The Unapologetic Mathematician

The Category of Braids

Before I jet off for a week of knot homology — where I’m looking forward to seeing Scott Morrison, Jacob Rasmussen, and Ben Webster of the Secret Blogging Seminar — I thought I’d finally start tying in (sorry) this category theory stuff with knot theory.

As I said yesterday, the analogue of commutativity for a monoidal category is called a braiding. The name comes from a deep connection between braided monoidal categories and braid groups. Specifically, there is a (strict) braided monoidal category of braids: $\mathcal{B}raid$. The objects of this category are the natural numbers, and $\hom_{\mathcal{B}raid}(m,n)$ is the group $B_n$ if $m=n$ and empty otherwise. We think of a braid with $n$ strands as taking a collection of $n$ points and shuffling them around, and we draw a diagram like this:

As a side note, I always intend diagrams like these to be read from bottom to top, though other authors go the other way.

First, the monoidal structure. We don’t need any of the structural isomorphisms because we’re making a strict monoidal structure, but we do need a bifunctor $\underline{\hphantom{X}}\otimes\underline{\hphantom{X}}$. On the objects it’s just addition. For morphisms we need a way of taking a braid with $m$ strands and one with $n$ strands and making a braid with $m+n$ strands. We’ll just stand both of our braids next to each other like this:

Now we need a braiding. That is, for each pair of objects $(m,n)$ we need a morphism (a braid) from $m+n$ to $n+m$ that “commutes” with all the other morphisms. We’ll just pass the $m$ strands from the left over the $n$ strands from the right. Here’s what I mean for $m=3$ and $n=2$:

This is natural because if we stick on a braid below the leftmost three strands on the bottom, the Reidemeister moves will let us pull it up and over the other two strands until it’s sitting on top of the three strands at the right on the top. Similarly, a braid can be pulled along the undercrossing strands. Since our monoidal structure is strict, the hexagon identities degenerate into triangles, and the proof is just the following diagram:

So we indeed have a strict braided monoidal category. It turns out to be extremely important because of the following theorem of Joyal and Street:

If $\mathcal{M}$ is a braided monoidal category with underlying category $\mathcal{M}_0$ (forget the monoidal structure and the braiding to just have a regular old category), then the category of braided monoidal functors from $\mathcal{B}raid$ to $\mathcal{M}$ is equivalent to $\mathcal{M}_0$.

This is sort of like a coherence theorem for braided monoidal categories: for each natural isomorphism built from $\alpha$, $\lambda$, $\rho$, and $\beta$ there is an “underlying braid”, and two such isomorphisms are equal if and only if they have the same underlying braid. I’ll defer the proof of this for now, but you should think about it a bit. The details aren’t that bad, but the basic idea just leaps out at you after a bit.