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There is a result in Cartan Eilenberg (XII 6.4) that says that if $G$ is a finite group and $D$ a divisible abelian group with trivial $G$-action then for any $G$-module $M$ the cup product $$\widehat{H}^{r-1}(G,Hom(M,D)) \times \widehat{H}^{-r}(G,M) \longrightarrow \widehat{H}^{-1}(G,D)$$ induces an isomorphism $$\widehat{H}^{r-1}(G,Hom(M,D)) \longrightarrow Hom(\widehat{H}^{-r}(G,M),D)$$ for all $r \in \mathbb{Z}$

I wanted to know if this result would hold for non-Tate cohomology and homology groups, i.e., if instead of having a finite group $G$, we used an arbitrary group $G$ and instead of using cup products we used cap products, would the result still hold for all $r \in \mathbb{Z}$ or maybe for some $r$?

Thank you

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  • $\begingroup$ This is a comment. But I'm not able to comment on your question, so I'm posting these questions as answer: 1. What do you mean by "non-Tate cohomology and homology groups" ? 2. How is $\hat{H}^{-r}(G,M)$ defined if $r> 0$ and $G$ is an arbitrary group ? $\endgroup$
    – Demin Hu
    Commented Dec 9, 2012 at 21:37
  • $\begingroup$ Non-Tate surely means the usual homology and cohomology of a group. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 9, 2012 at 22:22
  • $\begingroup$ For arbitrary groups neither the OP not I talked about Tate cohomology, btw. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 10, 2012 at 0:56
  • $\begingroup$ Yeah by non-Tate I just ment usual homology and cohomology groups, this is why I was wondering about cap-products as they seem to work more generally than the cup-product. $\endgroup$
    – Chris Birkbeck
    Commented Dec 10, 2012 at 15:56

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The isomorphism, at least, exists. Take $M\leftarrow Y$ and $\mathbb Z\leftarrow X$ projective resolutions of $M$ and of $\mathbb Z$ as $G$-modules. The double complex $\hom_G(X,\hom_{\mathbb Z}(Y,D))$ gives rise, as usual, to two spectral sequences. Computing first cohomology with respect to the differential induced by that of $Y$ and then by that of $X$, we get $H^\bullet(G,\hom(M,D))$. On the other hand, rewriting the complex as $\hom_{\mathbb Z}(Y\otimes_G X,D)$ using standard adjunctions, and then computing homology first with respect to the differential induced by that of $X$ and then by that of $Y$ gives the other side of your isomorphism. Convergence then does the trick.

That the isorphism is given by cap products should follow from general nonsense...

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    $\begingroup$ This does what you want for half the $r$; I guess something similar will take care of the other half. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 8, 2012 at 19:21
  • $\begingroup$ (And, as usual, this can be rephrased in the language of derived categories and friends —I am old-fashioned that way... :-) ) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 8, 2012 at 19:28
  • $\begingroup$ Thank you very much, I need to compute this in detail for r=2, can you recomend a good source to read up on spectral sequences, given that I dont know much about them $\endgroup$
    – Chris Birkbeck
    Commented Dec 9, 2012 at 1:13
  • $\begingroup$ MacCleary's book has a pretty straightforward exposition; since you are already reading Cartan-Eilenberg, that's also readable (this is a somewhat unpopular opinion, though!) Weibel has a nice exposition too. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 9, 2012 at 1:31
  • $\begingroup$ You say the isomorphism exists. But which isomorphism ? All your complexes ($X, Y$ and the various hom's thereof) are positive. Can you please explain how the complex $\hom_{\mathbb Z}(Y\otimes_G X,D)$ leads to cohomology in negative degrees ? $\endgroup$
    – Demin Hu
    Commented Dec 9, 2012 at 22:50

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