TITLE: Elliptic Curves with equal trace of Frobenius Values
QUESTION [8 upvotes]: 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?

REPLY [11 votes]: 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 $|tr(V)| \leq n$, $|tr(W)| \leq N$:
$0 = \int_G tr(V) \overline{tr(W)} = \int_X tr(V) \overline{tr(W)} + \int_Y tr(V) \overline{tr(W)} = \int_X tr(V) \overline{tr(V)}+ \int_Y tr(V) \overline{tr(W)} = 1- \int_Y tr(V) \overline{tr(V)} + \int_Y tr(V) \overline {tr(W)} \geq 1 - 2 p n^2$