Matrix coefficients of a compact quantum group Let $(A, \Delta)$ be a $C^*$-algebraic compact quantum group (in the sense of Woronowicz).
Definition: A corepresentation matrix of $(A, \Delta)$ is a matrix $a=(a_{i,j}) \in M_n(A)$ such that
$$\Delta(a_{i,j}) = \sum_k a_{ik}\otimes a_{kj}$$
for all $i,j$.
The matrix $a$ is called non-degenerate (unitary) when $a$ is invertible (unitary) in the $C^*$-algebra $M_n(A)$.
Question:
If $A_0$ is the set of matrix entries of unitary corepresentation matrices of $(A, \Delta)$ and $A_1$ is the set of corepresentations (with no restrictions) of $(A, \Delta)$, then do we have $A_0 = A_1$?
Clearly $A_0 \subseteq A_1$, but I'm not sure if the converse inclusion holds. Any help will be appreciated!
 A: In general, the converse inclusion $A_1 \subseteq A_0$ does not hold.
As the counterexample below shows, it is not a good idea to define a corepresentation matrix as in the question. To really be considered as a corepresentation matrix, one should require $a$ to be invertible as an element in $M_n(A)$. Below is an example of a C$^*$-algebraic compact quantum group $(A,\Delta)$ containing an orthogonal projection $p \in A$ that is nontrivial ($p \neq 0$ and $p \neq 1$) and that satisfies $\Delta(p) = p \otimes p$. This pathological behavior can only happen in situations where the Haar state is not faithful.
Let me therefore start with a positive result showing that in the reduced case (i.e. the case where the Haar state is faithful), the $*$-algebra $A_0$ of coefficients of finite dimensional unitary corepresentations is highly canonical and, in particular, contains all coefficients of corepresentation matrices as in the question, by applying the following proposition to the $*$-algebra generated by the elements $a_{ij}$.
Proposition. Let $(A,\Delta)$ be a C$^*$-algebraic compact quantum group and assume that the Haar state on $A$ is faithful. Denote by $A_0$ the $*$-algebra of coefficients of finite dimensional unitary corepresentations. Let $A_1 \subseteq A$ be any $*$-subalgebra satisfying $\Delta(A_1) \subseteq A_1 \otimes_{\text{alg}} A_1$.
Then $A_1 \subseteq A_0$ and equality holds if and only if $A_1 \subseteq A$ is dense.
Proof. Denote by $h$ the Haar measure on $A$. By the Schur orthogonality relations, we can fix a complete set of irreducible and inequivalent unitary corepresentations $u_\alpha$ and bases for their underlying Hilbert spaces such that the matrix coefficients $u_{\alpha,i,j}$ satisfy
$$h(u_{\alpha,i,j}^* u_{\beta,k,l}) = \begin{cases} F_{\alpha,i} > 0 &\;\;\text{if $\alpha = \beta$, $i = k$, $j = l$,} \\ 0 &\;\;\text{otherwise.}\end{cases}$$
Define the map $\Phi : A \to A \otimes A \otimes A$ by $\Phi = (\Delta \otimes id) \circ \Delta$. We thus have
$$(h \otimes id \otimes h)((u_{\alpha,i,k}^* \otimes 1 \otimes u_{\alpha,l,j}^*) \Phi(a)) = F_{\alpha,l} \, h(u_{\alpha,i,j}^* a) \, u_{\alpha,k,l}$$
for all $a \in A$.
Denote by $V(\alpha) \subseteq A$ the linear span of $u_{\alpha,i,j}$. The previous formula shows that either $A_1$ is orthogonal to $V(\alpha)$ (w.r.t. the scalar product given by $h$), or $V(\alpha) \subseteq A_1$.
Given $a \in A_1$, we can take a finite-dimensional vector space $V \subseteq A$ such that $\Phi(a) \in V \otimes V \otimes V$. So if $a$ is not orthogonal to $V(\alpha)$, we have $V(\alpha) \subseteq V$. Therefore, $a$ is orthogonal to all but finitely many $V(\alpha)$. It follows that $a$ is contained in the linear span of finitely many $V(\alpha)$. This means that $a \in A_0$. Combined with the previous paragraph, we also find that $A_1 = A_0$ if we moreover assume that $A_1 \subseteq A$ is dense.
Counterexample. Let $G$ be any nonamenable group. Denote by $A_u = C^*(G)$ the universal C$^*$-algebra and denote by $A_r = C^*_r(G)$ the reduced C$^*$-algebra. Denote by $\lambda : A_u \to A_r$ the regular representation and by $\varepsilon : A_u \to \mathbb{C}$ the trivial representation. Write $\pi = \lambda \oplus \varepsilon$. Put $A = \pi(A_u)$. Since the trivial representation is not weakly contained in the regular representation, $A = A_r \oplus \mathbb{C}$. Since $\pi \otimes \pi$ is weakly contained in $\pi$ (actually, contained in a multiple of $\pi$), there is a unique comultiplication $\Delta$ on $A$ such that $\Delta \circ \pi = (\pi \otimes \pi) \circ \Delta_u$. Then, $(A,\Delta)$ is a C$^*$-algebraic compact quantum group.
Denote by $p = (0,1) \in A$ the natural projection. I prove that $\Delta(p) = p \otimes p$.
Denote by $\pi_\lambda : A \to A_r$ and $\pi_{\varepsilon} : A \to \mathbb{C}$ the homomorphisms satisfying $\pi_\lambda \circ \pi = \lambda$ and $\pi_\varepsilon \circ \pi = \varepsilon$. Then,
$$(\pi_\lambda \otimes \pi_\lambda) \circ \Delta \circ \pi = \Delta_r \circ \lambda \quad , \quad (id \otimes \pi_\varepsilon) \circ \Delta \circ \pi = \pi = (\pi_\varepsilon \otimes id) \circ \Delta \circ \pi \; .$$
Viewing $A_r \subseteq A$ (and noting that this inclusion is not unital), it follows that
$$\Delta(a) = \Delta_r(a) + a \otimes p + p \otimes a \quad , \quad \Delta(p) = p \otimes p \;\;,$$
for all $a \in A_r$.
The $*$-algebra $A_0$ of coefficients of finite dimensional unitary corepresentations of $(A,\Delta)$ is given by the group algebra $\pi(\mathbb{C}[G])$. In particular, the restriction of $\pi_\lambda$ to $A_0$ is injective. But $\pi_\lambda(p) = 0$. Thus, $p \not\in A_0$.
