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Motivation

In the Klein disk model of the hyperbolic plane, the points are the interior of the disk, and the lines in $H^2$ correspond to lines intersecting the interior.

a tiling in the Klein disk model

Similarly, the Euclidean plane can be modeled by the interior of a hemisphere of $S^2$ (or $\mathbb {RP}^2$ minus a line) so that lines in $\mathbb R^2$ are the intersections of geodesics of the sphere with the hemisphere.

In both cases, the angles aren't preserved, but the orderings of points on lines are preserved.

Definitions

For any open convex set in $\mathbb R^2$, consider the nonempty intersections of lines with the set as lines in an ordered geometry with the induced ordering from $\mathbb R^2$. Two such geometries are equivalent if there is a bijection between them preserving lines and the ordering on each line.

  1. Which convex open sets produce geometries equivalent to $H^2$?

  2. Which pairs of convex open sets produce equivalent geometries?

Either line-segment preserving maps are quite flexible, or else there should be ways to recover much of the information about convex sets from their incidence geometries.

Some weak results

You can distinguish the interior of a triangle from $H^2$ (or any other bounded convex open set) through the incidence relations. In the triangle, there are three lines such that every other line intersects at least one of the three. Any three lines through the vertices work. In $H^2$, you can always find a line disjoint from any finite collection of lines.

Similarly, if a line segment makes up part of the boundary of a set in the plane, then the incidence geometry is not $H^2$.

The incidence relation plus ordering is enough to construct ideal points of the boundary of the set. These correspond to maximal sets of rays so that for any two disjoint rays $R_1$ and $R_2$ in the set, the set of points $p$ so that for some $x_1 \in R_1$ and $x_2 \in R_2$, $p$ is between $x_1$ and $x_2$, is a triangle subgeometry.

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Surely if one convex set is a projective transformation of another, then they define equivalent geometries. In particular, the convex side of a parabola gives $H^2$. –  Konrad Swanepoel Feb 3 '10 at 21:41
    
Yes, that's correct, and a triangle is equivalent to a half-strip. –  Douglas Zare Feb 3 '10 at 22:01
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1 Answer

up vote 5 down vote accepted

Convex sets give the same geometries if and only if they are projectively equivalent. In particular, it is only conics that give $H^2$.

It is more natural to work in the projective plane $P^2(\mathbb{R})$. Then we define a set to be convex if its intersection with any line is empty or connected. We are given two open convex sets $C_1$ and $C_2$ in the plane with a bijection $\phi:C_1\to C_2$ which is order preserving in the sense that $a$ and $b$ separates $c$ and $d$ (on some projective line) if and only if $\phi(a)$ and $\phi(b)$ separates $\phi(c)$ and $\phi(d)$ (on some projective line).

We would like to prove that $\phi$ is a projective transformation.

The following attempt uses the theorem of Desargues together with the fundamental theorem of projective geometry.

Claim. $\phi$ can be extended to the whole of $P^2(\mathbb{R})$ such that lines are mapped to lines.

Consider any point $x\notin C_1$. We locate $\phi(x)$ by using the Theorem of Desargues as follows. Choose three lines through $x$ that intersect $C_1$ in three connected sets $c_1$, $c_2$, $c_3$. Choose points $a_i$ and $b_i$ on chord $c_i$. Then the triangles $\triangle a_1 a_2 a_3$ and $\triangle b_1 b_2 b_3$ are in perspective, so by the theorem of Desargues, the intersection points $p_{ij}$ of the lines $a_i a_j$ and $b_i b_j$ are collinear. This can be done in such a way that the $p_{ij}$ are all in $C_1$. For instance, the $c_i$ has to be chosen sufficiently close together, and each triangle has to be chosen so that its points are "almost collinear", with the two lines of collinearity intersecting inside $C_1$.

Then this picture of the triangles in perspective, without the point $x$, can be transferred to $C_2$ using $\phi$. All incidences are preserved, so by the converse of the theorem of Desargues, the three connected sets $\phi(c_i)$ lie on concurrent lines. Define $\phi(x)$ to be the point of concurrency. It is easy to see that the definition is independent of which chords through $x$ are used.

We are halway there. It remains to prove that the extended $\phi$ preserves collinearity.

So let $x,y,z$ be collinear in $\mathbb{R}^2$. We would like to show that $\phi(x), \phi(y), \phi(z)$ are collinear.

If at least two of $x,y,z$ are in $C_1$, it is clear that their images will also be collinear. So assume without loss of generality that $y,z\notin C_1$.

The case $x\in C_1$ is simple: the chord of $C_1$ through $x,y,z$ maps to the chord of $C_2$ through $\phi(x)$ and $\phi(y)$, and also to the chord through $\phi(x)$ and $\phi(z)$. Thus $\phi(x),\phi(y),\phi(z)$ are collinear.

If on the other hand, $x\notin C_1$, we again use the theorem of Desargues. Find triangles inside $C_1$ such that the points in which their corresponding sides intersect, all lie on the line through $x,y,z$. (As before, it is easy to see that this is possible.) By the converse of Desargues, the triangles are in perspective. Transfer this picture with $\phi$ to the plane in which $C_2$ lives. We again get two triangles in perspective, and by Desargues, $\phi(x)$, $\phi(y)$, $\phi(z)$ are collinear.

We have shown that lines are mapped onto lines. By (a very special case of) the fundamental theorem of projective geometry, $\phi$ is a projective transformation.

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Nice proof. Thanks! –  Douglas Zare Feb 6 '10 at 2:57
    
I tried to find a reference, but failed. My guess is that this should be about 100 years old. Hilbert introduced his metric on these geometries in 1895. –  Konrad Swanepoel Feb 6 '10 at 8:23
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