The boundaries of Sato-Tate, part I

Persiflage 2018-04-09

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.