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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeDec 23rd 2011

In the course I notice the following remnant discussion, which hereby I move from there to here

Eric: Some day this should hopefully tie into the beautiful stuff on Leinster measure (blog).

+– {: .query} Eric: $d$ is also the exterior derivative and $d\mu$ is a volume form. Is there a nice way to say this that is consistent with the above? Update: The Usenet discussion probably discusses this, but I’m too lazy to read the whole thing right now (past my bedtime!).

Toby: As a volume form is not, in general, the exterior derivative of anything, you cannot interpret the ‘$\mathrm{d}$’ in (eq:excessive) as an exterior derivative. You can do this for the ‘$\mathrm{d}$’ in (eq:Leibniz), of course, because that is on the real line, where the volume form is the exterior derivative of the identity function $(x \mapsto x)$. But in general, an absolutely continuous Radon measure $\mu$ on an oriented smooth $n$-dimensional manifold $X$ defines an $n$-form on $X$ (and vice versa), so if you call the form $\mu$ as well, then you want to use (eq:simple). (The exterior deriviative of the volume form, of course, would be zero!)

I'm actually halfway through writing an article differential form where I will address some of this. (I guess that I'm going through old Usenet posts of mine; I am using the conversation that you and I had with John in this old thread as reference for some of it!)

Eric: I look forward to it! By the way, that Usenet discussion was a nice blast from the past :)

Toby: I should note that, even given what I wrote above, there is still a slight clash of notation between measure theory and differential topology. To fix this, the $\mathrm{d}x$ in (eq:full) could be replaced with $|\mathrm{d}x|$. This has to do with the whole the-absolute-value-of-an-$n$form-is-an-$n$pseudoform and integration-of-$n$pseudoforms-is-more-fundamental-than-integration-of-$n$forms issue. I referred to this clash of notation in our Usenet conversation here.

Eric: It’s starting to come back to me now. Yeah, the measure is really a pseudo $n$-form and we settled on the notation $|dx|$ for that. We should at least give a nod to that idea I think in the above.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJun 30th 2012

I noticed that Wikipedia’s entry on meaure spaces does not mention morphisms of measure spaces. Then I discovered that the $n$Lab’s measure space does neither. That’s a major omission for a category-theoretic wiki.

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeJul 1st 2012

I’m having trouble coming up with a notion of morphism that reproduces the correct notion of isomorphism and allows the following four maps to be morphisms (much less (co)product diagrams when relevant):

• $\empty \to \mathbb{R}$,
• $\mathbb{R} \to \mathbb{R} + \mathbb{R}$,
• $\mathbb{R} \to \pt$,
• $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$.

Restricting attention to localisable measure spaces and taking morphisms to be equal if equal almost everywhere doesn’t help. But it allows us to dualise the question and ask:

• What is the appropriate notion of morphism between (commutative) $W^*$-algebras each of which is equipped with a weight?

Another analogous question is this:

• What is the appropriate notion of morphism between (smooth $n$-dimensional) manifolds each of which is equipped with an $n$-pseudoform?
• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJul 1st 2012
• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJul 1st 2012

We could just mention some possible choices that come up in practice. Such as the one used in Gromov’s article.

(I can’t do it right now, only have a minute online…)

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeJul 2nd 2012

I think that Gromov’s is a measuarable function (possibly modulo equality almost everywhere) that preserves measure. That category has neither an initial nor terminal object, but it has the correct core.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeAug 2nd 2012

Still doing some catch-up after months of inactivity…

Re #4: you might be right about “no definitive answer”, but it seems to me that Dmitri Pavlov’s was a very interesting answer, which I’d like to digest further. I am struck by the duality with commutative von Neumann algebras.

As Dmitri says, if one proceeds naively and considers measure-preserving functions, one doesn’t get a category with good co/completeness properties, just as in the case of manifolds with a volume form. But I idly wonder whether one can say anything interesting about the category of measure spaces as a monoidal (not cartesian monoidal) category…

• CommentRowNumber8.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2012

@Toby: The morphisms you identified in comment 3 can all be considered as morphisms between measurable locales = measurable spaces = opposite commutative von Neumann algebras.

There is also a good (from the nPOV) notion of morphism that incorporates weights / measures into it. In the language of von Neumann algebras (not necessarily commutative) such a morphism is a pair (f,T), where f: A\to B is a morphism of von Neumann algebras and T is an operator valued weight associated to f.

Neither Wikipedia nor nLab seem to have any articles about operator valued weights, which seems like a major omission to me given their importance in the field of von Neumann algebras.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeAug 3rd 2012

Dmitri: of course we would like it very much if you recorded your insights on operator-valued weights in the nLab…

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeAug 3rd 2012

Exposing my ignorance for a moment, and returning to the duality that struck me in #7, I notice that in the article on von Neumann algebras, this duality is called Gelfand-Naimark duality, with a link that directs to Gelfand spectrum. But the article on Gelfand spectrum does not discuss this formulation of Gelfand-Naimark.

Indeed, Google led me to a discussion at the Noncommutative Geometry blog, where Gelfand-Naimark seems to connote, for different people, different theorems. The commutative Gelfand-Naimark theorem I take as referring to a duality between commutative $C^\ast$-algebras and compact Hausdorff spaces; this is sometimes called just the Gelfand-Naimark theorem, and also sometimes called Gelfand duality as here in the nLab. (See also Johnstone’s Stone Spaces, p. 161.) The Gelfand-Naimark theorem I thought was a statement about general $C^\ast$-algebras: that they can be realized as algebras of bounded operators on a Hilbert space. And now we have it as a duality statement about commutative von Neumann algebras.

So, there doesn’t seem to be complete consensus on nomenclature. I for one would be happy to consider ’Gelfand-Naimark’ as referring to a body of related results, referring variously to commutative $C^\ast$-algebras, noncommutative $C^\ast$-algebras, commutative von Neumann algebras, etc. – this would seem to be the approach being tacitly or unconsciously adopted at the nLab. But all this should be discussed and clarified.

• CommentRowNumber11.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2012

@Todd: I guess I should first edit the von Neumann algebra article. Right now it doesn't even mention morphisms of von Neumann algebras, which is ridiculous from the nPOV.

• CommentRowNumber12.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2012

@Todd: Since the commutative Gelfand-Neumark duality was apparently introduced in its current form by Gelfand and Neumark in their original paper http://www.mathnet.ru/links/54a391a1d5335904acac6283dc6d6c48/sm6155.pdf, is there any reason we should drop Neumark's name from the commutative Gelfand-Neumark duality? Perhaps one of the earlier papers by Gelfand already contains the statement?

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeAug 3rd 2012

@Dmitri: I don’t personally know the history. Johnstone cites in his historical notes, page 166, two papers of Gelfand which I interpret him as claiming contain the essence of what we are here calling Gelfand duality:

• I.M. Gelfand, On normed rings, Dokl. Akad. Nauk USSR 23, 430-432

• I.M. Gelfand, Normierte Ringe, Mat. Sb. 9 (51), 3-24.

(On the spelling of Naimark/Neumark, we had a discussion about it starting here. Was it you who was arguing vigorously at MO that the spelling really ought to be Neumark, as it is a case not of transliteration but of how M.N. chose to spell his name in relevant publications?)

• CommentRowNumber14.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2012

@Todd: I have just looked into these papers myself and it seems that Satz 10 and Satz 10' in Gelfand's Normierte Ringe essentially state the commutative duality theorem. Link: http://www.mathnet.ru/links/9ee7dd923849a78d694ec7957f4da6e0/sm6046.pdf.

So I guess we should talk about Gelfand duality for (commutative) C-algebras (commutative C-algebras are contravariantly equivalent to compact regular spaces/locales) and (commutative) von Neumann algebras (commutative von Neumann algebras are contravariantly equivalent to measurable spaces/locales) and the Gelfand-Neumark theorem for C-algebras (every C-algebra has a faithful representation on a Hilbert space) and von Neumann algebras (every von Neumann algebra has a faithful (normal) representation on a Hilbert space).

Of course, Gelfand and Neumark only considered the case of C*-algebras, but I don't know any better names for the case of von Neumann algebras.

Concerning the names, I don't think it makes any sense to refer to the paper that is signed "I. Gelfand and M. Neumark" in any way other than Gelfand-Neumark.

Some authors intentionally distort the names when they cite this particular paper, which I find completely ridiculous (all bibliographic references must reproduce authors' names exactly as they are written in the paper).

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeAug 3rd 2012

For the purposes of this discussion, I’ll go along with the spelling ’Neumark’. Certainly I understand your reasoning, but I prefer to wait before deciding to edit.

Your suggestion for how to name these four distinct theorems doesn’t seem bad, but who first established the duality between commutative von Neumann algebras and localizable measurable spaces/locales? If it was X and not I.M. Gelfand, it seems that “Gelfand-X duality” could be a reasonable solution for this case. Possibly a similar proposal for the case of general von Neumann algebras.

• CommentRowNumber16.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2012
• (edited Aug 3rd 2012)

@Todd: This is precisely the topic of my question on MathOverflow: http://mathoverflow.net/questions/23408/reference-for-the-gelfand-neumark-theorem-for-commutative-von-neumann-algebras

The names of Jacques Dixmier and Irving Segal should certainly be mentioned in connection with Gelfand duality for commutative von Neumann algebras. The relevant papers (both from 1951) are

Dixmier, J. Sur certains espaces considérés par M. H. Stone. Summa Brasil. Math. 2 (1951). 151–182.

Segal, I. E. Equivalences of measure spaces. Amer. J. Math. 73 (1951). 275–313.

Dixmier's paper is mostly about the fact that commutative von Neumann algebras correspond to hyperstonean spaces under the usual Gelfand duality for commutative C*-algebras.

Segal's paper introduces localizable measurable spaces and essentially proves that equivalence classes of bounded measurable functions on such spaces form a von Neumann algebra. His exposition was later improved by John Kelley:

Kelley, J. L. Decomposition and representation theorems in measure theory. Math. Ann. 163 (1966) 89–94.

There are also some folklore facts, for example the fact that the projection functor maps commutative von Neumann algebras to complete Boolean algebras fully faithfully. I would expect such a result to be known already to Murray and von Neumann, but I cannot find a reference.

• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeAug 4th 2012

@Dmitri #16: thanks for this! Stupid question: by ’projection functor’, do you mean a functor that assigns to a commutative von Neumann algebra the Boolean algebra of idempotent elements?

• CommentRowNumber18.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 4th 2012

@Todd: Not just idempotent, but self-adjoint idempotent.

• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeAug 4th 2012

Ah right; thanks.

• CommentRowNumber20.
• CommentAuthorTobyBartels
• CommentTimeAug 5th 2012

@Dmitri #8: Sure, measurable spaces are no problem. Measurable spaces form a nice category, but measure spaces don’t; that’s what we’re getting here.

It’s yet possible that there could be a nice $\mathcal{M}$-category of measure spaces, where the tight morphisms are simply morphisms of measurable spaces that preserve measure exactly (so the thin isomorphisms are the correct isomorphisms of measure spaces), while the loose morphisms are something more general. I’m thinking of the way in which $Hilb$ (with short linear maps as morphisms) isn’t very nice, but $Hilb_TVS$ (with bounded linear maps as morphisms) is fairly nice (even nicer than its supercategory $TVS$) but has too many isomorphisms; so we combine these into a single $\mathcal{M}$-category.

Maybe the morphisms equipped with weights can serve as the loose morphisms here, actually giving us some sort of graded category?

• CommentRowNumber21.
• CommentAuthorTobyBartels
• CommentTimeAug 8th 2012
• (edited Aug 8th 2012)

It occurs to me that one could think of a (localisable) measurable space as a Banach space (the space of finite complex-valued measures) equipped with a commutative $C^*$-algebra structure (or a $*$-algebra structure with the property of being commutative and $C^*$) on its Banach dual (the algebra of bounded measurable functions). Now the morphisms are going the right way (since measures push forward while functions pull back, something that Jim Dolan said to me years ago without my realising its import). So the generating examples are the Banach space $\mathbb{C}$ with its dual the $*$-algebra $\mathbb{C}$ and the Banach space $L^1(\mathbb{R})$ with its dual the $*$-algebra $L^\infty(\mathbb{R})$.

Anyway, then a measure space is such an object together with an element of the Banach space (or of its extended positive cone, whatever). This makes it seem like the proper morphisms between such should preserve the measure exactly (not up to order on one side or the other, which is the obvious alternative).

• CommentRowNumber22.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 14th 2019

Corrected a typo.

• CommentRowNumber23.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 14th 2019
• (edited Jul 14th 2019)

The article claims:

Use R=]−∞,∞[ for a finite measure. Use ]−∞,∞] for a signed measure.

I am unaware of the source that defines signed measures in such an asymmetric way (i.e., as a difference μ−ν, where μ is a positive infinite measure and ν is a positive finite measure).

Typically, “signed measure” is a difference μ−ν, where μ and ν are positive infinite measures and μ or ν must be finite. This definition is used by Folland and Cohn in their textbooks.

• CommentRowNumber24.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 15th 2019

Corrected the definition of a signed measure. Added the definitions of semifinite and infinite measures. Mentioned the Radon-Nikodym theorem.

• CommentRowNumber25.
• CommentAuthorDmitri Pavlov
• CommentTimeJul 15th 2019