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slightly embarrassing; I asked this question because I thought that a statement equivalent to it had been proved in an unpublished manuscript of J-K-Yu, but when I looked again at the ms.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



The image of the point-pushing group in the hyperelliptic representation of the braid group
It's finite index by [Margulis' normal subgroup theorem.](http://groupprops.subwiki.org/wiki/Margulis%27_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 $Sp\_{2g}(\mathbb{Z})$. 


