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.

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