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

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-
I can't tell what your axioms might be. If you gave me two skew lines in $R^3$ and a compass and straightedge to wave about in midair, but no plane on which to draw, I'm not sure what I could accomplish. So, as I think you are probably envisioning a computer model of some kind, perhaps you could solve the $R^3$ problem and describe what was involved. Given enough detail on that I can probably solve your problem in $H^3$ or show it cannot always be done. – Will Jagy Oct 29 2010 at 22:22
@Will: Yes, this is a practical problem with drawing a diagram in a 3d computer model. I don't know if an exhaustive list of operations would be useful, but I can do things like draw lines between two points, circles or planes through three non-colinear points, spheres through 4 points, find intersections and I can do Euclidean isometries to objects (by distances and angles that I already know). – Henry Segerman Oct 29 2010 at 23:30
A solution to the $R^3$ version is not immediately apparent to me (although I think it should be possible), I'll have a think. For the Poincaré ball model embedded in $R^3$ there is more to work with: you have the endpoints of the $H^3$ geodesics to start from. – Henry Segerman Oct 29 2010 at 23:35

1
Now that I see what you are prepared to use, it is easy in $R^3.$ Take the two direction vectors for your skew lines and take their cross product, call that $N.$ Construct the plane normal to $N$ that contains one of the lines. Take two distinct points on the other line and drop their perpendiculars to the plane, thses being parallel to $N.$ Draw the line between the two image points, mark where it connects with the first line. Draw another line parallel to $N$ through that point. – Will Jagy Oct 30 2010 at 4:37
Ah yes, that works for $R^3$. – Henry Segerman Oct 30 2010 at 13:05
show 2 more comments

## 2 Answers

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....

-

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-

Yes, if we allow linear fractional transformations then this works. However, I don't know if they would really fit into the spirit of "ruler and compass construction", even in my somewhat loose interpretation: my natively Euclidean computer modelling program wouldn't do them for instance.  However, using this construction it would be possible to write a script to do the calculations. – Henry Segerman Jan 3 2011 at 0:42
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9495599269866943, "perplexity_flag": "head"}