The Unapologetic Mathematician

Mathematics for the interested outsider

The Braided Coherence Theorem

I find myself up earlier than expected this morning, so I thought I’d bang out the proof I’d promised of the “coherence theorem” for braided monoidal categories.

Recall that we’re considering a braided monoidal category \mathcal{M} with underlying “ordinary category” \mathcal{M}_0 — the same category, just forgetting that it’s braided and monoidal. Then the statement is that the category of braided monoidal functors from \mathcal{B}raid to \mathcal{M} is equivalent to \mathcal{M}_0.

To make our lives a bit easier, we’ll use the strictification theorem to pass from the (weak) monoidal category \mathcal{M} to a monoidally equivalent strict monoidal category \mathcal{S}, with underlying category \mathcal{S}_0. Now the braiding on \mathcal{M} induces (via the equivalence) one on \mathcal{S}. The new statement is that the category of strictly monoidal functors from \mathcal{B}raid to \mathcal{S} is isomorphic to \mathcal{S}_0. Then the result we really want (about \mathcal{M}) will follow.

One direction is easy: we take a functor F and evaluate it at the object 1 to get an object F(1)\in\mathcal{S}_0. A monoidal natural transformation \eta:F\rightarrow G has, in particular, the component \eta_1:F(1)\rightarrow G(1), which makes “evaluate at 1” a functor.

For the other direction, we’ll pick an object S of \mathcal{S}_0 and use it to construct a strictly braided monoidal functor from \mathcal{B}raid to \mathcal{S} that sends 1 to S, and then show that an arrow f:S\rightarrow S' in \mathcal{S}_0 induces a natural transformation between the functors built on S and S'. We’ll see that everything in sight is unique, so this construction actually provides an “on the nose” inverse functor to “evaluate at 1“.

So, we’ve got an object S and we set F_S(1)=S. Then for F to be monoidal we must set F(n)=S^{\otimes n}, since n is 1^{\otimes n} (remember that the “tensor power” is defined just like regular exponentiation: take the tensor product of n copies of the object in question). This completely defines the functor F_S on objects.

Now any morphism in \mathcal{B}raid is an element of some braid group B_n. Let’s look at simply crossing the left strand over the right in B_2: this is \beta_{1,1}, so to have a strictly braided functor we must set F_S(\beta_{1,1})=\beta_{S,S}. In B_n, the crossings of one strand over the one to its right generate the whole group. If we cross strand i over strand i+1 we’ll call the generator \sigma_i. In terms of the category \mathcal{B}raid, we can write \sigma_i as 1_{i-1}\otimes\beta_{1,1}\otimes1_{n-i-1}, so monoidality forces us to set F_S(\sigma_i)=1_{S^{\otimes i-1}}\otimes\beta_{S,S}\otimes1_{S^{\otimes n-i-1}}.

We do have to check that the relations of the braid group are satisfied. But if |i-j|>1 then we have F_S(\sigma_i\sigma_j)=F_S(\sigma_j\sigma_i) because the monoidal product in \mathcal{S} is a functor. And we see that F_S(\sigma_i\sigma_{i+1}\sigma_i)=F_S(\sigma_{i+1}\sigma_i\sigma_{i+1}) because this is exactly an instance of the triangle relation for the braiding on \mathcal{S}!

So once we pick the single object S=F_S(1) everything else about F_S is uniquely determined. If we have an arrow f:S\rightarrow S' then we must set the component \eta_1 of the induced natural transformation to be f:F_S(1)\rightarrow F_{S'}(1). Then monoidality forces \eta_n:F_S(n)\rightarrow F_{S'}(n) to be f^{\otimes n}. The construction is indeed a functor sending \mathcal{S}_0 to the category of strictly monoidal functors from \mathcal{B}raid to \mathcal{S}, and it is clearly inverse to evaluation at 1.

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July 6, 2007 - Posted by | Category theory, Knot theory

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