Completely isometric coaction of discrete quantum group is multiplicative?

Question

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!

Answer 1

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.