The image of the point-pushing group in the hyperelliptic representation of the braid group 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slightly embarrassing; I asked this question because I thought that a statement equivalent to it had been proved in an unpublished manuscript of J-K-Yu, but when I looked again at the ms.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.
 A: 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 $Sp_{2g}(\mathbb{Z})$. 
A: 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.
