Hyperbolic geometry - image of horoball

Question

This is a pretty basic question I suppose, but I want to check my intuition in hyperbolic geometry which is a field I've just started studying.

Let $A$ be a horoball in the hyperbolic disc $B^N$($=\mathbb{H}^N$), with tangent point $p\in S^N$. let $g$ be a conformal (i.e. oriention preserving) isometry of $B^N$.

I know that $g$ transforms hyperbolic balls to hyperbolic balls.

1. From that I believe I can deduce that $g$ also transforms horoballs to horoballs, right? Because horoball is after all a euclidean ball and therefore a hyperbolic ball.

2. In particular, if $g(p)=p$ then $g(A)=A$, right?

3. In the latter case, my intuition says that $g$ must be some sort of clockwise or anti-clockwise rotation towards the base point $p$. This is not very clear to me in that multidimensional case. For simplicity I am thinking of $\mathbb D=\mathbb B^2$ and a horoball at $(1,0)$ which is just a circle tangent at the point $(1,0)$. So I'm thinking that $g$ must be some flow in the direction of this cicle (just moving each point in the clockwise direction, each point in the correct amount so the orientation would be preserved). in particular $|g(x)|=|g^{-1}(x)|$ for $x$ the antipodal point to $(1,0)$ on this horocicle. Is this true at all?

Answer 1

I'm going to give an intrinsic/synthetic answer this question, i.e. an answer that not depend on any model of $\mathbb{H}^n$ (such as the upper half space model), but which instead uses only the intrinsic metric properties of $\mathbb{H}^n$.

Balls and horoballs in $\mathbb{H}^n$ each may be characterized by a metrically invariant property.

For balls, the metrically invariant property is simply the definition: $B \subset \mathbb{H}^n$ is a ball if and only if there exists $p \in \mathbb{H}^n$ and $r > 0$ such that $$B = \{q \in \mathbb{H}^n \,\bigm|\, d(p,q) \le r\} $$ This ball $B$ is denoted $B(q,r)$. (I'm doing closed balls, of course). From the fact that this definition is evidently invariant under isometries, it follows that if $B$ is a ball and $g$ is an isometry then $g(B)$ is a ball.

So far, so good.

In order to characterize horoballs by a metrically invariant property, we will need to use the above characterization of balls, but we'll need one more preliminary thing, namely a metrically invariant property that characterizes rays: a subset $R \subset \mathbb{H}^n$ is a ray if and only if there exists a finite geodesic segment $[a,b] \subset \mathbb{H}^n$ such that $R$ is the union of all finite geodesic segments $[a,c]$ such that $[a,b] \subset [a,c]$. This ray $R$ will be denoted $\vec{ab}$.

Now we're ready for the metrically invariant property that characterizes horoballs: a subset $H \subset \mathbb{H}^n$ is a horoball if and only if there exists a ray $\vec{ab}$ such that $H$ is the closure of the union of all balls $B = B(q,r)$ such that $q \in \vec{ab} - \{a\}$ and $r = d(a,q)$.

To summarize, since horoballs are characterized by a metrically invariant property, for every horoball $H$ and every isometry $g$ the image $g(H)$ is also a horoball.

Answer 2

Better say "center" than "origin".

$1.$ Yes and the action is transitive.

$2/3.$ For example, what happens if $p=\infty$ and $g(z)=z+i$?