The boundaries of Sato-Tate, part I

A caveat: the following questions are so obvious that they have surely been asked elsewhere, and possibly given much more convincing answers. References welcome!

The Sato-Tate conjecture implies that the normalized trace of Frobenius b_p \in [-2,2] for a non-CM elliptic curve is equidistributed with respect to the pushforward of the Haar measure of SU(2) under the trace map. This gives a perfectly good account of the behavior of the unnormalized a_p \in [-2 \sqrt{p},2 \sqrt{p}] over regions which have positive measure, namely, intervals of the form [r \sqrt{p},s \sqrt{p}] for distinct multiples of \sqrt{p}.

If one tries to make global conjectures on a finer scale, however, one quickly runs into difficult conjectures of Lang-Trotter type. For example, given a non-CM elliptic curve E over \mathbf{Q}, if you want to count the number of primes p < X such that a_p = 1 (say), an extremely generous interpretation of Sato-Tate would suggest that probability that a_p = 1 would be

\displaystyle{\frac{1}{4 \pi \sqrt{p}}},

and hence the number of such primes < X should be something like:

\displaystyle{\frac{X^{1/2}}{2 \pi \log(X)}},

except one also has to account for the fact that there are congruence obstructions/issues, so one should multiply this factor by a (possibly zero) constant depending one adelic image of the Galois representation. So maybe this does give something like Lang-Trotter.

But what happens at the other extreme end of the scale? Around the boundaries of the interval [-2,2], the Sato-Tate measure converges to zero with exponent one half. There is a trivial bound a_p  \le   t where t^2 is the largest square less than 4p. How often does one have an equality a^2_p = t^2? Again, being very rough and ready, the generous conjecture would suggest that this happens with probability very roughly equal to

\displaystyle{\frac{1}{6  \pi p^{3/4}}},

and hence the number of such primes < X should be something like:

\displaystyle{\frac{2 X^{1/4}}{3  \pi \log(X)}}.

Is it at all reasonable to expect X^{1/4 \pm \epsilon} primes of this form? If one takes the elliptic curve X_0(11), one finds a^2_p to be as big as possible for the following primes:

a_{2} = -2 \ge  -2 \sqrt{2} = -2.828\ldots,

a_{239} = -30 > -2 \sqrt{239} =  -30.919\ldots,

a_{6127019} = 4950   \le  2 \sqrt{p} = 4950.563\ldots,

but no more from the first 500,000 primes. That's not completely out of line for the formula above!

Possibly a more sensible thing to do is to simply ignore the Sato-Tate measure completely, and model E/\mathbf{F}_p by simply choosing a randomly chosen elliptic curve over \mathbf{F}_p. Now one can ask in this setting for the probability that a_p is as large as possible. Very roughly, the number of elliptic curves modulo p up to isomorphism is of order p, and the number with a_p = t is going to be approximately the class number of \mathbf{Q}(\sqrt{-D}) where -D = t^2 - 4p; perhaps it is even exactly equal to the class number H(t^2 - 4p) for some appropriate definition of the class number. Now the behaviour of this quantity is going to depend on how close 4p is to a square. If 4p is very slightly — say O(1) — more than a square, then H(t^2 - 4p) is pretty much a constant, and the expected probability going to be around 1 in p. On the other hand, for a generic value of p, the smallest value of t^2 - 4p will have order p^{1/2}, and then the class group will have approximate size p^{1/4 \pm \epsilon}, and so one (more or less) ends up with a heuristic fairly close to the prediction above (at least in the sense of the main term being around X^{1/4 \pm \epsilon}).

But why stop there? Let's push things even closer to the boundary. How small can a^2_p - 4p get relative to p? For example, let us restrict to the set S(\eta) of prime numbers p such that

\displaystyle{S(\eta):= \left\{p \ \left| \ p \in (n^2,n^2 + n^{2 \eta}) \ \text{for some} \ n \in \mathbf{Z} \right.\right\}}.

For such primes, the relative probability that a_p = \lfloor \sqrt{4p} \rfloor = 2n is approximately n^{\eta}/p \sim n^{2 \eta - 1}. So the expected number of primes with this property will be infinite providing that

\displaystyle{\sum \frac{n^{3 \eta}}{n^2 \log(n)}}

is infinite, or, in other words, when \eta \ge 1/3. So this leads to the following guess (don't call it a conjecture!):

Guess: Let E/\mathbf{Q} be an elliptic curve without CM. Is

\displaystyle{\liminf \frac{\log(a^2_p - 4p)}{\log(p)} = \frac{1}{3}?}

Of course, one can go crazy with even more outrageous guesses, but let me stop here before saying anything more stupid.

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3 Responses to The boundaries of Sato-Tate, part I

  1. This is a long long way from the Guess, but note that (at least) one can prove, for a non CM-curve E, that \liminf |a^2_p -4p| = \infty. If a^2_p - 4p = - D, then the eigenvalue of Frobenius generates the order \mathbf{Z}[\sqrt{-D}]. But, for fixed D, there are only finitely many j-invariants j_D in \overline{\mathbf{Q}} with CM by this order, and there cannot be a congruence j_E \equiv j_D \mod p for infinitely many primes unless there is an equality, which would contradict the assumption that E has CM.

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