9
$\begingroup$

[a repost from SE due to the lack of response]

Given a group $G$, let $A$ be a $G$-module and let $N\trianglelefteq G$. If I understand it correctly, the superscript "G/N" in the third term of the standard inflation-restriction exact sequence $$ 0\to H^1(G/N,A^N)\to H^1(G,A)\to H^1(N,A)^{G/N}\to H^2(G/N,A^N)\to H^2(G,A) $$ means the fixed points under the action of $G/N$ on the first cohomology group $H^1(N,A)$. But how is this action defined? Is there an elementary definition (in terms of cocycles) that does not refer to the Lyndon–Hochschild–Serre spectral sequence?

$\endgroup$
1
  • 1
    $\begingroup$ When in doubt, consult Ken Brown's Cohomology of Groups! $\endgroup$ Jul 30, 2015 at 16:17

1 Answer 1

13
$\begingroup$

You do not need LHS spectral sequence for this action. The functors $H^*(N,-)$ are the derived functors of $(-)^N:G\text{-mod}\rightarrow G/N\text{-mod}$, so they will carry a structure of $G/N$-modules. Explicitly, on the level of cocycles, it can be described as follows: if you have a one cocycle $f:N\rightarrow A$ and $g\in G$, then the action on $f$ will be given by $g\cdot(f)(n) = g\cdot f(g^{-1}ng)$

$\endgroup$
2
  • $\begingroup$ For this to be an action of $G/N$, $N$ must act trivially modulo $B^1(N,A)$. I do not see why this holds in general. Suppose $g\in N$. Then $(gf-f)(n)=g\cdot f(g^{-1}ng)-f(n)$. Why is this always of the form $ga-a$ for some $a\in A$? (Note that it is not assumed that $N$ is abelian. Otherwise $a=f(n)$, of course.) $\endgroup$
    – Anvita
    Jul 30, 2015 at 13:22
  • 2
    $\begingroup$ Assume that $g\in N$. Then by using the cocycle condition we have $gf(g^{-1}ng) - f(n) = gf(g^{-1}) + f(n) + nf(g) - f(n) = gf(g^{-1}) + nf(g)$. Now, since $f(gg^{-1}) = 0$ we have $f(g) + gf(g^{-1})) = 0$ so $gf(g^{-1}) = -f(g)$. This implies that the difference is $nf(g) - f(g)$ so we can take $a=f(g)$. $\endgroup$
    – Ehud Meir
    Jul 30, 2015 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.