# Cohomology of braid groups with coefficients in the group ring

Let $\mathbf B_n$ be the braid group on $n$ strings.

What is known about the cohomology of $\mathbf B_n$ with coefficients in its integral group ring: $H^*(\mathbf B_n;\mathbb Z \mathbf B_n)$?

The braid groups $B_n$ are Bieri-Eckmann duality groups of dimension $n-1$. It follows (either by definition or by a standard result depending on how you set things up) that $H^k(B_n;\mathbb{Z}[B_n])$ is $0$ for $k \neq n-1$ and is torsion-free for $k=n-1$.
Here is a brief description of how to see this. First, the group $B_n$ are of type F (i.e. they have compact Eilenberg-MacLane spaces). It follows that it is enough to find a finite-index subgroup of $B_n$ that is a duality group of dimension $n-1$. The pure braid group $PB_n$ do the job. To see that $PB_n$ is a duality group of dimension $n-1$, we use induction on $n$. The base case $n=1$ is trivial since $PB_1$ is the trivial group. For $n>1$, we will use the standard short exact sequence $$1 \longrightarrow F_{n-1} \longrightarrow PB_n \longrightarrow PB_{n-1} \longrightarrow 1,$$ where $F_{n-1}$ is the free group on $(n-1)$ generators and the map $PB_n \rightarrow PB_{n-1}$ comes from deleting the final strand. By induction $PB_{n-1}$ is a duality group of dimension $(n-2)$, and it is standard that $F_{n-1}$ is a duality group of dimension $1$. It follows that $PB_n$ is a duality group of dimension $(n-2)+1 = n-1$.
• Hi Andy, thank you for a very nice answer. What is the situation for the commutator subgroup $[B_n,B_n]$? – Jarek Kędra Oct 15 '17 at 10:15
• @JarekKędra: Sorry, I don't know much about the cohomology of $[B_n,B_n]$. Indeed, I don't even know what kind of finiteness properties it has, though I'm sure this must be in the literature somewhere. – Andy Putman Oct 15 '17 at 18:10