4 Amenability and non-positive curvature

4.1 CAT(0) spaces

Sectional curvature is defined for Riemannian manifolds. It was realized by Toponogov and Alexandrov that the sign of this curvature (for example non-negatvie) gives strong local properties for the distance associated to the Riemannian metric and actually that non-positive sectional curvature can be defined only with one inequality for the distance (if the space is simply connected). This is the notion of CAT(0) spaces. One can define more generally CAT($\kappa$) for $\kappa\leq0$. Intuitively, those are spaces with curvature bounded above by $\kappa$.

In a metric space $(X,d)$, a geodesic segment is the image of some isometric map $f\colon I\to X$ ($d(f(u),f(v))=|u-v|$ for all $u,v\in I$) where $I$ is a compact interval of $\mathbb{R}$. A metric space $(X,d)$ is geodesic if any two points are extremities of a geodesic segment. For two points $x,y\in X$, a midpoint is a point $m\in X$ such that $d(x,m)=d(y,m)=\frac{1}{2}d(x,y)$.

Definition 4.1 A CAT(0) space is complete metric space $X$ such that is

Geodesic.
For any $x,y,z\in X$ and $m$ midpoint of a geodesic segment between $x$ and $z$,

\[d(z,m)^2\leq 1/2(d(z,x)^2+d(z,y)^2)-1/4d(x,y)^2\]

Remark. If one replaces the less or equal sign by an equality sign, this is the classical parallelogram identity in $\mathbf{R}^2$.

For three points $x,y,z$ in a geodesic metric space $X$ a geodesic triangle $\Delta(x,y,z)$ is the union of three geodesic segments $[x,y], [y,z]$ and $[z,x]$. A comparison triangle $\Delta(\overline{x},\overline{y},\overline{z})$ is a triangle in $\mathbf{R}^2$ such that $d(\overline{x},\overline{y})=d(x,y)$, $d(\overline{x},\overline{z})=d(x,z)$ and $d(\overline{z},\overline{y})=d(z,y)$. For a point $p$ on a side of $\Delta(x,y,z)$, we denote by $\overline{p}$ the corresponding point on $\Delta(\overline{x},\overline{y},\overline{z})$

Proposition 4.1 Let $X$ a complete geodesic metric space. The following are equivalent:

$X$ is CAT(0).
For any geodesic triangle $\Delta(x,y,z)$ with comparison triangle $\Delta(\overline{x},\overline{y},\overline{z})$, for $p,q$ on sides of $\Delta$,

\[ d(p,q)\leq d(\overline{p},\overline{q}).\]

Proof. Let’s assume $X$ is CAT(0). The CAT(0) inequality means exactly that for $m$ midpoint of $x,y$ on geodesic segment $[x,y]$ between $x$ and $y$, we have $d(z,m)\leq d(\overline{z},\overline{m})$, repeating the argument, we have that for any point $p\in[x,y]$ with $d(x,p)=k/2^nd(x,y)$ with $k,n\in \mathbf{N}$, $d(z,p)\leq d(\overline{z},\overline{p})$. By continuity of the distance, for any point $p\in[x,y]$, $d(z,p)\leq d(\overline{z},\overline{p})$. Now, if $q\in[x,z]$, applying the above argument to the geodesic triangle $\Delta(x,p,z)$, we have for $q$d(p,q)\leq d(\overline{p},\overline{q})$.

Conversely, take $x,y,z\in X$ and $m$ midpoint of $x$ and $y$. Using a comparison triangle, we have $d(z,m)^2\leq d(\overline{z},\overline{m})^2= 1/2(d(\overline{z},\overline{x})^2+d(\overline{z},\overline{y})^2)-1/4d(\overline{x},\overline{y})^2.$

Lemma 4.1 In a CAT(0) space, geodesic segments are unique.

Proof. Let $S_1$, $S_2$ be two geodesic segments between two points $x,y$. Let $m_1$, $m_2$ be the midpoints on these segments. By the CAT(0) inequality, $m_1=m_2$. Repeating the argument for half segments, we see that $S_1$ and $S_2$ coincide on points with dyadic distance to $x$ and by continuity, they coincide for all points.

Lemma 4.2 Any CAT(0) space is contractible.

Proof. Let’s fix $x_0\in X$. For $x\in X$, we set $f_x\colon[0,1]\to[x_0, x]$, the constant speed parametrization of the geodesic segment $[x_0, x]$ such that $f_x(t)$ is the point at $(1-t)d(x_0,x)$ from $x_0$.

We define

$f\colon X\times t\to X$ by $f(x,t)=f_x(t)$. This is a deformation retraction from $X$ to $\{x_0\}$.

Example 4.1 Here are few examples:

Euclidean spaces and Hilbert spaces
trees with their geodesic distances
Any simply connected Riemannian manifold of non-positive sectional curvature
The space of definite positive matrices

Let $(X,d)$. For a bounded set $Y\subset X$, the circumradius of $Y$ is $r_0=\inf\{r>0,\ \exists x\in X \textrm{with} Y\subset B(x,r) \}$. A circumcenter is a point $x\in X$ such that $Y\subset \overline{B}(x,r_0)$.

Proposition 4.2 In a CAT(0) space, any bounded subset has a unique circumcenter.

Proof. Let $Y$ be a bounded set of a CAT(0) space. Let $(c_n,r_n)$ such that $Y\subset B(c_n,r_n)$ and $\lim r_n=r_0$. For any $n,m\in\mathbf{N}$ and $\varepsilon >0$, there is $y\in Y$ such that $d(\mu_{m,n},y)^2\geq r_0^2-\varepsilon$. By the CAT(0) inequality, we get that

\[d(c_n,c_m)^2\leq 4\left(\frac{r_n^2+r_m^2}{2}-r_0^2+\varepsilon\right)\] and thus $d(c_n,c_m)^2\leq 2\varepsilon$ for $n,m$ large enough. So this is a Cauchy sequence and by complete, it has a limit $c$. By passing to the limit, for any $y\in Y$, $d(c,y)\leq r_0$ and thus $c_0$ is a circumcenter. Uniqueness, follows again by applying the CAT(0) to the midpoint of two possible circumcenters.

Theorem 4.1 (Cartan fixed point theorem) Let $X$ be a CAT(0) space and $G$ be a group of isometries of $X$. If $G$ has a bounded orbit then $G$ has a fixed point.

Proof. Let $Y$ be a bounded orbit and $c$ its circumcenter. Since $g(Y)=Y$, for any $g\in G$ and $c$ is unique, we have that $g(c)=c$ for any $g\in G$. So we have a fixed point.

Corollary 4.1 Let $G$ be a bounded subgroup of $GL_n(\mathbf{R})$ then $G$ is conjugated to the subgroup of orthogonal transformations.

Proof. The group $GL_n(\mathbf{R})$ acts on the space of positive definite matrices by isometries via the formula $g\cdot M=^tg^{-1}Mg^{-1}$. This now a consequence of the above theorem.

Definition 4.2 Let $X$ be a CAT(0) space. A subset $C\subset X$ is convex if for $x,y\in C$, $[x,y]\subset C$.

Remark. For closed subspaces $C$, it suffices to prove that for any $x,y\in C$, the midpoint of $[x,y]$ belongs to $C$.

Proposition 4.3 In a CAT(0) space, balls are convex.

Proof. It suffices to prove the result for closed balls. For a closed ball $B$ of center $z$ and two points $x,y\in B$, the fact that the midpoint between $x$ and $y$ belongs to $B$ follows directly from the CAT(0) inequality.

Proposition 4.4 Let $X$ be a CAT(0) space. The distance function is convex, that if $c,c'\colon[0,1]\to X$ are two parametrizations of geodesic segments with constant speed then for all $t\in[0,1]$,

\[d(c(t),c'(t))\leq (1-t)d(c(0),c'(0))+td(c(1),c'(1)).\]

Proposition 4.5 Let $C$ be a closed convex subspace of a CAT(0) space. Then for $x\in X$, there is a unique $p\in C$ such that $d(x,p)=\inf_{c\in C} d(x,c)$. This point is called the projection of $x$ to $C$.

Proof. Let $(p_n)$ be a sequence of points in $C$ such that $d(x,p_n)\to\inf_{c\in C} d(x,c)=\ell$. For $n,m$, let $\mu_{m,n}$ be the midpoints of $p_n$ and $p_m$. Since $d(x,\mu_{m,n})\geq\ell$, we have $d(p_n,p_m)^2\leq 4\left(\frac{d(x,p_n)^2+d(x,p_m)^2}{2}-\ell^2\right)\to 0$. So $(p_n)$ is a Cauchy sequence. Its limit $p$ satisfies $d(x,p)=\inf_{c\in C} d(x,c)$. Such a point is unique by applying the CAT(0) inequality again.

Example 4.2 Let $C$ be the closed ball $\overline{B}(x_0,r)$ for some point $x$ in a CAT(0) space. Let us denote by $\pi(y)$ the projection on $C$ of a point $y\in X$. If $y\in \overline{B}(x_0,r)$ then $\pi(y)=y$ and otherwise $\pi(y)$ is the point on $[x_0,y]$ at distance $r$ from $x_0$.

4.2 The boundary at infinity

Definition 4.3 Let $X$ be a CAT(0) space. A geodesic ray is (the image of) an isometric embedding $c\colon\mathbf{R}_+\to X$. Two geodesic rays are asymptotic if they are at bounded Hausdorff distance one from another.

The boundary at infinity of $X$ is the set of equivalence classes of geodesic rays.

We often $c(\infty)$ or $\xi$ for a point in this boundary at infinity. We denote $\partial X$ for the boundary at infinity.

Example 4.3 The boundary at infinity of the euclidean space $\mathbf{R}^n$ is the set of parallel classes of half lines. This is a sphere at infinity.

Proposition 4.6 Let $X$ be a CAT(0) space, $x\in X$ and $\xi\in\partial X$. Then there is a unique geodesic ray $c\colon\mathbf{R}_+\to X$ such that $c(\infty)=\xi$ and $c(0)=x$.

Proof. Uniqueness follows from convexity of the metric.

Let $y=c(0)$ and let $a=d(x,y)$. We denote by $\sigma_t(s)$ the point on $[x,c(t)]$ at distance $s$ from $x$. We claim that $\sigma_t(s)$ is convergent to some $\sigma(s)\in X$ for fixed $s$ and $t\to\infty$ and $\sigma\colon\mathbf{R}^+$ is geodesic ray asymptotic to $c$.

The triangle inequality gives

\[t-a\leq d(x,c(t))\leq t+a.\]

Consider a comparison triangle $\Delta(\overline{x},\overline{c(t)},\overline{c(t')})$. Let $\alpha$ be the angle at $\overline{x}$. By the law of cosines we have

\[\cos(\alpha)=\frac{d(x,c(t))^2+d(x,c(t'))^2-d(c(t),c(t'))^2}{2d(x,c(t))d(x,c(t'))}\geq \frac{(t-a)^2+(t'-a)^2-(t-t')^2}{2(t+a)(t'+a)}.\]

Thus $\cos(\alpha)\to 1$ when $t,t'\to\infty$. Which means that $\alpha\to0$. So $d(\overline{\sigma_t(s)},\overline{\sigma_{t'}(s)})\to 0$ and thus $d(\sigma_t(s),\sigma_{t'}(s))\to 0$ for $t,t'\to\infty$. By completeness, there is a limit $\sigma(s)$ which satisfies $d(\sigma(s),\sigma(s'))=|s-s'|$ and $d(\sigma(s),c(\mathbf{R}^+))\leq a$.So, $\sigma$ and $c$ are asymptotic rays.

Definition 4.4 Let $(X_t)$ for $t\in\mathbf{R}_+$ be a collection of topological spaces with continuous maps $\pi_{r,s}\colon X_r\to X_s$ for $r>s$ such that for any $r,s,t$ with $r>s>t$, $\pi_{r,t}=\pi_{r,s}\circ \pi_{s,t}$. The inverse limit of this system is

\[\left\{(x_t)\in\prod_{t\in\mathbf{R}_+}X_t,\ \pi_{s,t}(x_s)=x_t, \forall s>t \right\}.\] Endowed with the induced topology from the product topology on $\prod_{t\in\mathbf{R}_+}X_t$.

Such an inverse limit is denoted $\underset{\leftarrow}{\lim}X_t$.

We denote $\overline{X}$ for $X\cup\partial X$. If we fix $x_0\in X$, we can identify any element of $\overline{X}$ with a unique geodesic segment or ray starting at $x_0$.

Let us fix some $x_0$ in a CAT(0) space $X$. For $s>r$, we define $\pi_{s,t}\colon \overline{B}(x_0,s)\to \overline{B}(x_0,r)$ to be the restriction of the projection map to $\overline{B}(x_0,r)$.

Lemma 4.3 There is a bijection between $\overline{X}$ and $\underset{\leftarrow}{\lim}\overline{B}(x_0,r)$.

Proof. To any point $y\in X$, we associate the collection $(y_r)_{r\geq0}$ where $y_r$ is the projection of $y$ on $\overline{B}(x_0,r)$. For $\xi\in\partial X$ we associate the collection $(c(r))_{r\geq0}$ where $c$ is the geodesic ray from $x_0$ to $\xi$ (parametrized with speed one). This defines an injective map $\overline{X}\to\underset{\leftarrow}{\lim}B(x_0,r)$.

Conversely, let $y=(y_r)_{r>0}$ be an element of $\underset{\leftarrow}{\lim}B(x_0,r)$. This means that for any $s>r$, $y_r=\pi_{s,r}(y_s)$. Assume that for some $s>r$, $y_s\neq y_r$, i.e. $y_s\notin \overline{B}(x_0,r)$. Then the map $t\mapsto y_t$ for $t\in [0,r]$ is an isometric parametrization of the segment $[x_0,x_r]$.

Let $t=\sup\{r>0,\ d(x_0,y_r)=r\}$. If $t=\infty$ then $r\mapsto y_r$ is a geodesic ray. If $t<\infty$, we have that for all $s,r>t$, $y_s=y_r$ and this point is necessarily $y_t$. So $y$ lies in the image of the map $X\to\underset{\leftarrow}{\lim}B(x_0,r)$.

Definition 4.5 The cone topology on $\overline{X}$ is the topology coming from the identification of $\overline{X}$ with $\underset{\leftarrow}{\lim}B(x_0,r)$.

Remark. This topology coincides with the metric topology on $X$ and a sequence $(x_n)$ of point of $X$ converges to $\xi\in\partial X$ if and only if for all $r>0$, the point at distance $r$ from $x_0$ on $[x_0,x_n]$ converges to the point at distance $r$ from $x_0$ on the geodesic ray from $x_0$ to $\xi$.

We say that a metric space is proper if closed balls are compact. In particular, it is locally compact. Recall that a compactification of a locally compact space $X$ is a compact topological space $Z$ such that there is a continuous bijection $i\colon X\to Z$ is a homeomorphism on its image such that $i(X)$ is an open dense subset of $Z$.

Proposition 4.7 Let $X$ be a proper CAT(0) space then $\overline{X}$ is a compactification of $X$.

Proof. Since $\overline{B}(x_0,r)$ is compact and Hausdorff for all $r\geq0$, the product $\prod_{r\geq0}\overline{B}(x_0,r)$ is compact Hausdorff and thus it suffices to prove that the inverse limit is closed which follows from the continuity of the projection maps $\pi_{s,r}$ (there are 1-Lipschitz).

Remark. The cone topology does not depend on the choice of base point $x_0$.

Recall that an isometry of a metric space $(X,d)$ is a bijection $g\colon X\to X$ such that $d(gx,gy)=d(x,y)$ for all $x,y\in X$.

Lemma 4.4 Any isometry $g$ of a CAT(0) space extends uniquely to a homeomorphism of $\overline{X}$

4.3 Horofunctions and Busemann functions

Let $X$ be a CAT(0) space. We endow the space of continuous functions on $X$, $\mathcal{C}(X)$ with the topology of uniform convergence on bounded subsets of $X$. The space of constant functions is a closed 1 dimensional subspace of $\mathcal{C}(X)$ and we denote by $\mathcal{C}_*(X)$ for the quotient space with the quotient topology. It can be identified with the subspace of $\mathcal{C}(X)$ of functions vanishing at some $x_0\in X$.

We have the embedding $X\to \mathcal{C}_*(X)$ mapping $x$ to $y\mapsto d(x,y)$. We denote by $\hat{X}$ the closure of its image in $\mathcal{C}_*(X)$.

Definition 4.6 A (class of) function in $\hat{X}\setminus X$ is called a horofunction. A sublevel set is called a horoball and a level set is called a horosphere.

Definition 4.7 Let $X$ be a CAT(0) space and $\xi\in\partial X$, The Busemann function associated to $\xi$ and vanishing at $x_0$ is

\[\beta_{\xi}(x,x_0)=\lim_{t\to\infty}d(x,c(t))-t.\]

Example 4.4 In a Hilbert case, Busemann functions are in bijection with unit vectors (or the projective space over the dual space)

Remark. This limit exists and the convergence is uniform on bounded sets.

Theorem 4.2 In a CAT(0) space, any horofunction is given by some Busemann function.

Lemma 4.5 Let $X$ be a CAT(0) space. Let $r,\varepsilon>0$ and $x_O\in X$ then there is $R>0$ such that for any $z\in B(x_0,r)$, $x\notin B(x_0,R)$ and $y\in[x_0,x]$ with $d(x_0,y)=R$ then

\[d(z,y)+d(y,x)<d(z,x)+\varepsilon.\]

Lemma 4.6 Let $x_0,x,y$ be points in a CAT(0) space such that $y\in [x_0,x]$ with $d(x_0,y)=\rho>0$ If $z\in X$ satisfies $d(x,z)=\rho$ then $d(x,z)-d(x,y)\geq \frac{d(y,z)^2}{2\rho}$.

Proof. Let $x_n$ be a sequence such $y\mapsto d(x_n,y)-d(x_n,x_0)$ converges uniformly on balls.If $(x_n)$ is bounded then $(x_n)$ converges to a point in $X$. Let’s consider the case where $x_n$ is unbounded. Let fix some $\rho>0$ and let $y_n\in[x_0,x_n]$ such that $d(x_0,y_n)=\rho$.Let’s fix $\varepsilon>0$. For $n,m$ large enough

\[|d(x_n,z)-d(x_n,x_0)-d(x_m,z)+d(x_m,x_0)|<\varepsilon\] for all $z\in B(x_0,\rho)$

Proposition 4.8 Let $h$ be a function on $X$ then $h$ is a horofunction if and only if

$h$ is convex
$h$ is 1-Lipschitz
$h$ as a unique minimum on each ball $\overline{B}(x_0,r)$ achieved at some $y$ on the sphere such that $h(y)=h(x_0)-r$.

4.4 Behaviour of individual isometries

Definition 4.8 Let $X$ be a CAT(0) space and $g$ an isometry of $X$. We define its translation length

\[\ell(g)=\inf_{x\in X}d(gx,x).\] The isometry is said to be elliptic if this infimum is a minimum and $\ell(g)=0$, it is hyperbolic if this is a minimum and $\ell(g)>0$. It is parabolic otherwise, i.e. the infimum is not a minimum.

We denote by $\mathrm{Min}(g)=\{x\in X,\ d(gx,x)=\ell(g)\}$. It is empty if and only if $g$ is parabolic.

Remark. It follows readily from the definition that an isometry of a CAT(0) space $X$ is elliptic if and only if it fixes a point in $X$.

Definition 4.9 Let $g$ be a hyperbolic isometry of a CAT(0) space. An axis $L$ for $g$ is a geodesic line $L\subset X$ that is invariant and on which $g$ acts by a translation of length $\ell(g)$.

Example 4.5 The isometries of $\mathbf{R}^2$ are rotations, symmetries, translation and gliding symmetry. The two first possibilities are elliptic and the two other are hyperbolic.

Example 4.6 The upper half-space model of the hyperbolic plane. It is $\mathbf{H}^2=\{z\in\mathbf{C},\ \Im(z)>0\}$ and the metric is $\frac{dx^2+dy^2}{y^2}$. In this model, geodesic lines are vertical half-lines starting at the real axis and half-circles with centers on the real axis.

The group $\mathrm{SL}_2(\mathbf{R})$ acts by homographies preserving the metric: The action of the matrix $\begin{bmatrix} a& b\\ c&d\end{bmatrix}$ on $z$ is $\frac{az+b}{cz+d}$.

The stabilizer of $i$ is $\mathrm{SO}_2(\mathbf{R})$, which are in particular all elliptic elements.

The subgroup of positive diagonal matrices $\begin{bmatrix}\lambda&0\\0&\lambda^{-1}\end{bmatrix}$ preserves the geodesic line $\Re(Z)=0$. A parametrization of this line with speed is given by $(t\mapsto ie^t)$. The point $ie^t$ is mapped to $\lambda^2ie^t=ie^{t+2\ln(\lambda)}$. So $\begin{bmatrix}\lambda&0\\0&\lambda^{-1}\end{bmatrix}$ acts by translation on this line with translation length $2\ln(\lambda)$. Since the projection to an invariant closed convex subspace reduced the translation length, all the elements of this diagonal group, except the identity, are hyperbolic isometries.

Let us denote by $\infty$ the point in $\partial \mathbf{H}^2$ corresponding to vertical geodesic rays. The stabilizer of $\infty$ is given by matrices of the form $\begin{bmatrix}\lambda&t\\0&\lambda^{-1}\end{bmatrix}$. Let us consider the ones with $\lambda=1$. So $\left\{\begin{bmatrix}\lambda&t\\0&\lambda^{-1}\end{bmatrix}, t\in\mathbf{R}\right\}$ is a subgroup which stabilizes each horizontal line $\{z\in\mathbf{H}^2,\ \Im(z)=h\}$ for some $h>0$. A point $z\in\mathbf{H}^2$ is translated horizontally to $z+t$ and the distance $d(z,z+t)\to0$ when $\Im(z)\to+\infty$. So all this elements are parabolic.

Lemma 4.7 Let $g$ be a hyperbolic isometry of a CAT(0) space then $g$ has an axis. Moreover the two end points of any axis are fixed points at infinity for $g$.

Proof. Consider $x\in\mathrm{Min}(g)$ and $m$ the midpoint between $x$ and $gx$. Consider the triangle $\Delta(x,gx,g^2(x))$. By a comparison triangle argument, $d(gm,m)\leq 1/2 d(x,g^2x)\leq d(x,gx)=\ell(g)$ and thus by definition of $\ell(g)$, $d(gm,m)=d(gm,gx)+d(gx,m)$ which implies that $[x,gx]\cup[gx,g^2x]$ is a geodesic segment. Repeating the argument, one proves that $[g^{-n}x,g^{-n+1}x]\cup\dots\cup[g^{n-1}x,g^{n}x]$ is the geodesic segment $[g^{-n}x,g^nx]$. So $\cup_{n\in\mathbf{N}} [g^{-n}x,g^nx]$ is a $g$-invariant line on which $g$ acts by translation of length $\ell(g)$. The two end points of this line are invariant by construction.

Lemma 4.8 Let $X$ be a CAT(0) space and $X_n$ be a nested sequence of non-empty bounded closed convex subspaces of $X$. Then $\cap_{n\in\mathbf{N}}X_n\neq\emptyset$.

Proof. Let $X_n$ be such nested sequence of bounded closed convex subspaces with circumradius $r_n$ and circumcentre $c_n$. We claim that $c_n$ is a Cauchy sequence.

First, observe that $c_n\in X_n$ (otherwise be projecting $c_n$ on $X_n$, we decrease all the distances). Fix $m\geq n$. Since $c_n$ is not the circumcenter of $X_m$, there is $x\in X_m$ such that $d(x,c_n)>r_m$. Let $p$ be the midpoint of $[c_n,c_m]$. By the CAT(0) inequality for $\Delta(c_n,c_m,x)$, we have

\[d(c_n,c_m)^2\leq 2(r_n^2+r_m^2)-4r_m^2.\]

Since $r_n$ is positive and non-increasing, it is convergent and thus $d(c_n,c_m)\to0$ when $n,m\to\infty$. Let $c$ be the limit of $(c_n)$. Since $c_m\in X_n$ for any $m\geq n$, $c\in X_n$ for any $n$ and thus $c\in\cap_{n\in\mathbf{N}} X_n$

Lemma 4.9 Let $X$ be a proper CAT(0) space and $X_n$ be a nested sequence of closed convex subspaces of $X$ such that $\cap X_n=\emptyset$. Assume that $g$ is an isometry such that $gX_n=X_n$ for each $n$. Then there is $\xi\in\cap_{n\in \mathbf{n}\partial X_n}$ that is $g$-invariant.

Proof. By the previous lemma, we know that $d(x,X_n)\to\infty$ (otherwise consider $C_n$ the intersection of $X_n$ with some very large ball to get that a contradiction with $\cap X_n=\emptyset$).

Let’s fix $x\in X$ and let $x_n$ to be the projection of $x$ on $X_n$. Since $X$ is proper, $\overline{X}$ is compact and up to extraction, we way assume that $[x,x_n]$ converges to a geodesic ray $[x,\xi)$ with $\xi\in \partial X$. Since $x_m\in X_n$, for all $m\geq n$, $\xi\in\partial X_n$ for all $n$.

Let $g$ be an isometry such that $gX_n=X_n$ for each $n$. Let $x\in X$ and let $x_n$ be the projection of $x$ on $X_n$. Since $gX_n=X_n$, we have that $gx_n$ is the projection of $gx$ on $X_n$. Since the projection to a closed convex subspace is 1-Lipschitz, we have that $d(x_n,gx_n)\leq d(x,gx)$. This implies that the Hausdorff distance between $[x,\xi)$ and $[gx,g\xi)$ is bounded by $d(x,gx)$ and thus the two geodesic rays are asymptotic i.e. $g\xi=\xi$.

Proposition 4.9 Let $X$ be a proper CAT(0) space and $g\in\mathrm{Isom}(X)$ be a parabolic isometry then $g$ has a fixed point at infinity.

Proof. It suffices to apply the previous lemma to $X_n=\{x\in X,\ d(gx,x)\leq \ell(g)+1/n\}$.

Definition 4.10 A splitting of a CAT(0) space $X$ is a pair of CAT(0) spaces $Y,Z$ such that $X=Y\times Z$ with distance \[d((y_1,z_1),(y_2,z_2))^2=d(y_1,y_2)^2+d(z_1,z_2)^2.\]

Proposition 4.10 (Flat strip Theorem) Let $c,c'$ be two geodesic lines of a CAT(0) space $X$. If these lines are asymptotic, i.e. the Hausdorff distance between their images is finite, the closed convex hull of their images is isometric to $[0,d]\times\mathbf{R}$.

Morever, if $X$ is the union of geodesic lines asymptotic to a geodesic line $c(\mathbf{R})$ then $X$ splits as $Y\times\mathbf{R}$ where $Y$ is the preimage of $c(0)$ under the projection to $c(\mathbf{R})$.

Definition 4.11 Let $g$ be an isometry of a CAT(0) space. It is a Clifford translation if $\mathrm{Min}(g)=X$.

Theorem 4.3 If $g$ is a non-trivial Clifford translation of $X$, then $X$ splits as a product $X=Y\times \mathbf{R}$, and $g(y,t) =(y,t+\ell(g))$ for all $t\in\mathbf{R}$ and $y\in Y$.

The space $X$ splits in a unique way as $H\times Y$ where $H$ is a Hilbert space and $Y$ has no non-trivial Clifford translation.

Remark. The factor $H$ in the above splitting is called the Euclidean de Rham factor.

The uniquess in the above statement implies that each isometry $g$ of $X$ can be written $g(h,y)=(g_H(h),g_Y(y))$ where $g_H$ is an isometry of $H$ and $g_Y$ is an isometry of $Y$.

4.5 Adams-Ballman theorem

Definition 4.12 A function $f\colon X\to \mathbf{R}$ is said to be affine, if $f\circ c$ is affine for any geodesic parametrisation $c\colon [a,b]\to X$.

Let $\xi$ be a point at infinity of a CAT(0) space $X$. The point $\xi$ is said to flat if $\beta_\xi$ is affine.

Example 4.7 Any point of $\partial \mathbf{R}^n$ is flat.

Lemma 4.10 Let $x,y,z$ be points in a CAT(0) space $X$ and $\xi\in\partial X$ then

\[\beta_\xi(x,z)=\beta_\xi(x,y)+\beta_\xi(y,z).\]

Proof. Let $x_n$ be a sequence of points converging to $\xi\in \partial X$ then for any bounded subset $B\subset X$, and $x\in B$, $\lim_{n\to \infty}d(x,x_n)-d(x_0,x_n)=\beta_\xi(x,x_0)$ and the convergence is uniform on $B$. Take $B=\{x,y,z\}$.

Since $d(x,x_n)-d(z,x_n)=d(x,x_n)-d(y,x_n)+d(y,x_n)-d(z,x_n)$, passing to the limit we get the result.

Proposition 4.11 Let $X$ be a proper CAT(0) without Euclidean factor. Assume that $\mathrm{Isom}(X)$ acts minimally on $X$ then there is no flat points in $\partial X$.

Proof. Let $\xi$ be a flat point. Since the action is minimal, no sublevel set of $\beta_\xi$ is invariant. So for any $x\in X$, we may find $(g_n)$ with $g_n\in G$ such that $\beta_\xi(x,g_nx)\to \infty$ (otherwise the closed convex hull of the orbit of $x$ would remain in a sublevel set). Since $\beta_\xi$ is affine, $-\beta_\xi$ is convex, not bounded from below and satisfies the characterization of Busemman function given in Proposition 4.8. So there is $\xi'\in\partial X$ such that $-\beta_\xi(\cdot,x_0)=\beta_{\xi'}(\cdot,x_0)$. Let fix $x\in X$ and consider the geodesic rays $c,c'$ to $\xi$ and $\xi'$. We claim that the union of their images is a geodesic line.

Let $t,t'>0$,

\[\begin{align} \beta_\xi(c(t),c'(t'))&=\beta_\xi(c(t),x)+\beta_\xi(x,c'(t'))\\ &=\beta_\xi(c(t),x)-\beta_{\xi'}(x,c'(t'))\\ &=t+t' \end{align}\]

Since $\beta_\xi$ is 1-Lipschitz, this implies that $d(c(t),c'(t'))=t+t'$ and thus the union of their images is a geodesic $L_x$.

For any $x,y$, $L_x$ and $L_y$ are at bounded distance and thus by Proposition 4.10, $X$ splits as $\mathbf{R}\times Y$ that is a contradiction.

Lemma 4.11 Let $X$ be a proper CAT(0) space and $G$ a group acting without fixed point at infinity then there is $Y\subset X$ $G$-invariant closed convex subset on which $G$ acts minimally.

Proof. By Zorn lemma, it suffices to prove that any chain (for the inclusion) $\mathcal{Y}$ of $G$-invariant closed convex subsets of $X$ has a non-empty intersection.

If some ball $B(x_o,r)$ intersects all $Y\in\mathcal{Y}$ then by Lemma 4.8, the intersection $\cap_{Y\in\mathcal{Y}}Y\neq\emptyset$.

So, to get a contradiction, assume this is not the case, for any $n\in\mathbf{N}$, we can find $X_n\in\mathcal{Y}$ such that $B(x_o,n)\cap X_n=\emptyset$. Now, we can apply Lemma 4.9, to get the existence of a fixed point at infinity. This is a contradiction.

Definition 4.13 A closed convex subspace $Y$ of a CAT(0) space is flat if it is isometric to some Hilbert space.

A flat subspace may have infinite or finite dimension. It may even have dimension zero. In that case, it is reduced to a point.

Theorem 4.4 (Adams-Ballmann theorem) Let $X$ be a proper CAT(0) space and let $G$ be an amenable topological group acting continuously by isometries. Either $G$ preserves a flat subspace of $X$ or $G$ fixes a point at infinity.

Proof. Assume that $G$ does not fix a point at infinity, by Lemma 4.11, there exists a minimal invariant subspace in $X$. So we may reduce to the case where $X$ is minimal.

Since $X$ is proper, $\overline{X}$ is compact. Since $G$ is amenable, there is an invariant probability measure $\mu$ on $\overline{X}$. If $\mu(X)>0$ then one can find a ball $B$ such that $\mu(B)>1/2\mu(X)$ which implies that for any $g\in G$, $gB\cap B\neq\emptyset$ (otherwise $\mu(gB\cup B)=2\mu(B)>\mu(X)$). So any point of $B$ has a bounded orbit and by Cartan fixed point theorem, there is a $G$-fixed point and thus a flat subspace of dimension 0.

If $\mu(X)=0$, then $\mu(\partial X)=1$. Fix $x_0\in X$, for any $\xi\in\partial X$, $|\beta_\xi(x,x_0)|\leq d(x,y)$ and thus the integral

\[f(x)=\int_{\partial X}\beta_\xi(x,x_0)d\mu(\xi)\]

is well-defined and gives a convex 1-Lipschitz function and thus an element of $\mathcal{C}(X)$.

For $g\in G$, we have

\[\begin{align*} f(gx)&=\int_{\partial X}\beta_\xi(gx,x_0)d\mu(\xi)\\ &=\int_{\partial X}\beta_{g^(-1)\xi}(x,g^{-1}x_0)d\mu(\xi)\\ &=\int_{\partial X}\beta_{\xi}(x,g^{-1}x_0)d\mu(\xi)\\ &=\int_{\partial X}\beta_{\xi}(x,x_0)d\mu(\xi)+\int_{\partial X}\beta_{\xi}(g^{-1}x_0,x_0)d\mu(\xi)\\ &=f(x)+f(g^{-1}x_0) \end{align*}\]

This shows that $f$ is quasi-invariant: Its image in $\mathcal{C}_*(X)$ is invariant.

Let $r=\inf_{x\in X}f(x)$, $r_n$ be a decreasing sequence converging to $r$ and let $X_n=f^{-1}((-\infty,r_n])$. If $\cap X_n=\emptyset$, we use Exercise 4.7 (up to considering $g^{-1}$ if $f(g^{-1}x_0)>0$), we conclude that there is a fixed point at infinity.

So we are reduced to the case $\cap X_n\neq\emptyset$ and this intersection is invariant (in that case, $f$ is actually invariant). By minimality, this intersection is $X$ itself, that is $f$ is constant.

But an integral of convex functions is constant if an only if almost all functions are affine. This implies that $\mu$-almost surely $\beta_\xi$ is affine i.e. almost all points at infinity are flat. This gives a contradiction with Proposition 4.11.

4.6 Geometry of the space of positive definite matrices with determinant 1

Let $\mathrm{SDP}_n(\mathbf{R})$ be the space positive definite matrices in $\mathrm{M}_n(\mathbf{R})$ with it is standard topology (i.e. identified with $\mathbf{R}^{n^2}$ as vector space). It can be identified with the space of scalar products on $\mathbf{R}^n$ and to the space of ellipsoids in $\mathbf{R}^n$ that gives three models for the same space.

This space is a dense open of the vector space of symmetric matrices $\mathrm{S}_n(\mathbf{R})$ (of dimension $n(n+1)/2$) and thus a differential manifold where the tangent space at each point is identified with $\mathrm{S}_n(\mathbf{R})$ . It is moreover a Riemannian manifold: at each point $S\in\mathrm{SDP}_n(\mathbf{R})$, the Riemannian metric is $\langle X,Y\rangle_S=\mathrm{Tr}(S^{-1}XS^{-1}Y)$.

Proposition 4.12 The sectional curvature of $\mathrm{SDP}_n(\mathbf{R})$is non-positive. The exponential map $\exp\colon\mathrm{S}_n(\mathbf{R})\to\mathrm{SDP}_n(\mathbf{R})$ is a diffeomorphism and geodesics through $I_n$ are of the form $t\mapsto\exp(tX)$ where $\mathrm{Tr}(X^2)=1$.

Corollary 4.2 The space $\mathrm{SDP}_n(\mathbf{R})$ is a CAT(0) space. The distance between $A\in\mathrm{SDP}_n(\mathbf{R})$ and $I_n$ is $\sqrt{\sum_{i=1}^n\ln(\lambda_i)^2}$ where $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$

The group $\mathrm{GL}_n(\mathbf{R})$ acts via the formula: $g\cdot S=gS^tg$. This action is transitive because any $S\in\mathrm{SDP}_n(\mathbf{R})$ has a symmetric square root $R$ i.e. $S=R^2=R^tR=R\cdot I_n$. One checks readily that the action is by isometry thanks to the trace property: $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$.

Definition 4.14 A symmetric space is a connected Riemannian manifold $M$ such that for any $p\in M$, there symmetry at $p$: an isometry $\sigma_p$ such that $\sigma_p(p)=p$ and $d_p\sigma_p=-\mathrm{Id}_p$.

Lemma 4.12 The space $\mathrm{SDP}_n(\mathbf{R})$ is a symmetric space.

Proof. By transitivity, it suffices to prove that there is symmetry at $I_n$. We claim that $\sigma\colon S\mapsto S^{-1}$ is the symmetry at $I_n$. Let’s compute the differential at $S$. Since $(S+H)^{-1}=S^{-1}(I_n+S^{-1}H)^-{1}=S^{-1}-S^{-2}H+o(H)$, we have $d_S\sigma(H)=-S^{-2}H$.

Thus \[\langle d_S\sigma(X),d_S\sigma(Y)\rangle_{S^{-1}}=\mathrm{Tr}(S^{-1}XS^{-1}Y)=\langle X,Y\rangle_S\]

and $\sigma_S$ is a symmetry.

A thorough study of the Riemann tensor shows the following characterization of flat subspaces.

Proposition 4.13 A subspace of $\mathrm{SDP}_n(\mathbf{R})$ containing $I_n$ is flat if and only if it is of the form $\exp{V}$ where $V$ is a subspace of $\mathrm{S}_n(\mathbf{R})$ such that any two elements commute.

Since commuting symmetric matrices are simultaneously diagonalisable, we get the following corollary.

Corollary 4.3 A maximal flat subspace of $\mathrm{SDP}_n(\mathbf{R})$ containing the identity is $\left\{\mathrm{diag}(\lambda_1,\dots,\lambda_n),\ \mathrm{with}\ \lambda_i>0\right\}$ and all maximal flat subspaces are of the form

\[S\left\{\mathrm{diag}(\lambda_1,\dots,\lambda_n),\ \mathrm{with}\ \lambda_i>0\right\}S\] for some $S\in\mathrm{SDP}_n(\mathbf{R})$ .

The following proposition identifies the Euclidean de Rham factor of $\mathrm{SDP}_n^1(\mathbf{R})$. Its proof relies on Riemannian geometry.

Proposition 4.14 The map

\[\begin{matrix} \mathbf{R}\times \mathrm{SDP}_n^1(\mathbf{R})&\to&\mathrm{SDP}_n(\mathbf{R})\\ (t,S)&\mapsto&e^{t/\sqrt{n}}S \end{matrix} \]

is an isometry.

So, the action of $\mathrm{GL}_n(\mathbf{R})$ on positive definite matrices with determinant 1, $\mathrm{SDP}_n^1(\mathbf{R})$ is given by the formula

\[ g\cdot A= (\det(g)^2)^{-1/n} \ gA^tg.\] From this formula, one readily sees that the action of homotheties on $\mathrm{SDP}_n^1(\mathbf{R})$ is trivial.

Proposition 4.15 The stabilizer of a point at infinity of $\mathrm{SDP}_n^1(\mathbf{R})$ is conjugated to a subgroup of blockwise upper triangular matrices.

Proof. A geodesic ray from $I_n$ is of the form $\rho\colon t\mapsto \exp(tX)$ where $X$ is a symmetric matrix of norm 1 with trace 0. Up to conjugating $X$, we may assume that $X=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_i>\lambda_{i+1}$ and the $\lambda_i$ are not all equal.

A element $g\in\mathrm{GL}_n(\mathbf{R})$ fixes the point at infinity corresponding to $\rho$ if $\rho$ $t\mapsto g\exp(tX)^tg$ remains at bounded distance from $\exp(tX)$, i.e. $\exp(-tX/2)g\exp(tX)^tg\exp(-tX/2)=(\exp(-tX/2)g\exp(tX/2))$ is bounded independently of $t$.

The $(i,j)$ coefficient of $\exp(-tX/2)g\exp(tX/2)$ is $e^{t(\lambda_j-\lambda_i)/2}g_{i,j}$. If $g_{i,j}\neq0$ then for $j<i$,$|e^{t(\lambda_j-\lambda_i)/2}g_{i,j}|\to\infty$. So if $g$ lies in the stabilizer of this point at infinity then $g_{i,j}=0$ for $i>j$. This shows that $g$ is blockwise upper triangular.

Lemma 4.13 The kernel of the action is given by homotheties.

Proof. The elements of the kernel fix all points at infinity and thus are diagonalisable in any base and thus are homotheties.

4.7 Amenable connected Lie groups

The goal of this section is to characterize amenable subgroups of $\mathrm{GL}_n(\mathbf{R})$ (for the induced topology) and then extend the result for general connected Lie groups. Since the closure of an amenable subgroup is amenable, it suffices to understand the closed ones. This is done in Theorem 4.5.

Definition 4.15 A direct sum decomposition of $\mathbf{R}^n$ is a collection of subspaces $(E_1,\dots, E_k)$ of $\mathbf{R}^n$ such that $\mathbf{R}^n=E_1\oplus\dots\oplus E_k$. It is non-trivial if $k\geq2$.

Lemma 4.14 Let $g\in\mathrm{GL}_n(\mathbf{R})$ that preserves some flat subspace of $\mathrm{SDP}_n^1(\mathbf{R})$ with positive dimension then there is a non trivial direct sum decomposition $\mathbf{R}^n=E_1\oplus\dots\oplus E_k$ and a permutation $\sigma\in S_k$ such that for any $i\in\{1,\dots,k\}$, $g(E_i)=E_\sigma(i)$.

Proof. We have seen that a flat subspace $A$ of $\mathrm{SDP}_n^1{\mathbf{R}}$ is collection of commuting positive definite matrices. Up to conjugation $g$,we may assume that $A$ contains the point $I_n$. In that case, $A=\{exp(ta),\ r\in\mathbf{R}, a\in A_0 \}$ where $A_0$ is a vector space of commuting symmetric matrices with zero trace. Since $A$ is not reduced to a point $A_0$ has positive dimension. So the elements in $A$ are simultaneous diagonalizable. Let $M$ in $A$ be an element with maximal number of distinct eigenspaces. Let $E_1,\dots, E_k$ be these eigenspaces. Because of commutation, these spaces are invariant by any element of $A$. If there is an element $N\in A$ such that $N|_{E_i}$ is not a homothety then $MN$ has at least 2 eigenspaces in restriction to $E_i$ and one can find an element of $A$ with stricly more eigenspaces than $M$. This is a contradiction and any element of $A$ is a homothety in restriction to $E_i$.

Now let $g$ that stabilizes $A$ so $g\cdot I_n=g^tg=\sum_{i=1}^k\lambda_ip_i$ where $p_i$ is the projection to $E_i$ and $\lambda_i>0$. Let $s$ be $\sum_{i=1}^k\sqrt{\lambda_i}p_i$ and $h=sg$. Then $h^th=I_n$ and $h$ is orthogonal. Moreover $s$ preserves $A$ so does $h$. Now, for any $M\in A$, $h\cdot M=hMh^{-1}$ and $h$ maps eigenspaces for $M$ to eigenspaces of $h\cdot M$ because this is the action by conjugation. If $M$ has distinct eigenvalues on the $E_i$ (i.e. it has $k$ distinct eigenvalues) then $h\cdot M$ has $k$ distinct eigenvalues and the eigenspaces of $h\cdot M$ are exactly the $E_i$’s. This means that $h$ maps each $E_i$ to some $E_j$ and thus there is a permutation $\sigma$ such that such $h(E_i)=E_{\sigma(i)}$. Now $g^{-1}(E_i)=h^{-1}s(E_i)=h^{-1}(E_i)=E_{\sigma^{-1}(i)}$ and thus for dimension reasons, $g(E_i)=E_{\sigma(i)}$.

Definition 4.16 A topological group $G$ is said to be solvable by compact if there is a closed normal solvable subgroup $S$ such that $G/S$ is compact (i.e $S$ is cocompact in $G$).

In that case, one also says that $G$ is a compact extension of a solvable group.

\[1\to S\to G\to K\to 1\]

Lemma 4.15 Let $G$ be a topological group and $G'$ a closed finite index subgroup of $G$. If $G'$ is solvable by compact then so does $G$.

Proof. Let $S$ be the closed normal subgroup of $G'$ that is solvable and cocompact. Then $S$ is a closed solvable that we may assume to be normal in $G$ up to passing to a finite index subgroup of $S$. The subgroup $G'/S$ has finite index in $G/S$ and is compact, so $G/S$ is a finite union of compact subsets and thus is compact.

Theorem 4.5 A closed amenable subgroup of $\mathrm{GL}_n(\mathbf{R})$ is a compact extension of a solvable group.

Let $G$ be a closed amenable subgroup of $\mathrm{GL}_n(\mathbf{R})$.

Let’s take a maximal direct sum decomposition $\mathbf{R}^n=E_1\oplus\dots\oplus E_k$ that is preserved by some finite index subgroup of $G'$ then let $G_i$ be the image of $G$ in $\mathrm{GL}(E_i)$. Assume that each $G_i$ has a normal solvable group $S_i$ such that $G_i/S_i$ is compact. Then the intersection $S$ of $\ker(G\to G_i/S_i)$ is a normal solvable closed subgroup and the image of $G$ is closed in $G_1\times\dots\times G_k$, so $G/S$ is compact since it is closed in $G_1/S_1\times\dots\times G_k/S_k$.

By maximality of the direct sum decomposition, no finite index subgroup of $G_i$ preserves a direct sum decomposition of $E_i$. So we are reduced to the case where $G$ has no finite index subgroup that preserves a non trivial direct sum decomposition.

By Adams-Ballmann theorem, we know that $G$ fixes a flat subspace (possibly reduced to a point) or a point at infinity. If $G$ fixes a point then $G$ is compact. If $G$ fixes a flat of dimension at least one then $G$ preserves a direct sum decomposition which is a contradiction.

If $G$ fixes a point at infinity then elements of $G$ are blockwise upper triangular, i.e there is $E_1\subset E_2\subset \dots E_k$ such that $E_i$ is $G$-invariant so $G$ induces a linear representation on $E_{i+1}/E_i$ without invariant subspace. Applying the same result to this space, we get that the image of $G$ in $\mathrm{GL}(E_{i+1}/E_i)$ is relatively compact. Since the intersection of the kernel is solvable, we get the result.

By Adams-Ballmannn theorem, we know that it fixes a point of $\mathrm{SDP}_n(\mathbf{R})$ (thus a closed subgroup of a compact group (i.e) compact), stabilizes a flat subspace

Theorem 4.6 Let $G$ be a connected Lie group then $G$ is a compact extension of a solvable group.

Proof. Let $G$ be such Lie group and $R$ be its radical, i.e. the largest normal connected solvable subgroup of $G$. Then $G/R$ is connected semisimple Lie group. Let $n$ be the dimension of this quotient and let’s consider the adjoint representation of $G/R\to\mathrm{SL}_n(\mathbf{R})$. The kernel of this map is the center of $G/R$, which necessarily discrete since $G/R$ is semisimple.

The image of $G/R$ via the adjoint representation, this is an amenable subgroup of $\mathrm{SL}_n(\mathbf{R})$ and thus solvable by compact. It then follows from Exercise 4.11 that $G$ is also solvable by compact.

4.8 Exercises

Exercise 4.1 Prove that the projection on a closed convex subspace $C$ of a CAT(0) space is 1-Lipschitz and that for any $x\in X$ and $y\in C$

\[d(x,y)^2\geq d(x,p)^2+d(p,y)^2\]

where $p$ is the projection of $x$ on $C$.

Exercise 4.2 Let $X$ be a CAT(0) space. Let $\xi\in\partial X$ and $(x_n)$ sequence of $X$ such that $x_n\to\xi$. Prove that the distance function to $x_n$ converges to $\beta_\xi$ in $\mathcal{C}^*(X)$.

Exercise 4.3 Prove that any isometry of $\mathbf{R}^n$ is elliptic or hyperbolic. Is it still true for a Hilbert space ? One may look at the example of $\ell^2(\mathbf{Z})$ and an isometry with non-trivial translation part and the shift map as linear part.

Exercise 4.4 Prove that any isometry of a tree is elliptic or hyperbolic.

Exercise 4.5 Let $g$ be some hyperbolic isometry of a CAT(0) space. Prove that any invariant geodesic line is an axis.

Exercise 4.6 Let $G$ be a topological group acting by isometries on a metric space $(X,d)$. Prove that the action is continuous (i.e. the map $(g,x)\mapsto gx$ is continuous from $G\times X$ with the product topology to $X$) if and only if for any $x\in X$, the orbit map $g\mapsto gx$ is continuous.

Exercise 4.7 Let $X_n$ be a nested sequence of closed convex subspaces of a proper CAT(0) space $X$ with empty intersection. The goal of this exercise is to prove that there is $\xi\in\partial X$ such that $g\xi=\xi$ for all $g$ isometry of $X$ such that $gX_n\subset X_n$ for all $n\in\mathbf{N}$.

Let $x_0\in X$ and $x_n$ be the projection of $x_0$ on $X_n$.

Prove that up to extraction we may assume that $x_n$ converges to some point $\xi\in \partial X$.
Consider the triangle $\Delta(x_0,x_n,gx_n)$ and denote $r_n=d(x_0,x_n)$, $s_n=d(x_0,gx_n)$ and $t_n=d(x_n,gx_n)$. Prove that $\left( d(x_0, g x_0) + d(gx_0, g x_n) \right)^2 \geq r_n^2 + t_n^2.$
Fix $r>0$. For $n$ large enough such that $r_n>r$ we denote by $\rho_n(r)$ the point on $[x_0,x_n]$ at distance $r$ from $x_0$ and by $\sigma_n(r)$ the point on $[x_0,gx_n]$ at distance $r$ from $x_0$. Prove that $t_n^2 \geq 2 r_n s_n \cdot \frac{1}{2 r^2} \cdot d(\rho_n(r), \sigma_n(r)).$
Deduce that $d(\rho_n(r), \sigma_n(r))\to 0$ when $n\to\infty$ and conclude that $\xi$ is $g$-invariant.

Exercise 4.8 Prove that for any two positive definite matrices $A,B$, the distance between $A$ and $B$ is

\[d(A,B)=\sqrt{\sum_{i=1}^n\left(\ln(\lambda_i/\mu_i)^2\right)}.\] where $A=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ and $A=\mathrm{diag}(\mu_1,\dots,\mu_n)$ in a common diagonalization basis.

Exercise 4.9 Give the formula for the symmetry at any $S\in\mathrm{SDP}_n(\mathbf{R})$.

Exercise 4.10 Prove that the kernel of the action of $\mathrm{GL}_n(\mathbf{R})$ is $\left\{\pm I_n\right\}$.

Exercise 4.11 Let $G$ be a topological group with a normal closed subgroup $S$ such that $G/S$ is solvable by compact. Prove that $G$ is solvable by compact.

Exercise 4.12 Let $G$ be a topological group and let $H$ be a closed subgroup. Prove that $H$ is cocompact, i.e. $G/H$ is compact if and only if there $K$ compact subspace of $G$ such that $G=KH$.