de Rham cohomology on finitely smooth manifolds

Question

In all of the places I've looked, de Rham cohomology is defined on $\mathcal{C}^\infty$ manifolds with $\mathcal{C}^\infty$ differential forms.

What about de Rham cohomology on $\mathcal{C}^r$ manifolds with $\mathcal{C}^r$ differential forms? Can de Rham cohomology still be defined? If a $\mathcal{C}^r$ manifold is homeomorphic to a $\mathcal{C}^\infty$ manifold, are the de Rham cohomologies isomorphic?

Specific question I am interested in The first de Rham cohomology of $S^1 \subset \mathbb{R}^2$ is isomorphic to $\mathbb{R}$. This means that any closed differential 1-form on $S^1$ is of the form $c \cdot \theta + d V$ where $c \in \mathbb{R}$, $\theta$ represents a basis element (e.g., the angle 1-form), and $V$ is a $\mathcal{C}^\infty$ function. Now, what if I have a space which is only $\mathcal{C}^r$ diffeomorphic to $S^1$. Is it true that any $\mathcal{C}^{r-1}$ closed differential 1-form is of the form $c\cdot \theta + dV$, where $\theta$ is $\mathcal{C}^{r-1}$ and $V$ is $\mathcal{C}^r$?

Update: As pointed out by Najib Idrissi, any $\mathcal{C}^r$ manifold $M$ is $\mathcal{C}^r$ diffeomorphic to a $\mathcal{C}^\infty$ manifold $N$. One could then define the de Rham cohomology of $M$ to be the de Rham cohomology of $N$. This certainly yields information about the topology of $M$ -- but I am interested in information about what types of closed differential forms may exist on $M$.

I think this is a relevant question: Do there exist closed $\mathcal{C}^r$ differential forms on $N$ which do not differ from some closed $\mathcal{C}^\infty$ form by an exact form?

Answer 1

In the specific case of $S^1$ the answer is positive. Namely, let $\omega$ be a $C^r$-smooth 1-form on $M=S^1$, $r\ge 0$. Then you can still integrate the form $\omega$ over $M$ and get a number $a=\int_M \omega$. Now, consider a new form, $\tau= \omega- a d\theta$. This form satisfies $\int_M \tau=0$. Now, fix some point $p\in M$ and consider the integrals $$ \int_{\gamma} \tau, $$ where $\gamma$ is a path in $M$ connecting $p$ to a point $z\in S^1$. Since $\int_M \tau=0$, this integral depends only on $z$ and not on $\gamma$. Thus, you obtain a $C^{r+1}$-smooth function $$ f(z)= \int_{p}^z \tau $$ and $df=\tau$. Hence, $\omega= a d\theta + df$.

This works because we are in dimension 1: Integration increases smoothness of a form. In higher dimensions, partial integration of $C^r$-smooth forms does not necessarily result in $C^{r+1}$-smooth forms. This means that you will have trouble proving the Poincare lemma with controlled smoothness. (At least, the standard proofs will not work.) I suspect that the Poincare lemma is simply false in your setting, namely, that there exist closed $C^r$-smooth forms $\omega$ on $R^n$ such that there does not exist $C^{r+1}$-forms $\eta$ such that $d\eta=\omega$.

Lastly, I still do not see why you are looking at this. One situation where one can effectively use Hodge - de Rham theory of low regularity forms is of Lipschitz manifolds, i.e. topological manifolds equipped with an atlas where the transition maps are locally Lipschitz. One can still define a de Rham complex in this situation by working with distributions. Unlike the $C^r$ problem, this is actually interesting since, while there are nonsmoothable topological manifolds, according to a theorem of Sullivan every topological manifold of dimension $\ne 4$ admits a unique Lipschitz structure. (In dimension 4 this is false.) Thus, one may develop an analytical approach to nonsmoothable topological manifolds using Lispchitz structure (or even compare different smooth structures on the same topological manifold using the uniqueness part of Sullivan's theorem). For instance, if I remember correctly, one can reprove Novikov's theorem about topological invariance of rational Pontryagin classes this way. Besides topology of manifolds, one can use low regularity forms in Lipschitz and quasiconformal analysis in domains in $R^n$.

Answer 2

Any $C^k$-manifold, with $k \ge 1$, has an essentially unique $C^\infty$ structure More precisely, by a theorem of Whitney any $C^k$ maximal atlas contains a $C^\infty$ atlas, and all such atlases are $C^\infty$ diffeomorphic.

By de Rham's theorem, de Rham cohomology is the same thing as singular cohomology with coefficients in $\mathbb{R}$, which is a homotopy invariant. A homotopy invariant is of course invariant under $C^\infty$-diffeomorphism, so I imagine that any kind of reasonable "de Rham complex" that you will manage to define will just give back singular cohomology with real coefficients. In any case I imagine that, in view of these two theorems, people didn't really try to work with these. (Though I'm not omniscient, so maybe it really exists in the literature.)

tl;dr: Any $C^k$-manifold, for $k \ge 1$, has an essentially unique $C^\infty$ structure. You can just define the de Rham complex of your manifold to be the de Rham complex of this $C^\infty$-manifold, any two choices giving homotopy equivalent complexes anyway.

Answer 3

I see that this question has already been answered, but I would just like to add that there is a way in which you can define a well-behaved complex of $C^k$ differential forms for $k \geq 1$.

The problem with taking the ordinary de Rham complex is that the derivative of a $C^k$ differential form is only necessarily a $C^{k-1}$ differential form. As such, we may naturally define $\Omega^r_{C^k}$ to be the space of $C^k$ differential $r$-forms whose exterior derivatives are also $C^k$. This is just my notation because I don't know of any standard notation for this space. Importantly, as the second exterior derivative of a differential form is zero, the exterior derivative of a $\Omega^r_{C^k}$ form is in $\Omega^{r+1}_{C^k}$. Note that, in dimensions greater than 1, these differential forms need not be $C^{k+1}$ since only their exterior derivatives need to be $C^{k}$, and not their partial derivatives.

The Poincare Lemma holds by the standard proof, and in fact, this space is a $C^{k+1}$ homotopy invariant. This space is well-defined for any $C^{k+1}$ manifolds, but not for $C^k$ manifolds, since it is impossible to tell whether a differential form is $C^k$ or not on a $C^k$ manifold.

I don't know a proof that the cohomology of this complex is naturally isomorphic to de Rham cohomology (on a smooth manifold), but I strongly suspect that it is.

Lastly, I want to add that you can go even further with this idea. If the manifold is merely Lipschitz, then you can still define $\Omega^r_{L^\infty}$ or $\Omega^r_{L^2}$, although I'm not very familiar with these. These are more well-studied, however, and are definitely naturally isomorphic to de Rham cohomology (on a smooth manifold).