Example: Take the irreps of $SU(2)$: $0,1/2,1,...$ (Spin notation.) The quantum dimensions are $1,q+1/q,q^2+q+1+1/q+1/q^2,...$. At $q=(-1)^{1/5}$ this evaluates to $1,\phi,0,...$ and you get the Fibonacci category, the Clebsch-Gordan series turning into fusion rules. (No point to formulate the "get" rigorously :-)
Obvious questions: 1. For a given fusion ring, what are the Lie algebras you can "get" it from, and 2. For a given Lie algebra, what fusion rings can you "get" by taking a subset of the irreps and setting $q=(-1)^{1/n}$? (E.g. Arxiv 1004.5456, where you "get" $SU(3)\rightarrow SU(3)_2, SU(3)_3/Z_3$, surely more are possible).
Since any Lie algebra has a Reshitikhine-Turaev invariant (oops, must I insert "semisimple"?), the construction including braiding, I assume the fusion rings to be braided. (Although it would be interesting if non-braided fusion rings are also a "get" from a more general object at $q=(-1)^{1/n}$.) As an amateur, I don't know if you can sharpen "braided" to "modular" or even "modular data", just call the relevant sharpest property "X" for a moment. Most interesting question 3: Does there exist fusion rings with the property "X" which are NOT possible to "get" from a Lie algebra? (Wouldn't rule that out a priori, I just say "Vogel plane"...)
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asked Jan 1, 2015 at 15:46
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$\begingroup$ Most probably you know much more about it than me, but I still dare to suggest that a promising place to look for cases for 3. would be fusion rings arising from representations of finite groups, or their "twistings" as in Freyd-Yetter (crossed G-sets). It is hard to imagine how could one associate any Lie algebra to such kind of data... $\endgroup$– მამუკა ჯიბლაძეCommented Jan 1, 2015 at 19:42
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$\begingroup$ Don't worry, I don't know Jack :-) Could you give an actual example for a fusion rule not associated with a Lie algebra (remember, braided. At least)? $\endgroup$– Hauke ReddmannCommented Jan 2, 2015 at 18:09
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