[Math] Langlands in dimension 2: the Yoshida conjecture

ag.algebraic-geometryarithmetic-geometryautomorphic-formslanglands-conjecturesmotives

Background:
One prominent part of the Langlands program is the conjecture that
all motives are automorphic.
It is of interest to consider special cases that are more precise, if less
sweeping. The idea is to formulate generalized modularity conjectures
that are as concrete as the Shimura-Taniyama-Weil conjecture
(now the elliptic modularity theorem). The latter gained much in
precision, for example, by Weil's experimental observation of the link between
the conductor of the elliptic curve and the level of the weight two
modular form. As a next step up the dimension ladder it is natural to consider
abelian surfaces over ${\mathbb Q}$, in which case one encounters

Yoshida's conjecture:
Any irreducible abelian surface $A$ defined over ${\mathbb Q}$ and with End$(A)={\mathbb Z}$ is modular in the sense that associated to each is a holomorphic Siegel modular
cusp eigenform $F$ of genus 2, weight 2, and some level $N$, such
that its spinor L-function $L_{\rm spin}(F,s)$ agrees with that of the
abelian surface
$$
L(H^1(A),s) ~=~ L_{\rm spin}(F,s).
$$

Questions:

  1. Has Yoshida's conjecture been proven for some classes of
    abelian surfaces over ${\mathbb Q}$?

  2. Are there lists of abelian surfaces and associated Siegel modular forms that extend the very useful lists constructed by Cremona and Stein for elliptic curves and their associated modular forms?

Best Answer

There's no need to require irreducibility. If $A=E_1 \times E_2$ is a product of elliptic curves the conjecture is true, by modularity of elliptic curves combined with Yoshida's lifting from pairs of classical cusp forms to Siegel modular forms of genus two. If $K$ is a real quadratic field and $E/K$ is a modular elliptic curve, then Yoshida's conjecture is true for the surface $A=\mathrm{Res}_{K/\mathbb{Q}}(E)$. This follows from a theorem of Johnson-Leung and Roberts; see arxiv 1006.5105. Presumably any individual $A$ can be done "by hand" using Faltings-Serre plus some serious computational cleverness.

Sorry if you already know all these things. :)

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