Timeline for Are all stabilizer groups of the co-adjoint action smooth?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 20, 2017 at 17:19 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 20, 2017 at 17:06 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 17, 2017 at 7:52 | comment | added | Uri Bader | Another approach which might work in all characteristics is to imitate the argument in the Bernstein-Zelevinsky appendix: change the algebraic structure in a way that does not change the $k$-points but makes things smoother. Unfortunately, not being an expert myself, I cannot advice you further without devoting too much time here. If something will come to mind I'll let you know. | |
| Feb 15, 2017 at 16:48 | comment | added | m07kl | @UriBader: Thanks for comments. I find the first Galois cohomology of a unipotent Group over characteristic 0 is trivial in Chap 2. Lemma 2.7 in Paltonov and Rapinchuk's book. But I can not work through the proof since I am not from this research area. Sorry. If you work out the proof please let me know. Please see Corollary 3.1.3 in projecteuclid.org/download/pdf_1/euclid.pja/1286198322 | |
| Feb 15, 2017 at 13:30 | comment | added | Uri Bader | (cont.) I would work through the proof. My guess is that it works verbatim, assuming the characteristic is high enough. Good luck. | |
| Feb 15, 2017 at 13:30 | comment | added | Uri Bader | m07kl, as remarked above, over algebraically closed field it is known that the orbits are closed for a unipotent group action on an affine variety. If you knew that the first Galois cohomology of a unipotent group is trivial it will also follow that this is the case for the action of the group of points over a local field. The vanishing of this Galois cohomology is known in characteristic 0. I am sure you could find this result (as well as an explanation of the general theory, in case you are not familiar with it) in the monumental book by Paltonov and Rapinchuk. | |
| Feb 14, 2017 at 20:27 | comment | added | m07kl | @user94041: Proposition 2.1 implies that U.x is closed in Zariski topology in X. I am asking for U(k).x is Hausdorff closed in X(k). They are different. An interesting example saying that: G.v is closed doesn't imply that G(k).v is closed is given in Section 7.1 content.algebraicgeometry.nl/2014-5/2014-5-025.pdf | |
| Feb 14, 2017 at 17:43 | comment | added | user94041 | I am not sure what you mean by "Hausdorff topology from $\mathfrak n^*$". Is that the one induced by some isomorphism of vector spaces with $k^N$? Proposition 2.1 in math.stanford.edu/~conrad/249BW16Page/handouts/unipgp.pdf proves that the orbits are closed in the Zariski topology. If the Hausdorff topology that you have in mind is finer than the Zariski topology on $\mathfrak n^*$, this gives an affirmative answer to Q1. | |
| Feb 14, 2017 at 16:06 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 13, 2017 at 13:58 | history | edited | m07kl |
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| Feb 11, 2017 at 3:58 | history | edited | YCor |
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| Feb 10, 2017 at 23:23 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 10, 2017 at 20:26 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 10, 2017 at 20:16 | history | edited | m07kl | CC BY-SA 3.0 |
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| Feb 10, 2017 at 20:07 | history | asked | m07kl | CC BY-SA 3.0 |