Differential forms and the de Rham cohomology - Part I

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1 Differential forms and the de Rham cohomology - Part I Paul Harrison University of Toronto October 30, 2009

2 I. Review Triangulation of Manifolds M = smooth, compact, oriented n-manifold. Can triangulate M, i.e. simplicial n-complex K, homeomorphism π : K M s.t. σ K coordinate neighbourhood (U,f ) near σ s.t. f 1 π is affine in σ. Write σ instead of π(σ). σ k K orientable by ordering vertices: σ k = q 0 q k. Given {q λ0,...,q λr } {q 0,...,q k }, τ = q λ0 q λr called r-face of σ. Inductively, orientation of τ i = q 0 q i 1 q i+1 q k agrees with orientation of σ iff i is even. Write F(σ) = set of faces of σ, boundary σ = σ\σ with appropriate orientation. k-chain A = a i σ k i Σ k is a formal sum of k-simplices. A = a i σ k i. For every simplex σ Σ k, there is a cosimplex σ Σ k defined by σ σ = 1, σ τ = 0 for τ σ. Define coboundary: ( A) B = A ( B).

3 Review Differential forms Λ k (T p M) = set of k-linear alternating functions on (T p M) k. Differential k-form ω Ω k (M) : p M ω(p) Λ k (T p M). Let f : M N be a smooth map, Df = tangent map of f. For ω Ω k (N) define f ω Ω k (M): (f ω)(p)(v 1,...,v k ) = ω (f (p))(df (p)(v 1 ),...,Df (p)(v k )) Cover compact n-manifold M with coordinate charts U i R n f i M, i = 1..k. Then partition of unity {φ i : M R} i=1..k s.t.: 1. All φ i are smooth 2. Supp (φ i ) U i i, where Supp (φ i ) = clos{p M : φ i (p) 0} 3. k i=1 φ i = 1 For n-form ω = αdx 1... dx n in coords x = (x 1,...,x n ), define V ω = U αdx 1...dx n. In general, define M ω = k i=1 U i φ i ω

4 II. Stokes Theorem Let M be a compact oriented n-manifold with boundary, ω an (n 1)-form on M. Then M ω = M dω. Proof. Let K = Supp (ω). Only need to prove for K V = f (U), for coordinate chart (U,f ), then use partition of unity. Write ω = n a j dx 1... dx j 1 dx j+1... dx n, j=1 n dω = ( 1) j 1 a j dx 1... dx n. x j j=1 Two cases to consider: f (U) M = and otherwise.

5 Stokes Theorem f (U) M = (1) f (U) M =. Then M ω = 0, so we must show ( ( 1) j 1 dω = a ) j dx 1... dx n = 0. x j M U WLOG, f 1 (K) int(q), where Q = {x H n : xj 1 x j xj 0} (H n = {x R n : x 0 0}), for xj 1, x0 j such that f 1 (K) int(q). Then, ( ( 1) j 1 a j )dx 1...dx n = ( 1) j 1 U x j = ( 1) j 1 [a j (x 1,...,xj 0,...,x n) Q a j (x 1,...,x 1 j,...,x n)]dx 1...dx j 1 dx j+1...dx n = 0 Q a j x j dx 1...dx n

6 Stokes Theorem f (U) M (2) f (U) M. Then ω M = a 1 (0,x 2,...,x n )dx 2... dx n Set Q as above, with x1 0 = 0. Then, dω = ( 1) j 1 M = = Q Q Q a j x j dx 1 dx n [ a1 (0,...,x n ) a 1 (x 1 1,...,x n) ] dx 2...dx n a 1 (0,x 2,...,x n )dx 2 dx n = M ω QED

7 III. Poincaré s Lemma M is contractible to p 0 M if smooth map C : M R M s.t. for all p M, C(p,1) = p and C(p,0) = p 0 Poincaré s Lemma. If M is contractible, ω Ω k (M) is closed (dω = 0) iff it is exact (ω = dη for some η). Proof. Let π : M R M be the projection map, ω = C ω. Note that ω can be written uniquely as ω = ω 1 + dt η, such that ω 1 (v 1,...,v k ) = 0 if some v i ker(dπ), and similarly for η.

8 Poincaré s Lemma Let i t : M M R, i t (p) = (p,t). Define I : Ω k (M R) Ω k 1 (M) as follows. For p M, v 1,...,v k 1 T p M, (Iω)(v 1,...,v k 1 ) = Claim: d(iω) = ω 1 0 [η(p,t)(di t (v 1 ),...,Di t (v k 1 ))] dt Sublemma. i 1 ω i 0 ω = d(iω) + I(dω). Note that I(ω 1 + ω 2 ) = I(ω 1 ) + I(ω 2 ), so enough to prove for ω = fdx i1... dx ik or ω = fdt dx i1... dx ik.

9 Poincaré s Lemma sublemma Case 1: ω = fdx i1... dx ik, we have dω = f t dt dx i 1... dx ik + terms not containing dt. ( 1 ) f I(dω)(p) = 0 t dt dx i1... dx ik =(f (p,1) f (p,0)) dx i1... dx ik =i 1 ω(p) i 0 ω(p) Case 2: ω = fdt dx i1... dx ik. Since i t takes M to a subspace of constant t, we have i1 ω = i 0 ω = 0, so we need d(iω) = I(dω).

10 Poincaré s Lemma sublemma n f dω = dx α dt dx i1... dx ik 1, x α=1 α (Idω)(p) = ( 1 ) f dt dx α dx i1... dx ik 1. α 0 x α On the other hand, {( 1 d(iω)(p) =d = α 0 ( 1 0 } fdt )dx i1... dx ik 1 ) f dt dx α dx i1... dx ik 1 x α

11 Poincaré s Lemma ω = (C i 1 ) ω = i 1 (C ω) = i 1 ω, 0 = (C i 0 ) ω = i 0 (C ω) = i 0 ω Since dω = 0, we get dω = C dω = 0. Setting α = Iω, ω = i1 ω = d(iω) = dα. QED

12 IV. The de Rham Chain Complex Define ( Int k (ω) ) (A) = A ω. Ω k 1 (M) Int k 1 d k 1 Ω k (M) Int k d k Ω k+1 (M) Int k+1 Σ k 1 k 1 Σ k k Σ k+1 Diagram commutes by Stokes Thm. Star St(σ) = σ F(τ) τ open in M, {St(p i)} cover M. Define subordinate partition of unity {φ i }, and define Φ k (q λ0 q λk ) = k! k ( 1) i φ λi dφ λ0 î dφ λk i=0

13 V. Elementary forms Theorem. Supp (Φ k σ) St(σ) Φ k A =dφ k 1 A Φ 0 ( q i ) =1 (1a) (1b) (1c) Furthermore, Φ k is a right inverse of Int k. Proof. (1a) follows immediately since Supp φ i St(q i ). For (1c), Φ 0 (q i ) = φ i, hence Φ 0 ( i q i) = 1. Now (1b). Let σ = q λ0 q λk, then dφ k (q λ0 q λk ) = (k + 1)!dφ λ0... dφ λk.

14 Sidenote: two important identities Recall ( σ) τ = σ ( τ) = 1 if σ τ and 0 otherwise. Hence q λ0 q λk = s q sq λ0 q λk, where is over s s.t. q s q λ0 q λk is simplex. In particular, 1 (k+1)! Φ (q λ0 q λk ) = 1 (k+1)! s Φ(q sq λ0 q λk ). Also note that d ( φ i ) = 0. Since φ i = k i=0 φ λ i + s λ i φ s, we have k dφ λi + dφ s = 0 s any λ i i=0 We use both these identities in the following.

15 Elementary forms Proof of (1b) = 1 (k + 1)! s = Φ(q s q λ0 q λk ) s { φ s dφ λ0... dφ λk k } ( 1) i φ λi dφ s dφ λ0 î dφ λk i=0 φ s dφ λ0... dφ λk + s any λ i k k ( 1) i φ λi dφ λj dφ λ0...î... dφ λk i=0 = j=0 s any λ i φ k dφ λ0... dφ λk + =dφ λ0... dφ λk = k φ λi dφ λ0... dφ λk i=0 1 (k + 1)! dφ(q λ 0 q λk )

16 Elementary forms Int Φ k = Id Now prove Φ k is right-sided inverse of Int k by induction. k = 0 consider vertex q j and covertex q i, so q i q j = δ ij. Note by definition Φ 0 (q i ) = φ i and φ i (q j ) = δ ij. Also note for function f, Int 0 (f ) q j = f (q j ). Hence Int 0 ( Φ 0 (q i ) ) q j = δ ij, i.e. Int 0 Φ 0 = Id. k > 0 will follow from (Int k Φ k σ) τ = τ Φ k σ = { 0 if τ σ 1 if τ = σ Suppose τ σ, then q i F(σ) with q i / F(τ), so φ i 0 in τ, proving for τ σ.

17 Elementary forms Int Φ k = Id If σ = τ, write σ = ψ Then, Φ k σ =! Φ k ψ = dφ k 1 ψ = σ σ σ σ Φ k 1 ψ =! ψ Φ k 1 ψ = 1, where! uses the case σ τ and the last equality uses inductive hypothesis. For σ K, Φσ called an elementary form. QED

18 VI. Stay tuned! Define H k Σ = ker( k )/im( k 1 ), Hk Ω = ker(d k)/im(d k 1 ), the differential and simplicial cohomology groups respectively. Define H Σ = H k Σ, H Ω = H k Ω. For h Hk, h H r, can define cup product h h H k+r under which H Σ and H Ω become rings. We will show the subcomplex (ker Int ) is acyclic (i.e. ker d k = imd k 1 when restricted to ker Int ) and use this to prove de Rham s theorem: Int is a ring isomorphism between H Ω and H Σ.