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A semisimple braided category with duals is called modular when a certain matrix $S$ is invertible. The components $S_{AB}$ are indexed by (isomorphism classes of) simple objects of the category and one computes $S_{AB}$ by colouring the Hopf link with (representants of) $A$ and $B$ and evaluates the resulting diagram. One can show that a category is modular iff there are no "transparent" objects (objects that braid trivially with every other object) besides the monoidal unit.

Is being modular a specific property, say in the category of braided categories? Is being a modular category equivalent to being the limit of some diagram or satisfying some diagram?

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The Drinfeld center $Z(C)$ of a braided category "contains" $C$ and $\bar C$ (the category with opposite braiding) and therefore also $C\boxtimes \bar C$ and you can show that the following is equivalent

1) $C$ is modular

2) $Z(C)$ is equivalent with $C\boxtimes \bar C$.

In other words, for a braided category $C$ there is a natural notion of a center $Z(C)$ and a natural embedding $C\boxtimes \bar C$ in $Z(C)$ which is an equivalence if and only if $C$ is modular.

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In general, taking a "center" of a higher category is looking at the endomorphisms of the identity functor. For example, if you think of a monoid as a 1-category, then the endomorphisms of the identity functor are exactly the center of the monoid. Thinking of a tensor category as a 2-category with one object this also gives the Drinfeld center.

If you think of your braided tensor category as a 3 category with one object and one morphism, then this "center" construction yields a symmetric tensor category which is exactly the subcategory of transparent objects! (This is sometimes called the "Mueger center" to distinguish it from the Drinfeld center.) So modularity is just saying that the center of the braided tensor category is trivial.

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    $\begingroup$ The 2-morphisms of the 3cat are the objects of the BTC and the 3-morophisms of the 3cat are the morphisms of the BTC. Vertical composition gives you the tensor product, and the Eckmann-Hilton argument gives you the braiding. The first place I know of the Mueger center being called a center is arxiv.org/abs/math/0111205 but as explained in the introduction the idea goes back further. $\endgroup$
    – Noah Snyder
    Feb 16, 2014 at 18:15
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    $\begingroup$ You just do EH... $x \otimes y \rightarrow (x \star 1) \otimes (1 \star y) \rightarrow (x \otimes 1) \star (1 \otimes y) \rightarrow x \star y$ and then do the same thing once more to end up at $y \otimes x$. See also cheng.staff.shef.ac.uk/degeneracy/eggclock.pdf $\endgroup$
    – Noah Snyder
    Feb 16, 2014 at 22:05
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    $\begingroup$ As a warmup you might want to think through why a 2-category with one object and one 1-morphism is automatically a commutative monoid by EH. $\endgroup$
    – Noah Snyder
    Feb 17, 2014 at 3:59
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    $\begingroup$ Sorry, I have said it incorrectly in my last comment. This direction is clear. My problem is going from a braided monoidal category to a 3-category. $\endgroup$
    – Manuel Bärenz
    Feb 19, 2014 at 14:35
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    $\begingroup$ @Turion: Ok. That direction is essentially a coherence theorem for braided monoidal categories. Making that rigorous is going to be tricky, since making anything about 3-categories rigorous is tricky, but morally it just comes down to Artin's presentation for the braid group. That is, you need to check that braided tensor categories have an action of braids on them. $\endgroup$
    – Noah Snyder
    Feb 19, 2014 at 18:26

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