# Cohomology $H^*(G,K)$ of wreath products

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 $p$.

What is $H^*(G,k)$?

If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$.

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.

Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that:

If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above.

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.

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Have you tried using the Lyndon-Hochschild-Serre spectral sequence? en.wikipedia.org/wiki/Lyndon–Hochschild–Serre_spectral_sequence – Alain Valette Jan 24 '12 at 10:08

Set $H^\ast(-) := H^\ast(-,k)$ and $S_a = Sym(a)$. Let $G$ be the wreath product that fits into the extension $$1 \to S_a^b \to G \to S_b \to 1.$$

Claim: $H^n(G) = 0$ for $1 \le n \le p-3.$

Proof: The LHS spectral sequence corresponding to the extension is $$E_2^{pq} = H^p(S_b, (H^\ast(S_a)^{\otimes b})^q)$$ where $$(H^\ast(S_a)^{\otimes b})^q = \oplus_{i_1 + ... + i_b = q} H^{i_1}(S_a) \otimes \cdots \otimes H^{i_b}(S_a).$$

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 $(H^\ast(S_a)^{\otimes b})^q = 0$.

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 $1 \le n \le p-3$.

Remarks: 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).

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.

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