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This may have been answered here somewhere, but I'm unable to find it.

Is 2D hyperbolic geometry is unique up to isomorphism (or up to whatever's appropriate)? I know there are at least four models -- Poincare disk, half-plane, etc., but I assume they're essentially the same in some sense?

Another way to put my question: suppose we keep Euclid's first four postulates, but replace Playfair's equivalent of the parallel posutulate ("Given a line L and a point P not on L, there exists at most one line through P that doesn't intersect L) by, "Given a line L and a point P not on L, there exist at least two lines through P that don't intersect L." Can we deduce from the five resulting axioms that at most one geometry that results?

Thanks for any help you can offer! References would especially be appreciated. I'm an analyst, so I know how to read math, but I'm not an expert in this area, so I'm looking for sources that are accessible to mathematicians generally.

Best Regards,

Bob

Robert Gethner
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  • You're going to have to define what you mean by 'isomorphic' here; there's a map between e.g. the Poincare and the Klein models, but it's not conformal. I think the best candidate is isomorphism in a model-theoretic sense, in that all of the statements that can be proved about lines, triangles, and points are the same? – Steven Stadnicki Jun 11 '18 at 15:37
  • You might want to have a look at Tarski's Axioms to see how this is usually done for Euclidean geometry; an offhand remark in https://math.stackexchange.com/questions/1477616/tarski-like-axiomatization-of-spherical-or-elliptic-geometry suggests that the poster considers the same 'easy' for hyperbolic geometry, but I haven't seen that independently anywhere. – Steven Stadnicki Jun 11 '18 at 15:43
  • "there's a map between e.g. the Poincare and the Klein models, but it's not conformal." hum, there is definitely a conformal map between the two, and this map is an isomorphism. in fact, every simply connected Riemann surface that is not the sphere or the plane is conformally equivalent to the unit disk – Albert Jun 11 '18 at 22:29
  • Glougloubarbaki: what are your assumptions? This does not seem to be the standard Riemann mapping theorem which deals with subsets of the Euclidean plane with Euclidean metric (or a sphere). For example, ellipsoid is not the sphere nor the plane, but it is not conformally equivalent to the unit disk; fish eye projection maps the plane to the unit disk, but it is not conformally equivalent to the unit disk with Euclidean metric. If you consider the Poincaré and Klein models without taking their metrics into account, then they are basically the same thing. – Zeno Rogue Jun 12 '18 at 08:46
  • Identity is a map between Poincaré and Klein disks which is conformal (in fact, isometric) with respect to Euclidean metric, but not with respect to the metric in the model. So it keeps angles on the picture, but the creatures living inside the world will consider the angles to be changing. There is also a map which is conformal (in fact, isometric) with respect to the metric in the model, but not with respect to the Euclidean metric. – Zeno Rogue Jun 12 '18 at 08:47
  • Yes, it is unique, this is what Lobachevsky proved. See my answer https://math.stackexchange.com/questions/2343323/is-there-a-way-to-prove-that-absolute-geometry-must-take-place-on-a-riemannian-m/2343748#2343748 – Moishe Kohan Dec 29 '18 at 21:20

1 Answers1

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The appropriate notion here is isometry: if points $a$ and $b$ are mapped to $a'$ and $b'$ respectively, the distance between points $d(a,b) = d'(a',b')$. $d$ and $d'$ here denote the represented distance, not the Euclidean distance between the points in the model. Geometry is about measuring distances, so to say.

Assuming four postulates of Euclid plus your variant of the Playfair's axiom, you have that the curvature is fixed and negative, but the specific value of curvature is not known. With curvature -1, a triangle with four angles 45 degrees each will have an area of $\pi/4$. If the curvature is, say, -2, this triangle will have an area of $\pi/8$; in general, the sum of angles of a triangle minus $\pi$ equals area times the curvature (or, for surfaces where the curvature is not constant, the integral of the curvature). The curvature is only the matter of scale (a larger sphere will have smaller (positive) curvature, but it is essentially the same shape).

When we fix curvature -1, all the common models (Poincaré, Klein, hyperboloid, half-plane) are isometric (my page lists the common models and several less common ones; but I have no references for how the postulates plus curvature fix the geometry). An useful analogy: cartographers use many projections of the surface of the sphere (stereographic, Mercator, etc.) but they all describe the same mathematical object on a flat 2D map, and since the sphere is not flat, none of them is perfect, and they have different advantages and disadvantages. The same is true about the models of hyperbolic geometry.

Surfaces of constant curvature need not be isometric to hyperbolic plane, because they can correspond to only a fragment of a plane (a disk is not isometric to the whole plane, even though both have curvature 0) or they can be wrapped (a cylinder is not isometric to the whole plane, even though both have curvature 0 -- this happens with the tractricoid aka pseudosphere, which is listed sometimes as a model of hyperbolic geometry). However, such cases do not satisfy the postulates.

The isometric mappings between the common models not only exist, but they are also given by simple formulas. For example, the mapping between the half-plane and the Poincaré disk is inversion, and Klein/Poincaré/Gans models are obtained from the hyperboloid model with a simple perspective transformation.

Zeno Rogue
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