# The Unapologetic Mathematician

## Cocommutativity

One things I don’t think I’ve mentioned is that the category of vector spaces over a field $\mathbb{F}$ is symmetric. Indeed, given vector spaces $V$ and $W$ we can define the “twist” map $\tau_{V,W}:V\otimes W\rightarrow W\otimes V$ by setting $\tau_{V,W}(v\otimes w)=w\otimes v$ and extending linearly.

Now we know that an algebra $A$ is commutative if we can swap the inputs to the multiplication and get the same answer. That is, if $m(a,b)=m(b,a)=m\left(\tau_{A,A}(a,b)\right)$. Or, more succinctly: $m=m\circ\tau_{A,A}$.

Reflecting this concept, we say that a coalgebra $C$ is cocommutative if we can swap the outputs from the comultiplication. That is, if $\tau_{C,C}\circ\Delta=\Delta$. Similarly, bialgebras and Hopf algebras can be cocommutative.

The group algebra $\mathbb{F}[G]$ of a group $G$ is a cocommutative Hopf algebra. Indeed, since $\Delta(e_g)=e_g\otimes e_g$, 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 $B$ is not only monoidal, but it’s also symmetric! Indeed, given representations $\rho:B\rightarrow\hom_\mathbb{F}(V,V)$ and $\sigma:B\rightarrow\hom_\mathbb{F}(W,W)$, we have the tensor product representations $\rho\otimes\sigma$ on $V\otimes W$, and $\sigma\otimes\rho$ on $W\otimes V$. To twist them we define the natural transformation $\tau_{\rho,\sigma}$ to be the twist of the vector spaces: $\tau_{V,W}$.

We just need to verify that this actually intertwines the two representations. If we act first and then twist we find \begin{aligned}\tau_{V,W}\left(\left[\left[\rho\otimes\sigma\right](a)\right](v\otimes w)\right)=\tau_{V,W}\left(\left[\rho\left(a_{(1)}\right)\otimes\sigma\left(a_{(2)}\right)\right](v\otimes w)\right)\\=\tau_{V,W}\left(\left[\rho\left(a_{(1)}\right)\right](v)\otimes\left[\sigma\left(a_{(2)}\right)\right](w)\right)\\=\left[\sigma\left(a_{(2)}\right)\right](w)\otimes\left[\rho\left(a_{(1)}\right)\right](v)\end{aligned}

On the other hand, if we twist first and then act we find \begin{aligned}\left[\left[\sigma\otimes\rho\right](a)\right]\left(\tau_{V,W}(v\otimes w)\right)=\left[\sigma\left(a_{(1)}\right)\otimes\rho\left(a_{(2)}\right)\right]\left(w\otimes v\right)\\=\left[\sigma\left(a_{(1)}\right)\right](w)\otimes\left[\rho\left(a_{(2)}\right)\right](v)\end{aligned}

It seems there’s a problem. In general this doesn’t work. Ah! but we haven’t used cocommutativity yet! Now we write $a_{(1)}\otimes a_{(2)}=\Delta(a)=\tau_{B,B}\left(\Delta(a)\right)=\tau_{B,B}\left(a_{(1)}\otimes a_{(2)}\right)=a_{(2)}\otimes a_{(1)}$

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.

November 19, 2008 -

## 2 Comments »

1. […] 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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2. […] on in a way that is natural in to recover , and now we know this is equivalent to putting a cocommutative cogroup structure on […]

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