The Unapologetic Mathematician

Mathematics for the interested outsider

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.

November 12, 2008 Posted by | Algebra, Category theory, Representation Theory | 3 Comments

   

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