Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 11F (Geometry)

For a continuously differentiable curve $\gamma (t)=x(t)+iy(t)$ in the Poincare half-plane model, its hyperbolic length isThe hyperbolic lines are the geodesics: vertical Euclidean lines and Euclidean semicircles whose centres lie on the real axis.

The unique hyperbolic line through distinct $z,w\in H$ can be carried by an upper-half-plane Möbius transformation to the imaginary axis. Such transformations are isometries by hypothesis, so it is enough to consider endpoints $i{y}_0$ and $i{y}_1$. Every joining curve satisfiesEquality holds for the monotone vertical parametrizationThus the appropriately parametrized hyperbolic-line segment $[z,w]$ attains the infimum defining $\rho (z,w)$.

Now let $l,m$ have positive infimum distance. They neither intersect nor share an ideal endpoint: an intersection would give distance zero directly, while lines sharing an ideal endpoint approach one another arbitrarily closely near that endpoint. Hence they are ultraparallel. An upper-half-plane isometry puts them into the formThis is the normal form underlying the common perpendicular of ultraparallel hyperbolic lines. Put $z=e^{u+i\theta }$. Sincethe metric becomesEvery path from $l$ to $m$ therefore hasEquality requires $\theta =\pi /2$ and monotone $u$, so it is attained on the imaginary axis, which meets both semicircles orthogonally. Its intersection points $ir$ and $iR$ realize the distance

For a nonattained example, take the vertical lines $l=\{iy:y>0\}$ and $m=\{1+iy:y>0\}$. They share the ideal endpoint at infinity. The horizontal segment at height $y$ has hyperbolic length $1/y$, so their infimum distance is zero, but distinct points of $H$ always have positive distance. Hence the infimum is not attained.