Timeline for Do representations of real semisimple algebraic group have to be algebraic?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2015 at 18:12 | vote | accept | Jerry | ||
| Oct 1, 2015 at 1:18 | comment | added | Venkataramana | @grghxy: I did not take offence at all; I simply wanted to write that this was why I did not pass to the connected compotent of identity in the simply connected case. | |
| Sep 30, 2015 at 23:50 | comment | added | grghxy | @Venkataramana: I was aiming my comments to Jerry, rather than to you; sorry if that wasn't clear. (I was aware that you know extremely well everything I have been writing in my comments.) | |
| Sep 30, 2015 at 7:36 | comment | added | Venkataramana | Yes, I am aware that $G(\mathbb R)$ is connected, if $G(\mathbb C)$ is simply connected and semi-simple. | |
| Sep 30, 2015 at 7:31 | comment | added | grghxy | One other point: probably you are tacitly assuming (as is everyone giving comments/answers) that the original semisimple group in question is connected (in the sense of algebraic groups), as without that Lie algebras lose a bit of their control on the situation. Even over $\mathbf{R}$ there are subtleties if the group of $\mathbf{R}$-points is disconnected for the analytic topology, but it is an important (and remarkable) theorem of Cartan that such disconnectedness never happens when the given (connected!) semisimple group is simply connected (in the sense of algebraic groups!). | |
| Sep 30, 2015 at 7:28 | comment | added | Venkataramana | since G is semi-simple, this analytic isomorphism exists only as a covering;(see the example of $SL_3(\mathbb R)$ in my answer). I think there are no other counterexamples, because of the "rigidity" of semisimple group representations | |
| Sep 30, 2015 at 7:25 | comment | added | grghxy | The standard counterexample without the "simply connected" hypothesis in Venkataramana's answer is the inverse of the analytic isomorphism ${\rm{SL}}_n(\mathbf{R}) \rightarrow {\rm{PGL}}_n(\mathbf{R})$ for odd $n > 1$ (i.e., take the semisimple group of interest to be ${\rm{PGL}}_n$). Also, for an alternative treatment as a consequence of general procedures over any field of characteristic 0, see the answer by user27056 at mathoverflow.net/questions/114974/… | |
| Sep 30, 2015 at 5:57 | comment | added | GH from MO | I agree with Venkataramana, this question seems very appropriate. If it gets closed, I will vote to re-open it. | |
| Sep 30, 2015 at 5:36 | comment | added | Venkataramana | I do not understand why this was closed: it is well known and classical (but is not really discussed at length in many books) but definitely non-trivial. | |
| Sep 30, 2015 at 3:17 | review | Close votes | |||
| Sep 30, 2015 at 7:40 | |||||
| Sep 30, 2015 at 3:00 | answer | added | Venkataramana | timeline score: 7 | |
| Sep 30, 2015 at 2:53 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Sep 30, 2015 at 2:13 | history | asked | Jerry | CC BY-SA 3.0 |