Timeline for algebraic groups and their Lie algebras
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2013 at 17:51 | vote | accept | Andriy Regeta | ||
| Mar 3, 2013 at 17:51 | vote | accept | Andriy Regeta | ||
| Mar 3, 2013 at 17:51 | |||||
| Mar 3, 2013 at 17:51 | vote | accept | Andriy Regeta | ||
| Mar 3, 2013 at 17:51 | |||||
| Dec 2, 2012 at 18:31 | comment | added | anon | This is not a research level question. The answer is complicated, depending on the hypotheses, but can be found in books and in online notes. Briefly, over a field of characteristic zero, the category of representations of a semisimple Lie algebra is equal to that of the associated simply connected semisimple algebraic group, but otherwise the categories usually differ (e.g., for $\mathbb{G}_a$). | |
| Nov 30, 2012 at 16:09 | comment | added | Jim Humphreys |
@Andriy: Chevalley's classification shows that in the semisimple case (working over the complex field) the Lie groups and algebraic groups turn out to be (almost) the same. But this does not translate directly into results on their irreducible finite dimensional representations, though they too eventually turn out to be (almost) the same and are classified by highest weights, etc. See for example Jantzen's book Representations of Algebraic Groups, though he is motivated mostly by harder characteristic $p$ problems.
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| Nov 30, 2012 at 16:05 | comment | added | user27056 | @Andriy: Your necessity comment beyond the connected semisimple case above is not quite true: things work well for unipotent groups too (as noted at the end of my answer below). | |
| Nov 30, 2012 at 15:56 | answer | added | Jim Humphreys | timeline score: 6 | |
| Nov 30, 2012 at 15:45 | answer | added | user27056 | timeline score: 5 | |
| Nov 30, 2012 at 13:48 | answer | added | Marc Palm | timeline score: 2 | |
| Nov 30, 2012 at 13:43 | comment | added | Andriy Regeta | Say, field has characteristic zero and algebraically closed (say, complex numbers). V is just a finitely dimensional representation i.e., V is a finitely dimensional vector space over our field (with the structure of the representation of $G$ and respectively of $L$). | |
| Nov 30, 2012 at 13:30 | comment | added | Marc Palm | Algebraic groups over what field/ring? What kind of representations over which kind of $V$? There are some results available, but you need to be more specific. Best, Marc | |
| Nov 30, 2012 at 13:29 | comment | added | Andriy Regeta | Yes, sure, I forgot to add simply connectedness, It seems to me, it is always necessary (not only in semisimple case). | |
| Nov 30, 2012 at 13:10 | comment | added | Jason Starr | They are not always equivalent without some further hypothesis, e.g., in the semisimple case you should require that $G$ is simply connected. | |
| Nov 30, 2012 at 12:46 | history | asked | Andriy Regeta | CC BY-SA 3.0 |