monodromy representations of families of hyperelliptic jacobians

Question

Let $k$ be a number field , $f \in k[x]$ of degree $2g$ with distinct roots, and $X$ the complex plane with the roots of $f$ removed. Then define the abelian scheme $J \rightarrow X$ where for each $t \in X$, $J_{t}$ is the jacobian of the hyperelliptic curve $y^{2} = f(x)(x - t)$. Then there is a monodromy representation of the topological fundamental group $\rho: \pi_{1}(X) \rightarrow \mathrm{Sp}_{2g}(\mathbb{Z})$ which preserves the Riemann form on the homology elements of the fiber over the basepoint.

I know that the image of this representation is the subset of $\mathrm{Sp}(\mathbb{Z})$ whose image modulo 2 is the identity but don't know of any "elementary" proof of it. In particular, I've seen it claimed that for genus $g = 1$, the image of the representation is simply $\Gamma(2) \cap \mathrm{SL}_{2}(\mathbb{Z})$. Can anyone tell me how to prove it for this elliptic curve case, in an intuitive way? I'm looking for something similar to http://rigtriv.wordpress.com/2010/02/25/monodromy-representations/, which I can't quite follow.

I would very much appreciate any kind of argument, but especially a relatively intuitive, visual one.

Answer 1

I think I do have an informal argument in the $g = 1$ case that can be obtained by drawing pictures of one copy of the Riemann sphere with $\infty$ and three other points chosen, and drawing homology loops and looking at what would happen to them if one of the finite points is rotated around the others. But this is very rough, and I'd still appreciate someone else's insight.

Answer 2

I believe the following link answers your question (this is apparently an unpublished result of J.K Yu)

The image of the point-pushing group in the hyperelliptic representation of the braid group group

Answer 3

This isn't actually true for $g=1$. $-I$ is an element of $\Gamma(2) \cap SL_2(\mathbb Z)$ which satisfies the equation $a \cdot a = 1$. No element in the fundamental group of the complex plane minus two points satisfies this, so for it to be hit the map must have some kernel. But it's easy to check that the map has no kernel. Indeed, the complex plane minus two points has universal cover the upper half plane, which is the moduli space for elliptic curves with a pair of generators for their homology, and all the deck transformations lie in the group which acts by changing the generators and thus acts nontrivially on homology.

But it is true up to $\pm 1$. Here is a slightly offbeat answer, which is intuitive to me but might not be to anyone else. The glib summary is "because $\mathbb P^1$ minus $3$ points is the moduli space of elliptic curves with full level $2$ structure."

Suppose you have an element in $SL_2(\mathbb Z) \cap \Gamma(2)$. I am not sure how this differs from just $\Gamma(2)$. This is a mapping class of the torus. It is easy to see that you can realize this by a family of complex tori, i.e., elliptic curves, over the cirle. Take a lattice in $\mathbb C$ and move the generators around so they form a new pair of generators for the lattice that realizes your chosen element in $\Gamma(2)$.

Since the mapping class is in $\Gamma(2)$, the three $2$-torsion points will be preserved by this mapping class, so we can uniquely identify and label them.

Embed each elliptic curve in the family in a Weierstrauss form of type $y^2=f(x)(x-t)$ such that the first $2$-torsion point is the first root of $f$, the second two-torsion point is the second root of $f$ and the third $2$-torsion point is $t$. This embedding is possible and is unique up to the automorphism $y \to \pm y$. This gives a function, $t$, from the circle to the complex plane minus the roots of $f$. If you follow the family $y^2=f(x) (x-t)$ and the original family around this loop simultaneously, you see they remain the same up to $\pm y$, so they give the same element of $SL_2(\mathbb Z)$ up to $\pm 1$.