In arxiv:0909.3140 the $G$-extensions of a fusion category $\mathcal{D}$ are classified via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$. A crucial part of the classification is Theorem 7.7 that shows that equivalence classes of such functors are in 1-1 correspondence with equivalence classes of $G$-extensions of $\mathcal{D}$. However in the proof they never show that the tensor categories they obtain from such functors are actually rigid. I was also unable to find this statement anywhere else in the paper, it is not even mentioned. Is there some obvious proof of this that I am missing or are they in fact classifying not necessarily rigid $G$-extensions? (The latter would however contradict Definition 2.1, which defines a G-extension to be in particular a fusion category, hence rigid.)
1 Answer
Great question! Yes, this is missing from the original paper and is certainly needed for the main result. It is true though. There's two different proofs currently in the literature, Deshpande and Mukhopadhyay Corollary 2.11 and Davydov and Nikshych proof of Theorem 8.5.
You can also prove it using an argument I learned from Theo Johnson-Freyd, namely given an object $X$ in $C_g$ you can realize $C_g$ as $A$-mod in $C_e$ with $X$ corresponding to $A$ itself, then $C_{g^{-1}}$ can be identified with mod-$A$ and the dual object will be $A$ again and the evaluation and coevaluation maps are given by multiplication and the unit of the algebra structure on $A$.
As to your last two sentences, rigidity is pretty essentially to the whole setup, otherwise the graded parts don't need to be invertible and then you leave the world of homotopy theory. For example, take the semisimple category with two objects 1 and X with tensor product defined by $X \otimes X = 0$ (aka the universal counterexample to all questions involving rigidity), this is $\mathbb{Z}/2\mathbb{Z}$ graded but the odd part is not invertible as a bimodule over the even part.
-
2$\begingroup$ I should say the info from the first paragraph came from Dmitri and Victor when I emailed them this exact question a month ago. $\endgroup$ Commented Jun 30, 2021 at 18:56
-
$\begingroup$ Thanks, that clears things up! When you identify $C_g$ with $A$-mod, how do you get the $C_e$-left module structure on $C_g$? I guess it should be something like $c \in C$ acting from the left as $c^*$ from the right (using rigidity of $C$). But that would require an identification of $C_g$ with $C_g^{op}$. These two categories are equivalent but not canonically, so I guess it would boil down to an arbitrary choice of non-degenerate trace on each $C_g$? $\endgroup$ Commented Jul 1, 2021 at 16:03
-
$\begingroup$ Thinking about it again that wouldn't work, because the left and right action wouldn't commute. $\endgroup$ Commented Jul 1, 2021 at 16:10
-
1$\begingroup$ You use invertibility to identify $C_e$ with A-mod-A. Without invertibility you’re right you couldn’t get an action in the other side. $\endgroup$ Commented Jul 1, 2021 at 17:38