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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.

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3 Answers

up vote 13 down vote accepted

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 $B_n$.

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The key term to google here is "Artin group". –  David Speyer Mar 30 '11 at 17:38
    
David, Qiaochu: Thanks. I had been aware of the existence of Artin groups associated with Coxeter groups, but somehow hadn't connected the dots by noticing that the signed symmetric groups have a Coxeter presentation. –  Vipul Naik Mar 30 '11 at 18:45
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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 $X_n/(Z_2^n \rtimes S_n)$.

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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.

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Also known as the hyper-octohedral group. –  Scott Carter Mar 30 '11 at 20:39
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