Timeline for Elementary question about Langlands decomposition
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2017 at 16:47 | vote | accept | Sasha | ||
| Feb 15, 2017 at 23:54 | comment | added | Jim Humphreys | Maybe references to the Borel-Tits theory of abstract homomorphisms don't help in your situation, but here is a more focused account by Tits: ams.org/mathscinet-getitem?mr=0379749 (aimed at Lie groups, with some examples but without all details of proof). | |
| Feb 15, 2017 at 21:29 | answer | added | nfdc23 | timeline score: 9 | |
| Feb 15, 2017 at 20:38 | comment | added | Sasha | Thank you, I will think about your comments. I guess that my question is because I want to work with algebraic groups, and understand things like Langlands classification. Then I want to be assured that if I start with an algebraic real reductive group, then the "$M$" from Langlands decomposition $P=MAN$ of a parabolic is also algebraic, so that I can use "induction". Otherwise, if I still insist to work with algebraic groups only, I will have to work with $MA$ and say things like "$L^2$ modulo center" etc. | |
| Feb 15, 2017 at 20:26 | comment | added | nfdc23 | Unfortunately still "no". The problem is that $G(\mathbf{R})$ can be disconnected in complicated ways. Its component group is an elementary abelian 2-group that can be quite large. When it is nontrivial then $G(\mathbf{R}) \rightarrow G(\mathbf{R})/G(\mathbf{R})^0 = (\mathbf{Z}/(2))^n$ composed with any $(\mathbf{Z}/(2))^n \twoheadrightarrow \mathbf{Z}/(2) = \{1,-1\}\subset \mathbf{R}^{\times}$ is nontrivial but kills the Lie algebra. This cannot be $\ker|\chi|$ for algebraic $\chi$, since if $\ker|\chi|$ is open then $|\chi|=1$ (as $\mathbf{R}_{>0}$ has no open subgroup of finite index). | |
| Feb 15, 2017 at 20:17 | history | edited | Sasha | CC BY-SA 3.0 |
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| Feb 15, 2017 at 20:15 | comment | added | Sasha | @nfdc23 : Oh, I see, of course you are correct. One still has the meaningful question though, whether the algebraic ones are "enough" (is the intersection of preimages of $\{ 1 , -1\}$ under the algebraic ones the same as under all). | |
| Feb 15, 2017 at 19:41 | comment | added | nfdc23 | No: compose any such "algebraic" one with the sign character on $\mathbf{R}^{\times}$, or instead (and even worse in general) compose it with $x \mapsto |x|^{1/29}$. | |
| Feb 15, 2017 at 19:31 | comment | added | Jim Humphreys | For $G$ semisimple, which is probably the key case to consider here, you should look at the classic paper by Borel and Tits ams.org/mathscinet-getitem?mr=0316587 | |
| Feb 15, 2017 at 19:11 | history | asked | Sasha | CC BY-SA 3.0 |