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(1) Let $\mathcal F$ be a spherical fusion tensor category. Then Müger showed that $R=\bigoplus_{H\in\mathrm{Irr}(\mathcal F)} H\boxtimes H^\mathrm{op}$ canonically has the structure of a Frobenius algebra in $\mathcal F \boxtimes \mathcal F^\mathrm{op}$ and that the category of bimodules ${}_R(\mathcal F \boxtimes \mathcal F^\mathrm{op})_R$ has a braiding which makes it a modular tensor category equivalent to $Z(\mathcal F)$. Further, the "dual" Frobenius algebra is $\hat Q=I(1)$, where $I$ is the induction functor, ie the adjoint of the forgetful functor $F\colon Z(\mathcal F)\to \mathcal F$.

(2) Conversely, let $\mathcal C$ be a modular tensor category, then it is the center of a fusion category iff there is a Lagrangian algebra $A\in\mathcal C$, i.e. a commutative (Frobenius) algebra, such that $\dim A^2=\dim\mathcal C$. Namely, the right modules $\mathcal C_A$ have a tensor structure and $$Z(\mathcal C_A)=\mathcal C\cong\mathcal C\boxtimes \overline{\mathcal C_A^0}\cong \mathcal C\boxtimes \mathrm{Vect}\cong \mathcal C$$

Further, I conclude ${}_A\mathcal C_A\cong \mathcal C_A \boxtimes \mathcal C_A^\mathrm{op}$, because $\mathcal C_A^0\cong\mathrm{Vect}$.

Question: Let $A\in\mathcal C$ be a Lagrangian algebra. I guess it is true that $A$ and $\hat A$ by (2) are in of the form as in (1), ie $A\cong I(1)$ and $\hat A\cong R$. Is there any reference for such a result, which is citable?

Note: A special case when $\mathcal C \cong Z(\mathcal C_0)$ for $\mathcal C_0$ a unitary modular tensor category is contained implicitly in the subfactor literature http://arxiv.org/abs/math/0107127

I am particular interested and most comfortable with the case of unitary tensor categories, so feel free to add the right adjectives.

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  • $\begingroup$ $C_A$ is a left $\mathcal C=Z(\mathcal C_A)$ module category. Then I guess, because $A$ is commutative, $A=[A,A]$ the (action) internal hom of $A$ seen as the trivial right $A$-module. If the action of $Z(\mathcal C_A)$ on $\mathcal C_A$ is simply $F(\,\cdot)\otimes -$, then this would show that $A \in \mathcal C$ is the full center of the tensor unit $A\in\mathcal C_A$, which is well-known to coincide with $I(1)$. $\endgroup$
    – Marcel Bischoff
    Jun 12, 2015 at 16:06
  • $\begingroup$ As I understand it, these lagrangian algebras all look like this from the right perspective. If you realize your initial category as $\mathcal Z(\mathcal C_A)$, then $A=I(\mathbf 1)$ and any other lagrangian algebra will look like $B=\alpha(I(\mathbf 1))$ for some $\alpha\in\mathsf{Aut}\big(\mathcal Z(\mathcal C)\big)$. Thus $B$ looks like the canonical algebra adjusted by $\alpha$ from the perspective of $\mathcal C_A$, while $A$ Looks like the canonical algebra adjusted by $\alpha^{-1}$ from the perspective of $\mathcal C_B$. $\endgroup$
    – Sean Sanford
    Jun 11, 2022 at 23:43

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