reading my course I stumbled upon this fact:
Let $f\in\text{Isom}(\mathbb{H}^n)$ be an hyperbolic function (exactly 2 fixed points on $\partial\mathbb{H}^n$ and no fixed point in the interior). Suppose these fixed points are $p$ and $q$ and let $\ell$ be the unique geodesic from $p$ to $q$.
Then the geodesic is preserved under the action of $f$, i.e. $f(\ell)=\ell$.
This seems intuitive but yet I cannot prove it.
I class the only justifications were:
- Isometries preserve angles.
- Since $p$ and $q$ on the border are fixed so is the geodesic from one to the other.
But how can we show that if $x\in\ell$ then $f(x)\in\ell$. I cannot find a way to tackle this problem.
Edit:
What I tried since: Let's work in the upper half-space $\mathcal{U}^n$ where the distance is given by
$$
\cosh^{-1}(d_{\mathcal{U}^n}(x,y))=1+\frac{\|x-y\|^2}{2\cdot x_n\cdot y_n}\qquad x,y\in\mathcal{U}^n
$$
Then since $f$ has to be such that $f(0)=0$ and $f(\infty)=\infty$ and has to preserve the distance, we can (can we ?) say that $f$ is of the form:
$$
f(z)=\lambda z
$$
If this correct then $x\in\ell\implies x=a\cdot e_n$.
So $f(x)=\lambda a \cdot e_n \in\ell$. cqfd?
Edit n°2:
I found something else. Assuming the same conditions that in previous edit. Then since $f(\infty)=\infty$, $f\in\text{Stab}(\infty)$ and we showed that $\text{Stab}(\infty)\simeq\text{Sim}(\mathbb{R}^{n-1})$.
So we can in a way say that $f$ is in the similarity group of $\mathbb{R}^{n-1}$ but every $\psi\in\text{Sim}(\mathbb{R}^{n-1})$ can be uniquely written as
$$
\psi=\lambda\cdot g+ v\qquad\lambda\in\mathbb{R}_{>0},g\in O(n-1),v\in\mathbb{R}^{n-1}
$$
Since $f(0)=0$ then here $v=0$ and since it preserves $+\infty$ we can argue that $g=id$. Hence $f(z)=\lambda z$ ?
I'm sure there's an easier way out there, I would be thankfull if someone could point it out to me :)
