Isometries in CAT(0) spaces

In connection with a course on metric geometry, I came across two interesting exercises about isometries in CAT(0) spaces. CAT(0) spaces are essentially generalization of manifolds with non-positive curvature, but without invoking any smoothness or curvature explicitely.

Definition

We say a (complete) metric space $(X,d)$ is a CAT(0) space if the following holds: \begin{equation*} \forall x \quad \forall y \quad \exists m \text{ such that } \forall z: \frac{d^2(x,z) + d^2(y,z)}{2}- \frac{d^2(x,y)}{4} \geq d^2(z,m). \end{equation*} It is not difficult to show that $m$ is the (unique) midpoint of $x$ and $y$, i.e. $2d(x,m)=2d(y,m)=d(x,y)$. In other words the inequality tells that the midpoint lies somehow closer to any $z$ than its average distance to the two endpoints.

Definition

A map $f:X \rightarrow X$ is called an isometry if for any $x,y\in X$ we have $d(x,y)=d(f(x),f(y))$. The collection of all isometries on $X$ is denoted by $Isom(X)$.

Definition

We now define the norm of an isometry $f$ as $|f|:= \inf_{x \in X} d(f(x),x)$.
When the infimum is not attained, the isometry is called parabolic, those are usually more difficult to deal with. If the infimum is attained and $|f|=0$, then the isometry is called elliptic, if $|f|>0$, then hyperbolic.
There is a very powerful theorem (which we will not prove) that elliptic isometries fix a point and hyperbolic isometries fix a geodesic.

Problem 1

Let $g,h \in Isom(X)$ commute and elliptic. Then there exists a common fixed point.

Since $g,h$ are elliptic, there exists $x_0,x_1$ (not necessarily equal!) such that $gx_0=x_0$ and $hx_1 = x_1$.
We want to show that there exists $z\in X$ such that $gz=z=hz$.
We set $A=\langle g, h \rangle = \{g^n h^m: n,m\in \mathbb{Z}\}$.
Thus we have the equivalence: $A$ fixes a point $z\in X$ $\Longleftrightarrow$ $gz=z$ and $hz =z$.
By the well known fixed point theorem: $A$ fixes a point $\Longleftrightarrow$ $A$ has a bounded orbit in $X$.
It suffices to show that the orbit $A x_1$ is bounded. Take $a\in A$, since $g$ and $h$ commute, without loss of generality, we can write $a=g^n h^m$ for some $n,m \in \mathbb{Z}$. Now \begin{equation*} \begin{split} d(ax_1,x_1) &= d(g^nh^m x_1, x_1) \leq d(g^nh^m x_1, x_0) + d(x_0,x_1) = d(g^nh^m x_1, x_0) + d(x_0,x_1) \\ &= d(g^nh^m x_1,g^n x_0) + d(x_0,x_1) = d(h^m x_1, x_0) + d(x_0,x_1) = d(x_1, x_0) + d(x_0,x_1) = 2d(x_0,x_1), \end{split} \end{equation*} where we use the triangle inequality, the fact that $g^n$ is an isometry in the 4th equality and $x_0,x_1$ are fixed points for $g^n$ and $h^m$ in the 3rd and 5th equality.
Hence the orbit $Ax_1$ is bounded, by interchanging $g,h$ and $x_0,x_1$ respectively, the orbit $Ax_0$ is bounded as well.

Problem 2

Let $g,h \in Isom(X)$ commute and hyperbolic. Then there exists a common fixed point in $\partial X$.

We can think of points in $\partial X$ as rays in $X$, quotiented out by the equivalence relation $\sim$ (i.e. $\sigma \sim \tau$ if $\sup_t(\sigma(t),\tau(t))<\infty$).
To prove the statement we want to show existence of a ray $\gamma$ such that $\gamma \sim g \gamma$ and $\gamma \sim h \gamma$, i.e. $\sup_t d(g \gamma(t), \gamma(t))<\infty$ and $\sup_t d(h \gamma(t), \gamma(t))<\infty$.

Claim: We can take $\gamma:[0,\infty) \rightarrow X$ defined as $\gamma(t) := \sigma(t)$, where $\sigma$ is the invariant geodesic of $g$ (that exists, since $g$ is hyperbolic) with $g\sigma(s) = \sigma(s+|g|)$.
Clearly $g\gamma \sim \gamma$, since $\sup_{t\geq 0} d(g\gamma(t), \gamma(t))=\sup_{t\geq 0} d(g\sigma(t), \sigma(t)) = d(\gamma(t+|g|),\gamma(t)) = |g|\leq \infty$. It remains to show that $h \gamma \sim \gamma$.
Using that $g^k$ is an isometry for $k \in \mathbb{N}$, we note \begin{equation*} d(\gamma(0), h \gamma(0)) =d(\sigma(0), h \sigma(0))=d(g^k\sigma(0), g^kh \sigma(0)) = d(g^k\sigma(0), hg^k \sigma(0)) = d(\sigma(k|g|), h \sigma(k|g|)) \quad \forall k\in \mathbb{N}. \end{equation*} We now lift the result that the distance stays constant for points of the form $t=|g|\mathbb{N}$ to all positive $t$.
Let $M:= \max_{t\in [0,|g|]}d(\sigma(t),h \sigma(t))$, which is finite, since $[0,|g|]$ is compact and $d(\cdot,\cdot)$ continuous.
Thus for any $t\geq0$, $d(\sigma(t), h\sigma(t))\leq M$, since $t=N |g|+ t'$ with $t'\in [0,|g|]$ and $N \in \mathbb{N}$.