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!




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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?



Relating different definitions of dual of a compact quantum group
This is simply a matter of differing conventions. The definition of a discrete quantum group arising as the dual of a compact quantum group, using the multiplicative unitary, uses the conventions of Locally Compact Quantum Groups. A really useful paper bridging between algebraic quantum groups and LCQGs is [Kustermans, van Daele, C\*-Algebraic Quantum Groups Arising from Algebraic Quantum Groups](https://www.zbmath.org/?q=an%3A1009.46038) (Also [arXiv:q-alg/9611023](https://arxiv.org/abs/q-alg/9611023)) In particular, look at Definition 2.16 in this paper, which defines
$$ \hat\Delta(x) = W(x\otimes 1)W^\* $$
which is exactly the tensor flip of $\Delta\_{\hat{\mathbb G}}$. In other words, the two definitions of the coproduct on the dual differ by a flip. (Note that this paper was written before the work of Kustermans and Vaes, and uses instead the slightly older axioms of Masuda, Nakagami, Woronowicz).


There is nothing "deep" going on here: one has at certain points certain choices, and just different sources choose different choices for probably historical reasons. I would agree that this can sometimes be frustrating.




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**Update:** The map $\Phi$ the OP defines is
$$ \Phi(\omega)\Lambda(a) = \Lambda(\omega\star a) = \Lambda((\iota\otimes\omega)\Delta(a)). $$
If we define the *right* multiplicative unitary $V$ as
$$ V(\Lambda(a)\otimes\Lambda(b)) = (\Lambda\otimes\Lambda)(\Delta(a)(1\otimes b)) $$
then a simple calculation shows that
$$ (\iota\otimes\omega)(V) \Lambda(x) = \Lambda\big( (\iota\otimes\omega)\Delta(x) \big) = \Phi(\omega)\Lambda(x). $$
(Usually $V$ would be defined using the right Haar weight and its associated GNS construction, but here $\mathbb G$ is compact, so there are no left/right issues).
So $\Phi(\omega) = (\iota\otimes\omega)(V)$. Performing now a slightly formal calculation,
$$ (\Phi\otimes\Phi)(\hat\Delta(\omega))
= (\iota\otimes\iota\otimes\hat\Delta(\omega))(V\_{13} V\_{24})
= (\iota\otimes\iota\otimes\omega)(V\_{13} V\_{23})
= (\iota\otimes\iota\otimes\omega)(V\_{12}^\* V\_{23} V\_{12})
= V^\* (1\otimes (\iota\otimes\omega)(V)) V
= V^\* (1\otimes \Phi(\omega)) V. $$
Here we used the Pentagonal equation. So if we use the right multiplicative unitary, then everything works as we expect. (Basically, another left/right "choice" happening.)


