# 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

1. […] 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. […]

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2. […] categories.  If you want to learn more about braidings, check out some of John Armstrong’s excellent posts over at The Unapologetic Mathematician, he is going into the nitty-gritty of it […]

Pingback by Chain complexes as Graded C[epsilon]-modules, part 3: Bicomplexes and Superalgebras « The Everything Seminar | July 4, 2007 | Reply

3. […] A closed category is a symmetric monoidal category where each functor has a specified right adjoint called an […]

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4. […] Braidings on Span 2-Categories Now that we can add a monoidal structure to our 2-category of spans, we want to add something like a braiding. […]

Pingback by Braidings on Span 2-Categories « The Unapologetic Mathematician | October 14, 2007 | Reply

5. […] I don’t think I’ve mentioned is that the category of vector spaces over a field is symmetric. Indeed, given vector spaces and we can define the “twist” map by setting and […]

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6. […] So what’s the tensor product? Well, we start with the vector space tensor product and try to find a cone. This should give an adjunction . So let’s read this as another logical statement. A linear map is positive (and thus in ) if Expanding this condition on , we get that is positive if But is the usual closure adjunction in the category of vector spaces, turning a function-valued function of one variable into a vector-valued function of two variables. And we want every positive map from to to correspond to exactly one in just this way. Thus the cone on that makes the tensor product for into a left adjoint to the exponential is that of all finite sums of tensor pairs of positive elements. That is, if with all the and positive in their respective cones. As an exercise, verify that this tensor product is also symmetric. […]

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