# Cohomological dimension of $\mathcal{B}_n$

What is the cohomological dimension of the braid group $\mathcal{B}_n$ on n-strands ? A reference would be appreciated.

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An obvious upper bound would be $2n$ for $B_n$, by looking at the classifying space. I think you could even take an upper bound of $n-1$. – Steve D Jun 29 '11 at 15:26
There is a tag already for cohomological-dimension, so I figured we may as well use it – David White Jun 29 '11 at 16:30

The other answers are correct, but I wanted to point out a quick way to see that $B_n$ has cohomological dimension $n-1$.

One obtains a lower bound of $n-1$ since $\mathbb{Z}^{n-1}$ is a subgroup of $B_n$. Take $n-1$ disjoint non-isotopic loops forming a pants decomposition of the $n$ punctured plane, then Dehn twists about these give a subgroup isomorphic to $\mathbb{Z}^{n-1}$.

For an upper bound, one may use the fact that the moduli space of $n$ points in $\mathbb{C}$ (normalized to have sum $=0$) is a Stein manifold of complex dimension $n-1$, and therefore has a spine of dimension $n-1$. The fundamental group of this space is $B_n$. This space is equivalent to the space of monic polynomials of degree $n$ with zero trace (coefficient of degree $n-1=0$) and non-zero discriminant, which is how one may see that it is Stein (actually, an affine variety). The fact that the moduli space is a $K(B_n,1)$ follows from Teichmuller theory, or one may pass to the finite-sheeted cover of $n$ marked points, and see that this is an iterated surface bundle, and therefore its universal cover is contractible.

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It's $n-1$.

See Harer, The virtual cohomological dimension of the mapping class group of an orientable surface, Inventiones Mathematicae, Volume 84, Number 1, 157-176.

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There is a more geometric(?) method to obtain an upper bound of the cohomological dimension. Recall that the braid group $B_n$ is the fundamental group of the complement of the complexification of an essential hyperplane arrangement $\mathcal{A}_{n-1}$ in an $(n-1)$-dimensional vector space $\mathfrak{h}_{n-1}$ divided out by the action of the symmetric group $\Sigma_n$ of $n$ letters. Namely $$B_n = \pi_1((\mathfrak{h}_{n-1}\otimes\mathbb{C} - \cup_{H\in \mathcal{A}_{n-1}} H\otimes\mathbb{C})/\Sigma_n).$$ Since the complement is known to be $K(\pi,1)$, it serves as the classifying space of $B_n$.

In general, for any real hyperplane arrangement $\mathcal{A}$, Salvetti constructed a cell complex $\mathrm{Sal}(\mathcal{A})$ which is homotopy equivalent to the complement of the complexification. Furthermore if $\mathcal{A}$ is essential in an $n$ dimensional vector space, then $\dim\mathrm{Sal}(\mathcal{A}) = n$.

Thus the cohomological dimension of $B_n$ is bounded by $n-1$.

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I also recommend M. Korkmaz' survey on low-dimensional homology of Mapping Class Groups:

Korkmaz, Mustafa(TR-MET) Low-dimensional homology groups of mapping class groups: a survey. (English summary) Turkish J. Math. 26 (2002), no. 1, 101–114. 57M05 (20J05 57M07 57N05)

It is in fact quite easy to find a subgroup of $P_n$ isomorphic to $\mathbb Z^{n-1}$. It is generated by the full twist of the first $k$ strands, where $k= 2,3,\dots, n$.