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Stream: theory: category theory

Topic: Free abelian group as a colimit

Brendan Murphy (Feb 21 2024 at 03:20):

Let $\mathcal{E}$ be a category with products, coequalizers, and countable coproducts. Assume some exactness properties, eg that $\mathcal{E}$ is an infinitary pretopos. There's a category $\operatorname{Ab}(\mathcal{E})$ of abelian group objects of $\mathcal{E}$ and a forgetful functor $\operatorname{Ab}(\mathcal{E}) \to \mathcal{E}$. Sometimes this admits a left adjoint, the internal free abelian group functor. Let $\mathsf{FI}$ be the category of finite sets and injections. Any pointed set $(S, s_0)$ defines a covariant functor $\mathsf{FI} \to \mathsf{Set}$ sending $n$ to $S^n$. An object $X$ of $\mathcal{E}$ defines a contravariant functor $\mathsf{FI}^{\mathrm{op}} \to \mathcal{E}$ by restriction. Can we define the free abelian group on $X$ as the weighted colimit of $n \mapsto X^n$ by the functor $n \mapsto \Z^n$?

I was thinking about the construction of the free abelian group explicitly as the quotient of the coproduct of $X^n \times \Z^n$ and realized this looks sort of like the quotient that shows up for geometric realization of simplicial sets.

Brendan Murphy (Feb 21 2024 at 03:27):

Also, part of the quotient we take is modding out each $X^n \times \Z^n$ by the $S_n$-action, which seems like it might be accounted for in a colimit over a diagram category whose objects' automorphism groups are $S_n$?

Brendan Murphy (Feb 21 2024 at 03:33):

There are structure maps $\Z^n \to \operatorname{Hom}_{\mathcal{E}}(X^n, \text{free abelian group on } X)$ making the free abelian group a weighted cocone

Brendan Murphy (Feb 21 2024 at 15:14):

@Patrick Nicodemus has convinced me we want the category of finite sets (or the skeleton on finite cardinals) rather than FI. Surjections should correspond to the distributive law (n+m)*x = n*x + m*x. This also means we can understand the functors Fin -> Set, Fin^op -> E via the universal property of E as classifying commutative monoid and cocommutative comonoid objects

Brendan Murphy (Feb 21 2024 at 16:49):

Is there hom-tensor sort of adjunction with the functor tensor product?

Brendan Murphy (Feb 21 2024 at 16:50):

That might give a nice proof of this

Brendan Murphy (Feb 21 2024 at 16:52):

Ah I think it's in section 4 of these notes https://www2.mathematik.tu-darmstadt.de/~streicher/FIBR/DiWo.pdf

Reid Barton (Feb 21 2024 at 17:20):

I think this is the kind of description you would get by unrolling the statement that the composition (forgetful functor) $\circ$ (free abelian group functor) commutes with filtered colimits, and therefore is the left Kan extension of its restriction to the subcategory of finite sets.

Brendan Murphy (Feb 21 2024 at 17:24):

That makes sense, but I don't see how it generalizes to other categories E? I guess if E is a sheaf topos then any object should be the directed colimit of its (Bishop-?)finite subobjects?

Kevin Arlin (Feb 21 2024 at 18:25):

The free abelian group monad exists on any locally presentable category and it commutes with filtered colimits at least if the category is locally finitely presentable, and then you can give a similar left Kan formula for it in terms of the finitely presentable objects of E. I'm not immediately sure how the monad behaves on locally $\kappa$-presentable categories for larger values of $\kappa.$ Note that this includes many Grothendieck topoi, like categories of sheaves on basically any interesting space; sheaves are $\kappa$-filtered colimits of $\kappa$-presentable objects for a sufficiently large $\kappa$ depending only on the space (not on the sheaf) but $\kappa$ could be badly infinite.

Kevin Arlin (Feb 21 2024 at 18:26):

If by Bishop-finiteness you mean isomorphism with some finite coproduct of terminals, or something like that, that's not the right concept in a general category. For instance in a topos with more than two truth values there will be subterminals that don't satisfy this, but they're extremely finite.

Graham Manuell (Feb 21 2024 at 18:52):

Why do you say subterminals are "extremely finite"? Most constructivists would not agree with this.

Kevin Arlin (Feb 21 2024 at 18:55):

I don't think Brendan's a constructivist, basically.

Brendan Murphy (Feb 21 2024 at 18:57):

I'm not, if I was I might have a better idea of the correct version of finitenes here :P

Graham Manuell (Feb 21 2024 at 18:57):

Surely constructivists are the ones we would ask about what a good notion of finiteness in a topos is?

Brendan Murphy (Feb 21 2024 at 18:57):

The category I'm interested in is actually condensed sets, which I believe is not kappa-presentable for any kappa?

Kevin Arlin (Feb 21 2024 at 18:58):

It's good to know which objects are finite in a category no matter what your philosophical background is!

Brendan Murphy (Feb 21 2024 at 18:58):

It does make sense that we'd want to replace finite objects with finitely presentable, but in that case I don't think we recover this coend/weighted colimit construction I was interested in

Kevin Arlin (Feb 21 2024 at 18:58):

Ugh, yeah, condensed sets are class-locally presentable IIRC and that makes all this stuff pretty creepy.

Kevin Arlin (Feb 21 2024 at 18:59):

But depending on what you mean by "in that case", no, you should still recover the same construction, you just need more variety in your exponents than natural numbers.

Brendan Murphy (Feb 21 2024 at 19:00):

Right, but I don't think we actually need the extra variety. It seems like the weighted colimit should give us the correct thing even just with natural number exponents

Brendan Murphy (Feb 21 2024 at 19:02):

I think I recall that the free monoid on an object can be constructed as a countable coproduct of tensor powers of that object in any monoidal category where (x) distributes over countable coproducts, and this is the same sort of construction

Kevin Arlin (Feb 21 2024 at 19:02):

Yeah, you might be able to recover this formula as that for the free model of the Lawvere theory of abelian groups.

Brendan Murphy (Feb 21 2024 at 19:03):

I was thinking along those lines but I wasn't sure how to relate the lawvere theory to the free monad on categories other than Set

Graham Manuell (Feb 21 2024 at 19:04):

Kevin Arlin said:

It's good to know which objects are finite in a category no matter what your philosophical background is!

I was just trying to tell you that you are using a notion of finiteness that is at odds with that what community usually considers. It seems odd to claim you know the 'true version of finiteness' and that the usual definitions used by people that actually study these things are wrong. But anyway, I'll leave it at this so as to not derail things further.

Kevin Arlin (Feb 21 2024 at 19:05):

Maybe the point is just that I'm talking in the external logic.

Kevin Arlin (Feb 21 2024 at 19:07):

(That is, I'm not disagreeing with people who study "these things", I think we're just talking about different things.)

Benedikt Peterseim (Feb 21 2024 at 19:23):

Hi Brendan, as it happens, I asked a MathOverflow question on essentially this. There are some very nice unpublished lecture notes lecture notes by Zhen Lin Low on (generalizations) of this. All you need for the forgetful functor from abelian group objects to be (finitary) monadic is that your base category is cartesian closed, well-powered, regular and countably complete -- this is definitely applicable to condensed sets. (The issue for me is that I feel there should be a "classic" (and published) reference for this, which I haven't been able to locate, unfortunately.)

Kevin Arlin (Feb 21 2024 at 19:25):

Nice reference, Benedikt, I think the link is broken, though. Here it is; Zhen references Theorem 2.4.21.

Brendan Murphy (Feb 21 2024 at 19:27):

These notes are very nice!

Benedikt Peterseim (Feb 21 2024 at 19:29):

I agree :)

Kevin Arlin (Feb 21 2024 at 19:30):

Curious to see if you find support for your specific formula in there!

Brendan Murphy (Feb 21 2024 at 19:30):

Yeah it didn't look like it when skimming, but I do expect that putting things in terms of lawvere theories will help

Benedikt Peterseim (Feb 22 2024 at 07:19):

This is a bit of an awkward question, but is anyone of you here on MathOverflow? David White kindly gave an answer to the linked question above that doesn't tackle the free abelian group object issue, however. Apparently I phrased my question so poorly that he now thinks I'm a troll for not accepting his answer. :/

Morgan Rogers (he/him) (Feb 22 2024 at 07:50):

Reading the discussion there I also do not understand why you didn't accept his answer. What is missing in your opinion?

Benedikt Peterseim (Feb 22 2024 at 07:58):

In which of the given references is there a proof that the forgetful functor $Ab(C) \to C$ is monadic (Zhen Lin's Teorem 2.4.10) and that the resulting monad is symmetric monoidal (with respect to the cartesian monoidal structure on $C$)? Most importantly, I don't see how the given references would provide a (non-hand-wavy, rigorous) proof of the example cases I provided in the question.

Benedikt Peterseim (Feb 22 2024 at 07:59):

Note also that Zhen Lin's theorem uses different assumptions than my question, I actually forgot where I got them from (maybe a different place in the same notes).

Benedikt Peterseim (Feb 22 2024 at 08:00):

I may have missed something, of course, in that case I'm really sorry.

Benedikt Peterseim (Feb 22 2024 at 08:06):

I think the confusion stems from the fact that what Hovey calls a module is a "module (aka action) of a monoid in a monoidal category".

Benedikt Peterseim (Feb 22 2024 at 08:06):

Ah, wait, abelian group objects are monoids, of course.

Benedikt Peterseim (Feb 22 2024 at 08:08):

I was always thinking about the general result for algebras for a Lawvere theory, that's why I didn't get it! Thanks Morgan, I just needed someone to talk to realize this.

Benedikt Peterseim (Feb 22 2024 at 08:12):

Oh, wait again. Abelian group objects or not modules for the monad $\mathbb{Z}\times (-)$ on $C$ (these are $\mathbb{Z}$-actions; whatever $\mathbb{Z}$ is supposed to mean here). What am I missing?

Morgan Rogers (he/him) (Feb 22 2024 at 08:13):

Benedikt Peterseim said:

I was always thinking about the general result for algebras for a Lawvere theory, that's why I didn't get it! Thanks Morgan, I just needed someone to talk to realize this.

Ha I didn't say anything really, I just wanted to understand where the confusion was. There is less room for such discussion on MO (although I am told it does happen sometimes).

Benedikt Peterseim (Feb 22 2024 at 08:13):

I'm still confused, unfortunately. :sweat_smile:

Morgan Rogers (he/him) (Feb 22 2024 at 08:18):

Benedikt Peterseim said:

Oh, wait again. Abelian group objects or not modules for the monad $\mathbb{Z}\times (-)$ on $C$ (these are $\mathbb{Z}$-actions). What am I missing?

This is true; equivalently, it's the category of sets equipped with an automorphism. I'm more used to thinking of RMod as the category of algebras for the monda $R \otimes -$ on the category of abelian groups.

Morgan Rogers (he/him) (Feb 22 2024 at 08:18):

I want to check my copy of Borceux but my tablet is out of juice.

Morgan Rogers (he/him) (Feb 22 2024 at 08:20):

However, as soon as free abelian groups exist, there is a functor $X \mapsto R \otimes F(X)$ (or rather, the underlying set of the right-hand side) which gives a monad.

Morgan Rogers (he/him) (Feb 22 2024 at 08:21):

(I think you figured that out based on your argument in the question)

Benedikt Peterseim (Feb 22 2024 at 08:22):

Okay, so what I'm looking for is a reference for the fact the forgetful functor from commutative monoid objects is monadic.

Benedikt Peterseim (Feb 22 2024 at 08:23):

I think this can be pieced together from the references given on this lab page.

Benedikt Peterseim (Feb 22 2024 at 08:24):

I think I got it now, I might write up the complete argument with all references in one place later here, in case anyone is interested.

Morgan Rogers (he/him) (Feb 22 2024 at 08:26):

Sure. As Zhen Lin says, the hard part is dealing with the coequalizers of monoids which you need in order to apply monadicity theorems, but I get the impression it's a matter of technical intricacy rather than there existing possible obstructions.

Benedikt Peterseim (Feb 22 2024 at 14:20):

This is what I found. This is all related to the topic of this thread in that the proof of the following lemma proceeds by constructing the free commutative monoid as a filtered colimit, more precisely as the copruduct over the "symmetric powers" of an object:

Lemma [Lemma 4.4.5 in Brandenburg, p. 67]. Let $C$ be a countably cocomplete symmetric monoidal category category. Then $\mathrm{CMon}(C) \to C$ has a left adjoint $S: C \to \mathrm{CMon}(C)$.

Remark. The cited work calls symmetric monoidal categories "tensor categories" and assumes cococompleteness of C, but only countable cocompleteness is used.

Benedikt Peterseim (Feb 22 2024 at 14:21):

Lemma (no reference found). Let $C$ be a countably cocomplete symmetric monoidal category category. Then $\mathrm{CMon}(C) \to C$ is monadic.
Proof sketch. Use the previous lemma and write down an explicit (quasi-)inverse to the comparison functor.

Corollary. Let $C$ be a countably cocomplete cartesian closed category. Then $S$, viewed as an endofunctor on $C$, acquires the structure of a symmetric monoidal monad (a.k.a. commutative monad).
Very-sketchy attempt at a proof sketch. $S$ should canonically be a $C$-enriched monad (whatever that is), and according to the lab page of "enriched monad", we get the structure of a strong monad, which we can check to be commutative, and which is then "the same" as a symmetric monoidal monad.

Corollary. Let $C$ be a finitely complete, countably cocomplete cartesian closed category. Then $\mathrm{CMon}(C)$ is a closed symmetric monoidal category.
Proof sketch. This follows from [Brandenburg, Corollary 6.5.4.], applied to the symmetric monoidal monad $S$ and using the previous lemma (i.e. that $Alg(S)\cong CMon(C)$).

Remark. [Brandenburg, Corollary 6.5.4.] assumes cocompleteness of $C$ (instead of just countable cocompleteness), but I don't think this is used in the proof.

Benedikt Peterseim (Feb 22 2024 at 14:22):

Corollary. Let $C$ be a finitely complete, countably cocomplete cartesian closed category. Then $R\mathrm{Mod}(C)$ is a finitely complete, countably cocomplete closed symmtric monoidal category. If $C$ is cocomplete (resp. regular), then so is $R\mathrm{Mod}(C)$.
Proof. This now follows from Davids White's answer to the linked MathOverflow question .

Conclusion. Unfortunately, the above is still quite sketchy for my taste, and I still don't know whether this has been treated in full detail in the literature.

John Baez (Feb 22 2024 at 16:31):

I think it would be a nice community service to write up clean proofs of these things and put them in an appendix to a paper that uses these facts.

Benedikt Peterseim (Feb 22 2024 at 16:31):

@Morgan Rogers (he/him) Do you now see why I’m not satisfied with David White’s answer to the linked question? If I understand him correctly, he wants to apply a theorem about modules (aka actions) over a commutative monoid in some monoidal category. This cannot work for the category of commutative monoids itself, since that’s (a priori) not a category of modules/actions.

Benedikt Peterseim (Feb 22 2024 at 16:31):

@John Baez I completely agree.

Benedikt Peterseim (Feb 22 2024 at 16:33):

I’m just surprised that this seems to not have happened yet concerning these “basic” things.

John Baez (Feb 22 2024 at 16:36):

I don't think you need to use commutative monads / symmetric monoidal monads to show that $\mathrm{CMon}(C)$ is closed symmetric monoidal if $C$ is a countably complete cartesian closed category. I think you can show it "directly" without excessive pain.

Better yet, $(\mathrm{CMon}(C), \otimes)$ should be closed symmetric monoidal if $(C, \otimes)$
is countably complete symmetric monoidal closed category whose tensor product distributes over countable colimits.

$C = \mathrm{AbGrp}$ with the usual tensor product of abelian groups satisfies the latter hypotheses, and commutative monoids in $(\mathrm{AbGp}, \otimes)$ are commutative rings so this example is important, but this tensor product is not cartesian.

Benedikt Peterseim (Feb 22 2024 at 16:48):

“Better yet, $(\mathrm{CMon}(C), \otimes)$ should be closed symmetric monoidal if $(C, \otimes)$
is countably complete symmetric monoidal closed category whose tensor product distributes over countable colimits.“

I think I see how this might work: one can probably use the construction of the free commutative monoid $S(X)$ in terms of “symmetric powers” directly to obtain a symmetric monoidal structure on the monad $S$, and from that we get a closed symmetric monoidal structure on $\mathrm{CMon}(C)$. I do think that any construction of the tensor product on $\mathrm{CMon}(C)$ will at least implicitly use the the symmetric monoidal monad structure on $S$.

John Baez (Feb 22 2024 at 16:59):

The tensor product of commutative monoids in $(C, \otimes)$ is just the tensor product of their underlying objects in $C$, made into a commutative monoid using "the obvious formula". The associator, symmetry, etc. in $\mathrm{CMon}(C,\otimes)$ are inherited from those in $(\mathrm{C}, \otimes)$. So I think it's easy to check "directly" that $\mathrm{CMon}(C,\otimes)$ is symmetric monoidal, with each axiom following from the corresponding axiom for $(C, \otimes)$.

Is the problem that you find this too long and boring, so you want to get the job done using a more high-powered theorem?

Benedikt Peterseim (Feb 22 2024 at 17:03):

Mmh, so for $C=\mathsf{Set}$, the tensor product of commutative monoids is just their cartesian product? This seems wrong.

Benedikt Peterseim (Feb 22 2024 at 17:04):

It’s “a tensor product” (namely, the coproduct in that category), but not “the tensor product”.

John Baez (Feb 22 2024 at 17:04):

In $(\mathrm{Set}, \times)$ the "tensor product" of commutative monoids that I'm talking about is the cartesian product of (ordinary) commutative monoids.

In $(\mathrm{AbGp}, \otimes)$ the "tensor product"of commutative monoids that I'm talking about is the usual tensor product of commutative rings.

John Baez (Feb 22 2024 at 17:07):

But if you want the usual tensor product of ordinary commutative monoids, yes you need some other construction.

John Baez (Feb 22 2024 at 17:07):

For that, I see why commutative monads come in handy.

Benedikt Peterseim (Feb 22 2024 at 17:09):

Yes, but the question is how to construct the tensor product of commutative monoids in general. For example, with the cartesian product, the category of commutative monoids is not closed, as far as I can tell. And even if it were, I want the tensor product that makes the sentence “abelian groups are modules/actions over $\mathbb{Z}$, viewed as a monoid in $\mathrm{CMon}(Set)$” true.

Benedikt Peterseim (Feb 22 2024 at 17:10):

(Responding to the first paragraph)

John Baez (Feb 22 2024 at 17:13):

Okay, I got distracted from your actual goal by my love of thinking of commutative rings as commutative monoids in $(\mathsf{AbGp}, \otimes)$ where the tensor product is simply that of abelian groups.

John Baez (Feb 22 2024 at 17:16):

To define the tensor product of commutative monoids that you want, and show $(\mathrm{CMon}(C),\otimes)$ is closed symmetric monoidal when $C$ is a countably complete cartesian closed category, it definitely will be helpful to show that there's a commutative monad on $C$ whose algebras are commutative monoids.

John Baez (Feb 22 2024 at 17:17):

Someone should have proved that already, but I don't know if anyone has.

Morgan Rogers (he/him) (Feb 22 2024 at 21:23):

Benedikt Peterseim said:

Morgan Rogers (he/him) Do you now see why I’m not satisfied with David White’s answer to the linked question? If I understand him correctly, he wants to apply a theorem about modules (aka actions) over a commutative monoid in some monoidal category. This cannot work for the category of commutative monoids itself, since that’s (a priori) not a category of modules/actions.

I do indeed understand. I added a comment to that effect, we'll see if he comes back to it.

Benedikt Peterseim (Feb 23 2024 at 10:04):

Benedikt Peterseim said:

“Better yet, $(\mathrm{CMon}(C), \otimes)$ should be closed symmetric monoidal if $(C, \otimes)$
is countably complete symmetric monoidal closed category whose tensor product distributes over countable colimits.“

I think I see how this might work: one can probably use the construction of the free commutative monoid $S(X)$ in terms of “symmetric powers” directly to obtain a symmetric monoidal structure on the monad $S$, and from that we get a closed symmetric monoidal structure on $\mathrm{CMon}(C)$. I do think that any construction of the tensor product on $\mathrm{CMon}(C)$ will at least implicitly use the the symmetric monoidal monad structure on $S$.

@John Baez I just realized that this does not actually work, you do need a cartesian monoidal category. The intuitive reason (intuitive for me, at least) is that in order to construct the symmetric monoidal structure $S(X) \otimes S(Y) \to S(X\otimes Y)$ on $S$, which is a sort of "form-the-product-of-finite-multisets" morphism, we need to copy data. You actually mentioned the category of commutative rings before (viewing commutative rings as commutative monoids in $(\mathrm{Ab}, \otimes)$), which is not a closed symmetric monoidal category (well, not in any obvious/useful way, at least). So that gives a (sort of) counter-example to the claim that $(\mathrm{CMon}(C), \otimes)$ is closed symmetric monoidal as soon as $(C, \otimes)$ is countably complete symmetric monoidal closed category whose tensor product distributes over countable colimits.

Morgan Rogers (he/him) (Feb 23 2024 at 15:10):

(or more generally a symmetric monoidal category with diagonal morphisms)

John Baez (Feb 23 2024 at 23:16):

Benedikt Peterseim said:

John Baez I just realized that this does not actually work, you do need a cartesian monoidal category. The intuitive reason (intuitive for me, at least) is that in order to construct the symmetric monoidal structure $S(X) \otimes S(Y) \to S(X\otimes Y)$ on $S$, which is a sort of "form-the-product-of-finite-multisets" morphism, we need to copy data.

For some reason I'm not seeing that, e.g. in the case of the free commutative monoid monad $S: \mathsf{AbGp} \to \mathsf{AbGp}$ this map does things like

$x_1 x_2 \otimes y_1 y_2 \mapsto (x_1 \otimes 1)(x_2 \otimes 1)(1 \otimes y_1)(1 \otimes y_2)$

John Baez (Feb 23 2024 at 23:16):

I feel I'm making some mistake, but maybe you can point it out.

John Baez (Feb 23 2024 at 23:18):

Is the point that I'm not allowed to act like an element of $S(X)$ is just a tensor product of elements of $X$, there are linear combinations too?

John Baez (Feb 23 2024 at 23:20):

I certainly agree that if $\mathsf{CommRing}$ were closed with respect to the usual tensor product of commutative rings (which happens to be the coproduct), there would need to be a conspiracy of ring theorists working to conceal this fact from me.

Brendan Murphy (Feb 24 2024 at 00:22):

Here's a proof that it isn't closed wrt tensor product: if it was then the tensor with a fixed comm ring would be a left adjoint, hence preserve colimits, and in the case of coproducts this means $A \otimes_\Z (B \otimes_\Z C) \cong (A \otimes_\Z B) \otimes_\Z (A \otimes_\Z C)$. But taking $A, B, C$ direct powers of $\Z$ this is absurd

Brendan Murphy (Feb 24 2024 at 00:26):

Oh, and the nlab has a proof that it's impossible to be closed monoidal wrt the coproduct unless you're a contractible groupoid

John Baez (Feb 24 2024 at 00:26):

Okay, thanks.

Benedikt Peterseim (Feb 24 2024 at 05:49):

John Baez said:

Benedikt Peterseim said:

John Baez I just realized that this does not actually work, you do need a cartesian monoidal category. The intuitive reason (intuitive for me, at least) is that in order to construct the symmetric monoidal structure $S(X) \otimes S(Y) \to S(X\otimes Y)$ on $S$, which is a sort of "form-the-product-of-finite-multisets" morphism, we need to copy data.

For some reason I'm not seeing that, e.g. in the case of the free commutative monoid monad $S: \mathsf{AbGp} \to \mathsf{AbGp}$ this map does things like

$x_1 x_2 \otimes y_1 y_2 \mapsto (x_1 \otimes 1)(x_2 \otimes 1)(1 \otimes y_1)(1 \otimes y_2)$

In the case of $S: \mathsf{Set} \to \mathsf{Set}$, this map does the following:
$(x_1 x_2, y_1 y_2) \mapsto (x_1, y_1) (x_1, y_2) (x_2,y_1) (x_2,y_2),$
writing multiplication in $S(-)$ as juxtaposition. If we try to write this down for $S: \mathsf{AbGp} \to \mathsf{AbGp}$ instead, we get,
$x_1 x_2 \otimes y_1 y_2 \mapsto (x_1 \otimes y_1) (x_1\otimes y_2) (x_2\otimes y_1) (x_2\otimes y_2),$
which shows that we need to copy stuff -- we won't get something linear. Hope that clears up the intuition part, in addition to Brendan's proof.