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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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up vote 5 down vote accepted

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
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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
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