Let $\mathbb{G}$ be a compact quantum group in the sense of Woronowicz. We can look at its associated dense Hopf$^*$-subalgebra $\mathbb{C}[\mathbb{G}]$. Hence, in the framework of multiplier Hopf $*$-algebras (as introduced by Van Daele) it has a dual $$\widehat{\mathbb{C}[\mathbb{G}]}.$$

On the other hand, in the literature the dual discrete quantum group is sometimes defined as follows: Let $\Lambda_h: C(\mathbb{G})\to L^2(\mathbb{G})$ be the GNS-map associated to the Haar state $h: C(\mathbb{G})\to \mathbb{C}$. Then there is a unique unitary $W_{\mathbb{G}}$ defined by $$W_{\mathbb{G}}^*(\Lambda_h(x)\otimes \Lambda_h(y)) = (\Lambda_h\odot \Lambda_h)(\Delta(y)(x\otimes 1))$$ for all $x,y \in \mathbb{C}[\mathbb{G}].$ We can then form $W_{\widehat{\mathbb{G}}}:= \Sigma W_{\mathbb{G}}^* \Sigma$ and define $\ell^\infty(\widehat{\mathbb{G}})$ to be the $\sigma$-weak closure of $$\{(\omega \otimes \iota)(W_{\mathbb{G}}^*): \omega \in B(L^2(\mathbb{G})) \}.$$ This is a von Neumann algebra and one proves that $$\Delta_{\widehat{\mathbb{G}}}(x) = W_{\widehat{\mathbb{G}}}^*(1\otimes x)W_{\widehat{\mathbb{G}}}$$ defines a comultiplication $\ell^\infty(\widehat{\mathbb{G}}) \to \ell^\infty(\widehat{\mathbb{G}}) \overline{\otimes} \ell^\infty(\widehat{\mathbb{G}}).$ One shows that we have a canonical $*$-isomorphism $$\ell^\infty(\widehat{\mathbb{G}}) \cong \text{$\ell^\infty$-$\bigoplus$}_{\gamma\in \operatorname{Irr}(\mathbb{G})} B(H_\gamma)$$ and this allows us to define $c_c(\widehat{\mathbb{G}})$ as the corresponding algebraic direct sum in $\ell^\infty(\widehat{\mathbb{G}}).$

It is easily verified that $$\Delta_{\widehat{\mathbb{G}}}(c_c(\widehat{\mathbb{G}})) \subseteq M(c_c(\widehat{\mathbb{G}})\odot c_c(\widehat{\mathbb{G}})).$$ Hence, I believe that somehow the spaces $\widehat{\mathbb{C}[\mathbb{G}]}$ and $c_c(\widehat{\mathbb{G}})$ must be the same (i.e. isomorphic as multiplier Hopf $*$-algebras). How can I show this?

I managed to show that the map $$\Phi: \widehat{\mathbb{C}[\mathbb{G}]}\to c_c(\mathbb{G}): \omega \mapsto (\Lambda_h(a) \mapsto \Lambda_h(\omega\star a))$$ is a well-defined $*$-isomorphism (where $\omega \star a:= (\iota \otimes \omega)(\Delta(a)))$. However, calculations suggests that this map does NOT preserve the coproducts! Is there a fix for this? Maybe change a comultiplication to its opposite or use the antipode to reverse the actions?

Thanks in advance for any help/hints!

EDIT: Below, it is claimed that $$\Phi: \widehat{\mathbb{C}[\mathbb{G}]}\to c_c(\widehat{\mathbb{G}})$$ preserves the coproduct if the domain has the coproduct $\widehat{\Delta}$ uniquely determined by $$\widehat{\Delta}(\omega_1)(1\otimes \omega_2)(x\otimes y) := (\omega_1\otimes \omega_2)((x\otimes 1)\Delta(y))$$ and where the codomain has the coproduct $\Delta_{\widehat{\mathbb{G}}}^{\text{op}}$, i.e. $x \mapsto W_{\mathbb{G}}(x\otimes 1)W_{\mathbb{G}}^*$.

I.e. we need to show that $$\Delta_{\widehat{\mathbb{G}}}^{\text{op}}(\Phi(\omega)) = (\Phi\otimes \Phi)(\widehat{\Delta}(\omega))$$ for which it suffices to show that $$\Delta_{\widehat{\mathbb{G}}}^{\text{op}}(\Phi(\omega)) (1\otimes \Phi(\omega')) = (\Phi\otimes \Phi)(\widehat{\Delta}(\omega)(1\otimes \omega'))$$ which is equivalent with $$(\Phi(\omega)\otimes 1)W_{\mathbb{G}}^*(1\otimes \Phi(\omega')) = W_{\mathbb{G}}^*(\Phi\otimes \Phi)(\widehat{\Delta}(\omega)(1\otimes \omega')).$$

However, if we evaluate the left hand side in the vector $\Lambda_h(x)\otimes \Lambda_h(y)\in L^2(\mathbb{G})^{\otimes 2}$ (where $x,y \in \mathbb{C}[\mathbb{G}]$), we get (using Sweedler notation) $$\Lambda_h(y_1x_1)\otimes \Lambda_h(y_3)\omega(y_2x_2)\omega'(y_4)$$ while if we evaluate the right hand side in the same vector, we get $$\Lambda_h(y_1x_1)\otimes \Lambda_h(y_2)\omega(x_2 y_2)\omega'(y_3).$$

These expressions should be equal but they are not. Where do I go wrong?

`\text{$\ell^\infty$-$\bigoplus$}`

or so. $\endgroup$3more comments