Timeline for How many ways are there to prove flag variety is a projective variety?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2010 at 16:26 | comment | added | Kevin McGerty | @Torsten: ahh sorry now I see what you meant, sorry I misread you! | |
| Apr 14, 2010 at 15:40 | comment | added | Torsten Ekedahl | @Kevin: What we need is the other direction, given a dominant weight there is an irreducible representation with that weight as highest weight. | |
| Apr 14, 2010 at 15:21 | comment | added | Kevin McGerty | I'm confused as to what Torsten means, an irrep. has to have a highest weight vector simply because it has finitely many weight spaces. | |
| Apr 14, 2010 at 4:31 | comment | added | Torsten Ekedahl | (cont'd) Hence, one should rather be looking for a proof that uses as few as possible of the properties of (semi-simple) algebraic groups as the properness of $G/B$ is used to establish such properties. | |
| Apr 14, 2010 at 4:29 | comment | added | Torsten Ekedahl | It seems to me that one of the quickest ways of proving the existence of an irrep with a highest weight vector is to use the properness of $G/B$. I guess it can be done in other ways (by for instance first constructing it over the Lie algebra and integrate). However, I think that this illustrates that maybe this is not such a good question: One wants to have the properness of $G/B$ as early as possible as it has many important consequences (conjugacy of Borel subgroups for one, but also that representations induced from $B$ are finite dimensional). | |
| Apr 14, 2010 at 4:04 | comment | added | Ben Webster♦ | Any closed subvariety of its complement would contain a closed orbit with a Borel fixed point, and thus also contain a highest weight vector. But that's impossible. | |
| Apr 14, 2010 at 3:54 | history | answered | Allen Knutson | CC BY-SA 2.5 |