Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 26I (Differential Geometry)

For $v\in T_p\Sigma$, let ${\gamma }_v$ be the maximal geodesic with ${\gamma }_v(0)=p$, ${\dot{\gamma }}_v(0)=v$. Then ${\exp }_p(v)={\gamma }_v(1)$ on the open star-shaped set of $v$ for which time $1$ exists.

At the north pole, for $v=(v_1,v_2,0)$ and $r=|v|$, ${\exp }_p(v)=\cos rp+(\sin r/r)v$. Polar coordinates give $(\sin r\cos \theta ,\sin r\sin \theta ,\cos r)$. They are one-to-one for $0<r<\pi$, covering the sphere minus the two poles as a coordinate chart, with the north pole added as the polar origin; the south pole is the cut locus.

A great-circle arc of angle $\arccos (p\cdot q)\le \pi$ joins the points. The spherical distance inequality follows by integrating the polar metric $d{s}^2=d{r}^2+{\sin }^{2}r,d{\theta }^2$, so every curve has length at least the radial change. If $q\ne -p$, the shorter great-circle arc is unique; equality forces constant angular coordinate and monotone radial motion, hence only a reparametrisation.

A geodesic still joins most pairs, but if the unique minimizing great-circle arc passes through the deleted north pole, that arc is unavailable; the complementary great-circle arc remains a geodesic, so a geodesic does exist. A length minimizer need not: for points whose shorter arc passes through the missing pole, curves detouring arbitrarily close to it approach the spherical distance but never attain it.