I was reading Bridson and Haefligers book Metric Spaces of Non-Positive Curvature and was having trouble understand a proof of a proposition.
In Chapter II, Proposition 6.7, it is stated that
Let $X$ be a complete CAT(0) space and $\gamma$ be an isometry of $X$. Then $\gamma$ is elliptic if and only if $\gamma$ has a bounded orbit.
They say this an easy consequence of the fact ($\S$2 Proposition 2.7):
If $Y\subseteq X$ is a bounded set, then there is a unique point $c_Y\in X$ such that $Y\subseteq \overline{B}(c_Y,r_Y)$.
Above, $r_Y$ is the radius of $Y$, that is, the infimum of the positive numbers $r$ such that $Y\subseteq B(x,r)$ for some $x\in X$. Also, an isometry is defined to be elliptic if it has a fixed point.
I just can't seem to put the two together... Does an isometry having a fixed point automatically make it have a bounded orbit? In the case it does have a bounded orbit, do I have to use Brower's fixed point theorem?
Any hints would be very much appreciated! Thank you!
If $x_0$ is a fixed point of the isometry $\gamma$, $\{x_0\}$ is the orbit of $x_0$, and is bounded. So yes, $\gamma$ has a bounded orbit. In fact, we have more than that: $d(\gamma^n\cdot y,x_0) = d(\gamma^n\cdot y,\gamma^n \cdot x_0)) = d(y,x_0)$ for $y\in X$ and $n\in \Bbb Z$, and the orbit of $y$ lies in the sphere $S(x_0,d(y,x_0))$. It follows that any orbit is bounded.
No fancy result to use here. Let $Y$ be a bounded orbit and $c_Y$ its centre. For $y\in Y$, it follows from $\gamma$ being an isometry that $d(\gamma\cdot c_Y, y) = d(c_Y,\gamma^{-1}\cdot y)\leqslant d(c_Y,Y) \leqslant r_Y$. Hence, $Y \subset \bar{B}(\gamma\cdot c_Y,r_Y)$. By uniqueness of the centre, $\gamma\cdot c_Y=c_Y$, and $\gamma$ has a fixed point, i.e. $\gamma$ is elliptic.