lecture notes on basic differential topologylerario/antonio_lerario/... · 2018. 10. 23. ·...

LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY

ANTONIO LERARIO

These are notes for the course “Advanced Geometry 2” for the Master Diploma in Mathemat-ics at the University of Trieste and at SISSA. These notes are by no mean complete: excellentreferences for the subject are the books [1, 3, 4, 5, 6], from which in fact many proofs are takenor adapted. The notes contain some exercises, which the reader is warmly encouraged to solve(sometimes part of a proof is left as an exercise). I apologize in advance for the many mistakesand imprecisions that the reader might find: I will greatly appreciate if he/she could point outany of them.

References

[1] R. Bott and L. W. Tu. Di↵erential forms in algebraic topology, volume 82 of Graduate Texts in Mathematics.Springer-Verlag, New York-Berlin, 1982.

[2] M. Cornalba. Note di geometria di↵ereziale. http://mate.unipv.it/cornalba/dispense/geodiff.pdf.[3] V. Guillemin and A. Pollack. Di↵erential topology. AMS Chelsea Publishing, Providence, RI, 2010. Reprint

of the 1974 original.[4] M. W. Hirsch. Di↵erential topology, volume 33 of Graduate Texts in Mathematics. Springer-Verlag, New York,

1994. Corrected reprint of the 1976 original.[5] J. M. Lee. Introduction to smooth manifolds, volume 218 of Graduate Texts in Mathematics. Springer, New

York, second edition, 2013.[6] J. W. Milnor. Topology from the di↵erentiable viewpoint. Princeton Landmarks in Mathematics. Princeton

University Press, Princeton, NJ, 1997. Based on notes by David W. Weaver, Revised reprint of the 1965original.

Date: October 16, 2018.

1

2 ANTONIO LERARIO

Schedule and plan of the course

Lessons on Wednesdays and Thursdays11am–1pm in room A134, located in SISSA main buildingThere is no lesson on Thursday, November 1st (national hoilday)

Lesson 01, 03/10 Di↵erentiable manifolds and smooth maps: definition and examples (spheres, projectivespaces and Grassmanians)

Lesson 02, 04/10 The tangent bundle and the tangent map; smooth maps and their di↵erentialLesson 03, 10/10 Submanifolds and how to produce them. Immersions, embeddings and regular value

theorem (notion of transversality).Lesson 04, 11/10 More examples.Lesson 05, 17/10 How abstract is the notion of manifold? Part 1: compact manifolds embed in Rn

Lesson 06, 18/10 Sets of measure zero and mini-SardLesson 07, 24/10 How abstract is the notion of manifold? Part 2: Weak Whitney embedding theoremLesson 08, 25/10 Transversality and Sard’s lemma; every closed set is the zero set of a smooth functionLesson 09, 31/10 Proof of Sard’s LemmaLesson 10, 07/11 Parametric transversality and applicationsLesson 11, 08/11 Parametric transversality and applications, continuationLesson 12, 14/11 The normal bundle of an immersionLesson 13, 15/11 Approximations of continuous functions and applications (e.g. homotopy groups of

spheres)Lesson 14, 21/11 Vector bundles: definitionLesson 15, 22/11 Examples of vector bundlesLesson 16, 28/11 Vector bundles and sections of vector bundles on projective spacesLesson 17, 28/11 Associated bundles, classification theoremLesson 18, 05/12 Di↵erential formsLesson 19, 06/12 The Mayer-Vietoris sequence and first applicationsLesson 20, 12/12 Orientation and integrationLesson 21, 13/12 Stokes theorem and Poincare LemmaLesson 22, 18/12 Examples: cohomology of spheres and projective spaces

note: the time, date and location of this lesson is flexibleLesson 23, 19/12 Kunneth formula and Poincare dualLesson 24, 20/12 Final applications (degree, invariant cohomology...)

LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 3

1. Differentiable manifolds

Definition 1 (Ck manifold). A manifold of dimension m and class Ck is a paracompact1

Hausdor↵ space M such that:

(1) for every point x 2 M there exists a neighborhood U of x and a continuous function : U ! Rm which is a homeomorphism onto an open subset of Rm (the pair (U, ) iscalled a chart);

(2) for every pairs of charts (U1, 1) and (U2, 2) such that U1 \ U2 6= ; the map

2 � 1|�1

U1\U2: 1(U1 \ U2) ! Rm

is a Ck map (for k = 0 we obtain topological manifolds, for k � 1 di↵erentiable manifolds,for k = 1 smooth manifolds and for k = ! analytic manifolds).

A collection of charts {(Uj , j)}j2J as above such thatS

j2J Uj = M is called a Ck atlas.

Remark 2 (Atlases and di↵erential structures). Two Ck atlases A = {(U↵, ↵)}↵2A and B ={(V� ,'�)}�2B for M , are said to be equivalent if their union A [ B is still a Ck atlas. A Ck-di↵erential structure on M is the choice of an equivalence class of Ck atlases. If we take theunion of all atlases belonging to a Ck-di↵erential structure, we obtain a maximal Ck atlas. Thisatlas contains every chart that is compatible with the chosen di↵erentiable structure. (There is anatural one-to-one correspondence between di↵erentiable structures and maximal di↵erentiableatlases.)

From now on we will assume that the atlas we work with is maximal, so that we will have allpossible charts available.

A simple way to enrich a given atlas is as follows. Given a chart : U ! Rm around a pointx 2 M (as in point (1) of Definition 1), and given a neighborhood V ⇢ U of x we can easilyconstruct a chart ' : V ! Rm by simply taking ' = |V . Note that in this way we can constructa chart (V,') around any point with V contractible: it is enough to take V = �1(BRm( (x), ✏))for ✏ > 0 small enough.

Example 3. If ' : M ! Rn be a homeomorphism, then M is an analytic manifold. In fact onecan cover M with the single chart (M,'), and ' � '�1 = idRm : Rm ! Rm is analytic.

Example 4 (Product manifolds). If M and N are smooth manifolds with respective atlases{(U↵, ↵)}↵2A and {(V� ,'�)}�2B , then M ⇥ N is naturally a smooth manifold with the atlas{(U↵ ⇥ V� , ↵ ⇥ '�)}(↵,�)2A⇥B .

Example 5 (Spheres). The unit sphere Sn = {x20

+ · · · + x2n = 1} ⇢ Rn+1 can be endowed

with the structure of a smooth manifold as follows. Consider the point e0 = (1, 0, . . . , 0) 2 Sn

and the two open sets of the the sphere defined by U1 = Sn\{e0} and U2 = Sn\{�e0}. Weproduce explicit and nice homeomorphisms (i.e. charts) 1 : U1 ! Rn and 2 : U2 ! Rn, calledstereographic projections (see Figure 1), as follows:

1(x0, . . . , xn) =1

1 � x0

(x1, . . . , xn) and 2(x0, . . . , xn) =1

1 + x0

(x1, . . . , xn).

1Recall that a paracompact space is a topological space X for which every open cover has a locally finiterefinement. More precisely: given an open cover U = {U↵}↵2A fo X, there exists another open cover V ={V�}�2B such that (i) for every � 2 B there exists ↵(�) 2 A such that V� ⇢ U↵(�) (i.e. V refines U); (ii) forevery x 2 X there exists a neighborhood Vx of x which intersects only finitely many elements of V (i.e. V islocally finite). It is worth noticing that if a Hausdor↵ space is locally Euclidean (i.e. if it satisfies condition (1) ofDefinition 1) and connected, then this space is paracompact if and only if it is second countable (i.e. its topologyhas a countable basis).

4 ANTONIO LERARIO

e0

p1

1(p1)

p2

1(p2)

Sn

Rn

Figure 1. The stereographic projection 1 : Sn\{e0} ! Rn.

It is easy to verify that �1

1: Rn ! Sn is given by:

�1

1(y1, . . . , yn) =

1

1 + kyk2

�kyk2 � 1, 2y1, . . . , 2yn

�.

This in particular allows to write the explicit expression for 2 � 1|�1

U1\U2: Rn\{0} ! Rn\{0}:

2 � �1

1(y) =

y

kyk2,

which is a smooth map. Hence {(U1, 1), (U2, 2)} is a smooth atlas for Sn and turns it into asmooth manifold. This is called the standard di↵erential structure on Sn.

Example 6 (Real projective spaces). The real projective space RPn can be endowed the struc-ture of smooth manifold as follows. Recall that RPn = (Rn+1\{0})/ ⇠, where p1 ⇠ p2 if andonly if there exists � 6= 0 such that p1 = �p2. We denote by [x0, . . . , xn] the equivalence classof (x0, . . . , xn) 2 Rn+1\{0} (the xj are called homogeneous coordinates). For every j = 0, . . . , nconsider the open set Uj ⇢ RPn defined by:

Uj = {[x0, . . . , xn] such that xj 6= 0},

together with the homeomorphism j : Uj ! Rn given by:

j([x0, . . . , xn]) =

✓x0

xj, . . . ,

cxj

xj, . . . ,

xn

xj

◆

(here the “hat” symbol denotes that this element has been removed from the list). The inverse �1

j : Rn ! RPn is given by:

�1

j (y0, . . . , byj , . . . , yn) = [y0, . . . , 1, . . . yn],

where the “1” is in position j. As a consequence, for every i 6= j we have:

i � �1

j (y0, . . . , byj , . . . , yn) =

✓y0

yi, . . . ,

cyi

yi, . . . ,

1

yi, . . .

yn

yj

◆,

which is a di↵eomorphism of Rn\{0} to itself.

Exercise 7. Prove that RP1 and S1 are homeomorphic.

Example 8 (real Grassmannians). The real Grassmannian G(k, n) consists of the set of allk-dimensional vector subspaces of Rn, endowed with the quotient topology of the map:

q : {M 2 Rn⇥k such that rk(M) = k} ! G(k, n), q(M) = span{columns of M}.


In other words, G(k, n) (as a topological space) can be considered as the quotient of the set ofn ⇥ k real matrices of rank k (viewed as a subset of Rn⇥k) under the equivalence relation:

M1 ⇠ M2 () there exists L 2 GL(k, R) such that M1 = M2L.

Observe that G(1, n) = RPn�1 and that the above definition mimics the equivalence relationv1 ⇠ v2 if and only if there exists � 2 GL(1, R) = R\{0} such that v1 = �v2.

We want to endow G(k, n) with the structure of a smooth manifold. For every multi-indexJ = (j1, . . . , jk) 2

�nk

we denote by M |J the k ⇥ k submatrix of M obtained by selecting the

rows j1, . . . , jk (in this way M |Jc denotes the (n � k) ⇥ k submatrix of M obtained by selectingthe complementary rows). For every such multi-index J we define the open set:

UJ = {[M ] 2 G(k, n) such that det(M |J) 6= 0}.

(Note that this set is well defined.) Mimicking again the definition for projective spaces, wedefine the manifold charts J : UJ ! R(n�k)⇥k by:

J([M ]) = (MM�1

J )|Jc .

The expression of the inverse of a matrix in terms of its determinant and its cofactor sshows thatfor every pair of indices J1, J2 2

�nk

the map J2 � �1

J1is smooth. In this way {(UJ , J)}J2{n

k}is a smooth atlas for the k(n � k)-dimensional manifold G(k, n).

Exercise 9. Fill in all the details in the previous definition of the smooth structure on theGrassmannian.

Example 10 (The Complex projective line). Recall that the complex line CP1 is defined as thequotient space (C2\{0})/ ⇠ where (z0, z1) ⇠ �(z0, z1) for every � 2 C\{0}. As we did for realprojective spaces, we denote by [z0, z1] the homogeneous coordinates of a point on CP1. Considerthe two open sets U0 = {z0 6= 0} and U1 = {z1 6= 0} together with the charts j : Uj ! C ' R2

for j = 0, 1 which are given by:

1([z0, z1]) =z0

z1

and 0([z0, z1]) =z1

z0

.

We have that 0 � �1

1(z) = 1

z , which is a holomorphic map C\{0} ! C\{0} and consequently,using the identification C ' R2, a smooth map R2\{0} ! R2\{0}. If we wanted, we could alsowork with j as a real map, as follows (however, as the reader will see, using the field structureof C ' R2 simplifies a lot the computations). Given (x0, y0, x1, y1) 2 R4, let us denote by(x0 + iy0, x1 + iy1) = (z0, y0) 2 C2. We can write 1 : U0 ! C ' R2 as

1([x0 + iy0, x1 + iy1]) =x0 + iy0

x1 + iy1

=x0x1 + y0y1

x21

+ y21

+ i · y0x1 � x0y1

x21

+ y21

which means that the real map 1 : U1 ! R2 is given by:

1([x0 + iy0, x1 + iy1]) =

✓x0x1 + y0y1

x21

+ y21

,y0x1 � x0y1

x21

+ y21

◆,

with inverse �1

1: R2 ! CP1 given by:

�1

1(x, y) = [1, x + iy].

In particular 0 � 1|�1

U0\U1is given by (x, y) 7! 1

x2+y2 (x, �y), which is indeed a smooth map

R2\{0} ! R2\{0}.The complex projective line CP1 is homeomorphic to S2 and in fact, as smooth manifolds, theyare indistinguishable (see Exercise 27).

6 ANTONIO LERARIO

x = z + ay x = z + by

z

a b

✏H

Az

✏

Figure 2. An open set of the type U3 (grey region) for the topology of thePrufer surface.

Exercise 11. Generalize the previous example and prove that CPn can be endowed with thestructure of a smooth 2n-dimensional manifold.

Example 12 (A non-paracompact “manifold”: the Prufer surface). Let H = {(x, y), y > 0} bethe positive half-space and for every z 2 R consider the set

Az = {(x, y, z) 2 R3 with y 0}(each Az is a closed half space). The Prufer surface is the set

P = H [ [

z2RAz

!

endowed with the topology generated by open sets of the three following types:

(1) open sets U1 ⇢ H for the standard euclidean topology;(2) open sets U2 ⇢ Az \ {y < 0} (so U2 is a euclidean open set in {(x, y, z), y < 0} ' H);(3) open sets of the form U3 = Az(a, b, ✏) \ Hz(a, b, ✏) where (see Figure 2):

Az(a, b, ✏) = {(x, y, z) 2 R3, a < x < b, �✏ < y 0}Hz(a, b, ✏) = {(x, y) 2 H, 0 < y < ✏, z + ay < x < z + by}.

The resulting topological space is Hausdor↵ and locally euclidean. To see this, for every z 2 Rlet Uz = Az [ H (an open set) and define the map fz : R2 ! Uz by:

fz(x, y) =

⇢(x, y, z) y 0

(z + xy, y) y > 0


11

12

X2 = R [ {11} [ {12}

Figure 3. The two-points compactification of the real line: a compact, non-Hausdor↵, C1 manifold.

The map fz is continuous with continuous inverse (check it!) which we denote by z : Uz ! R2.The “manifold” structure on P is then given by the atlas {(Uz, z)}z2R. Notice now that the setZ = {(0, 0, z), z 2 R} is an uncountable subset of P which inherits the discrete topology, henceP cannot be second countable.

Example 13 (A non-Hausdor↵ “manifold”: the lines with two origins). The circle S1 can beseen as the one point compactification of the real line. We might as well define the n-pointscompactification Xn of R as follows. As a point set Xn = R [ {11, . . . , 1n} endowed with thetopology generated by all the open sets of the form:

U = {1j} [ {A ⇢ R with R\A compact} for some j = 1, . . . , n.

For n � 2 the space Xn can be endowed with a structure satisfying the axioms for the definitionof smooth manifold, except for the Hausdor↵ condition (see Figure 3). The space X2\{0} issometimes called “the line with two origins”.

Exercise 14. Fill in all the details in the construction of the “manifold” structure for the abovenon-examples.

1.1. The tangent bundle.

Definition 15 (Tangent bundle). Let M be a smooth manifold of dimension m and {(U↵, ↵)}↵2A

be a smooth atlas for M . We define the tangent bundle of M as:

TM =

a

↵2A

U↵ ⇥ Rm

!/ ⇠,

where (x1, v1)↵1 ⇠ (x2, v2)↵2 if and only if J ↵2 (x)( ↵1 � �1↵2

)v2 = v1 (in other words, theJacobian of the coordinates change sends v2 to v1). We endow TM with the quotient topologyfrom the defining equivalence relation “⇠”. The tangent bundle is endowed with the structureof a smooth manifold as follows (see Definition 1).

First, for every ↵ 2 A we call TM |U↵ the set {[(x, v)↵]} ⇢ TM and observe that (by con-struction) TM |U↵ = TU↵ is an open subset of TM which is homeomorphic to U↵ ⇥ Rm (under

8 ANTONIO LERARIO

the identification map); we call '↵ : TM |U↵ ! U↵ ⇥ Rm the inverse of this homeomorphism.The manifold charts for TM are constructed as follows. We define the homeomorphism

�↵ : TM |U↵ ! ↵(U↵) ⇥ Rm ✓ Rm ⇥ Rm

to be the composition �↵ = ( ↵ ⇥ idRm) � '↵, i.e. the map given by [(x, v)↵] 7! ( ↵(x), v). Thefamily {(U↵ ⇥ Rm,�↵)}↵2A gives a family of charts for TM (called natural charts).

In order to prove that the above definition turns TM into a smooth manifold, we need to checkwhat is the regularity of the change of coordinates maps. For every pair of charts (U1 ⇥ Rm,�1)and (U2 ⇥ Rm,�2) such that (U1 ⇥ Rm) \ (U2 ⇥ Rm) 6= ;, we consider therefore the map:

(1.1) �2 � ��1

1: 1(U1 \ U2) ⇥ Rm ! 2(U1 \ U2) ⇥ Rm.

We see that this map is given by (y, v) 7! ( 2 � �1

1(y), Jy( 2 � �1

1)v) and it is smooth if M

itself was smooth. Notice indeed that if M is a Ck manifold, the tangent bundle TM is only aCk�1 manifold, since in (1.1) we are computing one more derivative (in particular the tangentbundle is not defined if M is only C0).

The map p : TM ! M defined by [(x, v)↵] 7! x is called the projection map and the fiberover a point is denoted by TxM = p�1(x) and called the tangent space to M at x. An importantproperty of the tangent space is that it carries the structure of an m-dimensional vector space.In fact, given a chart (U1, 1) containing x, calling by (TU1,�1) the corresponding natural chartfor TM , the map �1|TxM : TxM ! Rm is a bijection and can be used to induce the vector spacestructure on TxM ; this structure is independent of the chart since, if (U2, 2) is another chartcontaining x then �1 � (�2|TxM )�1 : Rm ! Rm is a linear automorphism.

Example 16. Since we can cover an open set U ⇢ Rm with the single chart = idU : U ! Rm,the tangent bundle to U is simply:

TU = U ⇥ Rm

with the projection map (x, v) 7! x.

Exercise 17. Only using the definition of tangent bundle, try to visualize TS1 and to provethat it is homeomorphic to S1 ⇥ R.

1.2. Smooth maps between manifolds and their di↵erential.

Definition 18 (Smooth map). A continuous map f : M ! N between two smooth manifolds issaid smooth if for every choice of charts (U, ) for M and (V,') for N such that f(U) ⇢ V (thesecharts will be called adapted) the function ' � f � �1 : (U) ! Rn is smooth. If f is smooth,for every x 2 M we pick a chart (U, ) = (U↵, ↵) containing x and a chart (V,') = (V� ,'�)containing f(x) (up to shrinking U we can assume these carts are adapted), and we define thedi↵erential of f to be the map df : TM ! TN given by:

(1.2) df : [(x, v)↵] 7! [(f(x), J (x)(' � f � �1)v)� ]

(applying the chain rule shows that the definition is independent of the charts). The restrictionof df to TxM is denoted by:

dxf = df |TxM : TxM ! Tf(x)N.

It is a linear map, called the di↵erential of f at x.

Exercise 19. Prove that the projection map p : TM ! M for the tangent bundle of a smoothmanifold is a smooth map.

Exercise 20. Prove that the map f : S1 ! RP1 given by (x0, x1) 7! [x0, x1] is smooth.


We immediately observe a couple of useful properties, which readily follows from the definition:

(1) if f : M ! N and g : N ! P are smooth maps between manifolds, then

dx(g � f) = df(x)g � dxf

(“the di↵erential of a composition is the composition of the di↵erentials”);(2) if f : Rm ! Rn is smooth, then dxf : TxM ' Rm ! Tf(x)N ' Rn is just the classical

di↵erential of a smooth map, i.t. the unique linear map such that:

f(x + v) � f(x) = dxfv + O(kvk2).

Remark 21 (Another definition of tangent space). For practical purposes it is often convenientto have an alternative equivalent definition of tangent space and of (1.2). First, we can defineTxM as the set of equivalence classes of smooth curves � : (�✏, ✏) ! M such that �(0) = x, withthe equivalence relation �1 ⇠ �2 if and only if for every chart (U, ) containing x we have:

d

dt( (�1(t)))

��t=0

=d

dt( (�2(t)))

��t=0

.

If (U, ) = (U↵, ↵) is a chart containing x, to every equivalence class [�] we associate theelement [(�(0), d

dt ( (�(t)))|t=0)↵]; viceversa, given [(x, v)↵] we can construct the curve �(t) = �1↵ ( ↵(x)+ tv), which is well defined on the interval (�✏, ✏) for ✏ > 0 small enough. Since these

two process are inverse to each other, this shows that the two definitions are equivalent.This alternative definition is particularly convenient for “computing” the di↵erential of a mapf : M ! N . In fact, given v 2 TxM we can write v = [�] for some � : (�✏, ✏) ! M with�(0) = x and dxfv = [f � �].

Exercise 22. There is also a third equivalent definition of tangent space, viewing tangent vectorsas derivations. The reader is warmly encouraged to check out also this definition (for exampleas it is done in [5, Chapter 3]), and to prove that it is equivalent to the one we have given here.

Remark 23 (The di↵erential in coordinates). Let f : M ! N be a smooth map. Given a pointx 2 M and adapted charts : U ! Rm and ' : V ! Rn (with U neighborhhod of x and Vnegighborhood of f(x)) there is a natural way to construct bases for TxM and Tf(x)N whichallow to write the matrix associated to the linear operator dxf in these bases. This is done asfollows. First, let

TxRm = Rm = span{e1, . . . , em}and define for every i = 1, . . . , m the vectors2

@

@xi= (d (x)

�1)ej 2 TxM.

Similarly, setting y = f(x), for j = 1, . . . , m we define:

@

@yj= (d'(y)'

�1)ei 2 TyN.

Denoting by A = (aij) the matrix representing dxf in the coordinates given by the bases{ @@xi

}i=1,...,m for TxM and { @@yj

}j=1,...,n for TyN , we have by definition:

dxf@

@xi=

nX

j=1

aij@

@yj.

2These are just symbols, but they have a useful and suggestive interpretation!

10 ANTONIO LERARIO

R

X

Figure 4. The set X = graph(x 7! |x|) ⇢ R2 is an analytic manifold, becauseit can be covered with a single chart : (x, |x|) 7! x, but is is not a smoothsubmanifold of R2.

Applying df(y)' to both sides of the previous equation, and using the definitions, we have:

nX

j=1

aijej =nX

j=1

aijdf(y)'@

@yj= df(y)'

0

@nX

j=1

aij@

@yj

1

A

= df(y)'

✓dxf

@

@xi

◆= df(y)'

�dxf(d (x)

�1)ej

�

= d (x)(' � f � �1)ei.

In the last line we have the usual di↵erential of a map '�f � �1 : Rm ! Rn, whose representingmatrix in coordinates is the Jacobian matrix J (x)(' � f � �1); the above chain of equalitiesproves that this Jacobiam matrix coincides with the matrix A.

Remark 24. Observe the following nice identity (which justifies the choice of the notation inthe previous Remark 23) for a smooth map f : Rm ! R:

dxf@

@xi=

@f

@xi(x).

Definition 25 (Immersion, embedding and di↵eomorphism). Let ' : M ! N be a smoothmap. We will say that ' is an immersion if for every x 2 M the di↵erential dx' is injective.We will say that ' is an embedding if it is an immersions and moreover it is a homeomorphismonto its image. We will say that a homeomorphism ' : M ! N is a di↵eomorphism if it is anembedding (note that in particular the inverse '�1 : N ! M is also a di↵eomorphism).

Exercise 26. Prove that there is no immersion of S1 ⇥ S1 into R2. On the other hand there isan immersion of S1 ⇥ S1\{one point} ! R2.

Exercise 27. Prove that S2 and CP1 are di↵eomorphic.

1.3. Submanifolds and how to produce them.

Definition 28 (Submanifold). Let M be a smooth manifold of dimension m. A subset X ⇢ M iscalled a submanifold of dimension k if every x 2 X belongs to the domain of a chart3 (U, ) for Msuch that U \ X = �1(Rk), where Rk ✓ Rm is a linear subspace. (Observe that a submanifoldis itself a manifold with charts of the form (U \ X, |U\X).) The di↵erence dim(M) � dim(X)is called the codimension of X in M .

Example 29. For ↵ � 0 let �↵ ⇢ R ⇥ R be the graph of the function x 7! |x|↵. For an integerk, if k < ↵ < k + 1 then �↵ is a submanifold of R2 which is Ck but not Ck+1.

3Recall that we assumed that we have a maximal atlas at our disposal, see Remark 2.


Before proceeding we observe two important features of the notion of submanifold.

(1) It has a local character: X ✓ M is a submanifold if and only if there is an open cover{Xi}i2I of X and there are open sets {Mi}i2I in M such that for every i 2 I the setXi ⇢ Mi is a submanifold of Mi.

(2) It is invariant under di↵eomorphism: if ' : M ! N is a di↵eomorphim, then X ✓ M isa submanifold if and only if '(X) ⇢ N is a submanifold.

Exercise 30. Work out the details showing that the notion of submanifold has local characterand is invariant under di↵eomorphisms.

Theorem 31. A subset X ✓ M is a submanifold if and only if it is the image of an embedding.

Proof. If X ✓ M is a submanifold, then the inclusion ◆ : X ! M is an embedding: ◆ is ahomeomorphism onto its image ◆(X) = X and the di↵erential dx◆ : TxX ! TxM is simply theinclusion (hence it is injective).

Suppose now that f : X ! M is an embedding. We need to prove that f(X) is a submanifoldof M . For every x 2 X let (U1, ) be a chart for M on a neighborhood of f(x) and (V1,') achart for X on a neighborhood of X. Since f is a homeomorphism onto its image, there existsan open set W1 ⇢ M such that f(V1) = f(X) \ W1. We set:

Uf(x) = U1 \ W1 and Vx = f�1(Uf(x)).

Observe that f(X) \ Uf(x) = f(Vx) and consider the diagram of maps:

Rn

Vx Uf(x)

Rm

'

f

Because the definition of submanifold has a local character, it is enough to prove that f(Vx) ⇢U(fx) is a submanifold for every x 2 X. Moreover, by the invariance under di↵eomorphism,f(Vx) ⇢ U(fx) is a submanifold if and only if (f(Vx)) ⇢ Rm is a submanifold. Note now that

� f � '�1 : '(Vx) ! Rm

is an embedding of '(Vx) ⇢ Rn into an open subset of Rm and � f � '�1('(Vx)) = (f(Vx)).Thus we are reduced to prove the statement in the special case X = '(Vx) ⇢ Rn is an opensubset and h = � f � '�1 : X ! Rm is an embedding.

We pick now a point p 2 h(X) and we realize (locally around p) the image h(X) as the graphof a function. To this end we will assume that p is the origin (up to translations, which willnot change the property of being a submanifold) and we pick x0 2 X such that p = h(x0). Weconsider the splitting:

Rm = L1 � L2 where L1 = im(dx0h)

(L2 is any complementary space, for example L2 = L?1

). Since h is an embedding, L1 ' Rm.We also consider the projections on the two factors p1 : Rm ! L1 and p2 : Rm ! L2 and set:

h1 = p1 � h and h2 = p2 � h.

In this way, in the coordinates given by the splitting Rm = L1�L2 we have h = (h1, h2). Observe

12 ANTONIO LERARIO

4 ANTONIO LERARIO

f

Figure 3. An injective immersion f : I ! R2 of an open interval which is notan embedding (the image of f is a closed subset of R2).

Observe that f(X) \ Uf(x) = f(Vx) and consider the diagram of maps:

Rn

Vx Uf(x)

Rm

'

f

Because the definition of submanifold has a local character, it is enough to prove that f(Vx) ⇢U(fx) is a submanifold for every x 2 X . Moreover, by the invariance under di↵eomorphism,f(Vx) ⇢ U(fx) is a submanifold if and only if (f(Vx)) ⇢ Rm is a submanifold. Note now that

� f � '�1 : '(Vx) ! Rm

is an embedding of '(Vx) ⇢ Rn into an open subset of Rm and � f � '�1('(Vx)) = (f(Vx)).Thus we are reduced to prove the statement in the special case X = '(Vx) ⇢ Rn is an opensubset and h = � f � '�1 : X ! Rm is an embedding.

We pick now a point p 2 h(X) and we realize (locally around p) the image h(X) as the graphof a function. To this end we will assume that p is the origin (up to translations, which willnot change the property of being a submanifold) and we pick x0 2 X such that p = h(x0). Weconsider the splitting:

Rm = L1 � L2 where L1 = im(dx0h)

(L2 is any complementary space, for example L2 = L?1). Since h is an embedding, L1 ' Rm.

We also consider the projections on the two factors p1 : Rm ! L1 and p2 : Rm ! L2 and set:

h1 = p1 � h and h2 = p2 � h.

In this way, in the coordinates given by the splitting Rm = L1�L2 we have h = (h1, h2). Observethat, dx0h1 : Tx0X ! Rn is invertible and by the Inverse Function Theorem there exists an openneighborhhod A ⇢ L1 and a smooth function ↵ : A ! X such that

h1(↵(a)) = a for all a 2 A.

We define now the function ' : A ! L2 by:

'(a) = h2(↵(a)).

Figure 5. An injective immersion f : I ! R2 of an open interval which is notan embedding (the image of f is a closed subset of R2).

that dx0h1 : Tx0X ! Rn is invertible, and by the Inverse Function Theorem, there exist an openneighborhood A ⇢ L1 and a smooth function ↵ : A ! X such that:

h1(↵(a)) = a for all a 2 A.

We define now the function ' : A ! L2 by:

'(a) = h2(↵(a)).

Since h is an embedding (in particular a homeomorphism onto its image), by possibly furthershrinking the neighborhood A, there exists a neighborhood B ⇢ L2 of zero such that:

h(X) \ (A ⇥ B) = h(↵(A)).

In particular in the neighborhood A ⇥ B of p 2 h(X) we have:

h(X) \ (A ⇥ B) = {(a,'(a)) | a 2 A}.

Consider now the map : A ⇥ B ! Rm given by:

(a, b) 7! (a, b � '(a)).

It is easy to see that is a di↵eomorphism on a neighborhhod U of p (since its Jacobian isnonvanishing at this point); moreover, by construction:

h(X) \ U = {(a, (a)) | a 2 A} \ U = {(a, b) 2 U | b = '(a)} = �1({0} ⇥ Rn�m) \ U.

Let us summarize what we have proved: for every p 2 h(X) there is a neighborhood U of p anda map : U ! Rm which is a di↵eomorphism (a chart) such that h(X) \ U = �1(Rn�m): thisis exactly the requirement for h(X) to be a submanifold of Rn. ⇤Exercise 32. Work out the details of the last step of the previous proof.

Exercise 33. Let G(2, 4) be the Grasmmanian of 2-planes in R4 and consider the map p :G(2, 4) ! RP5 defined by:

p([M ]) = [det(M |j1,j2), . . . , det(M |j3,j4)]

(the subscripts range over all possible pairs (i, j) 2�

4

2

, there are 6 such pairs.) Check that the

map p (called Plucker embedding) is an embedding. Denoting by [p12, p13, p14, p23, p24, p34] thehomogeneous coordinates of RP5, prove that the image of p is the quadric4 {p12p34 + p13p24 +

4This quadric has signature (3, 3) and is double-covered by S2 ⇥ S2; since �(S2 ⇥ S2) = 4, then the Eulercharacteristic of G(2, 4) is 2.


p14p23 = 0}.

More generally the Grassmannian G(k, n) embeds into RP(nk)�1 via an analogous Plucker em-bedding (though the description of the image of this embedding is more complicated). Can youfigure out the general construction?

Definition 34 (Regular value). Let f : M ! N be a smooth map. We will say that y 2 N is aregular value for f if for every x 2 f�1(y) the di↵erential dxf (a linear map from TxM to TyN)is surjective.

Remark 35. Observe that a regular value does not need to be a value: if y /2 im(f) then thereis no x in f�1(y) and the condition in the above definition is automatically satisfied.

Theorem 36. Let y 2 N be a regular value of a smooth map f : M ! N . Then f�1(y) is asmooth submanifold of M of dimension dim(M) � dim(N) (if nonempty).

Proof. By using local character and invariance under di↵eomorphisms of the definition of sub-manifold, we can reduce (as in the poof of Theorem 31) to the case M ⇢ Rm is an open subsetand N = Rn. In this case the conclusion follows from the Implicit Function Theorem. ⇤

Whenever a manifold M is described as the preimage of a regular value M = f�1(y), we willsay that the equation f = y is regular.

Exercise 37. Work out the details of the last step in the previous proof.

Remark 38. We observe that the two ways we have to produce submanifolds (i.e. usingembeddings or giving their equations) are essentially the two ways we have to exhibit vectorsubspaces of a vector space (i.e. as the span os some vectors, or as the set of solutions of asystem of independent linear equations).

Example 39 (Smooth projective hypersurfaces). Let F : Rn+1 ! R be a homogeneous polyno-mial of degree d. Since F is homogeneous, the following set Z(F ) ⇢ RPn is well defined:

Z(F ) = {[x0, . . . , xn] 2 RPn such that F (x0, . . . , xn) = 0}.

If the vector rF =⇣@F@x0

, . . . , @F@xn

⌘(the gradient of F ) is nonzero at every non-zero point of {F =

0} ⇢ Rn+1 (a non-degeneracy condition), then Z(F ) is a smooth submanifold of RPn. To see thiswe use the fact that being a submanifold is a local property, i.e. in order to prove that Z(F ) is asubmanifold, it is enough to cover RPn with the open sets RPn =

Snj= Uj defined in Example 6

and prove that Z(F ) \ Uj is a submanifold of RPn for every j = 0, . . . , n. Moreover, since beinga submanifold is invariant under di↵eomorphisms, it is enough to prove that j(Z(F ) \ Uj) is asubmanifold of Rn for every j = 0, . . . , n (the j : Uj ! Rn are the manifold charts for RPn).Now, if we define the function fj : Rn ! R by fj(y0, . . . , byj , . . . , yn) = F (y0, . . . , 1, . . . , yn), theset j(Z(F ) \ Uj) is given by the equation:

j(Z(F ) \ Uj) = {fj = 0}.

14 ANTONIO LERARIO

The condition rF |{F=0}\{0} 6= 0 tells that the equation {fj = 0} is regular (i.e. 0 is a regularvalue of fj). In fact if for some y 2 {fj = 0} we had rfj(y) = 0, then:

rF (y0, . . . , 1, . . . , yn) =

✓@F

@x0

, . . . ,@F

@xj, . . . ,

@F

@xn

◆ ��(y0,...,1,...,yn)

=

0, . . . , 0,

@F

@xj

��(y0,...,1,...,yn)

, 0, . . . , 0

!

= (0, . . . , 0, d · F (y0, . . . , 1, . . . , yn), 0, . . . , 0)

= 0

contradicting the non-degeneracy condition (in the last line we have used Euler’s identity forhomogeneous functions d · F (x) =

Pnj=0

xj@F@xj

(x).) We will review this example in Example 94

below.

Exercise 40. Prove that if y is a regular value of a smooth map f : S1 ! R, then f�1(y) haseven cardinality.

Exercise 41. Prove that there is no smooth function f : RP2 ! R such that RP1 = f�1(y),where y 2 R is a regular value.

Exercise 42. Let M(r; n, m) ⇢ Rn⇥m be the set of matrices of rank r. Prove that M(r; n, m)is a smooth submanifold of codimension (n � r)(m � r).

1.4. Every compact manifold embeds into some Euclidean space.

Theorem 43. Let M be a compact manifold. There exists an embedding ' : M ! Rn for some(possibly very large) n � dim(M).

Proof. We first claim that if M is compact there exists a finite atlas {(Uk, k)}`k=1such that

k(Uk) ⇢ B(0, 2) ⇢ Rm (here m = dim(M)) and

(1.3) M =[

k=1

int� �1

k (D(0, 1))�.

Let now � : Rm ! [0, 1] a bump function constructed as in Lemma 68 with the choice c1 = 1and c2 = 2, and for every k = 1, . . . , ` define the function �k : M ! [0, 1] by

�k(x) = �Uk(x) · �( k(x))

(we are “pulling back” the bump function � to a bump function on M using the chart k).Observe that (1.3) ensures that the sets {Dk = ��1

k (1)}`k=1cover M .

Define now for k = 1, . . . , ` the functions fk : M ! Rm by

fk(x) = �Uk(x) · �k(x) k(x)

(we are using the �k to extend the charts k : Uk ! Rm to the whole M).

We finally define the map ' : M ! R`(m+1) by:

'(x) = (f1(x),�1(x), . . . , f`(x),�`(x)).

We need to check that ' is a homeomorphism onto its image and that for every x 2 M thedi↵erential dx' : TxM ! T'(x)R`(m+1) ' R`(m+1) is an injective linear map. Observe first that' is injective: if x 6= y and y 2 Dk, then either (1) x 2 Dk, in which case k(x) 6= k(y), or (2)x /2 Dk and 1 = �k(y) 6= �k(x). Thus ' is injective and takes value in a Hausdor↵ space, M is


compact: hence ' is a homeomorphism onto its image. It remains to check that dx' is injectivefor every x 2 M . Note that

dx' = (dxf1, dx�1, . . . , dxf`, dx�`),

hence to prove that dx' is injective it is enough to prove that one of its components is injective.Since every x belongs to some Dk, then dxfk = dx k is injective because k is a di↵eomorphism.This concludes the proof. ⇤Exercise 44. Prove that the product of spheres M =

Qmi=1

Sni embeds into R1+Pm

i=1 ni .

1.5. The tangent bundle for embedded submanifolds. In the case M ✓ Rn the tangentbundle can be defined in an equivalent way as follows (observe that if M is compact by Theorem43 we can always assume it is embedded in some Rn):

(1.4) TM = {(x, v) 2 Rn ⇥ Rn | v 2 TxM}.

Exercise 45. Assuming that M ⇢ Rn is an embedded submanifold, using the definition (1.4),check that TM is a smooth submanifold of Rn ⇥Rn. In other words, check that the “equations”v 2 TxM defining TM ⇢ Rn ⇥ Rn are regular.

This definition is particularly useful because it allows to talk about the “length” of a tangentvector without introducing the notion of Riemannian manifold.

Proposition 46. The set T 1M = {(x, v) 2 TM | kvk = 1} (called the unit tangent bundle) is asmooth submanifold of TM of dimension 2 dim(M) � 1.

Proof. Consider the smooth function ⇢ : TM ! R defined by:

⇢(x, v) = kvk2.

(This function is smooth because it is given by the composition of two smooth functions.) ThenT 1M = ⇢�1(1) and to see that T 1M ⇢ TM is a submanifold it is enough to prove that 1 is aregular value for ⇢ and then use Theorem 36. To this end, let (x, v) 2 TM such that kvk2 = 1and consider the curve � : (✏, ✏) ! TM given by �(t) = (x, (1 + t)v). Then �(0) = (x, v) and

d

dt�(t)

��t=0

=d

dtktvk2

��t=0

=d

dt(1 + t)2

��t=0

= 2.

Hence im(d(x,v)⇢) � span{d(x,v)⇢�

ddt�(t)

��t=0

�} = R and d(x,v)⇢ is surjective. ⇤

Remark 47. Observe that if M ⇢ Rn is compact, then T 1M is compact, because it is closedand bounded: M is compact, hence contained in a bounded set K ⇢ Rn and

TM ⇢ K ⇥ B(0, 1)

which is bounded in Rn ⇥ Rn.

Example 48 (The group SO(3)). The group SO(3) is defined by:

SO(3) =�A 2 R3⇥3 such that AAT = and det(A) = 1

.

This is an example of a Lie group, i.e. a group G which is also a smooth manifold and suchthat the group multiplication µ : G ⇥ G ! G, given by µ(g1, g2) = g1g2, and the inverse map⌘ : G ! G, given by ⌘(g) = g�1, are both smooth. The smooth structure on SO(3) is given asfollows. We consider the map F : R3⇥3 ! Sym(3, R) ' R6 given by:

F (A) = AAT ,

16 ANTONIO LERARIO

vx

TxS2

T 1xS2

S2

Figure 6. The (embedded) unit tangent bundle to S2. The vectors v and xare orthogonal and of norm one; the map (x, v) 7! (x, v, x ^ v) gives a homeo-morphism T 1S2 ! SO(3).

and we observe that is a regular value of F (check it!), then we can use Theorem 36 anddeduce that O(3) = F�1( ) is a smooth submanifold of R9. since the determinant functiontakes constant values on each components of O(3), we see that SO(3) is a union of connectedcomponents of O(3), hence it is itself a smooth manifold. One important remark: O(3) consistsactually of just two components:

O(3) = SO(3) [ (� · SO(3)) .

The group SO(3) is homeomorphic to unit tangent bundle of S2 ⇢ R3. In fact we can define acontinuous, surjective and one-to-one map f : T 1S2 ! SO(3) as follows (see Figure 6):

(x, v) 7! (x, v, x ^ v).

Since T 1S2 is compact, then f is a homeomorphism onto its image.

Exercise 49. Prove that SO(3) is also homeomorphic to RP3. (This implies that T 1S2 ' RP3.)

Exercise 50. Using Exercise 49 and Example 48, prove that every vector field on the sphereS2 (i.e. a continuous assignment v : S2 ! TS2 with v(x) 2 TxS2, see Section 4.1) must vanishat some point.

Example 51 (The group SU(2)). The group SU(2) is defined by:

SU(2) =

⇢A =

✓z1 z2

z3 z4

◆such that AA

T= and det(A) = 1

�.

We can endow SU(2) the structure of smooth manifold using again Theorem 36. As a smoothmanifold SU(2) is di↵eomorphic to the sphere S3 and this can be seen as follows. Observe that

the condition AAT

= can be written as:✓z4 �z2

�z3 z1

◆=

1

det(A)

✓z4 �z2

�z3 z1

◆= A�1 = A

T=

✓z1 z2

z3 z4

◆,


which implies that A is of the form:

A =

✓z1 z2

�z2 z1

◆.

In particular the group SU(2) can be described as:

SU(2) =

⇢A =

✓z1 z2

�z2 z1

◆such that |z1|2 + |z2|2 = det(A) = 1

�,

which is a sphere S3 ⇢ C2 = {(z1, z2)}.The (embedded) tangent space to SU(2) at the identity is usually denoted by su(2) and itconsists of the vector space of matrices:

T SU(2) = {X 2 C2⇥2 such that X = XT

and tr(X) = 0}.

Writing a matrix X 2 su(2) as

X =

✓iu �v � iz

�v + iz �iu

◆, (u, v, z) 2 R3,

we see that the determinant det(X) = u2 + v2 + z2 is a positive quadratic form on su(2)and it induces on it a Euclidean structure. This allows to define a homomorphism of groups' : SU(2) ! O(3) (i.e. a group representation) via:

'(A) = (RA : X 7! AXAT).

The image of ' is SO(3) and, recalling that SO(3) ' RP3 (Exercise 49), we therfeore have:

' : SU(2) ' S3 ! RP3 ' SO(3).

The map ' is a group homomorphism with kernel {± } and topologically it is the double covermap.

Exercise 52. Try to prove the properties of ' : SU(2) ! SO(3) rigorously (im(') = SO(3),ker(') = {± }, ' : S3 ! RP3 is the covering space map), possibly referring to [?, Chapter 8].

1.6. Sets of measure zero. In this section we will define the notion of “sets of measure zero”,which introduces new ideas in the topic. We remark from the beginning that we will not definethe measure of a set, we will just define a special class of sets in a manifold: sentences like “theset A has measure zero” or “the measure of A is zero” mean that A belongs to this special classof sets (those “of measure zero”).

Let D = [a1, b1] ⇥ · · · ⇥ [am, bm] be a cube in Rm. We will set µ(D) =Qm

i=1|bi � ai|.

Definition 53 (Sets of measure zero). We will say that A ⇢ Rm has measure zero if for every✏ > 0 there exists a countable family of cubes {Dk}k2J (countable means that the cardinality ofthe index set J is either finite or countable) such that

A ⇢[

k2J

Dk andX

k2J

µ(Dk) ✏.

If M is a smooth manifold, we will say that A ⇢ M has measure zero if for every chart (U, )for M the set (A \ U) ⇢ Rm has measure zero.

Remark 54. Observe that Q ⇢ R has measure zero. In fact, let Q = {rk}k2N and for every✏ > 0 consider the cubes Dk = [rk � ✏

2k+1 , rk + ✏2k+1 ]. Then:

Q ⇢[

k2NDk and

X

k2Nµ(Dk) =

X

k2N

✏

2k+1= ✏.

18 ANTONIO LERARIO

Remark 55. If A ⇢ M has measure zero, then it cannot contain any open set: in fact if V ⇢ Ais open, then for some chart (U, ) we have that (U \ V ) ⇢ Rm is open and nonempty. Inparticular (U \ V ) it contains a cube D with µ(D) = � > 0. If we now try to cover (U \ V )with a countable collection of cubes {Dk}k2J , this collection must cover D and

Pk2J µ(Dk) � �.

In particular if A ⇢ M has measure zero, its complement is dense: let V ⇢ M be any open set,then V \ Ac 6= ;, otherwise V would be contained in A.

On the other hand the reader should keep in mind that if B ⇢ M is dense, the complementdoesn’t necessarily have measure zero. For example let Q \ [0, 2⇡] = {qk}k2N and set

B = {(cos ✓, sin ✓) 2 S1 | ✓ 2 Q \ [0, 2⇡]}.

Then B is dense in S1, but its complement does not have measure zero. (Can you prove thisrigorously?)

Exercise 56. For every ✏ > 0 construct an open and dense set A ⇢ Rm of Lebesgue measuresmaller than ✏ (this exercise requires measure theory).

Lemma 57. Let M be a smooth manifold and A ⇢ M be a subset such that A ⇢S

k2J Ak withJ countable and with each Ak ⇢ M of measure zero; then A itself has measure zero.

Proof. Since J is countable, we may assume J = {1, 2, . . .} ⇢ N. For simplicity let us just discussthe case A ⇢ Rm, leaving the manifold case to the reader.

Now, for every ✏ > 0 and for every k 2 J cover Ak with cubes {Dk,j}j2Jk withP

j2Jkµ(Dk,j)

✏2k+1 (such a cover exists because each Ak has measure zero). Then

A ⇢[

k2J

[

j2Jk

Dk,j andX

k2J

X

j2Jk

µ(Dk,j) X

k2J

✏

2k+1 ✏.

⇤Lemma 58. Let U ⇢ Rm be an open set and A ⇢ U be of measure zero. If f : U ! Rn is a C1

function, then f(A) has measure zero.

Proof. Observe first that, since Rm is a second countable space, then U =S

k2J int(B(xk, rk))where J is a countable set. Then U can also be written as:

U =[

k2J

[

n2NB

✓xk, rk � 1

n

◆,

i.e. U can be written as a countable union of closed (hence compact) balls. Relabeling, we canwrite:

U =[

k2NBk

where each Bk is a closed ball. Let us set Ak = A \ Bk; observe that Ak has measure zero (asubset of a set of measure zero has itself measure zero). We will prove that f(Ak) has measurezero for every k 2 N, and the result will follow from Lemma 57, since f(A) =

Sk2N f(Ak).

Thus let us fix k. Observe that, since Bk is compact and f is C1, then f |Bk is Lispchitz andthere exists Lk > 0 such that for all x, y 2 Bk:

kf(x) � f(y)k Lkkx � yk.

In particular, if C ⇢ Bk has diameter d, then f(C) has diameter at most d · Lk; hence thereexists Ck > 0 (which also depends on m, but this is also fixed) such that if D ⇢ Bk is a cube,then f(D) is contained in a cube D0 with µ(D0) Ckµ(D).


Fix now ✏ > 0 and take a countable cover of Ak with balls {Dj}j2I such thatP

j2I µ(Dj) ✏Ck

(this cover exists because we are assuming Ak has measure zero). By the above reasoning eachf(Dj) is contained in a D0

j with µ(D0j) Ckµ(Dj). In particular:

f(Ak) ⇢[

j2I

D0j and

X

j2I

µ(D0j)

X

j2I

Ckµ(Dj) Ck · ✏

Ck ✏,

which proves f(Ak) has measure zero. ⇤Remark 59. Observe that the previous Lemma implies that for a set A ⇢ M to have measurezero it is enough to check (A \ U) ⇢ Rm has measure zero for all charts (U, ) belonging to agiven atlas (i.e. we do not have to check it on all possible charts).

Lemma 60 (mini-Sard’s lemma). Let A and B be smooth manifold with dim(A) < dim(B) andg : A ! B be a C1 map. Then g(A) has measure zero in B and B\g(A) is dense.

Proof. Let {(Uk, k)}k2I be a countable atlas for A and {(Vj ,'j)}j2J be a countable atlas forB. It is enough tot check that for every k 2 I and j 2 J the measure of 'j(f(Uk) \ Vj) ⇢ Rb iszero. Let Ra be the codomain of k and Wk be its image. Consider the projection on the secondfactor map p : Rb�a ⇥ Wk ! Wk. Then 'j(f(Uk) \ Vj) is the image of {0} ⇥ Wk ⇢ Rb�a ⇥ Wk

under the C1 map:

� : Rb�a ⇥ Wkp�! Wk

�1

�! Ukf�! Vj

j�! Rb

(this is a map between open sets in Rb.) Since {0} ⇥ Wk has measure zero in Rb�a ⇥ Wk, thenthe result follows from Lemma 58. ⇤Example 61. For every n there is a continuous surjective map f : S1 ! Sn, which of coursewhen n > 1 cannot be C1 (this map is constructed using space-filling curves).

1.7. Whitney embedding theorem: weak form.

Theorem 62 (Weak Whitney embedding). Let M be a compact manifold of dimension m.There exists an embedding ' : M ! R2m+1.

Proof. By Theorem 43 we can assume M is already embedded into some Rq. We will show thatif q > 2m + 1 then there exists a projection Rq ! Rq�1 which restricts to an embedding ofM ! Rq�1. The result will follow iterating the argument until q = 2m + 1.

To make things more precise, let us put coordinates (x1, . . . , xq) on Rq and let us identifyRq�1 = {xq = 0}. For every unit vector v 2 Sq�1\{xq = 0} let us consider the linear mapfv : Rq ! Rq�1 that projects in the direction of v. We need to find a vector v 2 Sq�1\{xq = 0}such that fv|M is an embedding. To this end we need to verify:

(1) fv|M is injective;(2) for every x 2 M the di↵erential dx(fv|M ) is injective.

The first condition is satisfied if there is no pair (x, y) 2 M ⇥ M with x 6= y such that:

v =x � y

kx � yk .

We can rephrase this condition as follows. Let � be the diagonal in M ⇥ M ; then fv|M isinjective if and only if v does not belong to the image of the map h1 : (M ⇥M)\� ! Sq�1 givenby:

h1(x, y) =x � y

kx � yk .

20 ANTONIO LERARIO

We apply now Lemma 60 to the smooth function h1. Since

dim(M ⇥ M)\� = 2m < q � 1 = dim(Sq�1)

then im(h1) has measure zero in Sq�1 and there is a dense set D1 = Sq�1\im(h1) ⇢ Sq�1 of“good” choices for v (as long as we are concerned with injectivity of fv|M ).

The condition on the injectivity of the di↵erential is handled as follows. Observe first that,since fv is linear, the di↵erential of fv at a point x 2 Rq equals fv itself:

dxfv = fv.

Consequently the kernel of the linear map dxfv : TxRn ' Rn ! Rn equals span{v} (the linespanned by the direction of v, since by definition fv is the projection in the direction of v).Moreover, the di↵erential of fv|M at a point x 2 M equals:

dx(fv|M ) = (dxfv)|TxM

and ker dx(fv|M ) = TxM \ span{v}. Thus the di↵erential of fv|M at a point x 2 M is injectiveif and only if v /2 TxM . We can rephrase this condition as follows. Consider the smooth maph2 : T 1M ! Sq�1 given by:

h2(x, v) = v.

Then fv|M is an immersion if and only if v does not belong to the image of h2. We can use nowLemma 60 again, since:

dim(T 1M) = 2m � 1 < q � 1 = dim(Sq�1).

As a consequence there is a dense subset D2 = Sq�1\im(h2) of “good” choices for v (as long aswe are concerned with injectivity of dfv|M ).

Observe that since T 1M is compact (see Remark 47), then im(h2) is compact and D2 is notonly dense, but also open. In particular the intersection D1 \ D2 is nonempty (in fact dense inSq�1) and every v 2 D1 \D2 produces a map fv which restricts to an embedding of M ! Rq�1.

⇤Exercise 63. Prove that if M is a compact smooth manifold of dimension m, then there existsan immersion of M in R2m.

Exercise 64. Let M be a compact manifold of dimension m and h : M ! Rq be a smooth map,where q � 2m+1. Prove that for every ✏ > 0 there exists an embedding h✏ : M ! Rq such that:

maxx2M

kh(x) � h✏(x)k ✏.


�c2 �c1 c1 c2

Figure 7. A bump function

2. Appendix: basic tools

2.1. The inverse and the implicit function theorem. These are basic tools in Di↵erentialTopology, but we will not prove them here. There are many excellent references for thesetheorems, for instance a complete discussion can be found online at [2].

Theorem 65 (Inverse function theorem). Let U ✓ Rn be an open set and f : U ! Rn be adi↵erentiable function of class Ck, k � 1. If the Jacobian matrix Jf(x) is nonsingular at x0

there exists a neighborhood A of x0 such that f |A : A ! Rn is invertible; moreover f(A) is openand the inverse function g : f(A) ! A is also Ck.

In the above statement, note that the conclusion “f(A) is open” follows from the Invarianceof Domain Theorem; moreover, using the language of Definition 25 below, the function f |A is adi↵eomorphism.

Theorem 66 (Implicit function theorem). Let U ✓ Ra+b be an open set with coordinates (x, y) 2Ra ⇥ Rb and f : U ! Rb be a di↵erentiable function of class Ck, k � 1. If the matrix @f

@y (x, y)

(consisting of the partial derivatives of f with respect to the y variables) is nonsingular at (x0, y0),there exist neighborhoods A of x0 and B of y0 and a function ' : A ! B of class Ck such that:

{f = f(x0, y0)} \ (A ⇥ B) = graph(').

2.2. Bump functions.

Definition 67 (Bumb function). Let 0 < c1 < c2 be positive distinct numbers. A bumb functionis a C1 function � : Rn ! [0, 1] such that (see Figure 7):

(1) �(x) = 1 for kxk c1;(2) 0 < �(x) < 1 for c1 < kxk < c2;(3) �(x) = 0 for kxk � c2.

Lemma 68. Bumb functions exist for every 0 < c1 < c2.

Proof. Let ↵ : R ! R be the C1 function

↵(x) = e� 1x�{x>0}(x),

where �{x>0}(x) denotes the characteristic function of {x > 0}. Out of this function we construct� : R ! R by:

�(x) =

R c2

x ↵(t � c1)↵(c2 � t)dtR c2

c1↵(t � c1)↵(c2 � t)dt

.

Finally we set �(x) = �(kxk). (The reader can see [4, Section 2.2] for pictures explaining theconstruction step by step.) ⇤

22 ANTONIO LERARIO

2.3. Partitions of unity.

Definition 69 (Partition of unity). Let M by a smooth manifold and U = {Uj}j2J be an opencover of M . A partition of unity subordinated to U is a collection of smooth maps {⇢j : M ![0, 1]}j 2 J such that:

(1) for every j 2 J the support supp(⇢j) ⇢ Uj ;(2) the family {supp(⇢j)}j2J is locally finite;(3)

Pj2J ⇢j ⌘ 1 (the sum is well defined because of the previous condition).

In order to prove the existence of partitions of unity, we will need the following Lemma.

Lemma 70. Let U = {Uj}j2J be an open cover for M . Then there is a locally finite atlas{(V↵, ↵)}↵2A such that {V ↵}↵2A refines U , each ↵(V↵) ⇢ Rm is bounded and each V ↵ ⇢ Mis compact. Moreover the open cover V = {V↵}↵2A has a shrinking W = {W↵}↵2A, i.e. W isitself a cover and for every ↵ 2 A we have W↵ ⇢ V↵.

Proof. The proof is left as an exercise. Work out the details carefully! ⇤

Proposition 71. Let M be a smooth manifold. Every open cover U of M admits a subordinatepartition of unity.

Proof. Let U = {Uj}j2J be the open cover. We observe first that if we can construct a partitionof unity {�i}i2I subordinated to a refinement V = {Vi}i2I of U , then we can also construct apartition of unity subordinated to U . In fact let g : I ! J be such that for every i 2 I we havethe inclusion Vi ⇢ Ug(i) (the existence of such g is in the definition of refinement) and define forj 2 J the function:

⇢j(x) =X

i2g�1(j)

�i(x).

It is easy to check that {⇢j}j2J is a partition of unity subordinated to U (essentially to obtainthe ⇢j we collate together some of the �i).

Let now V be the open cover produced by Lemma 70 and W its shrinking. For every ↵ 2 Awe cover the compact set ↵(W↵) ⇢ Rm by finitely many closed balls contained in ↵(V↵):

↵(W↵) ⇢k↵[

k=1

B↵,k,

where B↵,k = B(x↵,k, r↵,k) for some points x↵,k 2 Rm and radii r↵,k > 0. For every ball B↵,k

let �↵,k : Rm ! [0, 1] be a bump function centered at x↵,k such that �↵,k(x) > 0 if and onlyif x 2 int(B↵,k) (it su�ces to take a bumb function �, as in Lemma 68, with c1 < r↵,k and

c2 = r↵,x and then translate it at x↵,k, i.e. �↵,k(x) = �(x � x↵,k)). Put now �↵ =Pk↵

k=1�↵,k

and define µ↵ : M ! [0, 1) by:

µ↵(x) = �V↵(x) · �↵( ↵(x)).

Each µ↵ is a smooth function, strictly positive on W↵ and with supp(µ↵) ⇢ V↵. We define thedesired partition of unity {⇢↵}↵2A subordinated to V by:

⇢↵(x) =µ↵(x)P�2A µ�(x)

.

⇤


Exercise 72 (Smooth Urysohn’s Lemma). Let A and B be two closed and disjoint submanifoldsof a manifold M . Prove that there is a smooth function f : M ! [0, 1] such that f |A ⌘ 0 andf |B ⌘ 1.

lecture notes on basic differential topologylerario/antonio_lerario/... · 2018. 10. 23. ·...

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