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!