Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 11E (Geometry)

Every orientation-preserving isometry of the Poincare half-plane model has the formwith matrices identified up to multiplication by $-I$. Thus $G\cong PSL(2,\mathbb{R})$.

For $z=x+iy\in \mathbb{H}$, the affine map $(z-x)/y$ sends $z$ to $i$, proving transitivity on points. Real translations and positive dilations act transitively on finite boundary points, while inversion $z\mapsto -1/z$ exchanges $0$ and $\infty$; hence the action on $\mathbb{R}\cup \{\infty \}$ is transitive. A hyperbolic line is determined by its unordered pair of boundary endpoints, and a real Möbius transformation can send any such pair to $\{0,\infty \}$. Therefore $G$ is also transitive on hyperbolic lines.

By point transitivity, it suffices to take $P=i$. The conditionis equivalent to $d=a$, $c=-b$, and the determinant condition becomes $a^2+b^2=1$. Thus the stabilizer consists ofmodulo $\pm I$. Writing $a=\cos \phi$, $b=\sin \phi$ identifies it with rotations of the tangent plane, and the doubled matrix angle removes the $\pm I$ quotient. Hence

An isometry sends the geodesic through $P,Q$ to the imaginary axis and sends the two points to $ir$ and $i/r$ for some $r>0$. The transformationbelongs to $PSL(2,\mathbb{R})$ and exchanges $ir$ with $i/r$. Conjugating it back gives an orientation-preserving isometry exchanging $P$ and $Q$.

For uniqueness, the quotient of two such isometries fixes both $P$ and $Q$. An orientation-preserving hyperbolic isometry fixing two distinct interior points fixes their connecting geodesic and both tangent directions there, hence is the identity. Therefore the exchanging isometry is unique; geometrically it is the hyperbolic half-turn about the midpoint of the segment $PQ$.

Use an orientation-preserving isometry to send $P$ to $i$ and $\ell$ to the imaginary axis, whose boundary points are $0,\infty$. A geodesic $m$ through $i$ meeting it at angle $\theta$ has boundary pointsIndeed $uv=-1$, so the semicircle with endpoints $u,v$ passes through $i$, and its tangent there makes angle $\theta$ with the vertical.

The cross-ratio is invariant under the normalizing Möbius transformation. With the suitable ordering $(u,v,\infty ,0)$ and the convention used here,