Why the De Rham cohomology just a topological invariant？

Question

I actually know that the De Rham cohomology is a topological invariant since it coincides with the regular case ,but I have no idea how the information of differential structure vanish because the construction of De Rham cohomology seems highly depending on the differential structure . Could any one help me figuring out the reason? Thanks!

Answer 1

One answer would be: sheaf cohomology. But do not run away at once, please, I will elaborate on that answer.

The main idea is the following: think about locally constant functions on open subsets of your space (with values in $\mathbb R$). These do not depend on the differential structure, right ? The locally constant functions on open sets of a space $X$ form a sheaf $\underline{\mathbb R}$ (no need for a precise definition at this point). Now the cohomology of the space with coefficients in $\underline{\mathbb R}$ (whatever that means) should (by definition) look only at the topology of your space. And that is what it does. In fact, it is (under good hypotheses) isomorphic to the singular homology of the space (with real coefficients).

What is the link with De Rham cohomology ? Well, in order to compute sheaf cohomology, you need what is called a resolution of your sheaf. It turns out that when your space is a manifold, you can construct such a resolution of $\underline{\mathbb R}$ using the differential structure: the De Rham complex ! And any resolution of a given sheaf computes the same homology, which is the sheaf homology I was talking about above. Thus, the differential structure allows us to construct a resolution, but the cohomology we compute is really the cohomology of the sheaf of locally constant functions, whose definition depends only on the topology of the space.

If you know about resolutions, you can see the link with @Elad's answer. The fact that the De Rham complex is a resolution of $\underline{\mathbb R}$ means that the sequence of sheaves: $0 \rightarrow \underline{\mathbb R} \rightarrow \Omega^0 \rightarrow \Omega^1 \rightarrow \cdots$ is exact, where $\Omega^0$ is the sheaf of functions and, in general, $\Omega^i$ is the sheaf of $i$-forms. Remark that exactness at $\Omega^0$ is exactly the fact that locally constant functions are those whose differential is trivial.

The fact that it is exact as a complex of sheaves comes from these facts:

• Exactness, for sequences of sheaves, is a local notion;
• Locally, $X$ is a ball;
• For a ball, the Poincaré lemma ensures exactness.

I hope that this answer is helpful (even if I suppose that much reading is needed in order to really understand it. I would recommend reading Bott and Tu's beautiful book for a start -- where they do not write about sheaves, but about Cech cohomology, which is basically the same, and whose definition does not depend on the differential structure in an equally obvious way).

Answer 2

It doesn't really vanish. You need the differential structure to define it. Once you do have a differential structure there is one crucial point to understand. The point is the celebrated poincare lemma. The Poincaré lemma states that if $B$ is an open ball in $\mathbf{R}^{n}$, any smooth closed $p$ -form $\omega$ defined on $B$ is exact, for any integer $p$ with $1 \leq p \leq n$. Thus you can always solve a differential equation locally in a manifold. Thus the obstruction of solving the equation $$d\eta =\omega,$$
is connected to a global problem. This hints that the de rham cohomology group is not affected by the differential structure that is of local nature.

Heuristicly this tells us that if we know $H^*(M,\mathbb{R})$ we should know $H^{*}_{dr}(M,\mathbb{R})$. The Heuristics for the other direction is that almost everything that you can do with continuous functions you can do with differential function by approximation.