Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 11E (Geometry)

The disc model isand the upper half-plane model isThe Cayley mapmaps $D$ bijectively to $\mathfrak{h}$. Since $C^{\prime }(z)=2i/(1-z{)}^2$ and $\mathrm{Im}C(z)=(1-|z{|}^2)/|1-z{|}^2$, direct substitution gives $C^{\ast }{g}_{\mathfrak{h}}=g_D$.

Representing $C$ by $K=\left(\begin{array}{cc}i& i\\ -1& 1\end{array}\right)$, the disc isometry corresponding to $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is represented, up to a nonzero scalar, by

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.