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: .
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.