In Fusion Categories and Homotopy Theory, ENO attatch a 3groupoid to a fusion category. In the case of A graded vector spaces they further compute it's truncation as an orthogonal group $O(A \bigoplus A^*)$. There are also computations by Grossman and Snyder for examples that come from the AsaedaHaagerup subfactor. But I have not seen any descriptions for quantum groups at roots of unity. If some of these are already taken care of by quantum group coincidences in other contexts, that would also be appreciated.
1 Answer
As far as I know, no one has written this up, but I think you should be able to find the BrauerPicard groupoid for quantum groups at roots of unity by the following techniques. Now that I've written it down, there are a lot of gaps which would need to be filled in, several of which are not straightforward. This might well be a good project for a graduate student paper.
I'm going to assume you care about the kinds of roots of unity that appear in ReshetikhinTuraev, the general case would be quite a bit harder because the categories wouldn't be modular, but could probably be done with a bit more thought and some time with Steve Sawin's tables.
(Also, I'm not going to compute the more detailed problem that Pinhas and I consider of classifying all other categories Morita equivalent to the given one. Such a classification would be far too difficult because you would need the classification of "quantum subgroups" which is known only for SU(2), SU(3), and SU(4).)
Since C is modular, $Z(C) = C \boxtimes C^{op}$ where $C^{op}$ denotes the same category with the opposite braiding. By ENO the BrauerPicard group is the same as the group braided autoequivalences of $Z(C)$. We compute the braided autoequivalences of $C \boxtimes C^{op}$ in two steps: first combinatorially we want to figure out what such an autoequivalence can do on objects, and second we need to consider the case of an autoequivalence whose underlying functor is the identity.
Let me first sketch how this would work for SU(2). The objects in $C \boxtimes C^{op}$ are generated by $V \boxtimes 1$ and $1 \boxtimes V$ for $V$ the standard 2dimensional representation. We want to figure out what object $\mathscr{F}(V \boxtimes 1)$ can be. The only objects of the same dimension as these objects are of the form $g^i V \boxtimes g^j$ or $g^i \boxtimes (g^j V)$ where $g$ is the invertible object at the far end of the Weyl alcove. Just having the right dimensions for $\mathscr{F}(V \boxtimes 1)$ and $\mathscr{F}(1 \boxtimes V)$ isn't enough, you also need the object to be quaternionic (i.e. have the right parity), to have the right selfbraiding with itself, and to "commute" (i.e. have the same over and under braidings) with each other. Sorting this out requires some annoying futzing modulo 4, which I haven't done. By the universal property of SU(2) (cf.) it is not hard to check that once you have the above conditions satisfied then there's a unique braided autoequivalences realizing this.
The universal property let us skip the usual second step, which is to check that there are no nontrivial "guage automorphisms", which are tensor autoequivalences whose underlying functor is the identity. (See Lemma 5.3 and Remark 5.6.) This follows from the fact that the TemperleyLieb planar algebra has no automorphisms. It could also be done directly by a cohomological calculation following Liptrap's thesis. In conclusion, the answer for SU(2) is trivial if there are an even number of objects and the group of order 2 otherwise.
One can calculate the BP group of SU(2) in another way, by using the classification of quantum subgroups of SU(2) to directly calculate the BP groupoid. Invertible bimodules over C correspond to a pair of an indecomposable module category M and a tensor equivalence $C \cong M_C^*$. The indecomposable module categories over SU(2) at roots of unity are given by ADE Dynkin diagrams with the same Coxeter number. The dual is equivalent to $C$ only for type A and for the odd type D's. The second piece of data is a torsor for the group of outer autoequivalences of C. By a similar calculation as in the previous paragraph (but without braidings), you get an outer autoequivalence of C when $V$ and $gV$ have the same parity (so the number of objects is odd). So when there is an even number of simple objects you get the trivial group, when the number is 1 mod 4 you get $\mathbb{Z}/2\mathbb{Z}$ coming from an outer autoequivalence interchanging $V$ and $g \otimes V$, when the number is 3 mod 4 you get a group of order $4$ coming from both the outer automorphism and $D_{\text{odd}}$. It's not obvious to me which order four group you get in the last case.
In general things will be similar, but harder. For the combinatorial step, the permutation has to preserve the dimensions and the selfbraidings, which should already get you down to very few possibilities (typically they should all of which should look roughly like replacing the smallest rep $V$ by $g \otimes V$ for $g$ some invertible object in a corner of the Weyl alcove and possibly moving it to the other component). In general this step might be a bit tricky, for example in type B and D you have to worry about where the spin representations go, which won't be determined by where the standard rep goes and so might give you some extra options. Also in the cases where the Dynkin diagram has symmetry (A_n, D_4, E_6) you have some extra cases to consider. You also need to check whether there's actually an autoequivalence realizing each combinatorial possibility.
Second, you need to figure out in how many ways each of these combinatorial options is realized. Equivalently, you just need to know what the "guage automorphisms" are, that is the braided autoequivalences which act triviailly on objects. This can be done by a kind of "cohomology"like calculation following Jesse Liptrap's thesis, or by a "diagrammatic" argument following what Pinhas and I do. That is, any gauge automorphism gives a very special kind of automorphism of the "planar algebra" attached to the tensor category. In the nonexceptional types this comes down to studying automorphisms of the HOMFLY and BMW planar algebras, but in the exceptional cases I'm not sure how to do it.
I suppose I should also say a little bit about the higher structure. $\pi_2$ is just the invertible objects $Z(C)$, which is just the product of the invertible objects in $C$ with the invertibles in $C^{\text{op}}$, so is easy to work out. $\pi_3$ is always $\mathbb{C}^\times$. For the full structure you need to know quite a bit more, like the action of $\pi_1$ on $\pi_2$ (which is just a calculation of where the invertible objects go), and the Postnikov kinvariants. This could ve quite tricky.

$\begingroup$ For SU(2) all the BP groupoids for quantum subgroups have now been calculated in Cain EdieMichell's Ph.D. thesis: arxiv.org/abs/1709.04721 $\endgroup$ Nov 12, 2017 at 20:19