Cocyles for abelian extensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:08:05Z http://mathoverflow.net/feeds/question/51582 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensions Cocyles for abelian extensions Sebastian Burciu 2011-01-09T20:26:03Z 2011-01-10T14:22:12Z <p>Suppose we have an abelian extension of Hopf algebras,</p> <p>$$k \rightarrow k^G \rightarrow A \rightarrow kF \rightarrow k.$$ </p> <p>According to the general theory there is a left action of $F$ on $G$ and a $2$-cocycle $\sigma:F\times F \rightarrow k^G$ such that $A=k^G$ #${}_{\sigma} kF$ as algebras.</p> <p>1)Is it true that $\sigma^n=\mathrm{id}$ for some $n\geq 1$? In other words can one choose $\sigma'$ in the same class with $\sigma$ such that if $\sigma'(f, h)=\sum_{a \in G}\sigma'_a(f,h)p_a$ then $$\sigma\;'_a(f,h)^n=\mathrm{id}$$ for all $a, f, h$. Here $p_a$ is the usual notation for dual basis of the group element basis . </p> <p>2) A little more general question (that implies the first question) is if the set of all $2$-cocycles of $F$ with values in $k^G$ finite? Equivalently the question is if $H^2(F, k^G)$ is finite with $F$ acting on $k^G$ via the action of $F$ on $G$.</p> <p>3)A related question for what groups $X$ with an $F$-action the group $H^2(F, X)$ is finite? For example $X=k^*$ gives the usual Schur multiplier. </p> http://mathoverflow.net/questions/51582/cocyles-for-abelian-extensions/51653#51653 Answer by César Galindo for Cocyles for abelian extensions César Galindo 2011-01-10T14:22:12Z 2011-01-10T14:22:12Z <p>Sebastian,</p> <p>Suppose that $k$ is an algebraic closed field of characteristic zero.</p> <p>Let $G$ be a finite group and $X$ a finite right $G$-set, so $k^X$ is left $G$-module. We want to see that $H^2(G,k^X)$ is a finite group. Let $X=\cup_{i=1}^n X_i$ where each $X_i$ is a transitive $G$-set, then as $G$-module $k^X= \bigoplus_{i=1}^n k^{X_i}$, and $H^2(G,k^X)= \bigoplus_{i=1}^n H^2(G,k^{X_i})$. Now by Shapiro's Lemma <a href="http://en.wikipedia.org/wiki/Shapiro%27s_lemma%20%22Shapiro%27s%20Lemma%22" rel="nofollow">link text</a>, $H^2(G,k^{X_i})\cong H^2(G_i,k^*)$, where $G_i$ is the stabilizer of an element $x_i \in X_i$ and $G_i$ acts trivially over $k^*$. Finally, since $k$ is algebraic closed, the Schur multiplers $H^2(G,k^{X_i})$ are finites.</p>