Cohomology $H^*(G,K)$ of wreath products - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:55:23Z http://mathoverflow.net/feeds/question/86520 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86520/cohomology-hg-k-of-wreath-products Cohomology $H^*(G,K)$ of wreath products Chris Bowman 2012-01-24T09:53:58Z 2012-01-24T16:25:47Z <p>Let $G = Sym(a) \wr Sym(b)$ be a wreath product of symmetric groups - I'm particularly interested in the Weyl group of type $B$, $Sym(2) \wr Sym(n)$. Let $k$ be a field of characteristic $p$.</p> <p>What is $H^*(G,k)$? </p> <p>If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$. </p> <p>If $i=1$ and $G$ is as above then $H^1(G, k)=0$, for all $a$ and $b$, unless we're in characteristic 2.</p> <p>Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that:</p> <p>If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above.</p> <p>Any ideas if this is true? It might be that it holds with fewer restrictions on $i$ and $G$ - the proof I know for the symmetric group case uses the Schur functor and tilting modules for $GL_n$. However, the result proved is FAR more general (it concerns all Specht modules) - so maybe this $GL_n$ approach isn't needed.</p> http://mathoverflow.net/questions/86520/cohomology-hg-k-of-wreath-products/86536#86536 Answer by Ralph for Cohomology $H^*(G,K)$ of wreath products Ralph 2012-01-24T14:45:11Z 2012-01-24T16:25:47Z <p>Set $H^\ast(-) := H^\ast(-,k)$ and $S_a = Sym(a)$. Let $G$ be the wreath product that fits into the extension $$1 \to S_a^b \to G \to S_b \to 1.$$</p> <blockquote> <p><strong>Claim:</strong> $H^n(G) = 0$ for $1 \le n \le p-3.$</p> </blockquote> <p><em>Proof:</em> The LHS spectral sequence corresponding to the extension is $$E_2^{pq} = H^p(S_b, (H^\ast(S_a)^{\otimes b})^q)$$ where $$(H^\ast(S_a)^{\otimes b})^q = \oplus_{i_1 + ... + i_b = q} H^{i_1}(S_a) \otimes \cdots \otimes H^{i_b}(S_a).$$</p> <p>Let $1 \le q \le p-3$. Then $i_j \le p-3$ for $j=1,...,b$ and not all $i_j$ can be zero. Hence $H^i(S_a)=0$ for $1 \le i \le p-3$ implies $(H^\ast(S_a)^{\otimes b})^q = 0$. </p> <p>Thus $E_2^{\ast,q}=0$ for $1 \le q \le p-3$ and $E_2^{p,0} = 0$ for $1 \le p \le p-3$. This shows $H^n(G) = 0$ for $1 \le n \le p-3$. </p> <p><strong>Remarks:</strong> 1) Even more is true for wreath products: $$H^\ast(G) \cong H^\ast(S_b,(H^\ast(S_a)^{\otimes b})$$ as graded rings (cf. Nakaoka: Homology of the Infinite Symmetric Group, Ann. of. Math. 73(1961),229-257, Theorem 3.3). </p> <p>2) Since the extension splits, $H^\ast(S_b)$ is a direct summand of $H^\ast(G)$. Hence the vanishing range for the cohomology of $H^\ast(G)$ stated above cannot be improved. </p>