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Pretty much every text about non-euclidean geometries talks about the various models by Beltrami, Riemann, Poincaré, Klein, and others which demonstrate that if euclidean geometry is consistent, then hyperbolic and elliptic geometry also are consistent. So far, so good. I think I understand these arguments.

However, there are many sources which claim that it was also proved (by Klein?) that hyperbolic geometry and euclidean geometry are equiconsistent. My understanding is that this means that if hyperbolic geometry is consistent, so is euclidean - which would place both on an equal footing.

My questions now are:

  1. Is this also true for elliptic geometry? Does the consistency of euclidean geometry follow from consistency of elliptic geometry?
  2. Where can accessible proofs for these equiconsistency results be found? (I mean proofs that not only show the well-known direction described in my first paragraph.)
Jade Vanadium
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Frunobulax
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  • Edited to add the "reference-request" tag. I'm familiar with the interpretation of hyperbolic geometry (within Euclidean) where the hyperbolic plane is represented by an open disk, and lines are represented by circular arcs ending perpendicular to the disk boundary. Elliptic geometry admits a similar interpretation, with a half-closed disk. We can interpret planar from elliptic by deleting a single line from the sphere, I think, but not sure. A similar trick should allow interpretation of planar geometry from within hyperbolic, but idk. – Jade Vanadium Jan 30 '25 at 03:01
  • If nobody who sees this can find a reference in the next few days, then I'll probably just work it out myself and post the constructions as a self-contained answer. – Jade Vanadium Jan 30 '25 at 03:17

1 Answers1

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Yes, any of those geometries can interpret any other, and the existence of an interpretation guarantees equiconsistency. I'll prove this for the four most common geometries: flat (Euclidean), hyperbolic, elliptic, and spherical (AKA double elliptic). Showing that flat geometry interprets all the others is actually quite simple, so I'll describe that in the paragraph below. The main thrust of my answer is showing the reverse, that every geometry can construct a disk model for flat geometry.

As you likely know, the Poincare disk model is the standard way to interpret hyperbolic geometry from a flat metatheory. For elliptic geometry, it's easiest to interpret spherical geometry first, and then simply identify antipodal points. To interpret spherical geometry, we can use the one-point compactification of the plane. The compactification is possible since flat geometry can construct a disk model of itself, hence any point on the disk boundary can be identified as the sphere's missing point. The geometry of the sphere can be defined in terms of stereographic projection, which isn't too hard to work out. Basically you just need to figure out which circles qualify as lines on the model sphere.


To model flat geometry in any other, first we select a distinguished origin $0$ and unit $1\neq 0$ that's not too far away. Let $C$ denote the corresponding unit circle, then our model plane will take its points from the interior of $C$ (defined relative to $0$). The lines of our model will be any arc segment staying strictly interior to $C$, and which has two antipodal termination points on $C$. Here, an "arc" can either be a line, circle, horocycle, or hypercycle.

This model does not preserve circles, but you can define circles in terms of line segment congruency. Any in-model line segment can be translated to the origin $0$ using an in-model parallelogram construction, as depicted by this desmos graph. The translation to the origin is necessarily an actual line segment in the metatheory, so we can say that two in-model line segments are congruent iff their translations to the origin are congruent in the metatheory. The parallelogram construction is provably unique by consequence of the parallel postulate, which can be verified about the model regardless of the metatheory's geometry.


If my construction is performed in flat Euclidean geometry, you get an interpretation of flat geometry within itself. The model is proven isomorphic to the universe using the map $p\mapsto p/(1-||p||^2)$, where those vector/arithmetic operations are defined relative to the distinguished points $0,1$ in the obvious ways. To verify that this is actually an isomorphism, we only need to verify that each in-model line is mapped to an actual line in the metatheory. That requires some tedious geometry/algebra, but it's not too hard.

If you interpret Hyperbolic geometry using the Poincare disk model, and subsequently perform my interpretation of flat geometry inside Poincare's disk, then you can easily verify that the inner model is isomorphic to the metatheory's flat universe. In fact, if you use the same origin for both constructions, then the inner model is the same as if you hadn't used Poincare's disk at all, since the arcs of Poincare's model are exactly the arcs of the metatheory.

If you interpret spherical/elliptic geometry using the compactification I mentioned earlier, and subsequently construct an inner model of flat geometry inside the spherical model, we again get an isomorphism between the inner model and the flat universe. The sphere model uses exactly the same arcs as the flat model, so the situation is roughly the same as if you'd just constructed the flat model inside itself twice.

In any case, we see that the model plane constructed in any given geometry is isomorphic to the ordinary plane. So, at least in principle, the models of the flat plane in a given geometry ought to actually obey all of Euclid's axioms. Provided our axiomatization of a given geometry is strong enough to perform basic reasoning about its arcs, it should also be strong enough to prove Euclid's axioms about its model plane, resulting in equiconsistency.

Jade Vanadium
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