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seen | Feb 10 '13 at 9:17 | |
stats | profile views | 371 |
Oct 7 |
awarded | Yearling |
Dec 1 |
comment |
algebraic groups and their Lie algebras
@Jim Humphreys: Thanks for the kind words. The argument I give also works without change if ${\rm{GL}}(V)$ is replaced with any linear algebraic $k$-group. I'm not sure how the argument in D-G may differ from this, as I have always avoided looking at anything in D-G whenever possible. |
Nov 30 |
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algebraic groups and their Lie algebras
@Andriy: Your necessity comment beyond the connected semisimple case above is not quite true: things work well for unipotent groups too (as noted at the end of my answer below). |
Nov 30 |
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algebraic groups and their Lie algebras
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Nov 30 |
answered | algebraic groups and their Lie algebras |
Nov 25 |
awarded | Nice Answer |
Nov 25 |
revised |
Is the normalizer of a reductive subgroup reductive?
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Nov 25 |
revised |
Is the normalizer of a reductive subgroup reductive?
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Nov 24 |
revised |
Is the normalizer of a reductive subgroup reductive?
added 357 characters in body |
Nov 24 |
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Is the normalizer of a reductive subgroup reductive?
@Ben Wieland: I imposed a connectedness condition in my answer because it seemed almost certain that Martin Orr would have wanted that ("everything" breaks down in the theory of linear algebraic groups when one drops connectedness). One should never say "reductive" without connectedness in positive characteristic, since too many things break down in such cases. |
Nov 24 |
revised |
Is the normalizer of a reductive subgroup reductive?
added 133 characters in body |
Nov 24 |
revised |
Is the normalizer of a reductive subgroup reductive?
added 516 characters in body |
Nov 23 |
answered | Is the normalizer of a reductive subgroup reductive? |
Nov 17 |
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Adelic description of moduli of $G$-bundles on a curve
Dear RK: By Steinberg's theorem (Serre's conjecture), any connected smooth linear algebraic group over a perfect field of cohomological dimension $\le 1$ has vanishing degree-1 Galois cohomology. By Tsen, this applies to the function field of your Riemann surface (viewed as an algebraic curve, say), so your principal bundle extends across the missing point by "gluing" it to the trivial bundle over a Zariski-open neighborhood of the puncture. Now comes the overkill reference in English: apply Theorem 3 (also see Remark 2(b)!!) in the short 1995 paper by Drinfeld and Simpson in MRL #2. :) |
Nov 17 |
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Adelic description of moduli of $G$-bundles on a curve
Dear Justin: You have the good fortune to be a graduate student at Harvard, so all you need to do is to talk to people in your department. (I don't know any references; I figured it out for myself by thinking carefully. Beauville-Lazslo is helpful to "algebraize" from the completed data.) Anyway, one should only consider triviality at the generic point (not Zariski-locally), and in practice it is only reasonable to consider $G$-torsors for the etale topology (this is why Serre "invented" it...). So basically, this stuff is "only" good for the simply connected case (or variants like GL$_n$). |
Nov 16 |
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Adelic description of moduli of $G$-bundles on a curve
This answer overlooks crucial hypotheses without which there is no such bijection (only an injection): one has to know that all $G$-bundles are trivial at the generic point and that all $G$-bundles over finite extensions of k are trivial (such as for finite $k$ and connected $G$). Those conditions are where the "almost everywhere integral" aspect of adelic points is logically relevant for surjectivity. So things are good for GL$_n$ but certainly not PGL$_n$, for example. And likewise for simply connected semisimple $G$ and finite $k$ one is in good shape by Harder's theorem, etc |
Nov 13 |
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Rigid Uniformization vs Grothendieck's Local Monodromy Theory
@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 |
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Rigid Uniformization vs Grothendieck's Local Monodromy Theory
@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 |
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Rigid Uniformization vs Grothendieck's Local Monodromy Theory
@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 |
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Rigid Uniformization vs Grothendieck's Local Monodromy Theory
@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). |