Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: differential rig categories

fosco (Jan 23 2021 at 12:23):

(Previously "differential 2-rigs", but there's been a little of a fight about the correct name :grinning: )

So, the idea is really simple, and I will lazily summarise it in the abstract of the draft-of-a-paper I am writing. (Jokes apart, I find it quite self-explanatory of the general motivation I had, when this project began)

image.png

Now, if $\mathcal C$ is a rig category, in the sense that a monoidal structure $\otimes$ distributes over coproducts, a derivation is just an endofunctor $\partial : \mathcal C \to \mathcal C$ with the property that

• $\partial(A\cup B) \cong \partial A \cup \partial B$

• $\partial(A\otimes B) \cong \partial A \otimes B \cup A\otimes\partial B$.

As a general rule, one knows to be after some deep Mathematics when a seemingly innocuous assumption yields unexpected or counterintuitive results . This is the case here, because from these very natural assumptions most of the theory of differential rings carries over, with one, eminent, painful exception: there is absolutely no reason why the derivative of the monoidal unit is zero (=the initial object of the rig).

Let me offend your intelligence by recalling that if if $R$ is a differential ring, with derivation $d$, then $d1=d(1\cdot 1)= d1 +d1$; now, since the additive monoid of $R$ is cancellative, we get $d1=0$. In fact, in differential semiring theory, the most you can get is that the derivative of $1$ is an additive idempotent. Now, some authors require $d1=0$ as an axiom of a differential semiring, essentially because if this is not required, $d$ is not well-defined: $d(a)=d(a\cdot 1) = da + a \cdot d1$, which in its own right is $da + a\cdot d1 + a\cdot d1$, which... on the other hand, there are examples of semirings where every object is an additive idempotent (posets with $\vee$, an eminent instance of which are co-Heyting algebras, that carry interesting operations that satisfy the axioms of a derivation.

When one moves to true categories things get way hairier, essentially because $\partial 1 = \partial 1 + \partial 1$ entails a whole lot of counterintuitive properties: I will list the ones that I found until now, in discussions with @Nathanael Arkor @Fabrizio Genovese (with which this ended up being a joint work) @Daniele Palombi and others (I get pretty pushy when there's mathematics I like and I can't get it right :grinning: )

• First, $\partial 1$ must be "empty or big", in the sense that -for example in the category of sets- a set $A$ such that $A+A\cong A$ is either empty or infinite.

• On the other hand, in the category of _finite_ sets, $A$ can only be empty; same in the category of finite dimensional vector spaces, where $d=\dim V = 2d$ has 0 (thus the zero space) as unique solution. Same, in every category with a decent choice of dimension $\mathcal C \to \mathbb N$. (What are examples of these categories? Do they have a name)?

• On another hand, sometimes $\partial 1 = \partial 1 + \partial 1$ is forced to have just trivial solutions due to the request that this equality is in some sense natural; for example, let $\partial$ be a derivation in a category of functors $\mathcal A \to Set$; then, $\partial 1$ is a functor $\mathcal A \to Set$ such that $F \cong F + F$, and naturally so; such functors are few, because no bijection $FA\cong FA + FA$ can be natural if the domain $\mathcal A$ is just a little bit nontrivial. I have asked this question for the exact purpose of establishing this obstruction result.

• In an abstract category of course you can't uniquely attach to an object information about how big it is; but you still have morphisms and the Yoneda lemma! So, let's stick $\partial 1$ in the hom-set $\hom(\partial 1,Z)$ for a generic object $Z$; the fact that $\partial 1\cong \partial 1 + \partial 1$ yields $\hom(\partial 1, Z) \cong \hom(\partial 1+\partial 1, Z)\cong \hom(\partial 1,Z) \times \hom(\partial 1, Z)$. Now _this_ is interesting, because $\hom(\partial 1,Z)$ can only be empty, a singleton or infinite. Let me sum up: a property of $\mathcal C$ that -at least a priori- has nothing to do with its size, the presence of a differential structure, imposes some conditions on the numerosity of some of its hom-sets; if a category is finite, thus, $\hom(\partial 1, Z)$ is either empty or a singleton, and in particular it must be a singleton when $Z=\partial 1$. Again, having a differential structure and being finite entails that $\partial 1$ is a "rigid" object, in the sense that the monoid of endomorphisms of $\partial 1$ is trivial.

• More on this: whenever there is an extension $\partial 1 \hookrightarrow A$ in $\mathcal C$, there must be a monomorphism $\hom(\partial 1, \partial 1) \hookrightarrow \hom(\partial 1, A)$, so that if the domain, endomorphisms of $\partial 1$, is infinite, the codomain, arrows $\partial 1 \to A$, can only be bigger. The endomorphisms of $\partial 1$ clearly are "something special"! And this is why I would like to get to know them better: let $M=\hom(\partial 1,\partial 1)$ for short; then, from $M\cong M^2 = M\times M$ one easily gets, by induction, that $M\cong M^n$ for every finite $n$. How many monoids with this property do you know, apart the terminal one?

• Let's go deeper in the white rabbit's hole: if $\partial 1 \cong \partial 1 + \partial 1$, this means that $\partial 1$ is naturally a coalgebra for the "leave it or double it" functor $S : A\mapsto A +A$, in such a way that there is a unique map

  $\begin{array}{ccc} \partial 1 &\cong& \partial 1 + \partial 1 \\ \downarrow && \downarrow \\ C &\cong& C + C \end{array}$

  between $\partial 1$ and the terminal coalgebra of $S$; but wait, in the category of topological space $C$ is the Cantor set! What just happened here? Is the map $\partial 1 \to C$ an epi, a mono in general?

So. All these questions are vague and somewhat alluding to the fact that I do not have a clear picture of what's going on: indeed, I have no idea which wa to turn: what am I after? "What is" the derivative of 1 in a differential ring, like, really? How shall one interpret each of the strange properties that I have outlined?

fosco (Jan 23 2021 at 12:54):

Another interesting idea related to the perks and shortcomings of being $\partial 1$ is the following: as an endofunctor, $\partial$ might have interesting fixed points, and there is a standard procedure to build its initial algebra and terminal coalgebra. One could legitimately call such an object "Eulerian" or "Neperian", for obvious reasons (the most reasonable is that "exponential object" already means something else).

Initial algebras are trivial, in that $\partial 0=0$ by using the Leibniz rule. On the other hand, the triviality of terminal coalgebras is governed by $\partial 1$, in the sense that (assuming the terminal object is the monoidal unit, that's what is called a semicartesian monoidal category)

$1\leftarrow \partial 1 \leftarrow \partial\partial 1 \leftarrow \partial\partial\partial 1 \leftarrow \dots$

and the first ordinal $\lambda$ for which the transition morphism $v : \partial^\lambda 1 \to \partial^{\lambda + 1} 1$ is invertible realises the terminal coalgebra. Now, what is this map? What is the result of applying $\partial$ many times to $1$? Well, since $\partial 1 = \partial 1 + \partial 1$, and $\partial$ is linear, $\partial\partial 1 = \partial (\partial 1 + \partial 1) = \partial \partial 1 + \partial \partial 1$, thus the entire cochain is made by coalgebras for the functor $\_ + \_$ (even more, by coalgebras whose coalgebra map is invertible: how many such algebras are there? What's their shape?).

Fabrizio Genovese (Jan 23 2021 at 13:36):

Yeah, I feel like $\partial 1$ sometimes gives us a very coarse way of representing cardinality within $\mathcal{C}$. In Set theory, we "measure:" size of sets by using injections. This is due to Hartog's lemma, which basicaly says that for any two sets $X,Y$ you have either an injection $X \to Y$ or the other way around. Bernstein's theorem instead says that if you have monos both ways, then $X,Y$ are in bijection. This is what allows us to order sets using cardinality.

In here, if $\partial 1$ is big, meaning that $Hom(\partial 1, \partial 1)$ is infinite, then a monomorphism $\partial 1 \to \partial Z$ guarantees that there is a mono $Hom(\partial 1, \partial 1) \to Hom (\partial 1, Z)$. Hence, the generalized elements of $Z$ of type $\partial 1 \to Z$ are infinite, and $Z$ is in some sense "big". I feel there is much more to say about this if one looks deep enough...

Antonin Delpeuch (Jan 23 2021 at 13:57):

Very nice! Can you tell a bit more about the context: why are you interested in differential rig categories? What motivated this definition?

Fabrizio Genovese (Jan 23 2021 at 14:06):

To my knowledge, literally nothing.

Fabrizio Genovese (Jan 23 2021 at 14:07):

I mean, I don't see any immediate motivating application. I think it started as a cool set of mathematically pertinent questions that Fosco had

Fabrizio Genovese (Jan 23 2021 at 14:07):

And you know, you ask questions, give more answers, connect the points, keep asking more questions, ...

Fabrizio Genovese (Jan 23 2021 at 14:08):

This is also why we do not have a precise direction for this at the moment. This work can evolve in many multiple ways and we literally have no real bias. For me, it's just a matter of understanding how much more gas there's in the tank, and where can we get from here :smile:

fosco (Jan 23 2021 at 14:55):

Antonin Delpeuch said:

Very nice! Can you tell a bit more about the context: why are you interested in differential rig categories? What motivated this definition?

I just find it a very natural notion and the many examples prove it is.

fosco (Jan 23 2021 at 14:56):

But I might be misunderstanding the question: are you asking if a particular example motivated the definition? Yes (combinatorial species) and no (I started to think about them after having drafted the first definition of a derivation on a category). Same with "Brzozowski derivatives", whose existence I didn't even suspected before :smile:

Antonin Delpeuch (Jan 23 2021 at 16:02):

Ah ok, I did not know about combinatorial species, that totally explains the motivation :)

Eric Forgy (Jan 23 2021 at 18:51):

Should cancellation also happen in a rig? :thinking:

It seems reasonable to require
${}$
$1\otimes M\cong 2\otimes M\implies M\cong 0$
${}$
even for a rig, where $2 = 1+1.$
${}$

fosco (Jan 23 2021 at 19:36):

Eric Forgy said:

Should cancellation also happen in a rig? :thinking:

It seems reasonable to require
${}$
$1\otimes M\cong 2\otimes M\implies M\cong 0$
${}$
even for a rig, where $2 = 1+1.$
${}$

[Ah, you just said what was 2, sorry] :grinning:

Same as Todd's comment below, yes.

Todd Trimble (Jan 23 2021 at 19:37):

Boolean algebras are rigs. But $x \vee x = x$ doesn't imply $x = 0$.

John Baez (Jan 23 2021 at 19:49):

Eric Forgy said:

Should cancellation also happen in a rig? :thinking:

Not necessarily. A rig is a monoid object in $(\mathsf{CommMon}, \otimes)$. To tweak that definition to require extra properties just because they hold in rings and you like them would be like drawing a picture of a deer in the background of the Mona Lisa just because you like pictures with deer in them. Or worse, actually.

Reid Barton (Jan 23 2021 at 20:25):

Is "endofunctor with the property that [isomorphisms]" shorthand for endofunctor equipped with isomorphisms satisfying some unstated coherence laws?

fosco (Jan 23 2021 at 21:56):

Reid Barton said:

Is "endofunctor with the property that [isomorphisms]" shorthand for endofunctor equipped with isomorphisms satisfying some unstated coherence laws?

To await a coherence law, is itself a coherence law.

John Baez (Jan 23 2021 at 22:00):

Is that some kinda Zen thing?

Fabrizio Genovese (Jan 23 2021 at 23:54):

It'a also a very popular meme in Italy, coming from a dumb TV ad where they said "To await for pleasure is itself pleasure", or something like that :grinning:

John Baez (Jan 23 2021 at 23:56):

Oh.

fosco (Jan 24 2021 at 10:45):

Well, Campari took the tagline from a minor German philosopher, so I would say it is the exact opposite of a Zen thing.

Even more so because another quote from this guy's work is

It is the mark of great people to treat trifles as trifles and important matters as important.
(Hamburgische Dramaturgie (1767 - 1769), Vierunddreißigstes Stück Den 25. August 1767)

whereas Laozi LXIII says

大小多少，
報怨以德。
圖難于其易，
為大于其細。
天下難事必作于易，
天下大事必作于細，
是以聖人終不為大，
故能成其大。
-
Enlarge the belittled, increase the lessened,
Reward condemnation with Virtue.
Complexity is drawn from simplicity,
Greatness is found in triviality.
Problematic complexities must be resolved in simplicity,
Great accomplishments must be built on trivialities,
Hence the master continues to be unconcerned with great deeds,
Therefore is capable of accomplishing greatness.

fosco (Jan 29 2021 at 11:45):

Someone has a better name than "Leibnizator" for the cell $\mathfrak l_{AB} : \partial (A\otimes B) \Rightarrow \partial A \otimes B + A \otimes \partial B$ here?

fosco (Jan 29 2021 at 11:45):

image.png

fosco (Jan 29 2021 at 11:45):

also, do these coherence conditions remind you of something else?

Jules Hedges (Jan 29 2021 at 11:50):

"Leibnizator" sounds like the right name for the cell version of the Leibniz law, it's not significantly weirder than "pentagonator"

Jules Hedges (Jan 29 2021 at 11:51):

Although, on the basis that to an English speaker, German names make good supervillain names, it makes me think of this.....

Jules Hedges (Jan 29 2021 at 11:52):

"And now, watch as I pull this lever on the Leibnizator, behold as it begins producing my army of Leibniz clones! And you, Goody Man, cannot stop me! Muahahahahaha!" / "Dr. Leibniz, you fiend! You won't get away with this!"

fosco (Jan 29 2021 at 12:05):

https://www.youtube.com/watch?v=Usrk5HR1U4I

Matteo Capucci (he/him) (Jan 29 2021 at 12:40):

+1 for Leibnizator

fosco (Feb 06 2021 at 11:12):

Update: next Tuesday I will give an informal talk to my colleagues here in taltech about this whole story.

Feel free to join! Here the coordinates of the event; the title of the event contains a direct link to the zoom room.

Hope I'll meet you there!

fosco (Feb 20 2021 at 10:40):

So, I am slowly making progress, and I have a new question.

fosco (Feb 20 2021 at 10:51):

Let's state it for rings for the moment; say $R$ is a differential ring and a linear topology that makes sense of infinite products like

$\prod_{i=1}^\infty A_i$

now, is there a way to prove the "infinite Leibniz rule" algebraically, without resorting to any analtyic technique? The rule is

$\displaystyle \partial \Big( \prod_{i=1}^\infty A_i \Big) = \sum_{n=1}^\infty \Big(\partial A_n \times \prod_{i\ne n} A_i\Big)$

and ideally, the proof would go as follows:

image.png

and now "one sees" that the claim holds. Now

1. I feel ashamed because I can't find a more formal argument: can you help me?
2. If you followed the whole story of this thread, you might suspect that I am interested in this result because $\partial$ is a derivation on the 2-rig $\mathcal R$: you're right! Any reason why a similar argument wouldn't hold, in a category, replacing equalities with isomorphisms?

John Baez (Feb 20 2021 at 17:24):

To make the argument more formal, you could say the infinite products and coproducts are "limits" of finite ones (in some sense of "limit", to be determined), and assume that the derivation $\partial$ can be passed through these "limits", and thus reduce the infinite Leibniz rule to the finite one.

fosco (Feb 20 2021 at 18:18):

Thanks, @John Baez !

The proof I know is for sequences of holomorphic functions: if $f_n$ is such a sequence, then

$f' = \sum_{n=1}^\infty f_n' \prod_{k\ne n} f_k$

fosco (Feb 20 2021 at 18:20):

the proof, however, takes the logarithm of the product and uses the logarithmic derivative rule: $d(\log f)=\frac{f'}{f}$

fosco (Feb 20 2021 at 18:21):

I was wondering if there's a proof for differential-topological rings that one can try to adapt: afaicu the fact that a differential $d$ on a ring $R$ lifts to a continuous operator really depends on many petty details.

John Baez (Feb 20 2021 at 23:49):

A lot of analysis is the study of "petty details" like this.

fosco (Feb 21 2021 at 08:51):

John Baez said:

A lot of analysis is the study of "petty details" like this.

"I have always disliked analysis."

P.J. Freyd

Todd Trimble (Feb 21 2021 at 13:19):

Where's that from, Fosco?

(Normally, mathematicians declaring they dislike a certain field doesn't improve my opinion of them.)

Fawzi Hreiki (Feb 21 2021 at 13:20):

His paper 'Algebraic Real Analysis' if I recall correctly.

Fawzi Hreiki (Feb 21 2021 at 13:21):

http://www.tac.mta.ca/tac/volumes/20/10/20-10abs.html

Fawzi Hreiki (Feb 21 2021 at 13:48):

It's worth reading the appendix which is where that quote appears. I don't think it is meant to be derogatory. And anyway, the appendix ends with "... but, at least, now I like analysis."

John Baez (Feb 21 2021 at 17:35):

Good: if you don't like something about a field, the noble path is to fix it.

Todd Trimble (Feb 21 2021 at 17:47):

Well, that's good! And more what I would expect from Freyd: a creative response rather than just "I don't like it". Thanks for adding that.

fosco (Feb 21 2021 at 18:16):

Todd Trimble said:

Well, that's good! And more what I would expect from Freyd: a creative response rather than just "I don't like it". Thanks for adding that.

Why you don't like Freyd? :grinning:

Are you preparing a paper that starts with "I have always disliked P.J. Freyd" and ends with "...but at least, now, I lke Peter Freyd"?

Todd Trimble (Feb 21 2021 at 18:41):

What?

fosco (Feb 21 2021 at 18:52):

I was just kidding: I know Freyd solely from his work, and I consider him one of the most brilliant category theorists ever; instead you seem of a different advice: why?

Nathanael Arkor (Feb 21 2021 at 18:54):

I think you may have misread Todd's message :)

Todd Trimble (Feb 21 2021 at 18:55):

Nope, I agree with you. I asked where he said that, so I could get a better idea of the context.

fosco (Feb 21 2021 at 18:55):

haha! Yes, thanks Nathanael: I misread: I thought you said "more than what I would expect from Freyd"

fosco (Feb 21 2021 at 18:57):

so, my bad! Hope the misunderstanding is solved :smile:

Todd Trimble (Feb 21 2021 at 18:58):

No problem -- yes, it's resolved.

fosco (Feb 28 2021 at 09:30):

I have just one last question left here, and then I can turn my attention into something else ;-) or different.

As we all know, if $R$ is a differential ring and $d : R \to R$ a derivation, there is a nice formula for the derivative of a multiplicative unit $u^{-1}$:

$du^{-1} = -\frac{du}{u^2}$

(if the ring is not commutative, $du = -u^{-1}.du.u^{-1}$). Interesting fact is that if $u$ is a multiplicative unit in a _semi_ ring, the fact that it also has an additive inveres comes for free using the Leibniz rule provided $d1=0$ )

fosco (Feb 28 2021 at 09:37):

What about categories now?!

• Well, having a "multiplicative inverse" means "being a dualisabile object" (on the left or on the right, let's say on the right from now on)
• Whereas, having an additive inverse is not a suitable condition (I am interested in bimonoidal category where the additive structure is coproduct: if $A + B=\varnothing$ in a category, then $A=B=\varnothing$...)

Let's study the first condition and see where it goes: let's say $B\in\mathcal C$ (monoidal) has a right dual $B^\text{r}$, this means that regarding $\mathcal C$ as a bicategory, $B$ has a right adjoint: this in turn means that

$\epsilon : B\otimes B^\text{r} \to I \qquad \eta : I \to B^\text{r} \otimes B$

(I the monoidal unit) satisfy the triangle identities. Let's concentrate on one of them:

$(B^\text{r} \otimes \epsilon)\circ (\eta \otimes B^\text{r}) =1_{B^\text{r}}$

and let's see f there any chance to find an expression for $\partial B^\text{r}$ in terms of $B,\partial B, B^\text{r}$.

fosco (Feb 28 2021 at 09:44):

In particular let's derive the triangle identity: if I'm not wrong one gets

$(\partial B^\text{r} \otimes \epsilon)(\partial\eta\otimes B^\text{r}) + (B^\text{r} \otimes \partial\epsilon)(\eta \otimes \partial B^\text{r})$

from this one gets diagrams

image.png

and I'm interested in the red arrows, that can be composed in a map $\partial I \otimes B^\text{r} \to \partial B^\text{r} \to B^\text{r} \otimes \partial I$. In a proper sense (I am willingly hiding some coherence under the carpet), the object- and morphism-wise sum of these maps is just the derivative of the composition of unitors

$\zeta : I\otimes B^\text{r} \to B^\text{r} \to B^\text{r} \otimes I$

This means that the square

image.png

commutes, if the vertical maps are leibnization isomorphisms, and the red upper row is the sum we were trying to compute.

What now? What do you think?