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\title{A remark on infinite discrete isometry groups}

\author{
W. Hebisch \\
{\small  Mathematical Institute, Polish Academy of Sciences, Wroc\l aw, Poland}
}
\date{December 2006}
\begin{document}
\maketitle
\begin{abstract}
We prove that arbitrary infinite discrete isometry groups of euclidean
space are closely related to crystallographic groups.
\end{abstract}

{\bf Introduction}.  Crystallographic groups are discrete isometry
groups of euclidean space containing large subgroup of translations.
This subgroup of translations appear naturally in applications, but
it is also natural to ask if different types of symmetry are
possible.  In particular, discrete isometry groups containing few
(or no translations) are of some interest.  The topic of
crystallographic groups is classical and was studied quite
extensively.  Also discrete group actions (frequently on
spaces more general than euclidean space) were (and are) subject
of intensive research.  So it is likely that the result contained in this
note are known.  However searching massive literature is
much harder than proving theorems below.  While I make no claim
concerning novelty of the results the exposition still may be of
some interest.  In particular we get easy proof of main part of
\ref{comlem} using (hard) theorem of Gromov.

Our result essentially says that to get an infinite discrete isometry
group of of euclidean space one takes a crystallographic group $G_1$
acting on a subspace.  Then one takes representation of $G_1$ via
isometries on the complementary subspace.  Finally one add a finite
extension.  Since the image of representation in the step two
is (in general) not closed, one has a lot of possible representations.
Still, the overall group structure look simple.

{\bf Acknowledgements.} This note was inspired by the question
Edwin Clark posted in {\tt sci.math.research}.
 
{\bf Notation and preliminaries}  By euclidean space we mean finite
dimensional vector space over ${\cal R}$ equipped with the scalar product.
Scalar product determines norm and metric.
We denote by $O(H)$ group of orthogonal transformations of $H$.  As
usual $O(n)$ (where $n$ is an positive integer) we denote
group of orthogonal transformations of ${\cal R}^n$.

It is well known that each isometry of an euclidean space is a
composition of translation and orthogonal transformation and
the isometry group of an euclidean space is a cross product
of translations with orthogonal group (translation being a normal
subgroup).

If $h_1$ is an isometry of $H_1$ and $h_2$ is an isometry of $H_2$
then the product map: $(x, y) \mapsto (h_1(x), h_2(y))$ is an
isometry of $H = H_1\oplus H_2$ (with the scalar product on $H$
being the sum of scalar products on $H_1$ and $H_2$.  Product
map gives an injective homomorphism from the product of isometry groups of
$H_1$ and $H_2$ into isometry group of $H$.  Not all isometries
are in the image of the product map: all orthogonal transformation
corresponding to product maps have $H_1$ and $H_2$ as invariant
subspaces.  In fact, if an orthogonal transformation corresponding
to an isometry of $H$ leave $H_1$ and $H_2$ invariant then the
corresponding map may be written as a product map.  

Given a subgroup $G$ of the isometry group of $H$ we say that
the action of $G$ splits if there are two orthogonal subspaces
$H_1$ and $H_2$ of $H$ such that $H = H_1\oplus H_2$ and each
isometry $g$ from $G$ can be written as a product of isometries
$g_1$ of $H_1$ and $g_2$ of $H_2$.  If the action of $G$ splits
we say that $g_1$ (respectively $g_2$) is the restriction of $g$
to $H_1$ (respectively $H_2$).   We denote by $\pi_{H_1}$
the group homomorphism mapping $g$ to $g_1$.

We say that a subgroup $S$ of $G$ is discrete if and only if subset
topology on $S$ is discrete.  For subgroups of isometry groups of 
euclidean space an equivalent condition is: intersection of
the $S$-orbit of any $x$ has finite intersection with any compact
set.

{\bf Results}

\begin{theo}
Let $G$ be an infinite discrete subgroup of the isometry group
of ${\cal R}^3$ which contains no translations.  $G$ contains
a finite index subgroup generated by a single element $g$. The
component of
   $g$ in $O(3)$ is a rotation by irrational angle, and the translational
   part along the axis of the rotation is nonzero.
\end{theo}
{\it Proof}: This follows immediately from \ref{tha} below.  In detail,
let $g \in G_1$.  Rotational part of $g$ must be a rotation by irrational angle
(otherwise some power of $g$ would give a nontrivial translation).
So $V$ is just axis of the rotation and is one dimensional.  Consequently,
$G_1$ is generated by a single element and $g$ generate subgroup of
finite index in $G_1$ (hence in $G$).
$\diamond$

\begin{theo}\label{tha}
Let $G$ be an infinite discrete subgroup of the isometry group
of $n$-dimensional euclidean space $H$ which contains no translations.
Then there exists
a subspace $V$
such that the action of $G$ splits into a direct sum of actions on
$V$ and orthogonal complement of $V$.  Moreover there exists
a normal subgroup $G_1\subset G$ of finite index such that the
restriction $\pi_V$ 
of $G_1$ to $V$ is one-to-one and the image of $\pi_V$ is a discrete
subgroup of the group of translation of $V$.  In particular $G_1$
is finitely
generated (of rank less then $n$), torsion free and commutative.
\end{theo}

{\it Proof}: Consider a quotient mapping $\psi$ from $M(H)$ into $O(H)$.
Let $G_{o}$ be the closure of $\psi(G)$ in $O(H)$ and let $T$ be
the connected component of the identity in $G_{o}$. 


\begin{lem}\label{finclo}
If $G_1$ and $G_2$ are subgroups of a Lie group $G_{Lie}$ and
$G_1$ is a subgroup if finite index in $G_2$ then closure of
$G_1$ in $G_{Lie}$ and closure of $G_2$ in $G_{Lie}$ have the
same connected component of the identity.
\end{lem}
{\it Proof}: Let $S_1$ be the closure of $G_1$ in $G_{Lie}$ and
$S_2$ be the closure of $G_2$ in $G_{Lie}$.  Since $G_1$ if of
finite index in $G_2$, there are $x_1,\dots, x_m$ such that
$G_2 \subset \bigcup_{i} x_iG_1$.  So also
$G_2 \subset S_0 = \bigcup_{i} x_i S_1$.  However $S_0$ is closed
(as a finite sum of closed sets), so 
$S_2 \subset \bigcup_{i} x_i S_1$.  Hence,  $S_2$ has the same
dimension as $S_1$.  Now, connected component of the identity
in $S_1$ is contained in the component of the identity in $S_2$.
Since both contained in the components are of the same dimension
they must be equal.
$\diamond$

\begin{lem}\label{disccom}
Let $S$ be a compact Lie group.
There exists a neighbourhood $U$ of the identity $id$ in $S$ such
that if $G$ is a discrete subgroup of $S$ and $g \ne id$ is an element
of $G$ closest to the $id$ then $g$ commutes with $G \cap U$
\end{lem}
{\it Proof}: commutator mapping $(x, y) -> x^{-1}yxy^{-1}$ is smooth
and has value $id$ at $(id, id)$ (where $id$ the identity matrix).
So, in some neighbourhood of the identity in $S$ we have
$d(x^{-1}yxy^{-1}, id) \leq C d(x, id)d(y, id)$
where $d$ is the Riemannian metric in $S$.  If $d(x, id)$ and $d(y, id)$
is small enough we have:
$d(x^{-1}yxy^{-1}, id) < \min(d(x, id), d(y, id)).$
We choose $U$ a Riemannian ball around $id$ such that the above inequality holds
when $x\in U$
and $y \in U$.  Now, if $g \notin U$ then $G\cap U = \{id\}$ and the
claim is trivial.  If $g in U$ and $y in G\cap U$ then
$y_0 = g^{-1}ygy^{-1} \in G$ and
$d(y_0, id) < (d(g, id)$ which (by definition of $g$) is possible only
if $y_0 = id$.  So also in this case $g$ and $y$ commute.
$\diamond$

\begin{lem}\label{comlem}
$T$ is commutative
\end{lem}
{\it Proof}: First, consider the case of finitely generated $G$.
Isometry group of euclidean space is of polynomial growth, so
also $G$ (being a closed subgroup) is of polynomial growth. By Gromow
theorem \cite{Gr} $G$
has nilpotent subgroup $G_1$ of finite index.
Consequently $\psi(G_1)$ and its closure is nilpotent.  By \ref{finclo}
$T$ is equal to the connected component of the identity in the closure
of $\psi(G_1)$, so also
is nilpotent.  However, connected nilpotent subgroup of a compact group
is commutative, which gives our claim for finitely generated $G$.

Let $T_{S}$ be the the connected component of the identity in the closure
of finitely generated subgroup $S$ of $G$, let $T_0$ be $T_{S}$
of maximal dimension and let $S_0$ be a fixed $S$ corresponding to $T_0$.
Note that $T_{G_1} \subset T_0$ for all
finitely generated $G_1$.  Namely, the sum of generators $G_1$ and $S_0$
generate a subgroup $G_2$ such that $T_{G_1} \subset T_{G_2}$
and $T_0 \subset T_{G_2}$.  Since dimensions of $T_0$ and $T_{G_2}$ are
equal (and both are connected) we have $T_0 = T_{G_2}$.  Now, maximality
of $T_0$ implies that $\psi(G)$ normalizes $T_0$.  Since $T_0$ is a
(multidimensional) torus is group of authomorphisms is discrete.
On the other hand inner automorphizm of $\psi(G)$ act trivially on $T_0$ 
if and only if it acts trivially on the Lie algebra of $T_0$.  However,
inner automorphizms of $O(H)$ corresponding to elements of $\psi(G)$ act
on the Lie algebra of $T_0$ via
orthogonal mappings, so they can give
only finitely many authomorphisms of $T_0$.  Hence, passing to
a subgroup of finite index we may assume that $\psi(G)$ centralizes
$T_0$.  Consequently, $T_0$ is a central subgroup of $T$. 

Let $T_1$ be the maximal central torus in $T$ and let $S = T/T_1$.
To finish the proof we need to show that $T_1 = T$ which is 
equivalent to $S$ being trivial.  We claim that nontrivial $S$ 
contradict maximality of $T_1$.  More precisely, we will show
that nontrivial $S$ contains a central one parameter subgroup
(such subgroup would allow to enlarge $T_1$).  Let $\eta$ be
the composition of $\psi$ and the quotient map from $T$ to $S$
(passing if necessary to a subgroup of finite index we may assume
that the image of $\psi$ is contained in $T$).  If there is
a finitely generated subgroup of $G_1$ such that $\eta(G_1)$ is
not discrete, then we can produce central one parameter subgroup
in the same way as in the fist stage of the proof.  So we
need only consider the case when $\eta(G_1)$ is discrete for all
finitely generated $G_1$.  Fix enumeration $j \mapsto x_j$ of 
$G = \{x_1, x_2, \dots\}$
and let $S_j$ be the subgroup of $S$ generated by 
$\eta(x_1), \dots, \eta(x_j)$.  Let $g_j \ne id$ be the element of $S_j$
closest to the identity and let $U$ be a neighbourhood of the identity
in $S$ given by \ref{disccom}.  $g_j$ converge to identity, so by
compactness of the sphere in the Lie algebra of $S$ we can choose natural
numbers $n_j$ such that the mappings $t \mapsto g_j^{[n_jt]}$ (where $[\cdot]$
denotes the integer part) converge to a one parameter subgroup $s(t)$ of $S$
when $j$ goes to infinity.
By \ref{disccom} for $j \geq i$ $g_j$ commutes
with $S_i\cap U$, so also $s(t)$ (as a limit) commutes with $S_i\cap U$.
Since the sum of $S_i$ is dense in $S$ $s(t)$ commutes with all elements
of $U$.  Since $S$ is connected $U$ generates $S$, so $s(t)$ is central.
$\diamond$


By the lemma $G$ contains a subgroup $G_1$ such that
$\psi(G_1) \subset T$.
Since $G$ contains no translations the kernel of $\psi$ is trivial
and $\psi$ is injective.  So $G_1$ is commutative.
We put $V = \{x \in H: \forall_{g \in T} gx = x \}$.  
Since $T$ as a connected component of the identity is a normal
subgroup of the closure of $\psi(G)$, also $\psi(G)$ normalizes $T$
and consequently $V$ is invariant under the action of $\psi(G)$.

This gives the first part of the theorem: the action of $G$ is 
a direct sum of actions on $V$ and its orthogonal complement.

In the sequel (if necessary replacing $G$ by current $G_1$) we
may assume that $\psi(G) \subset T$ and $G$ is commutative.  
Let $G_1$ be
a finitely generated subgroup of $G$ such that
$V = \{x \in H: \forall_{g \in \psi(G_1)} gx = x \}$.
Such subgroups exist: the set of fixed points of $\psi(G_1)$ is a
linear space.  If this set is bigger than $V$, then there is $x \notin V$
fixed by $\psi(G_1)$.  By definition of $V$ we can find an $g\in G$
such that $\psi(g)x \ne x$.  Adding this $g$ to $G_1$ we reduce the dimension
of the set of fixed points of $\psi(G_1)$. After finitely many
such steps we will get set of the same dimension as $V$, hence equal to $V$.
Let $g_1,\dots,g_n$ generate $G_1$ (we fix a set of generators) and
let $W$ be the orthogonal complement of $V$ in $H$.  We claim that there
exists an $\epsilon>0$ such that $\|\psi(g_k)x - x\| \geq \epsilon \|x\|$
for all $x\in W$.  Namely, it is enough to prove the inequality for
$x$ belonging to the unit sphere of $W$.  But then $\|\psi(g_k)x - x\|$
is a positive (since $\psi(g_k)x \ne x$) continuous function on a compact
set, so it is bounded from below by a positive $\epsilon$. 

Now, consider restriction to $W$ of an arbitrary $g\in G$.  We can
write $gx = R_gx + w_g$ where $R_g \in O(W)$ and $w_g \in W$.

\begin{lem}
$\|w_g\| \leq 2\max_{i=1,\dots, n} \|w_{g_i}\|/\epsilon$
\end{lem}
{\it Proof}: $G$ is now commutative, so $gg_i = g_ig$.  We write
$gg_ix = R_g(R_{g_i}x + w_{g_i}) + w_g = R_gR_{g_i}x + R_gw_{g_i} + w_g$,
$g_igx = R_{g_i}(R_gx + w_g) + w_{g_i} = R_{g_i}R_gx + R_{g_i}w_g + w_{g_i}$,
so 
$R_gw_{g_i} + w_g = R_{g_i}w_g + w_{g_i}$
and
$R_{g_i}w_g - w_g = R_gw_{g_i} - w_{g_i}$
Now $\|w_g\| \leq \|R_{g_i}w_g - w_g\|/\epsilon = 
\|R_gw_{g_i} - w_{g_i}\|/\epsilon 
\leq 2\max_{i=1,\dots, n} \|w_{g_i}\|/\epsilon$.
$\diamond$

Finally, consider kernel of $\pi_V$.  If $g \in \ker \pi_V$ then
$g$ restricted to $V$ is an identity.  We can identify the isometry
group of $W$ with the subgroup of isometry group of $H$ which
restricted to $V$ is an identity.  So, we can identify $\ker \pi_V$
with a discrete subgroup of the isometry group of $W$.  By the
lemma above this subgroup is contained in a bounded (hence compact) 
subset, so $\ker \pi_V$ is finite.  It follows that $\pi_V$ is
discrete.   Since $\pi_V(G)$ consists of translation
it is a free abelian group of rank not exceeding the dimension of $V$.
Since $\pi_V(G)$ is free abelian we can build the inverse map giving
us a subgroup $G_1$ of finite index in $G$ such that $\pi_V(G_1)$
is a discrete subgroup of translation group of $V$.  Passing if
necessary to a subgroup of finite index we may assume that $G_1$
is normal in our original $G$. 
$\diamond$

\begin{theo}
The conclusion of \ref{tha} remains valid for arbitrary discrete
subgroups of the isometry group of euclidean space $H$.
\end{theo}
{\it Proof}:  Again, let $\psi$ be the quotient map from isometry group
to $O(H)$.
Consider the subgroup $T$ of translations contained in
$G$.  $T$ is a normal subgroup, so corresponding vectors span a
subspace $W$ of $H$ invariant under the action of $\psi(G)$. 
Consequently, the action of $G$ splits into direct sum of actions
on $W$ and on its orthogonal complement $U$.  Each element of $T$
restricted to $U$ gives identity, so the restriction to $U$ gives
mapping from $G/T$ into isometry group of $U$.  Since $T$ is a
co-compact subgroup of isometry group of $W$ the restriction
$\pi_U$ considered as mapping from $G/T$ is closed and has finite kernel.
Also, the image of $\pi_U$ contains no translations. Namely, if
$\pi_U(g)$ is a translation then $gx = R_gx + u + w$ where $R_g \in O(W)$,
$u \in U$, $w \in W$.  Since $R_g$ normalizes lattice spanned by $T$
some power of $R_g$ equals identity $R_g^k = id$.  Consider now
$g^k = u' + kw$.  $g^k$ is a translation with respect to a vector
not contained in $W$.  This is a contradiction with definition of $W$.
Now we finish the proof applying \ref{tha} to $\pi_U(G)$.

\begin{thebibliography}{00}
\bibitem{Gr}M. Gromov. {\em Groups of polynomial growth and expanding maps.}
Publ. Ihes, 53, 1981, p.53-78.
\end{thebibliography}
\end{document}