States "absorbed" by a Haar idempotent on a compact quantum group Firstly, a small question of nomenclature. If $(S,\bullet)$ is a magma, is there good terminology to relate $a$ to $b$ when
$$a\bullet b=b=b\bullet a?$$
Can we say that $b$ absorbs $a$? Can we say that $a$ is $b$-invariant? Or $b$ is $a$-invariant?

Now, for my question.
Let $C(\mathbb{G})$ be an algebra of continuous functions on a compact quantum groups with comultiplication $\Delta$. There is no assumption that that $C(\mathbb{G})$ is any particular completion of the algebra of regular functions, so in particular we don't have that the Haar state is faithful (in fact in my application it will be in the case that the Haar state is not faithful).
Let $\mathbb{H}\subseteq \mathbb{G}$ be a quantum subgroup in the sense that there is a quotient map $\pi:C(\mathbb{G})\to C(\mathbb{H})$ such that:
$$\Delta_{C(\mathbb{H})}\circ \pi=(\pi\otimes \pi)\circ \Delta.$$
Where $h_{\mathbb{H}}$ is the Haar state on $C(\mathbb{H})$, the state $\phi_{\mathbb{H}}:=h_{\mathbb{H}}\circ \pi$ on $C(\mathbb{G})$ is called a Haar idempotent, in particular:
$$\phi_{\mathbb{H}}\star \phi_{\mathbb{H}}=\phi_{\mathbb{H}}\text{ where }\varphi_1\star \varphi_2=(\varphi_1\otimes \varphi)\Delta.$$
Consider a state $\varphi$ on $C(\mathbb{G})$ such that:
$$\varphi\star \phi_{\mathbb{H}}=\phi_{\mathbb{H}}=\phi_{\mathbb{H}}\star \varphi.$$
Is it the case that there exists a state $\varphi_0$ on $C(\mathbb{H})$ such that:
$$\varphi=\varphi_0\circ \pi?$$

I am happy to provide further definitions and facts around this question if required.
 A: Without making the assumption that $C(\mathbb{H})$ is the universal C$^*$-algebra of the compact quantum group $\mathbb{H}$, the answer is negative. Whenever $\mathbb{G}$ is not co-amenable, we could take $C(\mathbb{H}) = C_r(\mathbb{G})$ and $C(\mathbb{G}) = C_u(\mathbb{G})$. We define $\pi : C_u(\mathbb{G}) \to C_r(\mathbb{G})$ as the canonical quotient homomorphism. When $h$ denotes the Haar state on $C_r(\mathbb{G})$, the state $h_1 = h \circ \pi$ is the Haar state on $C_u(\mathbb{G})$. Therefore, the equality $\varphi \ast h_1 = h_1 = h_1 \ast \varphi$ holds for all states $\varphi$ on $C_u(\mathbb{G})$. Since $\pi$ is not faithful, there are states on $C_u(\mathbb{G})$ that do not factor through $\pi$.
However, if we assume that $C(\mathbb{H})$ is the universal C$^*$-algebra of $\mathbb{H}$, the answer is positive. The result and proof go as follows. It combines some analytic and algebraic considerations. As far as I know, such a result is not available in the literature, but other users might know a reference!
Proposition. Let $\mathbb{G}$ and $\mathbb{H}$ be compact quantum groups and $\pi : C(\mathbb{G}) \to C(\mathbb{H})$ a surjective unital $*$-homomorphism satisfying $(\pi \otimes \pi) \circ \Delta = \Delta \circ \pi$. Assume that $C(\mathbb{H})$ is the universal enveloping C$^*$-algebra of the Hopf algebra $\operatorname{Pol}(\mathbb{H})$. Denote by $h$ the Haar state on $\mathbb{H}$ and write $h_1 = h \circ \pi$. Let $\varphi$ be a state on $C(\mathbb{G})$. Then the following statements equivalent.

*

*There exists a state $\psi$ on $C(\mathbb{H})$ such that $\varphi = \psi \circ \pi$.


*We have that $\varphi \ast h_1 = h_1 = h_1 \ast \varphi$.


*We have that $\varphi \ast h_1 = h_1$.
Proof. The implications $1 \Rightarrow 2 \Rightarrow 3$ are trivial. Assume that 3 holds. Denote by $c_0(\widehat{\mathbb{G}})$ the $c_0$ direct sum of the matrix algebras $B(H_\alpha)$, $\alpha \in \operatorname{Irrep}(\mathbb{G})$. We denote by $\ell^\infty(\widehat{\mathbb{G}})$ the $\ell^\infty$ direct sum. Denote by $W \in M(C(\mathbb{G}) \otimes c_0(\widehat{\mathbb{G}}))$ the corresponding direct sum of unitary representations of $\mathbb{G}$. We similarly define $V \in M(C(\mathbb{H}) \otimes c_0(\widehat{\mathbb{H}}))$ for the compact quantum group $\mathbb{H}$. The morphism $\pi$ dualizes to a morphism $\widehat{\pi} : \ell^\infty(\widehat{\mathbb{H}}) \to \ell^\infty(\widehat{\mathbb{G}})$ satisfying $(\pi \otimes \text{id})(W) = (\text{id} \otimes \widehat{\pi})(V)$.
Denote by $p_\varepsilon \in c_0(\widehat{\mathbb{H}})$ the minimal central projection that corresponds to the trivial representation of $\mathbb{H}$. Write $q = \widehat{\pi}(p_\varepsilon)$. Note that $(h_1 \otimes \text{id})(W) = q$. Define $T \in \ell^\infty(\widehat{\mathbb{G}})$ by $T = (\varphi \otimes \text{id})(W)$. Since $(\Delta \otimes \text{id})(W) = W_{13} W_{23}$ and $\varphi \ast h_1 = h_1$, we get that $T = T q$. Write $R = W(1 \otimes q) - (1 \otimes q)$. Because
$$R^* R = 2 (1 \otimes q) - (1 \otimes q) W (1 \otimes q) - (1 \otimes q) W^* (1 \otimes q) \; ,$$
we conclude that $(\varphi \otimes \text{id})(R^* R) = 0$. Since $\varphi$ is a state on a C$^*$-algebra, it follows that
$$(\varphi \otimes \text{id} \otimes \text{id})(W_{12} R_{13}) = 0 \; .$$
This means that
$$(\varphi \otimes \text{id} \otimes \text{id})(W_{12} W_{13}) \, (1 \otimes q) = T \otimes q \; .$$
Write $T$ as the direct sum of the matrices $T_\alpha \in B(H_\alpha)$ for $\alpha \in \operatorname{Irrep}(\mathbb{G})$. So, for all $\alpha,\beta,\gamma \in \operatorname{Irrep}(\mathbb{G})$ and for all $X \in \operatorname{Mor}_\mathbb{G}(\alpha,\beta \otimes \gamma)$ and $Y \in \operatorname{Mor}_\mathbb{H}(\varepsilon,\gamma)$, we get that
$$T_\alpha \, X^*(1 \otimes Y) = X^*(1 \otimes Y) \, T_\beta \; .$$
By taking linear combinations, the same holds if $\gamma$ is any finite dimensional unitary representation of $\mathbb{G}$, not necessarily irreducible.
We now deduce that $T_\alpha \, Z = Z \, T_\beta$ for all $\alpha,\beta \in \operatorname{Irrep}(\mathbb{G})$ and $Z \in \operatorname{Mor}_\mathbb{H}(\beta,\alpha)$. To prove this, choose solutions of the conjugate equations for $\alpha,\beta \in \operatorname{Irrep}(\mathbb{G})$, given by $t \in \operatorname{Mor}_\mathbb{G}(\varepsilon,\overline{\beta} \otimes \beta)$ and $s \in \operatorname{Mor}_\mathbb{G}(\varepsilon,\beta \otimes \overline{\beta})$. Writing $\gamma = \overline{\beta} \otimes \alpha$, $X = s \otimes 1$ and $Y = (1 \otimes Z)t$, we get that $X^*(1 \otimes Y) = Z$ and the claim follows.
This means that $T = \widehat{\pi}(S)$ for some $S \in \ell^\infty(\widehat{\mathbb{H}})$.
Every $a \in \operatorname{Pol}(\mathbb{G})$ is of the form $a = (\text{id} \otimes \omega)(W)$ where $\omega$ is a uniquely determined, ``finitely supported'' functional on $\ell^\infty(\widehat{\mathbb{G}})$. Note that $\pi(a) = 0$ if and only if $\omega \circ \widehat{\pi} = 0$. Since $(\varphi \otimes \text{id})(W) = T = \widehat{\pi}(S)$, we conclude that there is a well defined linear functional $\psi : \operatorname{Pol}(\mathbb{H}) \to \mathbb{C}$ such that $\psi(\pi(a)) = \varphi(a)$ for all $a \in \operatorname{Pol}(\mathbb{G})$.
Then $\psi(1) = \varphi(1) = 1$ and, because $\pi : \operatorname{Pol}(\mathbb{G}) \to \operatorname{Pol}(\mathbb{H})$ is surjective, $\psi(b^* b) \geq 0$ for all $b \in \operatorname{Pol}(\mathbb{H})$. Since $C(\mathbb{H})$ is the universal C$^*$-algebra of the compact quantum group $\mathbb{H}$, it follows that $\psi$ uniquely extends to a state on $C(\mathbb{H})$ that we still denote as $\psi$. By construction, $\varphi = \psi \circ \pi$.
