TITLE: Dimension of de Rham Cohomology groups?
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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)