Timeline for Getting the Weyl dimension formula geometrically
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2010 at 23:25 | comment | added | Faisal | Victor: The clarification you seek can be found in the link posted by Mike. (Apologies for the delayed response!) | |
| Jul 14, 2010 at 14:36 | answer | added | David E Speyer | timeline score: 3 | |
| Jul 14, 2010 at 10:23 | answer | added | Bugs Bunny | timeline score: 1 | |
| Jul 13, 2010 at 13:49 | answer | added | David E Speyer | timeline score: 1 | |
| Jul 13, 2010 at 11:42 | answer | added | Jim Humphreys | timeline score: 4 | |
| Jul 13, 2010 at 7:15 | comment | added | Victor Protsak | If I remember correctly, you can get full Weyl character formula from equivariant $K$-theory and the dimension by "forgetting" the $T$-action. | |
| Jul 13, 2010 at 7:12 | comment | added | Victor Protsak | Can you, please, clarify where the RHS of your formula "lives", i.e. the meaning of the exponential functions $e^\alpha$ and $e^\lambda$ when $\alpha, \lambda\in\mathfrak{t}^*$ (together with the identifications made)? | |
| Jul 13, 2010 at 3:56 | comment | added | Faisal | Mike, that looks--at least to me--like the Borel-Hirzebruch argument. In particular, David only gets the dimension of the irrep after using the Weyl char formula. I've done some searching and have found a paper by Bernhard Koeck where he derives the Weyl char formula using some sophisticated version of Riemann-Roch. I was hoping something more elementary would suffice for the dimension formula. | |
| Jul 13, 2010 at 3:36 | comment | added | Mike Skirvin | You can think of the Weyl Character Formula as geometric because it can be proved using Borel-Weil-Bott and Riemann-Roch. David Speyer wrote a nice answer in regards to this, found here mathoverflow.net/questions/11422/… | |
| Jul 13, 2010 at 0:57 | history | asked | Faisal | CC BY-SA 2.5 |