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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?

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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 $G_0$.

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  • $\begingroup$ Thanks! With this interpretation, is the canonical map $L^\infty(G, \lambda)\to B(L^2(G, \lambda))$ an isometric $*$-morphism such that its image is a von Neumann algebra? For example, can we show that $L^\infty(G, \lambda)$ is its own commutant inside $L^2(G)$? $\endgroup$
    – Andromeda
    Commented Apr 10, 2023 at 16:19
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    $\begingroup$ Yes! That is right. $\endgroup$
    – Stefaan Vaes
    Commented Apr 10, 2023 at 16:23
  • $\begingroup$ @StefaanVaes Does one really need this re-interpretation of $L^\infty$ in this case? It seems to me that the Haar measure is always localizable and for localizable measure spaces the usual $L^\infty$ is a maximal abelian von Neumann subalgebra of $B(L^2)$ by the results from Segal's article Equivalences of Measure Spaces. Am I missing some measure-theoretic subleties? $\endgroup$
    – MaoWao
    Commented Apr 19, 2023 at 10:07
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    $\begingroup$ In Segal's paper, the notions of being measurable and being a null set, etc, are all defined locally (see Definition 2.2 in the paper). The following example is useful to keep in mind. Let $G = \mathbb{R} \times \mathbb{R}_d$ be the direct product of $\mathbb{R}$ with the usual topology and $\mathbb{R}$ with the discrete topology. Then $X = \{0\} \times \mathbb{R}$ is a Borel set of infinite Haar measure (by outer regularity), but nevertheless $X$ is a (locally) null set. The multiplication operator by $1_X$ on $L^2(G)$ is the zero operator. $\endgroup$
    – Stefaan Vaes
    Commented Apr 19, 2023 at 12:09
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    $\begingroup$ If $\lambda$ denotes the Haar measure on $G$ and if $F \in L^\infty(G)^+$ (in the sense of e.g. Folland), then $\varphi(f)$ is defined as the supremum of $\int_K F(g) d\lambda(g)$ over all compact subsets $K \subset G$. Equivalently, you could take the supremum over all Borel sets of finite measure, or all $\sigma$-finite Borel sets. I would not recommend to change the definition of the Haar measure. After all, the theory of Radon measures on locally compact spaces is well developed. $\endgroup$
    – Stefaan Vaes
    Commented May 5, 2023 at 14:36
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As an alternative to Stefaan Vaes' answer, I believe that one can take Fremlin's approach and work with the usual $L^\infty$ spaces (classes of a.e. equal measurable, essentially bounded functions). This however requires (as in Fremlin's book) to consider Haar measure defined on larger $\sigma$-algebra of Haar-measurable sets (in fact this measure is the completion of Haar measure restricted to Borel sets). Then Haar measure is regular, strictly localisable, $L^{\infty}(G,\mu)$ is a von Neumann algebra, and weight can be defined simply as $\varphi(f)=\int_G f d \mu$ for $f\in L^{\infty}(G,\mu)^+$. With this definition, and appropriately defined comultiplication, one gets locally compact quantum group. There is however a number of technical details to be checked.

One should also be careful that different authors use term ``regular measure'' with different meaning.

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