Ultraparallel theorem

In hyperbolic geometry, the ultraparallel theorem states that every pair of ultraparallel lines in the hyperbolic plane has a unique common perpendicular hyperbolic line.

Construction of ultraparallel lines in the Klein model

In the Klein model of the hyperbolic plane, two ultraparallel lines correspond to two non-intersecting chords. The poles of these two lines are the respective intersections of the tangent lines to the unit circle at the endpoints of the chords. Lines perpendicular to line A are modelled by chords such that when extended, the extension passes through the pole of A, and vice-versa. Hence we draw the unique line between the poles of the two given line, and intersect it with the unit disk; the chord of intersection will be the desired ultraparallel line. If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter is perpendicular as well in the hyperbolic space, and so we draw a line through the polar of the other line intersecting the diameter at right angles to get the ultraparallel line.

Proof in the Poincaré half-plane model

Let

$a<b<c<d$

be four distinct points on the abscissa of the Cartesian plane. Let $p$ and $q$ be semicircles above the abscissa with diameters $ab$ and $cd$ respectively. Then in the upper half-plane model HP, $p$ and $q$ represent ultraparallel lines.

Compose the following two hyperbolic motions:

$x\to x-a\,$

${\mbox{inversion in the unit semicircle}}\,$.

Then $a\to \infty$, $b\to (b-a)^{-1}$, $c\to (c-a)^{-1}$, $d\to (d-a)^{-1}$.

Now continue with these two hyperbolic motions:

$x\to x-(b-a)^{-1}\,$

$x\to \left[(c-a)^{-1}-(b-a)^{-1}\right]^{-1}x\,$

Then $a$ stays at $\infty$, $b\to 0$, $c\to 1$, $d\to z$ (say). The unique semicircle, with center at the origin, perpendicular to the one on $1z$ must have a radius tangent to the radius of the other. The right triangle formed by the abscissa and the perpendicular radii has hypotenuse of length ${\begin{matrix}{\frac {1}{2}}\end{matrix}}(z+1)$. Since ${\begin{matrix}{\frac {1}{2}}\end{matrix}}(z-1)$ is the radius of the semicircle on $1z$, the common perpendicular sought has radius-square

${\frac {1}{4}}\left[(z+1)^{2}-(z-1)^{2}\right]=z\,$.

The four hyperbolic motions that produced $z$ above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius ${\sqrt {z}}$ to yield the unique hyperbolic line perpendicular to both ultraparallels $p$ and $q$.