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.

  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Apr 1 '15 at 8:47
  • $\begingroup$ That was part of my line of thinking; is there any data that could lead to a form of a UCT? $\endgroup$ – user44191 Apr 1 '15 at 16:02
  • $\begingroup$ Say we were working over a field; would knowing $H_.(G, W)$ for each simple representation $W$ be enough? Each indecomposable? $\endgroup$ – 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.

  • $\begingroup$ I don't have immediate access to Brown's book; would you mind expanding with the proof? $\endgroup$ – user44191 Apr 1 '15 at 4:14
  • $\begingroup$ 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. $\endgroup$ – Andy Putman Apr 1 '15 at 4:17
  • $\begingroup$ This isn't a UCT, though. You're only swapping $M$-coefficients for $M^\ast$-coefficients. $\endgroup$ – Chris Gerig Apr 1 '15 at 5:51
  • 3
    $\begingroup$ @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. $\endgroup$ – Andy Putman Apr 1 '15 at 5:55

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