Question

This is probably something well known (either in the affirmative or in the negative) but I couldn't get this information easily:

Braid group:Symmetric group::?:Signed symmetric group

By "signed symmetric group" I mean the wreath product of the cyclic group of order two by the symmetric group with its usual action as a symmetric group. Equivalently, it is the group of n by n matrices under multiplication where every row and every column has exactly one nonzero entry and that entry could be +1 or -1.

I don't see an obvious way to generalize, nor do I see a reason why no analogue can exist.

Answer 1

There are braid groups attached to every Coxeter group which are obtained by forgetting that the generators in the standard presentation square to the identity but keeping the other relations. I believe the signed symmetric group is the Coxeter group of type $B_n$.

Answer 2

There is an analog of the braid group for each Coxeter group. The signed symmetric group is the Coxeter group of type B: as for the usual braid group, a presentation of it can be obtained from the standard Coxeter presentation by removing the torsion relation. Hence, it is generated by $\tau, \sigma_1,\dots, \sigma_{n-1}$ and relations

$ \tau \sigma_1 \tau \sigma_1 = \sigma_1 \tau \sigma_1 \tau $

$ \tau \sigma_i=\sigma_i\tau$ if i > 1

$\sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_{i} \sigma_{i+1}\ i=1,\dots,n-2 $

$ \sigma_i \sigma_j = \sigma_j \sigma_i \text{ if } |i-j| \geq 2 $

There is also a topological definition of this group: Coxeter groups are finite reflection groups, that is finite subgroups of $GL_n(\mathbb{R})$ generated by reflections. For the type B, the corresponding hyperplane are defined by equation $z_i-z_j=0$ (as for the symmetric group), $z_i+z_j=0$ and $z_i=0$.

Hence let $X_n={(z_1,\dots,z_n) \in (\mathbb{C}^{\times})^n, z_i \neq \pm z_j}$, and define the pure braid group of type B as the fundamental group of $X_n$. The full braid groupe of type B is the fundamental group of the quotient space $X_n/(Z_2^n \rtimes S_n)$.

Answer 3

To expand a bit on Qiaochu's answer: The signed symmetric group has a presentation with generators $s_i=(i,i+1)$ and $t$, which is the element that is -1 in the first coordinate and 1 in the others (in terms of signed permutation matrices, this is $\operatorname{diag}(-1,1,1,\dots,1)$).

The $s_i$'s satisfy the usual braid relations $$s_is_{i+1}s_i=s_{i+1}s_is_{i+1}$$ as well as $s_i^2=1$. You can check that there are two different ways of writing $\operatorname{diag}(-1,-1,1,\dots,1)$ in terms of these generators: $$s_1ts_1t=ts_1ts_1;$$ you also have that $t^2=1$.

If you forget about the relations that say that the squares of generators are 1, you get the braid group of type B. This braid group also has a topological intepretation as braids in a cylinder: see this paper.