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Stream: deprecated: id my structure

Topic: derivations for k-linear categories

Matteo Capucci (he/him) (Sep 27 2023 at 13:08):

The usual definition of K-derivation on a K-algebra $A$ (i.e. a K-linear map $D:A \to A$ such that $D(ab)=D(a)b+aD(b)$) seems to be readily generalizable to K-linear categories, by saying that a derivation $D:\cal A \to A$ of such categories is given by a function on objects, locally linear maps on morphisms, and instead of functoriality requiring that $D(f;g)=D(f);g + f;D(g)$.
Notice this directly generalizes the traditional definition since a K-algebra is a one-object K-linear category.

Has anyone ever heard of these? I didn't finid anything on Google but I'd be suprised if I'm the first to think about these things

Matteo Capucci (he/him) (Sep 27 2023 at 15:33):

You can even define the generalized version of derivations in which the codomain is just an $A$-bimodule: replace $A$-bimodule with 'two-sided fibration over $\cal A$' (of course, a suitably enriched in $K\bf Mod$), i.e. a two-sided fibration $\cal A \leftarrow M \to A$.

Matteo Capucci (he/him) (Sep 27 2023 at 15:34):

Then replace Leibniz with

$$D(fg) = f^*Dg + g_*Df$$

Matteo Capucci (he/him) (Sep 27 2023 at 15:35):

To have this typecheck, $D$ has to be a section of both sides of the two-sided fibration

Matteo Capucci (he/him) (Sep 27 2023 at 15:36):

Alternatively, you can replace two-sided fibration with bifibration, which makes being a section easier

Graham Manuell (Sep 27 2023 at 20:26):

It seems like such a derivation can send isomorphic objects to non-isomorphic objects. Indeed, if D is zero on morphisms, there is no restriction on what it can do to objects. Is this the intention?

Matteo Capucci (he/him) (Sep 27 2023 at 21:46):

Yeah these are not functors so there's no expectation for them to be isomorphism-preserving

James Deikun (Sep 28 2023 at 09:22):

Not sure what the map on objects is even getting you here though.

James Deikun (Sep 28 2023 at 09:26):

My instinct would be to say "a derivation $D$ riding a functor $F : \mathcal{A} \to \mathcal{B}$ is given by local linear maps $D_{x,y} : \mathcal{A}(x,y) \to \mathcal{B}(F(x),F(y))$ satisfying $D(f;g) = F(f);D(g) + D(f);F(g)$".

James Deikun (Sep 28 2023 at 09:35):

This gives them the structure of a $K$-module. But maybe they should actually have the structure of an $\mathcal{A}$-presheaf or profunctor from $\mathcal{A}$ to $\mathcal{A}$ or something, and I'm less sure how to define them so that that happens.

James Deikun (Sep 28 2023 at 09:38):

This is the kind of thing where it wouldn't hurt to have an application in mind ...

Matteo Capucci (he/him) (Sep 28 2023 at 11:07):

James Deikun said:

Not sure what the map on objects is even getting you here though.

What do you mean?

Matteo Capucci (he/him) (Sep 28 2023 at 11:09):

James Deikun said:

My instinct would be to say "a derivation $D$ riding a functor $F : \mathcal{A} \to \mathcal{B}$ is given by local linear maps $D_{x,y} : \mathcal{A}(x,y) \to \mathcal{B}(F(x),F(y))$ satisfying $D(f;g) = F(f);D(g) + D(f);F(g)$".

Uhm why a functor? I was moving along the correspondence 'category : two-sided fibration = ring : bimodule'

Matteo Capucci (he/him) (Sep 28 2023 at 11:12):

I believe your definition is subsumed by mine, anyway. You use the functor $F$ to change base for either $\cal B = B =B$ or $\cal B \leftarrow B^\downarrow \to B$, though I don't have time to check rn

Matteo Capucci (he/him) (Sep 28 2023 at 11:20):

James Deikun said:

This is the kind of thing where it wouldn't hurt to have an application in mind ...

Yeah indeed... I'm trying to understand whether such a notion could help shed light on some constructions in differential categories, chiefly the Faà di Bruno comonad. The point is that one can see the chain rule as saying that taking derivatives is a section of a fibration, for instance a section of the codomain fibration $B^\downarrow \to B$ (I'm simplyfing). But then when you take higher derivatives, the chain rule has to be combined with the Leibniz rule.
This comes from a subtlety: chain rule in first-year calculus is defined as $D(f;g)(x) = D(g)(f(x)) \cdot f(x)$, i.e. uses multiplication, but in differential geometry (third-year calculus?) multiplication seems to disappear: $d(f;g)(x,v) = dg_{f(x)}(df_x(v))$!

The resolution of the conundrum, of course, is that multiplication in first-year calculus is actually the application of a linear map to a vector, like in third-year calculus. This, however (and here's where I might be mistaken) suggests that the Leibniz rule is about the derivative of applications of linear maps, hence suggesting that it might be captured by an horizontal categorification of the notion of derivation in differential algebra.

James Deikun (Sep 28 2023 at 11:26):

Matteo Capucci (he/him) said:

James Deikun said:

Not sure what the map on objects is even getting you here though.

What do you mean?

I mean I think the map on objects in your definition is losing you structure without getting any useful flexibility in return.

Matteo Capucci (he/him) (Sep 28 2023 at 11:30):

What structure is it losing?

Matteo Capucci (he/him) (Sep 28 2023 at 11:31):

Or better, why do you think adding a functorial action too would improve the situation? :thinking:

James Deikun (Sep 28 2023 at 11:41):

Matteo Capucci (he/him) said:

Uhm why a functor? I was moving along the correspondence 'category : two-sided fibration = ring : bimodule'

Besides the notion of a derivation into a bimodule, there's a notion of a derivation riding an algebra homomorphism; this is the higher equivalent of that. I think derivation into a bimodule does subsume it though. But I think introducing this idea clarifies why you don't want the object map; a "plain" derivation should be riding the identity functor.

I think the proper equivalent here of a bimodule derivation is a derivation riding a K-linear profunctor $P : \mathcal{A} \: ⇸ \: \mathcal{A}$ where $D_{x,y} : \mathcal{A}(x,y) \to P(x,y)$ where $D(f ; g) = f ; D(g) + D(f) ; g$ where the $;$ on the right is the action of the profunctor. This subsumes the above definition when $P = \mathcal{A}(F,F)$.

James Deikun (Sep 28 2023 at 11:42):

Matteo Capucci (he/him) said:

What structure is it losing?

It's losing K-linear structure on the space of derivations.

John Baez (Sep 28 2023 at 11:51):

I like derivations taking values in a bimodule of a ring or algebra, and I like @James Deikun's idea of generalizing these to derivations riding $\mathsf{V}$-enriched profunctor $P: \mathsf{A} \to \mathsf{A}$, i.e. $P : \mathsf{A} \otimes \mathsf{A}^{\text{op}} \to \mathsf{V}$. It looks like we only need $\mathsf{V}$ to be $\mathsf{CommMon}$-enriched to make sense of the addition in the formula for a derivation, and also the linearity of the derivation.

John Baez (Sep 28 2023 at 11:52):

But we could take $\mathsf{V} = \mathsf{AbGp}$ or $\mathsf{V} = \mathsf{Vect}_k$ for some field $k$ if we want things to be more familiar.

John Baez (Sep 28 2023 at 11:52):

I feel I may be making a level slip somewhere, but I hope not.

Matteo Capucci (he/him) (Sep 28 2023 at 12:21):

James Deikun said:

Matteo Capucci (he/him) said:

Uhm why a functor? I was moving along the correspondence 'category : two-sided fibration = ring : bimodule'

Besides the notion of a derivation into a bimodule, there's a notion of a derivation riding an algebra homomorphism; this is the higher equivalent of that. I think derivation into a bimodule does subsume it though. But I think introducing this idea clarifies why you don't want the object map; a "plain" derivation should be riding the identity functor.

I see, now it's clear! I wasn't aware of the linear-riding definition!

Matteo Capucci (he/him) (Sep 28 2023 at 12:23):

James Deikun said:

I think the proper equivalent here of a bimodule derivation is a derivation riding a K-linear profunctor $P : \mathcal{A} \: ⇸ \: \mathcal{A}$ where $D_{x,y} : \mathcal{A}(x,y) \to P(x,y)$ where $D(f ; g) = f ; D(g) + D(f) ; g$ where the $;$ on the right is the action of the profunctor. This subsumes the above definition when $P = \mathcal{A}(F,F)$.

This sounds good, although note that profunctors are discrete two-sided fibrations :) which means we agree!

Matteo Capucci (he/him) (Sep 28 2023 at 12:24):

John Baez said:

It looks like we only need $\mathsf{V}$ to be $\mathsf{CommMon}$-enriched to make sense of the addition in the formula for a derivation, and also the linearity of the derivation.

Indeed, that's actually the setting I'm interested in (I guess these would be '$\N$-linear' categories?)

James Deikun (Sep 28 2023 at 12:30):

Matteo Capucci (he/him) said:

This sounds good, although note that profunctors are discrete two-sided fibrations :) which means we agree!

This part actually tends to fail in the enriched setting.

Matteo Capucci (he/him) (Sep 28 2023 at 12:31):

Ah uhm right, there you need to use two-sided codiscrete cofibrations instead...

Matteo Capucci (he/him) (Sep 28 2023 at 12:33):

I'm more interested in this remark though:
James Deikun said:

It's losing K-linear structure on the space of derivations.

I'm having a tough time seeing how derivations $D:\mathcal A \to P$ (where $P$ is a K-linear profunctor) would form a K-linear category, unless we still make them into an algebra, i.e. a one-object K-linear category. This works but it's a bit weird..?

James Deikun (Sep 28 2023 at 12:36):

Well, with this definition the structure that derivations have is just that they live in $\mathsf{V}$ (possibly up to size). Which is at least something!

John Baez (Sep 28 2023 at 12:37):

James didn't say derivations formed a category, just that you can add them and multiply them by elements of $K$. And I think that should be true if we define them correctly (in the $K$-linear case).

James Deikun (Sep 28 2023 at 12:40):

There might also be a nice structure where composition of endoprofunctors induces a heterogeneous composition on derivations, but I'd have to explicitly work it out to be 100% sure.

James Deikun (Sep 28 2023 at 12:41):

(I'm only, like, 60% sure.)

James Deikun (Sep 28 2023 at 16:29):

Usually when I think about higher derivations, btw, the cleanest framework seems to be that of Hasse-Schmidt derivations. You can generalize Hasse-Schmidt derivations from their usual context to a bimodule context, with an appropriate notion of power series algebra.

James Deikun (Sep 28 2023 at 17:27):

Here we can define the power series category $P[[t]]$ generated by a $\mathsf{V}$-enriched profunctor $P : \mathsf{A} \to \mathsf{A}$ (where $\mathsf{V}$ itself is $\mathsf{CommMon}$-enriched) as a category enriched in $\mathbb{N}$-graded objects of $\mathsf{V}$, where:

• $P[[t]](x,y)_n = P^n(x,y)$ ($P^n$ is the $n$-fold self-composition of $P$)
• $(f \circ g)_n = \sum_{k = 0}^n f_k \circ g_{n-k}$ ($\circ$ on the right is the semi-formal composition in $P^{n-k} \circ P^{k} \cong P^{n}$)

James Deikun (Sep 28 2023 at 17:29):

Then a Hasse-Schmidt derivation into $P$ is just a functor from $\mathsf{A}$ to $P[[t]]$!

James Deikun (Sep 28 2023 at 17:38):

The higher-order Leibniz rules fall out of this formalism really naturally; I wonder if something similar can happen with higher-order chain rules.