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Brendan Murphy (Aug 21 2023 at 01:44):Given an ordinary category I can endow the hom-sets with the structure of an abelian group and ask if composition is bilinear, or endow them with the structure of pointed sets and ask if composition preserves the point. In these cases, I'm fairly sure that we should get a category enriched over abelian groups or over pointed sets. The structure at play here seems to be that we have an adjunction where the right adjoint is faithful (but not full) and is lax monoidal. Also in both cases I've given, this forgetful functor is actually corepresented by the unit object. Maybe that's important?
Is there some general result about when we can endow an ordinary category with an enrichment by "reflecting" back along a functor like this? If so, could someone point me to where it's written down? I don't think the stuff on nlab about change of enriching category applies here, but I apologize if I'm missing something obvious
Patrick Nicodemus (Aug 21 2023 at 02:43):Some of your comments may be distractors as they are just stating things that hold in general.
Every monoidal category has a unit object . The Hom functor is a lax monoidal functor to Sets. In this sense there is always a forgetful functor from V-Cat to Cat. It may lose a lot of information as in the case where we want to view metric spaces as categories enriched over R^{+}\cup {\infty}, or very little.
It may not always be faithful; It is faithful precisely when 1 is a generator. it need not always have a left adjoint. If it does, it will necessarily send any set X to the coproduct of |X| many copies of the unit object.
If 1 is a generator of V then there are good theorems in general about gluing together objects in V out of copies of 1.
Patrick Nicodemus (Aug 21 2023 at 02:45):image.png
What I meant by "good theorems about gluing objects together out of copies of 1" is that there are adjoint functor theorems concerning generators
Patrick Nicodemus (Aug 21 2023 at 02:45):this is from Borceux's handbook of categorical algebra.
Brendan Murphy (Aug 21 2023 at 02:46):Thanks for the clarification. Essentially what I'm asking for is conditions that allow me to lift along that forgetful functor V-Cat -> Cat.
Patrick Nicodemus (Aug 21 2023 at 02:50):Ah, ok. I understand what you mean then.
Patrick Nicodemus (Aug 21 2023 at 02:57):https://ncatlab.org/nlab/show/commutative+algebraic+theory
Perhaps section 3 of this page is relevant to you. It says that if T is a monad on a monoidal category (such as Sets with Cartesian product) and it is a commutative monad, then the category of algebras for that monad inherit a monoidal structure (such as the tensor product or the smash product of pointed sets).
So, if your monoidal category V is monadic over Sets, and this monoidal structure on V is the one inherited from the monoidal structure, then lifting C to a V-cat is the same as constructing T-algebra structures on each Hom set compatible with the monoidal product of T-algebras.
Patrick Nicodemus (Aug 21 2023 at 02:58):That is the most immediate generalization of Ab and pointed sets I can think of. Although monadicity is a strong assumption. On the other hand you appear to be exactly looking for cases where it is sensible to talk about V as "a set equipped with additional structure."
Brendan Murphy (Aug 21 2023 at 02:58):ah yeah, that's great!
Brendan Murphy (Aug 21 2023 at 02:58):yeah yeah i am, the only case I had in mind I didn't mention was dg-categories and Ch(R) is still monadic over Set
Brendan Murphy (Aug 21 2023 at 03:01):although hm, you won't get the koszul sign rule will you...
Brendan Murphy (Aug 21 2023 at 03:04):maybe there's a trick where you go Set -> Ab -> R-Mod -> GrR-Mod -> Ch(R-Mod) and introduce the sign rule at the level of graded modules? assuming that the forgetful functor Ch(A) -> GrA is monadic, which I haven't checked
Patrick Nicodemus (Aug 21 2023 at 03:06):I don't know much about the Koszul sign rule. There cannot be many symmetry isomorphisms on chain complexes which actually commute with the differentials. Isn't it kind of forced on you?
But I agree it's plausible that you might have to work at the level of a different monadic adjunction.
Patrick Nicodemus (Aug 21 2023 at 03:07):What do you mean by the Koszul rule in simple terms here?
Brendan Murphy (Aug 21 2023 at 03:07):I think you're flipped around, it's the symmetry isomorphism that determines the differentials on various constructions
Patrick Nicodemus (Aug 21 2023 at 03:08):And you're worried that the monadicity theorem for Ch(Mod) you're thinking of gives the wrong tensor product on chain complexes?
Brendan Murphy (Aug 21 2023 at 03:09):by the koszul sign rule I mean the symmetry isomorphism on graded modules (or chain complexes) where if a has degree n and b has degree m then σ(a ⊗ b) = (−1)ⁿᵐ b ⊗ a
Brendan Murphy (Aug 21 2023 at 03:09):yep
Brendan Murphy (Aug 21 2023 at 03:09):if there's some formal algebraic construction i would expect it to just do σ(a ⊗ b) = b ⊗ a
Patrick Nicodemus (Aug 21 2023 at 03:09):And does that commute with the differentials?
Because the differentials are part of the object, so they should too be determined at the formal level.
Patrick Nicodemus (Aug 21 2023 at 03:10):Like, the differential on T(A)\otimes T(B) should be given by abstract nonsense if Ch(Mod) is a commutative algebraic theory over sets.
Brendan Murphy (Aug 21 2023 at 03:10):Thinking
Brendan Murphy (Aug 21 2023 at 03:13):so the differential we usually use is d^{C ⊗ D}_k(a ⊗ b) = d^C_n(a) ⊗ b + (-1)^n a ⊗ d^D_m(b), where k = n + m and a ∈ Cₙ, b ∈ Dₘ
Brendan Murphy (Aug 21 2023 at 03:14):I'm pretty sure that if you include no signs the result won't satisfy d^2 = 0
Brendan Murphy (Aug 21 2023 at 03:14):you'll get a term like x + x when it should be x - x
Patrick Nicodemus (Aug 21 2023 at 03:14):This also seems true to me.
Brendan Murphy (Aug 21 2023 at 03:14):I'm not sure how this comes from abstract nonsense though
Brendan Murphy (Aug 21 2023 at 03:14):Maybe I should think through whether Ch(R) -> GrMod(R) is monadic
Brendan Murphy (Aug 21 2023 at 03:16):so a chain complex is a graded module C equipped with a map d : C -> C[1], where C[1] has the grading shifted down by 1
Brendan Murphy (Aug 21 2023 at 03:16):satisfying the axiom d[1] ∘ d = 0
Brendan Murphy (Aug 21 2023 at 03:18):what would the free chain complex on a graded module M be? I guess my instinct is that we just add a shifted copy of M to itself. F(M) = M (+) M[1]. the differential is then (x, y) |-> (0, x)
Patrick Nicodemus (Aug 21 2023 at 03:19):Well, I personally have my own theoretical explanation for the Koszul convention.
I don't wish to appear as though I have a one track mind, but hear me out.
First of all, when you say "you have it backwards, the symmetry isomorphism determines the choice of differential on the tensor product" I prefer to disagree. The tensor product of chain complexes is the one that makes Dold-Kan strong monoidal, because it is inherited across that equivalence from the Day convolution of simplicial Abelian groups.
The augmented simplex category is a monoidal category, but it is not a symmetric monoidal category.
However, surprisingly enough, the free Ab-enriched category - which is also a monoidal category - can be equipped with a symmetry isomorphism. It is easy to see that this isomorphism is uniquely determined by the isomorphism (1 + 1) \cong (1+1).
By Yoneda, this passes to a symmetry on the Day convolution monoidal structure of presheaves, which is carried across the Dold-Kan correspondence, and that is how we get the symmetry convention on the tensor product.
Brendan Murphy (Aug 21 2023 at 03:19):lmao
Brendan Murphy (Aug 21 2023 at 03:20):my theoretical explanation for the symmetry isomorphism is about orientation or signs of permutations
Brendan Murphy (Aug 21 2023 at 03:20):the matrix representing a swap R^n x R^m -> R^m x R^n has determinant (-1)^nm
Patrick Nicodemus (Aug 21 2023 at 03:20):It is helpful here to think of a chain complex as a graded module which is a module over where x is in degree -1 and is concentrated in degree 0.
Patrick Nicodemus (Aug 21 2023 at 03:21):I borrowed this helpful insight from Alexander Shmakov.
Of course, modules over a ring are monadic over Abelian groups, where the monad is ; and so I would imagine that graded algebras are monadic over graded rings.
Patrick Nicodemus (Aug 21 2023 at 03:25):Anyway I think both ways of doing this will be successful. I still think doing this over Sets will give you the correct answer, as the Koszul convention is necessary to get the differential right, and the differential is determined by abstract nonsense.
Brendan Murphy (Aug 21 2023 at 03:28):I wasn't actually intending to talk about anything DG-y in my notes but now I feel like I should work out what we're talking about...
Brendan Murphy (Aug 21 2023 at 03:28):on the other hand my oral exam is in 46 days and I'm already sidetracked enough already
Brendan Murphy (Aug 21 2023 at 03:29):I just need to make sure I can get the signs right in front of a blackboard
Patrick Nicodemus (Aug 21 2023 at 03:30):I can give you a simple intuitive answer as to where the alternating signs come from "categorically".
The tensor product of chain complexes is cocontinuous in both arguments. If you take some simple chain complexes like S^n concentrated in a single degree, it's not obvious how to extrapolate the sign convention, but if you care about cocontinuity then T(X)\otimes T(Y) should be a colimit of copies of T(S^n) and T(S^m). And colimits involve gluing, and gluing in Abelian categories involves taking kernels, and that involves identifying f with negative g.
That's where the signs come from in my perspective, the gluing/quotienting involved in forming the tensor product.
Brendan Murphy (Aug 21 2023 at 03:31):taking cokernels?
Patrick Nicodemus (Aug 21 2023 at 03:33):Yes. Apologies
Patrick Nicodemus (Aug 21 2023 at 03:36):Actually like. If you just want to take the very simplest example:
concentrated in degrees 0 and 1.
This already has to have alternating sign conventions to get a well defined tensor product of this guy with itself
Patrick Nicodemus (Aug 21 2023 at 03:38):Then by cocontinuity this determines it for all chain complexes because and its higher tensor powers are dense in Ch(Ab) (in the Ab-enriched sense)
Steve Huntsman (Aug 21 2023 at 12:23):Brendan Murphy said:
Is there some general result about when we can endow an ordinary category with an enrichment by "reflecting" back along a functor like this?
This does not seem to answer your question as intended, but it does speak to the title of "easily checking that a category is enriched." Suppose someone hands you a finite digraph with data on the arcs and asks you to determine if it yields an enriched category. Pick a spanning polytree and pick any arc not in the polytree. Check that composition holds for that arc and the adjoining arcs on the polytree. Until/unless composition fails, repeat. This gives an algorithm linear in the number of arcs/putative morphisms. This can be slightly nontrivial because morphisms need not be invertible. Explicit examples along these lines are in https://arxiv.org/abs/2304.08334