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Let $K$ be a number field and $A/K$ an Abelian variety; let $l$ be a (rational) prime. Do there exist infinitely many primes $\mathfrak{p}$ of $K$ such that the Frobenius at $\mathfrak{p}$ acts semisimply on the integral Tate module $$T_l(A)\otimes \overline{\mathbb{Z}_\ell}?$$ By this I mean that there is a $\overline{\mathbb{Z}_\ell}$-basis of eigenvectors for the Frobenius action.

The answer is well-known to be yes if the integral Tate module is replaced with the rational Tate module (this goes back to Weil). If this is true, it should probably boil down to some Chebotarev argument, but I'm blanking at the moment.

Some remarks: As znt notes, the answer is "yes" if the image of the representation $G_K\to GL(A[\ell])$ contains a matrix with distinct eigenvalues, by Chebotarev.

Moreover, one can show that for any prime $\mathfrak{p}$ at which $A$ has good reduction, the Frobenius at $\mathfrak{p}$ acts semisimply on the $\ell$-adic Tate module for almost all $\ell$ (this follows from the theory of Frobenius tori). If the Mumford-Tate conjecture holds for $A$, I think it's enough to find any torus with good reduction at $\ell$ in the Mumford-Tate group of $A$, though I haven't written out the details. In any case, to find a counterexample it should be enough to find an Abelian variety over $K$ whose Mumford-Tate group contains no torus with good reduction at $\ell$.

One may also show the result holds if the image of the map $G_k\to GL(T_\ell(A))$ contains any element with distinct eigenvalues which is semi-simple over $\overline{\mathbb{Z}_\ell}$.

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    $\begingroup$ If there exists one element of Galois which has distinct eigenvalues mod $\ell$ then you're done by Cebotarev and Hensel. For example if $K=\Q$ and $\ell>2$ then use complex conjugation. $\endgroup$
    – znt
    Jun 30, 2016 at 23:17
  • $\begingroup$ @znt: yes, the interesting case is when the residual rep at $l$ is small. $\endgroup$
    – Daniel Litt
    Jun 30, 2016 at 23:23
  • $\begingroup$ @znt: I also am a bit confused about how you plan to use complex conjugation if $\dim(A)>1$. $\endgroup$
    – Daniel Litt
    Jun 30, 2016 at 23:32
  • $\begingroup$ Sorry -- my comment is erroneous as it stands -- was supposed to be about elliptic curves only but I failed to write "for example if $K=Q$ and $A$ is an elliptic curve and $\ell>2$...". Serves me right for posting late at night. Sorry. $\endgroup$
    – znt
    Jul 1, 2016 at 6:11
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    $\begingroup$ The Cebotarev argument for all $\ell \gg 0$ and every good $\mathfrak{p}$ away from $\ell$ is incorrect, since semisimple endomorphisms can have repeated eigenvalues. Counterexamples to the conclusion of that "distinct eigenvalues" conclusion are given by a product of two copies of the same (nonzero) abelian variety $A$ over $K$, and $K$-simple counterexamples are given by Weil restriction through a non-cyclic Galois extension. $\endgroup$
    – nfdc23
    Jul 3, 2016 at 14:28

1 Answer 1

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Probably not.

Let $E_1$ and $E_2$ be two distinct elliptic curves with identical $\ell$-torsion representations, and let $A$ be $E_1 \times E_2$ mod the diagonal $\ell$-torsion representation.

Then at any prime $\mathfrak p$ where Frobenius acts semisimply on $A[\ell]$, $E_1$ and $E_2$ are isogenous. Indeed, if the eigenvalues of Frobenius of $E_1$ and $E_2$ are distinct, an eigenbasis must consist two elements of $T_\ell (E_1)$ and two elements of $T_\ell(E_2)$, but there is no basis of $T_\ell(A)$, the lattice of elements of $T_\ell(E_1)+ T_\ell(E_2)$ where both parts are congruent mod $\ell$, consisting only of elements of $T_\ell(E_1)$ and $T_\ell(E_2)$. If the eigenvalues of Frobenius of $E_1$ and $E_2$ are equal, then they are isogenous.

Of course this is no problem as there are infinitely many such primes (https://arxiv.org/abs/1411.2914).

However, one can take a product of many such abelian surfaces for many different pairs of elliptic curves. The product is only semisimple if each pair of curves is isogenous. Standard heuristics suggest there should be only finitely many primes where all these pairs are isogenous (already 3 independent pairs should be sufficient).

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  • $\begingroup$ Very interesting (+1). One hint this approach gives as to what the correct statement should be is: the Frey-Mazur conjecture implies that for $\ell$ large in terms of the degree of the # field over which the $E_i$ are defined, a pair of elliptic curves with the same $\ell$-torsion rep are in fact isogenous. So this approach only can work for $\ell$ small w.r.t. the degree of the # field in question. So perhaps I should ask that $\ell$ be large in terms of the degree of the number field... $\endgroup$
    – Daniel Litt
    Oct 15, 2016 at 16:01
  • $\begingroup$ @DanielLitt Maybe $\ell$ needs to be large with respect to the degree as well, because you could also get into trouble with an equality of two small-dimensional representations of the $\ell$-torsion points of large-dimensional abelian varieties. $\endgroup$
    – Will Sawin
    Oct 15, 2016 at 20:07
  • $\begingroup$ @DanielLitt Regardless, because we're not going to prove the Frey-Mazur conjecture, nor some deep generalization of it, the strategy that makes sense to me is to identify some property of a closed subgroup of $GL_n(\mathbb Z_\ell)$ that implies a positive proportion of elements are $\ell$-adically semisimple, and decide when it makes sense to conjecture this property for large enough $\ell$. $\endgroup$
    – Will Sawin
    Oct 15, 2016 at 20:10
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    $\begingroup$ @DanielLitt I think a sufficient condition is that the $\ell$-adic monodromy group, viewed as an algebraic groups, has good reduction at $\ell$. Indeed every element of the identity component whose reduction mod $\ell$ is regular semisimple will act semisimply in this case. Moreover, any power of such an element will also act semisimply. Hence as long as there are any regular semisimple elements mod $\ell$ (I think this is always fine, but certainly for $\ell$ large enough), a positive-measure subset of any open subgroup will be semisimple. Because the image is open, we win. $\endgroup$
    – Will Sawin
    Oct 15, 2016 at 20:16
  • $\begingroup$ Yes, that's more or less what I was going for with the Mumford-Tate group in the original question -- this is supposed to be the same, more or less, as the $\ell$-adic monodromy group, if you believe the Mumford-Tate conjecture. I think there's probably some interesting things to be said about the natural integral structures on Mumford-Tate groups, their primes of bad reduction, etc... $\endgroup$
    – Daniel Litt
    Oct 16, 2016 at 17:31

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