Timeline for Is every connected reductive group over a local field already defined over a global field?
Current License: CC BY-SA 3.0
20 events
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| Mar 7, 2015 at 17:56 | comment | added | user74230 | @TimoRicharz: I don't know any reference, since I'd never heard of or thought about this result in such generality before seeing the question posed here. I'll email you at Bonn now. | |
| Mar 6, 2015 at 17:11 | comment | added | Timo Richarz | @user74230 Is there any reference for your result I can use in one of my preprints? If there is not, I kindly ask your permission to state your proof. In order to do so, it would be nice to have your full name. You can also contact me via email. Thanks for the argument and kind regards! | |
| Mar 5, 2015 at 16:31 | vote | accept | Timo Richarz | ||
| Mar 5, 2015 at 7:08 | comment | added | Dave Witte Morris | Correction: Instead of $\mathbb{Q}_p$, I should have said $K$, the local field (of characteristic 0) over which we know the kernel is defined. (Being defined over $K$ is the same as being invariant under $\mathrm{Gal}(\overline{K}/K)$, which is why I don't think Galois cohomology is needed.) | |
| Mar 5, 2015 at 5:43 | comment | added | Dave Witte Morris | Thanks, @YCor -- I didn't know whether $\mathbb{Q}$ would always work. For semisimple groups in characteristic 0, Borel-Harder says the universal cover of $\mathbf{G}$ can be defined over $\mathbb{Q}$. I think it is easy to see by Galois theory that the kernel of the isogeny to $\mathbf{G}$ is defined over some finite extension of $\mathbb{Q}$ that is contained in $\mathbb{Q}_p$ (since we are in characteristic 0), so I don't think any Galois cohomology is needed besides Borel-Harder. (PS The same argument can also glue a semisimple group to a torus, so the two cases can be treated separately.) | |
| Mar 4, 2015 at 23:23 | comment | added | Question Mark | @WillSawin: Thanks for the explanation. Now your approach makes more sense to me. | |
| Mar 4, 2015 at 23:11 | comment | added | Will Sawin | @user74230 I am saying precisely that your method avoids this space, allowing you to use the implicit function theorem, whereas my method doesn't work. I know better now than to argue with you. I never claimed that my method works - if it was a complete proof, I would post it as an answer - merely that it doesn't fail for the reason Question Mark says. | |
| Mar 4, 2015 at 23:07 | comment | added | user74230 | @WillSawin: I don't see why the "variety of all affine algebraic groups" under some bounded degree should have the requisite smoothness properties to apply the implicit function theorem (and the latter over what?). | |
| Mar 4, 2015 at 23:02 | comment | added | Will Sawin | @QuestionMark Just take the variety of all affine algebraic groups embedded in some affine space by equations of bounded degree. This is defined over $\mathbb Z$, and has a local point. The problem is that I don't know an appropriate continuity statement. As user74230 points out, if you avoid this enormous space of groups and just focus on the space of groups in $GL_n$ that are conjugate, over an algebraically closed field, to $H$, you get continuity using the implicit function theorem. | |
| Mar 4, 2015 at 21:12 | comment | added | YCor | @DaveWitteMorris: the result you mention proves a much stronger conclusion (in particular that adjoint/simply connected semisimple groups are $\mathbf{Q}$-definable). But this stronger conclusion is not true for arbitrary semisimple connected groups over $\mathbf{Q}_p$. For the question I would tend to believe that the material in Borel-Serre's 1964 paper "Theoremes de finitude en Cohomologie Galoisienne" answer the question (at least in char. 0). | |
| Mar 4, 2015 at 21:06 | comment | added | Question Mark | @WillSawin: As far as I can tell, either your variety parametrizing groups is defined over a local field and you have no global model, so it makes no sense to talk of global points, or, once you have deformed the coefficients that give "the associativity and other relations," you no longer have an initial local point to approximate by global points. My feeling is the same as that of user74230: this won't work (even to get a global model of $G$ as a group). | |
| Mar 4, 2015 at 20:36 | answer | added | user74230 | timeline score: 15 | |
| Mar 4, 2015 at 19:57 | comment | added | user74230 | Yes, by inner twisting one can reduce to the problem of algebraizing torsors for any smooth affine group over any henselian valued field whatsoever. This requires no input from number theory or serious algebraic geometry at all. I'll post a complete proof later, when I find time. I do not believe that the method suggested by Will can be turned into a proof; it seems too soft. | |
| Mar 4, 2015 at 19:05 | comment | added | Will Sawin | @QuestionMark - Deform the coefficients of the polynomials giving the group structure as well, and remember all the relations among the coefficients coming from the associativity and other relations. These variables and relations define an algebraic variety - take a point of that variety defined over a global field which approximates your local field point. This is easy because points defined over the algebraic closure of the global field are dense, both in the Zariski sense and over the local field. That's not the hard part. I don't know how to do the rest, so I can't post a complete answer. | |
| Mar 4, 2015 at 18:45 | comment | added | Dave Witte Morris | The easiest case, where $G$ is semisimple (and either adjoint or simply connected) and $K$ has characteristic 0, was proved by Borel and Harder [J. Reine Angew. Math. 298 (1978) 53-64], by showing the natural map of Galois cohomology sets $H^1(F, \mathrm{Aut}G) \to H^1(K,\mathrm{Aut}G)$ is surjective if $G$ is absolutely almost simple. (In fact, they allow any finite set of places of $F$ on the right-hand side, instead of only one completion.) Section 3 of the Borel-Harder paper seems to prove a similar result for groups of type $A$ in positive characteristic, and perhaps can be generalized. | |
| Mar 4, 2015 at 16:46 | comment | added | Question Mark | @WillSawin: I think you will have a hard time proving that your approximation is a group to begin with. Would you care to provide a complete argument? | |
| Mar 4, 2015 at 16:21 | answer | added | Jim Humphreys | timeline score: 2 | |
| Mar 4, 2015 at 15:18 | comment | added | Will Sawin | The connected reductive group is defined by some equations with coefficients in your local field. Take a global field dense in your local field and approximate them arbitrarily closely by elements of your global field. I think there will be a Krasner's Lemma phenomenon where if you get close enough the reductive group will be unchanged. | |
| Mar 4, 2015 at 14:23 | review | First posts | |||
| Mar 4, 2015 at 14:24 | |||||
| Mar 4, 2015 at 14:20 | history | asked | Timo Richarz | CC BY-SA 3.0 |