Let $M$ be a compact smooth orientable manifold of dimension $n$. I am looking for a simple proof that $H_{dR}^n(M) \cong \mathbb R$. Equivalently, an $n$-form which integrates to 0 is exact. I can show this via a rather indirect argument as follows: we know $H_{dR}^n(M) \cong H^n(M, \mathbb R)$, where $H^n$ denotes the singular cohomology. By the universal coefficient theorem (and the fact that $\mathbb R$ is a field) this is isomorphic to $Hom(H_n(M, \mathbb Z) , \mathbb R)$. From the (rather lengthy) proof in Section 3.3 of Hatcher's Algebraic Topology, we find that $H_n(M, \mathbb Z)$ is isomorphic to $\mathbb Z$, and so $Hom(H_n(M, \mathbb Z) , \mathbb R) \cong \mathbb R$. However, it seems like there should be a simpler way to prove this. Does anyone know of one?
Top deRham cohomology group of a compact orientable manifold is 1-dimensional
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2You probably want $M$ to be connected. – 2012-11-03
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3Integrate!! It is fairly straightforward to check that $\omega \mapsto \int_M \omega$ is the desired isomorphism. – 2012-11-03
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5I know that the isomorphism takes this form; the part I am having trouble with is seeing why it is injective. – 2012-11-03
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4I don't think there is a simple argument. (Standard proof of Poincare duality uses Mayer–Vietoris for induction by covering, I believe.) – 2012-11-05
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1@user15464: Maybe too late, but there is a nice elementary inductive proof in Spivak Vol 1, chap. 8. – 2017-10-26