## Equidistribution of Heegner Points

I saw a nice talk by Matt Young recently (joint work with Sheng-Chi Liu and Riad Masri) on the following problem.

For a fundamental discriminant $|D|$ of an imaginary quadratic field $F$, one has $h_D$ points in $X_0(1)(\mathbf{C})$ with complex multiplication by the ring of integers of $F$. Choose a prime $q$ which splits in $F = \mathbf{Q}(\sqrt{-|D|})$. One obtains a set of $2 h_D$ points in $X_0(q)(\mathbf{C})$, given explicitly as follows:

$\mathbf{C}/\mathfrak{a} \mapsto \mathbf{C}/\mathfrak{a} \mathfrak{q}^{-1}$

for $\mathfrak{a}$ in the class group and $\mathfrak{q}$ one of the two primes above $q$ in $F$. The complex points $X_0(q)(\mathbf{C})$ can be thought of as being tiled by $q+1$ copies of the fundamental domain $\Omega$ in the upper half plane.

Problem: How large does $D$ have to be to guarantee that every one of the $q+1$ copies of $\Omega$ contains one of the $2 h_K$ CM points by $\mathcal{O}_F$?

This is the question that Young and his collaborators answer. Namely, one gets an upper bound of the shape $|D| < O(q^{m + \epsilon})$ (with some explicit $m$, possibly 20), the point being that this is a polynomial bound. Note that this proof is not effective, since it trivially gives a lower bound on the order of the class group which is a power bound in the discriminant, and no such effective bounds are known.

I idly wondered during the talk about the following "mod-$p$" version of this problem. To be concrete, suppose that $p = 2$ (the general case will be similar). We now suppose that $D$ is chosen so that $2$ is inert in $F$. Then all the $h_K$ points in $X_0(1)(\overline{\mathbf{F}}_2)$ are supersingular, which means that they all reduce to the same curve $E_0$ with $j$-invariant $1728$. Now, as above, choose a prime $q$ which splits in $F$. The pre-image of $j=1728$ in $X_0(q)(\overline{\mathbf{F}}_2)$ consists of exactly $q+1$ points.

Problem: How large does $|D|$ have to be to ensure that these points all come from the reduction of one of the $2 h_K$ CM points by $\mathcal{O}_F$ as above?

Since $E_0$ is supersingular, we know that $\mathrm{Hom}(E_0,E_0)$ is an order in the quaternion algebra ramified at $2$ and $\infty$. In fact, it is equal to the integral Hamilton quaternions $\mathbf{H}$. If $E$ and $E'$ are lifts of $E_0$, then there is naturally a degree preserving injection:

$\mathrm{Hom}(E,E') \rightarrow \mathrm{Hom}(E_0,E_0) = \mathbf{H}.$

The degree on the LHS is the degree of an isogeny, and it is the canonical norm on the RHS.
In particular, if $E = \mathbf{C}/\mathfrak{a}$ and $E' = \mathbf{C}/\mathfrak{a} \mathfrak{q}^{-1}$, then one obtains a natural map:

$\psi_{\mathfrak{a}}: \mathfrak{q}^{-1} \simeq \mathrm{Hom}(E,E') \rightarrow \mathbf{H}$

preserving norms. The norm map on $\mathfrak{q}^{-1}$ is $N(x)/N(\mathfrak{q}^{-1})$. The image of the natural $q$ isogeny is simply $\psi_{\mathfrak{a}}(1)$, whose image has norm $q$. Hence the problem becomes:

Problem: If one considers all the $2 h_K$-maps:

$\psi_{\mathfrak{a}}: \mathfrak{q}^{-1} \rightarrow \mathbf{H}, \qquad \psi_{\mathfrak{a}}: \overline{\mathfrak{q}}^{-1} \rightarrow \mathbf{H},$

do the images of $1$ cover the $q+1$ elements of $\mathbf{H}$ of norm $q$?

Given a field $F$ in which $2$ is inert, it wasn’t obvious how to explicitly write down the maps $\psi_{\mathfrak{a}}$, but this problem does start to look similar in flavour to the original one. Moreover, to make things even more similar, in the original formulation over $\mathbf{R}$ one can replace modular curves by definite quaternion algebras ramified at (say) $2$ and $q$, and then the Archimidean problem now also becomes a question of a class group surjecting onto a finite set of supersingular points. In fact, this Archimedean analogue may well be *equivalent* to the $\mod 2$ version I just described! Young told me that his collaborators had mentioned working with various quotients coming from quaternion algebras as considered by Gross, which I took to mean the finite quotients coming from definite quaternion algebras as above. Hence, with any luck, they will provide an answer this problem.