Prove that isometries of the hyperbolic plane map horocycles to horocycles

Question

Working with the half-plane Poincaré model, I need to prove that isometries of the hyperbolic plane map horocycles to horocycles.

The definition I've been given of horocycle is that of a curve that's perpendicular to every geodesic in an "ultraparallel family". An "ultraparallel family" is the set of all the geodesics that have a point $(x,0)$ as one of their extremes.

I've tried using the knowledge of the form of the isometries in this model, which is

$$ \varphi(z) = \frac{az+b}{cz+d}, $$

but I haven't reached anything interesting. Up until now I had worked just with the disk model, so I'm at a loss. Thanks in advance for any help.

Answer 1

Having a more explicit description of horocycles helps here. In the disk model, horocycles are exactly the circles on the Riemann sphere contained in $\mathbb{D} = \{ z \mid |z| < 1\}$ and which are tangent to the circle $S = \{ z \mid |z| = 1\}$.

The isometries of $\mathbb{D}$ in this model are the Möbius transformations that send $\mathbb{D}$ and $S$ to themselves. It is a standard lemma in complex analysis is that Möbius transformations map circles on the Riemann sphere to circles on the Riemann sphere. So under a hyperbolic isometry, any horocycle is mapped to a circle on the Riemann sphere contained in $\mathbb{D}$ and tangent to $S$, hence another horocycle.

An addendum: with this definition of horocycles, a proof is to observe (a) ultraparallel families of geodesics are invariant under hyperbolic isometries, and (b) the property of being everywhere perpendicular to an ultraparallel family of geodesics is invariant under isometries. The first point follows from the description of isometries as Möbius transformations, the second from angle-preservation of isometries.