# Relating different definitions of dual of a compact quantum group

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?

• What is your reference for the dual in the multiplier Hopf case? It seems like the exact definition is going to be important in answering thus question. Apr 2, 2022 at 19:04
• @MatthewDaws Let's say that I follow the conventions in Van Daele's original paper "An algebraic framework for group duality". Apr 2, 2022 at 21:18
• $\ell^\infty-\bigoplus$ means the $\ell^\infty$ direct sum, right? If so, then it might be better to typeset it as $\text{$\ell^\infty$-$\bigoplus$}$ \text{$\ell^\infty$-$\bigoplus$} or so. Apr 2, 2022 at 22:09
• @LSpice That's correct! Thanks for the suggestion. I implemented it. Apr 2, 2022 at 22:11
• Where is your map $\Phi$ from? Because the analogous map, $\hat\pi$, in the paper I cite by Kustermans and van Daele, seems to not be equal to $\Phi$. Apr 3, 2022 at 19:52

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 (Also arXiv: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).
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.)
• Thanks for your answer. Does this mean that if we use $\Delta_{\widehat{\mathbb{G}}}^{\operatorname{op}}:= \Sigma \Delta_{\widehat{\mathbb{G}}}\Sigma$, then the map $\widehat{\mathbb{C}[\mathbb{G}]}\to c_c(\widehat{\mathbb{G}})$ preserves the coproducts? (So domain has the classical coproduct as in Van Daele's paper and the codomain has the flipped coproduct)? Apr 3, 2022 at 15:44
• I just saw your edit. Thanks for that! Actually, as follows from lemma 2.19 in the paper you mention, the 'correct' map is $\Phi(\omega)\Lambda(x) = \Lambda(x\star \omega S^{-1})$ so an easy fix with the antipode comes to the rescue :) Apr 7, 2022 at 18:02