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.