Lecture Notes on Basic Differential Topology

LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY

ANTONIO LERARIO

These are notes for the course “Advanced Geometry 2” for the Master Diploma in Mathematics at the University of Trieste and at SISSA. These notes are by no means complete: excellent references for the subject are the books [3,5,7,8,9], 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.

Date: November 4, 2020. 1 2 ANTONIO LERARIO

1. Differentiable manifolds

Definition 1 (m–dimensional Ck manifold). Let m N and k N , ω . A manifold M of dimension m and class Ck is a paracompact, Hausdorff∈ topological∈ space,∪ {∞ with} countably many connected components1 and such that: (1) for every point x M there exists a neighborhood U of x and a continuous function ∈ ψ : U Rm which is a homeomorphism onto an open subset of Rm (the pair (U, ψ) is called a→chart); (2) for every pairs of charts (U , ψ ) and (U , ψ ) such that U U = the map 1 1 2 2 1 ∩ 2 6 ∅ 1 m ψ2 ψ1 U− U : ψ1(U1 U2) R ◦ | 1∩ 2 ∩ → is a Ck map (for k = 0 we obtain topological manifolds, for k 1 differentiable manifolds, for k = smooth manifolds and for k = ω analytic manifolds).≥ ∞ S k A collection of charts (Uj, ψj) j J as above such that j J Uj = M is called a C atlas. { } ∈ ∈ Remark 2. In the second condition above, given two charts (U1, ψ1) and (U2, ψ2) such that 1 k U U = it is not enough to check that the map from ψ ψ − is C , but also that its 1 2 2 1 U1 U2 ∩ 6 k∅ ◦ | ∩ 3 inverse is C . For instance, let M = R with the two charts (U1, ψ1) = (R, x) and (U2, ψ2) = (R, x ). 1 3 ω 1 1/3 0 Then ψ ψ− = x is C but ψ ψ− = x is only C . 2 ◦ 1 1 ◦ 2 k Remark 3 (Atlases and differential structures). Two C atlases A = (Uα, ψα) α A and B = { k} ∈ k (Vβ, ϕβ) β B for M, are said to be equivalent if their union A B is still a C atlas. A C - { } ∈ ∪ differential 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-differential structure, we obtain a maximal Ck atlas. This atlas contains every chart that is compatible with the chosen differentiable structure. (There is a natural one-to-one correspondence between differentiable structures and maximal differentiable 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 ψ : U Rm around a point x M (as in point (1) of Definition1), and given a neighborhood V →U of x we can easily ∈ m ⊂ construct a chart ϕ : V R by simply taking ϕ = ψ V . Note that in this way we can construct → | 1 m a chart (V, ϕ) around any point with V contractible: it is enough to take V = ψ− (BR (ψ(x), )) for > 0 small enough. Example 4. If ϕ : M Rn is a homeomorphism, then M is an analytic manifold. In fact one → 1 m m can cover M with the single chart (M, ϕ), and ϕ ϕ− = id m : R R is analytic. ◦ R → Example 5. Given an open set U Rm, the single chart given by the inclusion U, Rm turns it into a smooth manifold. Why is⊆ the condition “with countably many connected components”→ in the above definition satisfied by U? (Try to prove it directly: an open set in Rm can have only countably many connected components). Example 6 (Product manifolds). If M and N are smooth manifolds with respective atlases (Uα, ψα) α A and (Vβ, ϕβ) β B, then M N is naturally a smooth manifold with the atlas { } ∈ { } ∈ × (Uα Vβ, ψα ϕβ) (α,β) A B. { × × } ∈ × n 2 2 n+1 Example 7 (Spheres). The unit sphere S = x0 + + xn = 1 R can be endowed with the structure of a smooth manifold as follows.{ Consider··· the point}e ⊂= (1, 0,..., 0) Sn and the 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 = {Uα}α∈A fo X, there exists another open cover V = {Vβ }β∈B such that (i) for every β ∈ B there exists α(β) ∈ A such that Vβ ⊂ Uα(β) (i.e. V refines U); (ii) for every x ∈ X there exists a neighborhood Vx of x which intersects only finitely many elements of V (i.e. V is locally finite). It is worth noticing that if a Hausdorff space is locally Euclidean (i.e. if it satisfies condition (1) of Definition 1) and connected, then this space is paracompact if and only if it is second countable (i.e. its topology has a countable basis), see the discussion you can find at this webpage https://math.stackexchange.com/questions/ 527642/the-equivalence-between-paracompactness-and-second-countablity-in-a-locally-eucl. There exist locally Euclidean spaces which are Hausdorff, paracompact but not second countable (for example R with the discrete topology); that is why we also add the condition that a manifold should have countably many connected components. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 3

e0 bc

b p1

n R b b

ψ1(p2) ψ1(p1)

b p2

n n Figure 1. The stereographic projection ψ1 : S e0 R . \{ } →

n n two open sets of the the sphere deﬁned by U1 = S e0 and U2 = S e0 . We produce explicit \{ n} \{− n} and nice homeomorphisms (i.e. charts) ψ1 : U1 R and ψ2 : U2 R , called stereographic projections (see Figure1), as follows: → → 1 1 ψ (x , . . . , x ) = (x , . . . , x ) and ψ (x , . . . , x ) = (x , . . . , x ). 1 0 n 1 x 1 n 2 0 n 1 + x 1 n − 0 0

1 n n It is easy to verify that ψ− : R S is given by: 1 → 1 1 2 ψ− (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 one to write the explicit expression for ψ2 ψ1 U− U : R 0 R 0 : ◦ | 1∩ 2 \{ } → \{ } 1 y ψ ψ− (y) = , 2 ◦ 1 y 2 k k n which is a smooth map. Hence (U1, ψ1), (U2, ψ2) is a smooth atlas for S and turns it into a smooth manifold. This is called the{ standard diﬀerential} structure on Sn.

Example 8 (Real projective spaces). The real projective space RPn can be endowed with the n n+1 structure of smooth manifold as follows. Recall that RP = (R 0 )/ , where p1 p2 if and \{ } ∼ ∼ only if there exists λ = 0 such that p1 = λp2. 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 ∈ \{ } 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 ψj : Uj R given by: → x0 cxj xn ψj([x0, . . . , xn]) = ,..., ,..., xj xj xj (here the “hat” symbol denotes that this element has been removed from the list). The inverse 1 n n ψ− : R RP is given by: j → 1 ψj− (y0,..., ybj, . . . , yn) = [y0,..., 1, . . . yn], where the “1” is in position j. As a consequence, for every i = j we have: 6 1 y0 cyi 1 yn ψi ψj− (y0,..., ybj, . . . , yn) = ,..., ,..., ,... , ◦ yi yi yi yj which is a diﬀeomorphism of Rn 0 to itself. \{ } Exercise 9. Prove that RP1 and S1 are homeomorphic. Example 10 (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 . { ∈ } → { } 4 ANTONIO LERARIO

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: × k M1 M2 there exists L GL(R ) such that M1 = M2L. ∼ ⇐⇒ ∈ 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 λ GL(R) = R 0 such that v1 = λv2. ∼ ∈ \{ } 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 rows 1 k ∈ k |J × j1, . . . , jk (in this way M J c denotes the (n k) k submatrix of M obtained by selecting 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 { ∈ |J 6 } (Note that this set is well defined.) Mimicking again the definition for projective spaces, we define (n k) k the manifold charts ψJ : UJ R − × by: → 1 ψ ([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 ψ ψ− is smooth. In this way (U , ψ ) n 1 2 k J2 J1 J J J ∈ ◦ { } ∈ k is a smooth atlas for the k(n k)-dimensional manifold G(k, n). { } − Exercise 11. Fill in all the details in the previous definition of the smooth structure on the Grassmannian.

Example 12 (The Complex projective line). Recall that the complex line CP1 is deﬁned as the 2 quotient space (C 0 )/ where (z0, z1) λ(z0, z1) for every λ C 0 . As we did for real \{ } ∼ ∼ ∈ \{ } 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 ψj : Uj C R for j = 0, 1 which are given{ by:6 } { 6 } → '

z0 z1 ψ1([z0, z1]) = and ψ0([z0, z1]) = . z1 z0 1 1 We have that ψ0 ψ− (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 ' \{ } → \{ } 2 with ψj as a real map, as follows (however, as the reader will see, using the field structure of C R 4 ' simplifies a lot the computations). Given (x0, y0, x1, y1) R , let us denote by (x0 +iy0, x1 +iy1) = 2 2 ∈ (z0, y0) C . We can write ψ1 : U0 C R as ∈ → ' x0 + iy0 x0x1 + y0y1 y0x1 x0y1 ψ1([x0 + iy0, x1 + iy1]) = = 2 2 + i 2 − 2 x1 + iy1 x1 + y1 · x1 + y1 2 which means that the real map ψ1 : U1 R is given by: → x0x1 + y0y1 y0x1 x0y1 ψ1([x0 + iy0, x1 + iy1]) = 2 2 , 2 − 2 , x1 + y1 x1 + y1 1 2 1 with inverse ψ− : R CP given by: 1 → 1 ψ1− (x, y) = [1, x + iy]. 1 1 In particular ψ ψ − 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 31).

Exercise 13. Generalize the previous example and prove that CPn can be endowed with the structure of a smooth 2n-dimensional manifold. Example 14 (A connected, non-paracompact “manifold”: the Pr¨ufersurface). Let H = (x, y), y > { 0 be the positive half-space and for every z R consider the set } ∈ 3 Az = (x, y, z) R with y 0 { ∈ ≤ } LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 5

x = z + ay x = z + by

ǫ H bc z a b

Figure 2. An open set of the type U3 (grey region) for the topology of the Pr¨ufersurface.

(each Az is a closed half space). The Prüfersurface is the set ! [ P = H A ∪ z z ∈R 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 < x < b, < y 0 { ∈ − ≤ } H (a, b, ) = (x, y) H, 0 < y < , z + ay < x < z + by . z { ∈ } The resulting topological space is Hausdorff and locally euclidean. To see this, for every z R let 2 ∈ Uz = Az H (an open set) and define the map fz : R Uz by: ∪ → (x, y, z) y 0 f (x, y) = z (z + xy, y) y≤ > 0 2 The map fz is continuous with continuous inverse (check it!) which we denote by ψz : Uz R . → The “manifold” structure on P is then given by the atlas (Uz, ψz) z . Notice now that the set { } ∈R Z = (0, 0, z), z R is an uncountable subset of P which inherits the discrete topology, hence P cannot{ be second∈ countable.} Example 15 (A non-Hausdorff “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: ∪ {∞ ∞ }

U = j A R with R A compact for some j = 1, . . . , n. {∞ } ∪ { ⊂ \ } For n 2 the space X can be endowed with a structure satisfying the axioms for the deﬁnition of ≥ n smooth manifold, except for the Hausdorﬀ condition (see Figure3). The space X2 0 is sometimes called “the line with two origins”. \{ } 6 ANTONIO LERARIO

1 ∞b b

∞2

X2 = R 1 2 ∪ {∞ }∪{∞ } Figure 3. The two-points compactiﬁcation of the real line: a compact, non- Hausdorﬀ, C∞ manifold.

Example 16 (A paracompact, Hausdorff, “manifold” which is not second countable). Consider X = R with the discrete topology. Then, X is paracompact (given any cover U of X, the open cover V = x x is a locally-finite refinement of U), Hausdorff and locally Euclidean (each {{ }} ∈R point x X is open and homeomorphic to R0). However X is not second countable, as it has uncountably∈ many components. More generally, let Y be a connected manifold (a honest manifold, ` in the sense of Definition1) and consider X = t Yt, where each Yt is a copy of Y (just labeled ∈R by a point in t R). Endow X with the disjoint union topology, so that it has uncountably many connected components∈ and it is not second countable. However X is Hausdorff and paracompact. In fact let U = Uα α A be an open cover of X and for every t R consider the open cover { } ∈ ∈ Ut = Uα Yt α A of the component Yt. Since Yt is paracompact, there exists a locally finite { ∩ } ∈ refinement Vt = Vt,β β Bt of Ut. Then V = Vt,β, t R, β Bt is a locally finite open cover of X which refines U{. } ∈ { ∈ ∈ }

Exercise 17. Fill in all the details in the construction of the “manifold” structure for the above non-examples.

1.1. The tangent bundle.

Definition 18 (Tangent bundle). Let M be a smooth manifold of dimension m and (Uα, ψα) α A be a smooth atlas for M. We define the tangent bundle of M as: { } ∈ ! a m TM = Uα R / , × ∼ α A ∈ where (x1, v1)α1 (x2, v2)α2 if and only if x1, x2 Uα1 Uα2 (the two open sets overlap and ∼ ∈1 ∩ x1, x2 belong to their intersection), Jψ (x)(ψα ψ− )v2 = v1 (in other words, the Jacobian of the α2 1 ◦ α2 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 α A we call TM Uα the set [(x, v)α] TM and observe that (by construc- ∈ | { } ⊂ m tion) TM Uα = TUα is an open subset of TM which is homeomorphic to Uα R (under the | m × identification map); we call ϕα : TM Uα Uα R the inverse of this homeomorphism. The manifold charts for TM are constructed| as→ follows.× We define the homeomorphism m m m φα : TM Uα ψα(Uα) R R R | → × ⊆ × m to be the composition φα = (ψα idR ) ϕα, i.e. the map given by [(x, v)α] (ψα(x), v). The m × ◦ 7→ family (Uα R , φα) α A 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 check m what is the regularity of the change of coordinates maps. For every pair of charts (U1 R , φ1) m m m × and (U2 R , φ2) such that (U1 R ) (U2 R ) = , we consider therefore the map: × × ∩ × 6 ∅ 1 m m (1.1) φ2 φ− : ψ1(U1 U2) R ψ2(U1 U2) R . ◦ 1 ∩ × → ∩ × LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 7

1 1 We see that this map is given by (y, v) (ψ2 ψ1− (y),Jy(ψ2 ψ1− )v) and it is smooth if M itself 7→ k◦ ◦ k 1 was smooth. Notice indeed that if M is a C manifold, the tangent bundle TM is only a 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)α] x is called the projection map and the fiber over a → 1 7→ 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, ψ1) containing x, calling by (TU1, φ1) the corresponding natural chart for TM, m the map φ1 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, ψ2) is another chart containing x 1 m m then φ1 (φ2 TxM )− : R R is a linear automorphism. ◦ | → m m Example 19. Since we can cover an open set U R with the single chart ψ = idU : U R , the tangent bundle to U is simply: ⊂ → m TU = U R × with the projection map (x, v) x. 7→ Exercise 20. 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 differential. Definition 21 (Smooth map). A continuous map f : M N between two smooth manifolds is said smooth if for every choice of charts (U, ψ) for M and (→V, ϕ) for N such that f(U) V (these 1 n ⊂ charts will be called adapted) the function ϕ f ψ− : ψ(U) R is smooth. If f is smooth, ◦ ◦ → for every x M we pick a chart (U, ψ) = (Uα, ψα) containing x and a chart (V, ϕ) = (Vβ, ϕβ) containing f∈(x) (up to shrinking U we can assume these charts are adapted), and we define the differential of f to be the map df : TM TN given by: → 1 (1.2) df : [(x, v) ] [(f(x),J (ϕ f ψ− )v) ] α 7→ ψ(x) ◦ ◦ β (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 differential of f at x. Exercise 22. Prove that the projection map p : TM M for the tangent bundle of a smooth manifold is a smooth map. → 1 1 Exercise 23. Prove that the map f : S RP given by (x0, x1) [x0, x1] is smooth. → 7→ 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 differential of a composition is the composition of the differentials”); m n m n (2) if f : R R is smooth, then dxf : TxM R Tf(x)N R is just the classical differential→ 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 24 (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 γ :( , ) M such that γ(0) = x, with the − → equivalence relation γ1 γ2 if and only if for every chart (U, ψ) containing x we have: ∼ d d (ψ(γ1(t))) = (ψ(γ2(t))) . dt t=0 dt t=0

If (U, ψ) = (Uα, ψα) is a chart containing x, to every equivalence class [γ] we associate the element d 1 [(γ(0), dt (ψ(γ(t))) t=0)α]; viceversa, given [(x, v)α] we can construct the curve γ(t) = ψα− (ψα(x) + tv), which is well| defined on the interval ( , ) for > 0 small enough. Since these two process − 8 ANTONIO LERARIO are inverse to each other, this shows that the two definitions are equivalent. This alternative definition is particularly convenient for “computing” the differential of a map f : M N. In fact, given v TxM we can write v = [γ] for some γ :( , ) M with γ(0) = x and d →fv = [f γ]. ∈ − → x ◦ Exercise 25. 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 [8, Chapter 3]), and to prove that it is equivalent to the one we have given here. Remark 26 (The differential in coordinates). Let f : M N be a smooth map. Given a point → x M and adapted charts ψ : U Rm and ϕ : V Rn (with U neighborhood of x and V ∈ → → neighborhood 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 = (dψ(x)ψ− )ej TxM. ∂xi ∈ Similarly, setting y = f(x), for j = 1, . . . , m we deﬁne:

∂ 1 = (dϕ(y)ϕ− )ei TyN. ∂yj ∈ Denoting by A = (a ) the matrix representing d f in the coordinates given by the bases ∂ ij x ∂xi i=1,...,m ∂ { } for TxM and j=1,...,n for TyN, we have by deﬁnition: { ∂yj } n ∂ X ∂ d f = a . x ∂x ij ∂y i j=1 j

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

n n  n  X X ∂ X ∂ aijej = aijdf(y)ϕ = df(y)ϕ  aij  ∂y ∂y j=1 j=1 j j=1 j ∂ 1 = df(y)ϕ dxf = df(y)ϕ dxf(dψ(x)ψ− )ej ∂xi 1 = d (ϕ f ψ− )e . ψ(x) ◦ ◦ i 1 m n In the last line we have the usual differential of a map ϕ f ψ− : R R , whose representing ◦ 1◦ → matrix in coordinates is the Jacobian matrix Jψ(x)(ϕ f ψ− ); the above chain of equalities proves that this Jacobiam matrix coincides with the matrix◦A.◦ Remark 27. Observe the following nice identity (which justifies the choice of the notation in the previous Remark 26) for a smooth map f : Rm R: → ∂ ∂f dxf = (x). ∂xi ∂xi Definition 28 (Immersion, embedding and diffeomorphism). Let ϕ : M N be a smooth map. → We will say that ϕ is an immersion if for every x M the differential dxϕ is injective. We will say that ϕ is an embedding if it is an immersion and∈ moreover it is a homeomorphism onto its image. We will say that a homeomorphism ϕ : M N is a diffeomorphism if it is an embedding (note 1 → that in particular the inverse ϕ− : N M is also a diffeomorphism). → Exercise 29. Let f : M N be a smooth map. Prove that the set of points x M such that → ∈ the differential dxf has maximal rank is open in M. Exercise 30. 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 31. Prove that S2 and CP1 are diffeomorphic.

2These are just symbols, but they have a useful and suggestive interpretation! LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 9

Figure 4. The set X = graph(x x ) R2 is an analytic manifold, because it can be covered with a single chart7→ | ψ| :(⊂ x, x ) x, but is is not a smooth | | 7→ submanifold of R2.

3 Remark 32. Consider on R the two atlases A1 = (R, ϕ1 = id) and A2 = (R, ϕ2 : x x ) . It { } 1 { 7→ } 0 follows from Remark2 above that the two atlases A1 and A2 are not C -equivalent (but only C equivalent). However R with the differential structure induced by A1 and R with the differential structure induced by A are diffeomorphic, the diffeomorphism being the map x x1/3. 2 7→ 1.3. Submanifolds and how to produce them. Definition 33 (Submanifold). Let M be a smooth manifold of dimension m. A subset X M is called a submanifold of dimension k if every x X belongs to the domain of a chart3 (U, ψ)⊂ for M 1 k k m∈ such that U X = ψ− (R ), where R R is a linear subspace. (Observe that a submanifold ∩ ⊆ is itself a manifold with charts of the form (U X, ψ U X ).) The difference dim(M) dim(X) is called the codimension of X in M. ∩ | ∩ − α Example 34. For α 0 let Γα R R be the graph of the function x x . For an integer k, ≥ ⊂ × 2 k k+17→ | | 2 2 if k < α < k + 1 then Γα is a submanifold of R which is C but not C . In fact, let ϕ : R R α 1 → be the chart given by (x, y) (x, y x ). Then ϕ− (R 0 ) = Γα. 7→ − | | × { } 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 { } ∈ { } ∈ ∈ Xi Mi is a submanifold of Mi. (2) It is⊂ invariant under diffeomorphism: if ϕ : M N is a diffeomorphim, then X M is a submanifold if and only if ϕ(X) N is a submanifold.→ ⊆ ⊂ Exercise 35. Work out the details showing that the notion of submanifold has local character and is invariant under diffeomorphisms. Theorem 36. 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 ι : X M is an embedding: ι is a ⊆ → homeomorphism onto its image ι(X) = X and the differential dxι : 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, ψ) be a chart for M on a neighborhood of f(x) and (V1, ϕ) a chart for X on a neighborhood∈ of X. Since f 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)

ϕ ψ

Rn Rm

3Recall that we assumed that we have a maximal atlas at our disposal, see Remark3. 10 ANTONIO LERARIO 4 ANTONIO LERARIO

2 FigureFigure 5. 3.AnAn injective injective immersion immersionf f: I: I RR2 ofof an an open open interval interval which which is is not not an →→ 2 embeddingan embedding (the (theimage image of f ofis af closedis a closed subset subset of R of2).R ).

Because the definition of submanifold has a local character, it is enough to prove that f(V ) Observe that f(X) Uf(x) = f(Vx) and consider the diagram of maps: x ⊂ U(fx) is a submanifold∩ for every x X. Moreover, by the invariance under diffeomorphisms, ∈ m f(Vx) U(fx) is a submanifold if and only if ψf(f(Vx)) R is a submanifold. Note now that ⊂ Vx Uf(⊂x) 1 m ψ f ϕ− : ϕ(Vx) R ◦ ◦ → n m 1 is an embedding of ϕ(Vx) R into an openϕ subset of ψR and ψ f ϕ− (ϕ(Vx)) = ψ(f(Vx)). ⊂ ◦ ◦ n Thus we are reduced to prove the statement in the special case X = ϕ(Vx) R is an open subset 1 m ⊂ and h = ψ f ϕ− : X R is an embedding. ◦ ◦ → Rn Rm We pick now a point p h(X) and we realize (locally around p) the image h(X) as the graph ∈ ofBecause a function. the definition To this end of submanifoldwe will assume has that a localp is character, the origin it (up is eno tough translations, to prove that whichf( willVx) not ⊂ changeU(fx) the is a property submanifold of being for a every submanifold)x X. andMoreover, we pick byx0 theX invariancesuch that underp = h( dixff0).eomorphism, We consider ∈ ∈m f(Vx) U(fx) is a submanifold if and only if ψ(f(Vx)) R is a submanifold. Note now that the splitting:⊂ ⊂ m 1 m ψ f ϕ− : ϕ(Vx) R = L1 L2 where L1 =R im(dx0 h) ◦⊕ ◦ → is an embedding of ϕ(V ) n into an open subset of m and ψ f ϕ 1(ϕ(V )) = ψ(f(Vn )). (L2 is any complementaryx space,R for example L2 = L1⊥).R Since h is an embedding,− x L1 R x. We ⊂ m ◦ ◦ m n ' alsoThus consider we are the reduced projections to prove on the the two statement factors inp1 the: R specialL1 caseand Xp2 =: Rϕ(Vx)L2 andR is set: an open 1 m → → ⊂ subset and h = ψ f ϕ− : X R is an embedding. ◦ ◦ →h = p h and h = p h. 1 1 ◦ 2 2 ◦ We pick now a point p h(X) and we realize (locallym around p) the image h(X)asthegraph In this way, in the coordinates∈ given by the splitting R = L1 L2 we have h = (h1, h2). Observe of a function. To thisn end we will assume that p is the origin⊕ (up to translations, which will that dx0 h1 : Tx0 X R is invertible, and by the Inverse Function Theorem, there exist an open not change the property of being a submanifold) and we pick x0 X such that p = h(x0). We neighborhood A →L and a smooth function α : A X such that:∈ consider the splitting:⊂ 1 → m h1(α(a)) = a for all a A. R = L1 L2 where L1 = im(dx0 h) ⊕ ∈ We define now the function ϕ : A L2 by: 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 theϕ two(a) =factorsh2(α(pa1)):.R L1 and p2 : R L2 and set: → → Since h is an embedding (in particularh = p ah homeomorphismand h = p ontoh. its image), by possibly further 1 1 ◦ 2 2 ◦ shrinking the neighborhood A, there exists a neighborhoodm B L2 of zero such that: In this way, in the coordinates given by the splitting R = L1 ⊂L2 we have h =(h1,h2). Observe n ⊕ that, dx0 h1 : Tx0 X R is invertibleh(X and) ( byA theB) Inverse = h(α Function(A)). Theorem there exists an open → ∩ × Inneighborhhod particular inA the neighborhoodL1 and a smoothA functionB of p αh:(XA) weX have:such that ⊂ × ∈ → h(X)h1((αA(a))B =)a = for(a, ϕall(aa)) aA. A . ∩ × { ∈| ∈ } We define now the function ϕ : A Lm by: Consider now the map ψ : A B R 2 given by: × →→ (a,ϕ b()a)=(a,h2 b(α(aϕ))(a.)). 7→ − It is easy to see that ψ is a diffeomorphism on a neighborhhod U of p (since its Jacobian is nonvanishing at this point); moreover, by construction: 1 n m h(X) U = (a, ψ(a)) a A U = (a, b) U b = ϕ(a) = ψ− ( 0 R − ) U. ∩ { | ∈ } ∩ { ∈ | } { } × ∩ Let us summarize what we have proved: for every p h(X) there is a neighborhood U of p and a m ∈ 1 n m map ψ : U R which is a diffeomorphism (a chart) such that h(X) U = ψ− (R − ): this is → n ∩ exactly the requirement for h(X) to be a submanifold of R . Exercise 37. Work out the details of the last step of the previous proof. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 11

Exercise 38. 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 ),..., det(M )] |j1,j2 |j3,j4 (the subscripts range over all possible pairs (i, j) 4 , there are 6 such pairs.) Check that the map ∈ 2 p (called Plücker embedding) is an embedding. Denoting by [p12, p13, p14, p23, p24, p34] the homoge- 5 4 neous coordinates of RP , prove that the image of p is the quadric p12p34 + p13p24 + p14p23 = 0 . (n) 1 { } More generally the Grassmannian G(k, n) embeds into RP k − via an analogous Plücker embedding (though the description of the image of this embedding is more complicated). Can you figure out the general construction? Definition 39 (Regular value). Let f : M N be a smooth map. We will say that y N is a 1 → ∈ regular value for f if for every x f − (y) the differential dxf (a linear map from TxM to TyN) is surjective. ∈ Remark 40. Observe that a regular value does not need to be a value: if y / im(f) then there is 1 ∈ no x in f − (y) and the condition in the above definition is automatically satisfied. 1 Theorem 41. Let y N be a regular value of a smooth map f : M N. Then f − (y) is a smooth submanifold of∈M; if nonempty, it has dimension dim(M) dim(→N). − Proof. By using local character and invariance under diffeomorphisms of the definition of submanifold, we can reduce (as in the poof of Theorem 36) to the case M Rm is an open subset and n ⊂ 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 42. Work out the details of the last step in the previous proof. 1 Remark 43. Observe that if X = f − (y) with y a regular value of f : M N, then: → TxX = ker dxf.

One inclusion is simple: if v TxX, then there exists γ :( , ) such that γ(0) = x and [γ] = v. Then, since f(γ(t)) y is a constant∈ curve, it follows that − ≡ d fv = [f γ] = 0. x ◦ The other inclusion follows from the fact that, since the equation f = y is regular, T X and { } x ker dxf have the same dimension! Remark 44. We observe that the two ways we have to produce submanifolds (i.e. using embed- dings or giving their equations) are essentially the two ways we have to exhibit vector subspaces of a vector space (i.e. as the span of some vectors, or as the set of solutions of a system of independent linear equations). Example 45 (Smooth projective hypersurfaces). Let F : Rn+1 R be a homogeneous polynomial → of degree d. Since F is homogeneous, the following set Z(F ) RPn is well defined: ⊂ n Z(F ) = [x0, . . . , xn] RP such that F (x0, . . . , xn) = 0 . { ∈ } If the vector F = ∂F ,..., ∂F (the gradient of F ) is nonzero at every non-zero point of ∇ ∂x0 ∂xn F = 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 ) n n Sn is a submanifold, it is enough to cover RP with the open sets RP = j= Uj defined in Example n 8 and prove that Z(F ) Uj is a submanifold of RP for every j = 0, . . . , n. Moreover, since being ∩ a submanifold is invariant under diffeomorphisms, it is enough to prove that ψj(Z(F ) Uj) is a n n ∩ n submanifold of R for every j = 0, . . . , n (the ψj : Uj R are the manifold charts for RP ). n → Now, if we define the function fj : R R by fj(y0,..., ybj, . . . , yn) = F (y0,..., 1, . . . , yn), the set ψ (Z(F ) U ) is given by the equation:→ j ∩ j ψ (Z(F ) U ) = f = 0 . j ∩ j { j } 4This quadric has signature (3, 3) and is double-covered by S2 × S2; since χ(S2 × S2) = 4, then the Euler characteristic of G(2, 4) is 2. 12 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 ). In∇ fact|{ if for}\{ some} 6 y f = 0 we had f (y{) = 0, then:} j ∈ { j } ∇ j ∂F ∂F ∂F F (y0,..., 1, . . . , yn) = ,..., ,..., ∇ ∂x ∂x ∂x 0 j n (y0,...,1,...,yn) ! ∂F = 0,..., 0, , 0,..., 0 ∂x j (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 Pn ∂F homogeneous functions d F (x) = xj (x).) We will review this example in Example 107 · j=0 ∂xj below. Exercise 46. Let F : Rn+1 R as in the previous example be a homogeneous polynomial of → n+1 degree d and define the function f = F Sn . Prove that if for every x R 0 such that F (x) = 0 we have F (x) = 0, then the equation| f = 0 is a regular equation∈ on S\{n (i.e.} f = 0 is a smooth submanifold∇ 6 of Sn, of dimension n { 1 if nonempty).} Prove that the covering{ map} n n − n q : S RP restricts to a smooth covering map p f=0 : f = 0 Z(F ) RP . → |{ } { } → ⊂ 1 1 Exercise 47. Prove that if y is a regular value of a smooth map f : S R, then f − (y) has even cardinality. → 2 1 1 Exercise 48. Prove that there is no smooth function f : RP R such that RP = f − (y), → where y R is a regular value. ∈ Exercise 49. Let Z RPn be a smooth algebraic hypersurface defined by the regular equation ⊂ F = 0 with F : Rn+1 R a homogeneous polynomial of even degree. Prove that there exists a { } → smooth function ϕ : RPn R whose zero set is Z and such that the equation ϕ = 0 is regular → { } on RPn. n m Exercise 50. 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 51. Let M be a compact manifold. There exists an embedding ϕ : 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, ψk) k=1 such that m { } ψk(Uk) B(0, 2) R (here m = dim(M)) and ⊂ ⊂ ` [ 1 (1.3) M = int ψk− (D(0, 1)) . k=1 m Let now λ : R [0, 1] a bump function constructed as in Lemma 190 with the choice c1 = 1 and c = 2, and for every→ k = 1, . . . , ` define the function λ : M [0, 1] by 2 k → λ (x) = χ λ(ψ (x)) k Uk(x) · k (we are “pulling back” the bump function λ to a bump function on M using the chart ψk). Observe 1 ` that (1.3) ensures that the sets D = λ− (1) cover M. { k k }k=1 m Define now for k = 1, . . . , ` the functions fk : M R by → f (x) = χ (x) λ (x)ψ (x) k Uk · k k m (we are using the λk to extend the charts ψk : Uk R to the whole M). → We finally define the map ϕ : M R`(m+1) by: → ϕ(x) = (f1(x), λ1(x), . . . , f`(x), λ`(x)). We need to check that ϕ is a homeomorphism onto its image and that for every x M the `(m+1) `(m+1) ∈ differential dxϕ : TxM Tϕ(x)R R is an injective linear map. Observe first that ϕ is injective: if x = y →and y D , then' either (1) x D , in which case ψ (x) = ψ (y), or (2) 6 ∈ k ∈ k k 6 k LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 13 x / D and 1 = λ (y) = λ (x). Thus ϕ is injective and takes value in a Hausdorff space, M is ∈ k k 6 k compact: hence ϕ is a homeomorphism onto its image. It remains to check that dxϕ is injective for every x M. Note that ∈ dxϕ = (dxf1, dxλ1, . . . , dxf`, dxλ`), hence to prove that dxϕ is injective it is enough to prove that one of its components is injective. Since every x belongs to some Dk, then dxfk = dxψk is injective because ψk is a diffeomorphism. This concludes the proof. m Pm Q ni 1+ i=1 ni Exercise 52. Prove that the product of spheres M = i=1 S embeds into R .

1.5. The tangent bundle for embedded submanifolds. 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 51 we can always assume it is embedded in some Rn): n n (1.4) EM = (x, v) R R v TxM . { ∈ × | ∈ } Exercise 53. Assuming that M Rn is an embedded submanifold, using the definition (1.4), ⊂ check that EM is a smooth submanifold of Rn Rn. In other words, check that the “equations” n n × v TxM defining EM R R are regular. ∈ ⊂ × Exercise 54. Let ϕ : M Rn be an embedding with image N = ϕ(M). Prove that the map → TM T Rn given by [γ] [ϕ γ] is an embedding with image TN. In particular if M Rn, TM an EM→ can be naturally7→ identified.◦ ⊂

The previous definition is particularly useful because it allows to talk about the “length” of a tangent vector without introducing the notion of Riemannian manifold. Proposition 55. The set E1M = (x, v) EM v = 1 (called the unit tangent bundle) is a smooth submanifold of EM of dimension{ 2∈ dim(M|) k k1. } − Proof. Consider the smooth function ρ : EM R defined by: → ρ(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 E M = ρ− (1) and to see that E M EM is a submanifold it is enough to prove that 1 is a regular value for ρ and then use Theorem⊂ 41. To this end, let (x, v) TM such that v 2 = 1 and consider the curve γ :(, ) TM given by γ(t) = (x, (1 + t)v). Then∈ γ(0) = (x, v) andk k → d d d γ(t) = tv 2 = (1 + t)2 = 2. dt t=0 dtk k t=0 dt t=0 d Hence im(d(x,v)ρ) span d(x,v)ρ γ(t) = R and d(x,v)ρ is surjective. ⊃ { dt t=0 } Remark 56. 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 57 (The group SO(3)). The group SO(3) is defined by: 3 3 T SO(3) = A R × such that AA = 1 and det(A) = 1 . ∈ 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 η : G G, 1 × → → given by η(g) = g− , are both smooth. The smooth structure on SO(3) is given as follows. We 3 3 6 consider the map F : R × Sym(3, R) R given by: → ' F (A) = AAT , and we observe that 1 is a regular value of F (check it!), then we can use Theorem 41 and deduce 1 1 9 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), 14 ANTONIO LERARIO

1 2 Tx S

v x

2 TxS

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 homeomorphism T 1S2 SO(3). 7→ ∧ → hence it is itself a smooth manifold. One important remark: O(3) consists actually of just two components: O(3) = SO(3) ( 1 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 Figure6): → (x, v) (x, v, x v). 7→ ∧ Since T 1S2 is compact, then f is a homeomorphism onto its image.

Exercise 58. Prove that SO(3) is also homeomorphic to RP3. (This implies that T 1S2 RP3.) ' Exercise 59. Using Exercise 58 and Example 57, prove that every vector field on the sphere S2 2 2 2 (i.e. a continuous assignment v : S TS with v(x) TxS , see Section 3.1) must vanish at some point. → ∈ Example 60 (The group SU(2)). The group SU(2) is defined by: z z T SU(2) = A = 1 2 such that AA = 1 and det(A) = 1 . z3 z4 We can endow SU(2) the structure of smooth manifold using again Theorem 41. As a smooth manifold SU(2) is diffeomorphic to the sphere S3 and this can be seen as follows. Observe that T the condition AA = 1 can be written as: z4 z2 1 z4 z2 1 T z1 z3 − = − = A− = A = , z3 z1 det(A) z3 z1 z2 z4 − − 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 LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 15 of the vector space of matrices:

2 2 T T1SU(2) = X C × such that X = X and tr(X) = 0 . { ∈ − } Writing a matrix X su(2) as ∈ iu v iz 3 X = − − , (u, v, z) R , v + iz iu ∈ − − 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 ϕ : SU(2) O(3) (i.e. a group representation) via: → T ϕ(A) = (R : X AXA ). A 7→ The image of ϕ is SO(3) and, recalling that SO(3) RP3 (Exercise 58), we therfeore have: ' 3 3 ϕ : SU(2) S RP SO(3). ' → ' The map ϕ is a group homomorphism with kernel 1 and topologically it is the double cover map. {± } Exercise 61. Try to prove the properties of ϕ : SU(2) SO(3) rigorously (im(ϕ) = SO(3), → ker(ϕ) = 1 , ϕ : S3 RP3 is the covering space map), possibly referring to [2, 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 Qm Let D = [a1, b1] [am, bm] be a cube in R . We will set µ(D) = bi ai . × · · · × i=1 | − | Definition 62 (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 k J (countable means that the cardinality of the index set J is either finite or countable) such{ that} ∈ [ X A D and µ(D ) . ⊂ k k ≤ k J k J ∈ ∈ If M is a smooth manifold, we will say that A M has measure zero if for every chart (U, ψ) for ⊂ M the set ψ(A U) Rm has measure zero. ∩ ⊂ Remark 63. 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: { } ∈ k k − 2k+1 k 2k+1 [ X X Q Dk and µ(Dk) = = . ⊂ 2k+1 k k k ∈N ∈N ∈N Remark 64. 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, ψ) we have that ψ(U V ) Rm is open and nonempty. In particular ψ(U V ) it contains a cube D with µ(D) = δ > 0. If∩ we now⊂ try to cover ψ(U V ) with a countable ∩ P ∩ collection of cubes Dk k J , this collection must cover D and k J µ(Dk) δ. In particular if A M has measure{ zero,} ∈ its complement is dense: let V M be any∈ open set,≥ then V Ac = , otherwise⊂ V would be contained in A. ⊂ ∩ 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, 2π] = qk k and set ∩ { } ∈N 1 B = (cos θ, sin θ) S θ Q [0, 2π] . { ∈ | ∈ ∩ } Then B is dense in S1, but its complement does not have measure zero. (Can you prove this rigorously?)

Exercise 65. For every > 0 construct an open and dense set A Rm of Lebesgue measure smaller than (this exercise requires measure theory). ⊂ S Lemma 66. Let M be a smooth manifold and A M be a subset such that A k J Ak with J countable and with each A M of measure zero;⊂ then A itself has measure zero.⊂ ∈ k ⊂ 16 ANTONIO LERARIO

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 A with cubes D with P µ(D ) k k,j j Jk j Jk k,j ∈ { } ∈ ∈ ≤ 2k+1 (such a cover exists because each Ak has measure zero). Then [ [ X X X A D and µ(D ) . ⊂ k,j k,j ≤ 2k+1 ≤ k J j Jk k J j Jk k J ∈ ∈ ∈ ∈ ∈ Lemma 67. 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 S 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: ∈ [ [ 1 U = B x , r , k k − n k J n ∈ ∈N i.e. U can be written as a countable union of closed (hence compact) balls. Relabeling, we can write: [ U = Bk k ∈N where each B is a closed ball. Let us set A = A B ; observe that A has measure zero (a subset k k ∩ k k of a set of measure zero has itself measure zero). We will prove that f(Ak) has measure zero for S every k N, and the result will follow from Lemma 66, since f(A) = k f(Ak). ∈ ∈N 1 Thus let us fix k. Observe that, since Bk is compact and f is C , then f Bk is Lispchitz and there exists L > 0 such that for all x, y B : | k ∈ k f(x) f(y) L x y . k − k ≤ kk − k In particular, if C Bk has diameter d, then f(C) has diameter at most d Lk; hence there exists C > 0 (which also⊂ depends on m, but this is also fixed) such that if D B· is a cube, then f(D) k ⊂ k is contained in a cube D0 with µ(D0) C µ(D). ≤ k P Fix now > 0 and take a countable cover of Ak with balls Dj j I such that j I µ(Dj) C { } ∈ ∈ ≤ k (this cover exists because we are assuming Ak has measure zero). By the above reasoning each f(D ) is contained in a D0 with µ(D0 ) C µ(D ). In particular: j j j ≤ k j [ X X f(Ak) Dj0 and µ(Dj0 ) Ckµ(Dj) Ck , ⊂ ≤ ≤ · Ck ≤ j I j I j I ∈ ∈ ∈ which proves f(Ak) has measure zero. Remark 68. Observe that the previous Lemma implies that for a set A M to have measure ⊂ zero it is enough to check ψ(A U) Rm has measure zero for all charts (U, ψ) belonging to a given atlas (i.e. we do not have∩ to check⊂ it on all possible charts). Lemma 69 (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. → \ 5 Proof. Let (Uk, ψk) k I be a countable atlas for A and (Vj, ϕj) j J be a countable atlas for B. { } ∈ { } ∈ b It is enough tot check that for every k I and j J the measure of ϕj(f(Uk) Vj) R is zero. a ∈ ∈ ∩ ⊂ Let R be the codomain of ψk and Wk be its image. Consider the projection on the second factor b a b a map p : R − Wk Wk. Then ϕj(f(Uk) Vj) is the image of 0 Wk R − Wk under the C1 map: × → ∩ { } × ⊂ ×

−1 b a p ψ f ψj b φ : 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 67.

5The existence of such countable atlases follows from the assumption that both A and B are paracompact with countably many components, hence second countable LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 17

Example 70. 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 71 (Weak Whitney embedding). Let M be a compact manifold of dimension m. There exists an embedding ϕ : M R2m+1. → Proof. By Theorem 51 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 } ∈ \{ } 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: ∈ \{ } v|M (1) fv M is injective; (2) for| every x M the differential d (f ) is injective. ∈ x v|M The first condition is satisfied if there is no pair (x, y) M M with x = y such that: ∈ × 6 x y v = − . x y k − k We can rephrase this condition as follows. Let ∆ be the diagonal in M M; then fv M is injective × q 1 | if and only if v does not belong to the image of the map h :(M M) ∆ S − given by: 1 × \ → x y h (x, y) = − . 1 x y k − k We apply now Lemma 69 to the smooth function h1. Since q 1 dim(M M) ∆ = 2m < q 1 = dim(S − ) × \ − q 1 q 1 q 1 then im(h1) has measure zero in S − and there is a dense set D1 = S − im(h1) S − of “good” choices for v (as long as we are concerned with injectivity of f ). \ ⊂ v|M The condition on the injectivity of the differential is handled as follows. Observe first that, since q fv is linear, the differential of fv at a point x R equals fv itself: ∈ dxfv = fv. n n n Consequently the kernel of the linear map dxfv : TxR R R equals span v (the line ' → { } spanned by the direction of v, since by definition fv is the projection in the direction of v). Moreover, the differential of f at a point x M equals: v|M ∈ d (f ) = (d f ) x v|M x v |TxM and ker d (f ) = T M span v . Thus the differential of f at a point x M is injective x v|M x ∩ { } v|M ∈ if and only if v / TxM. We can rephrase this condition as follows. Consider the smooth map 1 q ∈1 h : T M S − given by: 2 → 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 now Lemma 69| again, since: 1 q 1 dim(T M) = 2m 1 < q 1 = dim(S − ). − − q 1 As a consequence there is a dense subset D2 = S − im(h2) of “good” choices for v (as long as we are concerned with injectivity of df ). \ v|M 1 Observe that since T M is compact (see Remark 56), then im(h2) is compact and D2 is not only dense, but also open. In particular the intersection D1 D2 is nonempty (in fact dense in q 1 ∩ q 1 S − ) and every v D1 D2 produces a map fv which restricts to an embedding of M R − . ∈ ∩ → Exercise 72. Prove that if M is a compact smooth manifold of dimension m, then there exists an immersion of M in R2m. 18 ANTONIO LERARIO

Exercise 73. Let M be a compact manifold of dimension m and h : M Rq be a smooth map, → q where q 2m + 1. Prove that for every > 0 there exists an embedding h : M R such that: ≥ → max h(x) h(x) . x M k − k ≤ ∈ LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 19

2. Transversality and Sard’s lemma

2.1. Transversality. Deﬁnition 74. Let f : M N be a smooth map between two smooth manifolds and A N be → ⊂ a smooth submanifold. We will say that f is transversal to A and write f t A if for every x M such that f(x) A we have: ∈ ∈ Tf(x)N = Tf(x)A + im(dxf). Moreover, if A, B are submanifolds of a given manifold N, we will say that A and B are transverse and write A t B if the inclusion ιA : A N is transverse to B (or equivalently the inclusion ι : B N is transverse to A), in other words→ for every y A B we have: B → ∈ ∩ TyN = TyA + TyB. (Note that A t B means that, if A and B intersect they do it in a transverse way; in particular two nonintersecting manifolds are transverse.) Exercise 75. Let C Rn be a closed set. Prove that there exists a smooth function f : Rn R whose zero set coincides⊂ with C. → Exercise 76. Let C Rn Rn+1 be a closed set. Prove that there exists a submanifold ⊂ ⊂ A Rn+1 such that A Rn = C (in particular a nontransverse intersection of manifolds can be very⊂ complicated, essentially∩ as complicated as any closed set). Exercise 77. Generalize Exercise 75: prove that if C is a closed set in a smooth manifold M, there exists a smooth function f : M R whose zero set coincides with C. (Hint: use partition of unities.) →

Observe that in the case A = y is a point f t A if and only if y is a regular value of f; 1 { } moreover in this case f − (y) is a submanifold of M by Theorem 41. Next Theorem generalizes this result. Theorem 78. Let f : M N a smooth map between two smooth manifolds and A N be a → 1 ⊂ submanifold such that f t A. Then f − (A) is a submanifold of M of codimension: 1 codimM (f − (A)) = codimN (A).

Proof. Let x M such that f(x) = y A. Since A is a submanifold of N, for every y A there exists a neighborhood∈ U N and a chart∈ (U , ψ) such that: ∈ y ⊂ y 1 a A Uy = ψ− (R ). ∩ n a n 1 n n a Let us write now R = R R − and let us consider the projection π : R R − . Denote by ϕ the composite map ϕ = π× ψ; then: → ◦ 1 A U = ϕ− (0). ∩ y 1 We now consider the open set Wx = f − (Uy) (a neighborhood of x in M) and the map: n a φ = ϕ f : W R − . ◦ → We claim that 0 is a regular value of φ. In fact, since f t A we have: n R = Tψ(y)ψ(A) + im(dx(ψ f)). ◦ Applying dyπ on both sides we obtain: n a n a n R − = Tφ(x)R − = dyπ(R ) = dyπ(Tψ(y)ψ(A)) + im(dxφ) = im(dxφ), where we have used the fact that dyπ(Tψ(y)ψ(A)) = 0 . This proves that 0 is a regular value for 1 1 { } 1 φ and φ− (0) = Wx f − (A) is a submanifold of M. Repeating this for every x f − (A) we 1 ∩ 1 ∈ get a cover of f − (A) with open sets of M such that in each of these open sets f − (A) Wx is a 1 ∩ submanifold of M, hence f − (A) itself is a submanifold of M. 1 1 The dimension of f − (A) equals dim(f − (A) Wx) = dim(M) dim(N) a, hence by Theorem 41 its codimension is: ∩ − − 1 1 codim (f − (A)) = dim(M) dim(f − (A)) = dim(N) a = codim (A). M − − N 20 ANTONIO LERARIO

2.2. Sard’s Lemma. Definition 79. Let f : M N be a Ck function, k 1. We say that x M is a critical point → ≥ ∈ for f is the differential dxf : TxM Tf(x)N is not surjective. We will denote by Crit(f) M the set of critical points of f.A critical→ value is a point y N which is the image of a critical⊂ point. ∈ We will need the following Lemma, whose proof is easy and is left as an exercise. n 1 n Lemma 80. Let I R and K I R − R be a closed set. Assume that for every t I the ⊂ n 1 ⊂ × ⊂ n 1 n 1 ∈ set Kt = K ( t R − ) has measure zero in t R − R − . Then K has measure zero in ∩ { } × { } × ' Rn. We will now prove the following result, which is fundamental in differential topology; the proof is taken from [9, Chapter 3]. Theorem 81 (Sard’s Lemma). Let f : M N be a smooth map. Then f(Crit(f)) (i.e. the set of critical values of f) has measure zero in N→. In particular the set of regular values of f is dense in N (and open if M is compact). Remark 82. The Theorem is not true if f is merely Ck with k < : in this case, to make it true, one has to add the requirement that k > max 0, m n . ∞ { − } Proof. First, let us observe that we can reduce to study the case of a map f : U Rn, where U is n 6 → an open subset of R . In fact we can find countable atlases (Uj, ψj) j of M and (Vj, ϕj) j of { } ∈N { } ∈N N such that for every j N we have f(Uj) Vj. Then, because we are only dealing with countably ∈ ⊂ many sets, thanks to Lemma 66 it will be enough to prove that f(Crit(f Uj )) has measure zero for 1 | every j N. But now observe that Crit(f Uj ) = Crit(f ψj− ) (because ψj is a diffeomorphism); ∈ 1 | 1 ◦ 1 moreover ϕj(Crit(f ψj− )) = Crit(ϕj f ψj− ) and if we prove the statement for ϕj f ψj− , ◦ 1 ◦ ◦ ◦ ◦ then it will also be proved for f ψ− (since ϕ is a diffeomorphism, it sends sets of measure zero ◦ j j to sets of measure zero). The proof is by induction on m, the case m = 0 being clearly true. Let us assume we have n m 1 n proved it for all function f : U R with U R − and let’s prove it for a function f : U R → ⊂ → with U Rm. ⊂ We denote for simplicity by C the set of critical points of f and for every k 1 let us define: ≥ C = x U all partial derivatives of f of order up to k vanish at x . k { ∈ | } Note that each Ck is closed (it is given by the vanishing of finitely many continuous functions, the partial derivatives of f up to order k) and:

C C1 C2 C3 ... ⊃ ⊃ ⊃ 1 Note that this chain of inclusions is potentially inﬁnite, for example for the function f(x) = e− x2 the origin belongs to C for every k 1. The proof consists now of three steps. k ≥ (1) First we show that for k large enough the set f(Ck) has measure zero. (2) Then we show that f(C C1) has measure zero. (3) Finally we prove that for\ every k 1 the set f(C C ) has measure zero. ≥ k\ k+1 From this the conclusion follows, since:

f(C) = f(C C1) f(C1 C2) f(Ck 1 Ck) f(Ck), \ ∪ \ ∪ · · · ∪ − \ ∪ and each piece has zero measure by the above three steps. Let us prove the ﬁrst step. Let S U be a cube whose sides have length δ > 0 (a ﬁxed constant). ⊂ The set Ck can be covered by countably many such cubes, again by Lemma 66 it is enough to prove that f(Ck S) has measure zero. If x Ck all partial derivatives of f up to order k vanish at x and we can∩ write: ∈ (2.1) f(x + h) = f(x) + h k+1R(x, h) k k where R(x, h) is a continuous function, whose absolute value is consequently bounded by a constant a > 0 on S (which is compact).

6The existence of such countable atlases again follows from the assumption that manifolds are paracompact with countably many components, hence second countable LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 21

m m Let us now cover S R with r many cubes of size δ/r. If S1 is a cube of this subdivision containing a point x ⊂C , then every point in S is of the form x + h with h √mδ/r. ∈ k 1 k k ≤ Consequently (2.1) implies that f(S1) is contained in a cube centered at f(x) with sides of length k+1 b = 2a(√mδ/r) (this is just a constant). It follows that f(Ck S) is contained in the union of at most rm cubes whose total volume v is bounded by: ∩ n m b n m (k+1)n v r = b r − . ≤ rk+1 n m (k+1)n If k is large enough (say k m/n) then b r − 0 as r and for every > 0 the set ≥ → → ∞ f(Ck S) is contained in the union of ﬁnitely many cubes of total volume smaller than , i.e. it is has measure∩ zero. This proves point (1) above.

Let us now prove point (2). Let x C C1: we will ﬁnd an open neighborhood V of x such that f(V (C C )) has measure zero.∈ Since\ C C can be covered by countably many such ∩ \ 1 \ 1 neighborhhoods (because Rm is second countable), this will prove point (2) (again because of Lemma 66). ∂f Since x / C1, then there is at least one partial derivative, say , which is nonzero at x. Let ∈ ∂x1 us consider the map h : Rm Rm given by: → h(x) = (f1(x), x2, . . . , xm). Since ∂f (x) = 0 this map is a diﬀeomorphism between two open sets V (a neighborhood of x) ∂x1 m6 and V 0 R . We will assume V 0 is of the form V 0 = I V 00, with I R. The critical values of ⊂ 1 × ⊂ f V are the same of g = f h− . We denote by (g1, . . . , gn) the components of g. This map, by construction,| can be written◦ as:

g(t, x2, . . . , xn) = (t, g2(t, x), . . . , gn(t, x)). m 1 n 1 In particular for every t I we obtain a smooth map gt : R − R − : ∈ → gt(x2, . . . , xm) = (g2(t, x), . . . , gn(t, x)).

Note that a point x V 00 is critical for gt if and only if (x, t) is critical for g: in fact the Jacobian matrix of g has the form:∈  1 0 0  ···   Jg(t, x) =  ∗  .  .   . Jgt(x)  ∗ In particular rk(Jg(t, x)) = rk(Jgt(x))+1 and g is a submersion at (t, x) if and only if rk(Jg(t, x)) = n, i.e. if and only if rk(Jgt(x)) = n 1, which is equivalent to gt being a submersion at x. By the − m 1 n 1 inductive hypothesis, the measure of the set of critical values of gt : V 00 R − R − is zero. ⊂ n 1 → Let us denote now by K the set of critical values of g and by Kt t R − the set of critical S ⊂ { } × values of gt. Then K = t I Kt, and since we have proved that for every t I the measure of ∈ ∈ Kt is zero, by Lemma 80 K itself has measure zero. Note that we can apply Lemma 80 because, up to shrinking the neighborhood V 0, we can assume K is closed: critical points are closed (they are deﬁned by the vanishing of continuous functions), and intersected with a small enough ball are compact; hence the image of their intersection with a small enough ball under a continuous function is compact, hence closed. Since C C1 can be covered by countably many open sets like V , then together with Lemma 66 this proves\ (2).

Let us finally prove point (3). Observe that Ck is defined by the vanishing of all partial derivatives of f up to order k. If x Ck Ck+1, then there exists one such partial derivatives, call it u, ∂u ∈ \ such that u C 0 but (x) = 0. We argue now in a similar way as in the previous point and | k ≡ ∂x1 6 define the diffeomorphism h : V V 0 (V is a neighborhood of x): → h(x) = (u(x), x2, . . . , xm). m 1 1 m 1 Since h(Ck V ) 0 R − , the map g = f h− has all its critical points on 0 R − . ∩ ⊂ { } × ◦ { } × Consider now g = g 0 m−1 )). Each point in h(Ck V ) is critical point for g; by the inductive |{ }×R ∩ hypothesis, g(Crit(g)) has measure zero and consequently also f((Ck Ck+1) V ) has measure zero. Since C C can be covered by countably many open sets like V\, then the∩ result follows from k\ k+1 Lemma 66. 22 ANTONIO LERARIO

2.3. Parametric transversality. Theorem 83 (Parametric transversality). Let F : V M N be a smooth map between smooth manifolds and A N be a submanifold such that F× A.→ For every v V consider the map ⊂ t ∈ fv : M N deﬁned by fv(x) = F (v, x). Then the set v V fv t A is dense in V (in fact its complement→ has zero measure). { ∈ | }

The meaning of this theorem is the following: if we interpret F : V M N as a family of × → maps parametrized by V , under the assumption that F t A, “most” of the maps fv are transverse to A. Remark 84. If a map F : M N is a submersion, then it is automatically transverse to all submanifolds A, N. That is why→ sometimes in order to verify that the conditions of the previous theorem are satisﬁed→ it is enough to check that F is a submersion.

1 Proof. Since F t A then W = F − (A) is a submanifold of V M by Theorem 78. Let p1 : V M V the projection onto the ﬁrst factor and π = p . We× claim that if v is a regular value × → 1|W of π then fv is transversal to A. To this end observe that if v is a regular value of π, then: 1 (2.2) T (V M) = T M + T (F − (A)). (v,x) × x (v,x) We now write:

Tfv (x)N = TF (v,x)N = T A + d F (T (V M)) (since F A) F (v,x) (v,x) (v,x) × t 1 = TF (v,x)A + d(v,x)F (TxM) + d(v,x)(T(v,x)(F − (A))) (because of (2.2)) 1 = T A + d F (T M) (because d (T (F − (A))) T A) F (v,x) (v,x) x (v,x) (v,x) ⊂ F (v,x) = Tfv (x)A + dxfv(TxM), which is the condition fv t A. The result follows now from Sard’s Lemma. Exercise 85. Using the notation of the previous proof prove that indeed v is a regular value of π if and only if fv t A. Exercise 86. Let A, B be two submanifolds of Rn. For every w Rn define the translate manifold ∈ Aw = a + w a A (the translation of A in the direction of w). Prove that there exists a dense { n | ∈ } set D R such that for every w D the manifolds Aw and B are transverse to each other. ⊂ ∈ Exercise 87. Generalizing the previous exercise, let G be a Lie group and A, B be submanifolds. For every g G define Ag = Lg(A) (the left translation of A in the direction of g). Prove that the set of g G∈such that A B is dense. ∈ g t Exercise 88. Let f : Rn Rn be a smooth function and K Rn be a compact set. Assume → ⊂ n n that n > 1. Prove that for every > 0 there exists a smooth function f : R R such that the → differential dxf is never zero (as a linear operator) and supx K f(x) f(x) . ∈ k − k ≤ Exercise 89. Let f : Rn Rm be a smooth function and K Rn be a compact set. Assume that → ⊂ n m m 2n. Prove that for every > 0 there exists an immersion f : R R (i.e. the differential ≥ n → dxf has rank n at every point x R ) such that supx K f(x) f(x) . ∈ ∈ k − k ≤ Exercise 90. A knot is a topological embedding ϕ : S1 R3 (if ϕ is smooth, we call it a smooth → knot). Let R2 R3 be a linear subspace and ϕ : S1 R3 be a smooth knot with image K (which, ⊂ 3 →3 2 3 2 in particular, is a submanifold of R ). For every v R R denote by πv : R R the orthogonal ∈ \ 3 2 → projection in the direction of v. Prove that there exists v R R such that: (1) πv K is an immersion; (2) every z π (K) has at most two preimages (i.e.∈∈ there\ are no triple-points)| and (3) ∈ v given z πv(K) with two preimages x1, x2, the images of the differentials dx1 (πv K ) and dx2 (πv K ) are transverse.∈ | | LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 23

Figure 7. The Moebius band has a structure of a vector bundle E S1. The 1 → tautological bundle τ1,2 RP and the Moebius bundle are equivalent. →

3. Vector bundles

Deﬁnition 91 (Vector bundle). A vector bundle of rank k is a triple (π, E, M) where E and M are smooth manifolds and π : E M −→ is a smooth map such that:

S 1 k (1) there is an open cover M = α A Vα and diﬀeomorphisms ψα : π− (Vα) Vα R such that the following diagram is commutative:∈ → ×

1 ψα π− (Vα) Vα R ×

π p1

Vα

(p1 is the projection on the first factor). The family (Uα, ψα) α A is called a vector bundle { } ∈ atlas, and each ψα is called a trivialization. k (2) whenever Vα Vβ = , there is a continuous map gαβ : Vα Vβ GL(R ) such that the 1∩ 6 ∅ k k ∩ → map ψα ψ− :(Vα Vβ) R (Vα Vβ) R is given by: ◦ β ∩ × → ∩ × (x, v) (x, g (x) v). 7→ αβ · 1 The fiber over x M is the vector space Ex = π− (x). Given X M the restriction of E is ∈ 1 ⊂ the vector bundle (π −1 , π− (X),X) (the vector space structure is well defined by condition |π (X) (2) above). The family gαβ α,β A is called the cocycle of the bundle and satisfies the interesting properties: { } ∈ (3.1) g (x)g (x) = g (x) x V V V , g (x) = 1. αβ βγ αγ ∀ ∈ α ∩ β ∩ γ αα Example 92 (The trivial bundle). The trivial rank-k bundle on M is simply E = M Rk with × π : M Rk M given by the projection on the first factor (in other words there is a single bundle × → trivialization: Vα = M and ψα = idM ).

1 Example 93 (The Moebius band). The Moebius band is the vector bundle on S = U1 U2 (the ` ∪ open cover deﬁned in Example7) constructed as follows. We have U1 U2 = I1 I2 (the disjoint union of two intervals) and the cocycle for the Moebius band is: ∩ g 1 and g 1. 12|I1 ≡ 12|I2 ≡ −

Example 94 (How to construct a vector bundle). Let us consider an open cover U = Uα α A of a smooth manifold M and for every α, β A such that U U assume we are given a{ smooth} ∈ ∈ α ∩ β 24 ANTONIO LERARIO

k function gαβ : Uα Uβ GL(R ) such that the family gαβ satisfies properties (3.1). Then we ∩ → { } can construct a vector bundle E with cocylce gαβ by setting: { } ! a k E = Uα R / , × ∼ α A ∈ where (x1, v1)α1 (x2, v2)α2 if and only if x1 = x2 and v1 = gα1α2 (x)v2. We endow E with the quotient topology∼ from the defining equivalence relation “ ” and the projection map π : E M is simply given by [(x, v)] x. ∼ → 7→ Example 95 (The tautological bundle). The tautological bundle τk,n is a vector bundle on the Grassmnannian G(k, n) which, as a topological space, is defined by: n k τk,n = (W, x) G(k, n) R x W R . { ∈ × | ∈ ' } n The projection π : τk,n G(k, n) equals the restriction to τk,n G(k, n) R of the projection → ⊂ × on the first factor G(k, n) Rn G(k, n). × → n n Exercise 96. Let RP = G(1, n+1) with the open cover U = Uj j=0 as in Example8. Prove that { } 1 τk,n Uj is trivial and find the cocycle of τk,n associated to the open cover U. Prove that τ1,2 RP is a| Moebius band (see figure7). → 2 1 Exercise 97. Let f : RP [1 : 0 : 0] RP be the smooth map: \{ } → f([x0 : x1 : x2]) = [x1 : x2]. 2 1 Prove that RP [1 : 0 : 0] has the structure of a line bundle over RP with projection f. Is it a trivial bundle? \{ } Exercise 98. Let Sym(2, R) R3 be the space of symmetric matrices and let Γ Sym(2, R) be the set: ' ⊂ x y Γ = A = such that det(A) = 1, tr(A) = 0 . y z − Prove first that Γ is a smooth submanifold of Sym(2, R) which is diffeomorphic to S1. Consider also the set: 2 E = (A, v) Γ R such that Av = v . { ∈ × } Prove that E has the structure of a line bundle over Γ with projection p :(A, v) A. Is it a trivial bundle? 7→ Hint: Observe the two identities: cos θ sin θ 1 1 π = , θ = sin θ cos θ sin θ sin θ 6 2 − 1+cos θ 1+cos θ cos θ sin θ sin θ sin θ 1 cos θ = 1 cos θ , θ = 0. sin θ cos θ −1 −1 6 − Definition 99 (Morphism of vector bundles). Let π : E M and π : E M be two vector 1 1 → 1 2 2 → 2 bundles. A morphism of vector bundles between E and E is a smooth map f : E E which 1 2 1 → 2 is linear on every fiber and which covers a smooth map f : M1 M2 (i.e. such that the following diagram is commutative): →

f E1 E2

π1 π2 f M1 M2

If the linear map f (E1)x :(E1)x (E2)f(x) is injective for every x M1, then f is called a monomoprhims; if it| is surjective for→ every x M , it is called an epimorphism.∈ If for every ∈ 1 x M1 the linear map f (E1)x :(E1)x (E2)f(x) is an isomorphism, then f is called a bundle map∈ . An equivalence of vector| bundles→ is a bundle map which covers a homeomorphism and an isomorphism of vector bundles is a bundle map which covers the identity. A rank-k vector bundle π : E M is called trivial if it is isomorphic to the trivial bundle → M Rk. An isomorphism E M Rk is called a trivialization of E. If the tangent bundle TM of a× smooth manifold is trivial,→M is× said parallelizable. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 25

Exercise 100. Prove that Lie groups are parallelizable. (In particular, for example, since both RP3 and S3 are Lie groups, then they are parallelizable.) Example 101 (Dual bundle). Given a vector bundle E π M which trivializes over an open E −→ p cover U = Uα α A of M and with cocycle gαβ , the dual bundle E∗ M is constructed as in { } ∈ ∗ { } −→ Example 94 by taking the cocyle gE defined for x U U by: { αβ } ∈ α ∩ β E∗ 1 T k g (x) = (gαβ(x)− ) GL(R ). αβ ∈ When E = TM, the dual bundle is denoted by T ∗M and called the cotangent bundle. Example 102 (Determinant bundle). Given a vector bundle E π M which trivializes over an E −→ p open cover U = Uα α A of M and with cocycle gαβ , the determinant bundle det(E) M is { } ∈ { } −→ the line bundle constructed as in Example 94 by taking the cocyle gdet(E) defined for x U U { αβ } ∈ α ∩ β by: det(E) g (x) = det(gαβ(x)) GL(R). αβ ∈ π k1 1 Example 103 (Direct sum of vector bundles). Given two vector bundles R , E1 M and π k2 2 7 → −→ R , E2 M which trivialize over the same open cover U = Uα α A of M and with cocycles → −→ { } ∈ gE1 and gE2 , the direct sum of E and E is the vector bundle E E M constructed as { αβ} { αβ} 1 2 1 ⊕ 2 → in Example 94 with the cocycle defined for x U U by: ∈ α ∩ β E1 ! E1 E2 gαβ(x) 0 k1+k2 gαβ⊕ (x) = E2 GL(R ). 0 gαβ(x) ∈

3.1. Sections of vector bundles. Deﬁnition 104 (Section of a vector bundle). Let E π M be a vector bundle. A section of E is a smooth map s : M E such that π(s(x)) = x for−→ every x M, i.e. such that the following diagram is commutative:→ ∈ E

π s

M The zero section is the section that associates to every point x M the zero vector in E . ∈ x Exercise 105. Let Rk , E M be a rank-one bundle. Prove that E is trivial if and only if admits a nowhere vanishing→ section→ s : M E. → Remark 106 (How to build a section of a vector bundle). Let E π M be a rank-k vector bundle −→ k and U = Uα α A be an open cover of M such that ψα : E Uα Uα R for every α A. Let also { } ∈ | ' × ∈ gαβ be the corresponding cocyle. Given a section s : M E, for every α A we can consider { } k → ∈ the map ψα s : Uα Uα R (the section in the trivialization), which has the following form: ◦ → × (3.2) ψα(s(x)) = (x, σα(x)), k for some smooth function σα : Uα R . Using the definition of the cocyle, we see that the family of maps σ satisfies the relation:→ { α} (3.3) σ (x) = g (x)σ (x) x U U . α αβ β ∀ ∈ α ∩ β k Conversely, given a family of smooth functions σα : Uα R such that (3.3) is satisfied, this family defines a section of E (i.e. the unique{ section which→ in} in the trivializaztions has the expression (3.2)). n Example 107 (Polynomials and vector bundles over projective spaces). Let U = Uj j=0 be the n { } open cover Uj = xj = 0 of RP . For every d Z consider the cocycle: { 6 } ∈ d xj gij([x0, . . . , xn]) = , xi, xj = 0. xi 6

7Indeed, by Corollary 114 there is an open cover of M such that every vector bundle trivializes over this cover. 26 ANTONIO LERARIO

n We denote by ORP (d) the vector bundle constructed with the procedure of Example 94 corresponding to this cocyle. An interesting fact is that, when d 0, every homogeneous polynomial ≥ n of degree d deﬁnes in a natural way a section of ORP (d), as follows. Given p R[x0, . . . , xn](d) n ∈ deﬁne the family of functions (σp)j : Uj R by { → }j=0 x0 xj xn (σp)j([x0, . . . , xn]) = p ,..., ,..., . xj xj xj Observe now that for every i, j and [x] U U we have: ∈ i ∩ j d xj x0 xj xn gij([x])(σp)j([x]) = p ,..., ,..., xi xj xj xj x x = p 0 ,..., n xi xi

= (σp)i([x]) (where in the second equality we have used the fact that p is homogeneous of degree d). In particular, following the discussion of Remark 106, we see that the family (σ ) n deﬁnes a section { p j}j=0 n σp of ORP (d). (Note also that the map p σp is linear.) The procedure of de-homogenization of a polynomial with respect to a variable can7→ be geometrically described as the procedure of looking at the polynomial as a section in the corresponding trivialization!

3.2. Metrics on vector bundles. π Definition 108 (Metric on a vector bundle). Let Rk , E M be a vector bundle. A smooth → −→ metric on E is a family hx x X where each hx is a symmetric, bilinear, positive definite quadratic { } ∈ form on Ex such that the map (x, y, z) hx(y, z) defined on (x, y, z) M E E p(y) = x = p(z) is smooth. (A smooth metric on TM7→ is called a Riemannian{ metric.)∈ × × | } Proposition 109. Every vector bundles admits a smooth metric.

k Proof. Let U = (Uα, ψα) α A be a vector bundle atlas and call p1 : Uα R Uα and p2 : k k { } ∈ × → Uα R R the projections on the two factors. Let ρα α A be a partition of unity subordinated × → { } ∈ to U (see Proposition 193) and denote by , the standard scalar product on Rk. For every α A and (x, w , w ) M E E such thath· ·i π(w ) = x = p(w ), define: ∈ 1 2 ∈ × |Uα × Uα 1 2 h (w , w ) = ρ (x) p (ψ (w )), p (ψ (w )) . α,x 1 2 α h 2 α 1 2 α 2 i Since the support of ρα is contained in Uα, this function extends to a smooth function on the set of (x, y, z) M E E such that π(y) = x = p(z) (when (x, y, z) / M E Uα EUα then h (y, z) =∈ 0). The× metric× h is defined by: ∈ × | × α,x { x} X hx(y, z) = ρα(x)hα,x(y, z). α The fact that the family is smooth is clear by construction (it is the sum of smooth maps); the fact that for every x the quadratic form hx : Ex Ex is positive definite follows from the fact that each ρ 0 and that for every x M there is at× least one α A with ρ (x) = 0. α ≥ ∈ ∈ α 6 Exercise 110. Let π : E M be a smooth vector bundle endowed with a smooth metric and E E be a sub-bundle. Define→ the orthogonal complement bundle to E in E by: 1 ⊆ 1 E⊥ = (x, v) X E such that π(v) = x and v (E )⊥ 1 { ∈ × ∈ 1 x } (where the orthogonal complement is taken with respect to the metric in Ex), together with the projection p = p1 E⊥ . Prove that p : E1⊥ M is a smooth vector bundle of rank rk(E1⊥) = | 1 → rk(E) rk(E ) and that E E E⊥. − 1 ' 1 ⊕ 1 3.3. Parametric transversality for sections of vector bundles. Partition of unity can be used to produce “generic” sections of a vector bundle, as follows. π Theorem 111. Let Rk , E M be a vector bundle, s : M E be a smooth section and Z be the image of the zero→ section.−→ We assume that E is endowed→ with a metric and we denote by v the norm of an element v E . For every > 0 there exists a section s : M E such that k k ∈ x → s t Z and supx M s(x) s(x) < . ∈ k − k LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 27

` Proof. Let (Uj, ψj) j=1 be a finite vector bundle atlas (such a finite atlas exists because M { } ` is assumed to be compact) and ρj j=1 be a partition of unity subordinated to U. Given the k{ } k trivialization ψj : E Uj Uj R we define the smooth map φj : M R E by: | → × × → 1 ρj(x)ψj− (x, v) if x Uj φj(x, v) = ∈ the zero element of Ex otherwise k ` Using the family of maps φj we define the evaluation map Fs : R × M E by: { } × → ` X Fs(v1, . . . , v`, x) = s(x) + φj(x, vj). j=1 k ` Observe first that for every p = (v1, . . . , v`) R × the map x F (p, x) is a section of E: ∈ 7→  `  X π(Fs(v1, . . . , v`, x)) = π s(x) + φ(x, vj) = x. j=1

We claim that Fs t Z. Since ψj is a diﬀeomorphism, it is enough to check that for every j = 1, . . . , ` the map ψ F is transversal to U 0 . To this end we write j ◦ α × { }  `  X ψj(Fs(v1, . . . , v`, x)) = x, σ(x) + ρj(x)vj , j=1 from which we see that for every w Rk ∈ d(0,...,0,x)Fs(w, . . . , w, 0) = (0, w), and consequently (ψj Fs) t Uj 0 . The result follows now from Theorem 83: the set of ◦ × { } k ` p = (v1, . . . , v`) such that x Fs(v1, . . . , v`, x) is transversal to Z is dense in R × . Choose now k ` 7→ 1 a sequence pn n R × such that Fs(pn, ) t Z and pn . Since Fs(pn, ) n converges { } ∈N ⊂ · k k ≤ n { · } ∈N uniformly to s, it follows that there exists n large enough such that Fs(pn, x) s(x) for all x M. k − k ≤ ∈

3.4. The classification theorem. Definition 112 (Pull-back bundle). Let π : E Y be a smooth vector bundle and f : X Y be a smooth map. We define the total space of a→ new vector bundle, called the pull-back bundle→ :

f ∗E = (x, e) X E f(x) = π(e) . { ∈ × | } Denoting by p : X E X the projection on the ﬁrst factor, the projection map for f ∗E is 1 × → p = p ∗ : f ∗E X. 1|f E → k k If (Uα, ψα) α A is a vector bundle atlas for E and denoting by p2 : Uα R R the projection { } ∈ 1 × → on the second factor, the vector bundle atlas for f ∗E is (f − (Uα), ϕα) α A, where: { } ∈ 1 ϕ (x, e) = (p , p ψ )(x, e) = (x, p (ψ (e))) (x, e) f − (U ) E α 1 2 ◦ α 2 α ∀ ∈ α × |Uα The cocyle of the pull-back bundle f ∗E is the family 1 1 k −1 −1 hαβ : E f (Uα) f (Uβ ) f − (Uα) f − (Uβ) R | ∩ → ∩ × given by hαβ(x) = gαβ(f(x)). Theorem 113. Let X be a smooth manifold and π : E X I be a vector bundle8. Then → × E X 0 and E X 1 are isomorphic vector bundles. | ×{ } | ×{ } Proof. The proof is based on the following two facts.

8Strictly speaking so fare we have only defined vector bundles when the base space is a smooth manifold, and in this Theorem we are instead taking M × I. This is easily fixed by extending the notion of smooth manifold to include manifolds with boundary, see Definition 186. 28 ANTONIO LERARIO

(1) A vector bundle π : E X [a, b] is trivial if there exists c [a, b] such that both the → × ∈ restriction E X [a,c] and E X [c,b] are trivial. In fact, given trivializations: | × | × k k ψ1 : E X [a,c] X [a, c] R and ψ2 : E X [c,b] X [c, b] R , | × → × × | × → × × k we can replace ψ2 with the trivialization ψ3 : E X [c,b] X [c, b] R defined by | × → × × composing ψ2 with the map: 1 (x, t, v) (x, t, ψ (ψ )− (x, c, v)). 7→ 1 2 (2) Given a vector bundle E X I, there exists an open cover Vα α A of X such that → × { } ∈ E Vα I is trivial for every α A. In fact, for every x X and for every t [0, 1] we| can× find an open neighborhood∈ U of (x, t) X ∈I such that E is∈ trivial. x,t ∈ × |Ux,t Using the compactness of [0, 1], we can find open neighborhoods Ux,1,...,Ux,m of x and

0 = t0 < t1 < < tm = 1 such that E Ux,j [tj− ,tj ] is trivial for every j = 1, . . . , m. The ··· | × 1 previous fact (1) implies that, after setting Ux = Ux,1 Ux,m, then E Ux [0,1] is trivial. ∩· · ·∩ | × Now the proof of the Theorem. In order to make the exposition more clear, let us first consider the case when X is compact. Applying fact (2) above, we can find an open cover Uα α A of X { } ∈ k such that E Uα I is trivial. By compactness of X we can extract a finite open cover U = Uj j=1 | × k { } of X such that E Uj I is trivial. We consider now a partition of unity ρj j=1 subordinated to U and we define for|j =× 1, . . . , k: { } θ = ρ + + ρ . j 1 ··· j Let Γ X [0, 1] be the graph of θ and E = E . We define for every j = 1, . . . , k a bundle j ⊂ × j j |Γj isomorphism hj : Ej Ej 1 as follows. Observe that → − c c Γj (Uj I) = Γj 1 (Uj I), ∩ × − ∩ × 1 we therefore set the map hj to be the identity outside π− (Uj I). Since over Uj I the bundle k × × is trivial, we have E Uj I (Uj I) R and we can set: | × ' × × hj(x, θj(x), v) = (x, θj 1(x), v) (x, θj(x), v) Ej. − ∀ ∈ The composition h1h2 hk gives therefore an isomorphism between E X 1 and E X 0 . ··· | ×{ } | ×{ }

When X is only paracompact, we still get an open cover U = Uα α A such that E Uα I is { } ∈ | × trivial. Using paracompactness we refine U to a countable, locally finite V = Vj j ; for every { } ∈N j N, since Vj Uα for some α A, we still get that E Vj I is trivial. Let now ρj j be a ∈ ⊂ ∈ | × { } ∈N partition of unity subordinated to V and for every j N define as before: ∈ θj = ρ1 + ... + ρj.

Again as before for every j N we get an isomorphism hj : Ej Ej 1, where Ej = E graph(θj ). Now the delicate point would∈ be the infinite composition h h →, however− near every point| x X 1 2 ··· ∈ only finitely many ρj’s are non zero, so there is a neighborhood where all but finitely many hj’s are the identity. Hence h is a well defined vector bundle isomorphism between E X 1 and E X 0 . | ×{ } | ×{ }

An immediate consequence, which will be much useful is the following. Corollary 114. Every vector bundle E over a contractible manifold M is trivial.

Proof. Since M is contractible, it means that the identity map f0 = idM is homotopic to a constant map f x . But then by the previous theorem: 1 ≡ 1 k E = id∗E f ∗E = M R . ' 1 × Proposition 115. Let π : E M be a vector bundle. Then there exists another vector bundle → π : Eb M such that: b → ` E Eb M R . ⊕ ' × LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 29

Proof. We only prove the statement under the additional assumption that M is compact – the proof is similar to the proof of Theorem 51. n Let U = Uj j=1 be an open cover of M such that for every j = 1, . . . , n we have: (1) E Uj is { } m | trivial; (2) there exists manifold charts ϕj : Uj R with ϕj(Uj) = B(0, 1) and M is covered by 1 1 n n → ϕ− (B(0, )) . Let also ρ be a family of bump functions with ρ supported on U . { j 2 }j=1 { j}j=1 j j For every j = 1, . . . , n define the map: m k φj : E Uj R R | → × which is the composition of the trivialization over U and the map ϕ id k . Out of these maps j j × R we build φ : E Rn(m+k+1) defined by: 7→ φ = ((ρ π)φ , ρ π, . . . , (ρ π)φ , ρ π). 1 ◦ 1 1 ◦ n ◦ n n ◦ It is easy to see that φ : E R` is an embedding (we have set ` = n(m+k+1)), and we can identify 9 →` ` ` E with a submanifold of R . Consequently TE M E TM is a sub-bundle of M R = T R M | ' ⊕ × | and, after endowing T R` with a riemannian metric, we can finally set:

Eb = (TM E)⊥ TM. ⊕ ⊕ (Here the orthgonal complement bundle (TM E)⊥ is deﬁned as in Exercise 110). ⊕ Theorem 116. Let p : τ G(k, `) be the tautological vector bundle. k,` → (1) Given a rank-k vector bundle π : E M there exist ` > 0 and a smooth map f : M → → G(k, `) such that E = f ∗τk,`. (2) Let E E be two isomorphic rank-k vector bundles over M. Then, by point (1), there 0 ' 1 exist smooth maps f : M G(k, ` ) and f : M G(k, ` ) such that E = f ∗τ and 0 → 0 1 → 1 0 0 k,`0 E1 = f1∗τk,`1 . There exists ` > `0, `1 such that the two maps f0, f1 become homotopic after composition with the natural inclusions G(k, ` ) , G(k, `). i →

Proof. The proof of point (1) uses Proposition 115: let Eb such that E Eb M R`. Then E can be viewed as a sub-bundle of a trivial bundle and the map f : M ⊕G(k,' `) is simply:× → k ` f(x) = Ex R R . ' ⊂

ì For the second point we proceed as follows. For i = 0, 1, let gi : E R be the map ì ì → Ei M R R (a linear injection on every fiber). Set ` = 2 max `0, `1 = 2`2 and → × → `0 `2 `1 `2 { } consider the two inclusions L0 : R , R and L1 : R , R as the first coordinates. Let now → → jeven : R`0 R` be the homotopy: t → jeven(x , . . . , x , 0,..., 0) = (1 t)(x , . . . , x , 0,..., 0) + t(0, x ,..., 0, x ). t 1 `2 − 1 `0 1 `2 odd even odd Defining similarly jt to be the homotopy to the odd coordinates, we see that both jt and jt are homotopies of linear injections. even odd ` Let nowg ˜0 = j1 g0 andg ˜1 = j1 g1. Then for i = 0, 1 we have thatg ˜i : Ei R is a linear injection such that◦ the map ◦ → 1 f˜ : x g˜ (π− (x)) G(k, `) i 7→ i i ∈ ˜ ˜ has the property Ei = fi∗τk,`. Moreover, for i = 0, 1 we have that fi : M G(k, `) is homotopic f → to M i G(k, ` ) , G(k, `). The final homotopy is simply: −→ i → f˜ = (1 t)f˜ + tf˜ . t − 0 1 Exercise 117. Prove that, up to isomorphism, there are only two line bundles on S1 (the tautological τ1,2 and the trivial bundle).

9When M is not compact, one can use the stronger embedding theorem [7, Theorem 2.14] and directly embed ` φ : E → R . From this point the proofs, both in the compact and the non-compact case, are identical. 30 ANTONIO LERARIO

Remark 118. It is possible to provide a uniform estimate on the ` from point (2) of the previous Theorem, i.e. it is possible to prove that there exists ` > 0 (which depends on k and m = dim(M)) such that if for two maps f , f : M G(k, `) we have f ∗τ f ∗τ , then the two maps f 0 1 → 0 k,` ' 1 k,` 0 and f1 are homotopic. We give here a sketch of the proof for the case k = 1; this proof is missing many details, which I warmly encourage the reader to ﬁll-in, and has the only scope of giving a ﬂavor of how nicely the techniques we have introduced so far combine together. We claim that if m+1 f0, f1 : M G(1, m + 2) RP pullback homotopic bundles, then they are homotopic. First, following the→ proof of the' previous Theorem 116, if we compose these maps with the inclusion j : G(1, m + 2) RPm+1 , G(m, 2m + 4) RP2m+3, then they become homotopic: ' → ' h = j f j f = h . 0 ◦ 0 ∼ ◦ 1 1 2m+3 Let us call H : M I RP the (smooth, by construction) homotopy between h0 = H( , 0) and × → b a · a b 1 h1 = H( , 1). We use now the fact that if RP RP is a subspace which is disjont from RP − − , ·a b a b⊂1 then RP RP deformation retracts to RP − − . We apply this with the choice a = 2m + 3 and \ b = m + 1 (we call B RPm+1 = RPb a complementary subspace in RP2m+3 to our original ' RPm+1): this says that there exists a retraction 2m+3 m+1 m+1 r : RP RP RP , \ → which is a homotopy equivalence. Now we apply Theorem 83 (Parametric Transversality) for the map F : O(2m + 4) RP2m+3 RP2m+3 given by: × → F (g, [v]) = [gv].

It is not difficult to verify that F is a submersion and consequently that there are elements g arbitrarily close to the identity 1 O(2m + 4) such that the map ∈ g H :(x, t) [g H(x, t)] ◦ 7→ m is transversal to B. Since codimRP2 +3 (B) = m + 2 and dim(M I) = m + 1, transversality means that: × (g H)(M I) B = . ◦ × ∩ ∅ In particular g H gives a homotopy between g h0 and g h1, which misses B. Since both f0, f1 ◦ m+1 ◦ ◦ m+1 have image in RP , for > 0 small enough and for i = 0, 1 the maps j fi : M RP 2m+3 2m+3 ◦ 2→m+3 ⊂ RP B and g hi : M RP B are homotopic as maps with values in RP B (this is because\ “close enough”◦ maps→ are homotopic\ – a statement that has to be justified carefully).\ Thus so far we have obtained homotopy of maps: 2m+3 j f0 g h0 g h1 j f1 : M RP B, ◦ ∼ ◦ ∼ ◦ ∼ ◦ → \ where the homotopies are as maps with value in RP2m+3 B. In particular this gives a homotopy 2m+3 \ 2m+3 f t between f 0 = j f0 : M RP B and f 1 = j f1 : M RP B. We finally compose ◦ → \ ◦ → \ now this homotopy with the retraction r : RP2m+3 B RPm+1 and we get a homotopy of maps: \ → m+1 r f t : M RP ◦ → such that r f = r j f = f and r j f = r f = f . ◦ 0 ◦ ◦ 0 0 ◦ ◦ 1 ◦ 1 1 3.5. Vector bundles on spheres. Consider a rank-k vector bundle on the sphere

Rk E π Sn + and let D and D− be, respectively, the upper and the lower hemisphere. Then, by Corollary 114 both E + and E − are trivial and there is a well defined map: |D |D + n 1 k g : D D− = S − GL(R ). ∩ → n + (This is simply the cocyle of the cover S = D D−; this is not an “open” cover, but the cocycle n 1 k ∪ k is still defined.) The map g : S − GL(R ) is called a clutching function. Denoting by Vect (X) the set of isomorphism classes of rank-→ k vector bundles over X and by [Y : Z] the set of homotopy classes of maps f : Y Z, there is a well defined map: → n 1 k k n γ :[S − : GL(R )] Vect (S ), → LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 31 which associates to every clutching function (i.e. cocycle) g the vector bundle Eg constructed as in Example 94.

n 1 k k n Exercise 119. Prove that γ :[S − : GL(R )] Vect (S ) is well deﬁned, i.e. that homotopic clutching functions deﬁne isomorphic vector bundles.→ (Hint: use Theorem 113.)

Since GL(Rk) is not connected, in order to classify vector bundles on Sn we restrict to the so called oriented bundles, i.e. those vector bundles E Sn such that the determinant bundle det(E) is trivial and a trivialization has been chosen (this→ last condition is equivalent to choosing an orientation of one ﬁber Ex0 over a ﬁxed point x0). 0 k When n = 1, there are just two homotopy classes of maps g : S GL(R ) such that g(x0) k → ∈ GL+(R ). These correspond to the trivial bundle and the direct sum of (k 1) many trivial line bundles with the Moebius bundle. −

Exercise 120. Consider a vector bundle of rank k > 1:

Rk E π .

Prove that there exists a section s : S1 E of this vector bundle that is never zero, and in → particular that E E1 span s where E1 is a rank-(k 1) vector bundle and span s is trivial rank-one bundle. (Hint:' ⊕ use Theorem{ } 111.) − { }

n 1 When n 2, the sphere S − is path connected and if for a clutching function we have g(x0) k ≥ n 1 k 10 n ∈ GL+(R ), then g(S − ) GL+(R ) . In particular all rank-k bundles over S (with n 2) are orientable and every vector⊂ bundle has precisely two orientations. ≥ k n We denote by Vect+(S ) the set of isomorphism classes of oriented, rank-k vector bundles on Sn, where the isomorphisms must preserve the orientation of the chosen ﬁber. For general k, n, the k n k n map Vect+(S ) Vect (S ) that forgets the orientation is at most two-to-one, in fact exactly two- to-one except on→ those vector bundles that have an automorphism which reverts the orientation; n 1 this can only happen when S − is not connected, i.e. when n = 1, for which Exercise 120 applies. k n k n When n > 1 the map Vect+(S ) Vect (S ) is two-to-one. In light of this observation, the classiﬁcation of vector bundles on S→n follows from next Theorem 121.

+ n 1 k k n Theorem 121. The map γ :[S − : GL+(R )] Vect+(S ), that associates to a map g : n 1 k → S − GL+(R ) the bundle Eg with clutching function g, is a bijection. →

n 1 k Proof. Observe first that if we take g : S − GL+(R ) then the corresponding bundle is oriented. The map γ+ :[g] [E ] is well defined by→ Exercise 119 and we wish to find an inverse for it. 7→ g The obvious candidate is the map [E] [gE] which associates to a vector bundle E over Sn the 7→ + clutching function that we obtain by trivializing the vector bundle over D and D−. The fact that this map is well defined is left as an exercise (the proof uses the properties of a cocycle and the k fact that GL+(R ) is path connected).

k Exercise 122. Prove that the inclusion SO(k) , GL+(R ) is a homotopy equivalence, and n 1 k → consequently that [S − : GL+(R )] = πn 1(SO(k)). (Hint: use the Gram-Schmidt procedure.) − 2 2 1 Exercise 123. Prove that there is a natural bijection θ : Vect (S ) Z = π1(S ). Under this + → identiﬁcation we have θ(TS2) = 2 and θ(τ C ) = 1 (here C R2 , τ C S2 CP1 denotes the 1,2 − ' → 1,2 → ' complex tautological bundle, seen as a real, rank-2 oriented vector bundle).

Exercise 124. Prove that every rank-2 bundle over Sn is trivial when n > 2.

10 k Here GL+(R ) denotes the set of invertible k × k matrices with positive determinant, i.e. one of the two k components of GL(R ). 32 ANTONIO LERARIO

3.6. The normal bundle. Deﬁnition 125 (Normal bundle). Let X Y be an embedded submanifold. Endow the vector p ⊆ bundle TY Y with a Riemannian metric (see Deﬁnition 108). The normal bundle NX of X −→ in Y is the vector bundle NX = TX⊥ X (see Exercise 110). Note that NX TY . → ⊂ |X Exercise 126. Let X Y be a submanifold given by regular equations X = f1 = = fc = 0 . Prove that the normal⊂ bundle NX is trivial. (We assume a metric on TY has{ been chosen··· so that} it makes sense to speak about normal bundle of a submanifold.) Use this result to give another proof that RP1 RP2 cannot be the regular zero set of a function on RP2. ⊂ Proposition 127. Let M Rn be a smooth embedded manifold and endow T Rn = Rn Rn with the standard vector bundle⊂ metric. Denote by N M = (x, v) NM v < (an open× set in { ∈ | k k } NM). If M is compact, there exists > 0 such that the map τ : N M Rn given by: → τ :(x, v) x + v 7→ is an embedding.

Proof. Let us ﬁrst observe that for every x M we have: ∈ (3.4) rk(d(x,0)τ) = n.

First, clearly the image of d(x,0) contains TxM. Moreover for every v TxM ⊥ let us consider the curve γ :( δ, δ) NM given by γ(t) = (x, tv). Then γ(0) = x and ∈ − →

d d τ(γ(t)) = (x + tv) = v. dt t=0 dt t=0 n Then the image of d(x,0) also contains TxM ⊥ and since R = TxM + TxM ⊥, then (3.4) follows. It is easy to see that the map τ is smooth and for every x M there exists a neighborhood ∈ U 0 M and 0 > 0 such that: x ⊂ x rk(d τ) = n for all (x, v) U 0 M such that x U and v . (x,v) ∈ x ∈ x k k ≤ x As a consequence, by the Inverse Function Theorem, there exist U U 0 and < 0 such that: x ⊂ x x x x n (3.5) τ N x Ux : N Ux R is an embedding. | → ` By compactness of M, we can extract a ﬁnite cover Uxi i=1 of M out of Ux x M and there x { } { } ∈ exists 0 < < min i such that 0 `{ 2 } 2 n 2 (3.6) τ N Ux : N Uxi R | i → is an embedding for every i = 1, . . . , `.

Let us now consider for every 0 < < 0 the compact set:

3 2 3 N M = (x, v) NM v N 2M. ∈ k k ≤ 2 ⊂

3 2 n We claim that for > 0 small enough τ 3 : N M R is a homeomorphism onto its |N 2 M → 3 image. Since N 2 M is compact, Rn is Hausdorﬀ and τ is continuous, it is enough to show that 3 2 n for > 0 small enough τ 3 : N M R is injective. Suppose by contradiction that there |N 2 M → are sequences (xn, vn) n and (yn, wn) n with { } ∈N { } ∈N (3.7) (x , v ) = (y , w ) n n 6 n n 3 and (x , v ), (y , w ) N 2n M such that n n n n ∈ τ(xn, vn) = τ(yn, wn). Up to subsequences, we can assume x x M and y y M. Then n → ∈ n → ∈ x = lim τ(xn, vn) = lim τ(yn, wn) = y, n n →∞ →∞ which means that (again, up to subsequences) the two sequences (xn, vn) n N and (yn, wn) n N x { } ∈ { } ∈ eventually lie in the same N Ux, on which τ is injective by (3.5), contradicting (3.7). LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 33

3 Thus there exists > 0 such that τ : N 2 M Rn is a homeomorphism onto its image. In 3 → particular, since N M N 2 M, then: ⊂ n τ N M : N M R | → is a homeomorphism onto its image. Consequently, because of (3.6), τ is an embedding. |N M Corollary 128. Let M Rn be a compact manifold. Then there is an open set U Rn containing M and a smooth retraction⊂ r : U M which is a homotopy equivalence. ⊂ → n Proof. Let > 0 such that τ N M : N M R is an embedding. Observe that | → U = τ(N M) is an open set containing M. Let p : NM NM be the smooth map (x, v) (x, 0) and deﬁne: → 7→ 1 r = τ p τ − : U M. ◦ ◦ → Observe that the map p : N M N M is homotopic to id , through the homotopy: → N M pt(x, v) = (x, tv), p0 = p, p1 = idN M . Consequently, denoting by i : M U the inclusion we have: → 1 1 r i = id and i r = τ p τ − τ id τ − = id , ◦ M ◦ ◦ ◦ ∼ ◦ N M ◦ U which proves both the inclusion i : M U and the retraction r : U M are homotopy equiva- → → lences.

3.7. An approximation result. Theorem 129. Let A and B be smooth compact manifold and f : A B be a continuous function. There is a smooth map g : A B which is homotopic to f. → → Proof. Let B Rn be an embedding given by Theorem 51; we identify B with its image under → 0 this embedding. Proposition 127 says that there exists an 0 > 0 such τ embeds N B onto a neighborhood U = τ(N 0 B) of B; Corollary 128 says that the retraction r : U B is a homotopy equivalence. → We claim that we there exists > 0 small enough such that for every y B the ball B(y, ) (a n ∈ ball in R ) is entirely contained in U. In fact, since U is open, then for every y B there is y > 0 such that B(y, 3 ) U. Since M is compact, there are y , . . . , y M such that:∈ y ⊂ 1 ` ∈ ` [ M B(y , 3 ). ⊂ k yk k=1 Let = min , . . . , . Then every y M belongs to some B(y , 3 ) and { y1 y` } ∈ k yk B(y, ) B(y , 3 ) U. ⊂ k yk ⊂ We can think at f as a continuous function f : A Rn with image in B. Using Theorem 195, n → we produce a smooth map f : A R such that → (3.8) max f(x) f(x) . x A k − k ≤ ∈ Of course the image of the map f is not in B, but (3.8) implies f (x) B(f(x), ), hence the ∈ image of f lies in U. Using the retraction r we define: g = r f : A B. ◦ → To see that g is homotopic to f we simply define the homotopy: g (x) = r(tf (x) + (1 t)f(x)), g = f, g = g. t − 0 1 (The homotopy is well defined because for every t [0, 1] the point tf(x) + (1 t)f(x) belongs to the convex set B(f(x), ) U, i.e. tf (x) + (1 t)∈f(x) belongs to the domain− of the map r.) ⊂ − 34 ANTONIO LERARIO

Exercise 130. Let A and B be two compact manifold. Let us assume that B Rn and let us put a metric on the set of continuous functions C0(A, B) by: ⊂ d(f, g) = max f(x) g(x) f, g C0(A, B). x A k − k ∈ ∈ Under this assumptions, reﬁne the conclusion of Theorem 129 and prove that for every continuous function f : A B and every > 0 there is a smooth map f : A B such that f is homotopic to f and d(f , f→) . → ≤ As a corollary we can deduce the following important theorem. Theorem 131. If k < n then π (Sn) = 0 . k { } Proof. Let f : Sk Sn be a continuous map. Then, by Theorem 129, f is homotopic to a smooth map g : Sk Sn.→Since k < n, by Lemma 69, the map g is not surjective; consequently it misses → a point and factors through a map Sk Rn Sn, hence g is homotopic to a constant map and → → the same is true for f. Exercise 132. Use the previous Theorem together with the results from Section 3.5 to prove that every vector bundle on S3 is trivial. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 35

4. Differential forms and cohomology

A note to the reader: the subject of this section requires a bit more to be digested, because the construction of diﬀerential forms is a bit involved. To complement this chapter, I suggest to also read [1, Section 7], which is an excellent source and might help developing the “intuition” on these objects.

4.1. Alternating forms. Let V be a n-dimensional vector space. We define the k-th exterior power of V ∗ (the dual space of V ) as: k (V ∗) = α : V V R multilinear and alternating . ∧ { | × ·{z · · × } → } k times k Elements of (V ∗) are called k-forms. ∧ n 11 Let now η1, . . . , ηn be a basis for V ∗ and I = i1, . . . , ik k be a multi-index. We define { }k { } ∈ η = η η (V ∗) as the k-form such that: I i1 ∧ · · · ∧ ik ∈ ∧ η (v ) η (v ) i1 1 ··· i1 k  . .  (4.1) ηI (v1, . . . , vk) = det  . .  . η (v ) η (v ) ik 1 ··· ik k We observe that to the same multi-index I we can associated different forms: for example, in principle, given I = 1, 2 we could consider η1 η2 and η2 η1 (note that in virtue of definition (4.1) these two multilinear{ } maps are different). To∧ resolve this∧ ambiguity we adopt the convention that multi-indices are always ordered and I = i < < i . { 1 ··· k} The following fact is elementary and is left as an exercise... n Proposition 133. Let k denote the set of all possible multi-indices I = i1 < < ik . The k k { n ··· } set ηI I n is a basis for (V ∗). In particular the dimension of (V ∗) is k . { } ∈{k} ∧ ∧ Proof. Try to prove it yourself!

We now make clear the meaning of the symbol “ ”. ∧ k ` Deﬁnition 134 (Wedge product). Let γ (V ∗) and ω (V ∗) be the forms (written in the above basis) as: ∈ ∧ ∈ ∧ X X γ = γ η η and ω = ω η η . I i1 ∧ · · · ∧ ik J j1 ∧ · · · ∧ j` I n J n ∈{k} ∈{ ` } k+` We deﬁne the form γ ω (V ∗) (called “the wedge of γ and ω”) as: ∧ ∈ ∧ X γ ω = γ ω η η η η . ∧ I J i1 ∧ · · · ∧ ik ∧ j1 ∧ · · · ∧ j` I,J

The wedge product of forms has the following properties, which eventually follows from (4.1):

k ` (1) (bilinearity) For every a1, a2, b1, b2 R and forms γ1, γ2 (V ∗) and ω1, ω2 (V ∗) we have: ∈ ∈ ∧ ∈ ∧ (a γ + a γ ) (b ω + b ω ) = a b γ ω + a b γ ω + a b γ ω + a b γ ω . 1 1 2 2 ∧ 1 1 2 2 1 1 1 ∧ 1 1 2 1 ∧ 2 2 1 2 ∧ 1 2 2 2 ∧ 2 k ` m (2) (associativity) For every γ (V ∗), ω (V ∗) and θ (V ∗) we have: ∈ ∧ ∈ ∧ ∈ ∧ γ (ω θ) = (γ ω) θ. ∧ ∧ ∧ ∧ k ` (3) (anticommutativity) For every γ (V ∗), ω (V ∗) we have: ∈ ∧ ∈ ∧ γ ω = ( 1)k`ω γ. ∧ − ∧

11 At this point ηi1 ∧ · · · ∧ ηik is just a symbol for denoting a special k-form. 36 ANTONIO LERARIO

(Notice in particular that from the last property it follows that γ γ = 0.) The wedge product turns ∧ dim(V ) M k ∗(V ∗) = (V ∗) ∧ ∧ k=0 into a graded algebra. Exercise 135. Check the details for the previous properties, using (4.1).

L Definition 136 (Pull-back). Let W V be a linear map between two finite-dimensional vector −→ k k spaces. We define the pull-back map L∗ : (V ∗) (W ∗) by: ∧ → ∧ k (L∗η)(w , . . . , w ) = η(Lw ,..., Lw ), η (V ∗), w , . . . , w W. 1 k 1 k ∈ ∧ 1 k ∈ Note that a simple computation applying the above definition shows that:

(4.2) L∗(η η ) = L∗η L∗η . 1 ∧ · · · ∧ k 1 ∧ · · · ∧ k

Let now e1, . . . , en be a basis for V , f1, . . . , fm be a basis for W and let η1, . . . , ηn and { } { } { n m } φ1, . . . , φm be the corresponding dual bases for V ∗ and W ∗ respectively. Let L R × be the matrix{ representing} L : W V in these bases and for every pair of indices I n∈ and J m → ∈ k ∈ k let LI,J be the submatrix of L obtained by extracting the rows with index in I and the columns with index in J. Proposition 137. With the above notations, for every I n we have: ∈ k X (4.3) L∗(ηI ) = det(LI,J )φJ . J m ∈{ k }

Proof. Given α = α , . . . , α , β = β , . . . , β m observe that (4.1) implies: { 1 k} { 1 k} ∈ k φ φ (f , . . . , f ) = δ , α1 ∧ · · · ∧ αk β1 βk αβ consequently in order to prove (4.3) it is enough to prove that L∗ηI (fj1 , . . . , fjk ) = det(LI,J ). To this end observe that:

L∗η (f , . . . , f ) = η η (Lf ,..., Lf ) I j1 jk i1 ∧ · · · ∧ ik j1 jk n n ! X X = η η L e ,..., L e i1 ∧ · · · ∧ ik bj1 b bjk b b=1 b=1 η (Pn L e ) η (Pn L e ) i1 b=1 bj1 b ··· i1 b=1 bjk b  . .  = det  . .  η (Pn L e ) η (Pn L e ) ik b=1 bj1 b ··· ik b=1 bjk b η (L e ) η (L e ) i1 i1j1 i1 ··· i1 i1jk i1  . .  = det  . .  η (L e ) η (L e ) ik ikj1 ik ··· ik ikjk ik = det(LI,J ).

4.2. Differential forms. Let U Rn be an open set and consider the linear coordinate functions ⊂ x1, . . . , xn : U R. The differentials dx1, . . . , dxn of these functions are smooth sections of n→ T ∗U U (R )∗ U. Given a function f C∞(U, R) and a multi-index i1 < . . . < ik , we can consider:' × → ∈ { } η = f(x)dx dx x i1 ∧ · · · ∧ ik k n which is a section of the trivial vector bundle U (R )∗ U. In other words, for every x U, k × ∧ k n→ ∈ ηx defines an element of (TxU)∗. A section of U (R )∗ U is called a k-differential form ∧ × ∧ → LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 37 on U and the space of all sections of this bundle is denoted by:    X  Ω∗(U) = η = fI dxi1 dxik , f C∞(U, R) ∧ · · · ∧ ∈  I n  ∈{k} k n = C∞ sections of U (R )∗ U . × ∧ → Let now W Rm and V Rn be open sets, ψ : W V be a smooth map and η Ωk(V ). We ⊂ ⊂ k→ k∈ m define the pull-back of η under ψ, denoted ψ∗η Ω (W ) as the section of W (R )∗ W ∈ × ∧ → given by x (x, d ψ∗η ). In other words: 7→ x ψ(x) ψ∗ηx(w1, . . . , wk) = ηψ(x)(dxψw1, . . . , dxψwk). Observe that, by Proposition 137, given I = i < < i we have: { 1 ··· k} X ψ∗(dx dx ) = det(Jψ(x) )dx dx . i1 ∧ · · · ∧ ik I,J j1 ∧ · · · ∧ jk J m ∈{ k } When ψ : W, V is the inclusion, ψ∗η is denoted by η and called the restriction of η to W . → |W m n Example 138. Let ψ : R R be a smooth map with components ψ = (ψ1, . . . , ψn) and 1 n → consider dxi Ω (R ). Then we have: ∈ m X ∂ψi (4.4) ψ∗(dx ) = (x)dx . i ∂x j j=1 j In practice combining (4.2) and (4.4) gives the simplest way to compute the pull-back of a differential form.

We turn now to the deﬁnition of the main object of this section.

Definition 139 (Differential form). Let M be a smooth manifold with atlas (Uα, ψα) α A.A 12 { k } ∈ differential k-form η on M is a collection ηα α A of differential forms ηα Ω (ψα(Uα)) such that for every α, β A with U U = we{ have:} ∈ ∈ ∈ α ∩ β 6 ∅ 1 (4.5) (ψ ψ− )∗η = η . α ◦ β α β We want to show now that there exists a vector bundle

k m k (R )∗ , (T ∗M) M ∧ → ∧ −→ such that diﬀerential k-forms are exactly sections of this bundle. In fact, given the atlas (Uα, ψα) α A k {m } ∈ for the m-dimensional manifold M, we deﬁne the cocyle gαβ : Uα Uβ GL( (R )∗) as the map: ∩ → ∧ m m 1 m gαβ(x) = the matrix with entries det J(ψα ψβ− )(x) , I,J . k × k ◦ I,J ∈ k

k We deﬁne (T ∗M) M to be the vector bundle with cocycle gαβ , using the construction of Example 94∧ . Recalling→ the discussion from Remark 106, it follows{ now} from (4.3) and (4.5) that k k a diﬀerential k-form is a section of (T ∗M) M. We denote by Ω (M) the space of all smooth sections of this bundle: ∧ →

k k Ω (M) = C∞ sections of T ∗M . { ∧ } A diﬀerential k-form η Ωk(M), evaluated at x M is therefore a multilinear, alternating map k ∈ ∈ η (T ∗M): x ∈ ∧ x ηx : TxM TxM R multilinear and alternating. | × ·{z · · × } → k times

12k is called the degree of the form η and denoted by deg(η). 38 ANTONIO LERARIO

Example 140. We have the following identities:

0 Ω (M) = C∞(M, R) 1 Ω (M) = sections of T ∗M { }   X  Ω2(M) = objects which in coordinates look like f (x)dx dx ij i ∧ j  i

Given η Ωk(M) and θ Ω`(M) the wedge η θ Ωk+`(M) is defined by the pointwise ∈ ∈ ∧ ∈ operation (η θ)x = ηx θx. The wedge product of differential forms satisfies the following properties: ∧ ∧

k (1)( C∞(M, R)-bilinearity) For every a1, a2, b1, b2 C∞(M, R) and forms γ1, γ2 Ω (M) and ω , ω Ω`(M) we have: ∈ ∈ 1 2 ∈ (a γ + a γ ) (b ω + b ω ) = a b γ ω + a b γ ω + a b γ ω + a b γ ω . 1 1 2 2 ∧ 1 1 2 2 1 1 1 ∧ 1 1 2 1 ∧ 2 2 1 2 ∧ 1 2 2 2 ∧ 2 (2) (associativity) For every γ Ωk(M), ω Ω`(M) and θ Ωn(M) we have: ∈ ∈ ∈ γ (ω θ) = (γ ω) θ. ∧ ∧ ∧ ∧ (3) (anticommutativity) For every γ Ωk(M), ω Ω`(M) we have: ∈ ∈ γ ω = ( 1)k`ω γ. ∧ − ∧ The vector space

dim(M) M k Ω∗(M) = Ω (M) k=0 together with the multiplication given by the wedge of forms is a graded algebra over C∞(M, R). The pull-back of a diﬀerential form ω Ωk(N) under a smooth map ψ : M N is deﬁned to ∈ → be the form ψ∗ω such that:

(ψ∗ω)x(w1, . . . , wk) = ηψ(x)(dxψw1, . . . , dxψwk).

4.3. The exterior differential. Let U Rm be an open set. We define now a linear operator d :Ωk(U) Ωk+1(U) by the following two⊂ conditions: → 0 (1) if f Ω (U) = C∞(U, R), then df coincides with the usual differential of functions; ∈ P m P m (2) if ω = I fI dxi1 dxik then dω = I dfI dxi1 dxik . ∈{ k } ∧ · · · ∧ ∈{ k } ∧ ∧ · · · ∧ Example 141. If ω = f(x, y)dx + g(x, y)dy Ω1(R2), then ∈ ∂f ∂f ∂g ∂g dω = dx dx + dy dx + dx dy + dy dy ∂x ∧ ∂y ∧ ∂x ∧ ∂y ∧ ∂f ∂g = dy dx + dx dy ∂y ∧ ∂x ∧ ∂g ∂f = dx dy. ∂x − ∂y ∧

The exterior diﬀerential satisﬁes the following properties. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 39

(1) d(γ ω) = dγ ω + ( 1)deg(γ)γ dω. ∧ ∧ − 13 ∧ To see that this is true let γ = fI dxI and ω = gJ dxJ and write: d(γ ω) = d(f g dx dx ) ∧ I J I ∧ J = df dx g dx + f dg dx dx I I ∧ J J I J ∧ I ∧ J = dγ ω f dx dg dx dx dx ∧ − I i1 ∧ J ∧ i2 ∧ · · · ∧ ik ∧ J = dγ ω + f dx dx dg dx dx dx ∧ I i1 ∧ i2 ∧ J ∧ i3 ∧ · · · ∧ ik ∧ J = dγ ω + ( 1)deg(γ)γ dω. ∧ − ∧ (2) d2 = 0 0 Pm ∂f To see this let ﬁrst f Ω (U), such that df = dxj. Then: ∈ j=1 ∂xj

 m  X ∂f d(df) = d  dxj ∂x j=1 j m X ∂f = d dx ∂x j j=1 j m m ! X X ∂2f = dx dx ∂x ∂x j ∧ i j=1 i=1 i j m X ∂2f = dx dx = ( ) ∂x ∂x j ∧ i ∗ i,j=1 i j In the last sum each pair of indices i, j 1, . . . , m appears two times: one time we see 2 2 ∈ { } ∂ f dx dx and the other ∂ f dx dx . However dx dx = dx dx and ∂xi∂xj j i ∂xj ∂xi i j i j j i 2 ∧2 ∧ ∧ − ∧ ∂ f = ∂ f , consequently ( ) = 0. ∂xj ∂xi ∂xi∂xj ∗ k P For a general ω Ω (U) with ω = I m fI dxI , using the previous point we get: ∈ ∈{ k }   X d(dω) = d  df dx   I ∧ I  I m ∈ k {}  X = d(dfI ) dxI + dfI d(dxI ) = 0. ∧ ∧ m | {z } | {z } I 0 0 ∈{ k } m n (3) Given ψ C∞(W, V ) with W R and V R , we have d ψ∗ = ψ∗ d. Starting∈ again with the case of⊂ functions we⊂ have: ◦ ◦

m X ∂(f ψ) (d ψ∗)f = d(f ψ) = ◦ dx ◦ ◦ ∂x j j=1 j m n ! X X ∂ψi ∂f = ψ dx ∂x ∂y ◦ j j=1 i=1 j j n m X ∂f X ∂ψi = ψ dx ∂y ◦ ∂x j i=1 i j=1 j

| ∗{z } ψ dyi n X ∂f = ψ∗ ψ∗dy = ψ∗df. ∂y · i i=1 i

13 m Here and below we use the convention that for a multi-index I = {i1 < ··· < ik} ∈ k the symbol dxI denotes the form dxi1 ∧ · · · ∧ dxik . 40 ANTONIO LERARIO

P Now, for a general form ω = I n fI dyI : ∈{k}   X (d ψ∗)ω = d  ψ∗f ψ∗dy ψ∗dy  ◦  I · i1 ∧ · · · ∧ ik  I n ∈ k  { }  X = d  ψ∗f d(ψ∗y ) d(ψ∗y )  I · i1 ∧ · · · ∧ ik  I n ∈{k} X = dψ∗f ψ∗dy ψ∗dy I ∧ i1 ∧ · · · ∧ ik I n ∈ k {}  X = ψ∗  df dx dx   I ∧ i1 ∧ · · · ∧ ik  I n ∈{k} = (ψ∗d)ω. We want to extend now the exterior differential to forms on a manifold. Let M be a smooth k k+1 k manifold and η Ω (M). We define dη Ω (M) to be the section of T ∗M which in the ∈ ∈ ∧ trivialization given by an atlas (Uα, ψα) α A is defined by: { } ∈ (dη) = d(η ) Ωk+1(ψ (U )). α α ∈ α α k In order to check that this definition is well posed, i.e. that dηα α A defines a section of T ∗M we need to verify that (3.3) is verified for the cocycle of this{ bundle.} ∈ Using the above properties∧ we see that: 1∗ 1∗ g dη = ψ ψ− dη = d ψ ψ− η = dη , αβ α α ◦ β α α ◦ β α β which proves dη Ωk+1(M) is well defined. ∈ Exercise 142. Prove that the exterior differential has the following important property (which k follows from the previous local properties): given f C∞(M,N) and ω Ω (N), then: ∈ ∈ d(f ∗ω) = f ∗dω.

4.4. De Rham cohomology. We introduce now the concept of De Rham cohomology of a smooth manifold M. This concept combines the diﬀerential structure of the manifold with its topology and gives a homotopy invariant of the manifold M which is of fundamental importance.

Definition 143 (De Rham cohomology). The differential complex (Ω∗(M), d) is called the De Rham complex fo M. A form ω Ωk(M) is called closed if dω = 0 and exact if there exists k 1 ∈ 2 η Ω − (M) such that dη = ω. The fact that d = 0 implies that every exact form is closed and consequently∈ we can define the k-th De Rham cohomology of M as the vector space: closed k-forms on M Hk(M) = { }. exact k-forms on M { } Example 144 (The cohomology of R). The De Rham complex in this case is simply:

0 d0 1 d1 0 Ω (R) Ω (R) 0. −→ −→ −→ The kernel of d0 equals the set of constant functions and consequently: 0 H (R) = ker(d0) R. ' 1 1 We have ker(d1) = Ω (R) and we claim that also im(d0) = Ω (R): in fact, given ω = g(x)dx ∈ Ω1(R), if we deﬁne the function: Z x f(x) = g(t)dt, 0 we see that df = ω and consequently: 1 H (R) = 0 . { } LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 41

Remark 145. Notice that it follows from the deﬁnition that if two manifolds are diﬀeomorphic, then they have the same De Rham cohomology (prove it with details!). We will prove indeed that they have the same cohomology even if they are just homotopy equivalent (this will be a consequence of Corollary 147.

4.5. Poincar´eLemma. In this section we introduce a basic tools for computing the cohomology of a manifold. Consider the two commutative diagrams:

M R Ω∗(M R) × × (4.6) π s π∗ s∗

M Ω∗(M) where the map π is the projection on the ﬁrst factor and s(x) = (x, 0) is the “zero section”. Next Theorem implies that the both π∗ and s∗ induce isomorphisms in cohomology. Theorem 146 (Poincar´eLemma). In the following diagram, induced by the diagrams in (4.6):

H∗(M R) × π∗ s∗

H∗(M) both the maps π∗ and s∗ are isomorphims.

Proof. We will prove that: 1 1 s∗π∗ = H∗(M) and π∗s∗ = H∗(M ), ×R which will imply the statement. The ﬁrst identity is simple, since:

s∗π∗ = (π s)∗ = (id )∗ = 1 ∗ . ◦ M H (M) 1 For the second identity we will construct a linear map K :Ω∗(M R) H∗− (M R) (called chain homotopy) such that: × → ×

(4.7) π∗s∗ 1 = (Kd dK). − ± ± If (4.7) is true, then for every [ω] H∗(M R) we have: ∈ × π∗s∗[ω] [ω] = [(Kd dK)ω] − ± ± = [dKω] dω=0± = [d(Kω)] = 0. d(Kω) is exact

For the proof of (4.7) we observe that every form ω Ω∗(M R) can be written in a unique way ∈ × as ω = ω1 + ω2 with:

ω = f (x, t)π∗β and ω = f (x, t)π∗β dt, 1 1 1 2 2 2 ∧ where f1, f2 C∞(M R) and β1, β2 Ω∗(M). We deﬁne: ∈ × ∈ Z t Kω1 = 0 and Kω2 = f(x, s)ds π∗β2. 0 | {z } C∞(M ) ∈ ×R We check now that (4.7) is true for both forms ω1 (“of type (1)”) and ω2 (“of type (2)”).

(1) Let ω1 = fπ∗β1 of the ﬁrst type. Then:

π∗s∗ω ω = π∗s∗(f (x, t)π∗β ) f (x, t)π∗β 1 − 1 1 1 − 1 1 = (f (x, 0) f (x, t))π∗β 1 − 1 1 = ( ) ∗1 42 ANTONIO LERARIO

(Kd dK)ω = (Kd dK)(f π∗β ) − 1 − 1 1 = Kd(f1π∗β1) Kω1=0 = K (∂ f π∗β + ∂ f dt π∗β + f π∗dβ ) x 1 ∧ 1 t 1 ∧ 1 1 1 = K(∂tf1dt π∗β1) K(type (1))=0 ∧ deg(ω ) = ( 1) 1 K(∂ f π∗β dt) − t 1 1 ∧ Z t deg(ω1) = ( 1) ∂tf1(x, s)ds π∗β1 by def. of K(type− (2)) 0 deg(ω ) = ( 1) 1 (f (x, t) f (x, 0))π∗β − 1 − 1 1 deg(ω) 1 = ( 1) − ( ). − ∗1 This proves (4.7) holds true on forms of the ﬁrst type. (2) Let now ω = f π∗β dt. We have: 2 2 2 ∧ π∗s∗ω ω = f (x, t)π∗β dt = ( ), 2 − 2 − 2 2 ∧ ∗2 because s∗dt = 0.

(Kd dK)ω = K (∂ f π∗β dt + f π∗dβ dt) + − 2 x 2 ∧ 2 ∧ 2 2 ∧ Z t d f2(x, s)ds π∗β2 − 0 Z t Z t = ∂xf2(x, s)ds π∗β2 + f2(x, s)ds π∗dβ2 0 0 Z t ∂xf2(x, s)ds π∗β2 f2(x, t)dt π∗β2+ − 0 ∧ − ∧ Z t f2(x, s)ds π∗dβ2 − 0 = f (x, t)dt π∗β − 2 ∧ 2 =( 1)deg(β2)( ) − ∗2 deg(ω) 1 =( 1) − ( ). − ∗2

Corollary 147. The De Rham cohomology of Rn vanishes except in dimension zero, where H0(Rn) = R.

n 1 Proof. This follows immediately using Theorem 146 with the choice M = R − : n n n 1 H∗(R ) = H∗(R R) H∗(R − ) H∗(point). × ' '···' Corollary 148 (Homotopy axiom for De Rham cohomology). Homotopic maps induce the same map in cohomology.

Proof. Let f , f C∞(M,N) be homotopic maps between smooth manifolds, i.e. f = F ( , 0) 0 1 ∈ 0 · and f1 = F ( , 1) for a smooth map F : M R N (here we have extended the homotopy in such a way that F· (x, t) f (x) for t 0 and F×(x, t→) f (x) for t 1). ≡ 0 ≤ ≡ 1 ≥ For every t R consider the commutative diagram ∈ F M R N ×

ft π st ,

M LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 43 where st(x) = (x, t). Since for every t R the maps s∗ and π∗ are inverse to each other in ∈ t cohomology, it follows that s1∗ = s0∗ and consequently

f0∗ = s0∗F ∗ = s1∗F ∗ = f1∗.

As a corollary we immediately see that if two smooth manifolds are homotopy equivalent, they have the same De Rham cohomology. Corollary 149. Homotopy equivalent manifolds have the same cohomology.

Proof. If two manifolds are homotopy equivalent in the C0-sense, then they are also homotopy equivalent in the C∞ sense (by a variation of the argument in Theorem 129 for noncompact manifolds, if two maps are continuously homotopic, they are homotopic to smooth maps that are smoothly homotopic) and we can apply the previous Corollary 148. Exercise 150. Using the fact that every manifold (even non-compact) embeds in some Rn, work out all the details for the proof Corollary 149. (This is a long, but rewarding, exercise...)

Exercise 151. Compute the De Rham cohomology of R2 point . \{ } Exercise 152. Compute the De Rham cohomolgy of the Moebius band M (hint: use the fact that π : M S1 is a line bundle...) → 4.6. Mayer-Vietoris sequence. Let M be the union of two open sets A B. Then we have a diagram of maps: ∪

iA (4.8) M A B A B , q ∩ iB where iA and iB are the inclusions in A and B respectively. The Mayer-Vietoris sequence of forms is the sequence obtained by taking the induced map at the level of diﬀerential forms (and taking the diﬀerence of the two maps on the right):

∗ ∗ j iB iA 0 Ω∗(M) Ω∗(A) Ω∗(B) − Ω∗(A B) 0 (1) ⊕ (2) ∩ (3)

ω (ω , ω ) |A |B

(α, β) β A B α A B | ∩ − | ∩ Proposition 153. The Mayer-Vietoris sequence of forms is exact.

Proof. We check exactness at every step.

(1) j is injective since a form ω Ω∗(M) is zero if and only if both its restrictions to A and B are zero. ∈

(2) clearly (iB∗ iA∗ )(ω A, ω B) = 0; moreover if α A B = β A B, then there exists ω − | | | ∩ | ∩ ∈ Ω∗(A B) such that ω = α and ω = β. ∪ |A |B (3) to check that every element η Ω∗(A B) is the diﬀerence of two forms we simply take a partition of unity ρ , ρ subordinated∈ ∩ to the cover A, B and deﬁne: { A B} { } α = ρ η Ω∗(A) and β = ρ η Ω∗(B). − B ∈ A ∈ With this choice we have (i∗ i∗ )(α, β) = η. B − A

As a corollary we derive the following Theorem, which is a second basic tool for computing the cohomology of a manifold. 44 ANTONIO LERARIO

Theorem 154. Let M = A B be an m-dimensional manifold with A, B two open sets. There exists a long exact sequence (called∪ the Mayer-Vietoris long exact sequence):

∗ Hq(A B) Hq(A) Hq(B) Hq(A B) Hq+1(A B) d ··· ∪ ⊕ ∩ ∪ ··· which starts with 0 H0(A B) and ends with Hm(A B) 0. → ∪ → · · · · · · → ∩ → Proof. This follows from Proposition 153 and Proposition 196 Example 155 (Cohomology of spheres). The sphere S0 consists of two points and consequently: 0 0 q 0 H (S ) = R R and H (S ) = 0 q > 0. ⊕ ∀ We want to show now that for n > 0 we have: for q = 0, n Hq(Sn) = R 0 otherwise 1 Let us start with the case of S . Considering the open cover U1,U2 from Example7 (i.e. the cover given by the upper and the lower hemispheres). We have:{ }

U1 R U2 and U1 U2 R R. ' ' ∩ ' q Consequently the Mayer-Vietoris long exact sequence for this open cover is: 0 H0(S1) H0(U ) H (U ) H0(U U ) H1(S1) 0 . 1 ⊕ 0 2 1 ∩ 2 Since S1 is connected it follows that H0(S1) R, hence the above sequence equals: ' 0 R R R R R H1(S1) 0 ⊕ ⊕ and the only unknown element is H1(S1). However the alternating sum between two zeroes of the dimensions of vector spaces in an exact sequences is zero (see Exercise 197) and: dim(H1(S1)) 2 + 2 1 = 0 − − which gives H1(S1) = R. For the sphere Sn we argue as follows. First, since Sn is connected (n > 0), again H0(Sn) = R. n Moreover, since both U1 and U2 are diﬀeomorphic to R , for n > 0 their cohomology vanishes by Corollary 147. Conseuqently for n > 1 the only nonzero piece of the long Mayer-Vietoris exact sequence for the open cover given by the upper and the lower hemispheres is:

n 1 n 1 n 1 n n H − (U ) H − (U ) = 0 H − (U U ) ' H (S ) 0 . 1 ⊕ 2 1 ∩ 2 n 1 Since U U is homotopy equivalent to a sphere S − , Corollary 148 implies that: 1 ∩ 2 n 1 n 1 n 1 H − (U U ) H − (S − ) 1 ∩ 2 ' and we conclude by induction. 2 Exercise 156. Using the Mayer-Vietoris sequence, compute the cohomology of R p1, p2 . \{ } Exercise 157. Using the Mayer-Vietoris sequence, prove that: n odd H0( Pn) = and Hn( Pn) = R R R R 0 n even Remark 158 (The generalized Mayer-Vietoris principle). There is a far generalization of the Mayer-Vietoris sequence to countably many open sets, which works as follows. Let U = Uj j J { } ∈ be a finite open cover (possibly with infinitely many open sets, but countably). For every k N we define the topological space: ∈ a Σ = U U k j0 ∩ · · · ∩ jk j <

∂0 ∂0 ∂1 M Σ0 Σ1 Σ2 ∂2 ··· ∂1 LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 45

where ∂i is the inclusion which ignores the i-th open set (e.g. ∂0 : Uj0 Uj1 Uj2 , Uj1 Uj2 ). Using the maps from this diagram we can form an exact sequence (which∩ generalizes∩ → the Mayer-Vietoris∩ sequence of forms):

r δ δ (4.9) 0 Ω∗(M) Ω∗(Σ ) Ω∗(Σ ) 0 1 ··· Here δ :Ω∗(Σ ) Ω∗(Σ ) is deﬁned as follows. First observe that: k → k+1 Y Ω∗(Σ ) = Ω∗(V V ) k j0 ∩ · · · ∩ jk j <

η = (ηj jk ) with ηj jk Ω∗(Vj Vjk ). 0··· 0··· ∈ 0 ∩ · · · ∩ We deﬁne δη Ω∗(Σ ) as: ∈ k+1 k+1 X (δη) = ( 1)iη . j0 jk+1 j0 jbi jk ··· − ··· ··· i=0 The sequence (4.9) is called the generalized Mayer-Vietoris sequence and it is exact (see [3, Proposition 8.5]). The trick that we used in point (3) of Propostion 153 for proving exactness here generalizes as follows: pick η = (ηj jk ) Ω∗(Σk) such that δ(η) = 0 and deﬁne 0··· ∈ τ = (τj jk− ) Ω∗(Σk 1) by: 0··· 1 ∈ − ` X τj jk− = ρjτjj jk− , 0··· 1 0··· 1 j=1 where ρ ` is a partition of unity subordinated to the open cover U. (Check that in the { j}j=1 case of two open sets this gives the same as we already did!). Out of the generalized Mayer- Vietoris sequence one can construct the so-called Mayer-Vietoris spectral sequence, which is a far generalization of the long exact Mayer-Vietoris sequence. I encourage the willing reader to elaborate on this by reading [3, Chapter 8].

4.7. The Mayer-Vietoris argument. The Mayer-Vietoris sequence can be used as a tool for proving statements by induction on the cardinality of “good” open covers of a manifold. This idea (combining the Mayer-Vietoris sequence with inductive proofs using “good” open covers) is sometimes called the Mayer-Vietoris argument. What is meant by “good” is the following. Definition 159 (Good cover). An open cover of an n-dimensional manifold is called a good cover if all finite nonempty intersections of open sets from it are diffeomorphic to Rn Remark 160. Every manifold has a good cover – this is proved using Riemannian geometry. We will take this statement for granted. Moreover the generalized Mayer-Vietoris sequence technique can be used to prove that the De Rham cohomology of M is isomorphic to the simplicial cohomology of the nerve of a good cover [3, Theorem 8.9]. Theorem 161. If a manifold has a finite good cover, then its cohomology is finite dimensional.

Proof. The proof is an interesting application of the above principle. First observe that from the exactness of the long Mayer-Vietoris sequence for the open cover M = A B it follows that the cohomology of M has dimension bounded by: ∪

(4.10) dim(H∗(M)) dim(H∗(A)) + dim(H∗(B)) + dim(H∗(A B)), ≤ ∩ and in particular if A, B and A B have ﬁnite-dimensional cohomology, then also A B does. Now we prove the statement of the∩ theorem by induction on the cardinality k of the good∪ cover. For k = 1 the manifold is diﬀeomorphic to Rn and the theorem is true. Assume now the theorem is proved for all manifolds with a good cover of cardinality k and let M = U U 1 ∪ · · · ∪ k+1 such that U k+1 is a good cover. Then we can take the cover of M given by the two open sets: { j}j=1 A = U U and B = U . 1 ∪ · · · ∪ k k+1 46 ANTONIO LERARIO

Both A and B have finite dimensional cohomology by induction; moreover U U ,...,U U { 1 ∩ k+1 k ∩ k+1} is a good cover of A B of cardinality k, so A B also has finite dimensional cohomology and the ∩ ∩ statement follows from (4.10). 4.8. Compact support. Given a smooth manifold M of dimension m it is also possible to restrict our attention only to forms with compact supports, i.e. we can consider k k Ω (M) = sections of T ∗M which have compact support . c { ∧ } Observe that the exterior differential preserves the property of having compact support and hence there exists a well-defined subcomplex:

0 d0 1 d1 dm−2 m 1 dm−1 m dm 0 Ω (M) Ω (M) Ω − (M) Ω (M) 0. −→ c −→ c −→ · · · −→ c −→ c −→ The cohomology of this complex is called the compactly supported De Rham cohomology and denoted by: compactly supported, closed k-forms on M Hk(M) = { }. c compactly supported, exact k-forms on M { } Example 162 (The compactly supported cohomology of R). In this case the complex for compactly supported forms is:

0 d0 1 d1 0 Ω (R) Ω (R) 0. −→ c −→ c −→ We see that ker(d0) equals the set of constant functions with compact supports, i.e. ker(d0) = 0 and consequently: { } 0 H (R) = 0 . c { } 1 For the cohomology Hc (R) the situation is already a bit more interesting. Consider the linear map: Z 1 :Ωc (R) R R → which takes a 1-form ω = g(x)dx and gives R g(x)dx. This map is clearly surjective (being nonzero) R and: 1 Ωc (R) (4.11) R. ker(R ) ' R · We claim that: Z (4.12) im(d0) = ker . R · In fact given f Ω0(R), then for a < 0 < b with absolute value large enough we have ∈ c Z Z ∂f df = (x)dx = f(b) f(a) = 0 R R ∂x − and df ker(R ). Viceversa, let ω = g(x)dx ker(R ). Then we can deﬁne the function R∈x R · ∈ R · f(x) = 0 g(t)dt which has the property of being compactly supported (because ω is compactly supported) and satisﬁes df = ω. From this, using (4.11) and (4.12) we see that: 1 1 ker(d 1 ) 1 1 Ωc (R) Ωc (R) Ωc (R) Hc (R) = | = R = R. im(d0 Ω0( )) im(d0 Ω0( )) ' ker( ) | c R | c R R · Exercise 163. Let A, B be two open sets in an m-dimensional manifold. There is a Mayer-Vietoris sequence of forms with compact support:

ι∗ σ 0 Ω∗(A B) Ω∗(A) Ω∗(B) Ω∗(A B) 0 , c ∩ c ⊕ c c ∪ where ι (α) = (α, α) and σ(α, β) = α + β. Prove that this sequence is exact and consequently that there∗ exists a− long exact sequence:

d Hq(A B) Hq(A) Hq(B) Hq(A B) ∗ Hq+1(A B) ··· c ∩ c ⊕ c c ∪ c ∩ ··· LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 47

0 m which starts with 0 Hc (A B) and ends with Hc (M) 0. This long exact sequence is called the−→Mayer-Vietoris∩ −→ long · · · exact sequence with· · · −→ compact support−→ . Exercise 164 (The compactly supported cohomology of Rn). The scope of this guided exercise is to generalize Example 162 and to prove that: q n R q = n H (R ) c ' 0 q = n 6 We will mimic the construction of Theorem 146. First we consider the diagram of maps:

Ωq+1(M R) c × π∗ e∗ ,

q Ωc(M) q where π and e are constructed as follows. Every form in Ωc(M R) can be written (using the notation∗ of the∗ proof of Theorem 146) as a unique linear combinations× of forms of two types:

ω = f (x, t)π∗β and ω = f (x, t)π∗β dt, 1 1 1 2 2 2 ∧ where now the involved forms have compact support. We define π to be zero on forms of the first type and we set: ∗ Z ∞ π (f2(x, t)π∗β2 dt) = f2(x, t)dt π∗β2. ∗ ∧ · −∞ Check that this defines π (this amounts to verify that the decomposition into sum of forms of the two types is unique and∗ that π maps to the space fo forms with compact support). For the ∗ 1 R definition of e we pick a form e = e(t)dt Ωc (R) such that e(t)dt = 1 and we set: ∗ ∈ R q e (η) = π∗η e, η Ωc(M). ∗ ∧ ∈ Using the strategy of Theorem 146 prove that π and e induce isomorphisms in cohomology. (Here q ∗ q 1 ∗ 1 one has to find an appropriate K :Ωc(M R) Ωc− (M R) such that e π = (Kd dK)...) × → × ∗ ∗ − ± ± n 4.9. Orientation and integration. Pick coordinate functions x1, . . . , xn : R R, which we → assume to be ordered. Let ω Ωn(Rn) and write it as: ∈ c ω = fdx dx , 1 ∧ · · · ∧ n n for some function f Cc∞(R , R). We define the integral of ω as: ∈ Z Z f(x)dx1 dxn = f(x)dx1 dxn. n n [R ] ∧ · · · ∧ R ··· On the right hand side we have the standard Riemann integral of a smooth, compactly supported function, while on the left hand side we have introduced the symbol “[Rn]” to stress that a ordering of the coordinates has been chosen (this notation will become more transparent soon). n n Let now ϕ C∞(R , R ) be a diffeomorphism and consider the form ∈ ϕ∗ω = (f ϕ) det(Jϕ) dy dy . ◦ · · 1 ∧ · · · ∧ n (This identity follows from Proposition 137.) From this we see that the integral of a top differential n n form ω Ωc (R ) is not invariant under diffeomorphisms, but only under diffeomorphisms that preserve∈ the orientation, i.e. which have everywhere positive Jacobian. In fact, using the above definition we have: Z Z ϕ∗ω = (f ϕ) det(Jϕ) dy1 dyn n n [R ] R ◦ · · ··· Z = sign(det(Jϕ)) (f ϕ) det(Jϕ) dy1 dyn n ◦ · | | · ··· ZR = sign(det(Jϕ)) fdx1 dxn n ··· ZR (4.13) = sign(det(Jϕ)) ω. n [R ] This discussion justifies the following definition. 48 ANTONIO LERARIO

Deﬁnition 165 (Orientable manifold). Let M be a smooth manifold with an atlas A = (Uα, ψα) α A. We say that M is orientable if for every α, β A such that V V = we have: { } ∈ ∈ α ∩ β 6 ∅ 1 (4.14) det(J(ψ ψ− )) > 0. α ◦ β In this case A is called an oriented atlas. We say that a smooth manifold is orientable14 if it admits an oriented atlas.

The manifolds Rn and Sn are both orientable: Rn because it can be endowed with a single chart (the identity), while for spheres one has to pay attention because the atlas constructed in Example 7 is not oriented.. Exercise 166. Prove that the sphere Sn is orientable, by building an oriented atlas.

Observe that the condition (4.14) can be reinterpreted in the language of vector bundles: since 1 (ψ ψ− ) is the cocycle of the tangent bundles, a manifold is orientable if and only if its tangent α ◦ β bundle is orientable (as a vector bundle15, in the sense of Section 3.5). Using the language of vector bundles we can rephrase the orientability condition as follows. Proposition 167. A smooth manifold M of dimension m is orientable if and only if the bundle m (T ∗M) M is trivial. ∧ →

Proof. Assume that M is orientable and let (Uα, ψα) α A be an oriented atlas. Let also ρα α A { } ∈ m { } ∈ be a partition of unity subordinated to this atlas. Fix coordinates x1, . . . , xm : R R and set: → m σ = ρ ψ∗ (dx dx ) Ω (M). α α α 1 ∧ · · · ∧ m ∈ This is a form on the whole M because we are multiplying ψ∗ (dx dx ) (which is a form α 1 ∧ · · · ∧ m on Uα) by the function ρα which is zero outside Uα. Consider now the form: X ω = σ Ωm(M). α ∈ α A ∈ m We claim that this form is never zero, i.e. it is a nowhere zero section of the line bundle (T ∗M), which will therefore be trivial by Exercise 105. In order to see that ω is never zero, it∧ is enough 1∗ to show that for every β A the form ω = ψ− ω Ω∗(ψ (U )) is never zero. To this end ∈ β β ∈ β β observe that: 1∗ ωβ = ψβ− ω

X 1∗ = ψβ− σα α A ∈ X 1∗ = ψ− ρ ψ∗ (dx dx ) β α α 1 ∧ · · · ∧ m α A ∈ X 1 1∗ = ρ ψ− ψ ψ− (dx dx ) α ◦ β · α ◦ β 1 ∧ · · · ∧ m α A ∈ X 1 1 = ρ ψ− det J(ψ ψ− ) dx dx α ◦ β · α ◦ β 1 ∧ · · · ∧ m α A ∈ = g (x)dx dx , β 1 ∧ · · · ∧ m P 1 1 where we have set gβ = α A ρα ψβ− det J(ψα ψβ− ) . Since this function is never zero, ∈ ◦ · ◦ it follows that ωβ is never zero. m m Assume now that the bundle (T ∗M) is trivial. Then there exists ω Ω (M) which is never ∧ ∈ zero – we will use this form to build an oriented atlas for M. Let U = (Uα, ψα) α A be an atlas { } ∈ for M (it might not be oriented yet). For every α A there exists a never vanishing function fα such that: ∈

ψ∗ (dx dx ) = f ω. α 1 ∧ · · · ∧ m α 14 2 As we will see, not all manifolds are orientable (for example RP is not orientable). 15We say orientable “as a vector bundle”, because TM as a smooth manifold is always orientable (i.e. TTM is orientable as a vector bundle). Try to prove this! LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 49

Now, if fα > 0 we set ψbα = ψα, otherwise if fα < 0 we replace with the chart: n ψbα = ϕ ψα : Uα R ◦ → where ϕ(x , . . . , x ) = ( x , x , . . . , x ) is a diﬀeomorphism which reverses the orientation. It is 1 m − 1 2 m now easy to check that the atlas (Uα, ψbα) α A is an oriented atlas. { } ∈ Let now ω , ω Ωm(M) be two nowhere zero forms. We say that they are equivalent ω ω 1 2 ∈ 1 ∼ 2 if ω1 = fω2 with f > 0. If M is orientable, an orientation for M is the choice of an equivalence class of nowhere vanishing forms (notice that on a connected manifold there are only two possible orientations). Once this choice has been made, then one can build an oriented atlas for M as in the proof of Proposition 167. The choice of the orientation class is denoted by [M]. Example 168 (RP2 is not orientable). We show in this example that RP2 is not orientable 2 2 by proving that (T ∗RP ) is not a trivial bundle. To this end let us ﬁrst consider the form ∧ ω Ω2(R3) given by: b ∈ ω = x dx dx x dx dx + x dx dx . b 1 2 ∧ 3 − 2 1 ∧ 3 3 1 ∧ 2 We restrict this form to the two sphere S2, i.e. we consider ω Ω2(S2) given by: ∈ ω = i∗ωb where i : S2 , R3 is the inclusion. Notice that, denoting by α : S2 S2 the antipodal map, we have: → →

(4.15) α∗ω = ω, − 1 1 1 3 3 because α = S2 and ∗ωb = ωb (here : R R is the map x x). Observe that 2 2 − | − − − → 2 2 7→ − ω Ω (S ) is never zero, hence it deﬁnes a trivialization of (T ∗S ). In other words, the ∈ 2 ∧2 2 map (x, t) tωx gives a trivializing diﬀeomorphism S R (T ∗S ) (we already knew this 7→ × → ∧ bundle is trivial since S2 is orientable). Assume now that RP2 is orientable. Then there exists a nowhere vanishing form η Ω2(RP2) and we can pull-back this form to S2 using the quotient 2 2 ∈ map q : S RP . The resulting form q∗τ is also never vanishing and consequently there exists → a nowhere zero function f : S2 R such that: → q∗η = fω. From the commutativity of the diagram S2 α S2

q q

2 RP we deduce that α∗q∗η = q∗η. However we have:

fω = q∗η = α∗(q∗η) = α∗(fω) = fω, − (the last identity by (4.15)), which contradicts the fact that f is never zero. Definition 169 (Integral of a form of degree m on an oriented m-dimensional manifold). Let M be an orientable and oriented manifold, with oriented atlas (Uα, ψα) α A, and consider the { } ∈ corresponding subordinated partition of unity ρα α A. We define the integral of an m-form τ Ωm(M) as the number: { } ∈ ∈ c Z X Z 1∗ τ = ψα− (ρατ). [M] α A ψα(Uα) ∈ In order to see that this definition does not depend on the choice of the partition of unity or the oriented atlas, we argue as follows. First observe that, once an orientation has been fixed and m an oriented atlas (Uα, ψα) α A has been chosen, then for every form σ Ω (Uα) the integral: { } ∈ Z Z ∈ 1∗ σ = ψα− σ n [Uα] R is well defined, i.e. it does not depend on the choice of the oriented chart. This follows from (4.13), since the transition function for two oriented charts belonging to the same orientation class have 50 ANTONIO LERARIO positive Jacobian. As a consequence, if now (Vβ, ϕβ) β B is another oriented atlas positively { } ∈ oriented with (Uα, ψα) α A, with corresponding partition of unity γβ β B, then: { } ∈ { } ∈ Z Z X 1 τ = (ψα− )∗(ρατ) [M] α A ψα(Uα) ∈ X Z = ρατ α A [Uα] ∈ X Z P = ραγβτ γβ =1 α A,β B [Uα] ∈ ∈ X Z = ραγβτ supp(ραγβ ) Uα Vβ ⊂ ∩ α A,β B [Uα Vβ ] ∈ ∈ ∩ X Z = γβτ. by symmetry β B [Vβ ] ∈ 4.10. Stokes’ theorem. In this section we work with orientable manifolds that can possibly have boundary (see Definition 186 below). We start with the following Lemma. Lemma 170. Let M be orientable with ∂M = . Then ∂M is orientable. 6 ∅

Proof. Let (Vα, ϕα) α A be an oriented atlas for M: { } ∈ 1 det J(ϕ ϕ− ) > 0 whenever V V = . α ◦ β α ∩ β 6 ∅

We claim that (Uα ∂M, ϕα Uα ∂M ) is an oriented atlas for ∂M (here Ab A denotes the { ∩ | ∩ }α Ab ⊆ index set labeling only the charts that intersect∈ ∂M).

Whenever Vα Vβ ∂M = we have that the map ϕαβ = ϕα ϕβ : W H is a diﬀeomorphism between an open∩ set ∩W H6 and∅ its image such that: ◦ → ⊆ . φαβ = ϕαβ W ∂H : W ∂H ∂H, | ∩ ∩ −→ m 1 and we want to prove that det(Jφαβ) > 0 (note that φαβ is a map between open subsets of R − , m 1 where m = dim(M)). Let us put coordinates (x, y) H = R − [0, ) such that H = y 0 ∈ × ∞ { ≥ } and ∂H = y = 0 . Observe that the condition ϕαβ(W ∂H) ∂H implies that, writing the { } (1) (2) m 1 ∩ ⊆ components ϕαβ = (ϕ , ϕ ) R − R, then for all j = 1, . . . , m 1 we have: αβ αβ ∈ × − ∂ϕ(2) ∂ϕ(2) αβ (x, 0) = 0 and αβ (x, 0) > 0. ∂xj ∂y

In particular the Jacobian matrix of ϕαβ at points of the form (x, 0) has the form:

∂φαβ ! ∂x (x) Jϕ (x, 0) = (2)∗ . αβ ∂ϕαβ 0 ∂y (x, 0)

(2) ∂ϕαβ Since ∂y (x, 0) > 0 and det(Jϕαβ) > 0, then it must also be det(Jφαβ) > 0. Deﬁnition 171 (Induced orientation on the boundary). Let M be an orientable and oriented manifold of dimension m with boundary ∂M. The induced orientation on ∂M is ( 1)m the orientation that one obtains considering the restriction to ∂M of an oriented atlas for M−. If [M] is the orientation on M, we denote by [∂M] the induced orientation on the boundary ∂M.

In other words, if the orientation on H is given by the class (in the sense of Section 4.9) of the m form dx1 dxm 1 dy, the induced orientation on ∂H is given by the class of ( 1) dx1 ∧· · ·∧ − ∧ m − ∧· · ·∧ dxm 1. The reason to introduce the ( 1) is to have a formula with no signs in Stokes’ Theorem. − − Theorem 172 (Stokes’ Theorem). Let M be an orientable and oriented manifold of dimension m 1 m, possibly with boundary. For every ω Ωc − (M) we have: Z∈ Z (4.16) dω = ω. [M] [∂M] LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 51

Proof. Let (Uα, ϕα) α A be an oriented atlas for M and ρα α A be a subordinated partition of unity. We will{ prove} that∈ for every α A we have: { } ∈ ∈ Z Z (4.17) d(ραω) = ραω, [Uα] [∂Uα] where ∂Uα is the oriented manifold Uα ∂M (and not the boundary of Uα in M!) with the orientation obtained by restricting the induced∩ orientation [∂M]. From (4.17) the conclusion (4.16) follows, since: ! Z Z X dω = d ραω [M] [M] α A ∈ Z X = d (ραω) [M] α A ∈ X Z = d (ραω) α A [M] ∈ X Z = d (ραω) supp(d(ραω)) Uα ⊂ α A [Uα] ∈ X Z = ραω (4.17) α A [∂Uα] ∈ X Z = ραω supp(ραω) Uα ⊂ α A [M] Z ∈ = ω. [∂M]

In order to prove (4.17) we observe that there are two types of charts in an altas (Uα, ϕα) α A for { } ∈ a manifold with boundary: (i) charts such that Uα ∂M = and (ii) charts such that Uα ∂M = ; we prove (4.17) separately for these two types. ∩ ∅ ∩ 6 ∅ m Let ϕ : Uα R be a chart of the ﬁrst type. Then, by deﬁnition: → Z Z Z 1∗ 1∗ d (ραω) = ϕα− d(ραω) = d ϕα− ραω . m [Uα] [ϕ(Uα)] [R ] 1 m 1 We can write the form ϕ− ∗ ρ ω Ω − (ϕ (U )) as: α α ∈ c α α m 1 X (4.18) ϕ− ∗ ραω = fα,j(x)dx1 dxdj dxm, α ∧ · · · ∧ ∧ · · · j=1 and consequently we have: m 1 X j+1 ∂fα,j (4.19) d ϕ− ∗ ρ ω = ( 1) (x)dx dx . α α − ∂x 1 ∧ · · · ∧ m j=1 j We integrate the summands of (4.19) one by one: Z Z j+1 ∂fα,j j+1 ∂fα,j ( 1) (x)dx1 dxm = ( 1) (x)dx1 dxm n m [R ] − ∂xj ∧ · · · ∧ R − ∂xj ··· Z Z ∞ ∂fα,j = (x)dxj dx1 dxdj dxm m−1 ∂xj ··· ··· R −∞ Z lim fα,j(R) fα,j( R) dx1 dxdj dxm m−1 R − − ··· ··· R Z →∞ = 0 dx1 dxdj dxm = 0, m− R 1 · ··· ···

∂fα,j where fα,j( R) = fα,j(x1,..., R, . . . , xm) and we have used the fact that (x) has compact ± ± ∂xj support. Since for an open set U of the ﬁrst type we have ∂U = , this proves (4.17) in this case. α α ∅ 52 ANTONIO LERARIO

Let us consider now the case of a chart ϕ : U H such that ϕ (U ) ∂H = . Observe first α α → α α ∩ 6 ∅ 1∗ that, writing ϕα− ραω as in (4.18), we have: Z Z m (4.20) ραω = ( 1) fα,m(x, 0)dx1 dxm, [∂Uα] − ∂H ··· where the ( 1)m comes from the definition of induced orientation. Proceeding as before we get: − Z Z m X j+1 ∂fα,j (4.21) d (ραω) = ( 1) (x)dx1 dxm 1dy − ∂x ··· − [Uα] H j=1 j (notice that we have used a different symbol “y” to denote the last coordinate) and again we integrate the summands one by one, i.e. we consider: Z j+1 ∂fα,j (4.22) ( 1) (x)dx1 dxm 1dy. H − ∂xj ··· − For j = m we can write the integral in (4.22) (up to its sign) as: Z 6 Z ∂fα,j ∂fα,j (x, y)dx1 dxm 1dy = (x, y)dx1 dxm 1dy − m− − H ∂xj ··· R 1 [0, ) ∂xj ··· Z × ∞ Z ∞ ∂fα,j = (x, y)dxj dx1 dxdj dxm 1dy m− − R 2 [0, ) ∂xj ··· ··· Z × ∞ −∞ (4.23) = 0 dx1 dxdj dxm 1dy = 0 m−2 [0, ) · ··· ··· − R × ∞ where we have used again the fact that ∂fα,j has compact support. For the last summand we have ∂xj instead: Z Z Z m+1 ∂fα,m m+1 ∞ ∂fα,m ( 1) (x, y)dx1 dxm 1dy = ( 1) (x, y)dy dx1 dxm 1 − m− − − H ∂y ··· − R 1 0 ∂y ··· Z m+1 = ( 1) fα,m(x, 0)dx1 dxm − m−1 − ··· Z R = ραω. (4.20) [∂Uα] Together with (4.21) and (4.23), this proves (4.17) also in this second case and concludes the proof. Exercise 173 (The volume form of Sn). Let us denote by ω Ωn(Sn) the form: ∈ n+1  X j+1 (4.24) ω = i∗  ( 1) xjdx1 dxdj dxn+1 − ∧ · · · ∧ ∧ · · · ∧ j=1 where i : Sn Rn+1 denotes the inclusion. This is called the volume form of Sn. Compute the integral of ω →on Sn, with respect to the orientation induced on Sn by viewing it as the boundary of the disk Dn+1 and orienting Dn+1 in the standard way as a subset of Rn+1 and show that: n+1 Z 2π 2 ω = n+1 . n [S ] Γ 2 Example 174. Let us consider again the form ω Ωn(Sn) defined in (4.24). We claim that Hn(Sn) is generated by [ω]. In fact we know that dω∈ = 0 (because it is a top form), hence its cohomology class is well defined. Moreover, by the previous Exercise 173, we have: Z ω > 0. [Sn] n n n 1 n Assume that [ω] = 0 in H (S ). Then ω = dη for some η Ω − (S ) and consequently, using Stokes’ ∈ Z Z Z

0 < ω = dη = η =n 0, n n Stokes n ∂S = [S ] [S ] [∂S ] ∅ which is a contradiction. LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 53

4.11. Poincaréduality. Let M be an orientable and oriented manifold without boundary and of q n q dimension m. Then we can define a linear map P : H (M) (Hc − (M))∗ as follows: given an q n q −→ element [α] H (M) the linear functional P ([α]) : H − (M) R is defined by: ∈ c → Z β α β. 7−→ [M] ∧ The fact that the map P is well defined follows from Stokes’ Theorem. Poincaréduality asserts that, when the cohomology of M has finite dimension, P is an isomorphism. Theorem 175 (Poincaréduality). Let M be an orientable and oriented manifold of dimension m, q n q with no boundary and with finite-dimensional cohomology. Then P : H (M) (H − (M))∗ is −→ c an isomorphism for every q N. ∈ Proof. We will prove the statement under the additional assumption that M has a finite good cover. Under this assumption the strategy of the proof is another example of application of the Mayer-Vietoris argument: we proceed by induction on the cardinality k of the good cover. (In other words: we will assume Poincaréduality is true for all manifolds with a good cover consisting of at most k elements and we prove it for manifolds with a good cover of cardinality k + 1.) The case k = 0 (the base of the induction) follows immediately from Exercise 164: because n n n Hc∗(R ) = Hc (R ) R, this vector space is generated by [f(x)dx1 dxn], where f n ' n ∧ · · · ∧ ∈ Cc∞(R , R) (the form f(x)dx1 dxn on [R ] cannot be exact by Stokes’ theorem, since its integral does not vanish). ∧ · · · ∧ k+1 For the inductive step we consider a manifold M with a good cover Vj j=1 and write it as M = A B where: { } ∪ A = V V and B = V . 1 ∪ · · · ∪ k k+1 We now “pair together” the long exact Mayer-Vietoris sequence for A, B and the dual of the Mayer-Vietoris sequence with compact support for A, B from Exercise{ 163}: { } Hq(A B) Hq+1(A B) Hq+1(A) Hq+1(B) ∩ ∪ ⊕ P PA∪B PA PB A∩B ⊕ n q n q 1 n q 1 n q 1 (H − (A B))∗ H − − (A B) ∗ H − − (A) ∗ H − − (B) ∗ c ∩ c ∪ c ⊕ c The bottom sequence (the dual of the sequence from Exercise 163) is exact since the dual of an exact sequence is still exact. These two long exact sequences give rise to diagram of maps which are all commutative, up to a sign. The only tricky part is to verify commutativity of the square:

∗ Hq(A B) d Hq+1(A B) ∩ ∪ PA∩B PA∪B , ∗ n q (d∗) n q 1 (H − (A B))∗ H − − (A B) ∗ c ∩ c ∪ where the map d∗ is the connecting homomorphism of the Mayer-Vietoris long exact sequence and (d )∗ is the dual of the connecting homomorphism of the long exact Mayer-Vietoris sequence for compact∗ supports. Given [ω] Hq(A B) we want to prove that the two linear maps ∈ ∩ n q 1 n q 1 (d )∗PA B[ω]: Hc − − (A B) R and PA Bd∗[ω]: Hc − − (A B) R ∗ ∩ ∪ −→ ∪ ∪ −→ n q 1 are the same (up to a sign). In other words we need to prove that for every [η] Hc − − (A B) we have: ∈ ∪ Z Z (4.25) ((d )∗PA B[ω]) [η] = ω d η = d∗ω η = (PA Bd∗[ω]) [η]. ∗ ∩ [A B] ∧ ∗ ± [A B] ∧ ± ∪ ∩ ∪ Let us recall now the deﬁnition of the involved connecting homomorphisms. Let ρA, ρB be a { q+1 } partition of unity subordinated to A, B . A representative d∗ω of d∗[ω] is a form in Ω (A B) such that (see Section 4.6): { } ∪

(4.26) d∗ω = d (ρ ω) and d∗ω = d (ρ ω) . |A − B |B A 54 ANTONIO LERARIO

q Since ω Ω (A B), it follows from (4.26) that d∗ω has support in A B. A representative d η ∈ ∩ n q ∩ ∗ of d [η] is a form in Ωc − (A B) such that (see Exercise 163): ∗ ∩ (4.27) ( d η, d η) = (d (ρAη) , d (ρBη)) . − ∗ ∗ Using this information, we can compute: Z Z ω d η = ω d (ρBη) [A B] ∧ ∗ (4.27) [A B] ∧ ∩ Z ∩ = ω dρB η dη=0 [A B] ∧ ∧ ∩ Z deg ω = ( 1) dρB ω η − [A B] ∧ ∧ Z ∩ deg ω = ( 1) d (ρBω) η = ( ). dω=0 − [A B] ∧ ∗ ∩ Similarly we have: Z Z d∗ω η = d (ρBω) η ∧ supp(d∗ω) A B − ∧ [A B] ⊂ ∩ [A B] ∪ Z∩ = dρBω η = ( ). suppdω=0 − [A B] ∧ ± ∗ ∩ This proves (4.25) and we have established that the diagrams are commutative. Poincar´eduality follows from the ﬁve’s lemma: in fact in the next commutative diagram (this is just the previous diagram where we have denoted by V the cohomology of A B and by W the direct sum of the cohomology of A and the one of B) ∩

V q W q Hq(A B) V q+1 W q+1 ∪ PA∩B PA PB P PA∩B PA PB ⊕ A∪B ⊕ n q n q n q n q 1 n q 1 (V − )∗ (W − )∗ (H − (A B))∗ V − − ∗ W − − ∗ c c c ∪ c c

PA PB and PA B are isomorphisms (by the inductive hypothesis, because A, B and A B have ⊕ ∩ ∩ a good cover of cardinality smaller than k), hence PA B is an isomorphism as well. ∪

4.12. Künneththeorem. Künneththeorem allows to compute the cohomology of a product of two manifolds as a tensor product of the two cohomologies. Observe first that we have a diagram:

M N ×

π1 π2

M N which allows to pull back forms on M and N to forms on M N. × Theorem 176 (K¨unneth). Let M and N be two smooth manifolds. There is an isomorphism

φ : H∗(M) H∗(N) H∗(M N) ⊗ −→ × which is given by:

(4.28) [α] [β] [π∗α π∗β]. ⊗ 7→ 1 ∧ 2 In particular for every q 0 we have: ≥ q M p q p (4.29) H (M N) = H (M) H − (N) × ⊗ p 0 ≥ (where we adopt the convention Hp = 0 for p < 0). LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 55

Proof. (sketch) We will prove the theorem under the additional assumption that M has a ﬁnite good cover, again using the Mayer-Vietoris argument. Since the reader should at this point be familiar with this technique we only sketch an outline, leaving the details to him/her. The base of the induction is just Theorem 146: M has a good cover consisting of just one open set, hence M Rm and consequently: ' m m H∗(R N) H∗(N) = H∗(R ) H∗(N) = R H∗(N). × ' ⊗ ⊗ k+1 For the inductive step, as before, we consider a manifold M with a good cover Vj j=1 and write it as M = A B where: { } ∪ A = V V and B = V . 1 ∪ · · · ∪ k k+1 Then we take the Mayer-Vietoris exact sequence of the pair A, B and, for every p we tensor it with Hp(N), obtaining many exact sequences (tensoring an{ exact} sequence with a vector space keeps exactness): Hq(A B) Hp(N) (Hq(A) Hq(B)) Hp(N) ··· ∪ ⊗ ⊕ ⊗ ··· We then take the direct sum over all p and combine all these sequences together into a single sequence which, after relabeling the indices, looks like:

Lq p p q Lq p p p q (H (A B) H − (N)) ((H (A) H (B)) H − (N)) ··· p=0 ∪ ⊗ p=0 ⊕ ⊗ ··· This exact sequence is naturally paired with the long Mayer-Vietoris exact sequence of the open cover A N,B N for M N: { × × } × Lq p q p Lq p q p n q (H (A B) H − (N)) ((H (A) H − (B)) H − (N)) ··· p=0 ∪ ⊗ p=0 ⊕ ⊗

φA∪B φA φB ⊕ Hq((A B) N) Hq(A N) Hq(B N) ··· ∪ × × ⊕ × where the vertical arrows are the corresponding K¨unnethmaps, as in (4.28). By the inductive hypothesis we know that all these maps – except φA B, are isomorphisms: if we prove that the previous diagram is commutative, the theorem follows∪ again by the ﬁve lemma. The only tricky part is again the commutativity of the piece of the diagram which contains the connecting homomorphisms:

∗ Lq p p q d Lq+1 p+1 q p (H (A B) H − (N)) H (A B) H − (N) p=0 ∩ ⊗ p=0 ∪ ⊗

φA∩B φA∪B ∗ Hq((A B) N) d Hq+1((A B) N) ∩ × ∪ × We leave to the reader the verification that also this piece of the diagram is commutative (one should take, in the definition of the bottom connecting homomorphism, the partition of unity ρ π , ρ π ...) { A 1 B 1} Example 177 (The cohomology of Sn Sn). The cohomology of Sn Sn can be computed using KünnethTheorem. In fact, using (4.29×), we have: × n q n n M q n p q n (4.30) H (S S ) H (S ) H − (S ) × ' ⊗ p=0 n 0 n n n and, using the fact that H∗(S ) = H (S ) H (S ), the only non-zero cohomology in (4.30) is: ⊕ H0(Sn Sn) H0(Sn) H0(Sn) × ' ⊗ Hn(Sn Sn) H0(Sn) Hn(Sn) Hn(Sn) H0(Sn) × ' ⊗ ⊕ ⊗ H2n(Sn Sn) Hn(Sn) Hn(Sn). × ' ⊗ Using the explicit formula (4.28) of the Künnethisomorphism, if we denote by [ω] Hn(Sn) the generator from Example 174, we have: ∈ 0 n n n n n 2n n n H (S S ) = [f 1] ,H (S S ) = [π∗ω], [π∗ω] ,H (S S ) = [π∗ω π∗ω] . × h ≡ i × h 2 1 i × h 1 ∧ 2 i 56 ANTONIO LERARIO

Exercise 178. Using Künnethformula, compute the cohomology of S1 S1 (k many factors). ×· · ·× Exercise 179. Compute the cohomology of S1 S1 minus a point. × 4.13. Invariant forms. Let M be a smooth manifold and G be a finite group acting freely on it. Then the quotient space M/G can be endowed the structure of a smooth manifold in such a way that the quotient map π : M M/G is a normal covering space and a smooth submersion. The De Rham cohomology of M/G→can be computed using the following theorem. Theorem 180. Let π : M M/G be the quotient under a free action of a finite group. Then the → map π∗ : H∗(M/G) H∗(M) is injective with image the vector space: → G H∗(M) = [ω] H∗(M) such that g∗[ω] = [ω] for every g G . { ∈ ∈ }

Proof. We first observe a preliminary fact: if a form ω Ω∗(M) satisfies g∗ω for all g G (such ∈ ∈ a form is called “invariant”), then it descends to a form ω Ω∗(M/G) such that π∗ω = ω (try to make this rigorous using the definition of a form as a bunch∈ of local sections gluing together nicely...)

We start by proving injectivity of π∗. Let [α] H∗(M/G) such that π∗[α] = 0, i.e. such that ∈ π∗α = dη for some η Ω∗(M). Then we can consider the form: ∈ 1 X η = g∗η, b G g G | | ∈ which is readily seen to be invariant and descends to a form η on M/G. The form η satisfies dα = η, hence [α] = 0. G Let us prove now that the image of π∗ equals H∗(M) . One inclusion is clear, since g∗π∗[α] = π∗[α] (for every g G we have π g = π). For the other inclusion, let [ω] H∗(M) such that ∈ ◦ ∈ g∗[ω] = [ω] for every g G. Then: ∈ g∗ω = ω + dηg 1 P for some form ηg. Consider the invariant form ωb = G g g∗ω. Then there is a closed form ω on | | M/G such that π∗ω = ωb and [ωb] = π∗[ω]. It remains to show that [ωb] = [ω]. To this end observe that:   1 X 1 X 1 X 1 X g∗ω ω = (g∗ω ω) = dηg = d  ηg , G − G − G G g g g g G | | | | | | | | ∈ and ω and ωb are co-homologous. n n n Example 181 (The cohomology of RP revised). Let G = idSn , α Z2 where α : S S is n n { }' → the antipodal map x x. Then RP = S /Z2 and 7→ − n n H∗(RP ) H∗(S )Z2 . ' We know from Example 174 that Hn(Sn) is generated by the volume form [ω] and consequently: n H∗(S ) = [f 1], [ω] . h ≡ i On the other hand an explicit computation shows that: n+1 α∗[f 1] = [f 1] and α∗[ω] = ( 1) [ω]. ≡ ≡ − Thus, using Theorem 180, we see that the cohomology of RPn can be identified with: * n+1 + n 1 + ( 1) H∗(RP ) [f 1], − [ω] ' ≡ 2 Example 182 (The cohomology of G(2, 4)). Recall from Exercise 38 that G(2, 4) can be embedded into RP5 through a map p : G(2, 4) RP5 whose image is: → p(G(2, 4)) = p p + p p + p p = 0 { 12 34 13 24 14 23 } 5 (here [p12, p13, p14, p23, p24, p34] denote the homogeneous coordinates of RP ). The quadric defining p(G(2, 4)) has signature (3, 3) and putting it in normal form, we can find coordinates [y0, . . . , y5] in RP5 such that: G(2, 4) q(y) = y2 + y2 + y2 y2 y2 y2 = 0 . '{ 0 1 2 − 3 − 4 − 5 } LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 57

Let us look at the manifold deﬁned by the same equation in S5: M = (y , . . . , y ) S5 q(y) = y2 + y2 + y2 y2 y2 y2 = 0 . { 0 5 ∈ | 0 1 2 − 3 − 4 − 5 } It is easy to see that M is a smooth submanifold of S5 (because the equation is regular) which double covers G(2, 4). The antipodal map α : S5 S5 restricts to a map τ : M M whose quotient is G(2, 4): → → π : M G(2, 4) is a normal covering space. → It follows from Theorem 180 that:

H∗(G(2, 4)) H∗(M)Z2 ' 6 2 where Z2 = idM , τ . The manifold M is given in R by the equations y = 2 (we take the sphere { } k k of radius √2 to make the notation simpler) and q(y) = 0. Using coordinates (w, z) R3 R3, these two equations are equivalent to: ∈ × M = w 2 = 1, z 2 = 1 , k k k k hence M S2 S2 R3 R3. The cohomology of M is given by K¨unnethTheorem: denoting by π , π : S2' S2× S⊂2 the× two projections and by [ω] H2(S2) the generator from Example 174, 1 2 × → ∈ 2 2 H∗(M) = H∗(S S ) = [f 1], [π∗ω], [π∗ω], [π∗ω π∗ω] . × h ≡ 1 2 1 ∧ 2 i The action of Z2 on M is given by: τ(z, w) = ( z, w), − − and consequently:

τ ∗[f 1] = [f 1], τ ∗[π∗ω] = [π∗ω], τ ∗[π∗ω] = [π∗ω], τ ∗[π∗ω π∗ω] = [π∗ω π∗ω]. ≡ ≡ 1 − 1 2 − 2 1 ∧ 2 1 ∧ 2 From this, using Theorem 180, we see that the cohomology of G(2, 4) is isomorphic to:

H∗(G(2, 4)) [f 1], [π∗ω π∗ω] . ' h ≡ 1 ∧ 2 i (Notice in particular that G(2, 4) is orientable.) Exercise 183 (The cohomology of a projective quadric). Generalizing the previous example, let q : Rn+1 R be a nondegenerate quadratic form of signature (a, b), a + b = n + 1. Compute the → cohomology of q = 0 RPn and q = 0 Sn. { } ⊂ { } ⊂ 1 1 Exercise 184. The Klein bottle K can be seen as (S S )/Z2, where Z2 = idS1 , τ and τ(z, w) = (1/z, w). Compute the cohomology of K. Is K ×orientable? { } − Exercise 185. Prove that a compact and connected manifold M of dimension n is orientable if and only if Hn(M) R and that M is not orientable if and only of Hn(M) = 0. (Hint: use the orientation double cover,' as deﬁned in [6, pag 234]...) 58 ANTONIO LERARIO

5. Appendix

5.1. Manifolds with boundary. In this section we introduce the notion of a manifold with boundary. To start with, define a half-space H Rn as: ⊂ n H = x R λ(x) 0 { ∈ | ≥ } where λ : Rn R is a linear map. The half-space is called proper if λ = 0, in which case the boundary of H→is the set: 6 ∂H = x H λ(x) = 0 . { ∈ | } The definition of a manifold chart can now be extended to be a map ϕ : U Rn which is a homeomorphism to an open subset of a half-space. → Definition 186 (Manifold with boundary). A manifold with boundary, of dimension m and class Ck is a paracompact Hausdorff space M such that: (1) for every point x M there exists a neighborhood U of x and a continuous function ψ : U H which∈ is a homeomorphism onto an open subset of a half-space H (the pair (U, ψ)→ is still called a chart); (2) for every pairs of charts (U , ψ ) and (U , ψ ) such that U U = the map 1 1 2 2 1 ∩ 2 6 ∅ 1 m ψ2 ψ1 U− U : ψ1(U1 U2) R ◦ | 1∩ 2 ∩ → is a Ck map16.

By the Invariance of Domain Theorem [6, Theorem 2B.3], a coordinate change cannot map an interior point of a half-space into a boundary point; consequently the set of boundary points of M (points which in some chart are mapped the boundary of a half-space), is well deﬁned and called the boundary of M.

5.2. The inverse and the implicit function theorem. These are basic tools in Diﬀerential 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 [4].

Theorem 187 (Inverse function theorem). Let U Rn be an open set and f : U Rn be a k ⊆ → differentiable function of class C , k 1. If the Jacobian matrix Jf(x) is nonsingular at x0 there ≥ n exists a neighborhood A of x0 such that f(A) is open and f A : A f(A) R is invertible; moreover 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 Invariance of Domain Theorem; moreover, using the language of Definition 28 below, the function f A is a diffeomorphism. |

Theorem 188 (Implicit function theorem). Let U Ra+b be an open set with coordinates (x, y) ⊆ ∂f ∈ Ra Rb and f : U Rb be a diﬀerentiable function of class Ck, k 1. If the matrix (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 ϕ : A B of class Ck such that: 0 0 → f = f(x , y ) (A B) = graph(ϕ). { 0 0 } ∩ × 5.3. Bump functions.

Deﬁnition 189 (Bumb function). Let 0 < c1 < c2 be positive distinct numbers. A bumb function n is a C∞ function λ : R [0, 1] such that (see Figure8): → (1) λ(x) = 1 for x c ; k k ≤ 1 (2)0 < λ(x) < 1 for c1 < x < c2; (3) λ(x) = 0 for x c .k k k k ≥ 2

Lemma 190. Bumb functions exist for every 0 < c1 < c2.

16 k A map f : W1 → W2 between open sets in half-spaces W1 ⊆ H1 and W2 ⊆ H2 is called C if it extends to a k ˜ n C map on a open neighborhood W1 ⊂ W1 ⊂ R . LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 59

b b b b c c c c − 2 − 1 1 2

Figure 8. A bump function

Proof. Let α : R R be the C∞ function → 1 α(x) = e− x χ x>0 (x), { } where χ x>0 (x) denotes the characteristic function of x > 0 . Out of this function we construct { } { } γ : R R by: → R c2 α(t c )α(c t)dt x − 1 2 − γ(x) = R c2 . α(t c1)α(c2 t)dt c1 − − Finally we set λ(x) = γ( x ). (The reader can see [7, Section 2.2] for pictures explaining the k k construction step by step.)

5.4. Partitions of unity.

Definition 191 (Partition of unity). Let M by a smooth manifold and U = Uj j J be an open { } ∈ cover of M.A partition of unity subordinated to U is a collection of smooth maps ρj : M [0, 1] J such that: { → }j ∈ (1) for every j J the support supp(ρ ) U ; ∈ j ⊂ j (2) the family supp(ρj) j J is locally finite; P { } ∈ (3) j J ρj 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 192. Let U = Uj j J be an open cover for M. Then there is a locally ﬁnite atlas { } ∈ m (Vα, ψα) α A such that V α α A reﬁnes U, each ψα(Vα) R is bounded and each V α M is { } ∈ { } ∈ ⊂ ⊂ compact. Moreover the open cover V = Vα α A has a shrinking W = Wα α A, i.e. W is itself { } ∈ { } ∈ a cover and for every α A we have W V . ∈ α ⊂ α Proof. The proof is left as an exercise. Work out the details carefully! Proposition 193. Let M be a smooth manifold. Every open cover U of M admits a subordinate partition of unity.

Proof. Let U = Uj j J be the open cover. We observe first that if we can construct a partition { } ∈ of unity λi i I subordinated to a refinement V = Vi i I of U, then we can also construct a partition{ of} unity∈ subordinated to U. In fact let g : I{ }J∈ be such that for every i I we have → ∈ the inclusion Vi Ug(i) (the existence of such g is in the definition of refinement) and define for j J the function:⊂ ∈ X ρj(x) = λi(x). i g−1(j) ∈ It is easy to check that ρj j J is a partition of unity subordinated to U (essentially to obtain the { } ∈ ρj we collate together some of the λi). Let now V be the open cover produced by Lemma 192 and W its shrinking. For every α A m ∈ we cover the compact set ψα(W α) R by finitely many closed balls contained in ψα(Vα): ⊂ k [α ψ (W ) B , α α ⊂ α,k k=1 m where Bα,k = B(xα,k, rα,k) for some points xα,k R and radii rα,k > 0. For every ball Bα,k m ∈ let λα,k : R [0, 1] be a bump function centered at xα,k such that λα,k(x) > 0 if and only if → 60 ANTONIO LERARIO x int(Bα,k) (it suffices to take a bumb function λ, as in Lemma 190, with c1 < rα,k and c2 = rα,x ∈ Pkα and then translate it at xα,k, i.e. λα,k(x) = λ(x xα,k)). Put now λα = k=1 λα,k and define µ : M [0, ) by: − α → ∞ µ (x) = χ (x) λ (ψ (x)). α Vα · α α Each µ is a smooth function, strictly positive on W and with supp(µ ) V . We define the α α α ⊂ α desired partition of unity ρα α A subordinated to V by: { } ∈ µα(x) ρα(x) = P . β A µβ(x) ∈ Exercise 194 (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 ≡ 5.5. A first approximation result. Partitions of unity are used to glue together locally defined object on a manifold (for example vector fields, Riemannian metrics...). Here is an important application of the existence of partitions of unity: we can approximate continuous functions f : M Rn with smooth functions (we refine this result in Theorem 129). → Theorem 195. Let M be a compact, smooth manifold and f : M Rn be a continuous function. n → Then, for every > 0 there is a smooth function f : M R such that: → max f(x) f(x) . x M k − k ≤ ∈ Proof. Let > 0 and for every z M let U be an obpen neighborhood of z such that ∈ z (5.1) f(z) f(y) for all y U . k − k ≤ ∈ z consider the open cover Uz z M and let ρz z M be a partition of unity subordinated to this cover. Define the smooth{ function:} ∈ { } ∈ X f(x) = ρz(x)f(z). z M ∈ Let us evaluate

X X f(x) f (x) = ρ (x)f(z) ρ (x)f(x) k − k z − z z M z M ∈ ∈ X = ρ (x)(f(z) f(x)) z − z M X∈ ρ (x) f(z) f(x) = ( ) (since ρ 0). ≤ z k − k ∗ z ≥ z M ∈ Now, either (1) x supp(ρz) Uz, in which case ρz(x) f(z) f(x) ρz(x) by (5.1) or (2) ρ (x) = 0, in which∈ case also ρ⊂(x) f(z) f(x) = 0 ρ k(x). Hence− k ≤ z z k − k ≤ z X ( ) ρ (x) = . ∗ ≤ z z M ∈ This concludes the proof.

5.6. Some homological algebra. We recall here some basic facts on homological algebra. A diﬀerential complex is a sequence of vector spaces with a diﬀerential d such that d2 = 0:

q 1 d q d q+1 C − C C ··· ···

d2=0 The condition that d2 = 0 implies that we can deﬁne the cohomology of the complex by

ker(d q ) Hq(C) = |C . im(d q−1 ) |C LECTURE NOTES ON BASIC DIFFERENTIAL TOPOLOGY 61

Given two diﬀerential complexes (A, dA) and (B, dB), a chain map is a linear map

f : A∗ B∗ such that fd = d f. −→ A B The condition fdA = dBf implies that there is a well deﬁned map:

f ∗ : H∗(A) H∗(B) −→ given by f ∗[α] = [f(α)]. An exact sequence of diﬀerential complexes f g 0 A B C 0 is called a short exact sequence. Proposition 196. A short exact sequence of diﬀerential complexes:

f g 0 A B C 0 in which f and g are chain maps, induces a long exact sequence between the cohomologies of these complexes:

f ∗ g∗ ∗ Hq(A) Hq(B) Hq(C) d Hq+1(A) ··· ···

Proof. The interesting part of the proof is the construction of the map d∗. This is done as follows. Pick [c] Hq(C), with c Cq. such that d c = 0 and consider the commutative diagram: ∈ ∈ C ......

f g 0 Aq+1 Bq+1 Cc+1 0

dA dB dC . f g 0 Aq Bq Cq 0

...... Since g is surjective, there exists b Bq such that g(b) = c. Consider db. By commutativity of the diagram, 0 = dg(b) = g(db) and∈ consequently db is in the kernel of g and, by exactness, there exists a Aq+1 such that f(a) = db. We set: ∈ d∗[c] = [a]. The rest of the proof is left as an exercise to the reader. Exercise 197. Consider the exact sequence of ﬁnite dimensional vector spaces:

0 V1 V2 Vm 1 Vm 0 ··· − Prove that Pm ( 1)k dim(V ) = 0. k=1 − k References [1] V. I. Arnold. Mathematical methods of classical mechanics, volume 60 of Graduate Texts in Mathematics. Springer-Verlag, New York, [1989?]. Translated from the 1974 Russian original by K. Vogtmann and A. Weinstein, Corrected reprint of the second (1989) edition. [2] M. Artin. Algebra. Prentice Hall, Inc., Englewood Cliffs, NJ, 1991. [3] R. Bott and L. W. Tu. Differential forms in algebraic topology, volume 82 of Graduate Texts in Mathematics. Springer-Verlag, New York-Berlin, 1982. [4] M. Cornalba. Note di geometria differeziale. http://mate.unipv.it/cornalba/dispense/geodiff.pdf. [5] V. Guillemin and A. Pollack. Differential topology. AMS Chelsea Publishing, Providence, RI, 2010. Reprint of the 1974 original. [6] A. Hatcher. Algebraic topology. Cambridge University Press, Cambridge, 2002. [7] M. W. Hirsch. Differential topology, volume 33 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1994. Corrected reprint of the 1976 original. [8] J. M. Lee. Introduction to smooth manifolds, volume 218 of Graduate Texts in Mathematics. Springer, New York, second edition, 2013. [9] J. W. Milnor. Topology from the differentiable viewpoint. Princeton Landmarks in Mathematics. Princeton University Press, Princeton, NJ, 1997. Based on notes by David W. Weaver, Revised reprint of the 1965 original. 62 ANTONIO LERARIO

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