The image of the point-pushing group in the hyperelliptic representation of the braid group - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T11:15:41Zhttp://mathoverflow.net/feeds/question/105048http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/105048/the-image-of-the-point-pushing-group-in-the-hyperelliptic-representation-of-the-bThe image of the point-pushing group in the hyperelliptic representation of the braid groupJSE2012-08-19T17:09:09Z2012-08-21T17:17:49Z
<p>Let $B_{2g+1}$ be the Artin braid group on $2g+1$ strands. There is a symplectic representation </p>
<p>$\rho: B_{2g+1} \rightarrow Sp_{2g}(\mathbf{Z})$</p>
<p>called the "hyperelliptic representation," which can be described as follows. The braid group is the fundamental group of the moduli space of configurations of 2g+1 points on the disc; each such configuration gives you genus-g surface which double covers the disk, ramified at those 2g+1 points and at the boundary; the representation is the usual monodromy action of the fundamental group on the homology of the fiber.</p>
<p>Alternately, we can think of $\rho$ as the specialization of the Burau representation to $t=-1$.</p>
<p>On the other hand, inside $B_{2g+1}$ there is a "point-pushing subgroup" H -- this can be thought of as the group of braids in which the first $2g$ strands stay fixed in place while the last strand is allowed to wind around the others. The group is thus naturally identified with the fundamental group of a disc with 2g punctures. It's a subgroup of the pure braid group, and it's the kernel of the Birman exact sequence.</p>
<p><strong>Question</strong>: What is the image $\rho(H)$ of the point-pushing subgroup in the hyperelliptic representation?</p>
<p>The image of the pure braid group under $\rho$ is the congruence subgroup $\Gamma(2)$, so $\rho(H)$ is a subgroup of that. It is known to be Zariski dense. Is $\rho(H)$ all of $\Gamma(2)$? Is it at least finite index? </p>
<p><strong>Update</strong>: OK, this is slightly embarrassing; I asked this question because I thought that a statement equivalent to it had been proved in an unpublished manuscript of J-K-Yu, but when I looked again at the ms. I though it was only proving something weaker. But now I see that Yu did prove this after all! However, I am very happy to know how to do it the way Agol explained below.</p>
http://mathoverflow.net/questions/105048/the-image-of-the-point-pushing-group-in-the-hyperelliptic-representation-of-the-b/105059#105059Answer by Agol for The image of the point-pushing group in the hyperelliptic representation of the braid groupAgol2012-08-19T22:23:49Z2012-08-20T01:27:16Z<p>It's finite index by <a href="http://groupprops.subwiki.org/wiki/Margulis%27_normal_subgroup_theorem" rel="nofollow">Margulis' normal subgroup theorem.</a> </p>
<p>Since $H \lhd P_{2g+1}$, then $\rho(H)\lhd \rho(P_{2g+1})$. Since $\rho(P_{2g+1})$ is finite index in $\rho(B_{2g+1})=\Gamma(2)$ (I'm taking your word for this),<br>
therefore $\rho(H)$ is either finite or finite-index in $\rho(P_{2g+1})$, and therefore in $\Gamma(2)$.
Since you also say that $\rho(H)$ is Zariski dense, it can't be
finite. Since finite-index subgroups of $Sp_{2g}(\mathbb{Z})$ have the congruence subgroup property,
I think that also means that $\rho(H)=\overline{\rho(H)}$, its
congruence closure in $Sp_{2g}(\mathbb{Z})$. </p>
http://mathoverflow.net/questions/105048/the-image-of-the-point-pushing-group-in-the-hyperelliptic-representation-of-the-b/105094#105094Answer by Dan Margalit for The image of the point-pushing group in the hyperelliptic representation of the braid groupDan Margalit2012-08-20T14:07:58Z2012-08-20T14:07:58Z<p>In fact it contains the level 4 subgroup. Using the lantern relation, you can show that the point-pushing subgroup contains all 4th powers of transvections (use the fact that the square of a Dehn twist about an odd curve acts trivially on the homology of the double cover), and Mennicke proved that these generate level 4. Also, level 4 mod level 2 is just sp(2g,Z_2), so it shouldn't be too hard to compute the exact image.</p>