Dimension of de Rham Cohomology groups?

Question

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?

Answer 1

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.

Answer 2

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)

Answer 3

If $M$ is orientable, then applying Poincare duality twice we find that $H^p_{dR}(M)\cong H^p_{dR}(M)^{**}$. Since no infinite-dimensional vector space can be isomorphic to its double dual vector space, it follows that $H^p_{dR}(M)$ is finite-dimensional.

If $M$ is nonorientable, then consider its orientation covering $\hat\pi:\hat M\to M$. The induced cohomology map $\hat\pi^*: H^p_{dR}(M)\to H^p_{dR}(\hat M)$ is injective by Lemma 17.33 in Lee's Smooth Manifolds book. Since a finite-sheeted covering map is a proper map, it follows from the compactness of $M$ that $\hat M$ is compact, and hence we can apply the preceding argument to $\hat M$.