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Stream: theory: category theory

Topic: Flat object in a symmetric monoidal category

Jean-Baptiste Vienney (Nov 15 2023 at 20:37):

In this paper: On Flat Semimodules over Semirings, they define:
Definition: A semimodule $M$ over a semiring is flat iff $- \otimes N$ preserves finite limits.
And the main result of the paper is:
Theorem: A semimodule $M$ is flat iff it is a filtered colomit of finitely generated free semimodules.

We could more generally define:
Definition: Let $(\mathcal{C},\otimes,I)$ be a finitely complete symmetric monoidal category. An object $M \in \mathcal{C}$ is flat iff $- \otimes M$ preserves finite limits.
Definition: Let $(\mathcal{C},\otimes,I)$ be a symmetric monoidal category with finite coproducts. An object $M$ is finitely generated free iff there is an isomorphism $M \cong I^{\oplus n}$ for some $n \ge 1$.
Then, we could maybe have:
Conjecture: Let $(\mathcal{C},\otimes,I)$ be a finitely complete symmetric monoidal category with finite coproducts and $M \in \mathcal{C}$, then $M$ is flat iff $M$ is a filtered colimit of finitely generated free objects.

Does anything like this is known?

I don't think that this conjecture is true in such a general setting, but maybe it works with additional requirements. I would be interested to know about such requirements which could make it true. When I look at the previous paper, it seems to be all very categorical, probably I should just try to understand the paper and everything could maybe be categorified.

There is a lot of material about flat functors on the nlab but it is never about flat functors in a monoidal category. But maybe we can recover what I want from this material. In this case the paper "On Flat Semimodules over Semiring" is a bit useless. Or there is a paper Purity and flatness in symmetric monoidal closed exact categories but they assume a lot more structure for their setting, it looks very complicated and I don't see that they talk about something like this (but I haven't read it and I probably would not be able to understand it properly quickly).

John Baez (Nov 16 2023 at 02:14):

I like your conjecture. I wish I understood this remark in the nLab, which might offer a slick way to prove this conjecture:

Theorem 2.5. A module is flat if and only if it is a filtered colimit of finitely generated free modules.

This observation (Wraith, Blass) can be put into the more general context of modelling geometric theories by geometric morphisms from their classifying toposes, or equivalently, certain flat functors from sites for such topoi.

Jonas Frey (Nov 16 2023 at 03:30):

The paper

Bulman-Fleming, S., & McDowell, K. (1978). Flat semilattices. Proceedings of the American Mathematical Society, 72(2), 228-232.

states that the two notions of flatness do not coincide in general. From the first page:

Screenshot_20231115_223033.png

Jonas Frey (Nov 16 2023 at 03:41):

In the context of classifying toposes, the term "flat model" is used for filtered colimits of (finitely generated) free models of an algebraic theory. Flat models of an algebraic theory are classified by the topos of presheaves on the Lawvere theory.

This topos is a subtopos of the presheaf topos on the "enveloping finite-limit theory" of the Lawvere theory, which classifies all models of the theory.

This concept of flatness doesn't make sense for general geometric theories, where there is no notion of free model, but generalizes at least to models of "Generalized algebraic theories" in Cartmell's sense (meaning filtered colimit of representable models here), and I use it in this sense in connection with the fat small object argument in my recent paper on clan duality.

Jonas Frey (Nov 16 2023 at 03:49):

Anyway, it's interesting to know that the notions coincide for semimodules!

Jean-Baptiste Vienney (Nov 16 2023 at 05:56):

Jonas Frey said:

The paper

Bulman-Fleming, S., & McDowell, K. (1978). Flat semilattices. Proceedings of the American Mathematical Society, 72(2), 228-232.

states that the two notions of flatness do not coincide in general. From the first page:

Screenshot_20231115_223033.png

Thanks!

Jean-Baptiste Vienney (Nov 16 2023 at 05:58):

The equivalence between a flat object and an $L$-flat object might still be verified by some class of monoidal categories which contains the categories of semimodules over a semiring but is not reduced to the class of these categories.

Jean-Baptiste Vienney (Nov 16 2023 at 05:59):

For instance, maybe that:
Conjecture 2: Let $(\mathcal{C},\otimes,I)$ be a finitely complete symmetric monoidal category enriched over commutative monoids (such that $- \otimes -$ is bilinear) with finite coproducts. Then for every $M \in \mathcal{C}$, $M$ is flat iff $M$ is $L$-flat.

Jean-Baptiste Vienney (Nov 16 2023 at 06:02):

Where you say that $M$ is $L$-flat if it is a filtered colimit of some $I^{\oplus n_\alpha}$.

Jean-Baptiste Vienney (Nov 16 2023 at 06:03):

Of course, it is still maybe false but at least categories of $M$-sets don't verify these conditions because they are not enriched over commutative monoids.

Jean-Baptiste Vienney (Nov 16 2023 at 06:05):

Monoidal categories can be quite different from a category of semimodules but symmetric monoidal categories enriched over commutative monoids are much closer.

Matteo Capucci (he/him) (Nov 16 2023 at 10:27):

Maybe it's helpful to notice that every object $M$ of a symmetric monoidal category $\cal C$ is a module over the commutative monoid ${\cal C}(I,I)$ of scalars in $\cal C$. What I mean exactly by module here is that there is, for every $M:\cal C$ a functor ${\bf B}{\cal C}(I,I) \to \cal C$ that picks $M$ and sends a scalar $s:I \to I$ to its conjugation by the right or left unitors (Idk if it matters which one you pick).

Matteo Capucci (he/him) (Nov 16 2023 at 10:30):

Moreover, if $\cal C$ has coproducts (I might need infinite ones though) then it is $\bf Set$-tensored so you can internalize the action of ${\cal C}(I,I)$ to a morphism $\sum_{s:{\cal C}(I,I)} M \to M$ in $\cal C$ itself. Maybe this helps too.

John Baez (Nov 16 2023 at 12:41):

At this point if I were trying to prove @Jean-Baptiste Vienney's conjecture, maybe with extra hypotheses, I'd first study Wraith and Blass' proof for modules of rings - they both write quite well! - and then study Yetsin Katsov's proof for semimodules of semirings. Then I'd try to copy what they did in greater generality, seeing what hypotheses I need.

John Baez (Nov 16 2023 at 12:47):

But without knowing those proofs I'm stuck.

Jean-Baptiste Vienney (Nov 16 2023 at 12:52):

Yes, sure ahah. I’ll try to do it!

Jean-Baptiste Vienney (Nov 16 2023 at 13:13):

Matteo Capucci (he/him) said:

Maybe it's helpful to notice that every object $M$ of a symmetric monoidal category $\cal C$ is a module over the commutative monoid ${\cal C}(I,I)$ of scalars in $\cal C$. What I mean exactly by module here is that there is, for every $M:\cal C$ a functor ${\bf B}{\cal C}(I,I) \to \cal C$ that picks $M$ and sends a scalar $s:I \to I$ to its conjugation by the right or left unitors (Idk if it matters which one you pick).

I know that $\mathcal{C}[I,I]$ is a monoid. Moreover it is commutative (in any monoidal category) and you get actions $\mathcal{C}[I,I]\times \mathcal{C}[A,B] \rightarrow \mathcal{C}[A,B]$. If $\mathcal{C}$ is enriched over commutative monoids, it turns out that $\mathcal{C}[I,I]$ is a moreover a semi-ring and every $\mathcal{C}[A,B]$ a $\mathcal{C}[I,I]$-semi-module. So monoidal categories enriched over commutative monoids are just monoidal categories enriched over semi-modules over an unspecified semi-ring and I love this class of categories!

Jean-Baptiste Vienney (Nov 16 2023 at 13:14):

But what is $\mathbf{B}\mathcal{C}[I,I]$?

John Baez (Nov 16 2023 at 13:16):

$\mathbf{B}$ of a monoid is the corresponding one-object category. If the monoid is commutative, this category is symmetric monoidal.

Jean-Baptiste Vienney (Nov 16 2023 at 13:18):

Oh, ok. And what is the tensor product?

John Baez (Nov 16 2023 at 13:18):

The only thing it possibly could be.

John Baez (Nov 16 2023 at 13:18):

There aren't a lot of binary operations here!

Jean-Baptiste Vienney (Nov 16 2023 at 13:19):

It is the composition = the product of the monoid?

John Baez (Nov 16 2023 at 13:20):

Yes.

Jean-Baptiste Vienney (Nov 16 2023 at 13:21):

I’m confused. So in this category, the tensor product is equal to the composition? It’s weird.

John Baez (Nov 16 2023 at 13:23):

Maybe it's weird, but it's inevitable and famous. Whenever you have a monoidal category $\mathcal{C}$ with unit object $I$, the hom-set $\mathcal{C}[I,I]$ has two binary operations: composition and the tensor product. But they are the same.

John Baez (Nov 16 2023 at 13:24):

Study the periodic table, my son! :upside_down:

.

John Baez (Nov 16 2023 at 13:25):

Those arrows pointing up and to the right are $\mathbf{B}$.

John Baez (Nov 16 2023 at 13:26):

A monoid is a one-object category.

A commutative monoid is a one-object monoidal category.

A commutative monoid is a one-object braided monoidal category.

A commutative monoid is a one-object symmetric monoidal category.

Etc.

John Baez (Nov 16 2023 at 13:27):

I explained this stuff here:

https://www.youtube.com/watch?v=X1PkkqDwf8Y

Jean-Baptiste Vienney (Nov 16 2023 at 13:29):

John Baez said:

Maybe it's weird, but it's inevitable and famous. Whenever you have a monoidal category $\mathcal{C}$ with unit object $I$, the hom-set $\mathcal{C}[I,I]$ has two binary operations: composition and the tensor product. But they are the same.

I remember this now but I didn’t know that you get a symmetric monoidal category with tensor product equal to the composition if you start from a monoid! I’m glad to learn this.

John Baez (Nov 16 2023 at 14:56):

If you start from a commutative monoid.

Matteo Capucci (he/him) (Nov 17 2023 at 09:41):

I was reading Callum Reader's thesis and found out my observation above is well-known (unsurprisingly)! See Section 5.2.

John Baez (Nov 17 2023 at 11:05):

Yes, I work a lot on $\mathsf{Vect}_k$-enriched symmetric monoidal categories $\mathcal{C}$, and for these, $\mathcal{C}[I,I]$ is an associative algebra over the field $k$, and this algebra acts on every object, so in some sense every object of $\mathcal{C}$ is a module of this algebra. But notice - and also in the more general situation you were talking about - objects of $\mathcal{C}$ are not sets, so these objects are "modules in $\mathcal{C}$" rather than modules in the usual sense. (Of course, objects in $\mathcal{C}$ may have underlying sets.)

Also, as you noted, the hom-sets of $\mathcal{C}$ are modules in the usual sense.