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 bijeciton $g\colon X\to X$ 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 $r,\varepsilon>0$ 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

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

Proof. We give the proof in the case of a proper space for simplicity.

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.

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 $\ell(g)$.

Lemma 4.6 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 $g$.

Lemma 4.7 Let $X$ be a CAT(0) space and $X_n$ be a nested sequence of bounded closed convex subspaces of $X$. Then $\cap_{n\in\mathbf{N}}X_n\neq\emptyset$.

Lemma 4.8 Let $X$ be a proper CAT(0) space and $X_n$ be a nested sequence of closed convex subspaces of $X$ such that $\cap X_n$. Assume that $g$ is an isometry such that for any $n$, $gX_n=X_n$. Then there is $\xi\in\cap_{n\in \mathbf{n}\partial X_n}$ that is $g$-invariant.

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

Proposition 4.10 Flat strip theorem

Proposition 4.11 $Min(\gamma)$

Definition 4.10 Clifford translations

Theorem 4.3 de Rham factor

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

Let $\mathrm{SDP}_n{\mathbf{R}}$ be the space positive definite matrices with determinant 1

Action of $\mathrm{SL}_n(\mathbf{R})$. It is transitive

Geodesics

Proposition 4.12 Stabilizers of points at infinity.

Proposition 4.13 Flat subspaces

4.6 Adams-Ballman theorem

Busemann functions, limits, cocycle relation

Proposition 4.14 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 $\partial X$.

Lemma 4.9 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.

4.7 Amenable connected Lie groups

Here we consider the action of $\mathrm{GL}_n(\mathbf{R})$ on positive definite matrices with determinant 1 $\mathrm{SDP}_n{\mathbf{R}}$ given by the formula

\[ g\cdot A= (\det(g)^2)^{-1/n} \ ^tg^{-1}Ag^{-1}.\]

Lemma 4.10 The kernel of the action is given by homotheties.

Definition 4.11 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 4.11 Let $g\in\mathrm{GL}_n(\mathbf{R})$ that preserves some flat subspace of $\mathrm{SDP}_n(\mathbf{R})$ 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 $\mathrm{SDP}_n{\mathbf{R}}$ 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 symmetrice matrices with zero trace. Since $A$ is not reduced to a point $A_0$ has positive dimension. So the element in $A$ are simultaneous diagonalizable. Let $M$ in $A$ be an element with maximal number of distinct eigenspaces. Let $E_1$,, $E_k$ 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 $g\cdot I_n=^tg^{-1}g^{-1}=\sum_{i=1}^k\lambda_ip_i$ 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 $^thh=I_n$ 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 $k$ disctinct 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, $g(E_i)=E_{\sigma(i)}$.

Lemma 4.12 Let $G$ be a topological 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 4.4 An 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 $\mathrm{GL}_n(\mathbf{R})$.

Lets 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 normal solvable closed subgroup and its the image of $G$ is closed in $G_1\times\dots\times G_k$, so $G/S$ is compact since it is close 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 4.5 Let $G$ be a connected Lie group then $G$ is a compact extension of a solvable group.

In that case one says that $G$ is solvable-by-compact. This means that we have a short exact sequence

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

where $S$ is a normal solvable closed subgroup and $K\simeq G/S$ is a compact 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.

4.8 Exercises

Exercise 4.1 Prove that the projection on a closed convex subspace is 1-Lipschitz.

Exercise 4.2 Let $g$ be some hyperbolic isometry of a CAT(0) space. Prove that any invariant geodesic line is an axis.