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Stream: event: Categorical Probability and Statistics 2020 workshop

Topic: Commutative Wstar-algebras

Tobias Fritz (Jun 09 2020 at 15:04):

I'd like to argue that the opposite of the category of commutative $W^*$-algebras, with Scott continuous (aka "normal") $*$-algebra homomorphisms as the morphisms, deserves to be explored further as a candidate for a convenient category of measurable spaces. This category is well-known to have a measure-theoretic flavour, but my impression is that the details of this are not yet as well-understood and appreciated as they perhaps deserve to be. As the "opposite" indicates, we're supposed to think of a $W^*$-algebra as an algebra of measurable functions (on an imaginary measurable space), and of a morphism as the "predicate transformer" or "Heisenberg picture" version of a measurable map.

Here are a few features of this category. @Robert Furber has worked on it extensively during his thesis, so he may be able to say more.

• Dropping the commutativity assumption provides an immediate generalization to the quantum setting. (@Arthur Parzygnat mentioned ongoing work on generalizations to the infinite-dimensional case. Does it involve $W^*$-algebras?)
• The category is well-known to be complete and cocomplete
• By reflecting a monoidal closure result of Kornell back to the commutative case by abelianization, it apparently follows that this category is even cartesian closed. Edit: as pointed out by Robert, I had misread Kornell's statement, which rather amounts to monoidal closure.
• It already comes with a supercategory of "stochastic maps" given by the normal positive unital maps between $W^*$-algebras. I suspect that the completeness properties of the category can be used to show that this supercategory is the Kleisli category of a monad which behaves like the monad of probability measures.
• In this category, every probability measure has a support, and more generally every stochastic map has a support, in the sense of a factorization through a smallest subobject. Of course, these supports will generally be very non-spatial, perhaps vaguely similar to double negation sublocales in locale theory.

There also are ways to describe this category directly without taking opposites, but the details here are notoriously subtle. @Robert Furber seems to understand this very well, so perhaps he can say more. There's also a recent preprint on this by Dmitri Pavlov.

Sam Staton (Jun 09 2020 at 19:44):

Hi Tobias Fritz, about the monoidal structure of commutative W*-algebras being cartesian. Probably you thought about this more than me. My worry is about a (co-)diagonal map for the tensor. In the category of localizable measurable spaces (or whatever they are called), is the diagonal of $\mathbb{R}\times \mathbb{R}$ not a designated null-set? Maybe my intuition is off. (Here I mean $\mathbb{R}$ with the Lebesgue null-sets, as a localizable measurable space.)

Tobias Fritz (Jun 09 2020 at 20:17):

Sam Staton said:

Hi Tobias Fritz, about the monoidal structure of commutative W*-algebras being cartesian. Probably you thought about this more than me. My worry is about a (co-)diagonal map for the tensor. In the category of localizable measurable spaces (or whatever they are called), is the diagonal of $\mathbb{R}\times \mathbb{R}$ not a designated null-set? Maybe my intuition is off. (Here I mean $\mathbb{R}$ with the Lebesgue null-sets, as a localizable measurable space.)

Good point! I don't know what this looks like from the Gelfand dual perspective in terms of localizable measurable spaces, but at the $W^*$-algebra level you have to be careful with the proper choice of tensor product, since there are at least two known possible options, and the usual spatial tensor product indeed suffers from that deficiency. The relevant tensor product is rather the "Dauns tensor product", as nicely explained in Section 5.1 of Duplicable von Neumann algebras by Kenta Cho and Bram Westerbaan. This is the coproduct in the category of commutative $W^*$-algebras.

Edit: I see now that Von Neumann Algebras form a Model for the Quantum Lambda Calculus, also by Kenta and Bram, has already done a quantum analogue of what I'm suggesting, by using monoidal closure to prove that $W^*$-algebras interpret the quantum lambda calculus.

John Baez (Jun 09 2020 at 22:10):

What happens when we take the Dauns tensor product of two copies of the commutative W*-algebra $L^\infty[0,1]$?

John Baez (Jun 09 2020 at 22:11):

I'm too lazy to look it up, but it sounds like we don't get $L^\infty[0,1]^2$.

(I'm using Lebesgue measure to define these $L^\infty$ spaces.)

Robert Furber (Jun 09 2020 at 22:42):

Tobias Fritz said:

• By reflecting a monoidal closure result of Kornell back to the commutative case by abelianization, it apparently follows that this category is even cartesian closed.

Tobias, when we talked about this yesterday, I said that Kornell proved (opposite) monoidal closure for the spatial tensor, which is _not_ the coproduct of commutative W-star-algebras. I hadn't read the paper in a while, and then you assured me that Kornell used the Dauns tensor, which is the coproduct of commutative W-star-algebras (because it is the "commuting coproduct" in the category of W-star-algebras). I agreed that if this were the case, then the opposite category of commutative W-star-algebras would be cartesian closed - but this was only because I didn't have access to the paper.

Kornell uses a tensor with a tilde ~ over it for the Dauns tensor (see just before Proposition 6.1), and uses a tensor with a line over it for the spatial tensor (Definition 7.1). He _clearly_ states that Dauns tensoring by L^\infty(S^1) does _not_ preserve limits (essentially because ergodic transformations exist). So Corollary 6.7 is that commutative W-star op is _not_ cartesian closed. It is monoidal closed for the spatial tensor. More information on what the spatial tensor is in a minute.

Robert Furber (Jun 09 2020 at 22:43):

John Baez said:

I'm too lazy to look it up, but it sounds like we don't get $L^\infty[0,1]^2$.

(I'm using Lebesgue measure to define these $L^\infty$ spaces.)

You are right. That is the spatial tensor. For example, the Dauns tensor, being a product, has a diagonal map.

Robert Furber (Jun 09 2020 at 22:59):

Tobias Fritz said:

There also are ways to describe this category directly without taking opposites, but the details here are notoriously subtle. Robert Furber seems to understand this very well, so perhaps he can say more. There's also a recent paper on this by Dmitri Pavlov.

In the current version of that paper, there is an error that I explained in the following link, and that Dmitri agreed is wrong and agreed to correct:
https://mathoverflow.net/questions/359825/image-of-probability-measures-under-measurable-mappings/359832#comment907656_359832

The problem is that we are not guaranteed to get a normal homomorphism from a measurable transformation, only a sigma-normal one. This is really the most difficult part of the deleted chapter of my thesis. The way of fixing it in the case of compact strictly localizable spaces is given here:
https://mathoverflow.net/a/361787/61785

Compact doesn't mean compact in the topological sense, by the way, but is a topology-free way of describing the good properties that Radon measures have. Compact strictly localizable complete measure spaces are the correct generalization of Rokhlin's standard Lebesgue spaces when all restrictions on cardinality are removed.

John Baez (Jun 09 2020 at 23:09):

Robert Furber said:

John Baez said:

I'm too lazy to look it up, but it sounds like we don't get $L^\infty[0,1]^2$.

(I'm using Lebesgue measure to define these $L^\infty$ spaces.)

You are right. That is the spatial tensor. For example, the Dauns tensor, being a product, has a diagonal map.

Good. I was mainly asking what W*-algebra we do get by taking the Dauns tensor product of $L^\infty[0,1]$ with itself. Does anyone around here know?

Robert Furber (Jun 09 2020 at 23:16):

John Baez said:

Good. I was mainly asking what W*-algebra we do get by taking the Dauns tensor product of L^\infty[0,1]L
∞
[0,1] with itself. Does anyone around here know?

You aren't going to like it. It doesn't have separable predual. I think it can't be proved to exist without AC.

Robert Furber (Jun 09 2020 at 23:21):

Maybe that was too curt. But it will take me a lot of effort to write down a description of it. I've had big problems with people asking me questions out of curiosity, I know how to answer them, I spend a lot of time writing an answer, and then they don't even give a shit.

Robert Furber (Jun 10 2020 at 00:05):

Here is Dauns's paper: https://www.jstor.org/stable/1996061?seq=1
He builds the tensor using the universal enveloping algebra and the projective C*-tensor. Kornell and others do it by verifying the solution set condition of the adjoint functor theorem.

Robert Furber (Jun 10 2020 at 00:43):

Sam Staton said:

My worry is about a (co-)diagonal map for the tensor. In the category of localizable measurable spaces (or whatever they are called), is the diagonal of $\mathbb{R}\times \mathbb{R}$ not a designated null-set? Maybe my intuition is off. (Here I mean $\mathbb{R}$ with the Lebesgue null-sets, as a localizable measurable space.)

This is why the Dauns tensor (which is the coproduct of commutative $W^*$-algebras, so the product when you have the opposite category) is not the same as the spatial tensor. Let me switch back to commutative $W^*$-algebras without taking the opposite category so I don't get confused.

Products are what you think they should be - they correspond to disjoint unions of measure spaces. Equalizers are a little bit funny, but they're funny in the right way - as described in Lemma 6.3 of Kornell's paper, the equalizer of an ergodic transformation and the identity is $\mathbb{C}$, so ergodic transformations look like they have no invariant subsets, as they should (their invariant subsets all have measure zero or complement of measure zero). However, the colimits come out funny. The coproduct is the Dauns tensor, and I showed that you get an unexpectedly large colimit of a sequence of finite-dimensional commutative $W^*$-algebras in my answer here: https://mathoverflow.net/a/284325/61785
(There I had to go through a faff to prove that the filtered colimit of commutative $C^*$-algebras is commutative - just ignore that and work with commutative $C^*$-algebras and $W^*$-algebras to begin with).

I think this is because we do not want the colimits that are there. The spatial tensor of $L^\infty(X,\mu)$ and $L^\infty(Y,\nu)$ does come out as $L^\infty(X \times Y, \mu \otimes \nu)$. The spatial tensor has a universal property as an "independent coproduct" (or product in the opposite category). Alex Simpson has described a version of this here: https://web.archive.org/web/20180919135310/http://homepages.inf.ed.ac.uk/als/Talks/cambridge14.pdf
linking to this answer:https://mathoverflow.net/a/120957/61785
Fremlin also describes a version of this in 325D here:
https://www1.essex.ac.uk/maths/people/fremlin/chap32.pdf

I think what is necessary is to always use "indepedent coproducts" (independent products viewed in the opposite category).

$L^1(X \times Y, \mu \otimes \nu)$ also has a universal property in Banach spaces, due to Grothendieck as the projective Banach space tensor of $L^1(X,\mu)$ and $L^1(Y,\nu)$.

Robert Furber (Jun 10 2020 at 04:27):

Tobias Fritz said:

• It already comes with a supercategory of "stochastic maps" given by the normal positive unital maps between $W^*$-algebras. I suspect that the completeness properties of the category can be used to show that this supercategory is the Kleisli category of a monad which behaves like the monad of probability measures.

This is strongly related to what Prakash (with Chaput, Danos and Plotkin) does with $L^\infty$ (he restricts to the case of a probability measure and uses only the positive cone, though as he mentioned in his talk you can go back and forth between cones and vector spaces with only a little difficulty). Conditional expectations are exactly what they should be. I consider this to be the right starting point, working with probability measures (on standard Borel spaces) because this is the case that matters for applications.

Actually, my original reason for considering Gelfand duality for $W^*$-algebras in my thesis was exactly to get a monad on measure spaces whose Kleisli category was dual to positive unital maps of $W^*$-algebras, as the $W^*$-analogue of what Bart and I did for $C^*$-algebras in https://doi.org/10.2168/LMCS-11(2:5)2015 . Around 2015, I thought I had proved this, modulo Gelfand duality of measure spaces (I didn't use the solution set condition to define the monad, but rather a construction involving the enveloping $W^*$-algebra). However, writing up Gelfand duality for measure spaces took up so much time I didn't have time to put this monad in my thesis, and then certain members of the thesis committee demanded that I delete the part on Gelfand duality for measure spaces anyway. That left such a bad feeling that I never really picked up any momentum to work on finishing it again (this is why only I say I thought I had proved it - not because I have serious doubts about the proof, but because I only consider myself to have actually proved something after I have written it up, because small gaps in the argument always need to be filled in during this process.).

I will tell you why a nontriviality arises in this approach. The Kleisli adjunction between $\mathbf{CHaus}$ and $\mathcal{K}\ell(\mathcal{R})$ (for $\mathcal{R}$ the Radon monad) becomes an adjunction $\mathbf{CC}^*(C(\mathcal{S}(A)),B) \cong \mathbf{CC}^*_{PU}(A,B)$ (where $\mathcal{S}$ is the state space and $C$ the algebra of continuous functions). So the adjunction that defines the monad $T$ that you want here is of the form $\mathbf{CW}^*(T(A),B) \cong \mathbf{CW}^*_{PU}(A,B)$. Consider the special case of $B = \mathbb{C}$. For any $W^*$-algebra $A$, $\mathbf{CW}_{PU}^*(A,\mathbb{C})$ is the normal state space (corresponding to probability density functions or "truly continuous" probability measures), while $\mathbf{CW}^*(A,\mathbb{C})$ is the "normal spectrum", which is the set of isolated points (points that are an open subset, as singletons) of the Gelfand spectrum of $A$. Under our adjunction $\mathbf{CW}^*(T(\mathbb{C}^2), \mathbb{C}) \cong \mathbf{CW}^*_{PU}(\mathbb{C}^2,\mathbb{C}) \cong [0,1]$, so the Gelfand spectrum of $T(\mathbb{C}^2)$ has a continuum of isolated points, i.e. it contains a discrete continuum as a subspace. This means that any faithful measure for $T(\mathbb{C}^2)$ is not $\sigma$-finite. Some people are misled by an analogy into thinking that $T(A)$ should be $\ell^\infty$ of the normal state space on $A$. I can show that this is not the case with an explicit example - $T(\mathbb{C}^2) \cong C([0,1])^{**}$. This is proved using the Yoneda lemma, by the following calculation using the universal property of the enveloping algebra and the fact that all maps out of a finite-dimensional $W^*$-algebra are normal. Let $B$ be a commutative $W^*$-algebra:

$\mathbf{CW}^*(C([0,1])^{**}, B) \cong \mathbf{CC}^*(C([0,1]), B) \cong \mathbf{CC}^*_{PU}(\mathbb{C}^2, B) \cong \mathbf{CW}^*_{PU}(\mathbb{C}^2,B) \cong \mathbf{CW}^*(T(\mathbb{C}^2), B)$

(I showed this example to Bram Westerbaan and it is included in one of his papers, I think. This is a simplification of the general construction (the general construction compensates for the fact that the third step only works in the finite-dimensional case).)

The previous paragraph is the entire reason I considered non-$\sigma$-finite measures at all (not being a standard Borel space comes along for the ride). It is just necessary if you want such a monad, by the universal property. Unfortunately people always cut me off and don't let me explain this - people seem to just want to believe I like making things difficult for no reason (or because I want to prove I am intelligent or something). It is impossible to defend against such an accusation - any attempt to show why it's not true can simply be wilfully misunderstood, and so my reputation just gets ruined with such people forever (they never ask for clarification of anything ever again, for example).

In this part of measure theory the difference between $\aleph_0, \aleph_1$ and $2^{\aleph_0}$ (the continuum) cannot be neglected, as certain blithe souls are wont to. This is why a special argument is needed to prove that $L^\infty(f)$ for a measurable map is normal and not just $\sigma$-normal. If there is a measure $\mu$ on $[0,1]$ with the power set $\sigma$-algebra that vanishes on singletons, and $\nu$ is the counting measure (this makes $L^\infty([0,1],\mathcal{P},\nu) \cong \ell^\infty([0,1])$, then the identity map $f : ([0,1], \mathcal{P},\mu) \rightarrow ([0,1],\mathcal{P}, \nu)$ is a measurable transformation (measurable and reflects nullsets) but $L^\infty(f)$ is $\sigma$-normal but not normal - the projections $(\chi_{\{x\}})_{x \in [0,1]}$ are all mapped to $0$ by $L^\infty(f)$ but their join is $1$. In the end, the right definitions rule out this example - but you have to know what the right definitions are first.

Since people here seem to be interested in it, I am prepared to write up my work in this area, but only if people understand that if you ask for a monad here, you might get something you don't like. That's not my fault. Don't shoot the messenger.

Sam Staton (Jun 10 2020 at 07:25):

Thank you, Robert Furber, for this explanation of the monad. I am trying to understand it by matching it up with my informal intuitions about localizable measurable spaces. I found a brief email chat with Matthijs Vákár from 2015, where we came up with the following worry. If there was a Giry monad on localizable measurable spaces, what would $G(X,M,N)$ be? would it be absolutely continuous measures? but what about the unit (Dirac?)? And now perhaps you are even pointing out that, "what would $G(2,2^2,\{\emptyset\})$ be?" doesn't have an especially nice answer, because what would the null-sets of $G(2,2^2,\{\emptyset\})$ be?
[Here I am writing $(X,M,N)$ with $M$ a $\sigma$-algebra and $N$ an ideal of null-sets. Also I may have discussed this with you in the past, and if so I am sorry if I forgot it.]

Tobias Fritz (Jun 10 2020 at 14:08):

Thank you @Robert Furber for the correction on the tensor product and elaborating on probabilistic Gelfand duality. That $C([0,1])^{**}$ comes up as the object representing normal positive unital maps out of $\mathbb{C}^2$ makes sense to me; I think of the formal dual of $C([0,1])^{**}$ as the object playing the role of the unit interval in this approach. So to me, this monad still seems like an interesting example worth having in the bestiary of probability monads, even if the lack of cartesian closure makes it less convincing.

Anyway, since the subtopic on $W^*$-algebras has quite a number of comments now, I'll make it into its own topic.

Over the past two days, I've said a number of embarrassing and premature things which I shouldn't have. Probably the mixture of excitement and exhaustion left over from the workshop has contributed to this. So I'll stop commenting for some time to cool down and relax.

Robert Furber (Jun 10 2020 at 15:58):

Sam Staton said:

I found a brief email chat with Matthijs Vákár from 2015, where we came up with the following worry. If there was a Giry monad on localizable measurable spaces, what would $G(X,M,N)$ be? would it be absolutely continuous measures? but what about the unit (Dirac?)? And now perhaps you are even pointing out that, "what would $G(2,2^2,\{\emptyset\})$ be?" doesn't have an especially nice answer, because what would the null-sets of $G(2,2^2,\{\emptyset\})$ be?

You are absolutely right to be worried about that. Really the problem is that the whole thing goes nonconstructive. The space $G(2,\mathcal{P}(2), \{\emptyset\})$ is isomorphic (I take it to be equal) to the Gelfand spectrum of $C([0,1])^{**}$. The null sets are the meagre sets (which happen to be the nowhere dense sets). Since $C([0,1])^{**}$ is a $W^*$-algebra, this $\sigma$-ideal does form the nullsets of a measure (proved to exist using Zorn's lemma).

In all the following discussion I am going to just use $X$ for a compact complete strictly localizable measure space, instead of repeating every time the $\sigma$-algebra and either the defining measure or the $\sigma$-ideal of null sets.

The thing I was calling $T$ earlier, the left adjoint to the inclusion $\mathbf{CW}^* \hookrightarrow \mathbf{CW}^*_{PU}$, defines a comonad $T : \mathbf{CW}^* \rightarrow \mathbf{CW}^*$, and by measure-Gelfand duality defines the monad $G : \mathbf{Meas} \rightarrow \mathbf{Meas}$.

Regarding the points of $G(X)$ versus the absolutely continuous measures. We have:

$\mathbf{Meas}(1, G(X)) \cong \mathbf{CW}^*(T(L^\infty(X)), \mathbb{C}) \cong \mathbf{CW}^*_{PU}(L^\infty(X), \mathbb{C}) = \mathcal{NS}(L^\infty(X))$

The last thing on the right is the normal states of $L^\infty(X)$, which each arise uniquely from integrating against "truly continuous" measures (in the case that $X$ is $\sigma$-finite, truly continuous is the same as absolutely continuous, in general absolutely continuous only implies $\sigma$-normal, not normal), or equivalently from the probability density functions $\mathcal{DF}(X)$, the set of elements of $L^1(X)$ that are positive with integral 1. In fact, I was using $[0,1] \cong \mathcal{D}(2) \cong \mathcal{NS}(\mathbb{C}^2)$ earlier - we have $T(\mathbb{C}^n) \cong C(\mathcal{D}(n))^{**}$ by essentially the same argument.

However, $\mathbf{Meas}(1,G(X))$ is not the points of $G(X)$, but only those points that have strictly positive measure (otherwise the inclusion map of a point won't be nullset-reflecting). If those were the only points, then we would have, for example, $C([0,1])^{**} \cong \ell^\infty([0,1])$. This is not so, and a common mistake by many respectable people who are much less careful than you (the others being thinking $C([0,1])^{**} \cong \mathcal{L}^\infty([0,1])$ or $L^\infty([0,1])$ with the Lebesgue measure -- all wrong):

$\mathcal{NS}(\ell^\infty([0,1])) \cong \ell^1([0,1]) \not\cong C([0,1])^* \cong \mathcal{NS}(C([0,1])^{**})$

The reason for the non-isomorphism is that $C([0,1])^*$ contains a copy of $L^1([0,1])$ with Lebesgue measure, whereas in $\ell^1([0,1])$ the unit ball is the $\sigma$-convex hull of its extreme points.

So $G(2)$ contains $[0,1] \cong \mathcal{DF}(2) = \mathcal{D}(2)$ as a discrete subset where each point has strictly positive measure, and then a lot of ultrafiltery fairy dust (to the tune of at least $2^{2^{\aleph_0}}$) to avoid $L^\infty(G(2))$ being equal to $\ell^\infty(G(2))$.

Robert Furber (Jun 10 2020 at 17:17):

Arthur Parzygnat said:

Robert Furber, I'm only beginning to explore this area and I am curious about this paragraph. What is the precise statement? If $(X,\Sigma,\mu)\xrightarrow{f}(Y,\Lambda,\nu)$ is a measure-preserving function between measure spaces, then the corresponding homomorphism $L^{\infty}(Y,\Lambda,\nu)\xrightarrow{L^{\infty}(f)}L^{\infty}(X,\Sigma,\mu)$ (obtained via pullback) need not be normal? If so, under what assumptions on the measure spaces guarantee normality? (Simply pointing me to a reference would be incredibly helpful for me).

Unfortunately I couldn't find any reference for this material, which is why I wrote the deleted chapter of my thesis :sad:. But since there is general interest in it, I will make the time to write this up (though I think I will solicit your opinions on the manuscript rather than sending it off for publication). I don't know a measure-preserving example, nor of a proof that no measure-preserving example exists - the example I know is nullset-reflecting but not measure-preserving.

What almost everybody does is to restrict to the case that $\mu$ and $\nu$ are $\sigma$-finite. In this case, $\sigma$-normal implies normal. This is because you can't have an uncountable almost disjoint family of sets of positive measure in these spaces, so all joins of projections reduce to countable joins. I would be using this approach too, but for the fact I mentioned elsewhere that you can't get a monad for probabilistic Gelfand duality for commutative $W^*$-algebras using only $\sigma$-finite spaces.

Strictly localizable spaces are the correct generalization of $\sigma$-finiteness to include all commutative $W^*$-algebras - these are disjoint unions of spaces of finite measure, with the adjustment to the $\sigma$-algebra that is required (arbitrary, not just countable, unions of measurable sets in these disjoint pieces must be measurable, and the measures must also add up correctly). So the problem occurs with spaces that are too wide (technical term - have large cellularity), or are too discrete.

The problem of non-normal maps $f : (X,\mu) \rightarrow (Y,\nu)$ cannot occur if $(X,\mu)$ is compact (not topologically, a measure-theoretic notion) and $(Y,\nu)$ is strictly localizable. I proved this by using strict localizability of $Y$ to reduce to the case where it is discrete, and then using compactness of $X$ to apply Fremlin's Lemma 451P or 451Q (depending on the version, as mentioned here: https://mathoverflow.net/a/361787 ). As it is already required to use the category of compact, complete and strictly localizable spaces for there to exist a measurable function $f : (X,\mu) \rightarrow (Y,\nu)$ for each normal *-homomorphism $L^\infty(Y,\nu) \rightarrow L^\infty(X,\mu)$, and all measure spaces occurring in practice are compact and strictly localizable (and all commutative $W^*$-algebras can be defined by such a measure) I didn't think this was any great loss.

Robert Furber (Jun 10 2020 at 17:37):

One addendum: Sometimes people think that if $(X,\mu)$ and $(Y,\nu)$ are $\sigma$-finite, and $g : L^\infty(Y,\nu) \rightarrow L^\infty(X,\mu)$ is a normal *-homomorphism, then there exists $f : (X,\mu) \rightarrow (Y,\nu)$ such that $g = L^\infty(f)$. This is true for standard Borel spaces and standard Lebesgue spaces, but not for $\sigma$-finite measure spaces, nor even for probability measure spaces.

Let $(X,\mu)$ be $1$ with its unique probability measure. Let $Y$ be your favourite uncountable set with the countable-cocountable $\sigma$-algebra (the $\sigma$-algebra consisting of countable sets and their complements). Define the measure $\nu$ on $Y$ to assign measure $0$ to countable sets and measure $1$ to cocountable sets. Then $L^\infty(X,\mu) \cong \mathbb{C} \cong L^\infty(Y,\nu)$. But there is no nullset reflecting map $(X,\mu) \rightarrow (Y,\nu)$, because each singleton in $Y$ has $\nu$-measure zero.

The reason for this is that $(Y,\nu)$ is not compact (in the measure-theoretic sense). What's worse is that we could take $Y = [0,1]$ and then $\nu$ is the restriction of Lebesgue measure to the countable-cocountable $\sigma$-algebra. This is what convinced me that the sub-$\sigma$-algebra definition of conditional expectation isn't quite the Right Thing™ - we should really be considering complete subalgebras of the measure algebra, not the $\sigma$-algebra (or sub-$W^*$-algebras, or doing conditional expectations in terms of homomorphisms).

Sam Staton (Jun 10 2020 at 19:40):

Thanks @Robert Furber, that's very informative, I appreciate it. Incidentally, standard probability spaces and null-set reflecting maps are one of Alex Simpson's index categories, so maybe his approach is a way to sidestep the "ultrafiltery fairy dust".

Arthur Parzygnat (Jun 10 2020 at 21:18):

@Robert Furber, woah, thank you. This is indispensable knowledge. Your comment about the lack of fullness of the functor $L^{\infty}$ even in the case of (arbitrary) probability spaces answers another question I had, so I am glad you also addressed this (and even provided a simple counter-example!). I would be immensely grateful if you could share a draft of your "lost" thesis chapter. To give you some context, I've been thinking about disintegrations (closely related to conditional expectations) and Bayesian inversion in an attempt to generalize some theorems we have from the finite-dimensional (non-commutative) case to the infinite-dimensional setting. But while we were working on this and analyzing the relationship to (commutative) conditional expectations in the measure-theoretic sense, we came across many questions of this sort (and my question in Panangaden's talk was related to this). The most general results I could find in this regard appear in Fremlin's work, which is a beast and would be a bit time-consuming for me to go through. Thanks again!
@Tobias Fritz Feel free to move this discussion with Robert to the stream on commutative Wstar-algebras (I will do it if I'm allowed, but I don't know how).

Paolo Perrone (Jun 10 2020 at 22:32):

"Algebraic probability", or "algebraic measure theory". I like the idea. (I have to do a lot of reading.)

Robert Furber (Jun 14 2020 at 14:07):

Sam Staton said:

Thanks Robert Furber, that's very informative, I appreciate it. Incidentally, standard probability spaces and null-set reflecting maps are one of Alex Simpson's index categories, so maybe his approach is a way to sidestep the "ultrafiltery fairy dust".

The problem is that a standard probability space has at most countably many atoms. Any measure space $(X,\mu)$ such that $C([0,1])^{**} \cong L^\infty(X,\mu)$ has a continuumful of atoms, as I explained above.

For any commutative $W^*$-algebra, $A$ there is a measure space $(X,\mu)$ that is a coproduct of points and measure spaces of the form $2^\kappa$ with its independent-fair-coin measure, for $\kappa$ an infinite cardinal, such that $L^\infty(X,\mu) \cong A$, so one can always get rid of the ultrafilters. But there is a "complementarity" between points and functions - in the aforementioned representation, one can easily describe the points, but the functions realizing a given normal homomorphism of $W^*$-algebras are defined nonconstructively. When using the Gelfand spectrum, the points are defined nonconstructively, but it is easy to define a continuous function realizing any given normal homomorphism of $W^*$-algebras, using ordinary Gelfand duality.

In any case, let $\Sigma$ mean the algebra of Lebesgue measurable sets on $[0,1]$, $\mu : \Sigma \rightarrow [0,1]$ Lebesgue measure and $\nu : \mathcal{P}([0,1]) \rightarrow [0,\infty]$ be the counting measure. I believe I have a proof that $C([0,1])^{**} \cong L^\infty( ([0,1], \nu) + ([0,1]^2, \mu \otimes \nu))$. This may even be a well-known fact to some people. I think I will include this as an example, if it turns out right.

Robert Furber (Jun 14 2020 at 14:23):

Arthur Parzygnat said:

Robert Furber, I would be immensely grateful if you could share a draft of your "lost" thesis chapter.

I am somewhat unsatisfied with it now, and I was writing a new version, but that is not close enough to being finished. I should be clear - the deleted chapter was only on non-probabilistic Gelfand duality for $W^*$-algebras and compact complete strictly localizable measure spaces. Since I have agreed to write up my work on probabilistic Gelfand duality for $W^*$-algebras, I will put a summary of the relevant facts about the non-probabilistic version. Another option is to look at Dmitri Pavlov's paper, posted earlier. As far as I can tell, there is no problem with the $\sigma$-finite case there, and he has agreed to fix the problem I showed him.

Tobias Fritz (Jun 14 2020 at 15:11):

Robert Furber said:

I believe I have a proof that $C([0,1])^{**} \cong L^\infty( ([0,1], \nu) + ([0,1]^2, \mu \otimes \nu))$.

That's very interesting, Robert! Every Borel probability measure on $[0,1]$ defines a normal state on $C([0,1])^{**}$, right? Is there an explicit description of how such a normal state acts on the right-hand side of your (conjectural?) isomorphism?

I think I can see what happens with the pure point part and the absolutely continuous part of a given measures. So by the Lebesgue decomposition, it's enough to consider the question for a singular continuous measure.

Edit: Hmm, by what you said in your preceding paragraph, I guess the proof of your isomorphism is too non-constructive for there to be a meaningful answer to my question. I guess it amounts to showing that both sides have the same number of points and $2^\kappa$ pieces for every $\kappa$, or something along these lines, right?

Arthur Parzygnat (Jun 14 2020 at 20:08):

Robert Furber said:

Arthur Parzygnat said:

Robert Furber, I would be immensely grateful if you could share a draft of your "lost" thesis chapter.

I am somewhat unsatisfied with it now, and I was writing a new version, but that is not close enough to being finished. I should be clear - the deleted chapter was only on non-probabilistic Gelfand duality for $W^*$-algebras and compact complete strictly localizable measure spaces. Since I have agreed to write up my work on probabilistic Gelfand duality for $W^*$-algebras, I will put a summary of the relevant facts about the non-probabilistic version. Another option is to look at Dmitri Pavlov's paper, posted earlier. As far as I can tell, there is no problem with the $\sigma$-finite case there, and he has agreed to fix the problem I showed him.

Right, basically my initial idea when he posted his paper was (in attempt to understand his paper) to go directly from his category CSLEMS (appropriately modified to include Markov kernels) to CVNA (also modified appropriately to include positive unital maps). But I'm not sure how much time I want to devote to this since my main interest in this is only to better understand conditional expectations, disintegrations, Bayesian inversion, etc. Since I expect this will take some time, I'm not really in a rush. Nevertheless, I'm personally interested in your summary of the relevant facts regarding the probabilistic version. Please keep me in mind when you have some of that available. :)

Robert Furber (Jun 15 2020 at 15:48):

Tobias Fritz said:

That's very interesting, Robert! Every Borel probability measure on $[0,1]$ defines a normal state on $C([0,1])^{**}$, right?

Yes, bijectively, in fact.

Edit: Hmm, by what you said in your preceding paragraph, I guess the proof of your isomorphism is too non-constructive for there to be a meaningful answer to my question. I guess it amounts to showing that both sides have the same number of points and $2^\kappa$ pieces for every $\kappa$, or something along these lines, right?

The proof goes by finding a set of $2^{\aleph_0}$ diffuse probability measures on $[0,1]$ with disjoint supports in $C([0,1])^{**}$, and then proving that there are not more than $2^{\aleph_0}$ probability measures on $[0,1]$ (many ways to do this), so that a maximal family of states with disjoint support in $C([0,1]))^{**}$ has cardinality $2^{\aleph_0}$. We make a maximal family starting with the atoms coming from singletons of $[0,1]$, followed by these diffuse measures, all of which have $L^\infty([0,1],\mu)$ isomorphic to $L^\infty([0,1],\lambda)$, with $\lambda$ being Lebesgue measure (by Rokhlin's classification of standard Lebesgue spaces). I do not construct a partition of unity of $C([0,1])^{**}$ by projections, and rely on Zorn's lemma to show such a thing exists (basically the commutative case of the proof in Takesaki volume II that every $W^*$-algebra admits a faithful normal semifinite weight).

It seems likely that this fact, or the equivalent fact that $C([0,1])^*$ is isometrically isomorphic to $\ell^1([0,1]) \oplus^1 \bigoplus^1_{x \in 2^{\aleph_0}} L^1([0,1])$ is well-known to the right people, but I'll include a proof of it in my write-up.

Of course, as I warned earlier, this is a great deal harder than working with $C([0,1])^{**}$ directly. I think Arveson or somebody said that early on people defined the universal von Neumann algebra of a $C^*$-algebra $A$ by finding a universal representation of $A$ and taking the bicommutant of $A$ in it, and the reason that Sakai's definition of $W^*$-algebra was the one that caught on was that you could use it to define $A^{**}$ as the universal $W^*$-algebra directly.

Robert Furber (Jun 15 2020 at 15:51):

@Arthur Parzygnat I have some time to work on it now, so I'll produce a write-up of it that will perhaps form (part of) a paper. I can't give a time scale, unfortunately, it's just done when it's done.

Tobias Fritz (Jun 15 2020 at 17:36):

That makes sense, thanks. I can see why it's harder than working with $C([0,1])^{\ast\ast}$ directly, but that isomorphism still provides some additional helpful intuition on what that beast looks like.