# Morita equivalent algebras in a fusion category

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.

Questions:

(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!

• Are you familiar with the answers to these questions for ordinary algebras? – Qiaochu Yuan Oct 19 '17 at 8:04
• @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? – kolaka Oct 19 '17 at 9:29
• 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). – Qiaochu Yuan Oct 19 '17 at 21:37
• 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. – Marcel Bischoff Oct 22 '17 at 13:02
• arxiv.org/abs/math/0111139 – Marcel Bischoff Oct 22 '17 at 13:06

• 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. – kolaka Oct 22 '17 at 9:03