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Apr 13, 2017 at 12:57 history edited CommunityBot
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May 5, 2011 at 17:22 comment added S. Carnahan Warning: the above comment uses an unorthodox definition of "characteristic polynomial".
May 5, 2011 at 17:12 comment added S. Carnahan I'd just like to point out that it might be easier to view your orbifold as a weighted $\mathbb{P}^2$, since you can change coordinates from the unordered triple of eigenvalues to the ordered triple of coefficients of the characteristic polynomial of Casimir (whose roots are the eigenvalues). I'm not sure how $\mathfrak{sl}_2$ fits in to this picture.
May 5, 2011 at 14:46 vote accept Noah Snyder
May 5, 2011 at 6:53 comment added DamienC @Noah Snyder: you are right. Thanks!
May 5, 2011 at 0:04 comment added Noah Snyder @DamienC: Those two "different" points for $\mathfrak{sl}_3$ are actually the same in the quotient $\mathbb{P}^2/S_3$. So that's a different issue.
May 4, 2011 at 20:26 answer added Bruce Westbury timeline score: 6
May 4, 2011 at 20:20 comment added DamienC Even more surprising: $\mathfrak{sl}_3$ and $\mathfrak{so}_8$ appear twice (they also appear in the exceptionnal series in math.tamu.edu/~jml/LMunivpub.pdf).
May 4, 2011 at 19:57 comment added Noah Snyder Perhaps I'm using the word "metric" wrong here. Vogel says "pseudo-quadratic" rather than "metric." The point is that you have (in addition to the bracket and the crossing) an adjoint pair of maps of representations $\mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathbf{1}$ and $\mathbf{1} \rightarrow \mathfrak{g} \otimes \mathfrak{g}$. But at any rate, its a bilinear form, not a Hermitian one.
May 4, 2011 at 19:36 comment added José Figueroa-O'Farrill But they are different as metric Lie algebras: the inner product is definite in $\mathfrak{so}_3$ and lorentzian in $\mathfrak{sl}_2$, assuming I'm understanding the question.
May 4, 2011 at 19:33 comment added Ben Webster Could it have something to do with the fact that over R, they are different?
May 4, 2011 at 19:19 history asked Noah Snyder CC BY-SA 3.0