Timeline for How Does a Borel Subgroup Know Which Weights Are Dominant
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 4, 2015 at 2:18 | comment | added | Keenan Kidwell | first act on the character being $\lambda$ by the longest Weyl group element to get something which is actually $\overline{B}$-dominant. | |
| Jul 4, 2015 at 2:17 | comment | added | Keenan Kidwell | Dear @David, I don't have my copy of Jantzen with me, but I think if you work in the algebraic category (algebraic induction, etc.), things are as follows. If you take a $B$-dominant weight and induce it from the opposite Borel $\overline{B}$ (relative to $B$) then what you get is a non-zero representation whose socle (in characteristic zero, just the whole thing due to semisimplicity) is the representation of your group $G$ with highest $B$-weight $\lambda$. If you induce from $\overline{B}$ a weight which is not $B$-dominant, then you get zero. If you want to induce from $B$, you should | |
| Nov 11, 2009 at 5:40 | comment | added | Rob Harron | To me, a root is a weight of a given maximal torus acting on the lie algebra of G. What my answer was trying to convey was that if you picked a different torus in B, then you get different roots and a notion of dominant on this new set of roots. Ben Webster's answer is clearly better though. | |
| Nov 11, 2009 at 5:12 | comment | added | Ben Webster♦ | I mean, simple roots. | |
| Nov 11, 2009 at 5:12 | comment | added | Ben Webster♦ | B/N acts on the Lie algebra n/[n,n] by the adjoint action. The weights of that action are the positive roots. | |
| Nov 11, 2009 at 4:43 | comment | added | Dinakar Muthiah | But the splitting is not unique. How do you get B/N to act on N? If you induce from a non-anti-dominant weight you'll get 0. | |
| Nov 11, 2009 at 4:25 | comment | added | David E Speyer | That said, I would be interested in knowing what happens when you induce a non-dominant weight form B to G. | |
| Nov 11, 2009 at 4:25 | comment | added | David E Speyer | B has a maximal abelian quotient T, given a short exact sequence 0 -> N -> B -> T -> 0. This sequence is semi-direct, and the positive weights are the weights of T acts on N. There are no choices here. | |
| Nov 11, 2009 at 4:22 | comment | added | Dinakar Muthiah | But without choosing a maximal T, how would you say which roots are dominant? | |
| Nov 11, 2009 at 4:14 | history | answered | Rob Harron | CC BY-SA 2.5 |