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Is there a Kunneth formula relating $H^i(k[G],A)\otimes_k H^i(k[G],B)$ and $H^i(k[G],A\otimes_k B)$ where $A\otimes_k B$ is given the diagonal $G$ action ?

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 It is quite possible that $H^0(G,A)=H^0(G,B)=0$ yet $H^0(G,A\otimes B)\neq0$ (for example, let $A$ be a non-trivial $1$-dimensional module and $B$ its dual) so one cannot expect something as simple as Künneth's formula for the general case. – Mariano Suárez-Alvarez Apr 16 2012 at 16:50
Although see: Greenblatt, R, Homology with local coefficients and characteristic classes. Homology, Homotopy Appl. 8 (2006), no. 2, 91–103, for a K\" unneth formula for a product action. – Mark Grant Apr 16 2012 at 17:38
Thanks very much! – unknown (google) Apr 16 2012 at 17:53
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 This question has already been asked and answered: mathoverflow.net/questions/75472/… – Chris Gerig Apr 17 2012 at 1:04
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 Kunneth relates $H^{\bulled}(G\times H, A\otimes B)$ with $H^{\bullet}(G,A)\otimes H^{\bullet}(H,B),$ if $A$ and $B$ are $k$-modules over groups $G$ and $H$ respectively. You can certainly take $G = H$ if you like. However, to relate $H^{\bullet}(G\times G,A \otimes_k B)$ to $H^{\bullet}(G,A\otimes_k B)$ is a special case of the general problem of relating cohomology of a group with coefficients in some module to cohomology of a subgroup with coefficients in the same module. In general it's hard to find a relation (it could be easier if $G$ were abelian, so that the diagonally embedded ... – Emerton Apr 17 2012 at 2:27
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