# Prove that the Euclid's parallel postulate is false on the hyperboloid model

Consider the two-dimensional case, i.e. of the hyperbolic place. Define our hyperboloid as the set of points $x=(x_1,x_2,x_3)$ in 3-space (note: Minkowski space, but not needed for this problem) that fulfill $x_1^2 - y_1^2 - z_1^2 = 1$ and $x_1>0$. A hyperbolic line is defined as the intersection of a plane that goes through the origin with the hyperboloid. Two distinct hyperbolic points (points on the hyperboloid) determine a hyperbolic line. Consider the drawing labelled [2] on this picture for clarification (from this website):

I would like to prove something that seems intuitively clear:

Given a hyperbolic line $h$ and a hyperbolic point $p$ not on $h$, there are infinitely many hyperbolic lines through $p$ that do not intersect $h$.

(note: this is the hyperbolic variant of Euclid's parallel postulate)

My attempt at a proof begins like this:

We have by definition that $h$ is the intersection of the plane $m_1$ determined by two distinct hyperbolic points and the origin with the hyperboloid. As $p$ is not contained by $h$, we have that $p$ cannot be contained by $m_1$ either, so any hyperbolic line spanned by $p$ is the intersection of a different plane $m_2$, determined by $p$, the origin and some point $s$, with the hyperboloid. $m_1$ and $m_2$ are thus distinct planes but because they meet at least at one point, the origin, they cannot be parallel and hence we know from Euclidean geometry that their intersection is a line $l$. Because $h$ is on the hyperboloid, if $l$ and $h$ intersect it means that the hyperbolic line spanned by $p$ intersects $h$ in a hyperbolic point, so we want to show that there are infinite ways to choose the point $s$ so that $h$ and $l$ do not intersect.

and at this point I was going to use a topological argument to show that we can select infinitely more points $s$ so that $h$ and $l$ do not intersect than not. But I could not do this, I think my method might be off and that algebra is a better way to go. Does anyone here know how to show this or can point me to a reference with a proof?

• I do not know if this comes from a course and if there is more you have to do in the future regarding hyperbolic geometry. Here is a book I used when I took a course of non euclidean geometry (in which hyperbolic geometry forms a BIG topic): "Introduction to Non-Euclidean Geometry" by Harold Wolfe. Published by Dover Publications Inc. A very inexpensive book in which I remember your question is answered in great detail. I just can't come up with the right approach anymore. (Too long ago) – imranfat Mar 18 '14 at 20:44
• @imranfat My question is related to a paper that I'm writing for my university. I found a copy of this book, are you sure that the author discusses this in particular? I do not see any mention of the hyperboloid model and I do not find it in the index. – Sid Mar 18 '14 at 21:03
• The book Euclidean and Non-Euclidean Geometry by Patrick Ryan treats hyperbolic geometry in your framework, and contains a proof that Playfair's axiom fails in the hyperbolic plane. (Can't give you an exact reference since I don't have the book at hand....) – Andrew D. Hwang Mar 18 '14 at 21:36
• @user86418 As far as I understand it, the parallel postulate (which is equivalent to Playfair's axiom in Euclid and Hilbert's framework) fails on the hyperbolic plane per definition, so what does Ryan prove? I am not so sure if this is relevant to my question since we do not know if there is hyperbolic geometry on the hyperboloid until we have proven this. – Sid Mar 18 '14 at 22:21
• Ryan's approach is to show that $H$, the upper half of the hyperboloid in $\mathbf{R}^{2,1}$, with lines being intersections of $H$ with suitable planes, satisfies Euclid's first four axioms but not Playfair's. His proof that Playfair's axiom is false in this model amounts to what you're asking. – Andrew D. Hwang Mar 19 '14 at 0:12

You may use the projective model (or Klein model). That is to say, projectivise everythng:

For every line $l$ trough $0$ consider its intersection with the plane $\pi=\{x_1=1\}$. Every point of the hyperboloid is mapped to a point of the unit disc in $\pi$. Since the geodesic of the hyperbolic plane are intersections of the hyperboloid with planes through the origin, they project to usual segments in the disc, connecting two points of the boundary. Let's call them chords.

Now, in a disc it is easy to se that given a chord $r$ and a point $p$ not in $r$ there are infinitely many chords through $p$ that do not intersect $r$.

a complete reference for hyperbolic geometry is the book of Ratcliffe "Foundations of hyperbolic manifolds". There you find the projective model well descripted.

• I have Ratcliffe's book (or do you mean "foundations of hyperbolic manifolds"? or is there another book I'm not aware of?), it's great but he does not go deep into axiomatics (and does not discuss this). Projecting onto the Klein model sounds interesting but when you do so, isn't it possible that the image of $p$ will be on the same line as the image of $h$ in the Klein model (if $p$ lies "beneath" $h$ on the hyperboloid for example)? – Sid Mar 18 '14 at 21:23
• yes sure, is "manifolds" not "geometry". Have you solved your problem? – user126154 Mar 19 '14 at 8:32
• No I haven't, I haven't found the proof in the literature that's been suggested and I have not been able to construct a proof myself so far. – Sid Mar 19 '14 at 13:58
• My answer is a proof: in the hyperboloid model, every point is determined by a line in $\mathbb R^3$ (trough the origin) and lines are intersection of $2$-planes with the hyperboloid. Now interserct everything with the plane $x_1=1$ and... as in the answer – user126154 Mar 19 '14 at 17:17
• I didn't see you had edited your answer, I understand you better now. So is there a bijection here, i.e. no two distinct points on the hyperboloid are mapped to the same point on the disc under your projection? If so then your solution works absolutely because it is very simple to show that the hyperbolic axiom is valid on the Klein model. – Sid Mar 19 '14 at 18:03