# How do we see the rank of the braid group?

The only presentation of the braid group that most people ever see is the standard Artin presentation

$$B_n=\langle σ_1,\cdots,σ_{n−1}|\ σ_iσ_j=σ_jσ_i\ \ (|i−j|>1),\ σ_iσ_{i+1}σ_i=σ_{i+1}σ_i σ_{i+1}\rangle.$$

It would be surprising if the rank of $B_n$ were less than $n−1$, as then there would be a simpler presentation. Fortunately, the rank of the braid group is in fact $n−1$. I'm curious about various ways to see this fact, since I don't know any.

The rank is in fact two, and so the rest of this question is irrelevant.

The only useful tool I know for bounding group ranks is the trivial lemma that if $G\twoheadrightarrow H$ is surjective, then $\mathrm{rank}(G)\ge \mathrm{rank}(H)$. Unfortunately I don't see how we can do much better than the trivial bound $\mathrm{rank}(B_n) \ge 2$ that $B_n$ being non-cyclic gives us. The only hope I have here is that for $n \ge 4$ we have a homomorphism $B_n \to S_n \times \mathbb{Z}$ (underlying permutation plus writhe) whose image might have rank 3.

Are there any nice surjections out of $B_n$ that would let us see a better bound for the rank?

I'm aware that $B_n = \mathrm{Mod}(D^2-\{n\text{ points}\},\partial D^2)$, and so there might be a minimality argument along the lines of the symplectic representation argument for $$Mod(\Sigma_g) \twoheadrightarrow \mathrm{Sp}(2g,\mathbb{Z}) \subset \mathrm{Aut}(H_1(\Sigma_g,\mathbb{Z})).$$ I don't see anything in, say, Farb-Margalit about this, so I'd appreciate a reference that talks about the rank calculation from the mapping-class or configuration space perspective.

• if by rank you mean minimal number of generators, then the braid group is two generated and hence has rank two. – Venkataramana Apr 19 '15 at 15:24
• Shucks, that it is. I suppose that means it's good I wasn't able to find a surjection onto a rank-3 group... – dvitek Apr 19 '15 at 17:05