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

  • 1
    Checking whether something is a braid group representation is straightforward; you just have to check the braid relations. Have you done that yet?2011-03-12
  • 0
    Hi... yes I checked and it works fine, but then why bothering with complicated representations like the Burau one, when this one is simpler....2011-03-12
  • 1
    Perhaps it is not faithful? (In fact, it shouldn't be faithful. The linearity of the braid groups was until recently an open problem so a faithful linear representation shouldn't be easy to write down.) I guess the Burau representations aren't faithful either...2011-03-12
  • 0
    I have to admit, I didn't check for faithfulness, it just arose while doing some other stuff with braid groups...2011-03-12
  • 0
    @Alexandre: as you surely expect, checking faithfulness is a difficult thing!2011-03-12
  • 0
    I'll have to ask an expert then... by the way, I was thinking that if my hypothesis concerning the wreath product is true, then Zm () Sn could be considered as "constrained" braid groups where the generators of the braid group have a limited number m of twists... does that make any sense ? Has this ever been studied ?2011-03-12
  • 0
    @Alexandre: regarding that, look at http://mathoverflow.net/questions/48849/transpositions-of-order-three: Gjergji mentions a paper which studies such things.2011-03-12
  • 0
    (By the way, I don't really see what matrices you have in mind. Your explanation that they are obtained by «swapping pairs of columns and adding» does not tell me anything... Could you write down explicitly a couple more?)2011-03-12
  • 2
    Your hypothesis is not correct: for n=3, your wreath product is solvable, but $B_3$ contains a non-abelian free subgroup.2011-03-12
  • 0
    Would be :$$\begin{bmatrix} 1&0&0 \\\\ 1&0&1 \\\\ 1&1&0 \end{bmatrix}$$ 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...2011-03-12
  • 0
    @Alexandre: if you add them to the body of the question, it is much easier to find them :)2011-03-12
  • 0
    Sorry for that... I corrected it, hope it will make more sense...2011-03-12
  • 0
    @Steve ... thanks for the answer. I had found a couple of elements from the wreath product that were satisfying the braid relations, but I wasn't looking at the big picture (this whole thing comes from rather applied stuff...)2011-03-12
  • 0
    @user8167: "but then why 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