A locally compact quantum group (in the sense of Vaes-Kustermans) consists of the data $(M, \Delta, \varphi, \psi)$ with $M$ a von Neumann algebra, $\Delta: M \to M \overline{\otimes} M$ a normal unital $*$-homomorphism satisfying coassociativity and $\varphi, \psi: M_+ \to [0, \infty]$ normal, semifinite, faithful weights satisfying left and right invariance.
One often writes $M= L^\infty(\mathbb{G})$ and refers to $\mathbb{G}$ as the locally compact quantum group.
The motivating example for the latter notation comes from the situation where $G$ is a locally compact group, $M= L^\infty(G)=L^\infty(G, \lambda)$, $\Delta: L^\infty(G)\to L^\infty(G)\overline{\otimes} L^\infty(G) \cong L^\infty(G\times G)$ given by $\Delta(f)(s,t) = f(st)$ and $$\varphi: L^\infty(G)\to \mathbb{C}: f \mapsto \int_G f d\lambda, \quad \psi: L^\infty(G)\to \mathbb{C}: f \mapsto \int_G fd\rho$$
where $\lambda$ is left Haar measure and $\rho$ is right Haar measure.
However, I see some problems. For example, for a general locally compact group $G$, it need not even be true that $L^\infty(G, \lambda)$ is a von Neumann algebra (acting on $L^2(G, \lambda)$) or it does not even need to be a $W^*$-algebra (if it is a $W^*$-algebra, then it is automatically true that integration is a normal weight). Of course, if $G$ is $\sigma$-compact, then everything works out nicely. So concretely, my question is:
If $G$ is a non $\sigma$-compact locally compact group, how exactly can we view it as a locally compact quantum group? Do we have to re-define $L^\infty(G, \lambda)$ as is usually done in harmonic analysis?
Best Answer
In the general, not necessarily $\sigma$-compact setting, one has to interpret $L^\infty(G,\lambda)$ as the von Neumann algebra of locally measurable functions that are bounded outside a locally null set. This is very well explained in Section 2.3 of Folland's "A course in abstract harmonic analysis".
Also note that as explained in Proposition 2.4 of that same book, every locally compact group $G$ admits a $\sigma$-compact subgroup $G_0 \subset G$ that is open. So measure theoretically, we are always in the quite tame situation of a (possibly uncountable) disjoint union of copies of $G_0$.