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: excellent references for the subject are the books [1, 3, 4, 5, 6], from which in fact many proofs are taken or 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 mistakes and imprecisions that the reader might find: I will greatly appreciate if he/she could point out any of them.
References [1] R. Bott and L. W. Tu. Di↵erential forms in algebraic topology,volume82ofGraduate 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.AMSChelseaPublishing,Providence,RI,2010.Reprint of the 1974 original. [4] M. W. Hirsch. Di↵erential topology,volume33ofGraduate Texts in Mathematics.Springer-Verlag,NewYork, 1994. Corrected reprint of the 1976 original. [5] J. M. Lee. Introduction to smooth manifolds,volume218ofGraduate Texts in Mathematics.Springer,New York, second edition, 2013. [6] J. W. Milnor. Topology from the di↵erentiable viewpoint.PrincetonLandmarksinMathematics.Princeton University Press, Princeton, NJ, 1997. Based on notes by David W. Weaver, Revised reprint of the 1965 original.
Date:October16,2018. 1 2 ANTONIO LERARIO
Schedule and plan of the course
Lessons on Wednesdays and Thursdays 11am–1pm in room A134, located in SISSA main building There is no lesson on Thursday, November 1st (national hoilday)
Lesson 01, 03/10 Di↵erentiable manifolds and smooth maps: definition and examples (spheres, projective spaces and Grassmanians) Lesson 02, 04/10 The tangent bundle and the tangent map; smooth maps and their di↵erential Lesson 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-Sard Lesson 07, 24/10 How abstract is the notion of manifold? Part 2: Weak Whitney embedding theorem Lesson 08, 25/10 Transversality and Sard’s lemma; every closed set is the zero set of a smooth function Lesson 09, 31/10 Proof of Sard’s Lemma Lesson 10, 07/11 Parametric transversality and applications Lesson 11, 08/11 Parametric transversality and applications, continuation Lesson 12, 14/11 The normal bundle of an immersion Lesson 13, 15/11 Approximations of continuous functions and applications (e.g. homotopy groups of spheres) Lesson 14, 21/11 Vector bundles: definition Lesson 15, 22/11 Examples of vector bundles Lesson 16, 28/11 Vector bundles and sections of vector bundles on projective spaces Lesson 17, 28/11 Associated bundles, classification theorem Lesson 18, 05/12 Di↵erential forms Lesson 19, 06/12 The Mayer-Vietoris sequence and first applications Lesson 20, 12/12 Orientation and integration Lesson 21, 13/12 Stokes theorem and Poincar´eLemma Lesson 22, 18/12 Examples: cohomology of spheres and projective spaces note: the time, date and location of this lesson is flexible Lesson 23, 19/12 Kunneth formula and Poincar´edual Lesson 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 M there exists a neighborhood U of x and a continuous function 2 : U Rm which is a homeomorphism onto an open subset of Rm (the pair (U, )is called! a chart); (2) for every pairs of charts (U , ) and (U , ) such that U U = the map 1 1 2 2 1 \ 2 6 ; 1 m 2 1 U U : 1(U1 U2) R | 1\ 2 \ ! is a Ck map (for k = 0 we obtain topological manifolds, for k 1 di↵erentiable manifolds, for k = smooth manifolds and for k = ! analytic manifolds). 1 k A collection of charts (Uj, j) j J as above such that j J Uj = M is called a C atlas. { } 2 2 k Remark 2 (Atlases and di↵erential structures). Two CS atlases = (U↵, ↵) ↵ A and = A { k} 2 B k (V ,' ) B for M, are said to be equivalent if their union is still a C atlas. A C - { } 2 A[B di↵erential structure on M is the choice of an equivalence class of Ck atlases. If we take the union of all atlases belonging to a Ck-di↵erential structure, we obtain a maximal Ck atlas. This atlas contains every chart that is compatible with the chosen di↵erentiable structure. (There is a natural one-to-one correspondence between di↵erentiable structures and maximal di↵erentiable atlases.) From now on we will assume that the atlas we work with is maximal, so that we will have all possible charts available. A simple way to enrich a given atlas is as follows. Given a chart : U Rm around a point x M (as in point (1) of Definition 1), and given a neighborhood V !U of x we can easily 2 m ⇢ construct a chart ' : V R by simply taking ' = V . Note that in this way we can construct ! | 1 a chart (V,') around any point with V contractible: it is enough to take V = (BRm ( (x),✏)) for ✏>0 small enough. Example 3. If ' : M Rn be a homeomorphism, then M is an analytic manifold. In fact one ! 1 m m can cover M with the single chart (M,'), and ' ' =id m : R R is analytic. R ! Example 4 (Product manifolds). If M and N are smooth manifolds with respective atlases (U↵, ↵) ↵ A and (V ,' ) B,thenM N is naturally a smooth manifold with the atlas { } 2 { } 2 ⇥ (U↵ V , ↵ ' ) (↵, ) A B. { ⇥ ⇥ } 2 ⇥ n 2 2 n+1 Example 5 (Spheres). The unit sphere S = x0 + + xn =1 R can be endowed { ··· }⇢ n with the structure of a smooth manifold as follows. Consider the point e0 =(1, 0,...,0) S n n 2 and the two open sets of the the sphere defined by U1 = S e0 and U2 = S e0 .We \{ } n \{ n } produce explicit and nice homeomorphisms (i.e. charts) 1 : U1 R and 2 : U2 R , called stereographic projections (see Figure 1), as follows: ! ! 1 1 (x ,...,x )= (x ,...,x ) and (x ,...,x )= (x ,...,x ). 1 0 n 1 x 1 n 2 0 n 1+x 1 n 0 0
1Recall that a paracompact space is a topological space X for which every open cover has a locally finite refinement. More precisely: given an open cover = U↵ ↵ A fo X,thereexistsanotheropencover = U { } 2 V V B such that (i) for every B there exists ↵( ) A such that V U↵( ) (i.e. refines ); (ii) for { } 2 2 2 ⇢ V U every x X there exists a neighborhood Vx of x which intersects only finitely many elements of (i.e. is 2 V V locally finite). It is worth noticing that if a Hausdor↵space is locally Euclidean (i.e. if it satisfies condition (1) of Definition 1)andconnected,thenthisspaceisparacompactifandonlyifitissecondcountable(i.e.itstopology has a countable basis). 4 ANTONIO LERARIO
e0
Sn
p1
Rn
1(p2) 1(p1)
p2
n n Figure 1. The stereographic projection 1 : S e0 R . \{ }!
1 n n It is easy to verify that : R S is given by: 1 ! 1 1 2 (y ,...,y )= y 1, 2y ,...,2y . 1 1 n 1+ y 2 k k 1 n k k 1 n n This in particular allows to write the explicit expression for 2 1 U U : R 0 R 0 : | 1\ 2 \{ }! \{ } 1 y (y)= , 2 1 y 2 k k n which is a smooth map. Hence (U1, 1), (U2, 2) is a smooth atlas for S and turns it into a smooth 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- n n+1 ture of smooth manifold as follows. Recall that RP =(R 0 )/ ,wherep1 p2 if and \{ } ⇠ ⇠ only if there exists = 0 such that p1 = p2. We denote by [x0,...,xn] the equivalence class n+16 of (x0,...,xn) R 0 (the xj are called homogeneous coordinates). For every j =0,...,n 2 \{ } n consider the open set Uj RP defined by: ⇢ U = [x ,...,x ] such that x =0 , j { 0 n j 6 } n together with the homeomorphism j : Uj R given by: ! x x x ([x ,...,x ]) = 0 ,..., j ,..., n j 0 n x x x ✓ j j j ◆ (here the “hat” symbol denotes that this element has beenc removed from the list). The inverse 1 n n : R RP is given by: j ! 1 j (y0,...,yj,...,yn)=[y0,...,1,...yn], where the “1” is in position j. As a consequence, for every i = j we have: b 6 1 y0 yi 1 yn (y ,...,y ,...,y )= ,..., ,..., ,... , i j 0 j n y y y y ✓ i i i j ◆ which is a di↵eomorphism of Rn 0 to itself. c \{ b} Exercise 7. Prove that RP1 and S1 are homeomorphic. Example 8 (real Grassmannians). The real Grassmannian G(k, n) consists of the set of all k-dimensional vector subspaces of Rn, endowed with the quotient topology of the map: n k q : M R ⇥ such that rk(M)=k G(k, n),q(M) = span columns of M . { 2 }! { } LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 5
In other words, G(k, n) (as a topological space) can be considered as the quotient of the set of n k n k real matrices of rank k (viewed as a subset of R ⇥ ) under the equivalence relation: ⇥ M1 M2 there exists L GL(k, R) such that M1 = M2L. ⇠ () 2 n 1 Observe that G(1,n)=RP and that the above definition mimics the equivalence relation v1 v2 if and only if there exists GL(1, R)=R 0 such that v1 = v2. ⇠ 2 \{ } We want to endow G(k, n) with the structure of a smooth manifold. For every multi-index J =(j ,...,j ) n we denote by M the k k submatrix of M obtained by selecting the 1 k 2 k |J ⇥ rows j ,...,j (in this way M c denotes the (n k) k submatrix of M obtained by selecting 1 k J the complementary rows). For| every such multi-index ⇥J we define the open set: U = [M] G(k, n) such that det(M ) =0 . J { 2 |J 6 } (Note that this set is well defined.) Mimicking again the definition for projective spaces, we (n k) k define the manifold charts J : UJ R ⇥ by: ! 1 ([M]) = (MM ) c . J J |J The expression of the inverse of a matrix in terms of its determinant and its cofactor sshows that n 1 for every pair of indices J ,J the map is smooth. In this way (U , ) n 1 2 k J2 J1 J J J 2 { } 2 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 the Grassmannian.
Example 10 (The Complex projective line). Recall that the complex line CP1 is defined as the 2 quotient space (C 0 )/ where (z0,z1) (z0,z1) for every C 0 . As we did for real \{ } ⇠ ⇠ 2 \{ } 1 projective spaces, we denote by [z0,z1] the homogeneous coordinates of a point on CP . Consider 2 the two open sets U0 = z0 =0 and U1 = z1 =0 together with the charts j : Uj C R for j =0, 1 which are given{ 6 by:} { 6 } ! ' z0 z1 1([z0,z1]) = and 0([z0,z1]) = . z1 z0 1 1 We have that 0 (z)= , which is a holomorphic map C 0 C 0 and consequently, 1 z \{ }! \{ } using the identification C R2, a smooth map R2 0 R2 0 . If we wanted, we could also ' \{ }! \{ } work with j as a real map, as follows (however, as the reader will see, using the field structure 2 4 of C R simplifies a lot the computations). Given (x0,y0,x1,y1) R , let us denote by ' 2 2 2 (x0 + iy0,x1 + iy1)=(z0,y0) C . We can write 1 : U0 C R as 2 ! ' x0 + iy0 x0x1 + y0y1 y0x1 x0y1 1([x0 + iy0,x1 + iy1]) = = 2 2 + i 2 2 x1 + iy1 x1 + y1 · x1 + y1 2 which means that the real map 1 : U1 R is given by: ! x x + y y y x x y ([x + iy ,x + iy ]) = 0 1 0 1 , 0 1 0 1 , 1 0 0 1 1 x2 + y2 x2 + y2 ✓ 1 1 1 1 ◆ 1 2 1 with inverse : R CP given by: 1 ! 1 1 (x, y)=[1,x+ iy]. 1 1 In particular is given by (x, y) (x, y), which is indeed a smooth map 0 1 U0 U1 x2+y2 | \ 7! R2 0 R2 0 . \{ }! \{ } The complex projective line CP1 is homeomorphic to S2 and in fact, as smooth manifolds, they are indistinguishable (see Exercise 27). 6 ANTONIO LERARIO
x = z + ay x = z + by
✏ H z a b
✏
Az
Figure 2. An open set of the type U3 (grey region) for the topology of the Pr¨ufer surface.
Exercise 11. Generalize the previous example and prove that CPn can be endowed with the structure of a smooth 2n-dimensional manifold. Example 12 (A non-paracompact “manifold”: the Pr¨ufer surface). Let H = (x, y),y >0 be { } the positive half-space and for every z R consider the set 2 3 Az = (x, y, z) R with y 0 { 2 } (each Az is a closed half space). The Pr¨ufer surface is the set
P = H A [ z z ! [2R endowed with the topology generated by open sets of the three following types: (1) open sets U H for the standard euclidean topology; 1 ⇢ (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\{U = }A (a, b, ✏) H (a, b, ✏) where (see Figure{ 2): }' 3 z \ z 3 Az(a, b, ✏)= (x, y, z) R ,a
11
12
X2 = R 1 2 [{1 }[{1 } Figure 3. The two-points compactification of the real line: a compact, non- Hausdor↵, C1 manifold.
2 The map fz is continuous with continuous inverse (check it!) which we denote by z : Uz R . ! The “manifold” structure on P is then given by the atlas (Uz, z) z . Notice now that the set { } 2R Z = (0, 0,z),z R is an uncountable subset of P which inherits the discrete topology, hence P cannot{ be second2 } countable. Example 13 (A non-Hausdor↵“manifold”: the lines with two origins). The circle S1 can be seen as the one point compactification of the real line. We might as well define the n-points compactification Xn of R as follows. As a point set Xn = R 1,..., n endowed with the topology generated by all the open sets of the form: [{1 1 }
U = j A R with R A compact for some j =1,...,n. {1 }[{ ⇢ \ } For n 2 the space X can be endowed with a structure satisfying the axioms for the definition n of smooth manifold, except for the Hausdor↵condition (see Figure 3). The space X2 0 is sometimes called “the line with two origins”. \{ } Exercise 14. Fill in all the details in the construction of the “manifold” structure for the above non-examples.
1.1. The tangent bundle.
Definition 15 (Tangent bundle). Let M be a smooth manifold of dimension m and (U↵, ↵) ↵ A be a smooth atlas for M.Wedefinethetangent bundle of M as: { } 2
m TM = U↵ R / , ⇥ ⇠ ↵ A ! a2 1 where (x1,v1)↵ (x2,v2)↵ if and only if J (x)( ↵ )v2 = v1 (in other words, the 1 ⇠ 2 ↵2 1 ↵2 Jacobian of the coordinates change sends v2 to v1). We endow TM with the quotient topology from the defining equivalence relation “ ”. The tangent bundle is endowed with the structure of a smooth manifold as follows (see Definition⇠ 1).
First, for every ↵ A we call TM U↵ the set [(x, v)↵] TM and observe that (by con- 2 | { }⇢ m struction) TM U↵ = TU↵ is an open subset of TM which is homeomorphic to U↵ R (under | ⇥ 8 ANTONIO LERARIO
m the identification map); we call '↵ : TM U↵ U↵ R the inverse of this homeomorphism. The manifold charts for TM are constructed| as! follows.⇥ We define the homeomorphism m m m ↵ : TM U↵ ↵(U↵) R R R | ! ⇥ ✓ ⇥ to be the composition ↵ =( ↵ idRm ) '↵, i.e. the map given by [(x, v)↵] ( ↵(x),v). The m ⇥ 7! family (U↵ R , ↵) ↵ A gives a family of charts for TM (called natural charts). { ⇥ } 2 In order to prove that the above definition turns TM into a smooth manifold, we need to check m what is the regularity of the change of coordinates maps. For every pair of charts (U1 R , 1) m m m ⇥ and (U2 R , 2) such that (U1 R ) (U2 R ) = , we consider therefore the map: ⇥ ⇥ \ ⇥ 6 ; 1 m m (1.1) 2 : 1(U1 U2) R 2(U1 U2) R . 1 \ ⇥ ! \ ⇥ 1 1 We see that this map is given by (y, v) ( (y),J ( )v) and it is smooth if M 7! 2 1 y 2 1 itself was smooth. Notice indeed that if M is a Ck manifold, the tangent bundle TM is only a k 1 C manifold, since in (1.1) we are computing one more derivative (in particular the tangent bundle is not defined if M is only C0).
The map p : TM M defined by [(x, v)↵] x is called the projection map and the fiber ! 1 7! over a point is denoted by TxM = p (x) and called the tangent space to M at x. An important property 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 chart m for TM, the map 1 TxM : TxM R is a bijection and can be used to induce the vector space | ! structure on TxM; this structure is independent of the chart since, if (U2, 2) is another chart 1 m m containing x then 1 ( 2 TxM ) : R R is a linear automorphism. | ! m m Example 16. Since we can cover an open set U R with the single chart =idU : U R , the tangent bundle to U is simply: ⇢ ! m TU = U R ⇥ with the projection map (x, v) x. 7! Exercise 17. Only using the definition of tangent bundle, try to visualize TS1 and to prove that 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 is said smooth if for every choice of charts (U, ) for M and (!V,') for N such that f(U) V (these 1 n ⇢ charts will be called adapted) the function ' f : (U) R is smooth. If f is smooth, ! for every x M we pick a chart (U, )=(U↵, ↵) containing x and a chart (V,')=(V ,' ) containing f2(x)(uptoshrinkingU we can assume these carts are adapted), and we define the di↵erential of f to be the map df : TM TN given by: ! 1 (1.2) df :[(x, v) ] [(f(x),J (' f )v) ] ↵ 7! (x) (applying the chain rule shows that the definition is independent of the charts). The restriction of df to TxM is denoted by: d f = df : T M T N. x |TxM x ! f(x) 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 smooth manifold is a smooth map. ! 1 1 Exercise 20. Prove that the map f : S RP given by (x0,x1) [x0,x1] is smooth. ! 7! LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 9
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 ! ! d (g f)=d g d f x f(x) x (“the di↵erential of a composition is the composition of the di↵erentials”); m n m n (2) if f : R R is smooth, then dxf : TxM R Tf(x)N R is just the classical di↵erential! of a smooth map, i.t. the unique' linear! map such that:' f(x + v) f(x)=d fv + O( v 2). x k k Remark 21 (Another definition of tangent space). For practical purposes it is often convenient to have an alternative equivalent definition of tangent space and of (1.2). First, we can define TxM as the set of equivalence classes of smooth curves :( ✏,✏) M such that (0) = x,with the equivalence relation if and only if for every chart (U,! ) containing x we have: 1 ⇠ 2 d d ( ( (t))) = ( ( (t))) . dt 1 dt 2 t=0 t=0 If (U, )=(U↵, ↵) is a chart containing x, to every equivalence class [ ] we associate the d element [( (0), 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 map f : M N. In fact, given v TxM we can write v =[ ] for some :( ✏,✏) M with (0) = x!and d fv =[f ]. 2 ! x Exercise 22. There is also a third equivalent definition of tangent space, viewing tangent vectors as derivations. The reader is warmly encouraged to check out also this definition (for example as 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 point ! x M and adapted charts : U Rm and ' : V Rn (with U neighborhhod of x and V 2 ! ! negighborhood of f(x)) there is a natural way to construct bases for TxM and Tf(x)N which allow to write the matrix associated to the linear operator dxf in these bases. This is done as follows. First, let m m TxR = R = span e1,...,em { } and define for every i =1,...,m the vectors2
@ 1 =(d (x) )ej TxM. @xi 2 Similarly, setting y = f(x), for j =1,...,m we define:
@ 1 =(d'(y)' )ei TyN. @yj 2
Denoting by A =(aij) the matrix representing dxf in the coordinates given by the bases @ @ i=1,...,m for TxM and j=1,...,n for TyN, we have by definition: { @xi } { @yj } @ n @ d f = a . x @x ij @y i j=1 j X
2These are just symbols, but they have a useful and suggestive interpretation! 10 ANTONIO LERARIO
X
R
Figure 4. The set X = graph(x x ) R2 is an analytic manifold, because it can be covered with a single chart7! | | :(⇢ x, x ) x, but is is not a smooth | | 7! submanifold of R2.
Applying df(y)' to both sides of the previous equation, and using the definitions, we have:
n n @ n @ a e = a d ' = d ' a ij j ij f(y) @y f(y) 0 ij @y 1 j=1 j=1 j j=1 j X X X @ A @ 1 = d ' d f = d ' d f(d )e f(y) x @x f(y) x (x) j ✓ i ◆ 1 = d (' f )e . (x) i 1 m n In the last line we have the usual di↵erential of a map ' f : R R , whose representing 1 ! matrix in coordinates is the Jacobian matrix J (x)(' f ); the above chain of equalities proves that this Jacobiam matrix coincides with the matrix A. Remark 24. Observe the following nice identity (which justifies the choice of the notation in the previous Remark 23) for a smooth map f : Rm R: ! @ @f dxf = (x). @xi @xi Definition 25 (Immersion, embedding and di↵eomorphism). Let ' : M N be a smooth ! map. We will say that ' is an immersion if for every x M the di↵erential dx' is injective. We will say that ' is an embedding if it is an immersions2 and moreover it is a homeomorphism onto its image. We will say that a homeomorphism ' : M N is a di↵eomorphism if it is an 1 ! embedding (note that in particular the inverse ' : 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 is ⇥ an 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.AsubsetX M is called a submanifold of dimension k if every x X belongs to the domain of a chart3 (U, )⇢ for M 1 k k m2 such that U X = (R ), where R R is a linear subspace. (Observe that a submanifold \ ✓ is 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 ↵ 0let ↵ R R be the graph of the function x x . For an integer ⇢ ⇥ 2 k 7!k+1 | | k,ifk<↵ 3Recall that we assumed that we have a maximal atlas at our disposal, see Remark 2. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 11 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 i I of X and there are open sets Mi i I in M such that for every i I the set { } 2 { } 2 2 Xi Mi is a submanifold of Mi. (2) It is⇢ invariant under di↵eomorphism: if ' : M N is a di↵eomorphim, then X M is a submanifold if and only if '(X) N is a submanifold.! ✓ ⇢ Exercise 30. Work out the details showing that the notion of submanifold has local character and 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 a ✓ ! homeomorphism onto its image ◆(X)=X and the di↵erential dx◆ : TxX TxM is simply the inclusion (hence it is injective). ! Suppose now that f : X M is an embedding. We need to prove that f(X) is a submanifold ! of M. For every x X let (U1, ) be a chart for M on a neighborhood of f(x) and (V1,')a chart for X on a neighborhood2 of X.Sincef is a homeomorphism onto its image, there exists an open set W M such that f(V )=f(X) W . We set: 1 ⇢ 1 \ 1 1 U = U W and V = f (U ). f(x) 1 \ 1 x f(x) Observe that f(X) U = f(V ) and consider the diagram of maps: \ f(x) x f Vx Uf(x) ' Rn Rm Because the definition of submanifold has a local character, it is enough to prove that f(V ) x ⇢ U(fx) is a submanifold for every x X. Moreover, by the invariance under di↵eomorphism, 2 m f(Vx) U(fx) is a submanifold if and only if (f(Vx)) R is a submanifold. Note now that ⇢ ⇢ 1 m f ' : '(Vx) R ! n m 1 is an embedding of '(Vx) R into an open subset of R and f ' ('(Vx)) = (f(Vx)). ⇢ n Thus we are reduced to prove the statement in the special case X = '(Vx) R is an open 1 m ⇢ subset and h = f ' : X R is an embedding. ! We pick now a point p h(X) and we realize (locally around p) the image h(X) as the graph of a function. To this end2 we will assume that p is the origin (up to translations, which will not change the property of being a submanifold) and we pick x0 X such that p = h(x0). We consider the splitting: 2 m R = L1 L2 where L1 =im(dx0 h) m (L2 is any complementary space, for example L2 = L1?). Since h is an embedding, L1 R . m m ' We also consider the projections on the two factors p1 : R L1 and p2 : R L2 and set: ! ! h = p h and h = p h. 1 1 2 2 m In this way, in the coordinates given by the splitting R = L1 L2 we have h =(h1,h2). Observe 12 ANTONIO LERARIO 4 ANTONIO LERARIO f 2 FigureFigure 5. 3.AnAn injective injective immersion immersion f : I R 2 ofof an an open open interval interval which which is is not not ! R 2 an embedding (the image of f is a closed! subset of 2). an embedding (the image of f is a closed subset of RR ). n thatObservedx0 h1 that: Tx0fX(X) RU is invertible,= f(V ) and and consider by the Inverse the diagram Function of maps: Theorem, there exist an open !\ f(x) x neighborhood A L1 and a smooth function ↵ : A X such that: ⇢ f ! h (↵(Vax)) = a forU allf(xa) A. 1 2 We define now the function ' : A L2 by: ! ' '(a)=h2(↵(a)). Since h is an embedding (in particular an homeomorphismm onto its image), by possibly further shrinking the neighborhood A, there existsR a neighborhoodR B L of zero such that: ⇢ 2 Because the definition of submanifold has a local character, it is enough to prove that f(V ) h(X) (A B)=h(↵(A)). x ⇢ U(fx) is a submanifold for every x X\ . Moreover,⇥ by the invariance under di↵eomorphism, In particular in the neighborhood A 2 B of p h(X)wehave:m f(Vx) U(fx) is a submanifold if and only if (f(Vx)) R is a submanifold. Note now that ⇢ ⇥ 2 ⇢ h(X) (A B)=1 (a, '(a)) ma A . \ f⇥ ' : '{(Vx) R| 2 } m ! Consider now the map : A nB R given by: m 1 is an embedding of '(Vx) ⇥R into! an open subset of R and f ' ('(Vx)) = (f(Vx)). ⇢ n Thus we are reduced to prove the statement(a, b) (a, in b the' special(a)). case X = '(Vx) R is an open 1 m 7! ⇢ subset and h = f ' : X R is an embedding. It is easy to see that is a di↵!eomorphism on a neighborhhod U of p (since its Jacobian is We pick now a point p h(X) and we realize (locally around p) the image h(X)asthegraph nonvanishing at this point);2 moreover, by construction: of a function. To this end we will assume that p is the origin (up to1 translations,n m which will noth change(X) U the= property(a, (a)) of beinga A a submanifold)U = (a, b) andU web = pick'(ax) =X such( 0 thatRp = h)(x ).U. We \ { | 2 }\ { 2 | 0}2 { }⇥ \0 Letconsider us summarize the splitting: what we have proved: for every p h(X) there is a neighborhood U of p and m 2 1 n m a map : U R which is am di↵eomorphism (a chart) such that h(X) U = (R ): this R = L1 L2 where L1 = im(dx0 h) ! n \ is exactly the requirement for h(X) to be a submanifold of R . m ⇤ (L2 is any complementary space, for example L2 = L1?). Since h is an embedding, L1 R . m m ' ExerciseWe also consider 32. Work the out projections the details on of the the two last factors stepp of1 : theR previousL1 and proof.p2 : R L2 and set: ! ! 4 Exercise 33. Let G(2, 4) be theh1 Grasmmanian= p1 h and ofh2 2-planes= p2 h. in and consider the map p : R G(2, 4) P5 defined by: m In this way,R in the coordinates given by the splitting R = L1 L2 we have h =(h1,h2). Observe ! n that, dx h1 : Tx X is invertible and by the Inverse Function Theorem there exists an open 0 0 R p([M]) = [det(M j1,j2 ),...,det(M j3,j4 )] neighborhhod A !L and a smooth function| ↵ : A X such| that ⇢ 1 !4 (the subscripts range over all possible pairs (i, j) 2 , there are 6 such pairs.) Check that the h1(↵(a)) = a for2 all a A. map p (called Pl¨ucker embedding) is an embedding. Denoting2 by [p12,p13,p14,p23,p24,p34]the 5 4 homogeneousWe define now coordinates the function of R'P: A, proveL2 by: that the image of p is the quadric p12p34 + p13p24 + ! { '(a)=h (↵(a)). 4This quadric has signature (3, 3) and is double-covered2 by S2 S2;since (S2 S2)=4,thentheEuler ⇥ ⇥ characteristic of G(2, 4) is 2. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 13 p14p23 =0 . } (n) 1 More generally the Grassmannian G(k, n)embedsintoRP k via an analogous Pl¨ucker em- bedding (though the description of the image of this embedding is more complicated). Can you figure out the general construction? Definition 34 (Regular value). Let f : M N be a smooth map. We will say that y N is a 1 ! 2 regular value for f if for every x f (y) the di↵erential dxf (a linear map from TxM to TyN) is surjective. 2 Remark 35. Observe that a regular value does not need to be a value:ify/im(f)thenthere 1 2 is no x in f (y) and the condition in the above definition is automatically satisfied. 1 Theorem 36. Let y N be a regular value of a smooth map f : M N. Then f (y) is a smooth submanifold of2M 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 subset n ⇢ and N = R . In this case the conclusion follows from the Implicit Function Theorem. ⇤ 1 Whenever a manifold M is described as the preimage of a regular value M = f (y), we will say 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. using embeddings or giving their equations) are essentially the two ways we have to exhibit vector subspaces of a vector space (i.e. as the span os some vectors, or as the set of solutions of a system of independent linear equations). Example 39 (Smooth projective hypersurfaces). Let F : Rn+1 R be a homogeneous polyno- ! mial of degree d.SinceF is homogeneous, the following set Z(F ) RPn is well defined: ⇢ n Z(F )= [x0,...,xn] RP such that F (x0,...,xn)=0 . { 2 } If the vector F = @F ,..., @F (the gradient of F ) is nonzero at every non-zero point of F = r @x0 @xn { 0 Rn+1 (a non-degeneracy⇣ condition),⌘ then Z(F ) is a smooth submanifold of RPn. To see this we}⇢ use the fact that being a submanifold is a local property, i.e. in order to prove that Z(F )isa n n n submanifold, it is enough to cover RP with the open sets RP = j= Uj defined in Example 6 n and prove that Z(F ) Uj is a submanifold of RP for every j =0,...,n. Moreover, since being \ S a submanifold is invariant under di↵eomorphisms, it is enough to prove that j(Z(F ) Uj)isa n n \ n submanifold of R for every j =0,...,n (the j : Uj R are the manifold charts for RP ). n ! Now, if we define the function fj : R R by fj(y0,...,yj,...,yn)=F (y0,...,1,...,yn), the set (Z(F ) U ) is given by the equation:! j \ j b (Z(F ) U )= f =0 . j \ j { j } 14 ANTONIO LERARIO The condition F F =0 0 = 0 tells that the equation fj =0 is regular (i.e. 0 is a regular value of f ). Inr fact|{ if for}\{ some} 6 y f =0 we had f ({y) = 0,} then: j 2{ j } r j @F @F @F F (y0,...,1,...,yn)= ,..., ,..., r @x0 @xj @xn ✓ ◆ (y0,...,1,...,yn) @F = 0,...,0, , 0,...,0 @xj (y0,...,1,...,yn) ! =(0,...,0,d F ( y ,...,1,...,y ), 0,...,0) · 0 n =0 contradicting the non-degeneracy condition (in the last line we have used Euler’s identity for n @F homogeneous functions d F (x)= xj (x).) We will review this example in Example 94 · j=0 @xj below. P 1 1 Exercise 40. Prove that if y is a regular value of a smooth map f : S R,thenf (y) has even cardinality. ! 2 1 1 Exercise 41. Prove that there is no smooth function f : RP R such that RP = f (y), ! where y R is a regular value. 2 n m Exercise 42. Let M(r; n, m) R ⇥ 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=1 such that m { } k(Uk) B(0, 2) R (here m =dim(M)) and ⇢ ⇢ ` 1 (1.3) M = int k (D(0, 1)) . k[=1 m Let now : R [0, 1] a bump function constructed as in Lemma 68 with the choice c1 =1 and c = 2, and for! every k =1,...,` define the function : M [0, 1] by 2 k ! (x)= ( (x)) k Uk(x) · k (we are “pulling back” the bump function to a bump function on M using the chart k). 1 ` Observe that (1.3) ensures that the sets D = (1) cover M. { k k }k=1 m Define now for k =1,...,` the functions fk : M R by ! f (x)= (x) (x) (x) k Uk · k k m (we are using the k to extend the charts k : Uk R 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 M the `(m+1) `(m+1) 2 di↵erential dx' : TxM T'(x)R R is an injective linear map. Observe first that ! ' ' is injective: if x = y and y Dk, then either (1) x Dk, in which case k(x) = k(y), or (2) x/D and 1 = 6 (y) = (x2). Thus ' is injective and2 takes value in a Hausdor6 ↵space, M is 2 k k 6 k LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 15 compact: hence ' is a homeomorphism onto its image. It remains to check that dx' is injective for every x M. Note that 2 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,thendxfk = dx k is injective because k is a di↵eomorphism. This concludes the proof. ⇤ m m ni 1+ i=1 ni Exercise 44. Prove that the product of spheres M = i=1 S embeds into R . P 1.5. The tangent bundle for embedded submanifolds.Q In the case M Rn the tangent bundle can be defined in an equivalent way as follows (observe that if M is compact✓ by Theorem 43 we can always assume it is embedded in some Rn): n n (1.4) TM = (x, v) R R v TxM . { 2 ⇥ | 2 } 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” n n ⇥ v TxM defining TM R R are regular. 2 ⇢ ⇥ This definition is particularly useful because it allows to talk about the “length” of a tangent vector without introducing the notion of Riemannian manifold. Proposition 46. The set T 1M = (x, v) TM v =1 (called the unit tangent bundle) is a smooth submanifold of TM of dimension{ 2dim(2 M|k) k 1. } Proof. Consider the smooth function ⇢ : TM R defined by: ! ⇢(x, v)= v 2. k k (This function is smooth because it is given by the composition of two smooth functions.) Then 1 1 1 T M = ⇢ (1) and to see that T M TM is a submanifold it is enough to prove that 1 is a regular value for ⇢ and then use Theorem⇢ 36.Tothisend,let(x, v) TM such that v 2 =1 and consider the curve :(✏,✏) TM given by (t)=(x, (1 + t)v).2 Then (0) = (x, vk) andk ! d d d (t) = tv 2 = (1 + t)2 =2. dt dtk k dt t=0 t=0 t=0 d Hence im(d(x,v)⇢) span d(x,v )⇢ (t) = R and d(x,v)⇢ is surjective. { dt t=0 } ⇤ Remark 47. Observe that if M Rn is compact, then T 1M is compact, because it is closed ⇢ and 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: 3 3 T SO(3) = A R ⇥ such that AA = and det(A)=1 . 2 This is an example of a Lie group, i.e. a group G which is also a smooth manifold and such that the group multiplication µ : G G G, given by µ(g1,g2)=g1g2, and the inverse map 1 ⇥ ! ⌘ : G G, given by ⌘(g)=g , are both smooth. The smooth structure on SO(3) is given as ! 3 3 6 follows. We consider the map F : R ⇥ Sym(3, R) R given by: ! ' F (A)=AAT , 16 ANTONIO LERARIO 1 2 Tx S v x 2 TxS S2 Figure 6. The (embedded) unit tangent bundle to S2. The vectors v and x are orthogonal and of norm one; the map (x, v) (x, v, x v) gives a homeo- morphism T 1S2 SO(3). 7! ^ ! and we observe that is a regular value of F (check it!), then we can use Theorem 36 and 1 9 deduce that O(3) = F ( ) is a smooth submanifold of R . since the determinant function takes constant values on each components of O(3), we see that SO(3) is a union of connected components of O(3), hence it is itself a smooth manifold. One important remark: O(3) consists actually 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 a continuous, surjective and one-to-one map f : T 1S2 SO(3) as follows⇢ (see Figure 6): ! (x, v) (x, v, x v). 7! ^ 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 sphere 2 2 2 2 S (i.e. a continuous assignment v : S TS with v(x) TxS , see Section 4.1) must vanish at some point. ! 2 Example 51 (The group SU(2)). The group SU(2) is defined by: z z T SU(2) = A = 1 2 such that AA = and det(A)=1 . z z ⇢ ✓ 3 4◆ We can endow SU(2) the structure of smooth manifold using again Theorem 36. As a smooth manifold SU(2) is di↵eomorphic to the sphere S3 and this can be seen as follows. Observe that T the condition AA = can be written as: z4 z2 1 z4 z2 1 T z1 z2 = = A = A = , z3 z1 det(A) z3 z1 z3 z4 ✓ ◆ ✓ ◆ ✓ ◆ LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 17 which implies that A is of the form: z z A = 1 2 . z z ✓ 2 1◆ In particular the group SU(2) can be described as: z1 z2 2 2 SU(2) = A = such that z1 + z2 =det(A)=1 , z2 z1 | | | | ⇢ ✓ ◆ 3 2 which is a sphere S C = (z1,z2) . The (embedded) tangent⇢ space{ to SU} (2) at the identity is usually denoted by su(2) and it consists of the vector space of matrices: 2 2 T T SU(2) = X C ⇥ such that X = X and tr(X)=0 . { 2 } Writing a matrix X su(2) as 2 iu v iz 3 X = , (u, v, z) R , v + iz iu 2 ✓ ◆ 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: ! T '(A)=(R : X AXA ). A 7! The image of ' is SO(3) and, recalling that SO(3) RP3 (Exercise 49), we therfeore have: ' 3 3 ' : SU(2) S RP SO(3). ' ! ' The map ' is a group homomorphism with kernel and topologically it is the double cover map. {± } 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 define the measure of a set, we will just define a special class of sets in a manifold: sentences like “the set A has measure zero” or “the measure of A is zero” mean that A belongs to this special class of sets (those “of measure zero”). m m Let D =[a1,b1] [am,bm]beacubeinR .Wewillsetµ(D)= bi ai . ⇥···⇥ i=1 | | Definition 53 (Sets of measure zero). We will say that A Rm has measureQ zero if for every ⇢ ✏>0 there exists a countable family of cubes Dk k J (countable means that the cardinality of the index set J is either finite or countable) such{ that} 2 A D and µ(D ) ✏. ⇢ k k k J k J [2 X2 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 k N and for every ✏>0 consider the cubes D =[⇢r ✏ ,r + ✏ ]. Then: { } 2 k k 2k+1 k 2k+1 ✏ Q Dk and µ(Dk)= = ✏. ⇢ 2k+1 k k k [2N X2N X2N 18 ANTONIO LERARIO Remark 55. If A M has measure zero, then it cannot contain any open set: in fact if V A ⇢ ⇢ is open, then for some chart (U, ) we have that (U V ) Rm is open and nonempty. In particular (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 k J , this collection must cover D and k J µ(Dk) . In particular if A M has measure{ zero,} 2 its complement is dense: let V M be2 any open set, then V Ac = , otherwise⇢ V would be contained in A. ⇢ P \ 6 ; On the other hand the reader should keep in mind that if B M is dense, the complement ⇢ doesn’t necessarily have measure zero. For example let Q [0, 2⇡]= qk k and set \ { } 2N 1 B = (cos ✓, sin ✓) S ✓ Q [0, 2⇡] . { 2 | 2 \ } Then B is dense in S1, but its complement does not have measure zero. (Can you prove this rigorously?) Exercise 56. For every ✏>0 construct an open and dense set A Rm of Lebesgue measure smaller than ✏ (this exercise requires measure theory). ⇢ Lemma 57. Let M be a smooth manifold and A M be a subset such that A k J Ak with ⇢ ⇢ 2 J countable and with each Ak M of measure zero; then A itself has measure zero. ⇢ S Proof. Since J is countable, we may assume J = 1, 2,... N. For simplicity let us just discuss { }⇢ the case A Rm, leaving the manifold case to the reader. ⇢ Now, for every ✏>0 and for every k J cover Ak with cubes Dk,j j J with µ(Dk,j) 2 { } 2 k j Jk ✏ (such a cover exists because each A has measure zero). Then 2 2k+1 k P ✏ A D and µ(D ) ✏. ⇢ k,j k,j 2k+1 k J j J k J j J k J [2 [2 k X2 X2 k X2 ⇤ 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. ⇢ ! m Proof. Observe first that, since R is a second countable space, then U = k J int(B(xk,rk)) where J is a countable set. Then U can also be written as: 2 S 1 U = B x ,r , k k n k J n ✓ ◆ [2 [2N i.e. U can be written as a countable union of closed (hence compact) balls. Relabeling, we can write: U = Bk k [2N where each B is a closed ball. Let us set A = A B ; observe that A has measure zero (a k k \ k k subset of a set of measure zero has itself measure zero). We will prove that f(Ak) has measure zero for every k N, and the result will follow from Lemma 57,sincef(A)= k f(Ak). 2 2N 1 Thus let us fix k. Observe that, since Bk is compact and f is C ,thenf BkSis Lispchitz and there exists L > 0 such that for all x, y B : | k 2 k f(x) f(y) L x y . k k kk k In particular, if C Bk has diameter d,thenf(C) has diameter at most d Lk;hencethere exists C > 0 (which⇢ also depends on m, but this is also fixed) such that if D· B is a cube, k ⇢ k then f(D) is contained in a cube D0 with µ(D0) C µ(D). k LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 19 ✏ Fix now ✏>0 and take a countable cover of Ak with balls Dj j I such that µ(Dj) { } 2 j I Ck (this cover exists because we are assuming A has measure zero). By the above reasoning2 each k P f(D ) is contained in a D0 with µ(D0 ) C µ(D ). In particular: j j j k j ✏ f(Ak) Dj0 and µ(Dj0 ) Ckµ(Dj) Ck ✏, ⇢ · Ck j I j I j I [2 X2 X2 which proves f(Ak) has measure zero. ⇤ Remark 59. Observe that the previous Lemma implies that for a set A M to have measure ⇢ zero it is enough to check (A U) Rm has measure zero for all charts (U, ) belonging to a given 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) and g : A B be a C1 map. Then g(A) has measure zero in B and B g(A) is dense. ! \ Proof. Let (Uk, k) k I be a countable atlas for A and (Vj,'j) j J be a countable atlas for { } 2 { } 2 b B. It is enough tot check that for every k I and j J the measure of 'j(f(Uk) Vj) R is a 2 2 \ ⇢ zero. Let R be the codomain of k and Wk be its image. Consider the projection on the second b a b a factor map p : R Wk Wk.Then'j(f(Uk) Vj) is the image of 0 Wk R Wk under the C1 map: ⇥ ! \ { }⇥ ⇢ ⇥ 1 b a p f j b : R Wk Wk Uk Vj R ⇥ ! ! ! ! b b a (this is a map between open sets in R .) Since 0 Wk has measure zero in R Wk,then { }⇥ ⇥ the result follows from Lemma 58. ⇤ Example 61. For every n there is a continuous surjective map f : S1 Sn, which of course when 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 that q q 1 if q>2m + 1 then there exists a projection R R which restricts to an embedding of q 1 ! M R . The result will follow iterating the argument until q =2m +1. ! q To make things more precise, let us put coordinates (x1,...,xq) on R and let us identify q 1 q 1 R = xq =0. For every unit vector v S xq =0 let us consider the linear map q { q 1 } 2 \{ } q 1 fv : R R that projects in the direction of v. We need to find a vector v S xq =0 such that! f is an embedding. To this end we need to verify: 2 \{ } v|M (1) fv M is injective; (2) for| every x M the di↵erential d (f )isinjective. 2 x v|M The first condition is satisfied if there is no pair (x, y) M M with x = y such that: 2 ⇥ 6 x y v = . x y k k We can rephrase this condition as follows. Let be the diagonal in M M;thenfv M is ⇥ q 1 | injective if and only if v does not belong to the image of the map h1 :(M M) S given by: ⇥ \ ! x y h (x, y)= . 1 x y k k 20 ANTONIO LERARIO We apply now Lemma 60 to the smooth function h1.Since q 1 dim(M M) =2m Then fv M is an immersion if and only if v does not belong to the image of h2. We can use now Lemma |60 again, since: 1 q 1 dim(T M)=2m 1 c c c c 2 1 1 2 Figure 7. A bump function 2. Appendix: basic tools 2.1. The inverse and the implicit function theorem. These are basic tools in Di↵erential Topology, but we will not prove them here. There are many excellent references for these theorems, 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 a k ✓ ! di↵erentiable function of class C ,k 1. If the Jacobian matrix Jf(x) is nonsingular at x0 n there exists a neighborhood A of x0 such that f A : A R is invertible; moreover f(A) is open and the inverse function g : f(A) A is also C| k. ! ! In the above statement, note that the conclusion “f(A) is open” follows from the Invariance of Domain Theorem; moreover, using the language of Definition 25 below, the function f A is a di↵eomorphism. | Theorem 66 (Implicit function theorem). Let U Ra+b be an open set with coordinates (x, y) ✓ @f 2 Ra Rb and f : U Rb be a di↵erentiable function of class Ck,k 1.Ifthematrix (x, y) ⇥ ! @y (consisting of the partial derivatives of f with respect to the y variables) is nonsingular at (x0,y0), there exist neighborhoods A of x and B of y and a function ' : A B of class Ck such that: 0 0 ! f = f(x ,y ) (A B) = graph('). { 0 0 }\ ⇥ 2.2. Bump functions. Definition 67 (Bumb function). Let 0 Lemma 68. Bumb functions exist for every 0 Proof. Let ↵ : R R be the C1 function ! 1 ↵(x)=e x x>0 (x), { } where x>0 (x) denotes the characteristic function of x>0 . Out of this function we construct { } { } : R R by: ! c2 x ↵(t c1)↵(c2 t)dt (x)= c2 . ↵(t c1)↵(c2 t)dt Rc1 Finally we set (x)= ( x ). (The readerR can see [4, Section 2.2] for pictures explaining the k k construction step by step.) ⇤ 22 ANTONIO LERARIO 2.3. Partitions of unity. Definition 69 (Partition of unity). Let M by a smooth manifold and = Uj j J be an open U { } 2 cover of M.Apartition of unity subordinated to is a collection of smooth maps ⇢j : M [0, 1] J such that: U { ! }j 2 (1) for every j J the support supp(⇢ ) U ; 2 j ⇢ j (2) the family supp(⇢j) j J is locally finite; { } 2 (3) j J ⇢j 1 (the sum is well defined because of the previous condition). 2 ⌘ In orderP to prove the existence of partitions of unity, we will need the following Lemma. Lemma 70. Let = Uj j J be an open cover for M. Then there is a locally finite atlas U { } 2 m (V↵, ↵) ↵ A such that V ↵ ↵ A refines , each ↵(V↵) R is bounded and each V ↵ M { } 2 { } 2 U ⇢ ⇢ is compact. Moreover the open cover = V↵ ↵ A has a shrinking = W↵ ↵ A, i.e. is V { } 2 W { } 2 W itself a cover and for every ↵ A we have W V . 2 ↵ ⇢ ↵ Proof. The proof is left as an exercise. Work out the details carefully! ⇤ Proposition 71. Let M be a smooth manifold. Every open cover of M admits a subordinate partition of unity. U Proof. Let = Uj j J be the open cover. We observe first that if we can construct a partition U { } 2 of unity i i I subordinated to a refinement = Vi i I of , then we can also construct a partition{ of} unity2 subordinated to . In fact letVg : I{ }J2 be suchU that for every i I we have U ! 2 the inclusion Vi Ug(i) (the existence of such g is in the definition of refinement) and define for j J the function:⇢ 2 ⇢j(x)= i(x). i g 1(j) 2X It is easy to check that ⇢j j J is a partition of unity subordinated to (essentially to obtain { } 2 U the ⇢j we collate together some of the i). Let now be the open cover produced by Lemma 70 and its shrinking. For every ↵ A V m W 2 we cover the compact set ↵(W↵) R by finitely many closed balls contained in ↵(V↵): ⇢ k↵ (W ) B , ↵ ↵ ⇢ ↵,k k[=1 m where B↵,k = B(x↵,k,r↵,k) for some points x↵,k R and radii r↵,k > 0. For every ball B↵,k m 2 let ↵,k : R [0, 1] be a bump function centered at x↵,k such that ↵,k(x) > 0 if and only if x int(B !) (it su ces to take a bumb function , as in Lemma 68,withc Exercise 72 (Smooth Urysohn’s Lemma). Let A and B be two closed and disjoint submanifolds of a manifold M. Prove that there is a smooth function f : M [0, 1] such that f A 0 and f 1. ! | ⌘ |B ⌘0 there exists an embedding h✏ : M R such that: ! max h(x) h✏(x) ✏. x M k k 2 LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 21