# Universal coefficient theorem for group homology and cohomology

I've been looking for any kind of universal coefficient theorem for group homology and cohomology, including dual universal coefficient theorems. However, the only things I can find are ones where the group action on the coefficients is trivial. As such, my question is:

Let $G$ be a group, $M$ be a $G$-module. Is there any universal coefficient theorem for $H^*(G, M)$ or $H_*(G, M)$?

My specific interest, as in Cohomology of lattice with coefficients in field of rational functions, is where $G = \mathbb{Z}^n$, but I'd love if there were a more general answer.

• The basic problem with asking for a UCT for a nontrivial action is: what data is it supposed to depend on? It certainly shouldn't depend only on $H_{\bullet}(G, \mathbb{Z})$ and on the underlying abelian group of $M$, for example. – Qiaochu Yuan Apr 1 '15 at 8:47
• That was part of my line of thinking; is there any data that could lead to a form of a UCT? – user44191 Apr 1 '15 at 16:02
• Say we were working over a field; would knowing $H_.(G, W)$ for each simple representation $W$ be enough? Each indecomposable? – user44191 Apr 3 '15 at 15:22

You're only finding UCT in the literature for trivial group actions, because there is no general UCT for nontrivial group actions:

The general Kunneth formula does not hold for arbitrary groups and actions. But it does hold a good amount of times, and I elaborated on this here: Kuenneth-formula for group cohomology with nontrivial action on the coefficient.

Now the UCT, which relates $H_*(G,M)$ to $H_*(G,\mathbb{Z})$, only follows from the Kunneth formula for trivial group actions. The Kunneth theorem considers the tensor product $C_*\otimes D_*$ of two chain complexes, and the special case for UCT is $D_*=M$ for some $R$-module. To guarantee that the images of the boundary maps are $R$-projective, some extra assumptions are needed (like $R$ is a PID). For group homology, we work with $F_*\otimes_{\mathbb{Z}G}M$ where $F_*$ is a free resolution of $\mathbb{Z}$ as a $\mathbb{Z}G$-module. We cannot take $R=\mathbb{Z}G$ (otherwise the assumption about the boundaries would imply that all homology groups are trivial), so we take $R=\mathbb{Z}$. But then $M$ must be trivial as a $\mathbb{Z}G$-module in order to express $F_*\otimes_{\mathbb{Z}G}M$ as $C_*\otimes_\mathbb{Z}M$. In this case, $F_*\otimes_{\mathbb{Z}G}M=(F_*\otimes_{\mathbb{Z}G}\mathbb{Z})\otimes_\mathbb{Z}M$ and we can apply the UCT.

Morally, in lieu of Qiaochu's remark you must ask: Given any data involving the $G$-action, how would the operators $\oplus,\otimes,\text{Tor},\text{Ext}$ encode such information? And as shown above, you can't use that information to pass from $M$ to $\mathbb{Z}$. For example, let $\mathbb{Z}_2$ act on $M=\mathbb{Z}_2\oplus\mathbb{Z}_2$ by swapping the generators of the summands. Where would this maneuver exist on the coefficient $\mathbb{Z}$ or on any homological object? We can't simply "forget" the action, because $H^1(\mathbb{Z}_2,M_\text{nontriv})=0$ while $H^1(\mathbb{Z}_2,M_\text{triv})=M$.

I do not know a good reference for this, but if $k$ is a field and $M$ is a $k[G]$-module and $M^{\ast} = Hom(M,k)$ is its dual, then there is a natural isomorphism from $H^k(G;M^{\ast})$ to $Hom(H_k(G;M),k)$. This can be proved exactly like Proposition 7.1 in Brown's book on group cohomology.

• I don't have immediate access to Brown's book; would you mind expanding with the proof? – user44191 Apr 1 '15 at 4:14
• Sorry, I don't have time to type it out (and my copy of the book is at my office anyway; I just copied the reference from one of my old papers which quoted it). His book is pretty standard; I would expect that any university library would have it. – Andy Putman Apr 1 '15 at 4:17
• This isn't a UCT, though. You're only swapping $M$-coefficients for $M^\ast$-coefficients. – Chris Gerig Apr 1 '15 at 5:51
• @ChrisGerig: As you pointed out in your answer, there is no general theorem. But the result in my answer is a shadow of the UCT that does hold, is often useful, and is not well-documented in the expository literature. – Andy Putman Apr 1 '15 at 5:55