classical geometry - danny calegari

8/9/2019 Classical Geometry - Danny Calegari

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CLASSICAL GEOMETRY LECTURE NOTES

DANNY CALEGARI

1. A CRASH COURSE IN GROUP THEORY

A group is an algebraic object which formalizes the mathematical notion which ex-presses the intuitive idea of symmetry . We start with an abstract denition.

Denition 1.1. A group is a set G and an operation m : G G G called multiplicationwith the following properties:(1) m is associative . That is, for any a,b,cG

,m(a, m (b, c)) = m(m(a, b), c)

and the product can be written unambiguously as abc.(2) There is a unique element eG called the identity with the properties that, forany aG, ae = ea = a(3) For any aG there is a unique element in G denoted a

1 called the inverse of asuch that

aa 1 = a1a = eGiven an object with some structural qualities, we can study the symmetries of that

object; namely, the set of transformations of the object to itself which preserve the structurein question. Obviously, symmetries can be composed associatively, since the effect of asymmetry on the object doesnt depend on what sequence of symmetries we applied to theobject in the past. Moreover, the transformation which does nothing preserves the structureof the object. Finally, symmetries are reversible performing the opposite of a symmetryis itself a symmetry. Thus, the symmetries of an object (also called the automorphisms of an object) are an example of a group.

The power of the abstract idea of a group is that the symmetries can be studied bythemselves, without requiring them to be tied to the object they are transforming. So forinstance, the same group can act by symmetries of many different objects, or on the sameobject in many different ways.

Example 1.2 . The group with only one element e and multiplication e e = e is calledthe trivial group . Example 1.3 . The integers Z with

m(a, b) = a + bis a group, with identity

0.

Example 1.4 . The positive real numbers R + with m (a, b) = ab is a group, with identity 1.

Example 1.5 . The group with two elements even and odd and multiplication given bythe usual rules of addition of even and odd numbers; here even is the identity element.This group is denoted Z / 2Z .

Example 1.6 . The group of integers mod n is a group with m(a, b) = a + b mod n andidentity 0. This group is denoted Z /n Z and also by C n , the cyclic group of length n .

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2 DANNY CALEGARI

Denition 1.7. If G and H aregroups, one can form the Cartesian product , denoted GH .This is a group whose elements are the elements of GH where m : (GH )(GH ) G

H is dened by

m((g1 , h1), (g2 , h 2)) = ( mG (g1 , g2), m H (h1 , h2))The identity element is (eG , eH ).

Example 1.8 . Let S be a regular tetrahedron; label opposite pairs of edges by A,B,C .Then the group of symmetries which preserves the labels is Z / 2Z

Z / 2Z . It is alsoknown as the Klein 4 group .

In all of the examples above, m(a, b) = m(b, a). A group with this property is calledcommutative or Abelian . Not all groups are Abelian!

Example 1.9 . Let T be an equilateral triangle with sides A,B,C opposite vertices a,b,cin anticlockwise order. The symmetries of T are the reections in the lines running fromthe corners to the midpoints of opposite sides, and the rotations. There are three possible

rotations, through anticlockwise angles 0, 2/ 3, 4/ 3 which can be thought of as e,,2

.Observe that 1 = 2 . Let r a be a reection through the line from the vertex a tothe midpoint of A. Then r a = r 1a and similarly for r b, r c . Then 1r a = r c butr a 1 = r a so this group is not commutative . It is callec the dihedral group D3 and has6 elements.

Example 1.10 . If P is an equilateral ngon, the symmetries are reections as above androtations. This is called the dihedral group Dn and has 2n elements. The elements aree,,2 , . . . , n 1 = 1 and r 1 , r 2 , . . . , r n where r 2i = e for all i, r i r j = 2( ij ) and1r i = r i1 . Example 1.11 . The symmetries of an equilateral gon (i.e. the unique innite 2valenttree) denes a group D, the innite dihedral group . Example 1.12 . The set of 2

2 matrices whose entries are real numbers and whose de-

terminants do not vanish is a group, where multiplication is the usual multiplication of matrices. The set of all 22 matrices is not naturally a group, since some matrices are notinvertible. Example 1.13 . The group of permutations of the set {1 . . . n }is called the symmetric groupS n . A permutation breaks the set up into subsets on which it acts by cycling the members.For example, (3, 2, 4)(5 , 1) denotes the element of S 5 which takes 1 5, 2 4, 3 2, 4 3, 5 1. The group S n has n! elements. A transposition is a permutation whichinterchanges exactly two elements. A permutation is even if it can be written as a productof an even number of transpositions, and odd otherwise.

Exercise 1.14. Show that the symmetric group is not commutative for n > 2. Identify S 3and S 4 as groups of rigid motions of familiar objects. Show that an even permutation isnot an odd permutation, and vice versa.

Denition 1.15. A subgroup H of G is a subset such that if h H then h1H , andif h1 , h2 H then h1h2 H . With its inherited multiplication operation from G, H isa group. The right cosets of H in G are the equivalence classes [g] of elements g Gwhere the equivalence relation is given by g1 g2 if and only if there is an hH withg1 = g2h.

Exercise 1.16. If H is nite, the number of elements of G in each equivalence class areequal to |H | , the number of elements in H . Consequently, if |G| is nite, |H | divides |G|.


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CLASSICAL GEOMETRY LECTURE NOTES 3

Exercise 1.17. Show that the subset of even permutations is a subgroup of the symmetricgroup, known as the alternating group and denoted An . Identify A5 as a group of rigid motions of a familiar object.

Example 1.18 . Given a collection of elements {gi} G (not necessarily nite or evencountable), the subgroup generated by the gi is the subgroup whose elements are obtainedby multiplying together nitely many of the gi and their inverses in some order.

Exercise 1.19. Why are only nite multiplications allowed in dening subgroups? Showthat a group in which innite multiplication makes sense is a trivial group. This fact is not as useless as it might seem . . .

Denition 1.20. A group is cyclic if it is generated by a single element. This justies thenotation C n for Z /n Z used before.

Denition 1.21. A homomorphism between groups is a map f : G1 G2 such thatf (g1)f (g2) = f (g1g2) for any g1 , g2 in G1 . The kernel of a homomorphism is the sub-group K G

1 dened by K = f 1

(e). If K = e then we say f is injective . If everyelement of G2 is in the image of f , we say it is surjective . A homomorphism which isinjective and surjective is called an isomorphism .

Example 1.22 . Every nite group G is isomorphic to a subgroup of S n where n is thenumber of elements in G. For, let b : G {1, . . . , n }be a bijection, and identify anelement g with the permutation which takes b(h) b(gh) for all h.Denition 1.23. An exact sequence of groups is a (possibly terminating in either direction)sequence

Gi G i+1 Gi +2 . . . joined by a sequence of homomorphisms h i : G i G i +1 such that the image of h i isequal to the kernel of h i +1 for each i.

Denition 1.24. If a, b

G, then bab1 is called the conjugate of a by b, and aba1b1is called the commutator of a and b. Abelian groups are characterized by the property thata conjugate of a is equal to a and every commutator is trivial.

Denition 1.25. A subgroup N G is normal , denoted N G if for any n N andgG we have gng1N . A kernel of a homomorphism is normal. Conversely, if N is normal, we can dene the quotient group G/N whose elements are equivalence classes

[g] of elements in G, and two elements g, h are equivalent iff g = hn for some n N .The multiplication is given by m([g], [h]) = [gh] and the fact that N is normal says this iswelldened. Thus normal subgroups are exactly kernels of homomorphisms.

Example 1.26 . Any subgroup of an abelian group is normal.

Example 1.27 . Z is a normal subgroup of R . The quotient group R / Z is also called thecircle group S 1 . Can you see why?

Example 1.28 . Let Dn be the dihedral group, and let C n be the subgroup generated by .Then C n is normal, and Dn /C n = Z / 2Z .

Denition 1.29. If G is a group, the subgroup G1 generated by the commutators in G iscalled the commutator subgroup of G. Let G2 be the subgroup generated by commutatorsof elements of G with elements of G1 . We denote G1 = [G, G ] and G2 = [G, G 1]. DeneG i inductively by G i = [G, G i1]. The elements of G i are the elements which can bewritten as products of iterated commutators of length i. If Gi is trivial for some i that


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4 DANNY CALEGARI

is, there is some i such that every commutator of length i in G is trivial we say G isnilpotent .

Observe that every Gi

is normal, and every quotient G/Gi

is nilpotent.

Denition 1.30. If G is a group, let G0 = G1 and dene Gi = [G i1 , G i1]. If G i istrivial for some i, we say that G is solvable . Again, every Gi is normal and every G/G i issolvable. Obviously a nilpotent group is solvable.

Denition 1.31. An isomorphism of a group G to itself is called an automorphism . The setof automorphisms of G is naturally a group, denoted Aut (G). There is a homomorphismfrom : G Aut (G) where g goes to the automorphism consisting of conjugation byg. That is, (g)(h) = ghg1 for any h G. The automorphisms in the image of arecalled inner automorphisms , and are denoted by Inn (G). They form a normal subgroup of Aut (G). The quotient group is called the group of outer automorphisms and is denote byOut (G) = Aut (G)/ Inn (G).

Denition 1.32. Suppose we have two groupsG, H

and a homomorphism : G Aut (H ). Then we can form a new group called the semidirect product of G and H

denoted G H whose elements are the elements of G H and multiplication is given bym((g1 , h1), (g2 , h2)) = ( g1g2 , h 1(g1)(h2))

Observe that H is a normal subgroup of G H , and there is an exact sequence

1 H G H G 1 Example 1.33 . The dihedral group Dn is equal to Z / 2Z C n where the homomorphism : Z / 2Z Aut (C n ) takes the generator of Z / 2Z to the automorphism 1 , where denotes the generator of C n . Example 1.34 . The group Z / 2Z R where the nontrivial element of Z / 2Z acts on R

by x x is isomorphic to the group of isometries (i.e. 11 and distance preservingtransformations) of the real line. It contains D as a subgroup.Exercise 1.35. Find an action of Z / 2Z on the group S 1 so that Dn is a subgroup of Z / 2Z S 1 for every n .

Example 1.36 . The group whose elements consist of words in the alphabet a,b,A,B subject to the equivalence relation that when one of aA,Aa,bB,Bb appear in a word, theymay be removed, so for example

aBaAbb aBbbab

A word in which none of these special subwords appears is called reduced ; it is clear thatthe equivalence classes are in 11 correspondence with reduced words. Multiplication isgiven by concatenation of words. The identity is the empty word, A = a1 , B = b1 . Ingeneral, the inverse of a word is obtained by reversing the order of the letters and changingthe case. This is called the free group F 2 on two generators , in this case the letters a, b.It is easy to generalize to the free group F n on n generators , given by words in lettersa1 , . . . , a n and their inverse letters A1 , . . . , A n . One can also denote the letters Ai bythe letters a1i .Exercise 1.37. Let G be an arbitrary group and g1 , g2 . . . gn a nite subset of G. Showthat there is a unique homomorphism from F n G sending a i gi .


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Example 1.38 . If we have an alphabet consisting of letters a1 , . . . , a n and their inverses,we can consider a collection of words in these letters r 1 , . . . , r m . If R denotes the subgroupof F

ngenerated by the r

iand all their conjugates, then R is a normal subroup of F

nand

we can form the quotient F n /R . This is denoted by

a1 , . . . , a n |r 1 , . . . , r mand an equivalent description is that it is the group whose elements are words in the a iand their inverses modulo the equivalence relation that two words are equivalent if theyare equivalent in the free group, or if one can be obtained from the other by inserting ordeleting some r i or its inverse as a subword somewhere. The a i are the generators andthe r i the relations . Groups dened this way are very important in topology. Notice that a presentation of a group in terms of generators and relations is far from unique.

Denition 1.39. A group G is nitely generated if there is a nite subset of G whichgenerates G. This is equivalent to the property that there is a surjective homomorphismfrom some F n to G. A group G is nitely presented if it can be expressed as A|R forsome nite set of generators A and relations R .Exercise 1.40. Let G be any nite group. Show that G is nitely presented.

Exercise 1.41. Let F 2 be the free group on generators x, y . Let i : F 2 Z be thehomomorphism which takes x 1 and y 1. Show that the kernel of i is not nitelygenerated.

Exercise 1.42. (Harder). Let i : F 2F 2 Z be the homomorphism which restricts oneither factor to i in the previous exercise. Show that the kernel of i is nitely generated but not nitely presented.

Denition 1.43. Given groups G, H the free product of G and H , denoted GH , is thegroup of words whose letters alternate between elements of G and H , with concatenation asmultiplication, and the obvious proviso that the identity is in either G or H . It is the uniquegroup with the universal property that there are injective homomorphisms iG : G GH and iH : H GH , and given any other group I and homomorphisms j G : G I and jH : H I there is a unique homomorphism c from GH to I satisfying c iG = j Gand c iH = j H .Exercise 1.44. Show that denes an associative and commutative product on groups upto isomorphism, and

F n = ZZ Z

where we take n copies of Z in the product above.

Exercise 1.45. Show that Z / 2ZZ / 2Z= D.

Remark 1.46 . Actually, one can extend to innite (even uncountable) products of groupsby the universal property. If one has an arbitrary set S the free group generated by S is the

free product of a collection of copies of Z , one for each element of S .Exercise 1.47. (Hard). Every subgroup of a free group is free.

Denition 1.48. A topological group is a group which is also a space (i.e. we understandwhat continuous maps of the space are) such that m : G G G and i : G G, themultiplication and inverse maps respectively, are continuous . If G is a smooth manifold (see appendix for denition) and the maps m and i are smooth maps, then G is called a Liegroup .


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6 DANNY CALEGARI

Remark 1.49 . Actually, the usual denition of Lie group requires that G be a real analyticmanifold and that the maps m and i be real analytic. A real analytic manifold is like asmooth manifold, except that the coordinate transformations between charts are requiredto be real analytic, rather than merely smooth. It turns out that any connected, locallyconnected, locally compact (see appendix for denition) topological group is actually aLie group.

2. M ODEL GEOMETRIES IN DIMENSION TWO

2.1. The Euclidean plane.

2.1.1. Euclids axioms.

Notation 2.1. The Euclidean plane will be denoted by E 2 .

Euclid, who taught at Alexandria in Egypt and lived from about 325 BC to 265 BC,is thought to have written 13 famous mathematical books called the Elements . In theseare found the earliest (?) historical example of the axiomatic method . Euclid proposed5 postulates or axioms of geometry, from which all true statements about the Euclideanplane were supposed to inevitably follow. These axioms were as follows:

(1) A straight line segment can be drawn joining any two points.(2) Any straight line segment is contained in a unique straight line.(3) Given any straight line segment, a circle can be drawn having the segment as radius

and one endpoint as center.(4) All right angles are congruent.(5) One and only one line can be drawn through a point parallel to a given line.

The terms point, line, plane are supposed to be primitive concepts, in the sense that theycant be described in terms of simpler concepts. Since they are not dened, one is notsupposed to use ones personal notions or intuitions about these objects to prove theorems

about them; one strategy to achieve this end is to replace the terms by other terms (Hilbertssuggestion is glass, beer mat, table ; Queneaus is word, sentence, paragraph ) or evennonsense terms. The point is not that intuition is worthless (it is not ), but that by provingtheorems about objects by only using the properties expressed in a list of axioms, the proof immediately applies to any other objects which satisfy the same list of axioms, includingcollections of objects that one might not have originally had in mind. In this way, ourordinary geometric intuitions of space and movement can be used to reason about objectsfar from our immediate experience. One important remark to make is that, by modernstandards, Euclids foundations are far from rigorous. For instance, it is implicit in thestatement of the axioms that angles can be added , but nowhere is it said what propertiesthis addition satises; angles are not numbers, neither are lengths, but they have propertiesin common with them.

2.1.2. A closer look at the fourth postulate. Notice that Euclid does not dene congru-ence. A working denition is that two gures X and Y in a space Z are congruent if thereis a transformation of Z which takes X to Y . But which transformations are allowed? Byincluding certain kinds of transformations and excluding others, we can drastically affectthe avor of the geometry in question. If not enough transformations are allowed, distinctobjects are incomparable and one cannot say anything meaningful about them. If too manytransformations are allowed, differences collapse and the supply of distinct objects to in-vestigate dries up. One way of reformulating the fourth postulate is to say that space is


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homogeneous : that is, the properties of an object do not depend on where it is placed inspace. Most of the spaces we will encounter in the sequel will be homogeneous.

2.1.3. A closer look at the parallel postulate. The fth axiom above is also known as the parallel postulate . To decode it, one needs a workable denition of parallel. The usualdenition is that two distinct lines are parallel if and only if they do not intersect. So thepostulate says that given a line l and a point p disjoint from l, there is a unique line l pthrough p such that l p and l are disjoint. Historically, this axiom was seen as unsatisfying,and much effort was put into attempts to show that it followed inevitably as a consequenceof the other four axioms. Such an attempt was doomed to failure, for the simple reasonthat there are interpretations of the undened concepts point, line, plane which satisfythe rst four axioms but which do not satisfy the fth. If we say that given l and p thereis no line l p through p which does not intersect l, we get elliptic geometry . If we say thatgiven l and p there are innitely many lines l p through p which do not intersect l, we gethyperbolic geometry . Together with Euclidean geometry, these geometries will be the mainfocus of this course.

2.1.4. Symmetries of E 2 . What are the allowable transformations in Euclidean geome-try? That is, what are the transformations of E 2 which preserve the geometrical propertieswhich characterize it? These special transformations are called the symmetries (also calledautomorphisms ) of E 2 ; they form a group , which we will denote by Aut (E 2). A symmetryof E 2 takes lines to lines, and preserves angles, but a symmetry of E 2 does not have topreserve lengths. A symmetry can either preserve or reverse orientation. Basic symmetriesinclude translations, rotations, reections, dilations . It turns out that all symmetries of E 2

can be expressed as simple combinations of these.

Exercise 2.2. Let f : E 2 E 2 be orientationreversing. Show that there is a unique linel such that f can be written as g r where r is a reection in l and g is an orientation preserving symmetry which xes l , in which case g is either a translation parallel to l or adilation whose center is on l. A reection in l followed by a translation parallel to l is alsocalled a glide reection .

Denote by Aut + (E 2) the orientationpreserving symmetries, and by Isom + (E 2) theorientationpreserving symmetries which are also distancepreserving.

Exercise 2.3. Suppose f : E 2 E 2 is in Aut + (E 2) but not in Isom + (E 2). Then there is aunique point p xed by f , and we can write f as r d where d is a dilation with center pand r is a rotation with center p.

Exercise 2.4. Suppose f Isom+ (E 2). Then either f is a rotation or a translation, and it

is a translation exactly when it does not have a xed point. In either case, f can be writtenas r 1 r 2 where r i is a reection in some line li . f is a translation exactly when l1 and l2are parallel.

These exercises show that any distancepreserving symmetry can be written as a prod-uct of at most 3 reections. An interesting feature of these exercises is that they can beestablished without using the parallel postulate . So they describe true facts (where rele-vant) about elliptic and about hyperbolic geometry. So, for instance, a distance preservingsymmetry of the hyperbolic plane can be written as a product r 1 r 2 of reections in linesl1 , l2 , and this transformation has a xed point if and only if the lines l1 , l2 intersect.Exercise 2.5. Verify that the group of orientationpreserving similarities of E 2 which xthe origin is isomorphic to C , the group of nonzero complex numbers with multiplication


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8 DANNY CALEGARI

as the group operation. Verify too that the group of translations of E 2 is isomorphic to Cwith addition as the group operation.

Exercise 2.6. Verify that the group Aut + (E 2) of orientationpreserving similarities of E 2 isisomorphic to C C where Cacts on C by multiplication. In this way identify Aut + (E 2)with the group of 2 2 complex matrices of the form

0 1

and Isom + (E 2) with the subgroup where | | = 1 .2.2. The 2sphere.

2.2.1. Elliptic geometry.

Notation 2.7. The 2sphere will be denoted by S2 .

A very interesting reinterpretation of Euclids rst 4 axioms gives us elliptic geom-etry. A point in elliptic geometry consists of two antipodal points in S2 . A line in ellipticgeometry consists of a great circle in S2 . The antipodal map i : S2 S2 is the map whichtakes any point to its antipodal point. A line or point with the interpretation above isinvariant (as a set) under i , so we may think of the action as all taking place in the quotientspace S2 /i . An object in this quotient space is just an object in S2 which is invariant asa set by i. Any two great circles intersect in a pair of antipodal points, which is a singlepoint in S2 /i . If we think of S2 as a subset of E 3 , a great circle is the intersection of thesphere with a plane in E 3 through the origin. A pair of antipodal points is the intersectionof the sphere with a line in E 3 through the origin. Thus, the geometry of S2 /i is equivalentto the geometry of planes and lines in E 3 . A plane in E 3 through the origin is perpen-dicular to a unique line in E 3 through the origin, and viceversa. This denes a dualitybetween lines and points in S2 /i ; so for any theorem one proves about lines and points in

elliptic geometry, there is an analogous dual theorem with the idea of line and pointinterchanged. Let d denote the transformation which takes points to lines and vice versa.Circles and angles make sense on a sphere, and one sees that the rst 4 axioms of Euclid

are satised in this model.As distinct from Euclidean geometry where there are symmetries which change lengths,

there is a natural length scale on the sphere. We set the diameter equal to 2 .

2.2.2. Spherical trigonometry. An example of this duality (and a justication of the choiceof length scale) is given by the following

Lemma 2.8 (Spherical law of sines) . If T is a spherical triangle with sidelengths A,B,C and opposite angles ,, , then

sin(A)

sin()=

sin(B )

sin( )=

sin(C )

sin( )Notice that the triangle d(T ) has side lengths ( ), ( ), ( ) and angles( A), ( B ), ( C ). Notice too that sin( t) t for small t , so that if T is avery small triangle, this formula approximates the sine rule for Euclidean space. Let S2t

denote the sphere scaled to have diameter 2t ; then the term sin( A)sin( ) in the spherical sine

rule should be replaced with t sin( t 1 A)

sin( ) . In this way we may think of E2 as the limit as

t of S2 .


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Exercise 2.9. Prove the spherical law of sines. Think of the sides of T as the intersectionof S2 with planes i through the origin in E 3 , intersecting in lines li in E 3 . Then the lengthsA,B,C are the angles between the l

iand the angles ,, are the angles between the

planes i .

2.2.3. The area of a spherical triangle. If L is a lune of S2 between the longitude 0 andthe longitude , then the area of L is 2 .

Now, let T be an arbitrary spherical triangle. If T is bounded by sides li which meetat vertices vi then we can extend the sides li to great circles which cut up S2 into eightregions. Each pair of lines bound two lunes, and the six lunes so produced fall into twosets of three which intersect exactly along the triangle T and the antipodal triangle i(T ). Itfollows that we can calculate the area of S as follows

4 = area (S2) = area (lunes ) 4 area (T ) = 4( + + ) 4 area (T )In particular, we have the beautiful formula, which is a special case of the GaussBonnettheorem:

Theorem 2.10. Let T be a spherical triangle with angles ,, . Then

area (T ) = + + Notice that as T gets very small and the area 0, the sum of the angles of T approach . Thus in the limit, we have Euclidean geometry in which the sum of the angles of

a triangle are . The angle formula for Euclidean triangles is equivalent to the parallelpostulate.

Exercise 2.11. Derive a formula for the area of a spherical polygon with n vertices interms of the angles.

Exercise 2.12. Using the spherical law of sines and the area formula, calculate the areaof a regular spherical n gon with sides of length t .

2.2.4. Kissing numbers the NewtonGregory problem. How many balls of radius 1 canbe arranged in E 3 so that they all touch a xed ball of radius 1? It is understood that theballs are nonoverlapping, but they may touch each other at a single point; guratively,one says that the balls are kissing or osculating (from the Latin word for kiss), and thatone wants to know the kissing number in 3dimensions.

Exercise 2.13. What is the kissing number in 2 dimensions? That is, how many disks of radius 1 can be arranged in E 2 so that they all touch a xed disk of radius 1?

This question rst arose in a conversation between Isaac Newton and David Gregory in1694. Newton thought 12 balls was the maximum; Gregory thought 13 might be possible.It is quite easy to arrange 12 balls which all touch a xed ball arrange the centers atthe vertices of a regular icosahedron. If the distance from the center of the icosahedronto the vertices is 2, it turns out the distance between adjacent vertices is

2.103, so this

conguration can be physically realized (i.e. there is no overlapping). The problem is thatthere is some slack in this conguration the balls roll around, and it is unclear whetherby packing them more tightly there would be room for another ball.

Suppose we have a conguration of nonoverlapping spheres S i all touching the centralsphere S . Let vi be the points on S where they all touch. The nonoverlapping condition isexactly equivalent to the condition that no two of the vi are a distance of less than 3 apart.If some of the S i are loose, roll them around on the surface until they come into contactwith other S j ; its clear that we can roll loose S i around until every S i touches at least


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10 DANNY CALEGARI

two other S j , S k . If S i touches S j 1 , . . . , S j n then join vi to vj 1 , . . . , v j n by segments of great circles on S . This gives a decomposition of S into spherical polygons, every edge of which has length

3. Its easy to see that no polygon has 6 or more sides (why?).

Let f n be the number of faces with n sides. Then there are 32 f 3 +2 f 4 +52 f 5 edges, since

every edge is contained in two faces. Recall Eulers formula for a polygonal decompositionof a sphere

faces edges + vertices = 2so the number of vertices is 2 + 12 f 3 + f 4 +

32 f 5

Exercise 2.14. Show that the largest spherical quadrilateral or pentagon with side lengths3 is the regular one. Use your formula for the area of such a polygon and the fact aboveto show that the kissing number is 12 in 3 dimensions. This was rst shown in the 19thcentury.

Exercise 2.15. Show what we have implicitly assumed: namely that a connected nonemptygraph in S 2 with embedded edges, and no vertices of valence 1 , has polygonal complemen-

tary regions. Remark 2.16 . In 1951 Schutte and van der Waerden ([8]) found an arrangement of 13 unitspheres which touches a central sphere of radius r 1.04556 where r is a root of thepolynomial

4096x16 18432x12 + 24576 x10 13952x8 + 4096 x6 608x4 + 32 x2 + 1This r is thought to be optimal.

2.2.5. Reections, rotations, involutions; SO (3) . By thinking of S2 as the unit sphere inE 3 , and by thinking of points and lines in S2 as the intersection of the sphere with lines andplanes in E 3 we see that symmetries of S2 extend to linear maps of E 3 to itself which xthe origin. These are expressed as 3 3 matrices. The condition that a matrix M inducea symmetry of S2 is exactly that it preserves distances on S2 ; equivalently, it preserves theangles between lines through the origin in E

3. Consequently, it takes orthonormal framesto orthonormal frames. (A frame is another word for a basis .)

Any frame can be expressed as a 3 3 matrix F , where the columns give each of thevectors. F is orthonormal if F t F = id. If M preserves orthonormality, then F t M t MF =id for every orthonormal F ; in particular, M t M = FF t = id. Observe that each of thesetransformations actually induces a symmetry of S2 ; in particular, we can identify the set of symmetries of S2 with the set of orthonormal frames in E 3 , which can be identied withthe set of 33 matrices M satisfying M t M = id. It is easy to see that such matrices forma group , known as the orthogonal group and denoted O(3) . The subgroup of orientationpreserving matrices (those with determinant 1) are denoted SO (3) and called the specialorthogonal group .

Exercise 2.17. Show that every element of O(3) has an eigenvector with eigenvalue 1 or

1. Deduce that a symmetry of S2 is either a rotation, a reection, or a product s

r where

r is reection in some great circle l and s is a rotation which xes that circle. (How is thislike a glide reection?) In particular, every symmetry of S2 is a product of at most threereections. Compare with the Euclidean case.

2.2.6. Algebraic groups. Once we have algebraized the geometry of S2 by comparing itwith the group of matrices O(3) we can generalize in unexpected ways. Let A denote the eld of real algebraic numbers. That is, the elements of A are the real roots of polynomialswith rational coefcients. If a, b

A and b = 0 then a + b, a b,ab, a/b are all in A (this


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is the dening property of a eld). There is a natural subgroup of O(3) denoted O(3, A)called the 3 dimensional orthogonal group over A which consists of the 33 matrices M with entries in A satisfying M t M = id. Observe that this, too is a group. If p = (0 , 0, 1)we can consider the subset S 2(A), the set of points in S2 which are translates of p byelements of O(3, A).

We think of S 2(A) as the points in a funny kind of space. Let

S 1(A) = S 2(A) {z = 0}and dene the set of lines in S 2(A) as the translates of S 1(A) by elements of O(3, A).

First observe that if q, r are any two points in S 2(A) thought of as vectors in E 3 then thelength of their vector crossproduct is in A . If q, r are two points in S 2(A) then togetherwith 0 they lie on a plane (q, r ). Then the triple

q, q q rq r

,q rq r

is an orthonormal frame with coordinates in A . If we think of this triple as an element of O(3, A) then the image of S 1(A) contains q and r . Thus there is a line in S 2(A) throughq and r . (Here denotes the usual cross product of vectors.)Exercise 2.18. Show that the set S 2(A) is exactly the set of points in S2

E 3 with coordinates in A.

Exercise 2.19. Explore the extent to which Euclids axioms hold or fail to hold for S 2(A)or S 2(A)/i . What if one replaces A with another eld, like Q ?

Let O(2, A) be the subgroup of O(3, A) which xes the vector p = (0 , 0, 1). Then if M O(2,

A) and N O(3,A), N ( p) = NM ( p). So we can identify S 2(A) with the

quotient space O(3, A)/O (2, A), which is the set of equivalence classes [N ] where N O(3, A) and [N ]

[N ] if and only if there is a M

O(2, A) with N = N M . In general,if F denotes an arbitrary eld, we can think of the group O(3, F ) as the set of 33 matriceswith entries in F such that M t M = id. This contains O(2, F ) naturally as the subgroupwhich xes the vector (0, 0, 1), and we can study the quotient space O(3, F )/O (2, F ) as ageometrical space in its own right. Notice that O(3, F ) acts by symmetries on this space,by M [N ] [MN ]. The quotient space is called a homogeneous space of O(3, F ).Exercise 2.20. Let F be the eld of integers modulo multiples of 2. What is the groupO(3, F )? How many points are in the space O(3, F )/O (2, F )?

In general, if we have a group of matrices dened by some algebraic condition, for

instance det( M ) = 1 or M t M = id or M t JM = J for J = 0 I

I 0etc. then we can

consider the group of matrices satisfying the condition with coefcients in some eld. This

is called an algebraic group . Many properties of certain algebraic groups are independentof the coefcient eld. An algebraic group over a nite eld is a nite group; such nitegroups are very important, and form the building blocks of most of nite group theory.

2.2.7. Quaternions and the group S3 . Recall that the quaternions are elements of the 4dimensional real vector space spanned by 1,i , j ,k with multiplication which is linear ineach factor, and on the basis elements is given by

ij = k,jk = i,ki = j, i 2 = j 2 = k2 = 1


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12 DANNY CALEGARI

This multiplication is associative. The norm of a quaternion, denoted by

a1 + a2 i + a3 j + a4k = ( a21 + a22 + a

23 + a

24)

1/ 2

is equal to the length of the corresponding vector (a1 , a2 , a 3 , a 4) in R 4 . Norms are mul-tiplicative. That is, = . The nonzero quaternions form a group undermultiplication; the unit quaternions, which correspond exactly the to the unit length vec-tors in R 4 , are a subgroup which is denoted S3 . Let be the set of quaternions of the form1 + ai + bj + ck. Then is a copy of R 3 , and is the tangent space to the sphere S3 of unitnorm quaternions at 1. The group S acts on by

z = 1zfor z . Since it preserves lengths, the image is isomorphic to a subgroup of SO (3,

R ),which can be thought of as the group of orthogonal transformations of . In fact, thishomomorphism is surjective . Moreover, the kernel is exactly the center of S3 , which is

1. That is, we have the isomorphism S3/ 1= SO (3, R ).Exercise 2.21. Write down the formula for an explicit homomorphism, in terms of standard quaternionic coordinates for S3 and matrix coordinates for SO (3, R ).

In general, the conjugation action of a Lie group on its tangent space at the identity iscalled the adjoint action of the group. Since the group of linear transformations of thisvector space is a matrix group, this gives a homomorphism of the Lie group to a matrixgroup.

2.3. The hyperbolic plane.

2.3.1. The problem of models.

Notation 2.22. The hyperbolic plane will be denoted by H 2 .

The sphere is relatively easy to understand and visualize because there is a very nicemodel of it in Euclidean space: the unit sphere in E 3 . Symmetries of the sphere extendto symmetries of the ambient space, and distances and angles in the sphere are what oneexpects from the ambient embedding. No such model exists of the hyperbolic plane. Bits of the hyperbolic planecan be isometrically (i.e. in a distancepreserving way) embedded, butnot in such a way that the natural symmetries of the plane can be realized as symmetries of the embedding. However, if we are willing to look at embeddings which distort distances,there are some very nice models of the hyperbolic plane which one can play with and geta good feel for.

2.3.2. The Poincar e Model: Suppose we imagine the world as being circumscribed by theunit circle in the plane. In order to prevent people from falling off the edge, we makethe edges very cold. As everyone knows, objects shrink when they get cold, so peoplewandering around on the disk would get smaller and smaller as they approached the edge,so that its apparent distance (to them) would get larger and larger and they could never

reach it. Technically, the length elements at the point (x, y ) are2dx

(1 x2 y2),

2dy(1 x2 y2)

or in polar coordinates, the length elements at the point r, are

2dr(1 r 2)

,2rd

(1 r 2)


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This is called the Poincar e metric on the unit disk, and the disk with this metric is calledthe Poincar e model of the hyperbolic plane . With this choice of metric, the length of aradial line from the origin to the point (r

1, 0) is

r 10 2dr(1 r 2) = log 1 + r1 rr 1

0

= log1 + r 11 r 1

If is any other path from the origin to (r 1 , 0) whose Euclidean length is l, then its lengthin the hyperbolic metric is

hyperbolic length of = l0 2dt(1 r 2( (t)))where r ( (t)) is the distance from the point (t) to the origin. Obviously, l r 1 andr ( (t)) t with equality if and only if is a Euclidean straight line. This implies that theshortest curve in the hyperbolic metric from the origin to a point in the disk is the Euclideanstraight line.

Notice that for a point p at Euclidean distance from the boundary circle, the ratio of the hyperbolic to the Euclidean metric is

21 (1 )2

1

for sufciently small .

Exercise 2.23. (Hard). Let E be a simplyconnected (i.e. without holes) domain in R 2bounded by a smooth curve . Dene a metric on E as follows. Let f be a smooth,nowhere zero function on E which is equal to 1dist ( p, ) for all p sufciently close to . Let the length elements on E be given by (dxf,dyf ) where (dx, dy) are the usual Euclideanlength elements. Show that there is a continuous, 1 1 map : E D which distorts thelengths of curves by a bounded amount. That is, there is a constant K > 0 such that for

any curve in E ,1K

length D (( )) length E () K length D (())Denition 2.24. The circle D is called the circle at innity of D , and is denoted S 1. Apoint in S 1 is called an ideal point .

Now think of the unit disk D as the set of complex numbers of norm 1. Let , betwo complex numbers with | |2 | 2| = 1 . The set of matrices of the formM =

form a group, called the special unitary group SU (1, 1). This is exactly the group of complex linear transformations of C 2 which preserves the function v(z, w) =

|z

|2

|w

|2

and have determinant 1. These are the matrices M of determinant 1 satisfying M t JM = J where J = 1 00 1

.

Exercise 2.25. Dene U (1, 1) to be the group of 22 complex matrices M with M tJM =

J with J as above, and no condition on the determinant. Find the most general form of amatrix in U (1, 1). Show that these are exactly the matrices whose column vectors are anorthonormal basis for the norm dened by v.


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14 DANNY CALEGARI

Now, there is a natural action of SU (1, 1) on D by

z

z +

z + Observe that two matrices which differ by 1 act on D in the same way. So the actiondescends to the quotient group SU (1, 1)/ 1 which is denoted by P SU (1, 1) for the projective special unitary group .

Observe that this transformation preserves the boundary circle. Furthermore it takeslines and circles to lines and circles, and preserves angles of intersection between them; inparticular, it permutes segments of lines and circles perpendicular to D .Exercise 2.26. Show that there is a transformation in P SU (1, 1) taking any point in theinterior of D to any other point.

Exercise 2.27. Show that the subgroup of P SU (1, 1) xing any point is isomorphic to acircle. Deduce that we can identify D with the coset space P SU (1, 1)/S 1 ; i.e. D is ahomogeneous space for P SU (1, 1).Exercise 2.28. Show that the action of P SU (1, 1) preserves the Poincar e metric in D 2 . Deduce that the shortest hyperbolic path between any two points is through an arc of acircle orthogonal to D or, if the points are on the same diameter, by a segment of thisdiameter.

2.3.3. The upper halfspace model. The upper halfspace , denoted H , is the set of pointsx, y

R 2 with y > 0. Suppose now that the real line is chilled, so that distances inthis model are scaled in proportion to the distance to the boundary. That is, in (x, y ) coordinates the length elements of the metric are

dxy

,dyy

Observe that translations parallel to the xaxis preserve the metric, and are therefore isome-tries. Also, dilations centered at points on the xaxis preserve the metric too. The lengthof a vertical line segment from (x, y1) to (x, y2) is

y 2y 1 dyy = log y2y1A similar argument to before shows that this is the shortest path between these two points.

The group of 22 matrices with real coefcients and determinant 1 is called the speciallinear group , and denoted SL (2, R ) or SL (2) if the coefcients are understood. Thesematrices act on the upper halfplane, thought of as a domain in C , by

z

z + z +

Again, two matrices which differ by a constant multiple act on H in the same way, so the

action descends to the quotient group SL (2, R )/ 1 which is denoted by P SL (2, R ) forthe projective special linear group .As before, these transformations take lines and circles to lines and circles, and preserve

the real line.

Exercise 2.29. Show that the action of P SL (2, R ) preserves the metric on H . Deduce that the shortest hyperbolic path between any two points in the upper halfspace is through anarc of a circle orthogonal to R or, if the points are on the same vertical line, through asegment of this line.


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Exercise 2.30. Find a transformation from D to H which takes the Poincar e metric onthe disk to the hyperbolic metric on H . Deduce that these models describe the samegeometry. Find an explicit isomorphism P SL (2, R )

=P SU (1, 1).

2.3.4. The hyperboloid model. In R 3 let H denote the sheet of the hyperboloid x2 + y2 z2 = 1 with z positive. Let O(2, 1) denote the set of 3 3 matrices with real entrieswhich preserve the function v(x,y,z ) = x2 + y2 z2 , and SO (2, 1) the subgroup withdeterminant 1. Equivalently, O(2, 1) is the group of real matrices M such that M t JM = J where

J =1 0 00 1 00 0 1

Then SO (2, 1) preserves the sheet H .

Denition 2.31. A vector v in R 3 is timelike if vt Jv < 0, spacelike if vt Jv > 0 andlightlike if vt Jv = 0 . The Lorentz length of a vector is (vt Jv )1/ 2 , denoted v , and can

be positive, zero, or imaginary. The timelike angle between two timelike vectors v, w is

(v, w) = cosh 1 vt Jw

v wCompare this with the usual angle between two vectors in R 3 :

(v, w) = cos 1 vt w

v wwhere in this equation denotes the usual length of a vector. Notice that H is exactly theset of vectors of Lorentz length i, just as S2 is the set of vectors of usual length 1 (this hasled some people to comment that the hyperbolic plane should be thought of as a sphere of imaginary radius). Since distances between points in S2 are dened as the angle between

the vectors, it makes sense to dene distances in H as the timelike angle between vectors.For two vectors v, wH the formula above simplies to

(v, w) = cosh 1(vt Jw )Exercise 2.32. Let K be the group of matrices of the form

cos() sin() 0

sin() cos() 00 0 1and A the group of matrices of the form

cosh( ) 0 sinh( )0 1 0

sinh( ) 0 cosh( )

Show that every element of SO (2, 1) can be expressed as k1ak2 for some k1 , k2K and

aA; that is, we can write SO (2, 1) = KAK . How unique is such an expression?

Notice the group K above is precisely the stabilizer of the point (0, 0, 1)H . Thus wecan identify H with the homogeneous space SO (2, 1)/K .

Exercise 2.33. Let K be the subgroup of P SL (2, R ) consisting of matrices of the formcos() sin()

sin() cos()and A the subgroup of matrices of the form s 00 s1 . Find an


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16 DANNY CALEGARI

isomorphism from SO (2, 1) to P SL (2, R ) taking K to K and A to A . (Careful! Theisomorphism K K might not be the one you rst think of . . .) Remark 2.34 . The isomorphisms P SU (1, 1) = P SL (2,

R ) = SO (2, 1)are known as

exceptional isomorphisms , and one should not assume that the matter is so simple in higherdimensions. Such exceptional isomorphisms are rare and are a very powerful tool, sincedifcult problems about one of the groups can become simpler when translated into aproblem about another of the groups.

After identifying P SL (2, R ) with SO (2, 1) we can identify their homogeneous spacesP SL (2, R )/K and SO (2, 1)/K . This identication of H with H shows that H is anequivalent model of the hyperbolic plane. The straight lines in H are the intersection of planes in R 3 through the origin with H . In many ways, the hyperboloid model of thehyperbolic plane is the closest to the model of S2 as the unit sphere in R 3 .

Exercise 2.35. Show that the identication of H with H preserves metrics.

Two vectors v, wR 3 are Lorentz orthogonal if vt Jw = 0 . It is easy to see that if v is

timelike, any orthogonal vector w is spacelike. If vH and w is a tangent vector to H atv, then vt Jw = 0 , since the derivative ddt v + tw should be equal to 0 at t = 0 (by thedenition of a tangent vector). For two spacelike vectors v, w which span a spacelike vectorspace, the value of v

t Jwv w < 1, so (v, w) is an imaginary number. The spacelike angle

between v and w is dened to be i(v, w). It is a fact that the hyperbolic angle betweentwo tangent vectors at a point in H is exactly equal to their spacelike angle. The proofof this fact is just that the symmetries of the space H preserve this spacelike angle, andthe total spacelike angle of a circle is 2 . Since these two properties uniquely characterizehyperbolic angles, the two notions of angle agree.

The fact that hyperbolic lengths and angles can be expressed so easily in terms of trigonometric functions and linear algebra makes the hyperboloid model the model of choice for doing hyperbolic trigonometry.

2.3.5. The Klein (projective) model. Let D1 be the disk consisting of points in R 3 withx2 + y2 1 and z = 1 . Then we can project H to D1 along rays in R 3 which pass throughthe origin.Exercise 2.36. Verify that this stereographic projection is 1 1 and onto.

This projection takes the straight lines in H to (Euclidean) straight lines in D1 . Thisgives us a new model of the hyperbolic plane as the unit disk, whose points are usualpoints, and whose lines are exactly the segments of Euclidean lines which intersect D1 .One should be wary that Euclidean angles in this model do not accurately depict the truehyperbolic angles. In this model, two lines l1 , l2 are perpendicular under the followingcircumstances:

If l1 passes through the origin, l2 is perpendicular to l1 if any only if it is perpen-dicular in the Euclidean sense. Otherwise, let m1 , m 2 be the two tangent lines to D 1 which pass through theendpoints of l1 . Then l1 and l2 are perpendicular if and only if l2 , m 1 , m 2 intersect

in a point.

Exercise 2.37. Verify that l1 is perpendicular to l2 if and only if l2 is perpendicular to l1 .

It is easy to verify in this model that all Euclids axioms but the fth are satised.


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The relationship between the Klein model and the Poincar e model is as follows: wecan map the Poincar e disk to the northern hemisphere of the unit sphere by stereographicprojection. This preserves angles and takes lines and circles to lines and circles. In thismodel, the straight lines are exactly the arcs of circles perpendicular to the equator. Thenthe Klein model and this (curvy) Poincar e model are related by placing thinking of theKlein disk as the at Euclidean disk spanning the equator, and mapping D1 to the upperhemisphere by projecting points along lines parallel to the zaxis. This takes lines in D1to the intersection of the upper hemisphere with vertical planes. These intersections arethe circular arcs which are perpendicular to the equator, so this map takes straight lines tostraight lines as it should.

Exercise 2.38. Write down the metric in D1 for the Klein model. Using this formula,calculate the hyperbolic distance between the center and a point at radius r 1 in D1 .

Exercise 2.39. Show that Pappus theorem is true in the hyperbolic plane; this theoremsays that if a1 , a 2 , a 3 and b1 , b2 , b3 are points in two lines l1 and l2 , then the six line

segments joining the a i to the bj for i = j intersect in three points which are collinear.2.3.6. Hyperbolic trigonometry. Hyperbolic and spherical geometry are two sides of thesame coin. For many theorems in spherical geometry, there is an analogous theorem inhyperbolic geometry. For instance, we have the

Lemma 2.40 (Hyperbolic law of sines) . If T is a hyperbolic triangle with sides of lengthA,B,C opposite angles ,, then

sinh( A)sin()

=sinh( B )sin( )

=sinh( C )sin( )

Exercise 2.41. Prove the hyperbolic law of sines by using the hyperboloid model and trying to imitate the vector proof of the spherical law of sines.

2.3.7. The area of a hyperbolic triangle. The parallels between spherical and hyperbolicgeometry are carried further by the theorem for the area of a hyperbolic triangle. We relaxslightly the notion of a triangle: we allow some or all of the vertices of our triangle to beideal points. If all three vertices are ideal, we say that we have an ideal triangle . Noticethat since every hyperbolic straight line is perpendicular to S 1, the angle of a triangle atan ideal point is 0.Theorem 2.42. Let T be a hyperbolic triangle with angles ,, . Then

area (T ) = Proof: In the upper halfspace model, let T be the triangle with one ideal point at and two ordinary points at (cos(), sin()) and (cos( ), sin( )) in Euclidean coordinates, where , are both / 2. Such a triangle has angles 0, , . The hyperbolicarea is

cos( )

x =cos( ) y=1 x 2 1y2 dy dx= cos( )x =cos( ) 11 x2 dx

= cos1(x)cos( )

cos( )=


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18 DANNY CALEGARI

In particular, a triangle with one or two ideal points satises the formula. Now, for anarbitrary triangle T with angles ,, we can dissect an ideal triangles with all angles 0into T and three triangles, each of which has two ideal points, and whose third angle is oneof , , . That is,

area (T ) = ( ( )) ( ( )) ( ( )) =

2.3.8. Projective geometry. The group P SL (2, R ) acts in a natural way on another spacecalled the projective line , denoted RP 1 . This is the space whose points are the lines throughthe origin in R 2 . Equivalently, this is is the quotient of the space R 2 0 by the equiva-lence relation that (x, y ) (x, y ) for any

R. The unit circle maps 21 to RP 1 ,so one sees that RP 1 is itself a circle. The natural action of P SL (2, R ) on RP 1 is the projectivization of the natural action of SL (2, R ) on R 2 . That is,

x

y= x + y

x + yWe can write an equivalence class (x, y ) unambiguously as x/y , where we write

when y = 0 ; in this way, we can naturally identify RP 1 with R . One sees thatin this formulation, this is exactly the action of P SL (2, R ) on the ideal boundary of H 2in the upper halfspace model. That is, the geometry of RP 1 is hyperbolic geometry at innity . Observe that for any two triples of points a1 , a2 , a 3 and b1 , b2 , b3 in RP 1 whichare circularly ordered, there is a unique element of P SL (2, R ) taking a i to bi . For a fourtuple of points a1 , a 2 , a 3 , a 4 let be the transformation taking a1 , a 2 , a 3 to 0, 1, .Then (a4) is an invariant of the 4tuple, called the crossratio of the four points, denoted[a1 , a 2 , a 3 , a 4]. Explicitly,

[a1 , a 2 , a 3 , a 4] =(a1 a3)(a2 a4)(a1 a2)(a3 a4)

The subgroup of P SL (2, R ) which xes a point in Ris isomorphic to the group of orientation preserving similarities of R , which we could denote by Aut + (R ). This groupis isomorphic to R + R . where R + acts on R by multiplication. We can think of RP 1 asthe homogeneous space P SL (2, R )/ R + R .

Projective geometry is the geometry of perspective . Imagine that we have a transparentglass pane, and we are trying to capture a landscape by setting up the pane and paintingthe scenery on the pane as it appears to us. We could move the pane to the right or left;this would translate the scene left or right respectively. We could move the pane closer orfurther away; this would shrink or magnify the image. Or we could rotate the pane andourselves so that the sun doesnt get in our eyes. The horizon in our picture is RP 1 , and thetransformations we can perform on the image is precisely the projective group P SL (2, R ).

2.3.9. Elliptic, parabolic, hyperbolic isometries. There are three different kinds of trans-

formations in P SL (2, R ) which can be distinguished by their action on S 1.Denition 2.43. A nontrivial element P SL (2,

R ) is elliptic, parabolic or hyperbolicif it has respectively 0, 1 or 2 xed points in S 1. These cases can be distinguished by theproperty that |tr ( )| is 2, where tr denotes the trace of a matrix representativeof .

An elliptic transformation has a unique xed point in H 2 and acts as a rotation aboutthat point. A hyperbolic transformation xes the geodesic running between its two ideal


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xed points and acts as a translation along this geodesic. Furthermore, the points in H 2

moved the shortest distance by the transformation are exactly the points on this geodesic.A parabolic transformation has no analogue in Euclidean or Spherical geometry. It

has no xed point, but moves points an arbitrarily short amount. In some sense, it islike a rotation about an ideal point. Two elliptic elements are conjugate iff they rotateabout their respective xed points by the same amount. This angle of rotation is equalto cos1(|tr ( )/ 2|). Two hyperbolic elements are conjugate iff they translate along theirgeodesic by the same amount. This translation length is equal to cosh1(|tr ( )/ 2|). If a, b are any two parabolic elements then either a and b are conjugate or a1 and b areconjugate.

Exercise 2.44. Prove the claims made in the previous paragraph.

Exercise 2.45. Recall the subgroups K and A dened in the prequel. Let N denote the

group of matrices of the form 1 t0 1 . Show that every element of P SL (2,R ) can be

expressed as kan for some kK , a A

and n N . How unique is this expression?This is an example of what is known as the KAN or Iwasawa decomposition .

Exercise 2.46. Consider the group SL (n, R ) of n n matrices with real entries and determinant 1. Let K be the subgroup SO (n, R ) of real n n matrices M satisfyingM t M = id. Let A be the subgroup of diagonal matrices. Let N be the subgroup of matrices with 1s on the diagonal and 0s below the diagonal. Show A is abelian and N isnilpotent. Further, K is compact (see appendix), thought of as a topological subspace of the space R n

2

of n n matrices. Show that there is a KAN decomposition for SL (n, R ).2.3.10. Horocircular geometry. In the Poincar e disk model, the Euclidean circles in Dwhich are tangent to S 1 are special; they are called horocircles and one can think of them as circles of innite radius . If the point of tangency is taken to be in the upperhalfspace model, these circles correspond to the horizontal lines in the upper halfplane.

Observe that a parabolic element xes the family of horocircles tangent to its xed idealpoint, and acts on each of them by translation.

3. T ESSELLATIONS

3.1. The topology of surfaces.

3.1.1. Gluing polygons. Certain computer games get around the constraint of a nitescreen by means of a trick: when a spaceship comes to the left side of the screen, itdisappears and reappears on the right side of the screen. Likewise, an asteroid whichdisappears beyond the top of the screen might reappear menacingly from the bottom. Thescreen can be represented by a square whose sides are labelled in pairs : the left and rightsides get one label, the top and bottom sides get another label. These labels are instructionsfor obtaining an idealized topological space from the at screen: the left and right sides

can be glued together to make a cylinder, then the top and bottom sides can be glued to-gether to make a torus (the surface of a donut). Actually, we have to be somewhat careful:there are two ways to glue two sides together; an unambiguous instruction must specify theorientation of each edge.

If we have a collection of polygons P i to be glued together along pairs of edges, we canimagine a graph whose vertices are the polygons, and whose edges are the pairs of edgesin the collection. We can glue in any order. If we rst glue along the edges correspondingto a maximal tree in , the result of this rst round of gluing will produce a connected


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20 DANNY CALEGARI

polygon. Thus, without loss of generality, it sufces to consider gluings of the sides of asingle polygon.

Denition 3.1. A surface is a 2dimensional manifold. That is, a Hausdorff topologicalspace with a countable basis, such that every point has a neighborhood homeomorphic tothe open unit disk in R 2 (see appendix for denitions). A piecewiselinear surface is asurface obtained from a countable collection of polygons by glueing together the edges inpairs, in such a way that only nitely many edges are incident to any vertex.

Exercise 3.2. Why is the niteness condition imposed on vertices?

The following theorem was proved by T. Rado in 1924 (see [6]):

Theorem 3.3. Any surface is homeomorphic to a piecewiselinear surface. Any compact surface is homeomorphic to a piecewiselinear surface made from only nitely many poly-gons.

Denition 3.4. A surface is oriented if there is an unambiguous choice of top and bot-tom side of each polygon which is compatible with the glueing. i.e. an orientable surfaceis twosided.

3.1.2. The fundamental group. The denition of the fundamental group of a surface requires a choice of a basepoint in . Let p be such a point.

Denition 3.5. Dene 1( , p) to be the space of continuous maps c : S 1 sending0 S 1 to p (here we think of S 1 as I/ 0 1). Two such maps c1 , c2 are calledhomotopic if there is a map C : S 1 I such that

(1) C (, 0) = c1().(2) C (, 1) = c2().(3) C (0, ) = p.Exercise 3.6. Show that the relation of being homotopic is an equivalence relation on1( , p).

In fact, 1( , p) has the natural structure of a topological space; with respect to thistopological structure, the equivalence classes determined by the homotopy relation are thepathconnected components.

The importance of the relation of homotopy equivalence is that the equivalence classesform a group :

Denition 3.7. The fundamental group of with basepoint p, denoted 1( , p), has aselements the equivalence classes

1( , p) = 1( , p)/ homotopy equivalence

with the group operation dened by

[c1] [c2] = [c1c2]where c1c2 denotes the map S

1 dened byc1c2(t) =

c1(2t) for t 1/ 2c2(2t 1) for t 1/ 2

where the identity is given by the equivalence class [e] of the constant map e : S 1 p, andinverse is dened by [c]1 = [ i(c)], where i(c) is the map dened by i(c)( t) = c(1 t).


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Exercise 3.8. Check that [c][c]1 = [e] with the denitions given above, so that 1( , p)really is a group.

For a piecewiselinear surface with basepoint v a vertex of , dene O1() to be thespace of polygonal loops from v to v contained in the edges of . Such a loop consistsof a sequence of oriented edges

= e1 , e2 , . . . , e n

where e1 starts at v and en ends there, and ei ends where ei +1 starts. Let r (e) denote thesame edge e with the opposite orientation. For a polygonal path (not necessarily startingand ending at v) let 1 denote the path obtained by reversing the order and the orientationof the edges in .

One can perform an elementary move on a polygonal loop , which is one of the fol-lowing two operations:

If there is a vertex w which is the endpoint of some ei , and is any polygonalpath beginning atw

, we can insert or delete

1 betweenei

andei +1

. That is,

e1 , e2 , . . . , e i , , 1 , e i+1 , . . . , e n e1 , e2 , . . . , e n If is a loop which is the boundary of a polygonal region, and starts and ends at avertex w which is the endpoint of some ei , then we can insert or delete between

ei and ei+1 . That is,

e1 , e2 , . . . , e i , , e i +1 , . . . , e n e1 , e2 , . . . , e nDenition 3.9. Dene the combinatorial fundamental group of , denoted p1( , v), to bethe group whose elements are the equivalence classes

O1( , v)/ elementary moves

With the group operation dened by [][ ] = [ ], where the identity is given by the

empty polygonal loop 0 starting and ending at v, and with inverse given by i( ) = 1.

Exercise 3.10. Check that the above makes sense, and that p1( , v) is a group.

Now suppose that is obtained by glueing up the sides of a single polygon P in pairs.Suppose further that after the result of this glueing, all the vertices of this polygon areidentied to a single vertex v. Let e1 , . . . , e n be the edges and

= e1i 1 e1i 2 . . . e1i mthe oriented boundary of P . Then there is a natural isomorphism

p1( , v) = e1 , . . . , e n | that is, p1 can be thought of as the group generated by the edges ei subject to the relationdened by . Notice that each of the ei appears twice in , possibly with distinct signs. If is orientable, each ei appears with opposite signs.

Example 3.11 . Let T denote the surface obtained by glueing opposite sides of a squareby translation. Thus the edges of the square can be labelled (in the circular ordering) bya,b,a 1 , b1 and a presentation for the group is

p1(T, v) = a, b|[a, b]It is not too hard to see that this is isomorphic to the group Z

Z .


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3.1.5. Covering spaces.

Denition 3.19. A space Y is a covering space for X if there is a map f : Y

X (called

a covering projection ) with the property that every point xX has an open neighborhood

x U such that f 1(U ) is a disjoint union of open sets U i Y , and f maps each U ihomeomorphically to U . The universal cover of a space X (if one exists) is a simply

connected space X which is a covering space of X .

An open neighborhood U of a point x of the kind provided in the denition is said tobe evenly covered by its preimages f 1(U ). If X is locally connected, we can assume thatthe open neighborhoods which are evenly covered are connected.

For a path I X we can nd, for each point p I an open neighborhood U p of p which is evenly covered. Since I is compact (see appendix) only nitely many openneighborhoods are needed to cover I ; call these U 1 , . . . , U n . If we let V 1 denote somecomponent of f 1(U 1) which maps homeomorphically to U 1 . Then there is a unique mapg : I U 1 V 1 such that fg = id. Moreover, there is a unique choice of V 2 from amongstthe components of f

1

(U 2) such that g can be extended to g : I (U 1U 2) V 1V 2with fg = id. Continuing inductively, we see that the choice of V 3 , V 4 , . . . , V n are alluniquely determined by the original choice V 1 .

Exercise 3.20. Modify the above argument to show that for every map g : I X anyevery pf

1g(0) there is a unique lift g : I Y such that g(0) = p and f g = g.Exercise 3.21. Suppose g1 , g2 : I X with g1(0) = g2(0) and g1(1) = g2(1) arehomotopic through homotopies which keep the endpoints of I xed. Let g1 be some lift of g1 . Show that the lift g2 of g2 with g1(0) = g2(0) also satises g1(1) = g2(1) , and thesetwo lifts are also homotopic rel. endpoints. (Hint: let G : I I X be a homotopybetween g1 and g2 . Try to lift G to a map G : I I X satisfying appropriateconditions.)

Let p

X be a basepoint, and let p

f 1( p). Then the projection f : Y

X induces a homomorphism f

: 1(Y, p) 1(X, p ). Let K 1(X, p ) denote the image

of f . Then a loop : S 1 X with []K can be lifted to a loop : S 1 Y , by

the argument of the previous exercise. Conversely, if two loops , represent the sameelement of K , then their lifts represent the same element of 1(Y, p). Thus we may identify1(Y, p) with the subgroup K .

Exercise 3.22. Show that for a space Z and a map g : Z X there is a lift of g tog : Z Y with f g = g if and only if g(1(Z ))K .Denition 3.23. A space is semilocally simplyconnected if every point p has a neigh-borhood U so that every loop in U can be shrunk to a point in a possibly larger openneighborhood V .

Informally, a space is semilocally simplyconnected if sufciently small loops in the

space are homotopically inessential.Theorem 3.24. Let X be connected, locally connected and semilocally simplyconnected.For any subgroup G 1(X, x ) there is a covering space X G of X and a point y f 1(x) such that f

(1(X G , y)) = G.

Sketch of proof: Let (X ) be the space of paths : I X which start at x. Weinduce an equivalence relation on (X ) where we say two paths 1 , 2 are equivalent if [ 12 1]G. Set X G = ( X )/ . Since X is semilocally simply connected, for


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24 DANNY CALEGARI

sufciently small paths 1 , 2 between points x1 , x2X the loop 1

2 1 is contractible.So any two paths which differ only by substituting 1 in one for 2 in the other will beequivalent in (X ), and this says that the equivalence classes of (X ) are parameterizedlocally by points in X ; that is, X G is a covered space of X . It can be veried that X G issimply connected, and that it satises the conditions of the theorem.

Exercise 3.25. Fill in the gaps in the sketch of the proof above.

Exercise 3.26. Show that the universal cover of a space X , it if exists, is unique, by usingthe lifting property.

3.1.6. Discrete groups.

Denition 3.27. Let be a group of symmetries of a space X which is one of Sn , E n , H nfor some n . is properly discontinuous if, for each closed and bounded subset K of X ,the set of such that (K ) K = is nite. acts freely if no has a xedpoint; that is, if ( p) = p for any p, then = id.

For a point pX , the subgroup of which xes p is called the stabilizer of p, and istypically denoted by ( p). If acts properly discontinuously, then ( p) is nite for any p.

Denition 3.28. A subgroup of a Lie group G is discrete if K is nite for anycompact subset K G.Exercise 3.29. Show that if G is a Lie group of symmetries of a space X as above, then asubgroup is discrete iff it acts properly discontinuously on X .

Suppose acts on X freely and properly discontinuously. We can dene a quotientspace M = X/ where two points x, y X are identied exactly when there is some such that (x) = y.

Theorem 3.30. If acts on X freely and properly discontinuously, where X is one of Sn , E n , H n for some n , then the projection X X/ is a covering space, and X is theuniversal cover of X/ .

Proof: Pick a point xX and let U be a neighborhood of x which intersects onlynitely many translates (U ). Then there is a smaller V U which is disjoint from all itstranslates. Then under the quotient map, each translate (V ) is mapped homeomorphicallyto its image. Thus X is a covering space of X/ , and since it is simplyconnected, it isthe universal cover.

3.1.7. Fundamental domains. Let act on X properly discontinuously, where X is oneof the spaces Sn , E n , H n .

Denition 3.31. A fundamental domain for the action of is a polygon P X such thatfor all p in the interior of P , ( p)

P =

unless = id, and such that the faces of P arepaired by the action of .

The translates of a fundamental domain are disjoint except along their boundaries.Moreover, these translates cover all of X . Thus they give a tessellation of X , whose sym-metry group contains as a subgroup. For generic fundamental domains, there are noaccidental symmetries, and the group of symmetries of the tessellation is exactly . Ingeneral, fundamental domains can be decorated with some extra marking which destroysany additional extra symmetries of a fundamental domain.


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Denition 3.32. Choose a point pX . The Dirichlet domain of centered at p, is theset

D = {qX such that d( p,q) d( p,(q)) for all }In dimension 2, 3, in the cases we are interested in, D will be a locally nite polygon inX whose faces are paired by the action of ; but in general, D might not be a polygon.

A Dirichlet domain is a fundamental domain for .

3.2. Lattices in E 2 .

3.2.1. Discrete groups in Isom (E 2). Perhaps the most important theorem about discretesubgroups of Isom (E n ) is Bieberbachs theorem, that such a group has a free abelian sub-group of nite index. In dimension two, this can be rened as follows:

Theorem 3.33. Let act properly discontinuously on E 2 by isometries. Then has asubgroup + of index at most 2 which is orientationpreserving. Moreover, + is one of the following:

+ is a subgroup of stab ( p) for some p. In this case, + = Z /n Z consists of powers of a single rotation. + is a semidirect product

+ = Z /n Z Z

where the Z factor is generated by a translation, and n is 1 or 2. In the second case, the conjugation action takes x x .

+ is a semidirect product + = Z /n Z (ZZ )

where the Z

Z factor is generated by a pair of linearly indpendent translations,and n = 1 , 2, 3, 4 or 6. If n = 4 , the Z

Z is conjugate to the group of translationsof the form z

z + n + mi for integers n, m . If n = 3 or 6 , the Z

Z is conjugateto the group of translations of the form z z + n + m 1+ i32 for integers n, m .

Proof: First, there is a homomorphism

o : Isom (E 2) Z / 2Zwhere o() is 0 or 1 depending on whether or not is orientation preserving. The in-tersection of with the kernel of o is + , and it has index at most 2. Next, there is ahomomorphism

a : Isom + (E 2) S 1given by the action of isometries on equivalence classes of parallel lines, where the imageis the angles of rotation. There is an induced homomorphism

a : + S 1The kernel K of a in + consists of a group of translations, and is therefore abelian.

Suppose = a() for some + , and nonzero. Then is a rotation, and therefore

xes some p. Since + is properly discontinuous, either + stab ( p) in which case +

is cyclic and consists of the powers of a xed rotation, or there is a (nonunique) closestimage q = p of p under some . Since q = ( p), no translate of p is closer to q than p. If

|| < 26 , then0 < d (q, (q)) = 2sin

2

d( p,q) < d ( p,q)


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26 DANNY CALEGARI

which is a contradiction. Since the same must also be true for each power of , the orderof is 6. If the order is 5, then 0 < d ((q), 1 1( p)) < d ( p,q), which is acontradiction. Hence the image a(+ ) is Z /n Z where n = 1 , 2, 3, 4 or 6. In any case, theimage is cyclic, and is therefore generated by a single a( ) where is a rotation, so + isa semidirect product.

The kernel K of a in + is a properly discontinuous group of translations. If K isnontrivial and the elements of K are linearly dependent, they are all powers of the element of shortest translation length. The image a(+ ) must preserve the translation of ; inparticular, this image must be trivial or Z / 2Z .

If K is nontrivial and the elements of K are linearly independent, they are generatedby two elements of shortest translation length. The image a(+ ) must preserve the set of nonzero elements of shortest length, so if this image is Z / 4Z the group K is generated bytwo perpendicular vectors of equal length. If the image is Z / 3Z or Z / 6Z , K is generatedby two vectors at angle 26 of equal length.

The two special lattices (i.e. groups of translations generated by two linearly indepen-dent elements) appearing in theorem 3.33 are often called the square and hexagonal latticesrespectively.

3.2.2. Integral quadratic forms. A good reference for the material in this and the nextsection is [2].

Denition 3.34. A quadratic form is a homogeneous polyonomial of degree 2 in somevariables. That is, a function of variables x1 , . . . , x n such that

f (1x1 , . . . , n xn ) =n

i =1

2i f (x1 , . . . , x n )

A quadratic form is integral if its coefcients as a polynomial are integers.

We will be concerned in the sequel with integral quadratic forms of two variables, suchas 3x2 + 2 xy 7y2 or x2 3xy .Notice for every quadratic form f (, ) there corresponds uniquely a symmetric matrixM f such that f (x, y ) = x y M f

xy . In particular, if f (x, y ) = ax

2 + bxy + cy2 then

M f =a b2b2 c

Denition 3.35. We say that a quadratic form f (x, y ) represents an integer n if there issome assignment of integer values n1 , n 2 to x, y for which

f (n1 , n 2) = n

Given an integral quadratic form, it is a natural question to ask what interger valuesit represents. Observe that there are some elementary transformations on quadratic formswhich do not change the set of values represented. If we substitute x xy or y yxwe get a new quadratic form. Since this substitution is invertible , the new quadratic formobtained represents exactly the same set of values as the original. For instance,

x2 + y2 x2 + 2 xy + 2 y2This substitution denes an equivalence relation on quadratic forms.


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The most general form of substitution allowed is a transformation of the form

x px + qy, y rx + syFor integers p,q,r,s . Again, since this substitution must be invertible, we should have

ps qr = 1. That is, the matrix p qr s is in SL (2, Z ). Since these forms are

homogeneous of degree 2, the substitution x x, y y does nothing. One caneasily check that every other substitution has a nontrivial effect on some quadratic form.

In particular, we have the following theorem:

Theorem 3.36. Integral quadratic forms up to equivalence are parameterized by equiva-

lence classes of matrices of the form ab2

b2 c

for integers a,b,c modulo the conjugation

action of P SL (2, Z ).

Observe that what is really going on here is that we are evaluating the quadratic form f

on the integral latticeZ

Z. The group SL (2,

Z) acts by automorphisms of this lattice,and therefore permutes the set of values attained by the form f .

There is nothing special about the integral lattice here; it is obvious that SL (2, Z ) acts

by automorphisms of any lattice L. In particular, if L = e1 , e2 then M =

SL (2, Z ) acts on elements of this lattice by

M re 1 + se2 (r + s )e1 + ( r + s)e23.2.3. Moduli of tori, continued fractions and P SL (2, Z ).

Denition 3.37. Let r be a real number. A continued fraction expansion of r is an expres-sion of r as a limit of a (possibly terminating) sequence

n1 , n 1 +1

m1, n 1 +

1

m1 +1

n 2, n 1 +

1

m1 +1

n 2 + 1m 2

, . . .

where each of the n i , m i is a positive integer.

A continued fraction expansion of r can be obtained inductively by Euclids algorithm.First, n1 is the biggest integer r . So 0 r n1 < 1. If r n1 = 0 we are done.Otherwise, r = 1r n 1 > 1 and we can dene m1 as the biggest integer r . So 0 r m1 < 1. continuing inductively, we produce a series of integers n1 , m 1 , n 2 , m 2 , . . .which is the continued fraction expansion of r . If r is rational, this procedure terminates ata nite stage. The usual notation for the continued fraction expansion of a real number r is

r = n1 +1

m1+1

n2+1

m2+1

n3+. . .

The following theorem is quite easy to verify:

Theorem 3.38. If n1 , m 1 , . . . is a continued fraction expansion of r , then the successiveapproximations

n1 , n 1 +1

m1, n 1 +

1m1 + 1n 2

, . . .

denoted r 1 , r 2 , r 3 , . . . satisfy

|r r i | |r p/q | for any integers ( p,q) , where q < the denominator of r i+1 .


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28 DANNY CALEGARI

Thus, the continued fraction approximations of r are the best rational approximationsto r for a given bound on the denominator.

Let T be a at torus. Then the isometry group of T is transitive (this is not too hardto show). Pick a point p, and cut T up along the two shortest simple closed curves whichstart and end at p. This produces a Euclidean parallelogram P . After rescaling T , we canassume that the shortest side has length 1. We place P in E 2 so that this short side is thesegment from 0 to 1, and the other side runs from 0 to z where Im (z) > 0. By hypothesis,

|z| 1. Moreover, if |Re (z)| 12 we can replace z by z + 1 or z 1 with smaller norm,contradicting the choice of curves used for the decomposition. Let D be the region in theupper halfplane bounded by the two vertical lines Re (z) = 12 , Re (z) = 12 and the circle|z| = 1 .

The group P SL (2, Z ) acts naturally on H as a subgroup of P SL (2, R ). The action thereis properly discontinuous. The action of P SL (2, Z ) permutes the sides of D . The element

1 10 1 pairs the two vertical sides, and the element

0 1

1 0preserves the bottom side,

interchanging the left and right pieces of it. In particular, D is a fundamental domain forP SL (2, Z ). The quotient is topologically a disk, but with two cone points of order 2 and3 respectively, which correspond to the points i and 1+ i32 respectively, whose stabilizersare Z / 2Z and Z / 3Z respectively.

This quotient is an example of an orbifold .

Denition 3.39. An ndimensional orbifold is a space which is locally modelled on R nmodulo some nite group.

A 2dimensional orbifold looks like a surface except at a collection of isolated points pi where it looks like the quotient of a disk by the action of Z /n i Z , a group of rotationscentered at pi . The point pi is a cone point , sometimes also called an orbifold point . The

nite group is part of the data of the orbifold. One can think of the orbifold combinatoriallyas a surface (in the usual sense) with a nite number of distinguished points, each of whichhas an integer attached to it. Geometrically, this point looks like a cone made from awedge of angle 2/n i .

We can dene an orbifold fundamental group o1 () for a surface orbifold. Thinkingof our orbifold as X/ for the moment where acts properly discontinuously but notfreely, the orbifold fundamental group of should be exactly . This means that a smallloop around an orbifold point pi should have order n i in o1 () . Note that we are beingcasual about basepoints here, so we are only thinking of these groups up to isomorphism.

In any case, the orbifold H 2 /PSL (2, Z ) should have orbifold fundamental group iso-morphic to P SL (2, Z ). There is an element of order 2 corresponding to the loop around

the order 2 point; a representative of this element in P SL (2, Z ) is 0 1

1 0 . There is an

element of order 3 corresponding to the loop around the order 3 point; a representativeof this element in P SL (2, Z ) is 1 1

1 0. Note that these elements have order 2 and 3

respectively in P SL (2, Z ), even though the corresponding matrices have orders 4 and 6respectively in SL (2, Z ).

Every loop in the disk can be shrunk to a point; it follows that every loop in the orbifoldH 2 /PSL (2, Z ) can be shrunk down to a collection of small loops around the two conepoints in some order. That is, the group P SL (2, Z ) is generated by these two elements.


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Theorem 3.40. A presentation for P SL (2, Z ) is given by

P SL (2, Z )= ,

| 2 , 3 = Z / 2Z

Z / 3Z

where representatives of and are the two matrices given above.

Proof: By the discussion above, all that needs to be established is that there are noother relations that do not follow from the relations 2 = id and 3 = id. That is, thereis a homomorphism : G P SL (2, Z ) sending , to the two matrices given; all weneed to check is that the kernel of this homomorphism consists of the identity element.

A general element of G = , | 2 , 3 is a product a 1 b1 a 2 b2 . . . a n bn whereeach of the a i , bi are integers. We reduce the a i mod 2 and the bi mod 3; after rewriting of this kind, we are left with a product of the form above where every a i = 1 and every bi is 1or 2. We write L = and R = 2 , so that every nontrivial element of G is of the formw, 1w,w, 1w where w is a word in the letters L and R . Furthermore, we have therelation ( 2)3

We show that no word w in the letters L and R is trivial in the group P SL (2, Z ). Asimilar argument works for elements of G of the other forms.

Now, (L) = 1 01 1 and (R) =1 10 1 in P SL (2, Z ). Suppose that

w = Lm 1 Rn 1 Lm 2 Rn 2 . . . L m k Rn k

where all the m i , n i are nonzero, say. Then we can calculate

(w) = p rq s

where pq

=1

m1+1

n1+1

m2+1

n 2+. . .

1mk

rs =

1m1+

1n1+

1m2+

1n2+ . . .

1mk +

1nk

where the notation is for a continued fraction expansion. That is, thealternating coefcientsm i , n i give the continued fraction expansions of pq and

rs . In particular, (w) = Id unless

w is the empty word, and is an isomorphism.

Notice actually that this method of proof does considerably more. We have shown thatevery element obtained by a product of positive multiples of L and R is nontrivial. It isnot true that the group generated by L and R is free, since LR 1 has order 3. In fact, Land R together generate the entire group P SL (2, Z ). But the group generated by L2 andR2 is free, since a fundamental domain for its action is the domain D bounded by the lines

Re(z) = 1 , Re (z) = 1and the semicircles

|z 1/ 2| = 1 / 2, |z + 1 / 2| = 1 / 2Let denote this subgroup of P SL (2, Z ). The domain D is a regular ideal hyperbolicquadrilateral; the quotient H 2 / is therefore topologically a punctured torus. By the argu-ment of the previous section, a presentation is

1(punctured torus )= , | That is, = Z

Z .


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30 DANNY CALEGARI

Another description of is the following: there is an obvious homomorphism :P SL (2, Z ) P SL (2, Z / 2Z ) given by reducing the entries mod 2. The image grouphas order 6 and the surjection is onto, so the kernel is a subgroup of index 6. Since thefundamental domain D can be made from 6 copies of D , it follows that the index of inP SL (2, Z ) is 6. Moreover, is certainly contained in the kernel of . It follows that isexactly equal to this kernel.

is sometimes also denoted by (2) (for reduction mod 2) and is of considerableinterest to number theorists, who like to refer to it as the principal congruence subgroup of level 2.

Notice that the domain D is obtained from two ideal triangles. The union of all thetranslates of D by (2) gives a tessellation of H 2 by regular ideal quadrilaterals; a sub-division of D into two ideal triangles gives a subdivision of H 2 into ideal triangles. If we choose the subdivision along the line Re (z) = 0 , the ideal triangulation T of H 2 soobtained admits reection symmetry along every edge. The 1skeleton of the dual celldecomposition to this ideal triangulation is the innite 3 valent tree . There is a natural ac-

tion of P SL (2, Z ) on this tree, where the elements of order 3 are the stabilizers of verticesand the elements of order 2 are the stabilizers of edges. This description of P SL (2, Z )as a group of automorphisms of a tree gives another way to see that it is isomorphic toZ / 2Z

Z / 3Z .For a rational point p/q we can consider the straight line l perpendicular to the real

axis given by Re (z) = p/q . As this line l moves from to p/q it crosses through manydifferent triangles of T , and therefore determines a word w in the letters R, L and theirinverses. By induction, it is easy to show that the word w is of the form

w = Lm 1 Rn 1 Lm 2 Rn 2 . . . L m k Rn k

where pq

=1

m1+1

n1+1

m2+1

n 2+. . .

1mk

An irrational point r determines an innite wordw = Lm 1 Rn 1 Lm 2 Rn 2 . . .

where

r =1

m1+1

n1+1

m2+1

n2+. . .

is an innite continued fraction expansion of r .Notice that this word w is eventually periodic exactly when r is of the form a + b for

rational numbers a, b.

3.3. Finite subgroups of SO (3) and S3 .

3.3.1. The fair dice. A die is a convex 3dimensional polyhedron. We can ask underwhat conditions a die is fair that is, the probability that the die will land on a given

side is 1/n where n is the number

classical geometry - danny calegari

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