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Oct 11, 2010 at 9:07 comment added François Brunault @Kevin Buzzard: It would be definitely worth trying the strategy you outline in the comments to David Hansen's answer. I think that computing some Frobenius on a genus 2 curve might not be too hard, but I don't know how many would be needed...
Oct 11, 2010 at 9:00 comment added François Brunault @Kevin Buzzard: I was assuming End(A)=Z because it is the generic case for an abelian surface (and seemingly the most difficult). Also, it was not clear to me when precisely one should expect a Siegel MF (I am not familiar with lifting techniques). But after all, in the classical GL_2 case, weight 2 newforms also give rise to the CM elliptic curves.
Oct 5, 2010 at 20:58 answer added Olivier timeline score: 5
Oct 5, 2010 at 20:09 comment added Kevin Buzzard @Francois: why would one demand End(A)=Z? If the endomorphisms are bigger then the Galois representation is smaller...aah...OK, I see: are you concerned that in this case the representation might not be cuspidal?
Oct 5, 2010 at 20:07 comment added Kevin Buzzard @Francois: If one can find the candidate Siegel modular form (like Poor/Yuen can do sometimes) then perhaps one can use the strategy I outline in the comments to David Hansen's answer to answer Q1.
Oct 5, 2010 at 15:23 history edited Laie
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Oct 4, 2010 at 15:01 comment added François Brunault One difficulty in the case $\operatorname{End}_{\overline{\mathbb{Q}}}(A)=\mathbb{Z}$ is that there is apparently no analogue of the Eichler-Schimura construction for Siegel eigenforms (see the introduction of the Brumer/Kramer article). So even for a single explicitly given abelian surface, I wouldn't know how to answer questions 1 and 2 (but maybe I'm wrong).
Oct 3, 2010 at 20:52 comment added Laie This is indeed part of the story. Thanks for catching this glitch, it's corrected now.
Oct 3, 2010 at 20:45 history edited Laie CC BY-SA 2.5
added 25 characters in body
Oct 3, 2010 at 10:46 comment added François Brunault Shouldn't there be the hypothesis End(A)=Z in the statement of Yoshida's conjecture?
Oct 2, 2010 at 0:07 comment added Emerton ... then modularity of elliptic curves. (One reason that regularity occurs as a hypothesis is that the motives/Galois representations that one can extract from the cohomology of Shimura varieties are regular.)
Oct 2, 2010 at 0:06 comment added Emerton One reason that proving modularity (in the classical, not $p$-adic sense) is harder for abelian surfaces than for elliptic curves is that the Hodge structure of an abelian surface is not regular. We say that a Hodge structure is regular if all $h^{p,q}$ are $\leq 1$. Most modularity theorems have as a hypothesis that the Hodge structure attached to the motive one is trying to show is modular is regular. These don't apply to an abelian surface, since for an abelian surface $h^{1,0} = h^{0,1} = 2$. So, from the point of view of current methods, this is an intrinsically harder question ...
Oct 1, 2010 at 20:07 history edited Laie
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Oct 1, 2010 at 19:31 comment added Laie For the applications I have in mind p-adic modularity unfortunately doesn't suffice.
Oct 1, 2010 at 17:54 comment added Olivier The answer to question 1 is yes: you want to look at the works of Tilouine, Genestier-Tilouine and most recently V.Pilloni. The article of J.Tilouine which appeared in Compositio Math 142 would seem to be a good starting point for you. Note that typically, these results show p-adic modularity, in the sense that they do not establish that the form is really classical, as you wish. The so-called classicity property is much harder, but might be within reach.
Oct 1, 2010 at 15:48 comment added Qiaochu Yuan +1 for the background!
Oct 1, 2010 at 15:29 answer added David Hansen timeline score: 8
Oct 1, 2010 at 15:00 history asked Laie CC BY-SA 2.5