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New page: positive cone, including the extended positive cone of a W*-module.
Now also ordered group. Pretty basic.
Now also ordered group.
Is there any good categorical elucidation of an “ordered group” or similar thing?
I’ve been toying with something that is also “translation-invariant” (a buzzword you should tag your “condition” with for the purpose of text searches).
I’ve had problems stating such a thing as just a category because it seems to need a set of objects, and two sets of arrows, and - where arrows compose within sets but not across.
In my example I have one dimension of arrows that are a preorder while the second dimension is a partial order and they obey the “translation-invariant” condition.
The structure of my example (2 arrow dimensions and a arbitrary constraint) strikes me as quite ad hoc, so I was wondering if there was any slick way to show that what seems to be its structure is really just a reflection of some deeper structure - maybe a beautiful comma category.
Well, an ordered group is just a monoidal category which happens to be both thin and grouplike.
This stuff has been added (which anyone should feel free to do).
A remark on extended positive cones: The predual of a von Neumann algebra (also referred to as the space of finite measures in the positive cone article) is not a W*-module because it is not complete in the ultraweak topology. See Blecher's paper http://dmitripavlov.org/scans/blecher.pdf and references therein for the relevant facts about W*-modules.
Morally, however, the claims are true. Given a von Neumann algebra M its L_p-spaces (here L_p := L^{1/p}) for real p have a positive part, which can be extended. For p=0 we obtain the usual positive cone of positive affiliated operators of M (note that this does not require M to be represented on a Hilbert space, in particular it is completely canonical). For p>0 we obtain the positive cone of unbounded positive p-densities. For p=1 this is precisely the space of (normal) weights (not necessarily semifinite or faithful) on M. If M is commutative, these are precisely unbounded positive measures on M, not necessarily semifinite. These cones for p>0 are all canonically homeomorphic to each other via the obvious power operations. The case of p=0 is very different. However, there is a polar duality between the extended cones for p=0 and p=1 (this is in fact one way to define them, see Takesaki's book).
The above already covers the example of unbounded measures, but it can be extended further to L_p-modules of Junge and Sherman if desired. For p=0 we obtain the usual W*-modules, for p=1/2 we obtain representations of M on Hilbert spaces. The case of modules can in fact be reduced to the case of algebras using the standard linking algebra construction.
I wrote a little bit about L_p-modules here: http://mathoverflow.net/questions/45871/subfactor-theory-and-hilbert-von-neumann-algebras/45964#45964
Thanks, I changed ‘W*-module’ to ‘module over a W*-algebra’ (and similar); I didn’t mean to include every hypothesis actually implied by that term.
If you can define the extended positive part of more general ordered TVSes, then that would be nice. Maybe we can use that every Hausdorff TVS is at least embedded in its double dual?
@Toby: The revised version of the statement can only be true if you allow for uncountable linear combinations (and in this case it is indeed true, because every weight on a von Neumann algebra can indeed be represented as an infinite sum of finite weights).
The double dual trick can only work for p<=1 (which is the case for weights), because the dual of L_p(M) for p>1 is trivial unless M has an atomic part.
uncountable linear combinations
I was trying to avoid this, because we don't yet have an ambient space to do this in. Formal finite linear combinations are easy enough, but formal limit points are harder to describe.
the dual of L_p(M) for p>1 is trivial
Ah, yes. One could use the algebraic dual, but it would be hard to make that give what we want.
Also, is it true that every measurable space in the classical sense (a set equipped with a σ-algebra) is equivalent to a localisable measurable space in the sense that they have the same measures? What if we broaden the class of measurable spaces by equipping each with a σ-ideal of null sets (but still not requiring localisability)?
The problem with allowing only finite linear combinations is that every finite linear combination of finite measures has sigma-finite support, and if the von Neumann algebra itself is not sigma-finite, then there are measures with non-sigma-finite support, e.g., faithful measures.
What if we broaden the class of measurable spaces by equipping each with a σ-ideal of null sets (but still not requiring localisability)?
Radon-Nikodym theorem fails for every nonlocalizable measurable space, so if by "the same measures" you mean in particular that the isomorphism has to preserve the conjugation action, then the answer is no.
I don't know what conjugation action you mean, but I don't intend to preserve the correspondence between L1 functions and measures.
For an example of what I mean, consider the real line with the Borel sets (and the empty set as the only null set), which is not localisable. However, as you have explained elsewhere, this can be replaced by a localisable measurable space which is rather larger (in the sense of elements). I thought (but now I see that you don't say this) that this space has the same measures as the Borel measures. (But otherwise I wouldn't think it an adequate replacement.)
The extended positive cone of the predual has the following algebraic structure: it has a zero element, addition, multiplication by positive scalars, and conjugation: if f is an element of the extended positive cone of the predual and a is an element of the extended positive cone of the original von Neumann algebra, then we can define afa*=afa, the conjugation of f by a. Of course, in the commutative case this is just aa*f=a^2f.
If the extended positive cones of the preduals of two measurable spaces are "the same", I interpret this in particular to mean that the algebraic structures identified above should be isomorphic. However, if the Radon-Nikodym theorem fails for one space, but not for the other one, then there is no such isomorphism.
As for Borel measures, it does seem to me (even though I don't have a proof) that the space of measures on the measurable space Y under discussion should be the space of Borel measures on the real line.
This is strongly related to the question of how (equivalence classes of) measurable subsets of Y are related to Borel subsets of the real line. One could probably use a Stieltjes-like construction to turn a Borel measurable subset of the real line into a continuous monotonous function on the real line, which in its turn can be identified with a unique endomorphism of Y, from which we could probably extract (an equivalence class of) a measurable subset of Y.
Technical note: Put dollar signs around mathematics to make it come out more or less like mathematics in LaTeX (really iTeX). But (which is perhaps the main difference between iTeX and mathematics in LaTeX) use spaces between unrelated letters. Thus $a f a^* = a f a$ produces ‘afa* = afa’. (I think of the Radon–Nikodym theorem as about the action (a, f) ↦ af, but I guess in the noncommutative case we need to use the conjugation action instead?)
These actions of L0 = L∞ on measures are precisely what I don't want to preserve, because L0(X) (where X is the real line equipped with the Borel sets and only the empty set as null) is ugly (since it's not a von Neumann algebra). So we replace X with Y, thereby replacing L0(X) with the von Neumann algebra L0(Y) but keeping M(X) (the space of finite measures on X) the same as M(Y) (which now is the predual of L0(Y) and justifiably also called L1(Y)). Right?
I care because I want to be able to start with V = M(X) = M(Y) = L1(Y) as an ordered vector space, say that this is being acted on by the von Neumann algebra A = L0(Y), and so construct the extended positive cone V̄+, without saying anything about X (or Y) directly.
@Toby: iTeX's output is completely unreadable in my browser. I'm using the latest version of Chromium, so the problem clearly lies with iTeX, and in fact it seems to be caused by extensive use of pixel distances in iTeX's CSS. Apparently, my laptop's display has much higher DPI than the DPI of the monitor on which iTeX's CSS was created, which results in numerous overlaps between letters and generally unreadable output. Here is what this page looks like in my browser: http://dmitripavlov.org/screenshot.png Thus I am forced to look at the TeX source of your comments to make sense out of them, and clearly I cannot use this functionality in my comments.
I guess in the noncommutative case we need to use the conjugation action instead?
That's correct, conjugation preserves the positive cone, but multiplication doesn't.
Right?
This seems plausible to me, but requires proof. Perhaps something along the lines of what I sketched above might work.
OK, so what I thought was true may be true but is not known. Fair enough.
Does the stuff on the nLab itself (like positive cone and measurable locale) also look like this on your browser? If not, it's probably a problem with CSS settings right here on the Forum, and Andrew can probably fix it if I tell him about it. In the meantime, I'll remove the iTeX from my own comment.
The nLab looks ok to me. No overlaps etc.
I've now started a discussion about the technical problem with iTeX.
(Also, I actually edited my comments like I said that I would.)
Also, my technical note for your comment #12 above is this: If you don’t want an asterisk to produce italics (technically emphasis), then put a backslash before it: \* produces ‘*’.
Also by the way, Dmitri, do you know a good term for a set X equipped with a σ-algebra M ⊆ P(M) equipped with a σ-ideal N ⊆ M? For the special case where N = {∅}, the standard term is ‘measurable space’; for the special case where M/N is complete, we may say ‘localizable measurable space’. But is there a good term for the general case, not conflicting with standard terminology?
@Toby: Why do you claim that the term ‘measurable space’ requires N={∅}? In my mind N can be any sigma-ideal in M. For the case N={∅} I use the term ‘measure space’.
The meaning of the term ‘measurable space’ that I was taught in school and that one finds online is nothing but a set with a σ-algebra. And then a measurable function is a function whose preimage operation respects this. This is what I’m calling the standard meaning.
For the generalised notion of measurable space as a set equipped with a σ-algebra equipped with a σ-ideal, we have (even in the complete case) a more complicated definition of measurable function, but if both σ-ideals are trivial, then this reduces to the standard meaning of measurable function. So the standard meaning of measurable space is equivalent to the generalised meaning where the σ-ideal is trivial.
For the case N={∅} I use the term ‘measure space’.
This doesn’t match any terminology that I’ve seen before.
@Toby: I guess you are right, but measurable spaces in the usual sense are almost never localizable if we assume N={∅}. Thus one is always forced to consider them together with some sigma-ideal of negligible sets, which can come from some measure, for example. This is what happens in all textbooks on measure theory, by the way. So the concept of a measurable space in the old sense is of little use unless it is accompanied by additional data.
Given the fact that measurable spaces in the old sense form a full subcategory of measurable spaces in the new sense, perhaps it makes sense to simply reuse the same term for the generalized notion.
@Toby: I just found a paper which seems to be very relevant to your original question: http://www.ams.org/mathscinet-getitem?mr=193488 (see also references therein).
Given the fact that measurable spaces in the old sense form a full subcategory of measurable spaces in the new sense, perhaps it makes sense to simply reuse the same term for the generalized notion.
Yes, but it might also be confusing. If there’s a good name for the generalised notion, then I would like to consider it. If there isn’t, oh well.
I just found a paper which seems to be very relevant to your original question
You mean the one at the end of comment #9? Yes, it seems relevant. I don’t immediately see that it answers the question, but I’ll try it.
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