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I am interested if (part of) Tate's theorem can be extended to Abelian Varieties over $\mathbb{C}(t)$. Specifically, let $A$ be an Abelian Variety of dimension $g$ over $\mathbb{C}(t)$, and consider the associated Galois representation on torsion $\rho:G\rightarrow GL_{2g}(\mathbb{Q}_l)$, where $G$ is the absolute galois group of $\mathbb{C}(t)$, and $l$ is a prime number. Since in the isotrivial case - that is, where all fibers are the same Abelian Variety - the representation has finite image, I impose the condition that $A$ is simple and non-isotrivial. Are either of the following true:

(1) The $\mathbb{Q}_l$-algebra generated by $\rho(G)$ is the commutator of the endomorphism algebra of $A$

(2) $\rho(G)$ acts irreducibly on $T_l(A)\otimes_{\mathbb{Z}_l}\mathbb{Q}_l$ where $T_l(A)$ denotes the Tate module of $A$.

My (unreliable) guess is that (1) is too much to hope for but (2) has a fighting chance.

Thanks!

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    $\begingroup$ I dont think that (2) holds. Look for instance at the generic fibre of a Shimura curve classifying abelian surfaces with quaternionic multiplication. The action of the quaternions is not scalar but it commutes with Galois and so this would imply that the generic fibre is not simple, which is not true (a Shimura curve is not the projective line, agreed, but it should be possible to manufacture a counterexample from this). The fact that (2) holds over finite fields is quite specific. $\endgroup$
    – Damian Rössler
    Commented Feb 15, 2013 at 22:58
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    $\begingroup$ ...on the other hand, I think that (1) holds. Indeed, the Tate conjecture holds over ${\bf C}(t)$ by work of Faltings (see the book on the Mordell conjecture edited by Faltings-Wustholz). The Tate conjecture says precisely that the ${\bf Q}_l$-algebra of endomorphisms of the Tate-module, which commute with Galois, is generated by the endomorphisms of the abelian variety. Now apply the bicommutant theorem (see for instance Lang, Algebra, XVII, Th. 3.2) and the fact that the action of Galois is semisimple (also proven by Faltings) to get (1). $\endgroup$
    – Damian Rössler
    Commented Feb 15, 2013 at 23:07
  • $\begingroup$ Thanks, I was unaware that Faltings settled the Tate Conjecture for $C(t)$. As for my (2), I just stated it poorly I realize, I should have asked for the action on rational homology,without going to the Tate module. I'm perplexed by your comment about finite fields though: Isn't (2) just as false there as stated? Take an Elliptic curve over a finite field whose endomorphism field splits over $l$. Doesn't that make the action reducible? $\endgroup$
    – jacob
    Commented Feb 16, 2013 at 5:07
  • $\begingroup$ You are right abut the fact that it is even wrong over finite fields, sorry. What is specific about finite fields is that you can read the presence of two distinct simple factors from the fact that the char. pol. of the Frobenius has two distinct factors over $\bf Q$. $\endgroup$
    – Damian Rössler
    Commented Feb 16, 2013 at 10:11
  • $\begingroup$ One more correction: Faltings proves the Tate conjecture over fields finitely generated over their prime fields so that does not include ${\bf C}(t)$ (but ${\bf Q}(t)$ for instance). It is not clear to me whether this implies (1) and (2) over ${\bf C}$. $\endgroup$
    – Damian Rössler
    Commented Feb 16, 2013 at 17:22

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