Timeline for Do representations of real semisimple algebraic group have to be algebraic?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2015 at 18:12 | vote | accept | Jerry | ||
| Sep 30, 2015 at 7:25 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Sep 30, 2015 at 5:54 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed a typo and introduced a display, fixed {\it ...} as well
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| Sep 30, 2015 at 5:52 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Sep 30, 2015 at 5:38 | comment | added | Venkataramana | @Jerry: The Lie algebra is semi-simple, and hence its representation theory is described in terms of highest weights, which essentially means that all representations are explicitly constructed, hence algebraic. | |
| Sep 30, 2015 at 5:29 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Sep 30, 2015 at 5:07 | comment | added | Vesselin Dimitrov | @KevinCasto: For homomorphisms of Lie groups, smoothness is automatic from continuity. This is proved in Proposition 1.2.22 of Terry Tao's book Hilbert's Fifth Problem and Related Topics. | |
| Sep 30, 2015 at 4:52 | comment | added | Kevin Casto | On "the other end of things", how do you get from knowing $\rho$ is continuous to knowing it's smooth? I know the story from there, but on the face of it you need smoothness to talk about tangent spaces and Lie algebras. Do you use something like smooth approximation...? | |
| Sep 30, 2015 at 3:48 | comment | added | Jerry | Sorry for being stupid. Could you please explain a bit why by going to the Lie algebra and coming back we know that the representation is indeed algebraic? Since this involves taking log and exp, which are not algebraic. Is there any extra problem when the group is not split? | |
| Sep 30, 2015 at 3:00 | history | answered | Venkataramana | CC BY-SA 3.0 |