Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 11E (Geometry)

Letbe reflection in the unit circle, and represent a Möbius transformation $g$ byA direct calculation shows that $JgJ$ is represented byThus $gJ=Jg$ exactly when $M$ and $\tau (M)$ differ by a nonzero scalar. Applying the conjugate-linear involution $\tau$ twice shows that this scalar has modulus one. Rescaling $M$ by a suitable complex scalar then makes $M=\tau (M)$. Consequently all commuting maps, and only those maps, have the formThis is the Möbius maps commuting with reflection in the unit circle classification.

For such a map,It follows that $|g(z)|<1$ whenever $|z|<1$ precisely whenThe inverse has the same property, so these and only these maps preserve the unit disc.

The Poincare disc model isIts geodesics through the origin are the Euclidean diameters. To see directly that the radial segment from $0$ to $r{e}^{i{\theta }_0}$ minimizes length, write an arbitrary joining curve as $z(t)=r(t)e^{i\theta (t)}$. Its hyperbolic length satisfiesThe radial segment has constant $\theta$ and monotone $r$, so equality holds. Henceand every radial diameter is length minimizing. Rotational symmetry and uniqueness for the geodesic equation show that these are all geodesics through $0$.

Given any geodesic and a point $p$ on it, a disc-preserving maptakes $p$ to the origin. By part (a), $T$ is a hyperbolic isometry and commutes with $J(z)=1/\overline{z}$. The transformed geodesic is a diameter, hence a generalized circle invariant under $J$. Its inverse image is therefore also a Generalized circle under a Möbius transformation invariant under $J$. Thus every hyperbolic geodesic is the part in $\mathbb{D}$ of a Euclidean line or circle preserved by reflection in the unit circle; equivalently, it is a diameter or a circle orthogonal to the unit circle. This is the Geodesics of the Poincare disc description.

Rotate the disc so that $P=r\in (0,1)$ lies on the positive real axis. From part (b), the perpendicular hyperbolic line $\ell$ is a Euclidean circle orthogonal to the unit circle, with centre $c>1$ on the real axis and radius $R=c-r$. Orthogonality of the two circles givesCombining the two equations yields

The radial distance formula from part (b) saysThe hyperbolic double-angle identities now giveThereforeThis is the Euclidean circle representing a perpendicular hyperbolic line.

Use the hyperboloid model with Minkowski inner productPut the vertex with angle $\theta$ at $v=(1,0,0)$, and place the endpoints of its adjacent sides of lengths $a$ and $b$ atThe geodesic through $P$ perpendicular to $vP$ has spacelike unit normalwhile the geodesic through $Q$ perpendicular to $vQ$ has spacelike unit normal

These two geodesics form the remaining two sides of the quadrilateral. Their angle equals the angle between their normals in the tangent plane at their intersection. The third right angle therefore givesDividing by $\cosh a\cosh b$ proves the Lambert quadrilateral identity