De Rham's Theorem states that $H_{dR}^k(M) \simeq H^k(M;\mathbf R)$ for all $k$. This is what it states. But I've been struggling to understanding what it's telling me. Here's my understanding so far:
$H^k(M)$ with any coefficient group/ring is, in general, impossible to compute from the definition.
- If $M$ is a CW complex, we can compute its cellular homology, which is isomorphic to $H_k(M;\mathbf Z)$. Then use the universal coefficient theorem to get homology/cohomology with any coefficient group/ring.
- If $M$ is a simplicial or $\Delta$-complex, we can compute its simplicial homology, which is again isomorphic to $H_k(M;\mathbf Z)$.
- If $M$ is a smooth manifold, and if we can compute its de Rham cohomology, then we get $H^k(M;\mathbf R)$ by de Rham's Theorem, which doesn't allow us to fully recover $H_k(M;\mathbf Z)$, but does place strong restrictions (tells us the Betti numbers).
If $H^k_{dR}(M)$ is too hard to compute from the definition, i.e., there's a bunch of differential equations we don't know how to solve, and $M$ is homotopy equivalent to a CW/simplicial complex, then we can compute $H_k(M;\mathbf Z)$, which gives us $H^k(M;\mathbf R)$ by UCT, which is $H^k_{dR}(M)$ by de Rham's Theorem.
- So now we have (partial) information about solutions of these differential equations on $M$ that we weren't able to solve directly/explicitly.
- This also tells us that whether or not certain differential equations have solutions depends on the homotopy type of the set on which they're defined, which is not at all obvious. (Should it be?)
Is this it? Or are there deeper implications that I'm missing?
