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Stream: theory: categorical probability

Topic: Opposites of categories of measurable spaces

Nathaniel Virgo (Oct 28 2022 at 04:14):

Here's a question that I'm asking out of curiosity.

$\mathbf{Set}^\mathrm{op}$ is isomorphic to the category of complete atomic Boolean algebras. On the other hand, a $\sigma$-algebra is something quite a lot like a complete atomic Boolean algebra. (I'm not completely sure but I think $\sigma$-algebras can be characterised as countably complete atomic Boolean algebras.)

Since the opposite of the category of complete atomic Boolean algebras is something interesting (namely $\mathbf{Set}$), I'm wondering whether the opposite of the category of measurable spaces can also be characterised concretely, and if so what it corresponds to.

Is that something that's known? And if it is, does anything interesting happen if we restrict to standard Borel spaces?

Morgan Rogers (he/him) (Oct 28 2022 at 06:15):

The category of measurable spaces is already the opposite, in some sense, since measurable maps are those whose inverse image function restricts to a Boolean algebra homomorphism

Nathaniel Virgo (Oct 28 2022 at 08:24):

That makes complete sense. I guess being the opposite is what makes them "spaces", in some sense. At least, I've heard people say that sort of thing before. I wish I had a better intuition for it.

Jean-Baptiste Vienney (Oct 28 2022 at 08:24):

Do you know in which measure it is possible to do probability theory or integration... in $Set^{op}$?

In the same way that the category $CRing^{op}$ is equivalent to the category of affine schemes while being easier to apprehend, maybe could it be easier to use $Set^{op}$ rather than the category of measurable spaces?

Are countably complete atomic Boolean algebras a specific type of complete atomic Boolean algebra or is it the contrary?

John Baez (Oct 28 2022 at 08:28):

I thought Nathaniel was aiming at a precise question, though I'm not quite sure what it was - either "what's the opposite of the category of countably complete boolean algebras?" or "what's the opposite of the category of measurable spaces?" I'm curious about these questions now.

Nathaniel Virgo (Oct 28 2022 at 08:30):

It was "what's the opposite of the category of measurable spaces?", but the other one sounds interesting too.

Tobias Fritz (Oct 28 2022 at 08:47):

That's a great question! I don't have an answer, but here are two closely related things:

• Dmitri Pavlov proved a categorical duality between the category of commutative von Neumann algebras and a certain category of "enhanced" measurable spaces, where being enhanced refers to having singled out a $\sigma$-ideal of null sets. This is a measure-theoretic analog of Gelfand duality. There had also been slightly earlier or concurrent work by @Robert Furber along the same lines.
• Fremlin's measure algebras take a dual approach from the start by developing everything in terms of $\sigma$-complete Boolean algebras.

I'm pretty sure that @Robert Furber will know more. I'd also be interested in a concrete dual description of the category of standard Borel spaces.

John Baez (Oct 28 2022 at 09:16):

I feel the category of commutative von Neumann algebras is a very beautiful approach to defining the "right" concept of measurable space, so I'm glad you proved that, @Tobias Fritz. By the time I finished writing my book on measurable 2-groups, I felt that working with these might be better than working with standard Borel spaces.

John Baez (Oct 28 2022 at 09:19):

So it sounds like you're hinting, though not stating, that the category of $\sigma$-complete boolean algebras is not equivalent to the category of commutative von Neumann algebras. They both seem like potentially "right" concepts of measurable space, so now I'm curious about their relation. Is there an equivalence or at least an adjunction between these categories, and what's it like?

Jean-Baptiste Vienney (Oct 28 2022 at 09:35):

This paper seems to be relevant to your question: Abelian von Neumann algebras, measure algebras and $L^{\infty}$-spaces

Tobias Fritz (Oct 28 2022 at 10:11):

@John Baez, it's not my result: as I mentioned, Gelfand duality for commutative von Neumann algebras is due to Dmitri Pavlov and Robert Furber.

I agree that the (opposite) category of commutative von Neumann algebras is a lot more appealing than the category of standard Borel spaces. However, I'm not at all convinced that it's the "right" category for categorical probability. An important question in this regard is, what is the "right" monoidal structure on it? There are two canonical choices, and both seem to be deficient:

• The usual "spatial" tensor product of von Neumann algebras is too small, since the multiplication map $\mathcal{A} \otimes \mathcal{A} \to \mathcal{A}$ of a commutative von Neumann algebra $\mathcal{A}$ is not continuous in the relevant topology (unless $\mathcal{A}$ is finite-dimensional). So intuitively, this gives a category of measurable spaces in which the usual diagonals aren't measurable. That sounds bad! And in particular one cannot put the machinery of Markov categories to work here.
• The Dauns tensor product, which actually is the coproduct of commutative von Neumann algebras. This is very much in the spirit of having a nice category, and Robert recently showed that this category supports a probability monad as well. However, this tensor products creates very large algebras, for which even the predual is typically not separable! So my suspicion is that the hom-sets in this category are too large to constitute a meaningful model of probability.

I've been wondering if one can relax the definition of von Neumann algebra a little bit, for example to AW*-algebra, and get something that resolves this dilemma. It's very tempting to think that there must be a category of commutative algebras that is both nice and actually works for the purposes of probability.

John Baez (Oct 28 2022 at 10:14):

Is there a way to turn a $\sigma$-complete boolean algebra into some sort of algebra of complex-valued functions, generalizing how we can form the algebra of bounded measurable functions on a measurable space?

Tobias Fritz (Oct 28 2022 at 10:58):

I don't know! I would bet that the forgetful functor from commutative von Neumann algebras to $\sigma$-complete Boolean algebras has a left adjoint, and that this can be obtained as an instance of the special adjoint functor theorem. Probably Robert can tell us something more definite.

Tobias Fritz (Oct 28 2022 at 11:01):

It's worth noting that that forgetful functor actually factors across the category of complete Boolean algebras (without the $\sigma$), since the projection lattice of a von Neumann algebra is complete, and normal $*$-homomorphisms preserve arbitrary directed suprema and infima (basically by definition).

John Baez (Oct 28 2022 at 11:06):

Do you feel this means we should look at some generalization of commutative von Neumann algebras whose projection lattice is just $\sigma$-complete? Or does the true completeness of the projection lattice seem like mainly a good thing?

Tobias Fritz (Oct 28 2022 at 11:45):

I wish I could tell! Two years ago I was playing around with some ideas where one keeps true completeness. But it seems hard to get a good intuition on this. It's frustrating that commutative von Neumann algebras are so well-behaved as a category, but still not quite satisfactory (to me at least).

John Baez (Oct 28 2022 at 11:51):

So the Dauns tensor product of two copies of $L^\infty[0,1]$ is not $L^\infty([0,1]^2)$?

John Baez (Oct 28 2022 at 11:53):

That would count as annoying.

Tobias Fritz (Oct 28 2022 at 12:03):

Right, it's much larger than that, and $L^\infty([0,1]^2)$ is a proper quotient of it. A good way to understand this is to note that the diagonal map $[0,1] \to [0,1]^2$ does not dualize to $L^\infty([0,1]^2) \to L^\infty([0,1])$, since the latter map is not well-defined. I think Robert once noted that the tensor product algebra does not even have separable predual, but I don't know how to prove that.

$L^\infty([0,1])$ is not what I'd want to be looking at from a categorical probability perspective anyway, since I'd want the unit interval to be represented by an algebra whose states correspond to the probability measures on $[0,1]$. That's not the case for $L^\infty([0,1])$, since its states are only those probability measures that are absolutely continuous with respect to the Lebesgue measure.

BTW, one more recent paper relevant to this whole thread is Foundational aspects of uncountable measure theory by Jamneshan and Tao. There's a lot in there, and in particular some discussion of categorical dualities. The sheer number of categories that appear in that paper make it a tough read.

John Baez (Oct 28 2022 at 19:51):

Okay, thanks!

Nathaniel Virgo (Nov 06 2022 at 00:44):

Is there any reason to think that the category of measurable spaces and the category of countably complete atomic Boolean algebras aren't exactly each others' opposites? It seems to me that they should be, but I don't trust myself not to have missed some important subtlety.

(Does $\sigma$-complete mean the same as what I was calling countably complete, i.e. having all countable meets and joins?)

Robert Furber (Nov 06 2022 at 02:08):

Nathaniel Virgo said:

Is there any reason to think that the category of measurable spaces and the category of countably complete atomic Boolean algebras aren't exactly each others' opposites? It seems to me that they should be, but I don't trust myself not to have missed some important subtlety.

(Does $\sigma$-complete mean the same as what I was calling countably complete, i.e. having all countable meets and joins?)

First, thanks to @Tobias Fritz for bringing this thread to my attention. I have some further replies to add after this one but I'll add them when I have the time.

Consider two sets $X,Y$ with different cardinality, and make them into measurable spaces with the indiscrete $\sigma$-algebra (so $(X, \{\emptyset,X\})$ for $X$). So they give isomorphic 2-element $\sigma$-Boolean algebras (this is how I prefer to abbreviate "$\sigma$-complete Boolean algebra", imagine that Booleanalgebra is a single word if it helps at all). But $X$ and $Y$ cannot be isomorphic as measurable spaces because the forgetful functor would give us an isomorphism of sets, contradicting their different cardinalities.

Another problem is that the $\sigma$-algebra of a measurable space has no requirement to be atomic, and in fact we need non-atomic ones if we want uncountable products. Let $X$ be an uncountable set, and $2$ be the discrete measurable space structure on the set $2$, and then $2^X$ is not atomic. The reason is that every measurable set $S$ only uses countably many of the generators $B_{x,b} = \{ f : X \rightarrow 2 \mid f(x) = b \}$ of the product $\sigma$-algebra, so if $S$ is not empty, there is some index $x \in X$ such that for both $S \cap B_{x,0}$ is non-empty and $S \cap B_{x,1}$ is non-empty, so $S$ cannot be an atom.

I could also have called this "the Baire $\sigma$-algebra of $2^X$ with the product topology".

Anyway, the duality between measurable spaces and $\sigma$-Boolean algebras was worked out by Sikorski. I wrote a long post about it, and then the Latex didn't compile and I accidentally deleted it while trying to edit it. The point is, it's not a categorical duality but a contravariant adjunction, but you can restrict to where the unit and counit are isomorphisms to get an equivalence. The analogue of "Set" is $\sigma$-perfect measurable space (one in which every 2-valued measure is the $\delta$-measure of a unit point) and the analogue of "CABA" was not given a name by Sikorski, but I have called a $\sigma$-spatial $\sigma$-Boolean algebra by analogy to locale theory, and then you do get an equivalence between these two things. Section III of this paper is the place where I stated these things (Sikorski didn't use category theory, and I think preceded the notion of adjunction). Nowadays I would define this adjunction by having $2$ be a dualizing object between measurable spaces and $\sigma$-Boolean algebras, and therefore using $\sigma$-homomorphisms into $2$ rather than working with $\sigma$-ultrafilters.

Robert Furber (Nov 06 2022 at 02:42):

Tobias Fritz said:

I'd also be interested in a concrete dual description of the category of standard Borel spaces.

I worked the following one out at the time I was writing the "Unrestricted Stone" article I linked earlier. The dual, under Sikorski duality, of standard Borel spaces is countably presented $\sigma$-Boolean algebras.

• $(X,\Sigma)$ standard Borel implies $\Sigma$ countably presented: Every standard Borel space is isomorphic to either $(X,\mathcal{P}(X))$ for a countable set $X$ or $(2^\omega,\mathcal{B}o(2^\omega))$. The second case is the free $\sigma$-Boolean algebra on $\omega$, so is countably presented, and in the first case you have a countable presentation by taking the atoms as generators and relations making $x \land y = 0$ for distinct atoms and the join of all the atoms equal to 1. This is a countable number of relations, so $\powerset(X)$ is countably presented.
• $A$ countably presented implies there exists a standard Borel space $(X,\Sigma)$ with $\Sigma \cong A$: Let $G$ be a countable set of generators of $A$, so the free $\sigma$-Boolean algebra on $G$ is $B = \mathcal{B}o(2^{G})$. Since $A$ is countably presented, there exists a countable set $R \subseteq B$ defining an ideal $I$ such that $B/I \cong A$. Since $R$ is countable, its join $S$ exists, and generates the same ideal $I = \langle S \rangle$, so $I$ is a principal ideal. So $B/I \cong \mathcal{B}o(2^{G} \setminus S)$. The space $2^{G} \setminus S$, as a Borel subspace of a standard Borel space, is a standard Borel space.

This also means that you can define $\sigma$-complete Boolean algebras as countable-limit-preserving functors from standard Borel spaces into $\mathbf{Set}$. I intended to use this in an article about my approach to defining random variables in a general setting, but it's on a lower priority behind making progress with other things (also some of the results proved in it were published independently by Jamneshan and Tao).

One thing to be careful about is that countably generated and countably presented are not the same here. A countably generated $\sigma$-spatial $\sigma$-Boolean algebra is the Borel algebra of a separably metrizable space, not necessarily completely metrizable (this distinction is very important in measure theory), and in general a countably generated $\sigma$-Boolean algebra might not have any $\sigma$-homomorphisms into $2$ at all, so isn't part of the duality.

Tobias Fritz (Nov 06 2022 at 08:10):

That's really neat! It sounds quite similar to the idempotent adjunction between topological spaces and locales, which restricts to the equivalence between sober spaces and locales.

To summarize, let me state the two duality theorems that Robert mentions in a way that unfolds the definitions a bit more. First, the following two categories are dually equivalent:

• The category of $\sigma$-perfect measurable spaces and measurable maps, where a measurable space $(X,\Sigma_X)$ is $\sigma$-perfect if the canonical map from elements of $X$ to $\sigma$-complete ultrafilters on $\Sigma_X$ is a bijection. This condition is analogous to sobriety of a topological space.
• The category of $\sigma$-spatial Boolean algebras having countable joins and meets, with Boolean algebra homomorphisms preserving those. Here, $\sigma$-spatial means that every two distinct Boolean algebra elements can be separated by a $\sigma$-complete ultrafilter.

Second, this duality restricts to a dual equivalence between the following two categories:

• The category of standard Borel spaces and measurable maps.
• The category of countably presented $\sigma$-Boolean algebras.

The definition of "countably presented $\sigma$-Boolean algebra" is so much nicer than the definition of standard Borel space: the latter is quite unsatisfying to a category theorist since it refers to structure (namely a metric) that then gets completely ignored in the definition of the morphisms. (Of course one could also define standard Borel via Kuratowski's theorem, but that feels at least as unsatisfying.)

I don't want to pester you with too many questions, but I can't hold back on these:

• There's a deep theorem of descriptive set theory which states that every monomorphism of standard Borel spaces is an isomorphism onto its image. Could the above duality facilitate a simpler proof of this? In those terms, the statement is that if $\mathcal{A} \to \mathcal{B}$ is an epimorphism of countably presented $\sigma$-Boolean algebras, then it induces an isomorphism $\mathcal{A} / \langle S \rangle \cong \mathcal{B}$ for a principal ideal $\langle S \rangle$ in $\mathcal{A}$. So it seems to me that the difficulty is to show that "epimorphism" means "surjective", and then this statement should follow easily.
• Do you know to what extent $\sigma$-perfect measurable spaces are as well-behaved for the purposes of probability as standard Borel spaces are? For example, is there a disintegration theorem for probability measures on product spaces?

Tobias Fritz (Nov 06 2022 at 08:14):

Addendum: I see now that the duality for standard Borel spaces also appears on p.2 of Ruiyuan Chen's paper A universal characterization of standard Borel spaces.