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>> No.11704460 [View]
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11704460

>>11704401
I may have simplified (co)homology a bit, there can be tensor products and other nastyish stuff, BUT there are the Künneth and universal coefficient theorems etc. to deal with those. It is not as bad as it may seem to be.

>>11704404
Now that you mentioned invariants, do you know what a group ring/algebra is (check it out if not!), and the same question for augmentation ideals? If you then look at the standard resolution [math] \cdots \to RG\otimes_R RG\otimes_R RG \to RG\otimes_R RG \to RG \to R[/math], and let's say [math]R=\mathbb{Z}[/math] so that we get the group ing [math]G[/math], and then we tensor that with a module [math]M[/math] over the group ring, then we get [math]H_*(G; M)[/math] out of the resulting complex, and [math]H_0(G; M) = M_G[/math] is the (quotient) module of coinvariants. If we instead apply the functor [math]\text{Hom}(-, M)[/math], we get [math]^*(G; M)[/math] with the zeroth cohomology being the submodule of invariants, [math]M^G[/math]. That's a connection, but you should check the dteails out yourself, I'm a bit tipsy.

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