TITLE: Geodesics and Distance in Hyperbolic Space
QUESTION [10 upvotes]: I am trying to understand geodesics and distance in hyperbolic space $H^n$ (Inverse image of $1$ under $f(x_0,x_1,\dots,x_n)=-x_0^2+x_1^2+\dots+x_n^2$ and with the inherited metric). More precisely, I want to show that the Riemann distance between two points $p,q$ is $d(p,q)=arccosh(-p\cdot q)$.
Easy I thought, just compute the geodesics between two points $p,q$ and evaluate the length.
However, I don't really understand how to do this and would appreciate some help. How do I get a nice analytical formula which I can just "evaluate" to obtain the distance? In all the books I've read there are only very geometric arguments which don't really produce formulas and I don't really understand how to go between the two...

REPLY [9 votes]: I will write you a sketch. First start with geometric reasoning to determine the geodesics in the hyperboloid model of hyperbolic space.

Let $O(1, n)$ denote the subgroup of linear transformations of $\mathbb{R}^{n+1}$ preserving the pseudo-Riemannian metric called the Lorentz metric $(\cdot, \cdot)$ of the . Let $O_+(1, n)$ denote the matrices $(a_{ij})_{i,j=0,\dots, n}$ in $O(1, n)$ with $a_{00} > 0$. Show that
$O_+(1, n)$ is a subgroup of $O(1, n)$. Show that elements of $O_+(1, n)$
restrict to isometries of $H^n$.

Determine the geodesics in the hyperboloid model of hyperbolic space:
Let $P$ be a two-dimensional linear subspace of $\mathbb{R}^{n+1}$ containing the
point $p \in H^n$. Show there is an element of $O_+(1, n)$ fixing $P$ and acting
non-trivially on each point of $\mathbb{R}^{n+1}\backslash P$. Conclude that all geodesics of $H^n$ through $p$ can be obtained as intersections $P \cap H^n$ for some $P$ as
above.
You will need to use:
Let $(M,g)$ be a Riemannian and let $\varphi:M \to M$ be an isometry. Let $p \in M$ such that $$\varphi(p)=p$$
and let $v \in T_p M$ such that:
$$d\varphi_p(v)=v$$
If $\gamma(t)=exp_p(tv)$ (the geodesic with velocity $v$ that start at $p$) then:
$$\varphi \circ \gamma(t)=\gamma(t)$$
for all $t$. You will also use the fact that geodesic in a given deirection is unique.
Klein disk model of hyperbolic space:
Let $y_1,\dots, y_n$, be the standard coordinates on $D^n$. Equip $D^n$ with metric defined by
$$g_{ij} =
\frac{\delta_{ij}}{1-|y|^2}+\frac{y_i y_j}{(1-|y|^2)^2}
$$


Define a map $h : H^n_1 \to D^n$ by
$$y_i=\frac{x_i}{x_0}$$
Show that $h$ is an isometry.

Show that the geodesics of part (2) are sent by $h$ to straight lines in
the Klein model. (The map $h$ is a stereographic projection along
lines through the origin, to $D^n \subset \{x_0 = 1\} \subset L^{n+1}$).
Now calculate the length of a straight line paramtized in arc length between the two points with the Klien metric. 

This is not the fastest way to see this but it connects the geometric reasoning to a formula and you can understant the arrcosh function. Maybe now you will even be satisfied with the $arrcosh$ formula without the calculation. Because you can understand what we are measuring.