Timeline for Inner product on $V_{-\rho}$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2016 at 13:52 | comment | added | Allen Knutson | 1. Okay, I'm missing something stupid. If I twist by the Cartan involution, then in what sense is the resulting inner product "invariant"? 2. Thanks, Jim; fixed. 3. I don't care about the reals, I was just presenting part (1) as motivational. | |
| Dec 3, 2016 at 13:43 | history | edited | Allen Knutson | CC BY-SA 3.0 |
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| Dec 3, 2016 at 2:25 | comment | added | Victor Protsak | Allen, @Ben: Right, contragredient preserves the highest weight property, whereas the dual maps highest weight modules to lowest weight, and vice-versa. | |
| Dec 3, 2016 at 1:40 | comment | added | Ben Webster♦ | @AllenKnutson The obvious one, yes. I think in the context of category O, people often use "contragredient" to mean the obvious dual twisted by the Cartan involution, so you end up back in category O. | |
| Dec 2, 2016 at 17:42 | comment | added | Jim Humphreys | @Allen: Starting with a semisimple Lie algebra over $\mathbb{C}$, note that a Verma module with an integral highest weight is irreducible just when the weight is "antidominant"; so $M(-\rho) = L(-\rho)$ is a basic example. Here the most standard symmetric bilinear form is the contravariant form, which is nondegenerate precisely in such cases (Jantzen, Shapovalov). But all of this is over $\mathbb{C}$, so it probably doesn't fit your situation. [P.S. The first question in your highlighted part lacks a word at the end.] | |
| Dec 2, 2016 at 15:51 | comment | added | Allen Knutson | Isn't its contragredient a lowest-weight, not highest-weight, representation? | |
| Dec 2, 2016 at 15:50 | history | edited | Allen Knutson | CC BY-SA 3.0 |
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| Dec 1, 2016 at 13:24 | answer | added | Ben Webster♦ | timeline score: 2 | |
| Dec 1, 2016 at 6:02 | comment | added | Victor Protsak | This representation is isomorphic to its contragredient. So being irreducible, it has a nondegenerate invariant form. Over the reals, you can ask whether this form is definite. I think that for $\frak{sl}_2$ the answer is affirmative, by an explicit calculation. | |
| Dec 1, 2016 at 4:05 | history | asked | Allen Knutson | CC BY-SA 3.0 |