# Cohomology with group ring coefficients and compact support

Let $$X$$ be a contractible space and let $$G$$ be a group acting freely, properly discontinuously, and cocompactly on $$X$$. We get an induced action on the singular chain complex $$C_{\bullet}(X)$$.

Define $$D^{\bullet}$$ to be the subcomplex of the singular cochain complex $$C^{\bullet}(X) = \text{Hom}(C_{\bullet}(X),\mathbb{Z})$$ whose $$n$$th term consists of all $$f\colon C_n(X) \rightarrow \mathbb{Z}$$ such that for all $$\sigma \in C_n(X)$$, the set $$\{\text{g \in G | f(g \cdot \sigma) \neq 0}\} \subset G$$ is finite.

Question: Is $$D^{\bullet}$$ the same as the singular cochain complex with compact support? Or at least is the cohomology of $$D^{\bullet}$$ the same as the cohomology of $$X$$ with compact support?

It is clear that every singular cochain with compact support lies in $$D^{\bullet}$$ (this just uses the fact that $$G$$ acts properly discontinuously).

Here's the reason I am interested in this question. We can identify $$D^{\bullet}$$ with the complex $$\text{Hom}_G(C_{\bullet}(X),\mathbb{Z}[G])$$, namely $$f \in D^{n}$$ is identified with the $$G$$-invariant map from $$C_n(X)$$ to $$\mathbb{Z}[G]$$ taking $$\sigma \in C_n(X)$$ to $$\sum_{g \in G} f(g^{-1} \sigma) g$$. The condition in the math display above is precisely what we need for this to be a finite sum for all $$\sigma$$. So the cohomology of $$D^{\bullet}$$ is the same as the cohomology of $$X/G$$ (a $$K(G,1)$$) with coefficients in $$\mathbb{Z}[G]$$.

Note: This would be a much easier question is I were using simplicial complexes and simplicial homology; however, for technical reasons I really want to use singular homology.

• Hi Laura, the complex $D^\bullet$ doesn't coincide on the nose with the compactly supported cochains. E.g. taking $\Bbb R$ with its action of $\Bbb Z$, the complex $D^\bullet$ contains the 1-dimensional cochain $f$ such that: $f(\sigma) = 1$ if $\sigma(t) = a t$ for some $a$, and $f(\sigma) = 0$ otherwise. I haven't been able to determine whether it still computes compactly supported cohomology; I suspect not, but haven't been able to solve it. – Tyler Lawson Mar 10 at 14:52
• Nothing to do with your question, but doesn't the (co)homology of the complex $D^\bullet$ rather compute the cohomology of $X/G$ with coefficients in a local system locally isomorphic to $\mathbf{Z}[G]$, instead of constant coefficients $\mathbf{Z}[G]$? – Johannes Huisman Mar 12 at 15:25
• @TylerLawson: That's a good example -- I thought I had worked out $\mathbb{Z}$ acting on $\mathbb{R}$, but looking back over my notes I had screwed it up. Still very interested if you ever figure this out. – Laura Mar 13 at 19:29
• @JohannesHuisman: That's what I meant, and I hope it was clear from the context. It would be perverse to take a trivial local system that also happened to be a module over the fundamental group... – Laura Mar 13 at 19:30
• @Laura This previous question appears to indicate that this might be VIII.7.5 in Ken Brown's "Cohomology of groups"? mathoverflow.net/questions/102654/… I don't currently have access to the book to check. – Tyler Lawson Mar 13 at 20:24