Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've read in several places now the claim that there are only three geometries in dimension 2, up to scaling: Euclidean, spherical, and hyperbolic. This fits nicely with the three cases for the sign of the curvature (zero, positive, negative), but how would you prove this claim? Indeed, I'm not sure exactly what would need to be proved -- what does it really mean? What do we require of geometries in order to consider them "the same"? Sorry if this question is a little nebulous.

share|improve this question
    
Those three are by far the most common, obtained by varying the parallel postulate. But to claim that those are the only three, you need to specify what you mean by "geometry" and "2 dimensions". If you mean "A definition of lines in the plane which statisfy Hilbert's incidence, betweenness, conruence and continuity axioms, as well as some form of the parallel postulate", then as far as I know there might be only three non-equivalent models. –  Arthur Oct 29 '13 at 9:59
add comment

1 Answer

Yes, at least for compact surfaces, there are only three model geometries upto scaling. You are on the right track when you note that this fits nicely with the notion of curvature.

In short, suppose we are looking at compact $2$--dimensional surfaces (to put it more technically, I am considering the class of compact $2$--dimensional manifolds). Now there is a beautiful theorem, called the Gauss Bonnet theorem, which relates the curvature with an invariant of the surface called the Euler characteristic.

If $S$ is a compact surface, and $\chi(S)$ is the Euler characteristic, $K$ is the curvature

$K = 2\pi\chi(S)$ (Eq1)

On the LHS of (Eq1), we have the curvature, and on the RHS of (Eq1) we have the Euler Characteristic. Now the Euler characteristic is simply an invariant of the surface -- its simple to calculate, pick any triangulation, add the vertices and faces and subtract the edges!

Anyway, if $K>0$ this means that the Euler characteristic is positive. The only surfaces with positive Euler characteristic are the spheres. The same is true for the cases of Euclidean and Hyperbolic curvatures.

I have simplified things a bit, and its possible to drop the constant curvature constraint, and even the topological constraints on the compactness. But hope this gives you a place to start,

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.