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Suppose $X$ is a smooth proper curve over $\mathbb{F}_p$ for some prime number $p$. Let $l\neq p$ be a prime, and suppose $L$ is a rank 2 local system over $X$ with coefficients in $\mathbb{Z}_l$ such that $L$ "looks like" it comes fro an elliptic curve. In paricular:

1) $L$ is of weight 1. I.e. for any $\mathbb{F}_{p^m}$ point of $X$, the Frobenius acting on $L$ has eigenvalues that are Weil numbers of absolute value $\sqrt{p^m}$, and

2) $L$ has coefficient field $\mathbb{Q}$. That is, the corresponding $\pi_1(X)$ representation has traces in $\mathbb{Q}$.

$\textbf{Question}$: Does $L$ come from the Tate module of an elliptic curve over $X$?

$\textbf{Things I know:}$ I know that by work of L.Lafforgue $L$ is motivic, but I don't know how to extract the above statement.

Also, It sees like Drinfelds work on elliptic modules might allow you to find $L$ in the jacobian of some modular curve, which would be enough. This would require $L$ to be Steinberg at one place - though I'm not entirely sure what this means. In any case, I don't quite understand the details well enough.

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    $\begingroup$ What if there is an abelian surface over $X$ with quaternion algebra multiplication by some quaternion algebra split at $l$. Then the sheaf of $l$-adic Tate modules of that surface splits into the sum of two rank two local systems. Each one clearly has eigenvalues of Frobenius Weil numbers of the correct absolute value, but they also both have coefficient field $\mathbb Q$ as they are isomorphic. But they don't come from an elliptic curve. $\endgroup$
    – Will Sawin
    Commented Mar 31, 2016 at 21:26
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    $\begingroup$ This explains why you won't be able to find it in an object like a modular Jacobian - if this quaternion multiplication surface is a factor of the modular Jacobian, then because its Hecke eigenvalues are rational, the quotient of the Hecke algebra acting on it must be just the center of the quaternion algebra, which violates multiplicity one. Drinfeld gets around this by finding $L \boxtimes L$ in the cohomology of a variety on $X \times X$. But this shows up in $H^2$, of course, and not $H^1$. $\endgroup$
    – Will Sawin
    Commented Mar 31, 2016 at 21:42
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    $\begingroup$ You would be better served, I would think, by trying to show the Steinberg condition. Assuming that your Galois representation has infinite image of inertia at one place, by Grothendieck's local monodromy theorem it is a unipotent representation tensored with a character. By twisting you can force the character to be trivial, which I think means Steinberg, and win. If it has finite image of inertia at each place then it can't come from an elliptic curve unless it's an Artin representation, in which case you presumably have no problem. So you just have to eliminate that case. $\endgroup$
    – Will Sawin
    Commented Mar 31, 2016 at 21:46
  • $\begingroup$ Thanks Will! Yeah, I came up with that counterexample, and am now interested in whether its the `only one': In other words, whether this always shows up in H^1 of some abelian variety. The galois approach seems reasonable, but there are many representations with finite image of inertia everywhere, right? $\endgroup$
    – jacob
    Commented Apr 1, 2016 at 3:10
  • $\begingroup$ Sure, but I suspect this is not known and hard to know. Drinfeld's method doesn't produce the abelian variety directly for the reason I sketched. So you are left with some lifting problem where you have $L \otimes L$ appearing in $H^2$ of a surface and you want to find $L$ in $H^1$ of something. If the surface were K3 we could apply Kuga-Satake, but that's not known in general. $\endgroup$
    – Will Sawin
    Commented Apr 1, 2016 at 3:17

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