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?
Best Answer
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.