# Does any Artin Braid group come from a Braid group of a root system?

An Artin Braid group of $$n$$-strands can be seen as the fundamental group of the $$n^{\text{th}}$$ (unordered) configuration space $$\pi_1(\mathbb C^n\setminus \cup H_{ij})/S_n)$$, where $$H_{ij}=Z(x_i-x_j)$$ is an hyperplane.

More generally, a Braid group of a root system is defined as follows: Fix a root system corresponding to a finite-dimensional Lie algebra $$\mathfrak g$$. The braid group $$\text{Br}$$ of a root system is defined as $$\pi_1((\mathfrak h\setminus \mathfrak h_\alpha)/W)$$, where $$\mathfrak h$$ is the Cartan Lie subalgebra, $$W$$ is the Weyl group and $$\mathfrak h_\alpha$$ are the root hyperplanes corresponding to all roots.

I know $$S_n$$ is the Weyl group of $$\mathfrak{sl}(n,\mathbb C)$$, but the Cartan Lie subalgebra of $$\mathfrak{sl}(n,\mathbb C)$$ has rank $$n-1$$, not $$\mathbb C^n$$. So an Artin Braid group $$B(n)$$ is not the Braid group of the root system of $$\mathfrak{sl}(n,\mathbb C)$$. My question is: does every Artin Braid group $$B(n)$$ come from the Braid group of a root system of some Lie group's Lie algebra?

• @QiaochuYuan I think $\mathfrak{sl}(n+1,\mathbb C)$ has Cartan Lie algebra $\mathbb C^n$ but Weyl group $S_{n+1}$. So the Braid group for $\mathfrak{sl}(n+1,\mathbb C)$ is $\pi_1((\mathbb C^n\setminus \cup H)/S_{n+1}))$ not the $Br(n)$, right?
– Kat
Commented Aug 1 at 22:17

There actually isn't an issue here, you get $$B_n$$ as the braid group of $$\mathfrak{sl}_n(\mathbb{C})$$. This is because $$\mathbb{C}^n \setminus H$$ has an $$S_n$$-equivariant deformation retraction onto the subspace of points such that $$x_1 + \dots + x_n = 0$$, given by
$$(x_1, \dots x_n) \mapsto (x_1, \dots x_n) - t \frac{x_1 + \dots + x_n}{n} (1, 1, \dots )$$
for $$t \in [0, 1]$$. So the two have the same fundamental group, even after the quotient by $$S_n$$, which is $$B_n$$.