The Congruence Subgroup Property does not imply Invariable Generation
T. Gelander and C. Meiri
It was suggested in \cite{KLS that for arithmetic groups Invariable Generation is equivalent to the Congruence Subgroup Property. In this paper we dismiss this conjecture by proving that certain %arithmetic groups possessing the CSP are not IG. arithmetic groups which possess the later property do not possess the first one. %We also prove that some countable simple groups, and in particular $\PSL(2,\BQ)$, are not invariably generated.
1505.06881.tex
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\begin{document}
\author{T. Gelander and C. Meiri} 

\address{Mathematics and Computer Science\\
Weizmann Institute\\
Rechovot 76100, Israel\\}
\email{tsachik.gelander@gmail.com}

\address{Department of Mathematics\\ Technion - Israel Institute of Technology\\
Haifa, 32000, Israel\\}
\email{chenm@technion.ac.il}

\date{\today}

%\title{A metric version of the Jordan--Turing theorem}

\title{The Congruence Subgroup Property does not imply Invariable Generation}

\maketitle

%\let\languagename\relax % because of bug in babel

\begin{abstract}
It was suggested in \cite{KLS} that for arithmetic groups Invariable Generation is equivalent to the Congruence Subgroup Property. In this paper we dismiss this conjecture by proving that certain %arithmetic groups possessing the CSP are not IG. 
arithmetic groups which possess the later property do not possess the first one. 
%We also prove that some countable simple groups, and in particular $\PSL(2,\BQ)$, are not invariably generated.
\end{abstract}

\section{introduction}

In an attempt to give a purely algebraic characterisation for the Congruence Subgroup Property  (CSP), it was asked in \cite{KLS} whether for arithmetic groups the CSP is equivalent to
Invariable Generation (IG).  This was partly motivated by the result of \cite{KLS} that the pro-finite completion of an arithmetic group with the CSP is topologically finitely invariably generated. Another result in the positive direction was obtained in \cite{Gel}, namely that discrete subgroups, and in particular arithmetic subgroups, of rank one simple Lie groups are not IG. Recall the famous Serre conjecture that an irreducible arithmetic lattice $\gC\le G$ in a semisimple Lie group has the CSP iff $\text{rank}_\BR(G)\ge 2$. 
In the current paper we give a negative answer to the  KLS question by producing various counterexamples of arithmetic groups in higher rank which are not IG. Some of our examples are known to possess the CSP. 



\subsection{Invariable generation} 
Recall that a subset $S$ of a group $\gC$ {\it invariably generates} $\gC$ if $\gC= \langle s^{g(s)} | s \in S\rangle$ for every choice of $g(s) \in \gC,s \in S$. One says that a group $\gC$ is {\it invariably generated}, or shortly IG if one of the following equivalent conditions holds:
\begin{enumerate}
\item There exists $S \subseteq \gC$ which invariably generates $\gC$.
\item  The set $\gC$ invariably generates $\gC$.
%It is obvious that $G$ is not IG iff it admits a proper subgroup $H<G$ which meets every conjugacy class. 
\item Every transitive permutation representation on a non-singleton set admits a fixed-point-free element.
\item $\gC$ does not have a proper subgroup which meets every conjugacy class.
\end{enumerate}
The main results of this paper are that certain lattices are not IG. 

\medskip

A matrix $g\in \GL_n(\BC)$ is {\it net} if the multiplicative group $A(g)$ generated by the eigenvalues of $g$ does not contain any nontrivial root of unity. A linear group $\gC\le \GL_n(\BC)$ is {\it net} if all its nontrivial elements are net.
It is well known that every finitely generated linear group admits a finite index net subgroup (see \cite[Theorem 6.11]{Rag}). The assumption that $\gC$ is net is not crucial but very convenient for our purpose.

\begin{thm}\label{thm:O(2,2)}
Let $Q$ be a rational quadratic form of signature $(2,2)$, let $\BG=\text{SO} (Q)$ and let $\gC$ be a finite index net subgroup in $\BG(\BZ)$. Then $\gC$ is not invariably generated.
\end{thm}

The group $\gC$ in Theorem \ref{thm:O(2,2)} is an irreducible lattice in the higher rank semisimple group $G=\SO(2,2)\cong\PSL(2,\BR)\times\PSL(2,\BR)$, and as such is super-rigid, possesses the normal subgroup property and the strong approximation theorem. By Kneser's theorem \cite{Kneser} $\gC$ has the CSP (when the form is isotropic this also follows from Raghunathan's theorem \cite{Ra1,Ra2}). In particular the groups obtained in this way are counterexamples to the question of \cite{KLS}.

We also prove a general theorem (see Theorem \ref{thm:regular}) that under a certain condition on $G$ every {\it regular} lattice in $G$ (see Section \ref{sec:regular} for the definition) is not IG. This produces a large family of examples, for instance:

\begin{exam}\label{exm:Witte}(See \cite[Proposition 6.61]{Witte})
Let $L$ be a cubic Galois extension of $\BQ$. Let $D$ be a central division algebra of degree $3$ over $\BQ$ which contains $L$ as a subfield. Let $\OO_D$ be an order in $D$. Then $\SL(1,\OO_D)$ is isomorphic to an anisotropic arithmetic lattice in $\SL(3,\BR)$.
\end{exam}

Being a lattice in $\SL(3,\BR)$, the group $\SL(1,\OO_D)$ possesses many rigidity properties such as those mentioned after Theorem \ref{thm:O(2,2)} and even more; for instance it has Kazhdan's property (T) and IRS-rigidity (the Stuck--Zimmer theorem). The corresponding lattice is co-compact. By Serre's conjecture, $\SL(1,\OO_D)$ is expected to possess the CSP, but this is currently unknown.

\begin{thm}\label{thm2}
The arithmetic group $\SL_1(\OO_D)$ is not IG.
\end{thm}

Theorem \ref{thm:regular} is more general and applies to many other example. 

\subsection{The motivation}

Apart from trying to better  understand the notion of invariable generation in infinite groups\footnote{The origin of this notion is from Galois theory, and it is well known that finite groups are always IG.}, our motivation arose from an attempt to understand better `thin subgroups' of arithmetic groups. For instance an important question in this theory is whether every element belongs to some thin subgroup. 

\begin{defn}
Let $\gC$ be an arithmetic group. Let us say that an element $\gc\in\gC$ is {\it fat} if every Zariski dense subgroup containing it is automatically of finite index, i.e. $\gc$ does not belong to any thin subgroup of $\gC$. 
\end{defn}

The existence and classification problems of fat elements are extremely interesting and may shed light on many problems concerning thin subgroups. Being fat is clearly invariant under conjugacy, hence the existence of such elements is closely related to invariable generation. Indeed, if $\gC$ admits an infinite index Zariski dense subgroup which meets every conjugacy class then there are no fat elements in $\gC$. Therefore our results imply that in certain arithmetic groups all elements are `thin':

\begin{cor}
The arithmetic groups from Theorem \ref{thm:O(2,2)} and Theorem \ref{thm:regular} do not admit fat elements.
\end{cor} 

This applies for instance to certain lattices in $\SL(3,\BR)$ (cf. Example \ref{exm:Witte}). However, our method does not apply to $\SL(3,\BZ)$. New tools are required in order to study  property IG or existence of fat elements for this group. 

\begin{ques}
Is $\SL(3,\BZ)$ invariably generated? What about $\SL(n,\BZ)$ for $n\ge 4$?
\end{ques}



\subsection{About the proofs}
As in \cite{W76} and \cite{Gel} the idea is to construct an independent set consisting of one representative of each non-trivial conjugacy class. We use of course the dynamics on the associated projective space $\BP$, where we construct such an infinite family of ping-pong partners. In the proof of Theorem \ref{thm:regular} we use the standard action on points in $\BP$. The dynamical picture in the proof of Theorem \ref{thm:O(2,2)} is more involved and we are led to consider the action on lines in $\BP$ rather than points. Yet the spirit of the proof is similar.

The main technical novelty of this paper is a method that allows us to conjugate certain elements, in particular $\BR$-regular elements, such that their dynamics will simulate unipotent dynamics --- in the sense that all non-trivial powers contract the complement of some small open set into that set. We first allow ourselves to use elements from the enveloping group $G$ and then apply Poincar\'{e} recurrence theorem in order to approximate the solution in $\gC$.

\subsection{Acknowledgment } The first author was partially supported by ISF-Moked grant 2095/15. The second author was partially supported by ISF grant 662/15.
%The reason for the assumption that $\gC$ is either a lattice or a dense subgroup is that in both cases we are able to approximate the dynamics of elements from $G$ by elements from $\gC$. Indeed, in the first case we apply Poincar\'{e}  recurrence theorem. 



%%%%%%%%%%%%%%%%%%


\section{$\BR$-regular lattices}\label{sec:regular}
Suppose that $G=\BG(\BR)$ is a Zariski connected linear real algebraic group with a faithful irreducible representation in a $d$-dimensional real vector space, with associated projective 
space $\BP\cong\BP^{d-1}(\BR)$.
%$G=\BG(\BR)\le\GL_d(\BR)$ be a Zariski connected linear algebraic group. 
We identify $G$ with its image in $\GL_d(\BR)$. 
An element $g\in G$ is {\it regular} if it is either a regular unipotent or all its eigenvalues are of distinct absolute value (in particular each has multiplicity one). A subgroup $\gC\le G$ is regular if every nontrivial element of $\gC$ is regular.
For an element $g\in G$ we denote by $v_g$ and $H_g$ the attracting point and repelling hyperplane of $g$ in $\BP$, assuming they are unique. 
Let us say that a point or a hyperplane is $G$-{\it defined} if it is of the form $v_g$ (resp. $H_g$) for some $g\in G$. In addition let us say that an incident pair $(v,H)$ is $G$-defined if there is a regular element $g\in G$ for which $v=v_g$ and $H=H_{g^{-1}}$. Let us also say that an ordered pair $(U,V)$ of open neighborhoods in $\BP$ is $G$-{\it applicable} if there is a $G$-defined incident pair $(v,H)$ with $v\in U$ and $H$ entirely contained in the interior of the complement of $V$. 
We shall further suppose the following:

\begin{asmp}\label{asmp:mani-pairs}
The set of $G$-defined incident pairs is not contained in a proper sub-variety of the set of incident pairs.
\end{asmp}

% In this section we prove:

\begin{thm}\label{thm:regular}
Let $G$ be a Zariski connected linear real algebraic group with a faithful irreducible representation on a $d$-dimensional vector space
which satisfies Assumption \ref{asmp:mani-pairs} and let $\gC\le G$ be a regular lattice. 
Then $\gC$ admits an infinite rank free subgroup which intersects every conjugacy class. In particular, $\gC$ is not invariably generated.
\end{thm}

\begin{rem}
For $G=\SL_d(\BR)$ with its standard $d$-dimensional representation Assumption \ref{asmp:mani-pairs} is obviously satisfied. Moreover it is easy to check that the lattices in Example \ref{exm:Witte} are regular, by construction. Thus Theorem \ref{thm2} is a special case of Theorem \ref{thm:regular}. 
\end{rem}


%To set up notations, let us suppose that  


%Let $\gC\le G$ be a regular lattice.

\subsection{The proof of Theorem \ref{thm:regular}}\label{sec:regular}

\begin{prop}[Main Proposition]\label{prop:main}
Let $C$ be a nontrivial conjugacy class in $\gC$, and let $(U,V)$ be a $G$-applicable pair of open sets in $\BP$. Then there is $\gc\in C$ such that $\gc^n\cdot V,~n\in\BZ\setminus\{ 0\}$ are all disjoint and contained in $U$.
\end{prop}

%To prove the proposition we shall need the following technical lemma:

\begin{lem}\label{lem:the-conjugating}
Let $\gc\in \gC\setminus\{1\}$, and let $(U,V)$ be a $G$-applicable pair of open sets in $\BP$.  Then there is $\gd\in\gC$ such that 
\begin{enumerate}
\item $H_{\gd^{-1}}\cap \overline{V}=\emptyset$,
\item $v_\gc,v_{\gc^{-1}}\notin H_\gd$, 
\item $v_{\gd^{-1}}\notin H_\gc\cup H_{\gc^{-1}}$,
\item $v_\gd\in U$, and
\item $\gc^n\cdot v_{\gd^{-1}}\notin H_\gd$ for all $n\ne 0$.
\end{enumerate}
\end{lem}

\begin{proof}
%Poincar\'{e}  recurrence.
As a first step, observe that there exists a regular $g\in G$ such that all the five conditions hold with $g$ instead of $\gd$. Indeed, as $(U,V)$ is $G$-applicable there is $g$ for which $(1)$ and $(4)$ are satisfied. 
Moreover, in view of Assumption \ref{asmp:mani-pairs}, for every $n\in\BN\setminus\{0\}$ the set of regular $g\in G$ such that 
$\gc^n\cdot v_{g^{-1}}\in H_g$ is nowhere-dense. Hence by the Baire category theorem and the fact that $G$ admits an analytic structure, we may slightly deform $g$ in order to guarantee that condition $(5)$ holds as well.
Since $G$ is Zariski connected and irreducible, the set of $h\in G$ such that either $(2)$ or $(3)$ do not hold for $g^h$ is nowhere-dense. Thus, up to replacing $g$ by a conjugate $g^h$ with $h$ sufficiently close to $1$, all the five conditions are satisfied.

Next note that $\gc^n\cdot v_{g^{-1}}$ tends to $v_\gc$ when $n\to\infty$ and to $v_{\gc^{-1}}$ when $n\to-\infty$. Thus if we pick a sufficiently small symmetric identity neighborhood $\gO\subset G$ and $\gep>0$ sufficiently small, then we have:
\begin{enumerate}
\item $\gO\cdot (H_{g^{-1}})_\gep\cap \overline{V}=\emptyset$,
\item $v_\gc,v_{\gc^{-1}}\notin \gO\cdot (H_g)_\gep$, 
\item $\gO\cdot (v_{g^{-1}})_\gep\cap (H_\gc\cup H_{\gc^{-1}})=\emptyset$,
\item $\gO\cdot (v_g)_\gep\subset U$, 
\item $\gc^n\gO\cdot (v_{g^{-1}})_\gep\cap \gO\cdot (H_g)_\gep=\emptyset$ for all $n\ne 0$, and
\item $\gO\cdot (v_g)_\gep\cap\gO\cdot (H_g)_\gep=\emptyset=\gO\cdot (v_{g^{-1}})_\gep\cap\gO\cdot (H_{g^{-1}})_\gep$.
\end{enumerate}

Finally we wish to replace $g$ by an element $\gd\in \gC$ with similar dynamical properties. 
%In case $\gC$ is dense in $G$ we may clearly pick any $\gd\in\gC$ as long as it is sufficiently close to $g$. Thus we may restrict to he second case where 
Since $\gC$ is a lattice, $G/\gC$ carries a $G$-invariant probability measure. Let $\pi:G\to G/\gC$ denote the quotient map. By the Poincar\'{e}  recurrence theorem we have that for arbitrarily large $m$, $g^m\cdot\pi(\gO)\cap\pi(\gO)\ne\emptyset$, which translates to $\gO g^m\gO\cap\gC\ne\emptyset$, since $\gO$ was chosen to be symmetric. 
 
Note that replacing $g$ by a power $g^m,~m>0$ does not change the attracting points and repelling hyperplanes, while for larger and larger $m$ the element $g^m$ is becoming more and more proximal. In particular, if $m$ is sufficiently large then 
%\begin{equation}
$$
 g^m\cdot (\BP\setminus (H_g)_\gep)\subset (v_g)_\gep,~\text{and}~g^{-m}\cdot (\BP\setminus (H_{g^{-1}})_\gep)\subset (v_{g^{-1}})_\gep.
$$
%\end{equation}
In view of Item (6), this implies that any element of the form $h=w_1g^mw_2$ with $w_1,w_2\in\gO$ is very proximal and satisfies 
$$
 v_h\in\gO\cdot(v_g)_\gep,~v_{h^{-1}}\in\gO\cdot(v_{g^{-1}})_\gep,~H_h\subset\gO\cdot (H_g)_\gep,~H_{h^{-1}}\subset\gO\cdot (H_{g^{-1}})_\gep.
$$
In particular, we obtain the desired $\gd$ by choosing any element from the nonempty set $\gC\cap\gO g^m\gO$.
\end{proof}
%
%By replacing $\gd$ with $\gd^m$ for a large $m$, we obtain that:
%
%\begin{cor}\label{cor:regular}
%Under the assumptions of Lemma \ref{lem:the-conjugating}, there is $\ga\in\gC$ such that 
%the sets $(\gc^{\ga})^n\cdot V,~n\in\BZ\setminus\{ 0\}$ are all disjoint and contained in $U$. 
%\end{cor}

\begin{proof}[Proof of Proposition \ref{prop:main}]
Pick $\gc'$ in $C$ and let $\gd$ be the element provided by Lemma \ref{lem:the-conjugating}. Items (2,3,5) of the lemma together with the contraction property of high powers of $\gc'$ guarantee that if $O$ is a sufficiently small neighborhood $v_{\gd^{-1}}$ then its $\langle\gc'\rangle$ orbit stays away from $H_\gd$, i.e. there is a neighborhood $W$ of $H_\gd$ disjoint from all $\gc'^n\cdot O,~n\in\BZ\setminus\{0\}$. In particular the $\langle\gc'\rangle$ translates of $O$ are all disjoint, i.e. $O$ is $\langle\gc'\rangle$-wondering (in the terminology of \cite{Gel}). If $k$ is sufficiently large then by Item (1) of the lemma $\gd^{-k}\cdot V\subset O$ while by Item (4), $\gd^k\cdot (\BP\setminus W)\subset U$. It follows that $\gc=\gd^k\gc'\gd^{-k}$ is our desired element.


\end{proof}


We shall also require the following:

\begin{lem}\label{lem:(U,V)-pairs}
There is a countable family of disjoint open sets $U_k,~k\in\BN$ such that if we let $V_k=\bigcup_{i\ne k}U_i$ for any $k\in\BN$ then the pairs $(U_k,V_k),~k\in\BN$
are all $G$-applicable.
\end{lem}

\begin{proof}
Let $g\in G$ be a regular element and set $(v,H)=(v_g,H_g)$. Pick $h\in G$ near the identity such that $h\cdot v\notin H$ and $v\notin h\cdot H$. Let $c:[0,1]\to G$ be an analytic curve with $c(0)=g$ and $c(1)=h$. Since the curve $c$ is analytic, the boundary conditions allow us to chose a sequence $t_k\to 1$ such that if we let $(v_k,H_k)=(c(t_k)\cdot v,c(t_k)\cdot H)$ and $(v_\infty,H_\infty)=(h\cdot v,h\cdot H)$, then
for all $i\ne j\le\infty$ we have $v_i\notin H_j$. In addition, the set of pairs 
$\{(v_k,H_k):k<\infty\}$ is discrete and accumulates to $(v_\infty,H_\infty)$. Thus, we may choose neighborhoods $U_k\supset v_k$ sufficiently small so that $\overline{U_i}\cap H_j=\emptyset,~\forall i\ne j$ and so that the only point in $\overline{\cup U_k}\setminus\cup\overline{U_k}$ is $h\cdot v$. It is evident that the sets $U_k,~k\in\BN$ satisfy the required property.
\end{proof}

The strategy of the proof of Theorem \ref{thm:regular} is as follows.  Let $(U_k,V_k),~k\in\BN$ be as in Lemma \ref{lem:(U,V)-pairs}.
Let $C_k$ be an enumeration of the non-trivial conjugacy classes in $\gC$. In view of Proposition \ref{prop:main}, we may pick $\gc_k\in C_k$ so that 
$$
 \gc_k^n\cdot V_k\subset U_k,~\forall n\in\BZ\setminus \{0\}.
$$

In order to complete the proof of Theorem \ref{thm:regular} we may address 
the following variant of the ping-pong lemma (cf. \cite[Proposition 3.4]{Gel}):

\begin{prop}\label{prop:pingpong}
Let $\gC$ be a group and $X$ a compact $\gC$-space. Let $\gc_k\in \gC,~k\in\BN$ be elements of $\gC$, and suppose there are disjoint subsets $U_k\subset X$ such that for all $k\ne j$ and all $n\in\BZ\setminus\{0\}$ we have $\gc_k^n\cdot U_j\subset U_k$. Then 
$$
 \gD=\langle \gc_k,~k=1,2,3,\ldots\rangle=*_k\langle\gc_k\rangle,
$$
is the free product of the infinite cyclic groups $\langle\gc_k\rangle,k=1,2,3,\ldots$.

Furthermore, the limit set $L(\gD)$ is contained in $\overline{\cup_k U_k}$.
\end{prop} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%\section{Lattices in product of convergence spaces}
%Let $G$ be a locally compact group and $X$ a minimal compact metrisable $G$-space of cardinality $\ge 3$. We shall say that the pair $(G,X)$ is rank one if any non-elliptic element $g\in G$ acts convergencely on $X$. Recall that $g$ is elliptic if $\langle g\rangle$ is relatively compact, and that $\langle g\rangle\act X$ is convergence if there are $g^-,g^+\in X$ such that for any pair of compact sets $g^-\notin A,g^+\notin B$ the set $\{n:g^n\cdot A\cap B\ne\emptyset\}$ is finite. We shall say that a locally compact group $G$ is rank one if there is a compact metrisable $G$ space $X$ such that $(G,X)$ is a rank one pair. 
%For example of rank one pairs we have:
%
%\begin{itemize}
%\item Let $G$ be a rank simple algebraic group over a local field, and let $X$ be the visual boundary of the associated symmetric space or Bruhat--Tits tree.
%\item Let $Y$ be a proper Gromov hyperbolic space, $X=\partial Y$ and $G=\text{Isom}(Y)$.
%\end{itemize}
%
%\begin{thm}
%Let $n\ge 2$ and let $G_i,~i=1,\ldots n$ be rank one locally compact groups. Let $\gC\le\prod_{i=1}^nG_i$ be an irreducible lattice. Suppose that for each $\gc\in\gC\setminus\{1\}$ the projection to one of the factors is non-elliptic. Then $\gC$ is not IG.
%\end{thm}
%







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Arithmetic groups of type $\SO(2,2)$}

%
%{\bf Assumptions:} Let  $q(\xb)$ be an integral quadratic form of signature $(2,2)$ and let $(\cdot,\cdot)$ be the corresponding quadratic form. 
%Let $\Gamma \le SO(q,\Z)$  be a finite index subgroup such that no root of unity other then 1 is an eigenvalue of an element of $\Gamma$. 
%
%\begin{lemma}\label{first lemma} Let $f$ be the characteristic polynomial of some $g \in \Gamma$.  Then $f$ is 
%reciprocal and one of the following two conditions hold:
%\begin{itemize}
%\item All the roots of $f$ are real. 
%\item All the ports of $f$ are non-real complex numbers of absolute values different than 1. 
%\end{itemize}
%\end{lemma}
%\begin{proof} We first show that $f$ is reciprocal . The free coefficient of $f$ is $1$ and $f$ is a monic
%polynomial of degree 4. Thus, in order to show that $f$ is reciprocal it is enough to prove that if $\alpha$ is a root of $f$ then so is $1/\alpha$.  Let 
%$v_1 \in \C^4$ be an eigenvector of $g$ with eigenvalue $\alpha$.  Denote  $V:=
%\{ x\in \C^4\mid (v_1,x)=0\}$ and let $v_2 \in \C^4$ be such that $(v_1,v_2)=1$. Then, $gv_2=(1/\alpha) v_2 +v_3$ for some $v_3\in V$. Moreover,
%$g$ preserves $V$ and acts on $\C^4/V$ as multiplication by $1/\alpha$. This implies that $1/\alpha$ is an eigenvalue of $g$.  
%
%We now show that it is not possible that $f$ has a real root  $\alpha$ with absolute values different than 1 and a non-real root $re^{i\theta}$.
%Assume otherwise, since $f$ is real and reciprocal $r=1$ and $f(x)=(x-\alpha)(x-\frac{1}{\alpha})(x-e^{i\theta})(x-e^{-i\theta})$ 
%where $\alpha$ and $\theta$ are real numbers, $|\alpha| \ne 1$ and  $\theta$ is not an integral multiply of $\pi$. Choose $v_1 \in \R^4$  such that
%$gv_1=\alpha v_1$.  Define 
%$U:=\{{ x}\in \R^4\mid (g-e^{i\theta})(g-{e^{-i\theta}}){ x}=0\}$,  $V:=\{{ x}\in \R^4\mid ({v_1},{ x})=0\}$
%and  $W:=\{{ x}\in \R^4 \mid (g-\alpha)(g-{1\over \alpha}){ x}=0\}$. The non-trivial subspace $U \cap V$ of $U$ is invariant under $g$
% so the assumption on $\theta$ implies that it must be equal to $U$. In particular, every element of $U$ is orthogonal to $v_1$.  Fix ${v_2}\in W \setminus \R{v_1}$ and let $\gamma \in \R$ be such that $g {v_2}={1\over\alpha} {v_2}+\gamma{v_1}$. Let ${ x}\in U$ be orthogonal to $v_3$, then
% \begin{equation}
%0=({v_2},{x})=(g{v_2},g{ x})=(\frac{1}{\alpha}{v_2}+\gamma{v_1},g{x})=(\frac{1}{\alpha}{v_2},g{x})=\frac{1}{\alpha} ({v_2}, g{x}).
% \end{equation}
% Thus, the assumption of $\theta$ implies that $({v_3},{ x})=0$ for every ${x}\in U$. It follows that $\R^4$ is an orthogonal sum of $W$ and $U$ and either 
%$W$ and $U$ both have signature $(1,1)$ or one of them is positive definite and the other is negative definite.  The second case is not possible since an orthogonal transformation of a definite space cannot have an eigenvalue with absolute value strictly greater than 1. The 
%%imaginary numbers  $i$ and $-i$ are the only possible non-real roots of an orthogonal transformation of a space of signature $(1,1)$, hence, if the first case happens then $\theta=\pi/2$ and $r=1$, a contradiction. 
%first case cannot happen since any orthogonal transformation of a quadratic space of signature $(1,1)$ is diagonalisable over $\R$.
%%This implies  that $\beta=-1/\alpha \ne \alpha$, $\gamma=0$ and 
%%\begin{itemize}
%%\item $({ v},{ v})=
%%(g{ v},g{ v})=\alpha^2({ v},{ v})$ and so $({ v},{ v})=0$;
%%\item $({ w},{ w})=
%%(g{ w},g{ w})=\alpha^{-2}({ w},{ w})$ thus $({ w},{ w})=0$;
%%\item$({ v},{ w})=
%%(g{ v},g{ w})=-({ v},{ w})$ hence $({ v},{ w})=0$.
%%\end{itemize}
%%This leads to a contradiction since $f$ is not degenerate. 
% 
% If $z \in \C\setminus \R $ is a root of  $f \in \Z[x]$ then so is $\bar{z}$.  Thus if $f(x)$ has a non-real root with absolute value greater than 1 then $re^{i\theta}$, $\frac{1}{r}e^{i\theta}$ and  $\frac{1}{r}e^{-i\theta}$ are the other roots of $f$. Finally, $f(x)$ cannot have a non-real root with 
%absolute value 1 since then all the roots of $f(x)$ must have absolute value 1 and thus they are roots of unity which contradicts the assumption on eigenvalues of elements in $\Gamma$.
%\end{proof}
%
%
%\begin{lemma}\label{tech lemma} Let $g \in \Gamma$ be a non-identity element and denote its eigenvalues by $\alpha,\beta,\frac{1}{\beta},\frac{1}{\alpha}$ where 
%$|\alpha|\ge |\beta| \ge 1$.  Then there exists a base $B=(v_1,v_2,v_3,v_4)$ of $\R^4$
%for which the representative matrices $[g]_B$ and  $[q]_B$ are as in Table 1.
%\end{lemma}
%\begin{proof} We start by proving that the Jordan blocks
%of elements in $\Gamma$ are of sizes $[1,1,1,1]$ or $[2,2]$ or $[3,1]$. Assume that  $g$ has Jordan blocks of sizes $[2,1,1]$. By replacing $g$ with $g^{-1}$ is necessary  we can assume that there exists a base
%$B=(v_1,v_2,v_3,v_4)$  of $\R^4$ such that one of
%$[g]_B=\diag(J_2(\alpha),\frac{1}{\alpha},\frac{1}{\alpha})$. For every $i \ne 2$ and every $m$ 
%\begin{equation}
% (v_2,{v_i})=(g^mv_2,g^m{v_i})=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^{\pm m}v_i)
% \end{equation} 
% so $v_1$ is orthogonal to $v_1$, $v_3$ and $v_4$. Moreover, $(v_2,v_1)=0$ since for every $m \ge 1$, 
%\begin{equation}
%(v_2,v_2)=(g^mv_2,g^m v_2)=\alpha^{2m}(v_2,v_2)+2m\alpha^{2m-1}(v_2,v_1)
%+m^2\alpha^{2m-2}(v_1,v_1).
%\end{equation}
%Thus, $v_1$ is orthogonal to $\R^4$,
% a contradiction.
%
%
% 
%Assume that $g$ has a unique Jordan block of size 4 and let $B=(v_1,v_2,v_3,v_4)$ be a base of $\R^4$ such that$[g]_B=J_4(1)$. Then for every $m \ge 1$:
%\begin{equation}\label{eq 1}
% (v_1,v_4)=(g^mv_1,g^mv_4)=(v_1,v_4+mv_3+\frac{m(m-1)}{2}v_2+\frac{m(m-1)(m-2)}{6}v_1)
%\end{equation}
%\begin{equation}\label{eq 2}
%(v_3,v_3)=(g^mv_3,g^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
%\end{equation}
%%\begin{equation}\label{eq 3}
%%(v_3,v_3)=(g^mv_3,g^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
%%\end{equation}
%From Equation \ref{eq 1} it follows that $v_1$ is orthogonal to $v_1$, $v_2$ and $v_3$. Equation
%\ref{eq 2} shows that $v_2$ is orthogonal to itself and to $v_3$. 
%%Finally, Equation \ref{eq 3} implies that  $v_2$ is orthogonal to $v_3$. 
%Thus, the 2-dimensional subspace spanned by $v_1$ and $v_2$ is orthogonal to the  3-dimensional linear subspace spanned by $v_1$, $v_2$ and $v_3$, a contradiction.  
%
%In the rest of the proof we check the existence of the required base by a case by case verification (Lemma \ref{first lemma} and the first part of the proof implies that the cases below are all the possible cases).
%\\ \\
%Case 1: Let $V:=\{x \mid (g^2-2r\cos\theta+r^2)x=0\}$ and $U:=\{x\mid (g^2-\frac{2}{r}\cos\theta+\frac{1}{r^2})x=0\}$. Then the restriction of $q$ to $V$ and $U$ is the zero form. There exists a base $v_1,v_2$ of $V$ such that
%$[q|_V]_{v_1,v_2}=\left(\begin{array}{cc}
%c & d \\
%-d & c
%\end{array}\right)$ where $c$ and $d$ are as in Table 1. Since $(\cdot,\cdot)$ is non-degenerate there exists a base $v_3,v_4$ of $U$ such that $[q]_{v_1,v_2,v_3,v_4}$ has the required form. Since $g$ belong to the orthogonal group of $q$, $[g]_{v_1,v_2,v_3,v_4}$ must have the required form. 
%\\ \\
%Cases 2 and 3 can be readily verified.   
%\\ \\
%Cases 4: Assume that the Jordan blocks of $g$ are of sizes $[2,2]$ and $\alpha \ne 1$. Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[g]_B$ has the required form. For every $m$:
%\begin{equation}\label{eq 5}
%(v_2,v_2)=(g^mv_2,g^mv_2)=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^m v_2+m\alpha^{m-1}v_1)\end{equation}
%so the restriction of $q$ to the span of$v_1$ and $v_2$ is zero and similarly the restriction of 
%$q$ the span of$v_3$ and $v_4$ is zero.  Moreover, for every $m$
%\begin{equation}\label{eq 5}
%(v_2,v_3)=(g^mv_2,g^mv_3)=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^{-m}v_3)
%\end{equation}
%so $(v_1,v_3)=0$. 
%
%Moreover, $\alpha^2(v_2,v_3)=-(v_1,v_4)$ since for every $m$
%\begin{equation}\label{eq 5}
%(v_2,v_4)=(g^mv_2,g^mv_4)=
%(v_2,v_4)+m\alpha^{-1}(v_1,v_4)+m\alpha(v_2,v_3). 
%\end{equation}
%Finally, by replacing $v_4$ with $v_4-\frac{(v_2,v_4)}{(v_2,v_3)}v_3$ we get that 
%$[q]_B$ has the required form up to scalar. In order to fix the scalar, we replace $v_3$ and $v_4$ by a scalar multiple. 
%\\ \\
%Case 5: Assume that the Jordan blocks of $g$ are of sizes $[2,2]$ and $g$ is unipotent. 
%Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[g]_B$ has the required form. For every $m$:
%\begin{equation}\label{eq 5}
%(v_2,v_2)=(g^mv_2,g^mv_2)=( v_2+mv_1, v_2+mv_1)\end{equation}
%so $v_1$ is orthogonal to $v_1$ and $v_2$ and similarly $v_3$ is  orthogonal to $v_3$ and $v_4$.  Moreover, for every $m$
%\begin{equation}\label{eq 5}
%(v_2,v_3)=(g^mv_2,g^mv_3)=( v_2+mv_1,v_3)
%\end{equation}
%so $(v_1,v_3)=0$. Since $q$ is non-degenerate, $v_1$ is not orthogonal to $v_4$ and $v_3$ is not orthogonal to $v_2$.
% By replacing $v_2$ with $v_2-\frac{(v_2,v_2)}{2(v_2,v_3)}v_3$ and $v_4$ with   $v_4-\frac{(v_4,v_4)}{2(v_1,v_4)}v_1$
% we can assume that $(v_2,v_2)=(v_4,v_4)=0$.
% By replacing $v_2$ with $v_2-\frac{(v_2,v_2)}{2(v_2,v_3)}v_3$ and $v_4$ with   $v_4-\frac{(v_4,v_4)}{2(v_1,v_4)}v_1$
% we can assume that $(v_2,v_2)=(v_4,v_4)=0$.
%Now, $(v_2,v_3)=-(v_1,v_4)$ since for every $m$
%\begin{equation}\label{eq 5}
%(v_2,v_4)=(g^mv_2,g^mv_4)=
%(v_2mv_1,v_4mv_3). 
%\end{equation}
%Finally, by replacing $v_4$ with $v_4-\frac{(v_2,v_4)}{(v_2,v_3)}v_3$ we get that 
%$[q]_B$ has the required form up to scalar.  In order to fix the scalar, we replace $v_3$ and $v_4$ by a scalar multiple.
%\\ \\
%Case 6:  Assume that the Jordan blocks of $g$ are of sizes $[3,1]$. Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[g]_B$ has the required form. 
%Then for every $m$:
%%\begin{equation}\label{eq 4}
% %(v_1,v_2)=(gv_1,gv_4)=(v_1,v_2+v_1)
%%\end{equation}
%\begin{equation}\label{eq 5}
%(v_2,v_2)=(g^mv_2,g^mv_2)=(v_2+mv_1,v_2+mv_1)
%\end{equation}
%\begin{equation}\label{eq 6}
%(v_3,v_2)=(g^mv_3,g^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_2+mv_1)\end{equation}
%\begin{equation}\label{eq 7}
%(v_3,v_3)=(g^mv_3,g^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
%\end{equation}
%
%Equation \ref{eq 5} implies that $(v_1,v_1)=0$ and $(v_1,v_2)=0$. Equation \ref{eq 6} implies that $(v_2,v_2)=-(v_1,v_3)$. In turn, Equation \ref{eq 7} implies that $2(v_2,v_3)=(v_1,v_3)$. By replacing $v_3$ with $v_3-\frac{(v_3,v_3)}{2(v_1,v_3)}v_1$ we can assume that $(v_3,v_3)=0$.
%Moreover, for every $m$ \begin{equation}\label{eq 8}
%(v_4,v_3)=(g^mv_3,g^mv_4)=(v_4,v_3+mv_2+\frac{m(m-1)}{2}v_1)
%\end{equation}
%so $(v_1,v_4)=0$ and $(v_2,v_4)=0$. By replacing $v_4$ with $v_4-\frac{(v_3,v_4)}{(v_1,v_3)}v_1$ we can assume that $(v_4,v_3)=0$.
%\end{proof}
%
%{\bf Notation:} For every linearly independent $u,w \in \R^4$ let $l_{u,w}$ be the projective line induced by the span of $u$ and $w$.
%\begin{lemma} Let $g \in \Gamma$ be a non-identity element.
%Let $B=(v_1,v_2,v_3,v_4)$ a base of $\R^4$ for which  
%$[g]_B$ and $[q]_B$ are as in Table 1. 
%Let $l_0$, $l_\infty$ and  $l_{-\infty}$ be the corresponding isotropic projective lines
%defined in Table 1,  Then for every $\e>0$ the following properties hold: 
%\begin{enumerate}
%\item For every non-zero $m  \in \Z$, $g^m(l_0)\cap l_0=\emptyset$.
%\item For all but finitely many positive values of $m \in \Z$, $g^m(l_0) \subseteq N_\e(l_\infty) \cup N_\e(l_{-\infty})$.
%\end{enumerate}
%\end{lemma}
%
%\begin{table}\label{table}
%\begin{center}
%\begin{tabular}{| l | l | l | l | l |}
%\hline
%&  & $[g]_B$& $[q]_{B} $& $l_0,l_{\pm\infty}$\\
%\hline
%& && &\\
%%%%
%$(1)$ & $\begin{array}{c} 
%\alpha=re^{i\theta}\\
%\beta=re^{-i\theta} \\ 
%r>1 \\
%\theta \not \in \pi\Z \\
%c:=r\cos(\theta) \\
%d:=r\sin(\theta)
%\end{array}$ & 
%$\left(\begin{array}{cccc}
%c& d  & 0 & 0 \\
% -d & c & 0 & 0 \\
%0 & 0 &\frac{c}{r^2} & \frac{d}{r^2} \\
%0 & 0 & \frac{-d}{r^2}   & \frac{c}{r^2} \\
%\end{array}\right)$ & 
%$\left(\begin{array}{cccc}
%0 & 0 & 1 & 0 \\
%0 & 0 & 0 & 1 \\
%1 & 0 & 0 & 0 \\
%0 & 1 & 0 & 0\\
%\end{array}\right)$ &
%$\begin{array}{c}
%l_0:=l_{{v_1+v_4,v_2-v_3}} \\
%l_\infty:=l_{{v_1,v_2}}\\
%l_{-\infty}:=l_{{v_3,v_4}}
%\end{array}$
%%%%
%\\ & && & \\
%\hline
%&& & &\\
% %%%
%$(2)$  &$\left.\begin{array}{c}
%\alpha > \beta \ge 1 \\
%\end{array}\right.$
% & 
%$\left(\begin{array}{cccc}
%\alpha &0  & 0 & 0 \\
% 0 & \beta & 0 & 0 \\
%0 & 0 &\frac{1}{\beta}& 0 \\
%0 & 0 & 0  & \frac{1}{\alpha} \\
%\end{array}\right)$& 
%$\left(\begin{array}{cccc}
%0 & 0 & 0 & 1 \\
%0 & 0 & 1 & 0 \\
%0 &1 & 0 & 0 \\
%1 & 0 & 0 & 0\\ 
%\end{array}\right)$ &
%$\begin{array}{c}
%l_0:=l_{{v_1+v_3,v_2-v_4}} \\
%l_\infty:=l_{{v_1,v_2}}\\
%l_{-\infty}:=l_{{v_3,v_4}}
%\end{array}$\\
% %%%
%&& && \\ \hline
%&& && \\
%%%%
%$(3)$ &
%$\begin{array}{c}
%\alpha = \beta >  1 \\
%\end{array}$
% & 
%$\left(\begin{array}{cccc}
%\alpha &0  & 0 & 0 \\
% 0 & \alpha& 0 & 0 \\
%0 & 0 &\frac{1}{\alpha}& 0 \\
%0 & 0 & 0  & \frac{1}{\alpha} \\
%\end{array}\right)$& 
%$\left(\begin{array}{cccc}
%0 & 0 & 0 & 1 \\
%0 & 0 & 1 & 0 \\
%0 &1 & 0 & 0 \\
%1 & 0 & 0 & 0\\
%\end{array}\right)$&
%$\begin{array}{c}
%l_0:=l_{{v_1+v_3,v_2-v_4}} \\
%l_\infty:=l_{{v_1,v_2}}\\
%l_{-\infty}:=l_{{v_3,v_4}}
%\end{array}$ \\
%%%%    
%&&& &\\ \hline
%&&& &\\
%%%%
%$(4)$ &
%$\begin{array}{c}
%\alpha = \beta >  1 \\
%\end{array}$
% & 
%$\left(\begin{array}{cccc}
%\alpha &1  & 0 & 0 \\
% 0 & \alpha& 0 & 0 \\
%0 & 0 &\frac{1}{\alpha}& 1 \\
%0 & 0 & 0  & \frac{1}{\alpha} \\
%\end{array}\right)$& 
%$\left(\begin{array}{cccc}
%0 & 0 & 0 & \alpha \\
%0 & 0 & -\frac{1}{\alpha}& 0 \\
%0 &-\frac{1}{\alpha}& 0 & 0 \\
%\alpha & 0 & 0 & 0\\
%\end{array}\right)$ &
%$\begin{array}{c}
%l_0:=l_{{v_1+\alpha v_3,\alpha v_2+v_4}} \\
%l_\infty:=l_{{v_1,v_2}}\\
%l_{-\infty}:=l_{{v_3,v_4}}
%\end{array}$\\
%%%% 
%&&& &\\ \hline
%&&& &\\
%$(5)$ &
%$\begin{array}{c}
%\alpha = \beta =  1 \\
%\end{array}$
% & 
%$\left(\begin{array}{cccc}
%1&1  & 0 & 0 \\
% 0 & 1& 0 & 0 \\
%0 & 0 &1& 1 \\
%0 & 0 & 0  &1 \\
%\end{array}\right)$& 
%$\left(\begin{array}{cccc}
%0 & 0 & 0 & 1 \\
%0 & 0 & -1 & 0 \\
%0 &-1& 0 & 0 \\
%1 & 0 & 0 & 0\\
%\end{array}\right)$ &
%$\begin{array}{c}
%l_0:=l_{{v_1+v_3,v_2+v_4}} \\
%l_\infty:=l_{{v_1,v_3}}\\
%l_{-\infty}:=l_{{v_1,v_3}}
%\end{array}$\\
%%%% 
%&&& &\\ \hline
%&&& &\\
%%%%
%$(6)$ &
%$\begin{array}{c}
%\alpha = \beta=1 \\
%\end{array}$
% & 
%$\left(\begin{array}{cccc}
%1 &1  & 0 & 0 \\
% 0 &1& 1 & 0 \\
%0 & 0 &1& 0 \\
%0 & 0 & 0  & 1\\
%\end{array}\right)$& 
%$\pm\left(\begin{array}{cccc}
%0 & 0 & 2 &  0\\
%0 & -2 & 1 & 0 \\
%2 & 1 & 0& 0 \\
%0 & 0 & 0 & 1\\
%\end{array}\right)$&
%$\begin{array}{c}
%l_0:=l_{{v_2+\sqrt{2}v_4,2v_3+v_2-\frac{1}{4}v_1}} \\
%l_\infty:=l_{v_1,v_2-\sqrt{2}v_4}\\
%l_{-\infty}:=l_{v_1,v_2-\sqrt{2}v_4}
%\end{array}$
% \\
%&&&& \\
%%%%       
%\hline
%\end{tabular}
%\end{center}
%\begin{center}
%Table 1
%\end{center}
%\end{table}


\begin{asmp} Let  $q(\xb)$ be an integral quadratic form of signature $(2,2)$ and let $(\cdot,\cdot)$ be the corresponding bilinear form. Let $\Gamma \le SO(q,\Z)$  be a finite index net subgroup. 
\end{asmp}

\begin{lemma}\label{first lemma} Let $f$ be the characteristic polynomial of some $g \in \Gamma$.  Then $f$ is 
self reciprocal and one of the following two conditions holds:
\begin{itemize}
\item All the roots of $f$ are real. 
\item All the roots of $f$ are non-real complex numbers of absolute values different than 1. 
\end{itemize}
\end{lemma}
\begin{proof} We first show that $f$ is reciprocal. The free coefficient of $f$ is $1$ and $f$ is a monic
polynomial of degree 4. Thus, in order to show that $f$ is reciprocal it is enough to prove that if $\alpha$ is a root of $f$ then so is $1/\alpha$.  Let 
$v_1 \in \C^4$ be an eigenvector of $g$ with eigenvalue $\alpha$.  Denote  $V:=
\{ x\in \C^4\mid (v_1,x)=0\}$ and let $v_2 \in \C^4$ be such that $(v_1,v_2)=1$. Then, $gv_2=(1/\alpha) v_2 +v_3$ for some $v_3\in V$. Moreover,
$g$ preserves $V$ and acts on $\C^4/V$ as multiplication by $1/\alpha$. This implies that $1/\alpha$ is an eigenvalue of $g$.  

The next step is to show that it is not possible for $f$ to have a real root  $\alpha$ with absolute values different than 1 as well as a non-real root $re^{i\theta}$.
Assume otherwise, since $f$ is real and reciprocal $r=1$ and $f(x)=(x-\alpha)(x-\frac{1}{\alpha})(x-e^{i\theta})(x-e^{-i\theta})$ 
where $\alpha$ and $\theta$ are real numbers, $|\alpha| \ne 1$ and  $\theta$ is not an integral multiple of $\pi$. Choose $v_1 \in \R^4$  such that
$gv_1=\alpha v_1$.  Define 
$U:=\{{ x}\in \R^4\mid (g-e^{i\theta})(g-{e^{-i\theta}}){ x}=0\}$,  $V:=\{{ x}\in \R^4\mid ({v_1},{ x})=0\}$
and  $W:=\{{ x}\in \R^4 \mid (g-\alpha)(g-{1\over \alpha}){ x}=0\}$. The non-trivial subspace $U \cap V$ of $U$ is invariant under $g$
 so that the assumption on $\theta$ implies that it must be equal to $U$. In particular, every element of $U$ is orthogonal to $v_1$.  Choose ${v_2}\in W $ to be an eigenvector with eigenvalue $\frac{1}{\alpha}$. Let ${ x}\in U$ be orthogonal to $v_2$, then
 \begin{equation}
0=({v_2},{x})=(g{v_2},g{ x})=\frac{1}{\alpha} ({v_2}, g{x}).
 \end{equation}
 Thus, the assumption on $\theta$ implies that $({v_2},{ x})=0$ for every ${x}\in U$. It follows that $\R^4$ is an orthogonal sum of $W$ and $U$ and either 
$W$ and $U$ both have signature $(1,1)$ or one of them is positive definite and the other is negative definite.  The second case is not possible since an orthogonal transformation of a definite space cannot have an eigenvalue with absolute value strictly greater than 1. The 
%imaginary numbers  $i$ and $-i$ are the only possible non-real roots of an orthogonal transformation of a space of signature $(1,1)$, hence, if the first case happens then $\theta=\pi/2$ and $r=1$, a contradiction. 
first case cannot happen since any orthogonal transformation of a quadratic space of signature $(1,1)$ is diagonalisable over $\R$.
%This implies  that $\beta=-1/\alpha \ne \alpha$, $\gamma=0$ and 
%\begin{itemize}
%\item $({ v},{ v})=
%(g{ v},g{ v})=\alpha^2({ v},{ v})$ and so $({ v},{ v})=0$;
%\item $({ w},{ w})=
%(g{ w},g{ w})=\alpha^{-2}({ w},{ w})$ thus $({ w},{ w})=0$;
%\item$({ v},{ w})=
%(g{ v},g{ w})=-({ v},{ w})$ hence $({ v},{ w})=0$.
%\end{itemize}
%This leads to a contradiction since $f$ is not degenerate. 
 
 If $z \in \C\setminus \R $ is a root of  $f \in \Z[x]$ then so is $\bar{z}$.  Thus if $f(x)$ has a non-real root $z=re^{i\theta}$ with absolute value greater than 1 then $re^{-i\theta}$, $\frac{1}{r}e^{i\theta}$ and  $\frac{1}{r}e^{-i\theta}$ are the other roots of $f$. Finally, $f(x)$ cannot have a non-real root with 
absolute value 1 since then the previous paragraph implies that all the roots of $f(x)$ must have absolute value 1. Thus they are roots of unity which contradicts the assumption about the eigenvalues of elements in $\Gamma$.
\end{proof}


\begin{lemma}\label{tech lemma} Let $\gc \in \Gamma$ be a non-trivial element and denote its eigenvalues by $\alpha,\beta,\frac{1}{\beta},\frac{1}{\alpha}$ where 
$|\alpha|\ge |\beta| \ge 1$.  Then there exists a base $B=(v_1,v_2,v_3,v_4)$ of $\R^4$
for which the representative matrices $[\gc]_B$ and  $[q]_B$ are as in Table 1.
\end{lemma}
\begin{proof} We start by proving that the Jordan blocks
of elements in $\Gamma$ are of sizes $[1,1,1,1], [2,2]$ or $[3,1]$. Assume that  $\gc$ has Jordan blocks of sizes $[2,1,1]$. By replacing $\gc$ with $\gc^{-1}$ if necessary  we can assume that there exists a base
$B=(v_1,v_2,v_3,v_4)$  of $\R^4$ such that 
$[\gc]_B=\diag(J_2(\alpha),\frac{1}{\alpha},\frac{1}{\alpha})$. For every $i \ne 2$ and every $m$ 
\begin{equation}
 (v_2,{v_i})=(\gc^mv_2,\gc^m{v_i})=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^{\pm m}v_i)
 \end{equation} 
 so $v_1$ is orthogonal to $v_1$, $v_3$ and $v_4$. Moreover, $(v_2,v_1)=0$ since for every $m \ge 1$, 
\begin{equation}
(v_2,v_2)=(\gc^mv_2,\gc^m v_2)=\alpha^{2m}(v_2,v_2)+2m\alpha^{2m-1}(v_2,v_1)
+m^2\alpha^{2m-2}(v_1,v_1).
\end{equation}
Thus, $v_1$ is orthogonal to $\R^4$,
 a contradiction.


 
Assume that $\gc$ has a unique Jordan block of size 4 and let $B=(v_1,v_2,v_3,v_4)$ be a base of $\R^4$ such that$[\gc]_B=J_4(1)$. Then for every $m \ge 1$:
\begin{equation}\label{eq 1}
 (v_1,v_4)=(\gc^mv_1,\gc^mv_4)=(v_1,v_4+mv_3+\frac{m(m-1)}{2}v_2+\frac{m(m-1)(m-2)}{6}v_1)
\end{equation}
\begin{equation}\label{eq 2}
(v_3,v_3)=(\gc^mv_3,\gc^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
\end{equation}
%\begin{equation}\label{eq 3}
%(v_3,v_3)=(g^mv_3,g^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
%\end{equation}
From Equation \ref{eq 1} it follows that $v_1$ is orthogonal to $v_1$, $v_2$ and $v_3$. Equation
\ref{eq 2} then shows that $v_2$ is orthogonal to itself and to $v_3$. 
%Finally, Equation \ref{eq 3} implies that  $v_2$ is orthogonal to $v_3$. 
Thus, the 2-dimensional subspace spanned by $v_1$ and $v_2$ is orthogonal to the  3-dimensional linear subspace spanned by $v_1$, $v_2$ and $v_3$, a contradiction.  

In the rest of the proof we exhibit the existence of the required base by a case by case verification (Lemma \ref{first lemma} and the first part of the proof imply that the cases below account for all possible cases).
\\ \\
Case 1: Let $V:=\{x \mid (\gc^2-2r\cos\theta\gc+r^2)x=0\}$ and $U:=\{x\mid (\gc^2-\frac{2}{r}\cos\theta\gc+\frac{1}{r^2})x=0\}$. Then the restriction of $q$ to $V$ and to $U$ is the zero form. There exists a base $v_1,v_2$ of $V$ such that
$[\gc]_{v_1,v_2}=\left(\begin{array}{cc}
c & d \\
-d & c
\end{array}\right)$ where $c$ and $d$ are as in Table 1. Since $(\cdot,\cdot)$ is non-degenerate there exists a base $v_3,v_4$ of $U$ such that $[q]_{v_1,v_2,v_3,v_4}$ has the required form. Since $\gc$ belong to the orthogonal group of $q$, $[\gc]_{v_1,v_2,v_3,v_4}$ must have the required form. 
\\ \\
Cases 2 and 3 can be readily verified.   
\\ \\
Cases 4: Assume that the Jordan blocks of $\gc$ are of sizes $[2,2]$ and $\alpha \ne 1$. Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[\gc]_B$ has the required form. For every $m$:
\begin{equation}\label{eq 5}
(v_2,v_2)=(\gc^mv_2,\gc^mv_2)=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^m v_2+m\alpha^{m-1}v_1)\end{equation}
so the restriction of $q$ to the span of $v_1$ and $v_2$ is zero and similarly the restriction of 
$q$ to the span of $v_3$ and $v_4$ is zero.  Moreover, for every $m$
\begin{equation}\label{eq 5}
(v_2,v_3)=(\gc^mv_2,\gc^mv_3)=(\alpha^m v_2+m\alpha^{m-1}v_1,\alpha^{-m}v_3)
\end{equation}
so $(v_1,v_3)=0$. 

Moreover, $\alpha^2(v_2,v_3)=-(v_1,v_4)$ since for every $m$
\begin{equation}\label{eq 5}
(v_2,v_4)=(\gc^mv_2,\gc^mv_4)=
(v_2,v_4)+m\alpha^{-1}(v_1,v_4)+m\alpha(v_2,v_3). 
\end{equation}
Finally, by replacing $v_4$ with $v_4-\frac{(v_2,v_4)}{(v_2,v_3)}v_3$ we get that 
$[q]_B$ has the required form up to a scalar. In order to remedy this, we replace $v_3$ and $v_4$ by a scalar multiple. 
\\ \\
Case 5: Assume that the Jordan blocks of $\gc$ are of sizes $[2,2]$ and $\gc$ is unipotent. 
Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[\gc]_B$ has the required form. For every $m$:
\begin{equation}\label{eq 5}
(v_2,v_2)=(\gc^mv_2,\gc^mv_2)=( v_2+mv_1, v_2+mv_1)\end{equation}
so $v_1$ is orthogonal to $v_1$ and $v_2$ and similarly $v_3$ is  orthogonal to $v_3$ and $v_4$.  Moreover, for every $m$
\begin{equation}\label{eq 5}
(v_2,v_3)=(\gc^mv_2,\gc^mv_3)=( v_2+mv_1,v_3)
\end{equation}
so $(v_1,v_3)=0$. Since $q$ is non-degenerate, $v_1$ is not orthogonal to $v_4$ and $v_3$ is not orthogonal to $v_2$.
 By replacing $v_2$ with $v_2-\frac{(v_2,v_2)}{2(v_2,v_3)}v_3$ and $v_4$ with   $v_4-\frac{(v_4,v_4)}{2(v_1,v_4)}v_1$
 we can assume that $(v_2,v_2)=(v_4,v_4)=0$.
By replacing $v_4$ with $v_4-\frac{(v_2,v_4)}{(v_2,v_3)}v_3$ we 
can assume that $(v_2,v_4)=0$
Now, $(v_2,v_3)=-(v_1,v_4)$ since for every $m$
\begin{equation}\label{eq 5}
(v_2,v_4)=(\gc^mv_2,\gc^mv_4)=
(v_2+mv_1,v_4+mv_3). 
\end{equation}
Thus, $[q]_B$ has the required form up to a scalar.  In order to remedy the situation, we replace $v_3$ and $v_4$ by a scalar multiple.
%Now, $(v_2,v_3)=-(v_1,v_4)$ since for every $m$
%\begin{equation}\label{eq 5}
%(v_2,v_4)=(\gc^mv_2,\gc^mv_4)=
%(v_2+mv_1,v_4+mv_3). 
%\end{equation}
\\ \\
Case 6:  Assume that the Jordan blocks of $\gc$ are of sizes $[3,1]$. Let $B:=(v_1,v_2,v_3,v_4)$ be a base such that $[\gc]_B$ has the required form. 
Then for every $m$:
%\begin{equation}\label{eq 4}
 %(v_1,v_2)=(gv_1,gv_4)=(v_1,v_2+v_1)
%\end{equation}
\begin{equation}\label{eq 5}
(v_2,v_2)=(\gc^mv_2,\gc^mv_2)=(v_2+mv_1,v_2+mv_1)
\end{equation}
\begin{equation}\label{eq 6}
(v_3,v_2)=(\gc^mv_3,\gc^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_2+mv_1)\end{equation}
\begin{equation}\label{eq 7}
(v_3,v_3)=(\gc^mv_3,\gc^mv_3)=(v_3+mv_2+\frac{m(m-1)}{2}v_1,v_3+mv_2+\frac{m(m-1)}{2}v_1)
\end{equation}

Equation \ref{eq 5} implies that $(v_1,v_1)=0$ and $(v_1,v_2)=0$. Equation \ref{eq 6} implies that $(v_2,v_2)=-(v_1,v_3)$. In turn, Equation \ref{eq 7} implies that $2(v_2,v_3)=(v_1,v_3)$. Since $q$ is non-degenerate, $(v_1,v_3)\ne 0$. By replacing $v_3$ with $v_3-\frac{(v_3,v_3)}{2(v_1,v_3)}v_1$ we can assume that $(v_3,v_3)=0$.
Moreover, for every $m$ \begin{equation}\label{eq 8}
(v_4,v_3)=(\gc^mv_3,\gc^mv_4)=(v_4,v_3+mv_2+\frac{m(m-1)}{2}v_1)
\end{equation}
so $(v_1,v_4)=0$ and $(v_2,v_4)=0$. By replacing $v_4$ with $v_4-\frac{(v_3,v_4)}{(v_1,v_3)}v_1$ we can assume that $(v_4,v_3)=0$.
\end{proof}



\subsection{Completing the proof of Theorem \ref{thm:O(2,2)}}
Let us now explain how to prove Theorem \ref{thm:O(2,2)}. The structure of the proof follows the exact same lines as the proof of Theorem \ref{thm:regular}, only that instead of using the action on the projective space, we use the action on the planes Grassmannian $X:=\text{Gr}(2,\BR^4)$.  
%Fix a metric $\dist_{\P^3}$ on $\P^3$ compatible with the topology. 
The elements of $\text{Gr}(2,\BR^4)$ can be viewed as lines in projective space $\P^3$. We define two functions 
$\ud,\ld:X \times X \rightarrow \R$ by:
$$
\ud(l_1,l_2):=\max\{\dist_{\P^3}(x_1,x_2) \mid x_1 \in l_1 \text{ and }x_2 \in l_2 \}
 \text { and }$$
$$
\ld(l_1,l_2):=\min\{\dist_{\P^3}(x_1,x_2) \mid x_1 \in l_1 \text{ and }x_2 \in l_2 \}.
$$
Note that $\ud$ is a metric on $X$.
For every $\e>0$ and $l \in X$ we define two open sets of $X$:
$$
(l)_\e:=\{ t \in X \mid \ud(l,t) < \e \} \text{ and } [l]_\e:=\{ t \in X \mid \ld(l,t) < \e \}.$$


For linearly independent $u,v \in \BR^4$, $l_{u,v} \in X$ denotes their projective linear span. 
An element $l \in X$ is called {\it isotropic} if the restriction  to the corresponding plane is the zero form. 
The next lemma follows from Lemma \ref{tech lemma} and Table 1:    
%For the purpose of the proof let us say that a $[l]\in\text{Gr}(1,\BR^4)$ is (unipotently) $G$-defined if it is the invariant line of some regular unipotent $u\in G$.

\begin{lemma}\label{lemma tab} 
Let $\gc \in \Gamma$ be a non-trivial element.
Let $B=(v_1,v_2,v_3,v_4)$ be a base of $\R^4$ for which  
$[\gc]_B$ and $[q]_B$ are as in Table 1. 
Let $l_0$, $l_\infty$ and  $l_{-\infty}$ be the isotropic elements of $X$
defined in Table 1. Then $$\lim_{n\rightarrow \infty}\gc^n \cdot l_0=\l_\infty \text{   and    }\lim_{n\rightarrow -\infty}\gc^n \cdot l_0=\l_{-\infty}$$ where the convergence is with respect to the metric $\ud$.
Moreover, if $\e>0$ is small enough then  for every $0 \ne m \in \Z $, $\gc^m\cdot (l_0)_\e \cap [l_0]_\e=\emptyset$.
\end{lemma}



\begin{lemma}\label{lemma con} 
Let $l_1,l_2 \in X$ be isotropic elements and let $\e>0$.
There exists $\delta \in \Gamma$ such that
$\delta \cdot (X \setminus [l_1]_\e) \subseteq (l_2)_\e$ and $\delta^{-1}\cdot (X \setminus [l_2]_\e) \subseteq (l_1)_\e$.
\end{lemma}
\begin{proof} Choose  an  isotropic element $l_3 \in (l_2)_\e$ and $\e^*>0$ such that the span of $l_1$ and $l_3$
is $\R^4$, $[l_3]_{\e^*} \subset [l_2]_\e$ and 
$(l_3)_{\e^*} \subset (l_2)_\e$. It is enough to prove the lemma for $l_1$, $l_3$ and $\e^*$ instead of $l_1$, $l_2$
and $\e$. For every $r>0$ there exists $g_r \in \SO(2,2)(\R)$ which preserves $l_1$ and $l_3$, acts on $l_1$ by multiplication by $r$ and 
on $l_3$ by multiplication by $1/r$. Denote $\bar{\e}:=\e^*/2$ and fix a large enough  $r$ such that for every $m>0$,
$g_r^m \cdot (X \setminus [l_1]_{\bar{\e}}) \subseteq (l_3)_{\bar{\e}}$ and $g_r^{-m} \cdot (X \setminus [l_3]_{\bar{\e}}) \subseteq (l_1)_{\bar{\e}}$. Let $U$ be a symmetric neighborhood of the identity in $\SO(2,2)(\R)$ such that for every 
$u \in U$ and every $i \in \{1,3\}$, $u\cdot (l_i)_{\bar{\e}} \subseteq (l_i)_{{\e^*}}$ and $u\cdot [l_i]_{\bar{\e}} \subseteq [l_i]_{{\e^*}}$. As in Section \ref{sec:regular}, by the Poincar\'{e}  recurrence theorem there exist $u_1,u_2 \in U$
and $m>0$ such that $\delta:=u_1g_r^mu_2 \in \Gamma$. Evidently, $\delta$ is the desired element.  
\end{proof}

\begin{cor}\label{alm cor} 
Let $l \in X$ be an isotropic projective line. Then for every $\e>0$ and every conjugacy class $C \ne \{\textrm{Id} \}$ of
$\Gamma$ there exists $\gamma \in C$ such that $\gamma \cdot (X \setminus [l]_\e) \subseteq (l)_\e$.
\end{cor}
\begin{proof} Choose $\gamma_0 \in C$. Lemma \ref{lemma tab} states that there exists an isotropic $l_0 \in X$
and $\e_0>0$ such that for every $0 \ne m \in \Z $, $\gc_0^m\cdot (l_0)_{\e_0} \cap [l_0]_{\e_0}=\emptyset$. Denote 
$\e_1:=\min(\e,\e_0)$. Lemma \ref{lemma con} implies that there exists $\delta \in \gamma$ such that 
$\delta \cdot (X \setminus [l_0]_{\e_1}) \subseteq (l)_{\e_1}$ and $\delta^{-1} \cdot (X \setminus [l]_{\e_1}) \subseteq (l_0)_{\e_1}$. The element $\gamma:=\delta \gamma_0 \delta^{-1}$ satisfies the requirement. 
\end{proof}

%
%\begin{lemma}[ping-pong]\label{ch ping} Let $(l_i)_{i \ge 1}$ be a sequence of isotropic elements of $X$ and let
%$(\e_i)_{i \ge 0}$ be a sequence of positive numbers such that $(l_i)_{\e_i} \cap [l_j]_{\e_j}=\emptyset$
%whenever $i \ne j$. Let $(\gamma_i)_{i \ge 1}$ be a sequence of elements of $\Gamma$ such that for every $i$,
%$\gamma (X \setminus [l_i]_{\e_i}) \subseteq (l_i)_{\e_i}$. Then the natural map from 
%$*_{i \ge 1}\langle \gamma_i\rangle$ to $\langle \gamma_i \mid i \ge 1\rangle$ is an isomorphism.
%\end{lemma} 
% 

Pick a sequences $(l_i)_{i \ge 1}$ of isotropic elements of $X$ and a sequence of positive numbers $(\e_i)_{i \ge 0}$ such that $(l_i)_{\e_i} \cap [l_j]_{\e_j}=\emptyset$
whenever $i \ne j$. By Corollary \ref{alm cor}, there are $(\gamma_i)_{i \ge 1}\in\gC$ representing all the non-trivial conjugacy classes such that for every $i$,
$\gamma_i \cdot (X \setminus [l_i]_{\e_i}) \subseteq (l_i)_{\e_i}$. It follows that Proposition \ref{prop:pingpong} is satisfied with $U_i=(l_i)_{\e_i},~i\in\BN$. Hence $\langle\gc_i:i\in\BN\rangle$ is a free group and in particular of infinite index in $\gC$. This completes the proof of Theorem \ref{thm:O(2,2)}.
\qed









%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%


\begin{table}\label{table}
\begin{center}
\begin{tabular}{| l | l | l | l | l |}
\hline
&  & $[\gc]_B$& $[q]_{B} $& $l_0,l_{\pm\infty}$\\
\hline
& && &\\
%%%
$(1)$ & $\begin{array}{c} 
\alpha=re^{i\theta}\\
\beta=re^{-i\theta} \\ 
r>1 \\
\theta \not \in \pi\Q \\
c:=r\cos(\theta) \\
d:=r\sin(\theta)
\end{array}$ & 
$\left(\begin{array}{cccc}
c& d  & 0 & 0 \\
 -d & c & 0 & 0 \\
0 & 0 &\frac{c}{r^2} & \frac{d}{r^2} \\
0 & 0 & \frac{-d}{r^2}   & \frac{c}{r^2} \\
\end{array}\right)$ & 
$\left(\begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
\end{array}\right)$ &
$\begin{array}{c}
l_0:=l_{{v_1+v_4,v_2-v_3}} \\
l_\infty:=l_{{v_1,v_2}}\\
l_{-\infty}:=l_{{v_3,v_4}}
\end{array}$
%%%
\\ & && & \\
\hline
&& & &\\
 %%%
$(2)$  &$\left.\begin{array}{c}
\alpha > \beta \ge 1 \\
\end{array}\right.$
 & 
$\left(\begin{array}{cccc}
\alpha &0  & 0 & 0 \\
 0 & \beta & 0 & 0 \\
0 & 0 &\frac{1}{\beta}& 0 \\
0 & 0 & 0  & \frac{1}{\alpha} \\
\end{array}\right)$& 
$\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 0 & 0\\ 
\end{array}\right)$ &
$\begin{array}{c}
l_0:=l_{{v_1+v_3,v_2-v_4}} \\
l_\infty:=l_{{v_1,v_2}}\\
l_{-\infty}:=l_{{v_3,v_4}}
\end{array}$\\
 %%%
&& && \\ \hline
&& && \\
%%%
$(3)$ &
$\begin{array}{c}
\alpha = \beta >  1 \\
\end{array}$
 & 
$\left(\begin{array}{cccc}
\alpha &0  & 0 & 0 \\
 0 & \alpha& 0 & 0 \\
0 & 0 &\frac{1}{\alpha}& 0 \\
0 & 0 & 0  & \frac{1}{\alpha} \\
\end{array}\right)$& 
$\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 0 & 0\\
\end{array}\right)$&
$\begin{array}{c}
l_0:=l_{{v_1+v_3,v_2-v_4}} \\
l_\infty:=l_{{v_1,v_2}}\\
l_{-\infty}:=l_{{v_3,v_4}}
\end{array}$ \\
%%%    
&&& &\\ \hline
&&& &\\
%%%
$(4)$ &
$\begin{array}{c}
\alpha = \beta >  1 \\
\end{array}$
 & 
$\left(\begin{array}{cccc}
\alpha &1  & 0 & 0 \\
 0 & \alpha& 0 & 0 \\
0 & 0 &\frac{1}{\alpha}& 1 \\
0 & 0 & 0  & \frac{1}{\alpha} \\
\end{array}\right)$& 
$\left(\begin{array}{cccc}
0 & 0 & 0 & \alpha \\
0 & 0 & -\frac{1}{\alpha}& 0 \\
0 &-\frac{1}{\alpha}& 0 & 0 \\
\alpha & 0 & 0 & 0\\
\end{array}\right)$ &
$\begin{array}{c}
l_0:=l_{{v_1+\alpha v_3,\alpha v_2+v_4}} \\
l_\infty:=l_{{v_1,v_2}}\\
l_{-\infty}:=l_{{v_3,v_4}}
\end{array}$\\
%%% 
&&& &\\ \hline
&&& &\\
$(5)$ &
$\begin{array}{c}
\alpha = \beta =  1 \\
\end{array}$
 & 
$\left(\begin{array}{cccc}
1&1  & 0 & 0 \\
 0 & 1& 0 & 0 \\
0 & 0 &1& 1 \\
0 & 0 & 0  &1 \\
\end{array}\right)$& 
$\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0 \\
0 &-1& 0 & 0 \\
1 & 0 & 0 & 0\\
\end{array}\right)$ &
$\begin{array}{c}
l_0:=l_{{v_1+v_2,v_3+v_4}} \\
l_\infty:=l_{{v_1,v_3}}\\
l_{-\infty}:=l_{{v_1,v_3}}
\end{array}$\\
%%% 
&&& &\\ \hline
&&& &\\
%%%
$(6)$ &
$\begin{array}{c}
\alpha = \beta=1 \\
\end{array}$
 & 
$\left(\begin{array}{cccc}
1 &1  & 0 & 0 \\
 0 &1& 1 & 0 \\
0 & 0 &1& 0 \\
0 & 0 & 0  & 1\\
\end{array}\right)$& 
$\pm\left(\begin{array}{cccc}
0 & 0 & 2 &  0\\
0 & -2 & 1 & 0 \\
2 & 1 & 0& 0 \\
0 & 0 & 0 & 1\\
\end{array}\right)$&
$\begin{array}{c}
l_0:=l_{{v_2+\sqrt{2}v_4,2v_3+v_2-\frac{1}{4}v_1}} \\
l_\infty:=l_{v_1,v_2-\sqrt{2}v_4}\\
l_{-\infty}:=l_{v_1,v_2-\sqrt{2}v_4}
\end{array}$
 \\
&&&& \\
%%%       
\hline
\end{tabular}
\end{center}
\begin{center}
Table 1
\end{center}
\end{table}
\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





%\section{Further remarks and questions}
%
%\begin{itemize}
%\item proof can be carried out for other local fields.
%\item trees, Burger--Mozes lattices
%\item to ask about $\SL(2,\BR)$ and $\SL(3,\BZ)$.
%\end{itemize}
%



\begin{thebibliography}{999999}

%\bibitem[BG95]{BG95} O.V. Bogopolskii, V.N. Gerasimov, Finite subgroups of hyperbolic groups, Algebra and Logic {34}, (1995), 343--345.

\bibitem[D92]{Dixson}
J. D. Dixon, Random sets which invariably generate the symmetric group. Discrete Math. 105 (1992), 25--39.

\bibitem[HNN49]{HNN}
G. Higman, B. H. Neumann and H. Neumann, Embedding theorems for groups, J. London
Math. Soc. 24 (1949), 247--254.

\bibitem[G15]{Gel}
T. Gelander, Convergence groups are not invariably generated, to appear in IMRN.

\bibitem[KLS11]{KLS11}
W. M. Kantor, A. Lubotzky and A. Shalev, Invariable generation and the Chebotarev
invariant of a finite group, J. Algebra 348 (2011), 302--314.

\bibitem[KLS14]{KLS} W.M. Kantor, A. Lubotzky, A. Shalev, Invariable generation on infinite groups, preprint.

%\bibitem[Os10]{Osin} D. Osin, Small cancellations over relatively hyperbolic groups and embedding theorems. Ann. Math. 172 (2010), 1--39.

\bibitem[KM79]{Kneser} M. Kneser, Normalteiler ganzzahliger Spingruppen, J. Reine und Angew. Math. 311/312(1979), 191--214.


\bibitem[R72]{Rag} M.S. Raghunathan, {\it Discrete Subgroups of Lie Groups}, 
Springer, New York, 1972.

\bibitem[R76]{Ra1} M.S. Raghunathan, On the congruence subgroup problem. Inst. Hautes Etudes Sci. Publ. Math. No. 46 (1976), 107--161.

\bibitem[R86]{Ra2} M.S. Raghunathan, On the congruence subgroup problem. II. Invent. Math. 85 (1986), no. 1, 73--117.

\bibitem[S14]{Sa}  P. Sarnak, Notes on thin matrix groups. Thin groups and superstrong approximation, Math. Sci. Res. Inst. Publ., 61, Cambridge Univ. Press, Cambridge, (2014), 343--362.

\bibitem[W76]{W76} J. Wiegold, Transitive groups with fixed-point-free permutations, Arch. Math. (Basel) 27
(1976), 473--475.

\bibitem[We77]{W77} J. Wiegold, Transitive groups with fixed-point-free permutations. II, Arch. Math. (Basel)
29 (1977), 571--573.

\bibitem[WM14]{Witte} D. Witte--Morris, Introduction to Arithmetic Groups.
\end{thebibliography}
\end{document}



1505.06881.extracted_bib
[BG95]{BG95 O.V. Bogopolskii, V.N. Gerasimov, Finite subgroups of hyperbolic groups, Algebra and Logic {34, (1995), 343--345.
1505.06881.extracted_bib
[D92]{Dixson J. D. Dixon, Random sets which invariably generate the symmetric group. Discrete Math. 105 (1992), 25--39.
1505.06881.extracted_bib
[HNN49]{HNN G. Higman, B. H. Neumann and H. Neumann, Embedding theorems for groups, J. London Math. Soc. 24 (1949), 247--254.
1505.06881.extracted_bib
[G15]{Gel T. Gelander, Convergence groups are not invariably generated, to appear in IMRN.
1505.06881.extracted_bib
[KLS11]{KLS11 W. M. Kantor, A. Lubotzky and A. Shalev, Invariable generation and the Chebotarev invariant of a finite group, J. Algebra 348 (2011), 302--314.
1505.06881.extracted_bib
[KLS14]{KLS W.M. Kantor, A. Lubotzky, A. Shalev, Invariable generation on infinite groups, preprint. %
1505.06881.extracted_bib
[Os10]{Osin D. Osin, Small cancellations over relatively hyperbolic groups and embedding theorems. Ann. Math. 172 (2010), 1--39.
1505.06881.extracted_bib
[KM79]{Kneser M. Kneser, Normalteiler ganzzahliger Spingruppen, J. Reine und Angew. Math. 311/312(1979), 191--214.
1505.06881.extracted_bib
[R72]{Rag M.S. Raghunathan, {Discrete Subgroups of Lie Groups, Springer, New York, 1972.
1505.06881.extracted_bib
[R76]{Ra1 M.S. Raghunathan, On the congruence subgroup problem. Inst. Hautes Etudes Sci. Publ. Math. No. 46 (1976), 107--161.
1505.06881.extracted_bib
[R86]{Ra2 M.S. Raghunathan, On the congruence subgroup problem. II. Invent. Math. 85 (1986), no. 1, 73--117.
1505.06881.extracted_bib
[S14]{Sa P. Sarnak, Notes on thin matrix groups. Thin groups and superstrong approximation, Math. Sci. Res. Inst. Publ., 61, Cambridge Univ. Press, Cambridge, (2014), 343--362.
1505.06881.extracted_bib
[W76]{W76 J. Wiegold, Transitive groups with fixed-point-free permutations, Arch. Math. (Basel) 27 (1976), 473--475.
1505.06881.extracted_bib
[We77]{W77 J. Wiegold, Transitive groups with fixed-point-free permutations. II, Arch. Math. (Basel) 29 (1977), 571--573.
1505.06881.extracted_bib
[WM14]{Witte D. Witte--Morris, Introduction to Arithmetic Groups.