# Our goal is now to describe an isomorphism h p dr m h

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Our goal is now to describe an isomorphism H p dR ( M ) = H p ( M, R ) . The map itself is given as follows: Suppose [ w ] H p dR ( M ), so ω Ω p ( M ) with d ω = 0. Suppose that σ : Δ p M is smooth with ∂σ = 0. We can then define I ([ ω ]) = Z Δ p σ * ω R . Note that we have not defined what R Δ p means, because Δ p is not even a manifold with boundary — it has corners. We can develop an analogous theory 68 7 De Rham’s theorem* III Differential Geometry of integration on manifolds with corners, but we can also be lazy, and just integrate over Δ × p = Δ p \ { codimension 2 faces } . Now ω | Δ * p does not have compact support, but has the property that it is the restriction of a (unique) p -form on Δ p , so in particular it is bounded. So the integral is finite. Now in general, if τ = a i σ i C p ( M ), we define I ([ ω ])( τ ) = X a i Z Δ p σ * i ω R . Now Stokes theorem tell us Z ∂σ ω = Z σ d ω. So we have Lemma. I is a well-defined map H p dR ( M ) H p ( M, R ). Proof. If [ ω ] = [ ω 0 ], then ω - ω 0 = d α . Then let σ H p ( M, R ). Then Z σ ( ω - ω 0 ) = Z σ d α = Z ∂σ α = 0 , since ∂σ = 0. On the other hand, if [ σ ] = [ σ 0 ], then σ - σ = ∂β for some β . Then we have Z σ - σ 0 ω = Z ∂β ω = Z β d ω = 0 . So this is well-defined. Lemma. I is functorial and commutes with the boundary map of Mayer-Vietoris. In other words, if F : M N is smooth, then the diagram H p dR ( M ) H p dR ( N ) H p ( M ) H p ( N ) F * I I F * . And if M = U V and U, V are open, then the diagram H p dR ( U V ) H p +1 dR ( U V ) H p ( U V, R ) H p ( U V, R ) δ I I δ also commutes. Note that the other parts of the Mayer-Vietoris sequence commute because they are induced by maps of manifolds. Proof. Trace through the definitions. 69 7 De Rham’s theorem* III Differential Geometry Proposition. Let U R n is convex, then U : H p dR ( U ) H p ( U, R ) is an isomorphism for all p . Proof. If p > 0, then both sides vanish. Otherwise, we check manually that I : H 0 dR ( U ) H 0 ( U, R ) is an isomorphism. These two are enough to prove that the two cohomologies agree — we can cover any manifold by convex subsets of R n , and then use Mayer-Vietoris to patch them up. We make the following definition: Definition (de Rham) . (i) We say a manifold M is de Rham if I is an isomorphism. (ii) We say an open cover { U α } of M is de Rham if U α 1 ∩ · · · ∩ U α p is de Rham for all α 1 , · · · , α p . (iii) A de Rham basis is a de Rham cover that is a basis for the topology on M . Our plan is to inductively show that everything is indeed de Rham. We have already seen that if U R n is convex, then it is de Rham, and a countable disjoint union of de Rham manifolds is de Rham. The key proposition is the following: Proposition. Suppose { U, V } is a de Rham cover of U V . Then U V is de Rham. Proof. We use the five lemma! We write the Mayer-Vietoris sequence that is impossible to fit within the margins: H p dR ( U ) H p dR ( V ) H p dR ( U V ) H p +1 dR ( U V ) H p dR ( U ) H p +1 dR ( V ) H p +1 dR ( U V ) H p ( U ) H p ( V ) H p ( U V ) H p +1 ( U V ) H p ( U ) H p +1 ( V ) H p +1 ( U V ) I I I I I I I This huge thing commutes, and all but the middle map are isomorphisms. So by the five lemma, the middle map is also an isomorphism. So done.  #### You've reached the end of your free preview.

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