21
$\begingroup$

Projections in an arbitrary commutative von Neumann algebra form a complete Boolean algebra. Moreover, a morphism of commutative von Neumann algebras induces a continuous morphism of the corresponding complete Boolean algebras.

Thus we have a fully faithful functor F from the category of commutative von Neumann algebras to the category of complete Boolean algebras and their continuous morphisms.

The category of complete Boolean algebras and their continuous morphisms is a full subcategory of the opposite category of the category of locales.

Thus the functor F can be seen as implementing the Gelfand-Neumark duality for commutative von Neumann algebras. However, to obtain a satisfactory statement of the duality we still need to characterize in topological terms objects in the essential image of F, which we call measurable spaces (or locales, think of a localic version of point-set measurable spaces).

What additional topological conditions do we need to impose on a complete Boolean algebra to ensure that it is the algebra of projections of some von Neumann algebra, i.e., a measurable space?

It is relatively easy to pin down non-topological conditions. For example, a complete Boolean algebra comes from a von Neumann algebra if and only if it admits sufficiently many normal positive measures.

The reason for requiring additional conditions to be topological is that the resulting definition of a measurable space should be easy to relate to other parts of general topology.

For example, consider the forgetful functor that sends a commutative von Neumann algebra to its underlying C*-algebra. Applying the Gelfand-Neumark duality to both sides we obtain the forgetful functor from the category of measurable spaces to the category of compact regular locales (or compact Hausdorff spaces, if we have the axiom of choice). A topological definition of a measurable space should allow for an explicit description of this forgetful functor in terms of open sets. Other potential applications include functors that send a locale (or a topological space) to its underlying measurable space, or a smooth manifold to its underlying measurable space. More speculatively, one could use this definition to replace ad hoc techniques of classical point-set measure theory with standard tools of general topology.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ Can you give any examples of complete Boolean algebras that cannot occur as the projection algebra of a commutative von Neumann algebra? (Say, using your "non-topological condition"?) $\endgroup$
    – Manny Reyes
    Jul 26, 2011 at 22:21
  • 3
    $\begingroup$ @Manny: The Boolean algebra of clopen sets of any compact regular extremally disconnected space (i.e., a stonean space) is a complete Boolean algebra, which does not come from a von Neumann algebra unless the original space was hyperstonean. In a hyperstonean space meager sets are rare (nowhere dense), which is usually not the case for stonean spaces. In fact, every stonean space canonically splits as a disjoint union of a hyperstonean space, a space that contains a dense meager set, and a space where every meager set is rare and the support of every measure is rare. $\endgroup$
    – Dmitri Pavlov
    Jul 28, 2011 at 12:16
  • $\begingroup$ This is probably not relevant, but what does it mean to admit ‘sufficiently many’ positive normal measures? (I assume that ‘normal’ here means that the rule that $\lim_n \mu(A_n) = 0$ when $A_n \searrow_n \varnothing$ applies for any net $A$, not just for an $\omega$-sequence.) $\endgroup$
    – Toby Bartels
    Oct 8, 2011 at 6:32
  • $\begingroup$ @Toby: Here “sufficiently many positive normal measures” means that for any x≠0 we can find a positive normal measure μ such that μ(x)=1. Equivalently we can say that the supremum of the supports of all normal positive measures equals 1. By a normal positive measure here I mean an additive map from the boolean algebra to the positive reals that preserves suprema of arbitrary sets. Your condition is equivalent to the preservation of suprema of arbitrary sets. $\endgroup$
    – Dmitri Pavlov
    Oct 8, 2011 at 12:41
  • $\begingroup$ @Toby: One reference for this notion is Takesaki's Theory of Operator Algebras I, Definition III.1.14, Theorem III.1.17, and Theorem III.1.18. $\endgroup$
    – Dmitri Pavlov
    Oct 8, 2011 at 12:43

1 Answer 1

Reset to default
3
$\begingroup$

A pair $(\mathcal{A}, \mu)$ of a $\sigma$-complete boolean algebra $\mathcal{A}$ and a functional $\mu : \mathcal{A} \to [0, \infty]$ is called a measure algebra if $\mu$ is strictly positive and countably additive on disjoint sequences. A measure algebra is semifinite if whenever $\mu(a) = \infty$ there exists a $b < a$ such that $0 < \mu(b) < \infty$. A measure algebra is localizable if it is complete and semifinite.

A measure algebra can be constructed from a measure space by taking the boolean algebra of equivalence classes of measurable sets modulo the null sets. In the reverse direction, by the Loomis-Sikorski Theorem, every $\sigma$-complete boolean algebra is isomorphic to the quotient of the $\sigma$-algebra $\{ A \bigtriangleup B : A \text{ clopen }, B \text{ meager } \}$ by the $\sigma$-ideal of meager sets. An ordinary measure can be defined on the quotient in the natural way, and the concrete measure algebra of the resulting measure space is isomorphic to the original measure algebra.

The preceding construction can also be used to show that the localizable measure algebras are precisely the Boolean algebras of projections of commutative von Neumann algebras. Note the similarities between the definitions of a localizable measure algebra and a normal semifinite weight on a von Neumann algebra.

Due to the semifiniteness of the measure, the problem of characterizing the Boolean algebras that are measure algebras can be reduced to the case of finite measure. Call a $\sigma$-complete boolean algebra $\mathcal{A}$ finitely measurable when there exists a functional $\mu : \mathcal{A} \to [0, \infty)$ making it a measure algebra. Then a complete boolean algebra $\mathcal{A}$ has a functional $\mu : \mathcal{A} \to [0, \infty]$ making it a localizable measure algebra precisely when the set $\{ a \in \mathcal{A} : \mathcal{A}_a \text{ is finitely measurable } \}$ is order-dense in $\mathcal{A}$, where $\mathcal{A}_a$ is the principal ideal generated by $a$.

Unfortunately, even in the finite measure case there is no great solution to this problem. There’s a characterization by Kelley that reduces it to the existence of a strictly positive finitely additive measure and a combinatorial condition (what he calls weakly countably distributive). He also characterizes the existence of a finitely additive measure in terms of intersection/covering numbers. Gaifman wrote a survey paper on this problem, and Jech proved a game-theoretic characterization.

A good reference for most of the facts mentioned about measure algebras is volume 3 of Fremlin’s tome Measure Theory, particularly chapters 32 and 39.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I'm familiar with Fremlin's book. He doesn't seem to mention Jech's paper, though. Jech mentions other results apart from the game-theoretic Theorem 4.6. In §1 he states five equivalent characterizations of measure algebras that seem to use only order-theoretic notions, which would seemingly fit my request for topological characterizations. I'm also curious to which extent these descriptions can be used to develop measure theory. $\endgroup$
    – Dmitri Pavlov
    Oct 4, 2016 at 18:20
  • 1
    $\begingroup$ Fremlin also wrote a review about this problem recently (possibly even after this answer was written), with comprehensive references. (Don’t mind that the linked page says it’s a reprint of something else, the review is a comment after the reprinted paper.) $\endgroup$
    – Alex Shpilkin
    Nov 9, 2018 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.