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Mar 11, 2015 at 7:15 answer added Sergei Ivanov timeline score: 10
Apr 19, 2012 at 10:42 vote accept Earthliŋ
Apr 18, 2012 at 20:19 answer added user1437 timeline score: 14
Apr 18, 2012 at 17:06 comment added Andy Putman There is a nice description of the twisted first cohomology group as crossed homomorphisms modulo principal crossed homomorphisms. But there really isn't a particularly nice description of the twisted first homology group.
Apr 18, 2012 at 13:11 comment added Jim Conant @s.barmeier: Sorry I got the wrong term in the complex. This is what happens when I answer before my morning coffee. My answer gave $H_0(G)$ which gives a quotient of $M$, namely the coinvariants $M_G$.
Apr 18, 2012 at 12:59 comment added Earthliŋ But $\mathbb Z[G]\otimes_{\mathbb Z[G]}M$ is just $M$, no? A quotient of $M$ does sound not too bad. Thanks for your answers, I'll ask my pencil for more insights.
Apr 18, 2012 at 12:48 comment added Jim Conant You'll get a certain quotient of $\mathbb Z[G]\otimes_{\mathbb Z[G]}M$ by the boundary operator. In the case of twisted coefficients the two parts of the tensor will be mixed up, so you won't have a nice factorization.
Apr 18, 2012 at 12:42 comment added Earthliŋ In Brown's book I can only find the statement with trivial coefficients... Do you have a page reference? Or are you suggesting I should calculate $H_1$ via the bar resolution, which I can't imagine has a straight-forward answer for an arbitrary action.
Apr 18, 2012 at 12:26 history edited Earthliŋ CC BY-SA 3.0
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Apr 18, 2012 at 12:18 comment added berl13 You will find the answer in Brown's book "Cohomology of groups". When the action is not trivial you have to work with twisted coefficients.
Apr 18, 2012 at 10:55 history asked Earthliŋ CC BY-SA 3.0