# The Unapologetic Mathematician

## Representations of Hopf Algebras I

We’ve seen that the category of representations of a bialgebra is monoidal. What do we get for Hopf algebras? What does an antipode buy us? Duals! At least when we restrict to finite-dimensional representations.

Again, we base things on the underlying category of vector spaces. Given a representation $\rho:H\rightarrow\hom_\mathbb{F}(V,V)$, we want to find a representation $\rho^*:H\rightarrow\hom_\mathbb{F}(V^*,V^*)$. And it should commute with the natural transformations which make up the dual structure.

Easy enough! We just take the dual of each map to find $\rho(h)^*:V^*\rightarrow V^*$. But no, this can’t work. Duality reverses the order of composition. We need an antiautomorphism $S$ to reverse the multiplication on $H$. Then we can define $\rho^*(h)=\rho(S(h))^*$.

The antiautomorphism we’ll use will be the antipode. Now to make these representations actual duals, we’ll need natural transformations $\eta_\rho:\mathbf{1}\rightarrow\rho^*\otimes\rho$ and $\epsilon_\rho:\rho\otimes\rho^*\rightarrow\mathbf{1}$. This natural transformation $\epsilon$ is not to be confused with the counit of the Hopf algebra. Given a representation $\rho$ on the finite-dimensional vector space $V$, we’ll just use the $\eta_V$ and $\epsilon_V$ that come from the duality on the category of finite-dimensional vector spaces.

Thus we find that $\epsilon_\rho$ is the pairing $v\otimes\lambda\mapsto\lambda(v)$. Does this commute with the actions of $H$? On the one side, we calculate

\begin{aligned}\left[\left[\rho\otimes\rho^*\right](h)\right](v\otimes\lambda)=\left[\rho\left(h_{(1)}\right)\otimes\rho^*\left(h_{(2)}\right)\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\otimes\rho\left(S\left((h_{(2)}\right)\right)^*\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\right](v)\otimes\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\end{aligned}

Then we apply the evaluation to find

\begin{aligned}\left[\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)=\lambda\left(\left[\rho\left(S\left(h_{(2)}\right)\right)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)\right)\\=\lambda\left(\left[\rho\left(h_{(1)}S\left(h_{(2)}\right)\right)\right](v)\right)=\lambda\left(\left[\rho\left(\mu\left(h_{(1)}\otimes S\left(h_{(2)}\right)\right)\right)\right](v)\right)\\=\lambda\left(\left[\rho\left(\iota\left(\epsilon(h)\right)\right)\right](v)\right)=\epsilon(h)\lambda(v)\end{aligned}

Which is the same as the result we’d get by applying the “unit” action after evaluating. Notice how we used the definition of the dual map, the fact that $\rho$ is a representation, and the antipodal property in obtaining this result.

This much works whether or not $V$ is a finite-dimensional vector space. The other direction, though, needs more work, especially since I waved my hands at it when I used $\mathbf{FinVect}$ as the motivating example of a category with duals. Tomorrow I’ll define this map.