As in the title, say $\lambda$ is some irrep of the symmetric group $S\_n$, and $Br\_n$ the braid group on $n$ strands, 



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> What is $H^\*(Br\_n, \lambda)$?
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H*(braid group, irrep of symmetric group) = ?
I claim that the dimension of $H^\*(Br\_n;\lambda)$ is twice the dimension of the subspace of $\lambda$ fixed by the subgroup $S\_2\leq S\_n$.


Indeed, there is a spectral sequence
$$ H^p(S\_n;H^q(PBr\_n;\lambda)) \Longrightarrow H^{p+q}(Br\_n;\lambda). $$
As a $PBr\_n$-module, $\lambda$ is just a sum of finitely many copies of $\mathbb{Q}$ with trivial action. This means that $H^q(PBr\_n;\lambda)=H^q(PBr\_n)\otimes\lambda$, which is a $\mathbb{Q}[S\_n]$-modules and thus injective over $\mathbb{Z}[S\_n]$. This means that the spectral sequence collapses to an isomorphism
$$ H^\*(Br\_n;\lambda) = (H^\*(PBr\_n)\otimes\lambda)^{S\_n}, $$
so you just want to know the isotypical pieces of $H^\*(PBr\_n)$. The classifying space of $PBr\_n$ is the configuration space of distinct $n$-tuples in $\mathbb{C}$, and there is an elegant argument based on this and the Lefschetz fixed point theorem to determine the character of the $S\_n$-action on $H^\*(PBr\_n)$. The answer is that if we take two copies of the trivial representation of $S\_2$ and induce up to $S\_n$, we get $H^\*(PBr\_n)$. The claim now follows by Frobenius reciprocity.


This tells you the total dimension of $H^\*(Br\_n;\lambda)$, but not the degrees of the generators. For that, you would need to look in more detail at the structure of $H^\*(PBr\_n)$. There are generators $\alpha\_{ij}\in H^1$ for $1\leq i,j\leq n$ with $i\neq j$, subject only to the relations $\alpha\_{ij}=\alpha\_{ji}$ and $\alpha\_{ij}^2=0$ and
$$ \alpha\_{ij}\alpha\_{jk} + \alpha\_{jk}\alpha\_{ki} + \alpha\_{ki}\alpha\_{ij} = 0. $$


