I have a project I have to do. In order to do it I need to investigate this book.
In page 94 they defined hyperbolic isometry on a metric s.t it possesses no fixed point in the tree.
After that they claim that such isomtery possesses 2 fixed point in the boundary of a tree.
How could it be ? How it possesses no fixed point in the tree but 2 fixed point in the boundary of a tree ? Is it a typo ?
were they meant to elliptic isometry ? I having a little difficult to understand the proof
There is no typo, a hyperbolic isometry on a tree fixes two points at infinity. For instance, you may think about the unit translation on the real line: it has no fixed point on the real line itself, but it fixes the two points at infinity.
Some hints to prove that a hyperbolic isometry $g$ on a tree $T$ fixes exactly two points at infinity:
Let $x \in T$ satisfying $d(x,gx)= \inf \{ d(y,gy) \mid y \in T \}$, and let $\gamma_0$ be a geodesic from $x$ to $gx$. Prove that $[g]=\bigcup\limits_{n \in \mathbb{Z}} g^n \cdot \gamma_0$ defines a bi-infinite geodesic on which $g$ acts by translation.
Deduce that $g$ fixes at least two points at infinity.
Let $\xi \in \partial T$ be a point at infinity, and let $\gamma$ be geodesic ray from $x$ to $\xi$. Then $g \cdot \xi$ is the point at infinity associated to the infinite ray $g \cdot \gamma$. Prove that $g \cdot \xi \neq \xi$ if $\gamma$ is not included into $[g]$.
Conclude that the only points of $\partial T$ fixed by $g$ are the two endpoints of $[g]$.