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 $S(u(\pi)_{ij}) = u(\pi)_{ji}^{*}$ ?


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    $\begingroup$ Yes! Maybe look at Timmermann's book, "An invitation to quantum groups and duality.", as it's an good introduction, and nicely motivates from the analytic theory from algebraic considerations. But what you ask is true for general compact quantum groups. $\endgroup$ – Matthew Daws Nov 19 '13 at 20:30
  • $\begingroup$ @MatthewDaws : thank you ! I looked in this book and I found it on page 71 : proposition 3.1.7 v) (e5). $\endgroup$ – Sebastien Palcoux Nov 19 '13 at 22:46
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    $\begingroup$ The argument is just the uniqueness of the inverse of $u(\pi) \in M(d(\pi),\mathbb{C}) \otimes \mathcal{A}$. $\endgroup$ – Sebastien Palcoux Nov 19 '13 at 23:09
  • $\begingroup$ Sure! But why the antipode behaves in this way is perhaps interesting (and, in some sense, forms the basis for the definition of the antipode in more complicated analytic quantum group settings). $\endgroup$ – Matthew Daws Nov 20 '13 at 8:37
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    $\begingroup$ Awesome! Timmermann's book page 77 : the tensor product of corepresentation operators $X$ and $Y$, i.e. $X \boxtimes Y = X_{[13]}.Y_{[23]}$, is a natural quantum generalization of the product $g.h$ in a group ! $\endgroup$ – Sebastien Palcoux Nov 20 '13 at 17:38

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