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In Walker and Wang's article about (3+1)-TQFTs from premodular categories, they say on page 14 that you can take a quotient of a premodular category $\mathcal{C}$ by its symmetric fusion subcategory $\mathcal{S_C}$ consisting of transparent objects. That quotient is promised to be modular.

I'm willing to believe this and I'd like to use it, but I can't find a mathematical reference for it, nor have I managed to work it out in all detail. Is this a known fact? Is it treated somewhere?

My attempt of making sense of it, is the following:

  • Take a premodular category $\mathcal{C}$. That's just a ribbon category which is fusion (finitely semisimple etc.).
  • Take all the transparent (trivially braiding) objects. They form a subcategory $\mathcal{S_C}$.
  • Form an "exact sequence" $0 \to \mathcal{S_C} \to \mathcal{C} \to \mathcal{Q_C} \to 0$. I have no idea what exactly an exact sequence of ribbon fusion categories is or whether we even have to consider this. I'm vaguely expecting something like this:
  • $\mathcal{Q_C}$ has the same objects as $\mathcal{C}$, but somehow the morphisms to and from the objects in $\mathcal{S_C}$ are identified with 0.

What I don't understand about it is:

  • How does it work exactly?
  • How do I prevent the monoidal unit to get killed? It's in $\mathcal{S_C}$ after all.
  • Are the simple objects in the quotient also simple in $\mathcal{C}$? Do they braid the same, i.e. is the quotient obviously modular?
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  • $\begingroup$ By the way, the certainly wrong approach is to take the category generated by the nontransparent objects. This is likely to contain a lot of transparent objects and won't be modular in general. $\endgroup$ – Manuel Bärenz Feb 25 '15 at 16:42
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Short answer: Look for papers on "deequivariantization". (I think the original references are by Müger and Brugières, but I am not sure whether they used the term "deequivariantization".)

Longer answer:

The procedure mentioned in the paper is actually a sort of predual to the deequivariantization procedure found in the literature, in the sense that $Rep(C/S) \cong Deeq(Rep(C))$. So it is very closely related to deequivariantization, but not exactly the same thing.

You attempt was on the right track. One starts with $C$, then adds isomorphisms between objects of $S$ and the trivial object. This has to be done carefully, of course. One uses the fact that $S \cong Rep(G)$ for some finite group $G$, and $Rep(G)$ has a fiber functor.

A fancier way of thinking of it goes as follows. Since $S$ is symmetric monoidal we can think of it as an $n$-category for any $n$, and in particular we can think of it as a 4-category. The 3-category $C$ then becomes a module 3-category for the 4-category $S$. We can construct another module 3-category $F$ for $S$ by using the isomorphism $S\cong Rep(G)$ and the fiber functor for $Rep(G)$. Now we can now define $$ C/S = C \otimes_S F $$

$F$ is actually a $S$-$G_4$ bimodule 3-category, where $G_4$ denotes the finite group $G$ thought of as a 4-category. It follows that $C/S$ has a $G$ action. (A $G$ action on a 3-category is the same thing as a $G_4$-module structure on that 3-category.)

I have a preprint which fills in the details of the above arguments, but it is not yet ready for the arXiv. If you go to this page and look at the slides from talks at Vienna and Princeton (Feb 2014), you can find some of the details.

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  • $\begingroup$ The article by Müger you mention must be arxiv.org/pdf/math/9812040v3.pdf, I guess. $\endgroup$ – Manuel Bärenz Feb 26 '15 at 16:54
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Kevin Walker Feb 26 '15 at 19:19
  • $\begingroup$ BTW: I had added a comment referring to this article of Müger, which I immediately deleted, because I was in a rush, had no time to read the question in detail and it seemed to me that the op implicitly knew the answer already: mathoverflow.net/questions/146315/… $\endgroup$ – Marcel Bischoff Feb 27 '15 at 16:27

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