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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!

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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
    Commented Mar 15, 2012 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
    Commented Mar 15, 2012 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
    Commented Mar 15, 2012 at 16:54
  • $\begingroup$ @Matthew Daws: Where can I find Banica's paper? $\endgroup$
    – m07kl
    Commented Mar 15, 2012 at 17:00
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    $\begingroup$ @MatthewDaws teobanica.wordpress.com/publications $\endgroup$
    – JP McCarthy
    Commented Sep 30, 2015 at 16:37

1 Answer 1

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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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  • $\begingroup$ 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
    Commented Mar 15, 2012 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
    Commented Mar 16, 2012 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
    Commented Mar 16, 2012 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
    Commented Jan 7, 2013 at 12:56

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