Suppose $M$ is a manifold on which the discrete group $G$ acts freely and properly discontinuously. Is it then true that $\Omega^p(M)$ is an acyclic $G$-module, in the sense that $H^k(G,\Omega^p(M)) = 0$ for $k>0$ ?
Clearly this holds if $G$ is finite, because then $|G|$ is invertible in $\Omega^p(M)$and it annihilates positive-degree cohomology. I would like to know if it's true for more general $G$. I'm ready to assume that $G$ is countable, so as to make it a discrete Lie group.
I'm also wondering about $C^p(X)$, the group of singular cochains of a space $X$ if the group $G$ acts freely on it : is this acyclic ? If not, under what conditions is it ?
The answer for $C^p(X)$ is yes, and so it suggests that it is so for $\Omega^p(M)$, though I don't have a proof for that yet.
let $M$ be a $G$-module, we have $\hom_G(M,C^p(X)) = (\hom(M, C^p(X))^G = (\prod_{\hom(|\Delta^p|,X)}(M, \mathbb Z))^G= \hom(C_p(X), \hom(M,Z))^G= \hom_G(C_p(X), \hom(M,\mathbb Z))$.
Now given that $G$ acts freely on $X$, it acts freely on $\hom(|\Delta^p|,X)$ as well, so that $C_p(X)$ is a free $G$-module, and in particular $\hom_G(C_p(X),-)$ is exact.
It follows that $\mathrm{Ext}^i_{\mathbb ZG}(M, C^p(X)) \cong \hom_G(C_p(X), R^i\hom(-,\mathbb Z)(M))$ where $R^i$ denotes the $i$th right derived functor.
Now note that as $C_p(X)$ is a free nonzero (except of $X=\emptyset$, but let's not worry about that here) $G$-module, this $\hom_G$ is zero if and only if $R^i\hom(-,\mathbb Z)(M) = 0$.
But note that this group is completely independent of $X$, it only depends on $G,M$. For acyclicity, we're only interested in $M=\mathbb Z$ so I'll restrict to that situation.
Now to compute $R^i\hom(-,\mathbb Z)(\mathbb Z)$ we want a $G$-projective resolution of $\mathbb Z$. So take a contractible space $EG$ on which $G$ acts freely, so that $C_\bullet(EG)\overset{\epsilon}\to \mathbb Z$ (the augmented singular chain complex of $EG$) is a free resolution of $\mathbb Z$. Then apply $\hom(-,\mathbb Z)$ so we get $C^\bullet(EG)$, and then take its cohomology: we get $H^*(EG)$. But $EG$ is contractible, so this is $0$ : $R^i\hom(-,\mathbb Z)(\mathbb Z) = 0, i>0$
In particular it follows that $\mathrm{Ext}^i_{\mathbb ZG}(\mathbb Z, C^p(X)) = 0, i>0$, or in other words, $C^p(X)$ is acyclic.