Consider the following fusion categorie $F(i)$ with integer parameter $i$. Simple objects are $1,a,A,B$ (where $a$ and $A$ are conjugates). Nontrivial fusion rules are $a\bigotimes{a}=A$ (and conjugate), $B\bigotimes{a}=B$ (and conjugate), $B\bigotimes{B}=1\bigoplus{a}\bigoplus{A}\bigoplus{i*B}$ (i.e. multiplicity $i$). The system is well known for $i=2$ (all quantum dimensions are integer then - offhand I can't give the reference but http://arxiv.org/abs/1401.7096 should be your friend) but the fusion rules work for any i.
Naturally the question arises, when is $F(i)$ braided? (Actually even modular.) No such luck: for $i=\infty$. I have no problem at all with thinking $\infty$ as integer, but you are mathematicians :-)
Question: Can you make any sense out of a fusion category with some infinity multiplicity(ies), maybe as limit to another object? (The twist values are well defined, BTW - 18th roots of unity.) As modular fusion categories are rare and this wacko won't show up in any Galois theory-based classification program, it might be of interest.
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asked Jun 9, 2015 at 15:30
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$\begingroup$ As Andre points out in an answer below, I think you are wrong about it working "for any $i$". Indeed, often the fusion ring perhaps can be defined, but for only certain $i$ does it lift to a category. A particularly simple case is that $X\otimes X = 1 \oplus i X$ admits an associator only when $i = 0$ or $1$, and the latter only when your ground field has a square root of $5$ (actually, the golden ratio is what's needed). It is braided only when your ground field has has a primitive tenth root of unity, if my memory is correct. $\endgroup$– Theo Johnson-FreydCommented Jun 9, 2015 at 23:55
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$\begingroup$ Owch! I begin to think that what I always called a "fusion category" is only a "based ring". (I check: Associative, commutative, 1 element, conjugate element. Uhm, can you give me a simple condition for fusion rules or Verlinde matrix that a based ring can be categorified?) $\endgroup$– Hauke ReddmannCommented Jun 10, 2015 at 10:51
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$\begingroup$ I'm afraid that there isn't any simple condition... But have a look at pages 36–40 of arxiv.org/pdf/0712.1377v4.pdf for what's needed to really give a modular tensor category as opposed to just a fusion ring. In particular, the $F$-matrices (also known as the "associator") are very important. $\endgroup$– André HenriquesCommented Jun 10, 2015 at 23:16
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$\begingroup$ So, loosely, a based ring is a category iff it has a consistent (pentagon rule etc.) set of 6j symbols? $\endgroup$– Hauke ReddmannCommented Jun 11, 2015 at 10:56
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$\begingroup$ That's exactly correct. The 6j symbols are the entires of the so-called $F$-matrices. They control associativity. If you want the category to be braided, that's another piece of data, which should satisfy its own axioms. $\endgroup$– André HenriquesCommented Jun 12, 2015 at 17:39
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1 Answer
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Such tensor categories are called "near-group categories": these are semisimple tensor categories whose fusion rule includes exactly one noninvertible simple object. In your case, the group is $\{1,a,A\}$, and is isomorphic to $\mathbb Z/3$.
The paper http://arxiv.org/pdf/1401.1879v2.pdf shows that there's only finitely many values of your parameter $i$ such that your fusion ring comes from a fusion category (excluding $i=\infty$, the maximum possible value is $i=6$). Thus, I would find it very surprising if there were a semisimple category (suitably generalised to allow for infinite direct sums of simple objects) that realises the case $i=\infty$.
answered Jun 9, 2015 at 16:04