Very strong multiplicity one for Hecke eigenforms

Question

In Invent. math. 116, 645-649 (1994) Dinakar Ramakrishnan proves a theorem which I understand to imply that the following statement (in light of the fact that elliptic curves over $\mathbb{Q}$ are modular):

Theorem: Let $E_1 / \mathbb{Q}$ and $E_2 / \mathbb{Q}$ be elliptic curves and suppose that the (mod $p$) reductions of $E_1$ and $E_2$ have the same number of points for a set of primes of Dirichlet density $> 7/8$. Then $E_1$ and $E_2$ are isogeneous.

Is $7/8$ expected to be sharp here? Could it be that $7/8$ can be replaced by any $a > 0$ with the conclusion in the above Theorem unaltered?

I'm interested in answers to these questions both in the level of generality in which Ramakrishnan works and in the special case that I give above.

Answer 1

Let $E$ and $F$ be two elliptic curves. For convenience, suppose that they do not have CM. For each $p$, there is a Galois representation

\[\rho: = \rho_E + \rho_F : G_{\mathbb{Q}} \to \mathrm{GL}_2(\mathbb{F}_p) \times \mathrm{GL}_2(\mathbb{F}_p)\]

given by the action of Galois on the $p$-torsion of $E$ and $F$. For sufficiently large $p$, the Galois groups $\mathbb{Q}(E[p])$ and $\mathbb{Q}(F[p])$ are $\mathrm{SL}_2(\mathbb{F}_p)$-extensions of $\mathbb{Q}(\zeta_p)$. If $p$ is big enough, $\mathrm{PSL}_2(\mathbb{F}_p)$ is simple. Thus, either:

1). These extensions are disjoint, and $\mathrm{im}(\rho) = \ker(\det(x)/\det(y))$.

2). These extensions intersect in a have a common $\mathrm{PSL}_2(\mathbb{F}_p)$ extension of $\mathbb{Q}(\zeta_p)$, and hence a $\mathrm{PGL}_2(\mathbb{F}_p)$ extension of $\mathbb{Q}$. This implies that their projective representations are the same. This implies that they are equal up to twisting by a character. Since their determinants are also the same, this character is either trivial or quadratic.

If $E[p] = F[p]$ infinitely often then $E$ is isogenous to $F$, using Falting's proof of the Tate conjecture.

If Case 1) occurs, one can count that the density of pairs of elements $(\sigma,\tau)$ in $\mathrm{im}(\rho)$ such that $\mathrm{Tr}(\sigma) = \mathrm{Tr}(\rho)$ is $\sim 1/p$. By Cebotarev, this implies that the number of primes $\ell$ such that $a(E,\ell) = a(F,\ell) \pmod{p}$ has density at most $\sim 1/p$. Hence, if 1) occurs infinitely many times, then $a(E,\ell) = a(F,\ell)$ for set of density zero.

If $E[p]$ is a non-trivial quadratic twist of $F[p]$ for infinitely many primes, looking at the ramification of $E$ and $F$ it is easy to see that $E[p] = F[p] \otimes \chi$ for some fixed quadratic character $\chi$ for infinitely many $p$. Yet then using Faltings again, $E$ is isogenous to the twist of $F$ by $\chi$.

PS: This is a pretty standard argument.

PPS: The argument requires the existence of Galois representations (EDIT: with big image!) attached to $E$ (and $F$), which do not exist in general (EDIT: for example, weight one forms, see Toby's answer). The argument basically works for any pair of classical modular forms $e,f$ of weight at least $2$.

Answer 2

$7/8$ is best possible. For $\mathrm{GL}_n$, you can't improve on $1-1/(2n^2)$, although I'm not sure if the result is known for $\mathrm{GL}_n$. Ramakrishnan discusses this further on page 442 of his article in the Motives volumes; the motivation is that the corresponding result is easy to prove for irreducible finite-dimensional representations of finite groups $G$, viz that if two such representations of dimension $n$ have identical traces outside a set of size less than $1/(2n^2)|G|$, they are isomorphic (the proof is to check that the inner product of their characters is necessarily nonzero). There are explicit examples to show that this is best possible. One can use then use automorphic induction to produce examples showing that the same bound is best possible for strong multiplicity one.

Answer 3

It is not expected that any $a>0$ would work. If $E$, $E'$ are quadratic twists of one another they have the same number of points at a set of primes of D. density $1/2$.

First sentence added by David Speyer, to explain (hopefully correctly) which part of the question is being addressed.

Answer 4

I believe the following statement is a folklore conjecture: Let pi₁ and pi₂ be a pair of cuspidal automorphic representations of GL2 whose local constituents agree at a set of places of positive Dirichlet density. Then either pi₁ is twist-equivalent to pi₂, or both pi's correspond to Artin representations.

It is now possible to prove this over Q in the case when one of the pi's is known to be algebraic (i.e. it comes from a holomorphic cusp form), but one must appeal to solved cases of the Artin conjecture which to me anyway does not seem a very satisfying argument.

Answer 5

Sorry, I meant to say that furthermore one of the pi's should be genuine in Shahidi's sense, or in otherwords not twist-equivalent to itself. Automorphic inductions of grossencharakters, as in Kevin's example, certainly violate this.