But in an interview on Monday night's The Kelly File, Ammon Bundy could find no quarter for even a little bit of his nonsense. The interview consisted mainly of Ammon Bundy trying to draw Kelly into an argument about the nature of federal authority through the use of "clever" rhetorical questions, but the real star of the show was Kelly's stony glare.
Kelly: Your brother, Ryan, apparently told the "Oregonian," that you are willing to kill and be killed, if necessary here. Do you agree with that?
Bundy: That statement was taken way out of context. I've already talked to my brother about that, and so no, i wouldn't agree with it, because it wasn't his statement.
Kelly: So you will not let this come to violence?
Bundy: Well, I have a family. I have six children, a beautiful wife. I have a wonderful home. I have a business, several employees. I'm not here to die. I'm here to defend my freedoms and my liberties, and will I? Yes. But it's not going to come to that.
Even in context, that doesn't sound so much like a walkback as it does overconfidence that the federal government will once again cave in. If they don't, Bundy is still saying he will "defend" his freedoms, and he's still holed up in federal property with a gang of armed men. Maybe the feds should send Megyn Kelly in after them.


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text:
about work contact 
A More Progressive Universe
Originally published in Gather Journal: Heroines Issue
Rebecca Sugar became a pioneer in 2013 with the launch of Steven Universe, the first show in Cartoon Network's 21-year history created by a woman. Although its titular character is a preteen boy, the show is one of TV's most feminist, thanks to the narrative significance of the Gems, members of a female-presenting alien race.
Steven is half-Gem, half-human; his alien mom sacrificed her corporeal form to birth him, leaving him in the care of her compatriots, Garnet, Amethyst, and Pearl, who act as Steven's "three moms." In short, this is not a nuclear family. Among the many taboos addressed, the most subtly transgressive is that of female caretakers who don't cook. We're used to seeing TV moms beatifically be-aproned, but the Gems don't need to eat, and resultantly raise Steven without routine family dinners or lessons on the finer points of nutrition. And yet food is one of his favorite things about being part human. Left to his own devices, he nourishes himself as any adolescent might, on chips and ice cream sandwiches and homemade cheese-puff sushi. The show acknowledges the value of a family meal (in "Together Breakfast" Steven insists popcorn-topped waffles be appreciated by all) while allowing unconventional food dynamics to be portrayed as a functional reality, rather than a familial failure. Steven cooks for himself—and that doesn't mean he's unloved. His three intergalactic moms just express how they care for him in other ways.
 All Work
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© Adrienne Matei 2017
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redpajama_set_name: RedPajamaCommonCrawl


text:
Amy Mbacke Thiam (Senegal, 10 de noviembre de 1976) es una atleta senegalesa, especialista en la prueba de 400 m, en la que ha logrado ser campeona mundial en 2001.

Carrera deportiva
En el Mundial de Edmonton 2001 ganó la medalla de oro en los 400 metros, por delante de la jamaicana Lorraine Fenton y la mexicana Ana Guevara.

Y dos años después, en el Mundial de París 2003 ganó la medalla de bronce en la misma prueba, con un tiempo de 49.95 segundos, tras las dos mismas atletas que en la ocasión anterior, pero esta vez Ana Guevara ganó el oro, y Lorraine Fenton, la plata.

Referencias

Atletas de Senegal
Senegalesas del siglo XXI
Deportistas femeninas de Senegal

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
Given a group $G$ one may ask the question:
\begin{question}
In what ways can $G$ act non-trivially on a Gromov hyperbolic metric space?
\end{question}
Many interesting groups can be fruitfully studied via some natural action on
a Gromov hyperbolic space.  Examples include the action of an
amalgam or HNN extension on its Bass-Serre tree, the action of a
Kleinian group on $\mathbb{H}^n$, and the action of the mapping class group of
a surface on the curve complex of that same surface.  Alternatively, one can
study the space of actions of a fixed group on some (fixed or
varying) Gromov hyperbolic metric space.  For example, the $PSL(2,\mathbb{C})$--character variety of a group
parameterizes the space of actions of a fixed
group on the hyperbolic space $\mathbb{H}^3$.  Analysis of the structure of
this variety has led to many remarkable theorems about $3$--manifolds
and their fundamental
groups (for an introduction, see \cite{shalen:handbook}).
A larger ``variety'' (in scare quotes because there
is unlikely to be any algebraic structure) would describe all
non-trivial actions on Gromov hyperbolic spaces, up to some
appropriate equivalence relation.  We give
suggestions for how to define this equivalence relation and topologize the
variety in Section \ref{s:actions}.  Briefly, the equivalence relation
is that generated by coarsely equivariant quasi-isometric embeddings.
This equivalence is coarser than that given by quasi-conjugacy (as in
for instance \cite{msw:quasiactI}), but finer than that given by
equivariant homeomorphism of limit sets.
In this paper we concentrate on cases in which the set of equivalence
classes is particularly simple.

Certain equivalence classes of actions on hyperbolic spaces cannot be
ruled out, or
even really analyzed using the tools of negative curvature.  These are
the \emph{actions with an invariant horoball} (see Section
\ref{ss:combhoro} for the definition, and Theorem \ref{t:horeq} for
some characterizations).  Actions with an invariant
horoball (in the sense used in this paper) are always elementary; they
include the trivial action
on a point and actions which preserve some horofunction.
A cobounded
action on an unbounded space never has an invariant horoball.

The variety of actions of an irreducible higher rank lattice is
expected to be very simple.
\begin{conjecture}\label{c:main}
If $\Gamma$ is an irreducible lattice in a higher rank Lie group (or
in a
nontrivial product of locally compact groups) $G$,
there there are finitely many Gromov hyperbolic $G$--spaces (up to
coarse equivalence) without
invariant horoballs.
\end{conjecture}
In the case where $G$ is a simple Lie group of rank at least $2$, the
only expected actions are those with an invariant horoball.  If $G$
has more than one direct factor, then $\Gamma$ projects densely to
each factor.  If the factors have rank one, then there will clearly be
non-elementary isometric actions of $\Gamma$ on rank one symmetric
spaces.  We discuss this case in Section \ref{section:rankone}.

For actions by lattices in nontrivial products of simple Lie
groups, the conjecture follows from rigidity results of Monod
and Monod-Shalom
\cite{monod:superrigid,monodshalom:jdg}
if one restricts attention
to Gromov hyperbolic spaces which are also either CAT$(0)$ spaces,
proper and cocompact spaces, or bounded valence graphs.
(Cf.
\cite{gelanderkarlssonmargulis} for CAT$(0)$ spaces.)
For an example
of an action of a lattice on a Gromov hyperbolic space which is
inequivalent to any action on a Gromov hyperbolic CAT$(0)$ space,
consider a lattice as in the Appendix to \cite{manning:qfa}, which has
Property (T), but admits a non-trivial pseudocharacter (or homogeneous
quasi-morphism).  This
pseudocharacter gives rise to a cobounded action on a space
quasi-isometric to $\mathbb{R}$, fixing both ends.  On the other hand, an
action on a CAT$(0)$ space preserving a point at infinity would give
(via the Busemann function) a homomorphism to $\mathbb{R}$, and  a cobounded such
action would give an unbounded homomorphism to $\mathbb{R}$.
Such a homomorphism is ruled out by Property (T).

\begin{theorem}\label{thm:cdr2}
Suppose that $G$ is a simple Chevalley-Demazure
group scheme of rank at least $2$, and let $\mathcal{O}$ be the ring
of integers of any number field.  Then any isometric action of
$G(\mathcal{O})$ on a Gromov hyperbolic geodesic metric space has
an invariant horoball.
\end{theorem}
Some remarks:
\begin{enumerate}
\item Some closely related results are proved by Karlsson and Noskov
  \cite[Sections 8.2 and 8.3]{karlssonnoskov}; part of our strategy is
  similar to theirs, and to that of Fukunaga in \cite{fukunaga:FA}.
\item The rank $\geq 2$ assumption is necessary, as the
  action of $SL(2,\mathcal{O})$ on $\mathbb{H}^3$ obtained from the inclusion
  $SL(2,\mathcal{O})\longrightarrow SL(2,\mathbb{C})$  never has an
  invariant horoball.
\end{enumerate}

The rank one case is discussed in Section
\ref{section:rankone}, where we apply a result of Carter, Keller and
Paige to show a weaker theorem for these groups.
\begin{theorem}\label{thm:cdr1}
Let $\mathcal{O}$ be the ring of integers of a number field, and
suppose that $\mathcal{O}$ has infinitely many units.  Suppose $X$ is
quasi-isometric to a tree.  Every action of $SL(2,\mathcal{O})$ on $X$
has a bounded orbit.
\end{theorem}

\subsection{Relative hyperbolicity and bounded generation}
If $G$ is a finitely generated
group, $S$ is a generating set for $G$, and
$\mc{P}=\{P_1,\ldots,P_n\}$ is a collection of
subgroups of $G$, then one can form the \emph{coned space}
$C(G,\mc{P},S)$ as follows:  Let $\Gamma(G,S)$ be the Cayley graph of
$G$ with respect to $S$.
The coned space $C(G,\mc{P},S)$ is obtained
from $\Gamma(G,S)$ by coning each left coset of an element of $\mc{P}$
to a point.  (Here the $0$--skeleton of $\Gamma(G,S)$ is implicitly
identified with $G$.)

Recall that a graph is \emph{fine} if every edge is contained in only
finitely many circuits (i.e. embedded cycles) of any bounded length.
\begin{definition}
If $G$ is a group with finite generating set $S$, and a collection of
subgroups $\mc{P}=\{P_1,\ldots,P_n\}$, then
$G$ is \emph{weakly hyperbolic relative to} $\mc{P}$ if
$C(G,\mc{P},S)$ is $\delta$--hyperbolic for some $\delta$.

If, moreover, $C(G,\mc{P},S)$ is \emph{fine}, then $G$ is
\emph{(strongly) hyperbolic relative to} $\mc{P}$.
\end{definition}

A special case of weak relative hyperbolicity is bounded generation.
The following definition is easily seen to be equivalent to the
standard one:
\begin{definition}
A group $G$ is \emph{boundedly generated} by a collection of
subgroups $\mc{P}=\{P_1,\ldots,P_n\}$ if the Cayley graph of $G$ with
respect to $\cup\mc{P}$ has finite diameter.

If each $P_i$ is cyclic, generated by $g_i$, we say that $G$ is
\emph{boundedly generated by} $\{g_1,\ldots,g_n\}$.
\end{definition}

A cobounded action on an unbounded Gromov hyperbolic space does not
have an invariant horoball.
It is thus a corollary of Theorem \ref{thm:cdr2} that these higher rank
$G(\mathcal{O})$ are not strongly relatively hyperbolic
with respect to any system of proper subgroups.
\begin{corollary}\label{corollary:notrelhyp}
If $L=G(\mc{O})$ is as in Theorem \ref{thm:cdr2}, and $L$ is weakly
hyperbolic relative to a system of subgroups
$\mc{P}$, then the coned space $C(L,\mc{P},S)$ has finite
diameter for any finite generating set $S$.

In particular, $L$ is not strongly hyperbolic relative to any system
of proper subgroups.
\end{corollary}
By possibly altering $\mc{P}$ to add some cyclic subgroups, the first
part of the corollary can be restated:  If $L$ is weakly
hyperbolic relative to a system of subgroups $\mc{P}$ which generate
$L$, then $L$ is boundedly generated by $\mc{P}$.
\begin{remark}
Corollary \ref{corollary:notrelhyp} also can be easily deduced from
\cite{karlssonnoskov} in some special cases, including $L=SL(n,\mathbb{Z})$
for $n>2$.  The second part of Corollary \ref{corollary:notrelhyp}
(about strong relative hyperbolicity) can be deduced from known
theorems in a number of ways, perhaps most straightforwardly by
combining a theorem of Fujiwara \cite{fujiwara:gromovhyperbolic} with
one of Burger and Monod \cite{burgermonod:gafa}.
\end{remark}

\subsection{Outline}
In Section \ref{s:preliminaries} we recall some definitions and basic
results, first from the theory of Gromov hyperbolic spaces, and second
from Chevalley groups over number rings.  In Section \ref{s:actions}
an equivalence relation amongst hyperbolic $G$--spaces is proposed, and
generalized combinatorial horoballs are introduced.  In Section
\ref{s:elementary}, we improve on the statement and proof of
Proposition 3.9 of \cite{manning:qfa}, and use the improved version to
characterize hyperbolic $G$--spaces with invariant horoballs.
In Section \ref{s:bigrank} we prove Theorem \ref{thm:cdr2}, and in
Section \ref{section:rankone} we prove Theorem \ref{thm:cdr1}.

\subsection{Acknowledgments}
The author thanks Benson Farb and Hee Oh for useful conversations, and
Nicolas Monod for helpful comments on an earlier version of this paper.
This work was partly supported by an NSF Postdoctoral Research
Fellowship (DMS-0301954).

\section{Preliminaries}\label{s:preliminaries}
\subsection{Coarse geometry}
\begin{definition}
If $X$ and $Y$ are metric spaces, $K\geq 1$ and $C\geq 0$,
a \emph{$(K,C)$--quasi-isometric embedding} of $X$ into $Y$ is a function
$q\colon\thinspace X\to Y$ so that
 For all $x_1$, $x_2\in X$
\[\frac{1}{K}d(x_1,x_2)-C\leq d(q(x_1),q(x_2))\leq Kd(x_1,x_2)+C\]

If in addition the map $q$ is \emph{$C$--coarsely onto} -- \emph{i.e.},
every $y\in Y$ is
distance at most $C$ from some point in $q(X)$ -- then $q$ is called a
\emph{$(K,C)$--quasi-isometry}.
The two metric spaces $X$ and $Y$ are then
said to be \emph{quasi-isometric} to one
another.  This is a symmetric condition.
\end{definition}
\begin{definition}
A \emph{$(K,C)$--quasi-geodesic} in $X$
is a $(K,C)$--quasi-isometric embedding $\gamma\colon\thinspace\mathbb{R}\to X$.
We will occasionally abuse
notation by referring to the image of $\gamma$ as a quasi-geodesic.
\end{definition}
\subsection{Gromov hyperbolic spaces}
For more details on Gromov hyperbolic metric spaces, see
\cite{bridhaef:book} and \cite{gromov:wordhyperbolic}.
A number of equivalent definitions of Gromov hyperbolicity are known.
For geodesic spaces, we will use the one based on the existence of
\emph{comparison tripods}.
Given a geodesic
triangle $\Delta(x,y,z)$ in
any metric space, there is a unique \emph{comparison tripod}, $T_\Delta$,
a metric tree so
that the distances between the three extremal points of the tree,
$\overline{x}$, $\overline{y}$ and $\overline{z}$ , are the same as
the distances between $x$, $y$ and $z$.
There is an obvious map $\pi\colon\thinspace \Delta(x,y,z)\to T_\Delta$ which takes $x$ to
$\overline{x}$, $y$ to $\overline{y}$ and $z$ to $\overline{z}$, and
which is an isometry on each side of $\Delta(x,y,z)$.
\begin{definition}\label{d:geodgromov}
A geodesic space $X$ is \emph{$\delta$--hyperbolic} if for any geodesic triangle
$\Delta(x,y,z)$ and any point $p$ in the comparison tripod $T_\Delta$,
 the diameter of $\pi^{-1}(p)$ is less than $\delta$.  If $\delta$ is
 unimportant we may simply say that $X$ is \emph{Gromov hyperbolic}.
\end{definition}
Gromov hyperbolicity (of geodesic spaces) is a quasi-isometry invariant.

The following proposition about stability of quasi-geodesics in Gromov
hyperbolic spaces
is well-known (see, e.g. \cite[III.H.1.7]{bridhaef:book}).
\begin{proposition}\label{p:quasistable}
Let $K\geq 1$, $C\geq 0$, $\delta\geq 0$.  Then there is some
$B=B(K,C,\delta)$, so that whenever $\gamma$ and $\gamma'$ are two
$(K,C)$--quasi-geodesics with the same endpoints in a
$\delta$--hyperbolic geodesic metric space $X$, then the Hausdorff
distance between $\gamma$ and $\gamma'$ is at most $B$.
\end{proposition}

We will occasionally need to deal with spaces which are not geodesic.
If $X$ is $\delta$--hyperbolic in the sense of Definition
\ref{d:geodgromov}, then it
satisfies the four point condition:  For all $p_1$, $p_2$, $p_3$,
$p_4\in X$,
\begin{equation}\label{fourpoint}
d(p_1,p_4)+d(p_2,p_3)\leq\max\{d(p_1,p_2)+d(p_3,p_4)\mbox{, }d(p_1,p_3)+d(p_2,p_4)+2\delta\}.
\end{equation}
Conversely, if a geodesic space satisfies \eqref{fourpoint}, then it
is $6\delta$--hyperbolic in the sense of Definition \ref{d:geodgromov}.
(For both these facts, and the below definition, see
\cite[III.H.1.22]{bridhaef:book} or \cite{gromov:wordhyperbolic}.)
\begin{definition}\label{d:fourpoint}
A (not necessarily geodesic) metric space $X$ is \emph{$(\delta)$--hyperbolic}
if it satisfies the condition \eqref{fourpoint} above.  If $\delta$ is
unimportant, we simply say that $X$ is \emph{Gromov hyperbolic}.
\end{definition}

In order to describe the boundary of a Gromov hyperbolic space, we
introduce the \emph{Gromov product} notation.
\begin{definition}
If $x,y$ and $z$ are points in a metric space with a metric
$d(\cdot,\cdot)$, then
\[ \gp{x}{y}{z}:= \frac{1}{2}(d(x,z)+d(y,z)-d(x,y)).\]
\end{definition}
We should remark that in the context of Definition \ref{d:geodgromov},
$\gp{x}{y}{z}$ is the distance from
$\overline{z}$ to the central vertex of the comparison tripod for a
geodesic triangle with vertices $x$, $y$ and $z$.
\begin{definition}
Let $p\in X$ where $X$ is a Gromov hyperbolic space.
A sequence of points $\{x_i\}$ in a Gromov hyperbolic space
\emph{tends to infinity} if
$\lim_{i,j\to\infty}\gp{x_i}{x_j}{p}=\infty$.
Two such sequences are \emph{equivalent}, written $\{x_i\}\sim\{y_i\}$, if
$\lim_{i,j\to\infty}\gp{x_i}{y_j}{p}=\infty$.
The boundary $\partial X$ is the set of equivalence classes of
sequences which tend to infinity.
\end{definition}
The Gromov product extends (by taking a lim sup) to sequences which
tend to infinity, and this allows convergence in $\partial X$ to be
defined, giving $\partial X$ a natural topology.

\subsection{Isometries of Gromov hyperbolic spaces}
Isometries of geodesic hyperbolic spaces can be classified into three
types.
\begin{definition}
Let $f\colon\thinspace X\to X$ be an isometry.  If $x\in X$, we let
$O_x=\{f^n(x)\mid n\in \mathbb{Z}\}$.
We say that $f$ is \emph{elliptic} if $O_x$ is bounded.  We say that
$f$ is \emph{hyperbolic} if
$n\mapsto f^{n}(x)$ is a quasi-isometric embedding of $\mathbb{Z}$ into $X$.
We say that $f$ is \emph{parabolic} if $O_x$ has a unique limit point
in $\partial G$.
\end{definition}
The following was observed by Gromov \cite[8.1.B]{gromov:wordhyperbolic}.
Although it is often stated with an extra hypothesis of
properness, this hypothesis is unnecessary (See, for example the
proof in \cite[Chapitre 9]{cdp}, where the extra hypothesis is given,
but not used).
\begin{proposition}\label{p:isometryclass}
Every isometry of a geodesic Gromov hyperbolic space is elliptic,
parabolic, or hyperbolic.
\end{proposition}

\begin{lemma}\label{l:commute}
Suppose that $G$ acts on the geodesic Gromov hyperbolic space $X$, and that
$p\in G$ acts parabolically, fixing $e\in \partial X$.  If $g\in G$
commutes with $p$, then $g$ also fixes $e$.
\end{lemma}
\begin{proof}
Let $e'=g(e)$.  We have $e'=gp(e)=pg(e)=p(e')$, so $p$ fixes $e'$.
Since $p$ is parabolic, it fixes a unique point in $\partial X$, and
so $e'=e$.
\end{proof}

\begin{definition}
Let $G$ be a finitely generated group, and let $g\in G$.  If $n\mapsto
g^n$ is a quasi-isometric embedding, we say that $g$ is
\emph{undistorted}.  Otherwise, $g$ is \emph{distorted}.
\end{definition}
The proof of the following observation is left to the reader.
\begin{lemma}\label{lemma:distortion}
Let the finitely generated group $G$ act by isometries on a Gromov
hyperbolic space $X$.  If $g\in
G$ acts hyperbolically on $X$, then $g$ is undistorted.
\end{lemma}

\subsection{Chevalley groups}\label{defs:chevalley}
In this section we recall the definition of a Chevalley group over a
commutative ring.
 (All rings are assumed to have $1 \neq 0$.)
The simplest example of a Chevalley group is
$SL(n,\mathbb{Z})$.  If one thinks of $SL(n,\mathbb{Z})$ as being the ``$\mathbb{Z}$--points
of $SL(n,\mathbb{C})$,'' then the Chevalley-Demazure group scheme
identifies what the ``$R$--points of $G$'' are, where
$R$ is now allowed to be an arbitrary ring, and $G$ an arbitrary
complex semisimple Lie group.  It turns out that this idea is not entirely
well-formed, until one fixes an embedding of $G$ into $GL(n,\mathbb{C})$ for
some $n$.  The following exposition is largely
adapted from \cite{abe:chevalley} and \cite{tavgen:bg}.

Let $\rho\colon\thinspace G\to GL(V)$ be a representation of a connected complex semisimple
Lie group into the general linear group of a complex vector space $V$
of dimension $n$.  We will assume that
$d\rho\colon\thinspace\mathfrak{g}\to\mathfrak{gl}(V)$ is faithful (Here
$\mathfrak{g}$ is the Lie algebra of $G$, $\mathfrak{gl}(V)$ the Lie
algebra of endomorphisms of $V$.
Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ and let
$\Phi$ be the (reduced) root system relative to $\mathfrak{h}$.
Let $\Delta$ be a
choice of simple roots.  Then there is a \emph{Chevalley basis}
for $\mathfrak{g}$ (see \cite{steinberg:notes} or \cite{chevalley:tohoku}) of
the form $\mathcal{B}=\{X_\alpha\ |\ \alpha\in\Phi\}\cup\{H_\alpha\ |\
\alpha\in\Delta\}$, so that the $H_\alpha$ generate $\mathfrak{h}$ (and
thus commute) and the structure constants are all integral.
In other words, the $\mathbb{Z}$--span of $\mathcal{B}$ is actually a Lie algebra over
$\mathbb{Z}$.
In particular, if $\{\beta - r_{\beta,\alpha} \alpha, \ldots \beta,
\ldots \beta + q_{\beta,\alpha} \alpha\}$ are all the roots on the
line $\{\beta + t\alpha\ |\ t\in\mathbb{Z}\}$, then:
\begin{enumerate}
\item $[X_\alpha, X_{-\alpha}] = H_\alpha$.
\item $[H_\alpha, X_\beta] = A(\alpha,\beta) X_\beta$, where
  $A(\alpha,\beta)$ is an integer determined by $\alpha$
  and $\beta$.
\item $[X_\alpha,X_\beta] = 0$ if $\alpha + \beta$ is not in $\Phi$;
  otherwise $[X_\alpha,X_\beta] = \pm (r_{\beta,\alpha}+1)X_{\alpha+\beta}$.
\end{enumerate}
Notice that the $X_\alpha$ are all ad-nilpotent, and that the
$H_\alpha$ are ad-semisimple.

It can be shown (see, e.g. \cite[\S27]{humphreys:introduction})
that $V$ contains
an ``admissible lattice'' $V_\mathbb{Z}$: This is a free $\mathbb{Z}$--module in $V$
which is invariant under
$(d\rho(X_\alpha))^m / m!$ for any $\alpha$ and any $m$.  Let
$\{v_1,\ldots,v_n\}$ be a basis for $V_\mathbb{Z}$.  In terms of this basis,
$\rho(g)$ can be written as an $n$ by $n$ matrix with complex
entries for any $g$.  Let $x_{ij}\colon\thinspace G\to \mathbb{C}$ be the function which
simply reads off the $ij$th entry of this matrix.  The $x_{ij}$
generate an affine complex algebra $\mathbb{C}(G)$.  Let $\mathbb{Z}(G)$ denote the
$\mathbb{Z}$--algebra with the same generators.

We can endow $\mathbb{Z}(G)$ with a Hopf algebra structure by defining
a comultiplication $\mu^*$ by
$\mu^*(x_{ij}) = \sum_k x_{ik}\otimes x_{kj}$, a counit $\epsilon$
by $\epsilon(x_{ij}) = \delta_{ij}$, and an antipode $s$ by
$s(f)(g) = f(g^{-1})$.
(That $s$ maps  $\mathbb{Z}(G)$ into itself uses the fact
that $\det(\rho(g))=\pm 1$ for every $g\in G$ -- this
 follows from semisimplicity.)
Since $\mathbb{Z}(G)$ is a Hopf algebra over $\mathbb{Z}$, it defines a functor from
rings to groups as follows:
For any ring $R$, let
$G(R) = \mathrm{Hom}_\mathbb{Z}(\mathbb{Z}(G), R)$.  (Note that the elements of
$G(R)$ are $\mathbb{Z}$--algebra homomorphisms, so they send $1$ to $1$.)
We define a group operation
$\bullet$ on $G(R)$ by $\rho\bullet\sigma = (\rho\otimes\sigma)\circ
\mu^*$.
(In other words, $\rho\bullet\sigma(x_{ij}) = \sum_k
\rho(x_{ik})\sigma(x_{kj})$.)

Some observations:
\begin{enumerate}
\item $G(\mathbb{C})$ can be naturally identified with the image of $\rho\colon\thinspace
  G\to GL(V)$ -- if $\rho$ is assumed to be faithful, then clearly
  $G(\mathbb{C})\cong G$ doesn't depend on $\rho$.  On the other hand, for $R$
  arbitrary, $G(R)$ does  depend  on $\rho$.
\item  The assignment $R\mapsto G(R)$ is a covariant functor from
  commutative rings to groups.  This functor is often called a
  ``Chevalley-Demazure group scheme.''
\item\label{zoft}  Any Hopf algebra gives such a functor.  An
  important example is
  $\mathbb{Z}[t]$, with a Hopf ($\mathbb{Z}$--)algebra structure so that:
\begin{enumerate}
\item The comultiplication satisfies $\mu^*(t)=t\otimes 1 + 1\otimes
  t$.
\item The counit satisfies $\epsilon(t)=0$.
\item The antipode satisfies $S(t)=-t$
\end{enumerate}
In this case the functor $R\mapsto \mathrm{Hom}(Z[t],R)$ is the
  ``forgetful'' functor, which takes a ring to its underlying Abelian
  group.
\item\label{contrafunctor} Since $G(R)$ is $Hom_\mathbb{Z}(\mathbb{Z}(G),R)$, any morphism of Hopf algebras
 $\mathbb{Z}(G)\to A$ where $A$ is
some other Hopf algebra, will give rise to a homomorphism of groups
 $Hom_\mathbb{Z}(A,R)\to G(R)$.
\end{enumerate}
\begin{definition}
Let $\alpha$ be a root.  Then since $d\rho(X_\alpha)$ is
nilpotent, the formal sum $\exp ( t X_\alpha ) = \sum_{m=1}^\infty
t^m \frac{d\rho(X_\alpha)^m}{m!} $ is a matrix with entries which are
polynomials in $t$.
Because $\frac{d\rho(X_\alpha)^m}{m!}$ preserves
$V_\mathbb{Z}$, each $x_{ij}(\frac{d\rho(X_\alpha)^m}{m!})$ is an integer.
Thus we get a map
\[\mathrm{ev}_\alpha\colon\thinspace \mathbb{Z}(G) \to \mathbb{Z}[t],\]
which sends $x_{ij}$ to the (integral) polynomial in $t$ which appears
in the $ij$'th place of the matrix $\exp( t X_\alpha)$.  The map
$\mathrm{ev}_\alpha$ is a morphism of Hopf algebras, where $\mathbb{Z}[t]$ is given a
Hopf algebra structure as in observation \eqref{zoft} above.
This morphism gives rise via observation \eqref{contrafunctor} to a
homomorphism of the additive group underlying $R$ into the group
$G(R)$
\[x_\alpha\colon\thinspace R\to G(R).\]
The image of this map is the \emph{root subgroup}
of $G(R)$ corresponding to $\alpha$.
\end{definition}

\begin{definition}
The subgroup of $G(R)$ generated by the root subgroups
is denoted $E(R)$.
\end{definition}

In this paper, we focus mainly on the special case that $R$ is the
ring of integers of a number field $k$ and $\Phi$ has rank at least
two.  In this case, $G(R)=E(R)$ by
a result of Matsumoto \cite{matsumoto:csp}.  By a result
of Carter and Keller \cite{carterkeller:slno} in case $\Phi=A_n$,
and  Tavgen$'$ \cite{tavgen:bg} in general,
$G(R)$ is boundedly generated by the root subgroups.

It should also be noted that in this special case, $G(R)$ is actually
an irreducible lattice in a semisimple Lie group.  Indeed, if $s$ and
$t$ are the number of real
and complex places of $k$, respectively, then $G(R)$ is a lattice in the
Lie group
\[ G(\mathbb{R})^s\times G(\mathbb{C})^t.\]

\section{Equivalence of actions}\label{s:actions}
We wish to study the variety of actions of a group $G$ on
Gromov hyperbolic spaces, up to some kind of coarse equivalence.  By a
{\em (Gromov) hyperbolic $G$--space}, we will always mean a Gromov
hyperbolic geodesic metric space, equipped with an isometric
$G$--action.

\begin{definition}
Let $X$ and $Y$ be Gromov hyperbolic $G$--spaces.  We say that $X$ and
$Y$ are {\em equivalent} if they lie in the same
equivalence class, under the equivalence relation generated by
coarsely equivariant quasi-isometric embeddings.
\end{definition}
The following proposition should serve to clarify this equivalence
relation.
\begin{proposition}If $X_1$ and $X_2$ are equivalent Gromov hyperbolic
  $G$--spaces, then there is
  a third Gromov hyperbolic $G$--space $V$ which coarsely equivariantly
  quasi-isometrically embeds in both $X_1$ and $X_2$.
\end{proposition}
\begin{proof}
The key claim is the following:
\begin{claim}\label{c:diamond}
If $V$ and $W$ are hyperbolic $G$--spaces which coarsely equivariantly
  quasi-isometrically embed in a third hyperbolic $G$--space $X$, then
  there is a fourth hyperbolic $G$--space $A$ which coarsely
  equivariantly quasi-isometrically embeds into $V$ and $W$.
\end{claim}
Before proving the claim, let us see how it implies the proposition.
If $X_1$ and $X_2$ are equivalent hyperbolic $G$--spaces, they
must be joined by a sequence of coarsely
equivariant quasi-isometric embeddings:
\cdlabel{long}{ & V_1\ar[dl]\ar[dr] & & \cdots\ar[dl]\ar[dr] & &V_n\ar[dl]\ar[dr] & &
  V_{n+1}\ar[dl]\ar[dr] & \\
X_1 & & Z_1 &\cdots & Z_{n-1} & & Z_n & & X_2}
By applying the claim with $V=V_n$, $W=V_{n+1}$, and $X=Z_n$, we
obtain a hyperbolic $G$--space $A$ which coarsely equivariantly
quasi-isometrically embeds into both $V_n$ and $V_{n+1}$, and hence
into both $Z_{n-1}$ and $X_2$.  We can thus shorten the sequence
\eqref{long}, unless it is a shortest possible such sequence,
\cd{ & V_1\ar[dl]\ar[dr] & \\
 X_1 & & X_2,}
in which case the proposition is verified.

\begin{proof} (of Claim \ref{c:diamond})
We first construct a $G$--space $A_1$ which coarsely equivariantly
quasi-isometrically embeds in both $V$ and $W$ and is Gromov
hyperbolic but not geodesic.  We then show that $A_1$ is
quasi-isometric to a geodesic $G$--space $A$.

Choose $\delta>0$ so that all of $V$, $W$, and $X$ are
$\delta$--hyperbolic spaces.
By hypothesis, we may choose $K\geq 1$ and $C\geq 0$ so that there are maps
$\phi\colon\thinspace V\to X$ and $\psi\colon\thinspace W\to Y$ which are $C$--coarsely equivariant
$(K,C)$--quasi-isometric embeddings.  We choose constants $J_0<J_1<J_2$:
Let
$J_0 = 2B(K,C,\delta)+2\delta$,
where $B(K,C,\delta)$ is the constant of quasi-geodesic stability from
Proposition \ref{p:quasistable}, let $J_1=J_0+2C$, and let $J_2=4 J_1$.

Let $A_0$ be the subset of $V\times
W$ given by $A_0=\{(v,w)\mid d(\phi(v),\psi(w))\leq J_0\}$, and let $A_1$
be the smallest $G$--equivariant subset of $V\times W$ containing
$A_0$.  We endow $A_1$ with the (pseudo)metric
\[
 d((v_1,w_1),(v_2,w_2))=d(\phi(v_1),\phi(v_2))+d(\psi(v_2),\psi(w_2)).
\]
\begin{subclaim}
$A_1$ is Gromov hyperbolic.
\end{subclaim}
\begin{proof}
Since $A_1$ is not a geodesic space, we must work with the four-point
definition \ref{d:fourpoint}.  We will show that
$A_1$ is $(2\delta+4J_1)$--hyperbolic in the sense of Definition
\ref{d:fourpoint}.
Let $\{p_i=(v_i,w_i)\mid i=1,\ldots,4\}$ be four points in $A_1$.
For each $i$, we write $\uu{v}_i$ for $\phi(v_i)$ and $\uu{w}_i$ for
$\psi(w_i)$. By
reordering the points if necessary, we can assume that
\[d(\uu{v}_1,\uu{v}_3)+d(\uu{v}_2,\uu{v}_4)\geq
d(\uu{v}_1,\uu{v}_2)+d(\uu{v}_3,\uu{v}_4),\]
as in Figure \ref{f:palimpsest}.
\begin{figure}[htbp]
\begin{center}
\begin{picture}(0,0)%
\epsfig{file=palimpsest.pstex}%
\end{picture}%
\setlength{\unitlength}{3947sp}%
\begingroup\makeatletter\ifx\SetFigFont\undefined%
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  \reset@font\fontsize{#1}{#2pt}%
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  \selectfont}%
\fi\endgroup%
\begin{picture}(3300,1939)(2551,-2744)
\put(5551,-2686){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\phi(v_3)$}%
}}}}
\put(2776,-961){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\phi(v_2)$}%
}}}}
\put(2626,-1186){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\psi(w_2)$}%
}}}}
\put(2551,-2011){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\phi(v_1)$}%
}}}}
\put(2626,-2236){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\psi(w_1)$}%
}}}}
\put(5851,-1411){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\phi(v_4)$}%
}}}}
\put(5776,-1186){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\psi(w_4)$}%
}}}}
\put(5776,-2461){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\psi(w_3)$}%
}}}}
\end{picture}%
\caption{Points in $X$.}
\label{f:palimpsest}
\end{center}
\end{figure}
There are then two cases.

In case
\[ d(\uu{w}_1,\uu{w}_3)+d(\uu{w}_2,\uu{w}_4)\geq
d(\uu{w}_1,\uu{w}_2)+d(\uu{w}_3,\uu{w}_4),\]
then
$\max\{d(p_1,p_3)+d(p_2,p_4),d(p_1,p_2)+d(p_3,p_4)\}=d(p_1,p_2)+d(p_2,p_4)$.
Applying $\delta$--hyperbolicity in $X$, we obtain
\begin{eqnarray*}
d(p_1,p_4)+d(p_3,p_2) & = &
d(\uu{v}_1,\uu{v}_4)+d(\uu{v}_2,\uu{v}_3)+d(\uu{w}_1,\uu{w}_4)+d(\uu{w}_2,\uu{w}_3)
\\
 & \leq & d(\uu{v}_1,\uu{v}_3)+d(\uu{v}_2,\uu{v}_4)+ 2\delta
+d(\uu{w}_1,\uu{w}_3)+d(\uu{w}_2,\uu{w}_4)+ 2\delta \\
& = & d(p_1,p_3)+d(p_2,p_4)+ 4\delta \\
& = & \max\{d(p_1,p_3)+d(p_2,p_4),d(p_1,p_2)+d(p_3,p_4)\} + 2
(2\delta) \\
& \leq & \max\{d(p_1,p_3)+d(p_2,p_4),d(p_1,p_2)+d(p_3,p_4)\} \\
& & + 2(2\delta+4 J_1)
\end{eqnarray*}
as required.

If on the other hand
\[ d(\uu{w}_1,\uu{w}_3)+d(\uu{w}_2,\uu{w}_4) <
d(\uu{w}_1,\uu{w}_2)+d(\uu{w}_3,\uu{w}_4),\]
then it is still true (since $d(\uu{w_i},\uu{v_i})\leq J_1)$) that
\begin{eqnarray*}
d(\uu{w}_1,\uu{w}_3)+d(\uu{w}_2,\uu{w}_4) & \geq &
d(\uu{v}_1,\uu{v}_3)+d(\uu{v}_2,\uu{v}_4) - 4J_1 \\
& \geq & d(\uu{v}_1,\uu{v}_2)+d(\uu{v}_3,\uu{v}_4) - 4J_1 \\
& \geq & d(\uu{w}_1,\uu{w}_2)+d(\uu{w}_3,\uu{w}_4) - 8J_1.
\end{eqnarray*}
We therefore obtain
\begin{eqnarray*}
d(p_1,p_4)+d(p_3,p_2) & = &
d(\uu{v}_1,\uu{v}_4)+d(\uu{v}_2,\uu{v}_3)+d(\uu{w}_1,\uu{w}_4)+d(\uu{w}_2,\uu{w}_3)
\\
& \leq & d(\uu{v}_1,\uu{v}_3)+d(\uu{v}_2,\uu{v}_4)+ 2\delta
+d(\uu{w}_1,\uu{w}_2)+d(\uu{w}_3,\uu{w}_4)+ 2\delta \\
& \leq & d(p_1,p_3)+d(p_2,p_4) + 2(2\delta+4 J_1) \\
& \leq & \max\{d(p_1,p_3)+d(p_2,p_4),d(p_1,p_2)+d(p_3,p_4)\}\\& & + 2
(2\delta+4 J_1),
\end{eqnarray*}
which finishes the proof of the Subclaim.
\end{proof}
Though the space $A_1$ is hyperbolic, and $G$ quasi-acts on $A_1$
via the diagonal action on $V\times W$, the space $A_1$ is not  geodesic,
and $G$ does not act by isometries.  Both of these issues can be fixed
at once, by replacing $A_1$ by an appropriate graph.
Specifically, we let $A$ be the graph with vertex set
$V(A)=A_1$, and with an edge between every pair of vertices $p$ and
$q$ so that there exists some $g$ with $d(gp,gq)\leq J_2:=4J_1$.  Clearly
this graph is a geodesic $G$--space.

Let $\iota\colon\thinspace A_1\to A$ be the map which takes a point to the
corresponding vertex.  We claim that $\iota$ is a quasi-isometry, and
so $A$ is a Gromov hyperbolic space.

Let $a=(v_a,w_a)$ and $b=(v_b,w_b)$ be points in $A_1$ so that
$d(\iota(a),\iota(b))=1$.  There is some $g$ with
\[d(ga,gb)=d(\phi(gv_a),\phi(gv_b))+d(\psi(gv_a),\psi(gv_b))\leq J_2.\]
Since $\phi$ and $\psi$ are both $C$--coarsely equivariant,
$d(\phi(gv_a),\phi(gv_b))$ differs  by
at most $2C$ from $d(\phi(v_a),\phi(v_b))$; similarly, $d(\psi(gw_a),\psi(gw_b))$ differs from $d(\psi(w_a),\psi(w_b))$  by
at most $2C$.  It follows that $d(ga,gb)$ differs by at most $4C$ from
$d(a,b)$, and so $d(a,b)\leq J_2+4C<\frac{3}{2}J_2$.  As this is true for every pair
of points connected by an edge, we deduce
\begin{equation}\label{lower}
 d(\iota(p),\iota(q)) \geq \frac{2}{3J_2} d(p,q)
\end{equation}
for any pair of points $p$, $q\in A_1$.

We now obtain the complementary bound to \eqref{lower}.  Suppose $p=(v_1,w_1)$
and $q=(v_2,w_2)$ are any two points in $A_1$.  We write $\uu{v_1}$
for $\phi(v_1)$ and so on as before.
We will show that
\begin{equation}\label{upper}
d(\iota(p),\iota(q))\leq  \frac{1}{J_1}d(p,q)+2
\end{equation}
by constructing a path joining $\iota(p)$ to $\iota(q)$ in $A$.  The
vertices of this path will be points in $A_0\subset V\times W$ so that the first
coordinate lies on a geodesic between $v_1$ and $v_2$, while the
second lies on a geodesic between $w_1$ and $w_2$.

If $d(p,q)\leq 2J_2 = 8J_1$, then \eqref{upper} is automatically
satisfied.  We may therefore assume that $d(p,q)\geq 2J_2$.  Because
$d(\uu{v}_1,\uu{w}_1)$ and $d(\uu{w}_1,\uu{w}_2)$ are at most $J_0$,
we have
$\min\{d(\uu{v}_1,\uu{v}_2),d(\uu{w}_1,\uu{w}_2)\}\geq J_2-J_0 >
3J_0$. It follows that
$\gp{\uu{v}_2}{\uu{w}_2}{\uu{v}_1}>\gp{\uu{w}_1}{\uu{w}_2}{\uu{v}_1}$, as in
Figure \ref{f:bar}.
\begin{figure}[htbp]
\begin{center}
\input{bar.pstex_t}
\caption{A pair of pairs of points in $X$, together with comparison tripods.
The dashed line on the left is the image of a geodesic in $V$; the one
on the right is the image of a geodesic in $W$.}
\label{f:bar}
\end{center}
\end{figure}
In fact,
$\gp{\uu{v}_2}{\uu{w}_2}{\uu{v}_1}-\gp{\uu{w}_1}{\uu{w}_2}{\uu{v}_1}>J_0\geq
J_1/2$ (since clearly $B(K,C,\delta)\geq C$).
We may thus choose real numbers
\[ \gp{\uu{w}_1}{\uu{w}_2}{\uu{v}_1}=t_0<t_1<\cdots<t_k =
\gp{\uu{v}_2}{\uu{w}_2}{\uu{v}_1}\]
satisfying $\frac{J_1}{2}<|t_{i+1}-t_i|\leq J_1$ for each $i$.
Note that $k \leq \frac{1}{J_1}d(p,q)$ for such a choice.

Let
$\gamma$ be a unit speed geodesic from $\uu{v}_1$ to $\uu{w}_2$.  For
each $i$ between $0$ and $k$, there are points $x_i'$ on a geodesic
between $\uu{v}_1$ and $\uu{w}_1$ and $y_i'$ on a geodesic between
$\uu{w}_1$ and $\uu{w}_2$ which are distance at most $\delta$ from
$\gamma(t_i)$.  Since $V$ and $W$ are geodesic spaces, there are
geodesics $[v_1,v_2]$ in $V$ and $[w_1,w_2]$ in $W$.  The maps $\phi$
and $\psi$ send these geodesics
to $(K,C)$--quasi-geodesics in $X$.  Applying quasi-geodesic stability
(Proposition \ref{p:quasistable}), there are points
$x_i\in [v_1,v_2]$, $y_i\in [w_1,w_2]$ so that $\phi(x_i)$ and
$\psi(y_i)$ lie within $B+\delta$ of $\gamma(t_i)$.  Since
$d(\phi(x_i),\psi(y_i))\leq 2B+2\delta=J_0$, the point $p_i=(x_i,y_i)$
lies in $A_0\subseteq A_1$.  Moreover, $d(p_i,p_{i+1})\leq
2(2(B+\delta)+J_1)<J_2$, and so $d(\iota(p_i),\iota(p_{i+1}))\leq 1$.

Finally, one notes that
\[
d(\uu{v}_1,\phi(x_0))+d(\uu{w}_1,\psi(y_0))  \leq  J_0+2\delta+2 B=2J_0<J_2
\]
and likewise for $d(\uu{v}_1,\phi(x_0))+d(\uu{w}_1,\psi(y_0))$, from
which it follows that $d(\iota(p),\iota(p_0))$ and
$d(\iota(q),\iota(p_k))$ are at most one.  Thus
\[ d(\iota(p),\iota(q))\leq k+2 \leq \frac{1}{J_1} d(p,q)+2.\]
It is obvious that $\iota$ is $1$--almost onto, and so $\iota$ is a
quasi-isometry.

It follows that $A$ is a Gromov hyperbolic geodesic $G$--space
which coarsely equivariantly, quasi-isometrically embeds into $V$ and
$W$.

\end{proof}
\end{proof}

\begin{remark}
We record some observations about this equivalence:
\begin{enumerate}
\item Let $\Gamma$ be a  group acting isometrically on $\mathbb{H}^2$.  This
  action extends in an obvious way to either $\mathbb{H}^3$ or $\mathbb{C}\mathbb{H}^2$.
  Although there is no quasi-isometric embedding either from $\mathbb{H}^3$ to
  $\mathbb{C}\mathbb{H}^2$ or vice versa, these actions are equivalent under the
  equivalence relation.
\item A Gromov hyperbolic $G$--space has a bounded orbit if and only if
  it is equivalent to the trivial $G$--space consisting of a single
  point.
\item If $X$ and $Y$ are equivalent Gromov hyperbolic $G$--spaces,
  $x\in X$ and $y\in Y$, then the
  limit sets $\Lambda(X)=\{e\in \partial X\cap \overline{Gx}\}$ and
  $\Lambda(Y)=\{e\in \partial Y\cap \overline{Gy}\}$ are equivariantly
  homeomorphic.
\item The equivalence is perfectly well-defined in the more general
  setting of {\em quasi-actions} on geodesic Gromov hyperbolic
  spaces.  Call a geodesic Gromov hyperbolic space with a
  $G$--quasi-action a {\em hyperbolic quasi-$G$--space}.  Every
  hyperbolic quasi-$G$--space is equivalent to some hyperbolic
  $G$--space.
\end{enumerate}
\end{remark}

Given a group acting on a hyperbolic $G$--space $X$ and some basepoint
$x_0\in X$, one obtains a metric on $G$ given by the formula
\[d(g,h) = d(x_0,g^{-1}hx_0).\]
This metric is obviously determined by its values on $\{1\}\times G$.
The compact-open topology on real-valued functions on $G$ thus induces
a topology on the set of pointed hyperbolic $G$--spaces.
The quotient topology on the space of equivalence classes
of Gromov hyperbolic $G$--spaces is not Hausdorff.   For example, if
$G$ is a surface
group, then all of Teichm\"uller space is identified to a single
point, whose closure contains many inequivalent actions of $G$ on
$\mathbb{R}$--trees.

\subsection{Combinatorial horoballs}\label{ss:combhoro}
Combinatorial horoballs of the simplest possible type were defined in
\cite{rhds}, and used as building blocks for complexes naturally
associated to relatively hyperbolic groups.  The point there as here
is that these spaces can be used to ``hide'' an action on a
non-hyperbolic space in an action on a hyperbolic space.  There is
some flexibility as to how this can
happen which is deliberately ignored in \cite{rhds}; here we give a
more general construction.

\begin{definition}\label{d:admissible}
Let $X$ be a graph, acted on by $G$, and suppose that
\[\{B_i\colon\thinspace  X^{(0)}\to 2^{X^{(0)}}\}_{i\in \mathbb{N}}\]
is a collection of functions.  We will say that $B_*$ is
\emph{admissible} if it
satisfies the following
four axioms:
\begin{enumerate}
\item Connectedness:  If $v\in X^{(0)}$, then
  $B_1(v)=\{w\in X^{(0)}\mid d_X(v,w)\leq 1\}$.
\item\label{ax:exp} Exponential growth: Let $v$, $w\in X^{(0)}$ and $n\in \mathbb{N}$.  If
  $v\in B_n(w)$, then $B_n(v)\subset  B_{n+1}(w)$.
\item Symmetry: Let $v$, $w\in X^{(0)}$ and $n\in \mathbb{N}$.
  If $v\in B_n(w)$, then $w\in B_n(v)$.
\item $G$--equivariance: If $w\in X^{(0)}$, $n\in \mathbb{N}$, and $g\in G$,
  then $g B_n(w)=B_n(gw)$.
\end{enumerate}
\end{definition}

\begin{definition}\label{d:choroball}
Let $B_*$ be a sequence of functions $\{B_i\colon\thinspace  X^{(0)}\to
2^{X^{(0)}}\}_{i\in \mathbb{N}}$.  The \emph{combinatorial horoball based on
  $X$ and $B_*$}, or $\mc{H}(X,B_*)$, is the graph defined as follows:
\begin{enumerate}
\item $\mc{H}(X,B_*)^{(0)} = X^{(0)}\times \mathbb{N}$.
\item If $n\in \mathbb{N}$ and $v\in X^{(0)}$, then $(v,n)$ is connected to
  $(v,n+1)$ by an edge (called a \emph{vertical edge}).
\item If $n\in \mathbb{N}$ and $v\in B_n(w)$, then $(v,n)$ is connected to
  $(w,n)$ by an edge (called a \emph{horizontal edge}).
\end{enumerate}
\end{definition}
We leave the following Lemma as an exercise.
\begin{lemma}\label{l:horohyp}
Let $G$ act on the graph $X$.  If $B_*$ is admissible, then
$\mc{H}(X,B_*)$ is Gromov hyperbolic, and the action of $G$ on $X$
induces an action of $G$ on $\mc{H}(X,B_*)$.
\end{lemma}

Because of the flexibility of this construction, a group
typically admits many inequivalent actions on horoballs.

\begin{definition}\label{d:ihoroball}
A Gromov hyperbolic $G$--space {\em has an invariant horoball} if it is
equivalent to a $G$--space of the form $\mc{H}(X,B_*)$, for some graph
$X$ with a $G$--action, and some family $B_*$ which is admissible.
\end{definition}
In the next section we give some other characterizations of $G$--spaces
with invariant horoballs (Theorem \ref{t:horeq}).

\section{Elementary actions and pseudocharacters}\label{s:elementary}
In this section we show how an elementary action by $G$ on a hyperbolic space
gives rise to a pseudocharacter on $G$ which ``picks out'' the
elements which act hyperbolically.  We then see that if no element
acts hyperbolically, then the action has an invariant horoball.

\subsection{The pseudocharacter coming from an elementary action}
Recall the following definitions.
\begin{definition}
An action of $G$ on a hyperbolic space $X$ is \emph{elementary} if it
is either equivalent to the trivial action on a point, or if the
induced action on $\partial X$ has a global fixed point.\footnote{This definition is slightly more restrictive than the
  usual one, which allows for a pair of points in $\partial X$ to be
  preserved.}
\end{definition}
\begin{definition}
A \emph{quasicharacter} (or quasi-morphism) on a group $G$ is a
real valued function $q$ on $G$ satisfying
\begin{equation}\label{defect}
|q(gh)-q(g)-q(h)|<C\mbox{, for all } g,h\in G.
\end{equation}
The \emph{defect} of a quasicharacter is the
smallest $C$ so that \eqref{defect} is satisfied.  A
\emph{pseudocharacter} (or homogeneous quasi-morphism) is a
quasicharacter $p$ which satisfies the additional condition
\begin{equation*}
  p(g^n)=n p(g)\mbox{, for all }g\in G, n\in \mathbb{Z}
\end{equation*}
\end{definition}
In this subsection we show how an elementary action on a hyperbolic
space gives rise to a pseudocharacter which is nonzero precisely on
the elements which act hyperbolically.
We begin by studying ``quasi-horofunctions'' on the space
$X$, corresponding to a fixed point at infinity.  A
quasi-horofunction restricted to an arbitrary orbit will give
a quasicharacter, which can then be homogenized to give the desired
pseudocharacter.

\begin{definition}
(cf. \cite[7.5.D]{gromov:wordhyperbolic})
Let $\mathbf{x}=\{x_i\}$ be a sequence tending to infinity in the geodesic
hyperbolic space $X$.  The \emph{quasi-horofunction coming from}
$\mathbf{x}$ is the function $\eta_{\mathbf{x}}\colon\thinspace X\to \mathbb{R}$ given by
\[ \eta_{\mathbf{x}}(a) = \limsup_{n\to\infty}(d(a,x_n)-d(x_0,x_n)). \]
\end{definition}

We use the following observation repeatedly:
\begin{observation}\label{o:fourdelta}
Let $A$, $B$, $C$ and $D$ be four points in the $\delta$--hyperbolic
space $X$. If $\gp{C}{D}{A}$ and $\gp{C}{D}{B}$ are both larger than $d(A,B)$,
then
\[ | (d(B,C) - d(A,C)) - (d(B,D) - d(A,D)) | \leq 4\delta. \]
\end{observation}
The observation \ref{o:fourdelta} implies in particular:
\begin{lemma}\label{l:approxbyterm}
If $a\in X$, and $\mathbf{x}=\{x_i\}$ tends to infinity in $X$, then
for all $n$ sufficiently large,
\[| \eta_{\mathbf{x}}(a)-(d(a,x_n)-d(x_0,x_n))| \leq 4\delta.\]
\end{lemma}

We now can describe the dependence of $\eta_{\mathbf{x}}$ on the sequence
$\mathbf{x}$.
\begin{lemma}\label{l:key}
Let $\mathbf{x}=\{x_i\}$ and $\mathbf{y}=\{y_i\}$ be two sequences of
points in the geodesic $\delta$--hyperbolic space $X$
which tend to the same point in $\partial X$.  For any point $a\in X$,
we have
\[ | \eta_{\mathbf{x}}(a)-\eta_{\mathbf{y}}(a)-\eta_{\mathbf{x}}(y_0)|
\leq 16\delta. \]
\end{lemma}
\begin{proof}
Since $\mathbf{x}$ and $\mathbf{y}$ tend to the same point at
infinity, we may choose $N$ so that
$(z,z')_\alpha> 2\thinspace \mathrm{diam}\{a,x_0,y_0\}$ for every $z$, $z'$ in
$\{x_i \mid i\geq N\}\cup \{y_i \mid i\geq N\}$.  Using Lemma
\ref{l:approxbyterm} three times, the quantity
\[|\eta_{\mathbf{x}}(a)-\eta_{\mathbf{y}}(a)-\eta_{\mathbf{x}}(y_0)|\]
differs by at most $12\delta$ from
\begin{equation}\label{easy}
| (d(a,x_N) - d(a,y_N)) - (d(y_0,x_N)-d(y_0,y_N))|.
\end{equation}
By Observation \ref{o:fourdelta}, the quantity \eqref{easy} is at most
$4\delta$.  The Lemma follows.
\end{proof}

Using Lemma \ref{l:key}, we deduce that an isometry of $X$ changes
$\eta_{\mathbf{x}}(a)$ by approximately the same amount, independent of
the $a\in X$ chosen:
\begin{proposition}\label{p:isom}
Let $X$ be a geodesic $\delta$--hyperbolic space, and suppose that
$\mathbf{x}=\{x_i\}$ tends to $e\in \partial X$.  Let $a$ be any point
in $X$.
If $g\in \mathrm{Isom}(X)$ fixes $e$, then
$\eta_{\mathbf{x}}(ga)$ differs from
$\eta_{\mathbf{x}}(a)+\eta_{\mathbf{x}}(gx_0)$ by at most $16\delta$.
\end{proposition}\label{p:trans}
\begin{proof}
First note that if $g\mathbf{x}$ is the sequence $\{gx_i\}$, then
\[\eta_{\mathbf{x}}(a) = \eta_{g\mathbf{x}}(ga).\]
But by Lemma \ref{l:key},
\[ | \eta_{\mathbf{x}}(ga)-\eta_{g\mathbf{x}}(ga)-\eta_{\mathbf{x}}(g x_0)|
\leq 16\delta. \]
\end{proof}

\begin{corollary}\label{c:quasimorphism}
Suppose $X$ is a $\delta$--hyperbolic space, and that $G$ acts on $X$
fixing $e\in \partial X$.  Let $\mathbf{x}=\{x_i\}$ be any sequence tending to
$e$.  The function $q_{\mathbf{x}}\colon\thinspace G\to \mathbb{R}$ defined by
$q_{\mathbf{x}}(g)=\eta_{\mathbf{x}}(g x_0)$ is a quasicharacter of
defect at most $16\delta$.
\end{corollary}
\begin{proof}
Let $g$, $h\in G$.  Using Proposition \ref{p:isom},
\begin{eqnarray*}
|q_{\mathbf{x}}(gh)-q_{\mathbf{x}}(g)-q_{\mathbf{x}}(h)| & = &
 |\eta_{\mathbf{x}}(ghx_0)-\eta_{\mathbf{x}}(gx_0)-\eta_{\mathbf{x}}(hx_0)| \\
& \leq & |
 \eta_{\mathbf{x}}(hx_0)+\eta_{\mathbf{x}}(gx_0)-\eta_{\mathbf{x}}(gx_0)-\eta_{\mathbf{x}}(hx_0)|+16\delta = 16\delta.
\end{eqnarray*}
\end{proof}

\begin{proposition}\label{p:pseudo}
Let $X$, $e$, $\mathbf{x}$, and $q_{\mathbf{x}}$ be as in the
statement of Corollary \ref{c:quasimorphism}, and let the
pseudocharacter $p_{\mathbf{x}}\colon\thinspace G\to \mathbb{R}$
be given by
\[p_{\mathbf{x}}(g) = \lim_{n\to\infty}\frac{q_{\mathbf{x}}(g^n)}{n}.\]
Then $p_{\mathbf{x}}(g)\neq 0$ if and only if $g$ acts hyperbolically on $X$.
\end{proposition}

\begin{proof}
First, we suppose that $p_{\mathbf{x}}(g)\neq 0$.  Without loss of generality we
assume that $p_{\mathbf{x}}(g)>0$.  Since
\[p_{\mathbf{x}}(g)=\lim_{n\to\infty}\frac{\eta_{\mathbf{x}}(g^n x_0)}{n}>0,\]
there exists some $N$ so that
$\eta_{\mathbf{x}}(g^n x_0)>\frac{1}{2}p_{\mathbf{x}}(g)n$ for all
$n\geq N$.

Let $a$ and $b$
be integers.  Choosing some sufficiently large $M$ and applying the triangle
inequality and Lemma \ref{l:approxbyterm}, we obtain a lower bound for
$d(g^a x_0,g^b x_0)$:
\begin{eqnarray*}
d(g^a x_0,g^b x_0) & = & d(g^N x_0,g^{N+|b-a|}x_0)\\
 & \geq & d(g^{N+|b-a|} x_0, x_M)-d(g^N x_0,x_M)\\
 & \geq & \eta_{\mathbf{x}}(g^{N+|b-a|})-\eta_{\mathbf{x}}(g^N) -
 8\delta\\
 & \geq & \frac{1}{2}p_{\mathbf{x}}(g)(N+|b-a|)-\eta_{\mathbf{x}}(g^N)-8\delta\\
 & = & \frac{1}{2}p_{\mathbf{x}}(g) |b-a| - (\eta_{\mathbf{x}}g^N+8\delta-\frac{1}{2}p_{\mathbf{x}}(g)N).
\end{eqnarray*}
On the other hand, $d(g^a x_0,g^b x_0)\leq |b-a| d(x_0,gx_0)$, so the map
$n\mapsto g^n(x_0)$ is a quasi-isometric embedding, and $g$ acts
hyperbolically.

Conversely, suppose that $g$ acts hyperbolically.
It
follows that there is some $\epsilon>0$ so that
\[d(g^n x_0,x_0)>\epsilon n\] for all $n$.
By
replacing $g$ with $g^{-1}$, we may suppose that $\{g^i x_0\}$ tends
to a point in $\partial X\smallsetminus\{e\}$ as $i\to\infty$.
Thus there is some $R$ so that
\[\gp{g^n x_0}{x_i}{{x_0}}<R\] for all positive $n$ and $i$.
Lemma \ref{l:approxbyterm} implies that for sufficiently
large $m$,
\begin{eqnarray*}
\eta_{\mathbf{x}}(g^n x_0) & \geq & d(g^n x_0,x_m)-d(x_0,x_m)-4\delta\\
& = & d(g^n x_0,x_0)-2 \gp{g^nx_0}{x_n}{{x_0}} - 4\delta \\
& \geq & \epsilon n - (2 R +4\delta).
\end{eqnarray*}
Since $q_{\mathbf{x}}(g^n)\geq \epsilon n - (2R+4\delta)$ for all $n>0$, we must
have $p(g)\geq \epsilon>0$.
\end{proof}

\begin{remark}
Proposition \ref{p:pseudo} was proved in \cite[Proposition 3.9]{manning:qfa} for the case of
quasi-trees.  The proof here is somewhat more efficient even in this
case.
\end{remark}

\subsection{Characterization of $G$--spaces with invariant horoballs}
\begin{theorem}\label{t:horeq}
Let $X$ be a Gromov hyperbolic $G$--space.  The following are
equivalent:
\begin{enumerate}
\item\label{ih} $X$ has an invariant horoball.
\item\label{enh} $X$ is elementary, and no element acts
  hyperbolically.
\item\label{onepoint} $X$ is equivalent to a hyperbolic $G$--space $Y$, so that
  $\#(\partial Y)\leq 1$.
\end{enumerate}
\end{theorem}
\begin{proof}
That \eqref{ih} implies \eqref{onepoint} is trivial.

We next assume \eqref{onepoint} and show \eqref{enh}.  If \eqref{enh}
holds for $Y$, it holds for $X$, so we may suppose that $X=Y$.  If
$\#(\partial X) = 1$ or $G$ has a bounded orbit in $X$, then clearly
$X$ is elementary.  The only case remaining is that $\partial X$ is
empty, but $Gx$ is unbounded for some $x\in X$.  We show that this
case does not occur.  Chose a sequence
$\{g_i\}$ in $G$ so that $\lim_{i\to\infty}d(g_i x,x) = \infty$.
Since $\partial X$ is empty,
$\liminf_{i,j\to\infty}(g_ix,g_jx)_x<\infty$.  It follows that there
are elements $g_m$, $g_n$ so that $d(g_mx,x)$ and $d(g_nx,x)$ are much
larger than $(g_mx,h_nx)_x$.  It can then be shown (see, for
example, \cite[Chapitre 9, Lemme 2.3]{cdp}) that $g_mg_n$ is hyperbolic.
It follows that $\partial X$ contains at least two points (the fixed
points of $g_mg_n$), contrary to assumption.

It remains to show that \eqref{enh} implies \eqref{ih}.
Let $X$ be a hyperbolic $G$--space so that the action of $G$ is
elementary.  If $X$ is equivalent to a point (i.e. if $Gx$ is bounded
for $x\in X$), then $X$ is also
equivalent to a ray, which is the combinatorial horoball based on a
point.  We therefore may assume that $Gx$ is unbounded for any $x\in
X$.  We will construct a combinatorial horoball which coarsely
equivariantly quasi-isometrically embeds in $X$.
Let $\delta>0$ be some number so
that $X$ is $\delta$--hyperbolic.
Let $e$, $\mathbf{x}$,
$q_{\mathbf{x}}$ and $p_{\mathbf{x}}$ be as in the statements of
Corollary \ref{c:quasimorphism} and Proposition \ref{p:pseudo}.

To build the combinatorial horoball, we first must start with a graph $Y$
on which $G$ acts.  Choose a finite generating set $S$ for $G$, and
let $C_0 = \mathrm{diam}(S x_0)$.
Let $V(Y)=G$, and connect $g$ to $h$ in $Y$ if $d(g x_0, h x_0)\leq C_0$.  It
is clear that $G$ acts on $Y$; in fact, $Y$ is a certain Cayley graph
for $G$.  We next define the functions $B_n\colon\thinspace V(Y)\to 2^{V(Y)}$.
Let $C_1 = 2C_0+20\delta$,
and let
\begin{equation}\label{bstar}
B_n(g) = \{ h\in G\mid d(hx_0,gx_0)\leq (2n+1)C_1\}.
\end{equation}
\begin{claim}
The sequence of functions $B_*$ in \eqref{bstar} is admissible in the sense
of Definition \ref{d:admissible}.
\end{claim}
\begin{proof}
The only axiom which is not obvious is \eqref{ax:exp}.  We must show
that if $a$ and $b$ are in $B_n(v)$, then $a\in B_{n+1}(b)$ (or
equivalently $b\in B_{n+1}(a)$).
Put another way,
we must show that if $d(ax_0,vx_0)$ and $d(bx_0,vx_0)$
are bounded above by $(2n+1)C_1$ then $d(a x_0,b x_0)\leq (2n+3)C_1$.

Because no element of $G$ acts hyperbolically, the pseudocharacter
$p_{\mathbf{x}}$ is identically zero.  An easy argument shows that
$|q_{\mathbf{x}}(g)|\leq 16\delta$ for all $g\in G$.
Using Lemma \ref{l:approxbyterm}, we can choose some large $n$ so that
\[| \eta_{\mathbf{x}}(z)-(d(z,x_n)-d(x_0,x_n))| \leq 4\delta\]
for $z\in \{ax_0,bx_0,vx_0\}$.  It follows that
\begin{equation}\label{diff}
|d(z_1,x_m)-d(z_2,x_m)|\leq 16\delta+4\delta\leq C_1
\end{equation} for $z_1$,
$z_2\in\{ax_0,bx_0,vx_0\}$.  The assertion to be proved is symmetric
in $a$ and $b$, so we may assume
that $\gp{x_m}{b}{v}\leq \gp{x_m}{a}{v}$.
We deduce:
\begin{eqnarray*}
d(a,b) & \leq & \gp{v}{x_m}{a}+[\gp{a}{x_m}{v}-\gp{b}{x_m}{v}]+\gp{v}{x_m}{b}+2\delta\\
& = & d(v,a) +(d(b,x_m)-d(v,x_m))+2\delta \\
& \leq & (2n+1)C_1 + C_1 + 2\delta \leq (2n+3)C_1.
\end{eqnarray*}
The first line follows from examining the comparison tripods
for the triangles $\Delta(a,v,x_m)$ and $\Delta(b,v,x_m)$;
the last follows from \eqref{diff}.
\end{proof}

Since $B_*$ is admissible, the combinatorial horoball
$H=\mathcal{H}_{B_*}(Y)$ is a hyperbolic $G$--space.  It remains
to construct a coarsely equivariant quasi-isometric embedding from $H$
to $X$.  It
suffices to define this map on the vertices of $H$.
For each $g\in G = V(Y)$ and each $n\in\mathbb{N}$, choose some
$i(g,n)$ so that $\gp{x_k}{x_l}{{gx_0}}\geq 2nC_1$ for all $k$, $l\geq
i(g,n)$.  Choose also some unit speed geodesic $\gamma_{g,n}$ starting
at $g x_0$ and ending at $x_{i(g,n)}$.  Any vertex of $H$ is a pair
$(g,n)$ where $n\in \mathbb{N}$ and $g\in G$.  We define a map $\phi\colon\thinspace
V(H)\to X$ by
\begin{equation*}
\phi(g,n) = \gamma_{g,n}(nC_1).
\end{equation*}
A number of choices were made in the definition of $\phi$ (namely, the
sequence $\mathbf{x}$, the numbers $i(g,n)$, and the geodesics $\gamma_{g,n}$).
However, so long as $x_0$ is unchanged, different
choices lead to a function which differs by at most $\delta$
from $\phi$.  In particular, we could replace $\mathbf{x}$ by
$\mathbf{x}'=\{x_i'\}$, where $x_0'=x_0$ and $x_i'=hx_i$ for some
fixed $h$ and for all $i\geq 1$.
It follows that the distance between $\phi(hg,n)$ and $h\phi(g,n)$ is
at most $\delta$ for any $h$, $g\in G$ and $n\in \mathbb{N}$, and so the map
$\phi$ is coarsely equivariant.

It remains to show that $\phi$ is a quasi-isometric
embedding.

Note that if $v$ and $w$ are two vertices in $H$ connected by a
vertical path, then
$C_1d_H(v,w)-\delta\leq d(\phi(v),\phi(w))\leq C_1d_H(v,w)+\delta$,
where $d_H$ is the distance in $H$.

We therefore assume that $v=(a,n)$ and $w=(b,m)$, where $a\neq b$.
There is a unique $k$ so that $(2k-1)C_1<d(a,b)\leq (2k+1)C_1$.  If
$\max\{m,n\}\geq k$, then $d(v,w)=|m-n|+1$; otherwise
$d(v,w)=2k-(m+n)+\frac{1}{2}\pm\frac{1}{2}$.  Let
$I \geq \max\{i(a,n),i(b,m)\}$, and observe that the points $\phi(v)$
and $\phi(w)$ lie within $\delta$ of geodesics joining $ax_0$ to $x_I$
and $bx_0$ to $x_I$, respectively (see Figure \ref{f:deep}).
\begin{figure}[htbp]
\begin{center}
\input{deep.pstex_t}
\caption{A pair of points in the image of $\phi$.}
\label{f:deep}
\end{center}
\end{figure}
Note that $\gp{x_I}{ax_0}{{bx_0}}$
and $\gp{x_I}{bx_0}{{ax_0}}$ differ from $\frac{1}{2}d(ax_0,bx_0)$ by at
most $\frac{C_1}{2}$.
There are a couple of cases to consider.

First, assume that one or both of $n$ and $m$ is at least $k$.
Without loss of generality assume that $n\geq k$.  Since
$d(ax_0,bx_0)\leq (2k+1)C_1$, we have $(x_I,bx_0)_a\leq kC_1+C_1$.
Since $d(\phi(v),ax_0)\geq k C_1$,
it follows that $\phi(v)$ is at most $C_1+\delta$ from the
geodesic joining $bx_0$ to $x_I$. Accounting for the possible
difference between $\gp{x_I}{ax_0}{{bx_0}}$ and $\gp{x_I}{bx_0}{{ax_0}}$, we deduce that the distance between $\phi(v)$ and
$\phi(w)$ differs from $|n-m|C_1=(d(v,w)+\frac{1}{2}\pm\frac{1}{2})C_1$ by at most $2C_1+\delta$.  It
follows that in case one of $n$ or $m$ is at least $k$, we have
\[ C_1 d(v,w)-2C_1+\delta \leq d(\phi(v),\phi(w))\leq
C_1d(v,w)+3C_1+\delta.\]

In case both $n$ and $m$ are strictly less than $k$, we may argue as
follows.  Since $\gp{ax_0}{x_I}{{bx_0}}$ and $\gp{bx_0}{x_I}{{ax_0}}$ are both
at least $\frac{1}{2}(d(ax_0,bx_0)-C_1)\geq (k-1)C_1\geq \max\{m,n\}
C_1$, it follows that both $\phi(v)$ and $\phi(w)$ are within $2\delta$
of the geodesic joining $ax_0$ to $bx_0$.  From this it follows that
$d(\phi(v),\phi(w))$ differs from $d(ax_0,bx_0)-(n+m)C_1$ by at most
$4\delta$.  But since $d(ax_0,bx_0)$ differs by at most $C_1$ from $k
C_1$, we deduce that
\[(2k-(m+n))C_1 -(C_1+4\delta) \leq d(\phi(v),\phi(w))\leq
(2k-(m+n))C_1 + (C_1+4\delta),\]
from which it immediately follows that
\[C_1 d(v,w) - (2C_1+4\delta)\leq d(\phi(v),\phi(w))\leq C_1 d(v,w)
+(2C_1+4\delta).\]

In particular, $\phi$ is a $(C_1,2C_1+4\delta)$--quasi-isometric
embedding from the combinatorial horoball $H$ into $X$, and the
theorem is established.
\end{proof}

\section{Rigidity in rank $\geq 2$}\label{s:bigrank}
The purpose of
this section is to establish Theorem \ref{thm:cdr2}, but we will begin
with some lemmas which hold in a slightly broader context.
We suppose that $G$ is a simple Chevalley-Demazure
group scheme of rank at least $2$, that $\Phi$ is a root system for
$G$, and that $R$ is some commutative unital ring containing $\mathbb{Z}$.

\begin{lemma}\label{l:phiprime}
If $\alpha$ and $\beta\in \Phi$,
then there is a $\Phi' = \mathrm{Span}(\Phi')\cap\Phi\subseteq \Phi$
containing $\alpha$ and $\beta$ so that
$\Phi'$ is
isomorphic to $A_1\times A_1$, $A_2$, $B_2$ or $G_2$.  If $\Phi'\cong
A_1\times A_1$, then $\alpha\neq -\beta$.
\end{lemma}
\begin{proof}
If $\alpha$ and $\beta$ are linearly independent, then
$\Phi'=\mathrm{Span}(\{\alpha,\beta\})\cap\Phi$ is a root system of
rank two, and we simply recall that such a root system is always
isomorphic to one of those listed.

If $\beta=-\alpha$, we assert that there must be some $\gamma\in \Phi$
so that
$\mathrm{Span}(\{\alpha,\gamma\})\cap\Phi$ is not equal to $A_1\times
A_1$.
Suppose that there is no such $\gamma$.  Then either
$\Phi=\overline{\Phi}\times\langle\alpha\rangle$, or
$\Phi=\langle\alpha\rangle$.
Because $G$ is simple $\Phi$ cannot split as a product; because $G$
has rank at least two, $\Phi\neq A_1$.
\end{proof}
For the following two lemmas, we refer to \cite{carter:book}.  Although
the proofs there are done assuming that $R$ is a field, this
assumption is unnecessary;  see also \cite{stein71} or
\cite{steinberg:notes}.
\begin{lemma}\label{l:steinberg}{\em (Steinberg commutator
    relations)}\cite[Theorem 5.2.2]{carter:book}
If $\alpha$, $\beta\in \Phi$ and $t$, $u\in R$, then
\[[x_\alpha(t),x_\beta(u)]=\prod_{i,j>0\mbox{, }i\alpha+j\beta\in \Phi }
x_{i\alpha+j\beta}(N_{\alpha,\beta,i,j}t_iu_j), \]
where the $N_{\alpha,\beta,i,j}\in \mathbb{Z}$ are integers which depend only
on the order in which the product is taken.
\end{lemma}
\begin{lemma}\label{l:weylconj}\cite[Lemma 7.2.1]{carter:book}
If $\alpha\in \Phi$, $w$ is an element of the Weyl group of $\Phi$,
and $t\in R$,
then
$x_{w(\alpha)}(t)$ is conjugate either to  $x_\alpha(t)$
or $x_\alpha(-t)$.
\end{lemma}

\begin{lemma}\label{l:distortion}
Let $X$ be a hyperbolic $G(R)$--space.
If $g = x_\alpha(t)$ for some $\alpha\in \Phi$ and $t\in R$, then $g$
does not act hyperbolically on $X$.
\end{lemma}
\begin{proof}
By Lemma \ref{l:phiprime}, there is a subset $\Phi'$ of $\Phi$
containing $\alpha$ which is either isomorphic to $A_2$, $B_2$ or
$G_2$.  In each case, we may apply Lemma \ref{l:steinberg} some number
of times to show that $g$ is distorted in $G$; the details of this are
left to the reader.  By Lemma
\ref{lemma:distortion}, $g$ cannot act hyperbolically on $X$.
\end{proof}

\begin{proposition}\label{p:allfixone}
Let $X$ be a hyperbolic $G(R)$--space,
let $r$, $s\in \Phi$, and let $\rho_1$, $\rho_2\in R$.
Suppose that $p=x_r(\rho_1)$ acts parabolically, fixing some $e\in
\partial X$.  Then $ge=e$ for any other root element $g=x_s(\rho_2)$.
\end{proposition}
\begin{proof}
If $r=s$ or if
$\langle r,s\rangle = A_1\times A_1$, then
$p$ and $g$ commute.  By Lemma \ref{l:commute}, $g(e)=e$, and we are
done.

Otherwise $r$ and $s$ are contained in a two-dimensional root system
$\Phi'\subset \Phi$ which is isomorphic to
$A_2$, $B_2$ or $G_2$, by Lemma \ref{l:phiprime}.

Each case requires a separate argument.

\begin{case}\label{c:a2}
$\Phi'\cong A_2$.
\end{case}
$A_2 = \{\lambda_i \mid i\in \mathbb{Z}_6\}$ contains six roots, arranged
hexagonally; the angle between $\lambda_i$ and
$\lambda_j$ is $\frac{|i-j|}{3}\pi$.  Suppose $r=\lambda_i$ and $s=\lambda_j$.
By Lemma \ref{l:weylconj}, $h = x_s(\rho_1)$ is conjugate to either
$p$ or $p^{-1}$, so $h$ is a parabolic, fixing some point
$f\in\partial X$.
In case $|i-j|=1$, then Lemma \ref{l:steinberg} implies that
$h$ and $p$ commute, and so $f=e$ by Lemma \ref{l:commute}.
In case $|i-j|>1$, one argues by induction on $|i-j|$ to the same
conclusion: $f=e$.

Since $g$ commutes with $h$, we must have $ge=e$, again by Lemma
\ref{l:commute}.

\begin{case}\label{c:b2}
$\Phi'\cong B_2$.
\end{case}

Let $\alpha$ be a short root, and $\beta$ a long root, so that
$\alpha$ and $\beta$ span $\Phi'\subset \Phi$, as in
Figure \ref{f:b_2}.
\begin{figure}[htbp]
\begin{center}
\begin{picture}(0,0)%
\includegraphics{b_2.pstex}%
\end{picture}%
\setlength{\unitlength}{4144sp}%
\begingroup\makeatletter\ifx\SetFigFont\undefined%
\gdef\SetFigFont#1#2#3#4#5{%
  \reset@font\fontsize{#1}{#2pt}%
  \fontfamily{#3}\fontseries{#4}\fontshape{#5}%
  \selectfont}%
\fi\endgroup%
\begin{picture}(4352,2006)(867,-3021)
\put(4972,-2047){\makebox(0,0)[lb]{\smash{{\SetFigFont{6}{7.2}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\alpha$}%
}}}}
\put(3455,-1448){\makebox(0,0)[lb]{\smash{{\SetFigFont{6}{7.2}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\beta$}%
}}}}
\put(2546,-2026){\makebox(0,0)[lb]{\smash{{\SetFigFont{6}{7.2}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\alpha$}%
}}}}
\put(882,-1238){\makebox(0,0)[lb]{\smash{{\SetFigFont{6}{7.2}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\beta$}%
}}}}
\end{picture}%
\caption{$B_2$ and $G_2$.}
\label{f:b_2}
\end{center}
\end{figure}

There are two subcases, depending on whether $r$ is a short or long
root.
\begin{subcase}
The parabolic $p=x_r(\rho_1)$, where $r$ is a short root of $\Phi'$.
\end{subcase}
Without loss of generality,
we may assume that $r=\alpha$.  If $s\in
\{\alpha,2\alpha+\beta,-\beta\}$, then $g=x_s(\rho_2)$ commutes with
$p$, by Lemma \ref{l:steinberg}.
Thus by Lemma \ref{l:commute} $ge=e$.

Suppose next that $s = \pm (\alpha+\beta)$.  Then Lemma
\ref{l:steinberg} implies
\begin{equation}\label{c1}
p g p^{-1} g^{-1} = x_{r+s}(N \rho_1\rho_2),
\end{equation}
where $N=N_{r,s,1,1}$ is an integer.  Since $r+s\in
\{2\alpha+\beta,-\beta\}$, we already know that
$h:=x_{r+s}(N\rho_1\rho_2)$ fixes $e$.  We rearrange \eqref{c1} to give
\begin{equation}
 h^{-1} p = gpg^{-1}.
\end{equation}
Since $p$ and $h$ both fix $e$, so must $gpg^{-1}$.  The element
$gpg^{-1}$ also fixes $g(e)$, since $p$ fixes $e$.
Since $p$ is
parabolic, it can only fix one point in $\partial X$, and so $g(e)=e$.

If $s\in \{\beta, -\alpha\}$, let $r' = \alpha+\beta$; if
$s=-2\alpha-\beta$ let $r' = -\alpha-\beta$.  In any case there is an
element of the Weyl group of $\Phi$ taking $r$ to $r'$; by Lemma
\ref{l:weylconj}, there is a
$p'= x_{r'}(\pm \rho_1)$ which is
conjugate to $p$ in $E(\Phi,R)$.  Since $p'$ is conjugate to $p$, it
is parabolic; by the argument of the previous paragraph, $p'$ has the
same fixed point as $p$.  If $s\in \{\beta, -2\alpha-\beta\}$, then
$g$ commutes with $p'$, and so $g(e)=e$ by Lemma \ref{l:commute}.
Finally, if
$s=-\alpha$, then we may apply the argument of the previous paragraph
again (with $p'$ and $r'$ in place of $p$ and $r$), to deduce that $g(e)=e$.

\begin{subcase}
The parabolic $p=x_r(\rho_1)$, where $r$ is a long root of $\Phi'$.
\end{subcase}
In this case, we may assume for instance that $r=2\alpha+\beta$.  If
$s\in\{2\alpha+\beta, \alpha+\beta, \alpha, \beta, -\beta\}$, then
Lemma \ref{l:steinberg} implies that $g=x_s(\rho_2)$ commutes with $p$,
and so $g(e)=e$ by Lemma \ref{l:commute}.

Suppose then that $s\in \{-\alpha,-2\alpha-\beta,-\alpha-\beta\}$.
If $s\in\{-\alpha,-2\alpha-\beta\}$, let $r'=\beta$; if $s=
-\alpha-\beta$, then let $r'=-\beta$.  In either case, there is an
element of the Weyl group taking $r$ to $r'$, and so there is a
parabolic $p' = x_{r'}(\pm \rho_1)$ conjugate to $p$ by Lemma
\ref{l:weylconj}.   By the previous paragraph, $p'$ has the same fixed
point as $p$ does.  Applying the previous paragraph with $p'$ and $r'$
in the place of $p$ and $r$ implies that $g(e)=e$ for $g =
x_s(\rho_2)$. This completes the proof of Case \ref{c:b2}.

\begin{case}\label{c:g2}
$\Phi'\cong G_2$.
\end{case}
Let $\alpha$ be a short root, and $\beta$ a long root, so that
$\alpha$ and $\beta$ span $\Phi'\subset \Phi$, as in
the right half of Figure \ref{f:b_2}.

Again there are two subcases, depending on whether $r$ is a short or long
root.

\begin{subcase}
The parabolic $p=x_r(\rho_1)$, where $r$ is a long root of $\Phi'$.
\end{subcase}
If $s\in \{-\beta, 3\alpha+\beta, \pm(3\alpha+2\beta)\}$, then Lemma
\ref{l:steinberg} implies that $g=x_s(\rho_2)$ commutes with $p$, and
so $g(e)=e$ by Lemma \ref{l:commute}.

Suppose that $s\in \{2\alpha+\beta,-\alpha-\beta\}$.  Lemma
\ref{l:steinberg} implies that
\begin{equation}
p g p^{-1} g^{-1} = x_{r+s}(N\rho_1\rho_2)=: h,
\end{equation}
for some integer $N$.  Exactly as in Case \ref{c:b2}, $h$ commutes
with $p$, and so $h(e)=e$.  Thus $e= h^{-1} p (e) = gpg^{-1}(e)$ and
the parabolic $gpg^{-1}$ fixes $e$.  Again since $gpg^{-1}$ also fixes
$g(e)$, we must have $g(e)=e$.

Using Lemma \ref{l:weylconj} repeatedly we discover that for every
short root $r'$ there is a parabolic element $p'=x_{r'}(\pm \rho_1)$
with $p'(e)=e$.  Since $g$ must commute with some such element,
$g(e)=e$ as well.
\begin{subcase}
The parabolic $p=x_r(\rho_1)$, where $r$ is a long root of $\Phi'$.
\end{subcase}
Without loss of generality, we may assume that $r = 3\alpha+2\beta$.

If the inner product of $s$ with $r$
is nonnegative, then $s\in\{ \pm \alpha, \beta, \alpha+\beta,
\alpha+2\beta,\alpha+3\beta, 3\alpha+2\beta\}$, and $g$ commutes with
$p$ by Lemma \ref{l:steinberg}, and so $g(e)=e$.

Otherwise, a (possibly
repeated) application of Lemma \ref{l:weylconj} implies that $g$
commutes with a parabolic $p' = x_{r'}(\pm \rho_1)$ for some long root
$r$ of $\Phi'$, and with $p'(e)=e$.  This again implies via Lemma
\ref{l:commute} that $g(e)=e$.  This completes the proof in Case \ref{c:g2}.
\end{proof}

We are now ready to give the proof of Theorem \ref{thm:cdr2}.
\begin{proof}
A result of Tavgen$'$ \cite{tavgen:bg} shows that $G(\mathcal{O})$ is
boundedly generated by its root subgroups.
The ring of integers $\mathcal{O}$ is finitely generated as an Abelian
group; choose generators $\mu_1,\ldots,\mu_k$.  It follows from
Tavgen$'$'s result that the set
\[ S=\{x_\alpha(\mu_i) \mid \alpha \in \Phi\mbox{, }1\leq i\leq k\}\]
boundedly generates $G(\mathcal{O})$.

Each of these generators acts hyperbolically, elliptically or
parabolically on $X$.
By Lemma \ref{l:distortion}, none can act hyperbolically.
If all of the root elements act elliptically, then it follows from bounded
generation that the orbit of a point under the action of $G$ must be
bounded.

We therefore may assume that some $x_\alpha(\mu_i)$ acts parabolically on
$X$, fixing a single point $e\in \partial X$.  It follows from
Proposition \ref{p:allfixone} that all the root subgroups will fix
this point $e$, and so $G(\mathcal{O})$ fixes $e$.

By Proposition \ref{p:pseudo}, the pseudocharacter
$p_\mathbf{x}\colon\thinspace G(\mathcal{O})\to \mathbb{R}$ determined by a sequence
$\mathbf{x}=\{x_i\}$ tending to $e$
is nonzero exactly on the hyperbolic elements.  Thus $p_\mathbf{x}(g)=0$
whenever $g$ lies in a root subgroup.

An elementary argument (see for example \cite[Proposition
  5]{kotschick:quasi}) shows that a pseudocharacter $p$ on a boundedly
generated group
is determined by its values on
the bounded generators;  thus $p\equiv 0$ on $G(\mathcal{O})$.  It
follows that no element of $G(\mathcal{O})$ acts hyperbolically on
$X$.
By Theorem
\ref{t:horeq}, the $G(\mathcal{O})$--space $X$ has an invariant horoball.
\end{proof}

\section{Remarks on rank one}\label{section:rankone}
One can also ask what actions rank one Chevalley groups have on
hyperbolic spaces.  If $\mc{O}$ is a number ring with finitely many
units, then $SL(2,\mc{O})$ is a lattice either in $SL(2,\mathbb{R})$ or
$SL(2,\mathbb{C})$.  In particular, it has a proper non-elementary action on
$\mathbb{H}^2$ or $\mathbb{H}^3$.  Moreover, such a group admits uncountably many
distinct pseudocharacters (AKA homogeneous quasi-(homo)morphisms) up
to scale \cite{fujiwara:gromovhyperbolic,bestvinafujiwara:mcg}.  Each
such ``projective pseudocharacter'' gives rise to a
quasi-action on $\mathbb{R}$; no two such are equivalent.  Moreover, these
often give rise to quasi-actions on more complicated trees
\cite{manning:cocycles}.
The groups $SL(2,\mc{O})$ where $\mc{O}$ has infinitely many units are
more rigid.
In this section we
apply the main result of \cite{manning:qfa} to the special case of
actions on quasi-trees (defined below), and
speculate on the general situation.

Recall that a group $G$ is said to have
Property \ensuremath{(\mathrm{FA})}\ if every action by $G$ on a simplicial tree $T$ has a
fixed point.
\begin{definition}
A \emph{quasi-tree} is a graph which is quasi-isometric to a tree.
\end{definition}
\begin{definition}
A group $G$ has property \ensuremath{(\mathrm{QFA})}\ if every action by $G$ on a quasi-tree
$X$ has a bounded orbit.
\end{definition}
\begin{remark}
This is differently worded than the definition in \cite{manning:qfa},
but easily seen to be equivalent.  Note that \ensuremath{(\mathrm{QFA})}\ implies \ensuremath{(\mathrm{FA})}, but
not vice versa.
\end{remark}
As quasi-trees are in particular Gromov hyperbolic spaces which admit
no parabolic isometries (see Section 3.2 of \cite{manning:qfa}),
Theorem \ref{thm:cdr2} implies that higher rank Chevalley
groups over number rings have property \ensuremath{(\mathrm{QFA})}.

We recall a definition and a theorem from \cite[Section 4]{manning:qfa}.

\begin{definition}
Let $G$ be a group, and let $g$ be an element of $G$.  We will say $g$
is a \emph{stubborn element of $G$} if for all
$H< G$ with $[G:H]\leq 2$, there exists some integer $k_H>0$ so that
$g^{k_H}\in [H,H]$.
\end{definition}

\begin{theorem}\label{thm:qfa}\cite[Theorem 4.4]{manning:qfa}
Let $G$ be a group which is boundedly generated by elements
$g_1,\ldots,g_n$, so
that for each $i$, $g_i$ is a stubborn element of $B_i$ for some
amenable $B_i < G$. Then $G$ has Property \ensuremath{(\mathrm{QFA})}.
\end{theorem}
Note that the above theorem was misstated slightly in
\cite{manning:qfa}; the
word ``amenable'' was inadvertently omitted.

Here's an easy lemma:
\begin{lemma}\label{lemma:finiteindex}
Let $H<G$ be a subgroup of finite index.  If $H$ has Property
\ensuremath{(\mathrm{QFA})}, then so does $G$.
\end{lemma}

Our methods in the higher rank case use heavily the bounded generation of
$G(\mathcal{O})$ established by Tavgen$'$ in \cite{tavgen:bg}
for Chevalley groups over rings of integers of algebraic
number fields.  There is an analogous result of Carter, Keller, and Paige
in rank $1$, at least for $SL(2,\cdot)$ and certain number
fields:
\begin{theorem}\label{thm:ckp2}
\cite{wittemorris:ckp}
For any integer $d>1$ there is an $r=r(d)$ so that the following is true.
Let $K$ be a number field of degree $d$ over $\mathbb{Q}$, and let
$\mathcal{O}$ be the ring of integers of $K$.  If the $\mc{O}$ has
infinitely many units, then:
\begin{enumerate}
\item every element of $E(2,\mathcal{O})$ is a product of at most $r$
  elementary matrices, and
\item the index of $E(2,\mathcal{O})$ in $SL(2,\mathcal{O})$ is at
  most $r$.
\end{enumerate}
\end{theorem}
In the above statement, $E(2,\mathcal{O})$ is the subgroup of
$SL(2,\mathcal{O})$ generated by the root subgroups (the strictly
upper triangular and strictly lower triangular matrices).
The following statement implies Theorem \ref{thm:cdr1}:
\begin{theorem}\label{thm:rankone}
If $\mathcal{O}$ is the ring of integers of an algebraic number field
and $\mathcal{O}$ has infinitely many units, then $SL(2,\mathcal{O})$
has Property \ensuremath{(\mathrm{QFA})}.
\end{theorem}
\begin{proof}
By Lemma \ref{lemma:finiteindex} it suffices to show that
$E(2,\mathcal{O})$ has property \ensuremath{(\mathrm{QFA})}.
If $\Lambda= \{\lambda_1,\ldots,\lambda_n\}$ is an integral basis
for the number field $\mathcal{O}$, then
Theorem \ref{thm:ckp2} implies that $E(2,\mathcal{O})$ is boundedly
generated by the $2n$ elements $\textMtwo{1}{\lambda_i}{0}{1}$ and
$\textMtwo{1}{0}{\lambda_i}{1}$.
\begin{claim}
For each $i\in\{1,\ldots,n\}$,
$\textMtwo{1}{\lambda_i}{0}{1}$ is a stubborn element of
$B=\textMtwo{*}{*}{0}{*}\cap E(2,\mathcal{O})$.
\end{claim}
\begin{proof}
On page 189 of \cite{carter:book}, Carter observes that
$\textMtwo{t}{0}{0}{t^{-1}}$ can be written as a product of elementary
matrices, for any invertible $t\in \mc{O}$, as follows.
If
$\lambda$ is any invertible element, we may write
\[\Mtwo{0}{\lambda}{-\lambda^{-1}}{0} =
\Mtwo{1}{\lambda}{0}{1}\Mtwo{1}{0}{-\lambda^{-1}}{1}\Mtwo{1}{\lambda}{0}{1},\]
and then note that
\begin{equation}\label{computation}
\Mtwo{t}{0}{0}{t^{-1}}=\Mtwo{0}{t}{-t^{-1}}{0}\Mtwo{0}{-1}{1}{0}.
\end{equation}
Carter's observation shows that $\textMtwo{\omega}{0}{0}{\omega^{-1}}$ is in
$E(2,\mc{O})$ for any unit $\omega$ in $\mc{O}$.
Computing the commutator of \textMtwo{\omega}{0}{0}{\omega^{-1}} and
\textMtwo{1}{\lambda}{0}{1} for $\omega$ a unit of $R$ and $\lambda\in R$
yields:
\begin{equation}\label{computation2}
\left[\Mtwo{\omega}{0}{0}{\omega^{-1}},\Mtwo{1}{\lambda}{0}{1}\right]
 = \Mtwo{1}{(1-\omega^2)\lambda}{0}{1}.
\end{equation}

By assumption, the group of units of $\mc{O}$ is infinite.
Dirichlet's units theorem (see, e.g. \cite[Appendix B]{stewarttall})
implies that we may
choose $\omega_0\in \mc{O}^*$ a unit of infinite order.
Let $H<B$ be a subgroup of index at most two.  Then $H$ must contain
$\textMtwo{\omega_0^2}{0}{0}{\omega_0^{-2}}$ and
$\textMtwo{1}{2 r}{0}{1}$ for all $r\in \mc{O}$.  It follows from the
computations \eqref{computation} and \eqref{computation2}
that $\textMtwo{1}{i}{0}{1}\in [H,H]$ for all
$i$ in the ideal $I$ generated by $2(1-\omega_0^2)$.  Let $N$ be the
order of $R/I$. (The number $N$ is also called the \emph{norm} of $I$;
that it is finite when $I\neq (0)$ is an elementary fact of algebraic
number theory;  see, e.g., \cite[Chapter 5]{stewarttall}.)
For any of the $\lambda_i$, we have
$\textMtwo{1}{\lambda_i}{0}{1}^N=\textMtwo{1}{N\lambda_i}{0}{1}\in [H,H]$, and
so $\textMtwo{1}{\lambda_i}{0}{1}$ is stubborn.
\end{proof}
It remains to observe that $B<E(2,\mathcal{O})$ is solvable, and hence
amenable.
We may now apply Theorem \ref{thm:qfa} to conclude that
$E(2,\mathcal{O})$ has Property \ensuremath{(\mathrm{QFA})}.
\end{proof}
\begin{remark}
It was already known \cite[p. 68]{serre:trees} that the groups covered
by Theorem \ref{thm:rankone} possessed property \ensuremath{(\mathrm{FA})}.
\end{remark}
\begin{remark}
Another proof of \ref{thm:rankone} may be given as follows:  First
show that every unipotent is distorted.  It follows that the bounded
generators cannot (quasi)-act hyperbolically.  It is shown in
\cite[Corollary 3.6]{manning:qfa} that there are no parabolic isometries of
quasi-trees, and so each of the bounded generators (quasi)-acts
elliptically.  It then follows from bounded generation that any orbit
is bounded.
\end{remark}

Finally, we speculate on the variety of hyperbolic $\Gamma$-spaces, for
$\Gamma=SL(2,\mathcal{O})$, where $\mathcal{O}$ is the
ring of integers of a number field $k$.
We have already remarked that $\Gamma$ is a lattice in
\[\prod_{i=1}^s SL(2,\mathbb{R})\times \prod_{i=1}^t SL(2,\mathbb{C}),\]
where $s$ and $t$ are the number of real and complex places
respectively.  Projection to some factor gives an isometric action
either on $\mathbb{H}^2$ or $\mathbb{H}^3$.  Call a hyperbolic $\Gamma$--space
\emph{standard} if it is equivalent to $\mathbb{H}^2$ or $\mathbb{H}^3$ with one of
these actions.

\begin{conjecture}\label{conjecture:realquadratic}
Every quasi-action by $\Gamma$ on a Gromov hyperbolic metric space
either has an invariant horoball or is standard.
\end{conjecture}

\def$'$} \providecommand\url[1]{\texttt{#1}{$'$} \providecommand\url[1]{\texttt{#1}}

redpajama_set_name: RedPajamaArXiv


Well, the show must go on, right? When Charlie Sheen started to have his epic freak-outs, CBS pulled Two And A Half Men from its lineup and, eventually, announced that the show was over despite being the top show on TV and having the cast members under contract. Well, not so fast my friend; Two And A Half Men may be coming back with a new leading man. Rumor has it that Ashton Kutcher is replacing Charlie Sheen on this hit CBS sitcom.
Dumped by Goddess Bree Olson and now dumped by his TV show, it's got to be tough to be Charlie Sheen right now. Yes, his stage show is still drawing crowds in spite of (or because of) his difficulties, but to see the show go on without you literally? Of course, how long the show survives without its main draw remains to be seen, but if anyone can help keep the show afloat, it's Kutcher.
Just hire Demi Moore for a guest-shot as one of Kutcher's love interests, then funnel in Laura Prepon and Mila Kunis, and boom, the show is winning again.


Posted by BroadwayReviewed & filed under Shows, Top Ten Broadway Songs, TV.
So one thing that people don't realize is that Broadway and Musical Theatre offers a huge selection of shows for you to enjoy for each holiday. When you go to a show you normally get taken to a new place where the show takes place and just like Halloween, there are costumes, there can be death, there can be parties and there can be tons of tricks and treats for the audiences. You can find princesses and fairies, witches and phantoms. There are devils and demons and even Biblical characters. If you are stuck with looking for something to do for Halloween, you may want to think about going to a broadway show or renting one to watch with friends at home and decorating your house like the stage or theatre. Here are my top ten broadway shows or musicals for Halloween.
1. The Rocky Horror Show.
Not only is The Rocky Horror Show perfect for Halloween, but they actually expect you to show up in character. You have costumes, aliens, romance and tons of singing and dancing. Get your Time Warp on with this cult classic and enjoy Halloween Transyvanian style with the cast of the Rocky Horror Show.
Sweeny Todd starts out in a dungeon and with an eerie and gruesome song. It then leads you through the life of the Demon Barber of Fleet Street. As you watch how he becomes a murderer and turns the local people in the town into meat pies then serves them to their friends you laugh and be grossed out as you watch and enjoy some fun songs and fabulous characters. Perfect for Halloween or just an odd night out on Broadway.
3. The Nightmare Before Christmas.
Although this show had never made it to Broadway, it is a musical that is all about the world and land of Halloween. Watch as Jack Skellington tries to bring together the worlds of Christmas and Halloween while the Boogey Man tries to kill Santa Claus. You'll laugh and be amazed with the imagination of Tim Burton and the amazing music of Tim Rice while you journey into a land of Halloween and Terror through the lives of a love stricken rag doll and a Skeleton who wants to find out if there is more to life than just Halloween.
4. The Wizard of Oz.
The Wizard of Oz is a classic musical movie that has delighted and amazed families for generations. You can sing a long with Dorothy, The Tinman, The Cowardly Lion, The Scarecrow, the Munchkins and even the Wizard. You get to watch the Wicked Witch of the West fight with Dorothy and you can watch as Glinda the good witch helps to save the day. Enjoy Halloween night with some of the most loved characters in Movie Musical History and visit the land of Oz with all of your friends and family.
If the Wizard of Oz doesn't do it for you, try the other side of the story and go to see Wicked on Broadway. Learn the real reason why Elphaba became the Wicked Witch of the West and how she and Glinda got to know each other and became the people they became. Learn about what happened in Oz before Dorothy blew and in and even find out how Dorothy had arrived in Oz. Perfect for Halloween, you can dress up as your favorite characters and enjoy this breathtaking show with amazing music and incredible special effects and dancing.
6. The Phantom of the Opera.
The Phantom of the Opera is the perfect spooky and romantic Broadway Show for Halloween. Cuddle up with the person you love or throw your own Mascaraed Ball and enjoy the classic love story from Andrew Lloyd Weber. Watch as Christine becomes the lead in her Ballet and Theatre and as the Phantom helps to train her to become the star she was meant to be. This love story is not only one of the most well known and well loved stories and shows in Broadway history but it is perfect for Halloween and for just you and your partner or even to share with a group of friends.
Moulin Rouge is one of the most popular Movie Musicals from the 90's. Nicole Kidman takes you back to France for a magical and Absinthe filled adventure through the life of a performer in a burlesque theatre and with a man who falls in love. Watch the cast perform amazing musical numbers in over the top costumes to some of the most fun and familiar rock and pop songs from the 90's.
No other show takes you into another land like Into the Woods. Watch as your favorite storybook characters come to life and interact with each other. Watch birds peck out people's eyes, dresses fall from trees and even wicked witches turn out to be good. Watch bakers fight Giants and princes and princesses fall in love. This classic show is perfect for people who are looking for a fun show to watch on Halloween.
This story of Dr. Frankenstein's family coming back to Transylvania is not only one of the most brilliant shows produced by Mel Brooks but one that one of my favorite Actors, Megan Mullally (Karen from Will and Grace), helped make famous on broadway. Roll in the hay with the locals in their old town and watch as the new Dr. Frankenstein brings a new monster to life. This classic show is a perfect one to watch as you get to enjoy seeing one of the most famous and classic Halloween Monsters come to life on stage.
The Addams Family is one of the most well known families from TV and is also now one of the most popular shows on Broadway. Watch as they interact and share their love of death and life with their audiences. The Addams Family Musical is the perfect show to go to on Halloween if you can get a ticket. Watch as Gomez, Morticia and everyone else from the Addams Family come to life and enjoy Halloween with an entirely new show to watch and tons of fun songs to enjoy as The Addams Family comes to life to help you enjoy Halloween.
There are a ton of other shows for you to enjoy on Halloween and if you have another personal favorite, please feel free to leave a comment below.
Posted by BroadwayReviewed & filed under Shows, The Addams Family, TV.
I can't believe that Halloween is next week. Tonight is the Rocky Horror Picture Show Episode of Glee and the annual Washington DC High Heel Drag Race. This weekend is the Miss Adams Morgan Drag Queen Beauty Pageant and with halloween comes halloween movies, tv shows and more importantly theatre. I already reviewed the Rocky Horror Picture Show but another show that has taken roots on Broadway which not only screams Halloween but is Halloween year round is The Addams Family Musical starring three of my very favorite Broadway Stars, Nathan Lane, Jackie Hoffman and Bebe Neuwirth.
The Addams Family Musical is a show that not only perfectly represents one of the most loved and memorable families in TV history, but when you combine some amazing set designs, costuming and some incredible talents, you get a show that no one on earth can possibly resist. You have Gomez who loves to romance his wife Morticia. You have Fester Addams who is a ladies man, the only problem is that regular non Addams Family Musical ladies haven't realized it yet. You have the kids Wednesday and Pugsly and one of my favorites Granny Addams. They even remembered to include Lurch and other secondary Addams Family characters in the cast.
The Addams Family Musical takes you on a journey into a new world with a love of death and everything dark where graveyards become playgrounds and poison becomes a seasoning for food like salt or pepper. Electric chairs can double as recliners and torture devices become play toys. The Addams Family Musical brings you into the world of the Addams Family and helps you to rediscover why so many generations have fallen in love with this strange and fabulous family of fun loving friends. The Addams Family Musical is perfect for Halloween but even if you are in NYC for the spring, the Addams Family Musical is still a perfect show for you to see.
You'll discover an entirely new selection of music, singing and dancing and you'll also get to watch your favorite Addams Family members come to life on stage. The Addams Family Musical is one of the top selling shows on Broadway right now and one that I highly recommend. I am hoping to get to see it on my next trip to New York since I have only seen small bits and pieces when they did a mini performance at Sidetracks in Chicago for Showtunes Night when the cast was there. If you have seen the Addams Family Musical in New York City, feel free to leave a comment below with how you loved it or hated it.
Posted by BroadwayReviewed & filed under Activism and Charity, Jonathan Larson, Showtunes, Songs, Tick Tick Boom.
So I was watching a video clip on gay bullying last night and it actually got me to start to cry. Not only has there been a huge uprising in gay bullying, but there are people out there making it seem like gay bullying and gay teen suicide which is also on a huge rise is a good thing. They are saying toughen up and that mainstream media shouldn't be making a big deal out of it which is something that I 100% disagree with. I was reading this article on gay teen bullying and gay teen suicide and thought that it is something that is not only worth mentioning but also including in a post here on BroadwayReviewed.com. It also got me to remember one of my favorite showtunes, Louder Than Words from Tick Tick Boom.
Broadway is the home to many gay people and a ton of talented young men and women who aspire to share their talents with the world. They can sing, write, dance, act, play a character and take audiences to another time or even another world and to lose even one straight or gay person to bullying is absolutely unacceptable. Think of all the talented people we lose every year to diseases like Breast Cancer, AIDS, Drugs and everything else. For someone who could grow up to entertain millions of people, who could be the person who cures any of the diseases above or even someone who can make one person smile when they are down to take their own life because they were being bullied for being gay.
There was this absolutely horrible person I got into a fight on facebook with because in my opinion he was supporting gay teen bullying and justifying gay teen suicide and saying that people shouldn't be making a big deal about it because they want to push their beliefs on others, I almost lost it. Instead I said that I'll be watching him from heaven, and yes Cher will be there when he goes to hell because in reality Christianity is about loving one another and not spreading hate and hurting others. Anyways, that whole back and forth fight made me remember a song from the Broadway Musical Tick Tick Boom by Jonathan Larson.
Louder Than Words from Tick Tick Boom is not only an amazing song but I think it is the perfect song for people who have grown up and gone through wanting to commit suicide because of being gay bullied or because they know what these kids are going through. In the article I linked from above, these kids have it much worse than we did when we were in high school trying to make it through the day and the reason I think that Louder Than Words from Tick Tick Boom is the perfect song for these kids is because it makes anyone who grew up and was bullied for being gay would be inspired to help prevent gay teen bullying and help to actually make a difference. The post goes through some things that you can do and what I like most about it is that it actually brings out a good point, this is one cause where one person can actually make a difference. Louder Than Words from Tick Tick Boom is not only an amazing song but it is also inspirational. I highly encourage you to listen to it and play it for other people and then find a way to help your community develop a safe haven or a safety program not only for gay teen youth, but to help stop bullying for all teens and kids in school.
Bullying isn't just calling people names anymore. It is posting pictures and videos of them on the internet that can last forever and it is about harassing them 24 hours a day instead of just during school hours. Gay teen bullying and gay teen suicide from gay teen bullying needs to stop and people need to start taking action to help prevent it. Below is Louder Than Words from Tick Tick Boom as well as the video that actually got me to cry and also a great clip of Martin Sheen ripping the Christianity argument to shreds. The article I linked to has the Ellen Video, the Fox news video and I highly recommend the Tim Gunn one as well. Being gay is not a sin and there is so much more to life than the assholes you may have to deal with when you are younger. Life gets better and you have to tough it through it. Being bullied is not fun and eventually you will be able to get away from it. If you are a gay teen, trust me, life gets better. You don't have to stay in your town for your entire life and you will meet thousands of people just like you. You will have rights and the bullying will end. Tough it out and remember that life does get better.


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Gangtok Travel Guide
Photo by Pralay on Pixabay // CC0
Situated in the Eastern Himalayas, Gangtok is a destination that mountain lovers won't want to miss. While the tourist trade is bustling these days, things weren't always so peaceful. A generation or two ago, India clashed with its neighbour China, creating security issues.
These days, however, relations are slowly thawing, making Gangtok no different than any other destination in its border regions.
Come check out our Gangtok travel guide as we cover the best things to do in Gangtok, India.
On your first day exploring in Gangtok region, stop by Baba Harbhajan Singh Memorial Temple first. Built to honour the memory of a soldier that drowned tragically in the 1960s, it is a serene place. Its existence is a touching tale of how much his fellow privates loved him.
A few days after his passing, he appeared in the dreams of his friends. He urged them to build a shrine to honour his memory. To this day, evidence of his existence in the paranormal realm persists. Chinese soldiers are rumoured to have seen him out on patrol, while his former bunk occasionally has wrinkled sheets.
Many believe Baba Harbhajan Singh provides blessings from beyond the grave. Pilgrims regularly leave bottles of water here for several days. On their return, they believe the water will impart good fortune on those who drink it.
Note that there are two temples with the same name in the area. Baba's old charges built the original temple up on the Old Silk Road. However, the terrain up there is rough, leading authorities to build one at a lower altitude in 1982. If you are out of shape, stick to the lower temple. The elevation in the Gangtok area is already high; if you push yourself too hard, medical problems can result.
Hanuman Tok is another noteworthy temple in Gangtok region. As the name suggests, this hall of worship was built to honour Lord Hanuman, the Hindu god of strength. According to local legend, Hanuman rested here on his journey to deliver medicinal herbs to Lakshman. Inside, you'll find an idol of Lord Hanuman, which replaced a sacred stone that had been there for centuries.
An easy-to-climb staircase grants access to the temple. This institution is situated at 7,200 feet, meaning most should be able to handle the climb without difficulty.
If you want to take in some Buddhist sights in Gangtok, make certain Ranka Monastery is on your list. Built in the Zurmang Kagyud tradition of Buddhism, its appearance differs from Buddhist temples elsewhere in the world.
However, its stunning exterior has made it a sought-after filming site for many Bollywood films. Once you see its prayer wheels, young monks, and sweeping mountain views, you'll understand why. Finally, have lunch in the onsite restaurant – locals say that Bollywood star Amir Khan ate here during a shoot.
Learn more about the Tibetan heritage of the Gangtok area by paying a visit to the Namgyal Institute of Tibetology. Here, you'll find a library, a museum, and a reference centre. They aim to engender an understanding of the Tibetan people who now call India home.
The museum contains many artifacts Tibetans brought with them from their former homeland. These include statues, artwork, manuscripts, and coins. Its library's shelves contain 60,000 volumes of Tibetan origin – one of the largest collections in the world.
If you plan on visiting, do note its unique schedule. It is open 10 am-4 pm Monday to Saturday. However, it closes on the second Saturday of each month, and Sundays.
The Gangtok region is one of great natural beauty. If you love mountainous landscapes, make the journey out to Tsomgo Lake. While you can find many lakes throughout India, this is one of the few that are glacial in origin.
In summer, alpine flowers line the shores of this small but serene lake. Keep your eyes open for Brahmini ducks trolling on its surface. On the way there and back, you may also see wildlife that includes Himalayan black bears and red pandas.
Geography nerds won't want to miss out on a trip to Nathula Pass. Until 1962, the Silk Road ran over this 14,000-foot high mountain pass. Geopolitical tensions put a stop to that, as a war between India and China sealed the border.
It took until 2006 for the post to reopen. Even now, restrictions on the passage of goods and people exist. From June to September, merchants from both sides host a market trading in Chinese and Indian products. Among the stalls, you'll find a generator-powered ATM, which is said to be the highest in the world.
Photographers looking to capture the best views in Gangtok Region will want to check out the Tashi View Point. On a clear day, the views of the snow-capped Kanchenjunga range will take your breath away. Photographers are advised to arrive before 5 am, as conditions get crowded as the morning progresses.


Note - despite the little story below, this is believed to be a French computer-generated video. It is, nonetheless, very clever and fun.
It takes about 3 minutes and 25 seconds to watch.
Turn your sound on for this.
Read this first (with skepticism), then watch.
See how all of the balls wind up in catcher cones.
This incredible machine was built as a collaborative effort between the Robert M. Trammell Music Conservatory and the Sharon Wick School of Engineering at the University of Iowa. Amazingly, 97% of the machines components came from John Deere Industries and Irrigation Equipment of Bancroft, Iowa. Yes, farm equipment!
It took the team a combined 13,029 hours of set-up, alignment, calibration, and tuning before filming this video but as you can see, it was WELL worth the effort. It is now on display in the Matthew Gerhard Alumni Hall at the University and is already slated to be donated to the Smithsonian.