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As I understand, on a pseudosphere, a surface of constant negative curvature, we can realize a part of the hyperbolic plane (but not the entire plane due to Hilbert's 1901 theorem) and use this for example to prove that hyperbolic geometry is just as consistent as Euclidean geometry. But how does this compare to elliptic two-dimensional geometry? Naturally, we can consider a sphere, which is a surface of constant positive curvature, but a sphere only admits spherical geometry and not elliptic geometry if we consider Euclid's postulates (for example there is not always a unique shortest line between two points). How can we then use the sphere to show that elliptic geometry is as consistent as Euclidean geometry? Does there exist some other surface which we can use to do so? Can someone clarify this for me?

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  • $\begingroup$ Is elliptic in the usual definition: en.wikipedia.org/wiki/Elliptic_geometry. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 5 '14 at 12:27
  • $\begingroup$ You have to clarify what axioms you want. Note also that the projective plane does not embed in 3-space. Furthermore, existence of an embedding in 3-space has nothing to do with consistency (hyperbolic plane cannot not embed with $C^2$-degree of smoothness!) $\endgroup$ – Moishe Kohan Feb 5 '14 at 12:28
  • $\begingroup$ @studiosus The axioms that I want are Euclid's postulates except that we replace the parallel postulate with "No line can be drawn through any point not on a given line parallel to the given line in a plane", as I understand that is the basic axiomatic system we have for elliptic geometry. As for "existence of an embedding in 3-space has nothing to do with consistency", isn't this what Beltrami did in his papers in 1860s when he was proving the consistency of hyperbolic geometry? $\endgroup$ – Sid Feb 5 '14 at 12:32
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    $\begingroup$ @Sid: Note that Euclid did not have a complete set of axioms in the modern sense; Hilbert was the first to give such a list. If one takes Hilbert + the axiom you stated then the space you get is the projective plane which does not embed in $R^3$ for topological reasons (but admits an immersion). As for Beltrami's work, it had nothing to do with isometric embedding. From post-Hilbert viewpoint, elliptic geometry is consistent since it is realized by the real projective plane with its homogeneous metric. No need to embed it in any particular space. $\endgroup$ – Moishe Kohan Feb 5 '14 at 13:08
  • $\begingroup$ @studiosus Thank you for your answers. What I have here is that Beltrami "realized" a piece of the hyperbolic plane on the psudeo-sphere which he used to prove the consistency of hyperbolic geometry. So this means something else than isometric embedding? I have an English translation of Beltrami's papers which I intend to read later, but if can explain what Beltrami did in a simple way I would be very grateful. $\endgroup$ – Sid Feb 5 '14 at 13:16
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Here is an extended version of my comments.

  1. Note that Euclid did not have a complete set of axioms in the modern sense; Hilbert was the first to give such a list. If one takes Hilbert's axioms and substitutes the existence of a parallel line axiom with the nonexistence of a parallel line axiom ("For every line $L$ and a point $p\notin L$ there is no line through $p$ which is disjoint from $L$") then one can show that the resulting geometry is isomorphic to the one of the real projective plane equipped with homogeneous Riemannian metric (you have the freedom to choose the constant curvature, but that's all). Then you realize that real projective plane does not embed in $R^3$ for topological reasons (but admits an isometric $C^1$-smooth immersion by Nash-Kuiper's theorem).

  2. As for Beltrami's work: Consistency of a geometry from (post) Hilbert viewpoint has nothing to do with existence of an (isometric) embedding in a particular space. For instance, for a Riemannian manifold to exist, it suffices to define it via an atlas of charts together with a Riemannian metric tensor. What Beltrami did was to embed isometrically a proper open subset of the hyperbolic plane in $R^3$. Since a proper open subset of $H^2$ violates axioms of hyperbolic geometry (it is not homogeneous!), existence of such an embedding does not prove consistency of hyperbolic geometry. (Note that Hilbert proved that $H^2$ does not admit a $C^2$-smooth isometric embedding in $R^3$.) What Beltrami (or Poincare, depending on what you read) realized is that you can simply write down the familiar expression for the hyperbolic metric tensor on the upper half-plane (or the unit disk) in order to establish existence of hyperbolic geometry. Lastly, if you really want to embed isometrically all these geometries (including the elliptic one) in some Euclidean space, just use Nash isometric embedding theorem. You can embed all 2-dimensional geometries in $R^{10}$ via $C^\infty$-smooth isometric embeddings as a corollary of Gunther's improvement on Nash's theorem (I am sure that 10 is still not optimal, I think, it can be reduced to 6): M. Günther, "Zum Einbettungssatz von J. Nash", Mathematische Nachrichten 144 (1989) p. 165-187.

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  • $\begingroup$ Let me see if I understand correctly: Beltrami did embed isometrically a proper open subset of the hyperbolic plane in $\mathbb{R}^3$ but this was not enough to prove the consistency of hyperbolic geometry for reasons you mentioned. However, he later realized that he could prove the consistency of hyperbolic geometry in a different way: by "writing down the familiar expression for the hyperbolic metric tensor on the upper half-plane". Is this correct? $\endgroup$ – Sid Feb 5 '14 at 13:48
  • $\begingroup$ I would delete the word "later" from this sentence, other than that, it is correct. $\endgroup$ – Moishe Kohan Feb 5 '14 at 13:51
  • $\begingroup$ Why did Beltrami embed a subset of the hyperbolic plane if that was not enough to prove the consistency of hyperbolic geometry? Did he have some other reason to do so? $\endgroup$ – Sid Feb 5 '14 at 13:58
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    $\begingroup$ Sid: You would have to read Beltrami's original papers to figure this out. My guess is that at the time Riemann's viewpoint on geometry was not yet widely known, so he was addressing the audience which was familiar with geometry of surfaces in 3-space but not with Riemann's abstract framework. Take a look at en.wikipedia.org/wiki/Eugenio_Beltrami for references (John Stillwell might also provided some information on this topic in his book, see the same link). $\endgroup$ – Moishe Kohan Feb 5 '14 at 14:04
  • $\begingroup$ One more thing: Hilbert did his foundational work 20-30 years after Beltrami's papers. Only after Hilbert we have a definitive treatment of "classical" geometries, including non-Euclidean ones (about 2200 years after Euclid!). $\endgroup$ – Moishe Kohan Feb 5 '14 at 14:09
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One common way to realize elliptic geometry is by taking the sphere but identifying antipodal points. So you say a point and the point directly opposite the first one are in fact one and the same in terms of your model. You use two spherical points to model a single elliptic one.

for example there is not always a unique shortest line between two points

The case where this breaks is if the two points you try to connect are in fact antopodal. In the above model, these would be considered the same point, and the axioms agree that there is no unique line connecting a point to itself. So at least this example works out just fine, and other problems will get resolved in a similar manner.

Most people read Euclid's axioms in such a way that elliptic geometry violates not only the fifth, but also the one about how you can infinitely extend a line. So in that sense, hyperbolic geometry is consistent with the first four axioms, while elliptic geometry is not. But this depends in part on how exactly you formalize Euclid's axioms.

If you follow the axiomatization used by Hilbert in his Grundlagen der Geometrie (Foundations of Geometry), then the part about extending a line segment, which would be axiom IV.1 in Hilbert's work, seems valid enough. But some of the axioms of order, axiom II.3 in particular, are violated. Between-ness on the sphere simply isn't the same as in the Euclidean or hyperbolic plane. So even in Hilbert's axiomatization, elliptic geometry violates more than just the axiom of parallels.

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  • $\begingroup$ So the sphere can be a model for elliptic geometry if you modify Euclid's axioms? When Hilbert made Euclidean geometry rigorous in his 1899 Grundlagen der Geometrie, is this what he did? $\endgroup$ – Sid Feb 5 '14 at 13:05
  • $\begingroup$ Also, I would like to ask you to clarify if you can: a commenter said that "existence of an embedding in 3-space has nothing to do with consistency". But as I understand it, embedding a part of the hyperbolic plane into Euclidean space was precisely what Beltrami did to prove the consistency of hyperbolic geometry. Is this a misunderstanding on my behalf? $\endgroup$ – Sid Feb 5 '14 at 13:07
  • $\begingroup$ @Sid: I must confess that I'll have to look more closely at how Hilbert formalized things, to see how he wrote the second axiom about extending lengths. In any case, both elliptic geometry on the sphere and hyperbolic geometry in a conic the way Beltrami did it work by reinterpreting terms. For elliptic you reinterpret the definition of points, for hyperbolic you reinterpret angles and distances. $\endgroup$ – MvG Feb 5 '14 at 17:39
  • $\begingroup$ @MvG I keep mulling over something thinking I must be missing something. I feel like the portion of the $3$D sphere above the $x-y$ plane satisfies all the incidence, order and angle congruence axioms. I am not totally sure if the segment congruence axioms are satisfied because allowing the principle of juxtaposition allows you to construct line segments that "run out of the model." Is this typically why this half-sphere model is not used? $\endgroup$ – rschwieb Feb 6 '14 at 18:50
  • $\begingroup$ @rschwieb: After looking at Hilbert's axioms, I agree that the segment congruences are the most obviously violated ones in that approach. There might be other violated axioms, but if there are, then reasons why they are violated are likely to be more subtile, particularly if you don't include the circle within the $x$-$y$ plane itself. $\endgroup$ – MvG Feb 6 '14 at 22:24

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