von neumann algebras and measurable spaces

Question

I've read some pages on links between von neumann (VN) algebras and measurable spaces (Spectra of $C^*$ algebras and Non-commutative geometry from von Neumann algebras?), but I can't get the following:

• VN algebras are C*-algebras
• C*-algebras are equivalent to compact separated topological spaces
• VN are equivalent to measurable spaces

How come then we can find measurable spaces which are not topological spaces if any VN algebra is a C*-algebra? I suspect my question to be silly but I don't have the answer.

Thanks for your help.

Answer 1

I highly recommend Segal's original paper Equivalences of Measure Spaces [American Journal of Mathematics Vol. 73, No. 2 (1951), pp. 275-313], where he introduced localizable spaces, since this was before the terminology took off. In it he shows that an arbitrary measure space has maximal abelian (i.e. strongly closed) $L^\infty$ algebra if and only if it is localizable. So there do exist measure spaces which for which $L^\infty(X)$ is not a von Neumann algebra. But (and it's a big but), if you're at all interested in integration, then the class of localizable measure spaces is really as large a class of interesting measure spaces as there is. The following excerpt explains why:

The class of measure spaces with these properties (we call such spaces "localizable") constitutes in some ways a more natural generalization of the $\sigma$-finite measure spaces, than the class of arbitrary measure spaces. In particular, for a measure space to be localizable is equivalent to the validity for the space of the conclusion of the Radon-Nikodym theorem, or alternatively to the conclusion of the Riesz representation theorem for continuous linear functionals on the Banach space of integrable functions. Every measure space is metrically equivalent (by which we mean there is a measure-preserving isomorphism between the $\sigma$-finite measure rings - roughly speaking this means the spaces are equivalent as far as integration over them is concerned) to a localizable space, and this latter space is essentially unique.

The point of view being held here is that measure theory is not about defining odd pathological measure spaces, but about all the cool stuff you can build on the nicer spaces: integration theory, probability theory, dynamical systems, stochastic processes, ergodic theory. And all that interesting stuff can be done, or is being worked out, in von Neumann algebras.

So von Neumann algebras capture the integration bit of measure theory.

Regarding your point in the comments: "since in vNT all what distinguishes measure theory from topology (mainly measurable spaces not being topological) is absent (we are restricted to LMS, which are topological)?" The category of localizable measure spaces and measurable maps is equivalent to the category of hyperstonean topological spaces and hyperstonean maps - this is not a subcategory of topological spaces and continuous maps - the maps between these spaces have to preserve an extra structure, namely a family of normal measures. (Just like the category of topological spaces is not a subcategory of sets, because again they have different morphisms, but there is a forgetful functor Top-->Set).

Answer 2

To make things simpler let us deal with unital $C^*$-algebras. Then the category of commutative (unital) $C^*$-algebras and unital $*$-homomorphisms is dual to the category of compact topological spaces and continuous maps. This is generally what is meant by $C^*$-algebra theory is "non-commutative topology" because we just drop the commutative assumptions. Now it is true that von Neumann algebras are $C^*$-algebras but in general they are not a particularly interesting case. The easiest way to see this is that for a commutative $C^*$-algebra, $A$, then by Gelfand's theorem we know that is it isomorphic to the continuous complex valued function on some compact space, call it $X$, (we need that $A$ is unital). However, it is a fairly easy exercise to show that $X$ is metrizable $\Leftrightarrow$ $A$ is separable for the norm topology. Also a von Neumann algebra is separable for the norm topology $\Leftrightarrow$ it is finite dimensional. It is natural to restrict to the separable case (so that you can get a handle on a dense subset and from there do analysis). So if you restrict to the norm topology viewing von Neuman algebras as $C^*$-algebras is not particularly useful.

However, there is another topology on von Neumann algebras for which many algebras are separable. This is called the untraweak topology and comes from the fact that the algebra is the dual of a banach space. In the commutative case, $L^\infty([0, 1])$, this topology amounts to convergence in measure. So the topology on the von Neumann algebra that makes it manageable remembers the measure structure of $[0, 1]$ NOT the topological structure of $[0, 1]$. With this in mind there is a similar equivalence of categories between commutative vNA with $*$-homomorphisms and measure spaces with measurable maps.

Further, because of this difference in topologies, the two theories have quite a different "flavor", much of it comes down to which version of the spectral theorem one is allowed to apply.

Finally, one more things worth pointing out is that it is common said that a von Neumann algebra is the non-commutative analogue of a measure space. This is not quite true. The correct statement is that a vNA is the analogue of a $\textit{measured}$ space. Recall that a measured space is is a space with a $\sigma$-algebra and a choice of measure 0 sets, but not a specific choice of measure. This is the proper analogue because you don't get the "measure" until you choose a state, or weight, or the algebra.