MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Have you tried using the Lyndon-Hochschild-Serre spectral sequence?–Hochschild–Serre_spectral_sequence – Alain Valette Jan 24 '12 at 10:08
up vote 5 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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