Norm of contragredient of unitary representations of compact quantum groups Maybe the answer to the following question is known but I am unable to find it in the literature. Anyway, let me begin my question by fixing some notations and terms.
Let $G = (A, \Delta)$ be a compact quantum group in the sense of Woronowicz. For a finite dimensional unitary representation $U \in B(H) \otimes A$ of $G$, the contragredient $U^c$ of $U$ is the representation $(j \otimes \operatorname{Id})(U^*) \in B\bigl(\overline{H}\bigr) \otimes A$ of $G$, where the carrier space $\overline{H}$ is the conjugate Hilbert space of the finite dimensional Hilbert space $H$, and $j : B(H) \to B\left(\overline{H}\right)$ the $\ast$-anti-isomorphism sending $T$ to $\overline{T^*}$, the latter denotes the operator $\overline{\xi} \mapsto \overline{T^* \xi}$ on $\overline{H}$. Define $c(G)$ to be the supremum of all $\| U^c \|$, where $U$ runs through all finite dimensional unitary representations (of course, irreducible ones suffice) of $G$.
It is clear that if $G$ is of Kac-type, then $c(G) = 1$, as $U^c$ in the above is always unitary. With some effort, I can prove that $c(G) = +\infty$ for $G = SU_q(2)$ with $-1 < q \ne 0 < 1$, but whether $c(G) = +\infty$ for general non-Kac type $G$ seems more delicate, which prompts me into asking the following
Question. Does $c(G) < +\infty$ imply $G$ being of Kac type?
As there are already many results concerning characterization of Kac type compact quantum groups, it may well be possible that this question is already settled, in which case, I appreciate a reference to the literature.
 A: Yes, $c(G) < +\infty$ implies that $G$ is of Kac type. As far as I know, this result is not in the literature, but can be proven as follows.
Let $h$ be the Haar state. For every irreducible unitary representation $u \in M_n(\mathbb{C}) \otimes A$, there is a unique positive invertible $Q_u \in M_n(\mathbb{C})$ with $\operatorname{Tr}(Q_u) = \operatorname{Tr}(Q_u^{-1})$ and giving the orthogonality relations
$$h(u_{ij} u_{kl}^*) = \delta_{i,k} \, (Q_u)_{jl} \, \operatorname{Tr}(Q_u)^{-1} \; .$$
Note that $G$ is of Kac type if and only if $Q_u = 1$ for all irreducible unitary representations $u$. This positive $Q_u$ also unitarizes $u^c$. First defining the matrix $u^c \in M_n(\mathbb{C}) \otimes A$ by $(u^c)_{ij} = (u_{ij})^*$, we get that $(Q_u^{1/2} \otimes 1) u^c (Q_u^{-1/2} \otimes 1)$ is a unitary representation that we denote as $\overline{u}$. This is the unitary contragredient of $u$.
Proposition. The following statements are equivalent.

*

*$G$ is of Kac type.


*There exists a $C > 0$ such that $\|Q_u\| \leq C$ for every irreducible unitary representation $u$.


*There exists a $C > 0$ such that $\|u^c\| \leq C$ for every irreducible unitary representation $u$.
Proof. When $u$ is an arbitrary finite dimensional unitary representation, we can still canonically define $Q_u$ by decomposing $u$ as a sum of irreducibles. We have $Q_{u \otimes v} = Q_u \otimes Q_v$. Also, $Q_{\overline{u}} = Q_u^{-1}$. For every unitary representation $u \in M_n(\mathbb{C}) \otimes A$, we denote $\dim u = n$ and $\dim_q u = \operatorname{Tr}(Q_u)$. The orthogonality relations then say that
$$(\mathord{\text{id}} \otimes h)((u^c)^* u^c) = \frac{\dim u}{\dim_q u} \, Q_u \; .$$
1 $\Rightarrow$ 2 is trivial.
2 $\Rightarrow$ 3. Since $Q_{\overline{u}} = Q_u^{-1}$, also $\|Q_u^{-1}\| \leq C$. By definition, $\|u^c\| \leq C$ for every irreducible unitary representation $u$.
3 $\Rightarrow$ 1. Assume that $G$ is not of Kac type. Fix $C > 0$. We construct an irreducible unitary representation $u$ with $\|u^c\| > C$.
Since $G$ is not of Kac type, we can choose an $n$-dimensional irreducible unitary representation $v$ such that $Q_v \neq 1$. Define $q_{\max}$ as the largest eigenvalue of $Q_v$. Since $Q_v \neq 1$, we get that $\operatorname{Tr}(Q_v) < q_{\max} \, n$. Take an integer $k \geq 1$ such that $(\operatorname{Tr}(Q_v)^{-1} \, q_{\max} \, n)^k > C^2$. Define $w$ as the $k$-fold tensor power of $v$. Then,
$$(\mathord{\text{id}} \otimes h)((w^c)^* w^c) = \frac{\dim w}{\dim_q w} \, Q_w = \Bigl(\frac{\dim v}{\dim_q v}\Bigr)^k \, Q_v^{\otimes k} \; .$$
By construction, the operator norm of the right hand side is strictly larger than $C^2$. The operator norm of the left hand side is the maximum of $\| (\mathord{\text{id}} \otimes h) ((u^c)^* u^c)\|$, where $u$ runs through the irreducible subrepresentations of $w$. So, there exists an irreducible $u$ such that $\| (\mathord{\text{id}} \otimes h) ((u^c)^* u^c)\| > C^2$. It follows that $\|u^c\| > C$.
