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We reference [EGNO] for the concept of a braided fusion category. Following the conventions in [JFR], let $\mathcal{C}$ denote a braided fusion category equipped with a braiding $\beta$, and let $\mathcal{B} \subset \mathcal{C}$ represent a fully faithful inclusion of braided fusion categories. The centralizer of this inclusion is defined as:

$$\mathcal{Z}_2 (\mathcal{B} \subset \mathcal{C}) := \{ y \in \mathcal{C} \ | \ \beta_{y,x} \circ \beta_{x,y} = {\rm id}_{x \otimes y} \ \forall x \in \mathcal{B} \}.$$

The Müger center $\mathcal{Z}_2 (\mathcal{B})$ of $\mathcal{B}$ is the centralizer of the identity inclusion. This center forms a symmetric fusion category. A braided fusion category $\mathcal{B}$ is termed nondegenerate if $\mathcal{Z}_2 (\mathcal{B})$ is equivalent to the trivial fusion category ${\rm Vec}$. A braided fusion category $\mathcal{B}$ is defined as slightly-degenerate if $\mathcal{Z}_2 (\mathcal{B})$ is equivalent to the fusion category ${\rm sVec}$ of $C_2$-graded finite-dimensional vector spaces. A slightly-degenerate braided fusion category $\mathcal{C}$ is referred to as split if it is equivalent to $\mathcal{D} \boxtimes {\rm sVec}$, where $\mathcal{D}$ is nondegenerate.

A fusion category $\mathcal{C}$ is called integral if the Frobenius-Perron dimension (${\rm FPdim}$) of each of its simple objects is an integer. An integral fusion category is inherently pseudo-unitary (i.e. ${\rm FPdim}(\mathcal{C}) = {\rm dim}(\mathcal{C})$), and consequently, spherical, as discussed in [EGNO]. A braided fusion category that possesses a spherical structure is (termed) premodular. If such a category is either nondegenerate or slightly-degenerate, it is denoted as modular or super-modular, respectively.

The classification of super-modular fusion categories began with the works cited in [BGHNPRW17, BGNPRW20, BPRZ21]. To date, very few non-split examples have been identified. The only known integral examples split as $\mathcal{D} \boxtimes \mathrm{sVec}^-$, which are non-positive (here positive means that ${\rm dim}(X) > 0$ for every simple object $X$). Perhaps the assumptions of integrality and positivity in a super-modular fusion category imply that it must necessarily split.

Question: Is there a non-split super-modular positive integral fusion category?


References

[BGHNPRW17] Bruillard, Paul; Galindo, César; Hagge, Tobias; Ng, Siu-Hung; Plavnik, Julia Yael; Rowell, Eric C.; Wang, Zhenghan. Fermionic modular categories and the 16-fold way. J. Math. Phys. 58 (2017), no. 4, 041704, 31 pp.
[BGNPRW20] Bruillard, Paul; Galindo, César; Ng, Siu-Hung; Plavnik, Julia Y.; Rowell, Eric C.; Wang, Zhenghan. Classification of super-modular categories by rank. Algebr. Represent. Theory 23 (2020), no. 3, 795--809.
[BPRZ21] Bruillard, Paul; Plavnik, Julia; Rowell, Eric C.; Zhang, Qing. On classification of super-modular categories of rank 8. J. Algebra Appl. 20 (2021), no. 1, Paper No. 2140017, 36 pp.
[EGNO] Etingof, Pavel; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor. Tensor categories. Mathematical Surveys and Monographs, 205. American Mathematical Society, Providence, RI, 2015. xvi+343 pp.
[JFR] Johnson-Freyd, Theo; Reutter, David. Minimal nondegenerate extensions. J. Amer. Math. Soc. 37 (2024), no. 1, 81--150.
[Mu] Müger, Michael. On the structure of modular categories. Proc. London Math. Soc. (3) 87 (2003), no. 2, 291--308.

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  • $\begingroup$ To make the question entirely self-contained (so rare and so nice!), could you please also explain what does "integral" mean? $\endgroup$ Apr 24 at 5:27
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    $\begingroup$ @მამუკაჯიბლაძე: Sure! Integral means that the Frobenius-Perron dimension of every simple object is an integer. I will add the definition. $\endgroup$ Apr 24 at 5:38

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Example 2.1 in [BGHNPRW17] that corresponds to the centralizer of a fermion in the twisted Drinfeld double of $SL(2,5)$is an integral non-split super-modular category. In fact, if we use Proposition 3.2, then we can easily see (since $\mathbb A_5$ is simple) that the example does not contain modular categories of dimension 4; hence, by Theorem 3.1, it is non-split.

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  • $\begingroup$ Since Z(Rep(A5)) is perfect, meaning it contains no non-trivial invertible objects, how can it possibly contains a fermion? $\endgroup$ Apr 24 at 13:13
  • $\begingroup$ Or maybe, you wanted to write "the Drinfeld double of SL(2,5)". $\endgroup$ Apr 24 at 16:20
  • $\begingroup$ In fact, a twisted Drinfeld double $D^{\omega}{\rm SL}(2,5)$ with $ord(\omega)=4$. $\endgroup$ Apr 24 at 16:55
  • $\begingroup$ Yes, I fixed the typo. $\endgroup$ Apr 24 at 17:21
  • $\begingroup$ To avoid any confusion, the example is $\mathcal{Z}_2 ({\rm sVec} \subset {\rm Rep}(D^{\omega}G))$, with $G={\rm SL}(2,5)$ and $ord(\omega)=4$. It is of course not the symmetric category ${\rm Rep}(G,z)$ first mentioned in your Example 2.1. $\endgroup$ Apr 24 at 17:43

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