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Suppose we have two elliptic curves over $\mathbb{Q}$ with trivial rational torsion. Is there some density $\delta$ such that if the trace of Frobenius values of the two elliptic curves are equal on a set of primes of density $\delta$ then the two curves are equal?

Does the situation change if we know the conductors of the curves are different?

Is the question easier if we have two given curves as opposed to two arbitrary curves?

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If $E$ is a CM curve and $E'$ is a quadratic twist of $E$ that is not isogenous to $E$ then $E$ and $E'$ have the same trace 75% of the time. That might be the largest possible. Torsion doesn't matter (e.g. you can take the CM discriminant to be $-11$), and the conductors need not be the same (indeed usually they aren't). –  Noam D. Elkies Jun 6 at 3:37

2 Answers 2

Let $E$ and $E^{\prime}$ be the two non-isogenous elliptic curves. By Rankin-Selberg theory one can show that for large $x$ we have $$ \sum_{p\le x} \frac{(a_E(p)-a_{E^{\prime}}(p))^2}{p} \frac{\log p}{p} \sim 2 \log x. $$ If $a_E(p)$ and $a_E^{\prime}(p)$ match on a set of density $\delta$, then using the Hasse bound the LHS is at most $$ 16 (1-\delta) \log x. $$ This proves that $\delta \le 7/8$. So if the Frobenii match on a set of density $7/8$ the two curves are the same (up to isogeny). A general version of this for $GL(2)$ was established by Dinakar Ramakrishnan. See also C.S. Rajan's papers 2, 3 on this topic, which give more general results.

Added: In fact Rajan shows in 3 (see Corollary 2 there) the following: Suppose $E_1$ is a non-CM elliptic curve and that $E_2$ is an elliptic curve with $a_{E_1}(p)=a_{E_2}(p)$ for a set of primes of positive density. Then $E_2$ is isogenous to a quadratic twist of $E_1$. He also notes that this result follows from work of Serre (see the references in Rajan's paper). In other words, in the $\ell$-adic situation one has much more precise results than for general automorphic forms.

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When you say the curves are "the same", you mean that they are isogenous. –  Jeremy Rouse Jun 6 at 2:57
    
@JeremyRouse: Thanks! Will edit to reflect that. –  Lucia Jun 6 at 2:57

Clearly Noam Elkies' answer is optimal for CM curves. For non-CM curves that are a quadratic twist of one another, the probability is $1/2$, because the probability that $a_p=0$ is $0$ by Chebotarev's density theorem.

For a non-CM and CM curve, or two non-CM curves that are not a twist of each other, let $G$ be the image of the Galois group in $GL_2(\mathbb Q_\ell) \times GL_2(\mathbb Q_\ell)$ acting on the Tate modules in both curve. The subset of $G$ consisting of elements that have the same trace in both representations is closed in the Zariski topology of positive codimension, hence has $\ell$-adic measure $0$, by Chebotarev's density theorem a zero-density set of primes live there.

Hence Noam Elkies' example is optimal, and we can take any $\delta> 3/4$ to imply equality up to isogeny.

In general, the Sato-Tate conjecture is true for sets which are closed in the Zariski topology, because those sets are also closed, and hence measurable, in the $\ell$-adic topology, and we know equidistribution for the $\ell$-adic measure by Sato-Tate.

The $7/8$ bound is attained if instead of representations of elliptic curves we allow two-dimensional Artin representations. Take an extension of Galois group $D_4 \times \mathbb Z/2$, and compare the two-dimensional irreducible representation of $D_4$ to its quadratic twist by $\mathbb Z/2$.

Edit: There is a purely algebraic proof of the $7/8$ upper bound. In general:

Let $V$ and $W$ be two distinct $n$-dimensional irreducible representations of a compact Lie group $G$. The probability that $tr(V)=tr(W)$ is at most $1-1/2n^2$.

Proof: Let $X$ be the locus where $tr(V)=tr(W)$ and let $Y$ be the locus where it is not. Let $p$ be the measure of $Y$. Then by Schur's lemma, the definition of $X$, Schur's lemma again, and the trivial estimate $|tr(V)| \leq n$, $|tr(W)| \leq N$:

$0 = \int_G tr(V) \overline{tr(W)} = \int_X tr(V) \overline{tr(W)} + \int_Y tr(V) \overline{tr(W)} = \int_X tr(V) \overline{tr(V)}+ \int_Y tr(V) \overline{tr(W)} = 1- \int_Y tr(V) \overline{tr(V)} + \int_Y tr(V) \overline {tr(W)} \geq 1 - 2 p n^2$

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