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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.

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By Gel'fand-Naimark, there is an isomorphism $L_\infty(\mu) \cong C(K)$ for some hyperstonean space $K$, so you should perhaps google for the term hyperstonean/hyperstonian then. One can even have `a formula' for that compact space: $K=\mbox{Stone space}(\wp(\Omega)/\mathcal{N}_\mu)$, where $(\Omega, \mathcal{F}, \mu)$ is your measure space and $\mathcal{N}_\mu$ denotes the σ-ideal of $\mu$-null sets. Please have a look on vol. 1 of Takesaki's book for more details. –  Tomek Kania Jul 24 '13 at 13:10
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To be hyperstonian is a topological property (a property of a topological space, not of a measure space). By the above, every localisable measure space gives rise to a hyperstonian (compact) space and vice-versa. Separably acting commutative vNAs exhaust the list: $\ell_\infty(\Gamma)$ where $\Gamma$ is finite or countably infinite, $L_\infty[0,1]$ or $L_\infty[0,1]\oplus \ell_\infty(\Gamma)$. –  Tomek Kania Jul 24 '13 at 13:34
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Of course, one has to be careful. People often make a blanket assumption that measure spaces are always localisable. I can't think of any interesting non-localisable measure spaces though. –  Tomek Kania Jul 24 '13 at 13:39
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No, I do not agree with the statement that vNA theory is just a specific case of C*-algebras theory. This might be of interest to you: people.maths.ox.ac.uk/grabowski/files/brown.pdf –  Tomek Kania Jul 24 '13 at 14:05
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@Tomek: "I can't think of any interesting non-localiszble measure spaces" --- how about one-dimensional Hausdorff measure on $\mathbb{R}^2$? –  Nik Weaver Jul 24 '13 at 14:44
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2 Answers 2

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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).

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Thanks for your answer Ollie. Your point of HypStone not being a subcategory of Top makes it easier to accept the idea that vNA are not a particular case of C*-algebras, but from your answer I would deduce that a better name for vNAs should be "non-commutative integration theory" and not "non-commutative measure theory" because of the pathological cases I was evoking. One more general question: how come topology is not considered a particular case of measure theory since you can get the primitive notion (open sets) from the generator of the Borel set, hence all other notions (continuity etc)? –  Issam Ibnouhsein Jul 27 '13 at 15:03
    
And by topology I mean HKB topology, not less rigid versions which clearly can't be reduced to measure theory. PS: I understand Segal's point of pathological cases not being essential, but this view of measure theory might turn out to be too restrictive (who knows how measure theory will develop) so I prefer to consider it in full generality, and explicitely point at the integration part of it when dealing with category equivalences like vNA, hyperstonian spaces etc –  Issam Ibnouhsein Jul 27 '13 at 15:17
    
If you're worried about excluding these pathological measure spaces, you should note the same problem also applies to $C^*$-algebras: we are dualising only compact Hausdorff spaces, not general topological spaces! I'm not sure what you mean by getting the open sets from the generator of the Borel set. The topology of a Borel space can't be recovered from its $\sigma$-algebra - for instance $\mathbb{R}$ and $\mathbb{R}^2$ with their standard topologies are not homeomorphic - however the corresponding measurable spaces are isomorphic. –  Ollie Margetts Jul 27 '13 at 16:32
    
Yes, you are right concerning C*-algebras, though the fact we are restricted is pretty obvious so it didn't really bother me, whereas vNAs got me confused with the subtelties about localizability, integration etc. But now its all clear. Concerning the second point, I thought we could recover the topology from the $\sigma$-algebra through the generator of the $\sigma$-algebra, but your examples of $\mathbb{R}$ and $\mathbb{R}^2$ made me understand I was wrong. Thanks for your answer which helped me on the "psychological" level, and to all others for their (technically) detailed answers! –  Issam Ibnouhsein Jul 27 '13 at 16:46
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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.

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Thanks Owen for this clarification. But then I don't see the difference between what you said and what I said at some point, quote "Ok so do we agree that vNA theory is just a specific case of C*-algebras theory, which can be studied from measure-theoretic perspective thanks to the topology inherited from the predual". I understand that vNA are not interesting with the C*-algebras norm topology, but being able to endow them with a nicer topology and find many results is just a nice feature of this particular subset of C*-algebras. Tomek not agreeing with this makes me feel something is wrong.. –  Issam Ibnouhsein Jul 24 '13 at 15:10
    
The main issue to me is this (and I am repeating myself): vNAs being equivalent to nice measured spaces (localizable), they are not general enough to be called "noncommutative measured spaces" since one can probably find noncommutative analogs for non localizable measured spaces, and no correspondent vNA will be found for them. –  Issam Ibnouhsein Jul 24 '13 at 15:16
    
Small quibble about "separability" - aren't ultrapowers (borh VN and Cstar) useful things to consider? –  Yemon Choi Jul 24 '13 at 18:32
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@Yemon: Yes, I did gloss over this. Though to be honest, ultrapowers are a convenient language to talk about sequences of separable algebras, and every use of ultrapowers that I know of fits into this role. –  Owen Sizemore Jul 24 '13 at 19:38
    
What is the precise meaning of "[the ultraweak] topology amounts to convergence in measure"? Surely you can't mean that the topologies are the same: the topology of convergence in measure is metrizable while the ultraweak topology is not metrizable except in the finite-dimensional case. –  Martin Jul 28 '13 at 12:09
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