Is there a ruler and compass construction of the common perpendicular of two geodesics in H^3?

Question

Assume we have two geodesics in the Poincaré ball model of $\mathbb{H}^3$, viewed as arcs intersecting the boundary of and contained in the Euclidean unit sphere in $\mathbb{R}^3$. Is there a ruler and compass construction in $\mathbb{R}^3$ to construct their common perpendicular geodesic?

It seems that there is in 2 dimensions, see here. However the obvious generalisation to 3 dimensions would give the common perpendicular to geodesic planes containing the original two geodesics, rather than between the geodesics themselves.

Answer 1

If you really want to use just straight-edge and compass, don't go into $3$-space at all! Coxeter points out the equivalence between the inversive plane and hyperbolic space in the paper The inversive plane and hyperbolic space. In this particular situation one can think of lines as pairs of points in the inversive plane. Two pairs of points correspond to perpendicular lines of hyperbolic space when the two pairs are harmonic. So the problem is:

Given two pairs of points $(a,b)$ and $(c,d),$ to construct using straightedge and compass the (unique) pair $(x,y)$ harmonic to both.

Fenchel asserts the existence of such a thing, and relates them to square roots, but without mentioning the geometric argument, so here goes.

We first start off easy, and assume that $a = \infty.$ We define the (two) geometric means of $(c,d)$ with respect to b to be given as follows. (Basically they are the complex square roots of $d$ where $b = 0$ and $c = 1$.) Bisect the angle $\angle c b d$ by a line $\ell.$ Let $C_c, C_d$ be the circles centered at $b$ through $c,d$ respectively. Intersect these with $\ell$ to get four points $p_c, p_c', p_d, p_d',$ so that $(p_c,p_d)$ separate $(p_c',p_d').$ Let $q$ be the midpoint of $p_c$ and $p_d.$ Draw the circle $D$ centered at $q$ through $p_c$ and $p_d.$ Draw the perpendicular $\ell_\perp$ to $\ell$ at $b,$ and intersect it with $D$ at a point $r$. Then draw the circle $E$ centered at $b$ through $r$, and intersect it with $\ell$ at the points $s_{(c,d)},t_{(c,d)}$. These are the geometric means of $(c,d)$ with respect to $b$.

Now, it should be that $(s_{(c,d)}, t_{(c,d)})$ is harmonic to both $(\infty, b)$ and $(c,d).$ (I haven't worked this out yet.) So that's what we were looking for. In the case that $a \neq \infty,$ just let $C$ be the circle centered at $a$ through $b,$ and denote by $I_C$ the inversion in $C.$ Then the points we're looking for are $I_C(s_{I_C(c,d)}, t_{I_C(c,d)}) = (\sigma_{(c,d)}, \tau_{(c,d)}).$ These give the endpoints for the common perpendicular to the hyperbolic lines with endpoints $(a,b)$ and $(c,d).$

It's probably all in Fenchel, anyway....

Answer 2

I remembered that the isometries for the upper half space model of $\mathbf H^3$ are given by the action of $PSL_2\mathbf C$ on the points at infinity (the $xy$-plane as the complex numbers). So, given four geodesic endpoints, move one to $0$ with a pure translation. We now have three complex numbers $A,B,C,$ where $A$ is the other endpoint of the geodesic with an endpoint at $0.$ Define a complex number $\gamma$ that solves $$ ABC \gamma^2 + 2 BC \gamma + (B+C-A) =0, $$ where you might as well pick $\gamma = 0$ if the constant term $B+C-A=0.$
Next, apply the linear fractional transformation $$ h(z) = \frac{z}{\gamma z + 1} $$ to the plane. The result is $$ h(A) = h(B) + h(C) $$
That is, the midpoint of $0$ and $h(A)$ is the same as the midpoint of $h(B)$ and $h(C).$ So the common orthogonal geodesic is the vertical ray through $h(A)/2$ and allowing the third coordinate to vary. Then map everything back to your originals.

Note that there is no answer if two of your original geodesic endpoints coincide. In that case, they lie in a common flat, and lines asymptotic at infinity do not share a perpendicular.

See if I can do the link the right way this time, Milnor's survey on the first 150 years of hyperbolic geometry can be downloaded.