# Dimension of de Rham Cohomology groups?

Is there a simple way to prove that the de Rham cohomology groups of a compact manifold $M$ have finite dimension as $\mathbb{R}$-vector spaces?

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## 2 Answers

There is a simple proof that uses the following concepts: de Rham’s Theorem, Leray’s Theorem and geodesic convexity.

• Firstly, use de Rham’s Theorem to interpret the de Rham cohomology groups of $M$ as the sheaf cohomology groups of the constant sheaf corresponding to $\mathbb{R}$.

• Equip $M$ with a Riemannian metric, and use the compactness of $M$ to pick a finite open cover $\mathcal{U} = \{ U_{1},\ldots,U_{n} \}$ of $M$, where each $U_{i}$ is a geodesically convex subset.

• Each $U_{i}$ has trivial de Rham cohomology, as it is homeomorphic to an open convex subset of $\mathbb{R}^{n}$. Also, the intersection of geodesically convex subsets is also geodesically convex.

• The open cover $\mathcal{U}$ thus satisfies the conditions for applying Leray’s Theorem, so one can compute the Čech cohomology groups corresponding to $\mathcal{U}$, which are simply the de Rham cohomology groups.

• However, the Čech complex consists of only finite-dimensional vector spaces, because $\mathcal{U}$ is finite and the sections of the constant sheaf over each $U_{i}$ is $\mathbb{R}$ by the connectedness of geodesically convex subsets.

• Therefore, the de Rham cohomology groups of $M$ are finite-dimensional.

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that's great! thanks –  Heitor Fontana Mar 20 '13 at 13:09

Another way (although this is definitely not as simple as Leonard's way) is to use the Hodge theorem, which states that if we denote by $\mathcal{H}^k$ the vector space of harmonic $k$-forms:

$\mathcal{H}^k(M) = \{ \alpha \in \Omega^{k}(M): \bigtriangleup\alpha = 0\}$

where $\Omega^k(M)$ denotes the $k$-forms on $M$ and $\bigtriangleup$ is the Laplacian then every class $[\alpha] \in H^{k}(M)$ has a unique harmonic representative. So we have a vector space isomorphism:

$\mathcal{H}^k(M) \cong H^{k}M$

and then we can use the fact that $\bigtriangleup$ is an elliptic operator, and the kernel of an elliptic operator on a compact manifold is always finite dimensional (in fact this is pretty much the contents of the first section of this wikipedia page)

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Just a small comment. If a manifold $M$ (compact or not) is covered by finit contactible opens such that any intersection of two opens is contractible then, from Mayer-Vietoris exact sequence the de Rham cohomology groups of $M$ are finite-dimensional. –  amine Jul 13 '13 at 18:00