# 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!

• Timmermann Th.5.3.11 and Th.5.4.1 close but not enough? Mar 27, 2021 at 8:24
• @JP McCarthy. Timmerman assumes corepresentations to be non-degenerate. So yes, close but not enough.
– user167952
Mar 27, 2021 at 10:17
• Ah yes I see this in Def. 5.2.6. Mar 27, 2021 at 10:43

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, $$p \not\in A_0$$.

• Thank you for the beautiful answer!
– user167952
Mar 27, 2021 at 15:46