# The Unapologetic Mathematician

## Representations of Bialgebras

What’s so great about bialgebras? Their categories of representations are monoidal!

Let’s say we have two algebra representations $\rho:A\rightarrow\hom_\mathbb{F}(V,V)$ and $\sigma:A\rightarrow\hom_\mathbb{F}(W,W)$. These are morphisms in the category of $\mathbb{F}$-algebras, and so of course we can take their tensor product $\rho\otimes\sigma$. But this is not a representation of the same algebra. It’s a representation of the tensor square of the algebra:

$\rho\otimes\sigma:A\otimes A\rightarrow\hom_\mathbb{F}(V,V)\otimes\hom_\mathbb{F}(W,W)\cong\hom_\mathbb{F}(V\otimes W,V\otimes W)$

Ah, but if we have a way to send $A$ to $A\otimes A$ (an algebra homomorphism, that is), then we can compose it with this tensor product to get a representation of $A$. And that’s exactly what the comultiplication $\Delta$ does for us. We abuse notation slightly and write:

$\rho\otimes\sigma:A\rightarrow\hom_\mathbb{F}(V\otimes W,V\otimes W)$

where the homomorphism of this representation is the comultiplication $\Delta$ followed by the tensor product of the two homomorphisms, followed by the equivalence of $\hom$ algebras.

Notice here that the underlying vector space of the tensor product of two representations $\rho\otimes\sigma$ is the tensor product of their underlying vector spaces $V\otimes W$. That is, if we think (as many approaches to representation theory do) of the vector space as fundamental and the homomorphism as extra structure, then this is saying we can put the structure of a representation on the tensor product of the vector spaces.

Which leads us to the next consideration. For the tensor product to be a monoidal structure we need an associator. And the underlying linear map on vector spaces must clearly be the old associator for $\mathbf{Vect}(\mathbb{F})$. We just need to verify that it commutes with the action of $A$.

So let’s consider three representations $\rho:A\rightarrow\hom_\mathbb{F}(U,U)$, $\sigma\rightarrow\hom_\mathbb{F}(V,V)$, and $\tau:A\rightarrow\hom_\mathbb{F}(W,W)$. Given an algebra element $a$ and vectors $u$, $v$, and $w$, we have the action

\begin{aligned}\left[\left[(\rho\otimes\sigma)\otimes\tau\right](a)\right]((u\otimes v)\otimes w)=\\\left(\left[\rho\left(\left(a_{(1)}\right)_{(1)}\right)\right](u)\otimes\left[\sigma\left(\left(a_{(1)}\right)_{(2)}\right)\right](v)\right)\otimes\left[\tau\left(a_{(2)}\right)\right](v)\end{aligned}

On the other hand, if we associate the other way we have the action

\begin{aligned}\left[\left[\rho\otimes(\sigma\otimes\tau)\right](a)\right](u\otimes(v\otimes w))=\\\left[\rho\left(a_{(1)}\right)\right](u)\otimes\left(\left[\sigma\left(\left(a_{(2)}\right)_{(1)}\right)\right](v)\otimes\left[\tau\left(\left(a_{(2)}\right)_{(2)}\right)\right](v)\right)\end{aligned}

Where we have used the Sweedler notation to write out the comultiplications of $a$. But now we can use the coassociativity of the comultiplication — along with the fact that, as algebra homomorphisms, the representations are linear maps — to show that the associator on $\mathbf{Vect}(\mathbb{F})$ intertwines these actions, and thus acts as an associator for the category of representations of $A$ as well.

We also need a unit object, and similar considerations to those above tell us it should be based on the vector space unit object. That is, we need a homomorphism $A\rightarrow\hom_\mathbb{F}(\mathbb{F},\mathbb{F})$. But linear maps from the base field to itself (considered as a one-dimensional vector space) are just multiplications by field elements! That is, the $\hom$ algebra is just the field $\mathbb{F}$ itself, and we need a homomorphism $A\rightarrow\mathbb{F}$. This is precisely what the counit $\epsilon$ provides! I’ll leave it to you to verify that the left and right unit maps from vector spaces intertwine the relevant representations.

November 11, 2008