# The Unapologetic Mathematician

## Braidings and Symmetries

So we’ve got monoidal categories that categorify the notion of monoids. In building up, we had to weaken things, refusing to talk about “equalities” of objects. Instead we replaced them with (natural) isomorphisms and then talked about equalities of morphisms. Now we’re ready to specialize a bit.

Many monoids we’re interested in are commutative: $ab=ba$ for all elements $a$ and $b$ of our monoid $M$. So what does this look like up at the category level? We learned our lesson with associativity, so we already know we should look for a natural isomorphism with components $\beta_{A,B}:A\otimes B\rightarrow B\otimes A$ to swap objects around the monoidal product.

Of course, there’s a coherence condition:

If we start with $(A\otimes B)\otimes C$ and want to get to $B\otimes (C\otimes A)$, we could pull $A$ past $B$, associate, and then pull $A$ past $C$. We could also associate, pull $A$ past $B\otimes C$, and associate again. And, naturally, we want these two to be the same. There’s another coherence condition we’ll need, but first let’s notice something.

Since $\beta_{A,B}:A\otimes B\rightarrow B\otimes A$ is an isomorphism, there’s an inverse $\beta_{A,B}^{-1}:B\otimes A\rightarrow A\otimes B$. Of course, we also have $\beta_{B,A}:B\otimes A\rightarrow A\otimes B$ sitting around. Now the obvious thing is for these two to be the same, but actually that’s not what we want. In fact, there are very good reasons to allow them to be different. So in general we’ll have two different ways of pulling $A$ past $B$, either with a $\beta$ or with a $\beta^{-1}$, and both of them should satisfy the hexagon identity shown above. That’s our second coherence condition.

We call such a natural transformation satisfying (both forms of) the hexagon identity a “braiding”, and a monoidal category equipped with such a “braided monoidal category”, for a very good reason I’ll talk about tomorrow (hint). Now if by chance the braiding $\beta$ is its own inverse, we call it a “symmetry”, and call the category a “symmetric monoidal category”.

We say that a monoidal functor $F:\mathcal{C}\rightarrow\mathcal{D}$ is braided (symmetric) if its monoidal structure plays well with both braidings. That is, if $F_{(B,A)}\circ\beta_{F(A),F(B)}=F(\beta_{A,B})\circ F_{(A,B)}$. Notice that if $F$ is strictly monoidal this just says that $F(\beta_{A,B})=\beta_{F(A),F(B)}$.

Both of these kinds of categories give back commutative monoids when we decategorify. This tells us that even though commutativity is relatively straightforward down at the level of sets, categorification lets us tease apart a subtle distinction and get a better insight into what’s “really going on”.

We know that any category with finite products (or coproducts) can use them as a monoidal structure. As an exercise, use the universal properties of products to come up with a braiding, then show that it’s symmetric.

Another exercise: show that any braiding satisfies $\rho_A=\lambda_A\circ\beta_{A,I}$ and $\lambda_A=\rho_A\circ\beta_{I,A}$. That is, when we have a braiding the left and right units are determined by each other and the braiding.

July 2, 2007 Posted by | Category theory | 6 Comments