I added Doberkats discussion of barycenter maps which is how most people think about probability monads. Using the theory of barycenters requires some kind of compactness which is not a requirement of Polish spaces, however there are P-algebras which do satisfy that condition.

I also edited the proof of the Lemma concerning the fact that P(2)xP(2) is not isomorphic to P(3). By showing P(3) is a quotient of P(2)xP(2) that result is more obvious, and that proof generalized to the more general case quite easily.

Kirk Sturtz

]]>Fixed the broken link to Doberkats article, and put back Paolo’s characterization of the fundamental idea behind Doberkats representation of algebras.

Kirk Sturtz

]]>Added back the link to Doberkats paper, and showed the equivalence between his representation of algebras as convex spaces which satisfy some conditions, and the Eilenberg-Moore representation of algebras.

Kirk Sturtz

]]>Added corrections for the section on Algebras for the Giry monad which address the error pointed out by Paolo Perrone which result in showing the explicit construction of a G-algebra on Polish Spaces which is not free. Paolo, if you see any improvements to the arguments please make changes as needed. The section on the Kleisi category is woefully lacking. (But I confess having no interest in that.)

I still need to add back the reference to Doberkats article in the noted section.

Kirk Sturtz

]]>Paolo, I finally understood your concern.

Given an arbitrary space $X$ which is of cardinality $c$, we know there exists a unique map $h:GX \rightarrow X$ but how do we know $h=\mu_Y$ for some space $Y$? The concern is with the homeomorphism $X \rightarrow G(2)$ preserving the (super)convex space structure. I need to think about this.

]]>In the category of Polish spaces with measurable maps I agree with you - there probably is a a non-free algebra h:G([0,1] x [0,1]) ---> [0,1]x[0,1]. I suspect there are even non-free algebras h:G([0,1]) ----> [0,1] in Polish spaces with measurable maps. But not in the category of Polish spaces with continuous maps.

The fact that [0,1]x[0,1] is not "free" as a convex space has nothing to do with the terms "free/ non-free" when discussing algebras. When I say an (EM) algebra is free I am simply saying that it is a morphism mu_X: G(GX) ---> GX for some space X. If an algebra h:GY ----> Y is not a morphism of the form mu_X, i.e, there exists no space X such that Y=GX, then it is not-free. (I think we're talking apples and oranges :)

To ease your mind note that in the proof you can replace the space [0,1] with the space [0,1]x[0,1] and nothing changes. ]]>

Hi Kirk.

I do agree that superconvex spaces have been much overlooked in the past, and should be used more.

I still think that [0,1] x [0,1] is not free. The reason is that [0,1] x [0,1] is isomorphic to [0,1] in the category of Polish spaces (and measurable maps), but not in the category of Giry algebras. Said more precisely, given the componentwise algebra structure on [0,1] x [0,1] which you mentioned, we have that any isomorphism of measurable spaces [0,1] x [0,1] -> [0,1] is not a morphism of algebras. (Not even of convex or superconvex spaces).

The situation is similar for the category of groups: the group Z is free (over one generator), and *as a set* it is isomorphic to Z x Z, since they are both of countable cardinality. However, the group structures are not isomorphic (and the second one is not a free group).

To prove that [0,1] x [0,1] is not free as a convex space it suffices to show that its universal property fails. The universal property boils down to saying that every point is a unique convex combination of extreme points. Now the extreme points of [0,1] x [0,1] are the vertices of the square: (0,0), (0,1), (1,0), and (1,1). But now the point (1/2, 1/2) can be expressed both as 1/2 (0,0) + 1/2 (1,1) and as 1/2 (0,1) + 1/2 (1,0). (Geometrically, the square is not a simplex.)

Also, I think that in the next few days I should give a thorough makeover to the Giry monad article, a ton of research has been made since it was written.

]]>I believe that for any closed subspace of a real vector space the algebras on that (sub)space are going to be equivalent to the Kleisi category. Those categories of algebras are all subcategories of the category of superconvex spaces. In the few cases for where we know the algebras they all appear to be equivalent to the Kleisi category. Generally that property is unfortunately not considered in the derivation of the algebras, e.g. Doberkat work. This brings up a good point that should be considered for clarification on the nLab page for monads for proability measures - which categories of algebras have non-free algebras? I would like to add the case for Polish spaces with measurable maps which does have non-free algebras. Those algebras are the usual suspects G(n) ----> n which are not continuous. I don't know the algebras yet but there is an obvious subcategory of superconvex spaces which should at least lead to further understanding. (Polish spaces are countably generated so ... what are the countably generated superconvex spaces?) ]]>

Isn’t the closed unit square a non-free algebra?

]]>I rewrote the section on Algebras. The main change is that I wanted to point out why there are no non-free algebras for the case of Polish spaces. That proof points out explicitly why we need to think of the Giry monad via superconvex spaces - every algebra and every morphism of algebras is necessarily a countably affine map. You can’t prove the fact that the category of algebras is equivalent to the Kleisi category of the Giry monad without that fact which is why Doberkat didn’t recognize that his description of the algebras were the free algebras.

Kirk Sturtz

]]>added a few references (starting here) by C. Okay et al on using the Giry monad extended to simplicial sets to characterize quantum contextuality via simplicial homotopy theory

]]>added publication data for this item:

- Tom Avery,
*Codensity and the Giry monad*, Journal of Pure and Applied Algebra**220**3 (2016) 1229-1251 [arXiv:1410.4432, doi:10.1016/j.jpaa.2015.08.017]

Creating link to new page for super convex spaces.

Kirk Sturtz

]]>I suggest that if you request an author hyperlink, as in

- Ruben Van Belle,
*Probability monads as codensity monads*(2021) $[$arXiv:2111.01250$]$

that you immediately go and create that author page (just a brief page: pointer to the author’s web presence plus the reference).

Because otherwise nobody will do, typically, in which case the broken link looks bad.

]]>I added the reference to Ruben Van Belle’s article which can be viewed, in hindsight, as naturally suggesting the two constructions (dense and codense functors) which arise when considering super convex spaces. The explanation in the main article (on $G$-algebras) is the easiest way to motivate the constructions and explain why the countable set $\mathbb{N}$ is sufficient”.

Kirk Sturtz

]]>In explaining why we are naturally led to super convex spaces I added the relatively short proof that every G-algebra specifies a super convex space and, likewise, that every map of G-algebras is a countably affine map.

Incidentally, I suspect that Doberkats category may have the Klesi category = category of Algebras because continuous map do not permit discrete spaces so everything is embeddable into a vector space which yields the algebras as $\mu$ = averages over all probability measures.

Kirk Sturtz

]]>Corrections/Modifications. In the section on Algebras over the Giry monad: (1) Doberkat’s example is the free algebra on $\mathbf{2} =\{0,1\}$. It is rather useless as an example precisely because it is free. This leads to the second correction, (2) There seems to be a fundamental misunderstanding why one necessarily needs a concrete representation. I hope the added paragraph explains the necessity (concerning existence of non free algebras). (3) I updated my reference to my latest work - which gives the simplest most direct way to show the existence of non free algebras using the support of a probability measure (=added Lemma) which I wish I understood 10 years ago!)

Kirk Sturtz

]]>I’ll guess what is meant is that you take the inverse images of measurable sets of $[0, 1]$, which gives you a $\sigma$-algebra on $G(X)$.

]]>In the formula for the multiplication of the Giry monad (why is it called the “counit”?), what is P(X) in the integral? P(X) is only defined well below this formula. It looks like the formula should actually say G(X), not P(X).

The relevant change is Revision 23 by K. Sturtz on October 30, 2018.

]]>Between #21 and #28, there was discussion of additions by Sturz. He’s now brought out a new paper on Giry algebras over measurable spaces. Any views?

- Kirk Sturz,
*The existence and utility of Giry algebras in probability theory*, (arXiv:2006.08290)

With all this attention being paid to categorical probability theory now, surely some expert out there could help.

]]>I corrected the monadic part of the history.

]]>I have taken the liberty of turning the footnote into an item in the list if references – because that’s what it is and how you are using it in the text. Then I replaced the unspecified “in a video comment” with actual pointer to that reference.

(Am relieved to hear that the Pentagon doesn’t disagree with the category of probabilistic mappings.)

]]>Added a transcript of Lawvere’s comments.

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