mg.metric geometry - Ordered geometries from convex subsets of the plane

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 $phi:C_1to C_2$ 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 $xnotin C_1$. 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 $y,znotin C_1$.

The case $xin C_1$ 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, $xnotin C_1$, 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.