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We know that every symmetric fusion category (SFC) gives rise to data $N^{ij}_k$ that describe the fusion of simple objects: $i\times j = N^{ij}_k k$, and the data $\theta_i =\pm 1$ that describe the twist of simple objects.

My question is what are the conditions on the data $N^{ij}_k,\theta_i$ such that the data correspond to a SFC?

Any conditions beyond those for fusion category are welcome. (One may try to classify finite groups via a classification of SFCs.)

One also can ask a related (and more general) question: what are the conditions on the data $N^{ij}_k,\theta_i$ such that the data correspond to a pre-modular braided fusion category?

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    $\begingroup$ I think @GjergjiZaimi posted a similar question before about giving all the finitely many categorifications, but I don't see it right now. $\endgroup$
    – AHusain
    Commented Oct 10, 2015 at 9:17

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The information in the $\theta$'s is very weak, for example if a FC admits a symmetric structure with $\theta_i=-1$ for at least one $i$, it also admits a symmetric structure with $\theta\equiv 1$. Even, it is possible to have several non-equivalent symmetric structures with $\theta\equiv 1$ over a fixed fusion category.

On the other hand, since every SFC is equivalent to the representation category of a (super)-group a necessary condition over the fusion algebra $N_k^{ij}$ is that Frobenius-Perron dimension of every simple object is an integer.

Let us call a fusion algebra without non-trivial fusion subalgebra a simple fusion algebra. Since quotient groups of a finite group $G$ are in biyective correspondence with fusion subalgebras of $K_0(Rep(G))$, a condition on the fusion data of symmetric fusion category is that every simple fusion subalgebra have to be realized (in a unique way) as the fusion algebra of the SFC of representation of (a unique) simple group (the fusion data can be read from the character table). Also, you can read from your fusion data if it corresponds to a nilpotent group, see http://arxiv.org/abs/math/0610726 in this case using group cohomology is possible to find obstructions to the existence of a SFC that realize your fusion data.

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  • $\begingroup$ Thank you very much for the comment. Here is my understanding from what you wrote: Only a simple fusion algebra corresponds to a unique SFC (of the reps. of a simple finite group). Non-simple fusion algebra may correspond to several SFC's. In this case $N^{ij}_k$ + additional data specify a SFC. What is this additional data? $\endgroup$ Commented Oct 12, 2015 at 22:27
  • $\begingroup$ Additional data correspond to F-matrices (or 6j-symbols) and R-matrices (in this case symmetric R-matrices). You could take Definition 3.10 in arxiv.org/pdf/1305.2229v1.pdf and change the condition (iv) for $\widehat{S_{a,b}}=1$ for all $a,b$. $\endgroup$ Commented Oct 12, 2015 at 22:36
  • $\begingroup$ Thanks. We know that unlike $\theta_i$, F-matrices and R-matrices are not "gauge invariant". I wonder, do we know any gauge invariant data? The conditions on F-matrices and R-matrices are very hard to solve. $\endgroup$ Commented Oct 12, 2015 at 22:49
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    $\begingroup$ @Xiao-Gang Wen: Symmetric structures over the abstract FC $Rep(G)$ are in correspondence with pair $(A,\rho)$, where $A$ is an abelian normal subgroup of $G$ and $\rho:A\times A\to U(1)$ is a skew-symmetric bilinear form ad $G$-invariant, see Example 2.1 of arXiv: math/0605731, or arXiv: q-alg/9706007 Then for a simple group there is a unique symmetric structure over $Rep(G)$. By Deligne's theorem every ($\theta\equiv 1$) SFC is the category of representation of a unique (up to isomorphism) group. So, if you have SFC such that its fusion data is simple it corresponds to a unique simple group. $\endgroup$ Commented Oct 15, 2015 at 16:19
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    $\begingroup$ @Xiao-Gang Wen: There is an other result from group theory that say "every simple group is uniquely determined by its character table", it implies that simple groups are determined by its fusion data (I do not remember an exact reference ). This result uses the classification of finite simple groups. $\endgroup$ Commented Oct 15, 2015 at 16:34

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