Timeline for Calculating 6j-symbols (aka Racah-Wigner coefficients) for quantum groups
Current License: CC BY-SA 2.5
13 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 10, 2010 at 1:04 | comment | added | Chris Schommer-Pries | I added the tqft tag since I think it is of interest to people interested to tqfts. | |
| Mar 10, 2010 at 1:03 | history | edited | Chris Schommer-Pries |
added tag: tqft
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| Mar 9, 2010 at 21:05 | history | edited | Bruce Westbury | CC BY-SA 2.5 |
Link added
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| Mar 9, 2010 at 9:38 | vote | accept | Bruce Westbury | ||
| Feb 24, 2010 at 10:14 | comment | added | Bruce Westbury | Here is a reference where this is done. MR2169040 Gliske, S. ; Klink, W. H. ; Ton-That, T. Algorithms for computing generalized ${\rm U}(N)$ Racah coefficients. Acta Appl. Math. 88 (2005), no. 2, 229--249. | |
| Feb 23, 2010 at 21:21 | comment | added | Greg Kuperberg | Allen: Yes, you can define a numerical 6j-symbol given any system of canonical bases of $\text{Inv}(A \otimes B \otimes C)$, where "canonical" is taken in the informal sense. More formally, you can (a) use Lusztig's dual canonical bases; or (b) simply ask for any self-consistent set of formulas. A consistent set of formulas would define the bases with respect to which they are correct. | |
| Feb 22, 2010 at 9:06 | comment | added | Bruce Westbury | Allen, The decomposition of an exterior power with any representation is unique. The question was phrased to avoid this problem. I defined the 6j-symbols as the components of a functor. In general this would mean specifying bases in vector spaces and there is no consensus on how to do this (but there are several ways it can be done). You put a vector at each vertex of the tetrahedron graph to define a scalar. | |
| Feb 20, 2010 at 15:05 | comment | added | Allen Knutson | In SL(2) the decomposition of 2-fold tensor products is unique, prompting one to compare the decomposition of $(A \otimes B) \otimes C$ with $A \otimes (B\otimes C)$. But for other groups the decomposition isn't unique, without some canonical basis input perhaps. So it's never been clear to me how to define 6j symbols, much less compute them. | |
| Feb 19, 2010 at 17:24 | answer | added | Pavel Etingof | timeline score: 14 | |
| Feb 19, 2010 at 15:22 | comment | added | Bruce Westbury | I agree. Kirillov and Reshetikhin stated the result. Masbaum and Vogel proved it by showing both sides satisfied the same linear recurrence relations. Frenkel and Khovanov derive the result. The Frenkel and Khovanov method is the only one which looks as though it could be applied in other examples. | |
| Feb 19, 2010 at 14:43 | comment | added | Greg Kuperberg | I think that the result for sl(2) is due to Kirillov and Reshetikhin, and that these other references are later, although they may give new arguments. | |
| Feb 19, 2010 at 8:31 | history | asked | Bruce Westbury | CC BY-SA 2.5 |