Wednesday, November 19, 2008

Gross Zagier, Hilbert Class Polynomials, and Rank 2 Curves with 3 Sha

Today I gave a talk on the Gross-Zagier formula in the this seminar. I stated the main formula over the Hilbert class field, then explained exactly how to use it to deduce a formula over the quadratic imaginary field K. The proof involves using a summation trick and projecting onto eigencomponents of the Hecke algebra. I also talked about the relationship between L(f,chi,s), L_A(f,s), and the ranks of the eigencomponents of J(H)_C.

Right after my talk, several of us went to Microsoft Research, where we had lunch, then Drew Sutherland gave a superb talk on his very impressive multi-modular algorithm for computing Hilbert Class Polynomials modulo one cryptographic sized prime P. The main advantages of his multi-modular method are that it is very easy to parallelize, is extremely memory efficient compared to all other known approaches, and in several situations in the algorithm he can lower the constant in the time complexity by a big factor by using explicit models for X_1(N) (and some other tricks). Very nice.

When I got home, Mark Watkins had sent me a table of 630 elliptic curves of prime conductor and rank 2 that have Shafarevich-Tate groups of order divisible by 9 (see this Sage worksheet). The smallest-conductor example E we know of is y^2 + x*y = x^3 - x^2 + 94*x + 9, which has conductor "only" 53295337. The field K=QQ(sqrt(-7)) satisfies the Heegner Hypothesis, and E^K has rank 1. Also, an hour computation shows that Sha(E^K)_an = 1. I double checked that indeed Sha(E/QQ)_an = 9. Of course the torsion subgroup and Tamagawa numbers are trivial (this follows automatically from the Neumann-Setzer classification). So one expects that the Heegner-Kolyvagin subgroup [which I've never "publicly defined"] of E(K)=Z x Z x Z is a subgroup of index 3 = sqrt(#Sha). I wonder if it is possible to somehow compute -- at least conjecturally(!) -- which subgroup it is? Possibly, yes, by finding the subgroup of E(QQ)/3 E(QQ) that maps to a subgroup of Sel^(3)(E/QQ) that is orthogonal to the inverse image of Sha(E/Q)[3]. I wonder if complexity-wise such a computation is impossibly hard or reasonable?

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