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!



Matrix coefficients of a compact quantum group
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, $p \not\in A\_0$.


