5
$\begingroup$

Let $G=N\rtimes T$ and let $A$ be a $G$-module with a trivial $G$-action. The action of $T$ on induce a natural action of $T$ on the second cohomology group of $N$. Denote by $H^2(N,A)^T$ the $T$-invariant cohomology classes in $H^2(N,A)$.

For $A=\mathbb{C}^*$ and $(|N|,|A|)=1$ it is shown (e.g. in "The Schur Multiplier" by G. Karpilovsky) that $$H^2(G,\mathbb{C}^*)=H^2(N,\mathbb{C}^*)^T\times H^2(T,\mathbb{C}^*).$$

My question is: is this result is true when we replace $\mathbb{C}^*$ with $A$ and if not what is the generalization?

The proof given in Karpilovsky does not work. However, in http://link.springer.com/article/10.1007%2FBF01181625#page-1, Tahara deals with similar questions but I couldn't derive the desired result from there.

I will appreciate any help.

$\endgroup$
3
$\begingroup$

I guess, the answer you're seeking is the Hochschild-Serre spectral sequence. This is a spectral sequence with $E_2$-terms $$ E_2^{p,q}=H^p(T,H^q(N,V)) $$ which abuts to $H^{p+q}(G,V)$ for any $G$-module $V$. To derive this you should view $H^0(N,.)$ as a functor from $Mod(G)$ to $Mod(T)$ which is then followed by the functor $H^0(T,.)$ to the category of abelian groups. This pair of functors gives you a Grothendieck spectral sequence as described in Lang's book on Algebra. That's the one. Now as you are interested in $H^2$ only, there are not too many differentials around, so one should be able to do explicit computations here.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

From the spectral sequence mentioned by Anton one can derive an exact sequence in low degrees $$ 0\to E_2^{1,0}\to E^1\to E_2^{0,1}\to E_2^{2,0} \to \mathrm{ker}\left[E^2\to E_2^{0,2}\right]\to E_2^{1,1}\to E_2^{3,0}\ , $$ where $E^n=H^n(G,V)$ for $n=1,2$. I have found this exact sequence on page 38 of Sansuc's paper (without proof). Maybe this is what you need.

Note also that in order to construct the spectral sequence and the exact sequence in low degrees, you need only the short exact sequence $$ 1\to N\to G\to T\to 1.$$ From the splitting of this short exact sequence you can extract additional information about the exact sequence in low degrees.

| cite | improve this answer | |
$\endgroup$

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.