TITLE: Similarity transformations in $H^n$
QUESTION [0 upvotes]: 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.

REPLY [2 votes]: 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 if $f : \mathbb{H}^2 \to \mathbb{H}^2$ satisfies the similarity identity 
$$d(f(x),f(y)) = kd(x,y)
$$
for some constant $k>0$ then $f(T)$ is a triangle and 
$$\text{Area}(f(T)) = k^2 \text{Area}(T)
$$
Similarities preserve angles, and so from the Gauss-Bonnet theorem it follows that $\text{Area}(f(T))=\text{Area}(T)$, implying that $k=1$.
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