# Is every separable algebra in a modular tensor category Morita equivalent to a commutative one?

Separable algebras in modular tensor categories are interesting algebraic structures, which have received significant attention because of their connection to conformal field theories. My understanding is that it is only the Morita class of the algebra which is important for determining the conformal field theory. Also, in the more general case of a fusion category, it is known by a result of Ostrik that up to Morita equivalence, separable algebras correspond to module categories for the fusion category.

Over ordinary vector spaces, every separable algebra is Morita equivalent to a commutative one, which of course makes the Morita equivalence classes easier to study. So this motivates my question: is every separable algebra in a modular tensor category Morita equivalent to a commutative one? If not, what about restricting to symmetric separable algebras?

I am working always over $\mathbb{C}$ here, and assuming that the categories involved are semisimple and with a finite number of isomorphism classes of simple objects.

• You must be assuming that we are working over an algebraically closed field. The quaternions are a separable algebra over the reals, but they are not Morita equivalent to a commutative algebra. – Chris Schommer-Pries Jun 24 '14 at 10:57
• Yes, MTCs and fusion categories are usually taken to be over $\mathbb{C}$ by defintion, and I was assuming this---I will edit the question to clarify this point. – Jamie Vicary Jun 24 '14 at 14:18
• I'm not sure about your second question, since symmetric is pretty restrictive. You can't get a counterexample from su(2). It seems implausible to me that the answer could be yes, but it also might be hard to find a counterexample. – Noah Snyder Jun 24 '14 at 15:25
• One good reason to be interested in the symmetric case is that this is what is relevant for the FRS description of RCFT. – Jamie Vicary Jun 25 '14 at 9:04
• It is necessary that the modular invariant is of type I, if this helps – Marcel Bischoff Sep 12 '14 at 17:29

It is worth noting that module categories over a modular tensor category do have a classification in terms of commutative algebras. Namely, you pick two commutative algebras A and B plus a braided equivalence between $\mathrm{Rep}^0(A)$ and $\mathrm{Rep}^0(B)$. This is proved in Davydov-Nikshych-Ostrik Cor 3.8, though the idea goes back at least to Ocneanu.
• Thanks for this interesting answer. I am having difficulty pushing this argument through my brain. Suppose I have a commutative monoid object $A$ in a modular tensor category, and module actions $f:A \otimes X \to X$ and $g:A \otimes Y \to Y$. How are you defining the tensor product module? The only way I know how to do it is if $A$ is equipped with a bialgebra structure. – Jamie Vicary Jun 24 '14 at 14:30
• I am also confused for another reason: I thought separable algebras have semisimple module categories. Are you saying there's a semisimple $\mathbb{C}$-linear category that does not admit a monoidal structure? That would stun me. Surely I have missed something here. – Jamie Vicary Jun 24 '14 at 14:35
• Thanks. Can you clarify which algebra you have in mind in super vector spaces. Do you mean the exterior algebra on $I \oplus F$, where $F$ is the fermion? I don't see that this is separable. – Jamie Vicary Jun 24 '14 at 15:47