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).


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  • $\begingroup$ Maybe this arxiv.org/abs/0812.1546 paper on explicit Haar states on SU_q(N) is useful? Also, it could be of interest to state what the the answers for your question for SUq(2) are, if they are known. $\endgroup$ – Jan Jitse Venselaar Mar 15 '12 at 14:07
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    $\begingroup$ If you understand the irreducible corepresentations, then you understand the Haar state, as $h(1)=1$, and $h(a)=0$ for any other matrix coefficient $a$ of an irreducible corep. This is worked out in Timmermann's book (section 6.4) or in Banica's paper ams.org/mathscinet-getitem?mr=1378260 But maybe that's not "explicit" enough for you. $\endgroup$ – Matthew Daws Mar 15 '12 at 14:27
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    $\begingroup$ @Jan Jitse Venselaar: there is an injective unital $*$-rep on the Hopf $*$-algebra, which allows one to expree the Haar state on $SU_q(2)$ for $q\neq 1$. The Haar state is not tracial by e.g. the fact I mentioned above. It is faithful. see Example 3.1.6 "A Survey of $C^*$-Algebraic Quantum Groups. I" by Johan Kustermans and Lars Tuset $\endgroup$ – m07kl Mar 15 '12 at 16:54
  • $\begingroup$ @Matthew Daws: Where can I find Banica's paper? $\endgroup$ – m07kl Mar 15 '12 at 17:00
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    $\begingroup$ @MatthewDaws teobanica.wordpress.com/publications $\endgroup$ – JP McCarthy Sep 30 '15 at 16:37

Have you seen arXiv:math/0511253?



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.

(I do not have sufficient reputation to post this as a comment.)

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  • $\begingroup$ Thanks! This paper gives explicit formula for $A_u(1_n)$ and $A_o(1_n)$. They are both tracial, since $\kappa^2=id$. Do you know any thing about faithfulness? $\endgroup$ – m07kl Mar 15 '12 at 16:41
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    $\begingroup$ I believe (but I do not at this precise moment have access to the relevant papers, so please check for yourself) that $A_u(1_n)$ and $A_o(1_n)$ are not co-amenable for $n \geq 3$, which would imply that the Haar state is not faithful. $\endgroup$ – Tom Cooney Mar 16 '12 at 9:04
  • $\begingroup$ thanks! For universal compact quantum groups, Co-amenability is equivalent to faithfulness of the Haar state (see Thm 3.6 "Co-Amenability of Compact Quantum Groups" by E. B´edos G.J. Murphy L. Tuset). $A_o(n)$ is coamenable only for $n = 2$, while $A_u(n)$ is not coamenable for any $n$. $\endgroup$ – m07kl Mar 16 '12 at 9:41
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    $\begingroup$ It might be useful to stress that the Haar state is always faithful on the dense Hopf *-algebra spanned by the coefficients of the finite-dimensional corepresentations of a compact quantum groups (which is the *-algebra generated by the $u_{ij}$ in these examples) and - by construction - on the reduced $C^*$-algebra. $\endgroup$ – Uwe Franz Jan 7 '13 at 12:56

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