5
$\begingroup$

Let $\mathbb{G}$ be a compact quantum group (in the sense of Woronowicz) with discrete dual $\widehat{\mathbb{G}}$ which we view as a von Neumann algebraic locally compact quantum group (in the sense of Vaes-Kustermans). Let us denote its function algebra by $(\ell^\infty(\widehat{\mathbb{G}}), \hat{\Delta})$.

Consider a von Neumann algebra $M$ and a unital completely isometric (normal) map $$\alpha: M \to M \overline{\otimes} \ell^\infty(\widehat{\mathbb{G}})$$ satisfying the coaction property $$(\alpha\otimes \iota) \alpha = (\iota\otimes \hat{\Delta})\alpha.$$ Is it true that $\alpha$ is automatically multiplicative?

Of course, if $\Gamma$ is a discrete group, then a completely isometric map $$\alpha: M \to M\overline{\otimes}\ell^\infty(\Gamma)= \prod_{g\in \Gamma} M$$ is automatically multiplication-preserving because the map $\alpha$ is then a direct product of unital completely isometric maps $\alpha_g: M \to M$ which are automatically $C^*$-isomorphisms (by a result by Choi).

I tried to apply the same trick in this case: $$\ell^\infty(\widehat{\mathbb{G}})\cong \prod_{\gamma \in \operatorname{Irr}(\mathbb{G})} B(H_\gamma)$$ and the map $\alpha$ then breaks down as a collection of maps $$\alpha_\gamma: M \to M_{n_\gamma}(M): m \mapsto [u_{ij}^\gamma \rhd m]$$ where $U^\gamma = [u_{ij}^\gamma]$ is the irreducible representation $\gamma$ and $\rhd: \mathcal{O}(\mathbb{G})\odot M \to M$ the induced left module structure. If we can show that these maps are multiplication-preserving, then we are done. This, in turn, is equivalent with showing that their images are $C^*$-algebras, but neither of these claims are clear to me. On the level of left $\mathcal{O}(\mathbb{G})$-modules, the multiplicativity means $$g\rhd (mn)= (g_{(1)}\rhd m)(g_{(2)}\rhd n)$$ or in terms of matrix coefficients $$u_{ij}^\gamma\rhd (mn) = \sum_{k=1}^{n_\gamma} (u_{ik}^\gamma\rhd m)(u_{kj}^\gamma\rhd n).$$

Is the multiplicativity of the coaction $\alpha$ somehow automatic? I am starting to believe this isn't true, but I was not able to find a counterexample. Thanks in advance for your help!

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Yes, such a map $\alpha$ is automatically multiplicative and thus defines an action of $\widehat{\mathbb{G}}$ on $M$.

As in the question, denote by $\alpha_\gamma : M \to M \otimes B(H_\gamma)$ the components of $\alpha$, for any irreducible unitary representation $\gamma$ of $\mathbb{G}$. Fix an irreducible representation $\gamma$. It suffices to prove that $\alpha_\gamma$ is multiplicative.

Since $\alpha$ is unital completely isometric, $\alpha$ is also completely positive. Thus, all $\alpha_\gamma$ are unital completely positive (ucp).

We first prove that $\alpha_\varepsilon(x) =x$ for all $x \in M$. By the coaction property, $\alpha_\gamma \circ \alpha_\varepsilon = \alpha_\gamma$ for all $\gamma$. So, if $\alpha_\varepsilon(x)=0$, it follows that $\alpha(x)=0$ and thus $x=0$ because $\alpha$ is supposed to be isometric. Since $\alpha_\varepsilon(\alpha_\varepsilon(x)-x) = 0$ for all $x \in M$, it follows that $\alpha_\varepsilon(x) =x$ for all $x \in M$.

Let $\rho$ be the contragredient of $\gamma$ and choose morphisms $t \in \operatorname{Mor}(\varepsilon,\rho \otimes \gamma)$ and $s \in \operatorname{Mor}(\varepsilon,\gamma \otimes \rho)$ such that $t^* t = 1$ and $(s^* \otimes 1)(1 \otimes t) = 1$. Define the ucp map $$\theta : M \otimes B(H_\gamma) \to M : \theta(x) = (1 \otimes t^*)(\alpha_\rho \otimes \text{id})(x) (1 \otimes t) \; .$$ By the coaction property and the fact that $\alpha_\varepsilon = \text{id}$ proven above, $\theta(\alpha_\gamma(x)) = x$ for all $x \in M$. Fix a unitary $u \in \mathcal{U}(M)$. Since $\theta(\alpha_\gamma(u)) = u$ is a unitary and $\|\alpha_\gamma(u)\| \leq 1$, we find that $\alpha_\gamma(u)$ belongs to the multiplicative domain of $\theta$. We have that $\alpha_\gamma(u)^* \alpha_\gamma(u) \leq \alpha_\gamma(u^*u) = 1$. Applying $\theta$ and using that $\alpha_\gamma(u)$ belongs to the multiplicative domain of $\theta$, we find that $\theta(1-\alpha_\gamma(u)^* \alpha_\gamma(u)) = 0$. Below I will prove that $\theta$ is faithful. So, we conclude that $\alpha_\gamma(u)^* \alpha_\gamma(u) = 1$ for every unitary $u \in \mathcal{U}(M)$. This implies that $\alpha_\gamma$ is multiplicative.

It remains to prove that $\theta$ is faithful. Assume that $x \in M \otimes B(H_\gamma)$ such that $\theta(x^* x) = 0$. Then, $(\alpha_\rho \otimes \text{id})(x)(1 \otimes t) = 0$. Apply $\alpha_\gamma \otimes \text{id} \otimes \text{id}$ to conclude that $$(1 \otimes s^* \otimes 1) ((\alpha_\gamma \otimes \text{id})\alpha_\rho \otimes \text{id})(x) (1 \otimes 1 \otimes t) = 0 \; .$$ Using the coaction property of $\alpha$ and the fact that $\alpha_\varepsilon = \text{id}$ as proven above, the left hand side of the above expression equals $$x (1 \otimes s^* \otimes 1)(1 \otimes 1 \otimes t) = x \; .$$ So $x = 0$ and the faithfulness of $\theta$ is proven.

$\endgroup$
1
  • $\begingroup$ Thanks so much for your splendid answer! $\endgroup$
    – J. De Ro
    Commented Feb 3, 2023 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.