I’ve been away from this blog for too long–partially it’s because most of my expository energy has gone into preparing a collection of notes on algebraic geometry (to help me learn the subject). Someday I’ll post them.

Today, however, I’d like to talk about a clever proof I learned recently.

The following result is neat:

**Theorem**: Let be a compact smooth manifold. Then the de Rham cohomology groups are finite-dimensional.

I’m pretty sure it follows from Hodge theory and the finite-dimensionality of the harmonic forms. However, I learned a neat elementary proof that I’d like to discuss.

By de Rham’s theorem, we can compute the de Rham cohomology groups as the sheaf cohomology groups for denoting the constant sheaf associated to the group . Now, pick a Riemannian metric on . Each point has a neighborhood such that any two points in are joined by a unique geodesic contained in —such a neighborhood is called geodesically convex. It is clear that a geodesically convex neighborhood is homeomorphic (via the exponential map) to a convex set in , which has trivial de Rham cohomology, and also that the intersection of two geodesically convex sets is geodesically convex.

So pick a finite cover of by geodesically convex sets . Then on every intersection , the sheaf has trivial cohomology because this intersection is geodesically convex, hence diffeomorphic to a convex set in . In particular, the cover satisfies the hypotheses of Leray’s theorem. We can apply Cech cohomology with this cover to compute , or equivalently the de Rham cohomology.

But there are finitely many sets in this cover, and the sections of the sheaf over each of these sets is just the abelian group by connectedness of anything geodesically convex. So the Cech complex consists of finite-dimensional vector spaces; its cohomology thus consists of finite-dimensional vector spaces.

I learned this from Bott and Tu’s *Differential Forms in Algebraic Topology*, which appears to be a really fun read.