Let $\mathcal{A}$ be a Kac algebra (Hopf C*-algebra), with comultiplication $\delta$, counit $\epsilon$ and antipode $S$.

The following definition comes from this paper (p51-52) of Izumi-Longo-Popa :

A finite dimensional unitary corepresentation $\pi$ is a pair of a finite dimensional Hilbert space $H_\pi$ and a linear map $\Gamma_\pi :H_\pi \longrightarrow H_\pi \otimes \mathcal{A}$ satisfying $$(\Gamma_\pi \otimes id) \cdot \Gamma_\pi =(id \otimes \delta)\cdot \Gamma_\pi$$ and the following unitarity condition: if $\{ e(\pi)_i\}$ is an orthonormal basis of $H_\pi$ and $$\Gamma_\pi(e(\pi)_j)=\sum_i e(\pi)_i\otimes u(\pi)_{i,j},$$ then $u(\pi)=(u(\pi)_{i,j})$ is unitary as an element in $M(d(\pi), \mathbb{C})\otimes \mathcal{A}$, where $d(\pi)$ is the dimension of $H_\pi$.

(...)

Let $\Xi$ be a complete system of representatives of the irreducible corepresentations of $\mathcal{A}$. Then the linear span of $\{u(\pi)_{i,j}\}_{1\leq i,j \leq d(\pi), \; \pi \in \Xi}$ is a dense in $\mathcal{A}$ in weak topology.

We deduce from the definition of corepresentation that $$ \delta(u(\pi)_{ij}) = \sum_{k} u(\pi)_{ik} \otimes u(\pi)_{kj}$$

We add the assumption that $\mathcal{A}$ is finite dimensional, then $$card \{u(\pi)_{i,j}\}_{1\leq i,j \leq d(\pi), \; \pi \in \Xi} = dim(\mathcal{A})$$ and so $\{u(\pi)_{i,j}\}_{1\leq i,j \leq d(\pi), \; \pi \in \Xi}$ is a linear basis of $\mathcal{A}$ (is it true in the $\infty$-dimensional case ?).

Now, by definition of the counit

$$\sum_{k} \epsilon(u(\pi)_{ik})u(\pi)_{kj} = \sum_{k} u(\pi)_{ik} \epsilon(u(\pi)_{kj}) = u(\pi)_{ij}$$

So, by linear independence: $\epsilon(u(\pi)_{ij}) = \delta_{ij}$.

But by definition of the antipode

$$\sum_{k} S(u(\pi)_{ik})u(\pi)_{kj} = \sum_{k} u(\pi)_{ik} S(u(\pi)_{kj}) = \epsilon(u(\pi)_{ij})\mathbf{1} = \delta_{ij}\mathbf{1} $$
Now by unitarity of $u(\pi)$, we also have
$$\sum_{k} u(\pi)_{ki}^{*}u(\pi)_{kj} = \sum_{k} u(\pi)_{ik} u(\pi)_{jk}^{*} = \delta_{ij}\mathbf{1} $$

From the closeness the these equalities, I'm led to ask:

Question: Is it true that $S(u(\pi)_{ij}) = u(\pi)_{ji}^{*}$ ?

whythe antipode behaves in this way is perhaps interesting (and, in some sense, forms the basis for thedefinitionof the antipode in more complicated analytic quantum group settings). $\endgroup$ – Matthew Daws Nov 20 '13 at 8:37