# When modular tensor categories are equivalent?

I asked this question at math stack exchange math stack exchange but I haven't got any answer yet there.

I would like to know when we say that two modular tensor categories are equivalent.

Is it true that two modular tensor categories are equivalent if they are equivalent as monoidal categories? Or do we require more conditions?

Similarly, I would like to know the definitions of equivalence as ribbon/ braided/premodular/ categories.

I saw these equivalences in several papers but I could not find definitions of them in the papers.

• They need to be equivalent in the enriched sense and as ribbon categories. So in addition to the monoidal structure an equivalence must respect the linear structure on morphisms, the braidings, and the twists. – Qiaochu Yuan Mar 9 '16 at 0:30

A tensor category includes the information of a tensor product, which is something that takes objects and returns objects. This means that a tensor functor can't just "preserve tensor product" it needs to have higher data which gives isomorphisms between $F(X \otimes Y) \rightarrow F(X) \otimes F(Y)$ satisfying some naturality conditions.
But braided and ribbon structures are things that take objects and return morphisms. So they're up one dimension higher. This makes life easier. For example a braiding is an iso $\sigma_{X,Y}: X\otimes Y \rightarrow Y \otimes X$ and a monoidal functor is braided if $\mathcal{F}(\sigma_{X,Y})=\sigma_{\mathcal{F}(X),\mathcal{F}(Y)}$. This is just a condition, not a structure. Similarly, a spherical structure (which is equivalent to a ribbon structure) assigns to every object a morphism $f_X: X \rightarrow X^{**}$, and a functor preserves ribbon structure if $\mathcal{F}(f_X) = f_{\mathcal{F}(X)}$. Again this is just a condition not something complicated.