# Symmetries of module categories over the category of representations of quantum $sl(2)$

The category $\mathcal{C}_l$ of tilting modules of the quantum group $U_q(sl_2)$ quotiented out by the modules of zero quantum dimension has a natural structure as a semisimple monoidal category when the quantum parameter $q$ is a root of unity. The module categories of $\mathcal{C}_l$ (or equivalently, quantum subgroups of $U_q(sl_2)$) are classified by ADE diagrams. See for example, section 6 of Ostrik.

Let $K_0(\mathcal{C_l})$ be the Grothendieck ring of $\mathcal{C}_l$. The action of $\mathcal{C}_l$ on a module category by tensor product on the objects defines a based module $M$ over $K_0(l)$. The generating element of $K_0(\mathcal C_l)$ acts as a matrix $A$ on $M$, where $A$ is the adjacency matrix of the corresponding ADE Dynkin diagram.

The ADE Dynkin diagrams have graph symmetries: the A-type and E_6 graphs can have the order of the vertices reversed, the D-type graphs can switch the two short legs, and the D_4 graph exhibits a full $S(3)$ symmetry. All of these symmetries induce based module symmetries on the corresponding based module $M$.

In what ways can the symmetries on the based modules be extended to symmetries of the corresponding module categories? Can you construct module functors from the module category to itself that induce this symmetry on the based modules? If not, is there some lesser structure than a module category but greater than a based module that can be preserved under these symmetries?

Let $\mathcal C$ be the $\mathrm{SU(2)}_k$ modular tensor category and let $\mathcal M$ be a module category. Ostrik showed in the same article the OP mentioned that there is an algebra $A$ in $\mathcal C$, such that $\mathcal M\cong \mathcal C_A$ as (left) module categories. Let $\mathcal D=\mathrm{Fun}_\mathcal{C}(\mathcal M,\mathcal M)$ be the dual category, which is equivalent with the bimodule category ${}_A\mathcal C_A$.

Now by definition every invertible object in $\mathcal D$, i.e. every object in the Picard group $\mathrm{Pic}(\mathcal D)$, gives a symmetry of the module category $\mathcal M$ and also of its based ring. Then it is easy to check in the literature (see below) that all symmetries of $G=A,D,E$ Dynkin diagrams correspond to invertible elements of $\mathcal D$. Namely, one simply has to count that there are exactly $|\mathrm{Aut}(G)|$ invertible objects, *edit*: (see Noah Snyder's comment) acting "faithful" on the Grothendieck ring. Thus all symmetries of the Dynkin diagram are symmetries of corresponding module category

Much of the data of the dual categories for $\mathrm{SU}(2)_k$ has been calculated in the subfactor literature. See for example Section 6.2 of Jens Böckenhauer, and David E. Evans, Modular invariants, graphs and $\alpha$-induction for nets of subfactors. III., Comm. Math. Phys. 205 (1999), no. 1, 183–228. and Section 6 of Jens Böckenhauer, David E. Evans, and Yasuyuki Kawahigashi, Chiral structure of modular invariants for subfactors, Comm. Math. Phys. 210 (2000), no. 3, 733--784..

Remark. The OP ask for quantum $sl_2$, but by Proposition 8.2.6 in Fröhlich, Jürg(CH-ETHZ-P); Kerler, Thomas(1-HRV) Quantum groups, quantum categories and quantum field theory. Lecture Notes in Mathematics, 1542. Springer-Verlag, Berlin, 1993. viii+431 pp. ISBN: 3-540-56623-6 it follows that all modular tensor categories of $\mathrm{SU(2)}_k$ type are braided equivalent.

• You need to be a little careful, just counting the automorphisms isn't enough as you might have that some of them act trivially on the Grothendieck group. – Noah Snyder Jul 22 '15 at 16:04
• Thanks Noah, that's right, still in the case at hand one can verify that this does not happen. Do you have an example where this actually happens? I was thinking that the main result in arxiv.org/abs/1004.4725 might imply that this can actually not happen (just some physical intuition about CFT, though). – Marcel Bischoff Jul 22 '15 at 16:26
• If your category is Vec(G) for a finite group G, then any cohomology class $H^2(G,\mathbb{C}^\times)$ will give you a tensor autoequivalence of Vec(G) whose underlying functor is trivial. (You just use the cohomology class to define the natural tranformation $1_{gh} = \mathrm{id}(1_g \otimes 1_h) \rightarrow \mathrm{id}(1_g \otimes 1_h) = 1_{gh}$.) – Noah Snyder Jul 22 '15 at 17:06

Marcel's answer is great if you're familiar with the literature, but let me try to give a more direct argument.

Recall that for $A_n$, $D_{2n}$,$E_6$, and $E_8$ the module category actually is itself a tensor category with the action given by applying a tensor functor and then tensoring on the left. So suppose we have $\mathcal{F}:\mathcal{C} \rightarrow \mathcal{D}$ a tensor functor making $\mathcal{D}$ into a module category for $\mathcal{C}$. Clearly tensoring on the right by any object in $\mathcal{D}$ gives a module endofunctor, and tensoring on the right with an invertible object gives a module autoequivalence. For the 2-fold symmetry of $A_n$, the 3-fold symmetry of $D_4$ and the $2$-fold symmetry of $E_6$ it's easy to see that the Dynkin diagram automorphism is just given by tensoring on the right with an invertible object.

The harder case to deal with is the 2-fold symmetries of $D_{2n}$ and of $D_{2n+1}$. In the former case, there's no invertibles around to tensor with, and in the latter case the module category doesn't come from a tensor product.

In the odd case there's a nice trick. If $\mathcal{C}$ is a braided tensor category and $\mathcal{M}$ is a module category, then you can use the braiding to make tensoring with an object $X$ into a module endofunctor. If you take $X$ to be the nontrivial invertible object in quantum SU(2), then a quick combinatorial check shows that tensoring with it implements the diagram involution. In the even case, the above trick doesn't work since tensoring with the nontrivial invertible object in quantum SU(2) fixes the two ends of the fork.

In both the odd and even cases for type D, everything is governed by $\mathbb{Z}/2\mathbb{Z}$ so you can just do the calculation directly. Recall that $D_{2n}$ is realized as a module category as the category of right modules for the algebra $1+X$ (where X is the nontrivial invertible object). It's easy to see that the two ends of the fork correspond to two different module structures on the middle object. As Marcel said, any module endofunctor is then given by tensoring on the right with some A-A bimodule. Typically the only invertible A-A bimodules have underlying object A itself ($D_4$ has extras, we'll just not use them), and then it's just a $\mathbb{Z}/2\mathbb{Z}$ fact that there are exactly two such bimodules. We want to know if tensoring with the nontrivial one switches or leaves fixed the two mod-A structures on the middle object. This can be rephrased just in terms of the group $G=\mathbb{Z}/2\mathbb{Z}$, and so you can see that yes it switches them. I feel like there should be a cleaner way to see this though.