TITLE: The Haar state on compact quantum groups $A_u(Q)$ and  $A_o(Q)$
QUESTION [8 upvotes]: Let $Q\in GL_n(\mathbb{C})$. The free unitary quantum group is the universal $C^*$-algebra $A_u(Q)$ with generators $u_{ij},1\leq i,j\leq n$ and relations making $u=(u_{ij})$ as well as $Q\bar{u}Q^{-1}$ unitary, where $\bar{u}=(u_{ij}^*)$. The comultiplication is defined by
\begin{align}
\Phi(u_{ij})=\sum_k u_{ik}\otimes u_{kj}.
\end{align}
$(A_u(Q),\Phi)$ is a compact quantum group.
Let $Q\in GL_n(\mathbb{C})$ such that $Q\bar{Q}=\pm 1$. The free orthogonal quantum group is the universal $C^*$-algebra $A_o(Q)$ with generators $u_{ij},1\leq i,j\leq n$ such that $u=Q\bar{u}Q^{-1}$ is unitary. As above $(A_o(Q),\Phi)$ is a compact quantum group.
It is well-known that every compact quantum group admits a unique Haar state. My question is that 

What is the expliciit expression of the Haar state on $A_u(Q)$ and  $A_o(Q)$?
Are they faithful?

By the way, it is known for $SU_q(2)$ and $SU_q(2)\cong A_o(Q)$ for
$Q=\begin{pmatrix}
0& |q|^{1/2} \newline
-\text{sign}(q)|q|^{-1/2}&0
\end{pmatrix}.$
I just find out that the Haar state is tracial if and only if the coinverse $\kappa$ on the dense Hopf $*$-algebra satifies $\kappa^2=id$ (see "compact quantum groups" Thm 1.5 by Woronowicz).
Thanks!

REPLY [7 votes]: Have you seen arXiv:math/0511253?
INTEGRATION OVER COMPACT QUANTUM GROUPS
TEODOR BANICA AND BENOIT COLLINS 
Abstract. We find a combinatorial formula for the Haar functional of the orthogonal and unitary quantum groups. As an application, we consider diagonal coefficients of the fundamental representation, and we investigate their spectral measures.
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