Timeline for Can a simple lie algebra be determined by weights of its representation?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2012 at 10:50 | vote | accept | Dmitry Vaintrob | ||
| Jun 11, 2012 at 10:50 | answer | added | Dmitry Vaintrob | timeline score: 0 | |
| Jun 11, 2012 at 10:20 | comment | added | Dmitry Vaintrob | Thanks Victor! These are indeed counterexamples to my (intended) question. | |
| Apr 13, 2011 at 23:55 | history | edited | Dmitry Vaintrob | CC BY-SA 3.0 |
added words "weight datum"
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| Mar 24, 2011 at 22:05 | comment | added | Jim Humphreys | The original tag 'weights' is usually applied to other senses of the word, so I edited this and also added 'simple' to clarify which class of Lie algebras you are discussing. Aside from that, I agree with other comments that the question needs to be better focused since there are problems with the formulation. Some added motivation might help too, since the finite dimensional representations of simple Lie algebras (over the complex numbers) have been so well studied from many viewpoints. | |
| Mar 24, 2011 at 22:01 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
added 7 characters in body; edited tags
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| Mar 24, 2011 at 21:46 | comment | added | Victor Protsak | The fundamental representation of $\mathfrak{g}=\mathfrak{sl}_n$ and its dual cannot be distinguished in this way; the same applies to any non-self-dual representation of $\mathfrak{g}$ and its dual. Similarly, the defining representations of $\mathfrak{sp}_{2n}$ and $\mathfrak{o}_{2n}$ are indistinguishable, and these Lie algebras are not the Langlands duals of each other, because they are of types $C$ and $D.$ | |
| Mar 24, 2011 at 21:29 | comment | added | ARupinski | By "linear combination of points of $\mathbb{Z}^n$" are you referring to the full set of weights of $V$ or some subset? | |
| Mar 24, 2011 at 21:12 | history | asked | Dmitry Vaintrob | CC BY-SA 2.5 |