Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 14B (Geometry)

In the upper half-plane model, hyperbolic lines are vertical Euclidean rays and semicircles centred on the real axis, equivalently circle arcs meeting the real boundary orthogonally. A semicircle with endpoints $r<s$ is mapped to the imaginary axis byIts determinant is one, and its real endpoints map to $0,\infty$. A vertical line $x=a$ is mapped to the imaginary axis by $z\mapsto z-a$. Thus every hyperbolic line can be carried to a fixed one; composing one such map with the inverse of another proves transitivity of $\mathrm{PSL}(2,\mathbb{R})$ on the lines.

Reflection in $x=a$ is $R_a(z)=2a-\overline{z}$. Reflection in the unit semicircle is $R_{\circ }(z)=1/\overline{z}$. The latter fixes the semicircle pointwise, is an involution, and preserves the hyperbolic metric because $|dw|=|dz|/|z{|}^2$ and $\mathrm{Im}w=\mathrm{Im}z/|z{|}^2$. The vertical reflection similarly preserves $|dz|/\mathrm{Im}z$. Their composition in the specified order is

Parallel hyperbolic lines have one common ideal endpoint and no interior intersection. Move the common endpoint to infinity by a Möbius transformation. The two lines are then distinct vertical lines $x=a,x=b$. Their hyperbolic reflections have productwhose $m$th power is $z+2m(a-b)$. Since $a\ne b$, no positive power is the identity.

Ultraparallel hyperbolic lines have no common endpoint or intersection. To see their common perpendicular explicitly, first map one line to the imaginary axis. The other has two boundary endpoints on the same side, say $0<r<s$ after possibly reflecting. It is a semicircle of centre $(r+s)/2$ and radius $(s-r)/2$. The circle centred at zero with radius $\sqrt{rs}$ meets it orthogonally, since $((r+s)/2{)}^2=rs+((s-r)/2{)}^2$, and also meets the imaginary axis orthogonally. It supplies the common perpendicular. Map that perpendicular to the imaginary axis. Orthogonality then makes both lines semicircles centred at zero, of distinct radii $r,s$. This normal form can also be obtained by a real Möbius transformation carrying the endpoints of the common perpendicular to $0,\infty$. Their reflections are $z\mapsto r^2/\overline{z}$ and $z\mapsto s^2/\overline{z}$, so their product isNo positive power of this dilation is the identity because $r^2/s^2>0$ and $r\ne s$. Thus the product has infinite order in both nonintersecting cases. Geometrically the ultraparallel product is a translation along the common perpendicular through twice the separation of the lines.

If two hyperbolic lines intersect, move their intersection to the centre of the Poincaré disk model. Lines through the centre are diameters. Reflection in a diameter making Euclidean angle $\alpha$ with the real axis is $z\mapsto e^{2i\alpha }\overline{z}$. The composition of two such reflections is the rotation $z\mapsto e^{2i(\alpha -\beta )}z$. The Poincaré disk model is conformal, so the Euclidean angle difference is the hyperbolic intersection angle $\theta$, up to sign and replacing it by $\pi -\theta$, neither of which changes rationality of $\theta /\pi$.

The rotation has finite order exactly when $m\theta \in \pi \mathbb{Z}$ for some positive integer $m$, equivalently $\theta /\pi \in \mathbb{Q}$. If $\theta /\pi =r/s$ in lowest terms, its order is $s$. Conversely, the nonintersecting cases in part(a) have infinite order. Therefore