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Background

Let $E$ be an elliptic curve over $\mathbb{Q}$ and let $G_{\mathbb{Q}}$ be the absolute Galois group $Aut(\overline{\mathbb{Q}})$. For any positive integer $n$ the $n$-torsion subgroup $E\[n\](\overline{\mathbb{Q}})$ is stable under the $G_{\mathbb{Q}}$-action. Since $E\[n\](\overline{\mathbb{Q}})$ is isomorphic to $(\mathbb{Z}/n\\mathbb{Z})^2$ one gets a continuous (with respect to the profinite topology on the left and the discrete on the right) homomorphism $$ \overline{\rho_{E,\, n}}\colon G_{\mathbb{Q}} \to GL_2(\mathbb{Z}/n\mathbb{Z}) $$ which one calls the mod $n$ representation associated to $E$. As $n$ varies these are compatible and taking limits gives representations $\rho_{E,\ell^{\infty}}$ and $\rho_E$ with values in $GL_2(\mathbb{Z}_l)$ and $GL_2(\widehat{\mathbb{Z}})$ which one calls respectively the $\ell$-adic and adelic representations associated to $E$.

Alternatively, $\rho_{E,n }$ is isomorphic to the representation induced by the action of $G_{\mathbb{Q}}$ on the etale cohomology $H^1_{\text{et}}(E_{\overline{\mathbb{Q}}}; \mathbb{Z}/n\mathbb{Z})$; the description of $\rho_{E,n}$ via torsion generalizes to give representations $\rho_{A,n}$ for higher dimensional abelian varieties, but for a general variety one must instead use cohomlogy.

Serre famously proved that for $E$ an elliptic curve with $End(E) = \mathbb{Z}$, $\rho_E(G_{\mathbb{Q}})$ has finite index in $GL_2(\widehat{\mathbb{Z}})$. In particular for $\ell$ large $\rho_{l^{\infty}}$ is surjective; how large one must take $\ell$ depends on $E$.

Conjecture

This last fact does not depend on $E$. I.e. there exists a constant $N$ such that for every $E/\mathbb{Q}$ with $End(E) = \mathbb{Z}$ and $\ell \geq N$, the mod $\ell$ representation $\overline{\rho_{E,\ell}}$ is surjective; equivalenty there is an upper bound (independent of $E$) on the index of $\rho_E(G_{\mathbb{Q}})$.

By a recent paper of Bilu and Parent, one knows that the image is either surjective or contained in a non-split cartan subgroup.


My Question

Why do people expect this to be true? Does it follow from other believable conjectures? Is there some heuristic that predicts this?

One fact is that one can phrase this question as the failure of various modular curves (e.g. $X_{ns}(l)$, which have increasingly large genus) to have non-trivial rational points. Is there a geometric reason that one expects these modular curves to have no non-trivial rational points?

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Is the reason people expect it to be true simply that analogous results for the curves X_0(N) are true, by Mazur's theorem? I don't see why a random bunch of curves should only have the points on them that you can think of but I don't think that this is the source of the conjecture. –  Kevin Buzzard Nov 23 '09 at 9:27

3 Answers 3

I'm not sure there's a reason for these particular curves to have no points, but I think people expect in general that a "random" high-genus curve over Q has no points; so in a situation like this one one might guess that the curves have "only the points they have to have." In order to heuristicize about this I suppose you'd start by guessing what proportion of genus-g curves of height at most H had rational points (but already this is a bit sticky, since in large genus curves over Q are presumably concentrated on some unknown proper closed locus in the general-type M_g) and then ask whether this sum converges for the X_{ns}(ell)!

Edit: Also, of course, this is true when Q is replaced by the function field K of a curve over C a finite field, so long as you require that E is not isotrivial. (This hypothesis is something like the exclusion of CM curves in the number field case.) This follows from the fact that the modular curves X parametrizing various level structures have growing genus; X(K) corresponds to maps C -> X, and all such maps are constant once g(X) is large enough. Indeed, a theorem of Poonen shows that the X even have increasing gonality, which means that a yet stronger statement holds in the function field case: for each d, there is an M(K,d) such that: for all extensions K'/K of degree d, all non-isotrivial elliptic curves E/K', and all p > M(K,d), the Galois representation to Aut(E[p]) is surjective.

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Since you asked "Why do people expect this to be true?" and nobody mentioned it so far, part of the reason is "numerical evidence". Drew Sutherland (of MIT) very recently computed a massive amount of new data about images of Galois for over 100 million elliptic curves. See the notes from this talk or contact Sutherland.

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The conjecture is known for semistable curves. There the only non-surjective $\overline{\rho_p}$ come from isogenies and they have been treated by Mazur's theorem on $X_0(N)$.

When looking at the proof of Serre's result one sees that pairs $(E,p)$ for which the representation is not surjective have to be very special. For instance if the $j$-invariant of $E$ is not integral one can exclude all but finitely many $p$ immediately. My intuition is that it is very hard for $\overline{\rho_p}$ not to be surjective when there is no isogeny of degree $p$. For curves with integral $j$-invariant but without cm it is much harder to say anything general.

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