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

${\displaystyle a<b<c<d}$ 

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

Compose the following two hyperbolic motions:

${\displaystyle x\to x-a\,}$ 

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

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

Now continue with these two hyperbolic motions:

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

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

Then ${\displaystyle a}$  stays at ${\displaystyle \infty }$ , ${\displaystyle b\to 0}$ , ${\displaystyle c\to 1}$ , ${\displaystyle d\to z}$  (say). The unique semicircle, with center at the origin, perpendicular to the one on ${\displaystyle 1z}$  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 ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(z+1)}$ . Since ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(z-1)}$  is the radius of the semicircle on ${\displaystyle 1z}$ , the common perpendicular sought has radius-square

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

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