Is "being a modular category" a universal or categorical/algebraic property? 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?
 A: 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.
A: 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.
