## Cocommutativity

One things 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 extending linearly.

Now we know that an algebra is commutative if we can swap the inputs to the multiplication and get the same answer. That is, if . Or, more succinctly: .

Reflecting this concept, we say that a coalgebra is *co*commutative if we can swap the outputs from the comultiplication. That is, if . Similarly, bialgebras and Hopf algebras can be cocommutative.

The group algebra of a group is a cocommutative Hopf algebra. Indeed, since , we can twist this either way and get the same answer.

So what does cocommutativity buy us? It turns out that the category of representations of a cocommutative bialgebra is not only monoidal, but it’s also symmetric! Indeed, given representations and , we have the tensor product representations on , and on . To twist them we define the natural transformation to be the twist of the vector spaces: .

We just need to verify that this actually intertwines the two representations. If we act first and then twist we find

On the other hand, if we twist first and then act we find

It seems there’s a problem. In general this doesn’t work. Ah! but we haven’t used cocommutativity yet! Now we write

Again, remember that this doesn’t mean that the two tensorands are always equal, but only that the results after (implicitly) summing up are equal. Anyhow, that’s enough for us. It shows that the twist on the underlying vector spaces actually does intertwine the two representations, as we wanted. Thus the category of representations is symmetric.

[…] let’s say we have a group . This gives us a cocommutative Hopf algebra. Thus the category of representations of is monoidal — symmetric, even — […]

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