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Kadison and Ringrose define normal states $\omega$ of a von Neumann algebra $A$ as such that $\omega(H_\alpha)\to \omega(H)$ for each monotone increasing net of operators $H_\alpha$ with least upper bound $H$ (definition 7.1.11)

Let $A$ be a commutative von Neumann algebra and $NS(A)$ be its set of normal characters, and let us endow $NS(A)$ by some natural topology, for example, by the the weak topology generated by elements of $A$. Did anybody try to describe the topological properties of $NS(A)$? As far as I understand, usually this space is not compact, but from the construction of the von Neumann envelope it follows that such spaces are "natural covers" for all Hausdorff compact spaces. So I wonder how this picture can be explained from the topological point of view.

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    $\begingroup$ For $A=L^\infty[0,1]$ don't you get that $NS(A)$ is empty? $\endgroup$
    – Matthew Daws
    Commented Apr 5, 2012 at 17:20
  • $\begingroup$ Yes, I have corrected: USUALLY NS(A) is not compact. Thank you. $\endgroup$
    – Sergei Akbarov
    Commented Apr 5, 2012 at 18:21

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If by a normal character you mean a normal morphism of C*-algebras A→C, then every commutative von Neumann algebra canonically decomposes as a product of its atomic and diffuse parts, the atomic part canonically decomposes as a product ∏i∈IC, and the set of normal characters is canonically isomorphic to I. In geometric terms, every measurable space is a disjoint union of its atomic and diffuse parts, and the atomic part is a disjoint union of points, which can be identified with the normal characters of A. The only natural topology on the set of normal characters is the discrete topology, in particular the weak topology induced by A is discrete.

To answer the other question, if we take all characters (not necessarily normal), i.e., the spectrum of the underlying C*-algebra, then the resulting space is hyperstonean. Furthermore, for the von Neumann envelope of a commutative C*-algebra A the resulting map of compact Hausdorff spaces C*-Spec(W*-env(A))→C*-Spec(A) is the hyperstonean cover of C*-Spec(A).

Alternatively, one can invoke the Gelfand-Neumark theorem for commutative von Neumann algebras, which states in particular that the opposite category of the category of commutative von Neumann algebras is equivalent to the category of measurable locales, which in its turn embeds fully faithfully in the category of locales, which is very similar to the category of topological spaces (and contains Hausdorff topological spaces as a full subcategory). Thus the spectrum of a commutative von Neumann algebra is a locale, and bounded functions on this locale are precisely the elements of the original commutative von Neumann algebra. Of course, the argument above proves that the interesting part of this locale (the one that corresponds to the diffuse part of the original von Neumann algebra) has no points (but it is highly nontrivial anyway), in particular it is nonspatial, i.e., does not come from a topological space and thus provides an example of a pointfree / pointless topological space. Arguably this fact can be seen as yet another argument for replacing topological spaces by locales in mathematics.

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  • $\begingroup$ "If by a normal character you mean a normal morphism of C*-algebras" -- yes, of course. Discrete topology? So the topological explanation is trivial... That is interesting. But where is it written? By the way, in your link I did not find a mentioning of Gelfand-Neumark. $\endgroup$
    – Sergei Akbarov
    Commented Apr 5, 2012 at 19:37
  • $\begingroup$ The facts mentioned in the first two paragraphs one can find in Takesaki's Theory of Operator Algebras I, in particular Section III.1. The Gelfand-Neumark theorem for commutative von Neumann algebras is discussed in the last two paragraphs of the “Definitions” section in the link above. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 6, 2012 at 5:32
  • $\begingroup$ Dima, apparently, I am doing something wrong, but I can't understand the text in your link because of the abundance of phrases "Math Processing Error". What should I do to look at the formulas? $\endgroup$
    – Sergei Akbarov
    Commented Apr 6, 2012 at 6:09
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    $\begingroup$ @Dmitri: Takesaki's III.1 is 18 pages long. In addition, the text is quite difficult for non-specialists, and one should be quite sophisticated in those tricks to see that Corollary 1.13 indeed implies what I need. For 2 days I was trying to understand your hints, and finally I had to ask another specialist, Julia Kuznetsova. Only after her explanations I understood what one can have in mind here. My opinion is that each work, including helping people, can be done punctually and agreeably for those who need help. Or carelessly and disagreeably for them. And it is Julia who did it punctually. $\endgroup$
    – Sergei Akbarov
    Commented Apr 10, 2012 at 9:13
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    $\begingroup$ @Sergei: Instead of asking specific questions about the parts of the proof that are unclear to you, you immediately started to complain ("Stylistically it would be more precise to say..."). I cannot but remark that such nonconstructive position can hardly help you to achieve your goals. Comments on MathOverflow are not expected to be as precise as refereed papers. Some vagueness is always expected, and you are expected to ask further questions if something is unclear to you, instead of complaining and attempting to ridicule those who are trying to help you. $\endgroup$
    – Dmitri Pavlov
    Commented Apr 10, 2012 at 12:26

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