Timeline for Jordan decomposition in a classical group
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 28, 2010 at 5:25 | vote | accept | D. Savitt | ||
| Jul 7, 2010 at 11:43 | answer | added | JS Milne | timeline score: 2 | |
| Jul 1, 2010 at 16:24 | comment | added | D. Savitt | Victor: no, not overbearing at all, you were right that there was something more to be said to make the argument in the question work. (I realized this as well, just after I got on the plane, and it was annoying not to be able to fix it for 12 hours after that!) Anyhow, see my comments to Ryan's answer. If someone were bold enough to venture that the answer to my question is "no", I would probably accept it. | |
| Jul 1, 2010 at 8:45 | comment | added | Victor Protsak | David: I hope that at least you had a pleasant flight! I am sorry if my comment came across as overbearing: all I was trying to say was that it's the same issue in the Lie algebra case and the Lie group case. The main distinction is that the Lie algebra is cut out by $\textit{linear}$ conditions, but the technique of the proof (as in Ryan's answer) isn't too different. I also recommend looking at Procesi's book, 7.1.5 (and 7.1.4 may be relevant for your earlier Q). As an argument "by authority", since Procesi makes everything explicit for classical groups, maybe it's just not possible here. | |
| Jul 1, 2010 at 6:07 | history | edited | D. Savitt | CC BY-SA 2.5 |
added 81 characters in body
|
| Jul 1, 2010 at 6:04 | comment | added | D. Savitt | @Victor: naturally I'd omit a key detail right before getting on a 10-hour flight, so that it's been left hanging all day.... What you say is true, of course, but I don't think it's a serious problem. There's so much flexibility in the construction of the Jordan decomposition that I think you can always rig things so that the relevant polynomials are odd. I'll edit the question. | |
| Jul 1, 2010 at 5:37 | comment | added | Theo Johnson-Freyd | For comparison, in Mark Haiman's first-semester course on Lie theory (edited notes are on my website), we only proved the Lie algebra version. In Vera Serganova's second semester (unedited notes are on my website, and edited notes will be up by end of summer, I hope) we assert the group version without proof. | |
| Jun 30, 2010 at 23:36 | comment | added | Victor Protsak | There is something wrong with the "visibly true" condition: for the orthogonal Lie algebra, the condition is $JX=-X^t J,$ so $X$ is skew-symmetric (as opposed to symmetric) w.r.t. $J$ and $p(X)$ need not have this property (more precisely, it holds for $p$ odd, but not in general). So it's the same problem in the Lie algebra case as in the Lie group case. | |
| Jun 30, 2010 at 14:12 | comment | added | Boyarsky | @D.Savitt: Up to scaling conditions, all of the classical groups are characterized by preserving a line in a space of bilinear forms. So try to use Borel's argument directly in all of those cases, modulo the scaling aspects, and then take care of scalings via triviality of "algebraic characters" on unipotent points (which you can prove directly, since in characteristic zero) to take care of the trivial determinant condition when necessary. As for the trick for $g_u$, it seems that it could be unpleasant to prove by hand that the trick really works (e.g., $g_u$ commutes with $g$). | |
| Jun 30, 2010 at 14:08 | answer | added | Ryan Reich | timeline score: 2 | |
| Jun 30, 2010 at 13:34 | comment | added | D. Savitt | BTW, I'm not so pessimistic regarding the disconnectedness. For instance I think you're also fine for any element g such that g^N is in the image of exp for some N, since I believe g_u will be exp(1/N log (g^N)_u). | |
| Jun 30, 2010 at 13:29 | comment | added | D. Savitt | @Boyarsky ("Why do you prefer...?"): for a graduate course in Lie groups/Lie algebras where arguments invoking algebraic geometry would not be appropriate. [Though in fact, I think there's nothing wrong with seeing a "lowbrow" argument for some concrete groups before one does something more general -- you might understand the general case better if you can see how the argument is not so different from something you've already done by hand in specific cases -- as long as what you've does by hand is reasonably thoughtful.] | |
| Jun 30, 2010 at 13:03 | comment | added | Boyarsky | By the way, the case of finite $G$ is only interesting in positive characteristic, of course (as otherwise everything is semisimple). | |
| Jun 30, 2010 at 13:03 | comment | added | Kevin McGerty | Uniqueness of Jordan decompositions for endomorphisms means it is functorial. Can you use that to turn your conditions for classical groups into linear ones, e.g. looking at $\mathfrak g$ inside $\mathfrak{gl}(End(V))$ turns the condition $g^t = g^{-1}$ into $g(I) = I$, which then polynomials in $g$ will also satisfy. I guess this would really only be a spelling out of the general argument, but it would be explicit, which might be what you want? | |
| Jun 30, 2010 at 13:01 | comment | added | Boyarsky | Why do you prefer a case-by-case proof, specific moreover to the "classical" cases, instead of the straightforward proof in the general case (no semisimplicity hypotheses on the Lie algebra, etc.) as in Borel's book, which works over any algebraically closed field (and then any perfect field by Galois theory)? Recall that Borel's device is a description of $G$ as the stabilizer of a line in some representation of ${\rm{GL}}(V)$, which replaces the "description" with traces and bilinear forms? Trying to get it from the Lie algebra also makes no sense for disconnected $G$; consider finite $G$. | |
| Jun 30, 2010 at 12:36 | history | asked | D. Savitt | CC BY-SA 2.5 |