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Nov 13, 2012 at 6:30 comment added Will Sawin I'm asking for a sheaf satisfying the set of conditions but not coming from a rigid analytic variety. Of course, this is a bit unfair, as I haven't specified what the right set of conditions is. To get at the spirit of the question, they should be conditions that one can prove, without using rigid analytic methods, are satisfied for all abelian varieties.
Nov 13, 2012 at 6:25 comment added Will Sawin @xbnv: $R^1 \pi_* \mathbb Q_l$, etale category. I'm interested in seeing the difference between the monodromy and rigid-analytic picture. The specific claim I made was that any sheaf satisfying some natural set of conditions, come from a rigid-analytic group in this way. Thus reasoning sheaf-theoretically from those conditions and reasoning with rigid analysis would be equivalent, at least for solving problems one can state in terms of that sheaf.
Nov 13, 2012 at 5:48 comment added user27056 @Will: What is "the sheaf" (which functor, on which category)? There are examples of smooth connected proper rigid-analytic groups $X$ that do not arise from abelian varieties, even such $X$ with "complex multiplication" (in the sense of the endomorphism algebra of $X$ containing a CM field of degree twice the dimension of $X$). Such $X$ with CM have no analogue over the complex numbers, but would you regard them as being "fake abelian varieties"? I'm not sure what you're asking for.
Nov 13, 2012 at 5:00 comment added Will Sawin @xbnv: I'm only talking about things that can be stated in terms of properties of the sheaf. Can you construct a "fake abelian variety" in the form of a sheaf that looks like it comes from one, but actually does not, for rigid analytic reasons?
Nov 13, 2012 at 4:57 comment added user27056 @Davidac897: In situations where both viewpoints prove a common result it is reasonable to ask for a unified perspective, but the final paragraph of your question and the "Specific Question" go beyond this, into a realm that is disproved by the duality aspect with toric parts of the reductions. The rigid uniformization subsumes the monodromy operator. (As an aside, IMHO even for AV's admitting a principal polarization, it is a conceptual error to think about the orthogonality theorem without always keeping in mind the role of the dual, just like for Weil pairings in higher dimensions.)
Nov 13, 2012 at 4:38 comment added David Corwin I'm aware that the proof of the orthogonality theorem requires the dual AV, and I've read through the proof. But I still think that, regardless, the question is an interesting one - in the many cases where the two seem to give the same results, it's still interesting that there are two seemingly different proofs. As an aside, this isn't an issue for Jacobians, which are principally polarized (and which is what I happened to be using when I learned this).
Nov 13, 2012 at 4:12 comment added user27056 @Davidac897: It is not true that "if one fixes a polarization" then the role of the dual abelian variety can be suppressed. Plenty of abelian varieties do not admit a polarization with degree coprime to $\ell$, and in such cases you're not going to get the orthogonality theorem integrally (i.e., without inverting $\ell$) in the way you wish (with the dual hidden behind the polarization). The dual AV is a key ingredient in the proof of the orthogonality theorem. If you acquire more experience with abelian varieties then you will be in better position to understand the ideas in Expose IX.
Nov 13, 2012 at 4:08 history edited David Corwin CC BY-SA 3.0
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Nov 13, 2012 at 4:02 comment added user27056 @Will: What exactly is the meaning of your statement #1? There are statements seen via rigid uniformization for which there isn't any evident means of proof in terms of monodromy formalism; e.g., the canonical duality between character lattices of toric parts in the reductions (as noted in my comment). In particular, the rigid uniformization is so much richer than the monodromy operator (which only records inertial action, and so can be extracted from the rigid uniformization).
Nov 13, 2012 at 4:02 comment added David Corwin @xbnv: This should all be true if one fixes a polarization. See 2.5 of Expose IX.
Nov 13, 2012 at 3:54 comment added user27056 The generalization from elliptic curves to abelian varieties involves the dual abelian variety in an essential way that you appear to have completely missed (e.g., your statement of the orthogonality theorem of SGA7 is wrong). This causes you to overlook serious aspects such as that in case of semistable reduction, the toric parts of the reductions have canonically dual geometric character lattices; this fact lies far deeper than anything about the monodromy operator and is made vivid in rigid uniformization. I recommend more experience with abelian varieties before asking such questions
Nov 13, 2012 at 3:44 comment added Will Sawin 1. One possible answer would be "because we can find an analytic variety with any possible local monodromy group". vice versa is obviously trivial. Whether this is true depends on how you formulate the statement, and I'm pretty sure you can formulate it in a way such that it's true. 2. You mean "potentially good reduction", not "good reduction".
Nov 13, 2012 at 3:31 history edited David Corwin CC BY-SA 3.0
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Nov 13, 2012 at 3:21 history edited David Corwin CC BY-SA 3.0
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Nov 13, 2012 at 2:42 history asked David Corwin CC BY-SA 3.0