Is the de Rham complex independent of the smooth structure?

Question

The de Rham complex of a smooth manifold $M$ of dimension $n$ is the complex of differential forms $$\cdots\rightarrow 0\rightarrow\Omega^0(M)\rightarrow\Omega^1(M)\rightarrow\cdots\rightarrow\Omega^n(M)\rightarrow 0\rightarrow\cdots.$$ Its cohomology is a topological invariant, and in particular it does not depend on the smooth structure that is used to define the spaces $\Omega^k(M)$ and the differential $\Omega^k(M)\rightarrow\Omega^{k+1}(M)$. But do we know if the complex itself is independent of the smooth structure? I thought we might get somewhere if we find an identification of the $\Omega^0$ for different differentiable structures, but the best I can get from that is a chain homotopy equivalence, which is not surprising.

Answer 1

The de Rham-complex is a complex of $\mathbb{R}$-vector spaces. In fact, I claim

Proposition: Let $M$ and $M^{\prime}$ be non-empty smooth manifolds of the same positive dimension $n$ and assume that $H^k(M)\cong H^k(M^{\prime})$ for all $0\le k\le n$. Then, the de Rham-complexes of $M$ and $M^{\prime}$ are isomorphic.

(This, in particular, yields a positive answer to your question for non-empty manifolds of positive dimension. For the empty or $0$-dimensional manifolds, the answer is positive too, simply cause the smooth structure on such manifolds is unique.)

This boilds down to linear algebra, for which we need the

Lemma: If $M$ is a non-empty manifold of positive dimension, the spaces $\Omega^k(M)$ for $0\le k\le n$, $Z^k(M)=\ker d^k$ for $1\le k\le n$ and $B^k(M)=\mathrm{im}\ d^{k-1}$ for $1\le k\le n$ have dimension $\mathfrak{c}$.

Proof: First, the upper bound(s). The space $\Omega^k(M)$ is the space of smooth (in particular, continuous) sections of the vector bundle $\bigwedge^kT^{\ast}M\rightarrow M$. Since $M$ is second-countable and locally Euclidean, it is separable, i.e. has a countable, dense subset $X$. Then, the restriction $\Omega^k(M)\rightarrow\prod_{x\in X}\bigwedge^kT_x^{\ast}M$ is injective by continuity, but the latter is isomorphic to $\mathbb{R}^{\mathbb{N}}$ as a vector space, which has dimension $\mathfrak{c}$. The same upper bound holds for the spaces of closed and exact forms by monotonicity.

In the converse direction, it suffices by monotonicity to show the the space of exact $k$-forms ($1\le k\le n$) contains a linearly independent subset of size $\mathfrak{c}$. It suffices to work with forms that have compact support in a chart $U$ of $M$ for this. However, $U$ is diffeomorphic to an open subset of some $\mathbb{R}^n$ and if we take $\rho_r$ to be an appropriately chosen bump function whose support is precisely the closed disk of radius $r$, then the collection of $d^{k-1}(\rho_rdx_1\wedge\dotsc\wedge dx_{k-1})=\left(\sum_{i=k}^n\frac{\partial\rho_r}{\partial x_i}dx_i\right)\wedge dx_1\wedge\dotsc\wedge dx_{k-1}$ for $r>0$ is exactly that.$\blacksquare$

Now, recall that if you have two short exact sequences $0\rightarrow V^{\prime}\rightarrow V\rightarrow V^{\prime\prime}\rightarrow 0$ and $0\rightarrow W^{\prime}\rightarrow W\rightarrow W^{\prime\prime}\rightarrow 0$ of vector spaces and isomorphisms $V^{\prime}\stackrel{\sim}{\rightarrow}W^{\prime}$ and $V^{\prime\prime}\stackrel{\sim}{\rightarrow}W^{\prime\prime}$, then you can find an isomorphism $V\stackrel{\sim}{\rightarrow}W$ that turns these into an isomorphism of short exact sequences (I'd love drawing a diagram, but it's too inconvenient on MSE). This, we shall make repeated use of.

Proof of Proposition: Pick isomorphisms $h^k\colon H^k(M)\rightarrow H^k(M^{\prime})$ for $0\le k\le n$ (these are isomorphic by assumption) and $b^k\colon B^k(M)\rightarrow B^k(M^{\prime})$ for $0\le k\le n+1$ (these are isomorphic since they have the same dimension by the Lemma or for trivial reasons). Taking the short exact sequences $0\rightarrow B^k\rightarrow Z^k\rightarrow H^k\rightarrow 0$ for $M$ and $M^{\prime}$ respectively, we find a isomorphisms $z^k\colon Z^k(M)\rightarrow Z^k(M^{\prime})$ compatible with $b^k$ and $h^k$ for $0\le k\le n$. Then, taking the short exact sequences $0\rightarrow Z^k\rightarrow\Omega^k\rightarrow B^{k+1}\rightarrow 0$ for $M$ and $M^{\prime}$ respectively, we find isomorphisms $\omega^k\colon\Omega^k(M)\rightarrow\Omega^k(M^{\prime})$ compatible with $z^k$ and $b^{k+1}$ for $0\le k\le n$. Since the composite $\Omega^k\rightarrow B^{k+1}\rightarrow Z^{k+1}\rightarrow\Omega^{k+1}$ is the differential, the isomorphisms $\omega^k$ are compatible with the differentials by construction, hence define an isomorphism $\omega^{\bullet}\colon\Omega^{\bullet}(M)\rightarrow\Omega^{\bullet}(M^{\prime})$ of cochain complexes.$\blacksquare$

This was just linear algebra and not much topological insight. The take-away here should be that "very big" vector spaces are too flexible and cochain complexes of them don't have that much to offer besides their cohomology (something else they do offer, for example, is the dimension of $M$). However, this does not mean that differential forms are only interesting/useful to define cohomology! Rather than the de Rham complex of $M$, the algebraic object you really want is the differential graded algebra $\Omega^{\ast}(M)$. These have an incredibly rich theory that I'm not apt enough to explain. In fact, just the degree $0$ piece $\Omega^0(M)=C^{\infty}(M)$ as a commutative ring already determines both the topological space underlying $M$ and the smooth structure of $M$ (up to diffeomorphism).