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Stream: learning: questions

Topic: every object is a copower


view this post on Zulip Jules Hedges (Jun 08 2021 at 14:00):

Is there a name for categories in which every object is a coproduct of a set of copies of a fixed object?

view this post on Zulip Jules Hedges (Jun 08 2021 at 14:01):

It's a bit of a specific condition but quite a few important categories have this property: in Set we have XxX{}X \cong \coprod_{x \in X} \{ * \}, and from there I think it's inhereted to the kleisli category of any monad on Set, which includes such things as Rel

view this post on Zulip Jules Hedges (Jun 08 2021 at 14:02):

Another example is FVectKFVect_K, where every object is isomorphic to a direct product of a bunch of copies of KK

view this post on Zulip Nathanael Arkor (Jun 08 2021 at 14:03):

This is essentially the definition of (the opposite of) an algebraic theory / Lawvere theory.

view this post on Zulip Zhen Lin Low (Jun 08 2021 at 14:05):

Should we call such things co-Lawvere theories or Lawvere co-theories or co-Lawvere-theories?

view this post on Zulip Nathanael Arkor (Jun 08 2021 at 14:07):

Power and Shkaravska call them "Lawvere co-theories" in From Comodels to Coalgebras: State and Arrays.

view this post on Zulip Jules Hedges (Jun 08 2021 at 14:12):

Thanks. Looks like it's not really a well known thing, which is probably a sign that you can't do much interesting with it

view this post on Zulip Jules Hedges (Jun 08 2021 at 14:12):

For something I'm working on it turns out to be quite important

view this post on Zulip Nathanael Arkor (Jun 08 2021 at 14:14):

I think most people would just identify it with an algebraic theory, which is why the term "co-theory" doesn't appear frequently. Lawvere originally defined algebraic theories this way, but it's become more popular to define them in the dual sense instead.

view this post on Zulip Reid Barton (Jun 08 2021 at 14:28):

In this context, it might be more helpful to refer to the Kleisli category as the category of free algebras.

view this post on Zulip Morgan Rogers (he/him) (Jun 08 2021 at 14:30):

Jules Hedges said:

Thanks. Looks like it's not really a well known thing, which is probably a sign that you can't do much interesting with it

Hardly anyone is looking at it =/= it's not interesting :yum:
Maybe you'll find some applications of such categories which are completely different from the applications of Lawvere theories!

view this post on Zulip John Baez (Jun 08 2021 at 15:38):

I think most people will just look at these things through the looking-glass of "op" and call them Lawvere theories.

view this post on Zulip John Baez (Jun 08 2021 at 15:39):

But also remember this famous theorem from Lawvere's thesis: any of these things, namely the opposite of a Lawvere theory T, is also the category of finitely generated free models of T.

view this post on Zulip John Baez (Jun 08 2021 at 15:41):

Jules pointed out the example of FinSet, where every object is a finite coproduct of copies of the 1-element set. This is the category of finitely generated free models of the initial Lawvere theory, the "theory of sets".

view this post on Zulip John Baez (Jun 08 2021 at 15:43):

(Every set is free; sets have no nontrivial operations so no nontrivial relations are possible. Thus, a finitely generated free set is the same as a finitely generated set, which is just the same as a finite set!)

view this post on Zulip John Baez (Jun 08 2021 at 15:45):

He also pointed out the example of FinVect, where every object is a finite coproduct of copies of your chosen field K. This is the category of finitely generated free models of the Lawvere theory for K-modules.

view this post on Zulip John Baez (Jun 08 2021 at 15:46):

(Every K-module is free when K is a field. Thus, a finitely generated free K-modules is the same as a finitely generated K-module, which is just the same as a finite-dimensional vector space over K!)

view this post on Zulip John Baez (Jun 08 2021 at 15:48):

In both these examples everything is very simple because all models are free. But you can do more interesting examples, e.g. by taking T to be the theory of abelian groups, or groups.

view this post on Zulip Spencer Breiner (Jun 08 2021 at 15:56):

As far as naming, I would have thought the obvious answer would be "CROP" (or "CCCROP", if you want to be pedantic).

view this post on Zulip Jules Hedges (Jun 08 2021 at 17:24):

If you squint quite a lot it's sort of a form of discreteness, in the sense that every object is a "disjoint union" of "points"

view this post on Zulip Fawzi Hreiki (Jun 08 2021 at 18:42):

Lawvere theories used to be defined in this coproduct way (for reasons to do with embedding the theory into the category of algebras) but nowadays I think the standard is the product way (so that it meshes better with the rest of the hierarchy of doctrines).

view this post on Zulip Fawzi Hreiki (Jun 08 2021 at 18:47):

Jules Hedges said:

If you squint quite a lot it's sort of a form of discreteness, in the sense that every object is a "disjoint union" of "points"

Well there's something to be said about this, for example in the context of algebraic geometry. An Artinian local ring is basically just a field which may have nilpotents (i.e. infinitesimal extension) and every Artinian ring is a finite product of such Artinian local rings. So the idea is that the spectra of Artinian rings are disjoint sums of 'fuzzy points'. These spectra form an important subcategory of the category of spaces when you're doing anything related to deformation theory.

view this post on Zulip Fawzi Hreiki (Jun 08 2021 at 18:49):

However, there is of course more than just one Artinian local ring so we don't have the exact situation you asked for.

view this post on Zulip John Baez (Jun 08 2021 at 21:16):

Yeah, a category is extremely simple if every object is a coproduct of copies of a single object. Probably one reason people like sets and vector spaces is that this is true. (For vector spaces we need the axiom of choice to make it true.)

A fun question is: for which rings R is every module a coproduct of copies of R? It's true for fields, and then we get vector spaces. But what other kinds of rings? There's some discussion of this here, with the correct answer not very clearly marked.

view this post on Zulip Fawzi Hreiki (Jun 08 2021 at 21:27):

Its an interesting fact that every RR-module is free iff RR is a division ring. It just goes to show really how much more complicated modules are than vector spaces.