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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.

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    $\begingroup$ 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. $\endgroup$
    – Chris Schommer-Pries
    Commented Jun 24, 2014 at 10:57
  • $\begingroup$ 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. $\endgroup$
    – Jamie Vicary
    Commented Jun 24, 2014 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – Noah Snyder
    Commented Jun 24, 2014 at 15:25
  • $\begingroup$ One good reason to be interested in the symmetric case is that this is what is relevant for the FRS description of RCFT. $\endgroup$
    – Jamie Vicary
    Commented Jun 25, 2014 at 9:04
  • $\begingroup$ It is necessary that the modular invariant is of type I, if this helps $\endgroup$
    – Marcel Bischoff
    Commented Sep 12, 2014 at 17:29

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The answer to your first question is no. The category of modules over a commutative algebra has a tensor category structure such that the forgetful functor is a tensor functor, but the category of modules over noncommutative algebras need not. Since the category of modules is a Morita invariant this gives a negative answer. The usual name for this kind of situation is "Type II modular invariant" and the D_odd module categories over su(2) at roots of unity are the simplest examples. See Kirillov-Ostrik for more details on everything in this paragraph. In these examples you have a copy of super vector spaces inside su(2) at a root of unity and the algebra is the anti-symmetric one.

("Morally" super vector spaces with the group ring of Z/2 with each group element supported in its grade is the simplest answer to your question, but technically it's not modular. But you can find it inside modular categories.)

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.

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  • $\begingroup$ 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. $\endgroup$
    – Jamie Vicary
    Commented Jun 24, 2014 at 14:30
  • $\begingroup$ 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. $\endgroup$
    – Jamie Vicary
    Commented Jun 24, 2014 at 14:35
  • $\begingroup$ You're tensoring over A, not over the base field. It's just like how modules over an ordinary commutative algebra form a tensor category. For more details see arxiv.org/abs/math/0101219 $\endgroup$
    – Noah Snyder
    Commented Jun 24, 2014 at 14:43
  • $\begingroup$ Hopefully my edit clarified your second question. The issue is whether the module category has a tensor product that's compatible with the module category structure. $\endgroup$
    – Noah Snyder
    Commented Jun 24, 2014 at 14:55
  • $\begingroup$ 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. $\endgroup$
    – Jamie Vicary
    Commented Jun 24, 2014 at 15:47

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