The mathedu mailing list has a recent longish thread at


<http://www.nabble.com/Why-do-we-do-proofs--to25809591.html>


which discussed among other things whether we should teach triangles as labeled or unlabeled to high school students (this is a vast oversimplification of the thread). I have long been concerned with how we think (informally and formally) about mathematical objects, so naturally I started to consider how we think about triangles.


Consider circles. Most informal and formal descriptions involve an embedding into R^2, but they *can* be characterized as manifolds (even as Riemannian manifolds) of dimension 1 with specific properties, independent of any embedding. This sort of thing has turned out to be a major way to think about all sorts of spaces. So can we describe triangles in a similar way?


Unfortunately, manifolds are far removed from my usual mathematical work (category theory). What I *think* I understand is that there can be *piecewise* linear manifolds, even Riemannian ones. So perhaps we can say a triangle is a piecewise linear manifold of dimension 1 with certain properties. Now, I want to define a triangle so that it comes complete with information about the lengths of its sides and what the three angles are. Riemannian manifolds have a way to specify length and angles, and I can believe you can make the sides have specific lengths. But the angles? It seems to me that the tangent spaces (like those on a circle) result in all angles being 0 or pi, except at the corners where they don't exist. But I may not understand the situation correctly.


So my question is: Is there a known methodology that allows triangles to be characterized independent of embeddings in such a way that incorporates information about side lengths and angles?



Characterizing triangles unembeddedly
I'm not sure what you have in mind for circles--the only invariant of a connected compact Riemannian 1-manifold is its length; they have only extrinsic curvature (defined for a manifold embedded in Euclidean space) not intrinsic curvature. You'd have to consider Riemannian manifolds with some extra structure--perhaps just a function which is supposed to be the curvature of an embedding into R^2, and integrates to 2π. Then you could represent triangles similarly, with a curvature "function" which is the sum of three delta functions, one at each vertex. Of course the side lengths of a triangle determine its angles, so you might produce a "triangle" which cannot be embedded in R^2!


Interestingly for the case of 2-manifolds the situation is nicer. There is a theorem that a Riemannian 2-manifold of positive curvature homeomorphic to the sphere has a unique (up to rigid motion) isometric immersion in R^3 as a convex surface. There is an analogous statement for polyhedra but I am not sure of the precise statement. I think these results are due to Pogorelov and Alexandrov.


