Where does the generic triangle live?

This is a reformulation of my question Characterizing triangles unembeddedly.

Motivation 1: There is such a thing as a generic group. In category theory this is done by constructing "theory" of the group, which is a category in a certain doctrine. Functors (in that doctrine) to Set, or more generally to any topos, are groups. The barest such theory (as usually seen) is the Lawverean algebraic theory of groups. This theory is a category containing an object and operations making it a group object in that category, and the theory is the smallest such category that contains all finite limits. There are fancier ones; the fanciest is the classifying topos for groups, which is in some sense the initial topos-with-group object. Since in a topos, you have full-scale first order intuitionistic logic, the classifying topos for groups allows you to reason about the generic group inside the classifying topos and the theorems you prove will be true for all groups. (This is only an approximation of the actual situation.) In particular you can't prove it is abelian and you can't prove it isn't; the logic clearly does not have excluded middle.

Motivation 2: You can prove that a triangle that has two angles that are equal must be isosceles (has two sides that are equal). You can do this with Pappus' proof: Look at the triangle, flip it over the perpendicular from the odd angle to the other side, look at it again, and the side-angle-side theorem shows you that the "two" triangles are congruent, so two sides much be equal. This appears to me to be true without requiring the parallel postulate. So the theorem and the proof must be true not only in Euclidean 2-space but in any surface of constant curvature. (Here I am getting into territory I know very little about, so this particular motivation may be totally misguided.)

So what I want is a classifying space of some sort that contains the generic triangle in such a way that maps of the correct sort to any surface of constant curvature are triangles, and so that Pappus' proof can be carried out in the classifying space. The space doesn't have to be a topos or a category at all. I have no clue as to what sort of structure it would be.

Note 1: Even the Lawvere theory of groups has its own internal logic -- in this case equational logic. You certainly cannot prove the generic group is or is not abelian with equational logic.

Note 2: It does not seem reasonable to me that Pappus' proof would work in a surface with variable curvature. But maybe there is some trick to define "angle mod curvature" that would make it true.

Note 3 added 3 Dec 2009: One way of reformulating my question is: How do you give a suitably general definition of "triangle that allows Pappus' proof". Commenters who asked "which definition of triangle are you using" missed the point: I am asking for a definition. Mathematical research commonly consists of trying to find the right definition to make your intuitive proof work. Questions like that belong in MathOverflow and should not be criticized for not being "well formulated". (Of course many questions of this sort could have been solved by looking in Wikipedia or thinking for five minutes, and they deserve to be criticized.)

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I don't know that this is a well-defined question. From an Erlangen perspective, "triangle" means different things depending on whether you only care about properties up to isometry or affine transformations, for example. – Qiaochu Yuan Nov 18 '09 at 22:38
I want to know about any sort of work that might be considered an answer or might lead to an answer to this question. The specific work will no doubt have a specific definition of triangle. – SixWingedSeraph Nov 19 '09 at 1:58
It seems to me that by asking for a triangle for which Pappus proof holds, you are asking for an embedded triangle. (I read your other post.) Leinster's answer below gives a generic triangle, generic to one living in an affine space of 3 dimensions. <br> The category of all groups contains all groups. There are no others. Yet other than triangles in constant curvature spaces, there are going to be other triangles in non-constant curvature spaces. I think this post doesn't fully convey what your original question in the other post wants. – user2529 May 16 '10 at 10:14
Let me ask you this question: Where does the generic addition live? Would you say that its classifying space is some subset of ${\mathbb{N}}^{{\mathbb{N}}\times {\mathbb{N}}}$? – user2529 May 16 '10 at 10:17

3 Answers

There's some confusion about the question, so let me try answering a similar question that I think is strictly analogous to yours but easier. If you can confirm that the question I'm addressing really is analogous to yours, and that the answer is the kind of answer you're looking for, then maybe that will help prepare the ground for someone to answer your actual question.

So, let's pretend that you were interested in a much more rigid kind of geometry, so that a triangle consists of three joined-up straight line segments. There will be no curvature involved.

Formally, we might work with affine spaces, and define a triangle in an affine space $A$ to be an ordered triple of points in $A$. Of course, we imagine the three points joined up, but that won't be part of the formalism.

A generic triangle is an affine space $P$ equipped with a triangle $(p_1, p_2, p_3)$, with the property that for any affine space $A$ and triangle $(a_1, a_2, a_3)$ in $A$, there is a unique affine map $f: P \to A$ such that $$(f(p_1), f(p_2), f(p_3)) = (a_1, a_2, a_3).$$ The affine space $P$ is what you call the classifying space in your question.

It's easy to see that there does exist a generic triangle (and by the universal property, it's unique up to isomorphism). Indeed, we can take $P$ to be any two-dimensional affine space and $p_1, p_2, p_3$ to be any three affinely independent points of $P$.

Of course you know all this, Six Winged Seraph. But the point of saying it is to ask: is this the type of answer you want, in the more advanced setting that you describe?

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That is indeed analogous to my question. Thanks for the contribution. – SixWingedSeraph Dec 5 '09 at 0:33

The generic triangle lives in a subset of this function space $Axioms^{Languages}$.

What do I mean? First, let's us what is generic addition? Immediately, you would ask me, addition on which number system? The natural numbers, integers or what? Properly then, one can only say that for each (semi)ring $R$, addition is specified as some subset of $R^{R\times R}$.

Now I invoke this duality principle: Operations in number theory are analogous to objects in geometry. Objects in number theory are analogous to operations in geometry.

One of the reasons why this question by Six Winged Seraph is so difficult (and so interesting) is that we should think of the object "triangle" in geometry as analogous to the operation "addition" in number theory. We should not think of the "triangle" as analogous to the number "$n$". It is easy to say where the generic number "$n$" lives in: namely the set of natural numbers ${\mathbb{N}}$! (or whatever number system you might be considering) This is an easy question just as where does the generic rotations, dilations and reflections live in: namely the Lie group $Isom({\mathbb{R}}^2)$? (or ${\mathbb{H}}^2$ or ${\mathbb{S}}^2$ depending on which constant curvature space you are considering.)

The "triangle" question is hard because what you can pin down about a generic triangle is not what it is, but what you can do with it. That is, axioms. Using the duality principle again, we pin down what addition is by the axioms, not just axioms for addition, but also how multiplication interacts with it. Are you interested in statements like $23+34=57$, $n+n=2n$ etc? Only to the extent that these statements form part of the completed infinity of theorems from which can be generated by the axioms.

In a similar way, for each constant curvature plane, axioms that govern reflection, axioms that provide congruence criteria, Pappus' proof is generated from these axioms.

As a final refinement, one might also consider triangles in other non-constant curvature spaces. Just as addition can be considered for vector spaces or groups, not just rings. In this case, the full answer is that for any language in which triangles can be talked about, there is a corresponding axiomatization of what can be done with triangles.

Eg, the Pappus proof is for the first-order axiomatization of (non)-Euclidean geometry. There are different proofs for the plane as a Riemannian manifold (via integration) etc. For each of these axiomatizations, we have a different notion of triangle.

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As it happens, the klein hyperbolic models, the usual euclidean plane, and central projection for projective space all have geodesics that look like straight lines; on the other hand, these maps can't be conformal, so it's tricky talking about how they treat angles.

More generally, you can specify a map between two (small enough) geodesic triangles by requiring that it preserve two transverse geodesic partitions; but then what it does to any other geodesics is anyone's guess.

I rather suspect this isn't addressing your actual question, though.

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Alternatively; (spurred on by Tom's answer below), by the generic $T$ triangle let us mean the complete convex space generated by three points $A,B,C$, and satisfying Pappus' theorem in the most suitable sense. Then for any $\alpha,\beta,\gamma$ not colinear in any projective, affine, or hyperbolic space $X$, there exists a unique convex map from $T$ to $X$ mapping $A$ to $\alpha$, $b$ to $\beta$ and $c$ to $\gamma$. – some guy on the street Dec 3 '09 at 23:07
eep! not unique... many perspective shifts possible. – some guy on the street Dec 4 '09 at 1:54
Uniqueness is for pointed triangles; draw a bunch of pictures that look like perspective sketches of an infinite chequerboard; the Pappus' Theorem axiom gives that the proper diagonals are indeed lines; and then we can do a bisection search, finishing by continuity/completeness. – some guy on the street Dec 4 '09 at 2:22