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.)



If C is a fusion category over a field of nonzero characteristic and dim C = 0, is Z(C) ever fusion?
We eventually sorted this out, and it appears as (one direction of) Theorem 3.6.7. in [*Dualizable Tensor Categories*](http://arxiv.org/abs/1312.7188) (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.


