"Mini" fusion categories via 6j symbols - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:18:58Z http://mathoverflow.net/feeds/question/105151 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105151/mini-fusion-categories-via-6j-symbols "Mini" fusion categories via 6j symbols Hauke Reddmann 2012-08-21T11:31:27Z 2012-08-21T21:43:22Z <p>Just for fun, I set up the following scheme:<br> - A 6j symbol is everything that fulfils Biedenharn-Elliott. (Plus symmetry, orthogonality etc. if that doesn't follow from it anyway.)<br> - There are only a finite number $i$ of irreps (including $1$, the 1-dim "neutral"). Then I tried out which of the many possible Clebsch-Gordan expansions are mutally consistent with all the 6j equations via Biedenharn-Elliott. For two irreps I got only this: $1\bigotimes{X}=X (X=1,2), 2\bigotimes{2}=1+2.$<br> - Ha!<br> - Even I did see this before, it's the smallest Fibonacci fusion category. Now a quick research gave me a load of adjectives that come with fusion categories. Can you tell me which of them apply to my scheme?<br> - $i=3$ produces 3 "minis", with $i=4$ I get 6 (and I stopped here because of the dreaded combinatiorial explosion - are these already classified somewhere?).<br> - My main question, though, is whether you get a free parameter when you choose $i$ large enough. All my "minis" give only a set of fixed complex numbers for the values of the 6j symbols, the quantum dimension and the writhe normalizer (the minis all seem to be knot-theory compatible). The latter is a root of unity so I can speculate all the minis correspond to special values of the Jones polynomial. Or suchlike.</p> http://mathoverflow.net/questions/105151/mini-fusion-categories-via-6j-symbols/105162#105162 Answer by Bruce Westbury for "Mini" fusion categories via 6j symbols Bruce Westbury 2012-08-21T15:00:15Z 2012-08-21T15:00:15Z <p>Fusion categories are discrete (like finite groups) and you never have a "free parameter". This is an observation by Ocneanu.</p> http://mathoverflow.net/questions/105151/mini-fusion-categories-via-6j-symbols/105186#105186 Answer by Noah Snyder for "Mini" fusion categories via 6j symbols Noah Snyder 2012-08-21T18:57:16Z 2012-08-21T21:43:22Z <p>The number of non-isomorphic simple objects is called the <em>rank</em> of the fusion category.</p> <p>You've made an error somewhere, as in rank 2 you should also get an example where the nontrivial simple object squares to $1$. (In your notation $2 \otimes 2 \cong 1$.) </p> <p>For rank 2, Ostrik gave a complete classification (http://arxiv.org/abs/math/0203255). Ostrik's argument is somewhat indirect (via the Drinfel'd center). As far as I know, no one has given a direct classification in rank 2 via 6j symbols. The hard part is figuring out why there's no fusion categories with the fusion rule $X^2 = 1 + n X$ for $n>1$.</p> <p>For rank 3 if you only look at fusion categories that give knot invariants, then Ostrik has a complete classification (http://arxiv.org/abs/math/0503564).</p> <p>A great source of all information about "minis" that give knot invariants is Section 5.3 of Rowell-Strong-Wang (http://arxiv.org/abs/0712.1377). "Modular" means that you get knot invariants plus you have a certain "non-degeneracy" condition that's a bit hard to explain.</p>