i have a question about a "presentation" of the quantum $SU(n)$. Here presentation means the following. Let $(M,\Delta)$ be a quantum group in the sense of Kustermanns and Vaes. One can show that the von Neumann algebraic quantum group $SU_q(2)$, which will be denotes by $\mathscr{L}^{\infty}(SU_q(2))$ is $B(\ell^2(\mathbb{N}))\overline{\otimes}\mathscr{L}(\mathbb{Z})$ (where the tensor product denotes the von Neumann algebraic tensor product). Can one generalize this to $SU_q(n)$. Can one find an explicit expression for $\mathscr{L}^{\infty}(SU_q(n))$ as above?

Thank you very much

  • 1
    $\begingroup$ Small nitpick: by "is" you mean "is isomorphic as a von Neumann algebra". The point is that the coproduct of $B(\ell^2({\bf N}))\overline{\otimes} {\mathcal L}({\bf Z})$ is highly non-trivial! When one talks of "the quantum group ${\mathcal L}^\infty({\bf G})$'' it is usually understood (IMHO) that this refers not just to the underlying von Neumann algebra, but also the coproduct. $\endgroup$ – Yemon Choi Mar 14 '16 at 10:54
  • $\begingroup$ Yes, the coproduct is defined, but to prove that this is really a coproduct is not easy, you are right. $\endgroup$ – Zachary Gonzales Mar 14 '16 at 11:00
  • $\begingroup$ @YemonChoi: I know that one can represent $SU_q(2)$ faithfully on $\ell^2(\mathbb{N})\otimes\ell^2(\mathbb{Z})$ by some explicit formulas, but how to obtain the isomorphism? $\endgroup$ – Zachary Gonzales Mar 14 '16 at 11:10
  • $\begingroup$ For n=2 this is done at the C*-algebraic quantum group level in Woronowicz's original paper, although one needs to beware a typo: see mathoverflow.net/questions/210556/… Otherwise, for a more hands-on approach to the n=2 case, you could try this paper of Lance: ams.org/mathscinet-getitem?mr=1288151 $\endgroup$ – Yemon Choi Mar 14 '16 at 11:27

Theorem 3.1 of [2] is basically what you need. It implies $L^\infty(G_q) \simeq B(\ell^2(\mathbf{N})) \otimes L^\infty(T)$ for the maximal torus $T$ in $G$ (connected semisimple compact Lie group), and up to isomorphism the translation action of $T$ is just the standard one on the second factor.

As suggested there, Soibelman's work [1] on the classification of irreducible representations of $C(G_q)$ brings you most of the way for this. It implies that $C(G_q)$ is of type I and does not have finite dimensional irreducible representations of dimension $>1$, so any von Neumann algebraic closure has to be of the form $A_1 \oplus B(\ell^2(\mathbf{N})) \otimes A_2$ for some commutative von Neumann algebras $A_1, A_2$. Then you take into account of the torus action to show $A_1 = 0$ (edit: and that $A_2$ is diffuse) in the regular representation.

  1. Ya. S. Soibelman, Algebra of functions on a compact quantum group and its representations, Algebra i Analiz 2 (1990), no. 1, 190--212.
  2. Reiji Tomatsu, Product type actions of $G_q$, Adv. Math. 269 (2015), 162--196.
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