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Stream: deprecated: our papers

Topic: Toposes of Topological Monoid Actions

Morgan Rogers (he/him) (May 04 2021 at 08:41):

My paper on topological monoids is finally complete!

https://arxiv.org/abs/2105.00772

The purpose of this paper was to answer the basic questions about toposes of actions of topological monoids on sets. Its introduction gives a more detailed account, but here is an overview of the questions that were answered.

Are these categories of actions necessarily toposes? (Yes!)
What properties do they have?
Are they Grothendieck toposes; in particular, can we find natural sites for them? (Yes and yes!)
When are two such toposes equivalent; that is, when are two topological monoids Morita equivalent?
What information about the topological monoid can we recover from the topos if we are given its underlying discrete monoid? (We obtain a canonical coarsest topology on that monoid.)
What can we say about the topological properties of the representative monoids we get out of that process?
What about if we are only given the forgetful functor to the topos of sets? (We obtain a representing topological monoid built from the natural endomorphisms of the forgetful functor.)
Can we characterize these toposes? (Yes, in terms of their points.)
Do continuous monoid homomorphisms (or semigroup homomorphisms) produce geometric morphisms between these toposes? (Yes!)
If so, can we use these to characterize Morita equivalence between these toposes? (Alas, no!)

All questions and feedback are welcomed.

Ivan Di Liberti (May 04 2021 at 22:41):

I have a couple of questions.

1. For which kind of topological algebraic structure X (topological magmas, topological semigroups semigroups...) we have that Cont(X,Set) is a topos? Are groups and monoids somewhat special?

2. I am a bit confused by your Cor. 1.18. It is known that every topos with enogh points E is equivalent to Cont(G,Set), for some topological group G (this is really Galois(-Lascar) theory at the end of the day, G can be choosen to be the automorphism group of a saturated enough model). I must be superficial, but this sounds in contradiction with your 1.18 to me.

Morgan Rogers (he/him) (May 05 2021 at 07:01):

1. The proof that these are toposes in the monoid and group cases rests on the fact that the forgetful functor to Sets is (conservative and) the inverse image functor of a point of these toposes. I make the remark in the paper that you can extend a topological semigroup to a topological monoid (by freely adding an open identity element) without changing the category of continuous actions, so you get that the actions of those are toposes. For a topological magma, the fact that you have to map into the endomorphism monoid of a set to get an action means that you're going to factor through the associative reflection anyhow, so I'm pretty sure the same story will happen, but for less interesting reasons. Monoids are kind of the universal case of this, by virtue of actions on sets involving mapping into endomorphism monoids.
2. Every topos with enough points is the topos of actions of some topological groupoid. Asking that this be a group is a rather extreme restriction, since it forces the topos to be atomic.

Ivan Di Liberti (May 05 2021 at 08:02):

Thanks.

Morgan Rogers (he/him) (Feb 22 2023 at 12:11):

I said in another stream that I would comment on this now that this paper has been published in Compositionality, so it's about time I did that.

The main correction in the review process for this paper was that my proof of monadicity of 'powder monoids' was incorrect, and in the course of trying to correct it I came up with a proof that this is not the case, which might be interesting to recount (it's hidden away in Remark 4.36 of the paper, so you might not come across it otherwise).

There are a few ways to prove that a functor is not monadic. One is to provide a counterexample to the necessary and sufficient conditions of a monadicity theorem such as Beck's. However, while identifying a failure of conservativity or the existence of a right adjoint can be feasible, for a functor with both of these properties (such as a composite of monadic functors!) explicitly finding a counterexample of a coequalizer of a reflexive pair which fails to be reflected is a tricky business.

The alternative, which I used, was to show that the monad induced by the adjunction coincides with a monad whose algebras form a distinct category. That is, I showed that:

1. The category of zero-dimensional $T_0$ monoids is monadic over $\mathrm{Set}$.
2. The category of powder monoids is a proper subcategory of the category of zero-dimensional $T_0$ monoids (i.e. while every powder monoid is zero-dimensional and $T_0$, there is a zero-dimensional $T_0$ monoid which is not a powder monoid).
3. The monad induced by the forgetful functor from powder monoids coincides with the "free zero-dimensional $T_0$ monoid" monad.

Hence the forgetful functor from powder monoids cannot be monadic, since this would make every zero-dimensional $T_0$ monoid a powder monoid, a contradiction.

This is a similar strategy to that used to show that the forgetful topological spaces is not monadic over $\mathrm{Set}$, since the algebras for the monad induced by that functor are just sets; in that case, there are not enough algebras, rather than too many.

Morgan Rogers (he/him) (Feb 22 2023 at 12:23):

There are a number of open problems, conjectures and potential research directions in the conclusion of that paper. I am still too early in my career to have students under my sole responsibility, but if you are interested in collaborating or cosupervising a student on these topics, I would be very enthusiastic about exploring that possibility.

Ralph Sarkis (Feb 22 2023 at 12:25):

In 2., do you mean proper in a set-theoretical sense? Or is there a notion of "proper subcategory" that ensures the two categories are not equivalent?

Morgan Rogers (he/him) (Feb 22 2023 at 12:26):

I mean that the inclusion functor of the subcategory is not an equivalence (specifically, I show that a particular monoid is not isomorphic to a powder monoid)

Mike Shulman (Feb 22 2023 at 21:28):

Kudos for having the courage to publicly emphasize a mistake!

I have also found that often the best way to show that a functor is not monadic is to identify the algebras for its induced monad as being a different category.

Morgan Rogers (he/him) (Feb 23 2023 at 10:05):

Mike Shulman said:

Kudos for having the courage to publicly emphasize a mistake!

Thanks! It's not my first and will not be my last :sweat_smile: