Similarity transformations in $H^n$

Question

Why are similarity transformations necessarily isometries in $H^n$. Note that a similarity transformation is a bijective map $f:H^n\rightarrow H^n$ such that $d_{H^n}(f(x),f(y)))=kd_{H^n}(x,y)$, where $k>0$. Intuitively I can see why because of the distortion when you scale in $H^n$, but what is a rigorous proof of this. I would really appreciate if something could give a detailed explanation. Thanks.

Answer 1

The fact that a similarity is an isometry can be seen by application of the Gauss-Bonnet theorem, which says that the area of any triangle $T=\triangle(a,b,c) \in \mathbb{H}^2$ is equal to $2\pi$ minus the total exterior angle: $$\text{Area}(T) = 2 \pi - \biggl((\pi - \angle a) + (\pi - \angle b) + (\pi - \angle c)\biggl) $$

So suppose now that $f : \mathbb{H}^n \to \mathbb{H}^n$ satisfies the similarity identity $$d(f(x),f(y)) = kd(x,y) $$ for some constant $k>0$. Choose any isometrically embedded hyperbolic plane $\mathbb H^2 \hookrightarrow \mathbb H^n$ and any geodesic triangle $T \subset \mathbb H^2$. It follows that $f(T) \subset f(\mathbb H^2)$ is a geodesic triangle in an isometrically embedded hyperbolic plane $f(\mathbb H^2) \hookrightarrow \mathbb H^n$ and that $$\text{Area}(f(T)) = k^2 \text{Area}(T) $$ Similarities preserve angles, and so from the Gauss-Bonnet theorem it follows that $\text{Area}(f(T))=\text{Area}(T)$, implying that $k=1$.