Moulton plane

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The Moulton plane. Lines sloping down and to the right are bent where they cross the y-axis.

The Moulton plane is an example for an affine plane in which Desargues' theorem does not hold. It is named after the American astronomer Forest Ray Moulton . The points of the Moulton plane are simply the points in the real plane $\mathbb R^{2}$ and the lines are the regular lines as well with the exception, that for lines with a negative slope the slope doubles when they pass the Y-axis.

[edit] Formal Definition

The Moulton plane is an incidence structure $\mathfrak M=\langle P, G,\textrm I\rangle$ , where $P$ denotes the set of points, $G$ the set of lines and $I$ the incidence relation „lies on“:

$P:=\mathbb R^{2}$

$G:=(\mathbb R \cup \{\infty\}) \times \mathbb R,$

$\infty$ is just a formal symbol for $\not\in\mathbb R$ . It is used to describe vertical lines, which you may think of as lines with an infinitely large slope.

The incidence relation is defined as follows:
For $p=(x,y)\in P$ and $g=(m,b)\in G$ we have

$p\,\textrm I\,g\iff\begin{cases} x=b&\textrm{falls}\,m=\infty\\ y=mx+b&\textrm{falls}\,m\geq 0\\ y=mx+b&\textrm{falls}\,m\leq 0, x\leq 0\\ y=2mx+b&\textrm{falls}\,m\leq 0, x\geq 0. \end{cases}$

[edit] Application

The Moulton plane shows that affine planes in which Deargues' theorem does not hold do actually exist. As a consequence you have , that the associated projective plane, into which an affine plane can be extended, is non desarguesian as well. Since for $P G (2, F)$ Desargues' theorem actually does hold, this means, that there exist projective planes which are non isomorphic to $P G (2, F)$ . In other words not all projective planes can be described via the canonical construction ( $P (V)$ ) over a 2 dimensional vector space V.

[edit] References

Moulton,F.R.: A simple non-desarguesian plane geometry: Trans. Amer. Math Soc. 3 (1902),192-195
Beutelspacher, A/ Rosenbaum,U. : Projektive Geometrie: Vieweg (1992)