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!



Completely isometric coaction of discrete quantum group is multiplicative?
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.


