The Unapologetic Mathematician

Mathematics for the interested outsider

Braided Monoidal Categories with Duals

As one last example along these lines, let’s throw all these structures in together. We start with a monoidal category, and we want it to have both a braiding and duals. Naturally, they’ll have to play well together.

Now I’ll want to go back and tweak one of the things I said about duals. I insisted that the duality functor satisfy (M^*)^*\cong M, but we can actually weaken it slightly. What I defined before we can call a “right dual”, since it provides a pairing M\otimes M^*\rightarrow\mathbf{1}. Instead of insisting that (M^*)^*\cong M, we can put in “left duals” by hand. This will be another contravariant functor M\rightarrow {}^*M, and arrows \overline{\epsilon_M}:{}^*M\otimes M\rightarrow\mathbf{1} and \overline{\eta_M}:\mathbf{1}\rightarrow M\otimes{}^*M satisfying the “mirror images” of the identities for \eta and \epsilon.

What does this all have to do with a braiding? Well we can use a braiding to turn a left dual into a right dual and vice versa. Let’s say we’ve got (right) dual maps \eta_M:\mathbf{1}\rightarrow M^*\otimes M and \epsilon_M:M\otimes M^*\rightarrow\mathbf{1}, along with braiding maps \beta_{M,N}:M\otimes N\rightarrow N\otimes M. Now we can define left dual maps by \overline{\eta_M}[\beta_{M,M^*}^{-1}\circ\eta_M]:\mathbf{1}\rightarrow M\otimes M^* and \overline{\epsilon_M}[\epsilon_M\circ\beta_{M^*,M}]:M^*\otimes M\rightarrow\mathbf{1}. This will automatically satisfy the left dual axioms. You can either try to show this now using the algebra or hold off a bit until we have a better tool to attack such identities.

Okay, so a braided monoidal category with duals has a braiding and has right duals. Since we define left duals in terms of these structures they’ll automatically play well together. From these we can further build a morphism for each object called the “balancing”. It’s defined by \theta_M=(1_M\otimes\epsilon_M)\circ(\beta_{M,M}\otimes1_M)\circ(1_M\otimes\overline{\eta_M}). We call an object M “unframed” if \theta_M=1_M.

I know that there’s a lot of stuff here and it’s hard to remember it all. We need a better way to think about these things. Just like we had for braided categories and categories with duals, we have a diagrammatic rendering of the free braided monoidal category with duals on one object.

First I’ll deal with the free braided monoidal category with duals on one self-dual, unframed object. That is, we start with one object and say it’s its own (left and right) dual and for whom the balancing is the identity. Then we build all the other objects as tensor powers of this one. Then we throw in the braiding morphisms and the duality morphisms and insist they satisfy all the relevant relations. And what we get is the category \mathcal{T}ang of tangles. Here’s an example:
Sample Tangle

As usual we read this from bottom to top. Notice that this means we can read off the algebraic notation just by going top to bottom, left to right, to get (\beta_{1,1}\otimes\epsilon_1)\circ(1_1\otimes\beta_1\otimes1_1)\circ(\beta_1\otimes\eta_1):2\rightarrow2. The unframed condition tells us that we can untwist that loop to the right and still have the same tangle.

What if we didn’t ask that our generator be unframed? Then we get the category \mathcal{F}r\mathcal{T}ang of framed tangles, which differs only in that we can’t untwist loops like the one above. It’s like working with knot diagrams where all we can use are Reidemeister moves 2 and 3.

What if our generator is unframed, but isn’t self-dual? Then we get the category \mathcal{OT}ang of oriented tangles. All we do now is label each strand of our tangle diagram with an orientation, like we did in the case of oriented Temperley-Lieb diagrams.

What if our generator is neither unframed nor self-dual? Then we get the category \mathcal{F}r\mathcal{OT}ang of framed, oriented tangles.

Each of these leads to a diagrammatic way of looking at the respective sorts of categories. For example, go back and take the definition of left duals in terms of right duals and write it out as a diagram of framed, oriented tangles. Then write down the condition it must satisfy as an equation between framed oriented tangles. Finally, verify that it satisfies the equation by finding a sequence of Reidemeister moves 2 and 3 from one side to the other (and invoke the definition of right duals at some point).

The category of tangles is very simply described in combinatorial and algebraic terms, but you might not yet have noticed how general it is. To illustrate this, I give with no further comment an example of a tangle from {}0 to {}0:
Tangle from 0 to 0

July 13, 2007 Posted by | Category theory, Knot theory | 2 Comments