Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You probably want $M$ to be connected. – Zhen Lin Nov 3 '12 at 23:30
Integrate!! It is fairly straightforward to check that $\omega \mapsto \int_M \omega$ is the desired isomorphism. – Matt Nov 3 '12 at 23:30
I know that the isomorphism takes this form; the part I am having trouble with is seeing why it is injective. – user15464 Nov 3 '12 at 23:41
I don't think there is a simple argument. (Standard proof of Poincare duality uses Mayer–Vietoris for induction by covering, I believe.) – Grigory M Nov 5 '12 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.