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%%% Title: Generic subgroups of Lie Groups
%%% Author: Joerg Winkelmann
%%% e-mail: jwinkel@member.ams.org
%%% AMS Subject Classifictaion: 22E40
%%% Keywords and Phrases: Lie groups, discrete subgroups, generic subgroups
%%% Abstract:
%%% We investigate properties of generically choosen finitely
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\documentclass[12pt]{amsart}
\usepackage{amsmath,amsthm,amsfonts,amssymb}
\begin{document} 
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\newtheorem*{definition}{Definition}
\newtheorem*{claim}{Claim}
\newtheorem{corollary}{Corollary}
\newtheorem*{Conjecture}{Conjecture}
\newtheorem*{SpecAss}{Special Assumptions}
\newtheorem{example}{Example}
\newtheorem*{remark}{Remark}
\newtheorem*{observation}{Observation}
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\title{%
Generic subgroups 
of Lie Groups
}
%\shorttitle{Generic subgroups}
\author {J\"org Winkelmann}
%
\begin{abstract}
We investigate properties of generically choosen finitely
generated subgroups of real Lie groups.
In particular we ask whether discreteness is generic.
\end{abstract}
\subjclass{AMS Subject Classification: 22E40}
% http://www.ams.org/msc/
%
\address{%
J\"org Winkelmann \\
Universit\"at Basel\\
Mathematisches Institut \\
Rheinsprung 21 \\
CH-4051 Basel\\
Switzerland}
\email{jwinkel@member.ams.org\newline\indent{\itshape Webpage: }%
http://www.math.unibas.ch/\~{ }winkel/
}
%
\maketitle
\section{introduction}
In this article we investigate generic subgroups of Lie groups.
In particular we ask the following question:
\begin{quote}
\sl
Given a Lie group $G$ and a  natural number $n$,
when is it true, that $n$ generically choosen elements
$g_1,\ldots,g_n$
will generate a discrete subgroup of $G$ ?
\end{quote}

More precisely, given a topological group $G$ we define a subset
$\Delta_n$ of the $n$-fold product space $G^n=G\times\ldots\times G$
by
\[
\Delta_k(G) =
\{ (g_1,\ldots,g_k)\in G^k: \left<g_1,\ldots,g_k\right>
\text{ is a discrete subgroup of $G$} \}
\]
A complete description of these sets $\Delta_k$ is available only
for rather special cases. It is easy to obtain such a description
for abelian connected Lie groups. For semisimple Lie groups 
already the description of $\Delta_2$ is rather
complicated.
A complete description has been achieved only for the case $SL(2,\R)$.
For $SL(2,\C)$ there are partial results, in particular the
famous inequality of J\o rgensen.

In this article we do not look for a complete description of
$\Delta_k$. Here we concentrate on the question:

\begin{quote}\sl
Given a Lie group $G$ and a number $k\in\N$,
under which conditions on $G$ and $k$ are
 $\Delta_k$ or $G^k\setminus\Delta_k$
are sets of measure zero (with respect to the product measure of
the Haar measure of $G$).
\end{quote}
(The sets $\Delta_k$ are always measurable, see prop~\ref{measurable}.)

For $\Delta_1$ we have precise criteria answering these questions
in dependance
 on the topological properties of
the Cartan subgroups of $G$.

\begin{theorem}\label{thm1}
Let $G$ be a connected real Lie group.

Then the set $\Delta_1$ has positive measure 
if and only if $G$ admits a Cartan subgroup with
non compact connected components.

The set $G\setminus\Delta_1$ has positive
measure if and only if $G$ admits a Cartan subgroup
with compact connected components.

If $G$ contains both Cartan subgroups with compact connected
components and Cartan subgroups with non compact components,
then both $\Delta_1$ and $G\setminus\Delta_1$ are sets of infinite
measure.
\end{theorem}

We discuss compactness resp.~non-compactness of Cartan subgroups
in some detail, because the size of $\Delta_1$ depends on these
properties.

Let $G$ be a connected real Lie group, $R$ its radical and $S=G/R$.
Cartan subgroups with compact connected
components are necessarily compact (prop.~\ref{csg-compact}).
The existence of a compact Cartan subgroup
implies
that the center of $S$ is finite (by prop.~\ref{infini})
and that
$G/G^\infty$ is compact (lemma~\ref{dersurj}).

For general $k$ there is a strict dichotomy depending on whether
$S$ is compact or not.

\begin{theorem}\label{thm2}
Let $G$ be a connected real Lie group, $R$ ist radical, $N$ its
nilradical
 and
$S=G/R$.

If $S$ is non-compact, then both  $\Delta_k$ and
$G^k\setminus\Delta_k$ are of infinite measure for all $k>1$.

If $S$ is compact, then there exists a natural number
 $\delta_G$ such that
$G^k\setminus\Delta_k$ has measure zero for all $k\le\delta_G$
and $\Delta_k$ has measure zero for all $k>\delta_G$.

Furthermore the number  $\delta_G$ has the following properties:
\begin{itemize}
\item
$\delta_G=0$ if $G$ is
compact.
\item
$\delta_G\le 1$ unless $G$ is nilpotent.
\item
$\delta_G\le\dim G/G'$.
\end{itemize}
\end{theorem}

We also derive some related results concerning density
of generic subgroups.
\begin{theorem}\label{z-dense}
Let $G$ be a connected semisimple linear algebraic group.
Then for every $k\ge 2$ there is a subset $Z_k\subset G^k$
of measure zero such that for every $g=(g_1,\ldots,g_k)\in 
G^k\setminus Z_k$ the group $\left<g_1,\ldots,g_k\right>$ 
generated by $g_1,\ldots,g_k$
is Zariski dense in $G$.
\end{theorem}


\begin{theorem}\label{dense}
Let $G$ be a connected semisimple real Lie group.

There exists an open neighbourhood $W$ of $e$ in $G$
and for every $k\ge 2$ a subset $Z_k\subset W^k$ of measure zero such that
$\left<g_1,\ldots,g_k\right>$ is dense in $G$ for all
$(g_1,\ldots,g_k)
\in W^k\setminus Z_k$.
\end{theorem}

In particular, for every connected semisimple Lie group $S$
there do exist two elements $g_1,g_2\in S$ such that $g_1$ and
$g_2$ generate a dense subgroup of $S$.
This may be regarded as a Lie analog for a theorem in the
theory of finite simple groups which states that every finite simple
group  is generated by two elements (\cite{AG}).


This paper is organized as follows:
First we provide some examples, and introduce some basic facts
and notations.
We show that $\Delta_k$ is always measurable.
We prove that under certain assumptions invariant sets of positive
measure are automatically of infinite measure.
Investigating Cartan subgroups we derive our above mentioned results
on $\Delta_1$. Subsequently we prove the theorem on $\Delta_k$ 
($k\ge 2$) using a variety of different techniques ranging from
Zassenhaus neighbourhoods over amenable groups to proximal elements.

\subsection{Proof of the main results}
The first two statements of Theorem~\ref{thm1} follow
from prop.~\ref{CSG-generic}, 
prop.~\ref{prop9} and prop.~\ref{prop10}.
If $G$ contains both Cartan subgroup with compact connected
components and Cartan subgroups with non compact
connected components, then prop.~\ref{csg-conj} implies
that $G/R$ is non-compact. This allows us to invoke
cor.~\ref{no-fin-vol} in order to conclude that
both $\Delta_1$ and $G\setminus\Delta_1$ have infinite measure.

Theorem~\ref{thm2}: For $S$ non compact the statement follows
from cor.~\ref{zhaus} and cor.~\ref{prox}
combined with cor.~\ref{no-fin-vol}. If $S$ is compact and
positive dimensional then $\Delta_k(G)$ is a set of measure
zero for all $k\ge 2$ by prop.~\ref{amen}.
Theorem~\ref{thm1} combined with prop.~\ref{csg-conj} implies that
either $\Delta_1(G)$ or $G\setminus\Delta_1(G)$ is a set of measure
zero for $S$ compact and positive dimensional.
If $G$ is solvable, but not nilpotent, then the assertions
of the theorem follows from prop.~\ref{csg-conj} combined
with prop.~\ref{solv}.
For nilpotent $G$ the theorem follows from prop.~\ref{nil1}
and prop.~\ref{nil2}. Finally, the assertion that $\delta_G=0$
for compact $G$ follows from cor.~\ref{tors}.

Theorem~\ref{z-dense} follows from prop.~\ref{z-dense2}.

For every connected semisimple Lie group $S$ the adjoint
representation $Ad$ has the property that $Ad(S)$ is linear algebraic
and the kernel (which is the center of $S$) is discrete.
Together with prop.~\ref{alg-dense}
and lemma~\ref{dense2} 
this implies theorem~\ref{dense}.

\section{Examples of $\Delta_k$}
For abelian and certain nilpotent connected Lie groups 
an explicit description
is easy.
\begin{example}
Let $G=(\R^d,+)$. Then $\Delta_k$ is the set of all
$(v_1,\ldots,v_k)$ such that
$\dim_{\Q}\left<v_1,\ldots,v_k\right>_{\Q}
=\dim_{\R}\left<v_1,\ldots,v_k\right>_{\R}$.
In particular $\Delta_k$ is of measure zero for $k>d$ and
$G^k\setminus\Delta_k$ is of measure zero for $k\le d$.
\end{example}
\begin{example}
Let $G=\left(S^1\right)^n$ with $S^1=\{z\in\C^*:|z|=1\}$,
let $E$ denote the set of all roots of unity,
i.e., $z\in E$ if $z^N=1$ for some $N\in\N$ and
let $G_{tors}=E^n$. Then $\Delta_k=G_{tors}^k$ for all $k\in\N$ and
$\Delta_k$ has measure zero for all $k\in\N$.
\end{example}
\begin{example}
Let $V$ be a real vector space with antisymmetric bilinear form
$B(\cdot,\cdot)$.
The associated Heisenberg group $G$ is $V\times\R$ as manifold with
the group structure given by $(v,x)\cdot(w,y)=(v+w, x+y+B(v,w))$.
Then the group generated by two elements $(v,x)$ and $(w,y)$
is
\[
\{ (nv+mw, nx+my+(nm+k)B(v,w): n,m,k\in\Z \}.
\]
Hence $G^k\setminus\Delta_k$ has measure zero for $k=1,2$.
On the other hand, if $(v_i,x_i)_{i\in I}$ is a family of elements
in $G$, then the group generated by these elements contains
$2B(v_i,v_j)$ for all $i,j\in I$.
This implies that
$\Delta_k$ is of measure zero for $k>2$.
\end{example}
\begin{example}
Let $\alg g$ be the four dimensional nilpotent Lie algebra
given by $[X_1,X_2]=X_3$ and $[X_1,X_3]=X_4$.

Then $\Delta_k$ is a set of measure zero for all $k\ge 2$
for the associated simply-connected Lie group.
\end{example}
Based on free Lie algebras, for every $k\in\N$ it is possible
to construct a nilpotent Lie group such that $\Delta_k$ is
of positive measure.

For semisimple Lie groups there are many partial results.
\begin{example}
There is a complete description of $\Delta_2$ for $SL(2,\R)$,
obtained by J.~Gilman (\cite{Gil}).
\end{example}
\begin{example}
For $SL_2(\C)$ there is the famous inequality of J\o rgensen
(\cite{J}):
If $(A,B)\in\Delta_2$ for $G=SL_2(\C)\simeq \tilde{SO}(3,1)$, then either
the inequality 
\[
| (\trace A)^2-4| + | \trace (ABA^{-1}B^{-1})-2 | \ge 1
\]
is fulfilled
or $A$ and $B$ generate a subgroup of very special kind, called ``elementary''
subgroup. It is
easily verified that the set of all $(A,B)$ generating an
elementary subgroup is a set of measure zero.
Thus J\o rgensen's inequality implies that $G^2\setminus\Delta_2$
is of positive measure for $G=SL_2(\C)$.
Results generalizing J\o rgensen's inequality
have been obtained for $SO(n,1)$ ($n$ arbitrary),
see \cite{CC},\cite{FFJ},\cite{Martin}.
\end{example}

\begin{example}
Let $G$ be a compact Lie group.
Every discrete subgroup of $G$ is finite.
Hence $\Delta_1(G)=G_{tors}=
\{g\in G: g^n=e\, \exists n \}$.

If $G$ is connected or nilpotent, then $\mu(\Delta_1(G))=\mu(G_{tors})=0$
(see corollary \ref{tors} and lemma \ref{nil-tors} below).

If $G$ is neither connected nor nilpotent, then $G_{tors}$
may be a set of positive measure.
For instance, consider the compact solvable group
 
\[
\left\{
\begin{pmatrix}
\lambda & 0 \\
0 & \lambda^{-1}
\end{pmatrix}: \lambda\in S^1
\right\}
\cup
\left\{
\begin{pmatrix}
0 & \lambda^{-1}\\
-\lambda & 0 
\end{pmatrix}: \lambda\in S^1
\right\}.
\]
This compact group has two connected components and $g^2=e$ for every
element $g$ in the connected component not containing the neutral
element.
\end{example}

\section{Infinity of volume}
In order to show that certain sets have infinite volume with
respect to Haar measure, we proceed as follows: We first show
that these sets have positive measure. Then we apply a proposition
to be deduced in this section which implies that often sets of
positive measure in Lie groups have automatically infinite measure if they
are invariant under conjugation.

\begin{lemma}
Let $V$ be a affine variety defined over $\R$, $U$ a unipotent group
 acting on $V$
morphically (also over $\R$) and $\mu$ a $U(\R)$-invariant 
probability
measure on $V(\R)$ of Lebesgue measure class.

Then $\mu$ is supported in the fixed point set $V^U(\R)$.
\end{lemma}
\begin{proof}
If $V^U\ne V$, then there exists an $\R$-regular function $f$ on $V$
such $f\not\in\R[V]^U$ but such that $U$ stabilizes the vector space
$M$ spanned by $f$ and $\R[V]^U$.
For every one-parameter-subgroup $\alpha_t$ of $U$
there is a $U$-invariant regular function $g=g_\alpha$ such that
$\alpha_t^*f=f+tg$.
Then $f/g$ is an $\R$-regular map from
$V_g=\{g\ne 0\}$ to $\R$ which is equivariant for $\R$ acting
on $V$ via $\alpha_t$ and on itself by addition.
Now every $U$-invariant subset of $V_g$ is mapped surjectively
onto $\R$. Hence $V_g$ can not carry any non-trivial 
$U$-invariant finite measure. Therefore $\mu$ must be supported
inside the intersection of the zero sets of all the $g_\alpha$.
This intersection is again an affine $U$-variety, but
of strictly smaller dimension  that $V$ (unless $V=V^U$).
Arguing by induction, we may therefore deduce that the support
of $\mu$ must be contained in $V^U$.
\end{proof}
% Funzt genauso fuer jeden lokalen Koerper k.
\begin{proposition}
Let $G$ be a connected real Lie group and $R$ its radical.
Assume that there exists a probability measure $\mu$ on $G$
which is invariant under $G$ acting on itself by conjugation.

Then $G/R$ is compact or $\mu$ is concentrated on $R$.
\end{proposition}
\begin{proof}
Assume that $G/R$ is not compact.
Then there exists a real simple Lie group $S_0$ with trivial center
and a surjective Lie group homomorphism
$\tau:G\to S_0$.
The measure $\mu$ on $G$ induces a probablity measure $\mu_0$ on $S_0$
via $\mu_0(A)=\mu(\tau^{-1}(A))$ for $A\subset S_0$.
Note that $S_0$ is linear algebraic, because it is simple and
center-free. Hence $S_0$ is an affine $\R$-variety.
Thus we may invoke the above lemma and conclude that 
the support of $\mu_0$ is concentrated at the intersection
of the fixed point sets of all unipotent subgroups of $S_0$
acting on $S_0$ by conjugation.
Since $S_0$ is simple and non-compact, it is generated by its
unipotent subgroups. Hence this intersection is simply the center
of $S_0$, i.e. $\mu_0$ is concentrated at $\{e\}$.
Thus the support of $\mu$ is contained in $R$.
\end{proof}

\begin{corollary}\label{no-fin-vol}
Let $G$ be a connected real Lie group $G$ with radical $R$,
$k\in\N$, $\mu$ the product measure of the Haar measures on $G^k$
and $E\subset G^k$ a measurable set which is invariant under
the diagonal $G$-action on $G^k$ via conjugation.
Assume that $G/R$ is not compact.

Then either $\mu(E)=0$ or $\mu(E)=\infty$.
\end{corollary}

\begin{proof}
Assume the contrary, i.e., the existence of such an invariant
set $E$ with $0<\mu(E)<+\infty$.
Since $E$ is invariant and of finite volume, it follows
that conjugation by an element of $G$ can not change the volume,
i.e., $G$ must be unimodular.
Hence the Haar measure is invariant under conjugation.

Now let $\pi_1:G^k\to G$ denote the projection onto the first factor.
We define a probability measure $\eta$ on $G$ by
$\eta(X)=\mu(\pi^{-1}(X)\cap E)$.
Note that $R\ne G$ and therefore $\mu(R\times G^{k-1})=0$, implying
$\eta(R)=0$
Thus $\eta$ is a probability measure on $G$ which is invariant
under conjugation and such that $\eta(R)=0$.
This contradicts the preceding proposition.
\end{proof}


\section{Preparations}
In this article, $F_n$ always denotes the free group with $n$ elements
$\xi_i$. There is a natural map $\alpha:F_n\times G^n\to G$
defined as follows:
To every element $x=(g_1,\ldots,g_k)\in G^k$ we associated a group
homomorphism $\phi_x:F_n\to G$ induced by $\phi_x:\xi_i\mapsto g_i$.
We define $\alpha(\xi,x)=\phi_x(\xi)$.
Furthermore, for every $R\in F_n$ we define
a continuous map $\zeta_R:G^k\to G$ via
$\zeta_R(x)=\phi_x(R)$.

\begin{proposition}\label{measurable}
Let $G$ be a topological group fulfilling the second axiom of
countability.

Then $\Delta_k(G)$ is measurable (with respect to the Borel
algebra generated by the topology) for all $k\in\N$.
\end{proposition}
\begin{proof}
Note that $x=(g_1,\ldots,g_k)\not\in\Delta_k$ iff the subgroup
generated by the $g_i$ is not discrete and that this condition
is equivalent to the property that $e$ is not an isolated point
in the group generated by the $g_i$.

Let $(V_j)_{j\in J}$ be a neighbourhood basis of the topology of $G$
at $e$. By assumption $J$ may be choosen to be countable.

Then
\[
G^k\setminus\Delta_k
=
\cap_{j\in J}\cup_{R\in F_n}\,\zeta_R^{-1}
\left( V_j\setminus\{e\}\right)
\]
Since both $J$ and $F_n$ are countable, it follows that
$\Delta_k$ is measurable.
\end{proof}

Subgroups of discrete groups are again discrete.
Hence $\Delta_{k+l}\subset\Delta_k\times G^l$ for all $k,l\in\N$,
implying the observation stated below.

\begin{observation}
If $\Delta_k$ is of measure zero, then $\Delta_l$ is of measure
zero for all $l\ge k$.
\end{observation}

Real Lie groups are intrinsically analytic, i.e., they admit
a structure as a real analytic manifold such that the
maps defining the group structure are real analytic.
Furthermore, every continuous group homomorphism 
between real analytic Lie groups is
automatically real analytic.
This is useful for our purposes, since the identity principle for
real analytic maps has the following consequence:
\begin{lemma}
Let $f:M\to N$ be a real analytic map between 
connected real analytic manifolds.
Assume that $f$ has maximal rank somewhere.

Then $f^{-1}(S)$ is of measure zero for every subset $S\subset N$ of
measure zero.
(where the measure is defined with respect to 
(otherwise arbitrary) everywhere positive volume forms on $M$ and
$N$.)
\end{lemma}
\begin{corollary}\label{shat}
Let $G$ be a connected Lie group, $S$ a subset of measure zero
and
\[
\hat S = \{ g\in G: g^n\in S\ \exists n\in\N\}.
\]
Then $\hat S$ is a set of measure zero.
\end{corollary}
\begin{proof}
For every natural number $n$ the map $\phi_n:g\mapsto g^n=g\cdot\ldots\cdot g$
is a real analytic map which has maximal rank at $e$.
Thus $\hat S=\cup_{n\in\N}\phi_n^{-1}(S)$ is a countable union of
sets of measure zero and therefore itself a set of measure zero.
\end{proof}
\begin{corollary}\label{tors}
Let $G$ be a connected Lie group.

Then $G_{tors}=\{g\in G: g^n=e\ \exists n\}$
is a set of measure zero.
\end{corollary}

\begin{lemma}\label{nil-comp}
Let $H$ be a connected nilpotent Lie group.
Then there exists a unique maximal compact subgroup $K\subset H$.
Furthermore
$K$ is central in $H$.
\end{lemma}
\begin{proof}
Let $K$ be a maximal compact subgroup. The adjoint representation
of $K$ on $\Lie(H)$ is completely reducible. Combined with
$Ad(g)$ being unipotent for all $g\in H$ this implies that $K$ is
central in $H$. Uniqueness of $K$ follows from $K$ being central,
because maximal compact subgroups in a connected Lie group
are all conjugate.
\end{proof}
% Remark: If G nilpotent with G/G^0 fin. gen., then there still
% exists a unique maximal compact subgroup which is normal.
\begin{lemma}\label{cncn}
Let $G$ be a connected Lie group and $H$ a connected normal compact
nilpotent Lie subgroup.

Then $H$ is central in $G$.
\end{lemma}
\begin{proof}
The complete reducibility of representations of compact groups
implies that there exists an $Ad(H)$-stable  
vector subspace $V\subset\Lie G$ such that $\Lie G=V\oplus \Lie H$
as a vector space. Since $V$ is $Ad(H)$-stable, it is clear
that $[V,\Lie H]\subset V$. On the other hand, $[V,\Lie H]
\subset\Lie H$, because $H$ is normal. Thus $[V,\Lie H]=\{0\}$.

Complete reducibility of representations of compact groups can 
also be used to deduce that a compact connected nilpotent Lie group
is necessarily commutative. Together with $[V,\Lie H]=\{0\}$
this implies that $H$ is central.
\end{proof}

\begin{lemma}\label{nil-tors}
Let $K$ be a compact nilpotent Lie group and $K_{tors}$
its set of torsion elements.

Then $K_{tors}$ is a set of measure zero.
\end{lemma}
\begin{proof}
For every every natural number $n\ge 2$ the set
\[
\{g\in K:g^n=e\}
\]
is a closed real analytic subset of $K$.
Therefore either $K_{tors}$ is a set of measure zero,
or $K_{tors}$ contains a whole connected component of $K$.

Let $C$ denote the connected component of the center of $K^0$.
Since $K$ is nilpotent, this is a positive dimensional group,
i.e. $C\simeq (S^1)^g$ with $g>0$.
Since $C$ is central in $K^0$, there is an action of the finite
group $K/K^0$ on $C$. Again using the fact that $K$ is nilpotent
it is clear that this action must be trivial, i.e., $C$ is central in
$K$. Let $\alpha\in C\setminus C_{tors}$.
Now for any $k\in K$ both elements $k$ and $k\alpha$ are contained
in the same connected component of $K$ and they cannot be
simultaneously torsion elements. Therefore no connected component
of $K$ is contained in $K_{tors}$.
\end{proof}

\section{Relations}
We start by introducing some notation.
\begin{definition}
Let $G$ be a group and $k$ a natural number.
An element $R\in F_k$ is called a
\begin{em} relation\end{em} for $v=(g_1,\ldots,g_k)$ 
if $\zeta_R(v)=e$.

An element $R\in F_k$ is called a
\begin{em}general relation\end{em} for $G$
if $\zeta_R\equiv e$.

The set of all general relations for $G$ is denoted by
$R_k(G)=R_k$.
\end{definition}

For example, $\xi_1\xi_2\xi_1^{-1}\xi_2^{-1}$ is a general relation
for every commutative group. Actually, a group $G$ is commutative
if and only if $R_k=F_k'$ for all $k$ where $F_k'$ denotes the
commutator group of $F_k$.
Similarily, properties like being $m$-step nilpotent or $m$-step
solvable can be translated in conditions on $R_k(G)$.

\begin{lemma}
Let $G$ be a group and $k$ a natural number. Then $R_k(G)$ is
a normal subgroup of $F_k$.
\end{lemma}
\begin{proof}
If $A,B\in F_k$, then $\zeta_{ABA^{-1}}(x)=
\zeta_A(x)\zeta_B(x)\zeta_{A^{-1}}(x)$.
Hence $\zeta_{ABA^{-1}}\equiv e$ if and only if $\zeta_A\equiv e$.

Therefore $R_k(G)$ is normal in $F_k$.
\end{proof}

We will show that generic $k$-tuples $(g_1,\ldots,g_k)$ have no
relations except the general relations of the ambient Lie group.

\begin{definition}\label{sigma}
Let $G$ be a group.
Then we define
$\Sigma_k$ as the set of all $(g_1,\ldots,g_k)\in G^k$ for which
there are more relations than the general relations
of the group $G$.
\end{definition}


\begin{proposition}
Let $G$ be a connected Lie group and $k\in\N$.

Then $\Sigma_k$ is a subset of $G^k$ of measure zero.
\end{proposition}

\begin{proof}
Let $S= F_k\setminus R_k(G)$.
Then $\zeta_A:G^k\to G$ is a non-constant real analytic map
for every $A\in S$. It follows that
\[
\Sigma_k=\cup_{A\in S}\, \zeta_A^{-1}(e)
\]
is a set of measure zero, because $S$ is a countable set and
$\zeta_A^{-1}(e)$ is of measure zero for every $A\in S$.
\end{proof}

Thus a generic finitely generated subgroup
of a connected Lie group  fulfills no
relations except the general relations of the ambient Lie group.
It is therefore useful to determine the general relations
for Lie groups.

\begin{lemma}
Let $G$ be a connected Lie group.
Then $G$ is
commutative resp.~nilpotent resp.~solvable
if and only if every subgroup with two generators has the
respective property.
\end{lemma}

\begin{proof}
The statement is trivial concerning commutativity. In respect to
solvability it follows from the ``Tits alternative''
(see \cite{Tits}).
Thus we only have to show, that given a non-nilpotent connected
Lie group $G$, there exists a subgroup with $2$ generators which
is not nilpotent. Since $G$ is not nilpotent, Ado's theorem implies
that there is an element $v$ in the Lie algebra $Lie(G)$ such that
$ad(v)$ is not nilpotent.
Let $W$ be an irreducible $ad(v)$-sub module of $Lie(G)$ on which
$ad(v)$ is not trivial.
Then either $W$ is real one-dimensional and $ad(v)(w)=\lambda w$
for $w\in W$ with $\lambda\in\R\setminus\{0\}$ or
$V$ is real two-dimensional.
In both cases it is easy to check that $\exp(v)$ and $\exp(w)$
generate a non-nilpotent group for every $w\in W$.
\end{proof}
\begin{corollary}
A connected Lie group $G$ is
solvable resp.\ nilpotent resp.\ commutative if and only if
$F_2/R_2(G)$ has the respective property.
\end{corollary}
In contrast, for non-solvable Lie groups there are no general
relations.
\begin{proposition}
Let $G$ be a connected Lie group and assume that $G$ is not
solvable. Then $R_k(G)=\{e\}$
for all $k\in\N$.
\end{proposition}
\begin{proof}
Consider the adjoint representation $Ad:G\to GL(\Lie G)$.
Since $G$ is a central extension of $Ad(G)$ by the center of $G$,
non-solvability of $G$ implies that $Ad(G)$ is also non-solvable.
Due to ``Tits-alternative'' it follows that for every $k\in\N$
the group $Ad(G)$ contains a free subgroup with $k$ generators.
This subgroup can be lifted to a subgroup of $G$ (because of its freeness).
It follows that $R_k(G)=\{e\}$.
\end{proof}
\begin{lemma}
Let $G$ be a positive-dimensional  Lie group.
Then $R_k(G)$ is contained in the commutator group $F_k'$ of $F_k$
for every $k$.
\end{lemma}
\begin{proof}
The Lie group $G$ must contain a one-parameter subgroup.
Such a one-parameter subgroup is isomorphic to $\R$ or $\R/\Z$.
Both contain subgroups isomorphic to $\Z^k$ for every $k$.
\end{proof}

\begin{corollary}\label{cor-cinq}
Let $G$ be a positive-dimensional abelian connected Lie group.

Then $R_k(G)=F'_k$ for every $k\in\N$.
\end{corollary}

\section{Cartan subgroups}
We recall the notion of Cartan subalgebras and Cartan subgroups
for arbitrary Lie groups.
As standard references we use 
\cite{Bou} and \cite{Enz}.

\begin{definition}
Let $\alg g$ be a finite-dimensional Lie algebra over a field $k$.
A Lie subalgebra $\alg h$ is called {\em Cartan subalgebra}
if it is nilpotent and equals its own normalizer
(i.e. $[x,a]\in\alg h\,\,\forall a\in h$ implies $x\in\alg h$.)

A \begin{em}Cartan subgroup\end{em}
 $H$ of a group $G$ is a maximal nilpotent subgroup
such that $N_G(I)/I$ is finite for every normal subgroup $I$ of $H$ with
$H/I$ finite.
\end{definition}
Since Cartan subgroups are maximal nilpotent, and nilpotency is
inherited by the closure of a subgroup in a topological group,
it is clear that Cartan subgroups are closed for any topology
compatible
with the group structure.
For instance, considering Zariski topology it follows that
in an algebraic group every Cartan subgroup is an algebraic
subgroup.

Every Lie algebra contains Cartan subalgebras.
More precisely, an element $x$ in a Lie algebra $\alg g$ is called
{\em regular} if the multiplicity of $0$ as root of the characteristic
polynomial of $ad(x)$ is minimal. 
Regular elements form a dense open
subset of the Lie algebra.
For every regular element $x$ the
vector subspace 
\[
\{v\in\alg g:ad(x)^n(v)=0\ \exists n\}
\]
is a Cartan subalgebra. 
Conversely, every Cartan subalgebra arises in this way.

For a connected Lie group $G$ every Cartan subgroup is a closed
Lie subgroup such that the corresponding Lie subalgebra of $\Lie(G)$
is a Cartan subalgebra. 
Conversely, given a Cartan subalgebra of the Lie algebra of a Lie
group
$G$, there always exists a Cartan subgroup $H$ of $G$ whose
Lie algebra is the given Lie subalgebra of $\Lie(G)$.

There is also a notion of \begin{em} regular
elements \end{em} for Lie groups:
An element $g$ in a Lie group $G$ is called regular,
if the multiplicity of $1$ as root of the characteristic
polynomial of $Ad(g)$ is minimal.
This definition implies immediately that the set of all non regular
elements in a connected Lie group constitutes a nowhere dense real analytic
subset. In particular this is a set of measure zero.
For a regular element $g$ in a connected 
Lie group $G$ the weight space of $1$
\[
{\mathbf g}_1 = \{v\in\Lie G: (Ad(g)-I)^N(v)=0\ \exists N\}
\]
is a Cartan subalgebra of $\Lie G$ and
the corresponding Cartan subgroup of $G$ can be described as the set
of
all elements $x\in G$ such that $Ad(x)$ stabilizes ${\mathbf g}_1$ and
preserves the weight space decomposition of $\Lie G$ with respect
to the $ad({\mathbf g}_1)$-action on $\Lie G$
(see
 \cite{Neeb}).
Evidently this Cartan subgroup contains the element $g$ with which we
started. Thus every regular element in a connected Lie group
is contained in a Cartan subgroup.
This implies the subsequent statement.
\begin{proposition}\label{CSG-generic}
Let $G$ be a connected Lie group and $W$ the union of all
Cartan subgroups of $G$. Then $G\setminus W$ is a set of measure
zero (with respect to the Haar measure of $G$).
\end{proposition}

\begin{lemma}\label{dersurj}
Let $G$ be a connected Lie group.
Let $G^k$ be the $k$-th derived group (i.e. $G^0=G$ and
$G^{k+1}=[G,G^k]$ for all $k\in\N$) and $G^\infty=\cap_kG^k$.

Then every Cartan subgroup of $G$ maps surjectively on $G/G^\infty$.
\end{lemma}
\begin{proof}
This is an immediate consequence of the fact for every Cartan
subgroup $H$ the associated Lie algebra $\Lie (H)$ can be
described as
\[
\{v\in\alg g:ad(x)^n(v)=0\ \exists n\}
\]
for some regular element $x\in G$.
\end{proof}

Cartan subgroups behave nice under central extensions.
\begin{lemma}\label{red-cen}
Let $G$ be a Lie group, $C$ a connected central subgroup of $G$ and $H$ an
arbitrary subgroup of $G$.

Then $H$ is a Cartan subgroup of $G$ if and only if the
following two conditions are fulfilled:
\begin{enumerate}
\item
$C\subset H$ and
\item
$H/C$ is a Cartan subgroup of $G/C$.
\end{enumerate}
\end{lemma}
\begin{proof}
Centrality of $C$ implies that $H$ is maximal nilpotent if and only
if $C\subset H$ and $H/C$ is maximal nilpotent in $G/C$.
Connectedness of $C$ implies that $C\subset I$ for every
subgroup of finite index $I$ of $H$ provided $C\subset H$.
Hence the assertion is an easy consequence of the definition.
\end{proof}
% This is true even without assuming $C$ to be connected.

\begin{proposition}\label{csg-conj}
Let $G$ be a connected Lie group and $R$ its radical.
Assume that $G/R$ is compact.

Then all the Cartan subgroups of $G$ are conjugate.
\end{proposition}
\begin{proof}
We may consider Lie algebras instead of Lie groups, since
there is a one-to-one correspondance between Cartan subgroups
of $G$ and Cartan subalgebras of $\Lie G$.
Let $\Lie(H_i)$ (with $i=1,2$) be Cartan subalgebras of $\Lie G$.
Then $(\Lie(H_i)+\Lie R)/\Lie R$ are Cartan subalgebras of 
$\Lie G/\Lie H$ (\cite{Dix}) and they are conjugate, because
$G/R$ is compact. Thus there is no loss in generality in
assuming $\Lie I=\Lie H_1+\Lie G=\Lie H_2+\Lie R$.
Now $\Lie H_i\subset\Lie I\subset\Lie G$ implies that
both $\Lie H_i$ are Cartan subalgebras in $\Lie I$.
On the other hand $\Lie I$ is solvable.
Therefore the Cartan subalgebras of $\Lie I$ are conjugate
(\cite{Bou}, VII. \$3, thm.~3).
Hence $\Lie H_1$ and $\Lie H_2$ are conjugate.
\end{proof}

\begin{lemma}\label{csg-pos}
Let $G$ be a connected Lie group and $H$ a Cartan subgroup.

Then $\cup_{g\in G}\,gH^0g^{-1}$ is a subset of $G$ of positive
measure.
\end{lemma}
\begin{proof}
By \cite{Bou}, Ch.VII, \S4, lemme~3 and prop.~7. this set contains
an open set.
\end{proof}

\begin{lemma}\label{nh}
Let $G$ be a connected Lie group and $H$ a Lie subgroup.
Assume that $\dim N_G(H^0)-\dim H>0$.

Then
\[
A=\cup_{g\in G}\, gHg^{-1}
\]
is a set of measure zero.
\end{lemma}
\begin{proof}
Note that $N_G(H^0)=\{g\in G:Ad(g)(\Lie H)=\Lie H\}$.
Hence $N=N_G(H^0)$ is closed (even if $H$ is not). Then
$\pi:G\to G/N$ is a locally trivial fiber bundle.
Let $(U_i)_i$ be a countable family of open subsets of $G/N$
which cover all of $G/N$ and such that there are sections
$\sigma_i:U_i\to G$. Let $M$ be the disjoint union of all $U_i$.
Then $M\times H$ is a manifold with countably many connected
components such that $\dim M<\dim G$ and such that
there exists a surjective differentiable map $f:M\to G$
with $f(M)=A$, namely $f(u,h)=\sigma_i(u)h\sigma_i(h)^{-1}$
for $u\in U_i$ and $h\in H$.
It follows that $A$ has measure zero.
\end{proof}

\section{Compactness of Cartan subgroups}
\begin{proposition}\label{infini}
Let $S$ be a connected semisimple Lie group and $H$ a Cartan
subgroup. Assume that the center of $S$ is infinite.

Then the connected components of $H$ are not compact.
\end{proposition}
\begin{proof}
Let $Z$ denote the center of $S$, let $S_0=S/Z$ and let $K_0$ be a maximal
compact subgroup of $S_0$. Then $K_0=A\cdot U$ where $U$ is
semisimple,
$A$ is central in $K_0$, and $A\cap U$ is finite.
Now $K_0$ is a deformation retract of $S_0$, hence
$\pi_1(S_0)\simeq\pi_1(K_0)$. As a compact semisimple group, $U$ has
a finite fundamental group. Therefore the image of the group
homomorphism $i_*:\pi_1(A)\to\pi_1(S_0)$ induced by the inclusion map
$i:A\to S_0$ is of finite index in $\pi_1(S_0)$. Since $\rho:S\to S_0$
is an infinite covering, it follows that there is a 1-torus 
$S^1\simeq A_1\subset A$ such that $\rho^{-1}(A_1)$ has a connected
component isomorphic to $\R$. 

Now, if $H\subset S$ would be a Cartan subgroup with compact
connected component, then $\rho(H)$ would have compact connected
components, too. In this case $\rho(H)$ would have to be a compact
Cartan subgroup of $S_0$ and be conjugate to a Cartan subgroup of
$K_0$.
Hence $\rho(H)$ would contain a conjugate of $A$ and therefore
$H$ would contain a closed subgroup isomorphic to $(\R,+)$.
Thus there can not exist a Cartan subgroup of $S$ with
compact connected components.
\end{proof}

\begin{proposition}\label{csg-compact}
Let $G$ be a connected Lie group and $H$ a Cartan subgroup.

Assume that $H$ has compact connected components.

Then $H$ is compact.
\end{proposition}

\begin{proof}
Every compact connected normal nilpotent Lie subgroup of $G$ is
central (lemma~\ref{cncn}) and therefore
contained in every Cartan subgroup
(lemma~\ref{red-cen}).
Hence
we may divide $G$ by any such
subgroup and therefore without loss of generality assume that $G$
contains no compact connected normal nilpotent Lie subgroup.

Let $Z$ denote the center of $G$. Its connected component
is contained in $H$, hence $Z^0$ is compact and by the assumption
we just made we derive $Z^0=\{e\}$, i.e., $Z$ is discrete.

Let $N$ be the nilradical of $G$ and $C$ a maximal compact subgroup
of $N$. Connectedness of $N$ implies connectedness of $C$.
Hence lemma~\ref{nil-comp} implies that $C$ is characteristic
in $N$ and therefore normal in $G$. Thus $C=\{e\}$, since we assumed
that $G$ contains no compact normal connected nilpotent Lie subgroup.
It follows that $N$ is simply-connected. Since every Cartan subgroup
maps surjectively on $G/G'$, compactness of a connected component
of a Cartan subgroup implies that $G/G'\simeq R/(G'\cap R)$ is compact.
Hence $N\subset G'$. Since $G'\cap R\subset N$ for every 
connected Lie group $G$, it follows that $N=G'\cap R$.

We will now discuss the adjoint representation of $G$.
Its kernel is the center $Z$ of $G$. 
Since the center is discrete, the fibers of $ad$ are discrete.
Now $Ad$ maps $G'\cap R$ to a unipotent subgroup
of $GL(\Lie G)$.
Thus $Ad(N)$ is simply-connected and closed in $GL(\Lie G)$.
This implies that $Z\cap N=\{e\}$ and that $ZN$ is closed in $G$.
Using compactness of $R/N$ it follows that $R\cap Z$ is finite.

Now let us consider the projection $\pi:G\to G/R$.
The Lie algebra of $\pi(H)$ is a Cartan subalgebra of $\Lie(G/R)$
(\cite{Dix}). Hence the semisimple Lie group $G/R$ contains
a Cartan subgroup with compact connected component.
By the preceding proposition it follows that the center of $G/R$
is finite. Therefore $\pi(Z)$ is finite and consequently
$Z$ itself is finite.

Because $H$ is a Cartan subgroup and $\ker(Ad)=Z$ is central,
it is clear that $Ad(H)$ is a maximal nilpotent subgroup of $Ad(G)$.

The group $U=Ad(G'\cap R)=Ad(N)$ is a unipotent subgroup
of $GL(\Lie G)$, hence algebraic. Let $A$ denote the normalizer
of $U$ in $GL(\Lie G)$. Then $A$ is algebraic and
$U\subset Ad(G)\subset A$. The quotient group $A/U$ is algebraic
and the projection $\rho:A\to A/U$ is an morphism of algebraic
groups. Now $\rho(Ad(G))$ is compact and therefore algebraic.
It follows that $Ad(G)=\rho^{-1}(\rho(Ad(G)))$ is algebraic, too.
The Zariski closure of the nilpotent group $Ad(H)$ in $Ad(G)$ is
likewise nilpotent. But $Ad(H)$ is maximal nilpotent. Hence $Ad(H)$
is an algebraic subgroup of $GL(\Lie G)$.
Together with the finiteness of $Z$ this implies that $H$ has only
finitely many connected components.
Thus compactness of the connected components of $H$ implies
compactness of $H$.
\end{proof}

\begin{remark}
For a Cartan subgroup $H$ with non-compact connected components
it is possible that $H$ has infinitely many connected components.
For instance $G_0=SL(2,\R)$ contains a Cartan subgroup $H$ isomorphic
to the multiplicative group $\R^*$. If $\pi:G\to G_0$ denotes the
universal covering, then $\pi^{-1}(H)$ is a Cartan subgroup
of $G$ which has infinitely many connected components, because
$H$ is simply connected while $\pi_1(G_0)\simeq\Z$ is infinite.
\end{remark}

\section{$\Delta_1$ and the structure of Cartan subgroups}

\begin{proposition}\label{prop9}
Let $G$ be a connected Lie group, $H$ a Cartan subgroup with
non compact connected components, and
$\Omega=\cup_{g\in G}\, gHg^{-1}$.

Then $\mu(\Omega\cap\Delta_1)>0$ and $\mu(\Omega\setminus\Delta_1)=0$.
\end{proposition}

\begin{proof}
Let $K$ be a maximal compact subgroup of $H^0$. Then $K$ is
normal in $H^0$ (lemma~\ref{nil-comp}) and
$S=\cup_{g\in G}\,gKg^{-1}$ is a set of measure zero
(lemma~\ref{nh}). Note that $\overline{\{g^n:n\in\Z\}}$ is compact
for all $g\not\in\Delta_1$. This implies that for every $g\in\Omega
\setminus\Delta_1$ there exists a natural number such that $g^n\in S$.
Now cor.~\ref{shat} implies that $\Omega\setminus\Delta_1$ has measure
zero. On the other hand $\Omega$ has positive measure by
lemma~\ref{csg-pos}. Thus $\Omega\cap\Delta_1$ has positive measure.
\end{proof}

\begin{proposition}\label{prop10}
Let $G$ be a connected Lie group, $H$ a Cartan subgroup
with compact connected components
and $\Omega=\cup_{g\in G}\, gHg^{-1}$.

Then $\mu(\Omega\cap\Delta_1)=0$ and $\mu(\Omega\setminus\Delta_1)>0$.
\end{proposition}

\begin{proof}
From prop.~\ref{csg-compact} it follows that $H$ is compact.
Hence for every $g\in\Omega$ the generated group $\{g^n:n\in\Z\}$
is contained in a compact subgroup of $G$. Thus
$\Omega\cap\Delta_1\subset G_{tors}$
implying that $\Omega\cap\Delta_1$ has measure zero
(corollary~\ref{tors}).
\end{proof}

\section{Zassenhaus neighbourhoods}
We recall the existence of ``Zassenhaus neighbourhoods''.
\begin{definition}
Let $G$ be a Lie group. 
An open neighbourhood 
$U$ of the neutral element $e$ 
is called \begin{em} Zassenhaus neighbourhood\end{em}
if the following assertion is true:

\begin{em}For every discrete subgroup $\Gamma$ of $G$ the intersection
$\Gamma\cap U$ is contained in a connected nilpotent Lie subgroup
of $G$.
\end{em}
\end{definition}

\begin{theorem}[Zassenhaus, see \cite{Zas},\cite{KM}]
Every Lie group contains Zassenhaus neighbourhoods.
\end{theorem}

This has the following consequence:
\begin{corollary}\label{zhaus}
Let $G$ be a  connected Lie group which is not nilpotent and $n\ge 2$.
Then there exists an open neighbourhood $W_k$ of $(e,\ldots,e)$
in $G^k$ and a subset $\Sigma_k\subset W_k$ of measure zero
such that $\left<x\right>$ is not discrete for any
$x=(g_1,\ldots,g_k)\in W_k\setminus\Sigma_k$.
\end{corollary}
\begin{proof}
Let $U$ be a Zassenhaus neighbourhood and $W_k=U\times\ldots\times U$.
Let $\Sigma_k$ be defined as in def.~\ref{sigma}.
Then $\left<x\right>$ is not nilpotent for $x\in
W_k\setminus\Sigma_k$,
since $\left<x\right>\simeq F_k/R_k(G)$ for $x\not\in\Sigma_k$
and $F_k/R_k(G)$ is not nilpotent for a non-nilpotent Lie group
$G$. 
On the other hand, $U$ being a Zassenhaus neighbourhood
implies that $\left<x\right>$ is nilpotent for $x\in W_k\cap\Delta_k$.
Hence $W_k\setminus\Sigma_k \subset G^k\setminus\Delta_k$.
\end{proof}

\section{Amenable Lie groups}
A topological group is called 
\begin{em} amenable \end{em} if it admits a ``left invariant mean''
(see \cite{Gr} for more details of this definition and basic properties
of amenable topological groups).
Compact and solvable topological groups are amenable as well
as extensions of solvable by compact topological groups.
Closed subgroups of amenable groups are amenable.
Free discrete groups are not amenable.

\begin{proposition}\label{amen}
Let $G$ be a connected Lie group, $R$ its radical and assume
that $G/R$ is compact, but not trivial.

Then $\Delta_k$ is of measure zero for all $k\ge 2$.
\end{proposition}
\begin{proof}
Since $G$ is an extension of a solvable group by a compact one,
it is clear that $G$ and every closed subgroup of $G$ must be
amenable.
On the other hand $R_k(G)=\{e\}$, because
$G$ is not solvable.  Since free groups are not amenable,
it follows that $\left<x\right>$ can not be closed in $G$ for
$x\not\in\Sigma_k$.
Therefore $\Delta_k$ is contained in $\Sigma_k$ and has measure
zero.
\end{proof}
\section{Solvable Lie groups}
\begin{proposition}\label{solv}
Let $G$ be a connected solvable Lie group.
Assume that $G$ is not nilpotent.

Then $\Delta_k$ is of measure zero for all $k\ge 2$.
\end{proposition}
\begin{proof}
Let $\Gamma$ be a non-nilpotent discrete subgroup of $G$.
Then its commutator group $\Gamma'$ is contained in the commutator
group $G'$ of $G$. Let $N$ denote the universal cover group of $G'$,
$\pi:N\to G'$ the natural projection, and
$\Gamma_1=\pi^{-1}(\Gamma')$.
Since $N$ is nilpotent and simply-connected, the exponential map
$exp:\Lie(N)\to N$ is a diffeomorphism. It is known that for
every discrete subgroup $\Gamma_1\subset N$ the preimage
$\exp^{-1}(\Gamma_1)\subset\Lie N$ spans a finite-dimensional
$\Q$-vector $W$ subspace of $\Lie N$. The $\Gamma$-action on $G$
by conjugation preserves $\Gamma'$. Hence there is an induced
action on $N$ and $\Lie(N)=\Lie(G')$, for simplicity denoted by $Ad$.
Evidently $Ad(\Gamma)$ stabilizes $W$.
Since $\Gamma$ is not nilpotent, the $Ad(\Gamma)$-action on
$W$ can not be unipotent, i.e., there must be an element
$\gamma\in\Gamma$ such that $Ad(\gamma)$ considered as $\Q$-linear 
endomorphism of the $\Q$-vector space $W$ has a non zero eigenvalue
$\lambda$. This number $\lambda$ is contained in an algebraic
extension field of $\Q$. As a consequence one of the eigenvalues
of $Ad(\gamma)$ considered as a $\R$-linear transformation of
$\Lie(N)$ must be a non-zero algebraic number.
However, the set of all algebraic numbers in $\C$ is countable,
hence this set has measure zero. From this fact it is easily deduced
that a generic finitely generated subgroup is not discrete.
\end{proof}

\section{Proximal elements}

Proximal elements where utilized by Tits in proving what is now
commonly called the ``Tits alternative''.
Their usage is based on a freeness condition. As explained
in \cite{JW} this criterion can be modified to check for
freeness and discreteness.

\begin{observation}
Let $X$ be a topological space, $G$ a topological group acting
continuously on $X$.

Assume that there exist families
of open subsets $(V_i^+)_{i\in I}$, $(V_i^-)_{i\in I}$
of $X$ with $i\in\{1,\ldots,k\}$
such that the closures of all these open sets are compact and
mutually disjoint. Furthermore let $p\in X$ be an element not
contained in the closure of any of these open sets.

Let $W$ denote the set of all $k$-tuples $x=(g_1,\ldots,g_k)\in G^k$
such that 
\begin{align}
& g_i(p)\in V_i^+  \\
& g_i^{-1}\in V_i^- \\
& g_i(\bar V_j^+\cup\bar V_j^-)\subset V_i^+ \\
& g_i^{-1}(\bar V_j^+\cup\bar V_j^-)\subset V_i^- \\
& g_i(\bar V_i^+)\subset V_i^+ \\
& g_i^{-1}(\bar V_i^-)\subset V_i^- 
\end{align}
for all $i,j\in I$, $i\ne j$.

Then 
\begin{enumerate}
\item
$W$ is an open subset in $G^k$,
\item
For every $x=(g_1,\ldots,g_k)\in G^k$ the group
$\left<g_1,\ldots,g_k\right>$ is a free and discrete subgroup of $G$.
\end{enumerate}
\end{observation}
The openness of $W$ follows form the simply fact that
\[
\{g\in G:g(K)\subset\Omega\}
\]
is open in $G$ for every compact subset $K\subset X$ and every
open subset $\Omega\subset X$.

Furthermore the construction ensures that $\gamma(p)$
is contained in the union $A$ of the closures of the sets $U_i^+$ and
$U_i^-$ with $i$ running through $I$
for every non-trivial expression of the form
\[
\gamma=g_{i_1}^{n_1}\cdot\ldots\cdot g_{i_N}^{n_N}.
\]
It follows that no such expression is trivial (because 
$\gamma(p)\in A\not\ni
p$) and no sequence of such expressions converges in $G$ to $e$
(because no sequence in $A$ converges to $p$, since $A$ is closed
and $p\not\in A$). Therefore the generated subgroup of $G$ is free
and discrete.

The following existence result is taken from an earlier paper
of the author, see \cite{JW}.
\begin{proposition}
Let $S$ be a connected non compact semisimple linear algebraic group
(defined over $\R$ or $\C$).

Then for every $k\in\N$
there exists an action of $S$ on a projective space $X$,
a point $p\in X$ and families $V_i^+$ and $V_i^-$ as required
in the above observation such that this open set $W$ is non empty.
\end{proposition}
\begin{corollary}\label{prox}
Let $G$ be a connected Lie group and $k\in\N$. Assume that $G$ contains
a non compact semisimple Lie subgroup.

Then there exists an open subset $U$ of $G^k$
such that $u_1,\ldots,u_k$ generate a free discrete subgroup of $G$
for all $u=(u_1,\ldots,u_k)\in G^k$.
\end{corollary}
\begin{proof}
Let $R$ denote the radical. Then $\pi:G\to G/R$ restricted to $S$
induces a group homomorphism $S\to\pi(S)$
with discrete fibers.
Compact semisimple Lie groups have finite fundamental groups,
therefore non-compactness of $S$ implies that $G/R$ is non compact.
It follows that $Ad(G/R)$ is a non compact semisimple linear
algebraic group.
Thus we can apply the preceding proposition to $Ad(G/R)$.
\end{proof}

\section{Nilpotent Lie groups}
\begin{proposition}\label{nil1}
Let $G$ be a connected nilpotent Lie group and assume that
$\Delta_k$ has positive measure.
Then $G^k\setminus\Delta_k$ has measure zero.
\end{proposition}
\begin{proof}
Let $K$ be a maximal compact subgroup of $G$. Then $K$ is central in
$G$
(lemma~\ref{nil-comp}) and $G/K$ is a simply-connected nilpotent 
Lie group. Hence $G/K$ admits a unique structure as a real
unipotent linear algebraic group.

Let $\Lambda=F_k/R_k(G)$ and $\Delta_k^*$ be the set of all
$(g_1,\ldots,g_k)\in G^k$ such that the group generated by the $g_i$
is isomorphic to $\Lambda$ and such that in addition
the group generated by the $g_i$ has trivial intersection with $K$. 
Then $\Delta_k^*\setminus\Delta_k$
is a set of measure zero.
If $\Delta_k$ has positive measure, then $\Delta_k^*$ is not empty,
and there is an embedding $i$ of $\Lambda$ into $U=G/K$ as
a discrete subgroup.
By Malcev theory there is a real algebraic subgroup $V$ of $U$ such that
$i(\Lambda)=V(\Z)$.
Furthermore every group homomorphism from $\Lambda\simeq V(\Z)$ to $U$
is induced by an algebraic group homomorphism of $V$ to $U$ which in
turn corresponds to a Lie algebra homomorphism from $\Lie V$ to $\Lie
U$.
Thus we obtain amap $\eta:F_k\to F_k/R_k(G)=\Lambda\to\Lie V$.
Choose a finite set $\alpha_1,\ldots,\alpha_d\in F_k$ such that the
$\eta(\alpha_i)$ constitute a vector space basis of $\Lie V$
(This can be done, since $i(\Lambda)$ is cocompact in $V$.).

Now for every $g=(g_1,\ldots, g_k)\in G^k$, we obtain a group
homomorphism from $\Lambda=F_k/R_k(G)$ to $U$ which is induced
from a Lie algebra homomorphism from $\Lie V$ to $\Lie U$.
Thus the subgroup of $U$ generated by the $g_iK$ in $U=G/K$
is discrete and isomorphic to $\Lambda$ if and only the
associated Lie algebra homomorphism is injective.
The latter is condition expressible in determinants of images
of the $\eta(\alpha_i)$.
Therefore the projection of $\Delta_k$ in $(G/K)^k$
contains the complement of a closed
real algebraic subvariety of $(G/K)^k$.
It follows that $G^k\setminus\Delta_k$ has measure zero.
\end{proof}

\begin{proposition}\label{nil2}
Let $G$ be a connected nilpotent Lie group and assume
that $\Delta_k$ has positive measure.

Then $\dim G/G'\ge k$.
\end{proposition}
\begin{proof}
We may divide $G$ by its maximal compact subgroup $C$
(which is normal in $G$, see lemma~\ref{nil-comp}) and
thereby assume that $G$ is simply-connected.
In this case $G$ carries the structure of a real unipotent
group in a natural way.

By assumption there is a discrete subgroup $\Gamma\subset G$
with $\Gamma\simeq F_k/R_k(G)$. Then
$\rank_{\Z}(\Gamma/\Gamma')\ge k$ by cor.~\ref{cor-cinq}.
Using ``Malcev theory'' (\cite{Malcev}) it follows that there
is a connected Lie subgroup $H\subset G$ such that $H/\Gamma$ is
compact and $\dim H/H'= \rank_{\Z}(\Gamma/\Gamma')\ge k$.

Now $G/G'$ is a real vector space, and, due to the genericity of
$\Gamma$
we may assume that the real vector subspace of $G/G'$ spanned
by the image of $\Gamma$ is of real dimension
$\min\{\dim G/G',k\}$.

Assume $\dim G/G' >k$. Then the image of $\Gamma$ generates $G/G'$
as a real vector space. It follows that $H$ maps surjectively
onto $G/G'$. However, for a subgroup $H$ of a nilpotent group $G$
the equality $HG'=G$ already implies $H=G$.
Thus 
\[
\dim H/H'=\dim G/G'>\rank_{\Z}(\Gamma/\Gamma')=\dim H/H'
\]
which is absurd.
\end{proof}

\section{Density results}
\begin{lemma}
Let $S$ be a connected semisimple linear algebraic group.
Let $\Omega$ denote the set of all elements $g\in S$ such that
the Zariski-closure of $\{g^n:n\in\Z\}$ is a Cartan subgroup
of $S$.

Then $S\setminus\Omega$ is a set of measure zero.
\end{lemma}
\begin{proof}
For every semisimple element $g$ of $S$ the
Zariski closure of $\{g^n:n\in\Z\}$ is a
commutative reductive algebraic group.
Cartan subgroups of semisimple linear algebraic groups
are connected commutative reductive algebraic groups.
Commutative reductive algebraic groups have only countably
many algebraic subgroups.
Every connected commutative reductive algebraic subgroup
of $S$ is conjugate to a subgroup of a Cartan subgroup of $S$.
Now fix a Cartan subgroup $T$ of $S$.
If $g$ is a semisimple element of $S$ such that the Zariski
closure of $\{g^n:n\in\Z\}$ is not a Cartan subgroup,
then a power $g^n$ is conjugate to an element in a 
proper algebraic subgroup of $T$. 
Now
lemma~\ref{shat} combined with lemma~\ref{nh} implies the statement
together with the fact that the set of all non-semisimple elements
of $S$ is contained in an algebraic subvariety of $S$.
\end{proof}


\begin{lemma}
Let $S$ be a connected semisimple linear algebraic group and
$T$ a Cartan subgroup.
Then there exists a subset $C_T$ of measure zero such that for
every $g\in S\setminus C_T$ the group generated by $T$ and $g$
is Zariski dense in $S$.
\end{lemma}
\begin{proof}
It is well-known for every non-zero weight the weight space of
$Ad(T)$ acting on $\Lie S$ is one-dimensional.
This implies that every Lie subalgebra of $\Lie S$ containing
$\Lie T$ is a direct sum of $Ad(T)$ weight spaces.
It follows that there exist only finitely many connected
Lie subgroups of $S$ containing $T$.

A semisimple Lie group has no normal subgroups except
for the products of its simple factors.
For this reason a connected Lie subgroup $H$ of $S$
containing $T$ is not normal in $S$ (unless $H=S$).
Hence $N_S(H)\ne S$ for such $H$.
Therefore
\[
C_T=\cup \{N_S(H^0) : \text{ $H^0$ connected, $T\subset H^0$} \}
\]
is a finite union of proper submanifolds of $S$ and thus a
set of measure zero.
Now $C_T$ contains every closed Lie subgroup $H$ with $T\subset H\subset
S$. Hence for every $g\in G\setminus C_T$ 
the subgroup of $S$ generated by $T$ and
$g$
is Zariski dense (in fact dense) in $S$.
\end{proof}

\begin{proposition}\label{z-dense2}
Let $S$ be a connected semisimple linear algebraic group.
Then there exists a subset $W\subset S\times S$ such that
$S\times S\setminus W$ is a set of measure zero and
for every $(g_1,g_1)\in W$ the subgroup of $S$ generated
by $g_1$ and $g_2$ is Zariski dense in $S$.
\end{proposition}
\begin{proof}
Let $T$ be a Cartan subgroup, $N$ its normalizer in $S$ and
$\Omega$ and $C_T$ as in the preceding lemmata.
Let $\pi:G\to G/N$ denote the natural projection and
let $\sigma:G/N\to G$ be a measurable section.
Let $T_0$ be the set of all elements in $T$ which are not contained
in any proper algebraic subgroup of $T$.
Then there is a measurable bijection
$\phi: G/N\times T_0\to\Omega$
given by
\[
\phi: (x,t)\mapsto \sigma(x)\cdot t\cdot \sigma(x)^{-1}.
\]
Let $\xi:\Omega\to G/N$ be defined by the composition of $\phi^{-1}$
with the projection on the factor $G/N$.
Define
\[
W=\{(g_1,g_2)\in S\times S: g_2\not\in\xi(g_1)\cdot
C_T\cdot\xi(g_1)^{-1}\}.
\]
Then $S\times S\setminus W$ is of measure zero, because $C_T$ is of
measure zero in $S$ and $\xi$ is a measurable map.
Furthermore $(g_1,g_2)\in W$ implies that
$\xi(g_1)g_1\xi(g_1)^{-1}$ generates a Zariski dense subgroup of $T$
and $\xi(g_1)g_2\xi(g_1)^{-1}\not\in C_T$.
Then $\xi(g_1)g_1\xi(g_1)^{-1}$ and $\xi(g_1)g_2\xi(g_1)^{-1}$
generate a Zariski dense subgroup of $S$. Since
\[
x\mapsto \xi(g_1)\cdot x \cdot \xi(g_1)^{-1}
\]
is an automorphism of $S$ as algebraic group, this is equivalent to
the assertion that $g_1$ and $g_2$ generate a Zariski dense subgroup
of $S$.
\end{proof}

\begin{lemma}
Let $G$ be a semisimple linear algebraic group.
Let $G=\Pi_{i\in I}G_i$ be the representation of $G$ as product
of its simple algebraic subgroups.
Assume that $H\subset G$ is a subgroup of $G$ which is dense
in the Zariski topology and such that $\pi_J(H)$ is not discrete
for any subset $J\subseteq I$ where $\pi_J$ denotes the natural
projection $\pi_J:G\to\Pi_{i\in J} Ad(G_i)$.

Then $H$ is dense in $G$ (in its Hausdorff topology).
\end{lemma}
\begin{proof}
Let $\bar H$ denote the closure of $H$ with respect to the
Hausdorff topology.
By assumption $H$ is not discrete. Hence $\dim \bar H>0$.
The connected component $\bar H^0$ is normalized by $H$.
However, the normalizer of $\bar H^0$ equals the set of all $g\in G$
for which $Ad(g)$ stabilizes the vector subspace $\Lie(\bar H)$
of $\Lie G$. For this reason, the normalizer of $\bar H^0$ in $G$
is an algebraic subgroup of $G$. Since $H$ is contained in this
normalizer, it follows that the normalizer equals the whole group
$G$, i.e., $\bar H^0$ is normal in $G$.
Thus $\bar H^0=\Pi_{i\in I\setminus K}G_i$
for some subset $K\subset I$.
But this implies that there is a morphism with finite fibers from
$\bar H/\bar H^0$ to $\Pi_{i\in K}Ad(G_i)$.
Since $\bar H/\bar H^0$ is discrete, it follows that $K$ must be
empty.
This implies $\bar H=G$.
\end{proof}
\begin{proposition}\label{alg-dense}
Let $S$ be a connected semisimple linear algebraic group.
Then there exists an open neighbourhood $W$ of $(e,e)$ in $S\times S$
and a set of measure zero $\Lambda\subset S\times S$
such that $g_1,g_2$ generate a dense subgroup of $S$ for all
$(g_1,g_2)\in W\setminus\Lambda$.
\end{proposition}
\begin{proof}
Choose an open neighbourhood $V$ of $e$ in $S$ in such a way
that $\pi_J(W)$ is contained in a Zassenhaus neighbourhood
of $S_J$ for all $J\subset I$. Let $\Lambda_1$ be the set of
all $(g_1,g_2)$ such that there exists an $i\in I$ such that
$\pi_1(g_1)$, $\pi_i(g_2)$ do not generate a free group in
$Ad(S_i)$. Let $\Lambda_2$ be the set of all  $(g_1,g_2)$ such that
the generated subgroup is not Zariski dense.
Let $\Lambda=\Lambda_1\cup\Lambda_2$.
\end{proof}

\begin{lemma}\label{dense2}
Let $G$ be a connected semisimple Lie group, $Z$ a discrete
central subgroup and $H$ an arbitrary subgroup in $S$.

Then $H$ is dense in $S$ if and only if $ZH$ is dense in $S$.
\end{lemma}
\begin{proof}
The closure $\bar H$ is evidently normalized by $H$ and therefore
normalized by $ZH$ and its closure $\overline{ZH}=S$, i.e., $\bar H$
is a closed normal subgroup of $S$. 
Now $S/\bar H$ is a semisimple Lie group, but the image of $Z$
in $S/\bar H$ is dense and central.
A topological group with a dense central subgroup is
necessarily commutative, because the center is closed. 
Thus we arrive at a contradiction
unless $\bar H=S$.
\end{proof}

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