Bound on distance of geodesics in metric space

Question

Bridson and Haefliger in their Metric Spaces of Non-Positive Curvature in proof of proposition II.1.4 make use of the following fact: given any positive $\ell<D_\kappa$ there is constant $C$ depending only on $\ell$ and $\kappa$ such that if $c,c':[0,1]\to M^2_\kappa$ are two lineary parametrised geodesic segments of length at most $\ell$ and if $c(0)=c'(0)$, then $$d(c(t),c'(t))\leq C d(c(1),c'(1))$$ for all $t\in[0,1]$. $M^2_\kappa$ is two dimensional hyperbolic, Euclidean or spherical space of curvature $\kappa$. They reference 3.20, but I don't see that this fact immediately follows from that proposition, and it uses Riemannian metric, that I'm not too familiar with, and trying to prove it some other way, it's clear there is such $C$ for fixed $c,c'$, but how do I find one such $C$ for all possible pairs of curves of lenth at most $\ell$?

Answer 1

The statement you are trying to prove can be explained in words as follows. Given two geodesics that start at the same point find an estimate of their distance at time $t$ relative to their distance at time $1$ for all $t \in [0,1]$.

You can look at the three scenarios $\kappa=0, \kappa=-1, \kappa=1$ separately, other values of $\kappa$ follow from scaling.

$\kappa=0$ corresponds to flat Euclidean space. Geodesics are straight lines. $C=1$ works for all values of $l$.

It is a well known fact in Riemannian geometry that geodesics in hyperbolic space always move away from each other, which here translates to $C=1$ for all $l$. Rigorously proving that requires some knowledge of Riemannian geometry.

For $\kappa=1$ the value of $C$ does depend on $l$. Picture two geodesics (meaning great circles) on the sphere starting at the south pole. At $l=\pi$ they meet again at the north pole, so you need to restrict to $l< \pi$. For $l< \pi/2$ the geodesics are moving apart so they reach their greatest distance at $t=1$ and $C=1$ works. For $\pi/2 < l < \pi$ the points of greatest distance are on the equator. At $t=1$ they are closer to each other but you can get an estimate of the distance on the equator relative to the distance at $t=1$. The closer $l$ is to $\pi$, the bigger $C$ will need to be.