# why there are no parabolic (on a paraboloid) non-euclidean geometry?

I have seen in many contexts that Euclidean geometry is called also "parabolic geometry".

As in many things in mathematics (conics, differential equations, algebraic equations) the terms: elliptical, parabolic, and hyperbolic refer to the conics with their corresponding names.

You could say that a plane is deformed paraboloid (can you?), but why is it that it is not important to consider geometry over a paraboloid?

I know Riemannian geometry considers geometry over general surfaces (manifolds) but there might be something uninteresting about parabolids that mathematicians do not like. What is it?

Thanks.

• "Parabolic" in the sense of axiomatic geometry does not refer to a paraboloid. Anyway, one partial answer is that, unlike a sphere or hyperbolic space, a paraboloid (endowed with the metric it inherits from Euclidean space) does not have constant curvature. (Also, NB neither of this notions should be confused with the branch of the field of geometric structures within differential geometry called "parabolic geometries", which is currently an active research area.) – Travis Nov 25 '15 at 16:03
• @Travis : I like what you say about "constant curvature" . Do you think the fact that there are no "parabollic trignometric functions" is related to the same thing? – Herman Jaramillo Nov 25 '15 at 16:13
• It seems that since circular trigonometrical functions are based on $x^2+y^2=1$, hyperbolic trignometrical functions in $x^2-y^2=1$. The parabolic would be based on $y-x^2=0$ which shows a problem in lost of symmetry and degeneracy (the zero in the right). Do this make sense? – Herman Jaramillo Nov 25 '15 at 16:16
• The real question is probably why hyperbolic geometry is called hyperbolic in the first place -- it is definitely not the geometry of any kind of recognizable hyperboloid embedded in Euclidean space. – Henning Makholm Nov 25 '15 at 16:21
• See here and linked questions for various interpretations on "parabolic trigonometric functions". – Henning Makholm Nov 25 '15 at 16:24

Hyperbolic geometry is not really geometry on a hyperboloid. It's geometry on an infinite surface of constant negative Gaussian curvature, something which cannot be represented even in 3D. You can model it using a sheet of a hyperboloid, but the metric you get isn't the normal 3D metric you'd intuitively expect.

Elliptic geometry is not the geometry on an ellipsoid either. While spherical geometry is what you get as geometry on the sphere, elliptic geometry is what you get from that if you identify antipodal pairs of points. It's the geometry on a surface of constant positive Gaussian curvature.

Just like the parabola is the singular limiting case between ellipse and hyperbola, the parabolic geometry is the limiting case between elliptic and hyperbolic geometry. And between constant positive and constant negative curvature, that limiting case is zero curvature.

Might be you could model parabolic geometry on a paraboloid using some strange metric, but why bother if you can have a flat plane using normal Euclidean metric, perfectly intuitive?

For a nice uniform way of looking at these different geometries, I suggest looking into Cayley-Klein metrics. Perspectives on Projective Geometry by Richter-Gebert has some nice chapters on this. Disclaimer: I'm working with that author, so I might be somewhat biased here.

• (+1) Hyperbolic geometry can be viewed as the geometry of (half of) the two-sheeted hyperboloid $x^{2} + y^{2} - z^{2} = -1$, $z > 0$, a.k.a., the "unit sphere of future-pointing timelike unit vectors" embedded in Minkowski space, however. :) – Andrew D. Hwang Nov 27 '15 at 0:43
• @AndrewD.Hwang: I know. That's what I meant when I said it's not the normal metric. You could look at this naively, and believe that between Minkowski space with signature $+,+,-$ and surface $x^2+y^2-z^2=-1$ on the one hand and Euclidean space with signature $+,+,+$ and surface $x^2+y^2+z^2=1$ you might have a space with metric $+,+,0$ and surface $x^2+y^2=0$ but unfortunately this kind of argumentation doesn't seem to be very fruitful, since the above is not a description of the Euclidean plane. – MvG Nov 27 '15 at 1:06
• Ah, I see. :) I'd thought you were addressing the misperception that a one-sheeted hyperboloid (or a hyperbolic paraboloid) in Euclidean space is a model of hyperbolic geometry. – Andrew D. Hwang Nov 27 '15 at 13:52
• @AndrewD.Hwang: In my original answer I was addressing just that misconception. Nevertheless, I was aware of how the concept of a Minkowski space can be used to resolve that problem for hyperbolic geometry (by making the “strange” metric something natural in that space), but not in a way that allows for a smooth and sensible transition from Elliptic geometry via some “flat” or “parabolic” border case. – MvG Nov 27 '15 at 21:33