5
$\begingroup$

I was wondering if $\mathbb{Z} \wr S_n$, where $\mathbb{Z}$ is the usual group of integers, $S_n$ the symmetric group on n elements and $\wr$ the wreath product of two groups, contains the braid group $B_n$.

I was also wondering if $n+1$-dimensional matrices of the form :

$\begin{bmatrix} 1&0&0 \\\\ 1&0&1 \\\\ 1&1&0 \end{bmatrix}$ for B2

$\begin{bmatrix} 1&0&0&0 \\\\ 1&0&1&0 \\\\ 1&1&0&0 \\\\ 0&0&0&1 \end{bmatrix}$ and $\begin{bmatrix} 1&0&0&0 \\\\ 0&1&0&0 \\\\ 1&0&0&1 \\\\ 1&0&1&0 \end{bmatrix}$ for B3

and so on... form a representation of the braid group $B_n$.

Thank you for your help

A. Popoff

  • 0
    @user8167: "but then wh$y$ bothering with complicated representations like the Burau one" -- because different representations highlight different properties of the group being represented. If you were just interested in having a single representation, you could always just take the trivial one.2014-01-29

1 Answers 1

5

It is certainly not true that $\mathbb{Z}\wr S_n$ contains $B_n$, for any $n>2$. Indeed, by definition, $\mathbb{Z}\wr S_n$ has a free abelian subgroup of finite (indeed, $n!$) index.

On the other hand, the kernel of the natural map of pure braid groups $PB_n\to PB_{n-1}$ obtained by forgetting a strand is the free group of rank $n-1$ (this follows from the Birman Exact Sequence, I suppose), so if $n>2$, $PB_n$ (and hence $B_n$) contains a non-abelian free group. If it were contained in $\mathbb{Z}\wr S_n$ then this non-abelian free group would have an abelian subgroup of finite index, which is absurd.

[Note: this answer is essentially the same as Steve D's comment.]

  • 1
    As Steve suggests, it is in general easy to check if a given map defines a representation (you just have to check the relations), and very difficult to check whether a given representation is faithful. Indeed, in the following paper, Bridson and I exhibit a sequence of representations with the property that no algorithm determines which are faithful: http://arxiv.org/abs/1003.5117 .2011-03-13