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Stream: learning: questions

Topic: groupoids of affine planes

Hugo Jenkins (Aug 09 2022 at 02:56):

I’ve been thinking about equivalence of categories and am unsure about the following. Exercise 1.5.viii in Riehl’s Category Theory in Context asks to show an equivalence between the groupoids (affine planes) and (projective planes with distinguished line), both having as maps the structural isomorphisms. It doesn’t seem to be defined in the text but I take these to be minimally axiomatized, with just

A1. $\exists !$ line through distinct points
A2. $\exists !$ line through given $p$ and parallel to given $m$

P1. A1
P2. $\exists !$ point in distinct lines

The (given) construction for the equivalence $\textbf{Aff}\to \textbf{Proj}^{|}$ is to delete the distinguished line and all points on it, and restrict the incidence relation. Somewhat intricate logic then seems needed to check A1 and A2. Functoriality by restriction is obvious.

Checking essential surjectivity requires you to know an inverse construction to this on objects: to the affine plane $A$ add a new point for every pencil of parallels, and one new line containing exactly those new points. We then need to check P1 and P2. Then for fullness, we need to know how to extend an affine map between the outputs of deletion, to the points which were thrown out: the map must take pencils to pencils, and pencils of the deletion are in 1-1 correspondence with the deleted points. Faithfulness is trivial.

My question is, is the point of the exercise to do all these (to me anyway) nontrivial geometric things? Am I missing something? It doesn't seem possible that abstract (but still elementary) category theory could avoid this.

Simon Burton (Aug 09 2022 at 03:07):

Surely you can just take whatever definition of Aff and Proj is most convenient, and just use that?

Simon Burton (Aug 09 2022 at 03:10):

What is p and m here: "A2. ∃!\exists !∃! line through given p and parallel to given m" ?

Hugo Jenkins (Aug 09 2022 at 20:56):

Simon Burton said:

What is p and m here: "A2. ∃!\exists !∃! line through given p and parallel to given m" ?

$p$ is a point and $m$ is a line. So the structure consists of sets $P$ and $L$ called the point set and line set, plus a relation $\epsilon \subset P \times L$. The axiom is that for all $p\in P$ and $m\in L$, there's a unique $l\in L$ such that $p \epsilon l \text{ and } l \parallel m$. Parallel is a defined relation on lines, $l || m$ if either they are equal, or share no common point.

Hugo Jenkins (Aug 09 2022 at 21:05):

Perhaps one shouldn't ask for a solution which doesn't ultimately rest on these technicalities/checks. But what I'm wondering is: is there some way people carry around Proj and Aff as groupoids in their heads, which makes this equivalence clear/foundational? The reference to the Erlangen program in the question also makes it sound like the groupoids themselves are the objects of study in some sense.

John Baez (Aug 09 2022 at 22:15):

My question is, is the point of the exercise to do all these (to me anyway) nontrivial geometric things? Am I missing something? It doesn't seem possible that abstract (but still elementary) category theory could avoid this.

It sounds like this is an exercise to help you understand the concept of "equivalence of categories" by looking at a concrete example. Two popular ways to show that $F: \mathsf{C} \to \mathsf{D}$ is an equivalence of categories are

1) to construct a functor $G : \mathsf{D} \to \mathsf{C}$ and show that $FG$ and $GF$ are naturally isomorphic to identity functors,

2) to show $F$ is essentially surjective, full and faithful.

John Baez (Aug 09 2022 at 22:19):

So, I always start by seeing which looks easier. I feel 1) is "better" if it's not too hard, because it's explicitly giving us a way to go back, while 2) just says there exists a way to go back: getting it may require the axiom of choice. (If you want to avoid the axiom of choice there's vastly more to be said about this, but I don't want to get into that here.)

John Baez (Aug 09 2022 at 22:21):

You decided to use method 2): was this because you thought method 1) would be harder?

Hugo Jenkins (Aug 10 2022 at 21:00):

John Baez said:

You decided to use method 2): was this because you thought method 1) would be harder?

Yes. But I see that of course it required the same ingredients: thinking of what would have been the inverse on objects in order to check essential surjectivity, and on maps to check fullness. The only things abstracted away by using method 2) were having to think about the natural transformations and functoriality of the inverse.

Simon Burton (Aug 11 2022 at 01:23):

I find the groupoid language confusing, especially here where it looks like these are just groups, ie. there is only one affine plane and we are considering its transformation group. So this to me looks like an exercise in group theory, and i have a guess about what the answer is.. But the groupoid language is supposed to apply more generally, when there is some kind of theory expressed in first order logic, and we get a groupoid of automorphisms of models of the theory. So statements like "consider the groupoid of finite dimensional vector spaces (over a field k) & invertible linear maps" makes more sense to me as a groupoid. And I think there is a version of Riehl's exercise in a more general situation like this.

Simon Burton (Aug 11 2022 at 01:54):

I'm slowly catching up here. It seems like you really are thinking of many different affine (&projective) planes, including finite planes, etc etc. I was making things easier for myself by just thinking of affine & projective spaces over a fixed field, like the real numbers.

Jan Pax (Aug 11 2022 at 08:35):

Hi John, could you please have a look at this: https://categorytheory.zulipchat.com/#narrow/stream/229199-learning.3A-questions/topic/n.2Et.2E.20commutativity.20square

Hugo Jenkins (Aug 11 2022 at 23:47):

Simon Burton said:

I find the groupoid language confusing, especially here where it looks like these are just groups, ie. there is only one affine plane and we are considering its transformation group. So this to me looks like an exercise in group theory, and i have a guess about what the answer is.. But the groupoid language is supposed to apply more generally, when there is some kind of theory expressed in first order logic, and we get a groupoid of automorphisms of models of the theory. So statements like "consider the groupoid of finite dimensional vector spaces (over a field k) & invertible linear maps" makes more sense to me as a groupoid. And I think there is a version of Riehl's exercise in a more general situation like this.

Ah yeah, I am thinking of the full collections of models of the theory. So they are very disconnected groupoids.

Hugo Jenkins (Aug 11 2022 at 23:51):

In general, though, it seems the idea of "studying groupoids of geometric spaces of various kinds" might have several meanings/levels, and I am curious which is being referred to.

-We could study the individual groupoids as abstract categories up to equivalence (or even isomorphism). In that case it's not intuitive to me that there should be much to say.
-We could study (still abstractly) the Cat-environment that these all create together--the functors and natural transformations which exist between them.
-We could study the situation when one has defined various groupoids as concrete categories of spaces; and ask what categorical properties hold of the various natural and concretely defined objects/constructions. (E.g. every affine plane arises from deleting the distinguished line from some projective plane--essential surjectivity of the functor in the abstract language).

What does the Erlangen program actually do? Moreover, is there any sense in which the first two (abstract) approaches are really geometry? Is it ever possible for them to "reconstruct" or "recognize" that their categories/functors/nats arise from such-and-such concrete construction, and essentially no other, up to some notion of essential sameness on the concrete side?

Simon Burton (Aug 12 2022 at 04:53):

Well, firstly, the Erlangen program deals with the case where the groupoid is a group. Have you thought about the Riehl exercise in this context?

John Baez (Aug 12 2022 at 18:31):

I think it can be useful to work with a groupoid of affine planes over a field $k$ even if they're all isomorphic, since choosing everybody's favorite one, namely $k^2$, automatically gives us extra structure, like an origin $(0,0)$, which we must then avoid using if we are trying to do constructions that use solely with the affine structure.

I.e., working with all affine planes is a way of "anonymizing" them, thus "blinding" ourselves in a very useful way.

John Baez (Aug 12 2022 at 18:32):

Of course we could also blind ourselves by saying "let $X$ be an affine plane", so sometimes working with the groupoid of all of them is overkill.

John Baez (Aug 12 2022 at 18:33):

But it can be useful, since a functor out of this groupoid is a systematic way of turning any affine plane into something in some other category.

John Baez (Aug 12 2022 at 18:34):

Let $A$ be the groupoid of affine planes over a field $k$ and let $P$ be the groupoid of projective planes over $k$ equipped with a distinguished line.

There's a functor from $A$ to $P$ given by "adding a projective line at infinity".

This functor takes a bit of work! I think I'd do it by taking any such affine plane $X$ and adding one point $p$ for each maximal collection of parallel lines in $X$, then declaring all lines in this collection intersect at $p$.

Check that you get a projective plane. Let the distinguished line the set of all points you've added - check that this is a projective line. I am too lazy to check such things unless I'm being paid at my usual hourly rate.

John Baez (Aug 12 2022 at 18:36):

And there's a more obvious functor from $P$ back to $A$ given by "removing the distinguished line".

John Baez (Aug 12 2022 at 18:43):

Once you have both functors set up, I think it's not hard to check that they form an equivalence.

If you start with an affine plane, add a projective line at infinity, then remove that line, you get back exactly where you started!

John Baez (Aug 12 2022 at 18:45):

If you take a projective plane with a distinguished line, remove that line, then add a new projective line at infinity, the projective plane you get is not equal to the original one, so you should check that it's naturally isomorphic.

John Baez (Aug 12 2022 at 18:46):

You can also try to do all this stuff for all affine planes, defined axiomatically somehow, and all projective planes with a distinguished line.

Simon Burton (Aug 13 2022 at 09:09):

John Baez said:

I think it can be useful to work with a groupoid of affine planes over a field $k$ even if they're all isomorphic, since choosing everybody's favorite one, namely $k^2$, automatically gives us extra structure, like an origin $(0,0)$, which we must then avoid using if we are trying to do constructions that use solely with the affine structure.

I.e., working with all affine planes is a way of "anonymizing" them, thus "blinding" ourselves in a very useful way.

Well, my favourite affine plane involves $k^3$, not $k^2$, especially because we are also talking about projective geometry. But this brings up another point. Klein's brilliant idea with the Erlangen program was that all these different ways of assembling geometries, via the syntax, is not the thing, the automorphism group is the thing. I think Klein called it the principle group. And then geometry is studied via group theory. Category theory is some big generalization of this idea, but that's where it starts.

I'm hoping that Hugo is still interested... & I still need to think more about this approach via predicates & first order logic.

Hugo Jenkins (Aug 13 2022 at 23:12):

John Baez said:

Let $A$ be the groupoid of affine planes over a field $k$ and let $P$ be the groupoid of projective planes over $k$ equipped with a distinguished line.

There's a functor from $A$ to $P$ given by "adding a projective line at infinity".

This functor takes a bit of work! I think I'd do it by taking any such affine plane $X$ and adding one point $p$ for each maximal collection of parallel lines in $X$, then declaring all lines in this collection intersect at $p$.

Check that you get a projective plane. Let the distinguished line the set of all points you've added - check that this is a projective line. I am too lazy to check such things unless I'm being paid at my usual hourly rate.

Right. I did just that for the exercise. But there’s no immediate abstract reason the categories should be equivalent just from the definitions, right? At least not without going into categorical logic.

Hugo Jenkins (Aug 13 2022 at 23:15):

John Baez said:

You can also try to do all this stuff for all affine planes, defined axiomatically somehow, and all projective planes with a distinguished line.

How generally do you think it’s true that an equivalence of categories expresses some sort of logical equivalence between the data/content defining the objects on either side? This seems to be a pattern. E.g. after doing this exercise, wouldn’t one feel justified in saying something like “an affine plane is the same as a projective plane with distinguished line”? The notions are the same, and the equivalence is our formalization of that.

Hugo Jenkins (Aug 13 2022 at 23:29):

Simon Burton said:

John Baez said:

I think it can be useful to work with a groupoid of affine planes over a field $k$ even if they're all isomorphic, since choosing everybody's favorite one, namely $k^2$, automatically gives us extra structure, like an origin $(0,0)$, which we must then avoid using if we are trying to do constructions that use solely with the affine structure.

I.e., working with all affine planes is a way of "anonymizing" them, thus "blinding" ourselves in a very useful way.

Well, my favourite affine plane involves $k^3$, not $k^2$, especially because we are also talking about projective geometry. But this brings up another point. Klein's brilliant idea with the Erlangen program was that all these different ways of assembling geometries, via the syntax, is not the thing, the automorphism group is the thing. I think Klein called it the principle group. And then geometry is studied via group theory. Category theory is some big generalization of this idea, but that's where it starts.

I'm hoping that Hugo is still interested... & I still need to think more about this approach via predicates & first order logic.

Ah, I see. So the exercise definitely shows in the single object case that the automorphism groups are isomorphic. And that is what we should look for as the sort of "semantic classifier"...

Can perhaps groups or more general categories actually do this for general first-order theories?

David Michael Roberts (Aug 14 2022 at 11:04):

@Simon Burton since there are non-Desarguesian planes, proving it just for field planes (affine or projective) is not going to prove the whole thing (similarly with talking about Erlangen stuff: the automorphism group doesn't necessarily act transitively on a non-Des. plane). It is an important special case, sure, but the proof doesn't reduce to that.

Simon Burton (Aug 14 2022 at 11:38):

@David Michael Roberts No where am I claiming that the proof reduces to group theory!

Simon Burton (Aug 14 2022 at 11:39):

Hugo specifically asks about the Erlangen program, and I am interested in this story as well, so that's why I'm talking about groups.

David Michael Roberts (Aug 14 2022 at 14:02):

@Simon Burton
That's fine, I think I might have misread your intent. In any case, it's worth forestalling any other readers from leaping to conclusions as I (perhaps falsely) imagine them doing.

John Baez (Aug 14 2022 at 17:55):

Hugo Jenkins said:

John Baez said:

Let $A$ be the groupoid of affine planes over a field $k$ and let $P$ be the groupoid of projective planes over $k$ equipped with a distinguished line.

There's a functor from $A$ to $P$ given by "adding a projective line at infinity".

This functor takes a bit of work! I think I'd do it by taking any such affine plane $X$ and adding one point $p$ for each maximal collection of parallel lines in $X$, then declaring all lines in this collection intersect at $p$.

Check that you get a projective plane. Let the distinguished line the set of all points you've added - check that this is a projective line. I am too lazy to check such things unless I'm being paid at my usual hourly rate.

Right. I did just that for the exercise. But there’s no immediate abstract reason the categories should be equivalent just from the definitions, right? At least not without going into categorical logic.

I think you're right. I think this was intended to be an exercise in 'thinking about geometry using the language of categories', not an exercise in 'pure category theory'.

John Baez (Aug 14 2022 at 17:59):

And I think this is very useful! When I think "there's a systematic way I can turn an affine plane into a projective plane with a distinguished line", I want to immediately guess that this is a functor from one category to another. And when I think "and there's a way to go back", I want to immediately guess that this functor is an equivalence.

John Baez (Aug 14 2022 at 18:01):

The reason it's useful is that a phrase like "there's a systematic way" or "there's a way to go back" is a bit informal - it's not exactly clear what it means. And category theory lets us make these ideas precise, so we can prove theorems using them.

For example, when we go back we may only get back our original object up to isomorphism, but ideally (it turns out) up to a natural isomorphism, etc.

Hugo Jenkins (Aug 16 2022 at 00:46):

David Michael Roberts said:

Simon Burton
That's fine, I think I might have misread your intent. In any case, it's worth forestalling any other readers from leaping to conclusions as I (perhaps falsely) imagine them doing.

Ah, good! I am forestalled. But if we throw out some cases, does this become true? For example, if $k$ and $k^\prime$ are nonisomorphic fields, is it possible that $\text{Aut}(P^2(k)) \cong \text{Aut}(P^2(k^\prime))$?

Or even that the projective planes $P^2(k)$ and $P^2(k^\prime)$ can be abstractly isomorphic? Pretty sure not, but I'm not positive. Right at the end of Chapter 2 of Artin’s Geometric Algebra, he outlines how one would coordinatize an abstract projective plane, and notes that obtaining a field (if you do it by deleting a line and then using the affine construction given in the text) requires the choice of a line to delete; and “a canonical field associated with the plane would have to be explained as an equivalence class of fields”. Surely he can’t mean that the fields can actually be nonisomorphic, though.

I've been meaning to work through this chapter for a long time. Probably the following two would explain exactly the group-theoretic approach Simon is alluding to.

John Baez (Aug 16 2022 at 05:15):

Hugo Jenkins said:

For example, if $k$ and $k^\prime$ are nonisomorphic fields, is it possible that $\text{Aut}(P^2(k)) \cong \text{Aut}(P^2(k^\prime))$?

I doubt it, but it's a good question. The automorphism group of a projective plane is usually called its collineation group, so that's a good buzzword to find information about it.

For example the linked Wikipedia article says the collineation group of $P^2(k)$ is called a projective semilinear group of $k^2$, and it's the semidirect product of the more familiar projective linear group $PGL(2,k)$ and the automorphism group of the field $k$.

I feel people must have studied whether you can reconstruct $k$ from the collineation group of $P^2(k)$, perhaps with the help of this stuff.

John Baez (Aug 16 2022 at 05:20):

Hugo Jenkins said:

Or even that the projective planes $P^2(k)$ and $P^2(k^\prime)$ can be abstractly isomorphic? Pretty sure not [....]

Yes, there's a known way reconstruct the field $k$ from the projective plane $P^2(k)$, up to natural isomorphism, so an isomorphism of $P^2(k)$ and $P^2(k^\prime)$ gives an isomorphism of fields $k \cong k'$.

John Baez (Aug 16 2022 at 05:23):

I think Hartshorne's book Foundations of Projective Geometry explains how you can reconstruct a field $k$ from the projective plane $P^2(k)$, and also how to take an axiomatically defined projective plane and tell if it's of the form $P^2(k)$. There are also other books that cover this material.

John Baez (Aug 16 2022 at 05:24):

Projective planes that come from fields in this way are called Pappian, because they obey Pappus' hexagon theorem.

John Baez (Aug 16 2022 at 05:26):

It's quite wonderful how this beautiful ancient theorem turns out to characterize which projective planes come from fields!