What is the discrete quantum group associated to a compact group?

Question

Let $G$ be a compact topological group. Then $G$ is a CQG with function algebra $C(G)$ and the usual comultiplication on $C(G)$. Is there an easy description of the dual discrete quantum group $\widehat{G}$?

In the case of commutative compact groups, I would hope that there is a connection with the usual dual of a compact group (in the sense of Pontryagin duality), but since the function algebra of the discrete quantum group $\widehat{G}$ is a von Neumann algebra, it won't be something nice like $C(\widehat{G})$.

Answer 1

I believe that really the question is being asked in the context of Locally compact quantum groups. This is a framework using the machinery of $C^*$ and von Neumann algebras, with (amoung many aims, and many different motivations) the aim of extend Pontryagin duality to all locally compact groups.

A locally compact quantum group is an operator algebra with a coproduct, which behaves something like a Hopf algbera. However, instead of specifying an antipode and counit, instead invariant weights (non-commutative analogues of the Haar measure) are specified, and then (densely defined) analogues of the counit and antipode emerge from the theory. Built into the definition is that any locally compact quantum group has a "dual quantum group", with the dual of the dual being (in this setup) equal to what you started with.

Given a locally compact group $G$, we consider the $C^*$-algebra $C_0(G)$, or the von Neumann algebra $L^\infty(G)$, with the usual coproduct, and the left/right Haar measures giving invariant weights. Then following the theory through will show that the dual group is represented by the group $C^*$-algebra $C^*_r(G)$ (or $C^*(G)$) and the group von Neumann algebra $VN(G)$. I am afraid I am not aware of a completely elementary account of this, but it is a nice exercise to prove it yourself. See also Enock, Schwartz, Kac Algebras which treats Kac algebras (an earlier axiomatisation) with lots of details on these examples.

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So with $G$ compact, the "discrete dual quantum group" is represented by $C^*(G) = C^*_r(G)$ or $VN(G)$. Concretely, we use the Haar measure to form $L^2(G)$ and then let $G$ act by (left) convolution, say the unitary translation operators $(\lambda_s)_{s\in G}$. Then $VN(G)$ is the von Neumann algebra generated by these, and one can show (not entirely trivially) that there is a unital normal $*$-homomorphism $\Delta:VN(G)\rightarrow VN(G)\overline\otimes VN(G)$ with $\Delta(\lambda_s) = \lambda_s\otimes\lambda_s$. Using classical Peter-Weyl theory, it follows that $VN(G)$ is a direct sum of matrix algebras, corresponding to irreducible unitary representations of $G$, and $\Delta$ encodes how the tensor product of irreducibles can be written as sums of irreducibles (i.e. the Fusion rules, and the explicit unitary isomorphisms resulting).

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Comments suggest that the OP really had in mind the case when $G$ is abelian. In this case, we can form the classical Pontryagin dual $\widehat G$ which is a genuine discrete group. The example to keep in mind if when $G=\mathbb T$ then $\widehat G=\mathbb Z$. With the Haar measure normalised in the usual way, the Fourier transform gives a unitary $$ \mathcal F : L^2(G) \rightarrow \ell^2(\widehat G). $$ Then conjugation by $\mathcal F$ gives that $$ VN(G) \rightarrow \mathcal B(\ell^2(\widehat G)); \quad x \mapsto \mathcal F x \mathcal F^* $$ is an isomorhism onto $\ell^\infty(\widehat G)$. Similar $C^*(G)$ is isomorphic to $c_0(\widehat G)$. Conversely, $VN(\widehat G)$ is isomorphic to $L^\infty(G)$ and $C^*(\widehat G)$ isomorhic to $C(G)$.

Answer 2

It's not my area, but I think the answer is very simple: when $G$ is compact abelian the W${}^*$ dual quantum group is just $L(G) \cong l^\infty(\hat{G})$.