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Classically, two rings $R$ and $S$ are Morita equivalent if and only if any of the following is true

  1. ($R$-Mod) $\simeq$ ($S$-Mod).
  2. $S \simeq Hom_R(M,M)$, where $M$ is a finitely generated projective generator in ($R$-Mod).

A tensor category is a categorical analogue of a ring. Two tensor categories $C$ and $D$ are said to be categorical Morita equivalent if there is an exact $C$-module category $M$ and a tensor equivalence [1, Definition 7.12.17]

$$ D^{op} \simeq C^\star_M. $$

This definition resembles the second condition in the classical case. Thus my question:

In this case, do $C$ and $D$ have equivalent categories of module categories?

Reference

  • [1] Tensor Categories-[Etingof, Gelaki, Nikshych, and Ostrik]
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    $\begingroup$ There’s some technical issues you need to be careful about. Firstly, the collection of module categories forms a 2-category! Second you need to be careful about how you build that 2-category (probably you want the right exact functors). $\endgroup$ Jan 22, 2021 at 16:00

1 Answer 1

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Yes, it is Theorem 7.12.16 in [1]. In fact these are 2-equivalent (as the categories of modules are 2-categories).

Theorem 7.12.16. Let $M$ be a faithful exact module category over $C$. The $2$-functor

$$ (7.36) \quad\quad N \mapsto Fun_{C}(M,N): Mod(C) \to Mod((C_M^\star)^{op}) $$

is a $2$-equivalence.

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  • $\begingroup$ Thanks for pointing that out. Tricky! It's stated before the definition of categorical Morita equivalence. I've added the statement in your answer. I need some time to review however, as there is an extra condition ("faithful") in the theorem. $\endgroup$
    – Student
    Jan 22, 2021 at 20:31

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