# How weird can Modular Tensor Categories be over non-algebraically closed fields?

I am trying to understand better the behaviour and character of modular tensor categories over non-algebraically closed fields. How weird can they be?

The reason I am interested in this is that my collaborators and I are investigating extended 3D tqfts. These are known to be related to Modular Tensor Categories via the Reshetikhin-Turaev construction. However most of the literature seems to be focused on the case where the MTC is defined over an algebraically closed field, usually the complex numbers.

I don't see a reason to restrict to this case and I can imagine that there could be some very interesting examples, and hence interesting invariants, in other cases to.

For example over field $k$, a finite semisimple linear category will have simple objects whose endomorphisms rings are division algebras over $k$. If $k$ is algebraically closed, then all we get are copies of $k$. If $k$ is not algebraically closed then it is more interesting as we can have objects with different division algebras as endomorphisms.

Can this happen in a modular tensor category? i.e. is there an example of a Modular Tensor Category such that the simple objects have different division algebras for their endomorphism rings? What if we drop the requirement End(1) = k?

How sticky can it get?

• An important point here is that the definition of modularity becomes a bit more complicated. The vector space associated to the torus is no longer generated in general by a basis given by the isomorphism classes of simple objects. So the S-matrix acts on a possibly larger space: the product of the centres of endomorphism algebras of the simple objects. – Jamie Vicary Jul 16 '14 at 14:18
• Concerning your: "What if we drop the requirement $End(1) = k$?" I have the feeling that one should define $k$ to be $End(1)$. So, by definition, you then always have $End(1) = k$, and the question is what happens when you take $k$ to be a ring that looks further and further from an algebraically closed field. – André Henriques Jul 16 '14 at 14:28
• Building on André's comment: I guess that a monoidal Ab-category is automatically $k$-linear for $k:=\text{End}(1)$. – Jamie Vicary Jul 16 '14 at 14:39
• Modularity should be thought of as being as far from symmetry as possible. In the symmetric case the S-matrix has rank 1, and in the modular case it has full rank. On the other hand, Rep(G) is contained in its center which is modular, so @DavidSpeyer's example still works. – Noah Snyder Jul 16 '14 at 20:34
• There's a bunch of great stuff related to Andre's comment in this short paper of Greg's. – Noah Snyder Jul 16 '14 at 20:42

## 2 Answers

An example of MTC is Drinfeld double of a finite group $G$ (over any field of characteristic zero). This category contains representation category of $G$ as a subcategory. So all endomorphisms rings that you can find in representations of finite groups, you can also find in MTC. For example the quaternions will show up in the Drinfeld double of the quaternion group.

• A recent result of Ehud Meir is that the Schur group constructed out of semisimple Hopf algebras is in fact the whole Brauer group, son one can get all CSAs if we allow for general Hopf algebras instead of just groups. – Mariano Suárez-Álvarez Jul 17 '14 at 1:55
• @VictorOstrik: what is the definition of modularity by which these representation categories are MTCs? – Jamie Vicary Jul 17 '14 at 8:50
• @JamieVicary: one definition would be that extension of scalars to algebraically closed field gives MTC. Another possibility is to require that Muger center is trivial. – Victor Ostrik Jul 17 '14 at 19:24
• @VictorOstrik: is it written down somewhere that these definitions are equivalent in this case, and that they imply the existence of a bona-fide 3d TQFT? – Jamie Vicary Jul 18 '14 at 8:56

Here is an beginning of an existence proof for anything nontrivial whatsoever.

Work over the field $\mathbb{R}$, and consider the algebra $A = \mathbb{R} \oplus \mathbb{H}$. Then $\text{Mod}(A)$ has two simple objects given by the $\mathbb{R}$ and $\mathbb{H}$ factors, which we name $I$ and $Q$ respectively. Then $\text{Hom}(I,I) = \mathbb{R}$ and $\text{Hom}(Q,Q) \simeq \mathbb{R}^4$. I think the following fusion algebra is realizable via a linear functor $\otimes : \text{Mod}(A) \boxtimes \text{Mod}(A) \to \text{Mod}(A)$, thanks to the fact that $\mathbb{H} \otimes_R \mathbb{H} = M_4(\mathbb{R})$ is Morita equivalent to $\mathbb{R}$:

$I \otimes I = I$

$I \otimes Q = Q$

$Q \otimes I = Q$

$Q \otimes Q = I \oplus I \oplus I \oplus I$

The hope is that this is a monoidal structure, and that $Q$ is a self-dual object. It looks very weird given those fusion rules! But note that $\text{Hom}(Q,Q) \simeq \text{Hom}(I,Q \otimes Q) \simeq \mathbb{R}^4$, so it is not ruled out that this could be a rigid monoidal structure on $\text{Mod}(A)$.

I have not checked whether this can be given an associator, so things could fail at that step. Finally, one could take the double to obtain a nontrivial modular tensor category.