# Is every elementary absolute geometry Euclidean or hyperbolic?

Absolute geometry is any one that satisfies Hilbert's axioms of plane geometry without the axiom of parallels. It is well-known that it is either the Euclidean or a hyperbolic plane. For an elementary version we also drop the (Cantor's) axiom of continuity, Greenberg calls such geometries Archimedean H-planes in his survey paper. We can still define a metric on them in the usual way. Repeatedly bisecting a picked "unit segment" and laying its pieces off of any other gives a binary fraction (possibly infinite) that is assigned as the segment's length. The distance between two points is defined to be the length of the line segment connecting them, unique by incidence. Incidence, order and congruence imply that perpendicular is shorter than oblique, so the triangle inequality holds. Without the continuity however Archimedean H-planes may not be metrically complete. Will their completions still always be Euclidean or hyperbolic? In other words, if we remove the axiom ensuring completeness, and then take the completion, do we end up with what we started with?

Here is why I am having doubts. It is easy to check by limit arguments that most axioms still hold in the completion, but not that two lines intersect at no more than one point, for example. With extra points added perhaps some lines intersect more than once. Also, there is an algebraic classification of Archimedean H-planes due to Pejas described in the Greenberg's paper (p.760). One of them, called the semi-elliptic plane, is quite peculiar. It satisfies the Lambert's hypothesis of the acute angle (in any quadrilateral with three right angles the fourth angle is acute), but any two non-intersecting lines in it have a unique common perpendicular. In other words, no two lines are asymptotically parallel. If the semi-elliptic plane is isometric to a subset of a hyperbolic one then we should be able to obtain it by removing some points from the latter. But if we remove all lines asymptotically parallel to some line in the hyperbolic plane, nothing will be left other than that line itself. On the other hand, if the completion is elliptic, as the name suggests, then how does it square with the hypothesis of the acute angle?

Is the metric completion of an Archimedean H-plane always Euclidean or hyperbolic? If so than what subplane of a hyperbolic plane is semi-elliptic? If not then what is the completion of the semi-elliptic plane?

Unfortunately, Pejas's original works weren't translated from German. Here is the passage from Greenberg describing the semi-elliptic plane.

"Pejas gave the following example of an Archimedean H-plane in which the acute angle hypothesis holds but which is not hyperbolic; it is an example of a semi-elliptic plane, defined by the property that any two parallel lines have a unique common perpendicular (in a hyperbolic plane, two asymptotically parallel lines have no common perpendicular, whereas two divergently parallel lines have a unique one): Let $K_0$ be an Archimedean ordered field with two distinct orderings $<$ and $<'$ (for example, $K_0 = \mathbb{Q}(\sqrt{2})$). Let $L$, $L'$ be the real closures of $K_0$ with respect to these orderings within a given algebraic closure. Set $K = L\cap L'$. Then $K$ is Pythagorean, Archimedean, and contains an element $k$ such that $k<0$ and $0<'k$. We take $k$ as metric constant and the points of $\mathcal{K}$ to be all $(x,y)$ in the affine plane over $K$ for which $k(x^2 +y^2)+1>0$. $\mathcal{K}$ is the interior of the "absolute conic" $x^2 +y^2=-k^{-1}$, which is empty because $\sqrt{-k}\notin K$. Since $\mathcal{K}$ is maximal, that conic is the locus of all ideal points, so asymptotic parallels do not exist and the plane is semi-elliptic."

• Do I understand correctly that $({*})$ in Archimedean H-plane you can perform all ruler-and-compass construction? If yes then Bolyai's construction (mathoverflow.net/questions/45237) gives an other parallel line. This will do the job if the completion of your H-plane is Lobachevskian. So if $({*})$ is correct then the answer is "NO". – Anton Petrunin Sep 11 '14 at 19:01
• Conifold, you might want to try Hartshorne's book on this as well. – Will Jagy Sep 11 '14 at 19:17
• for that matter, Marvin won a prize for this recent survey, which shows a fair amount of extra stuff that was added for his fourth edition , typing link again: maa.org/programs/maa-awards/writing-awards/… see if it works this time. Yup, good link. – Will Jagy Sep 11 '14 at 19:35
• @Anton Petrunin Not quite. You can perform ruler and compass constructions as long as you do not rely on every line passing through interior of a circle intersecting the circle. Bolyai's construction and many others that Will Jagy uses in the Squaring paper for example rely on this, Greenberg calls it the line-circle axiom. It can not be proved without continuity, and it does not hold in the semi-elliptic plane. – Conifold Sep 11 '14 at 21:17
• @Will Jagy I read this paper when you linked it under your previous answer (thanks!), but for semi-elliptic plane it references an Appendix to Greenberg's book, of which the paper I linked is an expanded version, and Pejas's papers in German. I am still interested in lengths constructible by elementary means, which I realized is not the same as by ruler and compass w/o the line-circle axiom, and now I am not even sure that Euclidean or hyperbolic metric relations have to be satisfied! – Conifold Sep 11 '14 at 22:05

Your answer is correct, except that in the example you gave, one has to adjoin far more numbers than you described in order to get to that ordered Pythagorean not-Euclidean field K.

That example is described on p.594 of the fourth edition of my book Euclidean and Non-Euclidean Geometries: Development and History (W.H. Freeman, 2008). There I called such a plane where parallel lines have a unique common perpendicular an HE-plane, abbreviating "halb elliptisch" from Pejas' classification article. Since you are only interested in the Archimedean case in order to get a natural metric once a unit segment is chosen, your HE-plane satisfies the acute angle hypothesis (by the Saccheri-Legendre theorem). You then describe an HE-plane accurately as the interior of a virtual circle in the affine plane over an ordered Archimedean Pythagorean not-Euclidean field K. An Archimedean ordered field is a subfield of R (up to isomorphism), so when you metrically complete it, you get R.

To make your argument completely rigorous, you would have to prove that the metric completion of an Archimedean H-plane is again an H-plane. Then, since it is complete, it must be either the real Euclidean or the real hyperbolic plane (in the HE-plane case it is the real hyperbolic). As I pointed out on p.594 of my book and as you indicated, if the line-circle axiom holds, then an Archimedean H-plane must be either Euclidean or hyperbolic, but its coordinate field could be any Euclidean subfield of R, such as the field of constructible numbers or the field of real algebraic numbers.

See also my March 2010 MONTHLY survey paper mentioned by Will Jagy entitled "Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries." Section 2 is all about Will Jagy's results about regular-polygoning circles in the hyperbolic plane (you can't always "square" them), and Section 3 is about the undecidability and consistency of elementary geometry.

If you email me at mjg0@pacbell.net I will send you my latest updating of that article.

(The terminology for all this is confusing. Yes, Janos Bolyai did introduce the term "absolute geometry" for the common part of real Euclidean and real hyperbolic geometries. I have argued - and it has generally been accepted by other writers - that "neutral geometry" is a better name, because one remains neutral about which parallel postulate to assume. I also argued that "absolute geometry" should be the name for Bachmann's geometry based on reflections, since it includes not only neutral geometry but also elliptic and other geometries - see p.588 of my book. Furthermore, even for neutral geometry, why restrict to real geometries? A model of Hilbert's incidence, betweenness and congruence axioms we now call a Hilbert plane or an H-plane for short. Pejas classified all H-planes. His classification is described on pp. 588ff of my book.)

• Got your email. Hope the OP (Original Post-er) notices this, (s)he will be notified. – Will Jagy Sep 17 '14 at 22:08
• @Marvin Greenberg Thank you for the answer. Sorry, I confused formally real fields with real closed fields, one still has to take closure after adjoining square roots. I was hoping to avoid checking the axioms for the completion because according to Pambuccian's paper (p.19 public.asu.edu/~pusunac/papers/budapest21.pdf) the congruence of segments in $H$-planes is defined by a formula that dictates the metric to be a restriction of the usual Euclidean or hyperbolic. So after completion we should directly get $\mathbb{R}^2$ or a Klein disk with a standard metric? – Conifold Sep 18 '14 at 1:25
• Originally I was wondering if there are $H$-planes where any two segments have a common measure, so that excluded non-Archimedean ones. With the line-circle axiom the answer is negative also, as follows from Will Jagy's answer, but $HE$ seems trickier. – Conifold Sep 18 '14 at 1:26

The answer is yes. The completion of a semi-elliptic plane is hyperbolic. The name probably comes from the fact that each pair of lines in it has either a point or a single perpendicular in common (in the elliptic plane they have both), but there is nothing elliptic about it metrically. Semi-elliptic planes are just hyperbolic planes thinned out so cleverly that no pair of asymptotic parallels remains. What of the paradox of removing all asymptotic parallels to a line and having nothing left? The catch is that some lines are not removed entirely. All their points are removed except one. It takes two points to form a line in geometry, so the plane does not "see" the lines that only have one point in it. What's unexpected is that most (uncountably many) lines are removed entirely and still the set of sole survivors is enough to support a geometry, in fact it is dense in the hyperbolic plane!

This is easiest to explain using the Klein model of a hyperbolic plane as the interior of a circle $D$ in $\mathbb{R}^2$. The lines are the chords of the circle, and the asymptotic parallels intersect on the boundary $\partial D$. Let $K$ be a proper subfield of $\mathbb{R}$ such that $K^2\cap\partial D=\emptyset$. Any two lines in $K^2$ intersect at a point with $K$ coordinates, so any line in $\mathbb{R}^2$ that intersects a $K^2$ line at a point in $\partial D$ has at most one point with $K$ coordinates. Everything so far works even with $K=\mathbb{Q}$, the rest in the construction of $K$ is to ensure that it is Pythagorean ($a,b\in K$ implies $\sqrt{a^2+b^2}\in K$), and that $K^2\cap\partial D=\emptyset$. Then $K^2\cap D$ is a semi-elliptic plane, and every semi-elliptic plane arises in this way. The metric is up to scale the restriction of the Klein-Beltramy metric on $D$ (logarithm of the cross-ratio), so the completion is the hyperbolic plane $D$ itself. Since according to Pejas's classification every Archimedean H-plane is Euclidean, hyperbolic or semi-elliptic (Theorem 3 in Greenberg's paper) the completion is always Euclidean or hyperbolic.

Since $K$ is only Pythagorean but not Euclidean (every positive element has a square root) there are lines in $K^2\cap D$ that pass through interior of circles without intersecting them in violation of the so-called line-circle axiom. This can also be seen geometrically. In any Archimedean H-plane that satisfies the hypothesis of the acute angle and the line-circle axiom a Bolyai's construction produces a line asymptotically parallel to any given one, which can not happen in $K^2\cap D$. The "metric constant" $k\in K$ is the Gaussian curvature of the completion, and since $\sqrt{-k}\notin K$ it can not always be normalized. In other words, not every hyperbolic plane can be thinned out into a semi-elliptic one, the standard one can't be among others. Pejas's example of $K$ is obtained by adjoining to $\mathbb{Q}(\sqrt{2})$ the square roots of all elements $x=a+b\sqrt{2}$ that are positive along with their conjugates $\overline{x}=a-b\sqrt{2}$, and taking a metric constant $k<0$ with $\overline{k}>0$, for instance $k=1-\sqrt{2}$. The equation of $D$ is then $x^2 +y^2<-k^{-1}$.

I'd like to thank Will Jagy for pointing me in the right direction in his previous answer