What is the cohomological dimension of the braid group $\mathcal{B}_n$ on n-strands ? A reference would be appreciated.
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$\begingroup$ 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 $n-1$. $\endgroup$– Steve DCommented Jun 29, 2011 at 15:26
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$\begingroup$ There is a tag already for cohomological-dimension, so I figured we may as well use it $\endgroup$– David WhiteCommented Jun 29, 2011 at 16:30
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6 Answers
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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.
answered Jun 29, 2011 at 15:21
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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)
Turkish JM is available for free...
answered Jun 29, 2011 at 15:42
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The homology of braid groups with rational and finite field coefficients (all with constant action) was computed by Cohen in the appendix to chapter III of "The homology of iterated loop spaces" LNM 533.
answered Jun 29, 2011 at 15:38
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I will sadly have to answer to this old post to elaborate on the lower bound part of Agol's answer. I would be grateful if someone could turn this into a comment.
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 $k= 2,3,\dots, n$.
