# Is there a way to prove that absolute geometry must take place on a Riemannian manifold?

Absolute geometry (as I know the term) is just (Euclidean geometry) $-$ (parallel postulate).

(It is sometimes also called neutral geometry, because it is "neutral" w.r.t. the parallel postulate.)

The only spaces I know of which satisfy its axioms are Euclidean spaces and hyperbolic spaces, both of which are obviously Riemannian manifolds.

Question: Are these the only possible metric spaces which can satisfy the axioms of absolute geometry? If so, is there a way to prove this?

Attempt: By Postulate 4 here, it follows that absolute geometry is a subset of metric geometry. In addition to the existence of a metric, it also requires the existence of lines and angles, in order for the remaining postulates to be defined.

Both lines and angles appear to be a concept of ordered geometry, thus we have to restrict to metric spaces with some notion of intermediacy, i.e. "betweenness".

One way to establish such a notion is by restricting to geodesic spaces, and then to say that the point $P$ lies between the points $A$ and $B$ if and only if it is on a geodesic connecting $A$ and $B$. (The uniqueness of geodesics seems unnecessary, for example elliptic geometry should also be an ordered geometry, I think, but elliptic geometry does not have unique geodesics.)

But is it really necessary, rather than just sufficient, to restrict to geodesic spaces, in order to have ordered geometry? Ordered geometry doesn't even seem like it requires the existence of a metric, so why would it require the existence of geodesics?

Moreover, I see no reason how or why to restrict further from geodesic spaces to Riemannian manifolds in order to satisfy the axioms of absolute geometry.

• Of course there is also elliptic geometry ... which is usually not excluded by the axioms of absolute geometry (in Euclid's sense), but I suppose it is excluded by the "ruler postulate" – Hagen von Eitzen Jul 1 '17 at 20:40
• @HagenvonEitzen This page says that elliptic geometry is inconsistent with the axioms of neutral geometry, although I don't quite understand the argument: web.mnstate.edu/peil/geometry/C2EuclidNonEuclid/7elliptic.htm Also, I vaguely remember the book by Agricola and Friedrich, which introduced me to the notion of absolute geometry, also saying that elliptic geometry is close to being an absolute geometry but not quite. As for whether elliptic geometry is an ordered geometry, I assume it is, but haven't been able to find a source explicitly confirming or denying the claim. – Chill2Macht Jul 1 '17 at 21:52
• In other words elliptic geometry might not only not be an absolute geometry, but it also might not even be an ordered geometry (although I think it probably most likely is, given that it is defined for geodesic spaces). I mostly just claimed that elliptic geometry is an ordered geometry so that someone could correct me in case that is false and I could ask for a reference as a follow-up. – Chill2Macht Jul 1 '17 at 21:59

This is what Lobachevsky proved. He did not prove existence of hyperbolic geometry, but he proved its "uniqueness", as we will see below. Lobachevsky established trig formulate for hyperbolic triangles, including the hyperbolic cosine formula, just assuming the negation of the Playfair's Axiom. Here and below, a hyperbolic plane is a neutral geometry satisfying the negation of the Playfair's Axiom. (I will try to find a modern reference, maybe it is in Moise.) Just for the record: each hyperbolic plane comes with a scalar parameter $\kappa$, its Gaussian curvature, which, for the purpose of this answer I assume to be equal to $-1$. Hyperbolic planes with different curvature are, of course, non-isometric. (Without referring to the notion of curvature which requires a Riemannian metric, I will just assume that the area of each ideal triangle equals $\pi$.) Given this, let's prove that every abstract hyperbolic plane $\Pi$ is isometric to the Poincare disk. Pick a base point $o\in \Pi$ and a reference ray $\rho$ emanating from $o$. This defines the "polar coordinates" on $\Pi$, namely $P(r,\theta)$, where $r$ is the distance from $P$ to $o$ and $\theta\in [0,2\pi)$, the angle between $oP$ and $\rho$. Now, do the same in the Poincare disk model $D$, where $0$ will denote the center and the ray $R=\{(x,0): x>0\}\cap D$ plays the role of $\rho$. Then, define a map $f: \Pi\to D$ by sending $P(r,\theta)\in \Pi$ to $Q(r,\theta)\in D$. It is clearly a bijection, let us prove that it is an isometry. Take two points $A, B\in \Pi$. Then their distance in $\Pi$ is given by $$cosh(|AB|)= cosh(|oA|)cosh(|oB|) - sinh(|oA|)sinh(|oB|) cos(\angle(AoB)).$$
By the construction, the map $f$ preserves all the quantities on the right-hand side of this equation (the distance to the origin and the angle). Since the hyperbolic cosine formula also holds in the Poincare plane, it follows that $f$ is an isometry.
• Holy **** this is an amazing answer. Wow. I want to make sure I understand all of it, or else I feel like it would be wasted on me. So did Lobachevsky prove, as part of the trig formulas, that each such space comes with a scalar parameter $\kappa$? (Maybe the formulas depend on $\kappa$ somehow?) Also, is it also true that a (metric space) isometry with a Riemannian manifold (whose metric space is induced by its Riemannian metric) implies that the former space is not only metric space isometric to the Riemannian manifold, but also Riemannian metric isometric? – Chill2Macht Jul 2 '17 at 11:28
• @Chill2Macht: Yes, all three: Bolyai, Gauss and Lobachevsky knew about this parameter. I suspect that Gauss also knew its relation to the Gaussian curvature (via Gauss-Bonnet formula). In the curvature $\kappa$ hyperbolic plane area of a triangle with angles $\alpha,\beta,\gamma$ equals $(-\kappa)(\pi - \alpha- \beta-\gamma)$. As for your last question, it does not make sense as stated since you need to have Riemannian metrics on both spaces to talk about Riemannian isometry and you have a metric space and a Riemannian manifold. However... – Moishe Kohan Jul 2 '17 at 11:57
• @Chill2Macht: This is nearly a tautology. A metric space isometric to a Riemannian metric space is itself Riemannian. Proof. Let $f: M\to N$ be an isometry of metric spaces, where $N$ is Riemannian. First, take the pull-back of the smooth atlas on $N$ to make $M$ into a smooth manifold (you just need $f$ to be a homeomorphism). Next, pull-back the Riemannian metric $h$ from $N$ to $M$, let $g$ denote the pull-back metric. Thus $f: (M,g)\to (N,h)$ is both a metric isometry and a Riemannian isometry. Therefore, the original metric on $N$ is the one which is defined via the Riemannian metric $g$. – Moishe Kohan Jul 10 '17 at 8:38
Following the pdf you linked to, for any two point $P$, we can find $A,B$ such that $P,A,B$ are not collinear. Then by distance from $P$ and angle with $PA$ (with orientation determined by $B$), we can map the plane to $\Bbb R^2$ - bijectively. Using the triangle inequality, one sees that this map is continuous. It is not too hard to show that the inverse is also continuous.