TITLE: What is the distance between two points on the Berger metric of the squashed three-sphere?
QUESTION [7 upvotes]: The Berger metric on a "squashed" three-sphere is given (in Euler coordinates) by 
4 $ds^2 = \lambda^2 (d \tau + \cos \theta d \phi)^2 + d \theta^2 + \sin^2 \theta d \phi^2$.
See for example Eq. 1.11 of this paper. This formula describes the geodesic distance between two infinitesimally separated points. 
My question: what is the geodesic distance between any two points on the Berger sphere? 
What is the distance as a function of the coordinates $\tau_1,\phi_1,\theta_1$ and $\tau_2,\phi_2,\theta_2$ of the two points?

REPLY [2 votes]: Geodesics in Berger sphere become circle arcs in the quotient
$$\tfrac12\cdot\mathbb{S}^2=\mathbb{C}\mathrm{P}^1=\mathbb{S}^3/\mathbb{S}^1.$$
Once you see it, all calculations should be easy. (But I do not want to calculate.)