Before asking my question, let me give the necessary background. Readers that are comfortable with the language of universal and reduced compact quantum groups may skip the following two sections.

Definition: A pair $(A, \Delta)$ is called $C^*$-algebraic compact quantum group if $A$ is a unital $C^*$-algebra and $\Delta: A \to A \otimes A$ is a unital $*$-morphism that is coassociative and satisfies the quantum density rules.

Definition: A pair $(A, \Delta)$ is called an algebraic compact quantum group if $(A, \Delta)$ is a Hopf$^*$-algebra together with a positive functional that is both left and right invariant.

Some constructions. Full details can be found in e.g. Timmerman's book "An invitation to quantum groups and duality".

(1) Given a $C^*$-algebraic compact quantum group $(A, \Delta)$, we can define the space of matrix coefficients of unitary finite-dimensional corepresentations. This becomes an algebraic compact quantum group for the induced comultiplication and the Haar state. Constructing the antipode and the counit takes some work.

(2) Given an algebraic compact quantum group $(A, \Delta)$, one can associate two C*-algebraic compact quantum groups:

(i) The universal one: This is the $C^*$-envelope of the space of matrix coefficients.

(ii) The reduced one: One forms the associated universal $C^*$-algebraic compact quantum group $(A_u, \Delta_u)$ as in (i) and considers the Haar state $h_u:A_u \to \mathbb{C}$ on this quantum group. One then considers the associated GNS representation $\pi_u: A_u \to B(H_u)$ with respect to $h_u$ and gives $A_r := \pi_u(A_u)$ the structure of a compact quantum group by extending the comultiplication $\Delta$ on $A$.

Now, let $(A, \Delta)$ be a $C^*$-algebraic compact quantum group and let $(A_0, \Delta_0)$ be the space of matrix coefficients, which has the structure of an algebraic compact quantum group.

With this algebraic compact quantum group, we can associate $(A_r, \Delta_r)$ as in $(ii)$ (thus first consider the universal compact quantum group and consider the GNS-image).

However, in the literature, the following construction also occurs:

One considers the Haar state on $(A, \Delta)$ and one associated the GNS representation $\pi_h: A \to B(H_h)$ to it. One then defines $\widetilde{A}_r = \pi_h(A)$ and shows that the comultiplication $\Delta:A_0\to A_0\odot A_0$ extends to a comultiplication on $\widetilde{A}_r$ (note that the Haar measure is faithful on $A_0$, so that $\pi_h$ is injective on $A_0$). Thus, this construction "skips the universal compact quantum group".

I have the impression that $A_r$ and $\widetilde{A}_r$ are used interchangeably, so they must be somehow related.

Question: How are these $C^*$-algebraic compact quantum groups related? Are they canonically isomorphic? I.e. do we have $\pi_u(A_u)\cong \pi_h(A).$

  • 1
    $\begingroup$ I have seen another version again which says start with $A_0$, which has a Haar state, and define a norm $\|f\|_r=\|\pi_h(f)\|$, and the completion of this is another reduced version. I think what you might need is that all these $A_0$ for all these completions are isomorphic and that all these completions are minimal in that $\|\cdot\|_r\leq\|\cdot\|$ for any other norm on $A_0$. $\endgroup$ – JP McCarthy Mar 20 at 22:41
  • $\begingroup$ @JPMcCarthy Thanks for your comment. Do you have an idea how we can show that these completions are isomorphic? $\endgroup$ – user167952 Mar 20 at 23:11
  • 1
    $\begingroup$ I'm looking for a reference but it is the $A_0$ that would be isomorphic - you would be showing all these norms are minimal/equal on $A_0$. $\endgroup$ – JP McCarthy Mar 20 at 23:19
  • $\begingroup$ Something along these lines should be in Timmermann. There is a kind of triangle commutative diagram in the chapter about CQGs. I cannot help further right now apologies $\endgroup$ – JP McCarthy Mar 20 at 23:24
  • $\begingroup$ @JPMcCarthy Thanks for the reference, I'm familiar with this book but I don't quite see how it helps. Anyways, thanks for your time and help. $\endgroup$ – user167952 Mar 20 at 23:31

Yes, $\widetilde{A}_r$ and $A_r$ are canonically isomorphic. This follows from the following result in which you could take as $(B,\Delta_B)$ the universal C$^*$-algebraic compact quantum group $(A_u,\Delta_u)$ associated with $(A_0,\Delta_0)$.

Proposition. Let $(B,\Delta_B)$ and $(A,\Delta_A)$ be C$^*$-algebraic compact quantum groups and let $\pi : B \to A$ be a surjective unital $*$-homomorphism satisfying $(\pi \otimes \pi) \circ \Delta_B = \Delta_A \circ \pi$. Let $B_0 \subset B$ be a dense Hopf $*$-subalgebra. Assume that the restriction $\pi|_{B_0}$ is injective. Denote by $h_B$ and $h_A$ the Haar states on $(B,\Delta_B)$ and $(A,\Delta_A)$, respectively. Denote by $\pi_B$ and $\pi_A$ the corresponding GNS-representations.

Then, $h_B = h_A \circ \pi$ and there is a unique $*$-isomorphism $\theta : \pi_B(B) \to \pi_A(A)$ satisfying $\theta \circ \pi_B = \pi_A \circ \pi$.

Proof. Since $\pi|_{B_0}$ is injective, the restriction of $h_A \circ \pi$ to $B_0$ is an invariant state on the Hopf $*$-algebra $B_0$. By uniqueness of the invariant state, $h_A \circ \pi$ and $h_B$ are equal on $B_0$. By density of $B_0 \subset B$, we get that $h_B = h_A \circ \pi$. Since $\pi$ is assumed to be surjective, the rest follows immediately.

  • $\begingroup$ Thank you professor Vaes. This is exactly what I'm looking for. $\endgroup$ – user167952 Mar 21 at 10:52
  • $\begingroup$ Maybe one additional question: we want to define $\theta(\pi_B(b)) = \pi_A(\pi(b))$. Why is this well-defined? If $\pi_B(b) = 0$, then $0=h_B(b) = h_A\circ \pi(b)= \langle \pi_A\pi(b)\xi_A, \xi_A\rangle$ but how can we conclude that $\pi_A(\pi(b)) = 0$? $\endgroup$ – user167952 Mar 21 at 13:23
  • 2
    $\begingroup$ Let $\pi : B \to A$ be any surjective $*$-homomorphism between unital C$^*$-algebras and $\omega$ a state on $A$. Put $\varphi = \omega \circ \pi$. Since $\omega(\pi(a)^* \pi(b)) = \varphi(a^* b)$, there is a canonical unitary identification between the GNS-Hilbert spaces of $\omega$ and $\varphi$. This unitary implements the isomorphism between $\pi_\omega(A)$ and $\pi_\varphi(B)$. $\endgroup$ – Stefaan Vaes Mar 21 at 13:38
  • $\begingroup$ Nice argument. Thanks again. $\endgroup$ – user167952 Mar 21 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy