text:
\section{Initial remark}
 The author observed just after completing this article that the so called $\pi$-convergence property (i.e. lemma $18$ of \cite{papasoglu2009boundaries}) is valid for non proper actions (the proof does not make use of this assumption). This would give other proofs for the lemma \ref{lem:split} and the proposition \ref{prop:fixflat}. More importantly it is possible that for the same reason the results (or some results) of \cite{guralnik2011atransversal} hold for cocompact non proper actions too. If so, this would supersed the propositions \ref{prop:geodcompflat} and \ref{prop:semisimlocalfinflat} below.

\section{Introduction}
A metric space $(X,d)$ is called \ensuremath{\textnormal{CAT}(0)}\ if it is geodesic and all its geodesic triangles are not fatter than there euclidean comparison triangle. For general reference on \ensuremath{\textnormal{CAT}(0)}\ spaces, see \cite{BridsonHaefliger201011} and \cite{Ballmann200402}. A geodesic is an isometry $\varphi:I\rightarrow X$ where $I$ is an interval of \ensuremath{\mathbf{R}}. If $I=\ensuremath{\mathbf{R}}^+$ (resp. $I=\ensuremath{\mathbf{R}}$) then $\varphi$ is called a geodesic ray (resp. geodesic line). We emphasis that a \ensuremath{\textnormal{CAT}(0)}\ space might admit no ray (take any bounded convex subset of $\ensuremath{\mathbf{R}}^n$, any ball of any arbitrary \ensuremath{\textnormal{CAT}(0)}\ space, ...). The Hausdorff distance between two subsets $A$ and $B$ of a metric space $X$ is defined by (for $Y\subset X$ and $r>0$, \ball{Y}{r} designates $\cup_{y\in Y} B(y,r)$  where the balls are those of $X$):
\begin{equation*}
  \disthaus{A}{B}=\inf_{r>0}\left\{B\subset\ball{A}{r}\textrm{ and }A\subset\ball{B}{r}\right\}.
\end{equation*}
We call two rays $\varphi$ and $\psi$ asymptotic if there images are at bounded Hausdorff distance. This is equivalent to saying that the function $t\mapsto d(\varphi(t),\psi(t))$ is uniformly bounded. Being asymptotic is an equivalent relation. The quotient set of all geodesic rays by this equivalent relation is written \bndry{X}, and the class of a ray $\varphi$ is $\varphi(\infty)$. If $X$ is complete (i.e. if every Cauchy sequence converges), one can find a representative of any class of asymptotic rays radiating from any given point $x\in X$, and this representative is necessarily unique. Since we will only be interested in proper spaces (i.e. spaces whose closed balls are compact), we remark that such spaces are complete. Suppose $x\in X$ is now fixed, and give the set of rays emanating from $x$ the topology of uniform convergence on compact sets. This gives a topology on \bndry{X}\ and one can show that this topology is independent of $x$. We write \coneb{X}\ for this topological space and call it the boundary at infinity (endowed with the cone topology if we want to be specific).

There is another description of the boundary in terms of horofunctions. Let us define $\Theta: X\rightarrow \ensuremath{\mathrm{C}\left(X,\R\right)/\R}; x\mapsto [y\mapsto d(y,x)]$ where \ensuremath{\mathrm{C}\left(X,\R\right)/\R}\ is the quotient of continuous functions on $X$ quotiented by the constants with the quotient topology induced from the compact open topology on the set of continuous functions. Then \bndry{X} can be canonically identified with the boundary of $\Theta(X)$ in \ensuremath{\mathrm{C}\left(X,\R\right)/\R}. Indeed if $\varphi$ is a ray in $X$ then the following function is well defined and $1$-lipschitz on $X$:
\begin{equation*}
  \label{eq:busedef}
  x\mapsto\lim_{t\rightarrow+\infty}d(x,\varphi(t))-t.
\end{equation*}
 We write \buse{[\varphi]}{\varphi(0)}{x} for this limit and call the associated function of $x$ the Busemann function based at $[\varphi]$ with reference point $\varphi(0)$. The class in \ensuremath{\mathrm{C}\left(X,\R\right)/\R}\ of this limit only depends on the asymptotic class $[\varphi]$ of the ray $\varphi$. When $X$ has non positive curvature, all the elements of the boundary of $\Theta(X)$ in \ensuremath{\mathrm{C}\left(X,\R\right)/\R}\ can thus be obtained. In general we obtain two different compactifications (see \cite{paeng2007tits} for example for the case of manifolds without conjugate points).

 The horosphere based at $p\in\bndry{X}$ through $x\in X$ is the level set through $x$ of any Busemann function based at $p$, i.e. it is $b^{-1}(b(x))$ for any such Busemann function. A (closed) horoball based at $p$ is any set of the form $b^{-1}(]-\infty,a])$ where $a\in\ensuremath{\mathbf{R}}$ and $b$ is a Busemann function based at $p$.

\coneb{X} is compact and second countable (it is readily seen that if $X$ is complete not a priori proper, if \coneb{X} is compact and second countable and if all geodesics in $X$ can be extended to geodesic rays, then  $X$ is itself must be proper). From now on, all the \ensuremath{\textnormal{CAT}(0)}\ spaces we will encounter will be proper (this is equivalent to being locally compact for geodesic spaces). Choose $x\in X$ and let $\varphi$ and $\psi$ be two rays emanating from $x$. Then $2\cdot\arcsin(d(\varphi(t),\psi(t))/2t)$ converges to a real as $t$ goes to $+\infty$. Moreover this real depends only on the asymptotic classes of $\varphi$ and $\psi$, and not on $x$. We write $\tangle{\varphi(\infty)}{\psi(\infty)}$ for this limit. This defines a metric on $\bndry{X}$ and we shall refer to it as the Tits angle. The topology induced by  this metric is always finer that the cone topology, that is to say the identity $i:\bndry{X}\rightarrow\bndry{X}$ induces a continuous bijection $\bar{i}:(X,\angle)\rightarrow\coneb{X}$. We define the Tits metric, and write $\titsm{\cdot}{\cdot}$, to be the inner metric associated to the Tits angle. Hence this metric is defined by:
\begin{displaymath}
  \forall \varphi,\psi\in\bndry{X},\ \titsm{\varphi}{\psi}=\inf_{\gamma}\ell(\gamma),
\end{displaymath}
where $\gamma:[0,1]\rightarrow\bndry{X}$ ranges  over all the continuous path in $\bndry{X}$ (for the topology induced by the Tits angle) that connect $\varphi$ and $\psi$, and $\ell(\gamma)$ is the length of $\gamma$ with respect to the Tits angle defined by:
\begin{displaymath}
  \ell(\gamma)=\sup_{n,\ (t_i)_{1\leq 0\leq n}} \sum_{i=0}^{n-1}\angle(\gamma(t_i),\gamma(t_{i+1})),
\end{displaymath}
where the supremum runs over all $n>0$ and all $0=t_0<t_1<\cdot<t_n=1$. One can show that the balls of radius $r<\pi$ are the same for the Tits angle and the Tits metric (Part \Roman{deux} proposition $9.21$  of \cite{BridsonHaefliger201011}), so that they generate the same topology. We will refer to this topology as the Tits topology, and abuse the vocabulary and call it the Tits topology of $X$.

 Let us briefly remind the classification of isometries in \ensuremath{\textnormal{CAT}(0)}\ spaces. An isometry of a \ensuremath{\textnormal{CAT}(0)}\ space $X$ is called elliptic if it has a fixed point, hyperbolic if it translates some geodesic, and parabolic if it is neither elliptic nor hyperbolic. Isometries naturally act on the boundary \bndry{X} and it can be shown that parabolic isometries fix at least one point in it (we make use of properness here). Non-parabolic isometries are called semi-simple. A geodesic translated by a hyperbolic isometry is called an axis of this isometry. Moreover if $Y\subset X$ is a convex subset stabilised by an isometry $\phi$, then $\phi$ and its restriction to $Y$ have same type. Now let $\phi$ be an isometry fixing a point $p$ at infinity: if $b$ is a Busemann function based at $p$ and $\phi^{*}b$ is the action of $\phi$ on $b$ then $b-\phi^{*}b$ is a constant that only depends of the class of $b$ (i.e. on $p$). This constant is called the Busemann character of $\phi$ at $p$.

We say that $X$ is Tits compact if the Tits topology is compact. For example, it is well known that $\ensuremath{\mathbf{R}}^n$ with the usual flat metric has a Tits boundary isometric to the standard metric sphere of radius one, and so is Tits compact. More trivially, bounded convex subsets of any \ensuremath{\textnormal{CAT}(0)}\ space (remark that there always exist such subspaces since the balls of \ensuremath{\textnormal{CAT}(0)}\ spaces are convex) have empty boundary, and hence are Tits compact. The following example shows that there exist non flat \ensuremath{\textnormal{CAT}(0)}\ spaces with the geodesic extension property (i.e. such that every geodesic is part of a geodesic line) that are Tits compact:
\begin{ex}\label{ex:conesphere}
  Take $r\ge 1$ and $n\in\ensuremath{\mathbf{N}}^*$. Let $\mathcal{S}_r$ be the metric sphere of radius $r$ and dimension $n-1$ and construct the euclidean cone $X$ over $\mathcal{S}_r$ (see section $5$ Part \Roman{un} of \cite{BridsonHaefliger201011} for the definition and general properties of cones over metric spaces). Since the curvature of $\mathcal{S}_r$ is bounded above by $1$ and $\mathcal{S}_r$ is simply connected, $\mathcal{S}_r$ is metrically a $\cat{1}$ space. Then, according to Berestovskii' s theorem (see Part \Roman{deux} proposition $3.14$ in \cite{BridsonHaefliger201011}), $X$ is a \ensuremath{\textnormal{CAT}(0)}\ space. It is easy to see that it is locally compact, has the geodesic extension property, and that its Tits boundary is isometric to $\mathcal{S}_r$. To prove the last affirmation, take $y,z$ on the boundary $\mathcal{S}_r$ of $X$. Then the rays asymptotic to $y$ and $z$ are respectively $t\mapsto (y,t)$ and $t\mapsto (z,t)$, and
  \begin{align*}
    \forall t>0,\ d_X((y,t),(z,t))/2t&=\frac{1}{\sqrt{2}}\sqrt{1-\cos\min\left(d_{\mathcal{S}_r}(y,z),\pi\right)}\\
    &=\frac{1}{\sqrt{2}}\sqrt{2\sin^2\bigg(\min\left(d_{\mathcal{S}_r}(y,z),\pi\right)/2\bigg)}\\
    &=\sin\bigg(\min\left(d_{\mathcal{S}_r}(y,z),\pi\right)/2\bigg),
  \end{align*}
so that $\titsm{y}{z}=d_{\mathcal{S}_r}(y,z)$. In particular $X$ is Tits compact. Let us now turn our attention to the isometries of $X$. Since the cone construction is functorial, every isometry of $\mathcal{S}_r$ extends to an isometry of $X$. This isometry is always elliptic since it fixes the apex of the cone (which we shall call $0$ henceforth). Reciprocally, it is easy to see that the isometries fixing $0$ are uniquely induced by isometries of the Tits boundary $\mathcal{S}_r$. If $r=1$, $X$ is the Hilbert space of dimension $n$, so that there are plenty of other isometries, including translations (Clifford translations in the terminology of \ensuremath{\textnormal{CAT}(0)}\ spaces) in any direction (remark that the actions at infinity of all those translations are trivial, so one cannot hope to recover any of them from its continuation to \titsb{X}). Let us now turn to the case where $r$ is bigger than $1$. Then the inspection of the geodesics in $X$ shows that $0$ is the only point where geodesics branch. So isometries of $X$ must fix $0$. The correspondence between isometries of the Tits boundary of $X$ (naturally isometric to the initial $\mathcal{S}_{r}^{n-1}$) and isometries of $X$ is then easily proved to be a isomorphism of topological groups. So the group of isometries of $X$ is a whole lie group of positive dimension. However, it is compact, so can be considered relatively small and with rather trivial dynamics.
\end{ex}
\begin{defi}
  A subspace $Y$ of a \ensuremath{\textnormal{CAT}(0)}\ space $X$ is geodesic if for all points $x,y\in Y$, the geodesic in $X$ joining those two points lies inside $Y$.
\end{defi}
\begin{defi}
  A (half) flat plane in $X$ is a convex subset isometric to a euclidean (half) plane.   A (half) flat strip in $X$ is a convex subset isometric to a euclidean (half) strip, i.e. to $[0,a]\times\ensuremath{\mathbf{R}}$ ($[0,a]\times\ensuremath{\mathbf{R}}^+$) for some $a>0$.
\end{defi}
The next lemma is  classical and can be found in every textbook on non positively curved spaces like \cite{BridsonHaefliger201011} or \cite{Ballmann200402} and is nicknamed the flat strip theorem:
\begin{lemma}
  Let $X$ be a \ensuremath{\textnormal{CAT}(0)}\ space a let $\ell$ be a geodesic line in $X$ (i.e. $\ell$ is a subset of $X$ isometric to the real line). We call another line $l$ parallel to $\ell$ if they lie at bounded Hausdorff distance from one another. Then if $\ell$ and $l$ are parallel they bound a totally geodesic flat strip in $X$ (i.e. there exists an isometry $\varphi:\ensuremath{\mathbf{R}}\times[0,a]\rightarrow X$ such that $\varphi(\ensuremath{\mathbf{R}}\times{0})=\ell$ and $\varphi(\ensuremath{\mathbf{R}}\times{a})=l$ where $a>0$). Moreover the set of parallel lines to $\ell$ form a totally geodesic subspace that decompose has a product $\ensuremath{\mathbf{R}}\times Y$ where $Y$ is a Hadamard space and the first factor corresponds to the parallel lines in $X$ to $\ell$ (i.e. the parallel lines to $\ell$ are exactly the $\ensuremath{\mathbf{R}}\times\{a\}$ where $a\in Y$).
\end{lemma}
 We say that a group $G$ acts cocompactly on a space $X$ if there exists a compact subset $K$ of $X$ whose translates by $G$ cover $X$. In the same context we say that $G$ acts properly if for every compact subset $C$ of $X$ the number of elements $g$ in $G$ satisfying $g\cdot C\cap C\neq\emptyset$ is finite (we shall in time modify this definition). Since the spaces we will be concerned with will be proper, this  is equivalent to the various usual other definitions of properness (for discrete groups). Finally an action is geometric if it is both cocompact and proper.

\section{Minimal subspaces}

We remind results that can be found in \cite{caprace2009isometry}. The reader should refer to this paper for details  and proofs.
\begin{defi}
  Let $X$ be a \ensuremath{\textnormal{CAT}(0)}\ space and $Y\subset X$ a convex subset. We say that $Y$ has full boundary (in $X$) if $\bndry{Y}=\bndry{X}$. Moreover, we say that $Y$ is boundary minimal if it is closed, convex, has full boundary, and is minimal for this property (the partial order considered is the inclusion).
\end{defi}
The existence of boundary minimal subset is not clear. However, we have:
\begin{prop}
  If a \ensuremath{\textnormal{CAT}(0)}\ space $X$ is cocompact, then it admits boundary minimal subsets and all those subspaces are isometric (they are even parallel but we will not need it). Moreover, one can canonically assign to $X$ a subspace $\can{X}$ that is boundary minimal and fixed by all isometries of $X$. This space is called the canonical boundary minimal subspace of $X$.
\end{prop}
\begin{rmk}
  \can{X} can be defined in greater generality but we shall not use it.
\end{rmk}
The product of two \ensuremath{\textnormal{CAT}(0)}\ spaces $X_1$ and $X_2$ is again a \ensuremath{\textnormal{CAT}(0)}\ space. The Tits boundary of the product depends only on the Tits boundary of each factor and can be constructed explicitly. The construction is called joining (we slightly modify the usual definition since with the usual definition of \cite{BridsonHaefliger201011} one should take the inner metric associated to the joining in order to get the Tits metric of the product). The join of two spaces $Y_1$ and $Y_2$ is written $Y_1\star Y_2$. Hence, almost by definition we have $\titsb{\left(X_1\times X_2\right)}=\titsb{X_2}\star\titsb{X_1}$. Reciprocally if the boundary of $X$ is a join and if $X$ is geodesically complete then $X$ is a product. Without assuming geodesic completness, we have:
\begin{prop}\label{prop:bndrysphere}
  Let $X$ be a cocompact \ensuremath{\textnormal{CAT}(0)}\ space. Then \can{X} is flat if and only if \titsb{X} is a metric unit sphere.
\end{prop}
\begin{proof}
Since \can{X} has full boundary in $X$, the sufficient part is trivial. We now suppose that \titsb{X} is a metric unit sphere of dimension $n$.

  We can assume that $X$ is itself boundary minimal. Decompose $X$ as a product $\ensuremath{\mathbf{R}}^k\times Y$ where $Y$ is a \ensuremath{\textnormal{CAT}(0)}\ space that admits no euclidean factor. This decomposition exists, is unique, and every isometry of $X$ respects the splitting so that $Y$ is also cocompact. It must also be boundary minimal since so is $X$. Suppose that $X$ is not flat. Then $Y$ is not compact by minimality of $X$.  Moreover $\titsb{Y}$ is a unit sphere (it is the set of points at constant distance $\pi/2$ from the subsphere \titsb{\ensuremath{\mathbf{R}}^k} of \titsb{X}).

 Since it is also cocompact it admits a geodesic line $\varphi$ (a non compact proper geodesic space always admits a ray. It is then easy to obtain a line by cocompactness and properness of $X$). Let $x$ be a point on $\varphi$. Now $\varphi(+\infty)$ must be the opposite point of $\varphi(-\infty)$ in the metric sphere \titsb{Y} since it is the only point that lies at distance not lower than $\pi$ from $\varphi(-\infty)$. We shall consider the two endpoints of $\varphi$ as the poles of the sphere. If the boundary has only two points, then $Y=\varphi(\ensuremath{\mathbf{R}})$ by minimality and $Y$ is flat, contradicting the hypothesis. Pick a point $y\in\bndry{Y}$ on the equator, and let $\psi$ be the ray from $x$ to $y$. We hence have $\angle(\varphi(\pm\infty),y)=\pi/2$ and:
  \begin{align*}
    \pi=\angle_{x}\left(\varphi(+\infty),\varphi(-\infty)\right)&\leq \angle_{x}\left(\varphi(+\infty),y\right)+\angle_{x}\left(y,\varphi(-\infty)\right)\\
    &\leq \angle\left(\varphi(+\infty),y\right)+\angle\left(y,\varphi(-\infty)\right)=\pi,
  \end{align*}
So that we have equality everywhere. In particular we have
\begin{displaymath}
\angle_{x}\left(\varphi(\pm\infty),y\right)=\angle\left(\varphi(\pm\infty),y\right)=\pi/2.
\end{displaymath}
 From this we deduce that the rays $\varphi(\ensuremath{\mathbf{R}}^+)$ and $\psi$ (resp. $\varphi(\ensuremath{\mathbf{R}}^-)$ and $\psi$) span a convex flat quadrant. In fact it is easy to see that those two quadrants form a convex half flat plane with edge $\varphi$ admitting $y$ as boundary point (maybe the easiest way to see this is to remark that we chose $x$ arbitrarily). Hence the set \para{\varphi} of parallel lines to $\varphi$ (which is a closed convex subset of $X$) has full boundary so that $Y=\para{\varphi}$ by minimality. But we know that \para{\varphi} splits as $\ensuremath{\mathbf{R}}\times Z$ where $Z$ is a \ensuremath{\textnormal{CAT}(0)}\ space. This contradicts the definition of $Y$ and $X$ must be flat.
\end{proof}
\begin{quest}
  Are there non flat boundary minimal \ensuremath{\textnormal{CAT}(0)}\ spaces with Tits boundary isometric to a metric sphere (without assuming geodesic completness of course)?
\end{quest}
\section{Geometric group actions and Tits compactness}\label{sec:geomact}
We only include this section for the curiosity of the reader since we will prove a stronger result in section \ref{sec:geomactbetter}.

 The reference for this section is \cite{guralnik2011atransversal} and the reader should refer to it for proofs and details. We will also use the notations and definitions of this article. The definition $4.2$ in this paper does not always make sense since the Tits metric could be infinite. However we will only make use of the Theorem $3.22$ (the folding lemma) so this will not be a problem for us. Recall theorem $3.22$ of this article:
\begin{thm}[Folding lemma]\label{thm:gural}
   Suppose $G$ acts geometrically on a \ensuremath{\textnormal{CAT}(0)}\ space $X$ and let $d$ denote the geometric dimension of $\titsb{X}$. Then for every $(d+1)$-flat $F_0\subseteq X$ there exist $\omega\in\beta G$ and a $(d+1)$-flat $F\subseteq X$ such that $\ultra{\omega}$ maps all of $\bndry{X}$ onto $\SS=\bndry{F}$, with $\bndry{F_0}$ mapped isometrically onto $\SS$.
\end{thm}
\begin{rmk}
  The flat $F_0$ exists, see the theorem C in \cite{kleiner1999local} and in particular the equivalent statements $3$ and $5$.
\end{rmk}
But we have:
\begin{lemma}
  Let $X$ be a Tits compact \ensuremath{\textnormal{CAT}(0)}\ space on which a discrete group $G$ acts. Then for every $\omega\in\beta G$, $x\mapsto\ultra{\omega}(x)$ is distance preserving on \bndry{X}.
\end{lemma}
\begin{proof}
  Let us pick $x,y\in\titsb{X}$. Then for all $n>0$ the set of all $g\in G$ such that $\titsm{g(x)}{\ultra{\omega}(x)}<1/n$ and $\titsm{g(y)}{\ultra{\omega}(y)}<1/n$ is not empty (it has $\omega$ probability $1$, because the Tits balls of radius $1/n$ around $g(x)$ and $g(y)$ are open in the cone topology), so we can fix such an element $h_n\in G$ for each $n$. We have for all $n\in\ensuremath{\mathbf{N}}$:
  \begin{equation*}
   |\titsm{\ultra{\omega}(x)}{\ultra{\omega}(y)}-\titsm{x}{y}|=|\titsm{\ultra{\omega}(x)}{\ultra{\omega}(y)}-\titsm{h_{n}(x)}{h_{n}(y)}|
  \end{equation*}
Since this is true for every $n$, $\ultra{\omega}$ is distance preserving.
\end{proof}
We finally conclude this section with the:
\begin{prop}
     Suppose $G$ acts geometrically on a Tits compact \ensuremath{\textnormal{CAT}(0)}\ space $X$. Then \titsb{X} is a metric unit sphere, \can{X} is flat, and $G$ is a Bieberbach group.
\end{prop}
\begin{proof}
  According to the proposition \ref{prop:bndrysphere} and Bieberbach' s theorem (see the Theorem $4.2.2$ in the beautiful book \cite{thurston1997three}), it is enough to show that \titsb{X} is a unit sphere. Theorem \ref{thm:gural} gives us a ultra-filter $\omega\in\beta G$ and a metric sphere $\SS\subset\titsb{X}$ such that $\ultra{\omega}$ maps \bndry{X} surjectively onto $\SS$. According to the previous lemma it is also distance preserving, so it is an isometry.
\end{proof}
\section{Dynamics of hyperbolic isometries and Tits compactness}\label{sec:hypcomp}

In section \ref{sec:geodcomp} we will have to distinguish between two possibilities: either the full isometry group fixes a point at infinity or it contains a hyperbolic element. This section will be useful in the later case.
\begin{lemma}\label{lem:split}
Suppose $X$ is a Tits compact \ensuremath{\textnormal{CAT}(0)}\ space that admits a hyperbolic isometry $a$ and choose an axis $c$ of $a$. Consider a ray $r$ emanating from $c$ and different from $c$ itself (we assume such a ray exists). Then $r$ lies in some half flat plane with edge $c$.
\end{lemma}
\begin{proof}
  Let's write $b$ for the restriction of $a$ to the boundary at infinity \bndry{X}, and consider the sequence of maps $\left(b^{-n}\right)_{n\in\ensuremath{\mathbf{N}}}$. This is a sequence of isometries for the Tits metric which is assumed to give rise to a compact topology. According to the Arzelà-Ascoli theorem, we can fix a series of strictly increasing integers $n_k$ such that $\left(b^{-n_k}\right)$ converges uniformly to a continuous map $b$ of \bndry{X}. Now this map must be distance preserving as a pointwise limit of such maps. But a distance preserving map of a compact metric space is always surjective, and so $b$ is a (surjective) isometry of \titsb{X}. This will prove useful later. Write $d$ for the minimum displacement of $a$ i.e. $d=\inf_{x\in X} d(x,a(x))$.

Consider  an asymptotic direction $x$ different from $c(\pm\infty)$. Then we know (see Part \Roman{deux} proposition 9.8 in \cite{BridsonHaefliger201011}) that the angles $\angle_{c(d\cdot n_k)}\left(c(+\infty),x\right)$ increase to the Tits angle $\angle(c(+\infty),x)$:
\begin{equation}\label{eq:part0}
  \lim_{k\rightarrow+\infty}\angle_{c(d\cdot n_k)}(c(+\infty),x)=\angle(c(+\infty),x).
\end{equation}
Now if we push everything forward in the left hand side of this equality via the isometry $a^{-n_k}$ we get:
\begin{equation}\label{eq:part1}
  \lim_{k\rightarrow+\infty}\angle_{c(0)}(c(+\infty),b^{-n_k}(x))=\angle(c(+\infty),x).
\end{equation}
But the sequence of maps $b^{-n_k}$ also converge pointwise to $b$ in the cone topology, so we have:
\begin{equation}\label{eq:part2}
  \lim_{k\rightarrow+\infty}\angle_{c(0)}(c(+\infty),b^{-n_k}(x))=\angle_{c(0)}(c(+\infty),b(x)).
\end{equation}
and since $b$ is an isometry that fixes $c(+\infty)$ (because all the $b^{-n_k}$ do) we also have:
\begin{equation}\label{eq:part3}
\angle(c(+\infty),x)=\angle(c(+\infty),b(x)).
\end{equation}
Combining, \eqref{eq:part1}, \eqref{eq:part2} and \eqref{eq:part3} we finally get:
\begin{equation*}
\angle_{c(0)}(c(+\infty),b(x))=\angle(c(+\infty),b(x)).
\end{equation*}
But this equality implies that $c(0)$, $c(+\infty)$ and $b(x)$ span an infinite flat triangle, and this for every $x$ (the case where $x$ is either of $c(+\infty)$ and $c(-\infty)$ being trivial). Now by surjectivity of $b$ we just proved that given any asymptotic direction $y\in\bndry{X}$, $c(0)$, $c(+\infty)$ and $y$ span an infinite flat totally geodesic triangle. We call $\Delta_y$ this triangle.

Now remark that the parametrization of $c$ was chosen arbitrarily, so the previous results still holds if we replace $c$ by $c^\prime:t\rightarrow c(t-1)$. In particular, whenever we have an asymptotic direction $y\in\bndry{X}$, the three points $c^\prime(0)=c(-1)$, $c^\prime(+\infty)=c(+\infty)$ and $y$ span convex geodesic flat triangle. We write $\Delta^\prime_y$ for this triangle. But then the euclidean triangle $\Delta^\prime_y$ contains asymptotic rays to $y$ emanating from $c(t)$ for every $t\geq 0$. Since those rays are uniquely determined and lie in $\Delta_y$, we conclude that $\Delta^\prime_y$ extends $\Delta_y$. Proceeding recursively, we then obtain a totally geodesic flat half plane $\Pi_y$ with border $c$ and $y$ on its boundary.

Now if $r$ is the ray of the proposition, it suffices to take $y=r(\infty)$.
\end{proof}
The same method as in the proof of proposition \ref{prop:bndrysphere} then gives:
\begin{prop}\label{prop:hypflat}
 Suppose $X$ is a cocompact Tits compact \ensuremath{\textnormal{CAT}(0)}\ space that admits a hyperbolic isometry. Then $\can{X}$ splits with a euclidean factor.
\end{prop}
\begin{proof}
  We can always assume that $X$ is boundary minimal. We use the notations of the above lemma. Then the set of parallel lines to $c$ (remark it closed and convex) has full boundary in $X$ (this is the lemma \ref{lem:split}) and splits with a \ensuremath{\mathbf{R}}\ factor. By minimality, so does $X$.
\end{proof}
\begin{rmk}\label{rmk:piconv}
  We could have used the $\pi$-convergence property, i.e. lemma $18$ of \cite{papasoglu2009boundaries}, to show that $\bndry{X}$ splits as a join $\mathcal{S}^{0}\star Y$. Let us sketch the proof. Since the limit points $n$ and $p$ (we use the notations of \cite{papasoglu2009boundaries}) are at distance not lower than $\pi$ apart here (since they are joined by the axis of the hyperbolic isometry), the $\pi$-convergence property easily shows that for each point $a\in\bndry{X}$, we have $d(n,p)=d(n,a)+d(a,p)$ (and in fact $d(n,p)=\pi$). Now we can use the lemma $3.19$ of \cite{guralnik2011atransversal} to conclude.
\end{rmk}
\section{Dynamics of parabolic groups and Tits compactness}

In this section we study \ensuremath{\textnormal{CAT}(0)}\ spaces that admit a cocompact action by a group of isometries fixing a point at infinity. We were not able to show the existence of hyperbolic elements. However, we construct a series of isometries $\left(g_n\right)_{n\in\ensuremath{\mathbf{N}}}$ that behaves much like iterates of a single hyperbolic isometry. Then a refinement of the argument used in section \ref{sec:hypcomp} will give us the desired flat factor.
\begin{prop}\label{prop:veryhyp}
  Let $X$ be a \ensuremath{\textnormal{CAT}(0)}\ space and $G$ a group of isometries that acts cocompactly on $X$ and fixes a point $p$ on the boundary \bndry{X}. Then there exists a series $\left(g_n\right)_{n\in\ensuremath{\mathbf{N}}}$ of isometries in $G$, a geodesic line $R$ and a real $d\geq0$ such that:
  \begin{enumerate}
  \item\label{it:geodline} $R(-\infty)=p$,
  \item\label{it:distatmost} For every integer $n$, $g_n(R)$ is parallel to $R$ and lies at distance at most $d$ to it,
  \item\label{it:limitpt} $\left(g_n^{\pm}(q)\right)_{n\in\ensuremath{\mathbf{N}}}$ converges to $R(\pm\infty)$ for every $q\in X$.
  \end{enumerate}
\end{prop}
\begin{proof}
  It is easy to construct a geodesic line $R$ satisfying \ref{it:geodline}. Now  fix $x=R(0)$ and $d>0$ such that the translates of the ball $\ball{x}{d}$ by all isometries of $G$ covers $X$.

Take $n\in\ensuremath{\mathbf{N}}$ and let us construct the isometry $g_n$ above. For each integer $k>0$, fix an isometry $h_{n,k}$ in $G$ that takes $R(k)$ within $d$ of $R(k+n)$. Consider a reparametrization $L_{n,k}$ of $h_{n,k}(R)$ such that $L_{n,k}(0)$ and $x$ belong to the same horosphere centered at $p$ (this is possible because these two lines are asymptotic to $p$). Classically, the function $f_{n,k}:t\mapsto d(R(t),L_{n,k})$ is increasing (it is convex with a finite limit at $-\infty$) and so:
\begin{figure}[!h]
  \centering
  \input{hnk.tex}
  \caption{Definition of $h_{n,k}$.}
  \label{fig:hnk}
\end{figure}
\begin{equation}\label{eq:lineconv}
\forall t\leq n+k,\ f_{n,k}(t)\leq f_{n,k}(n+k)\leq d(R(n+k),h_{n,k}(R(k)))\leq d.
\end{equation}
Moreover $h_{n,k}$ sends $x=R(0)$ to $L_{n,k}(\gamma_{n,k})$ where $\gamma_{n,k}$ is the image of $h_{n,k}$ under the Busemann character based at $p$, i.e. $\gamma_{n,k}=\buse{p}{y}{h_{n,k}(y)}$ for any $y\in X$. But then (Busemann functions are $1$-lipschitz):
\begin{align*}
  \gamma_{n,k}=\buse{p}{R(k)}{h_{n,k}(R(k))}&\leq d(R(k),h_{n,k}(R(k)))\\
  &\leq d(R(k),R(n+k))+d(R(n+k),h_{n,k}(R(k)))\\
  &\leq n+d,
\end{align*}
and so:
\begin{equation}\label{eq:majo}
  d(x,h_{n,k}(x))\leq \underbrace{d(x,L_{n,k}(0))}_{=f_{n,k}(0)}+\underbrace{d(L_{n,k}(0),L_{n,k}(\gamma_{n,k}))}_{=\gamma_{n,k}}\leq d+n+d.
\end{equation}
This bound \eqref{eq:majo} does not depend on $k$. By properness of $X$ we can hence take a converging subsequence of $\left(h_{n,k}\right)_{k\in\ensuremath{\mathbf{N}}}$. We choose $g_n$ as the limit of this series. Now the condition \ref{it:distatmost} of the proposition is obtained by letting $k$ go to infinity in \eqref{eq:lineconv}. Let us prove that the condition \ref{it:limitpt} also holds. The cocycle relation for Busemann functions gives:
\begin{equation*}
\buse{p}{x}{h_{n,k}(x)}=\buse{p}{x}{R(n+k)}+\buse{p}{R(n+k)}{h_{n,k}(x)},
\end{equation*}
and hence:
\begin{equation*}
\gamma_{n,k}=\buse{p}{x}{h_{n,k}(x)}\geq\buse{p}{x}{R(n+k)}-d(R(n+k),h_{n,k}(x))\geq n-d.
\end{equation*}
Going to the limit in $k$ then gives $\gamma_{n}=\buse{p}{x}{g_{n}(x)}\geq n-d$. But the pythagorean Theorem in the flat strip spanned by $R$ and $g_n(R)$ gives (here $\delta$ is the Hausdorff distance between $R$ and $g_n(R)$):
\begin{equation}\label{eq:cond3foward}
  d(x,g_n(x))=\sqrt{\gamma_{n}^2+\delta^2}\geq\gamma_{n}\geq n-d.
\end{equation}
We also have $d(R,g_{n}^{-1}(R))=d(R,g_n(R))=\delta$ and $\buse{p}{x}{g_{n}^{-1}(x)}=-\gamma_k$ so that we have the same bound with $g_n^{-1}$ instead of $g_n$:
\begin{equation}\label{eq:cond3backward}
  d(x,g_{n}^{-1}(x))=\sqrt{\gamma_{n}^2+\delta^2}\geq\gamma_{n}\geq n-d.
\end{equation}
The conditions \ref{it:limitpt} immediately follows from \eqref{eq:cond3foward} and \eqref{eq:cond3backward} and the fact that $g_n(x)$ and $g_{n}^{-1}(x)$ remain at distance at most $d$ from $R$.
\end{proof}
We will make use of the dynamics of those series. First we will need the following very handful lemma:
\begin{lemma}\label{lem:limtitspar}
  Let $X$ be a \ensuremath{\textnormal{CAT}(0)}\ space with a geodesic line $R$, $y\in\bndry{X}$ and $d>0$ be arbitrary. Suppose we have a sequence of lines $R_n$ that lie at distance at most $d$ from $R$ in the Hausdorff metric (in particular they are parallel to it). Then for any sequence of reals $t_k$  such that $R_k(t_k)$ diverges to $R(+\infty)$ , the angles $\angle_{R_k(t_k)}(R(+\infty),y)$ converge to the Tits angle $\angle(R(+\infty),y)$.
\end{lemma}
\begin{rmk}
  The case where the $R_n$ all coincide with $R$ is classical (see Part \Roman{deux} proposition $9.8$ of \cite{BridsonHaefliger201011}) and we will make use of it to simply the proof.
\end{rmk}
\begin{proof}
  As pointed out in the above remark, we already know that the angles $\angle_{R(t)}(R(+\infty),y)$ converge to $\angle(R(+\infty),y)$ when $t$ goes to $+\infty$. Let $\epsilon>0$ be arbitrary. We now fix $t$ such that  $\angle_{R(t)}(R(+\infty),y)\geq \angle(R(+\infty),y)-\epsilon/2$ and write $x$ for $R(t)$. Let us parametrize the problem (we will omit the subscript $k$ to simplify the notations since we will only consider one $k$ at a time) and define (see figure \ref{fig:layoutangles}):
  \begin{align*}
  \alpha&=\angle_{R_k(t_k)}(R(+\infty),y),&   \theta &=\angle_{R_k(t_k)}(y,x),\\
    z_2 &=\angle_{R_k(t_k)}(x,R(-\infty)),&     z_1  &=\angle_x(R_k(t_k),R(+\infty)),\\
    \varphi&=\angle_x(R_k(t_k),y), &=\beta=\angle_{x}(R(+\infty),y).
  \end{align*}
\begin{figure}[h]
  \centering
  \input{layoutofangles.tex}
  \caption{Layout of the angles.}
  \label{fig:layoutangles}
\end{figure}
Let $\delta\leq d$ be the Hausdorff distance between the two rays $R_k$ and $R$. Then euclidean geometry in the flat strip spanned by those two rays gives:
\begin{align*}
  \sin(z_1)=\sin(z_2)=\delta/d(R_k(t_k),x)\leq d/d(R_k(t_k),x)\xrightarrow[+\infty]{k} 0.
\end{align*}
So if we take $k$ large enough then we have $z_1,z_2\leq\epsilon/4$. We will assume this condition is satisfied in the following. But then:
\begin{equation}\label{eq:majorpi}
  \pi=\angle_{R_k(t_k)}(R(+\infty),R(-\infty))\leq \alpha+\theta+ z_2.
\end{equation}
In the infinite triangle with vertex $R_k(t_k)$, $x$ and $y$ the sum of the angles must not exceed $\pi$ so $\theta+\varphi\leq\pi$. But $\varphi\geq \beta-z_1$, so combining this with \eqref{eq:majorpi} gives:
\begin{equation*}
\theta+\beta-z_1\leq\pi\leq\alpha+\theta+z_2,
\end{equation*}
and  finally we have:
\begin{align*}
  \alpha\geq \beta-(z_1+z_2)\geq \angle(R(+\infty),y)-\epsilon/2-(\epsilon/4+\epsilon/4)\geq \angle(R(+\infty),y)-\epsilon.
\end{align*}
Since we always have $\alpha\leq\angle(R(+\infty),y)$ we just proved the lemma.
\end{proof}
The proof of the following proposition is a refinement of that of the lemma \ref{lem:split}.
\begin{prop}\label{prop:fixflat}
  Let $X$ be a \ensuremath{\textnormal{CAT}(0)}\ space with compact Tits topology. Suppose that a group $G$ fixes a boundary point $p\in\bndry{X}$ and acts cocompactly on $X$. Then $\can{X}$ admits a flat factor.
\end{prop}
\begin{proof}
 We can always assume that $X$ is boundary minimal. Fix a sequence of isometries $g_n$ and a ray $R$ as given by the proposition \ref{prop:veryhyp} above. Let also $x$ lie on $R$. Passing to a subsequence if necessary, we can assume that the isometries ${g_k}_{|\bndry{X}}^{|\bndry{X}}$ converge uniformly to a distance preserving map $b:\titsb{X}\rightarrow\titsb{X}$. Of course, $b$ fixes $R(+\infty)$. Fix $y\in\bndry{X}$ arbitrary. Then according to the lemma \ref{lem:limtitspar} above we have:
 \begin{equation*}
   \lim_{k\mapsto+\infty}\angle_{g_k(x)}(R(+\infty),y)=\angle(R(+\infty),y).
 \end{equation*}
If we push everything foward in the left hand side by $g_{k}^{-1}$ then we obtain:
\begin{equation}\label{eq:eq1}
   \lim_{k\mapsto+\infty}\angle_{x}(R(+\infty),g_{k}^{-1}(y))=\angle(R(+\infty),y).
\end{equation}
But we obviously have:
\begin{equation}\label{eq:eq2}
   \lim_{k\mapsto+\infty}\angle_{x}(R(+\infty),g_{k}^{-1}(y))=\angle_{x}(R(+\infty),b(y)),
\end{equation}
and since $b$ fixes $R(+\infty)$ we also have:
 \begin{equation}\label{eq:eq3}
   \angle(R(+\infty),y)=\angle(R(+\infty),b(y)).
 \end{equation}
Combining \eqref{eq:eq1}, \eqref{eq:eq2} and \eqref{eq:eq3} we get:
\begin{equation*}
  \forall y\in\titsb{X},\ \angle_{x}(R(+\infty),b(y))=\angle(R(+\infty),b(y)).
\end{equation*}
Now $b$ is surjective since every distance preserving map of a compact set is, so finally we obtain:
\begin{equation}
  \forall z\in\bndry{X},\ \angle_{x}(R(+\infty),z).=\angle(R(+\infty),z).
\end{equation}
The end of the proof is word for word that of the proposition \ref{prop:hypflat}.
\end{proof}
\begin{rmk}
  The author first thought that the $\pi$-convergence property failed here because the $g_n$ constructed in the proposition \ref{prop:veryhyp} need not act properly on $X$. However the author recently realised that the properness hypotheses in the lemma $8$ of \cite{papasoglu2009boundaries} is not used in the proof! This observation yields another proof (see the remark \ref{rmk:piconv}).
\end{rmk}
\section{Geodesically complete Tits compact cocompact spaces are flat}\label{sec:geodcomp}

We shall now use our results to prove the following:
\begin{prop}\label{prop:geodcompflat}
  Every geodesically complete cocompact and Tits compact space is flat.
\end{prop}
\begin{rmk}
  We suspect that the geodesic completeness is unnecessary (in section \ref{sec:geomact} above and in its improvement section \ref{sec:geomactbetter} bellow this hypothesis is exchanged with another one). The other hypotheses however cannot be dropped.
\end{rmk}
To prove this proposition we will make use of the following result of \cite{caprace2009isometry} (the result in their is actually a little better):
\begin{thm}\label{thm:capmon}
  A geodesically complete cocompact \ensuremath{\textnormal{CAT}(0)}\ space $X$ not reduced to a singleton admits hyperbolic isometries or admits a cocompact action by a group of isometries fixing a point at infinity.
\end{thm}
The proof of the above theorem is rather evolved since it makes use of the structure of locally compact groups with trivial amenable radical and that of totally disconnected locally compact groups (indeed it even makes use of the fact due to G. Willis and published in $1994$, \cite{willis1994structure}, that a compact subgroup of a locally compact totally disconnected group lies in a compact open subgroup. The existence of compact open subgroups, due to van Dantzig, was proved in $1936$ in \cite{dantzig1936zurtop}).
\begin{rmk}\label{rmk:capmon}
  The proof of the proposition $6.8$ in \cite{caprace2009isometry} lacks some details and we take the opportunity to warn the reader. The inequality therein $d(gy, y)\leq d(cg,x , y)\angle_{c_{g,x}}(gx, x)$ does not hold for any $y\in[c_{g,x},x]$ like written but for $y$ between $c_{g,x}$ and $x$ close enough to $c_{g,x}$ as soon as the angle $\angle_{c_{g,x}}(gx, x)$ is not zero. If it is zero, then we have $d(gy, y)\leq d(cg,x , y)/2n$ for $y\in[c_{g,x},x]$ close enough to $c_{g,x}$ so the rest of the argument works equally well in this case. We wish to thank P.E. Caprace for his quick answer concerning this point.
\end{rmk}
\begin{proof}[Proof of the proposition \ref{prop:geodcompflat}]
  Once again we decompose $X$ as $\ensuremath{\mathbf{R}}^k\times Y$ where $Y$ admits no euclidean factor. We want to show that $Y$ is reduced to a point. Suppose not. Then according to the theorem \ref{thm:capmon} above $Y$ either admits a hyperbolic isometry or a cocompact action by a group of isometries fixing a point at infinity. In either case it must admit a flat factor (remark that a geodesically complete cocompact \ensuremath{\textnormal{CAT}(0)}\ space is boundary minimal), despite its definition. So $Y$ is reduced to a point and $X$ is flat.
\end{proof}
\section{Proper semi-simple actions with locally finite stabilisers}\label{sec:geomactbetter}

We use the argument of P.E. Caprace and N. Monod in order to replace the geodesic completness hypothesis in the proposition \ref{prop:geodcompflat} by the existence of a proper semi-simple actions with locally finite stabilisers (definitions below). In this section, we say say that a topological group $G$ acting continuously on a metric space $(X,d)$ acts properly if for every closed ball $B$ in $X$, the set of all $g\in G$ satisfying $g(B)\cap B\neq\empty$ is compact in $G$ (this coincide with the previous definition when $G$ has a discrete topology). With say that $G$ acts with locally finite stabilisers if for every $x\in X$ and $r>0$, the stabiliser $G_{x}$ acts as a finite group of transformation on \ball{x}{r} (i.e. the quotient of $G_x$ by the pointwise fixator of \ball{x}{r} is finite). Then:
\begin{prop}
  Let $G\subset\isom{X}$ acts properly and cocompactly by semi-simple isometries with locally finite stabilisers on a \ensuremath{\textnormal{CAT}(0)}\ space without fixing a point on the boundary. Then it admits a hyperbolic isometry.
\end{prop}
\begin{proof}
  It is essentially that of the corollary $6.10$ in \cite{caprace2009isometry} (we only replace the use of van Dantzig' s theorem with the refinement of G. Willis). The only case that might cause problems is when $G$ is totally disconnected and we shall assume henceforth that it is. We only need to show that the proposition $6.8$ in there still holds. This is not clear at first since it is there made use of the smoothness of the action. We use the notations in the proof of the proposition $6.8$. Obviously the following still holds: if an isometry $g$ of any complete \ensuremath{\textnormal{CAT}(0)}\ space $B$ has order not bigger than $n$, then $\angle_{c_{g,x}}(gx,x)\geq 1/n$ (see also the remark \ref{rmk:capmon} above).

Like in the paper of P.E. Caprace and N. Monod, let us suppose the contrary. By cocompactness we obtain a sequence of isometries $g_n$ of $X$ that converge to an isometry $g$ and for each integer $n$ a point $x_n$ not fixed by $g_n$ such that $c_{g_n,x_n}$ converge to some $c\in X$ and $\angle_{c_{g_n,x_n}}(g_n x_n, x_n)\rightarrow 0$. Since the angle $\angle_{c_{g_n,x_n}}(g_n x_n, x_n)$ depends only on the geodesic segment $[c_{g_n,x_n},x_n]$ we can further assume that $d(c_{g_n,x_n},x_n)\leq 1$. Of course, $g$ fixes $c$, i.e. $g\in G_{c}$ and $G_c$ is compact. By Willis' theorem on the structure of totally disconnected locally compact groups in \cite{willis1994structure}, $G_c$ is contained in a compact open subgroup of $G$ : there exists a point $\hat{c}\in X$ such that $G_{\hat{c}}$ is open in $G$ and $G_c\subset G_{\hat{c}}$. In particular for $n$ large enough $g_n$ fixes $\hat{c}$. Let $d=2+d(\hat{c},c)$. By assumption the group $G_{\hat{c}}$ acts as a finite group of isometries on \ball{\hat{c}}{d} so up to taking a subsequence we can assume all the $g_n$ coincide on this ball (we write $g$ for this restriction) and have finite order $m$ (since $g$ lies in a finite group). Hence $\angle_{c_{g_n,x_n}}(g_n x_n, x_n)=\angle_{c}(g x_n, x_n)\geq 1/m$. This is the desired contradiction.
\end{proof}
\begin{rmk}
  If all the stabilisers are finite (and not merely locally finite) then the proof is easier. Indeed $G$ admits a compact open subgroup $K$ by the structure of totally disconnected locally compact group. But by compactness such a group lies in the stabilizator of a point so must be finite by hypothesis, hence discrete. We just found an open discrete subgroup of $G$: $G$ must itself be discrete. But then Swenson's result (see Theorem $11$ in \cite{swenson1999cut}), whose proof only uses basic \ensuremath{\textnormal{CAT}(0)}\ geometry, applies.
\end{rmk}
The same proof as the one of the proposition \ref{prop:geodcompflat} (after restricting ourselves to the canonical boundary minimal subspace) then gives:
\begin{prop}\label{prop:semisimlocalfinflat}
  Let $X$ be a Tits compact \ensuremath{\textnormal{CAT}(0)}\ space and suppose a topological group $G$ acts properly and cocompactly by semi-simple isometries with locally finite stabilisers on $X$. Then \can{X} is flat, or equivalently \titsb{X} is a unit metric sphere.
\end{prop}

\section{On the existence of hyperbolic isometries}

We include two results than show that the difficulty in finding hyperbolic isometries in a cocompact \ensuremath{\textnormal{CAT}(0)}\ space $X$ is related to this existence of flats in $X$. In particular a cocompact \ensuremath{\textnormal{CAT}(0)}\ hyperbolic space always admits hyperbolic isometries. We begin by the easiest one:
\begin{prop}
  Let $G$ be a proper group of isometries of a \ensuremath{\textnormal{CAT}(0)}\ space $X$ acting cocompactly and fixing a point $p$ on the boundary. Then either $G$ possesses hyperbolic elements or there exists a flat in $X$ with $p$ as boundary point.
\end{prop}
 Under the hypothesis of the proposition we know that there exists a geodesic line $R$ asymptotic to $p$. The parallel set to $R$ splits as $\ensuremath{\mathbf{R}}\times Y$ for some \ensuremath{\textnormal{CAT}(0)}\ space $Y$. Then:
\begin{lemma}\label{lem:coc}
  $Y$ is cocompact.
\end{lemma}
\begin{proof}[Proof of the lemma]
  Let $d>0$ be such that the translates by $G$ of any ball of radius $d$ covers $X$. Let $a,b\in Y$ be arbitrary, and note $\varphi_a$ and $\varphi_b$ geodesic lines asymptotic to $p$ corresponding to  $a$ and $b$ respectively and whose origin are on the same horosphere based at $p$. For all $k\in\ensuremath{\mathbf{N}}$ let $h_k$ be an isometry of $X$ that sends $\gamma_a(k)$ within $d$ to $\gamma_b(k)$. The same type of comparisons we used to construct $g_n$ in the proof of the proposition \ref{prop:veryhyp} show that $h_k(\gamma_a(0))$ remains uniformly bounded, and hence the series of isometries $h_k$ has a convergence subsequence. Let $h$ be a limit of such a subsequence. Then $h$ is an isometry that sends $R$ to a parallel line that lies within $d$ of $\gamma_a$. Hence $h$ stabilizes \para{R}, respects its product decomposition $\ensuremath{\mathbf{R}}\times Y$, and sends $a$ in $Y$ to within $d$ of $b$. Since $b$ is arbitrary we just proved the cocompactness of $Y$.
\end{proof}
\begin{lemma}\label{lem:isomexist}
  There exists an isometry of $X$ preserving the parallel set of $R$ and with non zero Busemann character at $p$.
\end{lemma}
\begin{proof}
  It is the same construction as in the previous lemma (with $a=b$ arbitrary) except we choose $h_k$ that sends $\gamma_a(k)$ to within $d$ of $\gamma_a(k+2d)$ to force the Busemann character to be positive.
\end{proof}
\begin{proof}[Proof of the proposition]
  Let $\ensuremath{\mathbf{R}}\times Y$ be the parallel set to $c$.  If $Y$ is bounded then it admits a circumcenter $c$. Let $h$ be an isometry given by the lemma \ref{lem:isomexist}. Then $h$ stabilises the line corresponding to $\ensuremath{\mathbf{R}}\times \{c\}$ in the decomposition of \para{R}. Since the Busemann character of $h$ is non trivial, $h$ acts as a non degenerates translation on this line (i.e. it the translation length is not zero). But then $h$ is hyperbolic.

 Suppose now that \para{R} is not bounded. Then it admits a geodesic line $L$ since it is cocompact by lemma \ref{lem:coc} and $L\times R$ is the desired flat.
\end{proof}
To simplify the proof of the next proposition, we begin with a definition and a lemma:
\begin{defi}
  Let $\theta$ and $\psi$ be two asymptotic rays in a \ensuremath{\textnormal{CAT}(0)}\ space. Then we define $\distinf{\theta}{\psi}$ by:
  \begin{equation*}
    \distinf{\theta}{\psi}=\lim_{t\rightarrow+\infty}d(\theta(t),\hat{\psi}(t)),
  \end{equation*}
where $\hat{\psi}$ is a reparametrization of $\psi$ such that $\psi(t)$ and $\theta(t)$ always lie on the same horosphere based at $\psi(+\infty)=\theta(+\infty)$ (hence it does not depend on the unit-speed parametrisation of $\theta$ and $\psi$, but might not depend solely on their asymptotic class).
\end{defi}
\begin{lemma}\label{lem:distinfdiv}
  Let $X$ be a cocompact \ensuremath{\textnormal{CAT}(0)}\ space. Suppose there exists for every $n\in\ensuremath{\mathbf{N}}$ two asymptotic rays $\theta_n$ and $\psi_n$ with $\distinf{\theta_n}{\psi_n}\geq n$. Then $X$ contains a flat.
\end{lemma}
\begin{proof}
  By cocompacity, it is enough to find flat half strips of arbitrary width. Take $n\in\ensuremath{\mathbf{N}}$ and the corresponding rays $\theta_n$ and $\psi_n$. For each $k\in\ensuremath{\mathbf{N}}$ let $h_{n,k}$ be an isometry that sends $\theta_n(k)$ to within a fixed distance of a fixed point. Up to passing to a subsequence we can assume that the rays $h_{n,k}(\theta_n)$ and $h_{n,k}(\psi_n)$ converge to asymptotic rays $\theta_{n}^\prime$ and $\psi_{n}^\prime$. In fact, the distance between $\theta_{n}^\prime$ and $\psi_{n}^\prime$ is constant equal to \distinf{\theta_n}{\psi_n}, so that they bound a half flat strip of width at least $n$ and we are done.
\end{proof}
Here comes the long expected proposition:
\begin{prop}
  Let $X$ be a cocompact \ensuremath{\textnormal{CAT}(0)}\ space. Then either it admits a hyperbolic isometry or it contains a flat.
\end{prop}
\begin{cor}
  Let $X$ be a Gromov hyperbolic cocompact \ensuremath{\textnormal{CAT}(0)}\ space. Then it admits a hyperbolic isometry.
\end{cor}
\begin{rmk}
  The corollary and the proposition are in fact equivalent statement by the theorem A of \cite{bridson1995existence}.
\end{rmk}
\begin{proof}
\begin{figure}[h]
  \centering
  \input{twotriangles.tex}
  \caption{When $h_n$ is elliptic.}
  \label{fig:twotriangles}
\end{figure}
The beginning of the proof is the same as that of the theorem $11$ in \cite{swenson1999cut}, i.e. we get a point $r\in X$ and a family of isometries $h_n\in G$ such that the angle $\varphi=\angle_{h_n(r)}(h^{2}_n(r),r)$ tends to $\pi$ at infinity. We assume moreover that $d(r,h_n(r))$ diverges (we only have to choose the $r_j$ and $r_i$ in the proof of the theorem $11$ of \cite{swenson1999cut} such that $d(r_i,r_j)\rightarrow+\infty$). Suppose that $X$ admits no hyperbolic isometry. We shall distinguish between two cases.

\textbf{An infinity of the $h_n$ are elliptic: }We can assume all the $h_n$ are elliptic and we fix a fixed point $x_n$ of $h_n$ for each $n$. Then the isosceles triangles $x_nh_n(r)h_{n}^2(r)$ and $x_nrh_n(r)$ are isometric and we label the angles according to the figure \ref{fig:twotriangles}. Now isosceles triangles in the euclidean space have acute base angles. By comparison, this must be the case in all \ensuremath{\textnormal{CAT}(0)}\ spaces, so that we have here $\alpha_n,\ \beta_n\leq\pi/2$. But then:
\begin{align*}
  \underbrace{\angle_{h_n(r)}(h^{2}_n(r),r)}_{\rightarrow\pi}\leq\underbrace{\alpha_n}_{\leq\pi/2}+\underbrace{\beta_n}_{\leq\pi/2},
\end{align*}
and so both $\alpha_n$ and $\beta_n$ must converge to $\pi/2$.

Let us pick $a>0$. Since $d(r,h_n(r))$ diverges, for sufficiently large $n$ we can take $s_n$ (resp $v_n$) on the geodesic joining $x_n$ and $h_n(r)$ (resp. on the geodesic joining $x_n$ and $r$) such that $d(s_n,v_n)=a$ and $d(s_n,x_n)=d(v_n,x_n)$. Let $\alpha_{n}^\prime=\angle_{s_{n}}(x_n,v_n)$ and $\beta_{n}^\prime=\angle_{v_n}(x_n,s_n)$. The item $(3)$ in the proposition $9.8$, Part \Roman{deux},  of \cite{BridsonHaefliger201011} shows that the sum  $\alpha_{n}^\prime+\beta_{n}^\prime$ is not lower that $\alpha_{n}+\beta_{n}$ and we  know that the latter tends to $\pi$. Since we always have $\alpha_{n}^\prime,\ \beta_{n}^\prime\leq\pi/2$ (the triangle $s_nx_nv_n$ is again isosceles), both $\alpha_{n}^\prime$ and $\beta_{n}^\prime$ must converge to $\pi/2$. But then by direct comparison with euclidean geometry, the isosceles triangle $s_nv_nx_n$  becomes arbitrary big. By cocompacity we can take a subsequence $n_k$ such that $s_{n}v_{n}x_{n}$ converges to some $svx$ where $s,v\in X$ and $x\in\bndry{X}$. By upper semi-continuity of the angle (see proposition $9.2$ Part \Roman{deux} in \cite{BridsonHaefliger201011}) we have $\angle_s(v,x)\geq\pi/2$ and $\angle_v(s,x)\geq\pi/2$. But the sum of those two angles cannot be bigger that $\pi$ so we have equality and $svx$ must in fact be a half flat strips of width $a$ as shown for example in \cite{BridsonHaefliger201011}, proposition $9.3$ Part \Roman{deux}. We just constructed half strips of arbitrary big width in $X$. It is then easy to construct a whole flat-plane by cocompacity.
\begin{figure}[h]
  \centering
  \input{twoinftriangles.tex}
  \caption{When $h_n$ is parabolic.}
  \label{fig:twoinftriangles}
\end{figure}

\textbf{An infinity of the $h_n$ are parabolic: }We can assume this time that all the $h_n$ are parabolic. It is shown in \cite{ballmann1985manifolds} that $h_n$ stabilises all the horospheres of some point $p_n$ on the boundary (this need not be the case of all its fixed points). We use the notations of the figure \ref{fig:twoinftriangles} and call $\theta_n$ (respectively $\psi_n$) the ray issuing from $h_n^{2}(r)$ (resp. $r$) and asymptotic to $p_n$. We can assume that the distances at infinity $\distinf{\theta_n}{\psi_n}$ are bounded since otherwise we can apply the lemma \ref{lem:distinfdiv} above. Let $d>0$ be such a bound. Let $a>d$ be chosen arbitrarily. From now on the construction is similar to the case where the $h_n$ are elliptic. Let $s_n$ (respectively $v_n$) be on the ray from $h(r)$ (resp. $r$)  to $p_n$ such that $d(s_n,v_n)=a$ and $s_n$ and $u_n$ both lie on the horosphere based at $p_n$ (so we have an ``infinite isosceles'' triangle). Let $\alpha_{n}^\prime=\angle_{s_{n}}(p_n,v_n)$ and $\beta_{n}^\prime=\angle_{v_n}(p_n,s_n)$. It is proved in the proof of the lemma $4.3$ of \cite{ballmann1985manifolds} (or in the proposition $9.8$ Part \Roman{deux} \cite{BridsonHaefliger201011}) that the sum  $\alpha_{n}^\prime+\beta_{n}^\prime$ is not lower that $\alpha_{n}+\beta_{n}$ and we  know that the latter tends to $\pi$. We still have $\alpha_{n}^\prime,\ \beta_{n}^\prime\leq\pi/2$ (to see this one can either take finite isosceles triangles that converge to $s_nv_np_n$ or use the fact that the horosphere based at $p_n$ through $s_n$ and $v_n$ is convex) and hence  $\alpha_{n}^\prime$ and $\beta_{n}^\prime$ both converge to $\pi/2$. Now by cocompacity we can assume that the triangle $s_nu_np_n$ converges to $sup$ where $s,u\in X$ and $p\in\bndry{X}$. By upper semi-continuity of the angle, $\angle_s(p,u)$ and $\angle_u(p,s)$ cannot be lower than $\pi/2$. Since there sum is lower than $\pi$ they must both be equal to $\pi/2$ and the triangle is in fact a half flat strip of width $a$ (again proposition $9.3$ Part \Roman{deux} of \cite{BridsonHaefliger201011}). We constructed flat strips of arbitrary big width: we conclude like in the previous case.
\end{proof}
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Health News Florida: Department of Health proposes cutting 1,600 jobs: News. Politics. Media
According to Health News Florida: Under fire from lawmakers, the Florida Department of Health has proposed a sweeping plan to reorganize — and shrink — its operations.
According to Health News Florida: #
Under fire from lawmakers, the Florida Department of Health has proposed a sweeping plan to reorganize — and shrink — its operations. Among other things, it would move the state out of the primary-care business. #
The recommendations, released in a 154-page report late Tuesday afternoon, call for cutting 1,608 department jobs and consolidating dozens of divisions and bureaus. One of the proposals would buck the powerful doctors' lobby by lifting a requirement that the department secretary be a physician. #
Health News Florida also indicates that the Department of Health report calls for the state to stop paying for primary-care services at county health departments to save about $22.3 million. #
The article also says the report coincides with some of Gov. Rick Scott's transition team recommendations, such as moving away from primary care and allowing a non-physician to serve as department secretary. #
It adds, "Scott's transition team went further than the report's recommendations and called for a merger of the department with the Agency for Health Care Administration." #
Other changes include: #
A reduction of the department's organizational chart, going from 11 divisions to six and 50 bureaus to 18.
Many programs would be moved to other state agencies, including the Department of Children and Families, the Department of Environmental Protection, and the Agency for Health Care Administration.
Other programs would be farmed out to private contractors or see their funding disappear.
Utility-backed renewable energy legislation returns in the Senate: News. Politics. Media
Union members decry Thrasher bill they say would 'silence' them: News. Politics. Media
'Attrition Through Enforcement' group: Rick Perry 'weak' on immigration
Numbers USA — an organization that supports Attrition Through Enforcement immigration policy and wants lower immigration levels — blames Rick Perry's results in the Florida straw poll on his weak stance on immigration enforcement.
McCollum, Bondi file motion in health care lawsuit
Florida Attorney General Bill McCollum, along with A.G.-elect Pam Bondi, announced the filing of a motion responding to…

redpajama_set_name: RedPajamaCommonCrawl


The first two discs in this 'Complete Symphonic Works' series (10/11, 11/11) were outstanding. This third is wholly exceptional. The presence of the overture In Autumn and the Old Norwegian Romance with Variations gives the programme a Beechamesque feel. But Aadland and his astonishingly well-integrated German ensemble – by this I mean that they are guided into a natural-sounding Nordic style – need fear nothing by way of competition, not even from the RPO's dream woodwind section.
Aadland's Romance – unlike Beecham's it is complete and not marginally reorchestrated – becomes a kind of Norwegian Enigma Variations avant la lettre, mixing a large degree of symphonic seriousness with wit and Griegian nostalgia. The demonstration-class recording showcases a weight and colour of orchestration that puts this Grieg score in the correct but rarely considered position of contemporary to Strauss's early orchestral masterworks. A piece that (sssh, even on the two Beecham recordings) can sound bloated and occasional claims a place here alongside, indeed anticipates, the disguised turn-of-the-century unnamed symphonies of Rachmaninov, Elgar and the like.
A similar seriousness but never overblown grandeur informs In Autumn. Aadland has already shown us in this cycle that he is good at correctly scaling miniatures both in joy and in grief. He seconds those achievements here in the characterisation of the Lyric Suite and the Sigurd Jorsalfar music, while the five minutes of Klokkeklang ('Bell-ringing') become a spooky shadow of the Bøyg's music in Peer Gynt (eagerly awaited in this series). A compulsively playable record, hugely recommended.


Mariuca's Perfume Gallery: Busy but Lazy!
I was talking to my friend Monica earlier and we both agree that the lazy bug is in the air! We're just feeling so sluggish nowadays. For me, that's really not good since I actually have a lot of work on my plate like the new life insurance quotes job I received yesterday. As usual, my deadline is tight and with the weekend just around the corner, it is taking a lot of effort for me to get started on my work.
When I first started working, I didn't know much about insurance at all. I knew that it was important and one should consider it during major life moments like when you're buying a new home or getting married and having a baby but that was about it. I failed to appreciate the importance of insurance in ensuring our comfort in practically all aspects of our lives.
Over the years, I worked on a lot of insurance products, which helped me understand the product better. And now, insurance plays an important part in my married life. When you're in it for the long-run with your other half, you have to make security and protection a main priority. B and I have several insurance plans and I'm not opposed to upgrading when necessary. It's better to be safe than sorry right?
Anyway, I am feeling so lazy despite having so many things to do! I need a pick-me-up to get me started on my new job. Right now, all I can think of is my warm cosy bed. More coffee perhaps?
yes! lazy bug is in the air!
Rizal u FC2 lah! don :'( :'( ok....hahahahahah!
Hahahahha always lazy la we both, but what to do cause it's Fri edi…weekend mode!
Wah Rizal 3 points edi, syabas!
That's terrific Ayie, I can't imagine having my fave coffee only once a week!!
Me too Mon! I adore coffee!
coffee doesn't work on me anymore Marzie! I can have like a dozen mugs and I still fall asleep even when I'm not sleepy. I think I may have developed some kind of a sleeping disorder.. hmm..
Insurance is one of those necessary evils.
Sorry about the laziness. Sometimes it is the weather and other times it is because of too much work.
Usually a nice sunny day will pull you out.
Insurance is that there must be something basically. However, the article to see you feel your feelings.


Animas Security LLC concentrates the business on physical security to include installation and usage of electronic surveillance systems. The surveillance systems emphasize either Apprehension or Prevention, the choice driving the type and installation of the video systems. Neighbor-to-neighbor disputes, shopkeeper theft prevention, physical security for installations and facilities are some of the targets for the company. Event security is also provided when requested. Investigations work continues as Animas Investigations requests. The two subsidiaries provide a location for firearms instruction and a framework for providing instruction in various skill sets useful to the preservation of individual safety, family safety, and consultation for home, business and vehicular safety. Provides expert witness testimony in matters of security.
The Wolverine School of Personal Safety operates the Red Dawn Range offering Concealed Handgun Permit training, a wide range of advanced shooting and unarmed skills training.
in 2018 Animas Investigations was subsumed under Animas Security as a primary Professional Server of Legal Process. Investigations are relegated to that required by Service of Process.
Name change from DC Systems and expansion of investigative services and training and security services and formalizing a previous and ongoing professional relationship with the company. Recent activities included a four-month security with the Sky Ute Casino in Ignacio, CO, several investigations including two homicide investigations for the Office of the Alternate Defense Council in Colorado. In addition, numerous investigations with emphasis on surveillance were conducted in conjunction with an investigative agency in New Mexico. The successful business was sold in 2009, maintaining a professional business relationship with the company.
The position required the principal to take responsibility and authority for the physical security of the TE-EPSS Academy compound, including vetting of indigenous personnel, supervision of construction of security barriers, and leading a security force of twenty-two Americans and seventy Iraqis. Security was intended to physically safeguard a large American presence and a much larger Iraqi student population in the face of probable attempts at infiltration or penetration of the perimeter. Physical security requirements led to the clearing outside the perimeter security fence of the debris of the war and years of Iraqi neglect. The requirement also included the safe transport of personnel from Taji to Baghdad, Iraq. The contract was ended by the U.S. Army Contracting Office after the first year.
Owner and Director of DCSystems of Durango, Colorado.
Conducted legal, private, and insurance investigations; provided Security Services and Training to individuals, corporations, and law enforcement entities. Specialized in threat development with regard to physical security of persons, groups, facilities, and their activities. Participated as a principle trainer of Federal Aviation Administration Air Marshals. Operated as Security Manager for oil explorations in Algeria where the threat was religious fundamentalism. In that regard the principal coordinated with the security officers of various country embassies; coordinated emergency evacuation plans for medical evacuation and wholesale evacuation; coordinated with and closely interacted with Algerian Military Forces for close-in armed security; advised the corporation on procedures and wrote security plan for facilities and personnel. Conducted Security Survey for oil explorations and discovery operations in Colombia where the threat was street crime and Marxist guerrillas and narco-traffickers: Surveyed physical surroundings of both field and urban locations for corporate and personnel security; devised, wrote and provided a comprehensive plan for physical and operational security; recommended measures for communications security. Conducted Vulnerability Survey for a chemical redistribution company in locations in Mexico City and Guayaquil, Peru: Surveyed sites for points of access by unauthorized persons, workplace hazards, financial and operating vulnerabilities and guard forces for competence and operating procedures. Currently certified as Firearms Instructor, Use of Force Instructor, Port and Ship Security Field Analyst.
Drug Enforcement Administration, Washington, D.C.
Office of the Deputy Chief of Staff of the Army for Operations, Pentagon, Washington, D.C.
Chief, Security Assistance Control Team, Grenada.
Student, Armed Forces Staff College, Norfolk, Virginia.
U.S. Army Foreign Area Officer Course, Ft Bragg, North Carolina.
U.S. Army Ranger School, Ft Benning, Georgia. .
Psychological Operations and Civil Affairs Officer, Tieu Atar, Republic of South Vietnam.
Teaching you how to prepare for and survive the unexpected!


Stopsley Community Primary School - Beatbox Collective!
Inspired by Young Voices, Ru (from Beatbox Collective) came to work with all the children providing them with some top beatboxing tips.
The children thoroughly enjoyed learning about the different sound effects that can be made by manipulating our voices in a number of ways. Ask your child to demonstrate 'boots' and 'cats'!
Not only did they benefit from the experience of a professional beatboxer, some of the children had the opportunity to put their beatboxing skills to test within a 1 on 1 battle with a friend.
They all did tremondously and benefited from exploring other avenues in which they could pursue their musical talents.


text:
Q: Change variables based on radio selection in Angulajs I am building a calculator in angularjs as a wordpress php plugin. Passing parameters as variables via a shortcode.
I have a jquery ui-slider with min, max and step values set by passed params:

weight_minimum}"
     max="{$this->weight_maximum}" step="{$this->weight_increment}"
     ng-model="weight">

I also added two radio buttons after the slider:

    {{ "Pounds" | translate }}
      <input type="radio" checked="checked" name="radio"
             ng-model="selection" value="lbs">

    {{ "Kilograms" | translate }}
      <input type="radio" name="radio" ng-model="selection" value="kg">

How can I change min/max values of the slider based on radio selection? I tried using ng-if but that is for showing/hiding html. I also tried adding an if statement in the app.controller with no success.
if ($scope.selection == 'lbs') {
  weight_maximum = $this->weight_maximum;
} else if ($scope.selection == 'kg') {
  weight_maximum = $this->weight_maximum / 2.2;
}

A: Use ng-change event example in the plunker
Code : html

     Pounds
      <input type="radio" checked="checked" name="radio" ng-change="changUnit()" ng-model="selection" value="lbs">

     Kilograms
      <input type="radio" name="radio" ng-model="selection" value="kg" ng-change="changUnit()">

Min : {{weight_minimum}}

Max : {{weight_maximum}}

slider value : {{weight}}

your controller :

    var app = angular.module('plunker', ['ui.slider']);

    app.controller('MainCtrl', function($scope) {

      $scope.selection = 'kg'
      $scope.weight_minimum = 10;
      $scope.weight_maximum = 1000;
      $scope.weight_increment = 1;
      $scope.weight = 50;

      $scope.changUnit = function() {
        if ($scope.selection == 'lbs') {
         $scope.weight_maximum = $scope.weight_maximum * 2.2;
        } else if ($scope.selection == 'kg') {
          $scope.weight_maximum =  $scope.weight_maximum / 2.2;
        }

      };
    });

redpajama_set_name: RedPajamaStackExchange


Kintavar Exploration Inc. (TSX-V: KTR) (FRANKFURT: 58V) is pleased to announce drilling results for the eastern portion of the Sherlock zone. Drilling in December focused on testing the four main horizons (A-D) which are stacked together in the Sherlock zone. Drill holes MS-37 to MS-39 were testing the south-eastern extensions of horizon A while hole MS-41 was primarily testing the north-eastern horizon C extensions. Drilling assays of hole MS-46, which was testing horizons B, C and D in the western extension, are pending.
"The eastern portion of the Sherlock zone was less worked with the only drilling and trenching to date being the Toby area. The December program validated our interpretation and fully opened up that area. Going from only horizon B to the East to now confirmed A, B and C horizons is a major success for our geological model and gives us the confidence to do another large step out to the East to add more mineralized material in what we believe is shaping up to be a large volume open pit potential project. Next steps would be confirming horizon D to the East and to keep extending the entire mineralized sedimentary sequence as we follow the main structure," comments Kiril Mugerman, President & CEO of Kintavar.


text:
Miami University Chamber Singers and Choraliers perform Oct. 15
The Miami University Chamber Singers and Choraliers, conducted by William Bausano, present their Fall Concert, 7:30 p.m. Friday, Oct. 15, in Hall Auditorium.
The Chamber Singers will open the concert with two pieces sung in German from the 19th century composers, Franz Schubert and Robert Schumann. The remainder of their portion of the concert will include works by contemporary American composers, Peter Durow and Robert Linn.
The portion of the concert of the Choraliers, Miami University's oldest and largest women's chorus, begins with a chorus by 17th century French composer, Jean Bapiste Lully, and is followed by works of contemporary American composers.
The choir was established in 1908 and "has grown from an informal 25-member society into a premier chorus composed of singers representing majors from every division in the university," according to the department of music website.
After a series of live-streamed concerts in past semesters, Professor of Music and Director of Choral Activities William Bausano said he is "very happy to be able to have person-to-person rehearsals." He also looks forward to "having the choirs sing for a live audience."
Admission to the event is free.
The event will also be live-streamed and the link is available through the Miami University Event Calendar found here.
briefsleft
Miami Police reports increase in fire alarms triggered
Local labor leader calls for solidarity with train workers
Talawanda physics class hosts pumpkin catapult demonstration
OPD officers return from Hurricane Ian relief efforts
Miami Hillel reports vandalism of sukkah at its building
Outdoor Pursuit Center leads Spooky bike tour across Oxford
Refugee and women's rights advocate to speak at Miami's Hamilton campus
Candidates for house of representatives face off in debate
MUPD responds to 15 false fire alarms in one week
Oxford trick-or-treating, upcoming Halloween events set


text:
Video: Porsche 911 GT3 Takes on M6 Competition Package in Drag Race
Home » Models » BMW M6 » Video: Porsche 911 GT3 Takes on M6 Competition Package in Drag Race
March 26, 2018 / 2 minutes read
The BMW M6 in Coupe guise is no longer available for purchase. The Germans pulled it out of production as they are readying the ground for the new M8 but one essential piece from it will continue to live on – the engine. The new M5 is using a revised version of the 4.4-liter V8 twin-turbo mill used on the Competition Package models. Unlike the standard M6, the Competition Package versions came with 600 PS and a bit more torque, towards the end of the production cycle and that made quite a difference.
Understandably so, considering the M6 was on the heavier side of things and could use all the horsepower and torque it could muster. However, being a rear-wheel drive model, it did have some trouble putting all of it down. In order to do so in the best possible way, one would have to get super sticky tires and then a super sticky surface to try them out on, something like a drag strip, the likes of the one in the video below.
In the footage we can see the M6 going up against a rather unusual rival, a Porsche 911 GT3 of the 911.1 assortment. I'm saying 'unusual' because the Porsche is more of a track fiend rather than a drag strip monster. Yet, the power to weight ratios are pretty close and both cars come with rear-wheel drive therefore this is an interesting battle. The GT3 was also ECU tuned, making closer to 500 HP at the time of the race, out of its 3.8-liter flat six engine.
The race result was also quite interesting and, as it is usually the case on professional drag strips, took into account only the times posted after the cars set off, so don't jump to conclusions just because the M6 took off too early. It should also be noted that the M6 apparently didn't use launch control for some reason, while the Porsche did.
competition package
More from BMW M6
5 responses to "Video: Porsche 911 GT3 Takes on M6 Competition Package in Drag Race"
Icebreakerr says:
M6 <3
Scott Biden says:
That's a really good time for the M6. I'm surprised he didn't use launch control.
Wout says:
No matter how much I love BMW but this is a stupid comparison. GT3 is a track car. It's hailed for it's cornering, sharp steering, feedback, grip, weight balance, etc. Straight line racing was never the intention. Compared to other Porsche models it's even under-powered.
Cheslyn Owen says:
This is so true, 100% agreed. We have to start comparing apples to apples. The M4 GTS is more a direct competitor to the GT3. And we all know how that went. BMW will have to do a brand new sports car from the ground up to compete with a 911 and Mercedes GTR. And not the basic 3/4 series as its platform to take on these cars.
disqus_8YchtRit8p says:
BMW do not look to other manufacturers for product planning, when have they ever done "a brand new sports car from the ground up"? M is across their range, never a dedicated model. 8 Series is already racing even before going against AMG in the marketplace & I'll wait on new M3/4 before considering a rear engine VAG, a layout so archaic every other manufacturer has abandoned it.


So sorry for your loss. Our hearts are with you.
I love you very much Great Grandma. You are with the creator now where we are all intended and I will always remember your big happy smile is shining down on us.
So honoured to have met such an amazing woman. You will be loved and missed my so many. So happy Lily was able to meet you.
Until we meet again, I will cherish all the beautiful memories of you. I love you so much Grandma.
Fly high my sweet Grandmother. I hope you and Grandpa are dancing among the stars. You will be so missed by so many. I am so happy that I had 40 years with you and Damian and Raven were blessed with meeting you. Rest in Paradise.
I love you Granny. I miss you so much. I'll always cherish all the memories I have. Your amazing smile will always light me up inside.


Issue for Jan. 9, 1976 called v.4, no. 27 but constitutes v.5, no. 1; issue for July 7, 1989 called v.18, no. 11 but constitutes v.18, no. 13.
ing his first six-year term in the VS.
day Jan. 22 at 10 a.m. on Ch. 12.
Jewish Federation over the years.
fe?^ rlgl.!*. -. to the coup*.
and elsewhere in the world.
Center, now Temple Beth Am.
ture at 7205 Royal Palm Blvd..
School to the rear of the Temple.
rael, will be featured on the Jan.
manager of Bank Leumi le-Israel, which sponsored the dinner.
He has been responsible for major u.
hk const icntious and intelligent work as a law.
Sacks, Sam Schwartz. Milton Trupin, Pan!
in advance, in the privacy of your home or office.
fund your prearranged funeral in advance.
supplied and paid for whenever needed.
roe quardian plans program offers Insurance Funded or*rr.n.~ ___.
Insurance Funded Prearranged F^aip" am.
stork journey Sunday Jan. 29.
omptly at 5:30 p.m. that day.
federation building at 8358 W.
rean Mile in Fort Lauderdale.
narily open from 10 to 6 p.m.
owing that Sunday night Jan.
iseum to catalogue the objects.
countries between 1948 and 1950.
be enriched by the exhibit"
Mrs. Levy and her husband.
and Harry Shiller to be honored.
Danny Tadmore will speak here.
h??2n*' P******* for the Jan. 22 and Jan.
Padek; Maxine and Al Stein.
second one on Jan. 29.
Center at 6501 W. Sunrise Blvd.
Jewish Center, 9101 NW 57th St.
ing and entertaining. Jerry Kaye.
held st the home of the Lerners.
what pleases the Reagan Administration.
they saw it, is good for Israel.
whether they want it to or not.
and judged guilty of the crime the following Aug. id.
wanted to go to his grave in peace.
filthy, perverted Jew of New York."
On Dec. 22,1983. only two weeks ago. Georgia1!
dam iXINCSrri SI, Miami. Fla M1M Ft.ooa137J.4aW "'"
something akin to a resolution.
yet not exactly like it.
in Beirut on Oct. 23.
him to hold his peace.
words, people simply ran away.
WO s. Or are they?
as in the 1928 failed campaip"
demonstrate to nns. *>l **!
was a political entity. *"
then of Floral Park, N.Y..
dore Herzl died July 3.1904.
Flax is an inveterate letter writer.
Howing the release of U.S.
fald arrived in Beirut on ^ lfl83, th" kmmt numb,r m represents s drop of 98Ipsrc.
Secretary of State for Human "^E^wlLmSbv*bmins.
Rights and Humanitarian Affairs numbers rstaassd by ADrams.
/. K. WEINRAUB & Co., Inc.
2000 City Hall Dr.. LauderhiU.
City Library will meet at 7 p.m.
day Jan. 18 and 25. Call 474-5880.
and Moshe Waldoks at 1 p.m.
going world affairs at 10 a.m.
the value of antiques at 7:30 p.m.
Oakland Park Blvd. Abraham J.
The shalom show Is funded, in part.
8358 W. Oakland Park Blvd.
years old Saturday Jan. 14.
ber Leadership Mission to Israel.
ty's Hollywood and Los Angeles.
being memorialized at 2 p.m.
National Fund of Broward. Dr.
1965 through 1981 ... JNF.
land Park Blvd.. phone 561-4812.
Concord Hotel in the Catskills.
Jan. 15 Happy Hour ("hour?"
Graenxweig. That's at 8 p.m.
Jan. 18 at Pine Island Ridge.
Tamarac Chapter at 1:30 p.m.
Jan. 19 at Tamarac JC.
Sam Tamir and Jack Kohn.
open a Broward outlet soon .
rabbinical students in tat .
All Publix Bakeries open at 8:00 A.M.
lie Monday Jan. 23 when Prof.
specialty food section of your favorite supermarket.
p.m., Temple Beth Am, Margate.
Jewish Center Men's Club: 9 a.m.
CONDO UJA RALLY: 10 a.m.
State of Israel Bonds: 830 a.m.
Temple Beth Torah: 6:45 p.m.
Circle of Yiddish Crabs: 10 a.m.
lunch Noon. 1 p.m. masting.
tury Village Chspter: 12:30 p.m.
ter. Wine and cheese $3.50.
Golds Meir Chapter: 1 p.m.
Whiting Hall, 6767 NW 57th St..
house, members $14; others $16.
Post and Auxiliary: 7:30 p.m.
$25. Shirlee and Group entertain.
Polynesian Gardens UJA: 7 p.m.
2 Meals daily included fi^fi *"
By Fran Rasumny Barrett, JJ).
Q. My wife has a regular^.
ting her an electric out.
fill out a detailed form.
January 21. 1M4 1:30 pm 4:30 p.m.
Contact: owner Director Morgan I. levy. C.C.D.
A Well Balanced Summer Program.
gion, as speaker from 1 to 3 p.m.
local, state and national issues.
by games at 7:30 p.m.
Gold Coast Section 11:30 a.m.
at Temple Beth Am, Margate.
Club on State Rd. 84.
increased from 300 to about 2500.
Annette Kay to co-chair the Jan.
l/we wish to contribute $.
the Cantorlal Tribute Concert on January 22,1084.
My check in the amount off----------------la enclosed.
at 6 p.m. followed by dinner.
presented at 8 p.m. Sunday Jan.
Steiner. Bruce Syrop, Ed Kourt.
by Ronald (' Chalker. at 8 p.m.
Saturday Feb. 4 at the Temple.
1099 Pine Island Rd.. Sunrise.
B not Mitzvah at the 8:15 p.m.
youth all over the world. "
Bar Mitzvah during the 5 p.m.
Lucas Barnett. son of Ms.
at Temple Beth Torah. Tamarac.
at Temple Beth Am. Margate.
came a Bat Mitzvah at the Dec.
At the Saturday morning Jan.
7 service at Ramat Shalom.
rid., he h won aw.
pm Saturday lim.lpm, Sunday am., I p.m BaaM Pta!
TEMPLE SMOLOM tStlAtiO). US BB lit* A**.. .
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text:
Datejust Model
Models Datejust
Datejust (61)
Datejust Lady (9)
Datejust 41 (3)
Datejust II (1)
Oyster Perpetual (48)
Air King (35)
Submariner (18)
Daytona (12)
Day-Date (8)
GMT Master (6)
Yacht-Master (4)
Deepsea / Sea-Dweller (3)
Cellini (1)
Oysterquartz (1)
Sky Dweller
Rolex Datejust Watches
The Rolex Datejust first appeared in 1945.
At the time, it was the world's first automatic wristwatch to display the date, behind a small window at the three o'clock position, and also the first to bear the word "Chronometer" on the dial.
These two characteristics, both exclusive to Rolex, were the inspiration for its name: "Date-just.
" The "Cyclops" date magnifier was added a few years later, allowing the wearer to magnify the date by two and a half times.
It rapidly became emblematic of Rolex's style.
Like all Oysters, the Datejust features a screw-fitted crown and back cover, ensuring that it remains waterproof, as well as a grooved bezel with striations of varying sizes.
The Datejust shows the date in a small cut-out window at the 3 o'clock position, with a magnifier attached to the glass as well as a flexible bracelet with five small, fluted links known as the Jubilee bracelet in what was, at the time, a departure from tradition for Rolex.
> Read full description
Rolex Datejust watch in gold and stainless steel Ref: 1601 Circa 1977
Rolex Datejust watch in stainless steel Ref: 1601 Circa 1970
Rolex Datejust watch in stainless steel Ref: 16014 Circa 1978
Rolex Datejust watch in stainless steel Ref: 126200 Circa 2021
Rolex Datejust watch in gold and stainless steel Ref: 16203 Circa 2000
Rolex Datejust watch in white gold Ref: 178279 Circa 2010
Rolex Datejust watch in gold and stainless steel Ref: 116234 Circa 1991
Rolex Datejust watch in stainless steel Ref: 1601-3 Circa 1966
History and quote
THE DATEJUST, INNOVATIVE MECHANICS SIGNED BY ROLEX
From a mechanical point of view, since its launch the Datejust has been an innovative watch. It is the first watch that combines a watertight case, automatic movement, a date box and the precision of a chronometer. Quality assurances can be found, as a selling point, directly on the dial of the watch, below the Oyster Perpetual Datejust officially certified chronometer.
The term Oyster
Refers to the watertight sealing of the case, which is comparable to that of an oyster.
The term Perpetual
It means that the watch has a mechanical movement with automatic winding, it draws energy from continuous natural movements of the arm.
The term Datejust
It might seem self-evident if we do not put it in the context of the time. In 1945, mechanical watch-making was still in the early stages of the automatic era. Finding how to get enough movement energy to make the date change instantly every night at midnight on the dot was a challenge that we do not think about today. For the time, it was a real technical breakthrough, far more spectacular than a date window that started to turn at 10pm and finished moving at 2 o'clock in the morning. It was not until 1955 that the calibration allowing the instant jump from one day to the next was finalised and adopted by Rolex.
On the watch face, the words "officially certified chronometer" specifies that the quality of the watch chronometer is certified by an independent body, which depends on the Federation of the Swiss Watch Industry, called the COSC.
STYLISTIC CODES OF THE DATEJUST WATCH
From a stylistic point of view, there are four key points about the Datejust: its Cyclops crystal, its bracelet, its face and its bezel.
Printed on a very small disc, to improve readability Rolex was quick to add a magnifying glass over the date window. This oversized eye has become a Rolex signature, dubbed by collectors as the Cyclops, a nickname later used by the manufacturer. It magnifies the actual size of the date by 2.5 times.
The Jubilee strap
Was launched in 1945 on the Datejust. This new strap composed of five-piece curved links was created as an alternative to the Oyster strap, with its flat three-piece links. On the second-hand market, it is not uncommon to find Datejust mounted on leather straps or on NATO fabrics. Its absence can bring a lower turnaround in the auction room.
The dial
Datejust's major interest, compared to the rest of the Rolex collection, is the diversity of watch dials in existence and the rarity of some of them. This had always been the piece that Rolex dial-makers had the most freedom with making. Aficionados, always wanting to seize greater ownership of the brand, have christened several specific dials, which has made the price of the watches that have them increase. The linen dial appears to have been woven, the pie pan with its outer edges resembles a pie mould, the Buckley dial, white with Roman numerals in black paint, pays homage, not to the singer… but to the Rolex representative from the East Coast of the United States, John Buckley! The markings at 6 o'clock can also alter the price of a Datejust. If the word Swiss is flanked by two T, or better still by two O (sigma dial), the watch you have in your hands is a rarity! Generally speaking, the most frequently seen colour of dial at auction is silver (comprising 24% of auctioned Datejust watches), followed by black (18%). Rare colours are bringing in a good price at auction, which is why it is really important to verify that the dial is original.
The bezel
Finally, the many bezels available for the Datejust are a detail that is the most likely to alter the price of a second-hand Datejust. Fluted, smooth, bark, crenellated… so many variants that add to the charm of the model.
A common thread between celebrities, politicians, intellectuals and the public at large, the Datejust will always be a popular watch. Choosing it will not only bring you joy but also the knowledge that you have invested in something that will always have outstanding worth. Don't the international correspondents always say that the Rolex is the most international currency there is?
THE PRICE OF THE SECOND-HAND DATEJUST ROLEX: A STORY OF REFERENCES
The Datejust is seen as a safe bet in the watch-making industry: as proof, its price has multiplied by 3.7 in 10 years, across all models! Each Datejust has its own unique characteristics, their value differs depending on their year of production, their material, their reference...
Datejust references with 4 digits
(the oldest references): have seen their worth increase by 2.6 times in 10 years. Among these references, the most common on the second-hand market is 1601. For a steel model, the price has risen 37% in 10 years. Another fact: if the watch is accompanied by its original box and papers, its value jumps immediately by 20%. Furthermore, if the watch is guaranteed, the price can be increased by about €500 for a Rolex. But rareness often also plays a part in the asking price! Of the 4 digit references, the Reference 4467 is the most rare. The first Datejust to come out, its price is spectacular, slightly above €14,000 in 2014.
5 digit references
(references called "quick date", which will take over from the 4 digit references): they have seen their worth increase twofold in 10 years. Of these, the reference 16030 is the most widespread.
(the most recent Datejust references): these are much rarer to find at auction as they are still available to purchase through the Rolex catalogue. Ref. 116203, for example, costs just over €4,000. Its original look and two-tone dial make it a most sought-after model by those who wish to stand out.
All the figures of this study emanate from the databases of LuxPrice-index© , who survey the results of more than 200 000 auctions from the main auction houses around the world across a 12 year span.
Analysis report
View quotes for watches Rolex Datejust thanks to almost 400,000 sales results decrypted by our expert
The history of Rolex Datejust
The Rolex Datejust first appeared in 1945. At the time, it was the world's first automatic wristwatch to display the date, behind
The Rolex Watches collections
Rolex Datejust Watches Rolex Oyster Perpetual Watches Rolex Air King Watches Rolex Submariner Watches Rolex Daytona Watches Rolex Explorer Watches Rolex Day-Date Watches Rolex GMT Master Watches


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