# Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes?

Let $C$ be a fusion category with simple objects $X_1,...,X_n$, and let $Y_1,...,Y_n$ be objects with each $Y_i$ isomorphic to $X_i$. Is there a monoidal auto-equivalence $F:C \rightarrow C$ which takes each $X_i$ to $Y_i$, and such that $F$ is naturally isomorphic (as a monoidal functor) to the trivial auto-equivalence?

It seems like this should be true since the structure of the fusion category is determined by $6j$-symbols and one should be able to get the same $6j$-symbols for every choice of representatives of simple objects by appropriately choosing representative morphisms between their tensor products. But how can one write down such a functor?

• I would be more interested in such $F$, which is actually not isomorphic to the trivial auto equivalence... – Marcel Bischoff Sep 17 '14 at 4:40
• Well, that sometimes exists and sometimes doesn't. For $Vec_G$ (with trivial associator), the autoequivalences which fix simple objects, modulo natural isomorphism, are parametrized by $H^2(G,k^*)$. – Pinhas Grossman Sep 17 '14 at 21:01
• Yes it is a two cohomology problem, Kawahigashi and Longo showed that this cohomology vanishes for many examples, but I am not aware of a non-group example where it doesn't. – Marcel Bischoff Sep 18 '14 at 17:52
• There is a fusion category Morita equivalent to the even parts of the Asaeda-Haagerup subfactor which has a non-trivial auto-equivalence which fixes all the isomorphism classes of simple objects. Moreover, this auto-equivalence restricts to the trivial auto-equivalence on the subcategory of invertible objects, so I think it's definitely a non-group example. (This category appeared in recent not-yet-published joint work of Noah, Masaki Izumi, and myself.) But anyway I'm sure there are many other examples as well. – Pinhas Grossman Sep 18 '14 at 22:28

Let's forget about the monoidal structure for a moment. $C$ is semisimple, and semisimplicity means that to give a functor out of $C$ is the same thing as specifying where it sends simple objects. So that means that saying $\mathcal{F}(X_i)=Y_i$ determines a unique functor $\mathcal{F}$. Now, what does it mean for this functor to be naturally isomorphic to the identity functor? Well, again by semisimplicity, to give a natural transformation between functors is just to give its components on simple objects. So fixing isomorphisms $\eta_i: X_i \cong Y_i$ determines a unique natural isomorphism $\eta: \mathcal{F} \rightarrow \mathcal{Id}$.

Now let's bring the monoidal structure back into the picture. We have a functor $\mathcal{F}$ and a natural transformation $\eta: \mathcal{F} \rightarrow \mathcal{Id}$. But $\mathcal{Id}$ has a canonical structure as a monoidal functor! So you can just use "transport de structure" to move the monoidal structure on $\mathcal{Id}$ across the natural isomorphism to determine a monoidal structure $\mathcal{F}$. In other words, there's no additional content to the monoidal statement. Once you know that $\mathcal{F}$ is naturally isomorphic via $\eta$ to the identity functor, then there's a unique way to make $\mathcal{F}$ into a monoidal functor such that $\eta$ becomes a monoidal natural transformation. This is completely the same argument as saying if you have a set X and a group G and a bijection $f: X \rightarrow G$ then there's a unique way to make X into a group such that $f$ is an isomorphism of groups.

• p.s. Let's talk sometime this week. I started looking at our draft again today. – Noah Snyder Sep 17 '14 at 3:18