Timeline for Centralizers of subtori in reductive groups, derived subgroups
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2020 at 2:44 | comment | added | LSpice | I know it's three years later and the question has been answered, but, if you ever look for a reference, this appears in Kac and Weisfeiler - Coadjoint action of a semi-simple algebraic group and the center of the enveloping algebra in characteristic $p$, Proposition 2.1. | |
| May 22, 2017 at 18:19 | comment | added | Mikhail Borovoi | Note that the centers are also described (not as nicely as in the cheat sheet of nfdc23, but equivalently) in Table 3 on page 298 of the book: A. L. Onishchik and E. B. Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag, Berlin, 1990. This book contains many other useful tables. | |
| May 21, 2017 at 11:22 | vote | accept | Tippy Tipper | ||
| May 21, 2017 at 10:34 | comment | added | Jason Starr | Dear all, nfdc23 shared with me the cheat sheet. Here is a LaTeX version: math.stonybrook.edu/~jstarr/papers/Centers.pdf | |
| May 20, 2017 at 14:46 | comment | added | nfdc23 | @JasonStarr: OK, I will do this later today. | |
| May 20, 2017 at 14:42 | comment | added | Jason Starr | @nfdc23. If you are serious, and if you e-mail me the scan, I can probably use TikZ to make a nice LaTeX version of the document. | |
| May 20, 2017 at 14:41 | comment | added | nfdc23 | @JasonStarr: How does one insert a scan of a hand-written document? I don't know how to create decorated Dykin diagrams in LaTeX. | |
| May 20, 2017 at 14:35 | answer | added | nfdc23 | timeline score: 4 | |
| May 20, 2017 at 13:16 | answer | added | Jim Humphreys | timeline score: 1 | |
| May 20, 2017 at 12:08 | comment | added | Jason Starr | @nfdc23. Will you give us the cheat sheet? | |
| May 20, 2017 at 11:58 | answer | added | Paul Levy | timeline score: 6 | |
| May 20, 2017 at 2:08 | comment | added | nfdc23 | @PaulLevy: Oops, I should have looked up my cheat sheet of centers for each Killing-Cartan type stuffed into in my copy of Bourbaki (to see when $a^{\vee}(-1)$ can land in the center for a simple positive $a$) before making my comment. To compensate for that blunder, let me point to a more specific reference: see Corollary 9.5.11 for the simply connected case and Example C.3.4 for the case of type B$_2$ (explicitly exhibiting ${\rm{SO}}_3={\rm{PGL}}_2$) in ams.org/open-math-notes/omn-view-listing?listingId=110663 I'll check the cheat sheet later. | |
| May 19, 2017 at 19:12 | comment | added | Paul Levy | I can't agree with nfdc23 - for example, for any short root in the root system of type $B_n$, the group we obtain this way is ${\rm SO}_3={\rm PGL}_2$. I think your reasoning is fine for simply-connected groups, but $\alpha^\vee(-1)$ can be contained in the kernel of the universal cover - only in type $B$, though, I think. | |
| May 19, 2017 at 17:33 | comment | added | Tippy Tipper | nfdc23 - Thank your comment, which I need to think about more. I have to say, I am surprised by your response. Based on the amount of time spent on reductive groups of semisimple rank one in the standard texts, I would have assume that $PGL_2$ would play more of a role in the theory. Perhaps I didn't ask the correct question, though. Based on your response, maybe my question should change to: as $[G_{\alpha}, G_{\alpha}] \cong SL_2$ (when $G \neq PGL_2$), when does the corresponding map $SL_2 \rightarrow G$ factor through $PGL_2$? Some condition on $\alpha$? | |
| May 19, 2017 at 17:30 | comment | added | Tippy Tipper | Mikhail Borovoi, thank you for your edits, of course what you wrote is what I meant to write. | |
| May 19, 2017 at 17:28 | history | edited | Tippy Tipper | CC BY-SA 3.0 |
added 38 characters in body
|
| May 19, 2017 at 15:53 | history | edited | Mikhail Borovoi | CC BY-SA 3.0 |
corrected the definition of the reductive group $G_\alpha$ of semisimple rank 1.
|
| May 19, 2017 at 14:17 | comment | added | nfdc23 | One loses nothing by assuming $G$ is semisimple. Then that commutator subgroup is always ${\rm{SL}}_2$ except for precisely when $G = {\rm{PGL}}_2$. Indeed, first one checks this when $G$ is simply connected using that the simple positive coroots relative to a basis of the root system are a basis of the cocharacter lattice precisely in the simply connected case (and that any root is a simple positive root relative to some basis of the root system). In general examine the kernel of the simply connected central cover $\widetilde{G}\rightarrow G$ (which identifies the root systems). | |
| May 19, 2017 at 13:51 | history | asked | Tippy Tipper | CC BY-SA 3.0 |