Let G be a finite group, H a subgroup and V a G-module. Then the embedding H in G induces a restriction map on $H^{n}(G,A)$ to $H^{n}(H,A)$. My question is that is there any long exact sequence which contains this map? And generally how to compute $H^n(G,A)$ effectively when n is small, like 0,1.
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$\begingroup$ The standard technique for computing $H^n(G,A)$ when $n = 0$ is just the definition: one looks for $G$-fixed elements in $A$. For $n = 1$, here is a common method: the $G$-action on $A$ gives a homo. $G \to Aut(A)$; let $H$ be the kernel. (So $H$-acts trivially on $A$.) Then one has the inflation-restriction sequence $0 \to H^1(G/H,A) \to H^1(G,A) \to Hom_{G/H}(H^{ab},A)$ (where the last term is written as $Hom$ rather than $H^1$ precisely because $H$ acts trivially on $A$). Now one hopes that the outer two terms are easier to compute, and can be pieced together to understand $H^1(G,A)$. $\endgroup$– EmertonCommented May 25, 2010 at 18:54
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2 Answers
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I think you're looking for the Hochschild-Serre spectral sequence. It's slightly more complicated than a long exact sequence, but you can extract the very concrete "inflation-restriction sequence" out of it.
Your general question is a little too general to get a good answer, though there are some good other questions on this site that you will probably find very helpful, e.g.,
Intuition for Group Cohomology
Essential theorems in group (co)homology
If you really only want $n=0$ and $n=1$, these are very concrete and addressed in any of the standard references for group cohomology. $H^0$ is the fixed-point functor, and $H^1$ is the group of "crossed homomorphisms" (which reduce to regular homomorphisms when the action is trivial).
answered May 25, 2010 at 3:33
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2$\begingroup$ Unless $H$ is normal how do you get a Hochschild-Serre s.s.? $\endgroup$ Commented May 25, 2010 at 3:55
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$\begingroup$ Ah, good catch. The way the question was phrased, it made it seem like even basic references would be helpful. $\endgroup$ Commented May 25, 2010 at 20:06
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There is a long exact sequence but I think it is largely useless: We have that for any $H$-module $B$ the cohomology $H^n(H,B)$ is equal to the cohomology $H^n(G,B^G_H)$ of the induced module $B^G_H$. When $B$ is the restriction of a $G$-module $A$ we have a surjective $G$-map $A^G_H \to A$ and an injective $G$-map $A\to A^G_H$. Taking the kernel (resp.\ cokernel) we get short exact sequences of $G$-module and then the desired long exact sequence. However, the cohomology of the kernel (cokernel) seems in general at least as difficult to compute as that of $A$ and $B^G_H$.
As for the efficient computation of low degree cohomology it depends on what you mean by efficient. There are computer algebra packages that compute for reasonable sized problems but I don't think they use methods that are that far from brute computation.
answered May 25, 2010 at 4:04
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$\begingroup$ Why $A^{G}_{H} \rightarrow A$ is a G-map? It seems just a H-map. $\endgroup$– user1832Commented May 27, 2010 at 0:43
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$\begingroup$ The map $A^G_H \to A$ is a little bit tricky to define (and works only because $H$ ahs finite index in $G$). It depends on the fact that the coinduced and induced modules are isomorphic, i.e., $A^G_H$ is both right and left adjoint to the restriction functor. (Think of the case of the permutation module $k[G/H]$ which has a $G$-map $k \to k[G/H]$ mapping $1$ to the sum of the elements of $G/H$ and a $G$-map $k[G/H] \to k$ mapping all the elements to $1$.) $\endgroup$ Commented May 27, 2010 at 9:36