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 theorybased classification program, it might be of interest.

$\begingroup$ 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 JohnsonFreydJun 9, 2015 at 23:55

$\begingroup$ 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 ReddmannJun 10, 2015 at 10:51

$\begingroup$ 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 HenriquesJun 10, 2015 at 23:16

$\begingroup$ So, loosely, a based ring is a category iff it has a consistent (pentagon rule etc.) set of 6j symbols? $\endgroup$– Hauke ReddmannJun 11, 2015 at 10:56

$\begingroup$ That's exactly correct. The 6j symbols are the entires of the socalled $F$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 HenriquesJun 12, 2015 at 17:39
1 Answer
Such tensor categories are called "neargroup 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 $i=\infty$.