De Rham Cohomology In this section we will see how to use differential forms to detect topological prop- erties of smooth manifolds. Question. Is there a smooth function F : U → R such that ∂F ∂F (1) = f , and = f ; i.e. dF = f dx + f dx ? ∂x1 1 ∂x2 2 1 1 2 2 ∂2F ∂2F Since ∂x2∂x1 = ∂x1∂x2 , we must have ∂f ∂f (2) 1 = 2 ; i.e. the form f dx + f dx is closed. ∂x2 ∂x1 1 1 2 2 The appropriate question is therefore whether F exists, assuming that f = (f1,f2) satisfies (2). Is condition (2) also sufficient? Example 1. Consider the function f : R2 → R2 given by  −x2 x1  f(x1,x2)= 2 2 , 2 2 x1 + x2 x1 + x2 Then (2) is satisfied. However, there is no function F : R2 \{0}→R that satisfies (1). Indeed, assume there were such an function F satisfying (2), then 2π d Z F (cos θ, sin θ)dθ = F (1, 0) − F (1, 0) = 0. 0 dθ which contradicts the fact that d ∂F ∂F F (cos θ, sin θ)= (− sin θ)+ cos θ dθ ∂x ∂y

= − f1(cos θ, sin θ) sin θ + f2(cos θ, sin θ) cos θ =1. 2 Theorem 1. Let U ⊂ R be open and star-shaped w.r.t. the point x0.For 2 any smooth function (f1,f2):U → R that satisfies (2), there exists a function F : U → R satisfying (1). 2 Proof. Assume w.l.o.g. that x0 =0∈ R . Consider the function F : U → R, 1 Z F (x1,x2)= (x1f1(tx1,tx2)+x2f2(tx1,tx2))dt. 0 Then one has 1 ∂F Z ∂f1 ∂f2 (x1,x2)= [f1(tx1,tx2)+tx1 (tx1,tx2)+tx2 (tx1,tx2)]dt ∂x1 0 ∂x1 ∂x1 and d ∂f1 ∂f1 tf1(tx1,tx2)=f1(tx1,tx2)+tx1 (tx1,tx2)+tx2 (tx1,tx2). dt ∂x1 ∂x2 Hence 1 ∂F Z  d  ∂f2 ∂f1  (x1,x2)= tf1(tx1,tx2)+tx2 (tx1,tx2) − (tx1,tx2) dt ∂x1 0 dt ∂x1 ∂x2 t=1

=[tf1(tx1,tx2)] = f1(x1,x2). t=0 ∂F Analogously, = f2(x1,x2).  ∂x2

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1 2

Example 1 and Theorem 1 suggest that the answer to Question depends on the topology of U. 1 1 ∞ Definition. HdR(U)={ω ∈A (U):dω =0}/{df : f ∈ C (U)}. Theorem 1 can be reformulated as follows. 1 2 Corollary. HdR(U)=0whenever U ⊂ R is star-shaped. Example 1 implies the following. 1 2 Corollary. HdR(R \{0}) =06 .

• As another basic example, we consider the two-torus T2 = S1 × S1. We parametrize points on the torus via the usual angular coordinates (θ, ψ), θ, ψ ∈ [0, 2π), of the two circles. (Note that strictly speaking, these coordinates give a coordinate chart only on a dense subset of T2.) • Loosely speaking, among all oriented closed surfaces, the torus is characterized by having one hole. In other words, among all such surfaces only the torus has exactly two “really different” noncontractible loops, a :[0, 2π] → T2,b:[0, 2π] → T2, given by a(t)=(t, 0),b(t)=(0,t). Here “really different” means that a and b cannot be deformed into a loop which travels n times around a and m times around b for some m, n ∈ Z. — Using the map (m, n) 7→ na + mb, we see that the group Z ⊕ Z appears naturally as an algebraic description of the one hole in the torus. This group is called the fundamental group of the torus. • We want to construct differential forms which detect the difference be- tween the loops a, b, but which are insensitive to smooth deformation of these loops. — As candidates, we take the one-forms dθ, dψ ∈A(T2), which satisfy

Z dθ = Z dψ =2π, Z dθ = Z dψ =0. a b b a Moreover, if F :[0, 1] × [0, 1] → T2, F (0,t)=a(t), is a smooth deformation of a to a loop a∗, a∗(t)=F (1,t), then by Stokes’ theorem,

Z dθ = Z dθ + Z d(dθ)=Z dθ. a a∗ F ([0,1]×[0,1]) a∗ Thus dθ, dψ satisfy the conditions we specify. • The key property of dθ used in the preceeding computation is that dθ is closed, i.e. d(dθ)=0. 3

However, for any f ∈ C∞(T2), dθ + df is also closed with

Z (dθ + df )=Z dθ + Z f = Z dθ, a a ∂a a and similarly for the integral over b. — Thus as far as integration over loops is concerned, the forms dθ and dθ + df are indistinguishable; similar remarks hold for dψ. — Since the exact one-forms {df : f ∈ C∞(T2)} are included in the closed one- forms, we can quotient out the ambiguity by setting the first cohomology group of the torus to be

1 2 1 2 ∞ 2 HdR(T )={ω ∈A (T ):dω =0}/{df : f ∈ C (T )}.

1 2 HdR(T ) is naturally a real vector space and so is a group only in trivial sense. • The basic information contained in this vector space is its dimension. 1 2 Proposition. dim HdR(T )=2 Proof. Let ω = αdθ + βdψ be a closed one-form. Then ∂α ∂β (3) = . ∂ψ ∂θ

This equation pulls back to R2, and we consider α, β as periodic functions on R2. Then (3) is the integrability condition, i.e. the necessary and sufficient condition, to solve the equations ∂f ∂f = α, = β ∂θ ∂ψ for f ∈ C∞(R2). This is equivalent to solving

df = ω in R2,

which is valid on T2 iff f defines a function on the quotient T2 = R2/(R ⊕ R). In other words f must be periodic:

f(θ +2πn, ψ +2πm)=f(θ, ψ), ∀n, m ∈ Z.

(i) If α, β are constants not both zero, then

f(θ, ψ)=αθ + βψ + C, for some constant C.

Hence f does not descend to a function on T2. 2 1 2 This shows that in this case ω =6 df on T , and thus dim HdR(T ) ≥ 2. (ii) To study what happens when α, β have no constant part, we write the Fourier expansion of α, β:

inθ imψ inθ imψ α(θ, ψ)= X an,me e ,β(θ, ψ)= X bn,me e , n,m∈Z n,m∈Z 4

and assume that a0,0 = b0,0 = 0. Under this assumption, we have

(4) Z α = Z α = Z β = Z β =0. a b a b

Lifting back to R2 and using the explicit solution

(θ ,0) (θ ,ψ ) Z 0 Z 0 0 f(θ0,ψ0)=f(0, 0) + α(θ, 0)dθ + β(θ0,ψ)dψ, (0,0) (θ0,0)

we obtain

f(θ0 +2πn, ψ0 +2πm) − f(θ0,ψ0) (θ +2πn,0) (θ ,ψ ) Z 0 Z 0 0 = α(θ, 0)dθ − β(θ0,ψ)dψ (θ0,0) (θ0,0) (θ +2πn,ψ +2πm) Z 0 0 + β(θ0 +2πn, ψ)dψ (θ0+2πn,0) (θ ,ψ +2π) (θ +2π,0) Z 0 0 Z 0 (5) =m β(θ0,ψ)dψ + n α(θ, 0)dθ, (θ0,ψ0) (θ0,0)

by the periodicity of α(θ, ψ), β(θ, ψ). – Since the integrals in (4) are unchanged under deformation of a, b to vertical and horizontal circles, respectively, the right-hand side of (5) vanishes. ⇒ f is periodic. ⇒ df = ω on T2. 1 2 (iii) By (i) and (ii), HdR(T ) is precisely two dimensional with basis [dθ], [dψ]. 

Definition. The kth de Rham cohomology group of a manifold M is given by

k 2 k k−1 HdR(T )={ω ∈A (M):dω =0}/{dθ : θ ∈A (M)}.

k 1 th The dimension β of HdR(M) is called the k Betti number. • Since the spaces of closed and exact forms are both infinite dimensional, it is nontrivial fact that βk < ∞ for all k whenever M is compact. 5

Definition. Let M be an n-dimensional C∞ manifold. A k-form ω ∈Ak(M) is called a closed form if dω =0, and an exact form if there exists a (k − 1)-form η such that ω = dη. • Since d ◦ d = 0, any exact form is a closed form. Definition. Let

Zk(M) =Ker{d : Ak(M) →Ak+1(M)} = the set of all closed k-forms, Bk(M) =Im{d : Ak−1(M) →Ak(M)} = the set of all exact k-forms.

• Both Zk(M) and Bk(M) are linear subspaces of Ak(M). Definition. (1) Let M be a smooth n-manifold. The quotient space

k k k HdR(M)=Z (M)/B (M)

is called the k-dimensional de Rham cohomology group of M. (2) For a closed k-form ω ∈Ak(M), we denote the class it represents in the de Rham k cohomology group by [ω] ∈ HdR(M), and call this the de Rham cohomology class represented by ω. (3) Call the direct sum n ∗ k HdR(M)=M HdR(M) k=0 the de Rham cohomology group of M.

•A∗(M) is equipped with a product structure defined by the wedge poduct. ∗ The product structure induced that on HdR(M) in the following way. k ` k If x ∈ HdR(M), y ∈ HdR(M) are represented by closed forms ω ∈ Z (M), η ∈ Z`(M), respectively, then we set

k+` xy =[ω ∧ η] ∈ HdR (M), where (1) since d(ω ∧ η)=dω ∧ η +(−1)kω ∧ dη, ω ∧ η is a closed form; and (2) if ω0 = ω + dξ, η0 = η + dτ, then

ω0 ∧ η0 =(ω + dξ) ∧ (η + dτ) =ω ∧ η + d((−1)kω ∧ τ + ξ ∧ η + ξ ∧ dτ).

Hence the product xy is determined independently of the choice of closed forms representing x, y.

H∗(M) equipped with the product structre is called the de Rham cohomology algebra. 6

• Consider a smooth map f : M → N. This induces the pullback map f ∗ : Ak(N) →Ak(M). The equation f ∗dω = df ∗ω yields

dω =0⇒ df ∗ω = f ∗ω =0,

and θ = dω ⇒ f ∗θ = f ∗dω = df ∗ω. Thus f ∗ takes closed forms to closed forms and exact forms to exact forms. Therefore f ∗ induces a map, also called f ∗, on de Rham cohomology,

∗ k k f : HdR(N) → HdR(M),

given by f ∗[ω]=[f ∗ω]. • We have 0 0 HdR(M)={f ∈A (M):df =0}, which means that f must be constant on each connected component of M. Thus, we obtain the following Proposition. If M has q components, we have

0 ∼ q HdR(M) = R and β0 = q. 7

• To explain the topological significance of the higher Betti numbers, we claim: if M is smoothly homotopy equivalent to N, then k ∼ k HdR(M) = HdR(N), where the homotopy eqivalene means that there exist maps f : M → N, g : N → M F :[0, 1] × M → M, G :[0, 1] × N → N, with F (0,x)=gf(x),F(1,x)=x, G(0,y)=fg(y),G(1,y)=y. To show the homotopy equivalence property, we first consider a smooth map f : M → N. This induces the pullback map f ∗ : Ak(N) →Ak(M). The equation f ∗dω = df ∗ω yields dω =0⇒ df ∗ω = f ∗ω =0, and θ = dω ⇒ f ∗θ = f ∗dω = df ∗ω. Thus f ∗ takes closed forms to closed forms and exact forms to rxact forms. Therefore f ∗ induces a map, also called f ∗, on de Rham cohomology,

∗ k k f : HdR(N) → HdR(M), given by f ∗[ω]=[f ∗ω].

Definition. Let it : M → [0, 1] × M be the map

it(x)=(t, x), ∀t ∈ [0, 1]. Define Φ:Ak([0, 1] × M) →Ak−1(M) as follows: any k-form ω on [0, 1] × M has the local coordinate expression

i1 ik j1 jk ω =fi1···ik dx ∧···∧dx + gj1···jk−1 (t, x)dt ∧ dx ∧···∧dx

=ω1 + dt ∧ η,

i1 ik j1 jk−1 where ω1 = fi1···ik dx ∧···∧dx and η = gj1···jk−1 (t, x)dx ∧···∧dx have no dt terms. Set 1 Z ∗ k−1 ∗ Φωp = it η(t, p)dt ∈A Tp M 0 1 Z j1 jk−1 = X ( gj1···jk−1 (t, x)dt)dx ∧···∧dx . 0 j1<···jk−1

Let dM , d[0,1]×M denote the exterior derivatives on M,[0, 1] × M, respectively. 8

∗ ∗ k Lemma 1. i1 − i0 = dM ◦ Φ+Φ◦ d[0,1]×M on A ([0, 1] × M). Proof. Since both sides of the equations are linear operators on Ak([0, 1] × M), it suffices to check two cases; namely, i1 ik I Case 1. ω = fi1···ik dx ∧···∧dx = fI (t, x)dx . J Case 2. ω = gJ (t, x)dt ∧ dx . We consider them seperately below. i1 ik I Case 1. ω = fi1···ik dx ∧···∧dx = fI (t, x)dx . In this case

∂f d ω = (garbage) + I dt ∧ dxI . [0,1]×M ∂t

Thus

dΦ(ω)=0, (∵ Φ(ω)=0) 1 Z ∂fI I Φ ◦ d[0,1]×M ω =( (t, x)dt)dx 0 ∂t I =[fI (1,x) − fI (0,x)]dx , ∗ ∗ I i1ω − i0ω =[fI (1,x) − fI (0,x)]dx .

J ∗ ∗ Case 2. ω = gJ (t, x)dt ∧ dx . In this case, since i1dt = i0dt = 0, we have

∗ ∗ i1ω = i0ω =0.

On the other hand,

∂g Φ ◦ d ω =Φ(− J dt ∧ dxα ∧ dxJ ) [0,1]×M ∂xα 1 ∂g − Z J α ∧ J = ( α dt)dx dx 0 ∂x 1 Z  J  dM Φω =d gJ (t, x)dt dx 0 ∂ 1 Z  α ∧ J = α gJ (t, x)dt dx dx , ∂x 0

and hence Φ ◦ d[0,1]×M ω + dM ◦ Φω =0. 

• Now assume that f0, f1 : M → N are smoothly homotopic maps with the homotopy given by F :[0, 1] × M → N,

F (0,x)=f0(x),F(1,x)=f1(x). Equivalently, we have F ◦ i0 = f0,F◦ i1 = f1. 9

This implies

∗ ∗ ∗ ∗ f1 − f0 =(F ◦ i1) − (F ◦ i0) ∗ ∗ ∗ ∗ =i1F − i0F =dΦF ∗ − Φdf ∗, by Lemma 1 =dΦF ∗ − ΦF ∗d.

If [ω] ∈ Hk(N), then

∗ ∗ ∗ ∗ ∗ f1 ω − f0 ω = dΦF ω − ΦF dω = dΦF ω,

and hence ∗ ∗ ∗ ∗ f1 [ω]=[f1 ω]=[f0 ω]=f0 [ω]. Thus, we obtain the following Proposition. Smoothly homotopic maps of manifolds induce identical maps on de Rham cohomology groups. Theorem 2. If M and N are smoothly homotopy equivalent manifolds of mani- k ∼ k folds, then HdR(M) = HdR(N). Proof. If M and N are smoothly homotopy equivalent via maps

f : M → N, g : N → M,

then fg is homotopic to the identity map on N. Hence the induced maps on de Rham cohomology satisfy

∗ ∗ ∗ ∗ g f =(fg) = (IdN ) =IdN

∗ ∗ and similarly f g =IdM . Thus

∗ k k f : HdR(N) → HdR(M)

is an isomorphism of de Rham cohomology groups for all k. 

• By induction on n, we obtain the following corollary. Corollary 4 (Poincar´elemma). The de Rham cohomology of Rn is trivial. That is, R,k=0 Hk(Rn)=Hk(one point)= 0,k>0. In other words, if ω ∈Ak(Rn)(k>0) is an arbitrary closed form, there exists a (k − 1)-form η such that ω = dη.