Singular sets and parameters of generalized triangle orbifolds
Michael Belolipetsky
We study groups generated by three half-turns in Lobachevsky space and their factor-orbifolds. This generalized triangle groups are directly connected with the arbitrary $2$-generator Kleinian groups. Our main result is a complete description of the singular sets of the generalized triangle orbifolds. We also give a method to obtain the parameters defining a generalized triangle group from the structure of the singular set of its factor-orbifold and illustrate it by examples.
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\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\Q{{\mathbb Q}}
\def\C{\mathbb C}
\def\R{{\mathbb R}}
\def\Hy{{\mathcal H}}
\def\S{{\mathcal S}}

\def\M{\rm{M}}
\def\SL{\rm{SL}}
\def\PSL{\rm{PSL}}
\def\tr{\rm{tr}}

\title{Singular sets and parameters of generalized triangle orbifolds}
\author{Michael Belolipetsky}

\begin{document}

\begin{abstract}
We study groups generated by three half-turns in Lobachevsky space and their
factor-orbifolds. This generalized triangle groups are directly connected with
the arbitrary $2$-generator Kleinian groups. Our main result is a complete
description of the singular sets of the generalized triangle orbifolds. We also
give a method to obtain the parameters defining a generalized triangle group
from the structure of the singular set of its factor-orbifold and illustrate it
by examples.
\end{abstract}

\maketitle

\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newenvironment{remark}{\noindent\normalsize {\bf Remark.}}{}
\newenvironment{remarkN}[1]{\noindent\normalsize {\bf Remark #1.}}{}

\section{ Introduction}

After the famous lectures of W.~Thurston \cite{Thur} the study of
$3$-dimensional manifolds was focused on geometric and especially on hyperbolic
manifolds and orbifolds. An orientable hyperbolic $3$-orbifold can be obtained
as a factor-space of the Lobachevsky space $\Hy^3$ by the action of a discrete
subgroup of the group of isometries $Isom^+\Hy^3= \PSL(2,\C)$. So the
hyperbolic $3$-orbifolds are connected with the discrete subgroups of
$\PSL(2,\C)$ known also as {\em Kleinian groups}. Since a general subgroup of
$\PSL(2,\C)$ is discrete if and only if each of its $2$-generator subgroups is
discrete (see e.g. \cite{Berdon}), the class of the $2$-generator Kleinian
groups takes on special significance. This groups and in particular arithmetic
$2$-generator Kleinian groups were studied by F.~Gehring, C.~Maclachlan,
G.~Martin, J.~Montesinos, A.~Reid and others (see \cite{BM}, \cite{MM3},
\cite{GM}, \cite{MM2}, \cite{ADV}, \cite{HLM1}, \cite{J}, \cite{MM1} and the
references therein). It appears, that the $3$-orbifolds corresponding to the
$2$-generator Kleinian groups may have a very complicated structure. In a
special case of the groups with real parameters the orbifolds were studied by
J.~Gilman~\cite{GGM}, E.~Klimenko~\cite{K1},\cite{K2}; and have been recently
completely classified by E.~Klimenko and N.~Kopteva (see \cite{KK}).

In this article we consider the groups of hyperbolic isometries generated by
three half-turns in $\Hy^3$. If the axes of the half-turns pairwise intersect
then the group is isomorphic to a Fuchsian triangle group, so the groups
generated by three half-turns can be considered as a generalization of the
Fuchsian triangle groups. The generalized triangle groups are directly
connected with the arbitrary $2$-generator Kleinian groups (see Section~5) but
have much more trackable geometric structure. Thus, since all three generators
of the groups have fixed points (axes) in $\Hy^3$, the fundamental groups of
underlying spaces of the corresponding $3$-orbifolds are always trivial, so the
underlying space of such an orbifold is a $3$-sphere $\S^3$ or a counterexample
to the Poincare conjecture.

Further we will always suppose that the underlying space is $\S^3$. In
Section~2 we describe singular structures of common generalized triangle
orbifolds, which are naturally fall into eight types (Theorem~2.4).

The set of the generalized triangle groups can be parameterized by three
complex numbers that correspond to the complex distances~\cite{Fen} between the
axes of the generators. In Section~3 we use W.~Fenchel's technics~\cite{Fen} to
study this parameterization and its connection with the matrix representation
of the group in $\SL(2,\C)$. Section~4 shows how to find the parameters of the
group for the eight types of singular structures from Theorem~2.4. The
connection between the generalized triangle groups and the $2$-generator
Kleinian groups and their usual parameterization is considered in Section~5.

Section~6 presents examples of common generalized triangle orbifolds with
different types of singularities and in Section~7 we consider an example for
the exceptional case.

{\it Acknowledgement.}  I would like to thank Professor Alexander D.~Mednykh
for interesting me in this problem and numerous helpful discussions.

\section{ Structure of the singular sets}

The underlying space of a generalized triangle group is $3$-sphere, so the
singular sets are knotted graphs in $\S^3$ with the orbifolds fundamental
groups generated by three involutions. While depicting the singular sets we
follow a common notation writing indexes of singularity near corresponding
components and omitting indexes $2$. We will use Wirtinger presentation of the
fundamental group of a knotted graph \cite{Fox} adopted to the group of
orbifold \cite{HQ}, so the elementary arcs of the graph will correspond to the
words in the group.

\begin{definition} {\em Common generalized triangle group} is a Kleinian group
generated by three halftuns such that the fundamental group of the complement
of its singular set in $\S^3$ has also rank $3$.
\end{definition}

This definition allows us in some cases to consider groups of the complements
of the singular graphs instead of the orbifold groups. This saves from certain
difficulties, which may arise because the considered groups have many
involutions.

\begin{lemma} The class of the common generalized triangle groups is
invariant under Reidemeister moves for graphs.
\end{lemma}

\begin{proof}
This lemma is obvious because the Reidemeister moves do not change the
fundamental group of the complement of the singular graph.
\end{proof}

Now suppose we have a knotted graph such that some three arcs of the graph
correspond to the three elements generating the fundamental group of the
complement of the graph in $\S^3$. Then corresponding three involutions
generate our generalized triangle group. The first thing we do is we draw a
circle that contains a plane projection of the singular graph and using the
Reidemeister moves draw the generating arcs outside the circle:

\begin{figure}[ht]
\psfig{file=fig1.eps, scale=.60} \caption{The singular set (except the
generating arcs) of generalized triangle group $<a,b,c>$ is in disc $D$.}
\end{figure}

Here we have to note that Lemma~2.1 is not enough to carry on this
procedure because it does not say anything about the generators of
the group, and what we want is to obtain the group generated by
the three elements that correspond to the selected singular arcs.
However, it is a matter of a simple case-by-case verification to
see that the generators can be "preserved" and we leave it as an
exercise.


Now let us see what singular structures are possible inside $D$.
The only restrictions we have are that the group should be
generated by $a$, $b$, $c$ and that the corresponding factor-orbifold
is hyperbolic.

\begin{lemma}
Non trivial knots on singular arcs inside $D$ are impossible.
\end{lemma}

\begin{proof}

\begin{figure}[h!]
\psfig{file=fig2.eps, scale=.75} %!!! was .75
\caption{An illustration for the proof of Lemma~2.2.}
\end{figure}

Suppose there is a non-trivial knot on a singular arc inside $D$. Let $X$ be
the complement to the singular set in $S^3$. Consider decomposition $X =
X_1\cup X_2$ such that $X_2$ contains the knot and $\partial X_2$ has precisely
$2$ punctures (see Figure~2).  Here $X_2$ could be obtained from the tubular
neighborhood of the singular set, so the decomposition is always possible.

We now apply Van Kampen theorem to $X = X_1 \cup X_2$. For $P\in X_1\cap X_2$:

$$\pi_1(X_1, P) = <a,b,c \mid \dots>, $$ $$\pi_1(X_2, P) = <w_1',w_2',x,\dots
\mid \dots>$$ and $x$ can not be obtained as a word of $w_1'$, $w_2'$ since the
knot is non-trivial.

By Van Kampen theorem $$\pi_1(X, P) = <a,b,c,w_1',w_2',x,\dots \mid w_1w_1',
w_2w_2' \dots> = <a,b,c,x,\dots \mid \dots>.$$ Here again $x$ can not be
obtained as a word of the other generators. This means that the rank of
$\pi_1(X)$ is greater then $3$, which contradicts the definition of the common
generalized triangle group.
\end{proof}


Since arcs of the graph can not be knotted with themselves,
the singular graph has a stratiform structure shown on Figure~3 with all
the vertices inside $D_1$ (we use the common notation for integer tangles
with $l_i$, $m_i$, $n_i\in\Z$ for $i=1,\dots,k$).

\begin{figure}[ht]
\psfig{file=scheme.eps, scale=.65} %!!! was .75
\caption{Structure of the singular set given by signature
\newline $S(l_1, m_1, n_1, l_2, m_2, n_2,\dots ,l_k, m_k, n_k)$.}
\end{figure}

\begin{lemma}
The singular graph in $D_1$ has no cycles.
\end{lemma}

\begin{proof}
Let $X$ be the common generalized triangle orbifold, and $\tilde{X}$ - the
$3$-manifold obtained as the complement to the singular set of $X$ in $S^3$. To
prove the lemma it is sufficient to consider the first homology group
$H_1(\tilde{X})$. The canonical projection $\tilde{p}: \pi_1(\tilde{X})\to
\pi_1(\tilde{X})^{\rm ab} \cong H_1(\tilde{X})$ maps non of the generators of
$\pi_1(\tilde{X})$ to the homology cycle that corresponds to the singular cycle
in $D_1$. Since $H_1(\tilde{X})\cong\Z^r$ with no any elements of finite order,
this immediately follows that $rank(H_1(\tilde{X})) > rank
(\tilde{p}(\pi_1(\tilde{X}))),$ which is impossible.

Note, that this argument does not work for the orbifold fundamental and
homology groups because the homology groups are generated by the elements of
the finite order (See example in Section~7).
\end{proof}

Lemma~2.3 makes it possible to push all remaining tangles out of
$D_1$ by drawing vertices through the tangles as it shown on
Figure~4.

\begin{figure}[htb]
\psfig{file=fig4.eps, scale=.75}
\caption{Drawing a vertex through a tangle.}
\end{figure}

\noindent So for the graph in $D_1$ we have:

 - it has no cycles;

 - it has no tangles;

 - the group of its complement is generated by (six) arcs coming
into $D_1$ from the outside.

Together with the assumption that the orbifold is hyperbolic this gives
all we need to classify singular sets of the common generalized triangle
groups. In the following theorem we summarize the results:

\begin{theorem} The structure of the singular set of a common generalized
triangle orbifold is determined by the signature
$S(l_1, m_1, n_1, l_2, m_2, n_2,\dots ,l_k, m_k, n_k)$ (Figure~3)
together with $D_1$ corresponding to one of the cases $A$~-~$H$ (Figure~5).

\begin{figure}[ht]
\psfig{file=centers.eps, scale=.75} \caption{Singular structures inside $D_1$
(here $t_i\in\N\cup\{\infty\}$).}
\end{figure}

Orbifolds obtained from $A$, $B$, $C\ ($with $1/t_1+1/t_2+1/t_3<1)$, $D\ ($with
$1/t_1+1/t_2<1/2$ and $1/t_2+1/t_3<1/2)$ are compact; while orbifolds obtained
from  $C\ ($with $1/t_1+1/t_2+1/t_3=1)$, $D\ ($with $1/t_1+1/t_2=1/2$ or
$1/t_2+1/t_3=1/2)$, $E$, $F$, $G$ and $H$ are non compact.
\end{theorem}

\begin{remarkN}1
We have dropped one more combinatorially possible case that can be
obtained from $G$ deleting the arc, which joins the vertices, because
corresponding groups are always generated by only two half-turns.
\end{remarkN}\medskip

\begin{remarkN}2 Some relatively simple singular structures from
Theorem~2.4 correspond to non-hyperbolic orbifolds (see Examples~6C, 6D). All
such geometric orbifolds can be found in \cite{Dun} and we do not specially
focus on non-hyperbolic case.
\end{remarkN}\medskip

\begin{remarkN}3 A generalized triangle orbifold is uniquely determined
by the signature and $D_1$ as it was shown above, however different
signatures can correspond to the same orbifolds.
\end{remarkN}

\section{ Matrix representation and parameters}

In order to obtain a matrix representation for a group generated by
three half-turns one has to fix the axes of the generators. We do
this by means of complex distances between the hyperbolic lines
(\cite{Fen}, p.~67-70). We define the {\it complex distance} between
two oriented hyperbolic lines as follows: its real part equals the
length of the common perpendicular to the lines and imaginary part
is given by the angle between the lines, taken from the first to the
second line respecting orientations.

We call three complex distances between the axes of the generators by
the {\it parameters} of the group $\Gamma=\langle a,b,c\rangle$.  Let us
remark that starting from this place by group $\langle a,b,c\rangle$
we mean the group with the given generators taken in a fixed order, so,
in fact, we work with a {\em marked groups} generated by $a, b, c$.
$$
par(\langle a,b,c\rangle)=(\mu (a,b),\ \mu (a,c),\ \mu (c,b))=(\mu_0,\ \mu_1,\ \mu_2).
$$
(Here $\mu(\ldots,\ldots)=x+iy$, $x\in \R^+$, $y\in [0;2\pi)$ --- the complex
distance between the oriented axes of the half-turns.)

\medskip
\begin{remark}
Since the directions of the half-turns are undefined the parameters are
defined up to orientation of the axes. We can canonically link the
orientation of a hyperbolic line with the matrix of the half-turn in
$\SL(2,\C)$ (see \cite{Fen}, p.~63). This means that fixing the directions
of the axes of generators equals choosing the inverse image after canonical
projection $p:\SL(2,\C)\to\PSL(2,\C)$ of our $\PSL(2,\C)$-subgroup in
$\SL(2,\C)$. Following common agreement we usually call this inverse image by
representation of our group in $\SL(2,\C)$.
\end{remark}

\begin{lemma}
Complex vector $par(\langle a,b,c\rangle)$ determines group
$\langle a,b,c\rangle$ uniquely up to conjugation in $\PSL(2,\C)$, i.e.
if $par(\langle a,b,c\rangle)=par(\langle a',b',c'\rangle)$ then there
exists $h\in\PSL(2,\C)$ such that $a'=hah^{-1}$, $b'=hbh^{-1}$,
$c'=hch^{-1}$.
\end{lemma}

\begin{proof}
Consider the axes of the half-turns $a, b, c$ in $\Hy^3$. We endow the axes by
orientations and join them in couples by common perpendiculars. The result
is a hyperbolic right-angled hexagon (Figure~6).

\begin{figure}[ht]
\psfig{file=hexagon.eps, scale=.75}
\caption{Axes of the half-turns $a, b, c$ together with their common
perpendiculars give a hyperbolic right-angled hexagon.}
\end{figure}

Vector $par(\langle a,b,c\rangle)$ represents three complex lengths of the
sides of our hexagon which are common perpendiculars of the given axes. By
theorem from \cite{Fen} (p.~94) three pairwise non-adjacent sides define
right-angled hexagon uniquely up to simultaneous change of orientations
of the other three sides. It means that any other hexagon defined by
another group with the same parameters can be translated to the given
one by a hyperbolic isometry. This isometry obviously defines the required
conjugation $h\in\PSL(2,\C)$.

It may seem that we have missed the possibility of changing of the
orientations of the axes but we have already remarked that the orientations
effect only representation of the group in $\SL(2,\C)$ but not isometries
themselves. So in this situation we can change the orientations as it is
needed.
\end{proof}

Next thing to do is to find out some convenient representation for
$\langle a,b,c\rangle$ in $\SL(2,\C)$ and describe it in terms of
$par(\langle a,b,c\rangle)$. We denote the matrices corresponding to
the half-turns in $a$, $b$, $c$ by capitals $A$, $B$, $C$.

Half-turn in oriented hyperbolic line $m$ is represented by a matrix
$M\in\SL(2,\C)$ of the form $(\begin{array}{cc} m_{11}&m_{12}\\m_{21}&-m_{11}\end{array})$,
which is characterized by $M^2=-I$ or equivalently $\tr(M)=0$.
This matrix $M$ is called {\it the normalized line-matrix} of line $m$.
By extending transformation $M$ to the absolute $\overline{\C}=\partial\Hy^3$
it is easy to find the set of its fixed points in $\overline{\C}$ denoted by
$\it{fix}(M)$. Geometrically the fixed points $\{z_1,\ z_2\}=\it{fix}(M)$ are the
ends of the arc orthogonal to $\C$ which presents line $m$ in the
upper-space model of $\Hy^3$.

After a suitable conjugation of the group $\langle A,B,C\rangle$ in $\SL(2,\C)$
one can suppose that $\it{fix}(A)=\{ 1/\beta ;-1/\beta\}$,
$\it{fix}(B)=\{ \beta ;-\beta\}$ ($\beta\in\C$ and $|\beta|\geq 1$).

We have:
$$
A=\left(\begin{array}{cc} 0&i/\beta\\i\beta&0\end{array}\right),\ \
B=\left(\begin{array}{cc} 0&i\beta\\i/\beta&0 \end{array}\right),\ \
C=\left(\begin{array}{cc} c_{11}&c_{12}\\c_{21}&-c_{11}\end{array}\right),
$$
where $-c_{11}^2+c_{12}c_{21}=1$ since $C\in\SL(2,\C)$.

To find $\beta$ and $c_{ij}$ in terms of $par(\langle a,b,c\rangle)$ we
use the following important formula, which allows to find the complex
distance between hyperbolic lines $m_1$, $m_2$ from the matrices
$M_1$, $M_2$ of the lines (\cite{Fen}, p.~68):
$$ cosh(\mu(m_1,m_2))=-\frac{1}{2} \tr\,(M_1M_2).$$

Denote $\rho_k=-2\,cosh\mu_k$ ($k=0$, $1$, $2$). Then
$$\rho_0=\tr(AB)=\tr\left(\begin{array}{cc} -1/\beta^2&0\\0&-\beta^2\end{array}\right);$$
$$\rho_1=\tr(AC)=\tr\left(\begin{array}{cc} ic_{21}/\beta&-ic_{11}/\beta\\ic_{11}\beta&ic_{12}\beta\end{array}\right);$$
$$\rho_2=\tr(BC)=\tr\left(\begin{array}{cc} ic_{21}\beta&-ic_{11}\beta\\ic_{11}/\beta&ic_{12}/\beta\end{array}\right).$$

We obtain system of equations on $\beta$ and $c_{ij}$:
$$
\left\{
\begin{array}{l}
{\displaystyle -1/\beta^2-\beta^2=\rho_0};\\
{\displaystyle ic_{21}/\beta+ic_{12}\beta=\rho_1};\\
{\displaystyle ic_{21}\beta+ic_{12}/\beta=\rho_2};\\
{\displaystyle c_{11}=i\sqrt{c_{12}c_{21}+1}}.
\end{array}
\right.
$$
Solving this equations we find out all parameters in our representation.
Let us remark that this system has not unique solution but it really does not
matter, because different solutions give representations of the groups
$\Gamma=\langle a,b,c\rangle$ which are conjugate in $\PSL(2,\C)$.
The last statement is a direct consequence of Lemma~3.1 since by the
construction the different solutions correspond to the same $par(\Gamma)$.
For definiteness we can always fix the analytic branches of the square
roots in the following formulas.

Now we can write down the representation of $\langle a,b,c\rangle$
in $\SL(2,\C)$:

$$
A=\left(\begin{array}{cc} 0&i/\beta\\i\beta&0\end{array}\right),\ \
B=\left(\begin{array}{cc} 0&i\beta\\i/\beta&0 \end{array}\right),\ \
C=\left(\begin{array}{cc} c_{11}&c_{12}\\c_{21}&-c_{11}\end{array}\right);
$$
\begin{equation}
\begin{array}{ll}
\phantom{wwww}&{\displaystyle
  \beta=\sqrt{\frac{-\rho_0+\sqrt{\rho_0^2-4}}{2}}},\\
\phantom{wwww}&{\displaystyle \vphantom{\sqrt{\frac{-\rho_0+\sqrt{\rho_0^2-4}}{2}}}
  c_{21}=\frac{\rho_1/\beta-\rho_2\beta}{i/\beta^2-i\beta^2},\ \
  c_{12}=\frac{-\rho_1\beta+\rho_2/\beta}{i/\beta^2-i\beta^2},\ \
  c_{11}=i\sqrt{c_{12}c_{21}+1};}\\
\phantom{wwww}&{\displaystyle \vphantom{\sqrt{\frac{-\rho_0+\sqrt{\rho_0^2-4}}{2}}}
  \rho_k=-2\,cosh(\mu_k)\ {\rm for}\ k=0,1,2}.
\end{array}
\end{equation}

To obtain the representation of $\langle a,b,c\rangle$ in $\PSL(2,\C)$ one has
to consider the image of the representation in $\SL(2,\C)$ under
canonical projection $p:\SL(2,\C)\to\PSL(2,\C)$.


\section{ Deducing parameters from the singular structure}

Knowing the structure of the singular set of a generalized triangle orbifold
one can obtain a presentation of its fundamental group and a representation of
the group in $\SL(2,\C)$. This representation is defined by complex parameters
$\rho_0$, $\rho_1$ and $\rho_2$, so it is possible to find the parameters,
which correspond to a given singular structure. In this section and in
Section~6 we show how it can be carried on for the common generalized triangle
groups.

Suppose we are given a singular set of a common generalized triangle orbifold
(Figure~3). Using Wirtinger algorithm we can start from the given three arcs
$a$, $b$, $c$ and passing through the tangles obtain words on $a$, $b$, $c$
corresponding to all the other arcs. It is obvious that until we come to the
central part of the diagram ($D_1$), we do not obtain any equations on the
generators, so {\it all the words in the presentation of the group come from
$D_1$} (Figures~3, 5). Equations defining parameters are then obtained by
taking traces of corresponding words in $\SL(2,\C)$. With the help of the well
known formulas
\begin{equation}
\begin{array}{l}
\tr(W_1 W_2) = \tr(W_2 W_1),\\
\tr(W_1 W_2 W_1^{-1}) = \tr(W_2),\\
\tr(W_1 W_2)=\tr(W_1)\tr(W_2)-tr(W_1^{-1} W_2),\\
\tr(H^{-1})=-\tr(H) \ \ \ {\rm if\ }H{\rm\ is\ a\ matrix\ of\ a\ half-turn}\\
\end{array}
\end{equation}
the traces of the words on $A$, $B$, $C$ can be expressed in terms of
$\tr(AB)=\rho_0$, $\tr(AC)=\rho_1$, $\tr(BC)=\rho_2$ and $\tr(ABC)$ (an
explicit expression for $\tr(ABC)$ in terms of $\rho_i$ can be obtained using
(1)). Since we have three complex parameters we always need three independent
equations to define them. In the remaining part of this section we will give
the presentations of the groups and corresponding equations for each of the
cases $A-H$\ from Theorem~2.4.

\bigskip
\psfig{file=cA.eps, scale=.75}
\bigskip
\hfill
\parbox[b]{0.7\textwidth}{
Group:\\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ w_1 w_2,\ w_3 w_4 \rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=2,\\ \tr(W_3 W_4)=2,\\ \tr(W_5 W_6)=2
\end{array}
\right.$

\smallskip
(We denote by $w_i$ words on $a$, $b$, $c$\ and by $W_i$~--- corresponding
matrix words in $\SL(2,\C)$, $i=1,\dots,6.$ )
}

It follows from \cite{Fox} that one of the words $w_1 w_2$, $w_3 w_4$,
$w_5 w_6$ in the fundamental group is always a corollary of two other.
However by the rigidity argument in the matrix group all the three words
are independent and give three independent equations on the parameters.
See Example~6A for the illustration.

\bigskip
\psfig{file=cB.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ w_1 w_2,\ (w_3 w_4)^t,\ (w_5 w_6)^t\rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=2,\\ \tr(W_3 W_4)=-2cos(\pi/t),\\ \tr(W_5 W_6)=-2cos(\pi/t).
\end{array}
\right.$
\bigskip
}

\bigskip
\psfig{file=cC.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ (w_1 w_2)^{t_1},\ (w_3 w_4)^{t_2},\ (w_5 w_6)^{t_3}\rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=-2cos(\pi/t_1),\\ \tr(W_3 W_4)=-2cos(\pi/t_2),\\ \tr(W_5 W_6)=-2cos(\pi/t_3).
\end{array}
\right.$
\bigskip
}

\bigskip
\psfig{file=cD.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ (w_1 w_2)^{t_1},\ (w_5 w_6)^{t_2},\ (w_1 w_2 w_3)^{t_3}\rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=-2cos(\pi/t_1),\\ \tr(W_5 W_6)=-2cos(\pi/t_3),\\
\tr(W_1 W_2 W_3)=-2cos(\pi/t_2).
\end{array}
\right.$
\bigskip
}

It can be seen that word $W_1 W_2 W_3$ has an odd length, which follows that in
this case we need the expression for $\tr(ABC)$. Actually, this is the only
case where we may need the expression.

\bigskip
\psfig{file=cE.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ (w_1 w_2)^2,\ (w_5 w_6)^2\rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=0,\\ \tr(W_5 W_6)=0,\\ \tr(W_3 W_4)=-2.
\end{array}
\right.$
\bigskip
}
%\end{figure}

The third equation follows from the fact that the multiple of two
half-turns with the parallel axes is a parabolic isometry, so $w_3 w_4$ is
a parabolic corresponding to the cusp. We choose sign $'-'$ for the trace
of the parabolic element because the axes of the half-turns have the same
directions.

\bigskip
\psfig{file=cF.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ w_1 w_2\rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=2,\\ \tr(W_3 W_4)=-2,\\ \tr(W_5 W_6)=-2.
\end{array}
\right.$
\bigskip
}

\bigskip
\psfig{file=cG.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ (w_3 w_4 w_5)^2 \rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_3 W_4 W_5)=0,\\ \tr(W_1 W_2)=-2,\\ \tr(W_3 W_4)=-2.
\end{array}
\right.$
\bigskip
}

Here, as in case $D$, we again have a word $W_3 W_4 W_5$ of an odd length, but
since $\tr(W_3 W_4 W_5)=0$ we can easily avoid using the explicit expression
for $\tr(ABC)$ in this case. Really, using formulae $(2)$ we can present
$\tr(W_3 W_4 W_5)$ as $\tr(ABC){\rm P}[\tr(AB),\\ \tr(AC), \tr(BC)]$ where
${\rm P[...]}$ is a polynomial, and since $\tr(ABC)\neq 0$ we obtain equation
$\rm P[\tr(AB), \tr(AC), \tr(BC)]=0$. This procedure can be used for case $H$
as well.

\bigskip
\psfig{file=cH.eps, scale=.75}
\hfill
\parbox[b]{0.7\textwidth}{
Group: \\
$\langle a,\ b,\ c \mid a^2,\ b^2,\ c^2,\ (w_1 w_2)^t,\ (w_1 w_2 w_3)^2 \rangle,$

\medskip
equations:

\smallskip
$\left\{
\begin{array}{l}
\tr(W_1 W_2)=-2cos(\pi/t),\\ \tr(W_5 W_6)=-2,\\
\tr(W_1 W_2 W_3)=0.
\end{array}
\right.$
\bigskip
}

\section{ Connection with the two-generator Kleinian groups}

Let $\Gamma_0=\langle f,g\rangle$ be a discrete subgroup of $\PSL(2,\C)$.
There is a natural way to associate to $\Gamma_0$ a generalized triangle
group $\Gamma$ which is equal to $\Gamma_0$ or contains it as a subgroup
of index $2$. It follows that such properties as discreteness, arithmeticity
and others of two-generator subgroups of $\PSL(2,\C)$ can be studied
using the generalized triangle groups.

The group $\Gamma$ is constructed as follows (see \cite{Gilm}):
If neither $f$ nor $g$ are parabolic then let $N$ be a common perpendicular
to the axes of $f$ and $g$. In case of parabolic generator corresponding
end of the line $N$ is the fixed point of the parabolic. Obviously,
this construction uniquely defines line $N$ for any Kleinian group
$\Gamma_0=\langle f,g\rangle$. Let $c$ be the half-turn about $N$. Then
there exist such half-turns $a$ and $b$ that $f=ac$ and $g=cb$
(\cite{Fen}, p.~47). So we obtain group $\Gamma=\langle a,b,c\rangle$
which contains $\Gamma_0$ as a subgroup of index at most $2$.

Visa versa starting from the group $\Gamma$ generated by three half-turns
$a$, $b$, and $c$ one can easily obtain its two generator subgroup
$\Gamma_0=\langle ab,ac\rangle$. Since $\Gamma=\Gamma_0\cup a\Gamma_0\
(a^2=id)$, index of $\Gamma_0$ in $\Gamma$ is equal to $2$ or $1$
depending on if $a$ is contained in $\Gamma_0$ or not.

The set of two-generator Kleinian groups is usually parameterized by three
complex numbers \cite{GM}: $$ \beta(f)=\tr^2(f)-4,\ \beta(g)=\tr^2(g)-4,\
\gamma(f,g)=\tr([f,g])-2. $$ A common computation with traces shows that this
parameters are connected with our parameterization for the corresponding
generalized triangle group by the following formulas: $$ \beta(f)=\rho_1^2-4,\
\beta(g)=\rho_2^2-4,\
\gamma(f,g)=\rho_0^2+\rho_1^2+\rho_2^2+\rho_0\rho_1\rho_2-4. $$

Since we have described all the singular sets of the generalized triangle
orbifolds we can at least theoretically present an algorithm, which using
equations from the previous section enumerates all the parameters
corresponding to the generalized triangle groups and discrete two-generator
subgroups of $\PSL(2,\C)$. The main practical difficulty is that the
complexity of the equations grows as the singular structures
become more complicated. The other point is that we have to cast away
the redundant solutions of the equations somehow. In the next section we
will show how this can be done for the examples considered, however we do
not know any general method to carry on this procedure.

Let us also note that the suggested algorithm shows only that our set of
parameters is semi-recursive: we can enumerate all the parameters of
discrete generalized triangle groups, but there is no way to decide
in a finite number of steps does a given triple of complex numbers define
a discrete subgroup of $\PSL(2,\C)$ or not.


\section{ Examples of the common generalized triangle orbifolds. }

Let us consider examples for different types of singularities from Theorem~2.4
and find parameters defining corresponding generalized triangle groups. Here we
will try to give examples, which deserve a particular interest and have
previously appeared in different situations. Detailed descriptions of some
other generalized triangle orbifolds can be found in \cite{Bel_m} and
\cite{Bel}.

\bigskip

\psfig{file=exA.eps, scale=.75}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
a_{-1} = cac^{-1}\\
b_{-1} = aba^{-1}\\
c_{-1} = bcb^{-1}\\
a_1 = b_{-1}^{-1}ab_{-1}\\
b_1 = c_{-1}^{-1}bc_{-1}\\
b_2 = b_{-1}^{-1}b_1 b_{-1}\\
a_{-1}^{-1}b_2 = ca^{-1}c^{-1}ab^{-1}a^{-1}bc^{-1}bcb^{-1}aba^{-1}\\
%b_{-1}^{-1}c_2 = ab^{-1}a^{-1}bc^{-1}b^{-1}ca^{-1}cac^{-1}bca^{-1}\\
%c_{-1}^{-1}a_2 = bc^{-1}b^{-1}ca^{-1}c^{-1}ab^{-1}aba^{-1}cab^{-1}\\
\end{array}
$
\bigskip

Matrix group:
\medskip

$\langle A,\ B,\ C \mid -A^2, -B^2, -C^2,\\
CA^{-1}C^{-1}AB^{-1}A^{-1}BC^{-1}BCB^{-1}ABA^{-1},\\
AB^{-1}A^{-1}BC^{-1}B^{-1}CA^{-1}CAC^{-1}BCA^{-1},\\
BC^{-1}B^{-1}CA^{-1}C^{-1}AB^{-1}ABA^{-1}CAB^{-1}
\rangle$
}
\bigskip

Let $\tr(AB)=t_0$, $\tr(AC)=t_1$, $\tr(BC)=t_2$.

By taking traces of the words we obtain a system of equations on $t_0$, $t_1$
and $t_2$\footnote{We use an original computer program $<$tracer$>$(C++, 242
lines) to calculate trace polynomials for generalized triangle groups. The
program is available on request.}:

$$\left\{
\begin{array}{l}
3t_0-t_0^3-2t_0t_1^2+t_0^3t_1^2+2t_0^2t_1t_2-t_0^2t_1^3t_2-2t_0t_2^2+t_0^3t_2^2-t_0^3t_1^2t_2^2-t_0^2t_1t_2^3
= 2,\\
3t_2-t_2^3-2t_2t_0^2+t_2^3t_0^2+2t_2^2t_0t_1-t_2^2t_0^3t_1-2t_2t_1^2+t_2^3t_1^2-t_2^3t_0^2t_1^2-t_2^2t_0t_1^3
= 2,\\
3t_1-t_1^3-2t_1t_2^2+t_1^3t_2^2+2t_1^2t_2t_0-t_1^2t_2^3t_0-2t_1t_0^2+t_1^3t_0^2-t_1^3t_2^2t_0^2-t_1^2t_2t_0^3
= 2.
\end{array}
\right.$$

We are interested only in complex solutions of this system and by the
symmetry of the singular graph all the three roots should be equal.
The only such solutions are (approximately):
$$ (t_0, t_1, t_2) = (0.662359 \pm 0.56228i, 0.662359 \pm 0.56228i,
0.662359 \pm 0.56228i).$$
So we have $\rho_0=\rho_1=\rho_2 = 0.662359\dots \pm 0.56228\dots i$.
The signs before the imaginary part correspond to the orientations of
the generating arcs and do not affect the presentation of the group
in $\PSL(2,\C)$.

The primitive polynomial of the root is $t^3-t+1$, using the
arithmeticity test for generalized triangle groups from \cite{Bel}
it can be easily verified that the orbifold is arithmetic. It is also
interesting to note (see \cite{MV1}) that the two-fold covering of this
orbifold is Weeks-Mateveev-Fomenko manifold~--- the hyperbolic manifold
of the smallest known volume.

(Of course, here we could use the symmetry of the singular graph
from the beginning and initially suppose $\tr(AB)=\tr(AC)=\tr(BC)$. It
would considerably simplify the calculations but leads to some loss
of generality.)

\bigskip

\psfig{file=exB.eps, scale=.75}
\hfill
\parbox[b]{0.4\textwidth}{
$\begin{array}{l}
a_1 = bab^{-1}\\
b_1 = cbc^{-1}\\
c_1 = aca^{-1}\\
a_{-1} = c_1^{-1}ac_1\\
b_{-1} = a_1^{-1}ba_1\\
\\
a_{-1}^{-1}a_1 = ac^{-1}a^{-1}ca^{-1}bab^{-1}\\
b_{-1}^{-1}c^{-1} = ba^{-1}b^{-1}ab^{-1}c^{-1}\\
b_1c_1 = cbc^{-1}aca^{-1}\\
\end{array}
$
\vspace{50pt}
}
\bigskip

\noindent Equations on $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$:
$$\left\{
\begin{array}{l}
t_0^2t_1^2+t_0t_1t_2-t_0^2-t_1^2+2 = 2,\\
t_0^2t_2+t_0t_1-t_2 = -2cos(\pi/n),\\
t_1^2t_2+t_0t_1-t_2 = -2cos(\pi/n).\\
\end{array}
\right.$$

We will consider only $n=3$. Up to the symmetry and change of orientation the
system has only one complex solution, which defines the complex place of the
orbifold group:
$$\rho_0 = \frac{i+\sqrt{3}}{2},\ \rho_1 = \frac{i-\sqrt{3}}{2},\ \rho_2=0.$$
The squares of $\rho_0$, $\rho_1$ are algebraic integers and by arithmeticity
test \cite{Bel} the group is arithmetic defined over field $\Q[\sqrt{-3}]$.

\noindent (For $n=2$ and $n=\infty$ the group has no any complex places, so the
orbifold is not hyperbolic. For large values of $n$ the group has more then one
complex places induced by $cos(\pi/n)\to cos(k\pi/n)$, $(k,n)=1$ and so it is
not arithmetic.)

\bigskip

\psfig{file=exC.eps, scale=.70}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
b_{-2} = b^{-1}aba^{-1}b\\
b_{-1} = aba^{-1}\\
b_1 = b_{-2}bb_{-2}^{-1}\\
b_2 = cb_1c^{-1}\\
a_1 = b_{-1}^{-1}ab_{-1}\\
c_1 = aca^{-1}\\
\\
c_1a_1 = acb^{-1}aba^{-1}\\
b_{-2}^{-1}c^{-1} = b^{-1}ab^{-1}a^{-1}bc^{-1}\\
b_2a^{-1} = \\
\ \ cb^{-1}aba^{-1}bab^{-1}a^{-1}bc^{-1}a^{-1}
\end{array}
$
\vspace{90pt}
}
\bigskip

\noindent Equations on $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$:
$$\left\{
\begin{array}{l}
t_0t_2+t_1 = 0,\\
t_0^2t_2+t_0t_1-t_2 = -2cos(\pi/n),\\
(t_0^3-3t_0)(t_0t_1t_2+t_1^2+t_0^2-2)+t_0t_1^2+t_1t_2-t_0 = 0.
\end{array}
\right.$$
Complex solution ($n>3$):
$$\begin{array}{l}
\rho_0^2 = \frac{1}{2}(1+4sin^2(\pi/n)(1\pm\sqrt{1-\frac{4}{1+4sin^2(\pi/n)}}),\\
\rho_1 = -2cos(\pi/n)\rho_0,\\
\rho_2 = 2cos(\pi/n).
\end{array}
$$
(Here $\rho_2$ is always a real number; for $n=2,3$ $\rho_0$ and $\rho_1$ are
also real, so corresponding orbifolds are not hyperbolic $3$-orbifolds of
finite volume. According to \cite{Dun} for $n=2$ the orbifold admits the
spherical structure and for $n=3$ it is Euclidean.)

As it was shown in \cite{RV} the considered orbifolds can be obtained
as the factor-orbifolds of Fibonacci manifolds $F(2n,n)$ by their full
groups of isometries. So using the arithmeticity test from \cite{Bel}
we can now easily find all arithmetic Fibonacci manifolds (see also
\cite{HLM}). The field of definition of the orbifold group $\Q[\rho_0^2,
cos^2(\pi/n)]$ has exactly one complex place only if $n=4$, $5$, $6$, $8$
$12$, $\infty $; it is a matter of direct verification that this cases
satisfy the remaining conditions of the arithmeticity test. Thus
the orbifolds and corresponding Fibonacci manifolds are arithmetic
for $n=4$, $5$, $6$, $8$ $12$, $\infty $.

\bigskip

\psfig{file=exD.eps, scale=.70}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
c_{-2} = ac_{-1}a^{-1}\\
c_{-1} = b^{-1}cb\\
b_{-1} = c_{-2}^{-1}bc_{-2}\\
a_1 = c_{-2}^{-1}ac_{-2}\\
b_1 = cbc^{-1}\\
c_1 = b_{1}cb_{1}^{-1}\\
\\
a_1b_{-1}^{-1} = ab^{-1}c^{-1}ab^{-1}cba^{-1}\\
c_{-2}^{-1}c_1 = ab^{-1}c^{-1}ba^{-1}cbcb^{-1}c^{-1}\\
a^{-1}a_1b_{-1}^{-1} = b^{-1}c^{-1}ab^{-1}cba^{-1}
\end{array}
$
\vspace{110pt}
}
\bigskip

\noindent Equations on $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$:
$$\left\{
\begin{array}{l}
t_0 = 0,\\
4t_2^2-t_1^2t_2^2-t_2^4+t_1^2-2 = 0,\\
4t_1^2-t_1^2t_2^2-t_1^4 = 4cos^2(\pi/n).\\
\end{array}
\right.$$
Equation $\tr(AB) = 0$ geometrically means that the axes of half-turns
$a$ and $b$ intersect and orthogonal in $\Hy^3$.

The system has exactly one pair of complex-conjugate roots for any
$n>2$ (for $n=2$ the orbifold is Euclidean), this roots give the
values of parameters $\rho_0 (=0),$ $\rho_1,$ $\rho_2$ defining the
orbifold group. Since the analytic formulas for the solutions are
rather complicated we do not give them here. One can see that the
field of definition $\Q[\rho_1^2, \rho_2^2]$ of the group  has
exactly one complex place iff $n=3$, $4$, $6$, and it is easy to
check the other conditions of the arithmeticity test \cite{Bel}
for this values of $n$. So we obtain, that the orbifolds are arithmetic
for $n=3$, $4$, $6$.

This groups and orbifolds first appeared in \cite{HKM2} as an infinite
one-parameter family extending one of the three regular tessellations of
$\Hy^3$. It is interesting to note that the corresponding family for the one of
the two remaining regular tessellations is also known \cite{HKM1} and consists
of Fibonacci manifolds considered in the previous example; and, finally, for
the third tessellation the question of constructing such a family is still
open.

\bigskip

\psfig{file=exE.eps, scale=.70}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
a_1 = bab^{-1} \\
b_1 = a_1ba_1^{-1} \\
c_1 = aca^{-1} \\
a_2 = ca_1c^{-1} \\
\\
c_1b^{-1} = aca^{-1}b^{-1} \\
a_2a^{-1} = cbab^{-1}c^{-1}a^{-1} \\
b_1c^{-1} = baba^{-1}b^{-1}c^{-1} \\
\end{array}
$
\vspace{136pt}
}
\bigskip

\noindent Equations on $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$:
$$\left\{
\begin{array}{l}
t_0t_1+t_2 = 0,\\
t_0t_1t_2+t_0^2+t_1^2-2 = 0,\\
-t_0^2t_2-t_0t_1+t_2 = -2.
\end{array}
\right.$$
Solution:
$$\rho_0^2 = \rho_1^2 = 1+i,\ \rho_2^2=2i.$$

This orbifold first appeared in \cite{Adams} as the third hyperbolic
$3$-orbifold with a non-rigid cusp. We see that the orbifold is
arithmetic with the field of definition $\Q[i]$ and so its group is
commensurable in $\PSL(2,\C)$ with Picard group $\PSL(2,\Z[i])$.

\bigskip

\psfig{file=exF.eps, scale=.75}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
a_1 = bab^{-1}\\
a_2 = ca_1c^{-1}\\
b_1 = cbc^{-1}\\
b_2 = a_2b_1a_2^{-1}\\
b_3 = b^{-1}b_2b\\
c_{-1} = b^{-1}cb\\
c_{-2} = b_3^{-1}c_{-1}b_3\\
b_{-1} = a_2^{-1}ba_2\\
\\
c_{-2}^{-1}a_2 = \\
\  b^{-1}cbab^{-1}a^{-1}b^{-1}c^{-1}baba^{-1}b^{-1}c^{-1}bcbab^{-1}c^{-1}\\
b_{-1}^{-1}c = cba^{-1}b^{-1}c^{-1}b^{-1}cbab^{-1}\\
a^{-1}b_3 = a^{-1}b^{-1}cbaba^{-1}b^{-1}c^{-1}b
\end{array}
$
\vspace{32pt}
}
\bigskip

\noindent Equations on $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$:
$$\left\{
\begin{array}{l}
t_0^2t_1^3t_2^2+t_0t_1^2t_2(2-2t_2^2+t_0^2(-2+3t_2^2))+t_0t_2(5-4t_2^2+t_2^4+t_0^4(-1+t_2^2)^2+\\
\ +t_0^2(-4+5t_2^2-2t_2^4))+t_1(1-3t_2^2+t_2^4+t_0^2(-3+7t_2^2-4t_2^4)+t_0^4(1-4t_2^2+3t_2^4)) = 2,\\
t_2+(t_1+t_0t_2)(t_0-t_1t_2-t_0t_2^2) = -2,\\
t_0+(t_1+t_2t_0)(t_2-t_1t_0-t_2t_0^2) = -2.\\
\end{array}
\right.$$
Complex solution:
$$\rho_1 = 2,\ \rho_0 = \rho_2 = \frac{-1\pm\sqrt{-3}}{2}.$$

This example appeared as an orbifold that has the sister-figure-eight knot as a
two-shit cover. As we see, the orbifold is arithmetic (by test \cite{Bel}) with the field of
definition $\Q[\sqrt{-3}]$.

\bigskip

\psfig{file=exG.eps, scale=.75}
\hfill
\parbox[b]{0.5\textwidth}{
$\begin{array}{l}
a_1 = bab^{-1}\\
b_1 = cbc^{-1}\\
c_1 = aca^{-1}\\
a_{-1} = c_1^{-1}ac_1\\
c_{-1} = a_1ca_1^{-1}\\
\\
a_1b_1c_1 = bab^{-1}cbc^{-1}aca^{-1}\\
b^{-1}c_{-1}^{-1} = ab^{-1}c^{-1}ba^{-1}b^{-1}\\
a_1b_1 = bab^{-1}cbc^{-1}
\end{array}
$
\vspace{32pt}
}
\bigskip

With our usual notation $t_0=\tr(AB)$, $t_1=\tr(AC)$, $t_2=\tr(BC)$
we obtain $\tr(A_1B_1C_1)=\tr(ABC)(t_0t_1t_2+1) = 0$. Since
$\tr(ABC)\neq 0$ it gives $t_0t_1t_2+1 = 0,$ and so we have the
following system of equations on the parameters:
$$\left\{
\begin{array}{l}
t_0t_1t_2 = -1,\\
t_0t_2^2+t_1t_2-t_0 = -2,\\
t_0^2t_2+t_1t_0-t_2 = -2.
\end{array}
\right.$$
Complex solution:
$$\rho_0 = \rho_1 = \rho_2 = \frac{1\pm\sqrt{-3}}{2}.$$
Since $\rho_0 = \rho_1 = \rho_2$ the orbifold has an order $3$ symmetry,
which cyclically changes the generators of the group. This symmetry can
be easily seen on the spacial representation of the singular graph but
is lost on the plane projection. According to \cite{Bel} the considered
orbifold is one of the three arithmetic non-compact generalized triangle
orbifolds with equal parameters (the remaining two orbifolds correspond
to Case C).

\bigskip

${\it H.}$ We have tried more then 20 different structures for this case but
did not succeed in finding any arithmetic examples. The central part structure
implies rather strong restrictions on the orbifold: it should be non compact
and it is essentially asymmetric. Let us leave as a challenge {\it to find an
arithmetic example for case $H$ or prove that it does not exist}.


\section{ An example of the exceptional generalized triangle orbifold.}

Here we will consider an example of a generalized triangle orbifold, whose
singular set does not correspond to any of the eight types of singularities
from Theorem~2.4. This orbifold appears to be a counterexample to the Lemma~2.3
if we do not suppose that the orbifolds considered there are common.

The example is well known, it is the Picard orbifold. It was first shown in
\cite{Brun} that the Picard group can be generated by only two elements or by
three involutions. Let us reproduce what happens.

\bigskip

\psfig{file=picard.eps, scale=.8}
\hfill
\parbox[b]{0.65\textwidth}{
$\begin{array}{l}
b_1 = cbc^{-1}\\
b_{-1} = aba^{-1}\\
\end{array}
$
\vspace{72pt} }
\bigskip

We won't find the matrix representation in $\SL(2,\C)$ for the Picard group
since it is well known and can be easily obtained from the above. Our goal is
only to show that this group can be generated by three involutions, so we will
make all the calculations in the group, which is a bit easier then working with
the matrices.

We have $$\Gamma = <a,b,c,x \mid a^2, b^2, c^2, x^2, (xac)^2, (xabc)^3,
(bax)^3, (abax)^2>.$$ $(xac)^2 = 1$ implies $acx = xca$;\newline $(abax)^2 =
1$implies $xaba = abax$.\newline

Now consider the word $xabc$: $$1 = (xabc)^3 = xabcxabcxabc = xabcxab\ aa\
cxabc = xabc\ abax\ xca\ abc = xabcabacbc;$$ $$x = abcabacbc.$$

So $\Gamma =<a,b,c \mid a^2, b^2, c^2, (abcabacbc)^2, (abcabacbcac)^2,
(abacb)^3, (acabacbc)^2>.$

\bigskip

Note, that in the notations of Lemma~2.3 here we have $H_1(\tilde{X})\cong
\Z^4$ (so the rank of $\pi_1(\tilde{X})$ is greater then 3 and the Picard group
can not be a common generalized triangle group by the definition) but, the same
time,  $H_1(X)\cong \Z_2\oplus\Z_2$ has rank $2$.

\bigskip

The considered example is non compact. One can obtain an infinite series of the
compact exceptional generalized triangle orbifolds by $(n;0)$ Dehn surgeries
on the cusp of the Picard orbifold.


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A.~W.~Reid, and L.~Siebenmann, de Gruyter, Berlin (1992), 1--15.

\bibitem{Bel_m} M.~Belolipetsky, Arithmetic properties of the groups
of isometries generated by three involutions (in Russian), {\sl Magister
Thesis}, Novosibirsk State University, Novosibirsk (1997), 30~pp.

\bibitem{Bel} M.~Belolipetsky, On arithmetic Kleinian groups generated
by three half-turns, Preprint~00-010, Universit\"at Bielefeld,
Bielefeld~(2000), 17~pp.

\bibitem{Berdon} A.~Berdon, The geometry of discrete groups,
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\bibitem{BM} R.~Brooks, J.~P.~Matelski, The dynamics of $2$-generator
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\bibitem{Brun} A.~M.~Brunner, A two-generator presentation for the Picard group.
{\sl Proc. of Amer. Math. Soc.},~115, N~1 (1992), 45--46.

\bibitem{MM3} M.~D.~E.~Condor, C.~Maclachlan, G.~Martin, E.~A.~O'Brien,
$2$-generator arithmetic Kleinian groups III, {\sl DVI preprint}, Department
of Mathematical Sciences, University of Aberdeen.

\bibitem{Dun} W.~Dunbar, Geometric orbifolds, {\sl Rev. Mat. Univ. Complutense
Madr.},~1 (1988), 67--99.

\bibitem{Fen} W.~Fenchel, Elementary geometry in hyperbolic space,
De Gruyter, 1989.

\bibitem{Fox} R.~Fox, A quick trip through knot theory, Topology of
$3$-manifolds, {\sl Proc. 1961 Top. Inst. Univ. Georgia}, Prentice Hall.

\bibitem{Gilm} J.~Gilman, A discreteness condition for subgroups
of $\PSL(2,\C)$, {\sl Contemp. Math.},~211 (1997), 261--267.

\bibitem{GM} F.~Gehring and G.~Martin, Commutators, collars and the
geometry of M\"obius groups, {\sl J.~d'Analyse Math.},~63 (1994), 175--219.

\bibitem{GGM} F.~Gehring, J.~Gilman, G.~Martin, Discrete groups with real
parameters, {\sl preprint}, 1998.

\bibitem{MM2} F.~Gehring, C.~Maclachlan, G.~Martin, Two-generator arithmetic
Kleinian groups II, {\sl Bull. Lond. Math. Soc.},~30, N~3 (1998), 258--266.

\bibitem{ADV} F.~Gehring, C.~Maclachlan, G.~Martin, A.~Reid,
Arithmeticity, Discretness and Volume, {\sl Trans.~Am.~Math.~Soc.},~349, N~9
(1997), 3611--3643.

\bibitem{HQ} A.~Haefliger, N.~D.~Quach, Une presentation de groupe
fundamental d'une orbifold, {\sl Asterisque},~116 (1984), 98--107.

\bibitem{HKM1} H.~Helling, A.~C.~Kim, J.~L.~Mennicke, A geometric study
of Fibonacci groups, {\sl J. Lie Theory},~8, N~1 (1998), 1--23.

\bibitem{HKM2} H.~Helling, A.~C.~Kim, J.~L.~Mennicke, Some honey-combs
in hyperbolic 3-space, {\sl Comm. in Algebra},~23 (1995), 5169--5206.
%(Vol.23, N14)

\bibitem{HLM1} H.~M.~Hilden, M.-T.~Lozano and J.~M.~Montesinos, A characterisation
of arithmetic subgroups of $\SL(2,\R)$ and $\SL(2,\C)$, {\sl Math. Nachr.},~159
(1992), 245--270.

\bibitem{HLM} H.~M.~Hilden, M.-T.~Lozano, J.~M.~Montesinos, The
arithmeticity of the figure-eight knot orbifolds, In: {\sl Topology'90},
edited by B.~Apanasov, W.~D.~Neumann, A.~W.~Reid, and L.~Siebenmann,
de Gruyter, Berlin (1992), 169--183.

\bibitem{J} T.~J\o rgensen, On discrete groups of M\"{o}bius transformations.
{\sl Amer. J. Math.},~78 (1976), 739--749.

\bibitem{K1} E.~Klimenko, Discrete groups in three-dimensional Lobachevsky
space generated by two rotations, {\sl Sib. Math. J.},~30 (1989), 95--100.

\bibitem{K2} E.~Klimenko, Some remarks on subgroups of $\PSL(2,\C)$,
{\sl Q~\&~A in General Topology},~8, N~2 (1990), 371--381.

\bibitem{KK} E.~Klimenko, N.~Kopteva, Discreteness criteria for
${\mathcal RP}$-groups, {\sl Preprint~00-061}, Universit\"at Bielefeld,
Bielefeld~(2000), 22~pp.

\bibitem{MM1} C.~Maclachlan, G.~Martin, $2$-generator arithmetic Kleinian
groups, {\sl J. Reine Angew. Math.},~511 (1999), 95--117.

\bibitem{MV1} A.~Mednykh, A.~Vesnin, Visualization of the isometry
group action on the Fomenko-Matveev-Weeks manifold,  {\sl J. Lie Theory},~8,
N~1 (1998), 51--66.
%Hyperbolic $3$-manifolds as $2$-fold coverings according to Montesinos,
%{\sl Preprint~95-010}, Universit\"at Bielefeld, Bielefeld (1995), 23~pp.

\bibitem{RV} A.~Rasskazov, A.~Vesnin, Isometries of hyperbolic Fibonacci
manifolds, {\sl Sib. Math. J.},~40 (1999), 9--22.

\bibitem{Thur} W.~Thurston, The geometry and topology of 3-manifolds,
Lecture Notes, Princeton University, 1980.

\end{thebibliography}

\bigskip
\bigskip
\bigskip

%\noindent Michael Belolipetsky

\noindent Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia

\noindent e-mail: \tt mbel@math.nsc.ru

\end{document}

math0103015.extracted_bib
\begin{thebibliography}{99}

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A.~W.~Reid, and L.~Siebenmann, de Gruyter, Berlin (1992), 1--15.

\bibitem{Bel_m} M.~Belolipetsky, Arithmetic properties of the groups
of isometries generated by three involutions (in Russian), {\sl Magister
Thesis}, Novosibirsk State University, Novosibirsk (1997), 30~pp.

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by three half-turns, Preprint~00-010, Universit\"at Bielefeld,
Bielefeld~(2000), 17~pp.

\bibitem{Berdon} A.~Berdon, The geometry of discrete groups,
Springer-Verlag, 1983.

\bibitem{BM} R.~Brooks, J.~P.~Matelski, The dynamics of $2$-generator
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\bibitem{Brun} A.~M.~Brunner, A two-generator presentation for the Picard group.
{\sl Proc. of Amer. Math. Soc.},~115, N~1 (1992), 45--46.

\bibitem{MM3} M.~D.~E.~Condor, C.~Maclachlan, G.~Martin, E.~A.~O'Brien,
$2$-generator arithmetic Kleinian groups III, {\sl DVI preprint}, Department
of Mathematical Sciences, University of Aberdeen.

\bibitem{Dun} W.~Dunbar, Geometric orbifolds, {\sl Rev. Mat. Univ. Complutense
Madr.},~1 (1988), 67--99.

\bibitem{Fen} W.~Fenchel, Elementary geometry in hyperbolic space,
De Gruyter, 1989.

\bibitem{Fox} R.~Fox, A quick trip through knot theory, Topology of
$3$-manifolds, {\sl Proc. 1961 Top. Inst. Univ. Georgia}, Prentice Hall.

\bibitem{Gilm} J.~Gilman, A discreteness condition for subgroups
of $\PSL(2,\C)$, {\sl Contemp. Math.},~211 (1997), 261--267.

\bibitem{GM} F.~Gehring and G.~Martin, Commutators, collars and the
geometry of M\"obius groups, {\sl J.~d'Analyse Math.},~63 (1994), 175--219.

\bibitem{GGM} F.~Gehring, J.~Gilman, G.~Martin, Discrete groups with real
parameters, {\sl preprint}, 1998.

\bibitem{MM2} F.~Gehring, C.~Maclachlan, G.~Martin, Two-generator arithmetic
Kleinian groups II, {\sl Bull. Lond. Math. Soc.},~30, N~3 (1998), 258--266.

\bibitem{ADV} F.~Gehring, C.~Maclachlan, G.~Martin, A.~Reid,
Arithmeticity, Discretness and Volume, {\sl Trans.~Am.~Math.~Soc.},~349, N~9
(1997), 3611--3643.

\bibitem{HQ} A.~Haefliger, N.~D.~Quach, Une presentation de groupe
fundamental d'une orbifold, {\sl Asterisque},~116 (1984), 98--107.

\bibitem{HKM1} H.~Helling, A.~C.~Kim, J.~L.~Mennicke, A geometric study
of Fibonacci groups, {\sl J. Lie Theory},~8, N~1 (1998), 1--23.

\bibitem{HKM2} H.~Helling, A.~C.~Kim, J.~L.~Mennicke, Some honey-combs
in hyperbolic 3-space, {\sl Comm. in Algebra},~23 (1995), 5169--5206.
%(Vol.23, N14)

\bibitem{HLM1} H.~M.~Hilden, M.-T.~Lozano and J.~M.~Montesinos, A characterisation
of arithmetic subgroups of $\SL(2,\R)$ and $\SL(2,\C)$, {\sl Math. Nachr.},~159
(1992), 245--270.

\bibitem{HLM} H.~M.~Hilden, M.-T.~Lozano, J.~M.~Montesinos, The
arithmeticity of the figure-eight knot orbifolds, In: {\sl Topology'90},
edited by B.~Apanasov, W.~D.~Neumann, A.~W.~Reid, and L.~Siebenmann,
de Gruyter, Berlin (1992), 169--183.

\bibitem{J} T.~J\o rgensen, On discrete groups of M\"{o}bius transformations.
{\sl Amer. J. Math.},~78 (1976), 739--749.

\bibitem{K1} E.~Klimenko, Discrete groups in three-dimensional Lobachevsky
space generated by two rotations, {\sl Sib. Math. J.},~30 (1989), 95--100.

\bibitem{K2} E.~Klimenko, Some remarks on subgroups of $\PSL(2,\C)$,
{\sl Q~\&~A in General Topology},~8, N~2 (1990), 371--381.

\bibitem{KK} E.~Klimenko, N.~Kopteva, Discreteness criteria for
${\mathcal RP}$-groups, {\sl Preprint~00-061}, Universit\"at Bielefeld,
Bielefeld~(2000), 22~pp.

\bibitem{MM1} C.~Maclachlan, G.~Martin, $2$-generator arithmetic Kleinian
groups, {\sl J. Reine Angew. Math.},~511 (1999), 95--117.

\bibitem{MV1} A.~Mednykh, A.~Vesnin, Visualization of the isometry
group action on the Fomenko-Matveev-Weeks manifold,  {\sl J. Lie Theory},~8,
N~1 (1998), 51--66.
%Hyperbolic $3$-manifolds as $2$-fold coverings according to Montesinos,
%{\sl Preprint~95-010}, Universit\"at Bielefeld, Bielefeld (1995), 23~pp.

\bibitem{RV} A.~Rasskazov, A.~Vesnin, Isometries of hyperbolic Fibonacci
manifolds, {\sl Sib. Math. J.},~40 (1999), 9--22.

\bibitem{Thur} W.~Thurston, The geometry and topology of 3-manifolds,
Lecture Notes, Princeton University, 1980.

\end{thebibliography}