Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $B_{2g+1}$ be the Artin braid group on $2g+1$ strands. There is a symplectic representation

$\rho: B_{2g+1} \rightarrow Sp_{2g}(\mathbf{Z})$

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.

Alternately, we can think of $\rho$ as the specialization of the Burau representation to $t=-1$.

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.

Question: What is the image $\rho(H)$ of the point-pushing subgroup in the hyperelliptic representation?

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?

Update: 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slightly embarrassing; I asked this question because I thought that a statement equivalent to it had been proved in an unpublished manuscript of J-K-Yu, but when I looked again at the ms.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.

share|improve this question
    
I'm fairly certain Fred Cohen has known the answer to your question for some time. So if you don't get a response here you might want to try writing him. Off the top of my head I don't know the answer, though. –  Ryan Budney Aug 19 '12 at 17:48
    
I remember Dan Margalit describing the answer to me at some point, though I don't remember what it was. He had worked it out with Tara Brendle as part of their work on the hyperelliptic Torelli group. You might want to ask him (he sometimes visits MO, but he might miss this question). –  Andy Putman Aug 19 '12 at 19:17
add comment

2 Answers 2

up vote 13 down vote accepted

It's finite index by Margulis' normal subgroup theorem.

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),
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 $Sp_{2g}(\mathbb{Z})$.

share|improve this answer
5  
AGOOOOOOOOOOOOOOOOOOL! </andrescantor> –  JSE Aug 20 '12 at 14:50
    
\overline is not visible on my browser. –  Anton Petrunin Aug 20 '12 at 15:47
    
Ian's argument also proves that the image of the point-pushing group is Zariski-dense in the Burau representation itself, right? –  JSE Aug 20 '12 at 16:15
    
If you know $Burau(P_n)$ is Zariski dense, and normal subgroups of Zariski dense groups are Zariski dense (or finite?), then this might be true. –  Ian Agol Aug 20 '12 at 17:24
add comment

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.

share|improve this answer
    
There is a famous paper of A.Borel (Comptes Rendus 1976) where he computes the index. –  Igor Rivin Aug 20 '12 at 14:59
    
Igor, what is the title of the paper? –  Dan Margalit Aug 20 '12 at 15:05
    
More importantly, what is the index? –  JSE Aug 20 '12 at 16:14
2  
Sorry, by Borel, I meant, of course, J. Tits: Tits, Jacques Systèmes générateurs de groupes de congruence. (French. English summary) C. R. Acad. Sci. Paris Sér. A-B 283 (1976), no. 9, Ai, A693–A695. –  Igor Rivin Aug 20 '12 at 16:46
2  
Actually, it is not too hard to see that the image is 2g-dimensional, and so the index is (2g^2+g)-(2g) = g(2g-1). –  Dan Margalit Aug 20 '12 at 20:10
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.