# Suppose C and D are Morita equivalent fusion categories, can you say anything about R I: C->Z(C)=Z(D)->D?

If C and D are (higher) Morita equivalent fusion categories, then the Drinfel'd centers Z(C) and Z(D) are braided equivalent. Given any fusion category C we have a restriction functor Z(C)->C (by forgetting the "half-braiding"), and adjoint to that an induction functor C->Z(C).

If C and D are Morita equivalent then you can compose the induction and restriction to get a functor C->Z(C)=Z(D)->D. (Actually now that I think about you may need to fix the Morita equivalence in order to actually identify Z(C) and Z(D)?) Is there anything nice one can say about this composition? If C=D then Etingof-Nikshych-Ostrik says that $R \circ I(V) = \sum_X X \otimes V \otimes X^*$.

The reason that I ask is that Izumi calculated the induction and restriction graphs for the Drinfel'd center of one of the even parts of the Haagerup subfactor, and I would like to understand the same picture for the other even part.

• What does the 'higher' mean in this context? – Mariano Suárez-Álvarez Feb 22 '10 at 5:59
• Well "Morita equivalence" concerns algebras and invertible bimodules. Here I was referring to tensor categories and invertible bimodule categories over them. This is one categorical level up, and so sometimes it's called a "higher Morita equivalence" (coined by Mueger I think?) and sometimes just called a "Morita equivalence." – Noah Snyder Feb 22 '10 at 6:49

## 1 Answer

I think the answer should depend on the particular choice of Morita equivalence between C and D. So let M be (bi)module category connecting C and D. My first guess would be that $R\circ I(V)=\sum_X{\underline Hom}(X,V\otimes X)$ (sum over simple objects of M; ${\underline Hom}$ is the internal $Hom$).