# de Rham cohomology on finitely smooth manifolds

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?

• The first problem is that there isn't a $C^r$ de Rham complex: the differential of a $C^r$ form will only be $C^{r-1}$ in general. – Qiaochu Yuan May 12 '16 at 2:13
• @QiaochuYuan I've noticed that problem. I thought that perhaps one could create a complex with $C^r$ functions, $C^{r-1}$ 1-forms, ... , $C^{1}$ $(r-1)$-forms, and then set all subsequent groups in the complex equal to zero. – Matthew Kvalheim May 12 '16 at 2:49
• Then you will get a Cohomology theory which does not have many standard properties, for instance, Poincaré lemma will fail and the open ball will have infinite dimensional Cohomology groups. – Moishe Kohan May 12 '16 at 3:05
• @studiosus Would you mind elaborating on why those things might be true? – Matthew Kvalheim May 12 '16 at 3:11
• Think of the case r=1. Then you set d=0 on all forms of degree >0. Then all forms of degree >1 are closed but not exact. Even in degree 1 you have too many closed forms which are not exact since they are not closed in the usual sense. – Moishe Kohan May 12 '16 at 3:21

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 $R^n$.
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.