% Hyperbolic surfaces with prescribed symmetry
% Daniel Allcock
% April 2005

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\begin{document}

\title[Prescribed symmetry groups]{Hyperbolic surfaces with prescribed
  infinite symmetry groups}
%
\author{Daniel Allcock}
\address{Department of Mathematics\\University of Texas at Austin\\Austin, TX 78712}
\email{allcock@math.utexas.edu}
\urladdr{http://www.math.utexas.edu/\textasciitilde allcock}
%
\date{April 17, 2005}
\thanks{Partly supported by NSF grant DMS-024512.}
%
\begin{abstract}
For any countable group $G$ whatsoever, there is a complete
hyperbolic surface whose isometry group is $G$.  The argument is
elementary. 
\end{abstract}

\subjclass[2000]{Primary 51M09, Secondary 20F99}

\maketitle


Greenberg proved in 1972 \cite{greenberg} that for any finite group
$G$ there is a closed hyperbolic 2-manifold having isometry group
isomorphic to~$G$.  Kojima proved the corresponding result for
hyperbolic 3-manifolds \cite{kojima}.  Following work of Long and Reid
\cite{long-reid},  Belolipetsky and
Lubotzky extended this to hyperbolic manifolds of arbitrary dimension
\cite{BL}.  Our purpose here is to give a generalization in a
different direction, which also provides an elementary proof of
Greenberg's original theorem.

\begin{theorem}
\label{thm}
For any countable group $G$, there is a complete hyperbolic
surface $X$ having isometry group isomorphic to $G$.  When $G$ is
finite, $X$ may be taken to be closed.
\end{theorem}

We build $X$ by gluing together pairs of pants.  A pair of pants is a
surface homeomorphic to the 2-sphere minus three open disks whose
closures are disjoint, equipped with a hyperbolic metric under which
the boundary curves are geodesic. (All boundary curves in this note
are assumed geodesic.)  Up to isometry, there is a
unique pair of pants whose boundary lengths are any three given
positive numbers.  The standard way to build pairs of pants is by
doubling right-angled hexagons; see \cite{thurston}.  This
construction makes the following lemma obvious.

\begin{lemma}
\label{lem}
There exists $\e>0$ such that if $P$ is a pair of pants with all its
boundary curves of length${}<\e$, then 
(1) the boundary curves are
the only closed geodesics on $P$ of length${}<\e$, and (2) any
geodesic interval in $P$ of length${}<\e$ with its endpoints in $\partial
P$ lies entirely in $\partial P$.
\qed
\end{lemma}

We call a closed geodesic on a hyperbolic surface short if it has
length${}<\e$, and very short if it has length${}<\e/2$.  The point of
the lemma is that on a hyperbolic surface obtained by gluing together
pairs of pants with all short boundary curves, the short curves are
exactly the curves along which the pairs of pants were glued.  (1)
deals with short closed geodesics that lie in a single pair of pants
and (2) rules out the existence of any others.

\begin{proof}[Proof of the theorem:]
Let $I$ be a set indexing a sequence of group elements $a_i$ that
generate $G$;  by
enlarging $I$ we may suppose $|I|\geq2$.  The Cayley graph $C$ of $G$
has automorphism group equal to $G$, and we will use $C$ as a sort of
framework on which to build $X$.  Explicitly, $C$ has vertex set
$G$, and for each $g\in G$ and each $i\in I$ there is a directed edge
from $g$ to $ga_i$ labeled `$i$'.  An automorphism of $C$ is an
automorphism of the underlying graph which preserves the directions
and labels of edges.  $G$ acts on $C$ by left-multiplication, and it
is easy to see that $G$ is all of $\aut C$.  (An automorphism  $g$ of
$C$ permutes the vertices in a manner commuting with
$G$'s right-multiplication action  on itself; since this action is
transitive, $g$ is completely determined by what it does to any one
vertex.) 

We choose positive numbers $\a_i$, $\b_i<\e/2$ for $i\in I$, all
distinct.  We build for each $i$ a compact hyperbolic surface $E_i$
with two boundary curves, of lengths $\a_i$ and $\b_i$.  We may
suppose that these are the only very short curves on $E_i$, say by
building $E_i$ from two pairs of pants, with boundary curves of
lengths $\a_i$, $\e/2$ and $\e/2$ (resp. $\b_i$, $\e/2$ and $\e/2$),
glued together along their boundary curves of length $\e/2$.  We call
$E_i$ the $i$th `edge surface'.  Next we build a complete hyperbolic
surface $V$ with $2\cdot|I|$ boundary curves, of lengths $\a_i$ and
$\b_i$, say by gluing a sequence of pairs of pants together along
boundary curves which are short but not very short.  The shortness of
the boundary curves of the pairs of pants assures us that the only very short
geodesics on $V$ are its boundary curves.  We may suppose that each
of the pairs of pants has boundary components of three distinct
lengths; this lets us arrange for $V$ to have trivial isometry group,
as follows.  $\isom(V)$ must permute the short geodesics of $V$, hence
permute the pairs of pants, hence carry to itself the unique one of
these pairs of pants with a boundary curve of length $\a_0$, $0$ being some fixed
element of $I$.  A pair of pants with boundary curves of three
distinct lengths has isometry group $\Z/2$, and by rotating the gluing
of an adjacent pair of pants, we can make sure that the nontrivial
isometry does not extend to an isometry of $V$.  We call $V$
the `vertex surface'.

Now we build $X$.  We take one copy of the vertex surface for each
vertex of $C$ and attach one copy of the edge surface $E_i$ for each
edge of $C$ labeled $i$.  We attach the length $\a_i$ (resp. $\b_i$)
boundary curve of each copy of $E_i$ to the length $\a_i$
(resp. $\b_i$) boundary curve of the copy of $V$ from which the edge
originates (resp. to which it points).  This may be done in a manner
compatible with the action of $G$ on $C$, so that $G\sset\isom X$.

To see that $G$ is all of $\isom X$, suppose $g:X\to X$ is an
isometry.  Then it permutes the very short geodesics of $X$ and hence
the components of their complement, which are the (interiors of the)
edge and vertex surfaces we used to build $X$.  It must carry
components with two boundary curves to other such components, so it
carries edge (resp. vertex) surfaces to edge (resp. vertex) surfaces.
Since the $\a_i$ and $\b_i$ are all distinct, it carries edges
labeled $i$ to other such edges, and preserves their directions.
Therefore $g$ acts on $C$; by composing $g$ with an element of $G$ we
may suppose that $g$ fixes a vertex of $C$.  Since we chose $V$ to
have no isometries, $g$ must be the identity.

It is clear that $X$ is complete;  it may be taken to be
compact when $G$ is finite by taking $I$ finite and $V$ compact.
\end{proof}

\begin{thebibliography}{99}

\bibitem{BL}
M. Belolipetsky and A. Lubotzky, Finite Groups and Hyperbolic
Manifolds, to appear in {\it Invent. Math.}

\bibitem{greenberg}
L. Greenberg,  Maximal groups and signatures, 
{\it Discontinuous groups and Riemann surfaces (Proc. Conf.,
  Univ. Maryland, College Park, Md., 1973)}, Ann. of Math. Studies,
No. 79, Princeton University Press, 1974, pp. 207--226

\bibitem{kojima}
S. Kojima, Isometry transformations of hyperbolic 3-manifolds, {\it 
 Topology and its Applications}  {\bf 29} (1988) 297--307

\bibitem{long-reid}
D. D. Long and A. W. Reid, On asymmetric hyperbolic manifolds, to appear
{\it Math. Proc. Camb. Phil. Soc.}

\bibitem{thurston}
W. Thurston, {\it Three-dimensional geometry and topology.},
ed. Silvio Levy. Princeton Mathematical Series, 35.
Princeton University Press, 1997
\end{thebibliography}

\end{document}