# Completely isometric coaction of discrete quantum group is multiplicative?

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!

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.

• Thanks so much for your splendid answer! Commented Feb 3, 2023 at 19:38