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If $C$ is a fusion category and $\dim(C) \neq 0$ (the latter is automatic in characteristic zero, but not in nonzero characteristic), then the Drinfel'd center $Z(C)$ is fusion. More generally, if $C$ is a fusion category and $M$ is a semisimple module category over $C$, then the dual of $C$ over $M$ is fusion if $\dim(C)$ is nonzero. But if $\dim(C)=0$ these results need not hold. For example $\operatorname{Vec}(G)$ in characteristic $p$ when $p | \#G$ acts on $\operatorname{Vec}$ and its dual is $\operatorname{Rep}(G)$ which is not semisimple. Similarly, $Z(\operatorname{Vec}(G))$ is not semisimple when $p | \#G$.

I want to know if $C$ and $Z(C)$ both being semisimple implies that $\dim(C) \neq 0$.

(Feel free to assume everywhere that the base field is algebraically closed if you'd like to.)

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  • $\begingroup$ This is far from an answer, but on page 3 of "From Subfactors to Categories and Topology II," Mueger suggests that "there is little hope of proving semisimplicity" of the center when the dimension is 0. Of course, this does nothing to rule out the possibly that the center might just happen to be semisimple in some anomalous cases. He also points out that in any case the center could never be modular, as modular categories necessarily have nonzero dimension. $\endgroup$
    – Evan Jenkins
    Commented Aug 11, 2011 at 5:56
  • $\begingroup$ That means any counterexample would have to be nonspherical, right? Hence there couldn't be any known counterexamples. (Or maybe I'm missing something about what modularity means in nonzero characteristic.) Probably that can be bumped up to a proof by looking at the sphericalization (at least outside of characteristic 2). $\endgroup$
    – Noah Snyder
    Commented Aug 11, 2011 at 6:23
  • $\begingroup$ It appears so. In the paper "Categorical centers and Reshetikhin-Turaev invariants," Bruguieres and Virelizier show that the center of a spherical category is (not necessarily semisimple) modular even when the dimension is noninvertible, although it seems that some essential pieces of the proof are relegated to a paper that is still forthcoming. $\endgroup$
    – Evan Jenkins
    Commented Aug 11, 2011 at 18:53
  • $\begingroup$ That said, your desired result is probably less hard than proving that all fusion categories are spherical. Drinfeld, Gelaki, Nikshych, and Ostrik have managed to prove quite a lot about not-necessarily-spherical categories (including that the center of a fusion category is nondegenerate, the nonspherical analogue of modular), albeit only over an algebraically closed field of characteristic zero. One might hope that combining the BV and DGNO viewpoints would give such a result for nonspherical categories with noninvertible dimensions, but I don't want to speculate too much about it. $\endgroup$
    – Evan Jenkins
    Commented Aug 11, 2011 at 19:01
  • $\begingroup$ I think we've worked out that Z(C) fusion implies $\mathrm{dim} C \neq 0$. I will post an answer soon when the proof is posted. $\endgroup$
    – Noah Snyder
    Commented Jun 8, 2013 at 4:23

1 Answer 1

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We eventually sorted this out, and it appears as (one direction of) Theorem 3.6.7. in Dualizable Tensor Categories (joint with Christopher Douglas and Chris Schommer-Pries). Note that (for C semisimple) separability is equivalent to semisimplicity of Z(C) (see Corollary 3.5.9.), so Theorem 3.6.7 does address exactly this question.

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