Let $\mathcal{C}$ be a braided $\mathbb{k}$-linear fusion category ($\mathbb{k}$ algebraically closed; if necessary to answer my question you can also assume $\mathcal{C}$ to be pivotal or even modular). Two algebras $A,B$ in $\mathcal{C}$ are called Morita equivalent if the categories $A\hbox{-}\mathsf{Mod}(\mathcal{C})$ and $B\hbox{-}\mathsf{Mod}(\mathcal{C})$ of (left or right) modules internal to $\mathcal{C}$ are equivalent.


(1) Is there any alternative, equivalent characterization of Morita equivalence of two algebras which circumvents showing that their module categories are equivalent?

(2) In particular, is there a way to "calculate" the Morita equivalence class of a given algebra?

(3) Is there anything special one can say about the Morita equivalence class of an algebra, e.g. some classification result, maybe by imposing additional properties on the algebra?

Please also refer to literature, thanks!

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    $\begingroup$ Are you familiar with the answers to these questions for ordinary algebras? $\endgroup$ Oct 19, 2017 at 8:04
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    $\begingroup$ @QiaochuYuan Roughly. For example, $A$ and $B$ are Morita equivalent if and only if there are bimodules $M$ and $N$ such that $M \otimes_B N \cong A$ and $N \otimes_A M \cong B$ as bimodules. I suppose that this also holds for internal algebras. But are there any other characterizations? Or any literature where such things are summarized? $\endgroup$
    – kolaka
    Oct 19, 2017 at 9:29
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    $\begingroup$ You can get much more specific: for ordinary algebras, $A$ and $B$ are Morita equivalent if and only if $B = e M_n(A) e$ for $e \in M_n(A)$ which is what is called a full idempotent (see the last bit of qchu.wordpress.com/2015/05/17/generators for details). $\endgroup$ Oct 19, 2017 at 21:37
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    $\begingroup$ In this case, the Morita equivalence class of $A$ (under some technical assumptions) can be easily calculated using the internal hom. Namely, Ostrik shows that if $M$ is an $A$-module then $[M,M]$ is Morita equivalent to $A$ and every algebra Morita equivalent to $A$ arises this way. $\endgroup$ Oct 22, 2017 at 13:02
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    $\begingroup$ arxiv.org/abs/math/0111139 $\endgroup$ Oct 22, 2017 at 13:06

1 Answer 1


in the algebra case, B=eMn(A)e "because" B=End_A (P) with P f.g. proyective, so, finding e is the same as give a presentation of P as a direct summand of A^n. Also, P=F(B) where F is the functor giving the equivalence between B-mod and A-mod, that is f.g. projective because B is so as B-module. In summary, "e" is so explicit as the functor..

  • $\begingroup$ Could you please elaborate your answer? If I understand correctly, explicitly finding $P$ and hence the full idempotent $e$ amounts to finding the equivalence $F$ between the module categories, but this is exactly what I wanted to circumvent. $\endgroup$
    – kolaka
    Oct 22, 2017 at 9:03
  • $\begingroup$ well, I can say "e" is so explicit as the module F(B)", (instead of "as F"), and maybe you can say something in a particular case, just because B is a particular B-module and so F(B). For instance, in the algebra case, if A is such that every f.g. projective is free, then every Morita equivalent ring to A is necesarily of the form Mn(A). $\endgroup$ Oct 22, 2017 at 18:11

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