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Stream: learning: questions

Topic: projective objects

John Baez (Feb 26 2023 at 18:26):

For any ring R, every projective R-module is a summand of a free R-module. This follows easily from the definition of projective object and the fact that every R-module M has an epimorphism p: F $\to$ M where F is free.

John Baez (Feb 26 2023 at 18:27):

How much does this generalize to other abelian categories, or maybe more general additive categories? This is a vague question: I'll try to make it a bit more precise but really I'm just fishing for any interesting information about this question.

John Baez (Feb 26 2023 at 18:28):

For example I'm interested in a class of symmetric monoidal additive categories called 2-rigs.

John Baez (Feb 26 2023 at 18:29):

In a symmetric monoidal additive category we could call an object of the form $I \oplus \cdots \oplus I$, where I is the unit for the tensor product, a 'finitely generated free object'.

John Baez (Feb 26 2023 at 18:30):

I believe these objects are all projective and thus their summands are projective too.

John Baez (Feb 26 2023 at 18:32):

When our category is the category of modules of a commutative ring, every finitely generated projective object is a summand of some object of the form $I \oplus \cdots \oplus I$.

John Baez (Feb 26 2023 at 18:33):

But what about more general symmetric monoidal additive categories?

John Baez (Feb 26 2023 at 18:34):

I suppose one not-very-exciting but perfectly reasonable answer would be: in general, we should define an object $M$ of a symmetric monoidal additive category to be finitely generated if there exists an epimorphism

$p: I \oplus \cdots \oplus I \to M$

John Baez (Feb 26 2023 at 18:36):

If this epimorphism splits then $M$ will be projective. (Right? Hmm, I seem to be able to prove this assuming $I$ is projective, but I don't know why it needs to be.)

John Baez (Feb 26 2023 at 18:38):

But also, if $M$ is projective this epimorphism will split!

John Baez (Feb 26 2023 at 18:39):

So, with this definition of 'finitely generated', finitely generated projective objects are the same as summands of objects of the form $I \oplus \cdots \oplus I$... modulo my parenthetical question.

John Baez (Feb 26 2023 at 18:45):

So, maybe a concrete formulation of my vague question is: under what conditions on a symmetric monoidal additive category is the unit for the tensor product projective?

John Baez (Feb 26 2023 at 18:50):

(I made a lot of progress just while asking the question here!)

Jean-Baptiste Vienney (Feb 26 2023 at 19:57):

Just a little fact that I like: in every symmetric monoidal additive category, $\mathcal{C}[I,I]$ is a commutative ring and every hom-set $\mathcal{C}[A,B]$ is a $\mathcal{C}[I,I]$-module. If ever it can be useful. That's something we get by assuming our category symmetric monoidal compared to just an additive or abelian category.

John Baez (Feb 26 2023 at 20:06):

We - that is @Joe Moeller, @Todd Trimble and I - definitely use the fact that $\mathcal{C}[I,I]$ is a commutative ring!

John Baez (Feb 26 2023 at 20:06):

We say a symmetric monoidal additive category is connected if this commutative ring has no idempotents other than 0 and 1.

John Baez (Feb 26 2023 at 20:07):

If you have a nontrivial idempotent you can use it to write $\mathcal{C}$ as a product $\mathcal{C}_1 \times \mathcal{C}_2$ in a nontrivial way, at least if you're in the situation where idempotents in $\mathcal{C}$ split (which is something we always assume).

John Baez (Feb 26 2023 at 20:08):

So, the connected case is a useful case to study at first.

Jean-Baptiste Vienney (Feb 26 2023 at 20:10):

Oh, great! Do you use it in Schur functors and categorified plethysm?

John Baez (Feb 26 2023 at 20:12):

No, we use it in our next paper, which is not done yet: "The splitting principle".

John Baez (Feb 26 2023 at 20:13):

This is about generalizing the splitting principle to more general 2-rigs.

Jean-Baptiste Vienney (Feb 26 2023 at 20:16):

Maybe you could do it as well by assuming that your hom-sets are just commutative monoids and not necessarily abelian groups. In that case $\mathcal{C}[I,I]$ is a semi-ring and the $\mathcal{C}[A,B]$ semi-modules

Jean-Baptiste Vienney (Feb 26 2023 at 20:22):

I guess it's less interesting for a mathematician but that is for CS people (they don't like negative numbers)

Jean-Baptiste Vienney (Feb 26 2023 at 20:28):

Well by looking at what it is that seems difficult to get rid of the negative numbers.

Jean-Baptiste Vienney (Feb 26 2023 at 20:31):

I'm excited by the day when we will be able to prove the Fermat theorem (which is something about positive integers!) without using the notion of negative number.

Jean-Baptiste Vienney (Feb 26 2023 at 20:33):

But they are everywhere in the usual math: groups, rings, (co)homology...

Jean-Baptiste Vienney (Feb 26 2023 at 20:35):

I remember that Grothendieck made such an observation in Récoltes et Semailles maybe, that we could have made maths without them, at a time where semi-rings were not used at all I guess.

Jean-Baptiste Vienney (Feb 26 2023 at 20:46):

I'm excited by things like "semiring schemes"

John Baez (Feb 26 2023 at 21:28):

Jean-Baptiste Vienney said:

Maybe you could do it as well by assuming that your hom-sets are just commutative monoids and not necessarily abelian groups. In that case $\mathcal{C}[I,I]$ is a semi-ring and the $\mathcal{C}[A,B]$ semi-modules

Right I'm actually interested in a more specialized situation, not a less specialized one. Of course I'm interested in everything, in a diffuse sort of way - but right now @Joe Moeller, @Todd Trimble and I are writing about '2-rigs', which is our short name for symmetric monoidal $k$-linear categories that are Cauchy complete, where $k$ has characteristic zero.

John Baez (Feb 26 2023 at 21:29):

So, I really want results about when the unit object in such a category is projective!

Jean-Baptiste Vienney (Feb 26 2023 at 21:30):

Sure, sorry for the digression!

John Baez (Feb 26 2023 at 21:30):

I also want to generalize algebraic geometry from commutative rings to commutative rigs, but I'm not really working on that now.

Jean-Baptiste Vienney (Feb 26 2023 at 21:34):

Maybe you could be interested by this: Tangent categories as a bridge between differential geometry and algebraic geometry

Jean-Baptiste Vienney (Feb 26 2023 at 21:35):

I think they say that what they did could work for semirings

John Baez (Feb 26 2023 at 21:35):

I'm interested in too many things. :cry: Please don't show me more.

Jean-Baptiste Vienney (Feb 26 2023 at 21:35):

Ahah okay

Reid Barton (Feb 26 2023 at 21:37):

John Baez said:

So, maybe a concrete formulation of my vague question is: under what conditions on a symmetric monoidal additive category is the unit for the tensor product projective?

I'm not sure what a general condition might be, but here's a case where it doesn't happen: take the category of sheaves of $k$-vector spaces on a space $X$ such as the unit interval.

John Baez (Feb 26 2023 at 21:41):

Hmm, thanks - I'm surprised by that, which means I don't understand the relation between sheaves, vector bundles, and projectiveness as well as I should! What do projective sheaves of $k$-vector spaces on the unit interval look like?

What about in algebraic geometry? For example, say we take the category of sheaves of $\mathcal{O}$-modules on a complex variety. Then the unit for the tensor product is projective, right? Or am I seriously confused?

Reid Barton (Feb 26 2023 at 21:46):

Quasicoherent sheaves?

Reid Barton (Feb 26 2023 at 21:48):

In the topological space example, you have the surjection $kU \oplus kV \to kX$ for any open cover $X = U \cup V$ (here $kU$ is the free vector space on the sheaf of sets $U \subset X$, etc.). I'm not 100% sure exactly how to prove it, but it seems that for, say, $U = [0, 2/3)$ and $V = (1/3, 1]$, this surjection doesn't split.

Reid Barton (Feb 26 2023 at 21:50):

In the algebraic geometry setting you mentioned, the unit object can also fail to be projective if the variety is not affine, because then there can be nonzero higher cohomology of a quasicoherent sheaf.

Reid Barton (Feb 26 2023 at 21:51):

And Hom from the unit object to a sheaf $\mathcal{F}$ is its global sections, so Ext from the unit object to $\mathcal{F}$ is its cohomology.

Reid Barton (Feb 26 2023 at 21:52):

I also wanted to say that if your category is not abelian, then there are several inequivalent notions of "projective" that stem from different notions of "epimorphism".

Patrick Nicodemus (Feb 26 2023 at 22:14):

A useful notion of projective is projective relative to a comonad. An object X is projective wrt the comonad G if the counit map GX->X has a section. Equivalently, it is expressible as a retract of any object in the image of G.
In your example perhaps you can pick an adjunction between your symmetric monoidal category C and some underlying category D, with F: C->D left adjoint to U:C->D and take G=FU. Perhaps a finitely generated free object in C can be defined relative the adjunction as well by identifying a class of finite objects in C and calling their images in D finitely generated and free.

If the unit of the tensor product is in the image of F that would work well.

John Baez (Feb 27 2023 at 01:53):

Reid Barton said:

In the algebraic geometry setting you mentioned, the unit object can also fail to be projective if the variety is not affine, because then there can be nonzero higher cohomology of a quasicoherent sheaf.

Wow, I'm suffering from some mental disconnect. On the one hand I knew about cohomology of sheaves and knew that the cohomology of the structure sheaf $\mathcal{O}$ of a projective variety is interesting. On the other hand, in a separate part of my brain, I knew that the cohomology of a projective object should be trivial.

(Perhaps this part of my brain was insulated from the other due to two completely different meanings of "projective"? :upside_down:)

John Baez (Feb 27 2023 at 01:54):

But then, getting in the way, there was some part of my brain that thinks invertible objects should be projective, and that line bundles correspond to invertible sheaves.

John Baez (Feb 27 2023 at 01:56):

The problem is that I was over-generalizing: invertible modules of rings are projective, but not invertible sheaves!

Graham Manuell (Feb 27 2023 at 15:07):

The dualisable objects in the monoidal category of R-modules are precisely the finitely generated projective modules, so perhaps dualisable objects are a good notion of "finitely generated projective objects" more generally?

John Baez (Feb 27 2023 at 23:42):

Thanks, that's nice! Is there any nice theorem saying that under some conditions on symmetric monoidal category with finite coproducts, every dualizable object is a summand of an object $I \oplus \cdots \oplus I$?

Reid Barton (Feb 28 2023 at 11:13):

I think the category of representations of a (let's say finite) group is a case where that doesn't happen. Because then the unit object is the trivial representation, but I think all the finite-dimensional representations are dualizable.

Graham Manuell (Feb 28 2023 at 14:05):

Also, in the category of suplattices, all the projective objects are dualisable, but they aren't all retracts of finite copowers of $I$.

John Baez (Feb 28 2023 at 21:49):

Right, but I was asking for conditions under which it is true. :upside_down:

John Baez (Feb 28 2023 at 21:50):

It's true for the category of vector bundles on a compact Hausdorff space.

John Baez (Feb 28 2023 at 21:56):

And more generally it's true for the category of $R$-modules for a commutative ring $R$, with its usual tensor product.

John Baez (Feb 28 2023 at 21:57):

But maybe there aren't lots of other situations where it's true?

Graham Manuell (Feb 28 2023 at 23:49):

If it is important that it really is a direct summand you probably want the category to be enriched over abelian groups. If it's enough for it to just be a retract, then any category of semimodules over a semiring should work. I don't know how much further it can be generalised.