Projective Geometry

Elisabetta Fortuna · Roberto FrigerioRita Pardini

Solved Problems and Theory Review

UNITEXT 104

UNITEXT - La Matematica per il 3+2

Volume 104

Editor-in-chief

A. Quarteroni

Series editors

L. AmbrosioP. BiscariC. CilibertoM. LedouxW.J. Runggaldier

More information about this series at http://www.springer.com/series/5418

Elisabetta Fortuna • Roberto FrigerioRita Pardini

Projective GeometrySolved Problems and Theory Review

123

Elisabetta FortunaDipartimento di MatematicaUniversità di PisaPisaItaly

Roberto FrigerioDipartimento di MatematicaUniversità di PisaPisaItaly

Rita PardiniDipartimento di MatematicaUniversità di PisaPisaItaly

ISSN 2038-5722 ISSN 2038-5757 (electronic)UNITEXT - La Matematica per il 3+2ISBN 978-3-319-42823-9 ISBN 978-3-319-42824-6 (eBook)DOI 10.1007/978-3-319-42824-6

Library of Congress Control Number: 2016946954

Translation from the Italian language edition: Geometria proiettiva. Problemi risolti e richiami di teoriaby Elisabetta Fortuna, Roberto Frigerio and Rita Pardini, © Springer-Verlag Italia, Milano 2011.All Rights Reserved.© Springer International Publishing Switzerland 2016This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilarmethodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exempt fromthe relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material contained herein orfor any errors or omissions that may have been made.

Cover illustration: Tito Fornasiero, Punto di fuga, 34 � 49 cm, watercolor, 2010.http://bluoltremare.blogspot.com. Reproduced with permission.

Printed on acid-free paper

This Springer imprint is published by Springer NatureThe registered company is Springer International Publishing AG Switzerland

Preface

Projective geometry topics are taught in many degree programs in Mathematics,Physics and Engineering, also because of their practical applications in areas suchas engineering, computer vision, architecture and cryptography. The literature onthis subject includes, besides the classical extensive treatises, by now very datedfrom the point of view of language and terminology, some modern textbooks.Among these we mention only ‘E. Casas-Alvero, Analytic Projective Geometry,EMS Textbooks in Mathematics (2014)’, whose approach is close to ours and towhich we refer the reader for bibliographical references.

This is not one further textbook, in particular it has not been conceived forsequential reading from the first to the last page; rather it aims to complement astandard textbook, accompanying the reader in her/his journey through the subjectaccording to the philosophy of “learning by doing”. For this reason we make noclaim to be systematic; rather we have the ambition, or at least the hope, not only toease and reinforce the understanding of the material by presenting completelyworked out examples and applications of the theory, but also to awaken thecuriosity of the reader, to challenge her/him to find original solutions and developthe ability of looking at a question from different perspectives. In addition, the bookpresents among the solved problems some classical geometric results, whose proofsare accessible within the relatively elementary techniques presented here. Hopefullythese examples will encourage some readers to learn more about the topics treatedhere and undertake the study of classical algebraic geometry.

Indeed the starting point of the original Italian version of this work has been ourexperience in teaching the Projective Geometry course in the undergraduateMathematics degree program in Pisa, which brought us to realize the difficulties thestudents encounter. However, the book contains also topics that, although usuallynot treated in undergraduate courses, can be useful and interesting for a readerwishing to learn more on the subject. Another feature of the text is that not onlycomplex hypersurfaces and algebraic curves are studied, as it is traditional, butconsiderable attention is paid also to the real case.

v

The first chapter of the book contains a concise but exhaustive review of thebasic results of projective geometry; the interested reader can find the proofs in anytextbook on the subject. The goal of this first part is to provide the reader with anoverview of the subject matter and to fix the notations and the concepts used later.The following three chapters are collections of solved problems concerning,respectively: the linear properties of projective spaces, the study of hypersurfacesand plane algebraic curves and, finally, conics and quadrics. In solving the prob-lems we have given preference neither to the analytic nor to the synthetic approach,but every time we have chosen the solution that seemed to us more interesting, ormore elegant or quicker; sometimes we have presented more than one solution. Thedifficulty level varies, ranging from merely computational exercises to more chal-lenging theoretical problems. The exercises that in our opinion are harder aremarked with the symbol , whose meaning is: “take it easy, get yourself a cup ofcoffee or tea, arm yourself with patience and determination, and you will succeed inthe end”. In other cases, besides giving the solution that is objectively the simplest,we have proposed alternative solutions that require longer or more involvedarguments, but have the merit of giving a deeper conceptual understanding orhighlighting connections with other phenomena that would not be evident other-

wise. We have marked these, so to say, “enlightening” solutions with the symbol .

We have tried to offer in this way a guide to reading the text; indeed, we believe thatthe best way of using it, once the basic notions given in the theory review have beenacquired, is to try to solve the problems presented on one's own, using the solutionsthat we give for checking and complementing one's personal work.

The necessary prerequisites are limited to some basic notions, mostly in linearalgebra, that are usually taught in the first year courses of the Mathematics, Physicsor Engineering degree programs.

We are extremely grateful to Ciro Ciliberto for believing in this project,encouraging us and constantly sharing with us his experience as mathematician,book author and editor. We also wish to thank Dr. Francesca Bonadei of SpringerItalia for suggesting us to produce an English version of our book and assisting usin its preparation.

Pisa, Italy Elisabetta FortunaApril 2016 Roberto Frigerio

Rita Pardini

vi Preface

Contents

1 Theory Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Standard Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Projective Spaces and Subspaces, Projective Transformations . . . . 2

1.2.1 Projective Spaces and Subspaces. . . . . . . . . . . . . . . . . . . . 21.2.2 Projective Transformations . . . . . . . . . . . . . . . . . . . . . . . . 31.2.3 Operations on Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.4 The Projective Linear Group . . . . . . . . . . . . . . . . . . . . . . . 41.2.5 Fixed Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.6 Degenerate Projective Transformations . . . . . . . . . . . . . . . 41.2.7 Projection Centred at a Subspace . . . . . . . . . . . . . . . . . . . 51.2.8 Perspectivities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Projective Frames and Homogeneous Coordinates . . . . . . . . . . . . . 51.3.1 General Position and Projective Frames . . . . . . . . . . . . . . 51.3.2 Systems of Homogeneous Coordinates . . . . . . . . . . . . . . . 61.3.3 Analytic Representation of a Projective

Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.4 Change of Projective Frames. . . . . . . . . . . . . . . . . . . . . . . 71.3.5 Cartesian Representation of Subspaces . . . . . . . . . . . . . . . 71.3.6 Parametric Representation of Subspaces . . . . . . . . . . . . . . 81.3.7 Extension of Projective Frames . . . . . . . . . . . . . . . . . . . . . 91.3.8 Affine Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3.9 Projective Closure of an Affine Subspace . . . . . . . . . . . . . 111.3.10 Projective Transformations and Change

of Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Dual Projective Space and Duality . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4.1 Dual Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4.2 Duality Correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4.3 Linear Systems of Hyperplanes . . . . . . . . . . . . . . . . . . . . . 131.4.4 Duality Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.5 Dual Projectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

vii

1.5 Projective Spaces of Dimension 1 . . . . . . . . . . . . . . . . . . . . . . . . . 151.5.1 Cross-Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5.2 Symmetries of the Cross-Ratio . . . . . . . . . . . . . . . . . . . . . 161.5.3 Classification of the Projectivities of P1ðCÞ

and of P1ðRÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5.4 Characteristic of a Projectivity. . . . . . . . . . . . . . . . . . . . . . 18

1.6 Conjugation and Complexification . . . . . . . . . . . . . . . . . . . . . . . . . 191.7 Affine and Projective Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . 19

1.7.1 Homogeneous Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 201.7.2 Affine and Projective Hypersurfaces . . . . . . . . . . . . . . . . . 211.7.3 Intersection of a Hypersurface with a Hyperplane . . . . . . . 231.7.4 Projective Closure of an Affine Hypersurface . . . . . . . . . . 241.7.5 Affine and Projective Equivalence of Hypersurfaces . . . . . 251.7.6 Intersection of a Hypersurface and a Line . . . . . . . . . . . . . 261.7.7 Tangent Space to a Hypersurface, Singular Points . . . . . . 271.7.8 Multiplicity of a Point of a Hypersurface . . . . . . . . . . . . . 281.7.9 Real Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.8 Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.8.1 First Notions and Projective Classification . . . . . . . . . . . . 311.8.2 Polarity with Respect to a Quadric . . . . . . . . . . . . . . . . . . 331.8.3 Intersection of a Quadric with a Line . . . . . . . . . . . . . . . . 351.8.4 Projective Quadrics in P

2ðKÞ and in P3ðKÞ . . . . . . . . . . . 35

1.8.5 Quadrics in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.8.6 Diametral Hyperplanes, Axes, Vertices . . . . . . . . . . . . . . . 431.8.7 Conics of R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.8.8 Quadrics of R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.9 Plane Algebraic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511.9.1 Local Study of a Plane Algebraic Curve . . . . . . . . . . . . . . 511.9.2 The Resultant of Two Polynomials . . . . . . . . . . . . . . . . . . 531.9.3 Intersection of Two Curves . . . . . . . . . . . . . . . . . . . . . . . . 541.9.4 Inflection Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561.9.5 Linear Systems, Pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . 561.9.6 Linear Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571.9.7 Pencils of Conics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

2 Exercises on Projective Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3 Exercises on Curves and Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . 107

4 Exercises on Conics and Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

viii Contents

About the Authors

Elisabetta Fortuna was born in Pisa in 1955. In 1977 she received her Diploma diLicenza in Mathematics from Scuola Normale Superiore in Pisa. Since 2001 she isAssociate Professor at the University of Pisa. Her areas of research are real andcomplex analytic geometry, real algebraic geometry, computational algebraicgeometry.

Roberto Frigerio was born in Como in 1977. In 2005 he received his Ph.D. inMathematics at Scuola Normale Superiore in Pisa. Since 2014 he is AssociateProfessor at the University of Pisa. His primary scientific interests are focused onlow-dimensional topology, hyperbolic geometry and geometric group theory.

Rita Pardini was born in Lucca in 1960. She received her Ph.D. in Mathematicsfrom Scuola Normale Superiore in Pisa in 1990; she is Full Professor at theUniversity of Pisa since 2004. Her area of research is classical algebraic geometry,in particular algebraic surfaces and their moduli, irregular varieties and coverings.

ix

Symbols

fxi , 1p0ðxÞ, 1K

�, 1ci;jðAÞ, 2Ai;j, 2PðVÞ, 2dimPðVÞ, 2PnðKÞ, 2

LðAÞ, 3LðS1; S2Þ, 3LðP1; . . .;PmÞ, 3PGLðVÞ, 4PGLðnþ 1;KÞ, 4πH , 5½P�R, 6φR, 6Hi, 8U0, 9j0 : Kn ! U0, 9Ui, 10ji : Kn ! Ui, 10PðVÞ�, 12AnnðWÞ, 13Λ1ðSÞ, 13f�, 15βðP1;P2;P3;P4Þ, 15σ : Cn ! C

n, 19σðLÞ, 19HC, 19SC, 19

xi

I þJ , 22mI , 22Vðf Þ, 22I \H, 23λ0P0 þ . . .þ λn�1Pn�1, 24Fðλ0P0 þ . . .þ λn�1Pn�1Þ, 24I \U0, 24I , 25IðI ; r;PÞ, 26, 27TPðIÞ, 28rF, 27SingðIÞ, 27, 28mPðIÞ, 28CPðIÞ, 29σðf Þ, 29σðIÞ, 29, 31VRðIÞ, 29½f �

R, 29

½f �C, 30

IC, 30, 31IR, 30signðAÞ ¼ ðiþ ðAÞ; i�ðAÞÞ, 32tPAX ¼ 0, 32AP, 33polQðPÞ, 33Q�, 34A�, 34A, 39MN , 39eX , 39Sðf ; gÞ, 53Risðf ; gÞ, 53Sðf ; g; xnÞ, 54Risðf ; g; xnÞ, 54degxn f , 53IðC;D;PÞ, 55HFðXÞ, 56HessFðXÞ, 56HðCÞ, 56Λd , 56K½X�d , 56jðCÞ, 142CPðJ Þ, 164

xii Symbols

Chapter 1Theory Review

Projective spaces. Projective transformations. Duality. Theprojective line. Affine and projective hypersurfaces. Affine chartsand projective closure. Conics and quadrics. Plane algebraiccurves and linear systems.

Abstract Review of basic results of projective geometry: projective spaces and theirtransformations, duality, affine and projective hypersurfaces, plane curves, conics andquadrics.

1.1 Standard Notation

Besides a number of special conventions introduced whenever necessary, throughoutthe text we shall use standard notation and symbols, as commonly found in theliterature. To avoid misunderstandings, however, we recall below some of the mostfrequent notations used in the sequel.

The symbols N, Z, Q, R, C shall denote the sets of natural, integer, rational, realand complex numbers, respectively. The complex conjugate a − ib of a complexnumber z = a + ib is denoted by z.

Wewill denote byK a subfield ofC andbyK[x1, . . . , xn] the ring of polynomials inthe indeterminates (or variables) x1, . . . , xnwith coefficients inK. For anypolynomialf ∈ K[x1, . . . , xn] we will denote by fxi the partial derivative of f with respect to xi; asimilar notation will be used for partial derivatives of higher order. For a polynomialin a single variable p(x) the derivative will also be denoted by p′(x).

Let us set K∗ = K \ {0}.

The identity map of a set A will be denoted by IdA, or simply by Id.The symbolM(p, q, K) shall denote the vector space of p×qmatrices with entries

in the field K and M(n, K) the space of square matrices of order n. The invertiblematrices of order n form the group GL(n, K) and the real orthogonal matrices oforder n the subgroup O(n) of GL(n, R). The identity matrix of GL(n, K) is denotedby I (or by In if it is important to specify its order).

© Springer International Publishing Switzerland 2016E. Fortuna et al., Projective Geometry, UNITEXT - La Matematica per il 3+2 104,DOI 10.1007/978-3-319-42824-6_1

1

2 1 Theory Review

Since themapKn → M(n, 1, K) associating to any vectorX = (x1, . . . , xn) ∈ K

n

the column

⎛⎜⎝x1...

xn

⎞⎟⎠ is bijective, we will write vectors of K

n either as n-tuples or as

columns.For any A ∈ M(p, q, K) we will write rk A for the rank of A and, in case p = q,

det A and tr A for the determinant and the trace of A, respectively. If A = (ai,j) ∈M(n, K), we will denote by ci,j(A) the square submatrix of order n − 1 obtainedby deleting the row and the column of A containing ai,j. Moreover we will writeAi,j = (−1)i+j det(ci,j(A)).

For any vector space V , we will denote by GL(V ) the group of linear automor-phisms V → V .

1.2 Projective Spaces and Subspaces, ProjectiveTransformations

1.2.1 Projective Spaces and Subspaces

If V is a finite-dimensional K-vector space, the projective space associated to V isthe quotient set P(V ) = (V \{0})/ ∼, where∼ is the equivalence relation on V \{0}defined by

v ∼ w ⇐⇒ ∃k ∈ K∗ such that v = kw.

The integer dim P(V ) = dim V − 1 is called the dimension of P(V ). If V = {0},we have P(V ) = ∅ and dim(∅) = −1.

From now on, unless otherwise specified, V will denote a K-vector space ofdimension n + 1 and P(V ) the n-dimensional projective space associated to it.

Let us denote by π : V \ {0} → P(V ) the quotient map and by [v] the equivalenceclass of the vector v ∈ V \ {0}.

The space P(Kn+1) is also denoted by Pn(K) and called the standard projec-

tive space of dimension n over K. For any (x0, . . . , xn) ∈ Kn+1 we will denote

[(x0, . . . , xn)] simply by [x0, . . . , xn].A projective subspace of P(V ) is a subset of the form P(W ), where W is a linear

subspace of V . Thus one hasP(W ) = π(W \{0}), and hence dim P(W ) = dimW−1.If W = {0}, the subspace P(W ) is empty and its dimension is −1. A projectivesubspace is called a projective line if its dimension is 1, a projective plane if itsdimension is 2, a projective hyperplane if its dimension is n − 1. Following theusual terminology, for any projective subspace S of P(V ) the integer codim S =dim P(V ) − dim S is called the codimension of S.

1.2 Projective Spaces and Subspaces, Projective Transformations 3

1.2.2 Projective Transformations

Let V andW be twoK-vector spaces. Amap f : P(V ) → P(W ) is called a projectivetransformation if there exists an injective linear map ϕ : V → W such that f ([v]) =[ϕ(v)] for every v ∈ V \ {0}. In this case we write f = ϕ.

If ϕ : V → W is a linear map inducing f , the set of linear maps from V to Winducing f coincides with the family {kϕ |k ∈ K

∗}; in other words, the linear mapwhich induces a projective transformation is determined only up to multiplicationby a non-zero scalar.

If f is induced by a linear isomorphism ϕ (and so, in particular, dim P(V ) =dim P(W )), we say that f is a projective isomorphism. Two projective spaces over afield K are said to be isomorphic if there exists a projective isomorphism betweenthem; this is the case if and only if they have the same dimension.

A projective isomorphism f : P(V ) → P(V ) is called a projectivity of P(V ), andan involution if, additionally, f 2 = Id. An involution f is said to be non-trivial iff �= Id.

If f : P(V ) → P(W ) is a projective transformation and H is a subspace of P(V ),then f (H) is a subspace of P(W ) having the same dimension as H and f |H : H →f (H) is a projective isomorphism.

Two subsets A,B of P(V ) are said to be projectively equivalent if there exists aprojectivity f of P(V ) such that f (A) = B.

1.2.3 Operations on Subspaces

Let S1 = P(W1) and S2 = P(W2) be projective subspaces of P(V ). As P(W1) ∩P(W2) = P(W1 ∩ W2), the intersection of two (as a matter of fact, of any numberof) projective subspaces is a projective subspace. The subspaces S1 and S2 are calledincident if S1 ∩ S2 �= ∅, while skew if S1 ∩ S2 = ∅.

For any non-empty subset A ⊆ P(V ) the subspace generated by A is definedas the projective subspace L(A) obtained as intersection of all subspaces of P(V )

containing A. If A = S1 ∪ S2 with S1 = P(W1), S2 = P(W2) projective subspaces,the subspace generated by A will be denoted by L(S1, S2) and called the join of S1and S2; of course one has L(S1, S2) = P(W1+W2). Wewill denote by L(P1, . . . ,Pm)

the subspace generated by the points P1, . . . ,Pm.

Proposition 1.2.1 (Grassmann’s formula) Let S1, S2 be projective subspaces ofP(V ). Then dim L(S1, S2) = dim S1 + dim S2 − dim(S1 ∩ S2).

Grassmann’s formula implies that two subspaces S1, S2 cannot be skew if dim S1+dim S2 ≥ dim P(V ), so that for instance two lines in a projective plane or a line anda plane in a three-dimensional projective space always meet.

4 1 Theory Review

1.2.4 The Projective Linear Group

The projectivities of P(V ) form a group with respect to composition, called theprojective linear group and denoted by PGL(V ).

Since the automorphism that induces a projectivity f ∈ PGL(V ) is determinedonly up to a non-zero scalar, the map GL(V ) → PGL(V ) sending the automorphismϕ to the projectivity f = ϕ induces a group isomorphism between the quotient groupGL(V )/∼ andPGL(V ), where∼ denotes the equivalence relation onGL(V ) definedas follows: ϕ ∼ ψ if and only if there exists k ∈ K

∗ such that ϕ = kψ.If dim V = n + 1, the group GL(V ) is isomorphic to the multiplicative group

GL(n + 1, K) consisting of invertible square matrices of order n + 1 with entriesin K. As a consequence PGL(V ) is isomorphic to the group PGL(n + 1, K) =GL(n + 1, K)/ ∼, where ∼ denotes the equivalence relation which identifies twomatrices A,B if and only if there exists k ∈ K

∗ such that A = kB.

1.2.5 Fixed Points

If f is a projectivity of P(V ), a point P ∈ P(V ) is a fixed point of f if f (P) = P.More generally we say that A ⊂ P(V ) is invariant under f (or f -invariant for short)if f (A) = A (of course a point lying in an invariant subset need not be a fixed point).

If S, S1, S2 are subspaces of P(V ) invariant under f , then f |S is a projectivity ofS, and S1 ∩ S2 and L(S1, S2) are f -invariant.

If f ∈ PGL(V ) is induced by the linear isomorphism ϕ and P = [v], then P is afixed point of f if and only if v is an eigenvector of ϕ. As a consequence the fixed-point set of a projectivity is the union of projective subspaces S1, . . . , Sm of P(V )

such that, for each j = 1, . . . ,m, the subspaces Sj and L(S1, . . . , Sj−1, Sj+1, . . . , Sm)

are skew.If K is algebraically closed, any projectivity of P

n(K) has at least one fixed point.Similarly, any projectivity of a real projective space of even dimension has at leastone fixed point.

1.2.6 Degenerate Projective Transformations

It is possible to extend the definition of projective transformation so to include trans-formations induced by linear maps which fail to be injective. If ϕ : V → W is anon-zero linear map between K-vector spaces and H = P(ker ϕ) ⊂ P(V ), the mapf : P(V ) \ H → P(W ) defined by f ([v]) = [ϕ(v)], for any [v] ∈ P(V ) \ H, iscalled the degenerate projective transformation induced by ϕ. As with projectivetransformations, the linear map ϕ is determined by f up to a non-zero multiplicativeconstant. Observe that, when ϕ is injective, the projective subspace H is empty and

1.2 Projective Spaces and Subspaces, Projective Transformations 5

f is the projective transformation induced by ϕ. If S = P(U) ⊆ P(V ) is a subspacenot contained inH, the restriction of f to S \ (S∩H) is a degenerate projective trans-formation to P(W ) and, if S ∩ H = ∅, it is a projective transformation. The imageunder f of S \ (S ∩ H) is a projective subspace, more precisely it is the projectivequotient of the subspace ϕ(U) ⊆ W and its dimension is dim S − dim(S ∩ H) − 1(cf. Exercise 28).

1.2.7 Projection Centred at a Subspace

Let S = P(U) and H = P(W ) be projective subspaces of P(V ) such that S ∩H = ∅and L(S,H) = P(V ). If we set k = dim S and h = dimH, Grassmann’s formula(cf. Proposition 1.2.1) implies that k + h = n − 1. For any P ∈ P(V ) \ H, thespace L(H,P) has dimension h + 1 and hence, using Grassmann’s formula again,dim(S ∩ L(H,P)) = 0. Therefore L(H,P) intersects S exactly at one point. Themap πH : P(V ) \ H → S which associates to P the point L(H,P) ∩ S is calledthe projection onto S centred at H. One can easily check that πH is a degenerateprojective transformation (cf. Exercise 29).

1.2.8 Perspectivities

Assume that r and s are two distinct lines in the projective plane P2(K) and let

A = r ∩ s. For any point O /∈ r ∪ s, the restriction to r of the projection onto scentred atO is a projective isomorphism f : r → s, called perspectivity centred at O.Clearly f (A) = A by construction, and the inverse isomorphism f −1 : s → r is aperspectivity centred at O, too.

More generally, assume that S1, S2 are two subspaces of an n-dimensional pro-jective space P(V ) with dim S1 = dim S2 = k, and let H be a subspace such thatH ∩ S1 = H ∩ S2 = ∅ and dimH = n − k − 1. Then the restriction to S1 of theprojection onto S2 centred at H is a projective isomorphism f : S1 → S2, calledperspectivity centred at H. As with perspectivities between lines in the projectiveplane, the inverse isomorphism f −1 : S2 → S1 is a perspectivity centred at H and therestriction of f to S1 ∩ S2 is the identity map.

1.3 Projective Frames and Homogeneous Coordinates

1.3.1 General Position and Projective Frames

The points P0 = [v0], . . . ,Pk = [vk] of the projective space P(V ) are said to beprojectively independent if the vectors v0, . . . , vk ∈ V are linearly independent.Thus the points P0, . . . ,Pk are projectively independent if and only if the subspace

6 1 Theory Review

L(P0, . . . ,Pk) has dimension k; moreover the largest number of projectively inde-pendent points in P(V ) is dim P(V ) + 1.

More generally, if dim P(V ) = n, we say that the points P0, . . . ,Pk are in generalposition if either they are projectively independent (when k ≤ n) or k > n and nosubset of the Pi’s consisting of n + 1 points is contained in a hyperplane of P(V ).

Any ordered set R = {P0, . . . ,Pn,Pn+1} consisting of n + 2 points in generalposition is called a projective frame of P(V ); the points P0, . . . ,Pn are called thefundamental points of the frame, while Pn+1 is called the unit point of the frame.

1.3.2 Systems of Homogeneous Coordinates

Let R = {P0, . . . ,Pn,Pn+1} be a projective frame of P(V ). For any u ∈ V \ {0}such that [u] = Pn+1 there exists a unique basis Bu = {v0, . . . , vn} of V such that[vi] = Pi for each i = 0, . . . , n and u = v0 + · · · + vn. Moreover, for any λ ∈ K

∗we have Bλu = {λv0, . . . ,λvn}.

Any basis Bu obtained as above is called a normalized basis of V and canbe used to define a system of homogeneous coordinates in P(V ): if P = [v]and if (x0, . . . , xn) are the coordinates of the vector v with respect to the lin-ear basis Bu, we say that (x0, . . . , xn) is an (n + 1)-tuple of homogeneous coor-dinates of P with respect to the frame R. These coordinates are uniquely deter-mined up to a non-zero scalar, hence by homogeneous coordinates of P we actuallymean the homogeneous (n + 1)-tuple [P]R = [x0, . . . , xn] ∈ P

n(K). In particu-lar, it turns out that [P0]R = [1, 0, . . . , 0], [P1]R = [0, 1, . . . , 0], . . . , [Pn]R =[0, 0, . . . , 1], [Pn+1]R = [1, 1, . . . , 1].

Once a projective frame (and hence a system of homogeneous coordinates) inP(V ) has been chosen, instead of [P]R = [x0, . . . , xn] we can simply write P =[x0, . . . , xn].

Choosing a projective frame R is equivalent to choosing a projective isomor-phism between P(V ) and P

n(K). Namely the map φR : P(V ) → Pn(K) defined by

φR(P) = [P]R turns out to be the projective isomorphism induced by the linearisomorphism φB : V → K

n+1 sending v ∈ V to its coordinates with respect to thebasis B, where B is any normalized basis associated toR.

In Pn(K) the frame {[1, 0, . . . , 0], [0, 1, . . . , 0], . . . , [0, 0, . . . , 1], [1, 1, . . . , 1]}

is called the standard projective frame; a corresponding normalized basis is thecanonical basis of K

n+1.

Theorem 1.3.1 (Fundamental theorem of projective transformations) Let P(V ) andP(W ) be projective spaces over the field K such that dim P(V ) = n ≤ dim P(W ).Assume that R = {P0, . . . ,Pn+1} is a projective frame of P(V ) and that R′ ={Q0, . . . ,Qn+1} is a projective frame of an n-dimensional subspace S of P(W ).Then there exists a unique projective transformation f : P(V ) → P(W ) such thatf (Pi) = Qi for each i = 0, . . . , n + 1. If S = P(W ), then f is a projectiveisomorphism.

1.3 Projective Frames and Homogeneous Coordinates 7

In particular, given two projective framesR andR′ of P(V ), there is exactly oneprojectivity ofP(V )mappingR toR′; moreover, the only projectivity ofP(V ) fixingn + 2 points in general position is the identity map.

1.3.3 Analytic Representation of a Projective Transformation

Having homogeneous coordinates allows us to give an analytic representation ofsubspaces and projective transformations. For instance, let f : P(V1) → P(V2) bea projective transformation induced by the injective linear map ϕ : V1 → V2, withdim P(V1) = n and dim P(V2) = m. Let R1 and R2 be projective frames in P(V1)

and P(V2), respectively, with associated normalized bases B1 and B2. Denote by Athe matrix associated to ϕ with respect to the bases B1 and B2. If P ∈ P(V1), assumethat [P]R1 = [x0, . . . , xn] and [f (P)]R2 = [y0, . . . , ym]. If we let X = (x0, . . . , xn)and Y = (y0, . . . , yn), then there exists k ∈ K

∗ such that kY = AX. The matrix A,called the matrix associated to the transformation f with respect to the frames R1

and R2, is determined up to a non-zero scalar, because both the linear map ϕ andthe normalized bases B1 and B2 are so.

1.3.4 Change of Projective Frames

Let R1 and R2 be two projective frames with associated normalized bases B1 andB2. Let A ∈ GL(n + 1, K) denote the coordinate change between the bases B1 andB2. If P ∈ P(V ), assume that [P]R1 = [x0, . . . , xn] and [P]R2 = [y0, . . . , yn]. Ifwe let X = (x0, . . . , xn) and Y = (y0, . . . , yn), then there exists k ∈ K

∗ such thatkY = AX . The matrix A, determined up to a non-zero scalar, is the one governingthe change of frame (or change of homogeneous coordinates) between R1 and R2.

If φR1 : P(V ) → Pn(K) and φR2 : P(V ) → P

n(K) are the projective isomor-phisms induced by the frames R1 and R2, the relation kY = AX represents incoordinates the projective transformation φR2 ◦ φR1

−1 : Pn(K) → P

n(K).

1.3.5 Cartesian Representation of Subspaces

Let S = P(W ) be a k-dimensional projective subspace ofP(V ). LetR be a projectiveframe in P(V ) and B a normalized basis of V associated to the frame. Since W isa linear subspace of V of dimension k + 1, it admits a Cartesian representation incoordinates. More precisely, there exists a matrix A ∈ M(n − k, n + 1, K) such thatrk A = n− k andW is the set of all vectors w ∈ V whose coordinates X with respectto the basis B satisfy the relation AX = 0. Then

8 1 Theory Review

S = {P ∈ P(V ) | AX = 0, where [P]R = [x0, . . . , xn] and X = (x0, . . . , xn)}.

The equations of the homogeneous linear system AX = 0 are called the Cartesianequations of the subspace S with respect to the frame R.

In this way S is regarded as the set of points in P(V ) whose homogeneous coor-dinates with respect to R satisfy a homogeneous linear system of n − k indepen-dent equations, where n − k is the codimension of S in P(V ). In particular, everyhyperplane of P(V ) can be represented by means of a homogeneous linear equationin n + 1 variables a0x0 + · · · + anxn = 0 where at least one coefficient ai is non-zero. The hyperplanes Hi of equation xi = 0 are called coordinate hyperplanes, orfundamental hyperplanes.

1.3.6 Parametric Representation of Subspaces

A subspace S = P(W ) of dimension k can also be represented parametrically.Namely, if we view W as the image of an injective linear map ϕ : K

k+1 → V ,then S is the image of the projective transformation f : P

k(K) → P(V ) induced byϕ. The components of the analytic representation of f (with respect to the standardprojective frame of P

k(K) and a projective frame R of P(V )) are called paramet-ric equations of the subspace S with respect to R. In this way the homogeneouscoordinates of the points of S are homogeneous linear functions of k+1 parameters.

In order to determine these equations explicitly it suffices to choose k + 2 pointsQ0, . . . ,Qk+1 of S in general position (so that S = L(Q0, . . . ,Qk)) and constructthe projective transformation f which maps the standard projective frame of P

k(K)

to the projective frame {Q0, . . . ,Qk+1} of S (the existence of f is guaranteed by theFundamental theorem of projective transformations, cf. Theorem 1.3.1). If [Qi]R =[qi,0, . . . , qi,n], then P lies in S if and only if there exist λ0, . . . ,λk ∈ K such that

⎧⎪⎨⎪⎩

x0 = λ0q0,0 + . . . + λkqk,0...

xn = λ0q0,n + . . . + λkqk,n

,

where [x0, . . . , xn] are homogeneous coordinates of P. This occurs if and only if thematrix

M =⎛⎜⎝q0,0 . . . qk,0 x0...

......

...

q0,n . . . qk,n xn

⎞⎟⎠

(whose first k + 1 columns are linearly independent) has rank k + 1. By a well-known linear algebra result, rk(M) = k + 1 if and only if M contains an invertible(k + 1) × (k + 1) submatrixM ′ such that the determinant of every (k + 2) × (k + 2)

1.3 Projective Frames and Homogeneous Coordinates 9

submatrix of M containing M ′ vanishes; since there are n − k such matrices, thisprovides a Cartesian representation of S with n − k equations.

1.3.7 Extension of Projective Frames

If S is a projective subspace ofP(V ) of dimension k, wemay choose k+1 projectivelyindependent points P0, . . . ,Pk in S and then extend this set to a projective frameof P(V ). In the system of homogeneous coordinates [x0, . . . , xn] thus induced, thesubspace S is given by the equations xk+1 = · · · = xn = 0 and the points of S havehomogeneous coordinates of the form [x0, . . . , xk, 0, . . . , 0]. The map

S � P = [x0, . . . , xk, 0, . . . , 0] → [x0, . . . , xk] ∈ Pk(K)

is a projective isomorphism and defines on S a system of homogeneous coordinates,called system of homogeneous coordinates induced on S.

1.3.8 Affine Charts

Consider the fundamental hyperplaneH0 = {x0 = 0} of Pn(K) and letU0 = P

n(K)\H0. The map j0 : K

n → U0 defined by

j0(x1, . . . , xn) = [1, x1, . . . , xn]

is bijective and its inverse map U0 → Kn is

j0−1([x0, . . . , xn]) =

(x1x0

, . . . ,xnx0

).

The pair (U0, j0−1) is called standard affine chart of Pn(K).

We can generalize the notion of affine chart of a projective space P(V ) of dimen-sion n starting with any hyperplane H.

To do that, let f : Pn(K) → P(V ) be a projective isomorphism such that

f (H0) = H (and hence f (U0) = UH , whereUH = P(V )\H); the map jH : Kn → UH

defined by jH = f ◦ j0 is bijective and the pair (UH , jH−1) is called an affine chartof P(V ). Sometimes, by abuse of terminology, one uses the word “chart” to indicateeither the subset UH or the bijection with K

n alone, if no confusion arises.For instance, let H be a hyperplane of P(V ) given by the equation

a0x0 + · · · + anxn = 0 in the system of homogeneous coordinates induced by aprojective frame R. Since not all ai are zero, we can assume a0 �= 0 for instance.Then the map f : P

n(K) → P(V )which associates to the point [x0, . . . , xn] ∈ Pn(K)

10 1 Theory Review

the point of P(V ) of coordinates[x0 − a1x1 − · · · − anxn

a0 , x1, . . . , xn]with respect

to R is a projective isomorphism such that f (H0) = H. In this case the mapjH = f ◦ j0 : K

n → UH is given by

jH(x1, . . . , xn) = f ([1, x1, . . . , xn]) =[1 − a1x1 − · · · − anxn

a0, x1, . . . , xn

]

and its inverse map UH → Kn by

jH−1([x0, . . . , xn]) =

(x1∑n

i=0 aixi, . . . ,

xn∑ni=0 aixi

).

If P ∈ UH , the components of the vector jH−1(P) =(

x1∑aixi

, . . . ,xn∑aixi

)are

called affine coordinates of the point P in the chart UH . Points in H are called pointsat infinity (or sometimes improper points) with respect to the chart UH , while pointsin UH are called proper points with respect to UH .

Choosing another coefficient ai �= 0 in the equation of H and proceeding in asimilar way, we obtain a different affine chart.

Being an “improper point” is of course a relative concept: a point may lie atinfinity with respect to a chart and be proper with respect to another one; even more,every point P can be considered as an improper point simply by choosing a chartUH

determined by a hyperplane H containing P.If P(V ) = P

n(K) we will denote by Ui = Pn(K) \ Hi the chart determined by

the fundamental hyperplane Hi = {xi = 0}; proceeding as above it turns out that themap ji : K

n → Ui

ji(y1, . . . , yn) = [y1, . . . , yi−1, 1, yi+1, . . . , yn]

is a bijection with inverse

ji−1([x0, . . . , xn]) =

(x0xi

, . . . ,xi−1

xi,xi+1

xi, . . . ,

xnxi

).

If i = 0 we recover the map j0 defined at the beginning of this subsection.Observe that U0 ∪ · · · ∪ Un = P

n(K); the family {(Ui, ji−1)}i=0,...,n is calledstandard atlas of P

n(K).The composition of ji with the natural inclusion ofUi inP

n(K) gives an embeddingof K

n in Pn(K); thus we can think of P

n(K) as the completion of Kn obtained by

adding the hyperplane at infinity Hi.A projectivity f of P(V ) leaves an affine chart UH = P(V ) \ H invariant (i.e.

f (UH) = UH ) if and only if f (H) = H. The projectivities that leave an affine chartinvariant form a subgroup of PGL(V ). Up to a change of homogeneous coordinateswe may assume that H has equation x0 = 0; if A is a matrix associated to f in this

1.3 Projective Frames and Homogeneous Coordinates 11

system of coordinates, then f leavesH invariant if and only if, up to non-zero scalars,A is a block matrix of the form

A =(1 0 . . . 0B C

),

where B ∈ Kn and C is an invertible square matrix of order n. Therefore, in affine

coordinates f acts on the chart UH mapping X ∈ Kn to the vector CX + B, that is,

the restriction of f to the chart is an affinity. In other words the subgroup of PGL(V )

of the projectivities leaving a hyperplane invariant is isomorphic to the group ofaffinities of K

n.

1.3.9 Projective Closure of an Affine Subspace

Let W be an affine subspace of Kn defined by the equations

⎧⎪⎨⎪⎩

a1,1x1 + · · · + a1,nxn + b1 = 0...

...

ah,1x1 + · · · + ah,nxn + bh = 0

and consider, for instance, the embedding of Kn into P

n(K) given by the map j0defined above. Then j0 transformsW into the subset of proper points (with respect tothe hyperplane H0) of the projective subspace W of P

n(K) defined by the equations

⎧⎪⎨⎪⎩

a1,1x1 + · · · + a1,nxn + b1x0 = 0...

...

ah,1x1 + · · · + ah,nxn + bhx0 = 0

.

The subspace W coincides with the smallest projective subspace of Pn(K) con-

taining W , and is called the projective closure of W (with respect to j0).

1.3.10 Projective Transformations and Changeof Coordinates

Let f : P(V ) → P(W ) be a projective transformation, with dim P(V ) = n anddim P(W ) = m. Assume that R1,R2 are projective frames in P(V ) and that S1,S2

are projective frames in P(W ). For any P ∈ P(V ), denote by X (respectively X ′) acolumn vector whose entries are homogeneous coordinates of P with respect to R1

(respectively R2). Moreover, denote by Y (resp. Y ′) a column vector whose entries

12 1 Theory Review

are homogeneous coordinates of the point f (P) ∈ P(W )with respect to S1 (resp. S2).If A is a matrix associated to f with respect to the framesR1 and S1, then Y = kAXfor some k ∈ K

∗. On the other hand if we denote by N (resp. M) the change ofcoordinates betweenR1,R2 (resp. between S1,S2), then X ′ = αNX and Y ′ = βMYfor some α,β ∈ K

∗. Therefore

Y ′ = βMY = βkMAX = (βkα−1)MAN−1X ′,

and hence f is represented by the matrix MAN−1 with respect to the frames R2 andS2.

In particular, if f ∈ PGL(V ) and we use the same frame of P(V ) both in thesource and in the target of f , then the matrices A,B representing f in two differentsystems of homogeneous coordinates are similar up to a non-zero scalar, i.e. thereexist M ∈ GL(n + 1, K) and λ ∈ K

∗ such that B = λMAM−1. For instance, if K isalgebraically closed, the projectivity f is represented by a Jordan matrix in a suitableframe.

Two projectivities f , g ofP(V ) are said to be conjugate if there exists a projectivityh ∈ PGL(V ) such that g = h−1 ◦ f ◦ h. Therefore f , g are conjugate if and only ifthe matrices representing f and g in a given system of homogeneous coordinates aresimilar up to a non-zero scalar.

1.4 Dual Projective Space and Duality

1.4.1 Dual Projective Space

Let V be a K-vector space and denote by V ∗ its dual space. The projective spaceP(V ∗), also denoted by P(V )∗, is called dual projective space. If dim V = n + 1,then P(V )∗ has dimension n and hence it is projectively isomorphic to P(V ).

If B is a linear basis of V (obtained for instance from a projective frame of P(V )),then the dual basis B∗ of V ∗ can be used to induce on P(V )∗ a system of dualhomogeneous coordinates. If [L] ∈ P(V )∗ is the equivalence class of a non-zerolinear functional L ∈ V ∗ and if L(x0, . . . , xn) = a0x0 + · · · + anxn in the system ofcoordinates induced by B, then L has coordinates (a0, . . . , an) with respect to thebasis B∗ of V ∗ and [L] has homogeneous coordinates [a0, . . . , an].

Since a0x0 + · · · + anxn = 0 represents a hyperplane of P(V ), the space P(V )∗can be identified in a natural way with the set of projective hyperplanes of P(V ), andhence it has a structure of projective space.Accordinglywe say that givenhyperplanesare independent (respectively, in general position) if the corresponding points ofP(V )∗ are projectively independent (resp. in general position).We call homogeneouscoordinates of a hyperplane the homogeneous coordinates of the correspondingpoint of P(V )∗; in this way the hyperplane of equation a0x0 + · · · + anxn = 0 hashomogeneous coordinates [a0, . . . , an].

1.4 Dual Projective Space and Duality 13

1.4.2 Duality Correspondence

Let S = P(W ) be a projective subspace of P(V ) of dimension k. Then the annihilatorAnn(W ) = {L ∈ V ∗ | L|W ≡ 0} is a linear subspace of V ∗ of dimension n − k.Consider the map δ, having the set of subspaces of P(V ) of dimension k as domainand the set of subspaces of P(V )∗ of dimension n− k− 1 as target, which associatesto the subspace S = P(W ) the subspace P(Ann(W )). The map δ, called dualitycorrespondence, is a bijection that reverses inclusions and satisfies

δ(S1 ∩ S2) = L(δ(S1), δ(S2)) δ(L(S1, S2)) = δ(S1) ∩ δ(S2).

When k = n− 1 we recover the one-to-one correspondence between hyperplanes ofP(V ) and points of P(V )∗; when k = 0 we get a one-to-one correspondence betweenpoints of P(V ) and hyperplanes of P(V )∗.

1.4.3 Linear Systems of Hyperplanes

Every projective subspace of P(V )∗ is called a linear system; as seen in Sect. 1.4.2,the points of a linear system are hyperplanes of P(V ). If L is a linear system, theintersection S of all hyperplanes in L is called the centre of the linear system and Lcoincides with the set of all hyperplanes of P(V ) containing S. This is why the linearsystem L is also called the linear system of hyperplanes with centre S and denotedΛ1(S).

Since δ reverses inclusions, a hyperplane H of P(V ) contains S if and only ifthe point δ(H) ∈ P(V )∗ belongs to δ(S); therefore Λ1(S) = δ(S) and hence, ifdim S = k, it follows that dimΛ1(S) = n − k − 1.

More explicitly, if we consider a Cartesian representation of S by means of aminimal set of n− k homogeneous linear equations, we may view S as the intersec-tion of n − k hyperplanes H1, . . . ,Hn−k of P(V ) such that δ(H1), . . . , δ(Hn−k) areprojectively independent points in P(V )∗. On the other hand, as we saw above,

Λ1(S) = δ(S) = δ(H1 ∩ · · · ∩ Hn−k) = L(δ(H1), . . . , δ(Hn−k))

which shows again that dimΛ1(S) = n − k − 1.When k = n− 2 the linear system Λ1(S) has dimension 1 and is called the pencil

of hyperplanes with centre S. For instance if n = 2 and S contains only one point P,then Λ1(S) is the pencil of lines of the projective plane P(V ) with centre P; if n = 3and S is a line, then Λ1(S) is the pencil of planes of the projective space P(V ) withcentre the line S.

If L is a projective hyperplane of P(V ), we denote by Λ1(S) ∩ UL the set ofintersections of the projective hyperplanes of the linear system Λ1(S) with the chartUL = P(V ) \ L; this set consists of affine hyperplanes and is called linear system of

14 1 Theory Review

affine hyperplanes. When the centre S is not contained in L (and hence L /∈ Λ1(S)),the affine hyperplanes ofΛ1(S)∩UL meet in the affine subspaceS∩UL andΛ1(S)∩UL

is called a proper linear system of affine hyperplanes. If S ⊂ L, instead, it is calledan improper linear system of affine hyperplanes.

In particular, ifΛ1(S) is a pencil (i.e. it has dimension 1), thenΛ1(S)∩UL is calleda proper pencil (or an improper pencil) of affine hyperplanes; any two hyperplanesof an improper pencil are parallel. When n = 2 we thus recover the notion of properpencil of lines (which meet at a point) and that of improper pencil of lines (consistingof all lines parallel to a fixed line).

In the case of a proper linear system, mapping a hyperplane H ∈ Λ1(S) to theintersection H ∩ UL defines a bijection between the projective linear system Λ1(S)and the proper affine linear system Λ1(S) ∩ UL. If instead S ⊂ L, the hyperplane Lbelongs to Λ1(S) but its intersection with UL is empty.

1.4.4 Duality Principle

Since P(V )∗ is projectively isomorphic to P(V ), the bijection δ defined in Sect. 1.4.2can be seen as a map which transforms subspaces of dimension k of P(V ) intosubspaces of “dual dimension” n−k−1 of P(V ). The properties of δ recalled aboveimply the following:

Theorem 1.4.1 (Duality principle) Let P(V ) be a projective space of dimension n.Let P be an assertion about subspaces of P(V ), their intersections, joins, anddimensions. Denote by P∗ the “dual” assertion obtained from P by replacingthe words “intersection, join, contained, containing, dimension” by “join, intersec-tion, containing, contained, dual dimension” respectively (the “dual dimension” isn − k − 1 if the dimension is k). Then P holds if and only if P∗ holds.

A proposition is said to be self-dual if it coincides with its dual statement. Anexample of a self-dual proposition is the following:

Theorem 1.4.2 (Desargues’ Theorem) Let P(V ) be a projective plane and A1,A2,

A3,B1,B2,B3 points of P(V ) in general position. Consider the triangles T1 and T2 ofP(V ) with vertices A1,A2,A3 and B1,B2,B3; one says that T1 and T2 are in centralperspective if there exists a point O in the plane, distinct from the Ai and Bi, suchthat all lines L(Ai,Bi) pass through O. Then T1 and T2 are in central perspective ifand only if the points P1 = L(A2,A3) ∩ L(B2,B3), P2 = L(A3,A1) ∩ L(B3,B1) andP3 = L(A1,A2) ∩ L(B1,B2) are collinear.

For a proof of Desargues’ Theorem see Exercise 12.

1.4 Dual Projective Space and Duality 15

1.4.5 Dual Projectivity

Let f be a projectivity of P(V ) represented by a matrix A ∈ GL(n + 1, K) in asystem of homogeneous coordinates. The map f sends the hyperplane H of equationtCX = 0 to the hyperplane f (H) of equation tC′X = 0 where C′ = tA−1C. In fact,

tCX = 0 ⇐⇒ tCA−1AX = 0 ⇐⇒ t(tA−1C)AX = 0 ⇐⇒ tC′AX = 0.

Then the map f∗ : P(V )∗ → P(V )∗ defined by f∗(H) = f (H) is represented by thematrix tA−1 in the dual system of homogeneous coordinates in P(V )∗. So f∗ is aprojectivity of P(V )∗, called dual projectivity.

A hyperplane H of equation tCX = 0 turns out to be invariant under f if and onlyif there exists λ �= 0 such that C = λtA−1C, i.e. tAC = λC, and therefore if and onlyif C is an eigenvector of the matrix tA.

1.5 Projective Spaces of Dimension 1

1.5.1 Cross-Ratio

Given four points P1,P2,P3,P4 of a projective line P(V ), where P1,P2,P3 aredistinct, the cross-ratio β(P1,P2,P3,P4) is defined as the number y1y0 ∈ K ∪ {∞},where [y0, y1] are homogeneous coordinates ofP4 in the projective frame {P1,P2,P3}of P(V ).

In particular one has β(P1,P2,P3,P1) = 0, β(P1,P2,P3,P2) = ∞ andβ(P1,P2,P3,P3) = 1. In this way for every P4 �= P2 the number β(P1,P2,P3,P4)

is the affine coordinate of P4 in the affine chart P(V ) \ {P2} with respect to the affineframe {P1,P3}.

If [λi,μi] are homogeneous coordinates of Pi for i = 1, . . . , 4 in a given systemof homogeneous coordinates, we have that

β(P1,P2,P3,P4) = (λ1μ4 − λ4μ1)

(λ1μ3 − λ3μ1)

(λ3μ2 − λ2μ3)

(λ4μ2 − λ2μ4)

(setting as usual 10 = ∞). By means of this formula we can extend the notion ofcross-ratio to the case when three of the four points, but not necessarily the first three,are pairwise distinct.

If λi �= 0 for each i and we consider the affine coordinates zi = μiλi

of the points

Pi, the cross-ratio is given by

β(P1,P2,P3,P4) = (z4 − z1)(z3 − z2)

(z3 − z1)(z4 − z2).

16 1 Theory Review

Observe that, if P1,P2,P3 are distinct points of a projective line, then for everyk ∈ K ∪ {∞} there exists a unique point Q such that β(P1,P2,P3,Q) = k.

Let FO denote the pencil of lines passing through a point O in a projective planeP(V ). The image of the pencil under the duality correspondence is a line in P(V )∗,and therefore FO has a natural structure of 1-dimensional projective space. Thus,given four lines in a projective plane P(V ) passing through a pointO, three of whichare pairwise distinct, we can define the cross-ratio of the four lines as the cross-ratioof the four collinear points of P(V )∗ corresponding to them.

Similarly, given a subspace S of codimension 2 in a projective space P(V ), wedenote by FS the pencil of hyperplanes centred at S, i.e. the set of hyperplanes ofP(V ) containing S. SinceFS is a 1-dimensional subspace of the dual projective spaceP(V )∗, given hyperplanes H1,H2,H3,H4 ∈ FS such that at least three of them aredistinct, the cross-ratio β(H1,H2,H3,H4) is well defined.

The main property of the cross-ratio is its invariance under projective isomor-phisms:

Theorem 1.5.1 Let P1,P2,P3,P4 be points of a projective line P(V ), with P1,P2,

P3 distinct, and let Q1,Q2,Q3,Q4 be points of a projective line P(W ), withQ1,Q2,Q3 distinct. Then there exists a projective isomorphism f : P(V ) → P(W )

such that f (Pi) = Qi for i = 1, . . . , 4 if and only if β(P1,P2,P3,P4) = β(Q1,Q2,

Q3,Q4).

This property can also be used to construct a projective isomorphism f betweentwo projective lines P(V ) and P(W ) sending three distinct points P1,P2,P3 of P(V )

to three distinct pointsQ1,Q2,Q3 ofP(W ).Namely, for every X ∈ P(V ), f (X)is nec-essarily the unique point in P(W ) such that β(P1,P2,P3,X) = β(Q1,Q2,Q3, f (X)).

1.5.2 Symmetries of the Cross-Ratio

Since the cross-ratio of four points of a projective line depends on the order of thepoints and since there are 24 ways of ordering four distinct points, a priori there are24 possible values that the cross-ratio of an unordered set of four points can take.From the analytic expression of the cross-ratio we deduce that it is invariant underthe following permutations

Id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3),

so that

β(P1,P2,P3,P4) = β(P2,P1,P4,P3) = β(P3,P4,P1,P2) = β(P4,P3,P2,P1).

As a consequence, in order to describe the values of the cross-ratio relative tothe 24 ordered quadruples obtained by arranging the points P1,P2,P3,P4, it suffices

1.5 Projective Spaces of Dimension 1 17

to consider the ordered quadruples where P1 appears in the first position. If we setβ(P1,P2,P3,P4) = k, it turns out that

β(P1,P2,P4,P3) = 1

k, β(P1,P4,P3,P2) = 1 − k,

β(P1,P4,P2,P3) = 1

1 − k, β(P1,P3,P4,P2) = k − 1

k,

β(P1,P3,P2,P4) = k

k − 1.

Therefore, by permuting the four points in all possible ways, the cross-ratio can takeat most 6 distincts values, namely

k,1

k, 1 − k,

1

1 − k,k − 1

k,

k

k − 1.

It follows that two sets {P1,P2,P3,P4} and {Q1,Q2,Q3,Q4}, each consist-ing of four distinct points of a projective line P(V ), are projectively equiva-lent if and only if, letting k = β(P1,P2,P3,P4), we have β(Q1,Q2,Q3,Q4) ∈{k, 1k , 1 − k, 1

1 − k , k − 1k , k

k − 1

}.

However for certain special values of k the distinct values of the cross-ratio offour points are in fact fewer than 6 (cf. Exercise 21).

If β(P1,P2,P3,P4) = −1 we say that the quadruple (P1,P2,P3,P4) forms aharmonic quadruple.

1.5.3 Classification of the Projectivities of P1(C)

and of P1(R)

Let f be a projectivity of P1(K) with either K = C or K = R.

If K = C, then there exists a system of homogeneous coordinates in P1(C)where

f is represented by one of the following matrices:

(a)

(λ 00 μ

)with λ,μ ∈ K

∗, λ �= μ;

(b)

(λ 00 λ

)with λ ∈ K

∗;

(c)

(λ 10 λ

)with λ ∈ K

∗.

18 1 Theory Review

Recall that thematrix representing a projectivity is determined up tomultiplication

by a non-zero scalar and that the matrix

(1 1

λ0 1

)is similar to the matrix

(1 10 1

)for

any λ ∈ K∗. Therefore we may assume λ = 1 in the matrices listed above.

If K = R and the eigenvalues of one (and hence of every) matrix associated tof are real, then in a suitable system of homogeneous coordinates of P

1(R) the pro-jectivity f is represented by one of the matrices (a), (b), (c). However it may happenthat the eigenvalues of the matrix associated to f are not real, but rather complexconjugate: a+ ib and a− ib, with b �= 0. If so, f is represented, in a suitable systemof homogeneous coordinates, by a matrix of type:

(d)

(a −bb a

)with a ∈ R, b ∈ R

∗,

called a “real Jordan matrix”. Also in this case, by choosing a suitable multiple ofthe matrix we may assume, for instance, a2 + b2 = 1.

If we examine the fixed-point sets in the four cases, we see the following possi-bilities:

(a) there exist exactly two fixed points, that is the points of coordinates [1, 0] and[0, 1]; in this case we say that f is a hyperbolic projectivity;

(b) all points are fixed and f is the identity map;(c) there is a unique fixed point [1, 0]; f is then called a parabolic projectivity;(d) there are no fixed points and then we say that f is an elliptic projectivity.

Therefore the four cases can be distinguished on the basis of the number of fixedpoints. In particular the previous matrices give explicit examples of projectivities ofP1(C) having one or two fixed points and examples of projectivities of P

1(R) wherethe fixed-point set has cardinality zero, one or two.

1.5.4 Characteristic of a Projectivity

Assume that K = C or K = R. Let f be a hyperbolic projectivity of P1(K) and

denote its two distinct fixed points by A and B (cf. Sect. 1.5.3). LetR be a projectiveframe in P

1(K) having A and B as fundamental points (so that [A]R = [1, 0] and[B]R = [0, 1]). Since A and B are fixed points, f is represented, in the system of

homogeneous coordinates induced by R, by a matrix of the form

(λ 00 μ

).

If P ∈ P1(K) \ {A,B} and [P]R = [a, b], then [f (P)]R = [λa,μb], so the

definition implies immediately that β(A,B,P, f (P)) = μλ(in particular it does not

depend on the point P). The value μλis called the characteristic of the projectivity.

Clearly the characteristic depends on the order of the fixed points: if we exchange

A and B, then the characteristic takes the value λμ .

1.6 Conjugation and Complexification 19

1.6 Conjugation and Complexification

The space Rn embeds in a natural way in C

n and can be characterized as the set ofpoints z = (z1, . . . , zn) ∈ C

n such that σ(z) = z, where σ : Cn → C

n denotes theinvolution defined by σ(z1, . . . , zn) = (z1, . . . , zn).

If a1z1 + · · · + anzn = c is the equation of an affine hyperplane L of Cn (where

at least one of the complex numbers a1, . . . , an is non-zero and c ∈ C), then σ(L) isthe hyperplane of equation a1z1 + · · · + anzn = c and it is called the conjugate of L.

If the numbers a1, . . . , an, c are real, the equation a1x1 + · · · + anxn = c definesan affine hyperplane H of R

n; as a matter of fact, the same equation defines in Cn

an affine hyperplane called the complexification of H and denoted HC. Proceedingin an analogous way one can define the complexification SC of an affine subspace Sof R

n of any dimension; SC is an affine subspace of Cn having the same dimension

as S such that σ(SC) = SC and SC ∩ Rn = S.

If we regard Rn as embedded in C

n, then any affinity of Rn extends to an affinity

of Cn. The converse is not true: for example, an affinity T(X) = MX + N of C

n ingeneral does not map points of R

n to points of Rn. This happens if and only if T

transforms n + 1 affinely independent points of Rn into n + 1 affinely independent

points of Rn, i.e. if and only if M is a matrix of GL(n, R) and N ∈ R

n. In this casethe restriction of T to R

n is an affinity of Rn.

Anologous considerations can be made in the projective case. Also Pn(R) is

naturally embedded in Pn(C) and may be seen as the set of points of P

n(C) thatadmit a representative in R

n+1 \ {0}. If, for the sake of simplicity, we still denoteby σ the involution P

n(C) → Pn(C) defined by σ([z0, . . . , zn]) = [z0, . . . , zn], then

Pn(R) coincides with the set of points P ∈ P

n(C) such that σ(P) = P. Similarlyone can define the conjugate σ(H) of a projective hyperplane H of P

n(C) and thecomplexification SC of a projective subspace S of P

n(R).Every projectivity of P

n(R) extends to a projectivity of Pn(C). Conversely, a

projectivity f of Pn(C) represented by a matrix A ∈ GL(n+ 1, C) transforms points

of Pn(R) into points of P

n(R) if and only if there exists λ ∈ C∗ such that λA ∈

GL(n + 1, R); in this case the restriction of f to Pn(R) is a projectivity of P

n(R).

1.7 Affine and Projective Hypersurfaces

Quadrics, plane algebraic curves and conics are the examples of hypersurfaces wewill focus on in what follows. To avoid useless repetitions, in this section we collectnotations, definitions and some basic results about hypersurfaces that will be studiedin greater detail and applied to the above special cases in subsequent sections.

20 1 Theory Review

1.7.1 Homogeneous Polynomials

From now on we assume that K = C or K = R. We denote the ring of polynomialswith coefficients in K in the indeterminates x0, . . . , xn by K[x0, . . . , xn] and thedegree of the polynomial F(x0, . . . , xn) by degF. We recall (cf. Sect. 1.1) that Fxidenotes the partial derivative of F with respect to xi; higher-order partial derivativesare denoted in a similar way.

A non-zero polynomial F ∈ K[x0, . . . , xn] is said to be homogeneous if all itsmonomials have the same degree (homogeneous polynomials are usually denotedby capital letters). Every non-zero polynomial can be written uniquely as a sum ofhomogeneous polynomials.

Homogeneous polynomials have many important properties; here we just recallthat:

(a) a non-zero polynomial F(x0, . . . , xn) is homogeneous of degree d if and only ifthe identity F(tx0, . . . , txn) = tdF(x0, . . . , xn) holds in K[x0, . . . , xn, t];

(b) (Euler’s identity) if F(x0, . . . , xn) is homogeneous, then the equality

n∑i=0

xiFxi = (degF)F

holds in K[x0, . . . , xn];(c) a polynomial that divides a homogeneous polynomial is itself homogeneous.

If F(x0, . . . , xn) ∈ K[x0, . . . , xn] is homogeneous, then the polynomialf (x1, . . . , xn) = F(1, x1, . . . , xn) is called the dehomogenized polynomial of F withrespect to x0. If F is homogeneous of degree d and x0 does not divide F, the deho-mogenized polynomial of F still has degree d.

If f (x1, . . . , xn) ∈ K[x1, . . . , xn] is a degree d polynomial, the polynomial

F(x0, . . . , xn) = xd0 f

(x1x0

, . . . ,xnx0

)

is called the homogenized polynomial of f with respect to x0. It is immediate to checkthat F is homogeneous of degree d and is not divisible by x0.

Homogenizing a polynomial f ∈ K[x1, . . . , xn] and then dehomogenizing thepolynomial thus obtained gives f back. On the other hand, if F is homogeneous,then the polynomial obtained by first dehomogenizing it with respect to x0 andthen homogenizing the result coincides with F if and only if x0 does not divide F.Therefore, there exists a one-to-one correspondence between degree d polynomialsof K[x1, . . . , xn] and homogeneous degree d polynomials of K[x0, . . . , xn] that arenot divisible by x0. Analogous definitions can be given for any other variable xi.

If F(x0, . . . , xn) and f (x1, . . . , xn) are polynomials that are obtained one fromanother by homogenizing and dehomogenizing (so that, in particular, x0 does notdivide F), then F is irreducible if and only if f is. More precisely, if F factors as

1.7 Affine and Projective Hypersurfaces 21

F = c Fm11 · . . . ·Fms

s , where the Fi are homogeneous irreducible polynomials and c isa non-zero constant, then f factors as f = c f m1

1 · . . . · f mss , where fi is the polynomial

obtained by dehomogenizing Fi, and conversely.If we consider homogeneous polynomials in two variables with complex coeffi-

cients, one has the following important result, which is the homogeneous version ofthe Fundamental theorem of algebra:

Theorem 1.7.1 (Factorization of complex homogeneous polynomials in two vari-ables) Let F(x0, x1) ∈ C[x0, x1] be a homogeneous polynomial of degree d > 0.Then there exist (a1, b1), . . . , (ad, bd) ∈ C

2 \ {(0, 0)} such that

F(x0, x1) = (a1x1 − b1x0) · . . . · (adx1 − bdx0).

The pairs (ai, bi) are uniquely determined up to the order and non-zeromultiplicativeconstants.

A homogeneous pair [a, b] ∈ P1(C) is called a root of multiplicity m of a homo-

geneous polynomial F(x0, x1) ∈ C[x0, x1] if m is the largest non-negative integersuch that (ax1 − bx0)m divides F(x0, x1).

Theorem 1.7.1 does not hold for real homogeneous polynomials in two variables:just consider the polynomial x20 + x21. However, since the only irreducible real poly-nomials in one variable are those of degree one and those of degree two with negativediscriminant, we have the following real version:

Theorem 1.7.2 (Factorization of real homogeneous polynomials in two variables)Let F(x0, x1) ∈ R[x0, x1] be a homogeneous polynomial of positive degree. Thenthere exist pairs (a1, b1), . . . , (ah, bh) ∈ R

2 \ {(0, 0)} and triples (α1,β1, γ1), . . . ,(αk,βk, γk) ∈ R

3 \ {(0, 0, 0)} with βj2 − 4αjγj < 0 such that

F(x0, x1) =h∏

i=1

(aix1 − bix0) ·k∏

j=1

(αjx

21 + βjx0x1 + γjx

20

).

The pairs (ai, bi) and the triples (αj,βj, γj) are uniquely determined up to the orderand non-zero multiplicative constants.

In particular every homogeneous polynomial of odd degree in R[x0, x1] has atleast a real linear factor and hence a root in P

1(R).

1.7.2 Affine and Projective Hypersurfaces

We define an equivalence relation on K[x1, . . . , xn] by declaring that polynomialsf , g ∈ K[x1, . . . , xn] are equivalent if and only if there exists λ ∈ K

∗ such thatf = λg; if this is the case, we say that f and g are proportional. An affine hypersur-face of K

n is defined as a proportionality class of polynomials of K[x1, . . . , xn] of

22 1 Theory Review

positive degree; an affine hypersurface is called an affine curve when n = 2 and anaffine surface when n = 3.

If f is a representative of the hypersurface I, we say that f (x1, . . . , xn) = 0 is anequation of the hypersurface and that the degree of f is the degree of I. If I = [f ]andJ = [g], we denote the hypersurface [f g] by I+J ; moreover, for every positiveinteger m, we denote mI = [f m].

The hypersurface I = [f ] is said to be irreducible if f is. If f = cf m11 · . . . · f ms

s withc a non-zero constant and the fi distinct irreducible polynomials, the hypersurfacesIi = [fi] are called the irreducible components of I and the integer mi is calledthe multiplicity of the component Ii. Every irreducible component of multiplicitymi > 1 is called a multiple component; hypersurfaces without multiple componentsare said to be reduced.

For any f ∈ K[x1, . . . , xn] we set

V (f ) = {(x1, . . . , xn) ∈ Kn | f (x1, . . . , xn) = 0}.

Since V (λf ) = V (f ) ∀λ ∈ K∗, we call the (well-defined) set V (I) = V (f ) the

support of the hypersurface I = [f ].As the fieldK is infinite, the Identity principle for polynomials ensures that a poly-

nomial f of K[x1, . . . , xn] satisfies f (a1, . . . , an) = 0 for every (a1, . . . , an) ∈ Kn

if and only if f = 0. As a consequence, the complement of any hypersurface isnon-empty.

While a hypersurface uniquely determines its support, the converse is not true ingeneral: for instance, the hypersurfaces of equations f = 0 and f m = 0, althoughdifferent, have the same supports. Actually, if K = C the correspondence betweenhypersurfaces and supports becomes a bijection in the case of reduced hypersurfaces,while if K = R there exist distinct reduced hypersurfaces with the same supports(for example the real plane curves of equations x21 + x22 = 0 and x41 + x42 = 0).Wewill sometimes abuse the notation and denote both a hypersurface and its support bythe same symbol.

Another remarkable phenomenon is that, while the support of a complex hyper-surface contains infinitely many points if n ≥ 2, if the field is not algebraicallyclosed there exist hypersurfaces whose support is finite (as in the case of the curveof R

2 of equation x21 + x22 = 0), or even empty (think of the curve of R2 of equation

x21 + x22 + 1 = 0).An affine hypersurface I is called a cone if there exists a point P ∈ I (called

vertex) such that, for every Q ∈ I, Q different from P, the line joining P and Qis contained in I. For instance, if f is a homogeneous polynomial then the affinehypersurface I = [f ] is a cone with vertex at the origin. The set of vertices of acone may be infinite (cf. Exercise 112); for example the vertices of the cone of R

3

of equation x1x2 = 0 are precisely the points of the x3-axis.Any proportionality class of homogeneous polynomials of positive degree in

K[x0, . . . , xn] is called a projective hypersurface of Pn(K). Similarly to what we

did in the affine case one defines the concepts of equation and degree of a projectivehypersurface, of irreducible hypersurface, irreducible component, multiplicity of a

1.7 Affine and Projective Hypersurfaces 23

component, cone. A hypersurface is said to be a plane curve when n = 2 and aprojective surface when n = 3.

The support of a projective hypersurface can also be defined in a similar way asthe set of points P = [x0, . . . , xn] ∈ P

n(K) such that F(x0, . . . , xn) = 0. Note thatthis is a good definition, since it depends neither on the choice of the representativeof I nor on the choice of the homogeneous coordinates of P; in fact, since F ishomogeneous, one has F(kx0, . . . , kxn) = kdF(x0, . . . , xn) for every k ∈ K, withd = degF. For instance, the support of a hypersurface ofP

1(K) is a (possibly empty)finite set.

Every hyperplane of Pn(K) is the support of an irreducible hypersurface of degree

1 (that we shall also call a hyperplane).Also in the projective case there is a bijective correspondence between reduced

projective hypersurfaces and their supports only when K = C (cf. Exercise 63 forthe case of curves).

The (affine or projective) hypersurfaces of degree 2, 3, 4 are called quadrics,cubics, quartics, respectively. The quadrics of P

2(K) are called conics.

1.7.3 Intersection of a Hypersurface with a Hyperplane

Let I be a hypersurface of Pn(K) of equation F(x0, . . . , xn) = 0, and let H be a

hyperplane not contained in I of equation xi = L(x0, . . . , xi, . . . , xn), where L is ahomogeneous polynomial of degree 1 that does not depend on the variable xi.

If Pi is the ith point of the standard projective frame of Pn(K) and Hi is the

coordinate hyperplane of equation xi = 0, then Pi does not lie in H ∪ Hi, andtherefore one can define the perspectivity f : Hi → H centred at Pi (cf. Sect. 1.2.8),i.e. the restriction of the projection πPi : P

n(K) \ {Pi} → H onto H centred at Pi

(cf. Sect. 1.2.7). Recall that the standard homogeneous coordinates of Pn(K) induce

homogeneous coordinates x0, . . . , xi, . . . , xn on Hi. In turn, the latter coordinatesinduce, via the projective isomorphism f , homogeneous coordinates on H that areusually still denoted by x0, . . . , xi, . . . , xn. It is immediate to check that, with thesechoices, the point of H of coordinates [x0, . . . , xi, . . . , xn] coincides with the pointof P

n(K) of coordinates [x0, . . . ,L(x0, . . . , xi, . . . , xn), . . . , xn]. The polynomial

G(x0, . . . , xi, . . . , xn) = F(x0, . . . , xi−1,L(x0, . . . , xi, . . . , xn), xi+1, . . . , xn)

is non-zero because H is not contained in I; in the coordinates of H just described,the equation G(x0, . . . , xi, . . . , xn) = 0 defines a hypersurface of H, that we denoteby I ∩ H and that has the same degree as I.

It is easy to check that the point [a0, . . . , an] of Pn(K) belongs to I ∩ H if and

only if ai = L(a0, . . . , ai, . . . , an) and G(a0, . . . , ai, . . . , an) = 0.Once a homogeneous coordinate system in H, and thus a projective isomorphism

between H and Pn−1(K), has been chosen, the previous considerations essentially

24 1 Theory Review

reduce the definition of the hypersurface I ∩H to that of a hypersurface in Pn−1(K)

(cf. Sect. 1.7.2).Anotherway of fixing a homogeneous coordinate system inH is obtained by using

a parametric representation of the hyperplane. Let P0, . . . ,Pn−1 be projectively inde-pendent points such thatH = L(P0, . . . ,Pn−1).We fix representatives (pi,0, . . . , pi,n)of the points Pi and denote by λ0, . . . ,λn−1 the corresponding coordinate system onH. From now on we will write λ0P0 + · · · + λn−1Pn−1 instead of

[λ0p0,0 + · · · + λn−1pn−1,0, . . . ,λ0p0,n + · · · + λn−1pn−1,n].

Then we can represent H parametrically as the set of points λ0P0 + · · · + λn−1Pn−1

as [λ0, . . . ,λn−1] varies in Pn−1(K).

We will also write F(λ0P0 + · · · + λn−1Pn−1) instead of

F(λ0p0,0 + · · · + λn−1pn−1,0, . . . ,λ0p0,n + · · · + λn−1pn−1,n).

This convention will always be tacitly used henceforth.The polynomial

G(λ0, . . . ,λn−1) = F(λ0P0 + · · · + λn−1Pn−1)

(which is non-zero as H is not contained in I) is homogeneous in λ0, . . . ,λn−1 ofthe same degree as F. Therefore, when H is not contained in I, the polynomialG(λ0, . . . ,λn−1) defines a hypersurface of H (endowed with the homogenous coor-dinates λ0, . . . ,λn−1) which is denoted by I ∩ H and has the same degree as I.

In a similar way one can see that the intersection of an affine hypersurface withan affine hyperplane not contained in it is a hypersurface of H.

1.7.4 Projective Closure of an Affine Hypersurface

Identify Kn with the affine chart U0 = P

n(K) \ {x0 = 0} by means of the mapj0 : K

n → U0 defined by j0(x1, . . . , xn) = [1, x1, . . . , xn].Let F be a homogeneous polynomial that defines the projective hypersurface I

of Pn(K) and assume that x0 is not the only irreducible factor of F. If f denotes the

polynomial obtained by dehomogenizingF with respect to x0, the affine hypersurfacedefined by f is called the affine part of I in the chartU0. Its support is the intersectionof the support of I with U0: hence we denote it by I ∩ U0. The affine part I ∩ U0

has the same degree as I if and only if x0 does not divide F.If π : K

n+1 \ {0} → Pn(K) is the canonical quotient map, the set π−1(I) ∪ {0}

is a cone of Kn+1 with vertex 0. The affine part I ∩ U0 can then be regarded as the

intersection of the cone π−1(I)∪ {0} with the affine hyperplane of Kn+1 of equation

x0 = 1.

1.7 Affine and Projective Hypersurfaces 25

Similarly, for every hyperplane H of Pn(K) one can define the affine part of I in

the chart UH = Pn(K) \ H.

Let I = [f ] be an affine hypersurface of Kn. If F(x0, . . . , xn) is the homogenized

polynomial of f with respect to x0, we say that the projective hypersurface I = [F]is the projective closure of I. Since by dehomogenizing F with respect to x0 we getf back, the affine part I ∩ U0 of I coincides with I. In addition, since x0 does notdivide F, the intersection of I with the hyperplane H0 = {x0 = 0} is a hypersurfaceof H0 having the same degree as I. For example, if n = 2 the projective closure ofthe curve I contains only finitely many points of the line x0 = 0, which are calledimproper points or points at infinity of I.

We remark that, if I = [F] is a projective hypersurface and x0 does not divide F,then I ∩ U0 = I.

In addition, if I is an affine hypersurface and H is an affine hyperplane of Kn not

contained in I, then I ∩ H = I ∩ H.

1.7.5 Affine and Projective Equivalence of Hypersurfaces

Recall that two subsets of Kn are called affinely equivalent if there exists an affinity

ϕ that transforms one into the other. As hypersurfaces are not in general determinedby their supports, we introduce a notion of affine equivalence of hypersurfaces basedon their equations.

Let I be an affine hypersurface of Kn of equation f (X) = 0, where X =

(x1, . . . , xn), and let ϕ(X) = AX + B be an affinity of Kn, with A ∈ GL(n, K)

and B ∈ Kn. We denote by ϕ(I) the affine hypersurface of equation g(X) =

f (ϕ−1(X)) = 0. This notation is consistent with the fact that ϕ transforms thesupport of I into the support of ϕ(I).

Two affine hypersurfaces I, J of Kn are said to be affinely equivalent if there

exists an affinity ϕ of Kn such that I = ϕ(J ). As we mentioned above, the supports

of affinely equivalent hypersurfaces are affinely equivalent.Projective equivalence of projective hypersurfaces can be defined in an analogous

way. Let I be a hypersurface of Pn(K) of equation F(x0, . . . , xn) = 0 and g a

projectivity of Pn(K). If X = (x0, . . . , xn) and N ∈ GL(n + 1, K) is a matrix

associated to g, then g acts by mapping the point of coordinates X to the point ofcoordinates NX. We denote by g(I) the projective hypersurface of equation G(X) =F(N−1X) = 0.

Two projective hypersurfaces I,J of Pn(K) are said to be projectively equivalent

if there exists a projectivity g of Pn(K) such that I = g(J ). If this is the case, one

can check that g transforms the support of J into the support of I, and thereforethe supports of projectively equivalent hypersurfaces are projectively equivalent sets.Moreover the degree, the number and themultiplicities of the irreducible componentsof a hypersurface are preserved by projective isomorphisms (hence by changes ofhomogeneous coordinates of P

n(K)).

26 1 Theory Review

Let I andJ be two hypersurfaces not having [x0] as an irreducible component andassume that g is a projectivity of P

n(K) such that I = g(J ). If the affine chart U0 isinvariant under g, then g is represented by a block matrix of the form (cf. Sect. 1.3.8)

N =(1 0 . . . 0B C

),

with B ∈ Kn and C an invertible square matrix of order n. The restriction of g to the

chart U0 is the affinity of Rn given by Y �→ CY + B and transforms J ∩ U0 into

I ∩ U0, while the restriction of g to the hyperplane at infinity H0 is the projectivityof H0 represented by the matrix C that transforms J ∩ H0 into I ∩ H0.

Analogously, if I and J are affinely equivalent hypersurfaces of Kn and ϕ is

an affinity of Kn such that I = ϕ(J ), we may regard ϕ as the restriction to U0

of a projectivity g of Pn(K) that fixes U0 and such that I = g(J ) and I ∩ H0 =

g|H0(J ∩ H0). Therefore, the projective closures and the parts at infinity of affinelyequivalent hypersurfaces are projectively equivalent.

1.7.6 Intersection of a Hypersurface and a Line

Let I be a projective hypersurface of degree d of Pn(K) having equation

F(x0, . . . , xn) = 0 and let r be a projective line.LetR andQ be distinct points of r. In accordancewith the considerations and nota-

tions established in Sect. 1.7.3, once we have chosen two representatives (r0, . . . , rn)and (q0, . . . , qn) of R and Q respectively, the line r is the set of points λR + μQ as[λ,μ] varies in P

1(K) and the intersection points of I and r are obtained by solvingthe equation

G(λ,μ) = F(λR + μQ) = F(λr0 + μq0, . . . ,λrn + μqn) = 0.

If r is contained in I, the polynomial G(λ,μ) vanishes identically; otherwise, itis homogeneous of degree d. Therefore, if K = C, by Theorem 1.7.1 the equationG(λ,μ) = 0 has precisely d roots [λ,μ] in P

1(C), counted with multiplicity. IfK = R, instead, the equation has at most d real roots, corresponding to the linearfactors of G(λ,μ).

Looking at the contribution of each root, if [λ0,μ0] is a root of multiplicity m ofthe polynomial G(λ,μ) then we say that I and r have multiplicity of intersection mat the corresponding point P = λ0R + μ0Q, and we write I(I, r,P) = m.

Note that themultiplicity of intersection iswell defined, because it does not dependon the choice of the two points on the line r.

We set by definition I(I, r,P) = 0 if P /∈ I ∩ r and I(I, r,P) = ∞ if r iscontained in I.

Concerning the notion of multiplicity just introduced, one can verify that:

1.7 Affine and Projective Hypersurfaces 27

(a) if g is a projectivity of Pn(K), then I(I, r,P) = I(g(I), g(r), g(P)), hence in

particular the multiplicity of intersection is preserved under projective equiva-lence;

(b) if I and J are projective hypersurfaces of Pn(K), then

I(I + J , r,P) = I(I, r,P) + I(J , r,P);

(c) if the line r is not contained in the projective hypersurfaceI of degree d ofPn(C),then

∑P∈r I(I, r,P) = d (that is to say, I and r intersect precisely in d points

counted with multiplicity). In the case of real hypersurfaces, even counting inter-sections with multiplicity, in general we can only say that

∑P∈r I(I, r,P) ≤ d.

Proceeding in a similar way, it is possible to define the multiplicity of intersectionof a hypersurface and a line at a point in the affine case, aswell. Indeed, if I = [f ] is anaffine hypersurface ofK

n and r is the line joining two points R,Q ∈ Kn parametrized

by t �→ (1− t)R+ tQ, we say that I(I, r,P) = m if P = (1− t0)R+ t0Q and t0 is aroot of multiplicity m of the polynomial g(t) = f ((1 − t)R + tQ).

As in the projective case, it is possible to check that the multiplicity I(I, r,P)

is well defined and does not depend on the choice of the points R,Q. In addition,I(I, r,P) = I(I, r,P) and therefore the multiplicity of intersection of a projectivehypersurface and a line at a point can be computed in affine coordinates in any affinechart containing the point.

1.7.7 Tangent Space to a Hypersurface, Singular Points

Let I = [F] be a projective hypersurface of Pn(K) of degree d. We say that a

projective line r is tangent to I at P if I(I, r,P) ≥ 2.It is possible to check that the union of the tangent lines to the hypersurface

I = [F] at P = [v] coincides with the projective subspace of Pn(K) defined by

Fx0(v)x0 + · · · + Fxn(v)xn = 0;

this space is called tangent space to I at P and is denoted by TP(I). This notion iswell defined, i.e. it does not depend on the choice of the representative of P, sinceevery first-order partial derivative of a homogeneous polynomial of degree d is eitherzero or homogeneous of degree d − 1.

A line r contained in I is tangent to I at each of its points, and thus r ⊆ TP(I)

for every P ∈ r.Denoting the usual gradient of F by ∇F, we will write ∇F(P) = 0 if the gradient

of F vanishes at any representative of P.The point P ∈ I is said a singular point of I if ∇F(P) = 0; otherwise it is said

non-singular or smooth. We will denote by Sing(I) the set of singular points of I. If

28 1 Theory Review

P is non-singular, then TP(I) is a hyperplane of Pn(K), otherwise it coincides with

Pn(K).A hypersurface is said to be non-singular or smooth if all its points are non-

singular, otherwise it is said to be singular.A projective hyperplane is said to be tangent to I at P if it is contained in TP(I):

if P is non-singular, then the only hyperplane tangent at P is the tangent space TP(I);if P is singular, then every hyperplane through P is tangent.

In analogy to the projective case, we say that the affine line r is tangent to theaffine hypersurface I at P if I(I, r,P) ≥ 2.

If P = (p1, . . . , pn), the union of the lines tangent to I = [f ] at P coincides withthe affine subspace TP(I) of K

n of equation

fx1(P)(x1 − p1) + · · · + fxn(P)(xn − pn) = 0,

which is called tangent space to I at P.The point P ∈ I = [f ] is called a singular point of I if ∇f (P) = 0, i.e. if

all first-order partial derivatives of f vanish at P; otherwise the point is called non-singular or smooth. We will denote by Sing(I) the set of singular points of I. IfP is non-singular, the tangent space TP(I) is an affine hyperplane of K

n, otherwiseit coincides with K

n. An affine hyperplane is said to be tangent to I at P if it iscontained in TP(I).

Finally we observe that:

(a) P is singular for I if and only if it is singular for I;(b) an affine line r is tangent to I at P if and only if r is tangent to I at P;(c) TP(I) = TP(I).

1.7.8 Multiplicity of a Point of a Hypersurface

Let I be a projective hypersurface of degree d of Pn(K) and let P be a point in

Pn(K). Letting r vary in the set of lines passing through P, and excluding the lines

contained in the hypersurface, if any, the multiplicity of intersection I(I, r,P) mayvary between 0 and d. The integer

mP(I) = minr�P I(I, r,P)

is called multiplicity of P for I (or also multiplicity of I at P). Since there is alwaysat least one line not contained in the hypersurface, one has that 0 ≤ mP(I) ≤ d; inaddition mP(I) = 0 if and only if P /∈ I.

One can check (cf. Sect. 1.7.6) that:

(a) if g is a projectivity of Pn(K), thenmP(I) = mg(P)(g(I)), and therefore the mul-

tiplicity of a hypersurface at a point is preserved under projective equivalence;

1.7 Affine and Projective Hypersurfaces 29

(b) if I and J are projective hypersurfaces of Pn(K), then

mP(I + J ) = mP(I) + mP(J )

(cf. Exercise 53);(c) the multiplicity mP(I) can also be computed in affine coordinates in any affine

chart containing P.

Working in an affine chart where P has affine coordinates (0, . . . , 0), the affinepart of I is defined by an equation f = fm + fm+1 + · · · + fd = 0, where every fi isa homogeneous polynomial of degree i in K[x1, . . . , xn], unless it is the zero poly-nomial, and fm �= 0. In this case every line through P has multiplicity of intersectionwith I at P greater than or equal to m and the lines r for which I(I, r,P) > m areprecisely those whose affine part is contained in the hypersurface CP(I) of K

n ofequation fm = 0, which is called the affine tangent cone to I at P (indeed, it is a conewith vertex P). The projective closure CP(I) of CP(I) is called projective tangentcone to I at P and its support coincides with the union of P and the projective linesthrough P such that I(I, r,P) > m. For example, the affine tangent cone at (0, 0) tothe curve of R

2 of equation x2 +y2 −x3 = 0 consists only of the point (0, 0) becausex2 + y2 = 0 contains no lines.

By interpreting f = fm + fm+1 + · · · + fd as the Taylor expansion of f centred atP = (0, . . . , 0), one sees immediately that P ∈ I is a point of multiplicity 1 for Iif and only if at least one of the first-order partial derivatives of f does not vanish atP, i.e. if and only if the point is non-singular. In this case sometimes one says that Pis a simple point. Instead, P is a point of multiplicity m > 1 if and only if f and allits partial derivatives of order smaller than m vanish at P but there exists at least onepartial derivative of order m that does not vanish at P.

If we do not work in an affine chart but rather we use a homogeneous equationF = 0 defining I, by Euler’s identity (cf. Sect. 1.7.1) P is a point of multiplicitym > 1 if and only if all partial derivatives of F of order m − 1 vanish at P and thereexists at least one partial derivative of order m that does not vanish at P.

1.7.9 Real Hypersurfaces

Extending what has been done in Sect. 1.6, for any polynomial f ∈ C[x1, . . . , xn]we denote by σ(f ) the polynomial obtained from f by conjugating each coefficient.The hypersurface σ(I) = [σ(f )] is called the conjugate of the affine hypersurfaceI = [f ] ofC

n. For every affine hypersurface I ofCn with support V (I) ⊆ C

n we canconsider the set VR(I) of the points of the support that are real; in symbols, VR(I) =V (I) ∩ R

n. The points of VR(I) are called the real points of the hypersurface.On the other hand, starting with any polynomial f ∈ R[x1, . . . , xn], we can con-

sider both the equivalence class [f ]R in R[x1, . . . , xn] and the equivalence class [f ]Cin C[x1, . . . , xn]. Therefore, for every affine hypersurface I = [f ]R of R

n with sup-

30 1 Theory Review

port V (I) ⊆ Rn we may consider the affine hypersurface IC = [f ]C of C

n, calledthe complexification of I; then one has VR(IC) = V (I).

For example, the support of the affine curve I = [x2 + y2]R in R2 consists only

of the point (0, 0), while the support in C2 of the curve IC = [x2 + y2]C is the union

of the lines x + iy = 0 and x − iy = 0. Similarly, the complex support of the curveof equation x2 + 1 = 0 is the union of the lines x+ i = 0 and x− i = 0, whereas thereal support is empty.

If g ∈ C[x1, . . . , xn], it may happen that [g]C contains real representatives, i.e.there exist α ∈ C

∗ and h ∈ R[x1, . . . , xn] such that g = α h. In this case one saysthat the affine hypersurface I = [g]C of C

n is a real hypersurface; if we denoteIR = [h]R, then I = (IR)C. Hence there exists a natural bijection between hyper-surfaces of R

n and real hypersurfaces of Cn.

If g ∈ C[x1, . . . , xn], the polynomial gσ(g) has real coefficients, so the hypersur-face [g] + σ([g]) is real.

If η(X) = MX + N is an affinity of Cn with M ∈ GL(n, R) and N ∈ R

n (sothat η maps points of R

n to points of Rn, cf. Sect. 1.6), then η transforms every real

hypersurface of Cn into a real hypersurface.

From the point of view of reducibility, we note that both the curve of equationx2 + y2 = 0 and that of equation x2 +1 = 0 are reducible when regarded as complexcurves, while they are irreducible when regarded as real curves.

In general every polynomial h ∈ R[x1, . . . , xn] has a factorization inC[x1, . . . , xn] of the form

h = cϕm11 · . . . · ϕms

s ψn11 · . . . · ψnt

t σ(ψ1)n1 · . . . · σ(ψt)

nt ,

where c is a real number, ϕ1, . . . ,ϕs are distinct polynomials with real coefficientsthat are irreducible in C[x1, . . . , xn], and ψ1, . . . ,ψt are polynomials with com-plex coefficients with the following properties: they are irreducible, pairwise distinctand not proportional to any polynomial with real coefficients. Therefore, the realhypersurface I = [h]C of C

n can have both real irreducible components (deter-mined by the real factors ϕ1, . . . ,ϕs of h) and non-real irreducible components;more precisely, if J is a non-real irreducible component of I of multiplicity m,then σ(J ) is also a non-real irreducible component of I of the same multiplicity.Since ψjσ(ψj) ∈ R[x1, . . . , xn] and it is irreducible over the reals, the irreduciblecomponents of the hypersurface IR = [h]R are given by the real irreducible factorsϕ1, . . . ,ϕs,ψ1σ(ψ1), . . . ,ψtσ(ψt). In this case, the points of V (ψj) ∩ V (σ(ψj)), ifany, are real points contributing to the support VR(IR).

In particular, there are only two possibilities for an irreducible hypersurface I ofR

n: either IC is irreducible, or there exists a complex irreducible hypersurface Jsuch that IC = J + σ(J ). In the latter case deg I = deg IC = 2 degJ , so thissituation can occur only in the case of hypersurfaces of even degree. For example, ifI is a real quadric (that is, if deg I = 2) that is reducible over C, then its irreduciblecomponents are either two real hyperplanes or two complex conjugate hyperplanes.

1.7 Affine and Projective Hypersurfaces 31

Analogous considerations can be made in the projective case. Thus for everyhomogeneous polynomial F ∈ C[x0, . . . , xn] the hypersurface σ(I) = [σ(F)] iscalled the conjugate of the projective hypersurface I = [F] of P

n(C).In addition, for every homogeneous polynomial F ∈ R[x0, . . . , xn] the hypersur-

face IC = [F]C of Pn(C) is called the complexification of the hypersurface I = [F]R

of Pn(R); the support of I in P

n(R) coincides with the set of real points of the sup-port of IC in P

n(C). Singular points P ∈ Pn(R) for I are singular for IC, too: more

precisely, mP(I) = mP(IC).A projective hypersurface I = [F] of P

n(C) defined by a homogeneous polyno-mial F ∈ C[x0, . . . , xn] is called a real projective hypersurface if F contains a realrepresentative G ∈ R[x0, . . . , xn]. If f is a projectivity of P

n(C) that maps points ofPn(R) to points of P

n(R) (so that there exists a matrix A ∈ GL(n + 1, R) represent-ing it, cf. Sect. 1.6), then f transforms every real hypersurface of P

n(C) into a realhypersurface.

In view of the previous discussion, by combining the inclusion of Rn in P

n(R)

and Pn(R) in P

n(C), for every polynomial f ∈ R[x1, . . . , xn] it can be useful torelate the geometric properties (such as, for instance, support and irreducibility)of the real affine hypersurface I = [f ]R with those of its projective closure I inPn(R) and of the complexification (I)C in P

n(C). For example, the polynomialx21 +x22 −1 ∈ R[x1, x2] defines an irreducible conic I in R

2 whose projective closureI inP

2(R) does not intersect the line at infinity in any real point; this happens becausethe complexification (I)C in P

2(C) intersects H0 in the hypersurface x21 + x22 = 0whose support consists of the complex points [1, i] and [1,−i] which are not real.

1.8 Quadrics

1.8.1 First Notions and Projective Classification

A projective hypersurface of Pn(K) of degree 2 is called a quadric; a quadric of

P2(K) is called a conic.If F(x0, . . . , xn) = 0 is the equation of a quadricQ of P

n(K), there exists a uniquesymmetric matrix A of order n + 1 such that

F(x0, . . . , xn) = tXAX,

where tX = (x0 x1 . . . xn

). In this case we say that the quadric is represented by the

symmetric matrix A.The quadricQ of equation tXAX = 0 is said to be non-degenerate if the matrix A

is invertible; the rank of A is called rank of the quadric (this notion is well definedsince A is determined byQ up to a non-zero multiplicative scalar). For instance, thequadrics of rank 1 are hyperplanes counted twice.

32 1 Theory Review

If P = [Y ] is a point ofQ = [F] where F(X) = tXAX, if we compute the gradientof F at Y we immediately see that ∇F(Y) = 2 tYA. Therefore the tangent spaceTP(Q) is defined by tYAX = 0 and Sing(Q) = {P = [Y ] ∈ P

n(K) | AY = 0}. Hencethe quadric is singular if and only if det A = 0, that is if and only ifQ is degenerate;furthermore the singular locus of Q is a projective subspace of P

n(K).The quadric Q is reducible (that is, the polynomial F is reducible) if and only if

A has rank 1 or 2. In particular, if Q is reducible and n ≥ 2, then it is degenerateand hence singular; on the other hand, there exist quadrics that are singular butirreducible, such as the quadric of P

3(C) of equation x21 + x22 − x32 = 0, for instance.If we intersect a quadric Q with a hyperplane H not contained in Q, we obtain a

quadric of H which is singular at a point P if and only if the hyperplane is tangentto Q at P (cf. Exercise 58). More precisely, when the quadric Q is non-degenerateand the hyperplane H is tangent to Q at P, the quadric Q ∩ H has rank n − 1 andSing(Q ∩ H) = {P} (for a proof of this result see Exercise 167).

Recall that any projectivity of Pn(K) transforms a quadric into a quadric

(cf. Sect. 1.7.5). More precisely, if Q has equation tXAX = 0 and the projectivity gis represented by an invertible matrix N , then g(Q) has equation tXtN−1AN−1X = 0.Since the matrices A and tN−1AN−1 have equal ranks, g transformsQ into a quadricof the same rank.

Two quadrics Q and Q′ of Pn(K) of equations tXAX = 0 and tXA′X = 0, respec-

tively, are projectively equivalent (cf. Sect. 1.7.5) if and only if there exists λ ∈ K∗

such that the matrices A and λA′ are congruent, i.e. there exists M ∈ GL(n + 1, K)

such that λA′ = tMAM. If K = C this latter fact occurs if and only if A and A′ arecongruent, while if K = R it occurs if and only if A is congruent to ±A′.

As a consequence, it is possible to classify quadrics up to projective equivalenceby using the classification of symmetric matrices up to congruence.

If K = C, the quadrics Q and Q′ are projectively equivalent if and only if A andA′ have the same rank.

If K = R, instead, A and A′ are congruent if and only if they have the samesignature (by signature of A we mean the pair sign(A) = (i+(A), i−(A)) wherei+(A) denotes the positivity index of the matrix A and i−(A) denotes its negativityindex; recall that i+(A) + i−(A) coincides with the rank of A). Since sign(−A) =(i−(A), i+(A)), up to exchanging A with −A (±A define the same quadric) we shallassume i+(A) ≥ i−(A) from now on. Having agreed on this convention, the realquadricsQ andQ′, represented by the symmetric matrices A and A′ respectively, areprojectively equivalent if and only if sign(A) = sign(A′).

From the previous considerations we obtain the following:

Theorem 1.8.1 (Projective classification of quadrics in Pn(K))

(a) Any quadric of Pn(C) of rank r is projectively equivalent to the quadric of

equationr−1∑i=0

x2i = 0.

1.8 Quadrics 33

(b) Any quadric of Pn(R) of signature (p, r − p), with p ≥ r − p, is projectively

equivalent to the quadric of equation

p−1∑i=0

x2i −r−1∑i=p

x2i = 0.

1.8.2 Polarity with Respect to a Quadric

Let Q be a quadric of Pn(K) of equation tXAX = 0 with A a symmetric matrix.

If P = [Y ] ∈ Pn(K), the equation tYAX = 0 defines a subspace of P

n(K) whichdoes not depend on the choice of the representative Y of P; therefore we will writetPAX = 0 henceforth, without specifying which representative of P has been chosen.Similarly, if P = [Y ], AP denotes the point having the vector AY as a representative.

For every P ∈ Pn(K) the subspace of P

n(K) of equation tPAX = 0 is denoted bypolQ(P) and called the polar space of P with respect to Q.

If P ∈ Q, then polQ(P) = TP(Q) (cf. Sect. 1.8.1); in particular, if P is singularfor Q, then polQ(P) = P

n(K).If P ∈ P

n(K) \ Sing(Q), the equation tPAX = 0 defines a hyperplane of Pn(K),

which is called the polar hyperplane of P with respect to Q. This hyperplane, thatwill often be denoted simply by pol(P), corresponds to the point of coordinates APin the dual space P

n(K)∗. Therefore we have a map

pol : Pn(K) \ Sing(Q) → P

n(K)∗

that associates the point AP ∈ Pn(K)∗ to the point P ∈ P

n(K) \ Sing(Q).For instance, if the fundamental point Pi = [0, . . . , 1, . . . , 0] is non-singular for

Q, the polar hyperplane of Pi has equation ai,0x0 + · · · + ai,nxn = 0.IfQ is a non-degenerate quadric (i.e. it is non-singular), the matrix A is invertible

and therefore the map pol : Pn(K) → P

n(K)∗ is a projective isomorphism. In thiscase for every hyperplane H of P

n(K) there exists a unique point, called the poleof H, having H as polar hyperplane with respect to Q. In particular the pole ofthe ith fundamental hyperplane Hi = {xi = 0} is the point [Ai,0, . . . ,Ai,n] (whereAi,j = (−1)i+j det(ci,j(A)), cf. Sect. 1.1).

Let us recall the main properties of polarity:

(a) (reciprocity) P ∈ pol(R) ⇐⇒ R ∈ pol(P)

(in particular for every P ∈ Pn(K) and for every R ∈ Sing(Q) one has that

P ∈ pol(R) and hence, by reciprocity, R ∈ pol(P), that is, pol(P) contains thesingular locus of the quadric);

(b) P ∈ pol(P) ⇐⇒ P ∈ Q;(c) if P /∈ Q, pol(P) is a hyperplane which intersects Q in the locus of points

of intersection between Q and the lines passing through P and tangent to thequadric.

34 1 Theory Review

Property (c) immediately implies that there are at most two tangents to a non-degenerate conic passing through a given point P of the projective plane P

2(K). Inparticular, if there are exactly two tangents passing through P, the polar of P is theline joining the two tangency points on the conic.

Of course in the real case the polar of a point not lying on the quadric might notintersect the quadric at all (for instance the polar of [0, 0, 1] with respect to the realconic x20 + x21 − x22 = 0 is the line x2 = 0 which does not meet the conic in any realpoint).

Two points P,R in Pn(K) are said to be conjugate with respect to the quadric Q

if tPAR = 0. In particular property (b) implies that the support of the quadric can beseen as the set of points of P

n(K) which are self-conjugate with respect to Q.We say that n + 1 projectively independent points P0, . . . ,Pn of P

n(K) are thevertices of a self-polar (n + 1)-hedron for Q if for any i �= j the points Pi and Pj

are conjugate with respect to Q (if n = 2 we speak of a self-polar triangle). Inthis case if Pi /∈ Sing(Q) it turns out that pol(Pi) = L(P0, . . . ,Pi−1,Pi+1, . . . ,Pn);this fact occurs for each Pi if Q is non-degenerate, which justifies the terminology“self-polar”.

If Pi = [vi] for i = 0, . . . , n and A is a symmetric matrix representing the quadricwith respect to the basis {v0, . . . , vn}, then the points P0, . . . ,Pn are the vertices of aself-polar (n+ 1)-hedron forQ if and only if {v0, . . . , vn} is an orthogonal basis forthe inner product onK

n+1 associated to the matrix A. As a consequence, representingQ bymeans of a diagonalmatrix is equivalent to choosing a projective frame inP

n(K)

whose fundamental points are the vertices of a self-polar (n+1)-hedron. In particularthe theorem of projective classification of quadrics (cf. Theorem 1.8.1) implies thatevery quadric has a self-polar (n + 1)-hedron; if the quadric has rank r, it containsexactly r points not lying on the quadric and n− r + 1 points belonging to Sing(Q).

IfQ is a non-degenerate quadric represented by an invertible matrix A, the imageofQ under the projective isomorphism polQ is a non-degenerate quadric, called dualquadric and denoted by Q∗, which is associated to the symmetric matrix A−1 withrespect to the dual coordinates. We can thus regard the support of the dual quadricas the set of the hyperplanes tangent toQ. By means of the usual identification withthe bidual, one has that Q∗∗ = Q.

IfQ is non-degenerate, the set of matrices representingQ∗ contains in particularthe adjoint matrix A∗ = (det A)A−1. Recall that A∗ is defined also when det A = 0(since it is the matrix whose (i, j) entry is the value Aj,i ∈ K defined in Sect. 1.1)and A∗ is non-zero if rk A = n. If Q is a quadric of P

n(K) of rank n, we cantherefore extend the previous notion by defining the dual quadric ofQ as the quadricrepresented by the matrix A∗. In this caseQ∗ has rank 1 and its support is the imageof the degenerate projective transformation pol.

1.8 Quadrics 35

1.8.3 Intersection of a Quadric with a Line

Assume thatQ is a quadric of Pn(K) of equation tXAX = 0 and r is a line. As already

observed (cf. Sect. 1.7.6), if r is not contained in Q, then Q ∩ r consists of at mosttwo points, possibly coinciding when the line is tangent. If r is not tangent to Q (inparticular it is not contained in the quadric), we say that the line r is secant to Q ifQ∩ r consists of two distinct points and that r is external toQ if the support ofQ∩ ris empty.

More precisely, if K = C no line can be external and, if r is not tangent, thenQ ∩ r consists of exactly two distinct points. IfQ and r are real and Q ∩ r containsat least one real point, then Q ∩ r consists of two real points, possibly coinciding;moreover, if a real line is tangent to a real quadric, then the point of tangency is real.

Explicitly, if P,R are two distinct points of r and we regard the line as the set ofpoints λP + μR as [λ,μ] varies in P

1(K), then

λP + μR ∈ Q ⇐⇒ tPAP λ2 + 2 tPAR λμ + tRAR μ2 = 0. (1.1)

In particular if P,R ∈ Q are conjugate with respect to Q, then r ⊆ Q. Moreover,if P ∈ Sing(Q) and R ∈ Q, then r is contained in Q; therefore any quadric with asingular point P is a cone with vertex P.

1.8.4 Projective Quadrics in P2(K) and in P

3(K)

In this section we study quadrics in projective spaces of low dimension in a moredetailed way.

As already said, quadrics in the projective plane are called conics. A conic isdefined by an equation of type

a0,0x20 + a1,1x

21 + a2,2x

22 + 2a0,1x0x1 + 2a0,2x0x2 + 2a1,2x1x2 = 0,

that is tXAX = 0 with

A =⎛⎝a0,0 a0,1 a0,2a0,1 a1,1 a1,2a0,2 a1,2 a2,2

⎞⎠ and X =

⎛⎝x0x1x2

⎞⎠ .

The conic is non-degenerate if the symmetric matrixA is invertible; it is called simplydegenerate if A has rank 2 and doubly degenerate if A has rank 1.

In contrast to what happens for quadrics of Pn(C) when n ≥ 3, a conic of P

2(C)

is reducible if and only if it is degenerate; in this case its irreducible componentsare two lines (which coincide when the conic is doubly degenerate). If K = R the

36 1 Theory Review

previous equivalence fails: for instance the conic of P2(R) of equation x20 + x21 = 0

is degenerate, yet the polynomial x20 + x21 is irreducible in R[x0, x1, x2].We can list the distinct projective canonical forms of conics of P

2(K) by special-izing the result of Theorem 1.8.1 for n = 2:

Theorem 1.8.2 (Projective classification of conics of P2(K))

(a) Every conic of P2(C) is projectively equivalent to exactly one conic in the fol-

lowing list:(C1) x20 + x21 + x22 = 0(C2) x20 + x21 = 0(C3) x20 = 0.

(b) Every conic of P2(R) is projectively equivalent to exactly one conic in the fol-

lowing list:(R1) x20 + x21 + x22 = 0(R2) x20 + x21 − x22 = 0(R3) x20 + x21 = 0(R4) x20 − x21 = 0(R5) x20 = 0.

As expected, in the complex case there is one model for each value of the rank;in the real case, instead, we find: two non-degenerate models ((R1) whose supportis empty and (R2) whose support contains infinitely many real points), two simplydegeneratemodels ((R3)whose real support consists of a single point and (R4)whichis the union of two distinct real lines), and one model (R5) of rank 1.

The conic C of equation tXAX = 0 is singular if and only if det A = 0, i.e. if andonly if C is degenerate. If K = C and C is simply degenerate, the only singular pointis the point where the two irreducible components of the conic meet; if C is doublydegenerate, all points of the curve are singular.

If K = R and C is simply degenerate, C may be reducible, with two distinct linesas irreducible components, or irreducible, in which case the complexified conic CChas two conjugate complex lines as irreducible components (cf. Sect. 1.7.9). In eithercase the two lines meet at a real point which is the only singular point of C. If K = R

and C has rank 1, the support is a line, all of whose points are singular for C.One obtains equations in canonical form by choosing in P

2(K) a projective framewhose fundamental points P1,P2,P3 are the vertices of a self-polar triangle. A tri-angle of this type always exists (cf. Sect. 1.8.2 or Exercise 170) and its constructionis very simple when n = 2.

To do that, if the conic is non-degenerate, we can choose a point P1 not belongingto C and consider the line r1 = pol(P1) (in particular P1 /∈ r1). Then we chooseP2 ∈ r1 \ C and denote r2 = pol(P2); by reciprocity P1 ∈ r2 and therefore r1 �= r2.Finally we take P3 = r1 ∩ r2 and denote r3 = pol(P3); then r3 = L(P1,P2). Byconstruction P1,P2,P3 are vertices of a self-polar triangle in which each edge is thepolar of the opposite vertex.

If the conic is simply degenerate, the polar of any point passes through the onlysingular point Q; if we choose a point P1 not on the conic and a point P2 �= Q

1.8 Quadrics 37

on pol(P1), then P1,P2,Q are vertices of a self-polar triangle. If the conic is doublydegenerate and its support is a line r (counted twice), in order to construct a self-polartriangle it suffices to take two distinct points on r and another point not in r.

The list of projectivemodels of quadrics of the three-dimensional projective spaceis longer:

Theorem 1.8.3 (Projective classification of quadrics of P3(K))

(a) Every quadric of P3(C) is projectively equivalent to exactly one quadric in the

following list:(C1) x20 + x21 + x22 + x32 = 0(C2) x20 + x21 + x22 = 0

(C3) x20 + x21 = 0

(C4) x20 = 0.

(b) Every quadric of P3(R) is projectively equivalent to exactly one quadric in the

following list:(R1) x20 + x21 + x22 + x32 = 0(R2) x20 + x21 + x22 − x32 = 0

(R3) x20 + x21 − x22 − x32 = 0

(R4) x20 + x21 + x22 = 0

(R5) x20 + x21 − x22 = 0

(R6) x20 + x21 = 0

(R7) x20 − x21 = 0

(R8) x20 = 0.

Observe that in the real case there are 3 distinct projective types of non-degeneratequadrics of P

3(R) (i.e. the models (R1), (R2) and (R3) in Theorem 1.8.3), two typesof rank 3 (the models (R4) and (R5)), two types of rank 2 (the models (R6) and (R7))and only one type (R8) of rank 1.

If A is a symmetric matrix of order 4 representingQ, we know thatQ is singularif and only if det A = 0. More precisely:

(a) if rk A = 3, Q has a unique singular point P and it is a cone with vertex P(the support may contain only the point P if K = R, as for the model (R4) ofTheorem 1.8.3);

(b) if rk A = 2, Q contains a line r of singular points; in the complex case Qis reducible and its support consists of two distinct planes intersecting in r; ifK = R the quadricQ can be either reduciblewith two distinct planes intersectingin r as irreducible components, or irreducible, in which case the irreduciblecomponents of the complexified quadric QC are two conjugate complex planesintersecting in the real line r (cf. Sect. 1.7.9);

(c) if rk A = 1, all points of Q are singular, the quadric is reducible and its supportconsists of a plane counted twice.

38 1 Theory Review

Concerning the intersection ofQwith a planeH of P3(K), recall thatH is tangent

to Q if H ⊆ TP(Q) for some P ∈ Q.If not tangent, H is said:

(a) external to Q if the conic Q ∩ H has an empty support (this can occur only ifK = R);

(b) secant to Q if the conic Q ∩ H is not empty.

In particular, if we look at the intersection ofQ with the tangent plane TP(Q) at anon-singular point P, we observe that, if TP(Q) � Q, thenQ∩TP(Q) is a degenerateconic which is singular at P (cf. Sect. 1.8.1). As a consequenceQ∩ TP(Q) is a conewith vertex P (cf. Sect. 1.8.3).

Therefore for any smooth pointP ofQwe can have one of the following situations:

(a) Q∩TP(Q) = TP(Q); in this caseQ is reducible, that is it consists of two planes;in particular Q ∩ TR(Q) = TR(Q) for every smooth point R of Q;

(b) Q∩TP(Q) = {P} (this can happen only if K = R, when the complexified conicof Q ∩ TP(Q) is the union of two complex conjugate lines meeting at P);

(c) Q∩TP(Q) is a degenerate conic which is the union of two (possibly coinciding)lines (for instance the intersection of the real cone of equation x20 + x21 − x22 = 0with the tangent plane at any smooth point is a line counted twice).

If Q is irreducible, case (a) in the latter list cannot occur; then we say that

(a) P is an elliptic point if Q ∩ TP(Q) = {P};(b) P is a parabolic point if Q ∩ TP(Q) is a line counted twice;(c) P is a hyperbolic point if Q ∩ TP(Q) consists of two distinct lines.

It can be checked (cf. Exercise 172) that in a quadric Q of P3(K) which is non-

degenerate and non-empty either all points are hyperbolic or they are all elliptic (thelatter casemayoccur only ifK = R); in this casewe say thatQ is ahyperbolic quadricor an elliptic quadric, respectively. It can also be checked (cf. Exercise 173) that everynon-singular point of an irreducible degenerate quadric of P

3(K) is parabolic: in thiscase, if there is at least a non-singular point, we say that Q is a parabolic quadric.

In the complex case, in which no elliptic point can exist, all non-degeneratequadrics are hyperbolic, while all quadrics of rank 3 are parabolic. If K = R thereare irreducible examples of each of the three types (cf. Exercise 174).

Since the notion of elliptic, hyperbolic or parabolic point is projective, meaningit is invariant under projectivities, one can investigate the nature of the points ofan irreducible quadric by using its projective model, or a model that is projectivelyequivalent to it and admits a simple equation (cf. Exercise 177). For instance thequadric of P

3(R) of equation x0x3 − x1x2 = 0 is projectively equivalent to the(hyperbolic) quadric x20 + x21 − x22 − x23 = 0.

1.8 Quadrics 39

1.8.5 Quadrics in Rn

An affine quadricQ of Rn, that is an affine hypersurface of R

n of degree 2, is definedby an equation of the form

tXAX + 2 tBX + c = 0,

where A is a non-zero symmetric matrix of order n, B is a vector of Rn, c ∈ R and

tX = (x1 . . . xn

). If we identify R

n with the affine chart U0 = Pn(R) \ {x0 = 0}

by means of the map j0 : Rn → U0 defined by j0(x1, . . . , xn) = [1, x1, . . . , xn], the

projective closure Q of the affine quadric Q is represented by the block matrix

A =(c tBB A

),

while the quadric at infinity Q∞ = Q ∩ H0 is represented by the matrix A. We saythat the affine quadricQ is degenerate if its projective closure is degenerate, and wecall rank of Q the rank of Q, that is the rank of the matrix A.

If Q has rank n, Q has only one singular point P and it is a cone of vertex P(cf. Sect. 1.8.3). In particular we say that Q is an affine cone if P ∈ U0 and that it isan affine cylinder if P ∈ H0.

Two affine quadrics Q and Q′ of Rn are affinely equivalent if there exists an

affinity ϕ of Rn such that Q = ϕ(Q′) (cf. Sect. 1.7.5). If ϕ(X) = MX + N , with M

an invertible matrix of order n and N ∈ Rn, then ϕ is the restriction to the chart U0

of the projectivity of Pn(R) represented by the block matrix

MN =(1 0 . . . 0N M

)

(cf. Sect. 1.3.8). Ifϕ transformsQ intoQ′, thenMN represents a projectivity ofPn(R)

that transformsQ intoQ′ andM represents a projectivity of H0 that transformsQ∞into Q′∞; therefore if two affine quadrics are affinely equivalent, their projectiveclosures and their quadrics at infinity are projectively equivalent (the converse impli-cation holds too, as we will see later on in this section).

The map ϕ(X) = MX + N is an isometry of Rn if and only if M ∈ O(n); the

quadrics Q and Q′ of Rn are said metrically equivalent if there exists an isometry

ϕ(X) = MX + N such that ϕ(Q) = Q′.

For every X = (x1, . . . , xn) ∈ Rn let X =

⎛⎜⎜⎜⎝

1x1...

xn

⎞⎟⎟⎟⎠ =

(1X

), that is denote by X

the vector of Rn+1 obtained from X by appending 1 to it as first coordinate. Then the

40 1 Theory Review

equation of Q can be written in the form tXAX = 0; moreover ϕ(X) =(

1ϕ(X)

)=

(1

MX + N

)= MNX. This is why from now on we will say that the affine quadric

Q of Rn is represented by the symmetric matrix A of order n+ 1 and that the affinity

ϕ of Rn is represented by the matrix MN .

So the quadric ϕ−1(Q) is represented by the matrix

tMNAMN =

⎛⎜⎜⎜⎝

1 tN0...

0

tM

⎞⎟⎟⎟⎠

(c tBB A

)(1 0 . . . 0N M

)=

=(

tNAN + 2 tBN + c t(tM(AN + B))tM(AN + B) tMAM

).

By using an affinity A and A transform to congruent matrices, so their signaturesremain unchanged and in particular their ranks are preserved. Now, the matrix of aquadric of R

n is determined up to a scalar α �= 0, hence either positive or negative.Since for every symmetric real matrix S we have sign(αS) = (i+(S), i−(S)) ifα > 0,while sign(αS) = (i−(S), i+(S)) if α < 0, up to multiplying the quadric’s equationby −1 (i.e. modifying A to −A and A to −A) we may assume i+(A) ≥ i−(A) andi+(A) ≥ i−(A). Henceforth we shall always adopt this convention, with the effectthat the above signatures now depend only on the quadric but not on the particularequation, and the pair (sign(A), sign(A)) becomes an affine invariant of the quadric.

The quadricQ of equation F(X) = tXAX+2 tBX+c = 0 is said to have centre atC ∈ R

n if F(C + X) = F(C − X) for every X ∈ Rn, equivalently if the polynomials

F(C +X) and F(C −X) coincide. Clearly the origin O of Rn is a centre forQ if and

only ifB = O. IfC is a centre forQ, the translation τ (X) = X+C satisfies τ (O) = CandO becomes a centre for the translated quadricQ′ = τ−1(Q). Therefore forQ′ alldegree one coefficients are zero, i.e. AC +B = 0, making C a solution of the systemAX = −B. Conversely, if the system AX = −B admits a solution C, the point Cmust be a centre for the quadric.

It can be checked that, if ϕ is an affinity, C is a centre forQ if and only if ϕ−1(C)

is a centre for the quadric ϕ−1(Q). Therefore the property of having a centre is affine(that is any quadric which is affinely equivalent to a quadric with centre has itself acentre).

Quadrics without centre are called paraboloids.

It turns out that rk A ≤ rk A ≤ rk A + 2 and that the system AX = −B admitssolutions (that is, the quadric has a centre) if and only if rk A ≤ rk A + 1, while Qis a paraboloid if and only if rk A = rk A + 2. In particular Q is a non-degenerateparaboloid if and only if det A �= 0 and det A = 0.

1.8 Quadrics 41

Recall (cf. Sect. 1.8.2) that, if Q is non-degenerate, the pole of the hyperplane atinfinity H0 with respect to Q is the point C = [A0,0,A0,1 . . . ,A0,n] ∈ P

n(R).IfQ is a non-degenerate paraboloid, then A0,0 = det A = 0; hence the pole of the

hyperplane at infinity H0 with respect to Q is an improper point, H0 is tangent to Qand therefore (cf. Sect. 1.8.1) Q ∩ H0 is degenerate (of rank n − 1). The converseimplication holds too: ifQ is non-degenerate andQ∩H0 is degenerate, then rk A =n+1 and det A = 0, so that rk A = n−1 = rk A−2 and henceQ is a non-degenerateparaboloid.

If Q is a non-degenerate quadric with centre, then rk A = n + 1 and rk A = n,hence the systemAX = −B has only one solution (that is, there exists only one centre)

which, by using Cramer’s rule, is given by the point C =(A0,1

A0,0, . . . ,

A0,n

A0,0

)∈ R

n. If

we regard C as a point of Pn(R), we have that C = [A0,0,A0,1 . . . ,A0,n]. Therefore

the centreC coincides with the pole of the hyperplane at infinityH0 with respect toQand this is a proper point (becauseA0,0 = det A �= 0). SinceQ∩H0 is non-degenerate,we may distinguish cases according to the nature of the intersection betweenQ andthe hyperplane at infinity:

(a) Q is called hyperboloid if non-degenerate, with centre and Q ∩ H0 is a non-degenerate quadric with non-empty support;

(b) Q is called ellipsoid if non-degenerate, with centre and Q ∩ H0 is a non-degenerate quadric with empty support.

By taking into account the property of being a quadricwith centre or not, and usingclassical results of linear algebra, it is possible to determine distinct canonical formsup to affine equivalence and up to metric equivalence for quadrics in R

n, includingdegenerate ones. If we restrict ourselves to using isometries,

(a) the Spectral theorem ensures that, after applying a linear isometry X �→ MX ofR

n (with M ∈ O(n)), we can assume that the symmetric matrix A is diagonalwith ai,i = 0 for each i = rk A + 1, . . . , n;

(b) if Q has a centre, by translating the origin to a centre we may assume B = 0;(c) ifQ has no centre, after applying an isometry (and multiplying the equation by a

non-zero constant, if necessary) we may assume B = (0, . . . , 0,−1) and c = 0.

Therefore we obtain:

Theorem 1.8.4 (Metric classification of quadrics ofRn) Assume thatQ is a quadric

of Rn of equation tXAX = 0 with p = i+(A) ≥ i−(A) and i+(A) ≥ i−(A). Let

r = rk A and r = rk A. Then:

(a) there exist λ1, . . . ,λr ∈ R with 0 < λ1 ≤ · · · ≤ λp and 0 < λp+1 ≤ · · · ≤ λr

such thatQ is metrically equivalent to the quadric defined by one of the followingequations:

(m1) x21 + λ2x22 + · · · + λpx2p − λp+1x2p+1 − · · · − λrx2r = 0

in this case we have λ1 = 1, r = r and sign(A) = sign(A);

42 1 Theory Review

(m2) λ1x21 + · · · + λpx2p − λp+1x2p+1 − · · · − λrx2r − 1 = 0

in this case we have sign(A) = (p, i−(A) + 1) (in particular r = r + 1);(m3) λ1x21 + · · · + λpx2p − λp+1x2p+1 − · · · − λrx2r + 1 = 0

in this case we have sign(A) = (p + 1, i−(A)) (in particular r = r + 1);(m4) λ1x21 + · · · + λpx2p − λp+1x2p+1 − · · · − λrx2r − 2xn = 0

in this case we have r = r + 2.

(b) In cases (m2), (m3), (m4) the numbers λ1, . . . ,λr are uniquely determined byQ.

For quadrics of type (m1) the numbers λ2, . . . ,λr are not uniquely determined:for instance the quadric of equation x21 − 2x22 = 0 is metrically equivalent to the

quadric of equation x21 − 12x

22 = 0 (it suffices to consider the isometry that exchanges

x1 with x2 and multiply the equation thus obtained by a suitable scalar).We arrive at a metric canonical form of type (m1), (m2) or (m3) ifQ has a centre,

of type (m4) if the quadric is a paraboloid.

If we do not restrict ourselves to using isometries in the process of simplifyingthe equation of the quadric, by applying a suitable affinity after the metric reductiondescribed above, it is possible to “normalize” the coefficients, so that the coefficientsof the equation belong to the set {1,−1, 0}. Therefore we have:Theorem 1.8.5 (Affine classification of quadrics of R

n) Assume thatQ is a quadricof R

n of equation tXAX = 0 with p = i+(A) ≥ i−(A) and i+(A) ≥ i−(A). Setr = rk A and r = rk A. Then Q is affinely equivalent to exactly one among thefollowing quadrics:

(a1) x21 + · · · + x2p − x2p+1 − · · · − x2r = 0,

in this case we have r = r and sign(A) = sign(A);(a2) x21 + · · · + x2p − x2p+1 − · · · − x2r − 1 = 0

in this case we have sign(A) = (p, i−(A) + 1) (in particular r = r + 1);(a3) x21 + · · · + x2p − x2p+1 − · · · − x2r + 1 = 0

in this case we have sign(A) = (p + 1, i−(A)) (in particular r = r + 1);(a4) x21 + · · · + x2p − x2p+1 − · · · − x2r − 2xn = 0

in this case we have r = r + 2.

We arrive at one of the first three models in the case of quadrics with centre, atthe fourth model in the case of a paraboloid.

For each model in Theorem 1.8.5 the pair (sign(A), sign(A)) is different andtherefore it allows us to distinguish the different affine types, so that two affinequadrics Q and Q′ of equations tXAX = 0 and tXA′X = 0, respectively, are affinelyequivalent if and only if (sign(A), sign(A)) = (sign(A′), sign(A′)). We express thisproperty by saying that the pair (sign(A), sign(A)) is a complete system of affineinvariants.

So, ifQ is projectively equivalent toQ′ andQ∞ is projectively equivalent toQ′∞,then sign(A) = sign(A′) and sign(A) = sign(A′) and hence Q and Q′ are affinelyequivalent. In other words, if the projective closures and the quadrics at infinity of

1.8 Quadrics 43

two affine quadricsQ andQ′ are projectively equivalent, thenQ andQ′ are affinelyequivalent.

In order to decide which is the affine canonical form of a quadric it is thereforesufficient to compute the signatures of A and A. As a matter of fact, Theorem 1.8.5implies that it is not always necessary to compute both signatures. One can startby computing the signature of A (hence rk A) and the rank of A: if rk A = rk Aor if rk A = rk A + 2 the equation of the affine model is determined. If insteadrk A = rk A + 1, it is necessary to compute the signature of A too.

Recall that the positivity index of a symmetric matrix coincides with the numberof positive eigenvalues of the matrix and hence with the number of positive roots ofits characteristic polynomial. Since all the roots of the characteristic polynomial ofa real symmetric matrix are real, the computation of the signatures of A and of Afrom their characteristic polynomials is immediate as a consequence of the followingresult:

Theorem 1.8.6 (Descartes’ criterion) If the roots of a polynomial p(x) with realcoefficients are all real, then p(x) has as many positive roots, counted with multi-plicity, as the number of sign changes in the sequence of non-zero coefficients of thepolynomial.

1.8.6 Diametral Hyperplanes, Axes, Vertices

In this section we shall only consider non-degenerate quadrics of Rn.

In Sect. 1.8.5 we saw that, ifQ is a non-degenerate quadric with centre, the centrecoincides with the pole of the hyperplane at infinity H0 with respect to Q. So whenworking with non-degenerate quadrics we can extend the notion of centre by callingcentre of a non-degenerate quadric Q the pole of the hyperplane at infinity H0 withrespect to Q. In particular the notion of centre can be used also for non-centredquadrics. By using the terminology thus introduced, it turns out that:

(a) the pole of H0 is proper, that is it lies in Rn, if and only ifQ is a non-degenerate

quadric with centre (in this case Q has as centre precisely the pole of H0);(b) the pole of H0 is improper, that is it belongs to the hyperplane at infinity H0,

if and only if Q is a non-degenerate paraboloid (and Q is tangent to H0 at thecentre of Q).

An affine hyperplane is called a diametral hyperplane of a non-degenerate quadricQ (diametral plane if n = 3, diameter if n = 2) if its projective closure is the polarhyperplane of an improper point. If R is the centre of the quadric, by reciprocitya hyperplane is diametral if and only if its projective closure passes through thecentre R. For instance ifQ is a non-degenerate paraboloid, all diametral hyperplanesare parallel to one line, because their projective closures pass through the impropercentre of Q.

44 1 Theory Review

If Q is a non-degenerate quadric of Rn and if the diametral hyperplane

polQ(P) ∩ Rn relative to a point P ∈ H0 is orthogonal to the direction lP deter-

mined by the point at infinity P, then polQ(P) is called principal hyperplane of Q.Moreover, we call axis of a non-degenerate quadric Q every line of R

n which isthe intersection of principal hyperplanes (if n = 2 the notions of principal hyperplaneand axis coincide). We call vertex of Q every point where Q intersects an axis (noconfusion is possible between this notion of vertex and that of vertex of a cone,because cones are degenerate quadrics).

It can be checked that, for every vertex V of a non-degenerate quadric Q of Rn,

the tangent hyperplane toQ at V is orthogonal to the axis through V (for a proof seeExercise 191).

The hyperplane polQ(P) ∩ Rn = {tPAX = 0} relative to P = [0, p1, . . . , pn] is

principal if and only if there exists λ �= 0 such that(p1 . . . pn

)A = λ

(p1 . . . pn

),

that is if and only if the vector (p1, . . . , pn) is an eigenvector of A relative to a non-zero eigenvalue. The eigenvectors v0 ∈ R

n of A relative to the eigenvalue 0, if any, donot correspond to any principal hyperplane, because in this case the polar hyperplanedetermined by the point P = [(0, v0)] is the hyperplane at infinity H0, which is notthe projective closure of any affine hyperplane.

If the quadric Q is non-degenerate and with centre, it can be proved (cf. Exer-cise 190) that there exist n pairwise orthogonal axes of the quadric that meet in thecentre of Q. Therefore the quadric can be represented in metric canonical form bychoosing in R

n a Cartesian coordinate system having the centre of Q as origin andn pairwise orthogonal axes of the quadric as coordinate axes.

If Q is a non-degenerate paraboloid, it can be proved (cf. Exercise 190) that Qhas only one axis (which is the intersection of n − 1 pairwise orthogonal principalhyperplanes ofQ) and only one vertex. Thus we arrive at the metric canonical formof the equation of Q by choosing in R

n a Cartesian coordinate system having asorigin the vertex V of Q and as coordinate axes pairwise orthogonal lines passingthrough V in such a way that the nth coordinate hyperplane is the tangent hyperplaneto Q at V and the remaining n − 1 coordinate hyperplanes are pairwise orthogonalprincipal hyperplanes of Q.

Given a point P = (a1, . . . , an) ∈ Rn and a real number η ≥ 0, the set of points

of Rn whose distance from P is equal to η is a quadric of equation

(x1 − a1)2 + · · · + (xn − an)

2 = η. (1.2)

More generally, we call sphere any quadric of Rn defined by Eq. (1.2) with η ∈ R.

This sphere is an ellipsoid with non-empty support if η > 0, an ellipsoid with emptysupport if η < 0; if η = 0 it is a degenerate quadric with support consisting of onlyone point.

A sphere is a quadric with exactly one centre; all hyperplanes through the centreare principal, all lines through the centre are axes and all points in the support arevertices (cf. Exercise 190).

The image of a sphere under an isometry is a sphere; in particular the property ofbeing a sphere is invariant under isometric coordinate changes.

1.8 Quadrics 45

1.8.7 Conics of R2

In this section, using the notation introduced so far, we examine in detail conics ofR2

both in the affine and in themetric setting, as wewill do later for quadrics ofR3. Since

R2 ⊂ C

2 ⊂ P2(C), it is useful to consider a conic C of R

2 not only as embedded inits projective closure in P

2(R), but also as embedded in the complexified conic CCin C

2 and in the projective closure of CC in P2(C) (cf. Sect. 1.7.9).

A doubly degenerate conic has as support a line consisting entirely of singularpoints.

A simply degenerate conic C can be either reducible (with two distinct lines asirreducible components) or irreducible (in this case the complexified conicCC has twocomplex conjugate lines as irreducible components, cf. Sect. 1.7.9). In both cases thetwo lines meet at a real point, which is the only singular point of C. If the intersectionpoint of these two lines is proper, the conic is an affine cone; if it is improper, theconic is an affine cylinder (cf. Sect. 1.8.5). More precisely, the conic is called a realcone or a real cylinder if the support is a union of lines, an imaginary cone if thesupport reduces to the unique singular point, an imaginary cylinder if the support isempty.

A non-degenerate conic C is called:

(a) parabola if it has no centre (in this case H0 is tangent to C),(b) hyperbola if it has a centre and C intersects H0 in two distinct points,(c) ellipse if it has a centre and the lineH0 is external to C (more precisely, it is called

a real ellipse if the support of C is not empty, an imaginary ellipse otherwise).

In R2 the classification obtained in Theorem 1.8.5 reads:

Theorem 1.8.7 (Affine classification of conics of R2) Every conic of R

2(x,y) is

affinely equivalent to exactly one among the following conics:(a1) x2 + y2 − 1 = 0 (real ellipse);(a2) x2 + y2 + 1 = 0 (imaginary ellipse);(a3) x2 − y2 + 1 = 0 (hyperbola);(a4) x2 − 2y = 0 (parabola);(a5) x2 − y2 = 0 (pair of distinct incident real lines);(a6) x2 + y2 = 0 (pair of complex conjugate incident lines);(a7) x2 − 1 = 0 (pair of distinct parallel real lines);(a8) x2 + 1 = 0 (pair of distinct parallel complex conjugate lines);(a9) x2 = 0 (pair of coincident real lines).

46 1 Theory Review

Table 1.1 Affine invariants for conics of R2

Model sign(A) sign(A)

(a1) (2, 0) (2, 1)

(a2) (2, 0) (3, 0)

(a3) (1, 1) (2, 1)

(a4) (1, 0) (2, 1)

(a5) (1, 1) (1, 1)

(a6) (2, 0) (2, 0)

(a7) (1, 0) (1, 1)

(a8) (1, 0) (2, 0)

(a9) (1, 0) (1, 0)

Observe that:

(a) for canonical forms (a1), (a2) or (a3) the origin is the centre of the conic C andthe axes x and y are diameters with points at infinity which are conjugate withrespect to C;

(b) for the parabola C of equation (a4) the origin O is a point of the conic, the axis xis the tangent line at O and the axis y is the line passing through O and throughthe (improper) centre of C;

(c) the model (a5) is a real cone, (a6) is an imaginary cone, (a7) is a real cylinder,(a8) is an imaginary cylinder; the doubly degenerate model (a9) is a cone forwhich all points of the support are vertices.

An affine line r is called an asymptote of a non-degenerate conic C if its projectiveclosure r is tangent to C at one of its improper points (that is, if r is the polar ofan improper point of the conic with respect to C). A real ellipse has no asymptotesbecause it has no improper points, a parabola has no asymptotes because the tangent atthe unique point at infinity is the line at infinity,while a hyperbola has two asymptotes.The asymptotes of a hyperbola, being the affine parts of the polars of the improperpoints, are diameters and they meet in the centre of the hyperbola.

By using the convention previously adopted (cf. Sect. 1.8.5) of choosing equations

such that i+(A) ≥ i−(A) and i+(A) ≥ i−(A) and of denoting X =⎛⎝1xy

⎞⎠, we know

that two conics of R2 of equations tXAX = 0 and tXA′X = 0 are affinely equivalent

if and only if sign(A) = sign(A′) and sign(A) = sign(A′).Table1.1 shows the values of these invariants for the affine models listed in The-

orem 1.8.7.We call circle every sphere of R

2, that is every conic of equation (x− x0)2 + (y−y0)2 = η. This circle is a real ellipse if η > 0 and an imaginary ellipse if η < 0; ifη = 0 it is a degenerate conic whose support consists of only one point. The image

1.8 Quadrics 47

of a circle under an isometry is a circle; in particular being a circle is an invariantproperty under isometric coordinate changes.

The complexification of a circle has I1 = [0, 1, i] and I2 = [0, 1,−i] as improperpoints; they are called cyclic points of the Euclidean plane. One can easily check thatany conic of R

2 whose complexification has the cyclic points as improper points isa circle.

The cyclic points are closely related to metric properties. For instance by associ-ating to every line through a point P ∈ R

2 the orthogonal line through P we definean involution in the pencil FP of lines through P. In an affine coordinate systemwhere P = (0, 0) we thus associate the line of equation bx + ay = 0 to the line ofequation ax − by = 0. Since the projective closures of these two lines meet the lineat infinity x0 = 0 at the points [0, b, a] and [0, a,−b], respectively, the involutiongiven by the orthogonality relation in FP induces the so-called absolute involution[0, b, a] �→ [0, a,−b] on the improper line. This latter map has no real fixed points,while the corresponding involution of (H0)C (which is projectively isomorphic toP1(C)) has the cyclic points I1 and I2 as fixed points.The name “absolute involution” originates from the relationship between this

involution and the polarity with respect to the quadric ofH0 of equation x21 + x22 = 0,classically called the absolute conic. Namely this polarity associates to the point[b, a] the hyperplane ofH0 of equation bx1 + ax2 = 0, whose support consists of thepoint [a,−b] only.

A line of C2 having a cyclic point as improper point is called an isotropic line;

exactly two isotropic lines pass through every point of C2 (and hence through every

point of R2).

A point F of R2 is called a focus of a non-degenerate conic C if the projec-

tive closures of the isotropic lines passing through F are tangent to the projectiveclosure of the complexification of C. A non-degenerate conic has one or two foci(cf. Exercise 194).

A line of R2 is called a directrix of C if its projective closure is the polar line of a

focus with respect to C.In the case of R

2, restricting our attention only to non-degenerate conics,Theorem 1.8.4 yields:

Theorem 1.8.8 (Metric classification of conics of R2) Every non-degenerate conic

of R2(x,y) is metrically equivalent to exactly one among the following conics:

(m1) x2

a2+ y2

b2− 1 = 0 with a ≥ b > 0 (real ellipse);

(m2) x2

a2+ y2

b2+ 1 = 0 with a ≥ b > 0 (imaginary ellipse);

(m3) x2

a2− y2

b2+ 1 = 0 with a > 0, b > 0 (hyperbola);

(m4) x2 − 2cy = 0 with c > 0 (parabola).

Using the canonical equations of non-degenerate conics listed in Theorem 1.8.8,we can compute explicit formulas for vertices, foci, axes and directrices of the non-degenerate metric models listed above (cf. Exercise 196).

48 1 Theory Review

Starting from the canonical metric equations it is also easy to check (cf. Exer-cise 197) that non-degenerate conics can be characterized in terms of metric condi-tions, as we already did for the circle:

(a) a parabola is the locus of points of the plane such that the distance to a givenpoint (the focus) equals the distance to a given line not passing through that point(the directrix);

(b) an ellipse is the locus of points of the plane such that the sum of the distances totwo given distinct points of the plane (the foci) is constant;

(c) a hyperbola is the locus of points of the plane such that the absolute value of thedifference of the distances to two given distinct points of the plane (the foci) isconstant.

For another metric characterization of non-degenerate conics see Exercise 198 andthe Remark following it.

1.8.8 Quadrics of R3

Recall that a quadricQ ofR3 is non-degenerate if and only if a matrix A representing

it is invertible. Moreover, Q is reducible if and only if A has rank 1 or 2.If rk A = 4, Q can be either a paraboloid (if it has no centre, i.e. if H0 is tangent

to Q), or a hyperboloid (if Q has a centre and H0 is secant to Q) or an ellipsoid(if Q has a centre and H0 is external to Q). We can further refine this terminologyaccording to the nature of points ofQ (recall that points of non-degenerate quadricscan be only elliptic or hyperbolic, cf. Sect. 1.8.4). Hence we speak of

(a) elliptic hyperboloid, or two-sheeted hyperboloid, if all its points are elliptic;(b) hyperbolic hyperboloid, or one-sheeted hyperboloid, if all its points are hyper-

bolic;(c) elliptic paraboloid if all its points are elliptic;(d) hyperbolic paraboloid, or saddle, if all its points are hyperbolic.

Note that all points of an ellipsoid Q are necessarily elliptic: if there were ahyperbolic point P for Q, then Q ∩ TP(Q) would be the union of two lines so thatQ ∩ H0 would contain real points (cf. Exercise 175 too).

We also say that an ellipsoid is real (resp. imaginary) if its support is non-empty(resp. empty).

If rk A = 3, we know that the projective closure Q is a cone having only onesingular point S. Therefore the quadric Q is either an affine cone (if S is a properpoint, which happens if and only if det A �= 0) or an affine cylinder (if S is an improperpoint, which happens if and only if det A = 0). More precisely, we speak of a realcone and of a real cylinder if the support is a union of lines, of an imaginary cone ifthe support only contains the unique singular point, of an imaginary cylinder if thesupport is empty. For a real cylinder, the terminology is sometimes further refinedaccording to the intersection ofQ with the plane H0. Thus ifQ is a real cylinder, we

1.8 Quadrics 49

say thatQ is a hyperbolic cylinder ifQ∩H0 is a pair of distinct real lines, an ellipticcylinder ifQ∩H0 is a pair of complex conjugate lines, a parabolic cylinder ifQ∩H0

is a line counted twice. Note how these names have nothing to do with the natureof points on the quadric since all non-singular points on a cylinder are parabolic,the quadric being degenerate and irreducible (cf. Exercise 173). The terminologyactually refers to the fact that non-degenerate plane sections are hyperbolas, ellipsesand parabolas, respectively.

In R3 the classification of Theorem 1.8.5 translates into:

Theorem 1.8.9 (Affine classification of quadrics of R3) Every quadric of R

3(x,y,z)

is affinely equivalent to exactly one among the following quadrics:(a1) x2 + y2 + z2 − 1 = 0 (real ellipsoid);(a2) x2 + y2 + z2 + 1 = 0 (imaginary ellipsoid);(a3) x2 + y2 − z2 − 1 = 0 (hyperbolic hyperboloid);(a4) x2 + y2 − z2 + 1 = 0 (elliptic hyperboloid);(a5) x2 + y2 − 2z = 0 (elliptic paraboloid);(a6) x2 − y2 − 2z = 0 (hyperbolic paraboloid);(a7) x2 + y2 + z2 = 0 (imaginary cone);(a8) x2 + y2 − z2 = 0 (real cone);(a9) x2 + y2 + 1 = 0 (imaginary cylinder);(a10) x2 + y2 − 1 = 0 (elliptic cylinder);(a11) x2 − y2 + 1 = 0 (hyperbolic cylinder);(a12) x2 − 2z = 0 (parabolic cylinder);(a13) x2 + y2 = 0 (pair of incident complex planes);(a14) x2 − y2 = 0 (pair of distinct incident real planes);(a15) x2 + 1 = 0 (pair of parallel complex planes);(a16) x2 − 1 = 0 (pair of distinct parallel real planes);(a17) x2 = 0 (pair of coincident real planes).

As already observed (cf. Sect. 1.8.5), by the convention of choosing equations suchthat i+(A) ≥ i−(A) and i+(A) ≥ i−(A), two quadrics of R

3 of equations tXAX =0 and tXA′X = 0 are affinely equivalent if and only if sign(A) = sign(A′) andsign(A) = sign(A′). Useful, although partial, information can be obtained from twoadditional affine invariants, such as the vanishing of the determinant of A and thesign of the determinant of A (since A is a matrix of even order).

Table1.2 shows the values of these invariants for the affine models listed in The-orem 1.8.9. The pictures of some of these models appear in Fig. 1.1.

When considering only non-degenerate quadrics, in order to distinguish the affinetype of Q it suffices to compute the determinants of A and A and the signature of A.In fact,

(a) Q is a real ellipsoid ⇐⇒ det A �= 0, det A < 0 and A is positive definite;(b) Q is an imaginary ellipsoid ⇐⇒ det A �= 0, det A > 0 andA is positive definite;

50 1 Theory Review

Fig. 1.1 Some quadrics listed in Theorem 1.8.9

(c) Q is a hyperbolic hyperboloid ⇐⇒ det A �= 0, det A > 0 and A is indefinite;(d) Q is an elliptic hyperboloid ⇐⇒ det A �= 0, det A < 0 and A is indefinite;(e) Q is an elliptic paraboloid ⇐⇒ det A = 0 and det A < 0;(f) Q is a hyperbolic paraboloid ⇐⇒ det A = 0 and det A > 0.

In the case ofR3 the classification obtained in Theorem 1.8.4 yields metric canon-

ical equations for quadrics of R3.

1.9 Plane Algebraic Curves 51

Table 1.2 Affine invariants for quadrics of R3

Model det(A) det(A) sign(A) sign(A)

(a1) �=0 <0 (3, 0) (3, 1)

(a2) �=0 >0 (3, 0) (4, 0)

(a3) �=0 >0 (2, 1) (2, 2)

(a4) �=0 <0 (2, 1) (3, 1)

(a5) 0 <0 (2, 0) (3, 1)

(a6) 0 >0 (1, 1) (2, 2)

(a7) �=0 0 (3, 0) (3, 0)

(a8) �=0 0 (2, 1) (2, 1)

(a9) 0 0 (2, 0) (3, 0)

(a10) 0 0 (2, 0) (2, 1)

(a11) 0 0 (1, 1) (2, 1)

(a12) 0 0 (1, 0) (2, 1)

(a13) 0 0 (2, 0) (2, 0)

(a14) 0 0 (1, 1) (1, 1)

(a15) 0 0 (1, 0) (2, 0)

(a16) 0 0 (1, 0) (1, 1)

(a17) 0 0 (1, 0) (1, 0)

1.9 Plane Algebraic Curves

The definitions and considerations about affine and projective hypersurfaces pre-sented in Sect. 1.7 apply in particular to hypersurfaces of the plane (i.e. to the casen = 2), namely to plane algebraic curves. These will sometimes be called simply“curves” from now on. In particular we have the notions of projective closure of anaffine curve, affine and projective equivalence of curves, multiplicity of a point of acurve, tangent line, singular curve, and so on.

Therefore, in this section we just highlight some results about the local study ofa curve at one of its points and the intersection of two curves.

1.9.1 Local Study of a Plane Algebraic Curve

Let C be a projective curve of degree d of P2(K), where K = C or K = R, and let

P be a point of P2(K).

If F(x0, x1, x2) = 0 is a homogeneous equation of C, then P is a simple pointof the curve if and only if not every first-order partial derivative of F vanishes at P(cf. Sect. 1.7.8). In this case the tangent line to C at P has equation

52 1 Theory Review

Fx0(P)x0 + Fx1(P)x1 + Fx2(P)x2 = 0.

A singular point P ∈ C (i.e. a point such that mP(C) > 1) is called a double point ifmP(C) = 2, a triple point if mP(C) = 3, an m-tuple point if mP(C) = m. The point Pis an m-tuple point of C if and only if all (m − 1)th derivatives of F vanish at P andthere is at least one mth derivative of F that does not vanish at P.

All points of a multiple irreducible component turn out to be singular. If C is areduced curve and C1, . . . , Cm are its irreducible components, then Sing(C) is a finiteset and, more precisely (cf. Exercises 54 and 56),

Sing(C) =m⋃j=1

Sing(Cj) ∪⋃i �=j

(Ci ∩ Cj).

A line r is called a principal tangent to C at P if I(C, r,P) > mP(C).At a simple point, the notions of tangent line and of principal tangent coincide.

If P is a singular point of multiplicity m, any line through P is a tangent, while theprincipal tangents are those contained in the tangent cone to C at P (cf. Sect. 1.7.8).The set of principal tangents to a curve C +D at P consists of the principal tangentsto C at P together with the principal tangents to D at P (cf. Exercise 53).

In order to study the curve locally at a point P, it can be useful to choose a chartin which P = (0, 0). In such a chart, if f is a polynomial that defines the affine partof C and if we write f as a sum of homogeneous terms, the multiplicity at the origincoincides with the degree m of the non-zero homogeneous part fm of f of minimaldegree. So the affine parts of the principal tangents at P are defined by the linearfactors of fm. In particular, there are at most m = mP(C) principal tangents at P.More precisely, if K = C by Theorem 1.7.1 there are m principal tangents (countedwith multiplicity); if K = R, instead, by Theorem 1.7.2 there are at mostm principaltangents, but there may be none. For example, the point [1, 0, 0] is a double pointof the real curve of equation x0x21 + x0x22 − x31 = 0 at which there are no principaltangents.

If K = C, the singular point P is said to be ordinary if there are precisely mP(C)

distinct principal tangents at P. A double point is called a node if it is ordinary, acusp if it is non-ordinary; more precisely, one speaks of an ordinary cusp if the onlyprincipal tangent at P has multiplicity of intersection exactly 3 with the curve at thepoint.

If K = R, a singular point P is said to be ordinary if it is ordinary for thecomplexified curve CC; in that case, since mP(CC) = mP(C), the complexified curveCC has mP(C) distinct principal tangent lines at P but it is possible that the numberof the real ones is strictly smaller than mP(C).

If, in some affine chart, the affine part of C has equation f (x, y) = 0 andP = (a, b)is non-singular, the line of equation

fx(P)(x − a) + fy(P)(y − b) = 0

1.9 Plane Algebraic Curves 53

is the affine part of the (principal) tangent to C at P and is called the affine tangentat P.

Finally, generalizing the notion already introduced for non-degenerate conics, anaffine line r is called an asymptote for an affine curve D if its projective closure r isa principal tangent to D at one of its improper points.

1.9.2 The Resultant of Two Polynomials

Let D be a unique factorization domain (we are interested in the cases D = K andD = K[x1, . . . , xn]). Let us consider two polynomials f , g ∈ D[x] of positive degreesm and p, respectively,

f (x) = a0 + a1x + · · · + amxm am �= 0

g(x) = b0 + b1x + · · · + bpxp bp �= 0.

We call Sylvester matrix of f and g the square matrix of order m + p

S(f , g) =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

a0 a1 . . . . . . am 0 . . . 00 a0 a1 . . . . . . am 0 . . .

. . .. . .

. . .

0 . . . 0 a0 a1 . . . . . . amb0 b1 . . . . . . bp 0 . . . 00 b0 b1 . . . . . . bp 0 . . .

. . .. . .

. . .

0 . . . 0 b0 b1 . . . . . . bp

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

whose first p lines are determined by the coefficients of f , while the following onesby the coefficients of g. The determinant of S(f , g) is called the resultant of f andg; in symbols, Ris(f , g) = det S(f , g).

The main property of the resultant is that f and g have a common factor of positivedegree in D[x] if and only if Ris(f , g) = 0. If D = C, then Ris(f , g) = 0 if and onlyif f and g have a common root.

If f , g ∈ K[x1, . . . , xn], we may choose an indeterminate, say xn, and regard f andg as polynomials in xn whose coefficients are polynomials in x1, . . . , xn−1, that is,f , g ∈ D[xn] with D = K[x1, . . . , xn−1]. Denote by S(f , g, xn) the Sylvester matrixand by Ris(f , g, xn) the resultant of f and g regarded as elements of D[xn], so thatRis(f , g, xn) ∈ K[x1, . . . , xn−1]. In addition, we denote by degxn f the degree of fwith respect to the variable xn.

Among the many important properties of the resultant, we recall only those thatare essential for our purposes.

54 1 Theory Review

Specialization property: Setting X = (x1, . . . , xn−1), let

f (x1, . . . , xn) = a0(X) + a1(X)xn + · · · + am(X)xmn am(X) �= 0

g(x1, . . . , xn) = b0(X) + b1(X)xn + · · · + bp(X)xpn bp(X) �= 0

and let R(x1, . . . , xn−1) = Ris(f , g, xn).

For any c = (c1, . . . , cn−1) ∈ Kn−1, one can alternatively:

(a) evaluate the polynomials f and g at X = c first, thus obtaining the polynomialsin one variable fc(xn) = f (c, xn) and gc(xn) = g(c, xn), and then compute theresultant of fc(xn), gc(xn) ∈ K[xn];

(b) compute the resultant R(x1, . . . , xn−1) first, and then evaluate it at X = c.

If for instance am(c) �= 0 and bp(c) �= 0, then degxn f (x1, . . . , xn) = deg fc(xn)and degxn g(x1, . . . , xn) = deg gc(xn), so that the Sylvester matrix of fc(xn) andgc(xn) coincides with the matrix S(f , g, xn) evaluated at X = c. As a consequence,

Ris(f (c, xn), g(c, xn)) = R(c),

which is to say that specializing at X = c commutes with computing the resultant.

Homogeneity property: LetF(x1, . . . , xn) andG(x1, . . . , xn) be homogeneous poly-nomials of degrees m and p, respectively. Set X = (x1, . . . , xn−1) and let

F(x1, . . . , xn) = A0(X) + A1(X)xn + · · · + Amxmn

G(x1, . . . , xn) = B0(X) + B1(X)xn + · · · + Bpxpn,

where every non-zero Ai (resp. Bi) is a homogeneous polynomial of degree m − i(resp. p − i) in K[x1, . . . , xn−1].

If Am �= 0 and Bp �= 0, the polynomial Ris(F,G, xn) is either homogeneous ofdegree mp in the variables x1, . . . , xn−1 or zero.

1.9.3 Intersection of Two Curves

One can generalize the notion of multiplicity of intersection of two curves at a point,which was previously definedwhen one of the curves is a line. There are several waysof defining this concept; below we shall recall the one that suits better the effectivecomputation of the multiplicity.

Let C and D be two projective curves of P2(K) of degrees m and d, respectively,

and without common components, and let P ∈ C ∩ D. Choose a homogeneouscoordinate system [x0, x1, x2] such that [0, 0, 1] /∈ C ∪D and such that P is the onlypoint of C ∩ D on the line joining P and [0, 0, 1].

1.9 Plane Algebraic Curves 55

If C = [F] and D = [G], denote by Ris(F,G, x2) the resultant of F and Gwith respect to the variable x2. Since F(0, 0, 1) �= 0 and G(0, 0, 1) �= 0, one hasdegx2 F = degF, degx2 G = degG; in addition, since C and D have no commoncomponent, the polynomial R(x0, x1) = Ris(F,G, x2) is homogeneous of degreemd(cf. Sect. 1.9.2, homogeneity property).

If P = [c0, c1, c2], the multiplicity of [c0, c1] as root of the polynomialRis(F,G, x2) is called the multiplicity of intersection of the curves C and D at Pand we denote it by I(C,D,P).

One can check either directly or indirectly, in any case with some effort, that thisdefinition does not depend on the homogeneous coordinate system. One can alsoverify that, in case one of the curves is a line, the multiplicity of intersection justdefined coincides with the one defined earlier.

In case C and D are curves in P2(R) and P is a point of P

2(R), it is apparent thatI(C,D,P) = I(CC,DC,P).

Let f and g be the polynomials obtained by dehomogenizing F andGwith respectto x0, which define the affine parts of the curves C and D in the chart U0. Since[0, 0, 1] /∈ C ∪ D, specializing to x0 = 1 does not lower the degrees of F and Gand, by the specialization property of the resultant, one has Ris(F,G, x2)(1, x1) =Ris(f , g, x2)(x1); hence the multiplicity of intersection of C and D at P can also becomputed from the equations of their affine parts.

Theorem 1.9.1 Let C and D be projective curves of P2(K) without common com-

ponents. ThenI(C,D,P) ≥ mP(C) · mP(D)

for every point P ∈ P2(K).

See Exercise 110 for a proof of Theorem 1.9.1.We say that the curves C and D are tangent at P if I(C,D,P) ≥ 2. Note that

the notion of curves tangent at a point can be characterized in terms of the tangentlines to the curves at the point, namely I(C,D,P) ≥ 2 if and only if the curves haveat least one common tangent at P. Indeed, if the two curves are not singular at P,then I(C,D,P) ≥ 2 if and only if the tangent lines to the two curves at P coincide(cf. Exercise 109). In case at least one of the curves, say C, is singular at P, then alllines through P are tangent to C and at least one of them is tangent also to D; in factin this case I(C,D,P) ≥ mP(C) · mP(D) ≥ 2.

Concerning the global intersection of two projective curves, we have the followingfundamental result:

Theorem 1.9.2 (Bézout’s Theorem) Let C and D be projective curves of P2(C)

of degrees m and d, respectively. If C and D have no common component, theyhave exactly md common points, counted with the corresponding multiplicity ofintersection (in particular C ∩ D �= ∅).

In the real case Bézout’s Theorem does not hold, and it may even happen that twoprojective curves of P

2(R) do not intersect at all (just think of the two distinct real

56 1 Theory Review

conics of equations x20 + x21 − x22 = 0 and x20 + x21 − 2x22 = 0). However one obtainsthe following “weak” form of Bézout’s Theorem as an immediate consequence ofTheorem 1.9.2:

Theorem 1.9.3 (real Bézout’s Theorem) Let C andD be projective curves of P2(R)

of degrees m and d, respectively. If C andD have more than md common points, theynecessarily have a common irreducible component.

1.9.4 Inflection Points

A point P of a projective curve C is called an inflection point, or a flex, if it issimple and I(C, τ ,P) ≥ 3, where τ denotes the (principal) tangent to C at P. In caseI(C, τ ,P) = 3 one has an ordinary inflection point. For example, all points of a lineare inflection points, irreducible conics have no inflection points and all inflectionpoints of an irreducible cubic are ordinary.

Inflection points on a projective curve C of equation F(X) = 0 are characterized(and hence can be determined) as the non-singular points P such that HF(P) = 0,whereHF(X) = det(Fxixj (X)). If F has degree d ≥ 2, the polynomialHF(X) is eitherhomogeneous of degree 3(d−2) or identically zero (for example ifF = x1x2(x1−x2),then HF(X) = 0).

The symmetric matrix HessF(X) = (Fxixj (X))0≤i,j≤2 is called the Hessian matrixof F and, if HF(X) = det(HessF(X)) is non-zero, the curve of equation HF(X) = 0is called the Hessian curve of F(X) = 0 and is denoted by H(C).

So, when the field is C, the number of inflection points of a projective curveof degree d ≥ 3 is either infinite or, by Bézout’s Theorem, at most 3d(d − 2); ifnon-singular, such a curve must anyhow have at least one inflection point.

We recall that, ifC is a curveof equationF(X) = 0 andg(X) = MX is a projectivityof P

2(K), then the curve D = g−1(C) has equation G(X) = F(MX) = 0. Onechecks that the Hessian matrices of F and G are related by the equality HessG(X) =tM HessF(MX)M, so that HG(X) = (detM)2HF(MX). Therefore, the polynomialHG(X) is non-zero if and only if HF(X) is non-zero and, if so, H(D) = g−1(H(C)).

1.9.5 Linear Systems, Pencils

For every d ≥ 1 we denote by Λd the set of projective curves of degree d ofP2(K). From the definition of curve it follows that Λd = P(K[X]d), where K[X]d =

K[x0, x1, x2]d denotes the vector space consisting of the zero polynomial and allhomogeneous polynomials of degree d in x0, x1, x2. A basis of this vector spaceconsists of themonomials xi0x

j1x

d−i−j2 of degree d, so thatK[x0, x1, x2]d has dimension

(d + 1)(d + 2)2 . As a consequence, Λd is a projective space of dimension

1.9 Plane Algebraic Curves 57

N = (d + 1)(d + 2)

2− 1 = d(d + 3)

2.

Hence lines form a projective space of dimension 2, conics a space of dimension 5,cubics of dimension 9 and so on. In the homogeneous coordinate system induced bythe basis of monomials, the coordinates of a curve C ∈ Λd are the coefficients of oneof the polynomials defining it.

A projective subspace ofΛd is called a linear system of curves of degree d; a linearsystemof curves of dimension 1 is called a pencil. A linear systemof curves of dimen-sion r can be represented parametrically starting with r + 1 independent points in it,[F0], . . . , [Fr], so every curve of the linear system has equation

∑ri=0 λiFi(X) = 0.

Alternatively, it can be represented in Cartesian form as an intersection of hyper-planes.

Points belonging to all curves of a linear system are called base points of thelinear system. For pencils of curves the base points are determined by intersectingany two distinct curves F0(X) = 0 and F1(X) = 0 of the pencil, since every curve ofthe pencil has an equation of the form λF0(X) + μF1(X) = 0 with [λ,μ] ∈ P

1(K).IfQ is not a base point of the pencil, there exists a unique curve of the pencil passingthroughQ, because by imposing that a curve pass throughQ one obtains an equationin λ,μ that has a unique solution [λ0,μ0] ∈ P

1(K).If two curves of degree d are tangent to the same line l at a point P, then all curves

of the pencil spanned by the two curves are tangent to l at P.Finally we recall that, if f is a projectivity of P

2(K) and C is a curve of degreed, then the curve f (C) has also degree d, so that f induces a map Λd → Λd that isin fact a projectivity. In particular, f transforms every linear system of curves into alinear system of the same dimension; for instance, f transforms the pencil generatedby two curves C1 and C2 of degree d into the pencil generated by the curves f (C1)and f (C2).

1.9.6 Linear Conditions

The equation of a hyperplane of Λd is a homogeneous linear equation in the coordi-nates of Λd , namely in the coefficients of the generic curve of Λd . Such an equationis called a linear condition on the curves of Λd .

For instance, imposing that a curve pass through a given point P is a linear con-dition. Since N hyperplanes of P

N (K) have always at least one point in common,

there is always at least one curve of degree d that satisfies up toN = d(d + 3)2 linear

conditions.In particular, since dimΛ2 = 5, there is at least one conic passing through 5

given points in P2(K). And if no 4 of the 5 points are collinear, there is exactly one

such conic. In fact, two distinct conics in P2(K) either have no common irreducible

component or they do. In the former case Bézout’s Theorem (cf. Theorem 1.9.2 orTheorem 1.9.3) tells that they meet at 4 distinct points at most. In the latter case

58 1 Theory Review

(both conics are degenerate) they share a line plus a point not on the line. This meanstwo conics can meet at 5 points, no 4 of which are collinear, only if they coincide.Hence the system of conics passing through 4 non-collinear points is a pencil. If the5 points satisfy the stronger condition that no 3 are collinear, then the only conicthrough them is necessarily non-degenerate.

Imposing that a point P have multiplicity at least r ≥ 1 for a curve of Λd isequivalent to imposing that all derivatives of order r − 1 of a polynomial defining

the curve vanish at P. This corresponds to imposing r(r + 1)2 linear conditions.

Of course, by imposing k linear conditions one obtains in general a linear systemof dimension greater than or equal toN−k; the dimension is exactlyN−k only if thelinear conditions that one imposes are independent. For example, the linear systemof projective conics that pass through 4 points lying on the same line r consists ofthe reducible conics that are the union of the line r and of another line; this set is inone-to-one correspondence with the space of lines of the projective plane and is alinear system of dimension 2 (cf. Exercise 96).

Other examples of linear conditions arise from tangency conditions at given pointsto the curves of Λd . For example, if r is a line of equation tRX = 0 and P ∈ r, thena curve C of Λd of equation F(X) = 0 is tangent to r at P if and only if the vectorsR = (r0, r1, r2) and (Fx0(P),Fx1(P),Fx2(P)) are proportional, i.e. if and only if thematrix

M =(

r0 r1 r2Fx0(P) Fx1(P) Fx2(P)

)

has rank 1. This is equivalent to the vanishing of the determinants of two 2 × 2submatrices of M, and therefore to two independent linear conditions (cf. Exer-cise 95). Note that if M has rank 1, and hence there exists k ∈ K such that(Fx0(P),Fx1(P),Fx2(P)) = k(r0, r1, r2), then automatically C passes through thepoint P = [p0, p1, p2] since

(degF)F(P) = p0Fx0(P) + p1Fx1(P) + p2Fx2(P) = k(p0r0 + p1r1 + p2r2) = 0.

A case in which the two linear conditions are satisfied is when the second row ofM is zero, namely when C is singular at P; this agrees with the fact that every linepassing through a singular point of a curve is tangent to the curve at that point.

1.9.7 Pencils of Conics

A pencil of conics is a linear system of conics of dimension 1. If C1 = [F1] and C2 =[F2] are two distinct conics of the pencil, then the conics of the pencil have equationλF1+μF2 = 0 as [λ,μ] varies in P

1(K). If F1(X) = tXA1X and F2(X) = tXA2X withA1,A2 symmetric 3 × 3 matrices, the generic conic Cλ,μ of the pencil has equationtX(λA1 + μA2)X = 0.

1.9 Plane Algebraic Curves 59

The conic Cλ,μ is degenerate if and only if D(λ,μ) = det(λA1 + μA2) = 0.Therefore, if the homogeneous polynomial of degree three D(λ,μ) is non-zero, thepencil contains three degenerate conics at most. If D(λ,μ) is the zero polynomial,instead, all conics of the pencil are degenerate; this happens, for instance, when theconics of the pencil consist of a given line and an arbitrary other line passing througha given point.

However there exist pencils all of whose conics are degenerate without having acommon component: just think of the pencil spanned by x20 = 0 and by x21 = 0.

Recall that the base points of a pencil, i.e. the points common to all conics of thepencil, can be determined by intersecting any two distinct conics C1, C2 of the pencil.Since every pencil contains at least one degenerate conic, we may assume that oneof the conics spanning the pencil is degenerate, and therefore is the sum of two lines;this simplifies the computation of the intersection points of C1 and C2.

For instance, let us consider the pencil of conics passing through 4 non-collinearpoints (cf. Sect. 1.9.6): if the 4 points are in general position, then the set of basepoints of the pencil consists exactly of the 4 points; if 3 of the 4 points belong to aline r, then the set of base points is the union of the line r and of the fourth point notlying on r (in particular, it is an infinite set).

As remarked above, it is possible that the locus of base points of a pencil isan infinite set (when all conics of the pencil have a common component) and inthis case all conics of the pencil are degenerate. If the pencil contains at least onenon-degenerate conic, then the set of base points is finite.

We now describe the types of pencils of conics containing at least one non-degenerate conic: let F be such a pencil and let C1 and C2 be two distinct conics ofF , with C1 degenerate.

If K = C then the conics C1 and C2 intersect at 4, possibly non-distinct, points.We will denote the quadruple of intersection points by writing any point P as manytimes as the value of I(C1, C2,P). Hence, according to the number of distinct basepoints and their multiplicities, we obtain the following types of pencils containing atleast one non-degenerate conic:

(a) C1 ∩ C2 = {A,B,C,D}: the points A,B,C,D turn out to be in general position;the pencil contains 3 degenerate conics, namely L(A,B) + L(C,D), L(A,C) +L(B,D) and L(A,D) + L(B,C);

(b) C1∩C2 = {A,A,B,C}: in this case the points A,B,C are not collinear; C1 and C2(hence all conics of the pencil) are tangent at A to a line tA that contains neitherB nor C; the pencil contains 2 degenerate conics, namely L(A,B)+L(A,C) andtA + L(B,C) (see also Exercise 121);

(c) C1 ∩ C2 = {A,A,B,B}: all conics of the pencil are tangent at A to a line tA notpassing through B and at B to a line tB not passing through A; the pencil contains2 degenerate conics, namely tA + tB and 2L(A,B) (see also Exercise 125);

(d) C1 ∩ C2 = {A,A,A,B}: in this case C1 and C2 intersect at A with multiplicity ofintersection 3 and are tangent at A to a line tA not passing through B; the pencilcontains only one degenerate conic, namely tA + L(A,B);

60 1 Theory Review

(e) C1 ∩ C2 = {A,A,A,A}: this case occurs when C1 and C2 intersect at A withmultiplicity of intersection 4 and are tangent to a line tA; the pencil contains onlyone degenerate (actually, doubly degenerate) conic: 2tA.

If K = R, it is possible that the intersection points of the conics C1 and C2, evenif counted with multiplicity, are fewer than 4, and possibly none (just think, forexample, of the conics of equations x2 + y2 = 1 and x2 − 4 = 0). Another case inwhich there are no base points is when C1 is a conic whose support consists of just onepoint P (this happens when C1 is irreducible and (C1)C has two complex conjugatelines meeting at P as irreducible components) and C2 does not pass through P.

On the other hand the complexified curves (C1)C and (C2)C do intersect at four,possibly not distinct, points, that turn out to be the complex base points of F . Since(C1)C and (C2)C are real curves, if A = [a0, a1, a2] is a complex base point of thepencil, then the point σ(A) = [a0, a1, a2] is a base point as well. With regard to this,note that the line L(A,σ(A)) admits an equation with real coefficients, i.e. it is a realline. If instead B and C are any two points of P

2(C) (so in general L(B,C) does notadmit a real equation), then an equation of the line L(σ(B),σ(C)) is obtained froman equation of L(B,C) by conjugating its coefficients; as a consequence, the conicL(B,C) + L(σ(B),σ(C)) is a real conic.

Therefore, in the real case, besides the five types of pencils containing at least onenon-degenerate conic listed above, there exist also the following additional types:

(f) (C1)C ∩ (C2)C = {A,σ(A),B,C} where A �= σ(A) and B,C are real points:the pencil F has two distinct real base points and contains only one degenerateconic, namely L(A,σ(A)) + L(B,C);

(g) (C1)C ∩ (C2)C = {A,σ(A),B,B} where A �= σ(A) and B is a real point: thepencil has only one real base point, C1 and C2 (hence all conics of the pencil) aretangent at B to a real line tB; the pencil contains two degenerate conics, namelyL(A,B) + L(σ(A),B) and L(A,σ(A)) + tB;

(h) (C1)C ∩ (C2)C = {A,σ(A),B,σ(B)} where A �= σ(A) and B �= σ(B): the pencilhas no real base point and contains three degenerate conics; one of them, namelyL(A,σ(A)) + L(B,σ(B)), is a pair of lines; the other two, namely L(A,σ(B)) +L(σ(A),B) and L(A,B)+L(σ(A),σ(B)), are conics having only one point eachas support;

(i) (C1)C ∩ (C2)C = {A,σ(A),A,σ(A)} where A �= σ(A): all complexifications ofthe conics of the pencil are tangent at A and σ(A), respectively, to two complexconjugate lines tA and tσ(A); the pencil contains two degenerate conics, namelytA + tσ(A) and 2L(A,σ(A)).

Chapter 2Exercises on Projective Spaces

Projective spaces and projective subspaces. Projective framesand homogeneous coordinates. Projective transformations andprojectivities. Linear systems of hyperplanes and duality. Theprojective line. Cross-ratio.

Abstract Solved problems on projective spaces and subspaces, projective transfor-mations, the projective line and the cross-ratio, linear systems of hyperplanes andduality.

Notation: Throughout the whole chapter, the symbol K denotes a subfield of C.

Exercise 1. Show that the points

[1

2, 1, 1

],

[1,

1

3,4

3

], [2,−1, 2]

of the real projective plane are collinear, and find an equation of the line containingthem.

Solution. The points[12 , 1, 1

],[1, 13 , 43

]are distinct, so the point [x0, x1, x2] is

collinearwith them if and only if the vectors (1, 2, 2), (3, 1, 4), (x0, x1, x2) are linearlydependent, i.e., if and only if

0 = det

⎛⎝1 3 x02 1 x12 4 x2

⎞⎠ = 6x0 + 2x1 − 5x2.

Therefore, an equation of the line containing[12 , 1, 1

],[1, 13 , 43

]is given by 6x0 +

2x1 − 5x2 = 0, and this equation is satisfied by the point [2,−1, 2].

© Springer International Publishing Switzerland 2016E. Fortuna et al., Projective Geometry, UNITEXT - La Matematica per il 3+2 104,DOI 10.1007/978-3-319-42824-6_2

61

62 2 Exercises on Projective Spaces

Exercise 2. Find the values a ∈ C for which the lines of equations

ax1 − x2 + 3ix0 = 0, −iax0 + x1 − ix2 = 0, 3ix2 + 5x0 + x1 = 0

of P2(C) are concurrent.

Solution (1). If

A =⎛⎝

3i a −1−ia 1 −i5 1 3i

⎞⎠ ,

then the given lines all intersect if and only if the homogeneous linear systemAX = 0admits a non-trivial solution. This occurs if and only if 0 = det A = −3a2 − 4ia − 7,i.e., if and only if either a = i or a = −7

3 i.

Solution (2). Via the duality correspondence (see Sect. 1.4.2), the given lines deter-mine three points in the space P

2(C)∗, and the coordinates of these points withrespect to the frame induced by the standard basis of (C2)∗ are given by [3i, a,−1],[−ia, 1,−i], [5, 1, 3i]. An easy application of the Duality principle shows that thesepoints are collinear if and only if the given lines are concurrent. Finally, the points[3i, a,−1], [−ia, 1,−i], [5, 1, 3i] are collinear if and only if the determinant of thematrix A introduced above vanishes (see Exercise1).

Exercise 3. In P3(R) consider the points

P1 = [1, 0, 1, 2], P2 = [0, 1, 1, 1], P3 = [2, 1, 2, 2], P4 = [1, 1, 2, 3].

(a) Determine whether P1,P2,P3,P4 are in general position.(b) Compute the dimension of the subspace L(P1,P2,P3,P4), and find Cartesian

equations of L(P1,P2,P3,P4).(c) If possible, complete the set {P1,P2,P3} to a projective frame of P

3(R).

Solution. (a) Let v1 = (1, 0, 1, 2), v2 = (0, 1, 1, 1), v3 = (2, 1, 2, 2), v4 =(1, 1, 2, 3) be vectors in R

4 such that Pi = [vi] for every i, and set

A =

⎛⎜⎜⎝1 0 2 10 1 1 11 1 2 22 1 2 3

⎞⎟⎟⎠ .

It is easy to check that det A = 0, so v1, v2, v3, v4 are linearly dependent. Therefore,the points P1,P2,P3,P4 are not in general position.

(b) The determinant of the submatrix given by the first three lines and the firstthree columns of A is equal to −1, so v1, v2, v3 are linearly independent. Therefore,by point (a) the dimension of the linear subspace spanned by v1, . . . , v4 is equal to 3,so L(P1,P2,P3,P4) = L(P1,P2,P3) and dim L(P1,P2,P3,P4) = 2. Moreover, the

2 Exercises on Projective Spaces 63

same argument as in the solution of Exercise1 shows that a Cartesian equation ofL(P1,P2,P3,P4) = L(P1,P2,P3) is given by

0 = det

⎛⎜⎜⎝1 0 2 x00 1 1 x11 1 2 x22 1 2 x3

⎞⎟⎟⎠ = −x0 − 2x1 + 3x2 − x3.

(c) By point (b), if we replace the last column of A with the vector (0, 0, 0, 1)we obtain an invertible matrix, so the vectors v1, v2, v3, (0, 0, 0, 1) provide a basisof R

4. The projective frame induced by this basis is given by the points P1, P2,P3, [0, 0, 0, 1], [3, 2, 4, 6]. Therefore, this 5-tuple of points extends P1,P2,P3 to aprojective frame of P

3(R).

Exercise 4. Let l ⊂ P2(K) be the line of equation x0 + x1 = 0, set U = P

2(K) \ land let α,β : U → K

2 be defined as follows:

α([x0, x1, x2]) =(

x1x0 + x1

,x2

x0 + x1

),

β([x0, x1, x2]) =(

x0x0 + x1

,x2

x0 + x1

).

Find an explicit formula for the composition α ◦ β−1, and check that this map is anaffinity.

Solution. Let us first determine β−1. Let β([x0, x1, x2]) = (u, v). Since x0 + x1 �= 0on U, we may suppose x0 + x1 = 1, so that

u = x0x0 + x1

= x0, v = x2x0 + x1

= x2, x1 = 1 − x0 = 1 − u.

Therefore, β−1(u, v) = [u, 1 − u, v], hence α(β−1(u, v)) = (1 − u, v), and α ◦ β−1

is obviously an affinity.

Exercise 5. For i = 0, 1, 2, let ji : K2 → Ui ⊆ P

2(K) be the map introduced inSect. 1.3.8.

(a) Find two distinct projective lines r, s ⊂ P2(K) such that the affine lines

j−1i (r ∩ Ui), j

−1i (s ∩ Ui) are parallel for i = 1, 2.

(b) Is it possible to find distinct lines r, s ⊂ P2(K) such that the affine lines

j−1i (r ∩ Ui), j

−1i (s ∩ Ui) are parallel for i = 0, 1, 2?

Solution. Let li be the projective line of equation xi = 0, i = 0, 1, 2. If r ⊂ P2(K)

is a projective line, then the set j−1i (r ∩ Ui) is an affine line if and only if r �= li.

Moreover, for any given distinct projective lines r �= li, s �= li, the affine linesj−1i (r ∩ Ui) and j−1

i (s ∩ Ui) are parallel if and only if the point s ∩ r belongs toli. Therefore, if r, s are projective lines such that r �= s, r /∈ {l1, l2}, s /∈ {l1, l2} and

64 2 Exercises on Projective Spaces

r ∩ s = [1, 0, 0], then r and s satisfy the condition described in (a): for example, onemay choose r = {x1 + x2 = 0}, s = {x1 − x2 = 0}.

Moreover, since l0 ∩ l1 ∩ l2 = ∅, the condition described in (b) cannot be satisfiedby any pair of distinct lines of P

2(K).

Exercise 6. Let A,B,C,D be points of P2(K) in general position, and set

P = L(A,B) ∩ L(C,D), Q = L(A,C) ∩ L(B,D), R = L(A,D) ∩ L(B,C).

Prove that P,Q,R are not collinear.

Solution. Since A,B,C,D are in general position, we can choose a system of homo-geneous coordinates in P

2(K) where

A = [1, 0, 0], B = [0, 1, 0], C = [0, 0, 1], D = [1, 1, 1].

An easy computation shows that

P = [1, 1, 0], Q = [1, 0, 1], R = [0, 1, 1].

Since det

⎛⎝1 0 11 1 00 1 1

⎞⎠ �= 0, the points P,Q,R are not collinear.

Exercise 7. LetR = {P0, . . . ,Pn+1} be a projective frame of P(V ) and let 0 ≤ k <

n + 1. Set S = L(P0,P1, . . . ,Pk), S′ = L(Pk+1, . . . ,Pn+1).

(a) Show that there exists W ∈ P(V ) such that S ∩ S′ = {W }.(b) Prove that {P0, . . . ,Pk,W } is a projective frame of S, and that {Pk+1, . . . ,

Pn+1,W } is a projective frame of S′.

Solution. (a) By the definition of projective frame, we have dim S = k, dim S′ =n − k and dim L(S, S′) = n, so Grassmann’s formula implies that dim(S ∩ S′) =dim S + dim S′ − dim L(S, S′) = 0, and this proves (a).

(b) In order to show that {P0, . . . ,Pk,W } is a projective frame of S it is sufficientto prove that dim L(A) = k for every subset A ⊆ {P0, . . . ,Pk,W } containing exactlyk + 1 points (if this is the case, then clearly L(A) = S).

Let A be such a subset. If W /∈ A, then dim L(A) = k, since the points of R arein general position. Let us now assume that W ∈ A. The points of R are in generalposition, so we have

dim L((A \ {W }) ∪ S′) = dim L((A \ {W }) ∪ {Pk+1, . . . ,Pn+1}) = n,dim L(A \ {W }) = k − 1 .

But dim S′ = n − k, so

dim(L(A \ {W }) ∩ S′) = (k − 1) + (n − k) − n = −1,

2 Exercises on Projective Spaces 65

i.e., L(A \ {W }) ∩ S′ = ∅. Since W ∈ S′, this implies that W /∈ L(A \ {W }), sodim L(A) = dim L(A \ {W }) + 1 = k. Therefore, {P0, . . . ,Pk,W } is a projectiveframe of S. In the very same way one can prove that {Pk+1, . . . ,Pn+1,W } is a pro-jective frame of S′.

Exercise 8. Let r, r′ ⊂ P3(K) be skew lines, and take P ∈ P

3(K) \ (r ∪ r′). Showthat there exists a unique line l ⊂ P

3(K) that containsP andmeets both r and r′. Com-pute Cartesian equations for l in the case when K = R, the line r has equations x0 −x2 + 2x3 = 2x0 + x1 = 0, the line r′ has equations 2x1 − 3x2 + x3 = x0 + x3 = 0,and P = [0, 1, 0, 1].Solution. Let S = L(r,P), S′ = L(r′,P). An easy application of Grassmann’s for-mula implies that dim S = dim S′ = 2. Moreover, S �= S′ because otherwise r andr′ would be coplanar, hence incident. It follows that dim(S ∩ S′) < 2. On the otherhand, dim(S ∩ S′) = dim S + dim S′ − dim L(S, S′) ≥ 2 + 2 − 3 = 1, so l = S ∩ S′is a line. Since l and r (respectively, r′) both lie on S (respectively, S′), we have thatl ∩ r �= ∅ (respectively, l ∩ r′ �= ∅). Therefore l satisfies the required properties.

Let now l′ be any line of P3(K) containing P and meeting both r and r′.

We have l′ ⊆ L(r,P) = S, l′ ⊆ L(r′,P) = S′, so l′ ⊆ S ∩ S′ = l, and l′ = l sincedim l′ = dim l.

Let us now come to the particular case described in the statement of the exercise.The pencil of planes centred at r has parametric equations

λ(x0 − x2 + 2x3) + μ(2x0 + x1) = 0, [λ,μ] ∈ P1(R).

By imposing that the generic planeof thepencil pass throughPweobtain 2λ + μ = 0,so an equation of S is given by −3x0 − 2x1 − x2 + 2x3 = 0. In the same way oneproves that S′ is described by the equation−3x0 + 2x1 − 3x2 − 2x3 = 0. The systemgiven by the equation of S and the equation of S′ provides the required equationsfor l.

Exercise 9. Let W1,W2,W3 be planes of P4(K) such that Wi ∩ Wj is a point for

every i �= j, and W1 ∩ W2 ∩ W3 = ∅. Show that there exists a unique plane W0 suchthat W0 ∩ Wi is a projective line for i = 1, 2, 3.

Solution.For i �= j setPij = Wi ∩ Wj, andW0 = L(P12,P13,P23) (therefore,Pij =Pji

for every i �= j). IfP12,P13,P23 were not pairwise distinct, thenW1 ∩ W2 ∩ W3 wouldbe non-empty, while ifP12,P13,P23 were pairwise distinct and collinear, then the linecontaining themwould be contained in eachWi. In any case, our hypothesis would becontradicted, soP12,P13,P23 are in general position, andW0 is a plane.Moreover, byconstruction W0 ∩ Wi contains the line L(Pij,Pik), {i, j, k} = {1, 2, 3}. On the otherhand, if dim(W0 ∩ Wi) > 1, then W0 = Wi, so Wi ∩ Wj = W0 ∩ Wj contains a linefor every j �= i, a contradiction. Therefore, W0 ∩ Wi is a line for every i = 1, 2, 3.

Let now W ′0 be a plane satisfying the properties described in the statement, and

let li = W ′0 ∩ Wi for every i = 1, 2, 3. Then each li is a line, and Wi ∩ Wj ∩ W ′

0 =(Wi ∩ W ′

0) ∩ (Wj ∩ W ′0) = li ∩ lj �= ∅ (the lines li, lj both lie on W ′

0, so they must

66 2 Exercises on Projective Spaces

intersect). It follows that Pij ∈ W ′0 for every i, j = 1, 2, 3, and W0 ⊆ W ′

0. SincedimW ′

0 = dimW0, we can conclude that W ′0 = W0.

Exercise 10. Let r1, r2, r3 be pairwise skew lines of P4(K) and suppose that no

hyperplane of P4(K) contains r1 ∪ r2 ∪ r3. Prove that there exists a unique line that

meets ri for every i = 1, 2, 3.

Solution (1).For every i, j ∈ {1, 2, 3}, i �= j, let Vij = L(ri, rj). An easy application ofGrassmann’s formula implies that dim Vij = 3 for every i, j. Since the lines r1, r2, r3are not contained in a hyperplane, we have L(V12 ∪ V13) = L(r1, r2, r3) = P

4(K), sodim(V12 ∩ V13) = 2, again by Grassmann’s formula. Moreover, if V12 ∩ V13 ⊆ V23,then the line r1 is contained in V23, and this contradicts the fact that r1, r2, r3 arenot contained in a hyperplane. It follows that the subspace l = V12 ∩ V13 ∩ V23 hasdimension one, so it is a projective line.

We now check that l meets each ri, i = 1, 2, 3, and that l is the unique line ofP4(K) with this property. If {i, j, k} = {1, 2, 3}, by construction l lies on the plane

Vij ∩ Vik , which contains also ri. Since two coplanar projective lines always intersect,we deduce that l ∩ ri �= ∅.Moreover, if s is any linemeeting each ri, i = 1, 2, 3, then itis easy to show that s ⊆ Vij for every i, j ∈ {1, 2, 3}, so s ⊆ l. But dim s = dim l = 1,hence s = l.

Solution (2). Let us show how the statement of Exercise10 can be deduced from thestatement of Exercise8.

If V23 = L(r2, r3), a direct application of Grassmann’s formula implies thatdim V23 = 3. Moreover, since r1, r2, r3 are not contained in a hyperplane, the line r1is not contained in V23, so r1 ∩ V23 = {P} for some P ∈ P

4(K). Now the statementof Exercise8 implies that there exists a unique line l ⊆ V23 that meets r2 and r3 andcontains P. Therefore, this line meets ri for every i = 1, 2, 3. Moreover, any line thatmeets both r2 and r3 must be contained in V23, so it can intersect r1 only at P. Itfollows that l is the unique line of P

4(K) that satisfies the required properties.

Note. It is not difficult to show that, by duality, the statement of the exercise isequivalent to the following proposition:

Let H1,H2,H3 be planes of P4(K) such that L(Hi,Hj) = P

4(K) for every i �= j,and H1 ∩ H2 ∩ H3 = ∅. Then, there exists a unique plane H0 such that L(H0,Hi) �=P4(K) for i = 1, 2, 3.

On the other hand, an easy application of Grassmann’s formula implies that, if S, S′are distinct planes of P

4(K), then L(S, S′) �= P4(K) if and only if S ∩ S′ is a line,

while L(S, S′) = P4(K) if and only if S ∩ S′ is a point. It follows that the statements

of Exercises9 and 10 are equivalent.

Exercise 11. Let r and s be distinct lines in P3(K) and let f be a projectivity of

P3(K) such that the fixed-point set of f coincides with r ∪ s. For every P ∈ P

3(K) \(r ∪ s) let lP be the line joining P and f (P). Prove that the line lP meets both r ands, for every P ∈ P

3(K) \ (r ∪ s).

2 Exercises on Projective Spaces 67

Solution.Wefirst recall that the lines r and s are skew (cf. Sect. 1.2.5). Therefore, thestatement of Exercise8 ensures that, for any given P ∈ P

3(K) \ (r ∪ s), there existsa line tP passing through P and meeting both r and s. In order to conclude it is nowsufficient to show that tP = lP.

Let A = tP ∩ r, B = tP ∩ s. Then f (P) ∈ f (L(A,B)) = L(f (A), f (B)) =L(A,B) = tP. Therefore, tP contains both P and f (P), and so it coincides with lP.

Exercise 12. (Desargues’ Theorem) Let P(V ) be a projective plane and A1,A2,A3,

B1,B2,B3 points of P(V ) in general position. Consider the triangles T1 and T2 ofP(V ) with vertices A1,A2,A3 and B1,B2,B3; one says that T1 and T2 are in centralperspective if there exists a point O in the plane, distinct from the Ai and Bi, suchthat all lines L(Ai,Bi) pass through O.

Prove that T1 and T2 are in central perspective if and only if the pointsP1 = L(A2,A3) ∩ L(B2,B3), P2 = L(A3,A1) ∩ L(B3,B1) and P3 = L(A1,A2) ∩L(B1,B2) are collinear. (cf. Fig. 2.1).

Solution. It is easy to check that the pointsP1,P2,P3 satisfy the following properties:they are pairwise distinct, they are distinct from any vertex ofT1 andT2, and the pointsA1,B1,P3,P2 provide a projective frame of P(V ). The point A2 belongs to the lineL(A1,P3), so it has coordinates [1, 0, a2] for some a2 �= 0, sinceA1 andA2 are distinctfrom each other. A similar argument shows that:

A3 = [a3, 1, 1], B2 = [0, 1, b2], B3 = [1, b3, 1],

where a3, b2, b3 ∈ K, b2 �= 0, a3 �= 1 and b3 �= 1.Let us set P′

1 = L(A2,A3) ∩ L(P2,P3) and P′′1 = L(B2,B3) ∩ L(P2,P3). The

pointsP1,P2,P3 are collinear if and only ifP′1 = P′′

1 (and in this caseP1 = P′1 = P′′

1 ).The lines L(A2,A3) and L(B2,B3) have equations a2x0 + (1 − a2a3)x1 − x2 = 0and (1 − b2b3)x0 + b2x1 − x2 = 0, respectively. So P′

1 = [1, 1, 1 − a2a3 + a2] and

B1

B3

A1

O

B2

P3

P2

P1

A3A2

Fig. 2.1 The configuration described in Desargues’ Theorem

68 2 Exercises on Projective Spaces

P′′1 = [1, 1, 1 − b2b3 + b2]. Therefore, our previous considerations imply thatP1,P2

and P3 are collinear if and only if the following equality holds:

a2(1 − a3) = b2(1 − b3). (2.1)

Let us now analyze the condition that T1 and T2 are in central perspective. The lineL(A1,B1) has equation x2 = 0, the lineL(A2,B2) has equation a2x0 + b2x1 − x2 = 0,and the line L(A3,B3) has equation (1 − b3)x0 + (1 − a3)x1 + (a3b3 − 1)x2 = 0.These three lines are concurrent if and only if the corresponding points of the dualprojective plane are collinear, i.e., if and only if

det

⎛⎝

0 0 1a2 b2 −1

1 − b3 1 − a3 a3b3 − 1

⎞⎠ = 0. (2.2)

We observe that, if this condition holds, then the point O belonging to the three lineshas coordinates [b2,−a2, 0]. Since a2 �= 0 and b2 �= 0, the point O is distinct fromany vertex of T1 and T2.Finally, the conclusion follows from the fact that conditions (2.1) and (2.2) are clearlyequivalent.

Note. The solution just described directly proves that the two conditions stated inthe exercise are equivalent. We have already noticed in Sect. 1.4.4 that Desargues’Theorem is an example of self-dual proposition, so, in fact, it is sufficient to provejust one implication, since the other one follows from the Duality principle.

Exercise 13. (Pappus’ Theorem) Let P(V ) be a projective plane, and let A1, . . . ,A6

be pairwise distinct points such that the lines L(A1,A2), L(A2,A3),…, L(A6,A1)

are pairwise distinct. Consider the hexagon of P(V ) with vertices A1, . . . ,A6, andsuppose that there exist two distinct lines r, s such that A1,A3,A5 ∈ r, A2,A4,A6 ∈ sand O = r ∩ s is distinct from each Ai.

Prove that the points where the opposite sides of the hexagon meet are collinear,i.e., that the points P1 = L(A1,A2) ∩ L(A4,A5), P2 = L(A2,A3) ∩ L(A5,A6) andP3 = L(A3,A4) ∩ L(A6,A1) lie on a projective line (cf. Fig. 2.2).

Solution. By hypothesis we have r = L(A1,A3) and s = L(A2,A4). Since r �= s andthe point O = r ∩ s is not a vertex of the hexagon, the points A1,A2,A3,A4 form aprojective frame. In the corresponding system of homogeneous coordinates of P(V )

the line r has equation x1 = 0, the line s has equation x0 − x2 = 0, and the point Ohas coordinates [1, 0, 1]. The point A5 lies on r and is distinct from O, from A1 andfrom A2, so it has coordinates [1, 0, a] for some a ∈ K \ {0, 1}. In the same way,the point A6 has coordinates [1, b, 1], where b ∈ K \ {0, 1}. The line L(A1,A2) hasequation x2 = 0 and the line L(A4,A5) has equation ax0 + (1 − a)x1 − x2 = 0, sothe point P1 = L(A1,A2) ∩ L(A4,A5) has coordinates [a − 1, a, 0]. Similar compu-tations show that P2 has coordinates [0, b, 1 − a] and P3 has coordinates [b, b, 1]. Itfollows that the points P1, P2 and P3 are collinear, since

2 Exercises on Projective Spaces 69

A5

A3

A2A4

P3

s

A1

r

P2

P1

A6

Fig. 2.2 The configuration described in Pappus’ Theorem

det

⎛⎝a − 1 a 00 b 1 − ab b 1

⎞⎠ = 0.

Exercise 14. Let A,A′,B,B′ be pairwise distinct non-collinear points of P2(K).

Prove that A,A′,B,B′ are in general position if and only if there exists a projectivityf : P

2(K) → P2(K) such that f (A) = B, f (A′) = B′, f 2 = Id.

Solution. Suppose that a projectivity f as in the statement exists. We first observethat f (B) = f (f (A)) = A, f (B′) = f (f (A′)) = A′. In particular, the lines L(A,B) andL(A′,B′) are invariant under f . Moreover, since A,A′,B,B′ are not collinear, thelines L(A,B) and L(A′,B′) are distinct and meet at a single pointO such that f (O) =f (L(A,B) ∩ L(A′,B′)) = L(A,B) ∩ L(A′,B′) = O. It is immediate to check that, ifA,A′,B,B′ were not in general position, than we would have O ∈ {A,A′,B,B′}, andthis would contradict the fact that no point in {A,A′,B,B′} is a fixed point of f . Wehave thus shown that A,A′,B,B′ are in general position, as desired.

Let us now prove the converse implication. Assume that the points A,A′,B,B′ arein general position. The Fundamental theorem of projective transformations ensuresthat there exists a (unique) projectivity f : P

2(K) → P2(K) such that f (A) = B,

f (A′) = B′, f (B) = A, f (B′) = A′. By construction, f 2 and the identity of P2(K)

coincide on A,A′,B,B′, so they coincide on the whole of P2(K), again by the Fun-

damental theorem of projective transformations. Therefore, f satisfies the requiredproperties. We also observe that any projectivity satisfying the conditions describedin the statement must coincide with f on A,A′,B,B′, so f is uniquely determined bysuch conditions.

Exercise 15. Determine a projectivity f : P2(R) → P

2(R) with the following prop-erties: if P = [1, 2, 1], Q = [1, 1, 1] and r, r′, s, s′ are the lines described by theequations

r : x0 − x1 = 0, r′ : x0 + x1 = 0s : x0 + x1 + x2 = 0, s′ : x1 + x2 = 0,

then f (r) = r′, f (s) = s′, and f (P) = Q. Is such a projectivity unique?

70 2 Exercises on Projective Spaces

Solution. First observe that P /∈ r ∪ s, Q /∈ r′ ∪ s′. Let P1 = r ∩ s = [1, 1,−2],Q1 = r′ ∩ s′ = [1,−1, 1], and let us choose points P2,P3 distinct from P on r, s,respectively: for example, we set P2 = [0, 0, 1], P3 = [−1, 1, 0]. It is easy to checkthat the pointsP1,P2,P3,P are in general position. In the sameway, ifQ2 = [0, 0, 1],then Q2 ∈ r′, and Q1,Q2,Q are in general position. If Q3 is any point of s′ distinctfrom Q1 and from s′ ∩ L(Q2,Q) = [1, 1,−1], then the quadruples {P1,P2,P3,P}and {Q1,Q2,Q3,Q} provide two projective frames of P

2(R). Therefore, there existsa unique projectivity f : P

2(R) → P2(R) such that f (Pi) = Qi for every i = 1, 2, 3

and f (P) = Q. Since r = L(P1,P2), s = L(P1,P3), r′ = L(Q1,Q2), s′ = L(Q1,Q3),we also have f (r) = r′, f (s) = s′. Since Q3 may be chosen in infinitely many ways,an infinite number of projectivities satisfy the required conditions.

Let us explicitly construct f in the case when Q3 = [1, 0, 0]. A normalizedbasis associated to the frame {P1,P2,P3,P} is given by {v1 = (3, 3,−6), v2 =(0, 0, 8), v3 = (−1, 1, 0)}, while a normalized basis associated to the frame{Q1,Q2,Q3,Q} is given by {w1 = (−1, 1,−1), w2 = (0, 0, 2), w3 = (2, 0, 0)}.Therefore, f is induced by the unique linear isomorphism ϕ : R

3 → R3 such that

ϕ(vi) = wi for i = 1, 2, 3. A straightforward computation shows that this iso-

morphism is given, up to a non-zero scalar, by the matrix

⎛⎝

14 −10 0−2 −2 0−1 −1 −3

⎞⎠, so

the required projectivity f may be explicitly described by the following formula:f ([x0, x1, x2]) = [14x0 − 10x1,−2x0 − 2x1,−x0 − x1 − 3x2].Exercise 16. Let r, s, r′, s′ be lines in P

2(K) such that r �= s, r′ �= s′, and letg : r → r′, h : s → s′ be projective isomorphisms. Find necessary and sufficient con-ditions for the existence of a projectivity f : P

2(K) → P2(K) such that f |r = g and

f |s = h. Show that, when it exists, such an f is unique.

Solution. Let us consider the points P = r ∩ s, P′ = r′ ∩ s′ (which are uniquelydefined since r �= s, r′ �= s′). Of course, if the required projectivity exists, then wemust have g(P) = h(P) (so necessarily g(P) = h(P) = P′). Let us show that thiscondition is also sufficient.

Let P1,P2 be pairwise distinct points of r \ {P}, let Q1,Q2 be pairwise distinctpoints of s \ {P}, and set P′

i = g(Pi), Q′i = h(Qi), i = 1, 2. It is immediate to check

that the quadruplesR = {P1,P2,Q1,Q2},R′ = {P′1,P

′2,Q

′1,Q

′2} are in general posi-

tion, so they define two projective frames of P2(K). Therefore, there exists a unique

projectivity f : P2(K) → P

2(K) such that f (Pi) = P′i, f (Qi) = Q′

i for i = 1, 2. Wenow show that this projectivity satisfies the required conditions.

We have f (r) = f (L(P1,P2)) = L(P′1,P

′2) = r′, and in the same way f (s) = s′.

Therefore, f (P) = f (r ∩ s) = r′ ∩ s′ = P′. So the projective transformations f |rand g coincide on three pairwise distinct points of r, hence on the whole line r.In the same way one proves that f |s = h.

Finally, the fact that f is unique readily follows from the Fundamental theoremof projective transformations: any projectivity that satisfies the required conditionsmust coincide with f on R, hence on the whole of P

2(K).

2 Exercises on Projective Spaces 71

Exercise 17. Let S, S′ be planes of P3(K) and r, r′ lines of P

3(K) such thatL(r, S) = L(r′, S′) = P

3(K), and let g : S → S′, h : r → r′ be projective isomor-phisms. Find necessary and sufficient conditions for the existence of a projectivityf : P

3(K) → P3(K) such that f |S = g and f |r = h. Show that, if such an f exists,

then it is unique.

Solution. An easy application of Grassmann’s formula shows that there exist pointsP,P′ ∈ P

3(K) such that r ∩ S = {P}, r′ ∩ S′ = {P′}. Of course, if the required projec-tivity exists, then we must have g(P) = h(P) = P′. We now show that this conditionis also sufficient.

We extend P to a projective frame {P,P1,P2,P3} of S, and we choose distinctpoints Q1,Q2 on r, such that Q1 �= P, Q2 �= P. We first show thatR = {P1,P2,P3,Q1,Q2} is in general position, so it is a projective frame of P

3(K).To this aim it is sufficient to check that R does not contain any quadruple of copla-nar points. Since L(P1,P2,P3) = S and Ql /∈ S, the points P1,P2,P3,Ql cannot becoplanar for l = 1, 2. Therefore, we may assume by contradiction that Pi,Pj,Q1,Q2

are coplanar for some i �= j. In this case, the lines L(Pi,Pj) ⊂ S, L(Q1,Q2) = rintersect in P = S ∩ r, hence P,Pi,Pj are collinear. But this contradicts the fact thatP,Pi,Pj are in general position on S. We have thus proved that R is a projectiveframe.

Let now P′i = g(Pi), Q′

j = h(Qj) for i = 1, 2, 3, j = 1, 2. The same argument asabove shows that P′

1,P′2,P

′3,Q

′1,Q

′2 are also in general position, so there exists

a unique projectivity f : P3(K) → P

3(K) such that f (Pi) = P′i, f (Qj) = Q′

j fori = 1, 2, 3, j = 1, 2. Let us show that this projectivity coincides with g on S andwith h on r. We have

f (S) = f (L(P1,P2,P3)) = L(P′1,P

′2,P

′3) = S′,

f (r) = f (L(Q1,Q2)) = L(Q′1,Q

′2) = r′,

so f (P) = f (r ∩ S) = r′ ∩ S′ = P′. Therefore, f |S and g coincide on the projectiveframe {P,P1,P2,P3} of S, so they coincide on the whole of S. In the same way oneproves that f |r and h coincide on P,Q1,Q2, hence on r.

Finally, the fact that f is unique is now obvious: any projectivity satisfying therequired properties must coincide with f on P1,P2,P3,Q1,Q2, hence on the wholeprojective space P

3(K), since {P1,P2,P3,Q1,Q2} is a projective frame of P3(K).

Note. Exercises16 and 17 illustrate particular cases of a general fact regardingthe extension of projective transformations defined on subspaces of a projec-tive space. In the projective space P(V ) let us consider the projective subspacesS1, S2, S′

1, S′2, and let us fix projective isomorphisms g1 : S1 → S′

1, g2 : S2 → S′2.

Suppose also that L(S1, S2) = P(V ), g1(S1 ∩ S2) = g2(S1 ∩ S2) = S′1 ∩ S′

2,and g1|S1∩S2 = g2|S1∩S2 . Then, there exists a projectivity f : P(V ) → P(V ) such thatf |Si = gi for i = 1, 2. Moreover, if S1 ∩ S2 �= ∅, then such a projectivity is unique.

Exercise 18. Let r, s be distinct lines of P2(K), let A,B be distinct points of

r \ s, and take C,D ∈ P2(K) \ (r ∪ s). Show that there exists a unique projectiv-

ity f : P2(K) → P

2(K) such that f (A) = A, f (B) = B, f (s) = s, f (C) = D.

72 2 Exercises on Projective Spaces

r

C

Q

D

B

s

A

PQ P

Fig. 2.3 The construction described in Exercise18

Solution. Since C,D /∈ s, each line L(C,A), L(C,B), L(D,A), L(D,B) meets sat exactly one point, so we can set P = s ∩ L(C,A), P′ = s ∩ L(D,A), Q = s ∩L(C,B), Q′ = s ∩ L(D,B) (cf. Fig. 2.3). As C,D /∈ r, the points P,P′,Q,Q′ lie ons \ r. Finally, we obviously have P �= Q, P′ �= Q′. It follows that the quadruples{A,B,P,Q} and {A,B,P′,Q′} define projective frames of P

2(K).Let now f be the unique projectivity of P

2(K) such that f (A) = A, f (B) = B,f (P) = P′, f (Q) = Q′. Thenwe have f (s) = f (L(P,Q)) = L(P′,Q′) = s.Moreover,f (C) = f (L(A,P) ∩ L(B,Q)) = L(A,P′) ∩ L(B,Q′) = D, so f satisfies the requiredproperties. Conversely, if g is a projectivity of P

2(K) satisfying the required prop-erties, then g(A) = A, g(B) = B, g(P) = g(L(A,C) ∩ s) = L(A,D) ∩ s = P′ andg(Q) = g(L(C,B) ∩ s) = L(D,B) ∩ s = Q′, so g = f .

K Exercise 19. In P3(R), let r be the line of equations x0 − x1 = x2 − x3 = 0, let

H,H ′ be the planes of equations x1 + x2 = 0, x1 − 2x3 = 0, respectively, and letC =[1, 1, 0, 0]. Compute the number of projectivities f : P

3(R) → P3(R) that satisfy the

following conditions:

(i) f (r) = r, f (H) = H ′, f (H ′) = H, f (C) = C;(ii) the fixed-point set of f contains a plane.

Solution.Let us first determine the incidence relations that hold among the subspacesdescribed in the statement. It is immediate to check that C ∈ r and C /∈ H ∪ H ′. Asa consequence, r ∩ H and r ∩ H ′ both consist of a single point: more precisely, aneasy computation shows that, ifA = r ∩ H andB = r ∩ H ′, thenA = [1, 1,−1,−1],B = [2, 2, 1, 1]. Moreover, sinceH �= H ′, the set l = H ∩ H ′ is a projective line, andthe fact that A �= B readily implies that the lines r and l are skew.

Let now f be a projectivity satisfying the required conditions, and let S be a planepointwise fixed by f . In order to understand the behaviour of f , we now determinethe possible positions of S, showing first that S must contain l. We have H ∩ S =f (H ∩ S) = H ′ ∩ S, so H ∩ S = H ′ ∩ S is a projective subspace of H ∩ H ′ = l. Asdim(H ∩ S) ≥ 1, we have l = H ∩ S = H ′ ∩ S, so in particular l ⊂ S, as desired.

Since l and r are skew, the intersection r ∩ S consists of one point. This pointis fixed by f , so in order to determine S it is useful to study the fixed-point set

2 Exercises on Projective Spaces 73

of the restriction of f to r. Recall that by hypothesis C ∈ r, while by construc-tion A,B ∈ r. We also have f (A) = f (r ∩ H) = r ∩ H ′ = B, and in the same wayf (B) = A. Since f (C) = C, if g : r → r is the unique projectivity such that g(A) = B,g(B) = A, g(C) = C, then f |r = g. Moreover, an easy computation shows that, ifr = P(W ), then g is induced by the unique linear isomorphism ϕ : W → W suchthat ϕ(1, 1,−1,−1) = (2, 2, 1, 1) and ϕ(2, 2, 1, 1) = (1, 1,−1,−1). Therefore,the fixed-point set of g (hence, of f |r) is given by {C,D}, where D = [1, 1, 2, 2].It follows that either r ∩ S = C or r ∩ S = D.

Then, let S1 = L(l,C) and S2 = L(l,D). Since C /∈ l, D /∈ l and l ⊂ S, if C ∈ Sthen S = S1, while ifD ∈ S then S = S2. Moreover, since as observed above the linesr, s are skew, we have L(S1, r) = L(S2, r) = P

3(R).Take i ∈ {1, 2}. As g fixes bothC = S1 ∩ r andD = S2 ∩ r, we may exploit Exer-

cise17 to show that there exists a unique projectivity fi : P3(R) → P

3(R) such thatfi|Si = IdSi and fi|r = g. Moreover, what we have shown so far implies that, if f is aprojectivity satisfying the conditions described in the statement, then f necessarilycoincides either with f1 or with f2. We observe that f1 �= f2, because otherwise thefixed-point set of f1 would contain S1 ∪ S2. Being the union of pairwise skew pro-jective subspaces (cf. Sect. 1.2.5), the fixed-point set of f1 would then coincide withthe whole of P

3(R), contradicting the fact that f1(A) �= A.In order to conclude it is now sufficient to show that both f1 and f2 satisfy the

required conditions. By construction, for i = 1, 2 we have fi(C) = C, fi(r) = r, andthe fixed-point set of fi contains the plane Si. Finally, we have fi(H) = fi(L(l,A)) =L(l,B) = H ′ and fi(H ′) = fi(L(l,B)) = L(l,A) = H, as desired.

Exercise 20. Let f : P1(K) → P

1(K) be an injective function such that

β(P1,P2,P3,P4) = β(f (P1), f (P2), f (P3), f (P4))

for every quadruple P1,P2,P3,P4 of pairwise distinct points. Show that f is a pro-jectivity.

Solution. Let P1,P2,P3 be pairwise distinct points of P1(K), and set Qi = f (Pi) for

i = 1, 2, 3. Since f is injective, the sets {P1,P2,P3} and {Q1,Q2,Q3} are projectiveframes of P

1(K). Therefore, there exists a projectivity g : P1(K) → P

1(K) suchthat g(Pi) = Qi for i = 1, 2, 3. Moreover, since the cross-ratio is invariant underprojectivities, for every P /∈ {P1,P2,P3} we have

β(Q1,Q2,Q3, g(P)) = β(g(P1), g(P2), g(P3), g(P)) = β(P1,P2,P3,P) == β(f (P1), f (P2), f (P3), f (P)) = β(Q1,Q2,Q3, f (P)).

As a consequence, f (P) = g(P) for every P �= P1,P2,P3. On the other hand, fori = 1, 2, 3 we have f (Pi) = g(Pi) by construction, so f = g, and f is a projectivity.

K Exercise 21. (Modulus of a quadruple of points) Let

A = {P1,P2,P3,P4}, A′ = {P′1,P

′2,P

′3,P

′4}

74 2 Exercises on Projective Spaces

be quadruples of pairwise distinct points of P1(C), and set

k = β(P1,P2,P3,P4), k′ = β(P′1,P

′2,P

′3,P

′4).

(a) Show that the sets A,A′ are projectively equivalent if and only if

(k2 − k + 1)3

k2(k − 1)2= ((k′)2 − k′ + 1)3

(k′)2(k′ − 1)2.

(b) Let G be the set of projectivities f : P1(C) → P

1(C) such that f (A) = A. Forevery k ∈ C \ {0, 1}, compute the number |G| of elements of G.

Solution. (a) As discussed in Sect. 1.5.2, the sets A,A′ are projectively equivalentif and only if k′ belongs to the set

Ω(k) ={k,

1

k, 1 − k,

1

1 − k,k − 1

k,

k

k − 1

}.

(Observe that k, k′ ∈ C \ {0, 1}, since Pi �= Pj and P′i �= P′

j for every i �= j.)

Consider the rational function j(t) = (t2 − t + 1)3

t2(t − 1)2. In order to prove (a) it is

sufficient to show that k′ ∈ Ω(k) if and only if j(k′) = j(k).Via a direct substitution, it is immediate to verify that j(k) = j(k′) if k′ ∈ Ω(k).Conversely, let us assume that j(k) = j(k′), and set q(t) = (t2 − t + 1)3 − j(k)t2

(t − 1)2. Of course we have q(k′) = 0. Moreover, q is a polynomial of degree 6admitting every element of Ω(k) as a root. Therefore, if Ω(k) contains 6 elements,then it coincides with the set of roots of q, and since q(k′) = 0 we may deduce thatk′ ∈ Ω(k), as desired. If we set ω = 1+i

√3

2 , then a direct computation shows that|Ω(k)| < 6 only in the following cases:

• when k ∈ {ω,ω}, and in this case Ω(k) = {ω,ω};• when k ∈

{−1, 2, 12

}, and in this case Ω(k) =

{−1, 2, 12

}.

Moreover, if k ∈{−1, 2, 12

}then j(k) = 27

4 and q(t) = (t + 1)2(t − 2)2(t − 1

2

)2,

while if k ∈ {ω,ω} then j(k) = 0 and q(t) = (t − ω)3(t − ω)3. In any case, Ω(k)coincides with the set of roots of q, and from the fact that q(k′) = 0 we can deducethat k′ ∈ Ω(k), as desired.

(b) Of course G is a group. If S4 is the permutation group of {1, 2, 3, 4}, thenfor any given f ∈ G there exists ψ(f ) ∈ S4 such that f (Pi) = Pψ(f )(i) for everyi = 1, 2, 3, 4. Moreover, the map ψ : G → S4 thus defined is a group homomor-phism. If ψ(f ) = Id, then f coincides with the identity on 3 distinct points of P

1(C),hence on the whole of P

1(C): therefore, the homomorphism ψ is injective, so wehave |G| = |Imψ| ≤ |S4| = 24. Let us now investigate which permutations of the Pi

are actually induced by a projectivity. To this aim, we will exploit some elementaryfacts about group actions on sets.

2 Exercises on Projective Spaces 75

Let us consider the map η : S4 × Ω(k) → Ω(k) which is defined as follows: forevery h ∈ Ω(k) and σ ∈ S4, if Q1,Q2,Q3,Q4 ∈ P

1(C) are such that β(Q1,Q2,Q3,

Q4) = h, then η(σ, h) = β(Qσ(1),Qσ(2),Qσ(3),Qσ(4)). The properties of the cross-ratio described in Sect. 1.5.2 imply that η(σ, h) does not depend on the choice of theQi; moreover, η(σ, h) ∈ Ω(k), so η is indeed well defined. Finally, it is immediateto check that η(σ ◦ τ , h) = η(σ, η(τ , h)), so η defines an action of S4 on Ω(k).

From the fundamental property of the cross-ratio (cf. Theorem1.5.1) we deducethat σ ∈ Imψ if and only if β(P1,P2,P3,P4) = β(Pσ(1),Pσ(2),Pσ(3),Pσ(4)). In otherwords, Imψ coincides with the stabilizer of k with respect to the action we have justintroduced; moreover, this action is transitive, since every element of Ω(k) is thecross-ratio of an ordered quadruple obtained by permuting P1,P2,P3,P4. We thushave |S4| = |Stab(k)| |Ω(k)| = |Imψ| |Ω(k)|, so

|G| = |Imψ| = |S4||Ω(k)| = 24

|Ω(k)| .

Then, it follows from (a) that |G| = 12 if k ∈ {ω,ω}, |G| = 8 if k ∈{−1, 2, 12

}and

|G| = 4 otherwise.

Exercise 22. Let f : P1(R) → P

1(R) be the projectivity defined by

f ([x0, x1]) = [−x1, 2x0 + 3x1].

(a) Determine the fixed-point set of f .(b) For P = [2, 5] ∈ P

1(R), compute the cross-ratio β(A,B,P, f (P)), where A andB are the fixed points of f .

Solution. The projectivity f is induced by the linear map ϕ : R2 → R

2 which is

represented with respect to the canonical basis of R2 by the matrix

(0 −12 3

). This

matrix is diagonalizable, and it admits (1,−1) and (1,−2) as eigenvectors relative tothe eingenvalues 1 and 2, respectively. It follows that A = [1,−1] and B = [1,−2]are the only fixed points of f .

We have seen in Sect. 1.5.4 that, since A = [v] where v is an eigenvector of ϕrelative to the eigenvalue 1 and B = [w] where w is an eigenvector of ϕ relative tothe eigenvalue 2, the value β(A,B,Q, f (Q)) does not depend onQ ∈ P

1(R) \ {A,B},and it is equal to 2/1 = 2. Therefore, β(A,B,P, f (P)) = 2.

Exercise 23. (Involutions of P1(K)) Let f be a projectivity of P

1(K). Recall that fis an involution if f 2 = Id, and that an involution f is non-trivial if f �= Id.

(a) IfM =(a bc d

)is a matrix associated to f , show that f is a non-trivial involution

if and only if a + d = 0.(b) Show that f is a non-trivial involution if and only if there exist two distinct points

Q1,Q2 that are switched by f , i.e., such that f (Q1) = Q2 and f (Q2) = Q1.

76 2 Exercises on Projective Spaces

(c) Suppose that f is a non-trivial involution. Show that f has exactly either 0 or 2fixed points, and that it has exactly 2 fixed points if K = C.

(d) Suppose that f fixes the points A,B. Show that f is a non-trivial involution if andonly if, for any given point P ∈ P

1(K) \ {A,B}, the equality β(A,B,P, f (P))

= −1 holds (i.e., the characteristic of f is −1, cf. Sect. 1.5.4).(e) Show that f is the composition of two involutions.

Solution. (a) First observe that, if f �= Id, then the minimal polynomial ofM cannothave degree 1 (so it is equal to the characteristic polynomial of M). Moreover, wehave f 2 = Id if and only if there exists λ ∈ K

∗ such thatM2 = λI . It follows that f isa non-trivial involution if and only if the minimal polynomial and the characteristicpolynomial ofM coincide and are equal to t2 − λ. Now the conclusion follows sincethe coefficient of t in the characteristic polynomial ofM is equal to −a − d.

(b) If f �= Id is an involution, it is sufficient to choose any point Q1 not fixed byf and set Q2 = f (Q1). Conversely, let Q1,Q2 be points that are exchanged by f , letP be any point of P

1(K) \ {Q1,Q2} and set P′ = f (P). Then

β(Q1,Q2,P′,P) = β(f (Q1), f (Q2), f (P′), f (P)) == β(Q2,Q1, f (P′),P′) = β(Q1,Q2,P′, f (P′)),

where the first equality follows from the invariance of the cross-ratio under projectiv-ities (cf. Sect. 1.5.1), while the second and the third one follow from the symmetriesof the cross-ratio (cf. Sect. 1.5.2). Then f 2(P) = f (P′) = P and f is an involution.

(c) Point (a) implies that, if M is a matrix associated to f , then the characteristicpolynomial of M is equal to t2 + detM, so either M does not admit any eigenvalue(when − detM is not a square in K) or M admits two distinct eigenvalues relativeto one-dimensional eigenspaces (when − detM is a square in K). Therefore, f hasexactly either 0 or 2 fixed points. Moreover, if K is algebraically closed, then Madmits two distinct eigenvalues, so f has exactly two fixed points.

(d) As observed in Sect. 1.5.4, the cross-ratio β(A,B,P, f (P)) is independentof the choice of P. More precisely, in a projective frame R having A and B as

fundamental points, f is represented by a matrix N =(

λ 00 μ

), λ,μ �= 0. If P ∈

P1(K) \ {A,B} and [P]R = [a, b], then [f (P)]R = [λa,μb], so β(A,B,P, f (P)) =

μλ. Therefore, β(A,B,P, f (P)) = −1 if and only if μ = −λ, i.e., if and only if

N =(

λ 00 −λ

). This condition is equivalent to the fact that N2 is a multiple of the

identity, i.e., to the fact that f 2 = Id.(e) If f = Id there is nothing to prove, so we suppose that there exists A ∈ P

1(K)

such that f (A) = A′ �= A, and we set A′′ = f (A′). If A′′ = A, then f switches A andA′, so it is an involution by point (b). Therefore, we can suppose that A′′ �= A, so thatnecessarily A′ �= A′′, because otherwise we would have f (A′) = A′ = f (A), whichcontradicts the fact that f is injective. As a consequence, being pairwise distinct,the points A,A′,A′′ define a projective frame of P

1(K), and by the Fundamentaltheorem of projective transformations there exists a projectivity g : P

1(K) → P1(K)

2 Exercises on Projective Spaces 77

such that g(A) = A′′, g(A′) = A′, g(A′′) = A. Since g switches A and A′′, g is aninvolution. Moreover, f ◦ g switches A′ and A′′, so it is an involution too. Therefore,f = f ◦ (g ◦ g) = (f ◦ g) ◦ g is the composition of two involutions.

Exercise 24. Let A,B be distinct points of P1(K). Show that there exists a unique

non-trivial involution of P1(K) having A and B as fixed points.

Solution (1). Take P ∈ P1(K) \ {A,B}. If f : P

1(K) → P1(K) is a projectivity such

that f (A) = A and f (B) = B, then it follows from point (d) of Exercise23 that f is anon-trivial involution if and only if β(A,B,P, f (P)) = −1, i.e., if and only if f (P) isthe unique point such that A,B,P, f (P) is a harmonic quadruple. Since a projectivityof P

1(K) is uniquely determined by the values it takes on A,B,P, this concludes theproof.

Solution (2). If we fix a projective frame of P1(K) having A and B as fundamental

points, every projectivity f : P1(K) → P

1(K) such that f (A) = A and f (B) = B is

represented in the induced coordinate system by a matrix N =(1 00 λ

), λ ∈ K

∗. It

is easily seen that N2 is a multiple of the identity if and only if λ = ±1. Therefore,the unique non-trivial involution having A and B as fixed points is the projectivity

represented by the matrix

(1 00 −1

).

Exercise 25. Let f : P1(K) → P

1(K) be a projectivity, and let A,B,C ∈ P1(K) be

pairwise distinct points such that f (A) = A, f (B) = C. Show thatA is the unique fixedpoint of f (i.e., f is parabolic, cf. Sect. 1.5.3) if and only if β(A,C,B, f (C)) = −1.

Solution. We endow P1(K) with the homogeneous coordinates induced by the

projective frame {A,B,C}. With respect to these coordinates, f is represented by

a matrix M of the form

(1 λ0 λ

)for some λ ∈ K

∗. Now, if λ = 1 the matrix M has

exactly one eigenvalue, and this eigenvalue has geometric multiplicity one, so f isparabolic; otherwise, M has two distinct eigenvalues, and it is hyperbolic. So weneed to show that β(A,C,B, f (C)) = −1 if and only if λ = 1. But the coordinatesof f (C) are [1 + λ,λ], so

β(A,C,B, f (C)) = β([1, 0], [1, 1], [0, 1], [1 + λ,λ]) = −λ,

and this concludes the proof.

Exercise 26. Let A1,A2,A3,A4 be points of P2(K) in general position, and set

P1 = L(A1,A2) ∩ L(A3,A4), P2 = L(A2,A3) ∩ L(A1,A4), r = L(P1,P2),

P3 = L(A2,A4) ∩ r, P4 = L(A1,A3) ∩ r.

Compute β(P1,P2,P3,P4).

78 2 Exercises on Projective Spaces

Solution (1). If we endow P2(K) with homogeneous coordinates such that A1 =

[1, 0, 0], A2 = [0, 1, 0], A3 = [0, 0, 1], A4 = [1, 1, 1], we easily obtain that P1 =[1, 1, 0] and P2 = [0, 1, 1]. So r = {x0 − x1 + x2 = 0}, hence P3 = [1, 2, 1] andP4 = [1, 0,−1]. It follows that, if r = P(W ), then a normalized basis ofW associatedto the projective frame {P1,P2,P3} is given by (1, 1, 0), (0, 1, 1). Since (1, 0,−1) =(1, 1, 0) − (0, 1, 1), it follows that the required cross-ratio is equal to −1.

Solution (2). We set t = L(A1,A3) and Q = t ∩ L(A2,A4) (cf. Fig. 2.4). Of courseA2 /∈ r ∪ t, A4 /∈ r ∪ t, so the perspectivity f : r → t centred at A2 and the perspec-tivity g : t → r centred at A4 are well defined. By construction we have f (P1) = A1,f (P2) = A3, f (P3) = Q, f (P4) = P4, so, being A1,A3,Q,P4 pairwise distinct, thepoints P1,P2,P3,P4 are pairwise distinct too. Moreover, again by construction wehave g(A1) = P2, g(A3) = P1, g(Q) = P3, g(P4) = P4. Being the composition oftwo projectivities, the map g ◦ f : r → r is a projectivity, so

β(P1,P2,P3,P4) = β(g(f (P1)), g(f (P2)), g(f (P3)), g(f (P4))) == β(P2,P1,P3,P4) = 1

β(P1,P2,P3,P4),

where the first equality is due to the invariance of the cross-ratio under projectiv-ities (cf. Sect. 1.5.1), while the last one is due to the symmetries of the cross-ratio(cf. Sect. 1.5.2). So β(P1,P2,P3,P4)

2 = 1, hence β(P1,P2,P3,P4) = −1 since thePi are pairwise distinct, as observed above.

Note. (Construction of the harmonic conjugate) The previous exercise suggestsa way to explicitly construct the harmonic conjugate of three pairwise distinctpoints P1,P2,P3 of a projective line r ⊆ P

2(K), i.e., the point P4 ∈ r such thatβ(P1,P2,P3,P4) = −1. To this aim, let us consider a line s �= r passing throughP and choose distinct points A1,A2 on s \ {P}. Let A4 = L(A2,P3) ∩ L(A1,P2) and

A1

Q

r

t

A2

P1P2 P4

A3

P3

A4

Fig. 2.4 Constructing a harmonic quadruple

2 Exercises on Projective Spaces 79

A3 = L(P1,A4) ∩ L(A2,P2), and let P4 = r ∩ L(A1,A3) (observe that A3,A4,Q4 areactually well defined). It is immediate to check that A1,A2,A3,A4 are in generalposition, and it is proven in Exercise26 that β(P1,P2,P3,P4) = −1.

Exercise 27. LetP,Q,R, S be points ofP2(K) in general position, let lP = L(Q,R),

lQ = L(P,R), lR = L(P,Q) and set

P′ = L(P, S) ∩ lP, Q′ = L(Q, S) ∩ lQ, R′ = L(R, S) ∩ lR.

Also let P′′ ∈ lP, Q′′ ∈ lQ, R′′ ∈ lR be the points that are uniquely determined by thefollowing conditions:

β(Q,R,P′,P′′) = β(R,P,Q′,Q′′) = β(P,Q,R′,R′′) = −1.

Show that P′′,Q′′,R′′ are collinear.

E Solution (1). Let T = L(Q′,R′) ∩ lP, W = L(T , S) ∩ lR, Z = L(T , S) ∩ lQ. Byapplying Exercise26 to the case when A1 = R, A2 = R′, A3 = Q′, A4 = Q, it is easyto verify that β(S,T ,W,Z) = −1 (cf. Figs. 2.4 and 2.5). Thanks to the symmetriesof the cross-ratio (cf. Sect. 1.5.2) we thus have β(W,Z, S,T) = β(S,T ,W,Z) =−1. Moreover, the perspectivity f : L(S,T) → lP centred at P maps W,Z, S,Tto Q,R,P′,T , respectively, so β(Q,R,P′,T) = −1. Since by hypothesis β(Q,R,

P′,P′′) = −1, we can conclude that T = P′′.Let now g : lQ → lR be the perspectivity centred at P′′. By construction we have

g(P) = P, g(R) = Q, and moreover g(Q′) = R′ since P′′ = T . Since the cross-ratiois invariant under projectivities, we then have

β(P,Q,R′, g(Q′′)) = β(g(P), g(R), g(Q′), g(Q′′)) ==β(P,R,Q′,Q′′) = 1

β(R,P,Q′,Q′′)= −1,

sog(Q′′) = R′′. Thedefinitionof perspectivity implies now thatR′′ = L(P′′,Q′′) ∩ lR:in particular, the points P′′,Q′′,R′′ are collinear.

Solution (2).By hypothesis the points P,Q,R, S define a projective frame of P2(K),

so we can choose homogeneous coordinates such that P = [1, 0, 0], Q = [0, 1, 0],R = [0, 0, 1], S = [1, 1, 1]. We thus have lP = {x0 = 0}, lQ = {x1 = 0}, lR = {x2 =0}, L(P, S) = {x1 = x2}, L(Q, S) = {x0 = x2}, L(R, S) = {x0 = x1}, hence P′ =[0, 1, 1], Q′ = [1, 0, 1], R′ = [1, 1, 0].

It is now easy to determine P′′,Q′′,R′′: a normalized basis induced by theprojective frame {Q,R,P′} of lP is given by v1 = (0, 1, 0), v2 = (0, 0, 1), so P′′ =[v1 − v2] = [0, 1,−1]. In the samewayoneobtainsQ′′ = [1, 0,−1],R′′ = [1,−1, 0].It follows that P′′,R′′,Q′′ all belong to the line of equation x0 + x1 + x2 = 0.

80 2 Exercises on Projective Spaces

P

Q

R

T

R

Q

SW

Q

R

P

Z

lQ

lR

lP

Fig. 2.5 The configuration described in Exercise27

Exercise 28. Let P(V ), P(W ) be projective spaces over the fieldK, letH,K be pro-jective subspaces of P(V ), and let f : P(V ) \ H → P(W ) be a degenerate projectivetransformation. Show that f (K \ H) is a projective subspace of P(W ) of dimensiondimK − dim(K ∩ H) − 1.

Solution. Let S,T be the linear subspaces of V such that H = P(S), K = P(T). Thedegenerate projective transformation f is induced by a linear map ϕ : V → W suchthat ker ϕ = S, and it is immediate to check that f (K \ H) = P(ϕ(T)). On the otherhand, the dimension of the linear subspace ϕ(T) is given by

dim T − dim(T ∩ S)= (dimK + 1)− (dim(K ∩ H)+ 1)= dimK − dim(K ∩ H),

so we finally have dim f (K \ H) = dimK − dim(K ∩ H) − 1.

Exercise 29. Let S,H be projective subspaces of the projective spaceP(V ) such thatS ∩ H = ∅ andL(S,H) = P(V ), and letπH : P(V ) \ H → S be the projection onto Scentred atH (cf. Sect. 1.2.7). Show that πH is a degenerate projective transformation.

Solution (1). Let n = dim P(V ), k = dim S, h = dimH. An easy application ofGrassmann’s formula shows that k + h = n − 1. Moreover, it is easy to check that,if P0, . . . ,Pk are independent points of S = P(U) and Pk+1, . . . ,Ph+k+1 are inde-pendent points of H, then the set {P0, . . . ,Ph+k+1} is in general position in P(V ), soit can be extended to a projective frame R = {P0, . . . ,Ph+k+1,Q} of P(V ).

Let us fix the system of homogeneous coordinates x0, . . . , xn induced byR. TheCartesian equations of H and S are given by x0 = · · · = xk = 0 and xk+1 = · · · =xn = 0, respectively (cf. Sects. 1.3.5 and 1.3.7). For any given P = [y0, . . . yn] /∈ H itis easy to check that the subspaceL(H,P) is the set of points [λ0y0, . . . ,λ0yk,λ0yk+1+λ1, . . . ,λ0yn + λh+1], where [λ0, . . . ,λh+1] ∈ P

h+1(K). So L(H,P) ∩ S is the point[y0, . . . , yk, 0, . . . , 0] and πH is the degenerate projective transformation induced by

2 Exercises on Projective Spaces 81

the linear map ϕ : V → U described in coordinates by the formula (x0, . . . , xn) �→(x0, . . . , xk, 0, . . . , 0).

Solution (2).LetW,U be linear subspaces of V such thatH = P(W ) and S = P(U).It is easy to check that the conditions S ∩ H = ∅, L(S,H) = P(V ) imply that W ∩U = {0}, W + U = V , respectively. So V = W ⊕ U, and the projection pU : V →U mapping every v ∈ V to the unique vector pU(v) ∈ U such that v − pU(v) ∈ Wis well defined. The map pU is linear, and we have ker pU = W . Therefore, for everyv ∈ V \ W , the line spanned by the vector pU(v) coincides with the intersectionof U with the subspace spanned by W ∪ {v}. Hence for every v ∈ V \ W we haveπH([v]) = [pU(v)], so πH is the degenerate projective transformation induced by pU .

Exercise 30. In P3(R), let us consider the plane T1 of equation x3 = 0, the plane T2

of equation x0 + 2x1 − 3x2 = 0, and the pointQ = [0, 1,−1, 1], and let f : T1 → T2be the perspectivity centred at Q. Find Cartesian equations for the image through fof the line r obtained by intersecting T1 with the plane x0 + x1 = 0.

Solution.Bydefinition of perspectivitywe have f (r)=L(Q, r) ∩ T2, so theCartesianequations of f (r) are given by the union of an equation of L(Q, r) and an equation ofT2. Moreover, r is defined by the equations x0 + x1 = x3 = 0, so the pencil of planesFr centred at r has parametric equations

λ(x0 + x1) + μx3 = 0, [λ,μ] ∈ P1(R).

By imposing that the generic plane of the pencil pass throughQ one obtains [λ,μ] =[1,−1], so L(Q, r) has equation x0 + x1 − x3 = 0. Therefore, f (r) is defined by theequations x0 + 2x1 − 3x2 = x0 + x1 − x3 = 0.

Exercise 31. Let r, s ⊂ P2(K) be distinct lines, set A = r ∩ s and let f : r → s be a

projective isomorphism. Prove that:

(a) f is a perspectivity if and only if f (A) = A.(b) If f (A) �= A, then there exist a line t of P

2(K) and two perspectivities g : r → t,h : t → s such that f = h ◦ g.

(c) Every projectivity p : r → r of r is the composition of at most three perspectiv-ities.

Solution. (a) Every perspectivity between r and s fixes the point A (cf. Sect. 1.2.8).Conversely, if the isomorphism f : r → s fixes A, then we choose distinct pointsP1,P2 ∈ r \ {A} andwe setQ1 = f (P1),Q2 = f (P2) (cf. Fig. 2.6). The linesL(P1,Q1)

and L(P2,Q2) are distinct andmeet atO /∈ r ∪ s. If g : r → s is the perspectivity cen-tred at O, we have g(A) = A, g(P1) = Q1, g(P2) = Q2. The Fundamental theoremof projective transformations now implies that f = g, so f is a perspectivity.

(b) Choose pairwise distinct points P1,P2,P3 ∈ r \ {A} in such a way that thepoints Q1 = f (P1), Q2 = f (P2), Q3 = f (P3) are distinct from A. Denote by M thepoint of intersection of the lines L(P1,Q2) and L(P2,Q1), by N the point of intersec-tion of the lines L(P1,Q3) and L(P3,Q1), and by t the line L(M,N) (cf. Fig. 2.6). It

82 2 Exercises on Projective Spaces

t

Q2

rs

Q3

Q1

P1

P2Q2

O

P3

Q1

P2

P1

N

M

A A

s r

Fig. 2.6 Statements a on the left and b on the right of Exercise31

is easy to check that the line t is distinct both from r and from s and does not containthe points P1 and Q1. Let us denote by g : r → t the perspectivity centred at Q1 andby h : t → s the perspectivity centred at P1. Since (h ◦ g)(Pi) = Qi for i = 1, 2, 3, asin the previous point the Fundamental theorem of projective transformations impliesthat f = h ◦ g.

When considering a projectivity p : r → r, in order to prove (c) it is sufficient toapply (b) to the composition of pwith any perspectivity h : r → t between r and anyline t �= r.

Exercise 32. (Parametrization of a pencil of hyperplanes) Let P(V ) be a projectivespace of dimension n and let H ⊂ P(V ) be a subspace of codimension 2. Denoteby FH the pencil of hyperplanes centred at H. If t is a line transverse to FH , i.e., ift ⊂ P(V ) is a line disjoint from H, then denote by ft : t → FH the map sending anypointP ∈ t to the hyperplaneL(P,H) ∈ FH . Prove that ft is a projective isomorphism(which is called the parametrization of FH via the transverse line t).

Solution. Choose in P(V ) a system of homogeneous coordinates such that Hhas equations x0 = x1 = 0 and the line t has equations x2 = · · · = xn = 0; in thisway x0, x1 is a system of homogeneous coordinates on t. In the dual systemof homogeneous coordinates a0, . . . , an on P(V )∗, the pencil FH has equationsa2 = · · · = an = 0, so a0, a1 is a system of homogeneous coordinates on FH . Withrespect to these coordinates, the map ft : t → FH is given by [x0, x1] �→ [x1,−x0],so it is a projective isomorphism.

Note. One can use the fact that the parametrization ft is a projective isomor-phism in order to provide an alternative definition of the cross-ratio of four hyper-planes S1, S2, S3, S4 belonging to a pencil FH , without referring to the notion ofdual projective space (cf. Sect. 1.5.1). If t is a transverse line, it is sufficient toset β(S1, S2, S3, S4) = β(P1,P2,P3,P4), where Pi = t ∩ Si, i = 1, . . . , 4: in fact,the invariance of the cross-ratio under projective isomorphisms and the fact thatft(Pi) = Si ensure that the value β(P1,P2,P3,P4) does not depend on the choice of t.

2 Exercises on Projective Spaces 83

We also observe that one can prove that the cross-ratio β(S1 ∩ t, S2 ∩ t,S3 ∩ t, S4 ∩ t) is independent of t without bringing the pencil FH into play. Tothis aim it is sufficient to note that, if t′ is any other transverse line, and P′

i =t′ ∩ Si, i = 1, 2, 3, 4, then the points P′

1,P′2,P

′3,P

′4 are the images of the points

P1,P2,P3,P4 via the perspectivity between t and t′ centred at H (cf. Sect. 1.2.8),so β(P′

1,P′2,P

′3,P

′4) = β(P1,P2,P3,P4).

When n = 2 we have thus proved that, if FO is the pencil of lines of P2(K)

centred at O, then the parametrization of FO obtained via the transverse line t isa projective isomorphism. Moreover, if the lines t1, t2 do not contain O, then theperspectivity between t1 and t2 centred at O is the composition f −1

t2 ◦ ft1 . Since thecomposition of projective isomorphisms is a projective isomorphism, this argumentprovides an alternative proof of the fact that every perspectivity between two linesof the projective plane is a projective isomorphism.

Exercise 33. Let r and H be a line and a plane of P3(K), respectively; suppose that

r � H, and let P = r ∩ H. Let Fr be the pencil of planes of P3(K) centred at r, and

let FP be the pencil of lines of H centred at P. Prove that the map β : Fr → FP

defined by β(K) = K ∩ H is a well-defined projective isomorphism.

Solution. Let s ⊆ H be a line such that P /∈ s, and let fr : s → Fr , fP : s → FP bethe maps defined by fr(Q) = L(r,Q), fP(Q) = L(P,Q) for every Q ∈ s. By con-struction, s does not contain P and is skew to r, so Exercise32 ensures that fr andfP are well-defined projective isomorphisms. It is now immediate to check that themap β coincides with the composition fP ◦ f −1

r , so β is a well-defined projectiveisomorphism.

Exercise 34. Let r, s ⊂ P2(K) be distinct lines, let A = r ∩ s, and let f : r → s be

a projective isomorphism such that f (A) = A. Let also

W (f ) = {L(P1, f (P2)) ∩ L(P2, f (P1)) |P1,P2 ∈ r, P1 �= P2}.

Prove that W (f ) is a projective line containing A.

E Solution (1). We have seen in Exercise31 that f is a perspectivity centred at O ∈P2(K) \ (r ∪ s). Let l = L(A,O), and let FA be the pencil of lines of P

2(K) centredat A: by construction r, s, l ∈ FA. In order to show that W (f ) is contained in aline passing through A it is sufficient to show that, as M varies in W (f ) \ {A}, thecross-ratio β(s, r, l,L(A,M)) does not depend on M: namely, if this is the case andk = β(s, r, l,L(A,M)), then we necessarily have W (f ) ⊆ t, where t is the uniqueline of FA such that β(s, r, l, t) = k.

So let P1,P2 be distinct points of r, and let

M = L(P1, f (P2)) ∩ L(P2, f (P1)) = L(P1, s ∩ L(O,P2)) ∩ L(P2, s ∩ L(O,P1)).

84 2 Exercises on Projective Spaces

Of course, if P1 = A or P2 = A then M = A, so we can suppose that P1,P2

are distinct from A. Then it is easy to check that M �= A. We now show thatβ(s, r, l,L(A,M)) = −1. As already observed, this implies that W (f ) is containedin the line t ∈ FA such that β(s, r, l, t) = −1.

Let w = L(O,M), and observe that A /∈ w, so w is transverse to the pencil FA. IfN = s ∩ w and Z = r ∩ w, then by the Note following Exercise32 we have

β(s, r, l,L(A,M)) = β(s ∩ w, r ∩ w, l ∩ w,L(A,M) ∩ w) = β(N,Z,O,M).

On the other hand it is easy to check that, if we apply the construction describedin the statement of Exercise26 to the case when A1 = P1, A2 = f (P1), A3 = P2,A4 = f (P2), then we get β(O,M,N,Z) = −1 (cf. Figs. 2.4 and 2.7). Thanks to thesymmetries of the cross-ratio (cf. Sect. 1.5.2) we then have β(N,Z,O,M) = −1, soβ(s, r, l,L(A,M)) = −1, as desired.We have thus proved that the inclusionW (f ) ⊆t holds.

Let us check that also the converse inclusion holds. For every P ∈ r \ {A}we haveL(P, f (A)) ∩ L(f (P),A) = A, so A ∈ W (f ). So let R ∈ t \ {A} and let v be any linepassing through R and distinct both from t and from L(O,R). Let P1 = v ∩ r, P2 =f −1(v ∩ s). Since R �= A, v �= t and v �= L(O,R), the points P1,P2 are distinct andthey are also distinct from A. Moreover L(P1, f (P2)) = v, so what has been provedabove implies that L(P1, f (P2)) ∩ L(P2, f (P1)) ∈ v ∩ t = {R}, and R ∈ W (f ). Wehave thus shown that t ⊆ W (f ), hence W (f ) = t, as desired.

E Solution (2). Let us fix a point P ∈ r \ {A}, take P0 ∈ r \ {A,P} and let M = L(P, f (P0)) ∩ L(P0, f (P)). We will show that, if t = L(A,M), then W (f ) = t.

We first prove the inclusion W (f ) ⊆ t. Observe that, if g : r → t is the perspec-tivity centred at f (P) and h : t → s is the perspectivity centred at P, then f and h ◦ gcoincide on A,P,P0, so f = h ◦ g (cf. Fig. 2.8). We easily deduce that for everyP1 ∈ r \ {P} we have L(P1, f (P)) ∩ L(P, f (P1)) = g(P1) ∈ t.

N

P2

s

A

lw

O

Z

M

f(P2)f(P1)

P1

r

t

Fig. 2.7 The configurations described in Solution (1) of Exercise34

2 Exercises on Projective Spaces 85

Let us nowprove that indeedL(P1, f (P2)) ∩ L(P2, f (P1)) ∈ t for everyP1,P2 ∈ r,P1 �= P2. This obviously holds if P1 = A or P2 = A, and also holds if P1 = P orP2 = P thanks to the previous considerations. Therefore, we may assume P1,P2 ∈r \ {A,P}. We consider the hexagon with vertices

P1, Q2 = f (P2), P, Q1 = f (P1), P2, Q = f (P).

By Pappus’ Theorem (cf. Exercise13 and Figs. 2.2, 2.8) the points

L(P1,Q2) ∩ L(P2,Q1), L(P,Q2) ∩ L(P2,Q), L(P,Q1) ∩ L(P1,Q)

are collinear. We have observed above that the second and the third of thesepoints are distinct and lie on the line t, so also the point L(P1,Q2) ∩ L(P2,Q1) =L(P1, f (P2)) ∩ L(P2, f (P1)) lies on t. We have thus proved that W (f ) ⊆ t.

We now come to the opposite inclusion. We observe that A ∈ W (f ) because forevery P ∈ r \ {A} we have L(P, f (A)) ∩ L(A, f (P)) = A. So let Q ∈ t \ {A}. We firstshow that f cannot be a perspectivity centred at Q. In fact, it is immediate to checkthat, if P1,P2 are distinct points of r \ {A}, then the points P1,P2, f (P1), f (P2) are ingeneral position, so the pointsA = L(P1,P2) ∩ L(f (P1), f (P2)),B = L(P1, f (P2)) ∩L(P2, f (P1)), L(P1, f (P1)) ∩ L(P2, f (P2)) are not collinear (cf. Exercise6). But wehave proved above that A ∈ t, B ∈ t, so if f were a perspectivity centred at Q,then we would have L(P1, f (P1)) ∩ L(P2, f (P2)) = Q ∈ t, and the points A,B,Qwould be collinear, a contradiction. So there exists a line v �= t containing Qand such that the points R1 = v ∩ r and R2 = f −1(v ∩ s) are distinct. We thushave L(R1, f (R2)) = v, hence L(R1, f (R2)) ∩ L(R2, f (R1)) ∈ t by the considerationsabove. Therefore L(R1, f (R2)) ∩ L(R2, f (R1)) = t ∩ v = Q, so Q ∈ W (f ). We havethus proved that t ⊆ W (f ).

t

P

P1

P2

t

Q

Q1

P0

P

M

f(P )

AA

Q2f(P0)

Fig. 2.8 Solution (2) of Exercise34: on the left, f is described as the composition of two perspec-tivities; on the right, the inclusion W (f ) ⊆ t as a consequence of Pappus’ Theorem

86 2 Exercises on Projective Spaces

Solution (3). By Exercise31, f is a perspectivity centred at O ∈ P2(K) \ (r ∪ s).

Take points B ∈ r, C ∈ s such that A,B,C,O are in general position, and endowP2(K) with the system of homogeneous coordinates induced by the projective frame

{A,B,C,O}. Then we have r = {x2 = 0}, s = {x1 = 0}. Let now P,P′ be distinctpoints in r \ {A}. We have P = [a, 1, 0], P′ = [a′, 1, 0] for some a, a′ ∈ K. Usingthat

f (P) = L(O,P) ∩ s = L([1, 1, 1], [a, 1, 0]) ∩ {x1 = 0},

one easily proves that f (P) = [1 − a, 0, 1]. In a similar way one gets f (P′) =[1 − a′, 0, 1]. We thus have L(P, f (P′)) = {x0 − ax1 + (a′ − 1)x2 = 0} andL(P′, f (P)) = {x0 − a′x1 + (a − 1)x2 = 0}, soL(P, f (P′)) ∩ L(P′, f (P))= [1 − a −a′,−1, 1]. It follows that, if P,P′ are distinct points in r \ {A}, then the pointL(P, f (P′)) ∩ L(P′, f (P)) lies on t \ {A}, where t is the line (through A) of equa-tion x1 + x2 = 0, and every point of t \ {A}arises in this way. On the other hand, forevery P ∈ r \ {A} one has L(P, f (A)) ∩ L(A, f (P)) = A, so W (f ) = t, as desired.

Exercise 35. Let r, s ⊂ P2(K) be distinct lines, set A = r ∩ s, and let f : r → s be

a projective isomorphism such that f (A) �= A. Let also B = f −1(A) ∈ r and set

W (f ) = {L(P, f (P′)) ∩ L(P′, f (P)) |P,P′ ∈ r, P �= P′, {P,P′} �= {A,B}}.

Prove that W (f ) is a projective line.

E Solution (1). Let us fix a point P1 ∈ r \ {A,B}. If P2,P3 are distinct points ofr \ {A,B,P1} and we set M = L(P1, f (P2)) ∩ L(P2, f (P1)), N = L(P1, f (P3)) ∩L(P3, f (P1)), t = L(M,N) (cf. Fig. 2.9), then it is easy to check that the pointsM,Nand the line t are well defined and that f = h ◦ g, where g : r → t is the perspectivitycentred at f (P1) and h : t → s is the perspectivity centred at P1. It readily followsthat

L(f (P1),P) ∩ L(P1, f (P)) = g(P) ∈ t ∀P ∈ r \ {P1}. (2.3)

In order to prove thatW (f ) ⊆ t, let us first show that t does not depend on the choiceof P1. If C = f (A) ∈ s, since f (A) �= A the points A,B,C are in general position.Together with our previous considerations, this implies that

B = L(P1,A) ∩ L(f (P1),B) = L(P1, f (B)) ∩ L(f (P1),B) ∈ tC = L(f (P1),A) ∩ L(P1,C) = L(f (P1),A) ∩ L(P1, f (A)) ∈ t.

(2.4)

Therefore, the line t contains both B and C, so that t = L(B,C). In particular, t doesnot depend on the choice of P1.

2 Exercises on Projective Spaces 87

NM

s

t

Br

A

P2

P1

P3

C

f(P2)

f(P1)

f(P3)

Fig. 2.9 The construction of the set W (f ) described in Solution (1) of Exercise35

Combining (2.3) with the fact that t is independent of P1 we deduce that

L(P, f (P′)) ∩ L(P′, f (P)) ∈ t ∀P ∈ r \ {A,B}, P′ ∈ r \ {P}.

Now the inclusion W (f ) ⊆ t readily follows from the fact that the expressionL(P, f (P′)) ∩ L(P′, f (P)) is symmetric in P,P′.

Let us now check that t ⊆ W (f ). We first observe that, by (2.4), the pointsB = t ∩ r,C = t ∩ s belong toW (f ). Let thenQ ∈ t \ {B,C}. If f (P) = L(P,Q) ∩ sfor every P ∈ r \ {A,B}, then f is the perspectivity centred at Q, so f (A) = A,against the hypothesis. Therefore, there exists P1 ∈ r \ {A,B} such that f (P1) �=L(P1,Q) ∩ s. Let P2 = f −1(L(P1,Q) ∩ s). By construction P2 �= P1. Let now Q′ =L(P1, f (P2)) ∩ L(P2, f (P1)). By construction Q′ ∈ L(P1, f (P2)) = L(P1,Q), andour previous considerations imply that Q′ ∈ W (f ) ⊆ t. It follows that Q′ = t ∩L(P1,Q) = Q, so Q ∈ W (f ). We have thus shown that t ⊆ W (f ), as desired.

Solution (2). Let us set C = f (A) ∈ s. The points A,B,C are not collinear, so wemay choose a system of homogeneous coordinates x0, x1, x2 such that A = [1, 0, 0],B = [0, 1, 0], C = [0, 0, 1]. We then have r = {x2 = 0}, s = {x1 = 0}, and we canendow r and s with the systems of homogeneous coordinates induced by x0, x1, x2.Since f (B) = A and f (A) = C, with respect to these coordinates the isomorphism f

is represented by the matrix

(0 λ1 0

), for some λ ∈ K

∗.

If P = [a, b, 0], P′ = [a′, b′, 0] are distinct points of r such that {P,P′} �= {A,B},we thus have f (P) = [λb, 0, a], f (P′) = [λb′, 0, a′]; so L(P, f (P′)) and L(P′, f (P))

have equationsba′x0 − aa′x1 − λbb′x2 = 0 andab′x0 − aa′x1 − λbb′x2 = 0, respec-tively. Since P �= P′ we have ab′ − ba′ �= 0, so the lines L(P, f (P′)) and L(P′, f (P))

meet at [0,λbb′,−aa′] (observe that we cannot have λbb′ = aa′ = 0 because{P,P′} �= {A,B}). If t is the line of equation x0 = 0, we can conclude thatW (f ) ⊆ t.

88 2 Exercises on Projective Spaces

On the other hand, take P = [λ, 1, 0] and P′ = [a′, b′, 0] with a′ �= λb′. ThenP �= P′ and {P,P′} �= {A,B}, and the previous computation shows that

L(P, f (P′)) ∩ L(P′, f (P)) = [0, b′,−a′].

This proves that every point of t, except possibly [0, 1,−λ], belongs to W (f ). Onthe other hand, of course we can choose distinct elements a, a′ ∈ K such that aa′ =λ2. If P = [a, 1, 0], P′ = [a′, 1, 0], then it is immediate to check that P �= P′ and{P,P′} �= {A,B}, and

L(P, f (P′)) ∩ L(P′, f (P)) = [0, 1,−λ].

We thus have t ⊆ W (f ), hence W (f ) = t, as desired.

Note. In Solution (1), after proving (2.3) one can conclude thatW (f ) = t by exploit-ing Pappus’ Theorem in the same spirit as in Solution (2) of Exercise34.

K Exercise 36. (a) Let A,A′,C,C′ be points of P1(K) such that A /∈ {C,C′} and

A′ /∈ {C,C′}. Prove that there exists a unique non-trivial involution f : P1(K) →

P1(K) such that f (A) = A′, f (C) = C′.

(b) Let A,B,C,A′,B′,C′ be points of P1(K) such that each quadruple A,B,C,C′

and A′,B′,C′,C contains only pairwise distinct points. Prove that there exists aunique involution f : P

1(K) → P1(K) such that f (A) = A′, f (B) = B′,

f (C) = C′ if and only if

β(A,B,C,C′) = β(A′,B′,C′,C).

(c) Let r ⊆ P2(K) be a projective line and let P1,P2,P3,P4 be points of P

2(K) \ rin general position. For every i �= j let sij = L(Pi,Pj), and set

A = r ∩ s12, B = r ∩ s13, C = r ∩ s14A′ = r ∩ s34, B′ = r ∩ s24, C′ = r ∩ s23

(cf. Fig. 2.10). Prove that there exists a unique involution f of r such thatf (A) = A′, f (B) = B′, f (C) = C′.

Solution. (a) The case when A = A′ and C = C′ has already been settled in Exer-cise24.

Let us now suppose A = A′, C �= C′ (the case when A �= A′, C = C′ being sim-ilar). If f satisfies the conditions of the statement, then necessarily f (A) = A′ = A,f (C) = C′, f (C′) = f (f (C)) = C. Moreover, by the Fundamental theorem of projec-tive transformations, there exists a unique projectivity mapping A,C,C′ to A,C′,C,respectively. By point (b) of Exercise23, this projectivity is an (obviously non-trivial)involution, and this concludes the proof in the case when A = A′, C �= C′.

We finally suppose A �= A′, C �= C′. Our hypothesis implies that the triplesA,A′,C and A,A′,C′ define two projective frames of P

1(K), so there exists a unique

2 Exercises on Projective Spaces 89

A

C

B

B

r

P1

A

P2

C

Z

P3

P4

W

Fig. 2.10 Exercise36, point (c): the case when Z /∈ r

projectivity f : P1(K) → P

1(K) such that f (A) = A′, f (A′) = A, f (C) = C′. More-over, by point (b) of Exercise23, the projectivity f is a non-trivial involution. Onthe other hand, any projectivity satisfying the required conditions necessarily mapsA,A′,C to A′,A,C′, respectively, so it must coincide with f .

(b) If an involution f with the required properties exists, then f (C′) = f (f (C)) =C, so

β(A,B,C,C′) = β(f (A), f (B), f (C), f (C′)) = β(A′,B′,C′,C)

thanks to the invariance of the cross-ratio with respect to projective transformations.We then suppose that β(A,B,C,C′) = β(A′,B′,C′,C), and we show that the

required involution exists (the fact that such involution is unique readily followsfrom the Fundamental theorem of projective transformations). By point (a), thereexists a non-trivial involution f : P

1(K) → P1(K) such that f (A) = A′, f (C) = C′.

Since f (C′) = C, we have β(A′,B′,C′,C) = β(A,B,C,C′) = β(f (A), f (B), f (C),

f (C′)) = β(A′, f (B),C′,C), where the first equality follows from the hypothesis,while the second one is due to the invariance of the cross-ratio with respect toprojective transformations. Since the points A′,B′,C′,C are pairwise distinct, thecondition β(A′,B′,C′,C) = β(A′, f (B),C′,C) implies that f (B) = B′, as desired.

(c)We first observe that the pointsA,B,C are pairwise distinct, because otherwisethe line r would contain P1, against the hypothesis. In a similar way one proves thatthe points A′,B′,C′ are also pairwise distinct, because otherwise the line r wouldpass through P2, P3 or P4, against the hypothesis.

Let us set W = s12 ∩ s34, T = s13 ∩ s24, Z = s23 ∩ s14. Since P1,P2,P3,P4 arein general position, the points W,T ,Z are not collinear (cf. Exercise6).

We first consider the case when r does not contain Z . Since C = r ∩ s14 andC′ = r ∩ s23, we have that C �= C′. Also observe that C′ �= A (since P2 /∈ r), andC′ �= B (since P3 /∈ r). In a similar way, since P4 /∈ r we have C �= A′ and C �= B′.Therefore, the quadruples A,B,C,C′ and A′,B′,C′,C contain only pairwise distinctpoints, so by point (b) we are left to show that β(A,B,C,C′) = β(A′,B′,C′,C).

90 2 Exercises on Projective Spaces

Let g : r → s23 be the perspectivity centred at P1 and h : s23 → r the perspec-tivity centred at P4. By construction, g maps the points A,B,C,C′ to the pointsP2,P3,Z,C′, respectively, and h maps P2,P3,Z,C′ to B′,A′,C,C′, respectively.We thus have

β(A,B,C,C′) = β(h(g(A)), h(g(B)), h(g(C)), h(g(C′))) == β(B′,A′,C,C′) = β(A′,B′,C′,C),

where the first equality is due to the invariance of the cross-ratio under projec-tive isomorphisms, while the last one is due to the symmetries of the cross-ratio(cf. Sect. 1.5.2). This concludes the proof in the case when r does not contain Z .

If Z ∈ r but T /∈ r, then we have B �= B′; a similar argument as above showsthat the quadruples A,C,B,B′ and A′,C′,B′,B contain only pairwise distinctpoints. By point (b), in order to conclude it is sufficient to show thatβ(A,C,B,B′) = β(A′,C′,B′,B). This equality can be proved as above by consid-ering first the perspectivity between r and L(P2,P4) centred at P1 and then theperspectivity between L(P2,P4) and r centred at P3.

Finally, if r passes both through Z and through T , then necessarilyW /∈ r, becauseotherwise W,T ,Z would be collinear. Then, a suitable variation of the previousargument allows one to conclude the proof also in this case.

Note. A quadrilateral of P2(K) is an unordered set of 4 points in general position

(called vertices) of the projective plane. A pair of opposite sides of a quadrilateral Qis a pair of lineswhose union contains the vertices ofQ (observe that any quadrilateralhas exactly three pairs of opposite sides).

Let r be a line that does not contain any vertex ofQ, and observe that the union ofeach pair of opposite sides ofQmeets r at exactly two points. Point (c) of Exercise36can be reformulated as follows: there exists an involution of r that switches the pointsof each pair determined on r by a pair of opposite sides of Q.

Exercise 37. Let P(V ) be a projective space of dimension n and let S1, S2 be distincthyperplanes of P(V ). Show that a projective isomorphism f : S1 → S2 is a perspec-tivity if and only if f (A) = A for every A ∈ S1 ∩ S2.

Solution (1). It is sufficient to show that, if f : S1 → S2 is a projective isomorphismthat is the identity on S1 ∩ S2, then f is a perspectivity. Let us choose points P1,P2 ∈S1 \ (S1 ∩ S2), set Q1 = f (P1), Q2 = f (P2) and denote by A the point at which theline L(P1,P2)meets the subspace S1 ∩ S2. Since f sends collinear points to collinearpoints and f (A) = A by hypothesis, the point A is collinear with Q1 and Q2. So thesubspace L(P1,P2,Q1,Q2) is a plane, and the lines L(P1,Q1) and L(P2,Q2) meetat a point O. It is immediate to check that O /∈ S1 ∪ S2; therefore, we can considerthe perspectivity g : S1 → S2 centred at O. The projective isomorphisms f and gcoincide on P1, P2 and on the whole of S1 ∩ S2. Therefore, the Fundamental theoremof projective transformations ensures that f = g.

Solution (2).We now give an analytic proof of the statement, by carrying out somecomputations in a suitably chosen system of homogeneous coordinates x0, . . . , xn

2 Exercises on Projective Spaces 91

A O1r2

S2

r1

l

S4

A

g1(l)

S2

O2

r2

lS5

g1(l)

O2S3

S3

S3S1

A

S2

Fig. 2.11 On the left: on the top, the construction of g1; on the bottom, the choice of l and theconstruction of S4. On the right: the choice of S5 and the conclusion of the proof

such that S1 has equation xn = 0 and S2 has equation xn−1 = 0. Using the fact that f isthe identity on the subspace of equations xn−1 = xn = 0, it is not difficult to show thatthe map f is described by the formula [x0, . . . , xn−1, 0] �→ [x0 + a0xn−1, . . . , xn−2 +an−2xn−1, 0, anxn−1], where a0, . . . an−2 ∈ K and an ∈ K

∗. Then, one can checkdirectly that f is the perspectivity centred at O = [a0, . . . , an−2,−1, an].Note. Exercise37 extends point (a) of Exercise31 (that provides a characterizationof perspectivities between lines in a projective plane) to the case of any dimension.

K Exercise 38. Let S1, S2 be distinct planes of P3(K). Prove that any projective iso-

morphism f : S1 → S2 is the composition of at most three perspectivities.

Solution. Let r1 = S1 ∩ S2 and take a point A ∈ S1 \ r1 such that the point A′ = f (A)

does not belong to r1 (cf. Fig. 2.11). Let S3 �= S2 be a plane passing through A′ andnot containing A, and choose a point O1 ∈ L(A,A′) \ {A,A′}. If π1 : S3 → S1 is theperspectivity centred at O1, then g1 = f ◦ π1 : S3 → S2 is a projective isomorphism.Moreover, A′ ∈ r2 = S3 ∩ S2 and g1(A′) = A′.

Let l �= r2 be a line contained in S3 and passing through A′. Since g1(A′) = A′, themap g1 transforms l into a line passing through A′. Let now S4 be the plane containingl and g1(l), so that l = S3 ∩ S4. Then Exercise31 implies that g1|l is a perspectivitycentred at a point O2 ∈ S4 such that O2 /∈ S3.

92 2 Exercises on Projective Spaces

r

sP

P

f(P )

P

f(P )Q

f(P )

lQ

t

t

Fig. 2.12 The construction described in the solution of Exercise39

Let now S5 be a plane containing g1(l) and distinct both from S2 and from S4,and let us consider the perspectivity π2 : S5 → S3 centred at O2. Then g1 ◦ π2 :S5 → S2 is a projective isomorphism that fixes the line g1(l) = S5 ∩ S2 pointwise.Now Exercise37 implies that there exists a perspectivity π3 such that g1 ◦ π2 = π3,so that f ◦ π1 ◦ π2 = π3. This implies the conclusion, since the inverse of aperspectivity is again a perspectivity.

Exercise 39. Let r, s ⊂ P3(K) be skew lines, and let f : r → s be a projective iso-

morphism. Prove that there exist infinitely many lines l such that f coincides withthe perspectivity centred at l.

Solution. Take P ∈ r, and let t = L(P, f (P)). We will show that, for every Q ∈t \ {P, f (P)}, there exists a line lQ with the following properties: lQ passes throughQ, lQ is skew both to r and to s, and f coincides with the perspectivity centred at lQ.WhenQ varies in t \ {P, f (P)}, the lines lQ are pairwise distinct, and this implies theconclusion.

Let us fix Q ∈ t \ {P, f (P)}, and choose P′,P′′ ∈ r \ {P}, such that P′ �= P′′. Weset t′ = L(P′, f (P′)), t′′ = L(P′′, f (P′′)) (cf. Fig. 2.12). If t′, t′′ were not skew, thenP′,P′′, f (P′), f (P′′) would be coplanar, so r and s would also be coplanar, a contra-diction. In the same way one proves that t is skew both to t′ and to t′′, so Q /∈ t′ ∪ t′′.Therefore, Exercise8 implies that there exists a unique line lQ that contains Q andmeets both t′ and t′′. Observe that no plane S can contain lQ ∪ r (or lQ ∪ s), becauseotherwise each of the lines t′, t′′ would contain at least two points of S, thus t′ and t′′would be coplanar, a contradiction. So lQ is skew both to r and to s. By construction,the perspectivity from r to s centred at lQ maps P to f (P), P′ to f (P′) and P′′ tof (P′′). By the Fundamental theorem of projective transformations, we deduce thatthis perspectivity coincides with f , whence the conclusion.

Note. An alternative solution of Exercise39 can be obtained by following the strat-egy described in the Note following Exercise 178.

2 Exercises on Projective Spaces 93

Exercise 40. Let P1 and P2 be distinct points of P2(K) and letF1,F2 be the pencils

of lines centred at P1 and P2, respectively. Let f : F1 → F2 be any function. Showthat the following facts are equivalent:

(i) f is a projective isomorphism such that f (L(P1,P2)) = L(P1,P2),(ii) there exists a line r disjoint from {P1,P2} and such that f (s) = L(s ∩ r,P2) for

every s ∈ F1.

Solution. (i) ⇒ (ii). Via the duality correspondence (cf. Sect. 1.4.2), the pencilsF1,F2 correspond to lines l1, l2 of the dual projective plane P

2(K)∗. Moreover,the projective isomorphism f : F1 → F2 induces the dual projective isomorphismf∗ : l1 → l2. The condition f (L(P1,P2)) = L(P1,P2) translates into the fact that f∗fixes the point at which l1 and l2 meet, and this implies in turn that f∗ is a perspectivity(cf. Exercise31). LetR ∈ P

2(K)∗ be the centre of this perspectivity, and let r ⊂ P2(K)

be the line corresponding to R via duality. Since R /∈ l1 ∪ l2 we have that P1 /∈ rand P2 /∈ r, and the fact that f∗(Q) = L(R,Q) ∩ l2 for every Q ∈ l1 implies thatf (s) = L(P2, r ∩ s) for every s ∈ F1, as desired.

(ii) ⇒ (i). For i = 1, 2, let gi : Fi → r be the map defined by gi(s) = s ∩ r.We have shown in Exercise32 and in the Note following it that the map gi is awell-defined projective isomorphism. It follows that f = g−1

2 ◦ g1 is a projectiveisomorphism. The fact that f (L(P1,P2)) = L(P1,P2) readily follows from thedefinition of f .

Exercise 41. In P2(R) consider the point P = [1, 2, 1] and the lines

l1 = {x0 + x1 = 0}, m1 = {x0 + 3x2 = 0},l2 = {x0 − x1 = 0}, m2 = {x2 = 0},l3 = {x0 + 2x1 = 0}, m3 = {3x0 + x2 = 0}.

Determine the points Q ∈ P2(R) for which there exists a projectivity f : P

2(R) →P2(R) such that f (P) = Q and f (li) = mi, i = 1, 2, 3.

Solution. First observe that the lines l1, l2, l3 (m1,m2,m3, respectively) belong tothe pencil FO centred at O = [0, 0, 1] (to the pencil FO′ centred at O′ = [0, 1, 0],respectively). Let r = L(O,P) = {2x0 − x1 = 0}.

Suppose now that f is a projectivity such that f (li) = mi for every i = 1, 2, 3. Thenf (O) = O′, and the dual projectivity associated to f induces a projective isomorphismbetween FO and FO′ . This isomorphism maps li to mi for i = 1, 2, 3. Therefore,if r′ = f (r), thenβ(l1, l2, l3, r) = β(f (l1), f (l2), f (l3), f (r)) = β(m1,m2,m3, r′) (cf.Sect. 1.5.1 for the definition and the basic properties of the cross-ratio of concurrentlines). We now fix on FO (on FO′ , respectively) a projective frame with respectto which the line of equation ax0 + bx1 = 0 (the line of equation ax0 + bx2 = 0,respectively) has homogeneous coordinates equal to [a, b]. With this choice, thelines l1, l2, l3, r ∈ FO have coordinates [1, 1], [1,−1], [1, 2], [2,−1], respectively, soβ(l1, l2, l3, r) = −9. On the other hand, the linesm1,m2,m3 ∈ FO′ have coordinates[1, 3], [0, 1], [3, 1], respectively, so if [a0, b0] are the coordinates of r′ in FO′ , thenwe have

94 2 Exercises on Projective Spaces

−9 = β([1, 3], [0, 1], [3, 1], [a0, b0]) = 1 · b0 − 3 · a01 · 1 − 3 · 3 · 3 · 1 − 0 · 3

a0 · 1 − b0 · 0 ,

hence [a0, b0] = [1, 27] and r′ = {x0 + 27x2 = 0}.We deduce that f (P) ∈ f (r) = r′.Moreover, since f is injective, we have f (P) �= f (O), so f (P) ∈ r′ \ {O′}.

Let nowQ ∈ r′ \ {O′}, and let us show that there exists a projectivity f : P2(R) →

P2(R) such that f (li) = mi for i = 1, 2, 3, and f (P) = Q. Let A1 ∈ l1 \ {O} and

A2 ∈ l2 \ ({O} ∪ (l2 ∩ L(A1,P)). By construction, the points O,A1,A2,P define aprojective frame of P

2(R). In a similar way, if B1 ∈ m1 \ {O} and B2 ∈ m2 \ ({O′} ∪L(B1,Q)), then the points O′,B1,B2,Q provide a projective frame of P

2(R). Letf : P

2(R) → P2(R) be the unique projectivity such that f (O) = O′, f (A1) = B1,

f (A2) = B2, f (P) = Q. Since f transforms lines into lines, we obviously havef (l1) = m1, f (l2) = m2, so we are left to show that f (l3) = m3.

The dual projectivity f∗ associated to f maps FO to FO′ and satisfies f∗(l1) = m1,f∗(l2) = m2, f∗(r) = r′, so thanks to the invariance of the cross-ratio under projectiveisomorphisms we get

β(m1,m2,m3, r′) = β(l1, l2, l3, r) == β(f∗(l1), f∗(l2), f∗(l3), f∗(r)) = β(m1,m2, f∗(l3), r′)

and f (l3) = m3, as desired.

Exercise 42. In P2(R) consider the lines l1, l2, l3 having equations x2 = 0,

x2 − x1 = 0, x2 − 2x1 = 0, respectively, and the line l4 having equationα(x0 − x1) +x2 − 4x1 = 0, whereα ∈ R. Consider also the linesm1,m2,m3,m4 having equationsx1 = 0, x1 − x0 = 0, x0 = 0, x1 − γx0 = 0, respectively, where γ ∈ R.

Find the values ofα and γ for which there exists a projectivity f ofP2(R) such that

f (li) = mi for i = 1, . . . , 4, and that transforms the line x0 = 0 into the line x2 = 0.Choose a pair of such values, and describe explicitly a projectivity satisfying the

required properties.

Solution (1). It is easy to check that the lines l1, l2, l3 all intersect at R = [1, 0, 0],while the lines m1,m2,m3,m4 all intersect at S = [0, 0, 1]. Therefore, a necessarycondition for the existence of f is that also the line l4 passes through R, i.e., thatα = 0. Moreover, if f exists, then the restriction of f to the line r = {x0 = 0} isa projective isomorphism between r and the line s = {x2 = 0}. We observe thatr intersects the lines li, i = 1, . . . , 4 at the points P1 = [0, 1, 0], P2 = [0, 1, 1],P3 = [0, 1, 2], P4 = [0, 1, 4], respectively, while s intersects the lines mi, i =1, . . . , 4, at the points Q1 = [1, 0, 0], Q2 = [1, 1, 0], Q3 = [0, 1, 0], Q4 = [1, γ, 0],respectively. Therefore, if f esists, then the restriction f |r : r → s is a projective iso-morphism such that f (Pi) = Qi for i = 1, . . . , 4, and the cross-ratioβ(P1,P2,P3,P4)

must coincide with the cross-ratio β(Q1,Q2,Q3,Q4). An easy computation nowshows that β(P1,P2,P3,P4) = 2

3 and β(Q1,Q2,Q3,Q4) = γγ − 1 . Therefore, by

imposing γγ − 1 = 2

3 , we obtain a second necessary condition for the existence of

f , namely γ = −2.

2 Exercises on Projective Spaces 95

We now prove that the conditions α = 0 and γ = −2 are also sufficient for theexistence of a projectivity f with the required properties. In fact, if γ = −2 thenthere exists a (unique) projective isomorphism g : r → s such that g(Pi) = Qi fori = 1, . . . , 4.Let us endow r, swith the systemsof homogeneous coordinates [x1, x2],[x0, x1], respectively, and let B be a 2 × 2 invertible matrix representing g with

respect to these coordinate systems. Then the matrix A =⎛⎝

00

B

1 0 0

⎞⎠ represents

(with respect to the standard homogeneous coordinates of P2(R)) a projectivity f

satisfying the required properties.Let us now construct g so that g([1, 0]) = [1, 0], g([1, 1]) = [1, 1], g([1, 2]) =

[0, 1]. In this way g(Pi) = Qi for i = 1, . . . , 3, so also g(P4) = Q4, due to the con-dition imposed on the cross-ratios. An easy computation shows that g is induced

e.g. by the matrix B =(2 −10 1

). Therefore, the projectivity P

2(R) induced by the

matrix A =⎛⎝0 2 −10 0 11 0 0

⎞⎠ satisfies the required properties.

We observe that, once the necessary conditions α = 0 and γ = −2 have beenestablished, one could also directly construct f as follows. Let U be a point inl3 \ {R,P3}, e.g.U = [1, 1, 2], and let V be a point inm3 \ {S,Q3}, e.g. V = [0, 1, 1].Since {R,P1,P2,U} and {S,Q1,Q2, V } are both projective frames of P

2(R), thereexists a unique projectivity f ofP2(R) thatmapsR,P1,P2,U to S,Q1,Q2, V , respec-tively. As r = L(P1,P2) and l3 = L(R,U), it easily follows that f (r)=L(Q1,Q2)=sand f (l3) = L(S, V ) = m3. Since P3 = r ∩ l3 and Q3 = s ∩ m3, one can concludethat f (P3) = Q3. Finally, the fact that f (P4) = Q4 follows from the conditionimposed on the cross-ratios.

Solution (2). If we consider a line of P2(R) as a point of P

2(R)∗, then the linesdescribed in the statement of the exercise correspond to the points

L1 = [0, 0, 1], L2 = [0,−1, 1], L3 = [0,−2, 1], L4 = [α,−α − 4, 1], R = [1, 0, 0]

M1 = [0, 1, 0], M2 = [−1, 1, 0], M3 = [1, 0, 0], M4 = [−γ, 1, 0], S = [0, 0, 1] ,

respectively. Therefore, we are looking for a projectivity g of P2(R)∗ such that

g(Li) = Mi for i = 1, . . . , 4, and g(R) = S. Since M1,M2,M3,M4 are collinear (asthey correspond to lines in a pencil), the points L1,L2,L3,L4 must be collinear too,and this occurs if and only if α = 0. Moreover, a necessary (and sufficient) conditionfor the existence of a projectivity g such that g(Li) = Mi for i = 1, . . . , 4 is thatthe cross-ratio β(L1,L2,L3,L4) coincides with the cross-ratio β(M1,M2,M3,M4);it can be checked that this occurs if and only if γ = −2.

96 2 Exercises on Projective Spaces

An easy computation (similar to the one carried out in solution (1)) shows that the

projectivity g associated e.g. to the matrix H =⎛⎝0 1 00 1 21 0 0

⎞⎠ is such that g(Li) = Mi

for i = 1, . . . , 4 and g(R) = S.We now come back to P

2(R) via duality. Thus, if we consider the points of

P2(R)∗ as lines of P

2(R), then the matrix tH−1 =

⎛⎜⎜⎝0 1 −1

20 0 1

21 0 0

⎞⎟⎟⎠ induces a projectivity

that satisfies all the required properties.

Exercise 43. (a) Let r, s be distinct lines of P2(K) passing through the point A.

Let B,C,D be pairwise distinct points of r \ {A}, and let B′,C′,D′ be pairwisedistinct points of s \ {A}. Prove that the lines L(B,B′), L(C,C′), L(D,D′) allintersect if and only if β(A,B,C,D) = β(A,B′,C′,D′).

(b) Let A,B be distinct points of a line r of P2(K). Let r1, r2, r3 be pairwise distinct

lines passing through A and distinct from r, and let s1, s2, s3 be pairwise distinctlines passing through B and distinct from r. Prove that the points r1 ∩ s1, r2 ∩s2, r3 ∩ s3 are collinear if and only if β(r, r1, r2, r3) = β(r, s1, s2, s3).

(c) Let A,B be pairwise distinct points of P2(K) and letFA (FB, respectively) be the

pencil of lines centred at A (at B, respectively). Let f : FA → FB be a projectiveisomorphism such that f (L(A,B)) = L(A,B). Prove that the set

Q =⋃s∈FA

s ∩ f (s)

is the union of two distinct lines.

Solution. (a) We denote by O the point at which the distinct lines L(B,B′) andL(C,C′) intersect; it is immediate to check thatO /∈ r ∪ s. LetϕO : r → s be the per-spectivity centred atO, which maps the points A,B,C to the points A,B′,C′, respec-tively. Since the cross-ratio is invariant under projective isomorphisms, we haveβ(A,B,C,D) = β(A,B′,C′,ϕO(D)). Therefore, β(A,B,C,D) = β(A,B′,C′,D′)if and only if ϕO(D) = D′, i.e., if and only if L(D,D′) passes through O.

(b) Via duality, the lines of the pencil centred at A (at B, respectively) form a lineof the dual projective plane. Using this, it is not difficult to show that the statementof (b) is just the dual statement of (a), so the conclusion follows from the Dualityprinciple (cf. Theorem1.4.1).

(c) By hypothesis f (L(A,B)) = L(A,B), so L(A,B) ⊆ Q. Let r1, r2 be distinctlines passing throughA and distinct fromL(A,B). Observe that ri �= f (ri) for i = 1, 2,and r1 ∩ f (r1), r2 ∩ f (r2) are distinct points. If t is the line joining the points r1 ∩ f (r1)and r2 ∩ f (r2), then point (b) implies that, for every r ∈ FA \ {L(A,B)}, the pointr ∩ f (r) belongs to t, and so Q ⊆ L(A,B) ∪ t. On the other hand, for every P ∈ t,if l = L(A,P) then f (l) is a line passing through B and such that f (l) ∩ l ∈ t, sof (l) ∩ l = P. This proves that every point of t belongs to Q, thus Q = L(A,B) ∪ t.

2 Exercises on Projective Spaces 97

Note. It is possible to prove point (c) via an argument based on the Duality principle.In fact, the statement of Exercise 40 implies that, if f : FA → FB is a projectiveisomorphism such that f (L(A,B)) = L(A,B), then there exists a line l ⊆ P

2(C) suchthat A /∈ l, B /∈ l, and f (s) = L(B, l ∩ s) for every s ∈ FA. This proves that, if s ∈ FA

is distinct from r = L(A,B), then s ∩ f (s) = l ∩ s, so in particular s ∩ f (s) ⊂ l. It isnow easy to show (as described above) that indeed Q = l ∪ r.

Exercise 44. (Invariant sets of projectivities of the projective plane) Let f be aprojectivity of P

2(K), where K = C or K = R. Find the possible configurations offixed points, invariant lines, axes (an axis is a pointwise fixed line) and centres (acentre is a fixed point such that every line passing through it is invariant) of f .

Solution. IfK = C, in a suitable system of homogeneous coordinates the projectivityf is represented by one of the following Jordan matrices (see also Sect. 1.5.3):

(a) A =⎛⎝1 0 00 λ 00 0 μ

⎞⎠ with λ,μ ∈ K \ {0, 1}, λ �= μ;

(b) A =⎛⎝1 0 00 1 00 0 λ

⎞⎠ with λ ∈ K \ {0, 1};

(c) A =⎛⎝1 0 00 1 00 0 1

⎞⎠;

(d) A =⎛⎝1 1 00 1 00 0 λ

⎞⎠ with λ ∈ K \ {0, 1};

(e) A =⎛⎝1 1 00 1 00 0 1

⎞⎠;

(f) A =⎛⎝1 1 00 1 10 0 1

⎞⎠.

Let now K = R. If a matrix (hence, every matrix) representing f has only real eigen-values, then in a suitable system of homogeneous coordinates of P

2(R) the projec-tivity f is represented by one of the matrices listed above. Otherwise, every matrixrepresenting f has one real and two conjugate non-real eigenvalues. In this case,there exists a system of homogeneous coordinates of P

2(R) with respect to which fis represented by the matrix

(g) A =⎛⎝a −b 0b a 00 0 1

⎞⎠ with a ∈ R, b ∈ R

∗.

Let us now analyze the cases listed above, focusing on the fixed-point set and onthe existence of invariant lines, axes and centres.

98 2 Exercises on Projective Spaces

Recall that P = [X] is a fixed point of f if and only if X is an eigenvector of A, anda line r of equation tCX = 0 is invariant under f if and only if C is an eigenvector ofthe matrix tA (cf. Sect. 1.4.5).

We set P0 = [1, 0, 0],P1 = [0, 1, 0],P2 = [0, 0, 1] and we denote by ri the lineof equation xi = 0 for i = 0, 1, 2.

Case (a). The eigenspaces of the matrix A are the lines generated by (1, 0, 0),(0, 1, 0), (0, 0, 1), respectively, so f has three fixed points, namely the pointsP0,P1,P2. Moreover, f admits three invariant lines, namely the lines r0, r1, r2. Weobserve that f |ri is a hyperbolic projectivity of ri for i = 0, 1, 2.

Case (b). The eigenvectors of the matrix A are of the form (0, 0, c) or (a, b, 0), soP2 is a fixed point and the line r2 consists of fixed points, i.e., it is an axis. Any linepassing through P2 has an equation of the form ax0 + bx1 = 0; since (a, b, 0) is aneigenvector of tA = A, every line passing through P2 is invariant, so P2 is a center.Summarizing, we have a centre P2 and an axis r2 such that P2 /∈ r2.

Case (c). In this case f is the identity, so every point is a centre and every line isan axis.

Case (d). The same argument as above shows that f has two fixed points, namelythe points P0 and P2, and two invariant lines, namely the lines r1 and r2. Therefore,there are no centres and no axes. We observe that f |r1 is a hyperbolic projectivity andf |r2 is a parabolic projectivity.

Case (e). The eigenvectors of A are of the form (a, 0, b), so the line r1 is an axis.Moreover, the eigenvectors of tA are of the form (0, a, b), so any line of equationax1 + bx2 = 0 is invariant; these lines are exactly the elements of the pencil centredat P0, so P0 is a center. In this case, the centre P0 belongs to the axis r1.

Case (f). It is readily seen that there exists a unique fixed point P0 and, by lookingat tA, a unique invariant line r2. We observe that the fixed point P0 belongs to theinvariant line r2, and f |r2 is a parabolic projectivity.

Case (g). Recall that this case occurs only when K = R and A has one real andtwo conjugate non-real eigenvalues. Therefore, P2 is the unique fixed point and r2is the unique invariant line; moreover, the invariant line does not contain the fixedpoint, and the restriction f |r2 is an elliptic projectivity.

Note. The solution of Exercise 44 shows that projectivities of P2(K) can be par-

titioned into 6 (when K = C) or 7 (when K = R) disjoint families, which can becharacterized according to the number of fixed points, invariant lines, centres andaxes, and to the incidence relations that hold among these objects. These familiesare classified by the Jordan form (when K = C) or by the “real Jordan form” (whenK = R) of the associated matrices.

Exercise 45. Let P = [1, 1, 1] and l = {x0 + x1 − 2x2 = 0} be a point and a line ofP2(R), respectively. Find an explicit formula for a projectivity f : P

2(R) → P2(R)

such that f (l) = l and that the fixed-point set of f coincides with {P}.

Solution. If g is the projectivity of P2(R) induced by the matrix B =

⎛⎝1 1 00 1 10 0 1

⎞⎠,

then [1, 0, 0] is the unique fixed point of g, and it is contained in the line {x2 = 0},

2 Exercises on Projective Spaces 99

which is invariant under g (cf. Exercise 44). Therefore, if h : P2(R) → P

2(R) isa projectivity such that h([1, 0, 0]) = P, h({x2 = 0}) = l, then the map h ◦ g ◦ h−1

gives the required projectivity.Since l = P(W ), where W ⊆ R

3 is the linear subspace generated by (1, 1, 1)and (2, 0, 1), such a projectivity h is induced for example by the invertible matrix

A =⎛⎝1 2 11 0 01 1 0

⎞⎠. As A−1 =

⎛⎝0 1 00 −1 11 1 −2

⎞⎠, we have ABA−1 =

⎛⎝3 1 −30 0 11 0 0

⎞⎠, so we can

set f ([x0, x1, x2]) = [3x0 + x1 − 3x2, x2, x0].Exercise 46. In P

2(R) consider the points

P1 = [1, 0, 0], P2 = [0, 1, 0], P3 = [0, 0, 1], P4 = [1, 1, 1],

Q1 = [1,−1,−1], Q2 = [1, 3, 1], Q3 = [1, 1,−1], Q4 = [1, 1, 1].

(a) Provide an explicit formula for a projectivity f of P2(R) such that f (Pi) = Qi

for i = 1, 2, 3, 4. Is such a projectivity unique?(b) Find all the lines of P

2(R) that are invariant under f .

Solution. (a) The points P1,P2,P3,P4 are in general position, so they define a pro-jective frame with associated normalized basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. In asimilar way, an easy computation shows thatQ1,Q2,Q3,Q4 define a projective framewith associated normalized basis {(1,−1,−1), (1, 3, 1), (−1,−1, 1)}. By the Fun-damental theorem of projective transformations, there exists a unique projectivity fsatisfying the required properties, and such an f is induced by the linear map defined

by the matrix B =⎛⎝

1 1 −1−1 3 −1−1 1 1

⎞⎠ .

(b) An easy computation shows that the characteristic polynomial of B is(2 − t)2(1 − t). The eigenspace ofB relative to the eigenvalue 2 is 2-dimensional andit is described by the equation x0 − x1 + x2 = 0, where (x0, x1, x2) are the standardcoordinates of R

3. The eigenspace of B relative to the eigenvalue 1 is 1-dimensional,and it is generated by the vector v = (1, 1, 1). Now it follows from Exercise 44 that,if r is the line of P

2(R) of equation x0 − x1 + x2 = 0 and P = [1, 1, 1] = [v] ∈P2(R), then r is pointwise fixed by f , and a line of P

2(R) distinct from r is f -invariantif and only if it passes through P.

Exercise 47. In P2(R) consider the points

P1 = [1, 0, 0], P2 = [0,−1, 1], P3 = [0, 0,−1], P4 = [1,−1, 2],

Q1 = [3, 1,−1], Q2 = [−1,−3, 3], Q3 = [−1, 1, 3], Q4 = [1,−1, 5].

(a) Construct a projectivity f of P2(R) such that f (Pi) = Qi for i = 1, 2, 3, 4. Is

such a projectivity unique?

100 2 Exercises on Projective Spaces

(b) Prove that there exist a point P and a line r such that P /∈ r, f (P) = P and r ispointwise fixed by f .

(c) Let s be a line passing through P and set Q = s ∩ r. Prove that the cross-ratioβ(P,Q,R, f (R)) is independent of s and of R, when s varies in the pencil of linescentred at P and R varies in s \ {P,Q}.

Solution. It is easy to check that the points P1,P2,P3,P4 define a projective frameof P

2(R) with associated normalized basis v1 = (1, 0, 0), v2 = (0,−1, 1), v3 =(0, 0, 1).Moreover, the pointsQ1,Q2,Q3,Q4 define a projective frame ofP2(R)withassociated normalized basis w1 = (3, 1,−1), w2 = (−1,−3, 3), w3 = (−1, 1, 3).Therefore, by the Fundamental theorem of projective transformations, there exists aunique projectivity f of P

2(R) such that f (Pi) = Qi for i = 1, 2, 3, 4, and this pro-jectivity is induced e.g. by the linear map ϕ : R

3 → R3 such that ϕ(vi) = wi for

i = 1, 2, 3. Since (0, 1, 0) = −v2 + v3 and −w2 + w3 = (0, 4, 0), the map ϕ is rep-

resented, with respect to the canonical basis of R3, by the matrix A =

⎛⎝

3 0 −11 4 1

−1 0 3

⎞⎠.

The characteristic polynomial of A is equal to (4 − t)2(2 − t). It is immediate tocheck that dim ker(A − 4I) = 2, so ϕ admits a 2-dimensional eigenspace V4 relativeto the eigenvalue 4, and a 1-dimensional eigenspace V2 relative to the eigenvalue 2.It follows that P = P(V2) and r = P(V4) are the point and the line requested in (b).An easy computation shows that P = [1,−1, 1] and that r has equation x0 + x2 = 0.

Let us now take Q ∈ r. We have seen in Sect. 1.5.4 that, since P = [v], wherev is an eigenvector of ϕ relative to the eigenvalue 2, and Q = [w], where w is aneigenvector ofϕ relative to the eigenvalue 4, for every R ∈ L(P,Q) \ {P,Q}we haveβ(P,Q,R, f (R)) = 4/2 = 2.

Exercise 48. Let r, s be distinct lines of P2(R) and set R = r ∩ s. Let A,B,C be

pairwise distinct points of r \ {R}, and let g : r → r be the unique projectivitysuch that g(A) = A, g(R) = R and g(B) = C. For every P ∈ P

2(R) \ r, set t(P) =L(B,P) ∩ s, and h(P) = L(C, t(P)) ∩ L(A,P).

(a) Prove that there exists a unique projectivity f : P2(R) → P

2(R) such thatf |P2(R)\r = h and f |r = g.

(b) Find the fixed points of f .(c) Show that f is an involution if and only if β(A,R,B,C) = −1 (Fig. 2.13).

Solution. If M,N are distinct points of s \ {R}, then the points A,B,M,N definea projective frame of P

2(R). Let us fix the system of homogeneous coordinatesinduced by this frame. Then A = [1, 0, 0], B= [0, 1, 0], r = {x2 = 0}, s= {x0 = x1},R = [1, 1, 0], and C = [1,β, 0] with β �= 0, 1.

(a) TakeP ∈ P2(R) \ r. ThenP = [a, b, c]with c �= 0. The line L(B,P) has equa-

tion cx0 − ax2 = 0, so t(P) = L(B,P) ∩ s = [a, a, c]. Therefore, the line L(C, t(P))

has equation −cβx0 + cx1 + a(β − 1)x2 = 0. Since L(A,P) has equation cx1 −bx2 = 0, we thus get h(P) = [a(β − 1) + b,βb,βc]. It follows that, on the setP2(R) \ r, the map h coincides with the projectivity f : P

2(R) → P2(R) represented

2 Exercises on Projective Spaces 101

P

C

h(P )

A

r

Bs

t(P )

R

Fig. 2.13 The construction described in Exercise48

(with respect to the chosen coordinates) by the invertible matrix

⎛⎝

β − 1 1 00 β 00 0 β

⎞⎠.

Therefore, in order to prove (a) we are left to show that f |r = g. However, sincef ([x0, x1, x2]) = [(β − 1)x0 + x1,βx1,βx2] for every [x0, x1, x2] ∈ P

2(R), we havef (A) = f ([1, 0, 0]) = [1, 0, 0] = A, f (R) = f ([1, 1, 0]) = [1, 1, 0] = R and f (B) =f ([0, 1, 0]) = [1,β, 0] = C, so f |r and g coincide on three pairwise distinct pointsof r, so that f |r = g.

(b) If P ∈ s \ {R}, then t(P) = P, and f (P) = h(P) = P. Moreover, we haveproved in point (a) that f (R) = R and f (A) = A. Therefore, every point of s ∪ {A}is fixed by f . Now, if f (M) = M for some M /∈ s ∪ {A}, then f is the identity(cf. Sect. 1.2.5), and this contradicts the fact that f (B) �= B. We have thus shownthat the fixed-point set of f is equal to s ∪ {A}.

(c) Since f (A) = A, f (R) = R and f (B) = C, Exercise23 implies that f 2|r =(f |r)2 = Idr if and only if β(A,R,B,C) = −1. Therefore, if f is an involution,then β(A,R,B,C) = −1. Conversely, let us suppose that β(A,R,B,C) = −1. Thenf 2|r = Idr . But f |s = Ids, so f 2|r∪s = Idr∪s. Since the fixed-point set of a projectiv-ity is given by the union of pairwise skew projective subspaces (cf. Sect. 1.2.5), wededuce that f 2 coincides with the identity on L(r, s) = P

2(R), and this concludes theproof.

Exercise 49. Let f : P2(K) → P

2(K) be a non-trivial involution.

(a) Show that the fixed-point set of f is given by l ∪ {P}, where l,P are a line and apoint of P

2(K), respectively, and P /∈ l.(b) Show that there exists an affine chart h : P

2(K) \ l → K2 such that

h(f (h−1(v))) = −v for every v ∈ K2.

Solution. (a)We first observe that the minimal polynomial of a linear endomorphismg of K

3 cannot be irreducible of degree 2. In fact, if this were the case, then g wouldnot have any eigenvalue. However, the characteristic polynomial of g would be divis-ible by the minimal polynomial of g. Therefore, having degree 3, the characteristic

102 2 Exercises on Projective Spaces

polynomial of g would be divisible by a linear factor, and this would imply that gadmits an eigenvalue, a contradiction.

Let now ϕ : K3 → K

3 be a linear map inducing f , and let m be the minimalpolynomial of ϕ. As f �= Id, we have that degm ≥ 2. Let us show that m is theproduct of two non-proportional linear factors. By hypothesis there exists λ ∈ K

∗such that ϕ2 = λ IdK3 . If λ is not a square in K then m, being a factor of t2 − λ,is irreducible of degree 2, and this contradicts what we have proved above. So letα ∈ K be a square root ofλ. Sincem divides t2 − λ = (t − α)(t + α) and degm ≥ 2,we have m = (t − α)(t + α). Also observe that, since α �= 0 and K ⊆ C, we haveα �= −α, so m is indeed the product of two non-proportional linear factors.

Therefore, K3 decomposes as the direct sum of two eigenspaces W1,W2 of ϕ

having dimensions 1 and 2, respectively. If P = P(W1), l = P(W2), then P /∈ l, andthe fixed-point set of f coincides with {P} ∪ l.

(b)After replacingϕbyα−1ϕor by−α−1ϕ,wemay suppose thatϕ|W1 = IdW1 andϕ|W2 = −IdW2 . Now, if v1 ∈ W1 \ {0} and {v2, v3} is a basis of W2, then {v1, v2, v3}is a basis of K

3. It is not difficult to show that, with respect to the homogeneouscoordinates of P

2(K) induced by {v1, v2, v3}, the map f is described by the for-mula f ([x0, x1, x2]) = [x0,−x1,−x2]. Moreover, with respect to these coordinateswe have l = P({x0 = 0}). After setting h : P

2(K) \ l → K2, h[1, x, y] = (x, y), we

finally have h(f (h−1(v))) = −v for every v ∈ K2.

Note. If K = C or K = R, Exercise 49 admits an easier solution that makes useof the Jordan form (or of the “real Jordan form”) (cf. Exercise 44).

K Exercise 50. Let f : P2(Q) → P

2(Q) be a projectivity such that f 4 = Id, f 2 �= Id.Compute the number of fixed points of f .

Solution (1). Letϕ : Q3 → Q

3 be a linear map inducing f , and letm, p ∈ Q[t] be theminimal and the characteristic polynomial of ϕ, respectively. Since the fixed-pointset of f coincides with the projection on P

2(Q) of the set of eigenvectors of ϕ, weanalyze the number and the dimensions of the eigenspaces of ϕ. To this aim, westudy the factorization of m and p.

Since f 4 = IdP2(Q), there exists λ ∈ Q∗ such thatϕ4 = λ IdQ3 , som divides t4 − λ

inQ[t]. Let us prove that λ is positive. If λwere negative, then t4 − λwould not haveany rational root, so ϕ would not have any eigenvalue. As a consequence, p wouldnot have any rational root so it would be irreducible, being of degree 3. By Hamilton-Cayley Theorem, we would have m = p, so m would have degree 3. Being dividedby m, the polynomial t4 − λ would be divided by a linear factor, thus admitting arational root. So λ would be positive, against our hypothesis. We have thus shownthat λ is positive.

Let us prove that p is the product of a linear factor and an irreducible factor ofdegree 2. This implies that ϕ has exactly one eigenspace, and that this eigenspaceis one-dimensional, so that f necessarily has exactly one fixed point. So let α ∈R

∗ be the positive fourth root of λ. If p were irreducible over Q, then Hamilton-Cayley Theorem would imply that m = p, so the polynomial t4 − λ = (t − α)(t +α)(t − iα)(t + iα) would be divided by a factor of degree 3 and irreducible over

2 Exercises on Projective Spaces 103

Q. By analyzing the cases α ∈ Q, α ∈ R \ Q, it can be easily shown that this isimpossible, so p is reducible, and it is divided by a linear factor. As a consequence, ϕadmits an eigenvalue, so α is rational. We now observe that t2 + α2 must divide m,because otherwise m would divide t2 − α2, and we would have f 2 = Id. Therefore,we may conclude that either p = m = (t − α)(t2 + α2) or p = m = (t + α)(t2 +α2), as desired.

Solution (2). Set g = f 2. We have seen in Exercise49 that the fixed-point set of g isgiven by {P} ∪ l, where P is a point and l is a line not containing P. Now, if a pointQis fixed by g, then g(f (Q)) = f 3(Q) = f (g(Q)) = f (Q), and f (Q) is also fixed by g.It follows that f ({P} ∪ l) ⊆ {P} ∪ l, and f (P) = P, f (l) = l, since f is a projectivity.So P is a fixed point of f .

Suppose now that Q �= P is another fixed point of f , and set s = L(P,Q). Ofcourse we have f (s) = s and, if s is endowed with homogeneous coordinates suchthat P = [1, 0], Q = [0, 1], then the map f |s can be represented by a matrix of the

form

(α 00 β

), α,β ∈ Q

∗. From f 4 = Id it follows that α4 = β4, so α2 = β2, and

g|s = f 2|s = Ids, which contradicts the fact that the fixed-point set of g is contained in{P} ∪ l, while the line s passes through P, so it contains points in P

2(K) \ ({P} ∪ l).This proves that P is the unique fixed point of f .

Note. Let H be the set of projectivities that satisfy the conditions described in thestatement. The solutions above show that any f ∈ H has exactly one fixed point, butthey do not exclude the possibility that H = ∅ (and, if this were the case, then anyanswer to the question “Howmany fixed points has f ?” would be correct!). However,it is easy to check that the map f : P

2(Q) → P2(Q) defined by f ([x0, x1, x2]) =

[x1,−x0, x2] belongs to the set H.

Note. It is easy to check that the solutions abovemay be adapted to deal with the casewhen the field Q is replaced by the field R. Therefore, the statement of Exercise50is still true if we replace P

2(Q) by P2(R).

On the contrary, every projectivity f of P2(C) such that f 4 = Id is induced by

a diagonalizable linear map, so it admits at least three fixed points. Moreover, inthis case the conditions f 4 = Id, f 2 �= Id are not sufficient to determine the numberof fixed points of f : if g, h : P

2(C) → P2(C) are the projectivities induced by the

matrices

⎛⎝1 0 00 1 00 0 i

⎞⎠,

⎛⎝1 0 00 −1 00 0 i

⎞⎠, respectively, then g4 = h4 = Id, g2 �= Id, h2 �= Id,

but g and h do not have the same number of fixed points.

Exercise 51. Let P1,P2,P3 be points of P2(K) in general position; let r be a line

such that Pi /∈ r for i = 1, 2, 3.

(a) Prove that there exists a unique projectivity f of P2(K) such that

f (P1) = P1, f (P2) = P3, f (P3) = P2, f (r) = r.

104 2 Exercises on Projective Spaces

A

r

B

Ms

P2

P3

Q

P1

N

Fig. 2.14 The construction described in the solution of Exercise51

(b) Prove that the fixed-point set of f is the union of a point M ∈ r and a lines ⊆ P

2(K) such that M /∈ s.

Solution. (a) Set A = L(P1,P2) ∩ r, B = L(P1,P3) ∩ r (cf. Fig. 2.14). It is immedi-ate to check that A,B,P2,P3 define a projective frame of P

2(K).If f is a projectivity satisfying the required properties, then f (L(P1,P2)) =

L(f (P1), f (P2)) = L(P1,P3), so f (A) = f (r ∩ L(P1,P2)) = r ∩ L(P1,P3) = B. Ina similar way one shows that f (B) = A so, as by hypothesis f (P2) = P3 andf (P3) = P2, the Fundamental theorem of projective transformations implies that therequired projectivity is unique, if it exists.

So let f : P2(K) → P

2(K) be the unique projectivity such that f (P2) = P3,f (P3) = P2, f (A) = B, f (B) = A. We have

f (P1) = f (L(A,P2) ∩ L(B,P3)) = L(B,P3) ∩ L(A,P2) = P1

andf (r) = f (L(A,B)) = L(f (A), f (B)) = L(B,A) = r,

so f satisfies the properties described in the statement.(b) IfM = L(P2,P3) ∩ r then

f (M) = f (L(P2,P3)) ∩ f (r) = L(P2,P3) ∩ r = M.

Moreover f (L(A,P3)) = L(B,P2) and f (L(B,P2)) = L(A,P3), so also the pointQ = L(A,P3) ∩ L(B,P2) is fixed by f . As L(A,P3) ∩ r = A, L(B,P2) ∩ r = B, wealso have Q /∈ r, and Q �= P1 because otherwise B would lie on L(P1,P2) andP1,P2,P3 would be collinear. So if s = L(Q,P1), then the point N = s ∩ r is welldefined. Since Q and P1 are fixed by f , we have f (s) = s, so f (N) = f (s ∩ r) =s ∩ r = N . The restriction of f to s fixes the three distinct points P1,Q,N , so itcoincides with the identity of s.

2 Exercises on Projective Spaces 105

Since P2,P3,A,B are in general position, the points M = L(P2,P3) ∩ L(A,B),P1 = L(P3,B) ∩ L(P2,A), Q = L(A,P3) ∩ L(B,P2) are not collinear (cf. Exer-cise6), so M /∈ s = L(P1,Q). Therefore, the fixed-point set of f contains the lines and the pointM ∈ r, which does not lie on s. Now, f cannot have other fixed points,because otherwise f would coincide with the identity on a projective frame of P

2(K),so it would be the identity, and this would contradict the fact that f (P2) = P3 �= P2.

Exercise 52. Let P1,P2,P3,P4 be points of P1(K) such that β(P1,P2,P3,P4) =

−1 and let (P1(K) \ {P4}, g) be any affine chart. Show that g(P3) is the midpoint ofthe segment with endpoints g(P1), g(P2), i.e., show that, ifαi = g(Pi) for i = 1, 2, 3,then α3 = α1 + α2

2 .

Solution (1). By definition of affine chart (cf. Sect. 1.3.8) we may endow P1(K)

with a system of homogeneous coordinates such that P1 = [1,α1], P2 = [1,α2],P3 = [1,α3], P4 = [0, 1]. Then it follows from Sect. 1.5.1 that

−1 = (1 − 0)(α2 − α3)

(α3 − α1)(0 − 1)= α2 − α3

α1 − α3,

hence α3 − α1 = α2 − α3 and α3 = α1 + α22 , as desired.

Solution (2). Thanks to the symmetries of the cross-ratio (cf. Sect. 1.5.2) we have

β(P2,P1,P3,P4) = β(P1,P2,P3,P4)−1 = (−1)−1 = −1 = β(P1,P2,P3,P4),

so there exists a projectivity f : P1(K) → P

1(K) such that f (P1) = P2, f (P2) = P1,f (P3) = P3, f (P4) = P4. As f (P4) = P4, the map h : K → K defined by h = g ◦f ◦ g−1 is a well-defined affinity. Therefore, there exist λ ∈ K

∗, μ ∈ K such thath(x) = λx + μ for every x ∈ K. Moreover, since f (P1) = P2 and f (P2) = P1, wehave h(α1) = α2, h(α2) = α1, so λα1 + μ = α2 and λα2 + μ = α1. It follows thatλ = −1 and μ = α1 + α2, so h(α3) = −α3 + α1 + α2. Moreover, from f (P3) = P3

we deduce that h(α3) = α3, so −α3 + α1 + α2 = α3, as desired.

Chapter 3Exercises on Curves and Hypersurfaces

Affine and projective hypersurfaces. Conjugation andcomplexification. Affine charts and projective closure.Singularities, tangent space and multiplicity. Intersection ofcurves in the projective plane. Plane cubics. Linear systems ofplane curves.

Abstract Solved problems on (real and complex) affine and projective hypersur-faces, tangent space and singularities, plane cubics, linear systems of plane curves.

Assumption: Throughout all the chapter, the symbol K will denote R or C.

Exercise 53. Let I, J be hypersurfaces of Pn(K), n ≥ 2, and let P be a point of

Pn(K). In addition, denote by CP(I), CP(J ) and CP(I + J ) the projective tangent

cones at P to I, J and I + J , respectively. Show that

mP(I + J ) = mP(I) + mP(J ), CP(I + J ) = CP(I) + CP(J ).

Solution. It is possible to choose homogeneous coordinates x0, . . . , xn ofPn(K) such

that P = [1, 0, . . . , 0]. Let yi = xix0 , i = 1, . . . , n, be the usual affine coordinates of

U0, and let f (y1, . . . , yn) = 0, f ′(y1, . . . , yn) = 0 be the equations of the affine partsI ∩ U0, J ∩ U0 of I, J , respectively. If d = mP(I) and d′ = mP(J ), then one has

f (y1, . . . , yn) = fd(y1, . . . , yn) + g(y1, . . . , yn),f ′(y1, . . . , yn) = f ′

d′(y1, . . . , yn) + g′(y1, . . . , yn),

where fd and f ′d′ are homogeneous of degrees d and d′ respectively, g is a sum of

monomials of degree greater than d and g′ is a sum of monomials of degree greaterthan d′. Note that fd = 0 and f ′

d′ = 0 are the equations of the affine tangent conesCP(I) and CP(J ), respectively.

Now the affine part (I + J ) ∩ U0 of I + J has equation

fd(y1, . . . , yn)f′d′(y1, . . . , yn) + h(y1, . . . , yn) = 0,

© Springer International Publishing Switzerland 2016E. Fortuna et al., Projective Geometry, UNITEXT - La Matematica per il 3+2 104,DOI 10.1007/978-3-319-42824-6_3

107

108 3 Exercises on Curves and Hypersurfaces

where fd f ′d′ is homogeneous of degree d + d′, and h is a sum of monomials of degree

greater than d + d′. It follows immediately that mP(I + J ) = d + d′ = mP(I) +mP(J ).

Moreover, from the equation of (I + J ) ∩ U0 written above one deduces thatthe affine tangent cone CP(I + J ) has equation fd f ′

d′ = 0, so that CP(I + J ) =CP(I) + CP(J ). The claim follows by passing to the projective closures.

Note. It is also possible to prove the equality concerning the multiplicities of P forI,J and I + J without resorting to a particular choice of coordinates. First of allwe observe that for every line r ⊆ P

n(K) one has

I(I + J , r,P) = I(I, r,P) + I(J , r,P) ≥ mP(I) + mP(J ),

so that mP(I + J ) ≥ mP(I) + mP(J ).Moreover, since the sum of the projective tangent cones to I and to J at P is

a hypersurface and the complement of the support of a hypersurface is non-empty(cf. Sect. 1.7.2), there exists a point Q ∈ P

n(K) that belongs neither to the projectivetangent cone to I at P nor the projective tangent cone to J at P. If r = L(P,Q), thenone has I(I, r,P) = mP(I), I(J , r,P) = mP(J ). Hence

mP(I + J ) ≤ I(I + J , r,P) = I(I, r,P) + I(J , r,P) = mP(I) + mP(J ).

The argument that we have just described also shows that the support of theprojective tangent cone to I + J at P coincides with the union of the supports ofthe projective tangent cones to I and to J at P.

Exercise 54. Let n ≥ 2 and let I be a hypersurface of Pn(K). Let I = m1I1 +

· · · + mkIk be the decomposition of I into irreducible components. Show that thepoint P is singular for I if and only if at least one of the following conditions holds:

(i) there exists j such that P ∈ Ij and mj ≥ 2;(ii) there exists j such that P ∈ Sing(Ij);(iii) there exist j �= s such that P ∈ Ij ∩ Is.Solution. The claim is an immediate consequence of Exercise 53. However we givean alternative proof.

Let F = 0 be an equation of I and let F = cFm11 · . . . · Fmk

k be the factorization ofF, where c ∈ K

∗, the Fi are mutually coprime irreducible homogeneous polynomialsand mi > 0 for every i = 1, . . . , k. By Leibniz’s rule

∇F(P) = cn∑

i=1

miFm11 (P) · . . . · Fmi−1

i (P) · . . . · Fmkk (P)∇Fi(P). (3.1)

If P belongs to at least two distinct components of I, or it belongs to a componentIj of multiplicitymj > 1, then all the terms of the right hand side of (3.1) vanish andtherefore P is singular for I. This case corresponds to conditions (i) and (iii).

3 Exercises on Curves and Hypersurfaces 109

If P belongs to only one component Ij of I and Ij is a component of I ofmultiplicity 1, by Eq. (3.1) one has ∇F(P) = c′∇Fj(P), where c′ = c

∏i �=j F

mii (P)

is a non-zero scalar. Hence in this case ∇F(P) = 0 if and only if ∇Fj(P) = 0, thatis if P ∈ Sing(Ij).Exercise 55. Let n ≥ 2. Show that:

(a) Two hypersurfaces I and J of Pn(C) have a non-empty intersection.

(b) If I is a smooth hypersurface of Pn(C), then I is irreducible.

Solution. (a) If I is defined by the equation F(x0, . . . , xn) = 0 and J is defined bythe equation G(x0, . . . , xn) = 0, we set:

F1(x0, x1, x2) = F(x0, x1, x2, 0, . . . , 0), G1(x0, x1, x2) = G(x0, x1, x2, 0, . . . , 0).

By Bézout’s Theorem there exists a point [a, b, c] ∈ P2(C) such that F1(a, b, c) =

G1(a, b, c) = 0. Thus the point P = [a, b, c, 0, . . . , 0] ∈ Pn(C) belongs to both I

and J , as required.

(b) Assume by contradiction that I is reducible and that consequently thereexist hypersurfaces I1 and I2 of P

n(C) such that I = I1 + I2. By (a) there existsP ∈ P

n(C) such that P ∈ I1 and P ∈ I2. By Exercise 54 the point P is singular forI, contradicting the assumption that I is smooth.

Exercise 56. Show that, if C is a reduced curve of P2(K), then Sing(C) is a finite set.

Solution. Let F(x0, x1, x2) = 0 be an equation of C and let d = degF. If d = 1, thenthe curve C is a line and therefore it has no singular points. So we can assume d ≥ 2.

Since C is reduced, we may write C = C1 + · · · + Cm, where the Ci are distinctirreducible curves. By Exercise 54 one has Sing(C) = ⋃

i Sing(Ci) ∪ ⋃i �=j(Ci ∩ Cj).

By Bézout’s Theorem the set Ci ∩ Cj is finite for every i �= j, since Ci and Cj areirreducible.

Hence to complete the proof it suffices to show that, ifC is irreducible, thenSing(C)

is finite. Since F �= 0, one has Fxj �= 0 for at least one index j ∈ {0, 1, 2}. The curveD defined by the equation Fxj = 0 has degree d − 1 and one has Sing(C) ⊆ C ∩ D.Since C is irreducible and has degree strictly greater thanD, the curves C andD haveno common components. By Bézout’s Theorem C ∩ D is a finite set, and thereforeSing(C) is also finite.

Exercise 57. Let I be a hypersurface ofPn(K)whose support contains a hyperplane

H ⊆ Pn(K). Show that H is an irreducible component of I. In particular, if I has

degree greater than 1, then I is reducible.

Solution. It is possible to choose homogeneous coordinates such thatH is defined bythe equation x0 = 0. Let F(x0, . . . , xn) = 0 be an equation of I. There exists a homo-geneous (possibly zero) polynomial F1(x0, x1, . . . , xn) and a homogeneous (possi-bly zero) polynomial F2(x1, . . . , xn) such that F(x0, . . . , xn) = x0F1(x0, . . . , xn) +F2(x1, . . . , xn). The fact that H ⊆ I implies that for every (a1, . . . , an) ∈ K

n we

110 3 Exercises on Curves and Hypersurfaces

have F(0, a1, . . . , an) = 0, so that F2(a1, . . . , an) = 0. Then F2 = 0 by the Identityprinciple of polynomials. So x0 divides F, namely H is an irreducible componentof I.Exercise 58. Let I be a hypersurface of P

n(K) and let P ∈ I be a point. Assumethat H ⊆ P

n(K) is a hyperplane passing through P and not contained in I. Provethat:

(a) mP(I ∩ H) ≥ mP(I); in particular, if I is singular at P, then I ∩ H is singularat P for any H.

(b) The hypersurface I ∩ H of H is singular at P if and only if H is contained in thetangent space TP(I).

(c) There exists a hyperplane H of Pn(K) such that I and I ∩ H have the same

multiplicities at P.

Solution (1). (a) If r is a line of H passing through P, then I(I, r,P) = I(I ∩H, r,P), so the result immediately follows from the definition of multiplicity of ahypersurface at a point.

(b)By definition, the hypersurfaceI ∩ H is singular atP if and only if I(I, r,P) =I(I ∩ H, r,P) ≥ 2 for any line r contained inH andpassing throughP.As the tangentspace TP(I) is the union of all lines that are tangent to I at P, the previous conditionis satisfied if and only H is contained in TP(I).

(c) Let r be a line through P such that the multiplicity of intersection I(I, r,P)

is minimal, and is therefore equal to the multiplicity of I at P. Then any hyperplaneH containing r has the required property.

Solution (2). (a) Let m = mP(I) ≥ 1. We can choose homogeneous coordinatessuch that P = [1, 0, . . . , 0] and H = {xn = 0}. In the affine coordinates yi = xi

x0 ,

i = 1, . . . , n, the affine part I ∩ U0 is defined by the equation f (y1, . . . , yn) = 0,where

f (y1, . . . , yn) = fm(y1, . . . , yn) + h(y1, . . . , yn),

fm is a non-zero homogeneous polynomial of degreem andh contains onlymonomialsof degree≥ m + 1. The affine coordinatesmentioned above induce affine coordinates(y1, . . . , yn−1) on the affine part H ∩ U0 of H. In these coordinates the affine part ofI ∩ H is defined by the equation g(y1, . . . , yn−1) = 0, where

g(y1, . . . , yn−1) = fm(y1, . . . , yn−1, 0) + h(y1, . . . , yn−1, 0).

From this we easily deduce that mP(I ∩ H) ≥ m = mP(I), as desired.

(b) The hypersurface I ∩ H is singular at P if and only if g(y1, . . . , yn−1) doesnot contain any linear term. This latter fact occurs if either m > 1 or m = 1 and thehyperplane f1(y1, . . . , yn) = 0 contains (and consequently coincides with) H ∩ U0.In the first case TP(I) = P

n(K) and so evidently H ⊆ TP(I). In the second caseTP(I) is the projective closure of the affine hyperplane f1(y1, . . . , yn) = 0, whichcoincides with H ∩ U0; therefore, also in this case we get that H ⊆ TP(I).

3 Exercises on Curves and Hypersurfaces 111

(c) Since fm is non-zero, by the Identity principle of polynomials there exists apoint Q = [1, v1, . . . , vn] ∈ U0 such that fm(v1, . . . , vn) �= 0. By applying the con-siderations made in (a) to a hyperplane H ⊆ P

n(K) containing the line L(P,Q), wesee that I ∩ H has multiplicity m at P.

Exercise 59. Given a hypersurface I ofPn(C), show that there exists a hypersurface

J of Pn(R) such that I is the complexification JC of J if and only if σ(I) = I.

Solution. If I is the complexification of a real hypersurface, then I can be defined bymeans of an equationF = 0withF ∈ R[x0, . . . , xn].Asσ(F) = F, also the conjugatehypersurface σ(I) is defined by F = 0, so it coincides with I.

Conversely, assume that I is a hypersurface such that σ(I) = I and I = [F]with F ∈ C[x0, . . . , xn]. The conjugate hypersurface σ(I) is defined by σ(F) = 0,so there exists λ ∈ C

∗ such that σ(F) = λF. Let us write F(x) = A(x) + iB(x), withA,B ∈ R[x0, . . . , xn]. Since F �= 0, up to replacing F by iF we may assume A �= 0,so that A defines a real hypersurface J . We have

A(x) = 1

2(F(x) + σ(F)(x)) = 1 + λ

2F(x),

so λ �= −1 and the polynomials F and A define the same hypersurface of Pn(C), i.e.

I = JC.

Exercise 60. Let r be a line of P2(C). Prove that:

(a) If r = σ(r), then r contains infinitely many real points.(b) If r �= σ(r), then the point P = r ∩ σ(r) is the only real point of r.

Solution. (a) If r = σ(r), by Exercise 59 the line r is the complexification of a realline s, so it contains infinitely many real points.

(b) If r and σ(r) are distinct lines, the point P = r ∩ σ(r) is a real point, becauseσ(P) belongs both to σ(r) and to σ(σ(r)) = r and hence it coincides with P. Inaddition, if Q ∈ r is a real point, then Q = σ(Q) ∈ r ∩ σ(r) = {P}, so P is the onlyreal point of r.

Exercise 61. Let I be an irreducible hypersurface of Pn(R) of odd degree. Show

that the complexified hypersurface IC is irreducible too.

Solution. LetJ be an irreducible component of IC. If σ(J ) �= J , the hypersurfacesJ and σ(J ) are distinct irreducible components of IC having the same degree. Sinceby hypothesis deg IC = deg I is odd, then there exists at least one component J ofIC such that σ(J ) = J . By Exercise 59 there exists a hypersurface K of P

n(R)

such that J = KC. Then K is a component of I and consequently K = I, as I isirreducible. It follows that IC = KC = J is irreducible, as required.

Exercise 62. If C denotes a cubic of P2(C), show that:

(a) If C is reducible and P ∈ C is a singular point, then C contains a line passingthrough P.

112 3 Exercises on Curves and Hypersurfaces

(b) If C has only one singular point P, and P is either a node or an ordinary cusp,then C is irreducible.

(c) If C has an inflection point P, then C is reducible if and only if C contains thetangent line at P.

(d) If C has finitely many singular points and finitely many inflection points, then Cis irreducible.

Solution. (a) If C is reducible, then C = Q + l, where l is a line and Q is a conic(which may be reducible too). Of course, if P ∈ l the result is trivial, so we maysuppose that P ∈ Q\l. Since P is singular for C, then P must be singular for Q too(cf. Exercise 54). In particularQmust be singular, so it is the union of two (possiblycoincident) lines, because every singular conic is reducible. It follows that in anycase P lies on a line contained in C.

(b)Assumeby contradiction thatC is reducible.As seen in (a),we haveC = Q + l,where l is a line passing throughP andQ is a (possibly reducible) conic. FurthermoreP, being a double point, is a smooth point for Q; the hypothesis that P is the onlysingular point of C readily implies that l ∩ Q = {P}. As a consequence l is thetangent line to Q at P, which easily implies that l is the only principal tangent toC at P (cf. Exercise 53). This yields a contradiction if P is a node of C. In additionI(C, l,P) = ∞, which leads to a contradiction if P is an ordinary cusp. Therefore Cis irreducible.

(c) Evidently C is reducible if it contains the tangent line at P. Conversely, assumethat C is reducible, say C = Q + l, where l is a line and Q is a (possibly reducible)conic. As the inflection point P is a non-singular point, then either P ∈ l\Q or P ∈Q\l. In the first case the tangent τP to C at P coincides with l, so τP is containedC. If instead P ∈ Q\l, then I(Q, τP,P) = I(C, τP,P) ≥ 3; sinceQ has degree 2, byBézout’s Theorem necessarily τP ⊆ Q ⊆ C.

(d) Assume by contradiction that C is reducible, and therefore C = l + Q, wherel is a line and Q is a conic (which may be reducible, too). If l ∩ Q is a finite set, allthe infinitely many points of l\Q are inflection points of C, a contradiction. Hencel and Q have infinitely many points in common, so Bézout’s Theorem implies thatl ⊆ Q. Then all points of l are singular for C, which again contradicts the hypotheses.It follows that C is irreducible.

Exercise 63. LetF(x0, x1, x2) = 0,G(x0, x1, x2) = 0be the equations of two reducedcurves of P

2(C). Prove that F and G define the same curve if and only if V (F) =V (G).

Solution. First of all recall that the polynomials F and G define the same curve ifand only if one of them is a scalar multiple of the other (and therefore they havein particular the same degree d), that is if and only if [F] = [G] in P(V ), whereV = C[x0, x1, x2]d is the vector space consisting of all homogeneous polynomialsof degree d and of the zero polynomial. Hence, if [F] = [G] then one obviously hasV (F) = V (G) (and this is the reason why the support of a curve is well defined!).

3 Exercises on Curves and Hypersurfaces 113

Assumenow thatV (F) = V (G). In addition, letF = F1F2 · . . . · Fk be the decom-position of F into irreducible factors; note that the Fi are pairwise coprime, because[F] is reduced. Fix i ∈ {1, . . . , k}. By Sect. 1.7.1 the polynomial Fi is homogeneous.Moreover, from the fact that V (F) ⊆ V (G) one deduces that V (Fi) is contained inV (G). Since V (Fi) consists of infinitely many points, Bézout’s Theorem implies thatthe curves defined by Fi and G have a common irreducible component, i.e. that Fi,being irreducible, divides G. Since this is true for every i and the Fi are pairwisecoprime, one can conclude that F divides G. On the other hand, by symmetry Gdivides F too, and this obviously implies that each of F and G is a scalar multiple ofthe other, as requested.

Note. Let F(x0, x1, x2) = 0, G(x0, x1, x2) = 0 be the equations of two curves ofP2(C) (possibly not reduced and possibly not of the same degree) such that

V (F) ⊆ V (G). The solution of Exercise 63 shows that every irreducible factor ofF divides G, so that every irreducible component of F is actually an irreduciblecomponent of G.

Exercise 64. If C is an irreducible curve of P2(R) of odd degree, prove that:

(a) The support of C contains infinitely many points.(b) If D is a curve of odd degree, then C ∩ D �= ∅.Solution. (a) Denote by d the degree of C and let r be a line of P

2(R) that isnot a component of C. The points of C ∩ r correspond to the roots of a non-zerohomogeneous polynomial g in two variables of degree d (cf. Sect. 1.7.6). As byhypothesis d is odd, by Theorem 1.7.2 the polynomial g has at least one real root, soC ∩ r is not empty. If we choose P ∈ P

2(R)\C, when r varies in the pencil of linescentred at P we find infinitely many points of C.(b) Of course, it is sufficient to consider the case when D is irreducible and distinctfrom C. Denote bym the degree ofD. Consider a system of homogeneous coordinatesx0, x1, x2 of P

2(R) such that the point of coordinates [0, 0, 1] belongs neither toC ∪ D nor to the union of the lines joining two points of C ∩ D. Let F(x0, x1, x2) = 0be an equation of C and G(x0, x1, x2) = 0 an equation of D and let R(x0, x1) =Ris(F,G, x2). In the chosen coordinates, the polynomial R(x0, x1) is non-zero andhomogeneous of degree dm; in addition, the irreducible factors of R(x0, x1) of degree1 are in one-to-one correspondence with the points of C ∩ D (cf. Sect. 1.9.3). As dmis odd, by Theorem 1.7.2 the polynomial R(x0, x1) has at least one irreducible factorof degree 1, which corresponds to a point Q ∈ C ∩ D.

Exercise 65. Assume thatC is an irreducible curve of degree d ofP2(R) that containsa non-singular point P. Prove that:

(a) If D is a curve of degree m such that md is even and I(C,D,P) = 1, then thereexists Q ∈ C ∩ D such that Q �= P.

(b) The support of C contains infinitely many points.

114 3 Exercises on Curves and Hypersurfaces

Solution. (a) Since C is irreducible by hypothesis and I(C,D,P) = 1, then C is nota component of D, so by the real Bézout’s Theorem the set C ∩ D is finite.

Assume that x0, x1, x2 is a system of homogeneous coordinates satisfying theproperties listed in the solution of Exercise64(b) and such that P = [1, 0, 0]; sothe points of C ∩ D are in one-to-one correspondence with the linear factors ofthe polynomial R(x0, x1) = Ris(F,G, x2), which is non-zero and homogeneous ofdegree dm. Recall that, by definition, the multiplicity of a factor of degree 1 in thefactorization of R(x0, x1) coincides with the multiplicity of intersection between CandD at the correspondingpoint (cf. Sect. 1.9.3).As I(C,D,P) = 1, thenR(x0, x1) =x1S(x0, x1), where S(x0, x1) is a homogeneous polynomial of degree dm − 1 which isnot divisible by x1. Since dm − 1 is odd, by Theorem 1.7.2 the polynomial S(x0, x1)has at least one irreducible factor of degree 1,which corresponds to a pointQ ∈ C ∩ Dsuch that Q �= P.

(b) Since a line of P2(R) contains infinitely many points, we can consider only

the case d > 1. Let D be a conic of P2(R) passing through P, such that D is non-

singular at P and the lines TP(D) and TP(C) are distinct. By Exercise 109 we haveI(C,D,P) = 1, so by part (a) there exists a point Q ∈ C ∩ D such that Q �= P.

LetP1,P2,P3 ∈ P2(R)\C be points such thatP,P1,P2,P3 are in general position.

The conics passing throughP,P1,P2,P3 forma pencilF = {D[λ,μ] | [λ,μ] ∈ P1(R)}

having {P,P1,P2,P3} as base-point set. Since imposing that a conic pass througha point and that it is tangent at that point to a given line are linear conditions(cf. Sect. 1.9.6), the set of conics ofF tangent to C atP (i.e. to TP(C) atP, cf. Exercise109) either coincides with F or contains at most one conic. The reducible conicsL(P,P1) + L(P2,P3) andL(P,P2) + L(P1,P3) belong toF and, asP,P1,P2,P3 arein general position, they are not both tangent to TP(C) in P. Therefore there existsat most one [λ0,μ0] ∈ P

1(R) such that D[λ0,μ0] is tangent to C at P. By part (a), forevery [λ,μ] �= [λ0,μ0] there exists a point Q[λ,μ] ∈ C ∩ D[λ,μ] such that Q[λ,μ] �= P.Since two distinct conics of the pencil meet exactly at P,P1,P2,P3 and P1,P2,P3 donot lie on C, when [λ,μ] varies in P

1(R)\{[λ0,μ0]} the points Q[λ,μ] are all distinct,so they form an infinite set.

Exercise 66. Analyse the singularities of the projective cubic C of P2(C) defined

by the equation

F(x0, x1, x2) = x30 + 2x20x2 + x0x22 + x21x2 = 0.

Solution. The singular points of C are given by the solutions of the system

⎧⎨⎩Fx0 = 3x20 + 4x0x2 + x22 = 0Fx1 = 2x1x2 = 0Fx2 = 2x20 + 2x0x2 + x21 = 0

.

From Fx1 = 0 one deduces x2 = 0 or x1 = 0. It is immediate to check that if x2 = 0then one has x0 = x1 = 0. If instead x1 = 0, then fromFx2 = 0 one deduces x0 = 0 orx0 = −x2. In addition, if x0 = x1 = 0 then from Fx0 = 0 one deduces x2 = 0, while

3 Exercises on Curves and Hypersurfaces 115

x0 = −x2 automatically implies Fx0 = 0. It follows that the only singular point of Cis given by [1, 0,−1].

Considering on U0 the coordinates u = x1x0 , v = x2

x0 , the affine part C ∩ U0 of Chas equation f (u, v) = 1 + 2v + v2 + u2v = 0. Furthermore, if τ : C

2 → C2 is the

translation defined by τ (u, v) = (u, v + 1), then the curve τ (C ∩ U0) has equationf (τ−1(u, v)) = v2 + u2v − u2 = (v − u)(v + u) + u2v = 0.Hence (0, 0) is an ordi-nary double point of τ (C ∩ U0). It follows that [1, 0,−1] is an ordinary double pointof C.Exercise 67. Consider the projective curve C of P

2(C) of equation

F(x0, x1, x2) = x0x22 − x31 + x0x

21 + 5x20x1 − 5x30 = 0

and the point Q = [0, 1, 0]. Check that C is non-singular and determine the pointsP ∈ C such that the tangent to C at P passes through the point Q.

Solution. Assume that (x0, x1, x2) is a solution of the system

⎧⎨⎩Fx0 = x22 + x21 + 10x0x1 − 15x20 = 0Fx1 = (x0 + x1)(5x0 − 3x1) = 0Fx2 = 2x0x2 = 0

.

FromFx2 = 0 one deduces that x0 = 0 or x2 = 0. In the former case fromFx1 = 0 onededuces x1 = 0, so that Fx0 = 0 allows one to conclude that x2 = 0, too. In the latter

case from Fx1 = 0 one deduces x0 = −x1 or x1 = 53x0. Together with the condition

x2 = 0, either of these equalities, when plugged in Fx0 = 0, implies x0 = x1 = 0. Ineither case we have shown x0 = x1 = x2 = 0, so that C is non-singular.

The tangent to C at the point [y0, y1, y2] has equation Fx0(y0, y1, y2)x0 +Fx1(y0, y1, y2)x1 + Fx2(y0, y1, y2)x2 = 0, and therefore it contains Q if and only ifFx1(y0, y1, y2) = 0. It follows that the points of C whose tangent contains Q areprecisely those determined by the solutions of the system

{F(y0, y1, y2) = y0y22 − y31 + y0y21 + 5y20y1 − 5y30 = 0Fx1(y0, y1, y2) = (y0 + y1)(5y0 − 3y1) = 0

which is satisfied by the points of coordinates [0, 0, 1], [1,−1, 2√2], [1,−1,−2

√2],

[3√3, 5√3, 2i

√10], [3√3, 5

√3,−2i

√10].

Note. We are going to show in Exercise 81 that, if C is a smooth curve of P2(C) of

degree greater than 1 and Q ∈ P2(C), then the set of lines tangent to C and passing

through Q is always finite and non-empty.Moreover, in the solution of Exercise 67 we observed that the points of the cubic

C whose tangent τP contains Q are the intersection points of C with the conic Qdefined by the equation Fx1 = 0. By Bézout’s Theorem the number of these points,counted with multiplicity, is equal to 6. Actually, it is not hard to check that the point

116 3 Exercises on Curves and Hypersurfaces

P = [0, 0, 1] is an inflection point and that I(C,Q,P) = 2, while the remaining 4points are not inflection points and C and Q intersect at these points with multiplic-ity 1.

Exercise 68. Denote by C the curve of C2 of equation f (x, y) = xy2 − y4 + x3 −

2x2y = 0. Determine:

(a) The improper points and the asymptotes of C.(b) The singular points of C, their multiplicities and their principal tangents, identi-

fying which of them are ordinary singular points.(c) The equation of the tangent to C at the point P = (4,−4).

Solution. (a) If we identify C2 with the affine chart U0 of P

2(C) by means of themap j0 : C

2 → U0 defined by j0(x1, x2) = [1, x1, x2], then the projective closure C ofC has equation

F(x0, x1, x2) = x0x1x22 − x42 + x0x

31 − 2x0x

21x2 = 0.

By computing the intersection between C and the line x0 = 0, it turns out that theonly improper point is P = [0, 1, 0].

With respect to the affine coordinates u = x0x1 , v = x2

x1 of the affine chart U1, the

point P has coordinates (0, 0) and the affine part C ∩ U1 has equation uv2 − v4 +u − 2uv = 0. So P is a simple point of C and the tangent to C ∩ U1 at P has equationu = 0. Therefore the tangent to C at P is the line x0 = 0; consequently C has noasymptotes.

(b) Recall that the singular points of C are the proper points that are singular forC. In order to determine the singular points of C, it suffices to solve the system

⎧⎨⎩Fx0 = x1x22 + x31 − 2x21x2 = x1(x2 − x1)2 = 0Fx1 = x0x22 + 3x0x21 − 4x0x1x2 = 0Fx2 = 2x0x1x2 − 4x32 − 2x0x21 = 0

which has as unique solution the point Q = [1, 0, 0], corresponding to (0, 0) ∈ C2.

Inspecting the equation of C we realize that (0, 0) is a triple point; since the homo-geneous part of degree 3 of f (x, y) is xy2 + x3 − 2x2y = x(x − y)2, we see that theprincipal tangents to C at the origin are the lines x = 0 and x − y = 0 (the latter onewith multiplicity 2). So the origin is a non-ordinary singular point.

(c) As Fx0(1, 4,−4) = 256, Fx1(1, 4,−4) = 128, Fx2(1, 4,−4) = 192, the equa-tion of the projective tangent to C at [1, 4,−4] is 4x0 + 2x1 + 3x2 = 0. It followsthat the tangent to C at P has equation 2x + 3y + 4 = 0.

Exercise 69. Consider the curve C of C2 of equation f (x, y) = x − xy2 + 1 = 0.

(a) Determine the singular points and the asymptotes of C.(b) Determine the inflection points of the projective closure of C, check that they

are collinear and compute the equation of the line containing them.

3 Exercises on Curves and Hypersurfaces 117

Solution. (a) If we identify C2 with the affine chart U0 of P

2(C) by means of themap j0 : C

2 → U0 defined by j0(x1, x2) = [1, x1, x2], the projective closure C of Chas equation F(x0, x1, x2) = x20x1 − x1x22 + x30 = 0.

In order to determine the singular points of C it is enough to observe that the onlysolution of the system

⎧⎨⎩Fx0 = x0(2x1 + 3x0) = 0Fx1 = (x0 − x2)(x0 + x2) = 0Fx2 = −2x1x2 = 0

is the homogeneous triple [x0, x1, x2] = [0, 1, 0]. It follows that P = [0, 1, 0] is theonly singular point of C, and that C has no singular point.

By intersecting C with the line at infinity x0 = 0, we easily see that the improperpoints of C are P and Q = [0, 0, 1] and that Q is a smooth point. As Fx0(0, 0, 1) =Fx2(0, 0, 1) = 0, the (principal) tangent to C at Q has equation x1 = 0. So x = 0 isthe equation of an asymptote of C.

In order to determine the principal tangents to C at P, we can use the affinecoordinates u = x0

x1 , v = x2x1 defined on the affine chartU1. In those coordinates P =

(0, 0) and the affine partC ∩ U1 has equation u2 − v2 + u3 = (u − v)(u + v) + u3 =0. So P is a double point, and the principal tangents to C ∩ U1 at P have equationsu + v = 0, u − v = 0. Therefore the principal tangents to C at P have equationsx0 + x2 = 0, x0 − x2 = 0, and C has as asymptotes the lines y = 1 and y = −1, inaddition to the line x = 0 found above.

(b) Since

HF(x0, x1, x2) = det

⎛⎝2x1 + 6x0 2x0 0

2x0 0 −2x20 −2x2 −2x1

⎞⎠ = 8(x20x1 − 3x0x

22 − x1x

22),

the inflection points of C are the simple points of C whose coordinates are solutionsof the system {

x20x1 − x1x22 + x30 = 0x20x1 − 3x0x22 − x1x22 = 0

.

If we subtract the second equation from the first one, easy computations show thatthe inflection points of C are the points [0, 0, 1], [−12, 9, 4i

√3], [−12, 9,−4i

√3].

These points lie on the line of equation 3x0 + 4x1 = 0.

Note. As shown in Exercise 62, a cubic having only one node and no other singularpoint is irreducible. Therefore, the fact that the inflection points of the projective clo-sure of C are collinear follows from a general result that we will show in Exercise 88.

Exercise 70. If C is the curve of R2 of equation f (x, y) = x3 − xy2 + x2y − y3 +

xy4 = 0, determine:

118 3 Exercises on Curves and Hypersurfaces

(a) The singular points of the projective closure C of C, computing, for each of thesepoints, the multiplicity, the principal tangents and the multiplicity of intersectionof C with each principal tangent.

(b) The points at infinity and the asymptotes of C.(c) All the lines of the pencil with centre [1, 0, 0] that are tangent to C at two distinct

points.

Solution. (a) The projective closure C of C has equation

F(x0, x1, x2) = x20(x31 − x1x

22 + x21x2 − x32) + x1x

42 = 0;

note that F = x20(x1 − x2)(x1 + x2)2 + x1x42.The singular points of C are characterized by the property that their homogeneous

coordinates annihilate the gradient of F, i.e. they are the solutions of the system

⎧⎨⎩Fx0 = 2x0(x1 − x2)(x1 + x2)2 = 0Fx1 = x20(x1 + x2)2 + 2x20(x1 − x2)(x1 + x2) + x42 = 0Fx2 = −x20(x1 + x2)2 + 2x20(x1 − x2)(x1 + x2) + 4x1x32 = 0

.

Hence the singular points are A = [0, 1, 0] and B = [1, 0, 0].We see from the equation of C that B is a triple point with the line τ1 of equation

x1 − x2 = 0 and the line τ2 of equation x1 + x2 = 0 (counted twice) as principaltangents, so that B is a non-ordinary triple point.

In the affine chartU0 a parametrization of τ1 ∩ U0 is given by themap γ : R → R2,

γ(t) = (t, t). Since γ(0) = B and the polynomial f (γ(t)) = t5 admits 0 as a root ofmultiplicity 5, one has I(C, τ1,B) = 5. Similarly, we have I(C, τ2,B) = 5.

In order to study the point A = [0, 1, 0], we work in the chart U1 with affinecoordinatesu = x0

x1 , v = x2x1 . In these coordinatesA = (0, 0) andC ∩ U1 has equation

u2(1 − v)(1 + v)2 + v4 = 0. Hence A is a non-ordinary double point (cusp) of C ∩U1, and the principal tangent to C at A is given by the projective closure τ3 of theaffine line u = 0, namely by the line τ3 = {x0 = 0}. In addition I(C, τ3,A) = 4, andtherefore A is a non-ordinary cusp.

(b) Intersecting C with the line x0 = 0 we find that the points A = [0, 1, 0] andP = [0, 0, 1] are the points at infinity of C. By what we have seen in (a), the point Adoes not give rise to an asymptote. We have also seen that P is a simple point of C;since ∇F(0, 0, 1) = (0, 1, 0), the tangent to C at P is the line of equation x1 = 0. Sowe find that the affine line of equation x = 0 is an asymptote of C.

(c) The lines of the pencil with centre B = [1, 0, 0] have equation ax1 + bx2 = 0as [a, b] varies in P

1(R).The line of the pencil corresponding to b = 0, i.e. the line x1 = 0, meets the

requirement, since it is tangent to C both at B and at P.So let us assume b �= 0, for instance b = −1, and see for which values of a ∈ R

the line x2 = ax1 is tangent to C at two distinct points. Substituting x2 = ax1 in theequation of the curve we get the equation

3 Exercises on Curves and Hypersurfaces 119

x31((1 − a)(1 + a)2x20 + a4x21) = 0

which shows that for every value of a the line intersects C at B = [1, 0, 0] withmultiplicity at least 3, as it is obvious since B is a triple point. Hence we have tosee for which values of a the remaining intersection points coincide but are differentfrom B.

The equation (1 − a)(1 + a)2x20 + a4x21 = 0 has two coincident solutions only ifeither a = 0 or (1 − a)(1 + a) = 0.

The value a = 0 is acceptable, since the corresponding line x2 = 0 is tangent toC at B (triple point) and at A (cusp).

On the other hand neither of the values a = 1 and a = −1 is acceptable since thecorresponding lines x2 = x1 and x2 = −x1 (that we denoted by τ1 and τ2) intersectC only at B with multiplicity 5.

Exercise 71. Consider the curve C of P2(C) of equation

F(x0, x1, x2) = x20x21 − x0x1x

22 − 3x41 − x20x

22 − 2x0x

31 = 0;

compute its singular points, their multiplicities and their principal tangents.Furthermore, say whether C is reducible.

Solution. The singular points of C are given by the solutions of the system

⎧⎨⎩Fx0 = 2x0x21 − x1x22 − 2x0x22 − 2x31 = 0Fx1 = 2x20x1 − x0x22 − 12x31 − 6x0x21 = 0.Fx2 = −2x0x2(x0 + x1) = 0

An easy computation shows that the singular points of C are P = [0, 0, 1], Q =[1, 0, 0], R = [1,−1, 2] and S = [1,−1,−2].

Now observe that P, R and S lie on the line r of equation x0 + x1 = 0. Hence onehas I(C, r,P) + I(C, r,R) + I(C, r, S) ≥ 2 + 2 + 2 = 6 > 4, so that by Bézout’sTheorem the line r is an irreducible component of C. In fact, if G(x0, x1, x2) =x0x21 − 3x31 − x0x22, then one has F(x0, x1, x2) = (x0 + x1)G(x0, x1, x2), and there-fore C is reducible.

Taking the coordinates u = x1x0 , v = x2

x0 on U0, the equation of the affine part

C ∩ U0 ofC becomes f (u, v) = u2 − uv2 − 3u4 − v2 − 2v3 = 0. Since in these coor-dinates Q corresponds to the origin, and the homogeneous part of smallest degreeof f is (u + v)(u − v), Q is an ordinary double point of C with principal tangents ofequation x1 + x2 = 0 and x1 − x2 = 0, respectively.

In order to compute the multiplicity and the principal tangents of the singularpoints of C that lie on r, we now exploit the information on the factorization of Fthat we have obtained. Since

Gx0 = x21 − x22, Gx1 = 2x0x1 − 9x21, Gx2 = −2x0x2,

120 3 Exercises on Curves and Hypersurfaces

if D is the curve of equation G = 0, the points P,R, S are simple points of D. SinceC = r + D, it follows that P,R, S are double points of C (cf. Exercise 53). Moreover,by computing the partial derivatives of G it is immediate to get the equations of thetangent lines rP, rR, rS to D at P,R, S. These lines are given by rP = {x0 = 0}, rR ={3x0 + 11x1 + 4x2 = 0}, rS = {3x0 + 11x1 − 4x2 = 0}. Then the principal tangentstoQ at P (at R, S, respectively) are given by r, rP (by r, rR, and by r, rS , respectively)(cf. Exercise 53).

Exercise 72. Consider the curve C of C2 of equation

f (x, y) = x2(y − 5)2 + y(bx + a2y) = 0,

where the parameters a, b vary in C.

(a) Determine the number of asymptotes of C.(b) Say whether the point (0, 0) ∈ C

2 is singular for C, compute its multiplicity andsay whether it is an ordinary point.

(c) Say whether there exist values of the parameters a, b ∈ C such that the curveC passes through the point (−1, 1) and it is tangent at that point to the line3x + 5y − 2 = 0.

Solution. (a) The projective closure C of C in P2(C) has equation

F(x0, x1, x2) = x21(x2 − 5x0)2 + x0x2(bx1 + a2x2) = 0.

Hence the points at infinity of C, i.e. the intersection points of C with the line x0 = 0,are P = [0, 1, 0] and Q = [0, 0, 1].

In order to determine the principal tangents to C atP, we use the affine coordinatesu = x0

x1 , v = x2x1 defined on the affine chart U1. With this choice of coordinates,

the point P corresponds to the point (0, 0) and the affine part C ∩ U1 has equationf1(u, v) = (v − 5u)2 + uv(b + a2v) = 0. Since f1 does not contain monomials ofdegree 1, P is singular. Moreover, the homogeneous part of f1 of degree 2 is givenby v2 + uv(b − 10) + 25u2: it is never identically zero, and it is the square of apolynomial of degree 1 if and only if b = 0 or b = 20. It follows that P is a non-ordinary double point if b = 0 or b = 20, and is an ordinary double point otherwise.Moreover, since the monomial u does not divide the quadratic homogeneous part off1, the line x0 = 0 is never one of the principal tangents to C at P. Therefore, thereare two asymptotes of C whose projective closures contain P for b �= 0 and b �= 20,and only one otherwise.

In order to study the point Q ∈ C, we now choose the affine coordinates s =x0x2 , t = x1

x2 defined on U2. With this choice of coordinates, Q corresponds to

the point (0, 0) and the affine part C ∩ U2 has equation f2(s, t) = t2(1 − 5s)2 +s(bt + a2) = 0.

If a �= 0, then the point Q is smooth, and the (principal) tangent to C at Q hasequation x0 = 0. In this case C has no asymptotes whose projective closure passesthrough Q.

3 Exercises on Curves and Hypersurfaces 121

If instead a = 0, then the quadratic homogeneous part of f2 is given by t(t + bs): itis always non-zero, and it is the square of a polynomial of degree 1 if and only if b = 0.It follows that P is a non-ordinary double point if b = 0, and it is an ordinary doublepoint otherwise. In addition, since the monomial s does not divide the quadratichomogeneous part of f2, the line x0 = 0 is never one of the principal tangents to C atP. Therefore, there are two asymptotes of C whose projective closures contain Q forb �= 0, and one otherwise.

In summary, the number of asymptotes of C is:

• 4 if a = 0 and b /∈ {0, 20};• 3 if a = 0 and b = 20;• 2 if a = b = 0, or a �= 0 and b /∈ {0, 20};• 1 if a �= 0, b ∈ {0, 20}.

(b) Since the equation of C contains no term of degree 1, (0, 0) is singu-lar. Moreover, the quadratic homogeneous part of the equation of C is given by25x2 + bxy + a2y2, that is always non-zero, and is equal to the square of a polyno-mial of degree 1 if and only if b = ±10a. Hence (0, 0) is a double point, and it isordinary if and only if b �= ±10a.

(c) Imposing f (−1, 1) = 0 one obtains b = a2 + 16. One has also fx(−1, 1) =b − 32, fy(−1, 1) = 2a2 − b − 8. So, the line of equation3x + 5y − 2 = 0 is tangentto C at (−1, 1) if and only if b = a2 + 16 and

0 = det

(b − 32 2a2 − b − 8

3 5

)= 8b − 6a2 − 136.

Since the system b − a2 − 16 = 8b − 6a2 − 136 = 0 has solutions a = ±2, b = 20,it follows that the line 3x + 5y − 2 = 0 is tangent toC at (−1, 1) if and only if a = ±2and b = 20.

Exercise 73. Consider the curve C of C2 of equation

f (x, y) = x3(x2 + a) + y(x3 − x2y + by + c) = 0,

where the parameters a, b, c vary in C.

(a) Compute the multiplicity of the point O = (0, 0) and the principal tangents to Cat O. When O is a simple point, determine the multiplicity of intersection at Obetween C and the tangent to C at O.

(b) Determine the improper points and the asymptotes of C.(c) Find the values of a, b, c forwhich the pointQ = (0, 2) is singular for C; for these

values, compute the multiplicity of Q for C, and say whether Q is an ordinarypoint.

Solution. (a) By reordering themonomials of f , one gets f (x, y) = cy + by2 + ax3 +yx2(x − y) + x5. Hence O is simple if and only if c �= 0. If this is the case, then the

122 3 Exercises on Curves and Hypersurfaces

tangent r to C at O has equation y = 0, and therefore the map γ : C → C2 given by

γ(t) = (t, 0) is a parametrization of r. Since f (γ(0)) = O, then I(C, r,O) is equalto the multiplicity of 0 as a root of f (γ(t)). Since f (γ(t)) = t3(t2 + a), one hasI(C, r,O) = 3 if a �= 0, and I(C, r,O) = 5 otherwise.

If instead c = 0 and b �= 0, the pointO is a non-ordinary double point, with y = 0as the only principal tangent. If c = b = 0 and a �= 0, thenO is a non-ordinary triplepoint, with the line of equation x = 0 as the only principal tangent. Finally, if b = c =a = 0, then O is a non-ordinary quadruple point, and the principal tangents to C atO have equations x = 0 (of multiplicity 2), y = 0 and x = y (each of multiplicity 1).

(b) The projective closure of C has equation

F(x0, x1, x2) = x31(x21 + ax20) + x0x2(x

31 − x21x2 + bx2x

20 + cx30) = 0.

Solving the system F(x0, x1, x2) = x0 = 0 one sees that P = [0, 0, 1] is the onlypoint at infinity of C. Now let u = x0

x2 , v = x1x2 be affine coordinates on U2. With

this choice of coordinates P coincides with (0, 0) and the affine part C ∩ U2 of theprojective closure of C has equation

g(u, v) = v3(v2 + au2) + u(v3 − v2 + bu2 + cu3) = 0.

The homogeneous component of g of smallest degree is u(bu2 − v2), so that if αis a square root of b, then the principal tangents to C at P intersect U2 in the affinelines of equations u = 0, v = αu, v = −αu (in particular, if b = 0 then P is notordinary, because the tangents v = αu, v = −αu coincide). Therefore, the principaltangents to C at P have equations x0 = 0, x1 = αx0, x1 = −αx0. It follows that, ifb = 0, then the curve C has only one asymptote, of equation x = 0, otherwise C hastwo asymptotes, of equations x = α and x = −α.

(c) In order for Q to be singular for C, one must have f (Q) = fx(Q) = fy(Q) = 0.Since f (0, 2) = 4b + 2c, fx(0, 2) = 0 and fy(0, 2) = 4b + c, this happens if andonly if b = c = 0. Therefore let us assume b = c = 0 and study the nature of thepoint Q. To this aim, observe that the translation τ : C

2 → C2 defined by τ (x, y) =

(x, y − 2) satisfies τ (Q) = (0, 0). Moreover

f (τ−1(x, y)) = x2(x(x2 + a) + (y + 2)(x − y − 2)),

so that the homogeneous component of f ◦ τ−1 of smallest degree is −4x2, and theorigin is a non-ordinary double point of τ (C). It follows that Q is a non-ordinarydouble point of C.Exercise 74. Consider the affine curve Ca,b of equation

f (x, y) = x3 − 2ay2 + bxy2 = 0

where the parameters a, b vary in C.

3 Exercises on Curves and Hypersurfaces 123

(a) Find the values of a, b for which the line at infinity is tangent to the projectiveclosure Ca,b of Ca,b and, for each of those values, say whether Ca,b is singular atthe point of tangency.

(b) Find a, b ∈ C such that Ca,b passes through the point (1, 2) with tangent linex − y + 1 = 0.

Solution. (a) The equation of Ca,b is

F(x0, x1, x2) = x31 − 2ax0x22 + bx1x

22 = 0.

The line at infinity is tangent to Ca,b at a point P if and only if x0(P) = F(P) =Fx1(P) = Fx2(P) = 0. Therefore, as Fx1 = 3x21 + bx22, Fx2 = −4ax0x2 + 2bx1x2, theline at infinity is tangent to Ca,b if and only if the system

⎧⎪⎪⎪⎨⎪⎪⎪⎩

x0 = 0

x31 − 2ax0x22 + bx1x22 = 0

3x21 + bx22 = 0

−4ax0x2 + 2bx1x2 = 0

has a non-trivial solution. It is easy to see that this fact occurs if and only if b = 0.In addition, if b = 0 the only non-trivial homogeneous solution is [0, 0, 1]. Finally,sinceFx0(0, 0, 1) = −2a, Ca,b is singular at the point of tangency if and only if a = 0.

(b) The curve Ca,b passes through (1, 2) with tangent x − y + 1 = 0 if and onlyif f (1, 2) = 0 and the vector (fx(1, 2), fy(1, 2)) is proportional to (1,−1). The firstcondition is equivalent to 1 − 8a + 4b = 0, while the second one is equivalent tofx(1, 2) = −fy(1, 2), i.e. to 3 − 8a + 8b = 0. The unique solution of the system con-

sisting of these two equations is a = −18 , b = −1

2 , so Ca,b passes through (1, 2)with

tangent x − y + 1 = 0 if and only if a = −18 , b = −1

2 .

Exercise 75. Let k ≥ 1 be an integer and consider the curve C of C2 of equation

f (x, y) = xky2 − x5 + a = 0, where the parameter a varies in C.

(a) For every k and a, determine the singular points of C and compute their multi-plicities. In addition, say whether these singularities are ordinary.

(b) For every k and a, determine the improper points and the asymptotes of C.(c) If k = 3, say which improper points of C are inflection points.

Solution. (a) The singular points of C are given by the solutions of the system

⎧⎪⎨⎪⎩

f (x, y) = xky2 − x5 + a = 0

fx(x, y) = kxk−1y2 − 5x4 = 0

fy(x, y) = 2xky = 0

.

124 3 Exercises on Curves and Hypersurfaces

The third equation implies that necessarily either x = 0 or y = 0. Though, if y = 0the second equation yields that x = 0, so every singular pointmust fulfil the conditionx = 0. Then from the first equation we deduce that no singular point exists if a �= 0.

Therefore we may suppose a = 0.If k > 1, all points of the form (0, y0), y0 ∈ C, are singular, while if k = 1 the

only singular point of C is O = (0, 0).At first let us investigate the nature of the point O.If k < 3, the homogeneous component of f of smallest degree is xky2, and hence

mO(C) = 2 + k. The point O is non-ordinary, because the principal tangent of equa-tion y = 0 has multiplicity 2.

If k = 3, we have f (x, y) = x3(y2 − x2), mO(C) = 5 and O is not ordinary,because the principal tangent of equation x = 0 has multiplicity 3.

If instead k > 3, the homogeneous component of f of smallest degree is −x5, soO is clearly a non-ordinary singular point of multiplicity 5.

In summary, for any value of k the point O is not ordinary of multiplicitymin{5, k + 2}.

As already remarked, if k = 1no additional singular point exists. So assume k > 1,and let us investigate the nature of the point (0, y0), with y0 �= 0. The translationτ : C

2 → C2 defined by τ (x, y) = (x, y + y0) fulfils the conditions τ (0, 0) = (0, y0)

andf (τ (x, y)) = xk(y + y0)

2 − x5 = y20xk − x5 + 2y0x

ky + xky2.

It follows that, if k < 5, the homogeneous component of smallest degree of f ◦ τ isy20x

k , so the point (0, y0) has multiplicity k for C and it is non-ordinary (because itsonly principal tangent is the line x = 0). If k > 5, the homogeneous component ofsmallest degree of f ◦ τ is −x5, so (0, y0) is a non-ordinary singular point for C ofmultiplicity 5. Finally, if k = 5 two cases can occur: if y0 �= ±1, the homogeneouscomponent of smallest degree of f ◦ τ is (y20 − 1)x5, so (0, y0) is a non-ordinarysingular point forC ofmultiplicity 5,while if y0 = ±1wehave f (τ (x, y)) = 2y0x5y +x5y2, and hence (0,±1) is a non-ordinary singular point for C of multiplicity 6.

In summary, if k > 1 each point of the form (0, y0) is a non-ordinary singularpoint, and its multiplicity is min{5, k}, except when k = 5 and y0 = ±1, because inthis case the multiplicity is 6.

(b) Let us study separately the cases k < 3, k = 3, k > 3.If k < 3, the projective closure of C of C has equation

F(x0, x1, x2) = x3−k0 xk1x

22 − x51 + ax50 = 0.

It readily follows that the only improper point of C is P = [0, 0, 1]. Using on U2

the affine coordinates u = x0x2 , v = x1

x2 , we see that the affine part C ∩ U2 of C has

equation u3−kvk − v5 + au5 = 0. Moreover, P has affine coordinates (0, 0). Sincethe principal tangents to C ∩ U2 at (0, 0) are evidently the lines of equations u = 0and v = 0, it follows that the principal tangents at P to C have equations x0 = 0 andx1 = 0. The first of these lines does not yield any asymptote, so the only asymptotefor C is the line x = 0.

3 Exercises on Curves and Hypersurfaces 125

If k > 3, the projective closure C of C has equation

F(x0, x1, x2) = xk1x22 − xk−3

0 x51 + axk+20 = 0.

So the improper points of C are P = [0, 0, 1] and Q = [0, 1, 0]. Using on U2 theaffine coordinates u, v defined in the previous paragraph in such a way that P =(0, 0), the affine part C ∩ U2 of C has equation vk − uk−3v5 + auk+2 = 0, whosehomogenous component of smallest degree is vk . Therefore the line v = 0 is theonly principal tangent to C ∩ U2 at (0, 0), and consequently the only asymptote of Cwhose projective closure passes through P is the line x = 0.

In order to study the asymptotes passing through Q, let us use on U1 the affinecoordinates s = x0

x1 , t = x2x1 . Thus Q is identified with the origin of C

2, and the

affine part C ∩ U1 of C has equation g(s, t) = t2 − sk−3 + ask+2 = 0. In the followinganalysis it may be useful to distinguish different cases according to the value of k.

If k > 5, the homogenous component of smallest degree of g(s, t) is t2; so theline t = 0 is the only principal tangent to C ∩ U1 at (0, 0), which easily implies thatthe only asymptote of C whose projective closure passes through Q is the line y = 0.

If k = 5, the homogenous component of smallest degree of g(s, t) is t2 − s2,hence the principal tangents to C ∩ U1 at (0, 0) are the lines t − s = 0 and t + s = 0;therefore there are two asymptotes of C whose projective closures pass through Q,that is the lines y − 1 = 0 and y + 1 = 0.

If k = 4, the homogenous component of smallest degree of g(s, t) is s; then theline s = 0 is the only principal tangent to C ∩ U1 at (0, 0) and so in this case thereexists no asymptote of C whose projective closure passes through Q.

Consider now the case k = 3. The equation of the projective closure of C is thenF(x0, x1, x2) = x31x

22 − x51 + ax50 = 0. The systemF(x0, x1, x2) = x0 = 0 is evidently

equivalent to the system x31(x21 − x22) = x0 = 0, so that the improper points of C are

P = [0, 0, 1],M = [0,−1, 1], N = [0, 1, 1]. Using onU2 the affine coordinates u, vintroduced above, the affine part C ∩ U2 of C has equation f (u, v) = v3 − v5 +au5 = 0. Moreover, P,M,N are identified with the points (0, 0), (0,−1), (0, 1) ofC

2, respectively. It follows immediately that the only principal tangent to C at P hasequation x1 = 0, so the only asymptote of C whose projective closure passes throughP has equation x = 0.

In order to determine the principal tangents to C atM and N , we compute the gra-dient of F. As Fx0 = 5ax40, Fx1 = 3x21x

22 − 5x41, Fx2 = 2x31x2, then ∇F(0,−1, 1) =

(0,−2,−2), ∇F(0, 1, 1) = (0,−2, 2), soM,N are non-singular for C, and the tan-gents to C at M and at N have equations x1 + x2 = 0 and x1 − x2 = 0, respectively.It follows that, if k = 3, the asymptotes of C are the lines x = y, x = −y, x = 0.

(c) Using the notation introduced in the final part of the solution of (b), first of allwe observe that P, being singular for C, cannot be an inflection point. We check that,instead, M and N are both inflection points of C. As seen above, the tangent τM tothe affine part C ∩ U2 at (0,−1) has equation v = −1, and so it can be parametrizedby means of the map γ : C → C

2 defined by γ(r) = (r,−1). Since γ(0) = (0,−1),then the multiplicity of intersection I(C, τM ,M) is equal to the the multiplicity of

126 3 Exercises on Curves and Hypersurfaces

0 as a root of the polynomial f (γ(r)). As f (γ(r)) = ar5, this multiplicity is greaterthan or equal to 5, so it is greater than 2. Moreover, as seen in the solution of (b),Mis non-singular for C, soM is an inflection point of it.

In the samewaywe get that alsoN is an inflection point of C: the tangent τN to C ∩U2 at (0, 1) is parametrized by η(r) = (r, 1), and f (η(r)) = ar5, so I(C, τN ,N) ≥ 5;this implies the thesis, since N is non-singular for C.Exercise 76. Find a cubic D of R

2 fulfilling the following conditions:

(i) (0, 1) is an ordinary double point with principal tangents of equations y =2x + 1 and y = −2x + 1;

(ii) the only improper points of D are [0, 1, 0] and [0, 0, 1];(iii) the line y = 5 is an asymptote for D.

In addition, say whether D has other asymptotes.

Solution. Consider the translation τ : (X,Y) → (x, y) = (X,Y + 1) such thatτ (0, 0) = (0, 1). Using the coordinates X,Y we must impose that the principal tan-gents have equations Y = ±2X. Therefore, the cubic must have an equation of theform

Y 2 − 4X2 + aX3 + bX2Y + cXY 2 + dY 3 = 0,

i.e., coming back to the coordinates x, y, an equation of the form

(y − 1)2 − 4x2 + ax3 + bx2(y − 1) + cx(y − 1)2 + d(y − 1)3 = 0.

By imposing that [0, 1, 0] and [0, 0, 1] be improper points, we get the conditionsa = 0 and d = 0. So the equation of the projective closure D of D is

F(x0, x1, x2) = x0(x2 − x0)2 − 4x0x

21 + bx21(x2 − x0) + cx1(x2 − x0)

2 = 0.

The projective closure of the line y = 5 has equation x2 − 5x0 = 0 and its improperpoint is [0, 1, 0]. Since ∇F(0, 1, 0) = (−4 − b, 0, b), the point [0, 1, 0] is smoothfor every b; if we impose that the line x2 − 5x0 = 0 be tangent at [0, 1, 0], we find theadditional condition b = 1. Finally observe that necessarily c = 0, because otherwiseD would have additional improper points distinct from [0, 1, 0] and [0, 0, 1]. Thuswe can conclude that the equation of D must be

(y − 1)2 − 4x2 + x2(y − 1) = 0,

which fulfils all the requested conditions.Finallywe easily check thatD has noother asymptotes, because the other improper

point [0, 0, 1] is smooth for D with tangent x0 = 0.

Exercise 77. Find a cubic D of P2(C) fulfilling the following conditions:

(i) [1, 0, 0] is an ordinary double point with principal tangents of equations x1 = 0and x2 = 0;

3 Exercises on Curves and Hypersurfaces 127

(ii) [0, 1, 1] is an inflection point with tangent x0 = 0;(iii) D passes through the point [1, 4, 2].In addition, say whether the cubic D is reducible.

Solution. In order that condition (i) be satisfied, in the affine chart U0 the equationof D ∩ U0 must be of the form

xy + ax3 + bx2y + cxy2 + dy3 = 0;

so D has equation

x0x1x2 + ax31 + bx21x2 + cx1x22 + dx32 = 0,

with a, b, c, d ∈ C not simultaneously zero.We want now to impose condition (ii), so we observe that in the chart U2, with

respect to the affine coordinates u = x0x2 , v = x1

x2 , the affine partD ∩ U2 has equation

f (u, v) = uv + av3 + bv2 + cv + d = 0.

Moreover, the affine part of the line r = {x0 = 0} has equation u = 0, while the point[0, 1, 1] has coordinates (u0, v0) = (0, 1). First of all we note that fu(0, 1) = 1, sothe point [0, 1, 1], provided it belongs toD, is non-singular and, in this case, condition(ii) is satisfied if and only if I(D, r, [0, 1, 1]) ≥ 3. Since the function γ : C → C

2,γ(t) = (0, t + 1) defines a parametrization of r ∩ U2 with γ(0) = (0, 1), the mul-tiplicity of intersection I(D, r, [0, 1, 1]) coincides with the multiplicity of 0 as aroot of the polynomial f (γ(t)) = a(t + 1)3 + b(t + 1)2 + c(t + 1) + d. Therefore,I(D, r, [0, 1, 1]) ≥ 3 if and only if

a + b + c + d = 3a + 2b + c = 3a + b = 0.

FinallyD passes through [1, 4, 2] if and only if 1 + 8a + 4b + 2c + d = 0. Sincethe only solution of the system

⎧⎪⎪⎨⎪⎪⎩

a + b + c + d = 03a + 2b + c = 03a + b = 01 + 8a + 4b + 2c + d = 0

is a = −1, b = 3, c = −3, d = 1, we obtain that the only cubic D satisfying therequested conditions has equation

x0x1x2 − x31 + 3x21x2 − 3x1x22 + x32 = 0.

As the point [0, 1, 1] ∈ D is an inflection pointwith tangent x0 = 0, byExercise 62the curve D is reducible if and only if it contains the line x0 = 0. By looking at the

128 3 Exercises on Curves and Hypersurfaces

equation of D we immediately see that this fact does not happen, and consequentlyD is irreducible.

Exercise 78. (a) Determine the equation of a cubic C of P2(C) passing through

[1, 6, 2], having at Q = [1, 0, 0] an inflection point with tangent x1 + x2 = 0and having at P = [0, 1, 0] a cusp with principal tangent x0 = 0.

(b) Say whether C has any additional singular points and inflection points.(c) Say whether C is irreducible.(d) Determine all the linesH of P

2(C) such that the affine curve C ∩ (P2(C)\H) hasno asymptotes.

Solution. (a) Let F(x0, x1, x2) = 0 be an equation of a cubic C with the requestedproperties. Set the affine coordinates u = x0

x1 , v = x2x1 on U1, and let f1(u, v) =

F(u, 1, v) be the equation of the affine part C ∩ U1 of C. Since P must be acusp of C with principal tangent x0 = 0, the origin (0, 0) ∈ C

2 must be a cusp ofC ∩ U1 with principal tangent u = 0, so that, up to non-zero scalars, f1(u, v) =u2 + au3 + bu2v + cuv2 + dv3 for some a, b, c, d ∈ C not all equal to 0. HenceF(x0, x1, x2) = x20x1 + ax30 + bx20x2 + cx0x22 + dx32.

So, with respect to the affine coordinates s = x1x0 , t = x2

x0 on U0, the equation

of the affine part C ∩ U0 of C is f0(s, t) = F(1, s, t) = s + a + bt + ct2 + dt3 = 0.In order for C to have an inflection point with tangent x1 + x2 = 0 at Q, thecurve defined by f0 must have an inflection point with tangent s + t = 0 at (0, 0).A parametrization γ : C → C

2 of this line is defined by γ(r) = (r,−r), and for thisparametrization one has γ(0) = (0, 0). Therefore, 0 must be a root of multiplicityat least 3 of the polynomial f0(γ(r)). Since f0(γ(r)) = a + (1 − b)r + cr2 − dr3,one must have a = c = 0, b = 1. Therefore f0(s, t) = s + t + dt3. Note that Q isautomatically non-singular for C, and so it is an inflection point. In addition, the con-dition [1, 6, 2] ∈ C implies that f0(6, 2) = 0, so 8 + 8d = 0 and d = −1, and finallyF(x0, x1, x2) = x20x1 + x20x2 − x32.

(b) It is immediate to check that the system

⎧⎨⎩Fx0 = 2x0x1 + 2x0x2 = 0Fx1 = x20 = 0Fx2 = x20 − 3x22 = 0

has (0, 1, 0) as the only non-trivial solution up to multiplication by non-zero scalars,and so P is the only singular point of C. Moreover, the determinant of the Hessianmatrix of F is given by

HF(X) = det

⎛⎝2x1 + 2x2 2x0 2x0

2x0 0 02x0 0 −6x2

⎞⎠ = 24x20x2.

It is immediate to check that the only non-zero homogeneous triples that satisfy thesystem F(x0, x1, x2) = HF(x0, x1, x2) = 0 are given by [0, 1, 0], [1, 0, 0], i.e., by the

3 Exercises on Curves and Hypersurfaces 129

coordinates of P and Q. Since P is singular, it follows that Q is the only inflectionpoint of C.

(c) Since C has a unique singular point and a unique inflection point, we deducethat C is irreducible from Exercise 62 (d).

(d) Note that the affine curve C ∩ (P2(C)\H) has no asymptotes if and only if forevery T ∈ C ∩ H there exists a unique principal tangent to C at T , and this tangentis equal to H. So let H be a line of P

2(C). Since C is irreducible and has degree 3,by Bézout’s Theorem the intersection C ∩ H consists of one, two, or three points. IfC ∩ H contains at least two points, the multiplicity of intersection of C andH is equalto 1 at at least one of these points, that we denote by S. It follows thatH is not tangentto C at S, so that S is non-singular and the (principal) tangent to C at S defines anasymptote of C ∩ (P2(C)\H). Therefore, if C ∩ (P2(C)\H) has no asymptotes, theremust be a point R ∈ P

2(C) such that C ∩ H = {R} and I(C,H,R) = 3. It followsthat either R is singular for C, or H is an inflection tangent to C at R. In the formercase, by what we have seen in (b) one must have R = P, and H must be the onlyprincipal tangent to C at P, henceH = {x0 = 0}. In addition, it is immediate to checkthat actually C ∩ {x0 = 0} = P, so that H coincides with the only principal tangentto C at the only point of C ∩ H. In the latter case, by what we have seen in (b) onemust have R = Q and H = {x0 + x1 = 0}. Bézout’s Theorem guarantees also in thiscase that C ∩ H = {Q}. In addition, the (principal) tangent to C at Q is precisely Hby construction. Summing up, the affine curve C ∩ (P2(C)\H) has no asymptotes ifeither H = {x0 = 0} or H = {x0 + x1 = 0}.Exercise 79. Let P,Q be distinct points of P

2(C) and let r be the line joining them.Determine the integers k for which there exists a quartic C of P

2(C) that satisfies thefollowing conditions:

(i) P is an inflection point of C with the line r as tangent;(ii) Q is an ordinary double point of C;(iii) C has k irreducible components.

Solution. If C is a quartic that satisfies properties (i) and (ii), then I(C, r,P) ≥ 3 andI(C, r,Q) ≥ 2. By Bézout’s Theorem, the line r must be an irreducible component ofthe quartic; more precisely, it must be a component of multiplicity 1, since P wouldbe singular otherwise. In particular, C must be reducible i.e. k ≥ 2.

On the other hand we see that for each of the values k = 2, 3, 4 we can find aquartic that satisfies properties (i) and (ii) and that has k irreducible components.

If k = 2, it is enough to take a quartic whose irreducible components are the liner and a non-singular (and therefore irreducible by Exercise 55) cubic not passingthrough P and passing through Q with a tangent different from r. In order to showthat such a cubic exists, we choose a homogeneous coordinate system x0, x1, x2 suchthatQ = [1, 0, 0] and P = [0, 1, 0], so that r has equation x2 = 0. The cubic definedin this coordinate system by the equation x0x22 − x31 − x1x20 = 0 has the requestedproperties.

130 3 Exercises on Curves and Hypersurfaces

If k = 3, it is enough to take a quartic whose irreducible components are r, anotherline s �= r passing through Q and an irreducible conic passing neither through P northrough Q.

If k = 4, it is enough to take a quartic whose irreducible components are r, anotherline s �= r passing throughQ and two additional distinct lines passing neither throughP nor through Q.

Exercise 80. Prove that, if C is an affine curve of C2 of degree n, then C has at most

n distinct asymptotes.

Solution. Let C be the projective closure of C in P2(C), and let r ⊆ P

2(C) be theline of equation x0 = 0. Since r is not a component of C (cf. Sect. 1.7.4), by Bézout’sTheorem r and C intersect at finitely many points P1, . . . ,Pk . Let ai be the number ofasymptotes of C whose projective closure passes through Pi. Of course, the integera = ∑k

i=1 ai is the total number of asymptotes ofC. In addition, since every asymptotepassing through Pi is a principal tangent to C at Pi, and the number of principaltangents to a curve at any of its points is bounded by the multiplicity of the point,for every i one has ai ≤ mPi(C).

Furthermore, from Bézout’s Theorem and from the fact that one has by definitionI(C, r,Pi) ≥ mPi(C) for every i = 1, . . . , k, we deduce that

n =k∑

i=1

I(C, r,Pi) ≥k∑

i=1

mPi(C).

Hence

a =k∑

i=1

ai ≤k∑

i=1

mPi(C) ≤ n,

and therefore the claim holds.

Exercise 81. Let C be a non-singular projective curve of P2(C) of degree n > 1

and, for every point P ∈ C, denote by τP the tangent to C at P. Given Q ∈ P2(C),

prove that the set CQ = {P ∈ C |Q ∈ τP} is non-empty and contains at most n(n − 1)points.

Solution.LetF = 0be an equationofC, and letQ = [q0, q1, q2]. The setCQ coincideswith the set of points whose homogeneous coordinates [x0, x1, x2] satisfy the system

{Fx0(x0, x1, x2)q0 + Fx1(x0, x1, x2)q1 + Fx2(x0, x1, x2)q2 = 0F(x0, x1, x2) = 0

.

At first let us show that the first equation of this system is not trivial. Otherwise,the polynomial q0Fx0 + q1Fx1 + q2Fx2 would be identically zero, and hence Fx0 , Fx1 ,Fx2 would be linearly dependent polynomials. By Bézout’s Theorem, since each Fxiis either homogeneous of degree n − 1 or zero, the set of solutions of the system

3 Exercises on Curves and Hypersurfaces 131

Fx0 = Fx1 = 0 is not empty as n > 1. On the other hand, because of the hypothesiswe are assuming by contradiction, the points corresponding to the solutions of theselatter two equations would annihilate Fx2 too, so they would be singular points of C.A contradiction, because C is non-singular by hypothesis.

The previous considerations imply that the set CQ coincides with the set of pointsof intersection between C and a curve of degree n − 1. Then, by Bézout’s Theorem,CQ is non-empty. Moreover, the curve C, which is non-singular, is irreducible byExercise 55. Using again Bézout’s Theorem, it follows that the cardinality of CQ isat most n(n − 1).

Note. A particular case of the general situation considered in this exercise isdescribed in Exercise 67.

Arguing in a similar but slightly subtler way, one can prove the more generalresult that, given an irreducible plane curve C of P

2(C) of degree n > 1 and a pointQ /∈ Sing(C), the set of lines passing through Q and tangent to C at some point isnon-empty and has cardinality ≤ n(n − 1).

Exercise 82. Assume that C is an irreducible quartic of P2(C) having 3 cusps. Show

that the three principal tangents to C at the cusps belong to a pencil.

Solution. Denote by A,B,C the three cusps of the quartic C and by τA, τB, τC theprincipal tangents at these points.Observe thatB /∈ τA because otherwise, byBézout’sTheorem, the line τA would intersect C in at least 5 points (counted with multiplicity)and consequently it would be a component of C, a contradiction. By the same reasonC /∈ τA, A /∈ τB and C /∈ τB. Therefore the lines τA and τB meet at a point D whichis distinct from A,B,C. On the other hand, the points A,B,C cannot lie on a line r,because otherwise r would be a component of C, contradicting the hypothesis. SoA,B,C,D are in general position.

Then we can choose a system of homogeneous coordinates where A = [1, 0, 0],B = [0, 1, 0],C = [0, 0, 1],D = [1, 1, 1]. In these coordinates τA has equation x1 −x2 = 0 and τB has equation x0 − x2 = 0.

A quartic having a cusp in A with principal tangent x1 − x2 = 0 has an equationof the form

x20(x1 − x2)2 + x0(ax31 + bx21x2 + cx1x22 + dx32)++ ex41 + fx31x2 + gx21x

22 + hx1x32 + kx42 = 0.

By imposing that C have a cusp in B with principal tangent x0 − x2 = 0, we obtainthe relations

a = 0, e = 0, f = 0, g = 1, b = −2.

Finally, by imposing that C have a cusp in C, we get

d = 0, k = 0, h = 0, c = ±2.

132 3 Exercises on Curves and Hypersurfaces

Therefore C has an equation of the form

x20x21 + x20x

22 + x21x

22 − 2x20x1x2 − 2x0x

21x2 + cx0x1x

22 = 0.

If c = 2, the equation of C would be (x0x1 − x0x2 − x1x2)2 = 0, so C would bereducible, in contradiction with the hypothesis. Hence c = −2 and then the principaltangent τC at C to C has equation x0 − x1 = 0; as a consequence the point D lies onthe third cuspidal tangent τC too.

K Exercise 83. Let P be a non-singular point of a curve C of P2(K) of equation

F(x0, x1, x2) = 0 and assume that the tangent τP to the curve at P is not contained inC. Setting m = I(C, τP,P), prove that:

(a) The Hessian polynomial HF of F is non-zero.(b) I(H(C), τP,P) = m − 2.(c) If m = 3 (i.e. if P is an ordinary inflection point of C), then I(H(C), C,P) = 1.

Solution. (a), (b) Recall at first (cf. Sect. 1.9.4) that one can check whether HF isthe zero polynomial or not working in any system of homogeneous coordinates ofP2(K); in addition, provided thatHF �= 0 and hence that the Hessian curve is defined,

I(H(C), τP,P) can be computed in any system of coordinates.So choose a system of homogeneous coordinates where P = [1, 0, 0] and τP

has equation x2 = 0; for the sake of simplicity, still denote by F(x0, x1, x2) = 0 anequation of C in this system of coordinates.

The affine part C ∩ U0 of the curve in the chartU0 has equation f (x, y) = 0, wheref (x, y) = F(1, x, y) is the dehomogenized polynomial of F with respect to x0, andthe tangent to C ∩ U0 at the origin has equation y = 0. It follows that

f (x, y) = xm ϕ(x) + yψ(x, y)

with ϕ ∈ K[x],ψ ∈ K[x, y],ϕ(0) �= 0 and ψ(0, 0) �= 0 (recall that m ≥ 2 by thedefinition of tangent line). Then we have

fx = xm−1h(x) + yψx(x, y)fy = ψ(x, y) + yψy(x, y)fxx = xm−2k(x) + yψxx(x, y)fxy = ψx(x, y) + yψxy(x, y)fyy = 2ψy(x, y) + yψyy(x, y)

(3.2)

with h(x) = mϕ(x) + xϕx(x) and k(x) = (m − 1)h(x) + xhx(x). In particularh(0) �= 0 and k(0) �= 0.

3 Exercises on Curves and Hypersurfaces 133

Let us now compute the dehomogenized polynomial HF(X) with respect to x0. Ifwe denote by d the degree of F, necessarily we have d ≥ 2 because by hypothesis Cdoes not contain the tangent line τP. Observe that by Euler’s identity we have

(d − 1)Fx0 = x0Fx0x0 + x1Fx0x1 + x2Fx0x2(d − 1)Fx1 = x0Fx0x1 + x1Fx1x1 + x2Fx1x2(d − 1)Fx2 = x0Fx0x2 + x1Fx1x2 + x2Fx2x2

dF = x0Fx0 + x1Fx1 + x2Fx2

which implies that

x0Fx0x0 = (d − 1)Fx0 − x1Fx0x1 − x2Fx0x2x0Fx0x1 = (d − 1)Fx1 − x1Fx1x1 − x2Fx1x2x0Fx0x2 = (d − 1)Fx2 − x1Fx1x2 − x2Fx2x2x0Fx0 = dF − x1Fx1 − x2Fx2 .

(3.3)

By using the relations (3.3) and the properties of the determinant, we obtain

x0HF(X) = det

⎛⎝x0Fx0x0 Fx0x1 Fx0x2x0Fx0x1 Fx1x1 Fx1x2x0Fx0x2 Fx1x2 Fx2x2

⎞⎠ =

= det

⎛⎝

(d − 1)Fx0 − x1Fx0x1 − x2Fx0x2 Fx0x1 Fx0x2(d − 1)Fx1 − x1Fx1x1 − x2Fx1x2 Fx1x1 Fx1x2(d − 1)Fx2 − x1Fx1x2 − x2Fx2x2 Fx1x2 Fx2x2

⎞⎠ =

= det

⎛⎝

(d − 1)Fx0 Fx0x1 Fx0x2(d − 1)Fx1 Fx1x1 Fx1x2(d − 1)Fx2 Fx1x2 Fx2x2

⎞⎠ =

= (d − 1) det

⎛⎝Fx0 Fx0x1 Fx0x2Fx1 Fx1x1 Fx1x2Fx2 Fx1x2 Fx2x2

⎞⎠ .

By using once more the relations (3.3) and the properties of the determinant weobtain

x20HF(X) = (d − 1) det

⎛⎝x0Fx0 x0Fx0x1 x0Fx0x2Fx1 Fx1x1 Fx1x2Fx2 Fx1x2 Fx2x2

⎞⎠ =

= (d − 1) det

⎛⎝dF (d − 1)Fx1 (d − 1)Fx2Fx1 Fx1x1 Fx1x2Fx2 Fx1x2 Fx2x2

⎞⎠ .

134 3 Exercises on Curves and Hypersurfaces

Thus, if we dehomogenize with respect to x0, we get

HF(1, x, y) = (d − 1) det

⎛⎝df (d − 1)fx (d − 1)fyfx fxx fxyfy fxy fyy

⎞⎠ . (3.4)

By replacing the relations (3.2) in (3.4) we obtain

HF(1, x, y) =

= (d − 1) det

⎛⎝d(xm ϕ + yψ) (d − 1)(xm−1h + yψx) (d − 1)(ψ + yψy)

xm−1h + yψx xm−2k + yψxx ψx + yψxy

ψ + yψy ψx + yψxy 2ψy + yψyy

⎞⎠ .

In order to compute the multiplicity of intersection at P = (0, 0) with the line ofequation y = 0, it is sufficient to compute HF(1, x, 0). Then we observe that

HF(1, x, 0) = (d − 1) det

⎛⎝d xm ϕ (d − 1)xm−1h (d − 1)ψ(x, 0)xm−1h xm−2k ψx(x, 0)ψ(x, 0) ψx(x, 0) 2ψy(x, 0)

⎞⎠ =

= xm−2 g(x)

with g(0) = −(d − 1)k(0)(ψ(0, 0))2. As k(0) �= 0 and ψ(0, 0) �= 0, then g(0) �= 0.This implies that HF is non-zero (so that the Hessian curve H(C) of C is defined),and also that I(H(C), τP,P) = m − 2.

(c) Ifm = 3, then I(H(C), τP,P) = 1 by part (b). It follows thatP is a smooth pointof H(C) and that the line τP is not tangent to H(C) at P. Therefore, by Exercise 109,the curves C and H(C) are not tangent at P, and hence I(H(C), C,P) = 1.

K Exercise 84. Let C be a reduced curve of P2(C) of equation F(x0, x1, x2) = 0. Show

that:

(a) If C is not a union of lines, than the Hessian polynomial HF(X) is non-zero.(b) If C is irreducible and has infinitely many inflection points, then C is a line.(c) If HF(X) �= 0, then the only common irreducible components of C and of the

Hessian H(C) are the lines contained in C.(d) If C is a union of lines all passing through one point, then HF(X) = 0.

Solution. (a) Since C is reduced, Sing(C) is a finite set (cf. Exercise 56). Moreover,by assumption there exists an irreducible component C1 of C which is not a line. Sothe claim can be proven by choosing a point P ∈ C1 which is non-singular for C andby using Exercise 83 (a).

(b) Assume by contradiction that C is not a line and denote by d ≥ 2 its degree.Let P ∈ C be a non-singular point, let τP be the tangent to C at P and let m =I(C, τP,P). Since C is irreducible, τP cannot be a component of the curve and som < ∞. By what we proved in Exercise 83, the polynomial HF(X) is non-zero andthe Hessian curve H(C) is such that I(H(C), τP,P) = m − 2. Since C has infinitely

3 Exercises on Curves and Hypersurfaces 135

many inflection points by assumption, one hasHF(Q) = 0 for infinitely many pointsQ ∈ C. Hence, by Bézout’s Theorem, C is an irreducible component of H(C) andtherefore I(C, τP,P) ≤ I(H(C), τP,P). So one obtains m ≤ m − 2, a contradiction.

(c) Any line contained in C is also contained in H(C), since all its points thatare non-singular for C (there are infinitely many of these, because C is reduced) areinflection points. Conversely, if C1 is a common irreducible component of C andH(C), then any point of C1 that is smooth for C is an inflection point (both for C1 andfor C). Since there exist infinitely many points with this property, C1 is a line by (b).

(d) If C is a union of lines all passing through one point, then up to a change ofcoordinates we may assume that they all pass through the point [1, 0, 0] and so theyhave an equation of the form ax1 + bx2 = 0. Then C is defined by a homogeneouspolynomial F not depending on x0; therefore the Hessian matrix of F has a zerocolumn and, as a consequence, its determinant is zero.

Exercise 85. Let C be the cubic of P2(K) defined by the equation:

F(x0, x1, x2) = x0x22 − x31 − ax1x

20 − bx30 = 0,

with a, b ∈ K. Show that:

(a) The point P = [0, 0, 1] is an inflection point of C.(b) C is irreducible.(c) C is non-singular if and only if g(x) = x3 + ax + b has no multiple roots or,

equivalently, iff 4a3 + 27b2 �= 0.(d) If K = C and C is non-singular, then there are precisely 4 lines passing through

P and tangent to C.Solution. (a) One has:

∇F(x0, x1, x2) = (x22 − 2ax0x1 − 3bx20,−3x21 − ax20, 2x0x2).

So it is immediate to check that P is a smooth point of C and that TP(C) is the linex0 = 0. Since F(0, x1, x2) = −x31, the line x0 = 0 intersects C at P with multiplicity3 and so P is an inflection point.

(b) Note first of all that it suffices to consider the case K = C, since if K = R andthe complexified curve CC of C is irreducible, then C is obviously irreducible too.

So assume K = C. Since P is an inflection point and the inflection tangent is theline x0 = 0 which is not contained in C, the curve is irreducible by Exercise 62.

(c) Since P is the only intersection point of C with the line at infinity x0 = 0,it is enough to examine the affine part C ∩ U0 of C. In the affine coordinates x =x1x0 , y = x2

x0 , the curve C ∩ U0 is defined by the equation f (x, y) = y2 − g(x) = 0,

where g(x) = x3 + ax + b. The singular points of C0 are the solutions of the system

2y = 0, g′(x) = 0, y2 − g(x) = 0,

136 3 Exercises on Curves and Hypersurfaces

i.e., they are the points (α, 0) with g(α) = g′(α) = 0. So C is singular if and onlyif g(x) has a multiple root α. This happens if and only if the resultant Ris(g, g′) ofthe polynomials g(x) and g′(x) = 3x2 + a vanishes (cf. Sect. 1.9.2). By definitionRis(g, g′) is the determinant of the Sylvester matrix:

S(g, g′) =

⎛⎜⎜⎜⎜⎝

b a 0 1 00 b a 0 1a 0 3 0 00 a 0 3 00 0 a 0 3

⎞⎟⎟⎟⎟⎠

,

and so it is equal to 4a3 + 27b2. Alternatively, one can observe that the pair ofequations g(x) = x3 + ax + b = 0, g′(x) = 3x2 + a = 0 is equivalent to the pair ofequations 2ax + 3b = 0, 3x2 + a = 0; then it is easy to check that the latter twoequations have a common solution if and only if 4a3 + 27b2 = 0.

(d)We have already checked that the line at infinity x0 = 0 is tangent to C atP. Theaffine part of the tangent line to C at a proper point R with affine coordinates(α,β)

has equation −g′(α)(x − α) + 2β(y − β) = 0, so that the point P belongs to TR(C)

if and only if β = 0 (and therefore g(α) = 0).So the proper points R such that P ∈ TR(C) are in one-to-one correspondence with

the roots of g(x) and the tangent lines to C at these points are distinct. Since C isnon-singular by assumption, the roots of g are distinct as explained in the solutionof (c). Since g has degree 3 and K = C, it follows from this argument that there arein all 4 lines that are tangent to C and pass through P.

K Exercise 86. (Weierstrass equation of a plane cubic) Let C be a non-singular irre-ducible cubic of P

2(K). Show that:

(a) C has at least one inflection point.(b) If P ∈ C is an inflection point, then there exists a system of homogeneous coor-

dinates x0, x1, x2 of P2(K) such that P has coordinates [0, 0, 1] and C is defined

by the equationx0x

22 − x31 − ax1x

20 − bx30 = 0,

where a, b ∈ K and 4a3 + 27b2 �= 0.(c) If K = C, there exists a system of homogeneous coordinates x0, x1, x2 of P

2(C)

such that P has coordinates [0, 0, 1] and C is defined by the equation

x0x22 − x1(x1 − x0)(x1 − λx0) = 0,

with λ ∈ C\{0, 1}.Solution. (a) First of all note that by Exercise 64 the support of C contains infinitelymany points also in the case K = R.

LetF(x0, x1, x2) = 0 be an equation of C and letHF(X) be theHessian polynomialof F. If HF = 0, then all the points of C, being smooth by assumption, are inflection

3 Exercises on Curves and Hypersurfaces 137

points (actually, using Exercises 61 and 84, one can show that this case does notoccur). If HF is non-zero, then it has degree 3; so, by Bézout’s Theorem if K = C,or by Exercise 64 if K = R, there is at least one point P ∈ C such that HF(P) = 0.Since C is non-singular, P is an inflection point.

(b) Choose homogeneous coordinates w0, w1, w2 such that P = [0, 0, 1] and theline TP(C) has equation w0 = 0. Since P is an inflection point, the curve C is definedby an equation of the form:

w0w22 + 2w2w0A(w0, w1) + B(w0, w1) = 0,

where A and B are homogeneous (or zero) polynomials of degrees 1 and 3,respectively. In the homogeneous coordinate system y0 = w0, y1 = w1, y2 = w2 +A(w0, w1) the equation of C takes the form y0y22 + C(y0, y1) = 0, where C is homo-geneous of degree 3. More precisely, since C is irreducible and therefore does nothave the line y0 = 0 as a component, one has C(y0, y1) = −cy31 + y0D(y0, y1), withc ∈ K

∗ and D either zero or a homogeneous polynomial of degree 2. Changing thecoordinate system to z0 = y0, z1 = y1

c , z2 = y2c2, the curve C is defined by an equa-

tion of the form z0z22 − z31 − αz21z0 − βz1z20 − γz30 = 0, withα,β, γ ∈ K. Finally, thechange of coordinates x0 = z0, x1 = z1 + α

3 z0, x2 = z2 transforms the equation of Cinto x0x22 − x31 − ax1x20 − bx30 = 0, with a, b ∈ K. Since in all the coordinate systemsthat we have used the point P has coordinates [0, 0, 1], we have obtained an equationof C of the requested form. In addition one has 4a3 + 27b2 �= 0 by Exercise 85.

(c) Consider the homogeneous coordinates y0, y1, y2 and the polynomialC(y0, y1)introduced in the solution of (b). Let c(y) = C(1, y), let α1,α2,α3 ∈ C be the rootsof c(y) and, for i = 1, 2, 3, let Qi = [1,αi, 0]. Note that the Qi are distinct, becauseif αi were a multiple root of c then the point Qi would be singular. There exists achange of homogeneous coordinates z0 = y0, z1 = μy0 + νy1, z2 = y2, with μ, ν ∈C, ν �= 0, such that in the coordinates z0, z1, z2 one hasQ1 = [1, 0, 0],Q2 = [1, 1, 0]andQ3 = [1,λ, 0], for a suitable λ �= 0, 1. In this coordinate system C is given by anequation of the form z0z22 − G(z0, z1) = 0. Since C ∩ {z2 = 0} = {Q1,Q2,Q3}, thereexists τ ∈ C

∗ such that G(z0, z1) = τz1(z1 − z0)(z1 − λz0). Let δ ∈ C be such thatδ2 = τ ; in the coordinates x0 = z0, x1 = z1, x2 = z2

δthe curve C is defined by the

equation x0x22 − x1(x1 − x0)(x1 − λx0) = 0

Exercise 87. Let D ⊂ P2(R) be an irreducible cubic. Show that D is non-singular

if and only if DC ⊂ P2(C) is non-singular.

Solution (1). Note that the singular points ofD are precisely the real singular pointsof DC. So, if DC is non-singular, then also D is non-singular.

Conversely, let us show that, if D is non-singular, then also DC is non-singular.So assume by contradiction that D is smooth and that there exists a singular pointP ∈ DC. The point Q = σ(P) is distinct from P and it is a singular point of DC, too.The line r = L(P,Q) intersects DC with multiplicity ≥ 2 both at P and at Q andtherefore is a component of DC by Bézout’s Theorem. Since r = σ(r), the line r isreal (cf. Exercise 59) and so it is a component of D, against the assumption.

138 3 Exercises on Curves and Hypersurfaces

Solution (2). As in Solution (1) we observe that, if DC is non-singular, then also Dis non-singular.

Assume now that the curve D is smooth. By Exercise 86 there exists a homo-geneous coordinate system x0, x1, x2 of P

2(R) in which D is defined by an equa-tion of the form F(x0, x1, x2) = x0x22 − x31 − ax1x20 − bx30 = 0 with a, b ∈ R suchthat 4a3 + 27b2 �= 0. Then the curve DC, being defined by the same equationF(x0, x1, x2) = 0 in the corresponding homogeneous coordinate system of P

2(C),is non-singular by Exercise 85.

Note. The assumption that D is irreducible cannot be removed. Indeed, if D =r + Q, where r is a line and Q is a non-degenerate conic such that Q ∩ r = ∅, thenthe curve D is non-singular but the complexified curve DC is singular at the pointsof rC ∩ QC (cf. Exercise 54).

Exercise 88. (Nodal complex cubics) If C is an irreducible projective cubic ofP2(C)

with one node, show that:

(a) C is projectively equivalent to the cubic of equation

x0x1x2 + x30 + x31 = 0.

(b) C has three inflection points, which are collinear.

Solution. (a) Observe at first that two curves are projectively equivalent if and onlyif there exist systems of projective coordinates where they are described by thesame equation. Denote by P the node of C and by r, s the principal tangents toC at P. Since the curve of equation x0x1x2 + x30 + x31 = 0 has a node at [0, 0, 1]with principal tangents of equations x0 = 0 and x1 = 0, on P

2(C) we can choosecoordinates y0, y1, y2 such that P = [0, 0, 1], r = {y0 = 0}, s = {y1 = 0} (we canreadily check that such a system of coordinates does exist). In the affine coordinatesu = y0

y2 , v = y1y2 defined on the affine chart U2 the equation of C ∩ U2 is of the

formuv + au3 + bu2v + cuv2 + dv3 = 0 for somea, b, c, d ∈ Cnot simultaneouslyzero. So C has equation y0y1(y2 + by0 + cy1) + ay30 + dy31 = 0. In addition, observethat, in order that C be irreducible, necessarily a �= 0, d �= 0.

Now it is not difficult to construct a new systemof coordinateswhereC is describedby the required equation.Namely, assume thatα, δ are cube roots ofa, d, respectively.As a �= 0, d �= 0, clearly α �= 0, δ �= 0. The matrix

⎛⎜⎜⎝

α 0 bαδ

0 δ cαδ

0 0 1αδ

⎞⎟⎟⎠

is invertible, so there exists a well-defined system of homogeneous coordinates

z0, z1, z2 on P2(C) such that z0 = αy0, z1 = δy1, z2 = y2 + by0 + cy1

αδ. As

3 Exercises on Curves and Hypersurfaces 139

z0z1z2 + z30 + z31 = αδy0y1y2 + by0 + cy1

αδ+ α3y30 + δ3y31 =

= y0y1(y2 + by0 + cy1) + ay30 + dy31,

in the system of coordinates z0, z1, z2 the curve C is defined by the equation given inthe statement, so the thesis follows.

(b) By using that projectivities preserve the multiplicity of intersection betweena line and a curve, it is immediate to check that, if C is a projective curve, P ∈ C andf : P

2(C) → P2(C) is a projectivity, then P is an inflection point of C if and only if

f (P) is an inflection point of f (C). Therefore, in order to prove statement (b) we mayassume that C is defined by the equation G(x0, x1, x2) = x0x1x2 + x30 + x31 = 0. Theequation of the Hessian curve of C is

HG(x0, x1, x2) = det

⎛⎝6x0 x2 x1x2 6x1 x0x1 x0 0

⎞⎠ = 2x0x1x2 − 6(x30 + x31) = 0.

It can be immediately checked that the system G(x0, x1, x2) = HG(x0, x1, x2) = 0is equivalent to x0x1x2 = x30 + x31 = 0, having as solutions the points [0, 0, 1],[1,−1, 0], [1,ω, 0], [1,ω2, 0], where ω �= −1 is a cube root of −1 in C. By con-struction, [0, 0, 1] is singular for C, while it is not difficult to check that [1,−1, 0],[1,ω, 0], [1,ω2, 0] are non-singular for C, so they are inflection points of C. Theseinflection points lie on the line x2 = 0, so they are collinear.

Note. Point (a) of the exercise ensures that two irreducible cubics ofP2(C), eachwithone node, are projectively equivalent. So, in particular, every complex irreduciblecubic with one node is projectively equivalent to the cubic of equation x0x22 − x31 −x0x21 = 0.

An explicit case of statement (b) is considered in Exercise 69.

Note. (Nodal real cubics) Denote by D an irreducible cubic of P2(R) having one

nodeP. Also the complexification C = DC has a node atP and, having an odd degree,it is irreducible by Exercise 61. Therefore, C has exactly three inflection points byExercise 88. Since C is the complexification of a real curve, the set of its inflectionpoints is invariant under conjugation. This observation implies that at least one of theinflection points of C is a real point Q, and consequently it is an inflection point ofD. Then, arguing as in Exercise 86, it is not difficult to show there exists a system ofprojective coordinates x0, x1, x2 of P

2(R) such that P = [1, 0, 0], Q = [0, 0, 1] andD is defined by one of the following equations:

(1) x0x22 − x31 + x0x21 = 0;(2) x0x22 − x31 − x0x21 = 0.

Note that the curves of P2(R) defined by the Eqs. (1) and (2) are not projectively

isomorphic, because the point P is the only singular point of both curves, but thetangent cones at P to the two curves are not projectively isomorphic. If we computethe Hessian explicitly, we can check that in case (1) the curve D has three inflectionpoints, while in case (2) only one inflection point exists.

140 3 Exercises on Curves and Hypersurfaces

Exercise 89. (Cuspidal complex cubics) If C is an irreducible cubic of P2(C) with

a cusp, show that:

(a) C has exactly one inflection point.(b) C is projectively equivalent to the cubic of equation x0x21 = x32.

Solution. (a)Denote byP the cusp of C and by r the principal tangent at that point. LetQ be another point of the cubic distinct fromP; necessarilyQ is non-singular, becauseotherwise the line L(P,Q) would meet the cubic in at least 4 points (counted withmultiplicity) and hence it would be an irreducible component of C, which instead isirreducible by hypothesis. Denote by s the tangent to C atQ. If s should pass throughP, then it would meet the cubic in at least 4 points (counted with multiplicity), againa contradiction. In particular r �= s, so the point R = r ∩ s is well defined. As thepoints P,Q,R are not collinear, we can choose a projective frame in P

2(C) havingthem as fundamental points, so that P = [1, 0, 0],Q = [0, 1, 0],R = [0, 0, 1].

In order that [1, 0, 0] be a cusp with tangent L(P,R) = {x1 = 0}, the equation ofC must be of the form

x0x21 + ax31 + bx21x2 + cx1x

22 + dx32 = 0,

with a, b, c, d ∈ C not simultaneously zero. The curve defined by the latter equationpasses through Q if and only if a = 0 and it has x0 = 0 as tangent at Q iff b = 0.Therefore, the equation of C is of the form

F(x0, x1, x2) = x0x21 + cx1x

22 + dx32 = 0

with d �= 0 because otherwise C would be reducible. Since we aim at determiningthe inflection points of C, observe that

HF(x0, x1, x2) = det

⎛⎝

0 2x1 02x1 2x0 2cx20 2cx2 2cx1 + 6dx2

⎞⎠ = −8x21(3dx2 + cx1).

Note that the line x1 = 0 intersects the curve C only at the singular pointP, coherentlywith the fact that P is singular and x1 = 0 is the (only) principal tangent to C. Asd �= 0, the line 3dx2 + cx1 = 0 is not a principal tangent and hence it intersects Cat P with multiplicity 2. It follows that it intersects C with multiplicity 1 at a pointT �= P, which necessarily is non-singular for C. (In fact, an easy computation showsthat T = [−2c3, 27d2,−9cd] and Fx0(T) �= 0). Therefore T is the only inflectionpoint of C.

(b) Observe at first that two curves are projectively equivalent if and only ifthere exist systems of projective coordinates where they are described by the sameequation.Observe also that the curve of equation x0x21 = x32 has a cusp at [1, 0, 0]withprincipal tangent of equation x1 = 0, one inflection point at [0, 1, 0] with tangent ofequation x0 = 0 and it passes through [1, 1, 1].

3 Exercises on Curves and Hypersurfaces 141

Denote by P, T the cusp and the inflection point of C, respectively; moreoverdenote by r the principal tangent to C at P and by s the tangent to C at the inflectionpoint T . If R = r ∩ s, as remarked in the solution of (a) the points P,T ,R are ingeneral position. In addition, as C is irreducible, the support of C is not containedin L(P,T) ∪ L(P,R) ∪ L(T ,R), so we can choose a point S ∈ C in such a way thatP,T ,R, S form a projective frame of P

2(C). Let us fix on P2(C) the homogeneous

coordinates y0, y1, y2 induced by the frame {P,T ,R, S}.As shown above, the equation of C is of the form

F(y0, y1, y2) = y0y21 + cy1y

22 + dy32 = 0

with d �= 0, and HF(y0, y1, y2) = −8y21(3dy2 + cy1). By imposing that C passthrough S and that T be an inflection point of C, we obtain the conditionsF(1, 1, 1) =HF(0, 1, 0) = 0, which imply that c = 0 and d = −1. Hence F(y0, y1, y2) = y0y21 −y32, and the thesis is proved.

Note. In particular the exercise proves that two irreducible cubics of P2(C), each

having one cusp, are projectively equivalent.Explicit cases of the property just proved are described in Exercise 78 and in

Exercise 105.

Note. (Cuspidal real cubics) Let D be an irreducible cubic of P2(R) having a cusp

P. The complexified curve C = DC also has a cusp at P and, having an odd degree, isirreducible by Exercise 61. So C has precisely one inflection point T by Exercise 89.Since C is the complexified curve of a real curve, the set of inflection points of Cis invariant under conjugation. It follows that T is a real point and therefore is aninflection point of D.

Since the support of D contains infinitely many points by Exercise 64, one canshow as in the solution of (b) the existence of a projective coordinate system x0, x1, x2of P

2(R) such that P = [1, 0, 0], T = [0, 1, 0], [1, 1, 1] ∈ D andD is defined by theequation x0x21 − x32 = 0. So the irreducible real cubics with a cusp are all projectivelyequivalent.

Finally note that the change of coordinates y0 = x0, y1 = x2, y2 = x1 transformsD into the cubic y0y22 − y31 = 0, which is in Weierstrass form.

K Exercise 90. Let C be a smooth irreducible cubic of P2(K) and let P1,P2 ∈ C be

two distinct inflection points. Show that:

(a) The line L(P1,P2) intersects C at a third point P3, different from P1 and P2,which is also an inflection point of C.

(b) There exists a projectivity g of P2(K) such that g(C) = C and g(P1) = P2.

Solution. (a) By Exercise 86 there exists a system of homogenous coordinatesx0, x1, x2 on P

2(K) such that P1 = [0, 0, 1] and C has equation

x0x22 − x31 − ax1x

20 − bx30 = 0,

142 3 Exercises on Curves and Hypersurfaces

where a, b ∈ K and 4a3 + 27b2 �= 0. Since P1 is the only point of C on the linex0 = 0, the point P2 has coordinates [1,α,β], with α,β ∈ K. Since deg C = 3 andP2 is an inflection point, the line TP2(C) intersects C only at P2, and therefore it doesnot pass through P1. A calculation analogous to those carried out in the solution ofExercise 85 (d) shows that this fact is equivalent to the condition β �= 0.

Now let f : P2(K) → P

2(K) be the projectivity defined by [x0, x1, x2] �→[x0, x1,−x2]. It is easy to check that f (P1) = P1 and f (C) = C. The point f (P2)

has homogeneous coordinates [1,α,−β] and so it is distinct from P2 and collinearwith P1 and P2, and therefore it is the third point P3 at which L(P1,P2) intersectsC. Moreover P3 = f (P2) is an inflection point, since it is the image of an inflectionpoint under a projectivity that preserves the curve C.

(b) Let now y0, y1, y2 be a system of homogeneous coordinates such that P3 =[0, 0, 1] and C has equation y0y22 − y31 − a′y1y20 − b′y30 = 0,with a′, b′ ∈ K (cf. Exer-cise 86). By what we have seen in the solution of (a), the coordinates of P1 and P2 inthis system are [1,α′,β′] and [1,α′,−β′], respectively, for some α′ ∈ K, β′ ∈ K

∗.It is now immediate to check that the projectivity g : P

2(K) → P2(K) defined by

[y0, y1, y2] �→ [y0, y1,−y2] has the requested properties.

Exercise 91. (Salmon’s Theorem) Let C be a smooth cubic of P2(C) and let P ∈ C

be an inflection point. Show that:

(a) There are precisely 4 lines r1, r2, r3, r4 tangent to C and passing through P.

(b) Let kP = β(r1, r2, r3, r4) and let j(C,P) = (k2P − kP + 1)3

k2P(kP − 1)2(note that j(C,P) is

independent of the ordering of the lines r1, r2, r3, r4 by Exercise 21); if P′ ∈ Cis another inflection point, then j(C,P) = j(C,P′).

Solution. (a) By Exercise 86 there exists a system of homogeneous coordinatesx0, x1, x2 such that P = [0, 0, 1] and the equation of C is in Weierstrass form. So theclaim follows from Exercise 85.

(b) If P′ ∈ C is an inflection point, then by Exercise 90 there exists a projectivityg of P

2(C) such that g(C) = C and g(P) = g(P′). If r′1, r

′2, r

′3, r

′4 are the lines tangent

to C and passing through P′, then one has g ({r1, r2, r3, r4}) = {r′1, r

′2, r

′3, r

′4}. So by

Exercise 21 one has j(C,P) = j(C,P′).

Note. (The j-invariant of a smooth cubic) Given a smooth cubic C of P2(C), we

have seen (cf. Exercise 86) that C has at least an inflection point P, and so one candefine j(C,P) as in Exercise 91. In addition, again by Exercise 91, j(C,P) does notdepend on the choice of the inflection point and therefore it is uniquely determinedby C. For this reason it is usual to write simply j(C). The complex number j(C) iscalled j-invariant, or modulus, of the cubic; indeed in Exercise 92 it is shown thatj(C) determines C up to projective equivalence.

K Exercise 92. Given a smooth cubic C of P2(C), let j(C) be the invariant defined in

Exercise 91 and in the Note following it. Prove the following statements:

3 Exercises on Curves and Hypersurfaces 143

(a) If Cλ is the cubic of equation:

x0x22 − x1(x1 − x0)(x1 − λx0) = 0,

with λ ∈ C\{0, 1}, then j(Cλ) = (λ2 − λ + 1)3

λ2(λ − 1)2.

(b) For every α ∈ C there exists a smooth cubic C of P2(C) such that j(C) = α.

(c) Two smooth cubics C and C ′ of P2(C) are projectively equivalent if and only if

j(C) = j(C ′).

Solution. (a) Arguing precisely as in Exercise 85 one checks that Cλ is smooth,that P = [0, 0, 1] ∈ Cλ is an inflection point and that the lines tangent to Cλ andpassing through P are the line r1 = {x0 = 0}, which is the tangent at P, andthe lines r2 = {x1 = 0}, r3 = {x1 − x0 = 0} and r4 = {x1 − λx0 = 0}. So one has

β(r2, r1, r3, r4) = λ, and therefore j(Cλ) = (λ2 − λ + 1)3

λ2(λ − 1)2.

(b) Fixα ∈ C; consider the polynomial p(t) = (t2 − t + 1)3 − αt2(t − 1)2. Sincep(t) has a positive degree, there exists λ ∈ C such that p(λ) = 0. Since p(0) =p(1) = 1, onehasλ �= 0, 1and so (λ2 − λ + 1)3

λ2(λ − 1)2= α. By (a) the cubicCλ of equation

x0x22 − x1(x1 − x0)(x1 − λx0) = 0 has modulus equal to α.

(c) Assume that C and C ′ are projectively equivalent and that g is a projectivity ofP2(C) such that g(C) = C ′. LetP ∈ C be an inflection point and let r1, r2, r3, r4 be the

lines passing through P and tangent to C; then P′ = g(P) is an inflection point of C ′and the lines r′

1 = g(r1), . . . , r′4 = g(r4) are the lines passing through P′ and tangent

toC ′. Since g induces a projective isomorphismbetween the pencil of lineswith centreP and the pencil of lines with centre P′, one has β(r1, r2, r3, r4) = β(r′

1, r′2, r

′3, r

′4),

and therefore, a fortiori, j(C) = j(C ′). So we have shown that j(C) is invariant underprojective isomorphisms.

Conversely, letC andC ′ be two smooth cubics such that j(C) = j(C ′) = α. ByExer-cise 86, there exist λ,λ′ ∈ C\{0, 1} such that C is projectively equivalent to the cubicCλ defined by the equation x0x22 − x1(x1 − x0)(x1 − λx0) = 0 and C ′ is projectivelyequivalent to the cubicCλ′ definedby the equation x0x22 − x1(x1 − x0)(x1 − λ′x0) = 0.Since projective transformations preserve j, by part (a) we have

(λ′2 − λ′ + 1)3

λ′2(λ′ − 1)2= j(C ′) = j(C) = (λ2 − λ + 1)3

λ2(λ − 1)2. (3.5)

Since projective equivalence is an equivalence relation, in order to conclude it isenough to show that Cλ and Cλ′ are projectively equivalent. Since the relation (3.5)holds, by Exercise 21 there exists a projectivity f of the line {x2 = 0} such that:

f {[1, 0, 0], [0, 1, 0], [1, 1, 0], [1,λ, 0]} = {[1, 0, 0], [0, 1, 0], [1, 1, 0], [1,λ′, 0]}.

144 3 Exercises on Curves and Hypersurfaces

In addition, since the permutations (12)(34), (13)(24) and (14)(32) preserve thecross-ratio of a quadruple P1,P2,P3,P4 of distinct points of a projective line(cf. Sect. 1.5.2), we may assume that [0, 1, 0] is fixed by f and so that f is ofthe form [x0, x1, 0] �→ [x0, ax0 + bx1, 0], where a, b ∈ C and b �= 0. Then, argu-ing as in the solution of Exercise 86 (c), one sees that the projectivity g of P

2(C)

defined by [x0, x1, x2] �→ [x0, ax0 + bx1, x2] transforms Cλ into a cubicD of equationx0x22 − τx1(x1 − x0)(x1 − λ′x0) = 0, where τ ∈ C

∗. Let δ ∈ C be such that τ = δ2;

the projectivity h defined by [x0, x1, x2] �→[x0, x1,

x2δ

]transforms D into Cλ′ .

Exercise 93. Prove that every smooth cubic C of P2(C) has exactly 9 inflection

points.

Solution (1).As C has degree 3 and it is irreducible by Exercise 55, for each inflectionpointP of C we have I(C,TP(C),P) = 3, i.e. all inflection points of C are ordinary. ByExercise 83 theHessian curveH(C) iswell defined and themultiplicity of intersectionof C with H(C) is equal to 1 at each inflection point of C. Since H(C) is a cubic,Bézout’s Theorem implies that C and H(C) meet exactly at 9 point, so C has exactly9 inflection points.

Solution (2).ByExercise 85 there exist homogeneous coordinates x0, x1, x2 ofP2(C)

such that C is defined by the equation F(x0, x1, x2) = x0x22 − x31 − ax1x20 − bx30 = 0,with a, b ∈ C and 4a3 + 27b2 �= 0. Aswe have shown in Exercise 85 and its solution,the point P = [0, 0, 1] is an inflection point of C and C intersects the line x2 = 0in three distinct points R1,R2,R3 such that P belongs to the tangent lines to C atR1,R2,R3. As C is smooth, and hence irreducible, then I(C,TRi(C),Ri) = 2 for i =1, 2, 3. In other words, the points R1,R2,R3 are not inflection points.

As P is the only improper point of C and it is an inflection point, it suffices tolook for inflection points only in the affine part C ∩ U0 of C. In the affine coordinatesx = x1

x0 , y = x2x0 , the curve C ∩ U0 is given by the equation f (x, y) = y2 − g(x) = 0,

where g(x) = x3 + ax + b. The Hessian matrix of F is:

HessF(X) =⎛⎝

−2ax1 − 6bx0 −2ax0 2x2−2ax0 −6x1 02x2 0 2x0

⎞⎠ .

If we compute the determinant of HessF(X) and we dehomogenize it with respectto x0, we obtain that the affine part H(C) ∩ U0 of the Hessian curve H(C) is givenby the equation h(x, y) = 3xy2 + 3ax2 + 9bx − a2 = 0. So the inflection points ofC ∩ U0 are the solutions of the system of equations

y2 − g(x) = 0, h(x, y) = 0.

By eliminating y2 in the second equation, we get the following system, which isequivalent to the previous one:

y2 − g(x) = 0, 3x4 + 6ax2 + 12bx − a2 = 0.

3 Exercises on Curves and Hypersurfaces 145

We are now going to show that the polynomial p(x) = 3x4 + 6ax2 + 12bx −a2 = 0 has distinct roots.We have p′(x) = 12x3 + 12ax + 12b = 12g(x), so p and p′have a common root if and only if p and g have a common root. Assume by contradic-tion that there exists α ∈ C such that p(α) = g(α) = 0: then the point R = [1,α, 0]belongs to the intersection of C with the line x2 = 0 and it is an inflection point, incontradiction with the observations made at the beginning. Therefore p(x) has 4 dis-tinct roots α1,α2,α3,α4, and none of them is a root of g(x); in correspondence witheach root αi we have two inflection points [1,αi,βi] and [1,αi,−βi] of C, whereβi �= 0 satisfies the relation β2

i = g(αi).In summary, the affine curveC ∩ U0 has 8 inflection points and soC has 9 inflection

points.

Note. (Configuration of the inflection points of a smooth complex cubic) If C is asmooth cubic of P

2(C) and P1, . . . ,P9 are the inflection points of C, by Exercise90 for each 1 ≤ i < j ≤ 9 the line L(Pi,Pj) contains a third inflection point. As aconsequence of this fact, it is not difficult to check that the set of lines joining twoinflection points of C has exactly 12 elements and the incidence relations amongthese 12 lines and the points P1, . . . ,P9 are the same that hold in the affine plane(Z/3Z)2. An explicit example of this correspondence is described in the Note fol-lowing Exercise 102.

K Exercise 94. (Inflection points of a smooth real cubic) Prove that every smooth andirreducible cubic C of P

2(R) has exactly three inflection points.

Solution. By Exercise 86 the curve C has at least one inflection point P and thereexists a system of homogeneous coordinates of P

2(R) such that P = [0, 0, 1] andthe affine curve C ∩ U0 is given by an equation of the form f (x, y) = y2 − g(x) = 0,where g is a monic real polynomial of degree 3 with distinct roots.

The polynomial g has at least one real root, because its degree is odd; so, up toperforming a change of affine coordinates of the form (x, y) �→ (x + c, y), we mayassume that g(0) = 0 and g(x) > 0 for x > 0. In other words, we may assume thatg(x) = x(x2 + mx + q) with m ≥ 0 and q > 0. Arguing exactly as in the solution ofExercise 93, one shows that the inflection points of C ∩ U0 are the solutions of thesystem:

y2 − g(x) = 0, p(x) = 3x4 + 4mx3 + 6qx2 − q2 = 0. (3.6)

Observe also that p′(x) = 12g(x). As p(0) = −q2 < 0 and p′(x) = 12g(x) > 0 forevery x > 0, there exists a unique α1 ∈ (0,+∞) such that p(α1) = 0. By con-struction g(α1) > 0, so there exists β1 ∈ R such that β2

1 = g(α1) and the pointsP1 = (α1,β1) and P2 = (α1,−β1) are inflection points of C ∩ U0.

If 0 is the only root of g(x), then g(x) < 0 for x ∈ (−∞, 0); in this case, sinceα1 is the only positive root of p(x), the system (3.6) has no solutions differentfrom P1 and P2. If instead g has roots λ1 < λ2 < 0 and if there exists a solutionQ = (α2,β2) of (3.6), then p(α2) = 0 and g(α2) = β2

2 > 0 (as explained in thesolution of Exercise 93, the polynomials p and g have no common roots). It follows

146 3 Exercises on Curves and Hypersurfaces

that λ1 < α2 < λ2 and that α2 is the only root of p in the interval (λ1,λ2), becausep(x) is increasing in this interval. Therefore in this case, besides P, P1 and P2, thereare exactly two additional inflection points corresponding to the pointsQ1 = (α2,β2)

andQ2 = (α2,−β2). In summary, we have shown that C can have either three or fiveinflection points.

Suppose that C has five inflection points. Then, by Exercise 90, the line L(P1,Q2)

intersects C in a third inflection point R and it can be readily checked that R does notcoincide with any of the points P, P1, P2, Q1 and Q2. Thus we get to a contradiction,and therefore C has exactly three inflection points.

Exercise 95. Let d ≥ 3; consider two distinct points P1,P2 ∈ P2(K) and, for i =

1, 2, a projective line li passing through Pi. Denote by Λd the space of projectivecurves of degree d, and let

F1 = {C ∈ Λd | l1 is tangent to C at P1},F2 = {C ∈ Λd | li is tangent to C at Pi, i = 1, 2}.

Show that Fi is a linear system, and compute its dimension.

Solution. On P2(K) consider homogeneous coordinates such that P1 = [1, 0, 0],

l1 = {x2 = 0}, and letF =

∑i+j+k=d

ai,j,kxi0x

j1x

k2 = 0

be the equation of the generic projective curve of degree d. As remarked in Sect. 1.9.6,the curve defined by F is tangent at [y0, y1, y2] to the line of equation ax0 + bx1 +cx2 = 0 if and only if ∇F(y0, y1, y2) is a multiple of (a, b, c). It follows that, inour situation, F represents a curve in F1 iff Fx0(1, 0, 0) = Fx1(1, 0, 0) = 0. Observe

now that∂xi0x

j1x

k2

∂x0(1, 0, 0) = d if i = d, j = k = 0, while it is zero otherwise. So

Fx0(1, 0, 0) = 0 iff ad,0,0 = 0. Similarly,∂xi0x

j1x

k2

∂x1(1, 0, 0) = 1 if i = d − 1, j = 1,

k = 0, and it is zero otherwise, so that Fx1(1, 0, 0) = 0 iff ad−1,1,0 = 0. Since thecoefficients ai,j,k give a system of homogeneous coordinates for Λd , the previousconsiderations imply that F1 is a subspace of Λd of codimension 2. As dimΛd =d(d + 3)

2 , we obtain that dimF1 = (d + 4)(d − 1)2 .

Suppose now that P1 /∈ l2, P2 /∈ l1. In this case, l1 and l2 meet at a point P3

which forms with P1,P2 a triple of points in general position. Then we can choosecoordinates such that P1 = [1, 0, 0], P2 = [0, 0, 1], P3 = [0, 1, 0], and hence l1 ={x2 = 0}, l2 = {x0 = 0}. So F defines a curve in F2 iff

Fx0(1, 0, 0) = Fx1(1, 0, 0) = 0, Fx1(0, 0, 1) = Fx2(0, 0, 1) = 0.

3 Exercises on Curves and Hypersurfaces 147

Arguing as above, one checks that these conditions are equivalent to ad,0,0 =ad−1,1,0 = a0,1,d−1 = a0,0,d = 0. Therefore F2 si defined by 4 independent linearconditions, and hence it has codimension 4 in Λd .

Suppose now P2 ∈ l1, P1 /∈ l2. In this case, we can choose coordinates suchthat P1 = [1, 0, 0], P2 = [0, 1, 0], l1 = {x2 = 0}, l2 = {x0 = 0}, and F ∈ F2 iffFx0(1, 0, 0) = Fx1(1, 0, 0) = 0, i.e. ad,0,0 = ad−1,1,0 = 0, and Fx1(0, 1, 0) = Fx2(0, 1, 0) = 0, i.e. a0,d,0 = a0,d−1,1 = 0. Also in this case F2 is defined by 4 inde-pendent linear conditions, so it has codimension 4 in Λd . Of course, the same resultholds also when P1 ∈ l2, P2 /∈ l1.

Finally, suppose l1 = l2 = L(P1,P2), and choose coordinates such that P1 =[1, 0, 0], P2 = [0, 1, 0]. Arguing as above, we obtain that F ∈ F2 iff ad,0,0 =ad−1,1,0 = a1,d−1,0 = a0,d,0 = 0. As d ≥ 3 we have d − 1 �= 1, so that also in thiscase F2 is defined by 4 independent linear conditions.

Therefore, in any case dimF2 = dimΛd − 4 = d2 + 3d − 82 .

Note. We briefly sketch two alternative methods for computing the dimension ofF1, that can also be used to compute dimF2.

Using the same notations and coordinates as in the above solution, if f (u, v) =F(1, u, v) is the equation of the affine part C ∩ U0 of the generic curve C ∈ Λd ,then f (u, v) = ad,0,0 + ad−1,1,0u + ad−1,0,1v + g(u, v), where g(u, v) is a sum ofmonomials all having degree greater than or equal to 2. Since the affine part l1 ∩ U0

of l1 has equation v = 0 and P1 has affine coordinates (0, 0), it follows that C passesthrough P1 with tangent l1 if and only if ad,0,0 = ad−1,1,0 = 0. This allows us toconclude that F1 has codimension 2 in Λd .

Alternatively, one can proceed as follows. As noted above, F represents a curveofF1 if and only if Fx0(1, 0, 0) = Fx1(1, 0, 0) = 0. Since the conditions that we havejust described are linear, it is clear thatF1 is a linear system of codimension at most 2.In order to show that codimF1 = 2, it suffices to produce a curve C1 ∈ Λd not passingthrough P1 = [1, 0, 0] and a curve C2 ∈ Λd passing through P1 having at P1 only onetangent different from l1 = {x2 = 0}. If this is the case, then, denoting byF3 the linearsystem consisting of the curves passing through P1, one has F1 � F3 � Λd , so thatdimF1 < dimF3 < dimΛd and dimΛ2 − dimF1 ≥ 2, as requested. Therefore, theproof can be concluded by setting, for instance, Ci = [Fi] with F1(x0, x1, x2) = xd0and F2(x0, x1, x2) = xd−1

0 x1.

Note in addition that the considerations made to compute the dimension of F1

apply also when d = 2; therefore the conics tangent to l1 at P1 form a linear systemof dimension 3. On the other hand, when d = 2 the dimension of the linear systemF2 depends on the position of the points P1,P2 with respect to the lines l1, l2; a caseis studied in Exercise 125.

Exercise 96. Let h, k ≥ 1 and let W be a linear system of curves of degree h inP2(K). Let C be a projective curve of degree k, and set

148 3 Exercises on Curves and Hypersurfaces

H = {C + C ′ | C ′ ∈ W}.

Show thatH is a linear system of curves of degree k + h, and that dimH = dimW .

Solution. Let V = K[x0, x1, x2]h (respectively V ′ = K[x0, x1, x2]h+k) be the vectorspace consisting of the zero polynomial and of the homogeneous polynomials ofdegree h (respectively h + k), and let F(x0, x1, x2) be a homogeneous polynomial ofdegree k that represents C. It is immediate to check that the function

ϕ : V → V ′, ϕ(G) = FG

is linear and injective. Moreover, if W = P(S), clearly one has H = P(ϕ(S)), andthe claim can be immediately deduced from this fact.

Exercise 97. Let Λ3 be the set of all cubics of P2(C), and let S ⊆ Λ3 be the set of

the cubics C that satisfy the following conditions:

(i) [1, 0, 0] is either a non-ordinary double point with principal tangent x2 = 0 or atriple point for C;

(ii) C passes through the points [0, 0, 1] and [1, 1,−1].Verify that S is a linear system and write down one of its projective frames.

Solution. Consider the curve C ∈ S of equation F(x0, x1, x2) = 0. Set affine coor-dinates u = x1

x0 , v = x2x0 on U0, and let f (u, v) = F(1, u, v). In order that C satisfy

(i), there must be a, b, c, d, e ∈ C not all zero and such that f (u, v) = av2 + bu3 +cu2v + duv2 + ev3, and so

F(x0, x1, x2) = ax0x22 + bx31 + cx21x2 + dx1x

22 + ex32 .

The fact that C passes through [0, 0, 1] and through [1, 1,−1] is equivalent to theconditions e = 0 and a + b − c + d − e = 0, hence the generic cubic of S has equa-tion

ax0x22 + bx31 + (a + b + d)x21x2 + dx1x

22 = 0, [a, b, d] ∈ P

2(C).

It follows thatS is a linear system of dimension 2, and a projective frame ofS is givenby the curves determined by the parametres [1, 0, 0], [0, 1, 0], [0, 0, 1], [1, 1, 1], thatis by the curves of equation

x0x22 + x21x2 = 0, x31 + x21x2 = 0,

x21x2 + x1x22 = 0, x0x

22 + x31 + 3x21x2 + x1x

22 = 0.

Exercise 98. Given two distinct points P and Q of P2(K), consider the set Ω of

cubics of P2(K) which are singular at Q and tangent at P to the line L(P,Q).

3 Exercises on Curves and Hypersurfaces 149

(a) Prove that Ω is a linear system and compute its dimension.(b) If R and S are points such that P,Q,R, S are in general position, show that

Λ = {C ∈ Ω | S ∈ C, C is tangent to the line L(R, S) at R}

is a linear system, compute its dimension anddetermine the number of irreduciblecomponents of every cubic of Λ.

Solution. (a) Observe that all cubics of Ω are reducible. Namely, if r = L(P,Q), forevery C ∈ Ω we have I(C, r,P) ≥ 2 and I(C, r,Q) ≥ 2, so by Bézout’s Theorem ris a component of C. Therefore Ω consists exactly of the cubics of the form r + Das D varies in the set of conics passing through Q (which must be singular); sincesuch conics form a linear system of dimension 4, it is easy to check (cf. Exercise 96)that Ω is a linear system of dimension 4 too.

(b) Assume that C = r + D is a cubic of Ω . As P,Q,R, S are in general position,neither S nor R lies on r; therefore C ∈ Λ if and only if the conic D, which passesthrough Q, passes through S too and is tangent at R to the line L(R, S). Then Dintersects L(R, S) at least in 3 points (counted with multiplicity), so L(R, S) is acomponent of D.

If t is any line passing through Q, the cubic r + L(R, S) + t satisfies the requiredconditions; thenΛ coincides with the set of cubics of the form r + L(R, S) + t wheret varies in the pencil of lines with centre Q. Therefore (cf. Exercise 96) Λ is a linearsystem of dimension 1.

Every cubic of Λ has 3 distinct irreducible components, except the cubic 2r +L(R, S) for which the line t of the pencil coincides with r.

Exercise 99. Consider in P2(C) the curve C of equation

F(x0, x1, x2) = 2x41 − x20x21 + 2x30x1 + 3x30x2 = 0

and the curve D of equation

G(x0, x1, x2) = ax21x22 + 2bx20x1x2 + 4x30x1 + cx30x2 = 0,

as a, b vary in C and c varies in C∗.

(a) Find the values of the parameters a, b, c for which the pencil generated by Cand D contains a curve G singular at the point P = [1, 0, 0]. For those values,determine the multiplicity and the principal tangents to G at P.

(b) Find the values of the parameters a, b, c for which the pencil generated by C andD contains a curve G such that its affine part G ∩ U0 has the line x − y = 0 asan asymptote and the corresponding point at infinity is non-singular for G.

Solution. (a) The affine part C ∩ U0 of C is defined by the equation f (x, y) =F(1, x, y) = 2x4 − x2 + 2x + 3y = 0 and the affine part D ∩ U0 of D is definedby the equation g(x, y) = G(1, x, y) = ax2y2 + 2bxy + 4x + cy = 0. The affine part

150 3 Exercises on Curves and Hypersurfaces

of the generic curve of the pencil generated by C and D has equation αf + βg = 0when [α,β] varies in P

1(C). The homogeneous linear term of αf + βg is equalto x(2α + 4β) + y(3α + cβ); so the pencil contains a curve which is singular atP = (0, 0) if and only if the system 2α + 4β = 3α + cβ = 0 has non-trivial solu-tions, that is iff c = 6. In addition, in this case the curve G of the pencil which issingular at P has equation 2f − g = 0. Since for c = 6 the first non-zero homoge-neous component of 2f − g is 2x(−x − by), then the origin is a double point forG, with principal tangents of equations x = 0 and x + by = 0. In particular, P isordinary for G if and only if b �= 0.

(b) The generic curve Gα,β of the pencil generated by C and D has equationHα,β = αF + βG = 0, [α,β] ∈ P

1(C).Assume that x − y = 0 is an asymptote of Gα,β ∩ U0, and assume that the point

at infinity of that line is non-singular for Gα,β . The projective closure of x − y = 0has equation x1 − x2 = 0, and intersects the improper line at [0, 1, 1]. A neces-sary and sufficient condition in order that [0, 1, 1] be a non-singular point of Gα,β

and the (principal) tangent to Gα,β at [0, 1, 1] have equation x1 − x2 = 0 is thatHα,β

x0 (0, 1, 1) = 0,Hα,βx1 (0, 1, 1) = −Hα,β

x2 (0, 1, 1) �= 0. An easy computation showsthat the first equation is always satisfied, while the second condition is equivalent to8α + 2aβ = −2aβ �= 0. It follows that the pencil generated by C and D contains acurve G satisfying the required conditions if and only if a �= 0. Moreover, if a �= 0then G = G−a,2, so that G has equation H−a,2 = −aF + 2G = 0.

Exercise 100. Consider the cubic D1 of P2(C) of equation

F(x0, x1, x2) = x1x22 − 2x0x1x2 − 6x20x2 + x0x

22 + 8x30 = 0.

(a) Check that D1 has exactly two singular points A and B.(b) Say whether D1 is irreducible.(c) Find a cubic D2 in P

2(C) such that all the cubics of the pencil generated by D1

and D2 have infinitely many inflection points, have A and B as singular pointsand pass through P = [−1, 2, 4].

Solution. (a) If we compute the gradient of the polynomial F

∇F = (−2x1x2 − 12x0x2 + x22 + 24x20, x2(x2 − 2x0), 2x1x2 − 2x0x1 − 6x20 + 2x0x2),

we see that the only singularities ofD1 are the points A = [0, 1, 0] and B = [1, 1, 2].(b)Any cubic having two singular points is surely reducible byBézout’s Theorem,

because the line joining the two singularities intersects the cubic in at least 4points (counted with multiplicity). In this case the line r = L(A,B) has equa-tion x2 − 2x0 = 0 and in fact the polynomial F factors as F = (x2 − 2x0)(x1x2 +x0x2 − 4x20). Therefore D1 is the union of the line r and the conic C of equationx1x2 + x0x2 − 4x20 = 0, which is non-singular and hence irreducible.

(c) As already observed in the solution of (b), every cubic which is singular bothat A and at B has the line r as a component. On the other hand, if we choose as D2

3 Exercises on Curves and Hypersurfaces 151

the sum of r and a conic passing through A, B and P and distinct from C, all cubicsof the pencil generated byD1 andD2 are singular at A and at B. For instance we cantakeD2 = r + L(A,P) + L(B,P), so that every cubic of the pencil generated byD1

and D2 is the sum of a line r and a conic of the pencil F generated by the conic Cand by the reducible conic L(A,P) + L(B,P).

Moreover, given a cubicD of this pencil, all points of r\{A,B} are inflection pointsofD unless r is a double component of the cubic.Namely, if the latter situation occurs,all points of r are singular and so they are not inflection points; however in this caseD consists, in addition to the double line r, of the tangent to C at P, so that all pointsof this line (except the intersection point with r) are inflection points of the cubic.

Exercise 101. Consider the points A = [1, 0,−1], B = [0, 1, 0], C = [1, 0, 0] andD = [0, 1, 1] of P

2(C). Consider also the lines r of equation x0 + x2 = 0, s ofequation x1 = 0 and t of equation 2x1 − 2x2 − x0 = 0. Denote by W the set ofprojective quartics Q of P

2(C) that satisfy the following conditions:

(i) A is a singular point;(ii) I(Q, r,B) ≥ 3;(iii) C is a singular point and I(Q, s,C) ≥ 3;(iv) the line t is tangent to Q at D.

(a) Say whether W is a linear system of curves and, if so, compute its dimension.(b) Say whether there exists in W a quartic having B as a triple point.(c) Write the equation of a quartic Q ∈ W such that the affine line t ∩ U0 is not an

asymptote of the affine part Q ∩ U0 of the quartic Q.

Solution. (a) Observe that the points A and B lie on the line r. If the quartic Qbelongs to W , then I(Q, r,A) ≥ 2 and I(Q, r,B) ≥ 3; so, by Bézout’s Theorem, rmust be an irreducible component ofQ. Moreover, the points A and C lie on the lines; since I(Q, s,A) ≥ 2 and I(Q, s,C) ≥ 3, in the same way we deduce that smust bean irreducible component of Q too. Therefore the quartic Q is necessarily given byr + s + C with C a curve of degree 2. A necessary and sufficient condition in orderthat this quartic satisfy all the conditions defining the setW is that the conic C passesthrough the point C (so that C is singular) and is tangent at D to the line t. So the setW consists of the quartics r + s + C, when C varies in the linear system of conicspassing through C and tangent atD to the line t. As these conics form a linear systemof dimension 2, then alsoW is a linear system of dimension 2 (cf. Exercise 96).

(b)Assumeby contradiction that there exists a quarticQ = r + s + C ∈ W havingB as a triple point. Then B has to be a double point for the conic C, i.e. C must havethe lines L(B,C) and L(B,D) as components. Since L(B,D) is distinct from the linet, the quartic is not tangent to t at D. This contradicts the hypothesis that Q ∈ W .

(c) Every quartic ofW is tangent to t at D. Therefore, the affine line t ∩ U0 is notan asymptote if and only if the pointD is singular and t is not a principal tangent to thecurve at that point. As L(C,D) �= t, all quarticsQ of the form r + s + L(C,D) + l,where l is a line passing through D and distinct from t, have the required property.If for instance we choose l = {x0 = 0}, then we haveQ = [x0x1(x0 + x2)(x1 − x2)].

152 3 Exercises on Curves and Hypersurfaces

Exercise 102. (Hesse pencil of cubics) Consider the pencil F = {Cλ,μ | [λ,μ] ∈P1(C)}, where Cλ,μ is the cubic of P

2(C) of equation

Fλ,μ(x0, x1, x2) = λ(x30 + x31 + x32) + μx0x1x2 = 0.

(a) Find the base points of F and show that all of them are non-singular for everyCλ,μ.

(b) Show that for every [λ,μ] ∈ P1(C) the Hessian curve H(Cλ,μ) of Cλ,μ is defined

and belongs to the pencil F , so that there is a map H : F → F which associatesH(Cλ,μ) to Cλ,μ.

(c) Prove that H : F → F has four fixed points, and that it is not a projectivity.(d) Find the inflection points of Cλ,μ when Cλ,μ is not a fixed point of H.

Solution. (a) Let ω = 1 + i√3

2 be a cube root of −1. If we consider the systemformed by the equations of C1,0 and of C0,1, it is immediate to check that the basepoints of F are

P1 = [0, 1,−1], P2 = [0, 1,ω], P3 = [0, 1,ω],P4 = [1, 0,−1], P5 = [1, 0,ω], P6 = [1, 0,ω],P7 = [1,−1, 0], P8 = [1,ω, 0], P9 = [1,ω, 0].

In addition, we have

∇Fλ,μ(x0, x1, x2) = 3λ(x20, x21, x

22) + μ(x1x2, x0x2, x0x1),

so that C1,0 is clearly non-singular (in particular no Pi is singular for C1,0). Sotake μ �= 0. For {i, j, k} = {1, 2, 3}, if yi = 0, yj �= 0, yk �= 0 it turns out that∂Fλ,μ

∂xi(y0, y1, y2) = μyjyk �= 0, which immediately implies that Pi is non-singular

for every cubic of F .

(b) The Hessian curve of Cλ,μ has equation

HFλ,μ= det

⎛⎝6λx0 μx2 μx1μx2 6λx1 μx0μx1 μx0 6λx2

⎞⎠ = −6λμ2(x30 + x31 + x32) + (216λ3 + 2μ3)x0x1x2 = 0.

As (λ,μ) �= (0, 0) clearly implies (−6λμ2, 216λ3 + 2μ3) �= (0, 0), we haveHFλ,μ

�= 0 and H(Cλ,μ) ∈ F for every Cλ,μ ∈ F .

(c) As seen in the previous part, H(Cλ,μ) = Cλ,μ if and only if [λ,μ] = [−6λμ2,

216λ3 + 2μ3], i.e. iff det(−6λμ2 216λ3 + 2μ3

λ μ

)= 0. This equation is equiv-

alent to λ(27λ3 + μ3) = 0, whose non-trivial solutions are [0, 1], [1,−3], [1, 3ω],[1, 3ω]. It follows that the fixed points ofH are the cubics C0,1, C1,−3, C1,3ω , C1,3ω . IfHwere a projectivity of the 1-dimensional projective space F , then it should coincide

3 Exercises on Curves and Hypersurfaces 153

with the identity map, since it has at least 3 fixed points. This contradicts the factthat H has exactly 4 fixed points, so H is not a projectivity.

(d) By the computations seen in (b), the intersection points between Cλ,μ andH(Cλ,μ) are given by the solutions of the system

{λ(x30 + x31 + x32) + μx0x1x2 = 0

−6λμ2(x30 + x31 + x32) + (216λ3 + 2μ3)x0x1x2 = 0.

As explained in (c), if C[λ,μ] is not a fixed point of H then

det

(−6λμ2 216λ3 + 2μ3

λ μ

)�= 0,

so that the previous system is equivalent to x30 + x31 + x32 = x0x1x2 = 0. The solutionsof this system exactly correspond to the base pointsP1, . . . ,P9 ofF , which, as seen in(a), are all non-singular for Cλ,μ. It follows that, ifH(Cλ,μ) �= Cλ,μ, then the inflectionpoints of Cλ,μ are exactly the base points of F .

Note. IfCλ0,μ0 is a fixed point ofH, then all non-singular points ofCλ0,μ0 are inflectionpoints. Since every cubic of F is reduced by part (a) of the Exercise, it follows fromExercise 84 that Cλ0,μ0 is the union of three lines. In fact, if [λ0,μ0] = [0, 1] thenthe cubic Cλ0,μ0 decomposes as the union of the lines of equation x0 = 0 (containingP1,P2,P3), x1 = 0 (containing P4,P5,P6), x2 = 0 (containing P7,P8,P9).

If [λ0,μ0] = [1,−3], then Cλ0,μ0 has equation x30 + x31 + x32 − 3x0x1x2 = 0, anddecomposes as the union of the lines of equation x0 + x1 + x2 = 0 (containingP1,P4,P7), ωx0 + ωx1 − x2 = 0 (containing P2,P6,P8), and ωx0 + ωx1 − x2 = 0(containing P3,P5,P9).

If [λ0,μ0] = [1, 3ω], then Cλ0,μ0 has equation x30 + x31 + x32 + 3ωx0x1x2 = 0, and

decomposes as the union of the lines of equation ωx0 − x1 − x2 = 0 (contain-ing P1,P5,P8), x0 − ωx1 + x2 = 0 (containing P2,P4,P9), and ωx0 + ωx1 − x2 = 0(containing P3,P6,P7).

Finally, if [λ0,μ0] = [1, 3ω], then Cλ0,μ0 has equation x30 + x31 + x32 +3ωx0x1x2 = 0, and decomposes as the union of the lines of equationωx0 − x1 − x2 =0 (containing P1,P6,P9), x0 − ωx1 + x2 = 0 (containing P3,P4,P8), and ωx0 +ωx1 − x2 = 0 (containing P2,P5,P7).

Now define a bijection Φ between the set {P1, . . . ,P9} and (Z/3Z)2:

P1 �→ (0, 0), P2 �→ (1, 0), P3 �→ (2, 0),

P4 �→ (0, 1), P5 �→ (2, 1), P6 �→ (1, 1),

P7 �→ (0, 2), P8 �→ (1, 2), P9 �→ (2, 2).

154 3 Exercises on Curves and Hypersurfaces

It is immediate to check, using the computations that we have just made, that threepoints Pi, Pj and Pk are collinear in P

2(C) if and only if Φ(Pi), Φ(Pj) and Φ(Pk)

are collinear in (Z/3Z)2, and that the four reducible cubics of the pencil correspondto the partition of the set of lines of (Z/3Z)2 in four subsets of parallel lines.

K Exercise 103. Let D and G be the cubics of P2(C) of equations x0(x21 − x22) = 0

and x21(x0 + x2) = 0, respectively. For [λ,μ] ∈ P1(C), consider the cubics Cλ,μ of

the pencil generated by D and G.(a) Determine the base points of the pencil.(b) Show that there exists a unique point P which is singular for all the cubics of

the pencil and, as [λ,μ] varies in P1(C), compute the multiplicity of P and the

principal tangents to Cλ,μ at P.(c) Determine all the reducible cubics of the pencil.

Solution. (a) Both generators of the pencil are reducible: D is the union of threedistinct lines, while G has two distinct irreducible components, one of which isdouble. The base points of the pencil are the points of D ∩ G; hence, consider-ing all the possible intersections of an irreducible component of D with an irre-ducible component of G, we find that the pencil has 5 base points, namely thepoints A = [1,−1,−1],B = [1, 1,−1],C = [0, 1, 0] (that lie on the line of equa-tion x0 + x2 = 0) and the points P = [1, 0, 0],Q = [0, 0, 1] (that lie on the doubleline x1 = 0 contained in G).

(b) The points which are singular for all the cubics of the pencil are precisely thepoints which are singular for both generators; looking at the five base points of thepencil, it is easy to see that only P = [1, 0, 0] is singular both for D and for G.

The generic cubic Cλ,μ of the pencil has equation λx0(x21 − x22) +μx21(x0 + x2) = 0. In order to determine the multiplicity of P and the principaltangents to Cλ,μ at P, we work in the affine chart U0 with the affine coordinatesx = x1

x0 , y = x2x0 . With respect to these coordinates the affine curve Cλ,μ ∩ U0 has

equationλ(x2 − y2) + μx2(1 + y) = (λ + μ)x2 − λy2 + μx2y = 0

and P has coordinates (0, 0). Since for [λ,μ] varying in P1(C) it is not possible that

λ and λ + μ vanish simultaneously, the homogeneous part (λ + μ)x2 − λy2 is neveridentically zero and so P is always a double point. In particular:

(b1) If λ = 0, P is a non-ordinary double point with principal tangent x1 = 0 (in thatcase we find again the cubic C0,1 = G which, as we already knew, has P as anon-ordinary double point since the double line x21 = 0 is a component of G).

(b2) If λ + μ = 0, P is a non-ordinary double point with principal tangent x2 = 0(in that case we obtain the cubic C1,−1 of equation x2(x0x2 + x21) = 0 whosecomponents x0x2 + x21 = 0 and x2 = 0 are mutually tangent at P).

(b3) If λ �= 0 and λ + μ �= 0, then P is ordinary double; denoting by α a square

root of λλ + μ

, the distinct principal tangents to the cubic at P are the lines of

equations x1 − αx2 = 0 and x1 + αx2 = 0.

3 Exercises on Curves and Hypersurfaces 155

(c) By what we have seen in Exercise 62, if Cλ,μ is reducible then it contains aline passing through P which of course will be a principal tangent at P.

As we have seen in part (b), P turns out to be non-ordinary only in the cases λ = 0or λ + μ = 0, namely in the cases corresponding to the reducible cubics G and C1,−1.

If P is ordinary double for a reducible cubic Cλ,μ, then by the considerations ofpart (b3) the line through P contained in Cλ,μ must be either the line x1 − αx2 = 0 orthe line x1 + αx2 = 0. On the other hand, since only even powers of x1 appear in theequation of Cλ,μ, we deduce immediately that Cλ,μ contains both these lines: indeed,if substituting x1 = αx2 in the equation of Cλ,μ we obtain the zero polynomial, thenthe same happens also substituting x1 = −αx2, and conversely.

Since α �= 0, these two lines pass neither through C = [0, 1, 0] nor through Q =[0, 0, 1]. Hence the line x0 = 0 must be the third irreducible component of the cubic,i.e. Cλ,μ must have equation x0(x21 − α2x22) = 0. Imposing that this cubic pass through

the points A = [1,−1,−1] and B = [1, 1,−1], we get α2 = 1, i.e. λλ + μ

= 1 and

so μ = 0. Therefore, besides the two reducible cubics found above, the only otherreducible cubic of the pencil is the cubic C1,0 = D.

It is also possible to conclude the proof of (c) in a slightly different way. Afterobserving that, whenever Cλ,μ is reducible and has P as an ordinary double point, thedegenerate conic of equation x21 = α2x22 must be contained in Cλ,μ, by substituting

x21 = λλ + μ

x22 in the equation of Cλ,μ one obtains the equationλμ

λ + μx32 = 0. In order

that this equation be identically zero, one must have λ = 0 (but in this case P is notan ordinary double point of Cλ,μ) or μ = 0, case that corresponds to the reduciblecubic C1,0 = D.

K Exercise 104. Consider the cubics D1 and D2 of P2(C) of equations

F1(x0, x1, x2) = x2(x21 − 2x0x1 + x22) = 0,F2(x0, x1, x2) = (x2 − x1)(x2 + x1)(x1 − x0) = 0,

respectively. Determine the base points and the reducible cubics of the pencil Fgenerated by D1 and D2.

Solution. The cubicD1 has as irreducible components the line r1 = {x2 = 0} and anirreducible conic; denote by r2, r3, r4 the lines of equations x2 − x1 = 0, x2 + x1 = 0and x1 − x0 = 0, respectively, which are the irreducible components of D2.

It can be easily computed that the base points of the pencil, i.e. the intersectionpoints betweenD1 andD2, are the pointsA = [1, 1,−1],B = [1, 1, 0],C = [1, 1, 1]and Q = [1, 0, 0]. Note that the points A,B,C are collinear and lie on the line r4;moreover, Q is a singular point both forD1 and forD2, so it is singular for all cubicsof the pencil.

156 3 Exercises on Curves and Hypersurfaces

Since the base locus ofF is a finite set, two distinct cubics ofF have no commoncomponents. Assume now that Cλ,μ is reducible. As Q is singular for Cλ,μ (becauseit is singular for every cubic of the pencil), there exists a line r contained in Cλ,μ andpassing through Q (cf. Exercise 62). Since Q /∈ r4, necessarily r �= r4.

If r = r1, then the previous observation implies that Cλ,μ = D1.If either r = r2 or r = r3, then Cλ,μ = D2 for the same reason.If r �= ri for i = 1, 2, 3, then Cλ,μ = r + G where G is a conic that necessarily

passes through the pointsA ∈ r4,B ∈ r4,C ∈ r4. HenceGmust be reducible andmustcontain the line r4. Again from the previous observation it follows that Cλ,μ = D2.

Therefore we obtain that the only reducible cubics of the pencil are the generatorsD1 and D2.

Exercise 105. Denote by C the curve of P2(C) of equation

F(x0, x1, x2) = x31 − 2x0x22 + x1x

22 = 0.

(a) Find the singular points of C, their multiplicities and their principal tangents. Inaddition, for each singular point of C, compute the multiplicity of intersectionbetween the curve and each principal tangent at that point.

(b) Find the inflection points of C.(c) Find a projective cubic D of P

2(C) such that the pencil F generated by C andD satisfies the following properties:

(i) all cubics of F have at least a non-ordinary double point or a triple point;(ii) the points [0, 0, 1] and [1, 1, 1] belong to the base locus of the pencil F .

Solution. (a) As∇F = (−2x22, 3x

21 + x22, 2x2(x1 − 2x0)),

the only singular point of C is P = [1, 0, 0]. If we set on U0 the affine coordinatesx = x1

x0 , y = x2x0 , the point P has coordinates (0, 0), while the affine part C ∩ U0 of C

has equation x3 − 2y2 + xy2 = 0. So P is a cusp with principal tangent r of equationx2 = 0. Moreover I(C, r,P) = 3.

(b) As

HF(X) = det

⎛⎝

0 0 −4x20 6x1 2x2

−4x2 2x2 2x1 − 4x0

⎞⎠ = −96x1x

22,

the only points of intersection between C and its Hessian are P and Q = [0, 0, 1].Since P is singular and Q is not, the only inflection point is Q.

(c) If we choose as D a cubic having P as a non-ordinary double point with thesame principal tangent as C, i.e. the line r = {x2 = 0}, then all cubics of the pencilgenerated by C and D fulfil property (i). If in addition D passes through the points[0, 0, 1] and [1, 1, 1], these points turn out to be base points of the pencil generatedby C and D. A cubic D with these properties is, for instance, the one defined by theequation x22(x0 − x1) = 0.

3 Exercises on Curves and Hypersurfaces 157

Note. In the solution of the previous exercise we proved, in particular, that C hasexactly one inflection point. This fact can also be deduced without performing anycomputation. Namely, C has an ordinary cusp as its unique singularity, so it is irre-ducible by Exercise 62. Then C has exactly one inflection point by Exercise 89.

K Exercise 106. Assume that C1 and C2 are curves of P2(K) of degree n that intersect

exactly at N distinct points P1, . . . ,PN . Let D be an irreducible curve of degreed < n passing through the points P1, . . . ,Pnd . Prove that there exists a curve G ofdegree n − d passing through the points Pnd+1, . . . ,PN .

Solution.Observe that Pj /∈ D for every j such that nd < j ≤ N : otherwise,D wouldhave more than nd points in common both with C1 and with C2 and consequently itwould be an irreducible component of both curves, while by hypothesis C1 and C2meet at finitely many points.

Denote by Q a point of D such that Q �= Pj for each j = 1, . . . , nd. In the pencilgenerated by C1 and C2 there is a curve W passing through Q; in particular W hasdegree n and passes through P1, . . . ,PN ,Q. Since W shares with D at least nd + 1distinct points and D is irreducible, then by Bézout’s Theorem D is an irreduciblecomponent of W , i.e. W = D + G, where G is a curve of degree n − d. As Wcontains the pointsP1, . . . ,PN andD containsP1, . . . ,Pnd but, as seen above, it doesnot contain any point among Pnd+1, . . . ,PN , these latter points must be contained inthe curve G.

K Exercise 107. (Poncelet’s Theorem) Assume that C is an irreducible projectivecurve of P

2(C) of degree n ≥ 3. Let r be a line that intersects C at n distinct pointsP1, . . . ,Pn and, for each i = 1, . . . , n, let τi be a line tangent to C at Pi. Denote byQ1, . . . ,Qk the additional points of intersection between C and the lines τ1, . . . , τn.Prove that Q1, . . . ,Qk belong to the support of a curve of degree n − 2.

Solution. First of all observe that each point Pi is simple for C: otherwise the line rwould meet the curve in more than n points (counted with multiplicity) and hence itwould be an irreducible component of C, which instead is irreducible by hypothesis.For the same reason r is not tangent to C at any Pi.

Consider the curve D of degree n defined by D = τ1 + · · · + τn and take Q ∈r\{P1, . . . ,Pn}. Assume that G is a curve of the pencil generated by C andD passingthrough Q. Since G has degree n and has at least n + 1 distinct points in commonwith r, then r is a component of G, that is G = r + G ′.

The curveG, like all the curves of the pencil generatedbyC andD, is tangent atPi tothe line τi; as r has not this property, thenG ′ has to pass through the pointsP1, . . . ,Pn.Consequently, as G ′ has degree n − 1, the line r is an irreducible component of G ′too, i.e. G = 2r + G ′′ with degG ′′ = n − 2. Then the points Q1, . . . ,Qk , which donot belong to r, must belong to the curve G ′′.

Note. If n = 3 the statement of Exercise 107 turns into the following result. If Cis an irreducible cubic of P

2(C), let r be a line meeting C in three distinct pointsP1,P2,P3 which are not inflection points of C and, for i = 1, 2, 3, let τi be a linetangent to C at Pi. If Qi is the additional point of intersection between τi and C, thenthe points Q1,Q2,Q3 are collinear.

158 3 Exercises on Curves and Hypersurfaces

Exercise 108. Consider the curves C and D of P2(C) defined by

F(x0, x1, x2) = x20 − x21 + x22 = 0, G(x0, x1, x2) = x0x1 − x22 + x21 = 0

respectively. For every P ∈ C ∩ D, compute I(C,D,P).

Solution. Note that [0, 0, 1] /∈ C ∪ D; so, if [q0, q1, q2] ∈ C ∩ D, then (q0, q1) �=(0, 0) annihilates the resultant

Ris(F,G, x2)(x0, x1) = det

⎛⎜⎜⎝

x20 − x21 0 1 00 x20 − x21 0 1

x21 + x0x1 0 −1 00 x21 + x0x1 0 −1

⎞⎟⎟⎠ = x20(x0 + x1)

2.

Since Ris(F,G, x2) �= 0, the curves C and D have no component in common. Ifx0 = 0, from F = G = 0 it follows that x22 − x21 = 0, a condition that identifies thepointsQ1 = [0, 1, 1] ∈ C ∩ D,Q2 = [0, 1,−1] ∈ C ∩ D. If insteadwe set x0 + x1 =0, then we find the point Q3 = [1,−1, 0] ∈ C ∩ D.

Observe now that Q1,Q2, [0, 0, 1] are collinear, while the line L(Q3, [0, 0, 1])does not contain any point of C ∩ D distinct from Q3. Therefore, the multiplic-ity of intersection I(C,D,Q3) is equal to the multiplicity of [1,−1] as a root ofRis(F,G, x2), that is 2.

By Bézout’s Theorem∑3

i=1 I(C,D,Qi) = 4; since of course I(C,D,Qi) ≥ 1 fori = 1, 2, then we can conclude that I(C,D,Q1) = I(C,D,Q2) = 1.

K Exercise 109. Consider two curves C,D in P2(K) without common components

and let P ∈ C ∩ D. Assume that P is non-singular both for C and for D, and denoteby τC, τD the tangents at P to C,D, respectively. Show that I(C,D,P) ≥ 2 if andonly if τC = τD.

Solution. Let Q /∈ C ∪ D ∪ τC be a point such that there exists no point of C ∩ Ddistinct from P and lying on the line joining P and Q. It is immediate to check thatthere are homogeneous coordinates such that P = [1, 0, 0], Q = [0, 0, 1] /∈ C ∪ Dand τC = {x2 = 0}. Let F(x0, x1, x2) = 0, G(x0, x1, x2) = 0 be the equations of Cand D, respectively, and denote by m the degree of F and by d the degree of G. Bydefinition I(C,D,P) is equal to the multiplicity of [1, 0] as a root of the resultantRis(F,G, x2).

As Q /∈ C, the coefficient of xm2 in F is not zero, so the degree of F with respectto the variable x2 ism. In particular, if f (x1, x2) = F(1, x1, x2) is the dehomogenized

3 Exercises on Curves and Hypersurfaces 159

polynomial ofF with respect to x0, then deg f = degF = m. Similarly, if g(x1, x2) =G(1, x1, x2), then deg g = degG = d. Moreover, since the specialization to x0 = 1does not lower the degrees of F andG, this specialization commutes with computingthe resultant (cf. Sect. 1.9.2). Therefore Ris(F,G, x2)(1, x1) = Ris(f , g, x2)(x1), sothat I(C,D,P) is equal to the order ord(Ris(f , g, x2)(x1)) (i.e. to the multiplicity of0 as a root of Ris(f , g, x2)(x1), or in other words to the largest i ≥ 0 such that xi1divides Ris(f , g, x2)(x1)).

Now, as τC = {x2 = 0}, up to multiplying f by a non-zero constant we have

f (x1, x2) = x2 + ϕ0(x1) + ϕ1(x1)x2 + · · · + ϕm−1(x1)xm−12 + axm2 ,

with ord (ϕ0(x1)) ≥ 2, ord (ϕ1(x1)) ≥ 1 and a �= 0. In the sameway, if τD = {αx1 +βx2 = 0} then

g(x1, x2) = αx1 + βx2 + ψ0(x1) + ψ1(x1)x2 + · · · + ψd−1(x1)xd−12 + bxd2 ,

with ord (ψ0(x1)) ≥ 2, ord (ψ1(x1)) ≥ 1 and b �= 0.The resultant Ris(f , g, x2)(x1) is given by the determinant of the Sylvester matrix

S(x1) =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

ϕ0(x1) 1 + ϕ1(x1) ϕ2(x1) . . . a 0 . . .

0 ϕ0(x1) 1 + ϕ1(x1) . . . . . . a . . ....

......

......

...

αx1 + ψ0(x1) β + ψ1(x1) ψ2(x1) . . . b 0 . . .

0 αx1 + ψ0(x1) β + ψ1(x1) . . . . . . b . . ....

......

......

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

For i, j = 1, . . . ,m + d, i < j, denote byDi,j the determinant of the 2 × 2 submatrixobtained by taking the first two columns and the ith and jth rows of S(x1), and denoteby D′

i,j the determinant of the corresponding cofactor, that is of the (m + d − 2) ×(m + d − 2) submatrix obtained by deleting in S(x1) the first two columns and theith and jth rows. By Laplace rule we get

det S(x1) =∑i<j

(−1)i+j+1Di,j(x1)D′i,j(x1).

We are going to estimate the order of det S(x1) by analysing how the varioussummands contribute to the previous summation. Since ϕ0(x1), ψ0(x1) are divisibleby x21, it readily follows that if i �= 1 or j �= d + 1 then we have ord(Dij(x1)) ≥ 2. Soord(Ris(f , g, x2)(x1)) ≥ 2 if and only if

ord(D1,d+1(x1)D′1,d+1(x1)) = ord(D1,d+1(x1)) + ord(D′

1,d+1(x1)) ≥ 2.

160 3 Exercises on Curves and Hypersurfaces

Now, all terms of

D1,d+1(x1) = ϕ0(x1)(β + ψ1(x1)) − (1 + ϕ1(x1))(αx1 + ψ0(x1)),

apart from αx1, have order at least two, so that ord(D1,d+1(x1)) ≥ 2 if and only ifα = 0.

Let us now check that D′1,d+1(0) �= 0. It is immediate to observe that D′

1,d+1(0)

is the determinant of the Sylvester matrix of the polynomials f (x2) = f (0, x2)x2 and

g(x2) = g(0, x2)x2 , which have no common roots: namely, since the line L(P,Q) does

not contain any point of C ∩ D different from P, then 0 is the only common root off (0, x2) and g(0, x2); on the other hand, τC has equation x2 = 0, so that the order off (0, x2) is equal to 1 and 0 cannot be a root of f . Therefore ord(D′

1,d+1(x1)) = 0, andI(C,D,P) = ord(Ris(f , g, x2)(x1)) ≥ 2 if and only if α = 0, i.e. iff τC = τD.

K Exercise 110. Assume that C and D are curves of P2(K) without common compo-

nents and let P ∈ C ∩ D. Prove that

I(C,D,P) ≥ mP(C)mP(D).

Solution. Let Q /∈ C ∪ D be a point such that there exists no point of C ∩ D distinctfrom P and lying on the line joining P and Q. One can readily check that there existhomogeneous coordinates such that P = [1, 0, 0] and Q = [0, 0, 1].

Denote byF(x0, x1, x2) = 0,G(x0, x1, x2) = 0 equations of C andD, respectively,with degF = d and degG = d′, and set m = mP(C), m′ = mP(D).

By definition I(C,D,P) is equal to the multiplicity of [1, 0] as a root of the resul-tant Ris(F,G, x2). Moreover, arguing as in the solution of Exercise 109 we obtainthat, if we set f (x1, x2) = F(1, x1, x2) and g(x1, x2) = G(1, x1, x2), the mentionedmultiplicity coincides with the order ord(Ris(f , g, x2)(x1)) of Ris(f , g, x2)(x1) (thatis with the multiplicity of 0 as a root of Ris(f , g, x2)(x1), or also with the largesti ≥ 0 such that xi1 divides Ris(f , g, x2)(x1)).

As Q /∈ C ∪ D, then

f (x1, x2) = ϕ0(x1) + ϕ1(x1)x2 + · · · + ϕd−1(x1)xd−12 + axd2 ,

g(x1, x2) = ψ0(x1) + ψ1(x1)x2 + · · · + ψd′−1(x1)xd′−12 + bxd

′2 ,

with a �= 0, b �= 0. In addition, the fact thatm = mP(C),m′ = mP(D) implies that ϕi

(resp. ψi) is divisible by xm−i1 (resp. by xm

′−i1 ) for each i = 0, . . . ,m (resp. for each

i = 0, . . . ,m′).

3 Exercises on Curves and Hypersurfaces 161

The resultant Ris(f , g, x2)(x1) is the determinant of the Sylvester matrix

S(x1) =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

ϕ0(x1) ϕ1(x1) ϕ2(x1) . . . . . . a 0 . . .

0 ϕ0(x1) ϕ1(x1) . . . . . . . . . a . . ....

......

......

......

...

ψ0(x1) ψ1(x1) ψ2(x1) . . . b 0 . . . . . .

0 ψ0(x1) ψ1(x1) . . . . . . b 0 . . ....

......

......

......

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

Consider now the matrix S(x1) obtained by multiplying by xm′−i+1

1 the ith rowof S(x1) for each i = 1, . . . ,m′, and by multiplying by xm−i+1

1 the (d + i)th row ofS(x1) for each i = 1, . . . ,m (since m ≤ d and m′ ≤ d′, the operations just describedactually make sense). Then we have

ord(det S(x1)) = ord(det S(x1)) + m(m + 1) + m′(m′ + 1)

2.

Moreover, since ord(ϕi(x1)) ≥ m − i (resp. ord(ψi(x1)) ≥ m′ − i) for each i =0, . . . ,m (resp. for each i = 0, . . . ,m′), then we obtain that, for each i = 1, . . . ,m + m′ the ith column of S(x1) is divisible by the monomial xm+m′−i+1

1 , so that

ord(det S(x1)) ≥ (m + m′)(m + m′ + 1)

2.

Therefore ord(det S(x1)) ≥ mm′, and hence I(C,D,P) ≥ mm′, as required.

K Exercise 111. Prove that, if two cubics of P2(K) intersect exactly at 9 points, then

every cubic passing through 8 of those 9 points passes through the ninth point too.

Solution. Denote by P1, . . . ,P9 the nine points where the two cubics C1 = [F1] andC2 = [F2] meet.

Some easy consequences immediately follow from the fact that the cubics intersectat finitely many points. First of all, however we choose four out of the nine points,they are not collinear, because otherwise the line joining themwould be an irreduciblecomponent both of C1 and of C2 by Bézout’s Theorem, so that the cubics would haveinfinitely points in common. Similarly, however we choose seven out of the ninepoints, they cannot lie on a conic.

Assume by contradiction there exists a cubic D = [G] passing through eight ofthe nine points, say P1, . . . ,P8, but not passing through P9.

The curveD does not belong to the pencil of cubics generated by C1 and C2, whichhas {P1, . . . ,P9} as base locus. Hence C1, C2 and D generate a 2-dimensional linearsystem L having {P1, . . . ,P8} as base locus; as a consequence, however we choosetwo points in P

2(K), there exist λ,μ, ν ∈ K not simultaneously zero such that thecubic [λF1 + μF2 + νG] passes through those two points.

162 3 Exercises on Curves and Hypersurfaces

We are now going to show that any three points chosen among P1, . . . ,P9 are notcollinear. Namely, assume by contradiction that, for instance, P1,P2,P3 lie on a liner and denote by Q the only conic passing through P4, . . . ,P8 (this conic is uniquebecause, as shown above, no four of the five points are collinear – cf. Sect. 1.9.6).Chosen two points A ∈ r\{P1,P2,P3} and B /∈ Q ∪ r, there exist λ,μ, ν ∈ K notsimultaneously zero such that the cubic W = [λF1 + μF2 + νG] passes throughP1, . . . ,P8,A,B. Since W intersects the line r at the four points P1,P2,P3,A, byBézout’s Theorem r is an irreducible component of W; hence W is the sum of rand a conic passing through P4, . . . ,P8 which, by the uniqueness recalled above,necessarily must be Q. On the other hand, we cannot have W = r + Q, becauseB ∈ W but B lies neither on r nor on Q.

Similar arguments allow us to prove that any six points chosen among P1, . . . ,P9

cannot lie on a conic. Namely, assume by contradiction that, for instance, P1, . . . ,P6

lie on a conic Q and denote by r the line joining P7 and P8. Arguing as above,we arrive at a contradiction by choosing a point A ∈ Q\{P1, . . . ,P6} and a pointB /∈ Q ∪ r, and by considering a cubic W of the linear system L passing throughP1, . . . ,P8,A,B.

The previous considerations allow us to get to a contradiction. Namely, considerthe line s = L(P1,P2) and the only conic G passing through P3, . . . ,P7. As shownabove, P8 /∈ s and P8 /∈ G. After choosing two distinct points A,B on s\{P1,P2},let W be a cubic of the linear system L passing through A and B. Then necessarilyW = s + G: namely, by Bézout’s Theorem s is an irreducible component of W;hence W is the sum of s and a conic passing through P3, . . . ,P7 which, by theuniqueness recalled above, necessarily must be G. On the other hand we cannot haveW = s + G, because P8 ∈ W but P8 /∈ s and P8 /∈ G.Exercise 112. If n ≥ 2, assume that I is a hypersurface of degree d of P

n(C) andlet P ∈ I be a point. Prove that:

(a) I is a cone with vertex P if and only if mP(I) = d.(b) If I is a cone, then the set X of vertices of I is a projective subspace of P

n(C).(c) If I is a cone with vertex P and Q ∈ Sing(I), then L(P,Q) ⊆ Sing(I).

Solution. (a) I is a cone with vertex P if and only if every line passing through Pand not contained in I intersects I only at P. Since such a line intersects I exactlyat d points counted with multiplicity (cf. Sect. 1.7.6), I is a cone with vertex P ifand only if for every line passing through P we have I(I, r,P) ≥ d, that is iff P is apoint of I of multiplicity d.

(b) If P1 �= P2 are vertices of I and R ∈ I, then L(P1,P2,R) ⊆ I: namely,P1 ∈ X implies that L(P1,R) ⊆ I, and P2 ∈ X implies that L(P2, S) ⊆ I for everyS ∈ L(P1,R); so L(P1,P2,R) = L(P2,L(P1,R)) ⊆ I. It follows that for every Q ∈L(P1,P2) and for every R ∈ I the line L(Q,R) is contained in I, so that everypoint of L(P1,P2) is a vertex of I. For every P1,P2 ∈ X, P1 �= P2, we so have thatL(P1,P2) ⊆ X, and therefore X is a projective subspace of P

n(C).Alternatively, we can argue as follows. By (a), P1 and P2 are points of I

of multiplicity d. We can choose homogeneous coordinates x0, . . . , xn on Pn(K)

3 Exercises on Curves and Hypersurfaces 163

such that P1 has coordinates [1, 0, . . . , 0] and P2 has coordinates [0, 1, . . . 0]. LetF(x0, . . . , xn) = 0 be an equation of I. Since P1 ∈ I, then

F(x0, . . . , xn) = xd−10 F1(x1, . . . , xn) + xd−2

0 F2(x1, . . . , xn) + · · · + Fd(x1, . . . , xn),

where Fi is either zero or homogeneous of degree i for i = 1, . . . , d. From theprevious expression it appears that P1 has multiplicity d for I if and only ifF1 = · · · = Fd−1 = 0, that is iff in F = Fd the variable x0 does not appear. In thesame way, since also P2 is a vertex, the variable x1 does not appear in F. It is nowimmediate to check that all points of the line L(P1,P2), which in the chosen systemof coordinates is described by x2 = · · · = xn = 0, are points of multiplicity d for I.

(c) Choose homogeneous coordinates such that P = [1, 0, . . . , 0], so that, as seenabove, I is defined by a homogeneous polynomial F which does not depend on thevariable x0. In particular Fx0 = 0, so the point Q of coordinates [a0, a1, . . . , an] issingular for I if and only if Fxi(a0, a1, . . . , an) = 0 for i = 1, . . . , n. In addition, fori = 1, . . . , n the polynomial Fxi is either zero or homogeneous and does not dependon the variable x0, so Fxi(a0, a1, . . . , an) = 0 iff Fxi(t, sa1, . . . , san) = 0 for every[s, t] ∈ P

1(C). It follows that if Q is singular for I, then all points of L(P,Q) aresingular for I.Note. Statements (b) and (c) of the previous exercise hold also on R, since theirproofs do not use properties of the complex numbers.

As regards (a), if K = R it is true that, if P ∈ I is a point of multiplicity d, thenI is a cone with vertex P, but the opposite implication is false. For instance, thecubic C of P

2(R) defined by the equation x0(x20 + x21 + x22) = 0 has as support theline r = {x0 = 0}, so it is a cone with vertex Q for every Q ∈ r. Though, it can beimmediately checked that all points of r are simple for C.Note. If I is a curve of degree d of P

2(C), then I is a cone with vertex P if and onlyif it is the sum of d (non necessarily distinct) lines passing through P. Namely, if Iis a cone with vertex P, from the definition it readily follows that the support of Iis the union of a family of lines passing through P. By applying Bézout’s Theorem(or Exercise 57), we easily deduce that every line contained in I is an irreduciblecomponent of the curve, so that I decomposes as the sum of d lines passing throughP (counted with multiplicity).

Therefore statement (a) implies that, if I is a curve of P2(C) of degree d and P

is a point of the curve of multiplicity d, then I decomposes as the sum of d linespassing through P (counted with multiplicity).

Exercise 113. If n ≥ 2 and H ⊂ Pn(K) is a hyperplane, consider a hypersurface J

of H of degree d and a point P ∈ Pn(K)\H. Prove that

X =⋃Q∈J

L(P,Q) ∪ {P} ⊆ Pn(K)

164 3 Exercises on Curves and Hypersurfaces

is the support of a hypersurface CP(J ) of degree d in Pn(K), called the “cone over

J with vertex P”, which verifies the following properties:

(i) CP(J ) is either irreducible or reduced if and only if J is;(ii) a point Q ∈ X\{P} is singular for CP(J ) if and only if the point L(P,Q) ∩ J is

singular for J .

Solution. Choose homogeneous coordinates x0, . . . , xn on Pn(K) in such a way that

P = [1, 0, . . . , 0] and H has equation x0 = 0. Note that these coordinates inducehomogeneous coordinates x1, . . . , xn onH, and letG(x1, . . . , xn) = 0 be an equationof J . Set F(x0, x1, . . . , xn) = G(x1, . . . , xn), and denote by CP(J ) the hypersurfaceof P

n(K) of equation F = 0. It is immediate to check that CP(J ) has degree d andit is a cone with vertex P.

Let us now check that the support of CP(J ) (which, thanks to the usual abuseof notation, we will simply denote by CP(J )) coincides with X. From the defi-nitions it follows that P ∈ X ∩ CP(J ). If the support of J is empty (and thereforeK = R), we have CP(J ) = {P} = X. Let nowQ ∈ J �= ∅. ThenQ = [0, b1, . . . , bn]for some [b1, . . . , bn] ∈ P

n−1(K), and in additionG(b1, . . . , bn) = 0.As the points ofL(P,Q)\{P} have coordinates [t, b1, . . . bn], t ∈ K and F(t, b1, . . . , bn)= G(b1, . . . , bn) = 0 for every t ∈ K, then L(P,Q) ⊆ CP(J ). This shows thatX ⊆ CP(J ).

Conversely, let Q ∈ CP(J )\{P} and set R = L(P,Q) ∩ H. If Q has coordinates[a0, . . . , an], the point R has coordinates [0, a1, . . . , an]. Since Q ∈ CP(J ), thenF(a0, . . . , an) = 0, so that G(a1, . . . , an) = 0. Therefore R belongs to the supportof J , and Q ∈ L(P,R) belongs to X. Since Q has been arbitrarily chosen, thenCP(J ) ⊆ X.

The hypersurface CP(J ) obviously verifies (i): if G = G1 · . . . · Gr is the decom-position of G into irreducible factors, if we set Fk(x0, x1, . . . , xn) = Gk(x1, . . . , xn),k = 1, . . . , r, thenF = F1 · . . . · Fr is the decomposition ofF into irreducible factors.

With regard to (ii), if a point Q �= P has coordinates [a0, a1, . . . , an], the pointR = L(P,Q) ∩ J has coordinates [0, a1, . . . , an]. Moreover we have Fx0 = 0 andFxi(a0, . . . , an) = Gxi(a1, . . . , an) for each i = 1, . . . , n. Therefore Q is singular forCP(J ) if and only if R is singular for J , as required.

Exercise 114. Consider a cubic surface S of P3(C) such that Sing(S) is a finite set.

Prove that:

(a) S is irreducible.(b) If P ∈ Sing(S) is a triple point, then S is a cone with vertex P and P is the only

singular point of S.Solution. (a) Assume by contradiction that S is reducible. Then S = H + Q, whereH is a plane and Q is a quadric, possibly reducible or non-reduced. By Exercise 54all points of the conicQ ∩ H are singular for S, against the assumption that Sing(S)

is a finite set (recall that, if n ≥ 2, then the support of a hypersurface Pn(C) contains

infinitely many points – cf. Sect. 1.7.2).

3 Exercises on Curves and Hypersurfaces 165

(b) By Exercise 112, if P ∈ Sing(S) is a triple point, S is a cone of vertex P. Inthis case, if there existed a pointQ �= P singular for S, by Exercise 112 (c) all pointsof the line L(P,Q) would be singular for S, contradicting the hypotheses.

Exercise 115. Let S be a cubic surface of P3(C) and let r be a line contained in

Sing(S). Prove that, if S is irreducible, then Sing(S) = r.

Solution. Assume by contradiction that there exists a point Q ∈ Sing(S)\r. Since Sis irreducible, the planeH = L(r,Q) is not contained in S (cf. Exercise 57), hence itintersectsS in a plane cubicC. Since r ⊆ C, byBézout’s TheoremonehasC = r + Q,with Q a conic of H. In addition, by Exercise 58 (a) one has r ⊆ Sing(C), so thatExercise 54 implies that r ⊆ Q. Then we deduce again from Bézout’s Theorem thatQ = r + s, where s is a line ofH. So C = 2r + s, andQ ∈ s, becauseQ /∈ r; thereforeQ is a smooth point of C. On the other hand, again by Exercise 58 (a), Q should be asingular point of C. So we have reached a contradiction, showing that Sing(S) = r.

Exercise 116. (a) Let S be a cubic surface of P3(C). Show that, if P is a singular

point of S, then there exists a line r ⊂ S such that P ∈ r.(b) Let T be a cubic hypersurface of P

4(C). Prove that for every P ∈ T there existsa line r such that P ∈ r and r ⊂ T .

Solution. (a) Let x0, x1, x2, x3 be a homogeneous coordinate system of P3(C) such

that P has coordinates [1, 0, 0, 0] and let yi = xix0 , i = 1, 2, 3, be the correspond-

ing affine coordinates on U0. Since S is singular at P by assumption, its affinepart S ∩ U0 is described by an equation of the form f (y1, y2, y3) = f2(y1, y2, y3) +f3(y1, y2, y3) = 0, where f2 is either the zero polynomial or is homogeneous of degree2 and f3 is either zero or homogeneous of degree 3 (however note that f2 and f3 cannotbe both zero). Given v ∈ C

3\{0}, the affine line rv passing through P and parallel tov is described in parametric form by rv = {tv | t ∈ C}, and it is immediate to checkthat rv is contained in S if and only if f2(v) = f3(v) = 0. If the polynomials f2 and f3are both not identically zero, then, being homogeneous, they define curves of P

2(C)

of degree 2 and 3, respectively. By Bézout’s Theorem, there exists at least one pointR = [v0] such that f2(v0) = f3(v0) = 0. On the other hand, in case f2 or f3 is identi-cally zero, the set of points of P

2(C) at which both f2 and f3 vanish is infinite, andtherefore it is still possible to choose v0 ∈ C

3\{0} such that f2(v0) = f3(v0) = 0. Inany case, the line rv0 is then contained in S and so is its projective closure r = rv0 .

(b) If P is a non-singular point of T , we denote by H the tangent hyperplaneTP(T ). If P is singular, instead, we denote byH any hyperplane containing P. IfH iscontained in T , then there exist infinitelymany lines passing throughP and containedin T . So we may assume that H is not contained in T . In this case S = T ∩ H is acubic surface of H and, by Exercise 58, is singular at P. So, by part (a), there existsa line r of H passing through P and contained in S ⊂ T .

Exercise 117. Let F(x0, x1, x2, x3) = x30 − x1x2x3, and consider the surface S ofP3(R) defined by the equation F(x0, x1, x2, x3) = 0.

166 3 Exercises on Curves and Hypersurfaces

(a) Determine the singular points of S and describe the tangent cone to S at each ofthese points.

(b) Determine the lines contained in S.(c) Say whether S is irreducible.

Solution. (a) Since ∇F(x0, x1, x2, x3) = (3x20,−x2x3,−x1x3,−x1x2), the singularpoints of S are P1 = [0, 1, 0, 0], P2 = [0, 0, 1, 0] and P3 = [0, 0, 0, 1]. Let us studythe nature of the singular point P1. An equation of S ∩ U1 in the affine coordinatesy0 = x0

x1 , y2 = x2x1 , y3 = x3

x1 is given by −y2y3 + y30 = 0. So P1 is a double point andthe projective tangent cone to S at P1 is x2x3 = 0. Similarly, one checks that P2

and P3 are double points whose tangent cones have equation x1x3 = 0 and x1x2 = 0,respectively.

(b) The intersection of S with the plane H0 = {x0 = 0} is the union of the threelines L(P1,P2), L(P1,P3) and L(P2,P3).

Now we check that there are no additional lines contained in S. Assume bycontradiction that r ⊆ S is a line not contained in H0 and let Q = r ∩ H0. Up topermuting the coordinates x1, x2, x3, we may assume that Q is on the line L(P2,P3).The plane K generated by r and by L(P2,P3) intersects S in a reducible cubic C,which is the union of r, L(P2,P3) and a line s (possibly not distinct from L(P2,P3)

and r).On the other hand, since L(P2,P3) ⊆ K andK �= H0, it is immediate to check that

K is defined by the equation λx0 + x1 = 0, for some λ ∈ R. So, eliminating x1, onesees that C is defined by the equation x0(x20 + λx2x3) = 0 (cf. Sect. 1.7.3 for a moredetailed description of the process of elimination of the variable x1). Therefore C isthe union of L(P2,P3) and of the conic of equation x20 + λx2x3 = 0, whose support,by what we have seen above, must contain the line r �= L(P2,P3). Since this conicis irreducible if λ �= 0 and is the line L(P2,P3) counted twice if λ = 0, we have acontradiction.

(c) Since S has degree 3, if it were reducible it would contain a plane, and soinfinitely many lines, contradicting the fact that S contains precisely three lines.

K Exercise 118. For λ ∈ C, consider the surface Sλ ⊂ P3(C) of equation

Fλ(x0, x1, x2, x3) = x40 + x41 + x42 + x43 − 4λx0x1x2x3 = 0.

(a) Determine the values λ ∈ C such that Sλ is singular.(b) For every value of λ as in (a), determine the multiplicity of the singular points

of Sλ.(c) Determine the values λ ∈ R such that Sλ has at least one real point.

Solution. (a) For i = 0, . . . , 3 denote the coordinate hyperplane of equation xi = 0by Hi. Note that, in the coordinates induced on H3 by the standard coordinatesof P

3(C), the curve Sλ ∩ H3 has equation G(x0, x1, x2) = x40 + x41 + x42 = 0 for anyλ ∈ C. Since∇G = 4(x30, x

31, x

32), the curve Sλ ∩ H3 is non-singular. By Exercise 58,

3 Exercises on Curves and Hypersurfaces 167

Sing(Sλ) ∩ H3 = ∅ for allλ ∈ C. A similar computation shows that Sing(Sλ) ∩ Hi =∅ for i = 0, 1, 2 and for all λ ∈ C. The gradient of Fλ is

∇Fλ = 4(x30 − λx1x2x3, x31 − λx0x2x3, x

32 − λx0x1x3, x

33 − λx0x1x2).

If ∇Fλ(x0, x1, x2, x3) = 0, multiplying the ith component of ∇Fλ by xi for i =0, 1, 2, 3 we get x4i = λx0x1x2x3. Multiplying the four relations thus obtained onededuces x40x

41x

42x

43(1 − λ4) = 0. Since, as remarked above, regardless of the value of

λ ∈ C no point of Sing(Sλ) belongs to a coordinate hyperplane, if Sλ is singular thenone has λ4 = 1, namely λ ∈ {1,−1, i,−i}. On the other hand, if λ ∈ {1,−1, i,−i}then the point P = [λ3, 1, 1, 1] satisfies ∇Fλ(P) = 0, and so it is singular for Sλ.

(b) Let us consider first of all the case λ = 1. By what we have seen in (a), nopoint of the coordinate hyperplaneH0 is singular for S1, hence it suffices to study theaffine part S1 ∩ U0 of S1, using the usual affine coordinates yi = xi

x0 , i = 1, 2, 3, onU0. The singular points of S1 are given by the solutions of the system of equations

y1y2y3 = 1, y31 = y2y3, y32 = y1y3, y33 = y1y2. (3.7)

Multiplying the second equation of (3.7) by y1 and substituting y1y2y3 = 1 oneobtains y41 = 1. Analogous computations give y42 = y43 = 1. It is now easy to checkthat the singular points of S1 are the points of affine coordinates (α,β,α3β3), forα,β ∈ {1,−1, i,−i}. Hence Sing(S1) consists of 16 points. One has (F1)x0x0 = 12x20and so (F1)x0x0(P) �= 0 for all P ∈ Sing(S1). This proves that all singular points ofS1 have multiplicity 2.

The singularities of the surfaces Sλ for λ = −1, i,−i can be analysed in the sameway.Alternatively, one can observe that, ifλ4 = 1, then the projectivity gλ : P

3(C) →P3(C)definedby [x0, x1, x2, x3] �→ [λx0, x1, x2, x3] transformsSλ intoS1. Therefore,

for λ4 = 1, Sλ is projectively equivalent to S1 and Sing(Sλ) consists of 16 points ofmultiplicity 2.

(c) Note that the projectivity of P3(C) defined by [x0, x1, x2, x3] �→ [−x0, x1,

x2, x3] maps real points to real points and transforms Sλ into S−λ for all λ ∈ R, sothat Sλ has real points if and only if S−λ has real points. So it is enough to considerthe case λ ≥ 0.

For λ = 1, it is immediate to check that S1 contains, for instance, the real pointsQ1 = [1, 1, 1, 1], Q2 = [1, 1,−1,−1].

Let us now study the intersections of Sλ with the line r = L(Q1,Q2) as λ variesin R. Since [0, 0, 1, 1] /∈ Sλ for every λ ∈ R, it is enough to consider the points ofthe form [1, 1, t, t], t ∈ R. One of these points belongs to Sλ if and only if g(t) =t4 + 1 − 2λt2 = 0. Let now λ > 1. Then one has g(0) = 1, g(1) = 2(1 − λ) < 0,and therefore there exists a real value t0 ∈ (0, 1) such that g(t0) = 0. So [1, 1, t0, t0] ∈Sλ, and Sλ has real points.

To conclude, we show that Sλ has no real points if 0 ≤ λ < 1. For every λ ∈ R

the curve Cλ = Sλ ∩ H0 is defined by x41 + x42 + x43 = 0, and so it does not containany real point. Therefore it is enough to determine the support of the real affine

168 3 Exercises on Curves and Hypersurfaces

hypersurface Sλ ∩ U0. Consider the affine coordinates yi = xix0 , i = 1, 2, 3, on U0

and let fλ(y1, y2, y3) be the polynomial obtained by dehomogenizing Fλ with respectto x0. Let Ω = {(a, b, c) ∈ R

3 | (b, c) �= (0, 0)}, fix (a, b, c) ∈ Ω and restrict fλ tothe line ra,b,c = {(a, tb, tc) | t ∈ R}. One obtains a polynomial gλ(t) = (b4 + c4)t4 −4λabct2 + a4 + 1. If abc ≤ 0, then gλ(t) ≥ a4 + 1 > 0 for every t ∈ R and so ra,b,cdoes not contain points ofSλ. Assume now abc > 0, set z = t2 and study the functionh(z) = (b4 + c4)z2 − 4λabcz + a4 + 1 for z ≥ 0. The minimum point of h is given

by z0 = 2λabcb4 + c4

. One has

(b4 + c4)h(z0) = (a4 + 1)(b4 + c4) − 4λ2(abc)2 ≥ (2a2)(2b2c2) − 4λ2(abc)2 == 4(abc)2(1 − λ2) > 0.

So for every choice of (a, b, c) ∈ Ω one has ra,b,c ∩ Sλ = ∅ if 0 ≤ λ < 1. Since theunion of the lines ra,b,c as (a, b, c) varies in Ω is the whole of R

3, it follows that Sλ

has no real points for 0 ≤ λ < 1.Summing up, we have proven that Sλ has real points if and only if |λ| ≥ 1.

We describe an alternativemethod to prove thatSλ has no real points if 0 ≤ λ < 1.As above, note that, if one fixes on the coordinate hyperplane H0 the homogeneouscoordinates induced by the standard frame of P

3(C), for all λ ∈ R the curve Sλ ∩ H0

is defined by x41 + x42 + x43 = 0, and so it does not contain any real point. Similarly oneshows that the homogeneous coordinates of any real point of Sλ are all non-zero. Solet x0, x1, x2, x3 be non-zero real numbers. From the inequality (x20 − x21)

2 ≥ 0 onededuces x40 + x41 ≥ 2x20x

21,and analogously one has x42 + x43 ≥ 2x23x

24. Adding these

inequalities one gets x40 + x41 + x42 + x43 ≥ 2(x20x21 + x22x

23). Moreover one has

x20x21 + x22x

23 = (|x0x1| − |x2x3|)2 + 2|x0x1x2x3| ≥ 2|x0x1x2x3|,

so that finally one has x40 + x41 + x42 + x43 ≥ 4|x0x1x2x3|. Since xi �= 0 for i = 0, . . . , 3,if 0 ≤ λ < 1 then

x40 + x41 + x42 + x43 ≥ 4|x0x1x2x3| > 4λx0x1x2x3.

It follows that Fλ(x0, x1, x2, x3) is positive for every [x0, x1, x2, x3] ∈ P3(R), so that

Sλ contains no real points.

K Exercise 119. Consider a planeH of P3(C), a non-degenerate conic C ⊂ H, a point

P ∈ C and let t ⊂ H be the tangent line to C at P. Let r and s be lines of P3(C) such

that:

(i) r and s are not contained in H;(ii) r ∩ (s ∪ C) = ∅;(iii) s ∩ C = P.

Let π : C → r be the map defined by π(Q) = L(s,Q) ∩ r if Q �= P and π(P) =L(s, t) ∩ r.

3 Exercises on Curves and Hypersurfaces 169

(a) Check that π is well defined and bijective.(b) Prove that

X =⋃Q∈C

L(Q,π(Q))

is the support of a cubic surface S.(c) Determine the singular points of S and the corresponding multiplicities, and say

whether S is reducible and whether it is a cone (Fig. 3.1).

Solution. (a) The restriction of π to C\{P} coincides with the restriction of theprojection onto r with centre s, and so it is well defined. Since s and r are skew byassumption, the plane L(t, s) does not contain r and so the point π(P) is well defined,too.

Let Q,Q′ be points of C such that π(Q) = π(Q′). If Q �= P and Q′ �= P, one hasL(s,Q) = L(s,π(Q)) = L(s,π(Q′)) = L(s,Q′), so that the line H ∩ L(s,Q) meetsC at P,Q,Q′. Since C is irreducible, we must have Q = Q′. Furthermore, if Q = Pand Q′ �= P, then the condition π(P) = π(Q′) would imply L(s, t) = L(s,π(P)) =L(s,π(Q′)) = L(s,Q′). The line t = H ∩ L(s, t) = H ∩ L(s,Q′), which is tangent toC, would then meet C also at the point Q′ �= P, contradicting again the irreducibilityof C. It follows that π is injective. The surjectivity of π is a consequence of the factthat, for any given point A ∈ r, the line H ∩ L(s,A) either is tangent to C at P, andin that case A = π(P), or intersects C at a point Q �= P, and in that case A = π(Q).

(b) It is possible to choose a projective frame {P0, . . . ,P4} of P3(C) such that

P0 = P, P2 ∈ C, P1 is the intersection point of t with the tangent line to C at P2,P3 ∈ s and P4 ∈ r. In the coordinate system induced by this frame the plane His defined by the equation x3 = 0, P has coordinates [1, 0, 0, 0], t is defined byx3 = x2 = 0, C is defined by the equation x0x2 − λx21 = 0 for some λ �= 0, and s

t

P2

P3P4

s

P0 = P r

P1

π(Q)

R

Q

Fig. 3.1 The configuration described in Exercise 119

170 3 Exercises on Curves and Hypersurfaces

is given by x1 = x2 = 0. Let [a, b, c, 0] be the coordinates of the point R = r ∩ H,and let us determine first the conditions imposed on a, b, c by assumption (ii). Sincer = L(P4,R), the lines r and s are skew if and only if there exists no non-zero linearcombination of (a, b, c, 0) and (1, 1, 1, 1) that satisfies the conditions x1 = x2 = 0,i.e. if and only if b − c �= 0. In addition, since r ∩ C = ∅, one has R /∈ C, so thatλb2 − ac �= 0.

Let Y ∈ P3(C)\s. The plane L(Y , s) intersects H in a line, and therefore it inter-

sects C, besides at P, at a second point Y ′, possibly not distinct from P. The pointsY ′ and π(Y ′) ∈ r are distinct because r ∩ C = ∅, and the point Y is on X if and onlyif Y , Y ′ and π(Y ′) are collinear.

If Y has coordinates [y0, y1, y2, y3], then the plane L(Y , s) has equation y2x1 −y1x2 = 0. Combining this equation and the equation of C, one sees that Y ′ has coor-dinates [λy21, y1y2, y22, 0]. In addition, if Y ′ �= P then one has π(Y ′) = L(s,Y ′) ∩ r =L(s,Y) ∩ r by definition, so that π(Y ′) is represented by the (unique up to multiples)non-zero linear combination of (a, b, c, 0) and (1, 1, 1, 1) that satisfies the equationy2x1 − y1x2 = 0. Then a simple computation shows that

π(Y ′) = [(a − c)y1 + (b − a)y2, (b − c)y1, (b − c)y2,−cy1 + by2]. (3.8)

If instead Y ′ = P, since L(s, t) has equation x2 = 0, one easily gets π(Y ′) = L(s, t) ∩r = [a − c, b − c, 0,−c]. Note also that in this case, since Y ′ = [λy21, y1y2, y22, 0] =[1, 0, 0, 0] = P, onemust necessarily have y2 = 0, implying y1 �= 0 since Y /∈ s. Onededuces that the formula (3.8), obtained in the case when Y ′ �= P, actually holds alsowhen Y ′ = P. So Y belongs to X if and only if

rk

⎛⎜⎜⎝y0 λy21 (a − c)y1 + (b − a)y2y1 y1y2 (b − c)y1y2 y22 (b − c)y2y3 0 −cy1 + by2

⎞⎟⎟⎠ ≤ 2. (3.9)

Note that the second and the third rows of the matrix in (3.9) are multiples of therow (1, y2, b − c) and they are both non-zero, because Y /∈ s. Hence condition (3.9)is equivalent to F(y0, y1, y2, y3) = 0, where

F(y0, y1, y2, y3) = det

⎛⎝y0 λy21 (a − c)y1 + (b − a)y21 y2 (b − c)y3 0 −cy1 + by2

⎞⎠ = (3.10)

= y3[(b − c)λy21 + (c − a)y1y2 + (a − b)y22] + (by2 − cy1)(y0y2 − λy21).

Therefore we have proven that a point Y ∈ P3(C)\s of coordinates [y0, . . . , y3]

belongs to X if and only if F(y0, . . . , y3) = 0, where F is a homogeneous poly-nomial of degree 3. So it follows that the support of the surface S defined by F(regarded from now on as a polynomial in x0, x1, x2, x3) coincides with the set Xon P

3(C)\s. Moreover it is evident that F(a0, 0, 0, a3) = 0 for all (a0, a3) ∈ C2,

3 Exercises on Curves and Hypersurfaces 171

so that s is contained in the support of S. Let us now show that s ⊆ X. One hasthat P ∈ L(P,π(P)) ⊆ X and, since Q ∈ s\{P}, the plane L(Q, r) intersects H ina line different from t, and therefore intersects C in at least one point Q1 �= P.So L(Q1,π(Q1)) = L(s,Q1) ∩ L(r,Q1) = L(s,Q1) ∩ L(r,Q) intersects s atQ. Thisproves that s is contained in X, and therefore that X coincides with the support of S.

(c) Using the equationF given in (3.10), it is immediate to check that all the pointsof the line s are singular for S. In order to compute their multiplicities, note thatFx1x1(x0, 0, 0, x3) = 2λ(b − c)x3. So the points Q ∈ s, Q �= P, verify Fx1x1(Q) �= 0and so have multiplicity 2. Moreover Fx1x2(P) = −c and Fx2x2(P) = 2b are not bothequal to zero because, as noted in the solution of (b), one has λb2 − ac �= 0, hencealso P has multiplicity 2.

Let us now show that S has no additional singular points. This verification canbe carried out analytically, by computing ∇F and showing that all the solutions of∇F = 0 satisfy x1 = x2 = 0. Otherwise one can argue synthetically, as follows.

LetA be a point ofS\s, letK = L(s,A), and consider the plane cubicS ∩ K . Sincethe support ofS ∩ K contains s, byBézout’s Theoremone hasS ∩ K = s + Q, whereQ is a conic in K . By Exercise 58 (a), the points of s are singular for S ∩ K sinces ⊆ Sing(S), and this implies easily that s ⊆ Q. Again by Bézout’s Theorem onehasQ = s + s′, where s′ ⊆ K is a line. More precisely, since r is contained in X andis disjoint from s, the support of the cubic S ∩ K = 2s + s′ cannot be contained in s,so that s′ �= s. In particular, the only singular points of S ∩ K are given by the pointsof s, hence the curve S ∩ K is smooth at A and, again by Exercise 58, S is smooth atA. So we have shown that Sing(S) = s.

Because all the points of Sing(S) have multiplicity 2, S is not a cone by Exer-cise 112.

Now let us show that S is not reducible. Assume by contradiction S = K + Q,where K is a plane and Q is a quadric. By Exercise 54, all the points of K ∩ Qare singular for S, therefore the support of K ∩ Q is contained in s. Then the curveK ∩ Q, being a conic of K , coincides with the double line 2s. In particular, s iscontained in K . Since we have shown above that the support of the intersection ofS with any plane containing s is the union of two distinct lines, we have reached acontradiction.

Note. It is natural to wonder how the situation described in Exercise 119 degenerateswhen the pointH ∩ r belongs to C. In that case it is easy to check that the polynomialF given in (3.10) turns out to be divisible by by2 − cy1, namely F vanishes on theplane L(s,H ∩ r). Then the cubic surface S defined by F splits as the union ofL(s,H ∩ r) and of a quadric Q, whose support coincides with the analogue of theset X defined here (cf. Exercise 186).

Chapter 4Exercises on Conics and Quadrics

Affine and projective conics and quadrics. Polarity, centres anddiametral hyperplanes. Conics and quadrics in Euclidean space.Principal hyperplanes, axes and vertices. Pencils of conics.

Abstract Solved problems on affine and projective conics and quadrics, polarity,conics and quadrics in Euclidean space, pencils of conics.

Notation: Throughout the whole chapter, the symbol K denotes R or C.

Exercise 120. (a) Find the equation of a conicC ofP2(C)passing through the pointsP1 = [1, 0, 1], P2 = [−1, 0, 0], P3 = [0, 1, 1], P4 = [0,−1, 0], P5 = [1, 3, 2].

(b) Check that C is non-degenerate and determine the pole of the line 5x0 + x1 −3x2 = 0 with respect to the conic C.

Solution. (a) The points P1, P2, P3, P4 are not collinear and hence the conics passingthrough these points form a pencil F (cf. Sect. 1.9.7) generated by the degenerateconics L(P1, P2) + L(P3, P4) and L(P1, P3) + L(P2, P4). Since

L(P1, P2) = {x1 = 0}, L(P3, P4) = {x0 = 0},

L(P1, P3) = {x0 + x1 − x2 = 0}, L(P2, P4) = {x2 = 0},

the generic conic Cλ,μ of F has equation

λx0x1 + μx2(x0 + x1 − x2) = 0.

This conic passes through P5 if and only if 3λ + 4μ = 0; choosing [λ,μ] = [4,−3],we get the conic of equation 4x0x1 − 3x2(x0 + x1 − x2) = 0.

(b) The conic C is represented by the matrix A =⎛⎝

0 4 −34 0 −3

−3 −3 6

⎞⎠. As det A �= 0,

the conic is non-degenerate.

© Springer International Publishing Switzerland 2016E. Fortuna et al., Projective Geometry, UNITEXT - La Matematica per il 3+2 104,DOI 10.1007/978-3-319-42824-6_4

173

174 4 Exercises on Conics and Quadrics

Since the map polC is a projective isomorphism (cf. Sect. 1.8.2), there existsexactly one point R = [a, b, c] such that polC(R) is the line r of equation 5x0 +x1 − 3x2 = 0. As polC(R) has equation tRAX = 0, that is

(4b − 3c)x0 + (4a − 3c)x1 + (−3a − 3b + 6c)x2 = 0,

the values a, b, c we are looking for are those for which there exists h �= 0 such that

(4b − 3c, 4a − 3c,−3a − 3b + 6c) = h(5, 1,−3).

The solutions of this linear system form the set {(t, 2t, t) | t ∈ C}; therefore the pointR = [1, 2, 1] is the requested pole.

Observe that an alternative way to determine R consists in choosing two distinctpoints M, N on r and taking R = polC(M) ∩ polC(N ): by reciprocity polC(R) =L(M, N ) = r .

Exercise 121. Assume that P1, P2, P3 are non-collinear points in P2(K) and let r be

a line passing through P1, which passes neither through P2 nor through P3. Considerthe subset of the space Λ2 of projective conics of P

2(K)

F = {C ∈ Λ2 | C passes through P1, P2, P3 and is tangent to r at P1}.

Show that F is a linear system and compute its dimension.

Solution (1). Imposing that a conic C is tangent to r at P1 corresponds to imposingtwo independent linear conditions (cf. Exercise 95 and the Note following it). Theconics ofF are obtained by imposing that they pass also through the points P2 and P3,which are two additional linear conditions, soF is a linear system of dimension≥ 1.

As a matter of fact F has dimension 1, i.e. it is a pencil. To prove that, assumeby contradiction that F has dimension at least 2 and choose a fourth point P4 /∈ r insuch a way that P1, P2, P3, P4 are in general position. Let F ′ be the linear systemof the conics of F that pass also through P4; since dimF ′ ≥ dimF − 1 ≥ 1, wecan choose two distinct conics C1, C2 in F ′. These conics meet in at least 5 pointscounted with multiplicity (because I (C1, C2, P1) ≥ 2 since C1, C2 are both tangent tothe line r at P1, cf. Sect. 1.9.3). As a consequence by Bézout’s Theorem C1 and C2contain a common line l and are both degenerate, that is C1 = l + r1 and C2 = l + r2.There are only two possibilities in order that the two previous conics be tangent to rat P1: either l = r or l and ri , i = 1, 2, meet at P1. The first situation cannot occur,because ri should pass through the points P2, P3, P4 which are not collinear. Also thesecond situation cannot occur, because at least two points among P2, P3, P4 shouldlie either on l or on ri and hence they would be collinear with P1. So we have founda contradiction, and therefore dimF = 1.

Solution (2). It is also possible to solve the exercise analytically. Let R be apoint of r not collinear with P2 and P3; choosing P1, P2, P3, R as a projec-tive frame, we have P1 = [1, 0, 0], P2 = [0, 1, 0], P3 = [0, 0, 1], R = [1, 1, 1] and

4 Exercises on Conics and Quadrics 175

r = L(P1, R) = {x1 − x2 = 0}. Let C be a conic of F and let M be a symmet-ric matrix that represents it. Since C passes through [1, 0, 0], [0, 1, 0], [0, 0, 1], the

entries on the principal diagonal ofM are necessarily zero, i.e.M =⎛⎝0 a ba 0 cb c 0

⎞⎠where

at least one among a, b, c ∈ K is non-zero. Therefore C has an equation of the formF(x0, x1, x2) = ax0x1 + bx0x2 + cx1x2 = 0.

Since∇F = (ax1 + bx2, ax0 + cx2, bx0 + cx1) and∇F(1, 0, 0) = (0, a, b), theline r is tangent to C at P1 if and only if b = −a. Therefore F consists precisely ofthe conics represented by matrices of the form

⎛⎝

0 a −aa 0 c

−a c 0

⎞⎠ , [a, c] ∈ P

1(K)

so F is a linear system of dimension 1, i.e. it is a pencil.

Note. The result holds also when the line r passes, for instance, through P2. Namelyin this case every conic C ∈ F meets r in at least 3 points counted with multiplicityand therefore by Bézout’s Theorem the line r is a component of C. As a consequencethe conics of F are exactly those of the form r + s when s varies in the pencil oflines centred at P3, so F is a pencil (cf. Exercise 96).

Exercise 122. In P2(R) consider the points

P1 = [0, 1, 2], P2 = [0, 0, 1], P3 = [2, 1, 2], P4 = [3, 0, 1].

Determine, if it exists, an equation of a conic passing through P1, P2, P3, P4 andtangent at P3 to the line r of equation x0 − x2 = 0.

Solution. Observe that the line r passes through the point P3 and that P1, P2, P3 arenot collinear. The conic we are looking for belongs to the pencil F of conics passingthrough P1, P2, P3 and tangent to r at P3 (cf. Exercise 121 and the following Note).It is generated by the degenerate conics r + L(P1, P2) and L(P1, P3) + L(P2, P3).

Since L(P1, P2) = {x0 = 0}, L(P1, P3) = {2x1 − x2 = 0} and L(P2, P3) ={x0 − 2x1 = 0}, the generic conic Cλ,μ of F has equation

λx0(x0 − x2) + μ(2x1 − x2)(x0 − 2x1) = 0.

The conic Cλ,μ passes through P4 if and only if 6λ − 3μ = 0; hence, choosing forinstance [λ,μ] = [1, 2], we find that the conic of equation

x20 − 8x21 + 4x0x1 − 3x0x2 + 4x1x2 = 0

satisfies the required properties.

176 4 Exercises on Conics and Quadrics

Exercise 123. Determine, if it exists, a non-degenerate conic C of P2(C) such that

(i) C passes through the points A = [0, 0, 1] and B = [0, 1, 1];(ii) C is tangent to the line x0 − x2 = 0 at the point P = [1, 1, 1];(iii) the polar of the point [2, 4, 3] with respect to C is the line 3x1 − 4x2 = 0.

Solution. The points A, B and P are not collinear. If r is the line of equationx0 − x2 = 0, a conic satisfying conditions (i) and (ii) belongs to the pencil generatedby the degenerate conics L(A, B) + r and L(A, P) + L(B, P) (cf. Exercise 121). Itis easy to compute that L(A, B) = {x0 = 0}, L(A, P) = {x0 − x1 = 0}, L(B, P) ={x1 − x2 = 0}, so the generic conic Cλ,μ of the pencil has equation

λx0(x0 − x2) + μ(x0 − x1)(x1 − x2) = 0,

and is therefore represented by the matrix

Aλ,μ =⎛⎝

2λ μ −λ − μμ −2μ μ

−λ − μ μ 0

⎞⎠ .

Therefore, if Cλ,μ is non-degenerate, the polar of the point [2, 4, 3] with respectto Cλ,μ has equation (λ + μ)x0 − 3μx1 + (2μ − 2λ)x2 = 0 and this line coincideswith the line of equation 3x1 − 4x2 = 0 if and only if λ + μ = 0. So, if we choosethe homogeneous pair [λ,μ] = [1,−1], we find the conic of equation

x20 + x21 − x0x1 − x1x2 = 0

which, being non-degenerate, satisfies the required properties.

Exercise 124. Consider the curve D of P2(C) of equation

F(x0, x1, x2) = x22 (x1 − x2)2 − x1(x1 + x2)x

20 = 0.

(a) Find the singular points of D and compute their multiplicities and principaltangents.

(b) Determine the dimension of the linear system F of the conics of P2(C) passing

through the singular points of D and tangent at Q = [1, 0, 0] to the line τ ofequation x1 = 0.

(c) Compute the points of intersection between D and C when C varies in F .

Solution. (a) The singular points D are the points whose homogeneous coordinatesannihilate the gradient of F . By solving the system

⎧⎨⎩

Fx0 = −2x0x1(x1 + x2) = 0Fx1 = 2x22 (x1 − x2) − x20 (x1 + x2) − x1x20 = 0Fx2 = 2x2(x1 − x2)2 − 2x22 (x1 − x2) − x1x20 = 0

4 Exercises on Conics and Quadrics 177

we find that the singular points of D are A = [0, 1, 0], B = [0, 1, 1] and Q =[1, 0, 0].

If we work in the affine chartU0 using the coordinates x = x1x0 , y = x2

x0 , the point

Q has coordinates (0, 0) andD ∩U0 has equation y2(x − y)2 − x(x + y) = 0. Thuswe see that Q is an ordinary double point for D where the principal tangents are theprojective closures of the lines x = 0 and x + y = 0, that is the lines x1 = 0 andx1 + x2 = 0.

If instead we work in the affine chart U1 using the coordinates u = x0x1 , v = x2

x1 ,we have A = (0, 0), B = (0, 1) and the affine part D ∩U1 of D has equation

v2(1 − v)2 − (1 + v)u2 = 0.

Thus we immediately see that A is an ordinary double point with principal tangentsthe projective closures of the lines v − u = 0 and v + u = 0, that is the lines x2 −x0 = 0 and x2 + x0 = 0.

In order to study the multiplicity of B, if we perform the translation (w, z) →(u, v) = (w, z + 1) that maps (0, 0) to B, we obtain the equation

(z + 1)2z2 − (z + 2)w2 = 0.

Therefore, B is an ordinary double point with principal tangents the projectiveclosures of the lines z − √

2w = 0 and z + √2w = 0, that is the lines x2 − x1 −√

2x0 = 0 and x2 − x1 + √2x0 = 0.

(b) The points A, B, Q are not collinear, A /∈ τ and B /∈ τ ; hence the set of conicspassing through A, B, Q and tangent to τ at Q is a linear system of dimension 1,that is a pencil (cf. Exercise 121).

(c) Since the conics τ + L(A, B) and L(Q, A) + L(Q, B) generate the pencil Fand have equations x1x0 = 0 and x2(x1 − x2) = 0, respectively, the generic conicCλ,μ of F has equation λx1x0 + μx2(x1 − x2) = 0.

Ifμ = 0, thenC1,0 ∩ D = {Q, A, B}. Thus, ifwe supposeμ �= 0,we can set t = λμ

and find the solutions of the system given by the equations of D and of Ct,1 = Cλ,μ.From the equation of Ct,1 we obtain the equality x2(x1 − x2) = −t x0x1. By squaringand substituting into the equation of D, one gets

x20 x1(t2x1 − (x1 + x2)) = 0.

In correspondence with the roots of the factors x0 and x1 we find the points Q, A, B.If instead t2x1 − (x1 + x2) = 0, if we substitute x2 = (t2 − 1)x1 into the equationof Ct,1 and if we divide by x1, we obtain the equation

t x0 + (t2 − 1)(2 − t2)x1 = 0,

having as unique homogeneous solution the pair [x0, x1] = [(1 − t2)(2 − t2), t].Thus it turns out that Ct,1 intersects D, besides at Q, A, B, only at the point of

178 4 Exercises on Conics and Quadrics

homogeneous coordinates [(1 − t2)(2 − t2), t, t (t2 − 1)]. This point coincides withQ if t = 0, with A if t = ±1, with B if t = ±√

2, otherwise it is distinct from Q, Aand B.

Note. Let f : P1(C) → P

2(C) be the map defined by

f ([λ,μ]) = [(μ2 − λ2)(2μ2 − λ2),λμ3,λμ(λ2 − μ2)].

The argument in the proof of part (c) of the previous exercise shows that the image off is contained inD. Since for every P ∈ D there exists a conic ofF passing throughP , it easily follows that f (P1(C)) = D. Therefore, the map f gives a parametriza-tion of D. Observe that such a parametrization is not injective, because f ([0, 1]) =f ([1, 0]) = Q, f ([1, 1]) = f ([1,−1]) = A, f ([√2, 1]) = f ([√2,−1]) = B.

Exercise 125. Let P1, P2 ∈ P2(K) be distinct points and let ri be a projective line

passing through Pi , i = 1, 2. Assume that P2 /∈ r1, P1 /∈ r2. Show that the set

F = {C ∈ Λ2 | ri is tangent to C at Pi , i = 1, 2}

is a linear system and compute its dimension.

Solution (1). The conics of F can be obtained by imposing tangency conditions totwo given lines at the points P1 and P2; this corresponds to imposing four linearconditions, so F is a linear system of dimension ≥1.

If C is a reducible conic in F , then C is tangent to r1 at P1 if and only if eitherr1 is an irreducible component of C or both irreducible components of the conicpass through P1. It follows that the only reducible conics in F , which must fulfil thetangency conditions both at P1 and at P2, are the conics r1 + r2 and 2L(P1, P2).

Let us now see that F has dimension 1. Assume by contradiction that F hasdimension at least 2 and choose a point P3 /∈ r1 ∪ r2 not collinear with P1 and P2.Then the set F ′ consisting of the conics of F passing also through P3 is a linearsystem of dimension≥1, so that we can choose two distinct conics C1, C2 in it. Theseconics meet in at least 5 points counted with multiplicity (because I (C1, C2, Pi ) ≥ 2for i = 1, 2, cf. Sect. 1.9.3). As a consequence, by Bézout’s Theorem C1 and C2are reducible and share a common line. Then, by the previous considerations, fori = 1, 2 either Ci = r1 + r2 or Ci = 2L(P1, P2), but both these conics pass throughP3. Therefore the hypothesis thatF has dimension at least 2 leads to a contradiction,and hence dimF = 1.

Solution (2). As for Exercise 121, we can give an analytic solution of the problem.Set Q = r1 ∩ r2. By hypothesis Q, P1, P2 are in general position, so we can

choose coordinates in such a way that Q = [1, 0, 0], P1 = [0, 1, 0], P2 = [0, 0, 1];consequently r1 = {x2 = 0}, r2 = {x1 = 0}. LetC be a conic ofF ; sinceC passes both

through P1 and through P2, it is represented by amatrixM =⎛⎝a b cb 0 dc d 0

⎞⎠with at least

4 Exercises on Conics and Quadrics 179

one among a, b, c, d ∈ K non-zero. Since pol(P1) has equation bx0 + dx2 = 0, theline r1 is tangent to C at P1 if and only if b = 0. Similarly, since pol(P2) has equationcx0 + dx1 = 0, the line r2 is tangent to C at P2 if and only if c = 0. Therefore, Fconsists precisely of the conics represented by matrices of the form

⎛⎝a 0 00 0 d0 d 0

⎞⎠ , [a, d] ∈ P

1(K)

and then F is a linear system of dimension 1, i.e. it is a pencil.

Exercise 126. Compute, if it exists, an equation of a non-degenerate conic C ofP2(C) such that:

(i) C is tangent to the line x0 + x2 = 0 at the point A = [1, 1,−1];(ii) C is tangent to the line x0 − x1 − x2 = 0 at the point B = [1, 1, 0];(iii) the polar of the point [2, 0, 1]with respect toC passes through the point [4, 2, 3].Solution. A conic fulfilling conditions (i) and (ii) belongs to the pencil generatedby the degenerate conics (x0 + x2)(x0 − x1 − x2) = 0 and 2L(A, B) (cf. Exercise125). Since L(A, B) has equation x0 − x1 = 0, the generic conic Cλ,μ of the pencilhas equation

λ(x0 + x2)(x0 − x1 − x2) + μ(x0 − x1)2 = 0

and is thus represented by the matrix

Aλ,μ =⎛⎝

2λ + 2μ −λ − 2μ 0−λ − 2μ 2μ −λ

0 −λ −2λ

⎞⎠ .

Since Aλ,μ

⎛⎝201

⎞⎠ =

⎛⎝

4λ + 4μ−3λ − 4μ

−2λ

⎞⎠, the point [2, 0, 1] is non-singular for every

Cλ,μ and the polar of [2, 0, 1]with respect to C is the line of equation (4λ + 4μ)x0 +(−3λ − 4μ)x1 − 2λx2 = 0. It contains the point [4, 2, 3] if and only if 4λ + 8μ = 0;therefore, in correspondence with the homogeneous pair [λ,μ] = [2,−1] we findthe conic of equation

x20 − x21 − 2x22 − 2x1x2 = 0

which, being non-degenerate, satisfies the required properties.

Exercise 127. In P2(C) consider the lines

r = {x2 = 0} s = {x1 − 2x2 = 0} t = {x1 = 0} l = {x0 − x2 = 0}

and the points P = r ∩ t , Q = t ∩ l, A = l ∩ r , R = [2,−1, 1].

180 4 Exercises on Conics and Quadrics

(a) Find an explicit formula for a projectivity f ofP2(C) such that f (r) = s, f (l) =

l, f (t) = t , f (R) belongs to the line of equation x0 + x1 = 0 and t is the onlyline in the pencil of lines centred at P that is invariant under f .

(b) Find, if it exists, a conic C of P2(C) tangent at P to t , tangent at A to l and such

that f (C) passes through the point [2,−2, 1].Solution. (a) It turns out that P = [1, 0, 0], Q = [1, 0, 1], A = [0, 1, 0] and B =l ∩ s = [1, 2, 1]. Moreover, we observe that the lines r, s, t pass through P whileP /∈ l.

If f is a projectivity such that f (r) = s, f (l) = l, f (t) = t , then f (A) = B,

f (P) = P, f (Q) = Q. Imposing the first two conditions, we see that f is repre-

sented by a matrix of the form M =⎛⎝1 b c0 2b d0 b e

⎞⎠ with b(2e − d) �= 0.

Also, we have that f (Q) = Q if and only if d = 0 and e = 1 + c, so that

M =⎛⎝1 b c0 2b 00 b 1 + c

⎞⎠. Since f (R) = [2 − b + c,−2b,−b + 1 + c], the point f (R)

belongs to the line of equation x0 + x1 = 0 if and only if c = 3b − 2; hence neces-

sarily M =⎛⎝1 b 3b − 20 2b 00 b 3b − 1

⎞⎠.

Let FP be the pencil of lines centred at P; the lines of this pencil have equa-tions αx1 + βx2 = 0, with [α,β] ∈ P

1(C). The pencil FP is a line of P2(C)∗ that

is invariant under the dual projectivity f∗ : P2(C)∗ → P

2(C)∗, which is representedby the matrix t M−1 in the system of dual homogeneous coordinates on P

2(C)∗(cf. Sect. 1.4.5). In the system of coordinates induced onFP the line αx1 + βx2 = 0has coordinates [α,β] and the restriction of f∗ to FP is represented by the matrix(3b − 1 −b

0 2b

). Then t , which has coordinates [1, 0], is the only line ofFP invariant

under f if and only if the eigenvectors of the latter matrix are precisely the non-zeromultiples of (1, 0), which happens if and only if b = 1.

So the only projectivity having the required properties is the one represented by

the matrix M =⎛⎝1 1 10 2 00 1 2

⎞⎠.

(b) The pencil of conics tangent to t at P and tangent to l at A is generated by thedegenerate conics l + t and 2r . Therefore, the generic conic Cλ,μ of this pencil hasequation

λx1(x0 − x2) + μx22 = 0.

Since f ([2,−1, 1]) = [2,−2, 1], f (Cλ,μ) passes through the point [2,−2, 1] if andonly if Cλ,μ passes through the point [2,−1, 1]; this occurs if and only if μ − λ = 0.Thus, if we choose the homogeneous pair [λ,μ] = [1, 1], we get the conic x1(x0 −x2) + x22 = 0 which satisfies all the required properties.

4 Exercises on Conics and Quadrics 181

Exercise 128. Given four distinct lines r1, r2, r3, r4 in P2(K), prove that there exists

at least one non-degenerate conic which is tangent to the four lines if and only ifr1, r2, r3, r4 are in general position.

Solution (1). Assume that there exists a non-degenerate conic which is tangent tor1, r2, r3, r4 and, for any i �= j , let Pi j = ri ∩ r j . Since for every point of P

2(K)

there exist at most two lines passing through that point and tangent to the conic (cf.Sect. 1.8.2), the line rh cannot pass through Pi j if h /∈ {i, j}. Therefore, any threelines chosen among r1, r2, r3, r4 are not concurrent, that is the four lines are in generalposition.

Conversely, let D be a non-degenerate and non-empty conic and denote byl1, l2, l3, l4 the lines tangent to D at four distinct points of D. As seen here above,the four lines are in general position. For any i �= j let Qi j = li ∩ l j . One can easilycheck that each of the quadruples P12, P23, P34, P41 and Q12, Q23, Q34, Q41 consistsof points in general position. Therefore, there exists a projectivity T of P

2(K) suchthat T (Q12) = P12, T (Q23) = P23, T (Q34) = P34 and T (Q41) = P41. This imme-diately implies that T (li ) = ri for i = 1, 2, 3, 4. Then the conic C = T (D) is tangentto r1, r2, r3, r4.

Solution (2). Arguing by duality, we can regard the lines r1, r2, r3, r4 as pointsR1, R2, R3, R4 of P

2(K)∗. The lines r1, r2, r3, r4 are in general position in P2(K) if

and only if the corresponding points in P2(K)∗ are in general position; in this case

there exists a non-degenerate conic passing through R1, R2, R3, R4 and then the dualconic is tangent to r1, r2, r3, r4 (cf. Sect. 1.8.2).

Note. The same argument used in Solution (2) proves that, if r1, r2, r3, r4, r5 are fivelines ofP2(K) in general position (that is the corresponding points R1, R2, R3, R4, R5

ofP2(K)∗ are in general position), then there exists exactly one non-degenerate conic

tangent to the five lines.

Exercise 129. Let A, B,C, D be points of P2(K) in general position and assume

that r is a line of P2(K) passing through D and such that A /∈ r , B /∈ r and C /∈ r .

Prove that there exists only one non-degenerate conic Q of P2(K) fulfilling the

following conditions:

(i) polQ(A) = L(B,C);(ii) polQ(B) = L(A,C);(iii) D ∈ Q and the line r is tangent to Q at D.

Solution. Observe that, if Q is a non-degenerate conic satisfying conditions (i) and(ii), then the pointC lies both on polQ(A) and on polQ(B); hence, by reciprocity, nec-essarily polQ(C) = L(A, B). As a consequence the points A, B,C are the verticesof a self-polar triangle for the conic.

In a suitable system of homogeneous coordinates in P2(K) we have that A =

[1, 0, 0], B = [0, 1, 0],C = [0, 0, 1], D = [1, 1, 1]. Then the non-degenerate con-ics Q having A, B,C as vertices of a self-polar triangle have equations of the formαx20 + βx21 + γx22 = 0, with at least one among α,β, γ non-zero (cf. Sect. 1.8.4).

182 4 Exercises on Conics and Quadrics

In order for Q to satisfy property (iii) too, the line r must be the polar of D withrespect to Q. This polar has equation ax0 + bx1 + cx2 = 0 with a + b + c = 0, asD ∈ r , and with a �= 0, b �= 0, c �= 0, as r does not pass through any of the pointsA, B and C . Instead, the line polQ(D) has equation αx0 + βx1 + γx2 = 0.

Therefore, we have polQ(D) = r if and only if the vectors (α,β, γ) and (a, b, c)are proportional. Hence the only conic having the required properties is the onedefined by the equation ax20 + bx21 + cx22 = 0.

Exercise 130. Denote by F the pencil of conics of P2(C) of equation

λ(4x0x1 − x22 − 4x21 ) + μ(x0x1 + x22 + 4x21 − 5x1x2) = 0 [λ,μ] ∈ P1(C).

(a) Find the degenerate conics and the base points of the pencil F .(b) Describe the projectivities f of P

2(C) such that f ([1, 1, 0]) = [1, 1, 0] andf (C) ∈ F for every C ∈ F .

Solution. (a) The generic conic Cλ,μ of the pencil is represented by the matrix

Aλ,μ =⎛⎜⎝

0 2λ + μ2 0

2λ + μ2 −4λ + 4μ −5

2μ

0 −52μ −λ + μ

⎞⎟⎠ .

Since det Aλ,μ =(2λ + μ

2

)2(μ − λ), the only degenerate conics of the pencil are the

ones corresponding to the homogeneous pairs [λ,μ] = [1, 1] and [λ,μ] = [−1, 4],i.e. the conics D1 and D2 of equations

x1(x0 − x2) = 0 and (2x1 − x2)2 = 0,

respectively. If we intersect D1 and D2, we get that the base points of the pencil areP = [1, 0, 0] and Q = [2, 1, 2].

(b) The conicD1 is simply degenerate and its irreducible components are the linesr1 = {x1 = 0} and r2 = {x0 − x2 = 0}, which meet at R = [1, 0, 1]. The conic D2

has the line L(P, Q) = {2x1 − x2 = 0} as an irreducible component of multiplicity2 and so it is doubly degenerate.

Recall (cf. Sect. 1.8.1) that every projectivity of P2(C) transforms a non-

degenerate (respectively, simply degenerate, doubly degenerate) conic into a non-degenerate (respectively, simply degenerate, doubly degenerate) conic.

Therefore, if f is a projectivity that transforms the conics of the pencil F intoconics of the same pencil, necessarily f (D1) = D1 and f (D2) = D2. In particularf must leave the line L(P, Q) invariant and must either leave the components r1, r2of D1 invariant or exchange them. In any case the point R = [1, 0, 1] where r1 andr2 meet must be invariant under f , while the points P and Q may be fixed or maybe exchanged by f .

4 Exercises on Conics and Quadrics 183

If we set S = [1, 1, 0], the points R, P, Q, S are in general position, so theirimages completely determine the projectivity f .

If f fixes the 4 points, then f is the identity map. The only other possibility is theprojectivity that fixes R, S and exchanges P and Q. One can easily check that thislatter projectivity is represented by the matrix

⎛⎝2 −1 −11 0 −12 −2 −1

⎞⎠ .

Finally note that, besides the identity map, also this second projectivity satisfiesthe properties required by the exercise: namely by construction f (D1) = D1 andf (D2) = D2, so that all the conics of the pencil generated by these two conics, i.e.of the pencil F , are transformed into conics of the same pencil.

Exercise 131. Assume that the points R, P1, P2 of P2(C) are not collinear and that

f is a projectivity of P2(C) such that f (R) = R, f (P1) = P2 and f (P2) = P1. Let

F be the pencil of conics tangent at P1 to the line r1 = L(R, P1) and tangent at P2 tothe line r2 = L(R, P2). Prove that f 2 = Id if and only if f (C) = C for every conicC ∈ F .

Solution. The pencilF is generated by the degenerate conics C1 = r1 + r2 and C2 =2L(P1, P2) and its only base points are P1 and P2 (cf. Exercise 125). The hypothe-ses immediately imply that f (r1) = r2, f (r2) = r1 and that the line L(P1, P2) isf -invariant; therefore, the degenerate conics C1 and C2 are f -invariant. Since finduces a projectivity of the space Λ2 of conics of P

2(C) (cf. Sect. 1.9.5), in par-ticular we obtain that f transforms the conics of the pencil F into conics of thesame pencil. Moreover, by the Fundamental theorem of projective transformationswe have that f (C) = C for every C ∈ F if and only if there exists another conicC3 ∈ F distinct from C1 and from C2 such that f (C3) = C3.

If f 2 = Id, the fixed-point set of f is the union of a line r and a point P /∈ r (cf.Exercise 49 and the following Note). If we had P = R, then r would intersect thelines r1 and r2 in two distinct points (each of them distinct from R) which would befixed under f , while the only fixed point lying on the two lines is R. Therefore, theline r of fixed points necessarily passes through R and, in particular, P �= R.

Let Q be a point of r \ L(P1, P2), Q �= R (cf. Fig. 4.1). Since Q is not a base pointof the pencil, there exists only one conic C3 ∈ F passing through Q; it is distinctboth from C1 and from C2 because Q /∈ C1 ∪ C2. The conic f (C3) belongs to F andpasses through f (Q) = Q; then necessarily f (C3) = C3 and so f (C) = C for anyC ∈ F .

Conversely suppose that f (C) = C for any C ∈ F . Since f (R) = R, f acts asa projectivity on the pencil of lines centred at R (which is isomorphic to P

1(C)).Since every projectivity of a complex projective line has at least one fixed point,there exists a line s passing through R which is invariant under f . By the hypothesesneither r1 nor r2 are invariant under f , so necessarily s �= r1 and s �= r2. Let C3be a non-degenerate conic of F (hence distinct from C1 and from C2) and denote

184 4 Exercises on Conics and Quadrics

s

C1

C2

C1

C2

C3

P2

P1

P2

P1

R R

r

C3 QQ2

Q1

Fig. 4.1 Left, if f 2 = Id, then f (C) = C for every C ∈ F ; right, if f (C) = C for every C ∈ F ,then f 2 = Id

by Q1, Q2 the points where C3 meets the invariant line s (cf. Fig. 4.1); one easilychecks that Q1 �= Q2. Observe that the points P1, P2, Q1, Q2 are in general position.Evidently f ({Q1, Q2}) = {Q1, Q2} and hence either f (Q1) = Q1 and f (Q2) = Q2

or f (Q1) = Q2 and f (Q2) = Q1. In both cases f 2 has P1, P2, Q1, Q2 asfixedpointsand thus f 2 = Id.

Exercise 132. Let C be a non-degenerate conic of P2(C). Let P be a point of C and

r the tangent line to C at P . Assume that R is a point of P2(C) such that R /∈ C ∪ r .

Prove that there exist exactly two projectivities f of P2(C) such that

f (C) = C, f (R) = R, f (P) = P.

Solution. Assume that f is a projectivity of P2(C) under which C, P and R are

invariant. Then eachof the two lines passing through R and tangent toC is transformedinto a line passing through R and tangent to C. If A and B are the points wherethe tangent lines to C passing through R meet the conic, then there are only twopossibilities: either f (A) = A and f (B) = B, or f (A) = B and f (B) = A.

In the first case f fixes four points in general position (that is A, B, P, R) andtherefore it is the identity map, which of course fixes the conic too (Fig. 4.2).

rA

R

B

C

P

Fig. 4.2 The configuration described in Exercise 132

4 Exercises on Conics and Quadrics 185

If f (A) = B and f (B) = A, the additional conditions f (P) = P and f (R) = Runiquely determine a projectivity f . In that case f 2 = Id, because f 2 fixes the pointsin general position A, B, P, R. Since C belongs to the pencil of conics tangent at Ato L(A, R) and tangent at B to L(B, R), Exercise 131 implies that f (C) = C.

Thus we have found exactly two projectivities with the required properties.

Exercise 133. If C is a non-degenerate conic of P2(K), let f be a projectivity of

P2(K) such that f �= Id and f (C) = C. Prove that:

(a) If f 2 = Id, then the fixed-point set of f is the union of a line r which is nottangent to C and its pole R with respect to C.

(b) If there exists a line of fixed points for f , then f 2 = Id.

Solution. (a) If f 2 = Id, the fixed-point set of f is the union of a line r and a pointP /∈ r (for a proof of this fact see Exercise 49 and the following Note).

Assume by contradiction that r is tangent to C, and let Q = r ∩ C (cf. Fig. 4.3). Forany B ∈ C \ {Q}, denote by rB the tangent to C at B. Clearly rB �= r , and so rB ∩ rconsists of only one point S. By construction, polC(S) = L(Q, B), so f ({Q, B}) ={Q, B}, since f (C) = C and f (S) = S. On the other hand, f (Q) = Q and f isinjective, so f (B) = B. Thus we have shown that f fixes all points of C, and hencef = Id, which contradicts the hypotheses. Therefore, r is not tangent to C.Set R = pol−1

C (r). Since f (r) = r and f (C) = C, it follows that f (R) = R.Moreover, since r is not tangent to C, we obtain that R /∈ r , and so necessarilyP = R.

(b) It is sufficient to consider the case K = C. Namely, if K = R, denote by Dthe complexified conic of C and by g the projectivity of P

2(C) induced by f . ClearlyD is non-degenerate, g(D) = D and, if r is a line consisting of fixed points of f , thecomplex line rC is a line of fixed points of g; moreover, the property g2 = IdP2(C)

implies that f 2 = IdP2(R).Thus assume thatK = C and that r ⊂ P

2(C) is a line of fixed points of f . Arguingas in the proof of (a), one shows that r is not tangent to C and that the pole R of

Q1

N

rB

R

S

B

Q

r

r

M

Q

s

Q2

Fig. 4.3 The configuration described in Exercise 133: on the left, the hypothesis that r is tangentto C leads to a contradiction; on the right, the solution of part (b)

186 4 Exercises on Conics and Quadrics

r with respect to C is fixed under f . Denote by M, N the points where r intersectsC and let s be any line passing through R and not tangent to C (cf. Fig. 4.3). Theline s is distinct from r , since R /∈ r , and it intersects r in a point Q which is fixedunder f . As a consequence s = L(R, Q) is f -invariant. If Q1, Q2 are the two pointswhere s intersects C, since f (C) = C we have that f ({Q1, Q2}) = {Q1, Q2}. So fmay either fix the points Q1, Q2 or exchange them: in both cases Q1, Q2 are fixedunder f 2. Since f 2 fixes the points M, N , Q1, Q2 which are in general position,then f 2 = Id.

Exercise 134. Consider the curve D of P2(C) of equation

F(x0, x1, x2) = x30 + x31 + x32 − 5x0x1x2 = 0.

Compute, if it exists, the equation of a conic tangent to the curve D at the pointsP = [1,−1, 0] and Q = [1, 2, 1] and passing through R = [1, 2,−3].Solution. First of all we can compute the tangents toD at the points P and Q. Since

∇F = (3x20 − 5x1x2, 3x21 − 5x0x2, 3x

22 − 5x0x1),

then ∇F(1,−1, 0) = (3, 3, 5) and ∇F(1, 2, 1) = (−7, 7,−7). Therefore the curveD is tangent at P to the line rP = {3x0 + 3x1 + 5x2 = 0} and at Q to the line rQ ={x0 − x1 + x2 = 0}. In order to find a conic which is tangent to D at the points Pand Q, so it is enough to find a conic which is tangent to rP at P and to rQ at Q (cf.Sect. 1.9.3).

The pencil of conics tangent at P to rP and at Q to rQ is generated by the conicsrP + rQ and 2L(P, Q); the generic conic of this pencil has therefore equation

λ(3x0 + 3x1 + 5x2)(x0 − x1 + x2) + μ(x0 + x1 − 3x2)2 = 0

as [λ,μ] varies in P1(C). The conic passes through R if and only if λ + 6μ = 0;

choosing the homogeneous pair [λ,μ] = [6,−1] we obtain the conic

17x20 − 19x21 + 21x22 − 2x0x1 + 54x0x2 − 6x1x2 = 0

which satisfies the required properties.

Exercise 135. In P2(C) consider the points P = [1,−1, 0], R = [1, 0, 0], S =

[0, 2, 1] and the conic C of equation F(x0, x1, x2) = 2x20 + 2x0x1 + 3x22 = 0. Find,if it exists, a non-degenerate conicD passing through R and S, tangent to C at P andsuch that polD(R) passes through the point [7, 3, 3].Solution. It is immediate to see that the point P belongs to the conic C; since∇F(1,−1, 0) = (2, 2, 0), the tangent to the conic C at P is the line r of equationx0 + x1 = 0.

4 Exercises on Conics and Quadrics 187

The conics passing through R and S and tangent to r at P formapencilF generatedby the degenerate conics L(P, R) + L(P, S) and L(R, S) + r (cf. Exercise 121).By easy computations we obtain that the pencilF consists of the conics of equations

λx2(x0 + x1 − 2x2) + μ(x0 + x1)(x1 − 2x2) = 0, [λ,μ] ∈ P1(C).

The generic conic Cλ,μ of F is thus represented by the matrix

Aλ,μ =⎛⎝

0 μ λ − 2μμ 2μ λ − 2μ

λ − 2μ λ − 2μ −4λ

⎞⎠ .

Since (1, 0, 0) /∈ ker Aλ,μ for any [λ,μ] ∈ P1(C), the polar of R with respect to Cλ,μ

is the line of equation μx1 + (λ − 2μ)x2 = 0; it passes through the point [7, 3, 3] ifand only if λ = μ. Thus we find the conic

x21 − 2x22 + x0x1 − x0x2 − x1x2 = 0

which is non-degenerate and therefore satisfies the requested properties.

Exercise 136. Assume that C is a non-degenerate conic of P2(C). Let P be a point

not lying on C and denote by r the polar of P with respect to C. Let s be a line passingthrough P and not tangent to C; denote by Q and R the points where s intersects C.

Show that r and s meet only at one point D, and that β(P, D, Q, R) = −1.

Solution (1). Since P /∈ C, the polar r does not pass through P; hence the lines rand s are necessarily distinct, so they meet only at one point D.

The point D cannot lie on the conic C: otherwise, since D is a point of the polarof P , the line L(D, P) = s would be tangent to the conic (cf. Sect. 1.8.2), whichcontradicts the hypotheses. Therefore, in particular D is distinct both from R andfrom Q (Fig. 4.4).

The polar polC(D) passes through P and intersects r at a point M such thatP, D, M are the vertices of a self-polar triangle for C. Then there exists a sys-tem of homogeneous coordinates in P

2(C) in which we have P = [1, 0, 0], D =[0, 1, 0], M = [0, 0, 1]. In these coordinates s has equation x2 = 0 and a matrix

associated to the conic is of the form

⎛⎝1 0 00 a 00 0 b

⎞⎠ with a and b non-zero, i.e.

x20 + ax21 + bx22 = 0 is an equation of the conic. If we compute the intersectionsbetween C and s, we find that Q = [α, 1, 0] and R = [−α, 1, 0] with α2 = −a.

In the system of coordinates x0, x1 induced on s we have P = [1, 0], D =[0, 1], Q = [α, 1], R = [−α, 1]. Therefore β(P, D, Q, R) = −1.

Solution (2). Since P /∈ C, the line polC(P) intersects C at two distinct points,say A and B. It is easy to check that A, B, P, Q are in general position, so that

188 4 Exercises on Conics and Quadrics

P

R

s

M

Q

BA r

D

Fig. 4.4 The configuration described in Exercise 136

there exists a system of coordinates of P2(C) in which we have P = [1, 0, 0], A =

[0, 1, 0], B = [0, 0, 1] and Q = [1, 1, 1]. If we impose that polC(A) = L(A, P) ={x2 = 0}, polC(B) = L(B, P) = {x1 = 0} and Q ∈ C, we easily obtain that C hasequation x20 − x1x2 = 0. Since s = L(P, Q) has equation x1 = x2, it follows thatR = [−1, 1, 1]. Moreover, since r = L(A, B) has equation x0 = 0, we get thatD = r ∩ s = [0, 1, 1]. Hence

β(P, D, Q, R) = β([1, 0, 0], [0, 1, 1], [1, 1, 1], [−1, 1, 1]) = −1.

E Solution (3). Arguing as in Solution (1), we show that there is exactly one point Dwhere r and s meet. Assume that A, B are defined as in Solution (2). The setF of theconics that are tangent at A to the line L(P, A) and tangent at B to the line L(P, B)

is a pencil having A and B as base points. Since R is distinct both from A and fromB, in the pencil there is exactly one conic passing through R: that conic is C.

Since the points P, Q, A, B and the points P, R, A, B are in general posi-tion, there exists exactly one projectivity f : P

2(C) → P2(C) such that f (P) = P ,

f (Q) = R, f (A) = A and f (B) = B. The map f leaves the lines L(P, A) andL(P, B) invariant and transforms C into a conic which is tangent to L(P, A)

at A, tangent to L(P, B) at B and passes through R. By the uniqueness prop-erty recalled above, f (C) = C. Moreover, f (s) = s, so that f (s ∩ C) = s ∩ C, thatis f ({Q, R}) = {Q, R}. As f (Q) = R, it follows that f (R) = Q. Finally, sincef (r) = r , we also have that f (D) = D. Hence

β(P, D, Q, R) = β( f (P), f (D), f (Q), f (R)) = β(P, D, R, Q) = β(P, D, Q, R)−1

(cf. Sect. 1.5.2). Then β(P, D, Q, R)2 = 1; since Q �= R, it follows that β(P, D,

Q, R) = −1.

4 Exercises on Conics and Quadrics 189

Exercise 137. Consider the following pencil of conics of P2(R):

2λx20 − (μ + λ)x21 + (μ − λ)x22 − 2μx0x1 − 2μx0x2 = 0, [λ,μ] ∈ P1(R).

(a) Find the degenerate conics and the base points of the pencil.(b) Show that there is exactly one line which is tangent to all conics of the pencil.(c) Determine all conics of the pencil that are parabolas in the affine chart U0.

Solution. (a) The generic conic Cλ,μ of the pencil is represented by the matrix

Aλ,μ =⎛⎝

2λ −μ −μ−μ −λ − μ 0−μ 0 μ − λ

⎞⎠ .

Since det Aλ,μ = 2λ3, the pencil contains exactly one degenerate conic D1 of equa-tion −x21 + x22 − 2x0x1 − 2x0x2 = 0, that is (2x0 + x1 − x2)(x1 + x2) = 0. Denoteby l1 and l2 the irreducible components of D1 having equations 2x0 + x1 − x2 = 0and x1 + x2 = 0 respectively; they meet at the point P = [1,−1, 1].

If we choose for instance the homogeneous pair [λ,μ] = [1, 0], we find the conicD2 of the pencil having equation 2x20 − x21 − x22 = 0. By intersecting the conics D1

and D2 spanning the pencil, we obtain that the base points are P = [1,−1, 1] andQ = [1, 1,−1].

(b) The irreducible conic D2 is tangent at the point P to the component l1 of D1;as a consequence, all conics of the pencil are tangent to the line l1 at P (cf. Sect. 1.9.5).

If r is a line which is tangent to all conics of the pencil, in particular r must betangent to the degenerate conic D1 = l1 + l2. Hence r must pass through P . On theother hand, every line passing through P and distinct from l1 is not tangent to D2.Therefore, the only line which is tangent to all conics of the pencil is l1.

(c) In the affine chart U0, using the affine coordinates x = x1x0 , y = x2

x0 , the conicCλ,μ ∩U0 has equation

2λ − (μ + λ)x2 + (μ − λ)y2 − 2μx − 2μy = 0.

It is a parabola if it is non-degenerate (that is λ �= 0) and its projective closureintersects the improper line only at one point. That occurs when λ2 − μ2 = 0 andtherefore for [λ,μ] = [1, 1] and [λ,μ] = [1,−1].Exercise 138. Consider the pencil of conics Cλ,μ of P

2(R) of equation

(λ + μ)x20 − μx21 − (λ + μ)x0x2 + μx1x2 = 0, [λ,μ] ∈ P1(R).

(a) Find the degenerate conics and the base points of the pencil.(b) Let r be the line of equation x2 = 0. Check that, for every P ∈ r , the pencil

contains exactly one conic Cλ,μ(P) passing through P .(c) Consider the map f : r → r defined by

190 4 Exercises on Conics and Quadrics

f (P) ={

(Cλ,μ(P) ∩ r) \ {P} if the conic Cλ,μ(P) is not tangent to rP otherwise

Check that f is a projectivity of r such that f 2 = Id.(d) Determine the homogeneous pairs [λ,μ] ∈ P

1(R) such that the affine part of theconic Cλ,μ in the affine chart U = P

2(R) \ {x0 + x2 = 0} is a parabola.Solution. (a) The generic conic of the pencil is represented by the matrix

Aλ,μ =⎛⎝2λ + 2μ 0 −λ − μ

0 −2μ μ−λ − μ μ 0

⎞⎠ .

Since det Aλ,μ = 2λμ(λ + μ), the degenerate conics of the pencil correspond to thehomogeneous pairs [0, 1], [1, 0] and [1,−1]. Therefore, they have equations

(x0 − x1)(x0 + x1 − x2) = 0, x0(x0 − x2) = 0, x1(x1 − x2) = 0.

Intersecting two of these degenerate conics, we find that the base points of the pencilare [0, 0, 1], [0, 1, 1], [1, 1, 1] and [1, 0, 1].

(b) As the line r does not contain any base point of the pencil, every point P ∈ rimposes a non-trivial linear condition on the conics of the pencil; hence there isexactly one conic of the pencil passing through P . Explicitly, if P = [y0, y1, 0],the only conic of the pencil passing through P is the conic Cλ,μ(P) defined by theequation

y21 x20 − y20 x

21 − y21 x0x2 + y20 x1x2 = 0.

(c) The points of intersection between Cλ,μ(P) and r are the points [x0, x1, 0]such that y21 x

20 − y20 x

21 = 0.We obtain only one intersection point if and only if either

y1 = 0 or y0 = 0; hence Cλ,μ(P) is not tangent to r provided that P �= P1 = [1, 0, 0]and P �= P2 = [0, 1, 0]. In that case the point of Cλ,μ(P) ∩ r distinct from P is thepoint [y0,−y1, 0].

So in the system of coordinates induced on r we have that f ([y0, y1]) =[y0,−y1] when [y0, y1] �= [1, 0] and [y0, y1] �= [0, 1]. On the other hand, by def-inition f (P1) = P1 and f (P2) = P2, so that f is represented in coordinates by[y0, y1] → [y0,−y1] on the whole line r . Thus it is evident that f is a projectiv-ity such that f 2 = Id.

(d) The affine conic Cλ,μ ∩U is a parabola when Cλ,μ is non-degenerate andCλ,μ ∩ {x0 + x2 = 0} consists of only one point (cf. Sect. 1.8.7). By intersecting Cλ,μ

with the line x0 + x2 = 0, we find the equation 2(λ + μ)x20 − μx0x1 − μx21 = 0,which admits only one solution if and only if μ(8λ + 9μ) = 0. Since the coniccorresponding to [λ,μ] = [1, 0] is degenerate, the only conic whose affine part isa parabola is the one obtained in correspondence with [λ,μ] = [9,−8], which hasequation

x20 + 8x21 − x0x2 − 8x1x2 = 0.

4 Exercises on Conics and Quadrics 191

Exercise 139. Denote by C the conic of P2(R) of equation

x20 + 2x21 + 2x0x2 − 6x1x2 + x22 = 0.

Check that C is non-degenerate and find the vertices of a self-polar triangle withrespect to C containing the line r = {x0 + x1 + x2 = 0}.Solution. The conic C is represented by the matrix

A =⎛⎝1 0 10 2 −31 −3 1

⎞⎠ ,

whose determinant is equal to −9; hence C is non-degenerate.Let P = [1,−1, 0] ∈ r \ C. Since (

1 −1 0)A = (

1 −2 4), the line polC(P)

has equation x0 − 2x1 + 4x2 = 0. Therefore, setting Q = polC(P) ∩ r , we haveQ = [−2, 1, 1], and so Q /∈ C. Finally, polC(Q) has equation x0 + x1 + 4x2 = 0,which implies that R = polC(P) ∩ polC(Q) = [4, 0,−1]. By construction, the tri-angle with vertices P, Q, R satisfies the requested property.

Exercise 140. Let F be a pencil of conics of P2(R). Prove that:

(a) F contains at least one degenerate conic.(b) F contains infinitely many non-empty conics.

Solution. (a) IfC1 andC2 are distinct conics ofF of equations tX AX = 0 and tX BX =0 respectively, the generic conic of the pencil F has equation tX (λA + μB)X = 0as [λ,μ] varies in P

1(R). Let G(λ,μ) = det(λA + μB). If G = 0, all conics of thepencil are degenerate. If G �= 0, then G is a real homogeneous polynomial of degree3, and so there exists at least one homogeneous pair [λ0,μ0] in P

1(R) such thatdet(λ0A + μ0B) = 0; so the corresponding conic of the pencil is degenerate.

(b) If all conics of F are non-empty, the statement is trivially true. So assumethat there exists at least an empty conic C1 in F and let C2 be another conic of thepencil. The conic C1 can be represented by a symmetric positive definite matrix,so that by the Spectral theorem we may choose a system of coordinates in P

2(R)

where C1 is represented by the identity matrix and C2 is represented by a diagonal

matrix

⎛⎝a 0 00 b 00 0 c

⎞⎠; without loss of generality wemay assume a ≤ b ≤ c. As C1 �= C2,

it cannot occur that a = b = c.At first consider the case b < c. The generic conic of the pencil is represented

by the matrix

⎛⎝

λ + μa 0 00 λ + μb 00 0 λ + μc

⎞⎠; if we exclude the case [λ,μ] = [1, 0],

corresponding to the empty conic C1, the conics of F \ {C1} are represented by the

192 4 Exercises on Conics and Quadrics

matrix At =⎛⎝t + a 0 00 t + b 00 0 t + c

⎞⎠ as t varies in R. If t ∈ (−c,−b) we have that

t + b < 0 and t + c > 0; therefore the symmetric matrix At is indefinite and henceevery conic of F having equation tX At X = 0 with t ∈ (−c,−b) is non-empty.

Using a similar argument we can prove the thesis when a < b.

Exercise 141. Assume that F is a pencil of conics of P2(K) containing a doubly

degenerate conic C1 and a non-degenerate conic C2. Prove that F contains at mosttwo degenerate conics.

Solution (1). First of all observe that it suffices to deal with the caseK = C. Namely,if K = R consider the pencil G of complex conics generated by (C1)C and (C2)C.Since G contains the complexifications of all conics of F , the number of degenerateconics of F is lower than or equal to the number of degenerate conics of G.

So assume K = C. By hypothesis C1 = 2r , where r is a line of P2(C). Since C2

is non-degenerate, r intersects C2 at two (possibly coincident) points A and B.If A �= B, consider the line τA tangent to C2 at A and the line τB tangent to C2

at B. The lines τA and τB are distinct and they meet only at one point M /∈ C2. Ifwe choose a point P ∈ C2 \ r , since the points M, A, B, P are in general position,there exists a system of homogeneous coordinates in P

2(C) in which we have M =[1, 0, 0], A = [0, 1, 0], B = [0, 0, 1], P = [1, 1, 1]; consequently r = L(A, B) hasequation x0 = 0. If we impose that C2 passes through A, B, P and that the polar ofM with respect to C2 is the line r = {x0 = 0}, we easily see that C2 has equationx20 − x1x2 = 0. Then the generic conic Cλ,μ of the pencil F generated by C1 and C2has equation tX Aλ,μX = 0 where

Aλ,μ = λ

⎛⎝1 0 00 0 00 0 0

⎞⎠ + μ

⎛⎝2 0 00 0 −10 −1 0

⎞⎠ =

⎛⎝

λ + 2μ 0 00 0 −μ0 −μ 0

⎞⎠ .

Since det Aλ,μ = μ2(λ + 2μ), the reducible conics of the pencil correspond to[λ,μ] = [1, 0] (which recovers the degenerate conic C1) and to [λ,μ] = [2,−1].Observe that the degenerate conic C2,−1 has equation x1x2 = 0 and hence it coin-cides with τA + τB .

If A = B, the line r is tangent to C2 at A. Choose P1 ∈ r \ {A} and two dis-tinct points P2, P3 of C2 \ r so that P1, P2, P3 are not collinear. As the pointsA, P1, P2, P3 are in general position, there exists a system of homogeneous coordi-nates in P

2(C) in which A = [1, 0, 0], P1 = [0, 1, 0], P2 = [0, 0, 1], P3 = [1, 1, 1].As a consequence, r has equation x2 = 0 and C2 is represented by a matrix of the

form

⎛⎝0 0 10 −2(1 + c) c1 c 0

⎞⎠ with c �= −1. Then the generic conic Cλ,μ of the pencil F

generated by C1 and C2 has equation tX Aλ,μX = 0 where

4 Exercises on Conics and Quadrics 193

Aλ,μ = λ

⎛⎝0 0 00 0 00 0 1

⎞⎠ + μ

⎛⎝0 0 10 −2(1 + c) c1 c 0

⎞⎠ =

⎛⎝0 0 μ0 −2μ(1 + c) μcμ μc λ

⎞⎠ .

Since det Aλ,μ = 2μ3(1 + c), the only reducible conic of the pencil is obtained for[λ,μ] = [1, 0], and so it is the conic C1 = 2r .

Solution (2). As observed in Solution (1), it suffices to deal with the case K = C.We have C1 = 2r , where r is a line of P

2(C); denote by A and B the two (possiblycoincident) points of intersection between C1 and C2.

If A �= B and if we denote by τA the line which is tangent to C2 at A and by τB theline which is tangent to C2 at B, for every conic C ofF we have that I (C, τA, A) ≥ 2and I (C, τB, B) ≥ 2. Therefore, if C is degenerate, the only possibilities are eitherC = 2r or C = τA + τB .

If A = B, then the line r is tangent to C2 at A. Moreover, the only base point ofF is A, so that, if C = l + l ′ is a degenerate conic of F , necessarily l ∩ C = {A},l ′ ∩ C = {A}. This latter fact can happen only if l = l ′ = r , and hence the onlydegenerate conic of the pencil is C1 = 2r .

Note. If K = R in both solutions of the Exercise we reduced to the complex caseby considering the pencil generated by the complexified conics of C1 = 2r and C2.

Alternatively, it is possible to adapt the two solutions to the case of a pencil ofreal conics arguing as follows. If the set r ∩ C2 is not empty, the considerationsmade in the previous solutions still hold. Therefore we need to deal only with thecase when r ∩ C2 is empty, that is the case when the complexification of r intersects(C2)C at two distinct and conjugate points, A and B = σ(A). In this case we showedin both solutions that the pencil of complex conics generated by (C1)C and (C2)Ccontains, besides (C1)C, only another non-degenerate conicQ = τA + τB , where τA(respectively τB) is the tangent to (C2)C at A (respectively at B). It follows thatσ(τA) = τσ(A) = τB , and hence σ(Q) = Q, so that, by Exercise 59, the conic Q isthe complexification of a real conic C. It is not difficult to prove, arguing for instanceas in the solution of Exercise 59, that C belongs to the pencil of real conics F .Therefore, if the set r ∩ C2 is empty, then F contains exactly two degenerate conics.

Exercise 142. (Cross-ratio of four points on a conic)

(a) Assume that A and B are two distinct points of a non-degenerate conic C ofP2(K). Denote by tA and tB the lines which are tangent to C at A and B respec-

tively, and by L(A, B) the line joining A and B. LetFA andFB be the pencils oflines with centres A and B respectively. Consider the mapψ : FA → FB definedby

(i) ψ(tA) = L(A, B);(ii) ψ(L(A, B)) = tB ;(iii) ψ(r) = L(B, Q) if r ∈ FA, r �= tA, r �= L(A, B), where Q is the inter-

section between r and C different from A (cf. Fig. 4.5). Prove that ψ is aprojective isomorphism.

194 4 Exercises on Conics and Quadrics

L(A, B)

tB

Q

Cr

A

tAB

ψ(r)

C

Fig. 4.5 If A and B are points of a conic C, then the conic C establishes a projective isomorphismbetween the pencils of lines FA and FB (cf. Exercise 142)

(b) Let P1, P2, P3, P4 be distinct points on a non-degenerate conic C of P2(K).

Define the cross-ratio of the four points as follows:

β(P1, P2, P3, P4) = β(L(A, P1), L(A, P2), L(A, P3), L(A, P4))

where A is a point of C different from each Pi . Prove that β(P1, P2, P3, P4) doesnot depend on the choice of A.

Solution. (a) Since the conic is non-degenerate, the lines tA and tB are distinct andhence they meet at a point C which does not lie on the conic. If R is a point on Cdifferent both from A and from B, it is easy to check that {A, B,C, R} is a projectiveframe (a real non-degenerate conic which is not empty contains infinitely manypoints, so that R exists also in the real case). In the induced system of homogeneouscoordinates we have A = [1, 0, 0], B = [0, 1, 0], C = [0, 0, 1], R = [1, 1, 1] andthe lines tA and tB have equations x1 = 0 and x0 = 0 respectively. If we impose thatC passes through [1, 1, 1] and has the lines x1 = 0 and x0 = 0 as polars at [1, 0, 0] and[0, 1, 0] respectively, we get immediately that the conic is represented by a matrix

of the form M =⎛⎝0 b 0b 0 00 0 −2b

⎞⎠ with b �= 0, so that C has equation x22 − x0x1 = 0.

Moreover, FA consists of the lines of equations a1x1 + a2x2 = 0 with [a1, a2] ∈P1(K), so that a1, a2 is a system of homogeneous coordinates for FA. Similarly, FB

consists of the lines of equations b0x0 + b2x2 = 0 with [b0, b2] ∈ P1(K), and b0, b2

is a system of homogeneous coordinates for FB .Let r be a line in FA, with r �= tA, r �= L(A, B). Then r has equation a1x1 +

a2x2 = 0 with a1 �= 0 and a2 �= 0. If we compute the intersections between r and Cwe find, besides the point A, the point Q = [a21, a22,−a1a2]. Then the line L(B, Q)

has equationa21x2 + a1a2x0 = 0; sincea1 �= 0, the line has equationa1x2 + a2x0 = 0and hence it has coordinates [a2, a1].

It follows that the restriction of the map ψ to FA \ {tA, L(A, B)} coincides withthe restriction toFA \ {tA, L(A, B)} of themap f : FA → FB defined in coordinatesby f ([a1, a2]) = [a2, a1], which is evidently a projective isomorphism.

4 Exercises on Conics and Quadrics 195

P2

P3

C

S

P1

C

Q2

Q3

Q1

T

f

Fig. 4.6 The construction of the projectivity required in Exercise 143 (a)

Observe that, in the system of coordinates chosen in FA, the line tA hascoordinates [1, 0] and L(A, B) has coordinates [0, 1]. Similarly, in the system ofcoordinates chosen in FB , L(A, B) has coordinates [0, 1] and tB has coordinates[1, 0]. Since f (tA) = f ([1, 0]) = [0, 1] = L(A, B) = ψ(tA) and f (L(A, B)) =f ([0, 1]) = [1, 0] = tB = ψ(L(A, B)), then ψ coincides with f and hence it is aprojective isomorphism.

(b) Let B be a point of C distinct from A and from each Pi . Then themapψ : FA →FB considered in (a) is a projective isomorphism such that ψ(L(A, Pi )) = L(B, Pi )for i = 1, . . . , 4. As the cross-ratio is invariant under projective isomorphisms,we have thatβ(L(A, P1), L(A, P2), L(A, P3), L(A, P4)) = β(L(B, P1), L(B, P2),L(B, P3), L(B, P4)) which implies the thesis.

Exercise 143. Consider two non-degenerate conics C and C ′ in P2(K). Let P1, P2,

P3, P4 be distinct points of C and Q1, Q2, Q3, Q4 distinct points of C ′.

(a) Show that there is exactly one projectivity f : P2(K) → P

2(K) such that f (C) =C ′ and f (Pi ) = Qi for i = 1, 2, 3.

(b) Show that there exists a projectivity f of P2(K) such that f (C) = C ′ and

f (Pi ) = Qi for i = 1, 2, 3, 4 if and only if β(P1, P2, P3, P4) =β(Q1, Q2, Q3, Q4), where β denotes the cross-ratio of four points on a conicdefined in Exercise 142.

Solution. (a) Let S = polC(P1) ∩ polC(P2), T = polC′(Q1) ∩ polC′(Q2). It is imme-diate to check that the quadruple P1, P2, P3, S forms a projective frame of P

2(C);the same holds for the quadruple Q1, Q2, Q3, T . Then, by the Fundamental the-orem of projective transformations, it is sufficient to prove that a proiectivityf : P

2(K) → P2(K) fulfils the required properties if and only if f (S) = T and

f (Pi ) = Qi for i = 1, 2, 3 (Fig. 4.6).If f satisfies the conditions described in the statement, then evidently f (Pi ) = Qi

for i = 1, 2, 3. Moreover, since projectivities preserve tangency conditions, we havethat f (polC(Pi )) = polC′(Qi ), and hence

f (S) = f (polC(P1) ∩ polC(P2)) = polC′(Q1) ∩ polC′(Q2) = T .

196 4 Exercises on Conics and Quadrics

Conversely assume that f (S) = T and f (Pi ) = Qi for i = 1, 2, 3. Then, fori = 1, 2 we have

pol f (C)(Qi ) = f (polC(Pi )) = f (L(Pi , S)) = L(Qi , T ) = polC′(Qi ),

so that both conics f (C) and C ′ pass through Q1 with tangent polC′(Q1), through Q2

with tangent polC′(Q2) and through Q3.Since Q1, Q2, Q3, T are in general position, there exists exactly one conic passing

through Q1 with tangent polC′(Q1), through Q2 with tangent polC′(Q2) and throughQ3 (cf. Exercise 125). This implies that f (C) = C ′, which proves the result.

(b) First of all assume there exists a projectivity f : P2(K) → P

2(K) such thatf (C) = C ′ and f (Pi ) = Qi for i = 1, 2, 3, 4. If O is any point of C distinct fromP1, P2, P3, P4 we have that f (O) = O ′ ∈ C ′ \ {Q1, Q2, Q3, Q4}. Furthermore, finduces a projectivity from the pencil of lines centred at O onto the pencil of linescentred at O ′, and this projectivity transforms ri = L(O, Pi ) into si = L(O ′, Qi )

for each i = 1, 2, 3, 4. Since the cross-ratio is invariant under projective trans-formations, we have β(r1, r2, r3, r4) = β(s1, s2, s3, s4), that is β(P1, P2, P3, P4) =β(Q1, Q2, Q3, Q4) by the definition of cross-ratio of points on a conic.

Conversely, assume that β(P1, P2, P3, P4) = β(Q1, Q2, Q3, Q4). As shown inpart (a), there exists a projectivity f : P

2(K) → P2(K) such that f (C) = C ′ and

f (Pi ) = Qi for i = 1, 2, 3.We are going to show that f (P4) = Q4, which concludesthe proof. As above let O ∈ C \ {P1, P2, P3, P4}, O ′ = f (O) ∈ C ′ and, for eachi = 1, 2, 3, 4, ri = L(O, Pi ), si = L(O ′, Qi ). As above, by the invariance of cross-ratio under projective transformations, we have

β(r1, r2, r3, r4) = β( f (r1), f (r2), f (r3), f (r4)) = β(s1, s2, s3, L(O ′, f (P4))).

Moreover, the hypothesis implies that β(r1, r2, r3, r4) = β(s1, s2, s3, s4), so thatL(O ′, f (P4)) = s4 = L(O ′, Q4). Hence

f ({O, P4}) = f (C ∩ L(O, P4)) = C′ ∩ L(O ′, f (P4)) = C′ ∩ L(O ′, Q4) = {O ′, Q4}.

As f (O) = O ′, it follows that f (P4) = Q4, and hence the thesis.

Exercise 144. Assume that A, B,C, D are points in general position in P2(K). Let

f : P2(K) → P

2(K) be the projectivity such that f (A) = B, f (B) = A, f (C) = Dand f (D) = C . Show that f (Q) = Q for every conic Q of P

2(K) passing throughthe points A, B,C, D.

Solution (1). The conics passing through the points A, B,C, D form a pencil Fcontaining 3 degenerate conics, that is C1 = L(A, B) + L(C, D), C2 = L(A, D) +L(B,C) and C3 = L(A,C) + L(B, D) (cf. Sect. 1.9.7). The hypotheses immedi-ately imply that each of the previous conics is invariant under f . As a consequencef acts on the pencil (which is projectively isomorphic to P

1(K)) as a projectivity

4 Exercises on Conics and Quadrics 197

having 3 fixed points and hence as the identity map; therefore, each conic passingthrough A, B,C, D is invariant under f .

E Solution (2). A different solution can be given by exploiting the notion of cross-ratio of four points on a conic (cf. Esercizio 142). As observed in Solution (1), thedegenerate conics C1, C2, C3 passing through A, B,C, D are evidently f -invariant,so it is sufficient to deal with the case whenQ is non-degenerate. If we denote by βQthe cross-ratio of four points on Q, using the symmetries of the usual cross-ratio ofpoints on a line it is easy to show that βQ(A, B,C, D) = βQ(B, A, D,C); therefore,using the result of Exercise 143 (b), there exists a projectivity g : P

2(K) → P2(K)

such that g(A) = B, g(B) = A, g(C) = D, g(D) = C and g(Q) = Q. Then theprojectivities g and f , which coincide on A, B,C, D, must coincide on the wholeP2(K), so that f (Q) = g(Q) = Q, as required.

K Exercise 145. Let C be a non-degenerate conic of P2(C).

(a) Denote by P, Q, R ∈ P2(C) the vertices of a self-polar triangle for C. Prove

that there exists a projectivity f of P2(C) such that f (C) = C, f (P) = Q,

f (Q) = P and f (R) = R.(b) Given two distinct points P, Q /∈ C such that the line joining them is not tangent

to C, prove that there exists a projectivity f of P2(C) such that f (C) = C and

f (P) = Q.

Solution. (a)Denote by A and B the pointswhere the line L(P, R) = pol(Q)meetsC(cf. Fig. 4.7). Since the vertices of a self-polar triangle do not belong to the conic andits edges are not tangent to the conic, the points A and B are distinct and differentboth from P and from R. Similarly denote by D and E the points where the lineL(Q, R) = pol(P) meets C; these points are distinct and different both from R andfrom Q.

Consider the projectivity f ofP2(C) such that f (A) = D, f (D) = A, f (B) = E

and f (E) = B (it exists because the points A, B, D, E turn out to be in generalposition). Then by Exercise 144 we have that f (C) = C.

D

A

R

BP

Q

E

C

Fig. 4.7 The construction described in the solution of Exercise 145 (a)

198 4 Exercises on Conics and Quadrics

Moreover, f (pol(Q)) = pol(P), f (pol(P)) = pol(Q), so that f (Q) = P andf (P) = Q. It follows that f (pol(R)) = pol(R) and hence f (R) = R.An alternative proof of statement (a) can be given as follows. Since P, Q, R

are in general position, there exists a projective frame of P2(C) having P, Q, R as

fundamental points. With respect to the homogeneous coordinates defined by this

frame, the conic C is represented by the matrix M =⎛⎝1 0 00 a 00 0 b

⎞⎠ with a, b ∈ C

∗.

Moreover, a projectivity f : P2(C) → P

2(C) satisfies the conditions f (P) = Q,f (Q) = P , f (R) = R if and only if it is represented by a matrix N such that N =⎛⎝0 c 01 0 00 0 d

⎞⎠ with c, d ∈ C

∗.

Observe that f (C) = C if and only if f −1(C) = C, i.e. if and only if tNMN = λMfor some λ ∈ C

∗. An easy computation shows that the latter condition is fulfilled ifand only if a2 = c2 and bd2 = ab. It follows that, if α ∈ C

∗ is a square root of a,

the projectivity associated to the invertible matrix

⎛⎝0 a 01 0 00 0 α

⎞⎠ satisfies the required

properties.(b) Consider first the case when Q ∈ polC(P). Then by reciprocity P ∈ polC(Q);

in addition, the lines polC(P) and polC(Q) are distinct, so that they meet at one pointR. The points P, Q, R turn out to be the vertices of a self-polar triangle for C; thenthe thesis follows from part (a).

If instead Q /∈ polC(P), let T = polC(P) ∩ polC(Q). Since T ∈ polC(P) andT /∈ C (which happens because L(P, Q) is not tangent to C), the points P and T aretwo vertices of a self-polar triangle for C. Then, as just proved, there exists a projec-tivity g of P

2(C) such that g(C) = C and g(P) = T . Similarly, since T ∈ polC(Q),there exists a projectivity h of P

2(C) such that h(C) = C and h(Q) = T . Then theprojectivity f = h−1 ◦ g fulfils the required properties (Fig. 4.8).

Note. Using Exercise 165, it is possible to show that the thesis of part (b) holds alsowhen the line passing through P and Q is tangent to C.

P

T

C

Q

Fig. 4.8 Exercise 145: how to derive part (b) from part (a) when Q /∈ polC(P)

4 Exercises on Conics and Quadrics 199

Exercise 146. Consider in P2(C) the points A = [1, 0, 0], B = [0, 1,−1] and C =

[1, 2,−3]. Find, if it exists, a non-degenerate conic C of P2(C) such that

(i) A, B,C are the vertices of a self-polar triangle for C;(ii) C passes through the point P = [1, 1, 1] and it is tangent at P to the line of

equation 3x0 − 4x1 + x2 = 0.

Solution. The conic C we are looking for has equation tXMX = 0 with

M =⎛⎝a b cb d ec e f

⎞⎠ .

The polar of the point Awith respect to C has equation ax0 + bx1 + cx2 = 0; in orderfor condition (i) to hold, necessarily this polar must coincide with the line L(B,C)

which has equation x0 + x1 + x2 = 0. Thus we find the first condition a = b = c,

so that M =⎛⎝a a aa d ea e f

⎞⎠ .

We now find further necessary conditions that a, d, e, f must fulfil if C satisfiesthe properties required in the statement of the exercise.

The polar of the pointC with respect to the conic has equation (a + 2d − 3e)x1 +(a + 2e − 3 f )x2 = 0. As the polar of A passes through C , by reciprocity the polarof C passes through A; in order to impose that polC(C) is the line L(A, B) it issufficient to impose that it passes through the point B. Thus we find the condition2d − 5e + 3 f = 0. By imposing these conditions we get, again by reciprocity, thatthe polar of B is the line L(A,C).

The conic C passes through P if and only if 5a + d + 2e + f = 0 and, if C isnon-degenerate, polC(P) coincides with the only tangent to C at P and it has equation3ax0 + (a + d + e)x1 + (a + e + f )x2 = 0. In order for this line to coincide withthe line r of equation 3x0 − 4x1 + x2 = 0, it is sufficient to impose that polC(P)

passes through another point of r distinct from P , for instance through [1, 0,−3].Thus we find the additional condition e + f = 0.

The linear system having as equations the conditions found above has as solutionsall multiples of the quadruple a = 1, d = −4, e = −1, f = 1. The conic associated

to the matrix M =⎛⎝1 1 11 −4 −11 −1 1

⎞⎠ is non-degenerate and satisfies all the required

conditions.

Exercise 147. Let C and D be two non-degenerate conics of P2(K) which meet at

4 distinct points.

(a) Show that no line in P2(K) is tangent to all conics of the pencil generated by C

and D.(b) Show that there exists a point Q ∈ P

2(K) such that polC(Q) = polD(Q) andthat the point Q lies neither on C nor on D.

200 4 Exercises on Conics and Quadrics

Solution. (a) Denote by A, B,C, D the four points where the conics C and Dmeet. Since the two conics are non-degenerate, the points A, B,C, D are in gen-eral position. The pencil F generated by C and D contains three degenerate conics(cf. Sect. 1.9.7), that is the conics

A1 = L(A, B) + L(C, D), A2 = L(A, D) + L(B,C), A3 = L(A,C) + L(B, D).

Assume by contradiction that there exists a line r tangent to all conics ofF . Since inparticular r is tangent to the degenerate conic A1, necessarily r must pass throughthe point M1 = L(A, B) ∩ L(C, D).

The line r is tangent also to the two degenerate conics A2 and A3. Hence rmust pass also through the points M2 = L(A, D) ∩ L(B,C) and M3 = L(A,C) ∩L(B, D). In particular the points M1, M2, M3 are collinear, which contradicts thefact that A, B,C, D are in general position (cf. Exercise 6). Thus (a) is proved.

(b) Recall that the map polC : P2(K) → P

2(K)∗ which associates to P ∈ P2(K)

the polar of P with respect to the non-degenerate conic C is a projective isomorphism(cf. Sect. 1.8.2). Therefore, the composite map pol−1

D ◦ polC is a projectivity ofP2(K)

and so it has at least a fixed point Q (cf. Sect. 1.2.5), that is a point such thatpolC(Q) = polD(Q) as required.

Assume by contradiction that Q belongs to C. Then the line polC(Q) is tangentto C at Q. Since polC(Q) = polD(Q), the point Q belongs also to the line polD(Q),so that Q lies also on the conic D and the line polD(Q) is tangent also to D at thepoint Q which belongs to both conics. Then the line is tangent to all conics of thepencil F , which contradicts part (a).

Exercise 148. Let P1, P2, P3, P4 be points of P2(K) in general position. Given

points O, O ′ in P2(K) such that P1, P2, P3, P4, O and P1, P2, P3, P4, O ′ are in gen-

eral position, let λ (resp. λ′) denote the cross-ratio of the four lines passing throughO (resp. O ′) and through P1, P2, P3, P4. Show that λ = λ′ if and only if there existsa conic passing through P1, P2, P3, P4, O, O ′.

Solution (1). For i = 1, 2, 3, 4 let ti = L(O, Pi ) and si = L(O ′, Pi ).Assume there exists a conic C passing through the points P1, P2, P3, P4, O, O ′;

necessarily this conic is non-degenerate. Then

λ = β(t1, t2, t3, t4) = β(s1, s2, s3, s4) = λ′

as an immediate consequence of Exercise 142 (b).Conversely suppose λ = λ′.We will prove that the points P1, P2, P3, P4, O, O ′ lie on a conic by showing ana-

lytically that the locusW of points Q such that P1, P2, P3, P4, Q are in general posi-tion and such that β(L(Q, P1), L(Q, P2), L(Q, P3), L(Q, P4)) = λ is contained ina conic passing through P1, P2, P3, P4.

4 Exercises on Conics and Quadrics 201

So let Q ∈ W . Since the points Pi are in general position, there exists a systemof homogeneous coordinates in which

P1 = [1, 0, 0], P2 = [0, 1, 0], P3 = [0, 0, 1], P4 = [1, 1, 1].

Then the line r = L(P1, P2) has equation x2 = 0; since Q /∈ r we have that Q =[y0, y1, y2] with y2 �= 0. We can compute the cross-ratio of the four lines L(Q, Pi )as the cross-ratio of their intersections with the transversal line r . We easily computethat

R3 = L(Q, P3) ∩ r = [y0, y1, 0] and R4 = L(Q, P4) ∩ r = [y2 − y0, y2 − y1, 0].

The hypothesis that the points P1, P2, P3, P4 are in general position implies that thepoints P1, P2, R3, R4 are distinct (and hence their cross-ratio is well defined and y2 −y0 �= 0, y2 − y1 �= 0). Therefore β(L(Q, P1), L(Q, P2), L(Q, P3), L(Q, P4)) =β(P1, P2, R3, R4)=β([1, 0], [0, 1], [y0, y1], [y2 − y0, y2 − y1])= y2 − y1

y1y0

y2 − y0.

As Q ∈ W , then y2 − y1y1

y0y2 − y0

= λ, that is Q belongs to the conic of equation

(x2 − x1)x0 − λx1(x2 − x0) = 0; one can immediately check that this conic passesthrough P1, P2, P3, P4.

E Solution (2). Like in Solution (1) one shows that, if the points P1, P2, P3, P4, O andO ′ lie on a conic C, then λ = λ′.

The opposite implication can be proved synthetically as follows. Let C be the(unique andnon-degenerate) conic passing through P1, P2, P3, P4, O and letC ′ be the(unique and non-degenerate) conic passing through P1, P2, P3, P4, O ′ (cf. Fig. 4.9).

Letψ : FO → FP1 denote the projective isomorphism defined in Exercise 142 (a),starting from the conic C and the points O and P1. If we denote by TP1(C) the tangent

P3

C

C

O

P1P2

O

P4

TP1 (C)

TP1 (C )

Fig. 4.9 Solution (2) of Exercise 148

202 4 Exercises on Conics and Quadrics

to C at the point P1, then

ψ(t1) = TP1(C) and ψ(t j ) = L(P1, Pj ) for j = 2, 3, 4.

Since the cross-ratio is invariant under projective isomorphisms, we have that

λ = β(TP1(C), L(P1, P2), L(P1, P3), L(P1, P4)).

Similarly, let ψ′ : FO ′ → FP1 denote the projective isomorphism defined in Exercise142 (a), starting from the conic C ′ and the points O ′ and P1. If we denote by TP1(C ′)the tangent to C ′ at the point P1, then

ψ′(s1) = TP1(C ′) and ψ′(s j ) = L(P1, Pj ) for j = 2, 3, 4

and henceλ′ = β(TP1(C ′), L(P1, P2), L(P1, P3), L(P1, P4)).

Since the lines L(P1, P2), L(P1, P3), L(P1, P4) are distinct, the hypothesis λ = λ′implies that TP1(C) = TP1(C ′), i.e. the conics C and C ′ have the same tangent r at P1.Since there exists only one conic passing through P1, P2, P3, P4 and with tangent rat P1, necessarily C = C ′.

E Solution (3). Like in Solution (1) one shows that, if the points P1, P2, P3, P4, O andO ′ lie on a conic C, then λ = λ′.

We are nowgoing to prove the opposite implication. Evidently ifO = O ′ the resultis trivially true. If O �= O ′, observe that the line L(O, O ′) cannot pass through anyof the Pi . Namely, suppose by contradiction that L(O, O ′) passes through at leastone of those points, for instance P1, and set r = L(P2, P3) and R = r ∩ L(O, O ′).The hypotheses immediately imply that R /∈ {P1, P2, P3, O, O ′}. If we compute λby using the transversal r , we have that

λ = β(R, P2, P3, L(O, P4) ∩ r).

Similarlyλ′ = β(R, P2, P3, L(O ′, P4) ∩ r).

From the hypothesis λ = λ′ it follows that L(O, P4) ∩ r = L(O ′, P4) ∩ r , that isO, O ′, P4 are collinear and hence also O, O ′, P1, P4 are collinear, which contradictsour hypotheses. Therefore, P1, P2, P3, P4, O, O ′ are in general position.

Denote by C the (unique and non-degenerate) conic passing through P1, P2, P3,O, O ′ (cf. Fig. 4.10) and let us prove that, if λ = λ′, then P4 ∈ C too, which provesthe thesis.

4 Exercises on Conics and Quadrics 203

O

P3

P4

s1s2

t4

C

P1

t1

t2

s3

O

P2

t3

s4

Fig. 4.10 Solution (3) of Exercise 148

Letψ : FO → FO ′ denote the projective isomorphism defined in Exercise 142 (a),starting from the conic C and the points O and O ′. Therefore, ψ(ti ) = si for i =1, 2, 3. Since the cross-ratio is invariant under projective isomorphisms, we have

β(t1, t2, t3, t4) = β(s1, s2, s3,ψ(t4)).

From the hypothesis λ = λ′, i.e. β(t1, t2, t3, t4) = β(s1, s2, s3, s4), it follows thatψ(t4) = s4.

By definition, t4 ∩ ψ(t4) is a point and belongs to C. But P4 ∈ t4 ∩ s4 = t4 ∩ ψ(t4)and hence P4 ∈ C.Exercise 149. Consider the pencil F of conics of P

2(C) of equations

λx1(x0 − x1 + x2) + μx2(2x0 − x1) = 0, [λ,μ] ∈ P1(C).

Let P = [1, 3, 1] and let r be the line x0 − x1 = 0. Check that the map f : F → rwhich associates to C ∈ F the point polC(P) ∩ r is well defined and it is a projectiveisomorphism.

Solution. The conic Cλ,μ of the pencil is represented by the matrix

Aλ,μ =⎛⎝

0 λ 2μλ −2λ λ − μ2μ λ − μ 0

⎞⎠ .

Since for every [λ,μ] inP1(C) the vector Aλ,μ

⎛⎝131

⎞⎠ =

⎛⎝

3λ + 2μ−4λ − μ3λ − μ

⎞⎠ is non-zero

(in other words P is non-singular for all conics in the pencil), then polCλ,μ(P) is a

line of equation tP Aλ,μX = 0 (cf. Sect. 1.8.2). Moreover, polCλ,μ(P) �= r for every

[λ,μ] ∈ P1(C), so that the map f is well defined.

204 4 Exercises on Conics and Quadrics

Easy computations show that the lines polCλ,μ(P) and r meet at the point f (P) =

[3λ − μ, 3λ − μ,λ − μ]. With respect to the homogeneous coordinates λ,μ on Fand x0, x2 on r , the map is thus represented by f ([λ,μ]) = [3λ − μ,λ − μ]. Asdet

(3 −11 −1

)�= 0, f is a projective isomorphism.

Exercise 150. Let C be a non-degenerate conic of P2(C); assume that r is a line not

tangent to C and s is a line not passing through the pole R of r with respect to C.(a) Check that the map f : s → r which associates to every point P ∈ s the point

polC(P) ∩ r is well defined and is a projective isomorphism.(b) If s = r , prove that f is a non-trivial involution and describe its fixed points.

Solution. (a) As R /∈ s, for every P ∈ s we have that P �= R and hence polC(P) �=polC(R) = r ; therefore the lines polC(P) and r meet only at one point and thus f iswell defined.

The map that associates to every point P ∈ s the line polC(P) is the restriction tos of the projective isomorphism polC : P

2(C) → P2(C)∗ (cf. Sect. 1.8.2). Then this

restriction is a projective isomorphism between s and a 1-dimensional subspace ofP2(C)∗, that is a pencil F of lines of P

2(C). The fact that f is well defined ensuresthat the line r does not belong to the pencil. Thus the map f coincides with thecomposition of the projective isomorphism polC |s : s → F with the parametrizationof F by means of the transversal r . Since this parametrization is an isomorphism(cf. Exercise 32 and the following Note), f is a projective isomorphism too.

(b) If s = r , for every P ∈ r we have that f (P) ∈ polC(P) and hence, by reci-procity, P ∈ polC( f (P)). As a consequence f ( f (P)) = polC( f (P)) ∩ r = P andso f is an involution. Furthermore, f (P) = P if and only if the line polC(P) passesthrough P . This latter fact occurs if and only if P lies on the conic. Then the fixed-point set of f is r ∩ C and hence, since r is not tangent to C, either it is empty or itcontains exactly two points. In any case f is different from the identity map.

Exercise 151. Assume that r ⊆ P2(C) is a projective line and f : r → r is a non-

trivial involution. Show that there exists a non-degenerate conic C not tangent to rand such that f (P) = polC(P) ∩ r for every P ∈ r .

Solution. The map f , being a non-trivial involution, has two fixed points A, B(cf. Exercise 23). Let C be any non-degenerate conic of P

2(C) passing through Aand through B; in particular this conic is not tangent to r = L(A, B). Denote byg : r → r the map which associates to every point P ∈ r the point polC(P) ∩ r ; asshown in Exercise 150, g is a non-trivial involution of r which fixes A and B. Sincethere exists exactly one non-trivial involution of r having A and B as fixed points(cf. Exercise 24), then f = g and hence the thesis.

Exercise 152. Assume that C is a non-degenerate conic of P2(C) and P ∈ P

2(C) isa point not lying on C. Let f : P

2(C) → P2(C) be a projectivity such that f (s ∩ C) =

s ∩ C for every line s passing through P . Prove that P is a fixed point of f and thatthe restriction of f to the polar of P with respect to C is the identity map.

4 Exercises on Conics and Quadrics 205

Solution. Denote by r1, r2 the two tangents to C passing through P; they meet theconic at the points A and B, respectively. Any line s passing through P and nottangent to C intersects the conic at two distinct points; since by hypothesis these twopoints either are fixed for f or are exchanged by f , the line s joining them is invariantunder f . Therefore, if s1, s2 are two lines passing through P and not tangent to C,we have f (s1) = s1, f (s2) = s2 and hence f (P) = f (s1 ∩ s2) = s1 ∩ s2 = P .

If we apply the hypothesis to the lines r1, r2, we get that also the points of tangencyA and B are fixed by f ; in particular the line polC(P) = L(A, B) is f -invariant. IfR is any point on polC(P) distinct both from A and from B, we have R = polC(P) ∩L(P, R); since both polC(P) and L(P, R) are f -invariant, then f (R) = R.

Exercise 153. Assume that C is a non-degenerate conic of P2(K). Let P1, P2, P3,

P4 be distinct points of C and for any i �= j let si j = L(Pi , Pj ). Show that the pointsA = s12 ∩ s34, B = s13 ∩ s24,C = s14 ∩ s23 are vertices of a self-polar triangle for C.

E Solution (1). Since A, B and C can be interchanged in the statement, it suffices toprove that polC(A) = L(B,C).

SinceC is non-degenerate, the distinct points P1, P2, P3, P4 are in general position.So let f be the only projectivity of P

2(C) such that

f (P1) = P2, f (P2) = P1, f (P3) = P4, f (P4) = P3.

The lines L(P1, P2) and L(P3, P4) turn out to be f -invariant, so that A is a fixed pointof f (cf. Fig. 4.11). Moreover, f (B) = f (L(P1, P3) ∩ L(P2, P4)) = L(P2, P4) ∩L(P1, P3) = B, that is also B is fixed by f . Arguing in a similar way one shows thatf (C) = C too. As a consequence the line L(B,C) is invariant under f .More precisely we can see that all points of L(B,C) are fixed. Namely the

points M = L(B,C) ∩ L(P1, P2) and N = L(B,C) ∩ L(P3, P4), obtained as inter-sections of invariant lines, are fixed points for f . It can be easily seen that the points

P1

P2

P3

C

A

N

C

B

P4

M

Fig. 4.11 The points A, B,C are vertices of a self-polar triangle for C

206 4 Exercises on Conics and Quadrics

B,C, M, N are distinct, so that the restriction of f to L(B,C), which is a projectivityof this line having four fixed points, is the identity map of L(B,C).

Since the points P1, P2, P3, P4 are in general position and fixed for f 2, thenf 2 = Id. Moreover, by Exercise 144, f (C) = C. Hence, by Exercise 133 (a), thefixed-point set of f is the union of a line not tangent to C and the pole of that line.It follows that the fixed point A is the pole of the line L(B,C) consisting of fixedpoints, i.e. polC(A) = L(B,C), as required.

E Solution (2). As observed at the beginning of Solution (1), it is sufficient to provethat polC(A) = L(B,C).

The conic C belongs to the pencil generated by the degenerate conicsD1 = s12 +s34 and D2 = s13 + s24. If tXM1X = 0 is an equation of D1 and tXM2X = 0 isan equation of D2, then there exist λ,μ ∈ C (not simultaneously zero) such thattX (λM1 + μM2)X = 0 is an equation of C. As A is singular for D1, then M1A = 0;similarly, since B is singular for D2, then M2B = 0. As a consequence tA(λM1 +μM2)B = λ tAM1B + μ tAM2B = 0, which proves that B ∈ polC(A). Arguing in asimilar way for the conics D1 and D3 = s14 + s23 (which has C as a singular point),one shows that C ∈ polC(A) and hence polC(A) = L(B,C).

Solution (3). Since C is non-degenerate, the points P1, P2, P3, P4 are in gen-eral position and hence they form a projective frame. In the induced system ofhomogeneous coordinates C is defined by an equation of the form x0x1 + ax1x2 −(1 + a)x0x2 = 0, with a ∈ K \ {0,−1}, and thus it is represented by the matrix⎛⎝

0 1 −1 − a1 0 a

−1 − a a 0

⎞⎠. Moreover, A = [1, 1, 0], B = [1, 0, 1] and C = [0, 1, 1].

Now it is immediate to check that the points A, B and C are pairwise conjugate withrespect to C, i.e. they form a self-polar triangle.

Exercise 154. Let A be a point not lying on a non-degenerate conic C of P2(K).

Consider two distinct lines r1 and r2 passing through A and secant to C. DenoteC ∩ r1 = {P1, P2} and C ∩ r2 = {P3, P4}. If S1 is the pole of L(P1, P3) and S2 is thepole of L(P2, P4) with respect to C, show that A, S1, S2 are collinear.

E Solution (1). Let B = L(P1, P3) ∩ L(P2, P4). By Exercise 153 (cf. Fig. 4.11), thepoints A and B are two vertices of a self-polar triangle for C; hence A ∈ polC(B).

Since B ∈ L(P1, P3), by reciprocity S1 ∈ polC(B); similarly S2 ∈ polC(B). ThenpolC(B) = L(S1, S2). Having already proved that also A belongs to polC(B) =L(S1, S2), then A, S1, S2 are collinear.

Solution (2). Since C is non-degenerate, the points P1, P2, P3, P4 are in general posi-tion and hence they form a projective frame. In the induced system of homogeneouscoordinates C is defined by an equation of the form x0x1 + ax1x2 − (1 + a)x0x2 = 0,

with a ∈ K \ {0,−1} and so it is represented by thematrix

⎛⎝

0 1 −1 − a1 0 a

−1 − a a 0

⎞⎠.

4 Exercises on Conics and Quadrics 207

In addition, L(P1, P3) has equation x1 = 0, L(P2, P4) has equation x0 − x2 = 0and A = [1, 1, 0]. Easy computations yield that S1 = [a, 1 + a, 1] and S2 = [a, a −1,−1]; since A, S1 and S2 belong to the line of equation x0 − x1 + x2 = 0, they arecollinear.

Exercise 155. Given a pencil of conics F of P2(C) containing at least one non-

degenerate conic, show that the following conditions are equivalent:

(i) there exist Q1, Q2, Q3 ∈ P2(C) that form a self-polar triangle for every conic

of F ;(ii) the base-point set of F consists of either four distinct points or two points each

counted twice (cf. Sect. 1.9.7).

Solution. Firstly observe that two points P, Q ∈ P2(C) are conjugate with respect

to every conic C ∈ F if and only if they are conjugate with respect to two distinctconics C1 and C2 of F . Namely, if Ai , i = 1, 2, is a symmetric matrix representingCi , the points P and Q are conjugate with respect to Ci if and only if tP Ai Q = 0 fori = 1, 2, and hence if and only if tP(λA1 + μA2)Q = 0 for every [λ,μ] ∈ P

1(C).As a consequence three points of the plane are the vertices of a self-polar triangle

for every conic of F if and only if they verify the same property for at least twoconics of the pencil.

Let us now show that condition (ii) implies condition (i), by examining separatelythe case when F has four distinct base points and the case when F has two basepoints each counted twice.

If F has four distinct base points P1, P2, P3, P4, then by Exercise 153 the pointsQ1 = L(P1, P2) ∩ L(P3, P4), Q2 = L(P1, P3) ∩ L(P2, P4) and Q3 = L(P1, P4) ∩L(P2, P3) are the vertices of a self-polar triangle for every non-degenerate conic ofF and so for infinitely many conics ofF . Then, because of the previous observation,Q1, Q2, Q3 are the vertices of a self-polar triangle for every conic of F .

If F has two base points P1 and P2 each counted twice, denote by C1 ∈ F thesingular conic 2L(P1, P2) and by C2 ∈ F any non-degenerate conic.

Let Q1 be the pole of the line L(P1, P2) with respect to C2. As L(P1, P2)intersects C2 at the two distinct points P1 and P2, the point Q1 does not belongto C2. Then (cf. Sect. 1.8.4 or Exercise 170) there exist Q2, Q3 ∈ P

2(C) suchthat Q1, Q2, Q3 are the vertices of a self-polar triangle for C2. By reciprocityQ2, Q3 ∈ polC2

(Q1) = L(P1, P2). Therefore, the points Q2 and Q3 are both sin-gular for C1, and hence they are conjugate to any point of the plane with respect toC1. Then Q1, Q2, Q3 are the vertices of a self-polar triangle both for C1 and for C2and consequently for any conic C ∈ F .

Suppose now that Q1, Q2, Q3 are points satisfying condition (i); we are going toshow that condition (ii) holds. In a system of homogeneous coordinates x0, x1, x2having Q1, Q2, Q3 as fundamental points, a conic C ∈ F is given by an equation ofthe form ax20 + bx20 + cx21 = 0, with a, b, c ∈ C not simultaneously zero. Observethat C is non-degenerate if and only if abc �= 0. Since by hypothesis F contains anon-degenerate conic, up to scaling the coordinates we may assume that F containsthe conic C1 of equation x20 + x21 + x22 = 0. If C2 ∈ F is a degenerate conic, up topermuting the coordinates we may assume that C2 has equation x20 + dx21 = 0 for

208 4 Exercises on Conics and Quadrics

some d ∈ C. It can be immediately checked that C1 ∩ C2 consists of four distinctpoints if d �= 0, 1 and consists of two points of multiplicity 2 if d = 0 or 1.

Note. IfF is a pencil of conics of P2(R), a sufficient condition for the existence of a

self-polar triangle for all conics ofF is the property thatF contains a conic C1 whichis non-degenerate and empty. Namely, if A1 is a symmetric matrix associated to C1,we may assume that A1 is positive definite. If C2 ∈ F is a conic distinct from C1 andrepresented by a symmetric matrix A2, by the Spectral theorem there is a changeof homogeneous coordinates on P

2(R) that simultaneously diagonalizes A1 and A2.Then the fundamental points of this projective frame are the vertices of a self-polartriangle for all conics of F .

Note that, since C1 is empty, the intersection set between (C1)C and (C2)C inP2(C)

consists of one or two pairs of conjugate points (cf. Sect. 1.9.7), and thus the pencil ofcomplex conics generated by (C1)C and (C2)C fulfils condition (ii) of the statement.

Exercise 156. If C and D are two distinct non-degenerate conics of P2(K), denote

byF the pencil of conics generated by C andD. For every P ∈ P2(K) let f (P) = Q

if polC(P) = polD(Q).

(a) Show that f defines a projectivity of P2(K).

(b) Show that P is a fixed point of f if and only if P is singular for a conic of thepencil F .

(c) What configurations of points in P2(K) may coincide with the fixed-point set

of f ?

Solution. (a) Let tX AX = 0 and tX BX = 0 be equations of C and D, respectively,with A, B symmetric invertible matrices of order 3. Since the line polC(P) hasequation tP AX = 0, then polC(P) = polD(Q) if and only if there existsα ∈ C \ {0}such that BQ = αAP , i.e. f (P) = Q = αB−1AP . As thematrix B−1A is invertible,f is well defined and it is a projectivity.(b) If P is a fixed point of f , there exists α ∈ C \ {0} such that P = αB−1AP ,

i.e. (B − αA)P = 0. Then P is a singular point for the conic tX (B − αA)X = 0 ofthe pencil F .

Conversely, if P is singular for some conic of the pencil, necessarily that conicis different both from C and from D, because each of these two conics is non-degenerate and hence non-singular. Then the conic having P as a singular point hasan equation of the form tX (B − αA)X = 0 for some α �= 0. So (B − αA)P = 0,that is P = αB−1AP , and hence P is a fixed point of f .

(c) Observe first that, as shown in Exercise 44, the fixed-point set of f maycoincide either with a finite set consisting of 1, 2 or 3 points, or with a line, or withthe union of one line and one point not lying on that line. We are now going toprove that each of these possible situations occurs in correspondence with a suitablechoice of the conics C and D, discussing separately the different cases according tothe number and the nature of the base points of F (cf. Sect. 1.9.7).

If F has 4 base points, then F contains exactly 3 degenerate conics, and each ofthem has exactly one singular point. Moreover, the singular points of these conicsare distinct, so that, as seen in (b), in this case f has exactly 3 fixed points.

4 Exercises on Conics and Quadrics 209

If instead we want f to have exactly 2 fixed points, it is sufficient to choose asC a non-degenerate conic passing through three independent points P1, P2, P3, todefine D′ = L(P1, P2) + L(P1, P3) and to choose as D any non-degenerate conic(of course distinct from C) belonging to the pencil F generated by C and D′. In thiscase P1, P2, P3 are the base points of the pencil, all conics of F are tangent at P1 tothe same line r and F contains two degenerate conics: one of them is D′, the otherone is the conic r + L(P2, P3). Hence the fixed points of f are P1 and r ∩ L(P2, P3).

In order to obtain only one fixed point we can argue as follows. Let C be a non-degenerate conic and A, B distinct points of C; denote by tA the tangent to C at A. IfD′ = tA + L(A, B), letD �= C be a non-degenerate conic of the pencil F generatedby C and by D′. Then the only degenerate conic of F is D′; so A is the only fixedpoint of f .

Assume now that C is a non-degenerate conic and tA is the tangent to C at a pointA ∈ C; letF be the pencil generated by C and 2tA. The only degenerate conic ofF is2tA, whose singular locus coincides with tA. In order to obtain a line of fixed pointsof f it is so sufficient to choose asD any non-degenerate conic ofF distinct from C.

Finally, given non-collinear points A, B,C ∈ P2(C), let F be the pencil gener-

ated by C ′ = L(A, B) + L(A,C) and D′ = 2L(B,C). Then C ′ and D′ are the onlydegenerate conics of F ; so, if C andD are two distinct and non-degenerate conics ofF , the fixed-point set of f coincides with L(B,C) ∪ {A}.

Thus we have proved that all configurations listed at the beginning of the solutionof (c) can be in fact obtained as the fixed-point set of f .

Exercise 157. Let A, B,C be points of P2(K) in general position. Given a projec-

tivity f : P2(K) → P

2(K) such that f (A) = B, f (B) = C and f (C) = A, provethat:

(a) There is no line r in P2(K) such that f (P) = P for every P ∈ r .

(b) f 3 = Id.(c) There is at least one non-degenerate conicQ passing through A, B,C and such

that f (Q) = Q.

Solution. (a) From the hypotheses it follows that

f (L(A, B)) = L(B,C), f (L(B,C)) = L(A,C), f (L(A,C)) = L(A, B).

(4.1)

If, by contradiction, there exists a line r of fixed points of f , this line cannot passthrough any of the points A, B, C . If we denote by D the point of intersectionbetween L(A, B) and r , then f (D) = D because D ∈ r . This gives a contradictionsince D = f (D) ∈ f (L(A, B)) = L(B,C) and D �= B.

(b) Let R be a fixed point of f (recall that all projectivities of P2(K) have a fixed

point, cf. Sect. 1.2.5). As observed in the solution of (a), this point cannot lie on anyof the lines L(A, B), L(B,C), L(A,C). Then the points A, B,C, R are in generalposition; since they are fixed for f 3, then f 3 = Id.

210 4 Exercises on Conics and Quadrics

(c) The setW of all conics passing through A, B,C is a linear systemof dimension2 and hence it is projectively isomorphic to P

2(K). The map f transforms everyconic of W into a conic of W and acts on the projective space W as a projectivity(cf. Sect. 1.9.5). Then in W there exists a fixed point of f , that is a conic Q suchthat f (Q) = Q. Necessarily Q is non-degenerate: namely, if it were degenerate, itshould contain a line passing through two points among A, B,C as an irreduciblecomponent, but then by (4.1) it should contain the three lines L(A, B), L(B,C) andL(A,C), which is not possible.

Exercise 158. Consider a non-degenerate conic C of P2(K) and let A, B ∈ C be

distinct points; denote by r1 the tangent line to C at A, by r2 the tangent line to C atB and let R = r1 ∩ r2.

(a) Show that for every X ∈ r1 \ {A} there is exactly one line rX passing through X ,tangent to C and different from r1.

(b) Let ψ : r1 → r2 be the map defined by

(i) ψ(A) = R;(ii) ψ(R) = B;(iii) ψ(X) = rX ∩ r2 for every X ∈ r1 \ {A}.Show that ψ is a projective isomorphism.

Solution. (a) Let X ∈ r1, X �= A. By reciprocity A belongs to polC(X), hencepolC(X) ∩ C �= ∅. In addition, polC(X) is not tangent to C because X /∈ C; there-fore, polC(X) intersects C, besides in A, in a second point PX such that the tangentline rX to C at PX passes through X . Since a point of P

2(K) lies on at most twolines tangent to a non-degenerate conic, the lines tangent to C and passing throughX are precisely r1 and rX . (Note that, if X = R, by construction we have PX = Band rX = r2).

(b) Since by construction the points A, B and R are not collinear, there exists inP2(K) a system of homogeneous coordinates in which A = [1, 0, 0], B = [0, 1, 0],

R = [0, 0, 1]. As a consequence, r1 has equation x1 = 0 and r2 has equation x0 = 0.Imposing that the conic passes through A and B and that the polar of R is the lineL(A, B) = {x2 = 0}, we obtain that C has equation tXMX = 0 with M of the form

M =⎛⎝0 a 0a 0 00 0 b

⎞⎠

with a, b ∈ K \ {0}.Let X = [y0, 0, y2] ∈ r1 \ {R, A} (so that y0 �= 0 and y2 �= 0). Thepoint PX where

the line rX is tangent to C is the point distinct from Awhere the line polC(X) intersectsthe conic. As polC(X) has equation y0ax1 + y2bx2 = 0, it turns out that PX hascoordinates [ay20 ,−2by22 , 2ay0y2]. Then the line rX = L(X, PX ) has equation

4 Exercises on Conics and Quadrics 211

det

⎛⎝

y0 ay20 x00 −2by22 x1y2 2ay0y2 x2

⎞⎠ = 0 that is 2by22 x0 − ay20 x1 − 2by0y2x2 = 0.

Hence ψ(X) = rX ∩ r2 = [0, 2by2,−ay0].With respect to the system of coordinates x0, x2 induced on r1 and to the system

of coordinates x1, x2 induced on r2, we have that ψ([y0, y2]) = [2by2,−ay0]; so therestriction of ψ to r1 \ {R, A} coincides with the restriction to r1 \ {R, A} of the

projective isomorphism f : r1 → r2 induced by the invertible matrix

(0 2b

−a 0

). On

the other hand it can be easily checked that f (A) = R and f (R) = B; therefore, fcoincides with ψ and hence ψ is a projective isomorphism too.

Note. Observe that the statement of the previous exercise is the dual statement ofExercise 142 (a).

Exercise 159. Assume that L is a linear system of conics of P2(C) of dimension 2

containing at least one non-degenerate conic. Prove that

B(L) = {P ∈ P2(C) | P ∈ C ∀C ∈ L}

is a finite set containing at most 3 points.

Solution. Let [F1], [F2], [F3] be three projectively independent conics of the linearsystem L, so that every conic of L has equation aF1 + bF2 + cF3 = 0 as [a, b, c]varies in P

2(C). Since L contains at least one non-degenerate conic, we can assumethat the conic [F1] is non-degenerate.

If P ∈ B(L), in particular P belongs to the supports of the conics [F1] and [F2].Since [F1] is non-degenerate, [F1] and [F2] meet in at most 4 points (cf. Sect. 1.9.7)and hence B(L) is a finite set containing at most 4 points. On the other hand, ifB(L) contained 4 points, these points would be in general position, so B(L) wouldcoincide with the set of base points of the pencil of conics F generated by [F1] and[F2] (cf. Sect. 1.9.7). Therefore, the conic [F3], passing through the base points ofF ,would belong to that pencil, and the linear system L would coincide with the pencilF , contradicting the hypothesis it has dimension 2.

K Exercise 160. (Pappus-Pascal’s Theorem) LetC be a non-degenerate conic ofP2(K)

and assume that P1, P2, P3, Q1, Q2, Q3 are distinct points of C. Show that the pointsR1 = L(P3, Q2) ∩ L(P2, Q3), R2 = L(P1, Q3) ∩ L(P3, Q1), R3 = L(P2, Q1) ∩L(P1, Q2) are collinear.

Solution (1). Consider the points A = L(P1, Q3) ∩ L(P3, Q2) and B = L(P1, Q2)

∩ L(P3, Q1) (cf. Fig. 4.12). As seen in Exercise 142, we can compute the cross-ratioof the points P3, P2, P1, Q2 on the conic as the cross-ratio of the lines joining thefour points with Q3. If we consider the intersections of the four lines with the lineL(P3, Q2), we get that

212 4 Exercises on Conics and Quadrics

Q3

R1

P2

P3

R2

B

Q2

Q1

P1

A

R3

Fig. 4.12 Pappus-Pascal’s Theorem

β(P3, P2, P1, Q2) = β(P3, R1, A, Q2).

If instead we compute the cross-ratio of P3, P2, P1, Q2 as the cross-ratio of the linesjoining the four points with Q1 and we consider the intersections of these lines withthe line L(P1, Q2), we get that

β(P3, P2, P1, Q2) = β(B, R3, P1, Q2).

Then β(P3, R1, A, Q2) = β(B, R3, P1, Q2), so there exists a projective isomor-phism f : L(P3, Q2) → L(P1, Q2) such that

f (P3) = B, f (R1) = R3, f (A) = P1, f (Q2) = Q2.

As Q2 is fixed for f , then f is the perspectivity centred at the point L(P3, B) ∩L(A, P1) = R2 (cf. Exercise 31). Since the perspectivity centred at R2 transformsR1 into R3, the points R1, R2, R3 are collinear.

Solution (2). The cubics D1 = L(P1, Q3) + L(P2, Q1) + L(P3, Q2) and D2 =L(P1, Q2) + L(P2, Q3) + L(P3, Q1) meet at the nine points P1, P2, P3, Q1, Q2,

Q3, R1, R2, R3. The first eight points belong to the cubic D = C + L(R1, R2). ByExercise 111, the cubic D has to pass through the point R3 too. Since R3 cannot lieon the irreducible conic C, then R3 ∈ L(R1, R2), which implies the thesis.

Solution (3). The cubics D1 = L(P1, Q3) + L(P2, Q1) + L(P3, Q2) and D2 =L(P1, Q2) + L(P2, Q3) + L(P3, Q1) meet at the nine points P1, P2, P3, Q1, Q2,

Q3, R1, R2, R3. The first six points belong to the irreducible cubic C; so by Exer-cise 106 the other three points R1, R2, R3 lie on a curve of degree 3 − 2 = 1, that ison a line.

Note. If in the statement of Pappus-Pascal’s Theorem we consider, instead of a non-degenerate conic C, a reducible conic union of two lines, we obtain Pappus’ Theorem(cf. Exercise 13).

4 Exercises on Conics and Quadrics 213

Exercise 161. Assume that C is a non-degenerate conic of P2(K), that A, B,C are

the vertices of a self-polar triangle for C and that r is a line passing through A andintersecting C at two distinct points P and Q. Prove that L(B, P) �= L(C, Q), andthat L(B, P) ∩ L(C, Q) ∈ C.

E Solution (1). First of all we observe that L(B, P) �= L(C, Q). Namely, otherwise thepoints B,C would lie on r , and hence A, B,C would be collinear, in contradictionwith the hypothesis that they are the vertices of a self-polar triangle for C.

In order to show that L(B, P) ∩ L(C, Q) ∈ C, we first investigate the caser = L(A, B) and r = L(A,C). If r = L(A, B) then Q ∈ L(B, P), so L(B, P) ∩L(C, Q) = Q ∈ C. Similarly if r = L(A,C) we have L(B, P) ∩ L(C, Q) =P ∈ C.

So we can assume that B /∈ r and C /∈ r . If L(B, P) were tangent to C, thenP ∈ polC(B) = L(A,C), and hence C ∈ L(A, P) = r , in contradiction with theassumption just made. Similarly one can show that L(B, Q) is not tangent to C.Therefore there are exactly two points R, S such that {R} = (L(B, P) ∩ C) \ {P},{S} = (L(B, Q) ∩ C) \ {Q} (cf. Fig. 4.13). As r �= L(A, B), then R �= Q and S �=P; in addition, the fact that B /∈ C easily implies that R �= S. Therefore P, Q, R, Sare distinct.

As proved in Exercise 153, the points A′ = L(P, Q) ∩ L(R, S) = r ∩ L(R, S),C ′ = L(P, S) ∩ L(Q, R) and B = L(P, R) ∩ L(Q, S) are the vertices of a self-polar triangle for C. Note that both A and A′ lie on r and also on polC(B). Moreover,polC(B) �= r because C ∈ polC(B), C /∈ r . It follows that A = A′, and hence C =polC(A) ∩ polC(B) = polC(A′) ∩ polC(B) = C ′. Therefore we have L(C, Q) =L(C ′, Q) = L(R, Q), so L(B, R) ∩ L(C, Q) = R ∈ C, as required.Solution (2). As observed at the beginning of Solution (1), the cases when eitherr = L(A, B) or r = L(A,C) can be easily discussed preliminarily. Thus assumethat B /∈ r and C /∈ r .

Since r is not tangent to C, it is immediate to check that the points A, B,C, Pform a projective frame ofP

2(K). Denote by x0, x1, x2 the homogeneous coordinatesinduced by that frame. As A, B,C are the vertices of a self-polar triangle for C,the equation of C has the form F(x0, x1, x2) = x20 + ax21 + bx22 = 0 with ab �= 0.

A

r

S

P

B RC

Q

Fig. 4.13 The configuration described in Exercise 161

214 4 Exercises on Conics and Quadrics

In addition, P ∈ C implies that b = −1 − a, so that F(x0, x1, x2) = x20 + ax21 −(1 + a)x22 . The line r = L(A, P) has equation x1 = x2, which easily yields Q =[1,−1,−1]. Therefore L(C, Q) and L(B, P) have equations x0 + x1 = 0 and x0 =x2, respectively. It follows that L(B, P) ∩ L(C, Q) = [1,−1, 1], which actuallybelongs to C.Exercise 162. (Chasles’ Theorem) Let C be a non-degenerate conic of P

2(K), andassume that P1, P2, P3 are distinct points ofP

2(K) such that polC(Pi ) �= L(Pj , Pk) if{i, j, k} = {1, 2, 3}. For i ∈ {1, 2, 3} let Ti = polC(Pi ) ∩ L(Pj , Pk), with {i, j, k} ={1, 2, 3}. Show that T1, T2, T3 are collinear.

Solution. If P1, P2, P3 lie on a line r , then evidently {T1, T2, T3} ⊆ r , which provesthe thesis. Therefore wemay assume that the points Pi are in general position, andwemay choose homogeneous coordinates inP

2(K) so that P1 = [1, 0, 0], P2 = [0, 1, 0],P3 = [0, 0, 1].

Denote by M a symmetric matrix representing C with respect to the coor-dinates just chosen, and denote by mi, j the entry of M lying on the i th rowand on the j th column. The line L(P2, P3) has equation x0 = 0, while polC(P1)has equation m1,1x0 + m1,2x1 + m1,3x2 = 0 (with m1,2, m1,3 not simultaneouslyzero because polC(P1) �= L(P2, P3)). Hence T1 = [0,m1,3,−m1,2]. Similarly T2 =[m2,3, 0,−m2,1] = [m2,3, 0,−m1,2], and T3 = [m3,2,−m3,1, 0] = [m2,3,−m1,3, 0].As (0,m1,3,−m1,2) − (m2,3, 0,−m1,2) + (m2,3,−m1,3, 0) = (0, 0, 0), the pointsT1, T2, T3 are collinear.

Note. If the points P1, P2, P3 lie on C, the line polC(Pi ) is the tangent to Cat Pi . If Q1, Q2, Q3 are points of C such that P1, P2, P3, Q1, Q2, Q3 are dis-tinct, the line polC(P1) can be seen as the limit of the line L(P1, Q2) when Q2

tends to P1. Similarly polC(P2) (resp. polC(P3)) is the limit of the line L(P2, Q3)

(resp. L(P3, Q1)) when Q3 tends to P2 (resp. when Q1 tends to P3). There-fore the points T1, T2, T3 can be obtained as limits of the points L(P1, Q2) ∩L(P2, Q1), L(P2, Q3) ∩ L(P3, Q2), L(P3, Q1) ∩ L(P1, Q3), respectively. Thusthe thesis of the exercise can be considered as a “limit version” of Pappus-Pascal’sTheorem (cf. Exercise 160).

Exercise 163. (Steiner construction) Let A and B be distinct points of P2(K) and

l = L(A, B); denote by FA (resp. FB) the pencil of lines centred at A (resp. B).Let f : FA → FB be a projective isomorphism such that f (l) �= l. Prove that Q ={r ∩ f (r) | r ∈ FA} is the support of a non-degenerate conic passing through A withtangent f −1(l) and through B with tangent f (l).

E Solution (1). The conics passing through A with tangent f −1(l) and through B withtangent f (l) form a pencil G (cf. Exercise 125).

Let s ∈ FA be a line different both from l and from f −1(l), and let P = s ∩ f (s) ∈Q. It is easy to check that A, B, P are in general position, that P /∈ f −1(l) and thatP /∈ f (l). Hence there exists exactly one conic C passing through P in the pencil Gand this conic is non-degenerate. In order to prove the thesis it is enough to showthat Q = C.

4 Exercises on Conics and Quadrics 215

Consider the map ψ : FA → FB defined by ψ( f −1(l)) = l, ψ(l) = f (l) and,if r ∈ FA \ {l, f −1(l)}, ψ(r) = L(B, Q), where Q denotes the intersection pointbetween r and C different from A. As shown in Exercise 142, ψ is a projectiveisomorphism and {r ∩ ψ(r) | r ∈ FA} = C.

Moreover, by construction we have ψ( f −1(l)) = l = f ( f −1(l)), ψ(l) = f (l),ψ(s) = f (s). Since ψ and f coincide on three distinct elements of FA, we deducethat ψ = f . Therefore Q = {r ∩ ψ(r) | r ∈ FA} = C, as required.

Solution (2). If C = f −1(l) ∩ f (l), it can be easily checked that the points A, B,Care in general position. Thus we can choose homogeneous coordinates on P

2(K)

in such a way that A = [1, 0, 0], B = [0, 1, 0], C = [0, 0, 1]. These coordinatesinduce on the pencil FA (resp. FB) homogeneous coordinates such that the lineax1 + bx2 = 0 (resp. cx0 + dx2 = 0) has coordinates [a, b] (resp. [c, d]).

By construction the lines l, f −1(l) in FA have coordinates [0, 1], [1, 0] respec-tively, while the lines l, f (l) inFB have coordinates [0, 1], [1, 0] respectively. There-fore in the chosen coordinates f can be represented by M =

(0 λ1 0

)for some

λ ∈ K∗. Hence, if r ∈ FA is the generic line of equation ax1 + bx2 = 0, then the

line f (r) has equation λbx0 + ax2 = 0. It follows that r ∩ f (r) = [a2,λb2,−λab],so that Q = {[a2,λb2,−λab] | [a, b] ∈ P

1(K)}.Since λ(a2)(λb2) = (−λab)2, the set Q is contained in the support of the conic

C of equation F(x0, x1, x2) = λx0x1 − x22 = 0. In addition, if T = [y0, y1, y2] ∈ Cthe following situations can occur:

(i) if y2 = 0, then either T = [1, 0, 0] = A = f −1(l) ∩ f ( f −1(l)) ∈ Q or T =[0, 1, 0] = B = l ∩ f (l) ∈ Q;

(ii) if y2 �= 0, then y0 �= 0 and, if a �= 0 is such that a2 = y0 and b = − y2λa , then

T = [a2,λb2,−λab] ∈ Q.

Therefore C ⊆ Q, and hence Q = C.In order to achieve the thesis it suffices to observe that C is non-degenerate, and

that ∇F(1, 0, 0) = (0,λ, 0), ∇F(0, 1, 0) = (λ, 0, 0), so that the tangents to C atA, B are actually f −1(l) and f (l).

Note. Exercise 43 (c) deals with the case when the line l is invariant under theisomorphism f ; also in that case the set Q is a conic, but a degenerate one.

K Exercise 164. Let C be a non-degenerate conic of P2(C) and assume that f is a

projectivity of P2(C) such that f (C) = C. Show that there exists P ∈ C such that

f (P) = P .

Solution.As seen in Sect. 1.2.5, there exists a point Q ∈ P2(C) such that f (Q) = Q.

IfQ ∈ C the thesis holds, hencewecan assume thatQ /∈ C. Theprojectivity f inducesa projectivity of the pencil FQ centred at Q, and also this projectivity has a fixedpoint, because FQ is a complex projective space. Then there is a line r ∈ FQ suchthat f (r) = r . If r is tangent to C and if P denotes the point of intersection betweenr and C, then f (P) = f (C ∩ r) = C ∩ r = P , and thus the thesis is proved.

216 4 Exercises on Conics and Quadrics

Therefore suppose now that r intersects C in two distinct points A, B. Thenf ({A, B}) = f (C ∩ r) = C ∩ r = {A, B}; so either f (A) = A and f (B) = B, orf (A) = B and f (B) = A. In both cases f 2(A) = A and f 2(B) = B. Let s =polC(Q) and denote by C, D the points of intersection between s and C (these pointsare distinct because Q /∈ C). As f (C) = C and f (Q) = Q, we have f (s) = s so that,arguing as above, f 2(C) = C and f 2(D) = D. Moreover, the points A, B,C, D aredistinct and, since they lie on a non-degenerate conic, they form a projective frameof P

2(C). Then the Fundamental theorem of projective transformations implies thatf 2 = Id. Taking into account the solution of Exercise 44, it is immediate to checkthat f is a projectivity of type (b); so there exists a line of fixed points of f . This lineintersects C in at least one point, and so C contains at least one point which is fixedfor f .

K Exercise 165. Let C be a non-degenerate conic of P2(K); assume that P is a point

of C and denote r = polC(P). Let g : r → r be a projectivity such that g(P) = P .Show that there is exactly one projectivity f : P

2(K) → P2(K) such that f (C) = C,

f (r) = r and f |r = g.

Solution. Let A1, A2 be distinct points of r \ {P}, and for i = 1, 2 denote by A′i the

point of intersection between C and the tangent to C different from r and passingthrough Ai , that is the point defined by the condition {A′

i } = (polC(Ai ) ∩ C) \ {P}(cf. Fig. 4.14). Moreover let B1 = g(A1) and B2 = g(A2); in a similar way, considerthe points B ′

1, B′2 ∈ C defined by {B ′

i } = (polC(Bi ) ∩ C) \ {P}.Let us now see that a projectivity f : P

2(K) → P2(K) fulfils the requirements of

the statement if and only if f (C) = C, f (P) = P , f (A′1) = B ′

1 and f (A′2) = B ′

2. Atthis point, the thesis is a consequence of Exercise 143 (a).

If f : P2(K) → P

2(K) satisfies the requirements of the statement, necessarilywe have f (C) = C and f (P) = g(P) = P . In addition f (Ai ) = g(Ai ) = Bi fori = 1, 2, so that f transforms L(Ai , A′

i ) into the tangent to f (C) = C passing throughf (Ai ) = Bi and distinct from f (r) = r . By construction this tangent coincideswith the line L(Bi , B ′

i ), so that f (A′i ) = f (L(Ai , A′

i ) ∩ C) = L(Bi , B ′i ) ∩ C = B ′

i ,as required.

Conversely assume that f (C) = C, f (P) = P , f (A′1) = B ′

1 and f (A′2) = B ′

2.Then f (r) = f (polC(P)) = pol f (C)( f (P)) = polC(P) = r . For i = 1, 2 similarlywehave f (L(Ai , A′

i )) = f (polC(A′i )) = polC(B ′

i ) = L(Bi , B ′i ), and hence f (Ai ) =

rP

A1

A2A1B1

B2B1

A2

B2

Fig. 4.14 The construction described in the solution of Exercise 165

4 Exercises on Conics and Quadrics 217

f (r ∩ L(Ai , A′i )) = r ∩ L(Bi , B ′

i ) = Bi . It follows that the projectivities f |r and gcoincide on the three distinct points P, A1, A2, and therefore they are equal as aconsequence of the Fundamental theorem of projective transformations.

Exercise 166. Denote by F a pencil of degenerate conics of P2(C) such that there

is no line which is an irreducible component of all conics in the pencil. Prove thatthe base-point set of F consists of a single point and that the pencil contains exactlytwo doubly degenerate conics.

Solution. First of all observe that F can contain at most two doubly degenerateconics. Namely, let C1 = 2r and C2 = 2s be two such conics in F , where r and s arelines. In a system of homogeneous coordinates where r has equation x0 = 0 and s hasequation x1 = 0, the generic conic Cλ,μ ∈ F is given by the equation λx20 + μx21 = 0,[λ,μ] ∈ P

1(C). It can be immediately checked that Cλ,μ is doubly degenerate if andonly if [λ,μ] ∈ {[1, 0], [0, 1]}, i.e. if and only if Cλ,μ ∈ {C1, C2}.

We are now going to show that F has exactly one base point. As observed above,we can choose as generators of the pencil two simply degenerate conics, C1 = r1 + r2and C2 = s1 + s2, where r1, r2, s1, s2 are distinct lines. Denote by A = r1 ∩ r2 thesingular point of C1 and by B = s1 ∩ s2 the singular point of C2. If A = B, the pencilhas exactly one base point, as required. Since by hypothesis C1 and C2 do not shareany component, if A and B are distinct, up to exchanging C1 with C2, we may assumethat A does not belong to the support of C2. Therefore, we can distinguish two casesdepending on whether B lies on the support of C1 or not.

In the first case we can suppose, for instance, that B belongs to r1. If P1 = r2 ∩ s1and P2 = r2 ∩ s2, we can immediately check that B, P1 and P2 are in general positionand that P1, P2 /∈ r1. All conics ofF pass through B, P1 and P2 and they are tangentat B to r1; hence, by Exercise 121, F coincides with the pencil of conics that satisfythese linear conditions. In other words we are in case (b) of Sect. 1.9.7 andF containsonly two degenerate conics, in contradiction with the hypothesis.

In the second case C1 and C2 meet at four points P1, P2, P3, P4 in general positionand hence F is the pencil of conics through P1, P2, P3, P4. Thus we are in case (a)of Sect. 1.9.7 and F contains only three degenerate conics, in contradiction with thehypothesis.

We have thus seen that F has only one base point A. Choose a system of homo-geneous coordinates x0, x1, x2 in P

2(C) such that A = [1, 0, 0] and [0, 1, 0] ∈ r1,[0, 0, 1] ∈ r2 and [1, 1, 1] ∈ s1. In this system of coordinates C1 is defined byx1x2 = 0 and C2 is defined by (x1 − x2)(x1 + αx2), with α �= 0,−1. The genericconic Cλ,μ of F is thus represented, as [λ,μ] varies in P

1(C), by the matrix

Mλ,μ =⎛⎝0 0 00 2μ λ + μ(α − 1)0 λ + μ(α − 1) −2μα

⎞⎠, whose rank is equal to 1 if and only

if λ2 + (α + 1)2μ2 + 2(α − 1)λμ = 0. This latter equality, considered as an equa-tion in [λ,μ] ∈ P

1(C), admits two distinct solutions, because (α − 1)2 − (α + 1)2 =−4α �= 0. Therefore F contains exactly two doubly degenerate conics, as required.

218 4 Exercises on Conics and Quadrics

Exercise 167. Let Q be a non-degenerate quadric of Pn(K) and let H be a hyper-

plane. Then Q ∩ H is a degenerate quadric of H if and only if H is tangent to Q ata point P . In addition, in that case the quadric Q ∩ H has rank n − 1 and P is itsonly singular point.

Solution. The first statement of the thesis easily follows from the fact that Q ∩ His a degenerate quadric of H if and only if it is singular and, by Exercise 58, thishappens if and only if the hyperplane H is tangent to Q at a point P .

In that case we can choose a system of homogeneous coordinates in Pn(K)

where H = {x0 = 0} and P = [0, . . . , 0, 1]. If in these coordinates Q has equationtX AX = 0, where A is a symmetric matrix, the quadricQ ∩ H is represented by thematrix c0,0(A) (cf. 1.1 for the definition of c0,0(A)). Since H is tangent to Q at P ,then pol(P) = H and hence an,1 = . . . = an,n = 0 and an,0 �= 0, i.e. the symmetricmatrix A is of the form

A =

⎛⎜⎜⎜⎜⎜⎝

a0,0 . . . an,0

a1,0 0... B

...

an−1,0 0an,0 0 . . . 0 0

⎞⎟⎟⎟⎟⎟⎠

where B denotes a symmetric matrix of order n − 1. As det A = −a2n,0 det B anddet A �= 0, it follows that det B �= 0 and so rk(c0,0(A)) = n − 1. As a consequenceSing(Q ∩ H) consists of a single point; since (0, . . . , 0, 1) ∈ K

n belongs to the kernelof c0,0(A), then Sing(Q ∩ H) = {P}.Exercise 168. LetQ be a quadric of rank r of P

n(K) and let P ∈ Pn(K) \ Q. Prove

that, if r = 1, then Q = 2 pol(P), and that, if r > 1, then the quadric Q′ = Q ∩pol(P) of pol(P) has rank r − 1.

Solution. Since P /∈ Q, the subspace pol(P) is a hyperplane not passing throughP . Then there are homogeneous coordinates in P

n(K) such that P = [1, 0, . . . , 0]and pol(P) = {x0 = 0}. In this system of coordinates Q is represented by a matrix

of the form A =(a0,0 00 C

), where C denotes a symmetric matrix of order n and

a0,0 ∈ K∗. Since r = rk(A) = rk(C) + 1, then C = 0 if and only if r = 1, and in

this case Q = 2 pol(P). For r > 1, the matrix C defines the quadric Q′ in pol(P),which consequently has rank r − 1.

Exercise 169. Consider a quadric Q of Pn(K) and let H ⊂ P

n(K) be a hyperplanenot contained in Q. Denote by Q′ the quadric Q ∩ H of H . Prove that polQ′(P) =polQ(P) ∩ H for every P ∈ H .

Solution. We can choose homogeneous coordinates x0, . . . , xn so that P = [1, 0,. . . , 0] and H = {xn = 0}. If A = (ai, j )i, j=0,...,n is a symmetric matrix represent-ing Q, then polQ(P) has equation

∑ni=0 a0,i xi = 0. In the coordinates x0, . . . , xn−1

induced on H , the quadric Q′ is defined by the matrix A′ = cn,n(A), obtained

4 Exercises on Conics and Quadrics 219

from A by deleting its last row and its last column (the matrix thus obtained isnon-zero because H � Q). Then polQ′(P) is the subspace of H defined by theequation

∑n−1i=0 a0,i xi = 0, so it coincides with H ∩ polQ(P).

K Exercise 170. (Construction of a self-polar (n + 1)-hedron) Let Q be a quadric ofrank r of P

n(K) and assume that P0, . . . , Pk ∈ Pn(K) \ Q are points such that Pi

and Pj are conjugate with respect to Q for any i �= j with 0 ≤ i, j ≤ k. Show that:

(a) P0, . . . , Pk are projectively independent and k + 1 ≤ r .(b) The hyperplanes pol(P0), . . . , pol(Pk) are independent and L(P0, . . . , Pk) ∩

pol(P0) ∩ · · · ∩ pol(Pk) = ∅.(c) There exist Pk+1, . . . , Pn ∈ P

n(K) such that P0, . . . , Pk, Pk+1, . . . , Pn are thevertices of a self-polar (n + 1)-hedron for Q.

Solution. (a) Observe that, for each i ∈ {0, . . . , k}, the subspace Si = L(P0, . . . ,Pi−1, Pi+1, . . . Pk) is contained in pol(Pi ). Since by hypothesis Pi /∈ Q, then pol(Pi )is a hyperplane and Pi /∈ pol(Pi ); in particular Pi /∈ Si . It follows that P0, . . . , Pkare projectively independent. Let us complete P0, . . . , Pk to a projective frame{P0, . . . , Pk, Qk+1, . . . , Qn+1} and let A be a matrix defining Q in the associated

system of homogeneous coordinates. Then A =(D tBB C

), where D is a symmet-

ric matrix of order k + 1, C is a symmetric matrix of order n − k and B is a matrix(n − k) × (k + 1). As the points P0, . . . , Pk do not belong toQ and are pairwise con-jugate, one can easily check that D = diag(d0, . . . , dk) is a diagonal matrix whoseelements di on the diagonal are all non-zero. Hence r = rk A ≥ rk D = k + 1.

(b) In the system of coordinates already used in the solution of (a) the point ofthe dual space P

n(K)∗ corresponding to pol(Pi ), i = 0, . . . , k, is given by the i-thcolumn of A. Having seen that the first k + 1 columns of A are independent, thenpol(P0), . . . , pol(Pk) are independent too. The subspace L(P0, . . . , Pk) is definedby xk+1 = · · · = xn = 0 and the subspace L(P0, . . . , Pk) ∩ pol(P0) ∩ · · · ∩ pol(Pk)is defined in L(P0, . . . , Pk) by the equations d0x0 = · · · = dkxk = 0, so it is empty.

(c) Denote by T the projective subspace pol(P0) ∩ · · · ∩ pol(Pk); by point (b)T has dimension n − k − 1 and does not intersect L(P0, . . . , Pk). By reciprocity,Sing(Q) is contained in the polar space of any point of P

n(K), so Sing(Q) ⊆ T .Moreover, Grassmann’s formula implies that dim L(L(P0, . . . , Pk), T ) = dim T +k + 1 = n, and therefore L(L(P0, . . . , Pk), T ) = P

n(K).Let us proceed by induction on m = r − k − 1, which is a non-negative integer

by part (a).If m = 0, let us prove that, however we choose projectively independent points

Pk+1, . . . , Pn ∈ T , the points P0, . . . , Pk, Pk+1, . . . , Pn form a self-polar (n + 1)-hedron.Asobserved above, L(P0, . . . , Pk, Pk+1, . . . , Pn) = L(L(P0, . . . , Pk), T ) =Pn(K), so the points P0, . . . , Pn are projectively independent. In addition, sincem =

r − k − 1 = 0, the projective subspaces Sing(Q) and T have the same dimension,so Sing(Q) = T . Consider i, j such that 1 ≤ i < j ≤ n. If j ≤ k (and hence i ≤ ktoo), the points Pi and Pj are conjugate by hypothesis, while Pj ∈ T = Sing(Q) ifj > k; so pol(Pj ) = P

n(K) and the points Pi , Pj are conjugate.

220 4 Exercises on Conics and Quadrics

Let now m > 0, and let us first prove that T � Q. Assume by contradictionthat T ⊆ Q and choose P ∈ T . Then L(P, P ′) ⊆ T ⊆ Q for every P ′ ∈ T ; as aconsequence L(P, P ′) ⊆ TP(Q) = pol(P) and T ⊆ pol(P), using the fact that P ′can be arbitrarily chosen. On the other hand, by reciprocity Pi ∈ pol(P) for eachi = 0, . . . , k, so pol(P) ⊇ T ∪ {P0, . . . , Pk} and pol(P) ⊇ L(T, L(P0, . . . , Pk)) =Pn(K). This implies that P ∈ Sing(Q), so T ⊆ Sing(Q) since P can be arbitrar-

ily chosen; therefore T = Sing(Q) by the observations made at the beginning ofthe proof of (c). Finally it follows that n − r = dim Sing(Q) = dim T = n − k − 1,contradicting the hypothesis m = r − k − 1 > 0.

We have thus seen that T \ Q is non-empty, so that it contains a point Pk+1.Then Pk+1 is conjugate with P0, . . . , Pk and, by the inductive hypothesis, there existPk+2, . . . , Pn ∈ P

n(K) such that P0, . . . , Pn are the vertices of a self-polar (n + 1)-hedron.

Exercise 171. Let Q be a quadric of Pn(K) and assume that P1, P2 ∈ P

n(K) \Sing(Q) are distinct points. Prove that pol(P1) = pol(P2) if and only if L(P1, P2) ∩Sing(Q) �= ∅.Solution (1). Let v1, v2 ∈ K

n+1 \ {0} be vectors such that Pi = [vi ], i = 1, 2. Sinceby hypothesis P1, P2 /∈ Sing(Q), for i = 1, 2 pol(Pi ) is the hyperplane defined bytvi AX = 0. Therefore, if pol(P1) = pol(P2) then there existsλ ∈ K

∗ such that Av1 =λAv2. The vector w = v1 − λv2 is non-zero, because v1, v2 represent distinct pointsof P

n(K) and so they are linearly independent. The point Q = [w] is singular forQ,because Aw = 0, and it lies on the line L(P1, P2).

Conversely, assume there existsQ ∈ L(P1, P2) ∩ Sing(Q) and letw ∈ Kn+1 \ {0}

be a vector such that Q = [w]. By hypothesis Q, P1 and P2 are distinct, so thereexist λ1,λ2 ∈ K

∗ such that w = λ1v1 + λ2v2. If we multiply the latter equality by

A on the left, we get 0 = Aw = λ1Av1 + λ2Av2, that is Av1 = −λ2λ1

Av2. Therefore

pol(P1) = pol(P2).Solution (2). Since pol : P

n(K) \ Sing(Q) → Pn(K)∗ is a (possibly degenerate)

projective transformation, if L(P1, P2) ∩ Sing(Q) = ∅ then the restriction of polto L(P1, P2) is a non-degenerate projective transformation, so it is injective, andhence pol(P1) �= pol(P2).

Conversely, if L(P1, P2) ∩ Sing(Q) �= ∅, then L(P1, P2) ∩ Sing(Q) is a point.As seen in Exercise 28, the image of L(P1, P2) \ Sing(Q) through the map pol is aprojective subspace of P

n(K)∗ having dimension

dim L(P1, P2) − dim(L(P1, P2) ∩ Sing(Q)) − 1 = 1 − 0 − 1 = 0,

i.e. it is a point. In particular it follows that pol(P1) = pol(P2).

Exercise 172. IfQ is a non-degenerate and non-empty quadric of P3(K), show that

one of the following statements holds:

4 Exercises on Conics and Quadrics 221

(i) for every P ∈ Q the setQ ∩ TP(Q) is the union of two distinct lines which meetat P;

(ii) for every P ∈ Q we have Q ∩ TP(Q) = {P}.(Recall that Q is said to be hyperbolic in case (i) and elliptic in case (ii), and thatcase (ii) can occur only if K = R – cf. Sect. 1.8.4.)

Solution. If P ∈ Q is a point of the quadric, since Q is non-degenerate, byExercise 167 the conic CP = Q ∩ TP(Q) is singular at P and has rank 2. So thesupport of CP either is the union of two distinct lines passing through P , and in thiscase P is called hyperbolic, or it consists only of the point P , and in this case P iscalled elliptic (cf. Sect. 1.8.4).

We are going to check that either all points of Q are hyperbolic (so statement (i)holds) or all of them are elliptic (and then we are in case (ii)). IfQ does not containany line, clearly case (ii) occurs. Therefore we can assume that Q contains a line r .For every point P ∈ r , the line r is contained in TP(Q), and hence also in CP , whichis consequently the union of two lines through P . If instead P ∈ Q \ r , there existsat least one point R ∈ r ∩ TP(Q). As R �= P and R ∈ CP , we can conclude that CP

is the union of two lines passing through P; so case (i) occurs.Let us show an alternative way to prove that Q cannot simultaneously contain

hyperbolic points and elliptic points. If Q contained an elliptic point P and alsoa hyperbolic point R, then Q ∩ polQ(P) = {P} and Q ∩ polQ(R) = r1 ∪ r2 withr1, r2 distinct lines meeting at R. As R /∈ polQ(P), then polQ(P) should intersectr1 ∪ r2 (and hence Q) in at least two points, contradicting the fact that P is elliptic.

Exercise 173. Let Q be a quadric of P3(K) of rank 3. Prove that, for every smooth

point P ∈ Q, the conic Q ∩ TP(Q) is a double line.(Recall that, in this case, ifQ \ Sing(Q) �= ∅, the quadricQ is called parabolic – cf.Sect. 1.8.4.)

Solution. The quadric Q, having rank 3, has only one singular point R, and ofcourse pol(R) = P

3(K). Let P ∈ Q be a smooth point. By reciprocity R ∈ pol(P) =TP(Q). Then, by Exercise 58, the conic Q ∩ TP(Q) is singular both at P and at R,and hence it coincides with 2L(P, R).

Exercise 174. Prove that:

(a) The quadric Q of P3(C) of equation x20 + x21 + x22 + x23 = 0 is hyperbolic.

(b) The quadric Q of P3(C) of equation x20 + x21 + x22 = 0 is parabolic.

(c) The quadric Q of P3(R) of equation x20 + x21 + x22 − x23 = 0 is elliptic.

(d) The quadric Q of P3(R) of equation x20 + x21 − x22 − x23 = 0 is hyperbolic.

(e) The quadric Q of P3(R) of equation x20 + x21 − x22 = 0 is parabolic.

Solution. First of all it is immediate to check that the quadric Q is non-degeneratein cases (a), (c) and (d), and that it has rank 3 in cases (b) and (e). In particular,in all cases the quadric Q is irreducible. Therefore, by Exercise 173, in cases (b)and (e) it suffices to check that Q contains a smooth point; as a matter of fact, in

222 4 Exercises on Conics and Quadrics

case (b) such check is unnecessary, because Q has only one singular point and thesupport of a complex quadric contains infinitely many points. Moreover, statement(a) follows from the fact that all non-degenerate quadrics of P

3(C) are hyperbolic(cf. Exercise 172). Instead, in cases (c) and (d) in order to decide whetherQ is eitherhyperbolic or elliptic, it is sufficient to choose a point P ∈ Q and determine thesupport of the conic Q ∩ TP(Q) (cf. Exercise 172).

(a) As recalled above, all non-degenerate quadrics of P3(C) are hyperbolic. How-

ever, choose for instance the point P = [1, 0, 0, i] ∈ Q, and let us check that theconic Q ∩ TP(Q) consists of two distinct lines which meet at P . The plane TP(Q)

has equation x0 + i x3 = 0, and the conic Q ∩ TP(Q) is defined in TP(Q) by theequation x21 + x22 = 0, whose support is the union of the lines x1 + i x2 = 0 andx1 − i x2 = 0.

(b) As observed before, it suffices to check that Q has at least a smooth point.Choose for instance P = [1, 0, i, 0] ∈ Q. Then TP(Q) is defined by the equationx0 + i x2 = 0, so P is non-singular. In addition, the conic Q ∩ TP(Q) is defined inTP(Q) by the equation x21 = 0, so in fact it is a double line.

(c) Choose P = [1, 0, 0, 1] ∈ Q. The plane TP(Q) has equation x0 − x3 = 0,and the conic Q ∩ TP(Q) is defined in TP(Q) by the equation x21 + x22 = 0, whosesupport contains only the point P .

(d) Choose P = [1, 0, 0, 1] ∈ Q. The plane TP(Q) has equation x0 − x3 = 0,and the conic Q ∩ TP(Q) is defined in TP(Q) by the equation x21 − x22 = 0, whosesupport is the union of the lines x1 + x2 = 0 and x1 − x2 = 0.

(e) Like in case (b), it is sufficient to check that Q has at least a smooth point.If P = [1, 0, 1, 0] ∈ Q, the tangent space TP(Q) is defined by x0 − x2 = 0, so thatP is smooth. In additon, the conic Q ∩ TP(Q) is defined in TP(Q) by the equationx21 = 0, so it is a double line.

Exercise 175. Assume thatQ is a non-degenerate and non-empty quadric of P3(R).

Prove that there exists a plane H of P3(R) external toQ if and only ifQ is an elliptic

quadric.

Solution.Assume thatQ is not elliptic. By Exercise 172, the quadricQ is hyperbolicand, given a point P ∈ Q, we have thatQ ∩ TP(Q) = r ∪ s, where r and s are distinctlines. As a consequence, for every plane H ofR3, the setQ ∩ H contains H ∩ (r ∪ s)and hence it is not empty. So no plane external to Q can exist.

Conversely, suppose that Q is an elliptic quadric. We can choose homogeneouscoordinates in which Q is defined by one of the equations listed in Theorem 1.8.3.Then by Exercise 174 we may assume that Q is defined by the equation x20 + x21 +x22 − x23 = 0. It is now immediate to check that the plane H of equation x0 = 2x3 isexternal to Q.

Exercise 176. Consider a quadric Q of P3(K) and assume that r, s, t ⊂ Q are dis-

tinct lines. Show that:

4 Exercises on Conics and Quadrics 223

(a) If r ∩ s = s ∩ t = r ∩ t = ∅, then Q is non-degenerate.(b) If r, s, t are coplanar, then Q is reducible.(c) If r, s, t meet at a point P and are not coplanar, then P is singular for Q.(d) If r and s are coplanar and r ∩ t = s ∩ t = ∅, then Q is reducible.

Solution. (a) Suppose by contradiction that Q is singular and consider a point P ∈Sing(Q). Since the lines r, s, t are pairwise skew,wemay assume that P /∈ r .As everyquadric having a singular point is a cone with that point as a vertex (cf. Sect. 1.8.3),Q is a cone of vertex P . So the plane H = L(P, r) is contained in the support ofQ.By Exercise 57, we haveQ = H + K , where K is a plane. Since r ∩ s = r ∩ t = ∅,the lines s and t are not contained in H . Then s, t ⊂ K and therefore s ∩ t �= ∅, acontradiction.

(b)Denote by H the plane containing r, s and t . If we had H � Q, then the supportof the conic Q ∩ H of H would contain three distinct lines, which is impossible.Therefore, H is contained in the support ofQ. By Exercise 57 H is a component ofQ, which is consequently reducible.

(c) Since the lines r, s, t are contained in Q, they are contained in TP(Q). ThenL(r, s, t) ⊂ TP(Q), so TP(Q) = P

3(K) because r, s, t are not coplanar. Then P issingular for Q.

(d) Denote by H the plane generated by r and s. If H is contained in the support ofQ, thenQ is reducible by Exercise 57. Otherwise the support of the conic C = Q ∩ His r ∪ s. Consider the point R = H ∩ t . By construction, R belongs to C, that is tor ∪ s, contradicting the hypothesis that r ∩ t = s ∩ t = ∅.

K Exercise 177. (Hyperbolic quadrics are ruled) Let Q be a non-degenerate hyper-bolic quadric of P

3(K). Show that the set of lines contained inQ is the union of twodisjoint families X1 and X2 such that:

(i) for every point P ∈ Q there exist a line of X1 and a line of X2 passing throughP;

(ii) any two lines of the same family are skew;(iii) if r ∈ X1 and s ∈ X2, then r and s meet at one point.

Solution (1).Assume that P is a point ofQ. By the definition of hyperbolic quadric,Q intersects the plane TP(Q) in a pair of distinct lines which meet at P , thereforefor any point ofQ there are two distinct lines r and s passing through it; by Exercise176 no other line contained in Q passes through P .

Given P0 ∈ Q, denote by r0 and s0 two lines passing through P0 and containedinQ. Let us define X1 as the set of lines r ⊂ Q such that r ∩ s0 is a point and X2 asthe set of lines s ⊂ Q such that s ∩ r0 is a point.

Suppose by contradiction there exists r ∈ X1 ∩ X2. By the definition of X1 andX2, the lines r , r0 and s0 are distinct and r ∩ r0 �= ∅, r ∩ s0 �= ∅. So either r , r0 ands0 pass through the point P0 or they are coplanar, contradicting Exercise 176.

Let us now show that for any point P of Q there is a line of X1 passing throughit. If P ∈ s0, the claim is true because there exists a line r �= s0 contained in Q and

224 4 Exercises on Conics and Quadrics

passing through P; by definition r ∈ X1. If P /∈ s0, let H = L(P, s0). The conicQ ∩ H contains s0 and the point P /∈ s0, so it is the union of s0 and a line r �= s0such that P ∈ r . The lines s0 and r , being coplanar, are incident, so r ∈ X1. In thesame way one shows that for any point ofQ there is a line of X2 passing through it.Observe now that, by Exercise 176 (b) and (c), for any point of Q there are exactlytwo lines contained in Q passing through that point; as a consequence X1 ∪ X2 isthe set of all lines contained in Q.

We are now going to show that any two lines of the same family are disjoint (thatis they are skew). If by contradiction there existed two incident lines r, s ∈ X1, thenthe three distinct lines s0, r, s either would be coplanar or would share a commonpoint. In both cases we would get to a contradiction with Exercise 176. The sameargument shows that any two distinct lines of X2 are skew.

Finally suppose by contradiction that r ∈ X1 and s ∈ X2 are skew. Choose a pointR ∈ s and consider the plane H = L(r, R). The conic Q ∩ H is the union of r anda line t passing through R (hence distinct from r ) and distinct from s (because r ands are skew). The line t does not belong to X1, because it intersects r ∈ X1 and it isdifferent from r , and it does not belong to X2, because it intersects s ∈ X2 and it isdifferent from s. We have so found a contradiction with the fact that X1 ∪ X2 is theset of all lines contained in Q.

Solution (2). Since all non-degenerate hyperbolic quadrics are projectively equiva-lent (cf. Theorem 1.8.3 and Exercise 174), it is sufficient to consider the quadric Qdefined by

det

(x0 x1x2 x3

)= x0x3 − x1x2 = 0. (4.2)

Namely, as pointed out in Sect. 1.8.4, Q is non-degenerate and hyperbolic: an easycomputation shows that the tangent plane to Q at the point [1, 0, 0, 0] is the plane{x3 = 0}which intersectsQ in the lines r0 = {x3 = x1 = 0} and s0 = {x3 = x2 = 0}.

Define X1 as the set of lines r[a,b] = {[λa,λb,μa,μb] | [λ,μ] ∈ P1(K)}, as

[a, b] varies in P1(K). In a similar way, define X2 as the set of lines s[a,b] =

{[λa,μa,λb,μb] | [λ,μ] ∈ P1(K)}, as [a, b] varies in P

1(K). Observe that theplanes of the pencil centred at s0 intersect Q, besides in s0, in a line of X1. Sim-ilarly, the planes of the pencil centred at r0 intersectQ, besides in r0, in a line of X2.

By using parametrizations, one can readily check that the lines of X1 and X2 arecontained in Q, that two lines of the same family are skew and that for every [a, b],[c, d] ∈ P

1(K) the lines r[a,b] and s[c,d] intersect at the point [ca, cb, da, db]. So thefamilies X1 and X2 are necessarily disjoint.

We now show that for every point of Q there are a line of X1 and a line of X2

passing through it, and that X1 ∪ X2 is the set of all lines contained in Q.Let P ∈ Q be a point of coordinates [α,β, γ, δ]. If, for instance, α �= 0, by using

Eq. (4.2) one can easily check that P belongs to the lines r[α,β] and s[α,γ]. If α = 0,one can argue in the same way but considering another coordinate of P . Thereforefor every point of Q there are a line of X1 and a line of X2 passing through it.Moreover, since every line r ⊂ Q is contained in TP(Q) for every P ∈ r and Q is

4 Exercises on Conics and Quadrics 225

hyperbolic, then every point P ∈ Q belongs exactly to two lines contained inQ. Asa consequence, X1 ∪ X2 is the set of all lines contained in Q.

Exercise 178. Assume that r and s are two lines of P3(K) such that r ∩ s = ∅ and

let f : r → s be a projective isomorphism. Prove that X = ⋃P∈r L(P, f (P)) is the

support of a non-degenerate hyperbolic quadric of P3(K).

Solution. If P0, P1 ∈ r are distinct points, set P2 = f (P0), P3 = f (P1). The pointsP0, . . . , P3 are in general position, so we can complete them to a projective frame{P0, . . . , P4}. In the homogeneous coordinates x0, . . . , x3 of P

3(K) induced by thisframe, r has equations x2 = x3 = 0, s has equations x0 = x1 = 0 and the projectiveisomorphism f is given by [y0, y1, 0, 0] → [0, 0, ay0, by1], with a, b ∈ K

∗. So X isthe set of points of coordinates [λy0,λy1,μay0,μby1], where [λ,μ], [y0, y1] vary inP1(K). It is easy to check that the points of X satisfy the equation

0 = ax0x3 − bx1x2 = det

(ax0 bx1x2 x3

), (4.3)

which defines a non-degenerate quadric Q of P3(K) of hyperbolic type.

Conversely, let R ∈ Q be a point of coordinates [c0, c1, c2, c3]. If c0 = c1 = 0,then R belongs to s = f (r), hence to X . Otherwise, let P ∈ r be the point ofcoordinates [c0, c1, 0, 0]. As Eq. (4.3) means precisely that [ac0, bc1] = [c2, c3] ∈P1(K), the point f (P) has coordinates [0, 0, c2, c3]. Therefore R lies on the line

L(P, f (P)) ⊆ X .

Note. Exercise 178 shows that it is possible to associate a hyperbolic quadric toevery projective isomorphism between two skew lines of P

3(K).Conversely, given a hyperbolic quadric Q, denote by X1, X2 the two rulings

of Q (cf. Exercise 177). If r, r ′, r ′′ are distinct lines of X1 and f : r → r ′ is theperspectivity centred at r ′′, then f does not depend on r ′′ and Q coincides withthe quadric associated to f . Namely, if P ∈ r , denote by sP the only line of X2

passing through P; for any r ′′ �= r, r ′, the line sP intersects both r ′ and r ′′, andf (P) = sP ∩ r ′. Thanks to this observation one can give an alternative proof ofExercise 39, which states that every projective isomorphism between skew lines ofP3(K) is a perspectivity, whose center can be chosen in infinitely many different

ways.

Exercise 179. Let r1, r2, r3 be pairwise skew lines of P3(K). Let X be the set of

lines of P3(K) such that s ∈ X if and only if s ∩ ri �= ∅ for i = 1, 2, 3. Prove that

Z = ⋃s∈X s is the support of a non-degenerate hyperbolic quadric of P

3(K) thatcontains r1 ∪ r2 ∪ r3.

Solution. Let f : r1 → r2 be the perspectivity centred at r3, and set Y = {L(P,

f (P)) | P ∈ r1}.Wewill first show that Y = X . Thanks to Exercise 178, this ensuresthat Z is the support of a non-degenerate quadric of P

3(K) of hyperbolic type. Thenwe will prove that r1 ∪ r2 ∪ r3 ⊆ Z .

226 4 Exercises on Conics and Quadrics

Let s ∈ X be a line. For i = 1, 2, 3, we set Qi = ri ∩ s. The plane H = L(Q1, r3)contains s = L(Q1, Q3). Since Q2 ∈ s, H coincideswith the plane L(Q2, r3). Hencewe have f (Q1) = Q2 and s = L(Q1, f (Q1)) ∈ Y . Therefore, we have X ⊆ Y .

Conversely, let P ∈ r1. By definition of perspectivitywe have f (P) = L(P, r3) ∩r2, so that the lines L(P, f (P)) and r3, being coplanar, do intersect. Therefore, fromthe fact that L(P, f (P)) ∩ r1 = P and L(P, f (P)) ∩ r2 = f (P) we deduce thatthe line L(P, f (P)) belongs to X . Since P can be arbitrarily chosen, this impliesthat Y ⊆ X .

In order to conclude, we are left to show that r1 ∪ r2 ∪ r3 ⊆ Z . Since X = Y , theset Z contains both r1 and r2, which are the domain and the image of f , respectively.Moreover, if R ∈ r3 then it is immediate to check that the plane L(r2, R) intersectsr1 at one point P . Therefore, being coplanar with r2, the line L(P, R) intersects eachof r1, r2, and r3, hence it belongs to X . We have thus shown that every point of r3belongs to a line of X , whence to Z .

Exercise 180. Let r1, r2, r3 be pairwise skew lines of P3(K). Prove that there exists

a unique quadric Q such that ri ⊂ Q for i = 1, 2, 3, and that Q is non-degenerate.

Solution. Exercise 179 implies that at least one quadricQ containing r1, r2, r3 exists.This quadric is non-degenerate of hyperbolic type and its support is equal to

⋃s∈X s,

where X is the set of lines of P3(K) that intersect each of r1, r2, r3. Moreover, by

Exercise 176 (a), no singular quadric can contain r1, r2, and r3.Suppose by contradiction thatQ andQ′ are distinct quadrics that contain r1, r2, r3,

and let A and A′ be symmetric 4 × 4 matrices that defineQ andQ′, respectively. Forevery [λ,μ] ∈ P

1(K) denote by Qλ,μ the quadric defined by the matrix λA + μA′.Every quadric Qλ,μ contains r1, r2, and r3. Moreover, if K = C then there existsat least one homogeneous pair [λ0,μ0] such that Qλ0,μ0 is degenerate, against theprevious discussion.

If K = R, thenQ is a real quadric containing r1, r2, and r3, and the complexifica-tion QC of Q is a complex quadric containing the complexifications of the lines r1,r2, and r3, so that QC is uniquely determined. This implies that also Q is uniquelydetermined.

Exercise 181. Consider the lines of P3(R) having equations

r1 = {x0 + x1 = x2 + x3 = 0}, r2 = {x0 + x2 = x1 − x3 = 0},r3 = {x0 − x1 = x2 − x3 = 0}.

Determine the equation of a quadric containing r1, r2 and r3.

Solution (1). It is easy to check that the lines ri are pairwise skew. Therefore, byExercise 180 there exists a unique quadricQ containing r1, r2 and r3, and this quadricis non-degenerate.

4 Exercises on Conics and Quadrics 227

Let

F1(x) = (x0 + x1)(x0 + x2), F2(x) = (x2 + x3)(x0 + x2),

F3(x) = (x0 + x1)(x1 − x3), F4(x) = (x2 + x3)(x1 − x3)

and, for i = 1, . . . , 4, let Qi be the quadric defined by the equation Fi (x) = 0.Each Qi has rank equal to 2, and contains both r1 and r2. Let us consider a linearcombination F = α1F1 + α2F2 + α3F3 + α4F4 of the polynomials Fi , where αi ∈R, and let us impose that F vanishes on r3. Since deg F = 2, to this aim it is sufficientto require that F(P1) = F(P2) = F(P3) = 0 for three distinct points P1, P2, P3 ∈ r3.Choosing the points P1 = [1, 1, 0, 0], P2 = [0, 0, 1, 1] and P3 = [1, 1, 1, 1], we getthe equations α1 + α3 = 0, α2 − α4 = 0, α1 + α2 = 0.

A non-trivial solution of this system is given by α1 = −1, α2 = α3 = α4 = 1.With this choice we have F(x) = −x20 + x21 + x22 − x23 , and this polynomial definesa quadric with the required properties.

Solution (2). By Exercise 179, the support of a quadric containing r1, r2 and r3 isgiven by the set

⋃s∈X s, where X is the set of lines that intersect each of r1, r2 and

r3. Moreover, it is possible to characterize the lines in X by studying the projectionsπi j : P

3(R) \ ri → r j on r j centred at ri , i, j ∈ {1, 2, 3}, i �= j . This will allow us toobtain an analytic formulation of the condition that a point belongs to a line in X ,thus getting an equation of the desired quadric.

A point Q ∈ P3(R) \ (r1 ∪ r2 ∪ r3) belongs to

⋃s∈X s if and only if π21(Q) =

π31(Q). Indeed, for a given line s ∈ X let us set Qi = s ∩ ri , i = 1, 2, 3. Thenfor every Q ∈ s \ {Q1, Q2, Q3} and for every i �= j ∈ {1, 2, 3} we have πi j (Q) =Q j . On the other hand, if Q ∈ P

3(R) \ (r1 ∪ r2 ∪ r3) and π21(Q) = π31(Q) = Q1,denote by s the line L(Q, Q1). Of course we have s ∩ r1 = Q1. Moreover, fori = 2, 3 we have s ⊂ L(Q, ri ) so that s and ri are incident, and s belongs to X .

Let us denote by [y0, y1, y2, y3] the coordinates of a point Q /∈ r2. The planeL(Q, r2) has equation (y3 − y1)(x0 + x2) + (y0 + y2)(x1 − x3) = 0 and the pointL(Q, r2) ∩ r1 has coordinates

[y0 − y1 + y2 + y3,−y0 + y1 − y2 − y3, y0 + y1 + y2 − y3,−y0 − y1 − y2 + y3].

Hence the projection π21 maps the point of coordinates [y0, y1, y2, y3] to the pointof coordinates

[y0 − y1 + y2 + y3,−y0 + y1 − y2 − y3, y0 + y1 + y2 − y3,−y0 − y1 − y2 + y3].

A similar computation gives the following expression of π31 in coordinates:

[y0, y1, y2, y3] → [y0 − y1,−y0 + y1, y2 − y3,−y2 + y3].

228 4 Exercises on Conics and Quadrics

Since r1 has equations x0 + x1 = x2 + x3 = 0, the points π21(Q) and π31(Q) areuniquely determined by the first and the third coordinates. Therefore, if Q /∈ (r2 ∪ r3)we have that π21(Q) = π31(Q) if and only if

rk

(y0 − y1 + y2 + y3 y0 + y1 + y2 − y3

y0 − y1 y2 − y3

)≤ 1,

i.e. if and only if

0 = det

(y0 − y1 + y2 + y3 y0 + y1 + y2 − y3

y0 − y1 y2 − y3

)= −y20 + y21 + y22 − y23 .

Then let Q be the quadric defined by the equation −y20 + y21 + y22 − y23 = 0. Byconstruction, every line s ∈ X has infinitely many points in common with Q, henceit is contained in Q. It follows that Q contains X , so Q contains r1, r2 and r3.

Exercise 182. (Polar of a line with respect to a quadric) LetQ be a non-degeneratequadric of P

3(K) and let r be a projective line. Prove that:

(a) When P varies in r , the planes pol(P) all intersect in a line r ′ (called the polarline of r with respect to Q).

(b) The polar of r ′ is r .(c) r = r ′ if and only if r is contained in Q.(d) If r and r ′ are distinct and incident, then r and r ′ are tangent to Q at the point

r ∩ r ′.(e) r is tangent to Q at a point P if and only if r ′ is tangent to Q at the same point

P .(f) If r is not tangent to Q, then there exist two distinct planes containing r and

tangent to Q if and only if r ′ is secant to Q (i.e. it intersects Q exactly at twodistinct points). In this case, the planes are tangent to Q exactly at the points ofQ ∩ r ′.

Solution. (a) The projective isomorphism pol : P3(K) → P

3(K)∗ transforms the liner into a line of P

3(K)∗, i.e. into a pencil of planes of P3(K) centred at a line r ′. In

particular, for any given points M and N of r , the planes pol(M) and pol(N ) aredistinct and their intersection coincides with the line r ′.

Statement (b) immediately follows from the reciprocity property of polarity.Hence the line r ′ is the locus of poles of the planes through the line r .

(c) Suppose r = r ′. Then for every point M ∈ r we have r ⊂ pol(M), henceM ∈ pol(M), so that M ∈ Q and r ⊂ Q. Conversely, if r ⊂ Q, then for every pointM ∈ r we have r ⊂ pol(M) = TM(Q), hence r = r ′.

(d) Denote by P the point where the distinct lines r and r ′ intersect, and letH = L(r, r ′) be the plane containing these lines. Since P ∈ r , by definition r ′ ⊂pol(P). Since P ∈ r ′, by (b) we have r ⊂ pol(P), hence pol(P) = H . In particular,P ∈ pol(P); hence P ∈ Q and L(r, r ′) = H = pol(P) = TP(Q). It follows that rand r ′ are tangent to Q at P .

4 Exercises on Conics and Quadrics 229

(e) If r = r ′, the conclusion trivially follows. Then we may suppose that r and r ′are distinct lines. If r is tangent to Q at P , then r is contained in the tangent planeTP(Q) = pol(P) so that for every point M ∈ r we have M ∈ pol(P). By reciprocityP ∈ ⋂

M∈r pol(M) = r ′, so r and r ′ are distinct and incident at P . Then by (d) alsor ′ is tangent to Q at P .

The converse implication immediately follows from what we have just provedand from (b).

(f) If r ′ is secant and intersectsQ at the distinct pointsM ′ and N ′, then by definitionof polar line and by (b) the planes TM ′(Q) = pol(M ′) and TN ′(Q) = pol(N ′) passthrough r .

Conversely, let us prove that if there exists at least one plane H passing throughr and tangent to Q, then r ′ is secant. Namely, if H is tangent to Q at a point M ,then M ∈ r ′, because r ′ is the locus of poles of the planes through r . Therefore,M ∈ Q ∩ r ′. Moreover, by (e) r ′ is not tangent to Q, hence it is secant.

Exercise 183. Find the planes of P3(C) that are tangent to the quadricQ of equation

F(x0, x1, x2, x3) = 2x21 + x22 − 2x1x2 − x23 + 2x0x1 = 0

and that contain the line r of equations x2 − x3 = 0, x0 + 3x1 − 3x2 = 0.

Solution (1). It is easy to check that the quadric Q is non-degenerate. The requiredplanes are the planes of the pencil F centred at r that intersect Q in a degenerateconic. The planes of F have equation

λ(x2 − x3) + μ(x0 + 3x1 − 3x2) = 0

where [λ,μ] varies in P1(C). Let us first discuss whether the plane H = {x2 −

x3 = 0} corresponding to the choice [λ,μ] = [1, 0] is tangent to the quadric or not.Using the homogeneous coordinates x0, x1, x2 on H , the conicQ ∩ H has equation

G(x0, x1, x2) = F(x0, x1, x2, x2) = 2x21 − 2x1x2 + 2x0x1 = 0.

Since this conic is degenerate, H is tangent to Q.After excluding H from the pencil F , the remaining planes Ht of the pencil are

described by the equation t (x2 − x3) + x0 + 3x1 − 3x2 = 0, where t varies inC. Weendow Ht with the homogeneous coordinates x0, x1, x2. Then, by substituting x0 =−3x1 + (3 − t)x2 + t x3 in F , we obtain that the conic Q ∩ Ht of the plane Ht hasequation −4x21 + x22 − x23 + (4 − 2t)x1x2 + 2t x1x3 = 0, so that it is represented by

the matrix At =⎛⎝

−4 2 − t t2 − t 1 0t 0 −1

⎞⎠ . Since det At = −4t + 8, the conic Q ∩ Ht

is degenerate (that is, Ht is tangent to Q) if and only if t = 2. Therefore, the onlyplanes containing r and tangent to Q are given by the plane H2 having equationx0 + 3x1 − x2 − 2x3 = 0, and the plane x2 − x3 = 0 previously found.

230 4 Exercises on Conics and Quadrics

Solution (2). The points R = [3,−1, 0, 0] and S = [0, 1, 1, 1] belong to r , andhence r ′ = pol(R) ∩ pol(S). We thus get that r ′ is the line of equations −x0 + x1 +x2 = 0, x0 + x1 − x3 = 0. By solving the system given by the equations of r ′ andthe equation ofQ, we obtain thatQ ∩ r ′ consists of the points M ′ = [1, 0, 1, 1] andN ′ = [1, 1, 0, 2]. By Exercise 182 and using thatQ is non-degenerate, we concludethat the required planes are the plane pol(M ′) and pol(N ′) having equation x2 − x3 =0 and x0 + 3x1 − x2 − 2x3 = 0, respectively.

Exercise 184. Let Q be a non-degenerate hyperbolic quadric of P3(R); let r be a

line that is not tangent to Q and let r ′ be its polar (cf. Exercise 182). Prove that:

(a) r is secant if and only if r ′ is secant.(b) There exist two planes of P

3(R) that pass through r and are tangent to Q if andonly if r is secant.

Solution. (a) If r ′ is secant and intersects the quadric at two distinct pointsM ′ and N ′,then by Exercise 182 (f) there exist two planes H1 and H2 that pass through r and aretangent toQ at the points M ′ and N ′, respectively. Let X1 and X2 be the families oflines contained in the hyperbolic quadricQ that are described in Exercise 177. SinceQ is hyperbolic, we have Q ∩ H1 = Q ∩ TM ′(Q) = m1 ∪ m2, where m1 ∈ X1 andm2 ∈ X2. In the same way Q ∩ H2 = Q ∩ TN ′(Q) = n1 ∪ n2, where n1 ∈ X1 andn2 ∈ X2. Observe that, since H1 ∩ H2 = r and r is not contained inQ, the four linesm1,m2, n1, n2 are distinct. Then Exercise 176 implies that there is no point belongingto three of these four lines. It follows from Exercise 177 that the lines m1 and n2intersect at a point of Q, and on the other hand m1 ∩ n2 ∈ H1 ∩ H2 = r , so that thepointm1 ∩ n2 belongs toQ ∩ r . In the same way the pointm2 ∩ n1 belongs toQ ∩ rand is distinct from m1 ∩ n2 because of the previous considerations. Therefore, r issecant.

The converse implication immediately follows from the fact that r is the polarof r ′.

(b) is an obvious consequence of (a) and of Exercise 182 (f).

Note. Let us consider the quadric Q and the line r defined in Exercise 183. In thatcase, both Q and r were defined by polynomials with real coefficients. Moreover, itis not difficult to check thatQ is hyperbolic, and that r is secant because it intersectsQ at the distinct real points M = [3, 0, 1, 1] and N = [0, 1, 1, 1]. Coherently withwhat we have just proved, in that exercise we found two planes that pass through rand are tangent to Q.

K Exercise 185. LetQ be a non-empty non-degenerate elliptic quadric of P3(R); let r

be a line that is not tangent toQ and let r ′ be the polar of r with respect toQ. Provethat:

(a) r is external if and only if r ′ is secant.(b) There exist two planes of P

3(R) passing through r and tangent toQ if and onlyif r is external.

4 Exercises on Conics and Quadrics 231

Solution. (a) Suppose that r ′ is secant and intersects the quadric at the distinctpointsM ′ and N ′. Then TM ′(Q) ∩ TN ′(Q) = pol(M ′) ∩ pol(N ′) = r ;moreover,Q ∩TM ′(Q) = {M ′} and Q ∩ TN ′(Q) = {N ′} because the quadric is elliptic. The lines rand r ′ are distinct, because r is not contained in the quadric, and they are not incident,since r is not tangent (cf. Exercise 182). Therefore, if there exists a point R ∈ Q ∩ r ,then necessarily R �= M ′, which contradicts the fact that Q ∩ r ⊂ Q ∩ TM ′(Q) ={M ′}.

Conversely, suppose that r is external. Then by Exercise 182 (e) the line r ′ is nottangent Q.

The complexificationQC is a non-degenerate complex quadric, so it is hyperbolic.Let X1 and X2 be the two families of complex lines contained inQC that are describedin Exercise 177. Observe that the conjugation σ preserves incidence relations, andrecall that lines of the same family are pairwise disjoint, while if li ∈ Xi , i = 1, 2,then l1 and l2 are incident. Therefore, σ either transforms each family of lines intoitself or switches X1 with X2. In order to understand which of these possibilitiesoccurs it is then sufficient to analyse the behaviour ofσ on a single line. Let P ∈ Q andH = TP(QC). Then σ(H) = H , hence the degenerate conic C = QC ∩ H = l1 ∪ l2,where l1, l2 are conjugate complex lines, is left invariant by σ. Since Q is elliptic,the lines l1, l2 cannot be real, so σ(li ) �= li for i = 1, 2. Therefore, σ switches theincident lines l1, l2, so that it switches X1 with X2.

Since in the complex case no line can be external to a projective quadric, let Mand N be the points of intersection between QC and rC. The complex points M andN are conjugate and r ′

Cis the intersection of the tangent spaces H1 = TM(QC) and

H2 = TN (QC), that are also conjugate to each other. We have QC ∩ H1 = QC ∩TM(QC) = m1 ∪ m2 withm1 ∈ X1 andm2 ∈ X2. In the samewayQC ∩ H2 = QC ∩TN (QC) = n1 ∪ n2 with n1 ∈ X1 and n2 ∈ X2. By the previous discussion we haveσ(m1) = n2, hence the lines m1 and n2 intersect at a real point R that belongs tor ′C

= H1 ∩ H2. Therefore R ∈ Q ∩ r ′, so r ′ is secant. In fact, the same argumentalso shows that m2 ∩ n1 belongs to Q ∩ r ′.

(b) is an obvious consequence of (a) and of Exercise 182 (f).

K Exercise 186. Let H be a plane of P3(C), let C ⊂ H be a non-degenerate conic,

let P, P ′ ∈ C be distinct points and t, t ′ ⊂ H the tangent lines to C at P and atP ′, respectively. Let r and s be skew lines of P

3(C) such that r ∩ H = P ′ ands ∩ H = P . Let π : C → r be the map defined by π(Q) = L(s, Q) ∩ r if Q �= Pand π(P) = L(s, t) ∩ r (Fig. 4.15).

(a) Prove that π is well defined and bijective.(b) Set

m = L(r, t ′) ∩ L(s, P ′), X =⋃

Q∈C\{P ′}L(Q,π(Q)) ∪ m,

and prove that X is the support of a non-degenerate quadric Q.

232 4 Exercises on Conics and Quadrics

s

r

t

R0

P

m

t

PH

C

Fig. 4.15 The configuration described in Exercise 186

Solution. (a) The proof is exactly the same as in the solution of Exercise 119 (a).(b) By Exercise 178, it is sufficient to show that there exists a projective isomor-

phism φ : r → s such that X = ⋃R∈r L(R,φ(R)).

Let us first define a projective isomorphismα : Fs → Fr , whereFs is the pencil ofplanes centred at s andFr is the pencil of planes centred at r . Setα(L(s, t)) = L(r, P)

andα(L(s, P ′)) = L(r, t ′). If K ∈ Fs is distinct both from L(s, t) and from L(s, P ′),then K intersects H in a line containing P and distinct both from t and from L(P, P ′).This line in turn intersects C at P and at a point QK distinct both from P and fromP ′. We then set α(K ) = L(r, QK ).

Let us check that α is a projective isomorphism. If FP and FP ′ are the pencils oflines of H with centre at P and P ′, respectively, we define ψ : FP → FP ′ by settingψ(t) = L(P, P ′), ψ(L(P, P ′)) = t ′, and ψ(l) = L(P ′, A) for every l ∈ FP , l �= t ,l �= L(P, P ′), where A is the point of intersection of l with C that is distinct from P .Exercise 142 implies that ψ is a well-defined projective isomorphism. Moreover, ifβs : Fs → FP , βr : Fr → FP ′ are defined by βs(K ) = K ∩ H , βr (K ′) = K ′ ∩ Hfor every K ∈ Fs , K ′ ∈ Fr , then βs and βr are well-defined projective isomorphisms(cf. Exercise 33). Finally, it is immediate to check that α = β−1

r ◦ ψ ◦ βs . Being thecompositionof projective isomorphisms, themapα is itself a projective isomorphism.

Let γ1 : Fs → r be the projective isomorphism defined by K → K ∩ r , andγ2 : Fr → s the projective isomorphism defined by K ′ → K ′ ∩ s (cf. Exercise 32)and let us set φ = γ2 ◦ α ◦ γ1

−1 : r → s. Being the composition of projective iso-morphisms, the map φ is a projective isomorphism.

4 Exercises on Conics and Quadrics 233

Let us now consider the sets of lines

Z1 = {L(R,φ(R)) | R ∈ r}, Z2 = {L(Q,π(Q)) | Q ∈ C \ {P ′}} ∪ {m}.

Since X = ⋃l∈Z2

l, in order to conclude it is sufficient to show that Z1 = Z2.The linem belongs to Z2 by definition. If R0 = L(r, t ′) ∩ s, thenm = L(R0, P ′),

so in order to prove thatm ∈ Z1 it is sufficient to show that φ(P ′) = R0. However, itreadily follows from the definitions that φ(P ′) = γ2(α(L(s, P ′)) = γ2(L(r, t ′)) =L(r, t ′) ∩ s = R0, as desired. Since π(P ′) = P ′, the map π restricts to a bijectionbetween C \ {P ′} and r \ {P ′}. For every R ∈ r \ {P ′}, by construction the pointsR, π−1(R) and φ(R) are collinear, and we have R �= π−1(R) and R �= φ(R). Thisreadily implies that Z1 \ {m} = Z2 \ {m}, so that Z1 = Z2.

Note. The construction described in Exercise 186 may be interpreted as a degen-eration of the construction of Exercise 119, where the case when the point r ∩ Hdoes not belong to C is considered. In that case, the analogue of the set X turns outto be the support of an irreducible cubic surface that, in the “limit” situation wherer ∩ C �= ∅, splits as the union of the plane L(s, H ∩ r) and the quadricQ describedhere.

Exercise 187. Let Q be a quadric of Rn having rank n and let A =

(c tBB A

)∈

M(n + 1, R) be a matrix representing Q. Prove that Q is an affine cone if and onlyif det A �= 0.

Solution. SinceQ has rank n, its projective closureQ has a unique singular point Pof homogeneous coordinates [z0, . . . , zn]; therefore, Sect. 1.8.3 implies that Q is acone whose vertex set is given by the point [z0, . . . , zn], so that Q is an affine coneif and only if z0 �= 0.

Let Z = (z1, . . . , zn) ∈ Rn and Z = (z0, z1, . . . , zn) ∈ R

n+1. The fact that P issingular forQ translates into the equation AZ = 0 or, equivalently, into the equationsAZ + z0B = 0 and tBZ + z0c = 0.

If z0 = 0, then Z �= 0 and AZ = 0, hence det A = 0.If z0 �= 0, since B = −z−1

0 AZ , the vector tB is a linear combination of the rowsof A, hence everyW ∈ R

n such that AW = 0 also satisfies tBW = 0. Let us supposeby contradiction that det A = 0 and let us denote by Z ′ = (z′

1, . . . , z′n) ∈ R

n a vectorsuch that Z ′ �= 0 and AZ ′ = 0. The point Q = [0, z′

1, . . . , z′n] ∈ P

n(R) is singularforQ and distinct from P , and this contradicts the fact that P is the unique singularpoint of Q.

Exercise 188. LetQ be a non-degenerate quadric ofRn , let H be a diametral hyper-

plane and let P = [(0, v)] ∈ Pn(R) be the pole of H with respect toQ, where v ∈ R

n .

(a) Prove that P ∈ H if and only if the vector v is parallel to H .(b) Suppose P /∈ H and denote by τH,v : R

n → Rn the reflection fixing H pointwise

and sending v to −v. Prove that τH,v(Q) = Q.

234 4 Exercises on Conics and Quadrics

Solution. (a) Let v = (v1, . . . , vn) and let a1x1 + · · · + anxn + b = 0 be an equationof H . The hyperplane H is defined by bx0 + a1x1 + · · · + anxn = 0, so that P ∈ Hif and only if a1v1 + · · · + anvn = 0, that is, if and only if v is parallel to H .

(b) By part (a), if P /∈ H then there exist affine coordinates x1, . . . , xn such thatv = (0, . . . , 0, 1) and H = {xn = 0}. In this system of coordinates, τH,v is describedby the formula (x1, . . . , xn) → (x1, . . . , xn−1,−xn). The quadric Q is then repre-

sented by a matrix of the form

⎛⎝

c tB 0B A 00 0 a

⎞⎠, where A ∈ M(n − 1, R), B ∈ R

n−1,

c ∈ R, and a ∈ R∗. It is immediate to check that the matrix defining τH,v(Q) is the

same as the one defining Q, hence τH,v(Q) = Q.

K Exercise 189. (Symmetries of quadrics of Rn) Let Q be a non-degenerate quadric

of Rn , let H ⊂ R

n be an affine hyperplane, and let τH : Rn → R

n be the orthogonalreflection with respect to H . Prove that:

(a) If the projective closure H of H is tangent to Q, then τH (Q) �= Q.(b) H is a principal hyperplane of Q if and only if τH (Q) = Q.

Solution. (a) Let P be the point where Q is tangent to H . If P is a proper point,then up to an isometric change of coordinates on R

n we may suppose that Phas coordinates (0, . . . , 0), and that H is defined by the equation y1 = 0. In thissystem of coordinates, Q is defined by f (y1, . . . , yn) = y1 + f2(y1, . . . , yn) = 0,where f2 is a homogeneous polynomial of degree 2. The orthogonal reflectionτH is given by (y1, . . . , yn) → (−y1, y2, . . . , yn). Therefore, the quadric τH (Q)

is defined by −y1 + f2(−y1, y2, . . . , yn) = 0. Suppose by contradiction that Q =τH (Q): then there exists λ ∈ R

∗ such that f (−y1, y2, . . . , yn) = λ f (y1, . . . , yn).Since f (−y1, y2, . . . , yn) = −y1 + f2(−y1, y2, . . . , yn), we must have λ = −1 andf2(−y1, y2, . . . , yn) = − f2(y1, y2, . . . , yn), that is, each monomial appearing in f2is of type y1y j , for some 2 ≤ j ≤ n. Hence f2 is divisible by y1, and the same is truefor f . However, this is impossible, since Q is non-degenerate, hence irreducible.

If P is a point at infinity, then we can suppose as above that H is the hyper-plane of equation y1 = 0, and that P is the point at infinity of the axis yn , that is,P = [0, . . . , 0, 1]. In this system of coordinates, Q is defined by f (y1, . . . , yn) =c + f1(y1, . . . , yn) + f2(y1, . . . , yn) = 0, where c ∈ R, the polynomial f1 is eitherzero or homogeneous of degree 1, and f2 = y1yn + g2(y1, . . . , yn−1), where g2is homogeneous of degree 2 in y1, . . . , yn−1. In this case τH (Q) is defined byf (−y1, y2, . . . , yn) = c − y1yn + f1(−y1, y2, . . . , yn) + g2(−y1, y2, yn−1) = 0.Then, as in the previous case, it is easy to check that if τH (Q) = Q then y1 dividesf and Q is reducible, against the hypothesis that Q is non-degenerate.

(b) Denote by w ∈ Rn \ {0} a vector orthogonal to H , and by R = [(0, w)] ∈

Pn(R) the point at infinity corresponding to the direction w. Of course we have

R /∈ H ; moreover, H is a principal hyperplane if and only if R is the pole of H withrespect to the projective quadric Q.

4 Exercises on Conics and Quadrics 235

If H is a principal hyperplane, then τH (Q) = Q by Exercise 188 (b). Conversely,suppose that τH (Q) = Q and let P be the pole of H . Denote by τH : P

n(R) →Pn(R) the projectivity induced by τH . The fixed-point set of τH is H ∪ {R}. On the

other hand, since τH (Q) = Q, we also have τH (P) = P . Since by (a) the projectivehyperplane H is not tangent to Q, the point P does not belong to H , hence P = Rand H is a principal hyperplane.

Exercise 190. Let Q be a non-degenerate quadric of Rn , and let A =

(c tBB A

)∈

M(n + 1, R) be amatrix representingQ. Let λ1, . . . ,λh be the non-zero eigenvaluesof A, where λi �= λ j if i �= j , let Vi be the eigenspace of A corresponding to theeigenvalue λi and set di = dim Vi , i = 1, . . . , h. Let also W be the set of principalhyperplanes of Q.

(a) Prove thatW is the union of h proper linear systemsW1, . . . ,Wh (cf. Sect. 1.4.3)such that dimWi = di − 1 for every i = 1, . . . , h.

(b) Prove that the dimension of the affine subspace

J =⋂H∈W

H

is equal to n − rk A.(c) Prove that, ifQ has a centre, then the centre ofQ is the intersection of n pairwise

orthogonal principal hyperplanes.(d) Suppose that Q is a paraboloid. Prove that Q has a unique axis, whose point

at infinity is contained in the projective closure of Q. Also prove that Q is theintersection of n − 1 pairwise orthogonal principal hyperplanes, and thatQ hasa unique vertex.

(e) Suppose thatQ is a sphere. Prove thatQ has a centre and that, if C is the centreofQ, then every plane passing throughC is principal, every line passing throughC is an axis, and every point of the support of Q is a vertex.

(f) Let n = 2. Prove thatQ admits at least one axis, thatQ has a centre if and onlyif it has at least two orthogonal axes, and thatQ is a circle if and only if it has atleast three pairwise distinct axes (and in this case,Q has a centre and every linepassing through its centre is an axis).

Solution. (a) A principal hyperplane of Q is the affine part of the polar hyperplanepol(P), where P = [(0, v)] and v is an eigenvector of A corresponding to a non-zeroeigenvalue (cf. Sect. 1.8.6). For every i = 1, . . . , h, let then Ki = Vi ∩ H0, whereH0 is the hyperplane at infinity, and let Wi be the set of hyperplanes H of R

n suchthat H = pol(P) for some P ∈ Ki . This shows that W = ⋃h

i=1 Wi . Moreover, ifv ∈ Vi \ {0} and X = (x1, . . . , xn) are the usual affine coordinates of R

n , then anequation of the principal hyperplane corresponding to v is given by tvAX + tBv = 0or, equivalently, by tvX = − 1

λi

tvB. It readily follows that, for every i = 1, . . . , h,

the linear system of projective hyperplanes pol(Ki ) does not contain the hyperplane

236 4 Exercises on Conics and Quadrics

at infinity H0, so that Wi is a proper linear system. Since pol : Pn(R) → P

n(R)∗ isan isomorphism, we finally have dimWi = dim Ki = di − 1 for every i = 1, . . . , h.

(b) For every i = 1, . . . , h, let Bi be an orthogonal basis of Vi , and let us set{v1, . . . , vm} = B1 ∪ . . . ∪ Bh . By the Spectral theorem A is similar to a diagonalmatrix, hence from the definition of the Vi one easily gets that m = d1 + . . . + dh =rk A. Moreover, the eigenspaces of A are pairwise orthogonal, hence vi is orthogonalto v j for every i �= j , i, j ∈ {1, . . . ,m}.

We have seen in the solution of (a) that the affine subspace J is defined by thelinear system ⎧⎪⎨

⎪⎩

tv1X = c1...

tvm X = cm

(4.4)

where c1, . . . , cm are real numbers. Since {v1, . . . , vm} is a set of linearly independentvectors, it readily follows that J is an affine subspace of R

n of dimension n −m = n − rk A. Also observe that, the vi being pairwise orthogonal, the system (4.4)describes J as the intersection of m pairwise orthogonal principal hyperplanes.

(c) Since the centre C of Q coincides with the pole of the hyperplane at infinity(cf. Sect. 1.8.6), if H is a principal hyperplane, then by reciprocity we have C ∈ H .Therefore, C ∈ ⋂

H∈W H = J . On the other hand, since Q has a centre, we haverk A = n (cf. Sect. 1.8.5), and part (b) implies that dim J = n − n = 0, so that J ={C}. Moreover, the solution of (b) also shows that J is the intersection of n pairwiseorthogonal principal hyperplanes.

(d) As observed in Sect. 1.8.5, since Q is a non-degenerate paraboloid we haverk A = n − 1. Therefore, by part (b) we have dim J = 1, and J is a line. Being theintersection of principal hyperplanes, J is an axis of Q. Moreover, if r is any axisof Q, by definition r is the intersection of principal hyperplanes, so that r ⊇ J , andr = J since they have the same dimension. Hence J is the unique axis ofQ and, asproved in the solution of (b), J is indeed the intersection of n − 1 pairwise orthogonalprincipal hyperplanes.

Let us now prove that the point at infinity of J belongs to Q. Since rk A =n − 1, there exists a non-zero vector v0 ∈ Ker A ⊆ R

n . Set P0 = [(0, v0)] ∈ H0.With the same notation as in the solution of (b), the Spectral theorem implies that v0is orthogonal to vi for every i = 1, . . . ,m = n − 1. From the explicit equations ofJ described in (b) it is now immediate to deduce that P0 is the point at infinity of J .Moreover, since Av0 = 0 we have P0 ∈ Q, as desired.

By definition of vertex, in order to conclude it suffices to prove thatQ ∩ J consistsof a single point. From Av0 = 0 we deduce that TP0(Q) = H0. Since of course J isnot contained in H0, we thus have I (Q, J , P0) = 1. It follows that J intersectsQ ata unique point V , which is the unique vertex of Q.

(e) By definition of sphere, Q has equation

(x1 − a1)2 + . . . + (xn − an)

2 = η

4 Exercises on Conics and Quadrics 237

for some η ∈ R, so that A = Id, tB = (−a1 . . . − an) and c = a21 + . . . + a2n − η.In particular, A being invertible, the quadricQ has a centre. Since the coordinates ofthe centre C of Q satisfy the system AX = −B, we also have C = (a1, . . . , an).

Since all vectors of Rn are eigenvectors for A, a hyperplane of R

n is principalif and only if it is equal to pol([0, v]) for some v ∈ R

n \ {0}, i.e. if and only if it isdescribed by the equation

v1x1 + v2x2 + . . . + vnxn = v1a1 + v2a2 + . . . + vnan

for some v = (v1, . . . , vn) ∈ Rn \ {0}. It readily follows that a hyperplane of R

n isprincipal if and only if it passes through C . Moreover, every line passing throughC is the intersection of n − 1 hyperplanes passing through C , so the axes of Q areprecisely the lines passing through C , and every point of the support ofQ is a vertexof Q.

(f) Let us first prove that for n = 2 the notions of axis and of principal hyperplanecoincide, so that the axes ofQ bijectively correspond to the linear subspaces spannedby an eigenvector of A corresponding to a non-zero eigenvalue (cf. Sect. 1.8.6).More precisely, the subspace spanned by an eigenvector v �= 0 determines an axisorthogonal to v. Since A is symmetric and rk A ≥ 1, the Spectral theorem implies thatQ admits at least one axis, and that Q admits two axes if and only if A is invertible,that is if and only if Q has a centre.

Recall that Q is a circle if and only if it is defined by an equation of the form(x − x0)2 + (y − y0)2 = c, where x0, y0, c ∈ R. This readily implies that Q is acircle if and only if A is a non-zero multiple of the identity, i.e. (since n = 2) if andonly if A admits three pairwise linearly independent eigenvectors corresponding tonon-zero eigenvalues. We can thus conclude thatQ is a circle if and only if it admitsthree distinct axes (and in this case, as we have seen in the solution of (e), Q has acentre and every line passing through the centre is an axis).

Exercise 191. Let Q be a non-degenerate quadric of Rn and let V be a vertex of

Q. Prove that there exists precisely one axis of Q containing V , and that this axis isorthogonal to the tangent hyperplane to Q at V .

Solution. Of course it is sufficient to show that, if r is an axis passing through V ,then r is orthogonal to TV (Q). By definition, r is the intersection of n − 1 principalhyperplanes whose projective closures are given by n − 1 projective hyperplanespolQ(P1), . . . , polQ(Pn−1), where Pi = [(0, vi )] ∈ H0 for i = 1, . . . , n − 1. Moreexplicitly, as we proved in the solution of Exercise 190 (b), if A is a symmetric matrixrepresentingQ, then for every i = 1, . . . , n − 1 the vector vi is an eigenvector of Acorresponding to a non-zero eigenvalue, and the principal hyperplane correspondingto Pi is described by an affine equation of the form tvi X = ci for some ci ∈ R. Inparticular, since dim r = 1, the vectors v1, . . . , vn−1 are linearly independent, so thatthe direction of r is spanned by the unique (up to multiplication by a non-zero scalar)vector vr ∈ R

n \ {0} such that tvivr = 0 for every i = 1, . . . , n − 1.

238 4 Exercises on Conics and Quadrics

Since V ∈ r = ⋂n−1i=1 polQ(Pi ), by reciprocity every Pi belongs to polQ(V ), i.e. to

the projective hyperplane tangent toQ at V . It readily follows that every vi belongsto the linear subspace associated to the affine hyperplane TV (Q). Being linearlyindependent, the vectors v1, . . . , vn−1 form a basis of the linear subspace associatedto TV (Q). From the fact that tvivr = 0 for every i = 1, . . . , n − 1we can then deducethat r is orthogonal to TV (Q), as desired.

Exercise 192. Let P1, P2, P3 ∈ R2 be non-collinear points. Prove that there exists

a unique circle C containing P1, P2, P3.

E Solution. If O is the point of intersection between the axes of the segments P1P2and P2P3, and ρ is the distance between O and P1, then an easy argument in planegeometry shows that the unique circle containing P1, P2 and P3 has centre in O andhas radius equal to ρ.

Here we offer an alternative solution, which is based on the characterization ofcircles described in Sect. 1.8.7. In fact, we recall that, if I1 = [0, 1, i] and I2 =[0, 1,−i] are the cyclic points of the Euclidean plane, then a conic C of R

2 is a circleif and only if the complexification CC of the projective closure C of C contains I1 andI2.

The points P1, P2, P3, I1, I2 of P2(C) are in general position: since P1, P2, P3

are projectively independent, the line L(I1, I2) = {x0 = 0} does not contain any Pi ,i = 1, 2, 3, and the lines L(Pi , Pj ), i, j ∈ {1, 2, 3}, intersect the line at infinity x0 = 0of P

2(C) in real points, which of course are distinct from I1 and I2. Therefore, thereexists a unique conic Q of P

2(C) such that P1, P2, P3, I1, I2 ∈ Q, and this conicis non-degenerate (cf. Sect. 1.9.6). The conjugate conic σ(Q) contains the pointsPi = σ(Pi ), i = 1, 2, 3, and the points I1 = σ(I2) and I2 = σ(I1). The uniquenessofQ implies thatσ(Q) = Q, hence there exists a conicD ofP

2(R) such thatQ = DC

(cf. Exercise 59). The affine part C = D ∩ R2 ofD is a circle passing through P1, P2

and P3.Let us now prove the uniqueness of C. If C ′ is a circle of R

2 containing P1, P2and P3, the complexification Q′ = C ′

C of the projective closure of C ′ is a conic ofP2(C) containing P1, P2, P3, I1, I2. Again by the uniqueness ofQ, we haveQ′ = Q,

whence C = C ′, that is, C is the unique circle of R2 passing through P1, P2 and P3.

Exercise 193. Let C be a parabola of R2 and let P be any point of C. Prove that

there exists an affinity ϕ : R2 → R

2 such that ϕ(P) = (0, 0) and ϕ(C) has equationx2 − 2y = 0.

Solution. Let τP ⊆ P2(R) be the projective line tangent to C at P , let Q ∈ P

2(R) bethe intersection point between τP and the line at infinity, and let R ∈ P

2(R) be thepoint at infinity of C. It is immediate to check that, if S ∈ R

2 ⊆ P2(R) is a point of

C distinct from P , then the points P, Q, R, S form a projective frame of P2(R). Let

f : P2(R) → P

2(R) be the projectivity such that f (P) = [1, 0, 0], f (Q) = [0, 1, 0],f (R) = [0, 0, 1], f (S) = [1, 1, 1]. Since f (Q) and f (R) lie on the line at infinity,f restricts to an affinity ψ of R

2.

4 Exercises on Conics and Quadrics 239

Let C ′ = ψ(C), and let g(x, y) = 0 be an equation of C ′. By construction C ′ passesthrough O = (0, 0), and the tangent to C ′ at O is the affine part of the line passingthrough O and [0, 1, 0], i.e. the line of equation y = 0. Therefore, C ′ is described byan equation of the form g(x, y) = y + ax2 + bxy + cy2, where at least one amonga, b, c is non-zero. Since R is the unique point at infinity of C, we also have that[0, 0, 1] is the unique point at infinity of C ′, so that b = c = 0, while the fact thatS ∈ C implies that (1, 1) ∈ C ′, so that a = −1. Therefore, C ′ has equationx2 − y = 0. The required affinity is then the composition of ψ with the affinity

η of R2 such that η(x, y) =

(x, y2

)for every (x, y) ∈ R

2.

Note. The previous exercise readily implies that the group of the affinities ofR2 that

leave a parabola C invariant acts transitively on C. This is clearly false if we considerisometries rather than affinities, because the vertex of C is fixed by every isometryof R

2 that leaves C invariant.

Exercise 194. Let C be a non-empty non-degenerate conic of R2. Prove that:

(a) If C has a centre but is not a circle, then C has two foci, each of which is distinctfrom the centre O of C.

(b) If C is a circle, then the center O of C is the unique focus of C.(c) If C is a parabola, then C has a unique focus.

Solution. Let I1 = [0, 1, i] and I2 = [0, 1,−i] be the cyclic points of the Euclideanplane, as usual. Let D = CC be the complexification of C, let D be the projectiveclosure of D, and recall that C is a circle if and only if I1, I2 ∈ D (cf. Sect. 1.8.7).

(a) Since C has a centre, the line at infinity x0 = 0 is not tangent to D (cf.Sect. 1.8.5). Moreover, as previously observed,D does not contain the cyclic points.Hence there are two distinct lines r1 and r2 of P

2(C) that pass through I1 and are tan-gent toD at the proper points Q1 = D ∩ r1, Q2 = D ∩ r2. Observe that if ri , i = 1, 2,passes through the centre O of C, then O ∈ ri = pol(Qi ), so that Qi ∈ pol(O) byreciprocity, which contradicts the fact that pol(O) is the line at infinity. Therefore,O /∈ r1 ∪ r2. The conjugate lines s1 = σ(r1) and s2 = σ(r2) pass through I2 and theyalso are tangent to D, since σ(D) = D. By Exercise 60, the points F1 = r1 ∩ s1and F2 = r2 ∩ s2 are precisely the real points belonging to r1 ∪ r2 ∪ s1 ∪ s2, so that{F1, F2} is the set of foci of C. As previously observed, we have O /∈ r1 ∪ r2, so thefoci F1 and F2 are distinct from O .

(b) If C is a circle, then I1 and I2 belong to D. Denote by r the line tangent to Dat I1. Since the line at infinity x0 = 0 intersects D at I1 and I2, the line r is distinctfrom {x0 = 0}. The conjugate line s = σ(r) is tangent to D at I2, hence it is distinctfrom r . The lines r and s, being the polars of points at infinity, intersect in the centreO of C. By Exercise 60, O is the unique real point of r ∪ s, so it is the unique focusof C.

(c) If C does not have a centre, then the conicD has precisely one point at infinity,which is real and distinct from I1 and I2. There are two lines that pass through I1and are tangent toD: the line at infinity x0 = 0 and another line r . The lines tangentto D and passing through I2 are the lines x0 = 0 and the conjugate line s = σ(r).

240 4 Exercises on Conics and Quadrics

By Exercise 60, the point F = r ∩ s is the unique real point of r ∪ s, hence it is theunique focus of C.Exercise 195. If C is a non-degenerate conic of R

2, prove that:

(a) If C has a centre and is not a circle, then one of the axes of C contains both fociof C.

(b) If C is a parabola, then the focus F of C belongs to the unique axis of C.

E Solution.The exercise can be readily solved analytically, by performing an isometricchange of coordinates that transforms C into its canonical form and by using theformulas for computing the foci of Exercise 196. We propose here a synthetic proof.

(a) By Exercise 194, C has two foci F1, F2, each distinct from the centre O of C.Moreover, by Exercise 190 (f), the conic C has exactly two mutually orthogonal axesr1 and r2. For i = 1, 2 denote by τi : R

2 → R2 the orthogonal reflection with respect

to the line ri . By Exercise 189, the isometries τ1 and τ2 (and hence τ = τ1 ◦ τ2 too)leave C invariant and therefore preserve the set {F1, F2}. As r1 is orthogonal to r2,the map τ is a central symmetry whose unique fixed point O = r1 ∩ r2 is the centreof C; so τ exchanges F1 and F2. It follows that, for instance, τ1 fixes F1 and F2 whileτ2 exchanges them. Since the fixed-point set of τ1 is r1, we have that F1, F2 ∈ r1.

(b) Note that by Exercise 190 (f) the conic C, which does not have a centre, has aunique axis r . By Exercise 194, C has only one focus F , which is therefore a fixedpoint for any isometry τ : R

2 → R2 such that τ (C) = C. If we apply this observation

to the orthogonal reflection with respect to r , we obtain that F belongs to r .

Note. If C is a non-degenerate conic with a centre and it is not a circle, the axiscontaining the foci of C is called focal axis. If C is an ellipse, the focal axis is alsocalled major axis, while if C is a hyperbola it is also called transverse axis. In bothcases, the focal axis intersects C at two vertices.

Exercise 196. In each of the following cases (a), (b), (c), (d), determine axes, ver-tices, foci and directrices of the non-degenerate conic C ofR

2 defined by the equationf (x, y) = 0:

(a) f (x, y) = x2

a2+ y2

b2− 1, with a > b > 0;

(b) f (x, y) = x2

a2+ y2

a2− 1, with a > 0;

(c) f (x, y) = x2

a2− y2

b2+ 1, with a > 0, b > 0;

(d) f (x, y) = x2 − 2cy, with c > 0.

Solution. (a) Since f (x, y) = f (−x,−y), the conic C has the origin O = (0, 0) ofR

2 as a centre. Moreover, C is represented by the matrix

A =⎛⎜⎝

−1 0 0

0 1a2

0

0 0 1b2

⎞⎟⎠ .

4 Exercises on Conics and Quadrics 241

As a �= b, the eigenspaces of the matrix A =⎛⎝

1a2

0

0 1b2

⎞⎠ relative to non-zero eigen-

values are the lines generated by (1, 0) and by (0, 1). Since the line polC([0, 1, 0])has equation x1 = 0 and the line polC([0, 0, 1]) has equation x2 = 0, the axes of Care the lines of equations x = 0 and y = 0. If we intersect these lines with C, we findthe vertices of C, which have coordinates

(a, 0), (−a, 0), (0, b), (0,−b).

Let us now compute the foci of C. If CC denotes the complexified conic ofC, we start by determining the lines of C

2 whose projective closures are tangentto CC and pass through one of the cyclic points I1 = [0, 1, i], I2 = [0, 1,−i]. Theconic CC has equation b2x21 + a2x22 − a2b2x20 = 0. If F1 is the pencil of lines ofP2(C) with centre I1, the set of lines of F1 distinct from the improper line coin-

cides with the set of lines rλ of equation λx0 + i x1 − x2 = 0, as λ varies in C. Ifwe substitute the equality x2 = λx0 + i x1 into the equation of CC, we obtain theequation

−(a2 − b2)x21 + 2λa2i x0x1 + a2(λ2 − b2)x20 = 0,

which has a double root if and only if (λa2i)2 + a2(a2 − b2)(λ2 − b2) = 0, that isiff λ = λ1 = −i

√a2 − b2 or λ = λ2 = i

√a2 − b2. So rλ is tangent to CC if and

only if either λ = λ1 or λ = λ2. Let now j = 1 or j = 2. By Exercise 60, the onlyreal point of rλ j is Fj = rλ j ∩ σ(rλ j ). In addition, as σ(CC) = CC, the line σ(rλ j ) =σ(L(Fj , I1)) = L(Fj , I2) is tangent to CC. Therefore the isotropic lines passingthrough Fj are tangent to CC, so that Fj is a focus of C. By using the equations ofrλ1 and rλ2 found above, it is now immediate to check that F1 and F2 have affinecoordinates

F1 =(√

a2 − b2, 0)

, F2 =(−

√a2 − b2, 0

).

Furthermore, being the only real points that belong to lines passing through I1 andtangent to CC, the points F1 and F2 are the only foci of C.

Finally, the directrices r1 and r2 of C are the affine parts of the lines pol(F1) andpol(F2), so they have equations

x = a2√a2 − b2

, x = − a2√a2 − b2

,

respectively.

(b) In this case C is a circle. Moreover, since f (x, y) = f (−x,−y), the centre ofC is the origin O = (0, 0) of R

2. By Exercise 190 (f), the axes of C are precisely thelines passing through O , and all points of the support of C are vertices.

242 4 Exercises on Conics and Quadrics

By Exercise 194 (b), the origin O is the only focus of C. Thus the polar of thefocus of C, since it coincides with the polar of the centre of C, is the improper line,and consequently C has no directrix.

(c) Proceeding as in part (a), we can easily show that C has the origin O = (0, 0)as a centre, and that the axes of C are the lines of equations x = 0 and y = 0. If weintersect these lines with C, we find the vertices of C, which have coordinates

(0, b), (0,−b).

In order to find the foci of C, we can proceed as in the solution of (a). Denote by CC

the complexified conic ofC, and consider the lines rλ of equationλx0 + i x1 − x2 = 0,with λ ∈ C. Since CC has equation b2x21 − a2x22 + a2b2x20 = 0, if we substitute theequality x2 = λx0 + i x1 into the equation of CC we obtain the equation

(a2 + b2)x21 − 2λa2i x0x1 + a2(b2 − λ2)x20 = 0,

which has a double root if and only if (λa2i)2 + a2(a2 + b2)(λ2 − b2) = 0, that isiff either λ = λ1 = √

a2 + b2 or λ = λ2 = −√a2 + b2. Therefore, rλ is tangent to

CC if and only if either λ = λ1 or λ = λ2. Proceeding as in the solution of (a), wefind that the foci of C coincide with the points rλ1 ∩ σ(rλ1) and rλ2 ∩ σ(rλ2), that iswith the points

F1 =(0,

√a2 + b2

), F2 =

(0,−

√a2 + b2

).

Finally, the directrices r1 and r2 of C are the affine parts of the lines pol(F1) andpol(F2), so they have equations

y = b2√a2 + b2

, y = − b2√a2 + b2

,

respectively.

(d) The conic C is represented by the matrix

A =⎛⎝

0 0 −c0 1 0

−c 0 0

⎞⎠ .

The only eigenspace of the matrix A =(1 00 0

)relative to a non-zero eigenvalue is

the line generated by (1, 0), so that C has a unique axis, of equation x = 0. If weintersect this line with C, we obtain the vertex of C, which coincides with the originO of R

2.Let us now determine the foci of C (coherently with what we proved in

Exercise 194 (c), wewill show that actually C has precisely one focus). The complex-

4 Exercises on Conics and Quadrics 243

ified conic CC of C has equation x21 − 2cx0x2 = 0. As C has no centre, the improperline is tangent to C. Therefore, since I1 /∈ CC and CC is non-degenerate, there existsexactly one line r ⊆ P

2(C) distinct from the improper line, containing I1 and tangentto CC. This line has equation λx0 + i x1 − x2 = 0 for some λ ∈ C. If we substitutethe equality x2 = λx0 + i x1 into the equation of CC, we obtain the equation

x21 − 2icx0x1 − 2λcx20 = 0,

which, as c �= 0, has a double root if and only if λ = c2 . Thus the line r has equation

cx0 + 2i x1 − 2x2 = 0; hence, as r ∩ σ(r) = [2, 0, c], the only focus of C is the point

F =(0,

c

2

).

Finally, the directrix of C is the affine part of the line pol(F), so it has equation

y = − c

2.

Exercise 197. (Metric characterization of conics) Assume that F1, F2 are distinctpoints ofR2 and that r is a line ofR2 such that F1 /∈ r ; denote by d the usual Euclideandistance.

(a) Given κ >d(F1, F2)

2 , let

X = {P ∈ R2 | d(P, F1) + d(P, F2) = 2κ}.

Show that X is the support of a real ellipse having F1 and F2 as foci. Moreover,if C is a real ellipse which is not a circle, show that the support of C coincideswith the locus of points of the plane for which the sum of the distances to thefoci of C is constant.

(b) Given κ ∈ R such that 0 < κ <d(F1, F2)

2 , let

X = {P ∈ R2 | |d(P, F1) − d(P, F2)| = 2κ}.

Show that X is the support of a hyperbola having F1 and F2 as foci. Show alsothat the support of every hyperbola C of R

2 coincides with the locus of points ofthe plane for which the absolute value of the difference of the distances to thefoci of C is constant.

(c) LetX = {P ∈ R

2 | d(P, F1) = d(P, r)}.

Show that X is the support of a parabola having F1 as focus and r as directrix.Show also that the support of every parabola C of R

2 coincides with the locus ofpoints of the plane that are equidistant from the focus of C and from the directrixof C.

244 4 Exercises on Conics and Quadrics

Solution. (a) Up to performing a change of coordinates by means of an isometry ofR

2, we may assume that the points F1 and F2 have coordinates (μ, 0) and (−μ, 0),respectively, withμ > 0. Thenμ < κ, and the point of the plane of coordinates (x, y)belongs to X if and only if

√(x − μ)2 + y2 +

√(x + μ)2 + y2 = 2κ, (4.5)

that is iff √(x − μ)2 + y2 = 2κ −

√(x + μ)2 + y2.

Squaring both sides of this equation, dividing by 4 and rearranging the terms, weobtain the equation

κ2 + μx = κ√

(x + μ)2 + y2.

Squaring again, after some simplifications we find the equation

x2

κ2+ y2

κ2 − μ2= 1, (4.6)

and one can check that Eq. (4.6) is actually equivalent to Eq. (4.5), so that, sinceκ > μ > 0, the set X is the support of a real ellipse.

Then from Exercise 196 (a) one can deduce that the foci of the ellipse of Eq. (4.6)are the points (

√κ2 − (κ2 − μ2), 0) = (μ, 0) = F1 and (−√

κ2 − (κ2 − μ2), 0) =(−μ, 0) = F2, as desired.

Assume now that C is a real ellipse which is not a circle. By Theorem 1.8.8 we can

choose Euclidean coordinates where C has equation x2

a2+ y2

b2= 1, with a > b > 0.

Let F1 = (√a2 − b2, 0), F2 = (−√

a2 − b2, 0) be the foci of C (cf. Exercise 196).The computations just carried out show that the set

{P ∈ R2 | d(P, F1) + d(P, F2) = 2a}

coincides with the support of C, which concludes the proof of (a).

(b) Up to performing a change of coordinates by means of an isometry of R2,

we may assume that the points F1 and F2 have coordinates (0,μ) and (0,−μ),respectively, with μ > 0. Then μ > κ, and the point of the plane of coordinates(x, y) belongs to X if and only if

∣∣∣√x2 + (y − μ)2 −

√x2 + (y + μ)2

∣∣∣ = 2κ, (4.7)

that is iff (√x2 + (y − μ)2 −

√x2 + (y + μ)2

)2 = 4κ2.

4 Exercises on Conics and Quadrics 245

This latter equation is equivalent to

x2 + y2 + μ2 − 2κ2 =√

(x2 + (y − μ)2)(x2 + (y + μ)2).

Squaring both sides of this equation, after several but trivial simplifications we getto the equation

x2

μ2 − κ2− y2

κ2+ 1 = 0, (4.8)

which is actually equivalent to Eq. (4.7). As μ > κ > 0, it follows that X is thesupport of a hyperbola.

Then from Exercise 196 (d) one can deduce that the foci of the hyperbolaof Eq. (4.8) are the points (0,

√κ2 + (μ2 − κ2)) = (0,μ) = F1 and

(0,−√κ2 + (μ2 − κ2)) = (0,−μ) = F2, as desired.

Assume now that C is a hyperbola of R2. By Theorem 1.8.8 we can choose

Euclidean coordinates where C has equation x2

a2− y2

b2+ 1 = 0, with a > 0, b > 0.

If F1 = (0,√a2 + b2), F2 = (0,−√

a2 + b2) are the foci of C (cf. Exercise 196),the computations just performed show that the set

{P ∈ R2 | |d(P, F1) − d(P, F2)| = 2b}

coincides with the support of C. Therefore, the support of every hyperbola coincideswith the locus of points of the plane for which the absolute value of the differenceof the distances to the foci of C is constant.

(c) Up to performing a change of coordinates by means of an isometry of R2, we

may assume that there exists μ > 0 such that F1 has coordinates (0,μ) and r hasequation y = −μ. So the point of the plane of coordinates (x, y) belongs to X if andonly if √

x2 + (y − μ)2 = |y + μ|,

that is iffx2 + (y − μ)2 = (y + μ)2.

This condition is in turn equivalent to the equation

x2 − 4μy = 0.

So, from Exercise 196 (d) one can deduce that X is the support of a parabola havingF1 as focus and r as directrix.

Assumenow thatC is a parabola ofR2. ByTheorem1.8.8we can chooseEuclidean

coordinates inR2 where C has equation x2 − 2cy = 0with c > 0. Since F =

(0, c2

)

and the line r of equation y = − c2 are the focus and the directrix of C respectively

(cf. Exercise 196), the computations just performed show that the set

246 4 Exercises on Conics and Quadrics

{P ∈ R2 | d(P, F) = d(P, r)}

coincides with the support of C. From this we deduce that the support of everyparabola of R

2 coincides with the locus of points of the plane that are equidistantfrom the focus and from the directrix.

Exercise 198. (Eccentricity of a conic) Assume that C is a non-degenerate conicof R

2 which is not a circle. Denote by F a focus of C and by r the correspondingdirectrix.

(a) Prove that F /∈ C and r does not intersect C.(b) Prove that there exists a positive real constant e, called eccentricity of the conic,

such thatd(P, F)

d(P, r)= e ∀P ∈ C

and check that e = 1 if C is a parabola, e < 1 if C is an ellipse, e > 1 if C is ahyperbola.

(c) If C is not a parabola, F1, F2 are the foci of C and V1, V2 are the vertices of Cthat lie on the line L(F1, F2) (cf. Exercise 195 and the Note following it), provethat

e = d(F1, F2)

d(V1, V2).

Solution. (a) Denote by I1 and I2 the cyclic points of the Euclidean plane; letD = CCbe the complexified conic of C and D its projective closure. As seen in the solutionof Exercise 194, the focus F is the intersection of a proper line r of P

2(C) passingthrough I1 and tangent to D at a proper point Q with the line σ(r) passing throughI2 and tangent toD at the point σ(Q). Moreover, Q �= σ(Q) and r �= σ(r). So thereare two distinct lines tangent to D and passing through F ; as a consequence F /∈ Dand in particular F /∈ C.

In addition, the fact that F = polD(Q) ∩ polD(σ(Q)) implies that polD(F) =L(Q,σ(Q)). Therefore, polD(F) intersects D at the non-real points Q and σ(Q),so that the directrix relative to F does not intersect C.

If C is a parabola, part (b) follows immediately from Exercise 197 (c), takinge = 1.

Let us now prove parts (b) and (c) when C is an ellipse which is not a circle. By

Theorem 1.8.8 we may assume that C has equation x2

a2+ y2

b2= 1, with a > b > 0.

By Exercise 196 (a) the foci of C are the points

F1 =(√

a2 − b2, 0)

, F2 =(−

√a2 − b2, 0

)

4 Exercises on Conics and Quadrics 247

and the equations of the directrices r1 and r2 relative to such foci are

x = a2√a2 − b2

, x = − a2√a2 − b2

,

respectively.Let now P = (x0, y0) ∈ C; since the coordinates of P satisfy the equation of C,

then y20 = b2 − b2

a2x20 . If for instance we take F = F1 and r = r1, then we have

d(P, F)2 =(x0 −

√a2 − b2

)2 + y20 =(1 − b2

a2

)x20 − 2

√a2 − b2 x0 + a2

and

d(P, r)2 =(x0 − a2√

a2 − b2

)2

= x20 − 2a2x0√a2 − b2

+ a4

a2 − b2.

Asa2 − b2

a2d(P, r)2 = d(P, F)2

it suffices to take

e =√a2 − b2

a

in order to have e2d(P, r)2 = d(P, F)2, and therefore d(P, F) = e d(P, r) for everypoint P ∈ C, as desired. It turns out that e < 1.

A similar computation shows that, choosing e as above, then we haved(P, F2)d(P, r2)

= e too.

We also observe that, again by Exercise 196, the vertices V1, V2 of C that lie on

the line L(F1, F2) are the points V1 = (a, 0), V2 = (−a, 0), so that e = d(F1, F2)d(V1, V2)

.

Finally let us prove parts (b) and (c) when C is a hyperbola. In this case we may

assume that C has equation x2

a2− y2

b2+ 1 = 0, with a > 0, b > 0. By Exercise 196

(c) the foci of C are the points

F1 =(0,

√a2 + b2

), F2 =

(0,−

√a2 + b2

),

and the equations of the directrices r1 and r2 relative to such foci are

y = b2√a2 + b2

, y = − b2√a2 + b2

,

respectively.

248 4 Exercises on Conics and Quadrics

Let now P = (x0, y0) ∈ C; since the coordinates of P satisfy the equation of C,then x20 = a2

b2y20 − a2. If for instance we take F = F1 and r = r1, then we have

d(P, F)2 = x20 +(y0 −

√a2 + b2

)2 =(1 + a2

b2

)y20 − 2

√a2 + b2 y0 + b2

and

d(P, r)2 =(y0 − b2√

a2 + b2

)2

= y20 − 2b2y0√a2 + b2

+ b4

a2 + b2.

Asa2 + b2

b2d(P, r)2 = d(P, F)2

it suffices to take

e =√a2 + b2

b

in order to have e2d(P, r)2 = d(P, F)2, and therefore d(P, F) = e d(P, r) for everypoint P ∈ C, as desired. In this case it turns out that e > 1.

A similar computation shows that, choosing e as above, one gets d(P, F2)d(P, r2)

= e.

Finally, again by Exercise 196, the vertices V1, V2 of C that lie on the line

L(F1, F2) are the points V1 = (0, b), V2 = (0,−b), so that e = d(F1, F2)d(V1, V2)

.

Note. In Exercise 197 we showed that, given a point F ∈ R2 and a line r not passing

through F , the set X = {P ∈ R2 | d(P, F) = d(P, r)} is the support of a parabola.

More in general, easy computations allow us to check that, given a positive real num-ber e, the set X = {P ∈ R

2 | d(P, F) = e d(P, r)} is the support of a non-degenerateconic which is an ellipse if e < 1, a hyperbola if e > 1 and evidently a parabola ife = 1.

Exercise 199. Consider the conic C of R2 of equation

5x2 + 5y2 − 10x − 8y + 8xy − 4 = 0.

(a) Find the affine canonical form of C.(b) Find the metric canonical form D of C and an isometry ϕ such that ϕ(D) = C.(c) Determine axes, vertices, foci and directrices of C.Solution. (a) Recall (cf. Sect. 1.8.5) that the pair (sign(A), sign(A)) is a completesystem of affine invariants, which determines the affine canonical form of the conic.

In the case we are studying the conic C is represented by the symmetric matrix

A =(c tBB A

)=

⎛⎝

−4 −5 −4−5 5 4−4 4 5

⎞⎠ .

4 Exercises on Conics and Quadrics 249

As det A = 9, trA = 10 and det A = −81, we immediately obtain that sign(A) =(2, 0) and sign(A) = (2, 1). So C is a real ellipse and its affine canonical form hasequation x2 + y2 = 1.

(b) In order to transform C into its metric canonical form, as a first step wediagonalize the matrix A by means of an orthogonal matrix. The eigenvalues of Aare 1 and 9 and an orthonormal basis of eigenvectors of A is formed by the vectors

v1 =(

1√2,− 1√

2

)and v2 =

(1√2, 1√

2

). Let ψ be the linear isometry of R

2 that

transforms the canonical basis into the basis {v1, v2}; if we consider ψ as an affinityand we use the notation fixed in Sect. 1.8.5, ψ is represented by the matrix

Mψ =

⎛⎜⎜⎝

1 0 0

0 1√2

1√2

0 − 1√2

1√2

⎞⎟⎟⎠ .

Therefore the conic C1 = ψ−1(C) is represented by the matrix

A1 = tMψAMψ =

⎛⎜⎜⎜⎝

−4 − 1√2

− 9√2

− 1√2

1 0

− 9√2

0 9

⎞⎟⎟⎟⎠ .

Solving the linear system A1

(xy

)= −B1, that is

(1 0

0 9

) (xy

)=

⎛⎝

1√29√2

⎞⎠, we

obtain (cf. Sect. 1.8.5) that the conic C1 has the point C1 =(

1√2, 1√

2

)as centre.

Thus we can eliminate the linear part of the equation by means of the translation

τ (X) = X + C1. If we set Mτ =

⎛⎜⎜⎝

1 0 01√21 0

1√20 1

⎞⎟⎟⎠, then the conic D = τ−1(C1) is rep-

resented by the matrix

A2 = tMτ A1Mτ =⎛⎝

−9 0 00 1 00 0 9

⎞⎠ .

Wehave so found that themetric canonical formD of C has equation x29 + y2 = 1

and that the isometry ϕ = ψ ◦ τ is such that ϕ(D) = C.

250 4 Exercises on Conics and Quadrics

(c) By Exercise 196 the conic D has the origin O = (0, 0) as a centre, has thelines of equations x = 0 and y = 0 as axes, the points (3, 0), (−3, 0), (0, 1) and(0,−1) as vertices; moreover, its foci are the points (2

√2, 0) and (−2

√2, 0), and

its directrices are the lines of equations x = 92√2and x = − 9

2√2.

Since the isometry ϕ transforms the centre, the axes, the vertices, the foci and thedirectrices of D into those of C, respectively, we readily obtain that

• the centre of C is the point C = ϕ(0, 0) = ψ(τ (0, 0)) = (1, 0);• the axes of C are the lines a1 = {x + y − 1 = 0} and a2 = {x − y − 1 = 0};• the vertices are the points

V1 = ϕ(3, 0) =(1 + 3√

2,− 3√

2

), V2 = ϕ(−3, 0) =

(1 − 3√

2,

3√2

);

V3 = ϕ(0, 1) =(1 + 1√

2,

1√2

), V4 = ϕ(0,−1) =

(1 − 1√

2,− 1√

2

)

• the foci of C are the points

F1 = ϕ(2√2, 0) = (3,−2), F2 = ϕ(−2

√2, 0) = (−1, 2);

• the directrices are the lines

d1 = {2x − 2y − 11 = 0}, d2 = {2x − 2y + 7 = 0}.

In addition we observe that, after determining the foci, the centre can be alsodetermined as the midpoint of the segment F1F2, an axis as the line L(F1, F2), theother axis as the line orthogonal to L(F1, F2) and passing through the centre, thevertices as the intersections of the axes with the conic.

Note. (Metric invariants of conics) If the exercise had not explicitly asked to deter-mine an isometry ϕ between the conic C and its metric canonical form D, we couldhave found an equation of D by looking at how the matrix associated to C changesunder an isometry.

Namely, if C has equation tX AX = 0 and if ϕ(X) = MX + N is an isometry

of R2 represented by the matrix MN =

(1 0 0N M

)with M ∈ O(2), we know (cf.

Sect. 1.8.5) that the conic ϕ−1(C) is represented by the matrix A′ = tMN A MN and,in particular, A′ = tM AM . As M is orthogonal, then A′ = M−1AM , i.e. A and A′

are similar matrices. So it is possible to determine a diagonal matrix A′ =(

λ1 00 λ2

)

similar to A by merely using the trace and the determinant of A, avoiding the explicitcomputation of the orthogonal matrix M . In addition, since det MN = ±1, thendet A′ = det A.

Recall that, if C is a non-degenerate conic, by means of an isometry, it is pos-sible to transform C into the conic represented either by a matrix of the form

4 Exercises on Conics and Quadrics 251

A′ =⎛⎝c 0 00 λ1 00 0 λ2

⎞⎠ if C has a centre, or by a matrix of the form A′ =

⎛⎝0 0 c0 λ1 0c 0 0

⎞⎠

if C is a parabola.In the first case one can determine c by using that det A′ = det A; then one can

determine the metric canonical form of C (cf. Theorem 1.8.8) by choosing a suitablemultiple of A′ and exchanging, if necessary, the coordinates via the linear isometry(x, y) → (y, x).

In the second case (when C is a parabola), we have det A′ = det A = −λ1c2 and

therefore c2 = −det Aλ1

. In addition, the matrices

⎛⎝0 0 c0 λ1 0c 0 0

⎞⎠ and

⎛⎝

0 0 −c0 λ1 0

−c 0 0

⎞⎠

are congruent, so that, if λ1 is positive (resp. negative), it suffices to choose as c

the negative (resp. positive) square root of −det Aλ1

in such a way that the metric

canonical form of C is the one represented by the matrix

⎛⎜⎝

0 0 cλ1

0 1 0cλ1

0 0

⎞⎟⎠.

In the particular case examined in the exercise we have trA = 10 and det A = 9,so that the characteristic polynomial of A is t2 − 10t + 9 and the eigenvalues ofA are 1 and 9. The conic C, having a centre, is metrically equivalent to the conic

represented by A′ =⎛⎝c 0 00 1 00 0 9

⎞⎠ where we can determine c by means of the condition

det A′ = det A, that is 9c = −81 and hence c = −9. Thus we find again that the

metric canonical form of C has equation x29 + y2 = 1.

We can now complete the previous observations by showing how the data requiredin part (c) of the exercise can be computed avoiding the use of an explicit isometrybetween the conic and its metric canonical form.

The centre C = (1, 0) can be readily computed by solving the system A

(xy

)=

−B.The eigenspaces of the matrix A are the lines generated by the vectors (1,−1) and

(1, 1). Since polC([0, 1,−1]) = {−x0 + x1 − x2 = 0} and polC([0, 1, 1]) = {−x0 +x1 + x2 = 0}, the axes of C are the lines of equations x − y − 1 = 0 and x + y − 1 =0.

Intersecting these lines with the conic, we obtain that the vertices of C are the

pointsV1 =(1 + 3√

2,− 3√

2

), V2 =

(1 − 3√

2, 3√

2

), V3 =

(1 + 1√

2, 1√

2

)and

V4 =(1 − 1√

2,− 1√

2

).

We now compute the foci of C as the points of intersection in R2 of the

isotropic lines whose projective closures are tangent to the complexification of C.(cf. Sect. 1.8.7).

252 4 Exercises on Conics and Quadrics

The affine lines ofC2 having the cyclic point I1 = [0, 1, i] as improper point have

equations y = i x − a with a ∈ C; the ones whose projective closures are tangent toCC correspond to the values a = 2 + 3i and a = −2 − i , so that we obtain the linesl1 = {y = i x − 2 − 3i} and l2 = {y = i x + 2 + i}.

The lines of C2 having the cyclic point I2 = [0, 1,−i] as improper point and

whose projective closures are tangent to CC are the conjugate lines of the linesl1, l2 (cf. Exercise 194), that is the lines l3 = σ(l1) = {y = −i x − 2 + 3i} and l4 =σ(l2) = {y = −i x + 2 − i}. Thus we find as foci the points

F1 = l1 ∩ l3 = (3,−2), F2 = l2 ∩ l4 = (−1, 2).

The directrices d1 and d2 are the affine parts of the lines polC(F1) and polC(F2); sothey are the lines of equations 2x − 2y − 11 = 0 and 2x − 2y + 7 = 0, respectively.

Exercise 200. Find the affine canonical form and the metric canonical form of theconic C of R

2 of equation x2 + y2 + 6xy + 2x + 6y − 2 = 0.

Solution. The conic C is represented by the symmetric matrix

⎛⎝

−2 1 31 1 33 3 1

⎞⎠ . This

matrix, which has a positive determinant but is not positive definite, has signature(1, 2). Since we adopted the convention to represent the conic bymeans of amatrix Asuch that i+(A) ≥ i−(A) and i+(A) ≥ i−(A), we multiply the equation of the conicby −1 so that it is represented by the matrix

A =⎛⎝

2 −1 −3−1 −1 −3−3 −3 −1

⎞⎠

for which sign(A) = (2, 1). As det A = −8, it follows at once that sign(A) = (1, 1).Therefore the conic C is a hyperbola and its affine canonical form has equationx2 − y2 + 1 = 0.

In order to determine the metric canonical form, we can use the considerationsmade in theNote followingExercise 199.As trA = −2, the characteristic polynomialof A is t2 + 2t − 8, so that the eigenvalues of A are 2 and −4. Therefore the conic is

metrically equivalent to the hyperbola represented by the matrix A′ =⎛⎝c 0 00 2 00 0 −4

⎞⎠

where c is determined by the condition det A′ = det A, that is−8c = −24 and hencec = 3. So the metric canonical form of C turns out to have equation 2

3 x2 − 4

3 y2 +

1 = 0.

Exercise 201. Consider the conic Cα of R2 of equation

x2 + αy2 + 2(1 − α)x − 2αy + α + 1 = 0

4 Exercises on Conics and Quadrics 253

with α ∈ R.

(a) Determine the affine type of Cα as α varies in R.(b) Find the values of α ∈ R such that Cα is metrically equivalent to the conic D of

equation 3x2 + y2 − 6x − 2y + 1 = 0.

Solution. (a) The conic Cα is represented by the matrix

Aα =⎛⎝1 + α 1 − α −α1 − α 1 0−α 0 α

⎞⎠ .

In order to determine the affine type of Cα it is sufficient to compute sign(Aα) andsign(Aα). Since

trAα = 1 + α, det Aα = α, det Aα = α2(2 − α),

we obtain:

• if α < 0, after multiplying the equation of Cα by −1 we have that sign(−Aα) =(1, 1) and sign(−Aα) = (2, 1), so Cα is a hyperbola;

• if α = 0, the matrix A0 has rank 1, so that C0 is doubly degenerate (in fact itsequation turns out to be (x + 1)2 = 0);

• if 0 < α < 2, then sign(Aα) = (2, 0) and sign(Aα) = (3, 0), so Cα is an imaginaryellipse;

• if α = 2, then sign(A2) = (2, 0) and sign(A2) = (2, 0), so C2 is a simply degen-erate conic whose real support consists of a single point;

• if α > 2, we have sign(Aα) = (2, 0) and sign(Aα) = (2, 1), and hence Cα is a realellipse.

(b) By the metric classification Theorem 1.8.8, the conics Cα andD are metricallyequivalent if and only if they have the same metric canonical forms. Thus one couldsolve the exercise by finding and comparing the canonical forms of Cα and D.

Though, in the Note following Exercise 199we saw that it is possible to determinethe metric canonical form of a non-degenerate conic provided we know trA, det Aand det A. Adapting those considerations to the problem of deciding whether twoconics C andD of equations tX AX = 0 and tX A′ X = 0, respectively, are metricallyequivalent, we observe that, if there exists a isometry ϕ(X) = MX + N that trans-formsD into C, then there is a real number ρ �= 0 such that tMN A MN = ρA′. SinceM is orthogonal, the matrix A is similar to the matrix ρA′ and hence trA = ρ trA′and det A = ρ2 det A′. In addition, as det MN = ±1, then det A = ρ3 det A′. If forinstance trA �= 0, then the numbers det A

(trA)2and det A

(trA)3turn out to be invariant under

isometries or metric invariants.On the other hand, if there exists ρ �= 0 such that trA = ρ trA′, det A = ρ2 det A′

and det A = ρ3 det A′, up to dividing by ρ the equation of C we may assume that thetriple (trA, det A, det A) coincides with the triple (trA′, det A′, det A′). As recalled

254 4 Exercises on Conics and Quadrics

above, if C andD are non-degenerate, then we can conclude that they have the samemetric canonical forms and hence by transitivity they are metrically equivalent.

Coming back to the particular case of the exercise, the conic D is represented bythe matrix

A′ =⎛⎝

1 −3 −1−3 3 0−1 0 1

⎞⎠

which has trA′ = 4, det A′ = 3 and det A′ = −9. It follows at once that D is non-degenerate and, more precisely, that it is a real ellipse.

So let us see for which values of α there exists ρ �= 0 such that trAα = ρ trA′,det Aα = ρ2 det A′ and det Aα = ρ3 det A′, i.e.

1 + α = 4ρ, α = 3ρ2, α2(2 − α) = −9ρ3.

With easy calculations we find that the only values that satisfy the three equationsare ρ = 1 and α = 3. Therefore, thanks to the previous considerations, the onlyconic of the family which is metrically equivalent to D is the conic C3 of equationx2 + 3y2 − 4x − 6y + 4 = 0.

Exercise 202. Verify that the conic C obtained by intersecting the plane H of R3 of

equation x + y + z = 0 with the quadric Q of R3 of equation

f (x, y, z) = xy − 2xz + yz + 2x − y + 2z − 1 = 0

is a hyperbola and find its centre and its asymptotes.

Solution. The points of H are of the form (x, y,−x − y) and the map ϕ : H →L = {z = 0} defined by ϕ(x, y,−x − y) = (x, y, 0) is an affine isomorphism thatallows us to use on H the system of affine coordinates (x, y) of L . In this system ofcoordinates the conic C = Q ∩ H has equation

g(x, y) = f (x, y,−x − y) = 2x2 − y2 + 2xy − 3y − 1 = 0.

Equivalently we may regard g(x, y) = 0 as the equation of the conic C ′ = ϕ(C) ofthe plane z = 0.

Since ϕ is an affine isomorphism, in order to determine the affine type of C itsuffices to determine the type of the conic C ′ represented by the symmetric matrix

A′ =(c′ tB ′B ′ A′

)=

⎛⎝

−2 0 −30 4 2

−3 2 −2

⎞⎠ .

As det A′ = −12 and det A′ �= 0, we realize immediately that C ′ is a hyperbola andso C is a hyperbola too.

4 Exercises on Conics and Quadrics 255

Since C ′ is a conic with centre and the notion of centre is invariant under affinities,also C is a conic with centre and, more precisely, the affine isomorphismϕ transformsthe centre of C into the centre of C ′.

Solving the linear system A′(xy

)=

(03

)we find that the centre of C ′ is the

point(12 ,−1

); therefore the centre of C is the point ϕ−1

(12 ,−1

)=

(12 ,−1, 12

).

Also the notion of asymptote is invariant under affinities, so that we may com-pute the asymptotes of C as the images of the asymptotes of C ′ through the affineisomorphism ϕ−1. Now, the points at infinity of C ′ are R = [0,√3 − 1, 2] andS = [0,−√

3 − 1, 2]. If we compute polC′(R) and polC′(S), it turns out that theasymptotes of C ′ are the lines of equations

4√3 x + (2

√3 − 6)y − 6 = 0 and 4

√3 x + (2

√3 + 6)y + 6 = 0

(which meet at the centre(12 ,−1

)). So the asymptotes of C are the lines of R

3 of

equations

{4√3 x + (2

√3 − 6)y − 6 = 0

x + y + z = 0

{4√3 x + (2

√3 + 6)y + 6 = 0

x + y + z = 0.

Exercise 203. For i = 1, . . . , 5 determine the affine type of the quadricQi of R3 of

equation fi (x, y, z) = 0, where:

(a) f1(x, y, z) = x2 + 3y2 + z2 + 2yz − 2x − 4y + 2;(b) f2(x, y, z) = 2x2 − y2 − 2z2 + xy − 3xz + 3yz − x − 4y + 7z − 3;(c) f3(x, y, z) = x2 + y2 − 2xy − 4x − 4y − 2z + 4;(d) f4(x, y, z) = 2x2 + y2 − 2z2 + 2xz − 10x − 4y + 10z − 6;(e) f5(x, y, z) = 4x2 + y2 + z2 − 4xy + 4xz − 2yz − 4x + 2y − 2z + 1.

Solution. (a) Using the convention adopted in Sect. 1.8.5, the quadric Q1 is repre-sented by the matrix

A1 =

⎛⎜⎜⎝

2 −1 −2 0−1 1 0 0−2 0 3 10 0 1 1

⎞⎟⎟⎠ .

We have det A1 = 2, det A1 = −2, sign(A1) = (3, 0) and sign(A1) = (3, 1). Thusfrom Table1.2 (cf. Sect. 1.8.8) we deduce that Q1 is a real ellipsoid.

(b) The matrix

A2 =

⎛⎜⎜⎝

−6 −1 −4 7−1 4 1 −3−4 1 −2 37 −3 3 −4

⎞⎟⎟⎠ ,

256 4 Exercises on Conics and Quadrics

which represents Q2, has rank 2. The line of equations x = 0, z = 0 intersectsQ2 precisely at the points M = (0,−1, 0) and N = (0,−3, 0), and in particu-lar it is not contained in the quadric. This fact ensures that the support of Q2

is the union of two distinct planes. Then these planes are the tangent planes toQ2 at M and N , respectively. Thus we find that the quadric decomposes intothe planes TM(Q2) = {x + y − 2z + 1 = 0} and TN (Q2) = {2x − y + z − 3 = 0}.Alternately we can compute the line of the singular points Sing(Q2) and then deter-mine the irreducible components of the quadric as the planes L(Sing(Q2), M) andL(Sing(Q2), N ).

(c) Let A3 be the symmetric matrix that representsQ3. Clearly we can determinethe affine type of Q3 by observing that rk A3 = 3 and det A3 = 0, so that Q3 is acylinder: namely, being degenerate, Q3 is either a cone or a cylinder, but it cannotbe a cone because of Exercise 187. More precisely, since the conic Q3 ∩ H0 hasequation x21 + x22 − 2x1x2 = (x1 − x2)2 = 0 and so it is a double line, we deducethat Q3 is a parabolic cylinder.

We can arrive at the same conclusion, for instance, by observing that the derivativewith respect to z of the polynomial f3 never vanishes and consequently no point ofQ3 is singular. In particular the point R = (0, 0, 2) is a smooth point of the quadric. Ifwe compute polQ3

([1, 0, 0, 2]), we obtain that the tangent plane TR(Q3) has equation2x + 2y + z − 2 = 0.

Let us now check the nature of the conic Q3 ∩ TR(Q3) proceeding as inExercise 202. The image of Q3 ∩ TR(Q3) on the plane z = 0 through the affineisomorphism (x, y,−2x − 2y + 2) → (x, y, 0) is the conic of equation

g3(x, y) = f3(x, y,−2x − 2y + 2) = (x − y)2 = 0.

This conic, and hence Q3 ∩ TR(Q3) too, is a double line, so that R is a parabolicpoint. Then (cf. Sect. 1.8.4)Q3 is necessarily a quadric of rank 3, that is either a coneor a cylinder. Having already observed that no point of the affine quadric is singular,clearly Q3 is a cylinder. Since its conic at infinity is a double line, the quadric is aparabolic cylinder.

(d) The symmetric matrix A4 that represents Q4 has determinant zero whiledet A4 �= 0. Then Exercise 187 ensures thatQ4 is a cone (in fact one can easily com-pute that Sing(Q4) = {[1, 1, 2, 3]} so that Q4 is a cone with vertex V = (1, 2, 3)).In order to decide whether it is a real or an imaginary cone, it is enough to seewhether the support of the quadric contains at least one real point in addition toV . For instance the line of equations x = z = 0 intersects Q4 at the points (0, y, 0)where y is a solution of the equation y2 − 4y − 6 = 0. Since this latter equation hastwo distinct real roots, we realize that Q4 is a real cone.

We could clearly get to the same result by computing that sign(A4) = (2, 1) andsign(A4) = (2, 1) and making use of Table1.2 of Sect. 1.8.8.

(e) The symmetric matrix A5 that representsQ5 has rank 1, so that the quadric isa double plane. Since this plane coincides with the singular locus of the quadric, inorder to determine it we can compute Sing(Q5), which turns out to be the plane ofequation 2x − y + z − 1 = 0.

4 Exercises on Conics and Quadrics 257

Exercise 204. Consider the quadrics Q1 and Q2 of R3 of equations

x2 + 3y2 + 3z2 + 2yz − 2x = 0 and x2 + y2 + 2z2 − 2xy + 3z − x − y + 1 = 0,

respectively. Verify that these quadrics are non-degenerate and determine their affinetypes and their principal planes.

Solution. The quadric Q1 is represented by the matrix

A1 =

⎛⎜⎜⎝

0 −1 0 0−1 1 0 00 0 3 10 0 1 3

⎞⎟⎟⎠ .

As A1 is positive definite and det A1 < 0, it follows that the quadric is a real ellipsoid.After realizing that A1 is positive definite, one can also reach the same conclusion

arguing as follows. The improper conic of Q1, being represented by the positivedefinite matrix A1, has no real points. So Q1 may only be either a real ellipsoid oran imaginary ellipsoid or an imaginary cone. But Q1 can be neither an imaginaryellipsoid nor an imaginary cone because the origin is a smooth point of the support.

In order to compute the principal planes ofQ1 we observe that A1 has eigenvalues1, 2, 4 and that the eigenspaces relative to them are the lines generated by the vectors(1, 0, 0), (0, 1,−1) and (0, 1, 1), respectively. If we compute the polars of the points[0, 1, 0, 0], [0, 0, 1,−1] and [0, 0, 1, 1] with respect to Q1, we find that the quadrichas the planes of equations x = 1, y − z = 0 and y + z = 0 as principal planes.

Examining the matrix

A2 =

⎛⎜⎜⎝

2 −1 −1 3−1 2 −2 0−1 −2 2 03 0 0 4

⎞⎟⎟⎠

one easily obtains that Q2 is an elliptic paraboloid (det A2 = 0 and det A2 < 0, cf.Sect. 1.8.8).

The matrix A2 has, in addition to the eigenvalue 0, the eigenvalue 4 with mul-tiplicity 2 and the eigenvectors relative to 4 are all the non-zero vectors of theform (a,−a, b). If we compute the polar of [0, a,−a, b] with respect to Q2, itturns out that the principal planes for Q2 are precisely the planes of equations4ax − 4ay + 4bz + 3b = 0 as the parameters a, b vary in R

2 \ {(0, 0)}.We can complete the study of the quadric Q2 by observing that the axis of the

paraboloid is the intersection of any two distinct principal planes, for instance thoseof equations x − y = 0 and 4z + 3 = 0; by intersecting the axis with Q2 we find

that the vertex of the paraboloid is the point(− 116 ,− 1

16 ,−34

).

258 4 Exercises on Conics and Quadrics

Note. The method based on the investigation of the conic at infinity used to identifythe affine type of the quadric Q1 can also be used to determine the affine typeof any non-degenerate quadric Q. Namely, if we denote the conic at infinity byQ∞ = Q ∩ H0, we have:

• if Q∞ is irreducible with an empty real support, then Q is either a real or animaginary ellipsoid (in order to decide whether it is real, it is enough to show theexistence of a real point in the support);

• ifQ∞ is irreducible with a non-empty real support, thenQ is either an hyperbolicor an elliptic hyperboloid (we can decide which is the case for instance by studyingthe nature of a point of the quadric);

• if Q∞ is a pair of distinct real lines, then Q is a hyperbolic paraboloid;• if Q∞ is a pair of complex conjugate lines, then Q is an elliptic paraboloid.

Exercise 205. Denote by Q the quadric of R3 of equation

f (x, y, z) = 2y2 − x2 + 2yz + 2z2 − 3 = 0.

Find all the lines of R3 contained inQ and passing through the point P = (1,

√2, 0)

of Q.

Solution. The quadric Q is represented by the symmetric matrix

A =

⎛⎜⎜⎝

−3 0 0 00 −1 0 00 0 2 10 0 1 2

⎞⎟⎟⎠ .

As det A = 9, the quadric is non-degenerate. More precisely we observe thatdet A = −3, sign(A) = (2, 1) and sign(A) = (2, 2). From Table1.2 of Sect. 1.8.8we deduce that Q is a hyperbolic hyperboloid.

Then, since P is a hyperbolic point, there are two distinct real lines contained inQ and passing through it; we can find them by identifying the components of thedegenerate conic Q ∩ TP(Q).

Easy computations yield that ∇ f = (−2x, 4y + 2z, 2y + 4z) and hence ∇ f (P)

= (−2, 4√2, 2

√2). So TP(Q) has equation −2(x − 1) + 4

√2(y − √

2) + 2√2

z = 0, that is x − 2√2 y − √

2 z + 3 = 0.The image of the degenerate conic C = Q ∩ TP(Q) on the plane x = 0 through

the projection p such that

(2√2 y + √

2 z − 3, y, z) → (0, y, z)

is the degenerate conic C ′ of equation

g(y, z) = f (2√2 y + √

2 z − 3, y, z) = −6y2 − 6yz + 12√2 y + 6

√2 z − 12 = 0

4 Exercises on Conics and Quadrics 259

represented, for instance, by the matrix

A =⎛⎝

4 −2√2 −√

2−2

√2 2 1

−√2 1 0

⎞⎠ .

Since we know that C ′ is the union of two distinct lines that meet at p(1,√2, 0) =

(√2, 0), we easily find that the components of C ′ are the lines of the plane x = 0 of

equations y − √2 = 0 and y + z − √

2 = 0. Therefore the irreducible componentsof C are the lines of R

3 of equations

{y − √

2 = 0x − 2

√2 y − √

2 z + 3 = 0

{y + z − √

2 = 0x − 2

√2 y − √

2 z + 3 = 0

which are the lines we looked for.

Exercise 206. Denote by d the Euclidean distance of R3. Given two skew lines

r, s ⊆ R3, consider the set

X = {P ∈ R3 | d(P, r) = d(P, s)}.

Moreover, let l ⊆ R3 be the only line that intersects both r and s and whose direction

is orthogonal both to the direction of r and to the direction of s, and set Q1 = r ∩ l,Q2 = s ∩ l.

(a) Show that X is the support of a hyperbolic paraboloid Q.(b) Show that l is the axis ofQ, and that the vertex ofQ coincides with the midpoint

of the segment having Q1 and Q2 as endpoints.(c) Show that Q is metrically equivalent to the quadric Q′ of equation x2 − y2 −

2z = 0 if and only if the directions of r and s are orthogonal and d(Q1, Q2) = 1.

Solution. (a) It is easy to check that, up to changing coordinates by means of isome-tries ofR

3, we may assume that r has equations x = y = 0. By performing a rotationaround r and a translation parallel to z (transformations that leave r invariant, it ispossible to transform s into a line of equations x − a = z − by = 0 for some a > 0,b ∈ R; so henceforth we will assume that r and s are defined by the equations justmentioned. Observe that, in the coordinates we have chosen, the line l has equationsy = z = 0 and intersects r (resp. s) at the point Q1 = (0, 0, 0) (resp. Q2 = (a, 0, 0)).

If P ∈ R3 has coordinates (x, y, z), clearly one has

d(P, r)2 = x2 + y2. (4.9)

In addition, since the line s can be parametrized through t → (a, t, bt), t ∈ R,the squared distance between P and s coincides with the minimum of (x − a)2 +(y − t)2 + (z − bt)2 where we let t vary in R. Then an easy computation shows that

260 4 Exercises on Conics and Quadrics

d(P, s)2 = (x − a)2 + y2 + z2 − (y + bz)2

1 + b2(4.10)

Thus, comparing (4.9) and (4.10) we get that X coincides with the support of thequadric Q of equation

y2 − z2 + 2byz + 2a(1 + b2)x − a2(1 + b2) = 0.

Let us now check that Q is a hyperbolic paraboloid. As Q is represented by thesymmetric matrix

A =

⎛⎜⎜⎝

−a2(1 + b2) a(1 + b2) 0 0a(1 + b2) 0 0 0

0 0 1 b0 0 b −1

⎞⎟⎟⎠ ,

the result follows from the fact that det A = 0 and det A = a2(1 + b2)3 > 0 (cf.Sect. 1.8.8).

(b) Recall that a hyperplane H ⊆ R3 is principal if and only if H = polQ(P)

for some P = [0, p1, . . . , pn] such that the vector (p1, . . . , pn) is a eigenvector forA relative to a non-zero eigenvalue (cf. Sect. 1.8.6). It is easy to check that A hasthe eigenvalues λ1 = 0, λ2 = √

1 + b2, λ3 = −√1 + b2. In addition, if b = 0 the

eigenspaces relative to λ2 = 1 and λ3 = −1 are generated by (0, 1, 0) and (0, 0, 1),respectively, so thatQ has two principal hyperplanes H1, H2, which have equationsy = 0 and z = 0, respectively. If instead b �= 0, the eigenspaces relative to λ2 andλ3 are generated by (0, b,

√1 + b2 − 1) and (0, b,−√

1 + b2 − 1), respectively, sothat the two principal hyperplanes H1, H2 of Q have equations

by +(√

1 + b2 − 1)z = 0, by −

(√1 + b2 + 1

)z = 0,

respectively. Since in both cases H1 ∩ H2 = l, the axis of Q coincides with l, asdesired. Furthermore, considering the system formed by the equations of l and theequation ofQ, we immediately obtain that the vertex V = Q ∩ l ofQ has coordinates(a2 , 0, 0

), and so it coincides with the midpoint of the segment having Q1, Q2 as

endpoints.

(c) Observe at first that d(Q1, Q2) = a, and that the directions of r and s areorthogonal if and only if b = 0. Suppose now that Q is metrically equivalent to thequadric Q′, which is represented by the matrix

B =

⎛⎜⎜⎝

0 0 0 −10 1 0 00 0 −1 0

−1 0 0 0

⎞⎟⎟⎠ .

4 Exercises on Conics and Quadrics 261

Arguing as in the Note following Exercise 199, the fact thatQ andQ′ are metricallyequivalent implies that det A = det B, and that the matrices A and B are similar.In particular, the eigenvalues of A must coincide with those of B, which are 0, 1and −1. As we have seen in the solution of part (b), this implies that b = 0, i.e.the directions of r and of s are orthogonal. As det B = 1 and det A = a2(1 + b2)3,when b = 0 the condition det A = det B and the fact that a > 0 imply that a = 1.Therefore d(Q1, Q2) = a = 1.

Conversely, if b = 0 and a = 1 the quadric Q has equation y2 − z2 + 2x −1 = 0. Ifϕ : R

3 → R3 is the isometry defined byϕ(x, y, z) =

(−z + 1

2 , x, y), then

ϕ(Q′) = Q, so that Q and Q′ are metrically equivalent.

Index

AAbsolute conic of the plane, 47Affine chart, 9

standard -, 9Affine part of a projective hypersurface, 24Asymptote, 46, 53Axis, 44

focal - , 240major - , 240transverse - , 240

CCentre

- of a conic, 40- of a linear system of hyperplanes, 13- of a non-degenerate quadric, 43- of a quadric, 40

Characteristic of a projectivity, 18Circle, 46Codimension of a projective subspace, 2Completion of Kn , 10Complexification

- of a projective hypersurface, 31- of a projective subspace, 19- of an affine hypersurface, 30- of an affine subspace, 19

Cone- over J with vertex P , 163affine -, 22, 39affine tangent -, 29imaginary -, 45, 48projective -, 23projective tangent -, 29real -, 45, 48

Conic, 23, 31- with centre, 40

absolute - of the plane, 47doubly degenerate -, 35eccentricity of a - , 246simply degenerate -, 35

Conicsmetric characterization of - , 243metric invariants of -, 250

Conjugate- of a projective hyperplane, 19- of an affine hyperplane, 19

Conjugation, 19Construction

- of a self-polar (n + 1)-hedron, 219- of the harmonic conjugate, 78Steiner - , 214

Coordinates- of a projective hyperplane, 12affine -, 10dual homogeneous -, 12homogeneous -, 6

Cross-ratio- of four hyperplanes, 16- of four lines, 16- of four points, 15- of four points on a conic, 193symmetries of the -, 16

Cubic, 23cuspidal complex - , 140cuspidal real - , 141modulus of a plane - , 142nodal complex - , 138nodal real - , 139Weierstrass equation of a plane -, 136

CurveHessian -, 56plane affine -, 22plane projective -, 23

© Springer International Publishing Switzerland 2016E. Fortuna et al., Projective Geometry, UNITEXT - La Matematica per il 3+2 104,DOI 10.1007/978-3-319-42824-6

263

264 Index

Cusp, 52ordinary -, 52

Cylinderaffine -, 39elliptic -, 49hyperbolic -, 49imaginary -, 45, 48parabolic -, 49real -, 45, 48

DDegree of a hypersurface, 22Descartes’ criterion, 43Diameter, 43Dimension of a projective space, 2Directrix, 47Duality

- correspondence, 13- principle, 14

EEccentricity, 246Ellipse, 45

imaginary -, 45real -, 45

Ellipsoid, 41real -, 48imaginary -, 48

Equation- of a hypersurface, 22Weierstrass - of a plane cubic, 136

EquationsCartesian - of a subspace, 8parametric - of a subspace, 8

Euler’s identity, 20

FFactorization of homogeneous polynomials,

21Flex, 56Focus, 47

GGrassmann’s formula, 3

HHarmonic conjugate, 78Harmonic quadruple, 17Hyperbola, 45

Hyperboloid, 41elliptic -, 48hyperbolic -, 48one-sheeted -, 48two-sheeted -, 48

Hyperplanecoordinate -, 8diametral -, 43fundamental -, 8polar -, 33principal -, 44projective -, 2tangent -, 28

Hypersurfaceaffine -, 21conjugate -, 29, 31degree of a -, 22irreducible -, 22projective -, 22real affine -, 30real projective -, 31reduced -, 22singular -, 28smooth -, 28

Hypersurfacesaffinely equivalent -, 25projectively equivalent -, 25

IInflection point, 56

ordinary -, 56Inflection points

- of a smooth complex cubic, 145- of a smooth real cubic, 145

Invariantj - of a smooth cubic, 142

Invariant set, 4Invariant sets of projectivities of the plane,

97Invariants

affine - of conics, 46affine - of quadrics, 49metric - of conics, 250, 253

Involution, 3absolute -, 47

Involutions of the projective line, 75Irreducible component

- of a hypersurface, 22multiple -, 22

LLine

Index 265

affine tangent -, 53external -, 35isotropic -, 47polar - , 228polar of a - with respect to a spacequadric, 228

principal tangent -, 52projective -, 2secant -, 35tangent -, 27, 28

Linear condition, 57Linear system

- of affine hyperplanes, 14- of curves, 57- of projective hyperplanes, 13base points of a -, 57improper - of affine hyperplanes, 14proper - of affine hyperplanes, 14

MMatrices

similar -, 12Matrix

- of a projective transformation, 7change of frame -, 7change of homogeneous coordinates -, 7Hessian -, 56real Jordan -, 18Sylvester -, 53

Modulus- of a plane cubic, 142- of a quadruple of points, 74

Multiplicity- of a point, 28- of an irreducible component, 22, 23

Multiplicity of intersection, 26, 55

NNode, 52Normalized basis, 6

PParabola, 45Paraboloid, 40

elliptic -, 48hyperbolic -, 48

Parametrization of a pencil of hyperplanes,82

Pencil- of conics, 58- of curves, 57

- of hyperplanes, 13- of lines, 13- of planes, 13Hesse - of cubics, 152improper - of affine hyperplanes, 14proper - of affine hyperplanes, 14

Perspectivity, 5Plane

diametral -, 43external -, 38projective -, 2secant -, 38tangent -, 38

Pointm-tuple -, 52- at infinity, 10double -, 52elliptic -, 38fixed -, 4hyperbolic -, 38improper -, 10inflection -, 56non-singular -, 27, 28ordinary singular -, 52parabolic -, 38proper -, 10simple -, 29singular -, 27, 28smooth -, 27, 28triple -, 52unit - of a projective frame, 6

Points- at infinity of an affine hypersurface, 25- in general position, 6base - of a linear system, 57conjugate -, 34cyclic -, 47fundamental - of a projective frame, 6improper - of an affine hypersurface, 25projectively independent -, 5real - of a hypersurface, 29

Polar of a linewith respect to a space quadric,228

Pole, 33Polynomial

dehomogenized -, 20homogeneous -, 20homogenized -, 20

Principleduality -, 14identity - of polynomials, 22

Projection centred at a subspace, 5Projective closure

266 Index

- of an affine hypersurface, 25- of an affine subspace, 11

Projective frame, 6standard -, 6

Projective isomorphism, 3Projective linear group, 4Projective space, 2

dual -, 12standard -, 2

Projective spacesisomorphic -, 3

Projective subspace, 2- generated by a set, 3Cartesian representation of -, 7parametric representation of -, 8

Projective subspacesincident -, 3join of -, 3skew -, 3

Projective transformation, 3analytic representation of a -, 7degenerate -, 4

Projectively equivalent sets, 3Projectivities

conjugate -, 12Projectivity, 3

- of the projective line, 17characteristic of a -, 18dual -, 15elliptic -, 18hyperbolic -, 18parabolic -, 18

QQuadric, 23, 31

- with centre, 40symmetries of a - of Rn , 234affine -, 39degenerate -, 39dual -, 34elliptic -, 38hyperbolic -, 38non-degenerate -, 31parabolic -, 38ruled - , 223

Quadricsmetrically equivalent -, 39projectively equivalent -, 32

Quartic, 23

RRank of a quadric, 31, 39Resultant, 53Root of a homogeneous polynomial, 21

SSaddle, 48Self-dual proposition, 14Self-polar (n + 1)-hedron, 34Self-polar triangle, 34Space

polar -, 33tangent -, 27, 28

Sphere, 44Standard atlas, 10Support

- of a projective hypersurface, 23- of an affine hypersurface, 22

Surfaceaffine -, 22projective -, 23

System of homogeneous coordinates, 6induced -, 9

TTangent curves, 55Theorem

Pappus’ -, 68Bézout’s -, 55Chasles’ - , 214Desargues’ -, 14, 67fundamental - of projective transforma-tions, 6

Pappus-Pascal’s - , 211Poncelet’s - , 157real Bézout’s -, 56Salmon’s - , 142

VVertex

- of a cone, 22- of a quadric, 44