
coordinates are $(x,y,z,t)=(2,-2,2,0)$. The previous discussion implies 
that $x^2(\chi_s)=z^2(\chi_s)=4$ and $y^2(\chi_s)$ is non-constant, hence a 
contradiction, because $x^2$, $y^2$ and $z^2$ are functions that do not depend 
on the lift from $PSL(2, \mathbb{C})$ to $SL(2, \mathbb{C})$.

The case $(x,y,z,t)=(2,2,-2,0)$ being analogous, we next assume that 
$(x,y,z,t)=(-2,-2,-2,0)$. We repeat the construction of the representation of 
$\rho_{0,s}$ as above, with the difference that now 
$$
\lambda_1(s)=\lambda_2(s)=\lambda_3(s)=\lambda_4(s),
$$ 
which implies that $x^2(\chi_s)=y^2(\chi_s)=z^2(\chi_s)$, leading again to a
contradiction.
\end{proof}









\section{A unique Conway sphere}
\label{section:thm1}

The aim of this section is to study the case when the link admits a unique
Conway sphere $C=C_1$. With the same notation as in the previous section, we 
shall denote by ${\mathcal O}_i$, $i=1,2$, the two orbifolds of the
decomposition. Since the Bonahon-Siebenmann family must be contained in
$\{C\}$, we see that the two orbifolds ${\mathcal O}_i$ are geometric. Note, 
moreover, that if both orbifolds are Seifert, their fibrations cannot
match, for else the base of the global fibration would be large
 and the link would then 
admit some other Conway sphere. In particular $C$ is precisely the only element
of the Bonahon-Siebenmann family.

\begin{Lemma}
\label{lemma:euclidean}
Assume ${\mathcal O}_i$ is a Seifert fibred orbifold, with Euclidean base. Then
the associated tangle consists of two straight vertical arcs plus an unknotted 
circle isotopic to the equator.
\end{Lemma}

\begin{proof}
The base of the fibration is a $2$-dimensional Euclidean orbifold. These are 
completely classified (see \cite{ThurstonBook}), and we can check that there 
are only two planar orbifolds with a unique boundary component (non planar 
bases cannot correspond to fibrations of the ball). The first one is obtained 
in the following way: quotient a ``pillowcase" by the reflection in a
great circle containing its four cone points and cut in half the resulting
orbifold. This way one obtains a rectangle with three mirrored sides and one 
boundary component, and two corner points (with angle $\pi/2$ and of dihedral 
type). The second one is obtained by quotienting a M\"obius band by a
reflection in a line orthogonal to its core. The result is a bigon with one 
mirrored side and one cone point of order $2$ in its interior (note that the
reflection fixes an extra point on the central curve of the M\"obius band). 
Both base orbifolds can also be seen as the quotient by two different 
reflections of a disc with two cone points of order $2$: the axis of the first 
reflection contains both cone points, while the second reflection exchanges
them. It is not hard to see that these bases correspond to the two different 
fibrations of the tangle described in the statement of the lemma, which are 
pictured in Figure \ref{fig:eucfib}. Note that the extra closed component of 
the tangle is a fibre of the second fibration, corresponding to the cone point 
of the base orbifold.
\end{proof}


\begin{figure}
\begin{center}
{
\includegraphics[height=4cm]{eucfib.eps}
}
\end{center}
   \caption{Two fibrations of the tangle with Euclidean base.}
\label{fig:eucfib}
\end{figure}



By putting a branching index $3$ on the closed component of the tangle of 
Lemma~\ref{lemma:euclidean} we get the following:

\begin{Corollary}
We can modify the branching indices of the orbifolds $\mathcal O_1$
and $\mathcal O_2$ so that:
\begin{itemize}

\item The singular components that meet the Conway sphere have branching index 
$2$;

\item The two resulting orbifolds, which will again be denoted by $\mathcal
O_1$ and $\mathcal O_2$, are either hyperbolic or Seifert fibred with 
hyperbolic base.
\end{itemize}
\end{Corollary}

\begin{Remark}\label{remark:ramification}
From now on $\mathcal O_i$, $i=1,2$, will denote the orbifold with underlying 
topological space $M_i$ and singular set $M_i\cap L$ with branching order $2$ 
on the components meeting the boundary and $3$ on the closed components. Remark 
that now $\mathcal O_i$, $i=1,2$, satisfies the conclusions of the above 
corollary. We shall denote by $\mathcal O_i'$, $i=1,2$, the compact orbifold 
obtained from $\mathcal O_i$ by drilling out open regular neighbourhoods of the 
two singular components of branching index $2$, i.e. those that meet the 
boundary.
\end{Remark}

Let $\chi_i\in X(\mathcal O'_i)$ be a lift of the character induced by the 
holonomy of the geometric structure of $\mathcal O_i$, chosen so that it
satisfies the compatibility condition established in 
Lemma~\ref{Lemma:liftsmatch}. Notice that Culler's theorem \cite{culler} allows 
to lift the representation of $\pi_1(M_i')$, and since $M_i'$ is obtained from 
$\mathcal{O}'_i$ by removing all components of order $3$, the lift extends from 
$M_i'$ to $\mathcal O'_i$, by changing the signs of the corresponding meridians
 as in the proof of Lemma~\ref{Lemma:liftsmatch}, 
if needed.

\begin{Lemma}\label{Lemma:dim3}
The character $\chi_i$ is a smooth point in $X(\mathcal O'_i)$. The traces of 
the two meridians and of a peripheral element of $\mathcal O_i$ define local 
coordinates in a neighbourhood of $\chi_i$.
\end{Lemma}

\begin{proof}

For a manifold admitting an irreducible representation, Theorem 5.6 in
Thurston's notes \cite{ThurstonNotes} states that the complex dimension of the
character variety is at least $-3e/2 +\theta$ where $e$ is the Euler characteristic
of the boundary of the manifold and $\theta$ the number of tori in the boundary.
If we fill some toric components with singular solid tori, in such a way that
the resulting orbifold still admits an irreducible representation, then
the same dimension bound still holds. This means that we can apply Thurston's
result to conclude that the variety of characters of $\mathcal O'_i$ has
dimension at least three. This gives a lower bound on the dimension.

To obtain an upper bound, we must work in the Zariski tangent space, i.e. the 
first cohomology group with coefficients in the Lie algebra twisted by the 
adjoint representation, that we denote by 
$H^1(\mathcal O'_i;\mathfrak{sl}(2,\mathbb C))$.

We first show that the variety of $PSL(2,\mathbb C)$-characters 
$X(\mathcal O_i,PSL(2,\mathbb C))$ is one-dimensional, locally parametrised by 
deformations of its boundary $\partial\mathcal O_i$ (since 
$ \pi_1(\mathcal O_i)$ has 2-torsion we must work in $PSL(2,\mathbb C)$ instead 
of $SL_2(\mathbb C)$). Moreover the Zariski tangent space at the holonomy 
character is also one dimensional, namely for the holonomy representation we 
have: $H^1(\mathcal O_i,\mathfrak{sl}(2,\mathbb C))\cong \mathbb C$.
In the hyperbolic case, this follows from the proof of Thurston's hyperbolic
Dehn filling theorem \cite{ThurstonNotes, KapovichBook} (see also
\cite{BoileauPorti} for the precise statement for orbifolds.) In the Seifert
fib