Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
What does the 'higher' mean in this context? –  Mariano Suárez-Alvarez 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
add comment

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 $Hom$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.