\section{Introduction}
It is one of the most important topics in the theory of Kleinian
group to study the topological structure of deformation spaces of
Kleinian groups. For such studies, it is indispensable to
guarantee that under some conditions, sequences of Kleinian groups
converge inside the deformation spaces. In other words, we need a
sufficient condition for a given sequence of Kleinian groups to
converge algebraically,  which should be as general as possible.
In particular, such a sufficient condition for quasi-conformal
deformations of geometrically finite Kleinian groups interests us
most. The quasi-conformal deformation space of a geometrically
finite Kleinian group $G$ is well understood by virtue of the work
of Ahlfors, Bers, Kra, Marden and Sullivan. To put it more
concretely, for a geometrically finite Kleinian group $G$, it is
known that there is a ramified covering map from the
Teichm\"{u}ller space of $\Omega_G/G$ to the quasi-conformal
deformation space of $G$, where $\Omega_G$ denotes the region of
discontinuity of $G$. Therefore, the sufficient condition for the
quasi-conformal deformation space should be expressed in terms of
sequences of the Teichm\"{u}ller spaces.

The first example of such a  sufficient condition for convergence is the result of  Bers in \cite{Be}, which shows that  the space of quasi-Fuchsian groups lying on the Bers slice is relatively compact.
On the other hand, in the process of proving the uniformisation theorem for Haken manifolds, Thurston proved
the
double limit theorem for quasi-Fuchsian groups and the compactness of deformation spaces
for acylindrical manifolds, in \cite{thui} and \cite{thuii} respectively.
These are
generalised to give a convergence theorem for general freely indecomposable Kleinian
groups in Ohshika \cite{OhZ} and \cite{OhI}.
The convergence in the deformation spaces for
freely decomposable groups is more complicated and is harder to understand.

In \cite{thb}, Thurston listed questions concerning Kleinian groups and 3-manifolds.
One of them asks how one can generalise the double limit theorem to the setting of Schottky groups.
This question was made into a more concrete conjecture using the notion of Masur domain, and then was generalised to function groups.
 Masur introduced in \cite{MaE} an open set in the projective  lamination space
of the boundary of a handlebody on which the mapping class group of the handlebody acts properly discontinuously.
This open set is what we call the Masur domain nowadays.
This notion is generalised by Otal \cite{chefjeune} to the exterior boundary of a compression body.
Thurston's conjecture is paraphrased as follows:
For a sequence in the Teichm{\"u}ller space converging in the
Thurston compactification to a projective lamination lying in the Masur domain of the
exterior boundary of a compression body $M$,
the corresponding sequence of convex cocompact representations in $AH(M)$ would converge after passing to a subsequence.
Otal in \cite{conti} first proved that Thurston's conjecture is true for rank-$2$ Schottky space provided that the limit lamination is arational, that is, any component of its complement is simply connected.
Canary in \cite{CanD} proved the conjecture for some special
sequences in Schottky space.
Ohshika in \cite{OhA} proved the conjecture for function
groups which are isomorphic to the free products of two surface groups under the same
assumption that the limit lamination is arational.
The strongest result in this direction
under the same assumption on the limit lamination was given by Kleineidam and Souto in
\cite{boches} without any other assumption on compression bodies.
Our main theorem, Theorem \ref{main} yields a
proof of this conjecture of Thurston in full generality without any extra assumption and generalises it to a
slightly larger set than the Masur domain.





We need to introduce some notions and notations to state our main theorem.
Consider a compact irreducible atoroidal $3$-manifold $M$ with boundary.
By Thurston's uniformisation theorem for atoroidal Haken manifolds, there is a
representation $\rho_0:\pi_1(M)\rightarrow \mathrm{Isom}({\mathbb H}^3)$ with the following properties :
$\rho_0(\pi_1(M))$ is geometrically finite, ${\mathbb H}^3/\rho_0(\pi_1(M))$ is homeomorphic to
$\mathrm{Int}(M)$, and any maximal parabolic subgroup of $\rho_0(\pi_1(M))$ is an Abelian
group of rank $2$.
Such a representation is said to {\it uniformise $M$}. Any quasi-conformal deformation of $\rho_0$  also uniformises $M$.
By the Ahlfors-Bers theory, the space $QH(\rho_0)$ of quasi-conformal deformations of $\rho_0$ up to conjugacy by  elements of $\mathrm{Isom}({\mathbb H}^3)$ is parametrised by the Teichm{\"u}ller space of the boundary of $M$.
More precisely, there is a (possibly ramified) covering map, called the {\it Ahlfors-Bers map} $\cal{T}(\partial_{\chi<0} M)\rightarrow
QH(\rho_0)$ whose covering transformation group is the group of isotopy classes of diffeomorphisms of $M$ which are homotopic to the identity.

The space $QH(\rho_0)$ is a subspace of the {\em deformation space} $AH(M)$.
 This deformation space $AH(M)$ is the space of discrete faithful representations $\rho:\pi_1(M)\rightarrow \mathrm{Isom}({\mathbb H}^3)$ up to conjugacy by elements of
$PSL_2({\mathbb C})$.
 It is endowed with the compact-open topology which is also called an {\em algebraic topology}.
In the main theorem, we shall consider sequences of representations given by sequences in the Teichm{\"u}ller
space
whose images under the Ahlfors-Bers map diverge in $QH(\rho_0)$ and give a sufficient condition for their convergence
in $AH(M)$.

Thurston introduced in \cite{thuiii} the notion of {\em doubly incompressible
curves}. This  can be extended to measured geodesic laminations in the following way.
\\
\indent
We say that a measured geodesic lamination $\lambda\in\cal{ML}(\partial M)$
is {\em doubly incompressible} if and only if \smallskip\\
\indent - $\exists\eta>0$ such that $i(\lambda,
\partial E)>\eta$ for any essential annulus, M{\"o}bius band or disc $E$.\smallskip\\
We denote by $\cal{D}(M)\subset\cal{ML}(
\partial M)$ the set of doubly incompressible measured geodesic laminations and by
$\cal{PD}(M)$ its projection in the projective lamination space $\cal{PML}(
\partial M)$.
It is not hard to see that $\cal{D}(M)$ contains the Masur domain (see \cite{masurext}).


Now we can state our main theorem.

\begin{thm}
\label{main}
Let $M$ be a compact irreducible atoroidal $3$-manifold with boundary, and $\rho_0 :
\pi_1(M) {\rightarrow} PSL(2,{\mathbb C})$ a geometrically finite representation that uniformises $M$.
Let $(m_n)$ be a sequence in the Teichm{\"u}ller space $\mathcal{T}(\partial M)$ which
converges in the Thurston compactification to a projective measured
lamination $[\lambda]$ contained in $\cal{PD}(M)$.
Let $q : \mathcal{T}(\partial M)\rightarrow QH(\rho_0)$ be the Ahlfors-Bers map, and suppose that
$(\rho_n:\pi_1(M){\rightarrow} G_n \subset PSL(2,{\mathbb C}))$ is a sequence of discrete faithful
representations corresponding to $q(m_n)$. Then passing to a subsequence, $(\rho_n)$
converges in $AH(M)$.
\end{thm}




The proof of Theorem \ref{main} proceeds as follows. By Theorems of Thurston \cite{thuii}
and Canary \cite{CanD}, the convergence of $m_n$ to $[\lambda]$ implies that there is a
sequence of weighted multi-curves $\lambda_n\in\cal{ML}(\partial M)$ such that
$l_{\rho_n}(\lambda_n)$ tends to $0$ and that the sequence $(\lambda_n)$ converges in
$\cal{ML}(\partial M)$ to a measured geodesic lamination whose projective class
is $[\lambda]$. Since $\lambda$ lies in $\cal{D}(M)$, the result comes from the following
theorem whose proof occupies the main part of this paper.

\begin{thm}
\label{th&}
Let $(\rho_n:\pi_1(M)\rightarrow \mathrm{Isom}({\mathbb H}^3))$ be a sequence of
discrete faithful representations that uniformise $M$ and let $(\lambda_n)\subset\cal{ML}(
\partial M)$ be a sequence of measured geodesic laminations such that
$(l_{\rho_n}(\lambda_n))$ is a bounded sequence and that $(\lambda_n)$ converges
in $\cal{ML}(
\partial M)$ to a measured geodesic lamination $\lambda\in\cal{D}(M)$. Then the sequence
$(\rho_n)$ has a compact closure in $AH(M)$, namely, any subsequence contains an
algebraically convergent subsequence.
\end{thm}

It should be noted that our result here is closely related to the Bers-Thurston density conjecture.
This conjecture states that every finitely
generated Kleinian group would be contained in the boundary
of the quasi-conformal deformation space of geometrically finite Kleinian groups without rank-$1$ maximal parabolic subgroups.
The conjecture was proved by combining the resolution  of the tameness conjecture by Agol and
 Calegari-Gabai, the ending lamination conjecture by Brock-Canary-Minsky, and
from some convergence theorems due to Thurston, Ohshika, Kleineidam-Souto and
Lecuire,  together with some topological argument due to  Ohshika and Namazi-Souto(See \cite{Ag}, \cite{CG}, \cite{thuiii}, \cite{OhA}, \cite{boches}, \cite{masurext}, \cite{Min}, \cite{OhP} and  \cite{Namazi}).
Although the convergence theorems  less general than ours were sufficient for this proof, we can simplify some of the argument there using our general result in this paper.

\section{Preliminaries}
\subsection{Deformation space}
Let $G$ be a finitely generated torsion-free Kleinian group, namely a (torsion-free and
finitely generated) discrete subgroup of $\mathrm{Isom}^+({\mathbb H}^3)$. We denote by
$({\mathbb H}^3/G)_0$ the non-cuspidal part of $({\mathbb H}^3/G)$, that is, the complement of
neighbourhoods of the cusps. We call an end of $({\mathbb H}^3/G)_0$ {\em geometrically finite}
when it has a neighbourhood intersecting no closed geodesics, and otherwise {\em
geometrically infinite}. The Kleinian group $G$ is geometrically finite if and only if
every end of $({\mathbb H}^3/G)_0$ is geometrically finite. Let $\Omega_G \subset S^2_\infty$
denote the region of discontinuity for the action of $G$ on the Riemann sphere
$S^2_\infty$. Each geometrically finite end corresponds to a component $\Sigma$ of
$\Omega_G/G$ in such a way that a neighbourhood of the end is compactified in the Kleinian manifold $({\mathbb H}^3\cup
\Omega_G)/G$ by adding $\Sigma$.

Let $M$ be a compact irreducible atoroidal manifold. Let $AH(M)$ denote the set of faithful
discrete representations from $\pi_1(M)$ to $PSL(2, {\mathbb C})$ modulo conjugacy. We endow
$AH(M)$ with the  topology induced from the representation space. We say that a
representation $\rho\in AH(M)$ has minimal parabolics if all of its maximal
parabolic
subgroups have rank $2$. The subspace of $AH(M)$ consisting of geometrically finite representations with
minimal parabolics is denoted by $CC(M)$. This space $CC(M)$ may contain several connected
components. The component consisting of representations $\rho$ for which there is a
homeomorphism from $\mathrm{Int}(M)$ to ${\mathbb H}^3/\rho(\pi_1(M))$ is denoted by $CC_0(M)$. The representations in $CC_0(M)$ are said to uniformise $M$.


For a Kleinian group $G$, if there is a quasi-conformal automorphism $f$ of
$S^2_\infty$ such that $f Gf ^{-1}$ is again a Kleinian group, then this group $f G
f^{-1}$ is said to be a {\em quasi-conformal deformation} of $G$. Quasi-conformal
deformations preserve the parabolic subgroups. Therefore a quasi-conformal deformation of a
Kleinian group with minimal parabolics has minimal parabolics. Furthermore
quasi-conformal deformations of a geometrically finite group are geometrically finite. By
the theory of Ahlfors-Bers, there is a ramified covering map from
$\mathcal{T}(\Omega_G/G)$ to the
space of quasi-conformal deformations of $G$ modulo conjugacy. Let $\partial_{\chi<0} M$
be the union\linebreak of the components of $\partial M$ which have negative Euler
characteristics. For\linebreak $G \in CC_0(M)$, there is a natural identification of $\Omega_G/G$ with $
\partial_{\chi<0} M$. The covering map $\mathcal{T}(
\partial_{\chi<0} M) \rightarrow CC_0(M)$ is called the {\em Ahlfors-Bers map}.

\subsection{${\mathbb R}$-trees}
An ${\mathbb R}$-tree $\cal{T}$ is a geodesic metric space in which any two points $x,y$ can be
joined by a unique simple arc. Let $G$ be a group acting by isometries on an ${\mathbb R}$-tree
$\cal{T}$. The action is {\em minimal} if there is no proper invariant subtree and {\em
small} if the stabilizer of any non-degenerate arc is virtually Abelian.

Morgan and Shalen \cite{mors1} made use of ${\mathbb R}$-trees to compactify  deformation
spaces. They used algebraic methods involving valuations, while the same result
has been obtained in Paulin
\cite{paulin} and
Bestvina \cite{best} using a more geometrical approach. In this paper we shall adopt the
point of
view of Kapovich-Leeb \cite{kl} (see also \cite[chapters 9 and 10]{kapo}). Let
$(\rho_n\subset
CC_0(M))$ be a sequence of representations such that no subsequence of $(\rho_n)$ converges
algebraically. Let $\Gamma \subset\pi_1(M)$ be a  set of generators and let
$\tilde x_n\subset{\mathbb H}^3$
be a point realising the minimum $\epsilon^{-1}_n$ on ${\mathbb H}^3$ of the function
$\max\{d(\tilde x,\rho_n(a)(\tilde x)),a\in \Gamma\}$ (see for example \cite{paulin} for
the existence of such a point). Since no subsequence of $(\rho_n)$ converges
algebraically,
$(\epsilon_n^{-1})$ tends to $\infty$. Choose a non-principal ultra-filter $\omega$ and
denote by $\epsilon_n{\mathbb H}^3$ the hyperbolic space ${\mathbb H}^3$ with the hyperbolic metric rescaled
by $\epsilon_n$. The ultra-limit $(X_\omega,x)=\omega-\lim (\epsilon_n{\mathbb H}^3,x_n)$ of the sequence
of rescaled spaces is defined as follows. Let $\Pi_{n}(\epsilon_n{\mathbb H}^3)$ be the infinite
product of the spaces $(\epsilon_n{\mathbb H}^3)$. We define a pseudo-distance $d_\omega$
on
$\Pi_{n}(\epsilon_n{\mathbb H}^3)$ by setting $$
d_\omega(y,z)=\omega-\lim d_{\epsilon_n{\mathbb H}^3}(\tilde y_n,\tilde z_n)$$
for any two points $y=(\tilde y_n)$ and $z=(\tilde z_n)$ lying in
$\Pi_{n}(\epsilon_n{\mathbb H}^3)$.

This function $d_\omega$ is a pseudo-distance in $\Pi_{n}(\epsilon_n{\mathbb H}^3)$ with
values in $[0,\infty]$ and we set
$(X_\omega,d_\omega)=(\Pi_{n}(\epsilon_n{\mathbb H}^3),d_\omega)/\sim$ where we identify points with
zero $d_\omega$-distance. Let $x=(\tilde x_n)$ denote the sequence of points $\tilde x_n$
defined above. The metric space $(X_\omega,x)$ is the set of points of $(X_\omega)$ with a
finite distance from $x$. This metric space is an ${\mathbb R}$-tree (cf. \cite{kl}). The action
of $\rho_n(\pi_1(M))$ on $\epsilon_n{\mathbb H}^3$ gives rise to an action of $\pi_1(M)$ on
$(X_\omega,x)$ by isometries. This action is small (cf. \cite{kl}). Let $\cal{T}$ be the
minimal invariant subtree of $X_\omega$ under this action. We say that
$(\rho_n)$
tends
to the action of $\pi_1(M)$ on $\cal{T}$ with respect to $\omega$. For $c\in\pi_1(M)$ let
us denote by $\delta_\cal{T}(c)$ the minimal translation distance of $c$ on $\cal{T}$.
Then
we have $\delta_\cal{T}(c)=\lim \epsilon_n^{-1} l_{\rho_n}(c)$, where $l_{\rho_n}(c)$ is the
length in ${\mathbb H}^3/\rho_n(\pi_1(M))$ of the closed geodesic in the free homotopy class of
$c$.

\subsection{Geodesic laminations}
\indent A {\em geodesic lamination} $L$ on a complete hyperbolic surface $S$ is a compact
set which is a disjoint union of complete embedded geodesics called {\em leaves}. It
is a classical fact that this definition is independent of a chosen hyperbolic metric on
$S$ (see \cite{meister2} for example). For a connected geodesic lamination $L$ which is
not a simple closed curve we denote by $\bar S(L)$ the smallest surface with compact geodesic
boundary containing $L$. Inside $\bar S(L)$ there are finitely many closed geodesics
(including the components of $
\partial\bar S(L)$) disjoint from $L$. These closed geodesics do not intersect each other
(cf. \cite{espoir}) and we denote by $
\partial'\bar S(L)\supset
\partial\bar S(L)$ their disjoint union. Removing a small tubular neighbourhood of
$\partial'\bar S(L)$ from $\bar S(L)$ we get an open surface $S(L)$. We call $S(L)$
the {\it surface embraced} by the geodesic lamination $L$ and $\partial'\bar S(L)$ the
{\it effective boundary} of $S(L)$. If $L$ is a simple closed curve, we define $S(L)$ to
be an annular neighbourhood of $L$ and we take $\partial'\bar S(L)=L$. When $L$ is not
connected, $S(L)$ is the disjoint union of the surfaces embraced by the connected
components of $L$.

We say that two geodesic measured laminations $L$ and $L'$ intersect
transversely if {\em at least} one leaf of $L$ intersects a leaf of $L'$
transversely.

 A {\em measured geodesic lamination} $\lambda$ is a geodesic lamination
$|\lambda|$ together with a transverse measure.
We always assume that $\lambda$ has $|\lambda|$ as its support.
We denote by
$\cal{ML}(S)$ the space of
measured geodesic laminations on $S$ endowed with the weak-$*$ topology.
To simplify the notations, we write $\cal{ML}(\partial M)$ instead of $\cal{ML}(\partial_{\chi<0} M)$ for a compact $3$-manifold $M$ with boundary.
The projective lamination space $\cal{PML}(\partial M)$ is defined to be
$(\cal{ML}(\partial M) -\{0\})/{\mathbb R}_+^*$ where $0$ stands for the measured lamination with empty support.
It should be noted that $\cal{ML}(\partial M)$ contains measured laminations whose restriction to some
component of $\partial M$ is empty.
The Teichm{\"u}ller space $\cal{T}(\partial M)$ denotes similarly
$\cal{T}(\partial_{\chi<0}M)$.
The boundary of the Thurston compactification of $\cal{T}(\partial M)$ is equal to
$\cal{PML}(\partial M)$.



\subsection{Train tracks and their realisations}

A {\em train track} $\tau$ in a hyperbolic surface $S$ is a  union of finitely
many rectangles
which meet each other only along non-degenerate segments contained in their
vertical sides. The
rectangles are called {\em branches}, and they are foliated by vertical segments called
{\em
ties}. A connected component of the intersection of the branches is called a {\em switch}.
The
branches are also foliated by horizontal segments, and a smooth arc which is a union of
horizontal segments is called a {\em rail} or a {\em train route}. A geodesic lamination is {\em carried} by
$\tau$ if it lies in $\tau$ in such a way that the leaves are transverse to the ties (see
\cite{Bouts} or
\cite{meister2} for more details about train tracks).

When $\tau$ is a train track and $\lambda$ is a measured geodesic lamination
whose support is carried by $\lambda$, we say that $\lambda$ is carried by $\tau$.
For a branch $b$ of $\tau$, we define the number $\lambda(b)$ to be the $\lambda$-measure
of a tie of $b$. This number does not depend on the choice of a tie in $b$.


Consider an action of $\pi_1(S)$ on an ${\mathbb R}$-tree $\cal{T}$ by isometries. A
measured lamination $\lambda$ is said to be {\em realised} in $\cal{T}$ if there is an
equivariant map (with respect to the action of $\pi_1(S)$) $\phi: {\mathbb H}^2 {\rightarrow} {\cal T}$ such that
the restriction of $\phi$ to any lift of a leaf of $\lambda$ in ${\mathbb H}^2$ is monotonic and
non-constant. A train track $\tau\in S$ is said to be realised in $\cal{T}$ if there is
an equivariant map $\phi: {\mathbb H}^2 {\rightarrow} {\cal T}$ which maps each lift of a branch of $\tau$ to a
non-trivial geodesic segment on ${\cal T}$ in such a way that each rail is mapped injectively,
and that each tie collapses to a point. By \cite{meister2}, $\lambda$ is realised in
$\cal{T}$ if and only if $\lambda$ is carried by a train track $\tau$ which is realised in
$\cal{T}$.

We say that a measured lamination $\lambda$ is {\em collapsed} by $\phi$ when
there is a train track $\tau$ carrying $\lambda$ such that every component of
$\tilde{\tau}$, the preimage of $\tau$ in ${\mathbb H}^2$, is mapped to a point by $\phi$. It is
straightforward that there exists an
equivariant map collapsing $\lambda$ if and only if the action of $i_*(\pi_1(S(\lambda))$
on $\cal{T}$ has a global fixed point.

\subsection{Compactification of $\tilde M$}

We denote by $\tilde M$ the universal covering of $M$ and by $p:\tilde M\rightarrow M$ the
covering projection. We compactify $\tilde M$ in the following way : endow $M$ with a
geometrically finite hyperbolic metric $\sigma$ with minimal parabolics and let us denote
by $N(\sigma)^{\rm thick}$ the complement in the convex core $N(\sigma)$ of
$\epsilon$-thin neighbourhood of
the cusps of $\sigma$ for some $\epsilon$ smaller than the Margulis constant. Let us choose
an isometry between the interior of $\tilde M$ and
${\mathbb H}^3$. Now we can consider $\tilde N(\sigma)^{\rm thick}$ as a closed subset of ${\mathbb H}^3$.
Since $\sigma$ is geometrically finite, there is a natural homeomorphism between $M$ and
$N(\sigma)^{\rm thick}$. Therefore we can regard $\tilde M$ as a closed subset of ${\mathbb H}^3$.
The
compactification $\overline{\tilde M}$ of $\tilde M$ is the closure of this closed subset
in the usual compactification of ${\mathbb H}^3$ by the unit ball. If we replace $\sigma$ by another
geometrically finite metric $\sigma'$ with minimal parabolics, it follows from
results of \cite{floyd} that we get a compactification which is homeomorphic to the one obtained with
$\sigma$. Therefore this definition is independent of the metric we chose. We
call this the Floyd-Gromov compactification of $\tilde M$.


\indent A {\em meridian} is a simple closed curve $c\subset
\partial M$ which bounds a disc in $M$ but not on $\partial M$. A compact surface
$\Sigma\subset
\partial M$ is incompressible if it contains no meridians. When we consider the
closure of a lift of an incompressible surface in $\overline{\tilde M}$, we have the
following :

\begin{lemma} \label{disque} Let $\Sigma\subset
\partial_{\chi<0} M$ be a compact connected incompressible surface which does not contain any
essential closed curve homotopic into $\partial_{\chi=0} M$.
Let $\hat\Sigma$ be the
universal covering of $\Sigma$, which is completed
in the usual way to a closed disc $\overline{\hat\Sigma}$. Then any lift of $\Sigma$ to $
\partial\tilde M$ is a disc whose closure in $\overline{\tilde M}$ is homeomorphic to
$\overline{\hat\Sigma}$ in an equivariant way.
\end{lemma}

\begin{proof}
This Lemma was proved in \cite{chefjeune}, see also \cite[Lemme 2.1]{espoir}.
\end{proof}


Let $\Sigma\subset
\partial M$ be a compact incompressible surface. Johannson and Jaco-Shalen defined a
characteristic submanifold $W$ relative to $\Sigma$ (cf. \cite{jojo} and \cite{jacs}). Notice that such a characteristic manifold was defined for any incompressible surface in the boundary of a Haken manifold.
Since we are considering atoroidal $M$ now, such a characteristic
submanifold is a disjoint union of
essential $I$-bundles over closed surfaces and essential solid tori. A disjoint
union $W$ of essential $I$-bundles and essential solid tori is a characteristic
submanifold if and only if it has the following two properties :\smallskip

\begin{itemize}
\item any essential $I$-bundle and any essential solid torus in $(M,\Sigma)$ can be homotoped in $W$;
\item no connected component of $W$ can be homotoped into another connected component of
$W$.\smallskip
\end{itemize}


By \cite{jojo} and \cite{jacs}, if $W$ and $W'$ are two  characteristic submanifolds
relative to $\Sigma$, then there is a diffeomorphism $\psi:M\rightarrow M$ isotopic to the
identity relative to $\partial M-\Sigma$ such that $\psi(W)=W'$ and that $\psi
(W\cap\Sigma)=W'\cap\Sigma$.


Such a characteristic submanifold can be found by looking only at $\overline{\tilde M}-\tilde
M$.

\begin{prop}[\cite{espoir}, \textsection $2$] \label{carac} Let $\Sigma$ and $\Sigma'\subset
\partial_{\chi<0} M$ be two compact, connected, incompressible surfaces which are disjoint
or equal and do not contain any essential closed curve which can be homotoped in $
\partial_{\chi=0} M$. Let $\tilde\Sigma\subset
\partial\tilde M$ (resp. $\tilde\Sigma'$) be a connected component of the preimage of
$\Sigma$ (resp. $\Sigma'$) and let $\Gamma\subset\rho(\pi_1(M))$ (resp. $\Gamma'$) be the
stabiliser of $\tilde\Sigma$ (resp. $\Gamma'$).

Then $\overline{\tilde\Sigma} \cap\overline{\tilde\Sigma}'$ is either empty or
equal to the limit set of $\Gamma \cap\Gamma'$.

In the latter case, if $\Gamma \cap\Gamma'$ is not cyclic, then it is the fundamental
group of an $I$-bundle which is a connected component of a characteristic submanifold of
$(M,\Sigma \cup\Sigma')$. If $\Gamma \cap\Gamma'$is cyclic, then it is a finite index
subgroup
of a solid torus which is a connected component of a characteristic submanifold of
$(M,\Sigma \cup\Sigma')$.
\end{prop}

\subsection{Geodesic laminations on compressible surfaces}
Let $M$ be a compact $3$-manifold with boundary and let $c\subset\partial M$ be a simple
closed curve. A {\em $c$-wave} is a simple arc $k$, with $k\cap c=
\partial k$ such that there is an arc $\kappa$ in $c$ with the simple closed curve
$k\cup\kappa$ bounding an essential disc in $M$.
In some literature, a $c$-wave is allowed to intersect $c$ in its interior.
A simple innermost argument shows that if there is a $c$-wave in this
generalised sense, there is one in our sense.

Let $L$ be a geodesic lamination on $
\partial_{\chi<0} M$ and let $c\subset
\partial M$ be a multi-curve.
In the following, we always assume that simple closed curves or multi-curves are
geodesics, hence there are no inessential intersection between them or with geodesic
laminations.
We
say that $L$ is in {\em tight position} with respect
to $c$ if $L$ contains no $c$-waves and if {\em every} leaf of $L$ intersects $c$
transversely.

The following Claim shows how to use cut-and-paste operations to construct a
meridian $m$ such that a given geodesic lamination contains no $m$-waves.

\begin{cl}   \label{limit}
Let $F\subset
\partial M$ be a compressible compact surface and let $\beta\subset F$ be a measured geodesic
lamination. Then either $\beta$ intersects transversely a meridian $m$ and
contains no $m$-waves, or there is a sequence of meridians $(m_i\subset F)$ converging
in the Hausdorff topology to a geodesic lamination which does not intersect $\beta$
transversely.
\end{cl}
\begin{proof}
If $\beta$ intersects no meridians transversely, then, since $F$ is compressible,
there is a meridian $m\in F$ such that $i(\beta,m)=0$. Setting $m_i=m$ for every $i$,
we get
the conclusion. Now we assume that $\beta$ intersects a meridian $m$ transversely. If
$\beta$ contains an $m$-wave $k$, let us ``cut $m$ along $k$'' to get a new meridian $m_1$
: let $\kappa$ be the closure of a connected component of $m-\partial k$ such that
$\int_{\kappa}d \beta\leq \frac{1}{2} i(\beta,m)$, and $m_1$ the simple closed geodesic
in the
free homotopy class of $k\cup\kappa$. We have $i(\beta , m_1)\leq\frac{1}{2} i(\beta,m)$
and $m_1$ is a meridian. If $\beta$ does not contains any $m_1$-waves, we are done. If
$\beta$ contains an $m_1$-wave $k_1$, then we cut $m_1$ along $k_1$ as above to get a new
meridian $m_2$ with $i(m_2,\beta)\leq\frac{1}{4} i(m,\beta)$. Repeating this operation,
we get either a meridian $m'$ such that $\beta$  contains no $m'$-waves or a
sequence of meridians $(m_i)$ such that $i(m_i,\beta)\longrightarrow 0$. In the latter
case,
we extract a subsequence such that $m_i$ converges in the Hausdorff topology to some
geodesic lamination $H$. Since $i(m_i,\beta)\longrightarrow 0$, we see that $H$ does not
intersect $\beta$ transversely.
\end{proof}



Let $L$ be a geodesic lamination on $
\partial_{\chi<0} M$. A leaf $l$ of $L$ is said to be {\em homoclinic} if a lift $\tilde l$
of $l$ to
the universal cover $\tilde M$ of $M$ contains two sequences of point $(\tilde x_n)$ and
$(\tilde y_n)$ such that the distance between $\tilde x_n$ and $\tilde y_n$ in $\tilde M$
is uniformly bounded whereas their distance measured on $\tilde l$ tends to $\infty$. By
Lemma \ref{disque} and by \cite[Affirmation 3.4]{espoir}, an incompressible surface cannot contain a homoclinic leaf.


Homoclinic leaves appear naturally in Hausdorff limits of sequences of meridians.
This is
illustrated by the following criterion of Casson whose proof can be found in
\cite{chefjeune} and \cite[Theorem B.1]{espoir}.

\begin{lemma}  \label{casson} Let $(m_n\subset
\partial M)$ be a sequence of meridians which converges to a geodesic lamination $H$ in the
Hausdorff topology. Then $H$ contains a homoclinic leaf.
\end{lemma}


\indent
A {\em simple half-geodesic} is an embedded half-line in $\partial M$ whose image is locally geodesic for some hyperbolic
metric on
$\partial_{\chi<0} M$.
 Let $\tilde l_+\subset\partial\tilde M$ be a half-geodesic and let $\Bar{\tilde l}_+$ be its closure in the Floyd-Gromov compactification of $\tilde M$.
We  say that $\tilde l_+$ {\em has a well-defined endpoint} if $\Bar{\tilde l}_+-\tilde l_+$ contains one point.
We  say that a geodesic $\tilde l\subset\partial\tilde M$ has two well-defined endpoints if $\tilde l$ contains two disjoint half geodesics each having a well-defined
 endpoint.
 Notice that we allow the two endpoints to be the same. Two
distinct leaves $\tilde l_1$ and $\tilde l_2$ of a geodesic lamination $\tilde L\subset\partial\tilde M$
are said to be {\it biasymptotic} if they both have two well-defined endpoints in $\overline{\tilde M}$
and the endpoints of $\tilde l_1$ coincide with those of $\tilde l_2$. A geodesic
lamination $A\subset
\partial M$ is said to be {\it annular} if the preimage of $A$ in $
\partial\tilde M$ contains a pair of biasymptotic leaves.\\


Annular laminations appear naturally as limits of sequences of annuli.

\begin{lemma} \label{annulus} Let $(A_n\subset
\partial M)$ be a sequence of (not necessarily embedded) essential annuli or M{\"o}bius
bands.
Suppose that $
\partial A_n$ converges in the Hausdorff topology to a geodesic lamination $E$. Suppose
also that there is a measured geodesic lamination $\alpha$ whose support is equal to $E$
such that
$i(\alpha,\partial A_n)\longrightarrow 0$. Then either $E$ is annular or $S(E)$ contains a
homoclinic leaf which does not intersect $E$ transversely.
\end{lemma}

\begin{proof}
The main argument of this proof comes from \cite[Lemme C.2]{espoir}.

Let $p:\tilde M\rightarrow M$ be the covering projection. Fix some complete hyperbolic
metric on $\partial_{\chi<0} M$, and let $S(E)$ be the surface embraced by $E$.

Assume first that $S(E)$ is incompressible. Let $\tilde S(E)\subset\partial\tilde
M$ be a lift of $S(E)$, and $\tilde F$ a fundamental domain for the action of
$\pi_1(S(E))$ on $\tilde S(E)$. Let $
\partial^1\tilde A_n$ and $\partial^2\tilde A_n$ be the boundary components of a
lift of $A_n$ to $\tilde M$ such that $\partial^1\tilde A_n$ intersects $\tilde F$. By
extracting a subsequence, we can assume that the sequence $(\partial^1\tilde A_n)$ converges to a lift
$\tilde e$ of a leaf $e$ of $E$. Since $A_n$ is an annulus or a M{\"o}bius band, the geodesic $\partial^2\tilde A_n$ has the same ends as
$\partial^1\tilde A_n$. By \cite[Affirmation 2.2]{espoir}, for any $\epsilon >0$, there are only finitely many translates of ${\tilde S}(E)$ whose limit sets have diameters greater than $\epsilon$. Therefore, after extracting a subsequence, all the geodesics $\partial^2\tilde A_n$ can be assumed to lie in the same lift $\tilde S(E)'\subset \partial\tilde M$ of $S(E)$. The ends of $\partial^2\tilde A_n$ converge to the ends of $\tilde e\subset \tilde S(E)$. By Lemma \ref{disque}, a geodesic in $\tilde S(E)'$ is uniquely determined by its two ends. Therefore
the sequence $(\partial^2\tilde A_n)$ converges to the lift $\tilde e'\subset \tilde S(E)'$ of a leaf
$e'$ of $E$ which has the same ends as $\tilde e$. If $\tilde S(E)=\tilde S(E)'$ then $
\partial^1\tilde A_n$ and $\partial^2\tilde A_n$ lie in $\tilde S(E)$. This contradicts
Lemma \ref{disque} since $\partial^1 A_n$ and $\partial ^2 A_n$ are not homotopic in
$S(E)$.
Hence we have $\tilde S(E)\neq\tilde S(E)'$ and $\tilde e\neq\tilde e'$. It follows that $E$ is an annular lamination.

Let us now consider the case when $S(E)$ is compressible. By Claim \ref{limit}
and the Casson's criterion (Lemma \ref{casson}), either there is a meridian $m\subset
S(E)$ such that $E$ does not contain any $m$-wave, or $S(E)$ contains a homoclinic leaf
which does not intersect $E$ transversely. Therefore we have only to consider the case
when
there is a meridian $m\subset S(E)$ such that $E$ intersects $m$ transversely and
contains no $m$-waves.

We set $
\partial^1 A_n=S^1\times\{0\}$ and $
\partial^2 A_n=S^1\times\{1\}$ where $A_n=S^1\times I$.
We shall first show that $E$ is in
tight position with respect to $m$. For this, we need to show that every leaf of $E$
intersects $m$. Since $E$ is the Hausdorff limit of $\partial A_n$, it has one or two
connected components.
Since there is a measured geodesic lamination whose support is equal to $E$, each
of these components is a minimal lamination. Therefore if a leaf of $E$ intersects $m$,
the same is true for all the leaves lying in the same connected component.
Thus if $E$ has
only one component, $E$ is in tight position with respect to $m$.

Now we assume that there are two connected components $E^1, E^2$ of $E$.
Since $E$ intersects $m$, we can assume that
$E^1$
intersects $m$. By extracting a subsequence, we can also assume that $E^1$ is the Hausdorff
limit of some $\partial^i A_n$, say $
\partial^1 A_n$ and that $E^2$ is the Hausdorff limit of $
\partial^2 A_n$. Note that $\partial^1 A_n$ contains
no
$m$-waves for sufficiently large $n$ (even if we allow an $m$-wave to have
self-intersection).
Otherwise taking a limit
in the Hausdorff topology of $m$-waves lying in $\partial^1 A_n$, we get either an $m$-wave
lying in $E$ contradicting the fact that $E$ is in tight position with respect to $m$, or
a sublamination of $E^1$ not intersecting $m$, contradicting the fact that $E^1$ is a
minimal lamination intersecting $m$. It follows from the fact that $
\partial^1 A_n$ contains no $m$-waves that the ends of any lift of $
\partial^1 A_n$ to $\tilde M$ are separated by a lift of $m$ (compare with
\cite[Affirmation 3.4]{espoir}). This implies that $
\partial^2 A_n$ intersects $m$. Since $E^2$ is the Hausdorff limit of $\partial^2 A_n$,
we see that $E^2$ also intersects $m$.

Thus we have proved that $E$ is in tight position with respect to $m$ whether
$E$ has  one or two connected components. Let us deduce that there
is a uniform upper bound $K$ on the lengths of the connected
components of $\partial A_n-m$. Assuming the contrary, we get a sequence of arcs
$k_n\subset
\partial A_n-m$ whose lengths tend to $\infty$. Extract a subsequence such that $(k_n)$
converges in the Hausdorff topology. The limit contains a sublamination of $E$ which does
not intersect $m$, contradicting the fact that $E$ is in tight position with respect to
$m$.

Let $e$ be a leaf of $E$ and let $\tilde e\subset\tilde M$ be a lift of $e$.
Denote by $(\tilde m_j)_{j\in{\mathbb Z}}$ the connected components of $p^{-1}(m)$ that $\tilde e$
intersects (in this order). Let $\tilde e[-t,t]$ be the segment of $\tilde e$ connecting
$\tilde m_{-t}$ to $\tilde m_{t}$. By assumption, $\partial A_n$ converges to $E$ in
the Hausdorff topology. Therefore, for sufficiently large $n$, a connected component
$\partial^1\tilde A_n$ of $p^{-1} (\partial A_n)$ intersects $\tilde m_{-t}$ and $\tilde
m_{t}$ and $\partial^1\tilde A_n [-t,t]$ converges to $\tilde e[-t,t]$ in the Hausdorff
topology. Let $\tilde A_n$ be the lift of $A_n$ whose boundary contain $\partial^1\tilde
A_n$. As we have seen above, $
\partial A_n$ does not contain any $m$-wave for sufficiently large $n$. Therefore both $\tilde
m_t$ and $\tilde m_{-t}$ separate the ends of $
\partial^1\tilde A_n$ (cf. \cite[Affirmation 3.4]{espoir}). It follows that
$\tilde m_t$ and $\tilde m_{-t}$ intersect $
\partial^2\tilde A_n$. Let $
\partial^2\tilde A_n [-t,t]$ be the segment of $
\partial^2\tilde A_n$ joining $\tilde m_t$ to $\tilde m_{-t}$. The length of $\partial^2\tilde A_n [-t,t]$ is less than
$2Kt$  where $K$ is the constant found in the previous paragraph. After extracting a subsequence, $(
\partial^2\tilde A_n [-t,t])$ converges in the Hausdorff topology to an arc $\tilde
e'[-t,t]\subset p^{-1}(E)$. Letting $t$ tend to $\infty$, we conclude that $\tilde e$ and
$\tilde e'$ have the same ends.

If $\tilde e=\tilde e'$, then $\partial^1\tilde A_n$
and $\partial^2\tilde A_n$ intersect an $\epsilon$-neighbourhood of $\tilde e$ for
sufficiently large $n$. Hence there are segments $\tilde I_n\subset
\partial\tilde M$ joining the two components of $\partial \tilde A_n$ with their projections $I_n$
such that $\int_{I_n}
d\alpha$ tends to $0$.
Since $\tilde I_n$ is homotopic to an arc on $\tilde A_n$ joining two boundary
components, $I_n$ is homotopic to an essential arc on $A_n$ joining two
boundary components. If
we cut $
\partial A_n$ along $
\partial I_n$ and glue two copies
of $I_n$, we get a closed curve which bounds a disc $D_n$ (not necessarily embedded). By
assumption, $i(\alpha,\partial A_n)$ and $\int_{I_n} d\alpha$ tend to $0$.
Therefore $i(\alpha,\partial D_n)$ tends to $0$. By the Loop Theorem (cf. \cite{hempel}), there is an
embedded disc $D'_n$ which is not parallel to $
\partial M$ such that $i(\alpha,
\partial D'_n)\longrightarrow 0$. Extracting a subsequence such that $(\partial D'_n)$
converges in the Hausdorff topology, we get a geodesic lamination $H\subset S(E)$ which
does not intersects $E=|\alpha|$ transversely. By Casson's criterion (Lemma \ref{casson}),
$H$ contains a homoclinic leaf.

When $\tilde e\neq\tilde e'$, by definition, $E$ is annular.
\end{proof}

As in the introduction, we say that a measured geodesic lamination
$\lambda\in\cal{ML}(\partial M)$ is {\em doubly incompressible} if and only if
:\smallskip\\
\indent - $\exists\eta>0$ such that $i(\lambda,
\partial E)>\eta$ for every essential annulus, M{\"o}bius band or disc $E$.\smallskip\\
We denote by $\cal{D}(M)\subset\cal{ML}(
\partial M)$ the set of doubly incompressible measured geodesic laminations and by
$\cal{PD}(M)$ its projection in the space $\cal{PML}(
\partial M)$ of projective measured laminations.\\

\indent Some properties of this set ${\cal D}(M)$ are discussed in \cite{masurext}. In
particular, we can deduce the following from \cite{espoir}.

\begin{lemma} \label{saintptot} Let $\lambda\in{\cal D}(M)$ be a measured geodesic
lamination and $l_+, l_-\subset \partial M$ two simple half-geodesics which do not intersect $|\lambda|$
transversely.
 Then any lift of $l_+$ (resp. $l_-$) to $\tilde M$ has a well-defined endpoint.

Furthermore, if a lift of $l_+$ has the same endpoint as a lift of $l_-$ then $l_+$ and $l_-$ are asymptotic on $\partial M$.
\end{lemma}

\begin{proof}
The first property, namely that any lift of $l_+$ (resp. $l_-$) to $\tilde M$ has a well-defined endpoint, can be deduced from the proofs of
 \cite[Lemme 3.1]{espoir} and \cite[Lemme 3.3]{espoir}.

The proof of the second property, namely that if a lift of $l_+$ has the same endpoint as a lift of $l_-$ then $l_+$ and $l_-$ are asymptotic on $\partial M$,
can be found in the paragraph (ii) in the proof of \cite[Lemme C3]{espoir}.
\end{proof}

From \cite{masurext}, we also get the following :

\begin{lemma} \label{anho}
Let $\lambda\in{\cal D}(M)$ be a measured geodesic lamination,
$A$ an annular geodesic lamination which is the Hausdorff limit of a sequence of
multi-curves, and $h$ a homoclinic leaf (of some geodesic lamination). Then the
support $|\lambda|$ of $\lambda$ intersects both $A$ and $h$ transversely.
\end{lemma}

\begin{proof} When $M$ is not a genus-$2$ handlebody, this is \cite[Lemma 3.5]{masurext}.
The case when $M$ is a genus $2$ handlebody is discussed in \cite{masurext} in the remark
following \cite[Lemma 3.5]{masurext}.
\end{proof}

Combining Claim \ref{limit}, Casson's criterion (Lemma \ref{casson}) and Lemma \ref{anho},
we get the following :
\begin{lemma}\label{tightposition} Let $\lambda \in \cal D(M)$, and let $S\subset
\partial
M$ be a compressible surface. Then there is a meridian $m$ in $S$ such that $S$ does not
contain any $m$-waves disjoint from $|\lambda|$.

\end{lemma}



\subsection{Some notations}
When $\lambda$ is a geodesic measured lamination, we denote by $|\lambda|$
the support of $\lambda$. For an arc $k$ whose intersections with $|\lambda|$ are
transverse, we will denote by $\int_k d\lambda$ the $\lambda$-measure of $k$.

Let $(u_n)$ and $(v_n)$ be two sequences of non-negative real numbers. We say
that $u_n$ is $o(v_n)$ if for any $\epsilon$ there is $N(\epsilon)$ such that for $n\geq
N(\epsilon)$, we have $u_n\leq\epsilon v_n$. We also write $u_n=o(v_n)$.

We will say that $u_n$ is $O(v_n)$ if there are $K,N>0$ such that for $n\geq
N$, we have $u_n\leq K v_n$.

We say that $u_n$ is $\Theta(v_n)$ if $u_n$ is $O(v_n)$ and $v_n$ is
$O(u_n)$.

\section{Realisations of doubly incompressible laminations}


Let $\pi_1(M)\curvearrowright\cal{T}$ be a small minimal action of $\pi_1(M)$ on
an ${\mathbb R}$-tree for a compact irreducible atoroidal $3$-manifold $M$.
Let $S$ be a connected
component of $\partial_{\chi<0} M$.
Using the map $i_*:\pi_1(S)\rightarrow\pi_1(M)$ induced
by the inclusion, we get an action of $\pi_1(S)$ on ${\cal T}$.
Therefore, if
$\lambda\in\cal{ML}(S)$ is a measured geodesic lamination, it makes sense to ask
whether or not it is realised in $\cal{T}$.
In this section, we shall discuss
this question for the connected components of a measured lamination lying in
$\cal{D}(M)$.

\begin{lemma}  \label{quivabien} Let $\pi_1(M)\curvearrowright\cal{T}$ be a
small minimal action of $\pi_1(M)$ on an ${\mathbb R}$-tree. Let $\lambda\in\cal{D}(M)$
be a measured geodesic lamination, and $\alpha$ a minimal sublamination of $\lambda$.
Then one of the following holds:

\begin{itemize}
\item the measured lamination $\alpha$ is realised in $\cal{T}$;
\item there is a sequence of train tracks $\theta_i$ each of which minimally
carries $\alpha$ and has the following properties:
\begin{itemize}
\item $\theta_i$ has only one switch $\kappa_i$
and $\kappa_i\supset\kappa_{i+1}$ for every $i$.
\item There are a point $p\in\cal{T}$, a sequence
$\eta_i\rightarrow 0$, and a sequence of $\pi_1(S)$-equivariant maps
$\phi_i:{\mathbb H}^2\rightarrow\cal{T}$ such that $\phi_i$ maps every branch of the
preimage of $\theta_i$ to a geodesic segment (which may be a point) with length
smaller than $\eta_i$ and such that a preimage of $\kappa_i$ is mapped to $p$
under
 $\phi_i$. Note that the number of branches of $\theta_i$ is uniformly bounded.
\end{itemize}
\end{itemize} Furthermore there is at least one component of $\lambda$ which
is realised in $\cal{T}$.
\end{lemma}

\begin{proof}
Notice that when $\alpha$ is collapsed in $\cal{T}$, it is easy to see that we are in the
second situation.

Let us first assume that $\alpha$ is a simple closed curve. If $\delta_\cal{T}(\alpha)=0$
then $\alpha$ is collapsed in $\cal{T}$, if $\delta_\cal{T}(\alpha)\neq 0$, then $\alpha$
is realised in $\cal{T}$.

Now assume that $\alpha$ is not a simple closed curve.
We cut $M$ along
a maximal family of essential discs disjoint from $S(\alpha)$. We denote by $N$ the
connected component of the resulting manifold that contains $\alpha$ on its boundary.
The surface $\partial N-S(\alpha)$ is incompressible in $N$. Let $\cal{T}_N$ be the
minimal
subtree of $\cal{T}$ that is invariant under the action of $\pi_1(N)$ regarded as a
subgroup
of $\pi_1(M)$. If $\cal{T}_N$ is trivial, then $\alpha$ is collapsed.
Therefore, from now on we assume
that $\cal{T}_N$ is not trivial. Let us denote by $F$ the connected component of $\partial
N$ containing $S(\alpha)$.
Since $\lambda$ lies in $\cal{D}(M)$, every component of $\partial S(\alpha)$ is essential
 in $M$; hence $S(\alpha)$ is incompressible on $F$.
In particular, the measured lamination $\alpha$ can be regarded as lying on
$F$.
Using the map $i_*:F\rightarrow
N$ induced by the inclusion,
we get an action of $\pi_1(F)$ on $\cal{T}_N$. Let $\cal{T}_F$ be the minimal subtree of
$\cal{T}_N$ that is invariant under the action of $\pi_1(F)$. We shall divide the
proof in two cases.

 \underline{First case} : $F$ is incompressible in $N$. Then the action of
$\pi_1(F)$ on
$\cal{T}_F$ is small. By Skora's theorem \cite{skora}, there are a measured geodesic lamination
$\beta$ on $F$
and an isomorphism $\phi:\cal{T}_\beta\rightarrow \cal{T}_F$ from the dual tree
$\cal{T}_\beta$ of $\beta$ to $\cal{T}_F$.
If $\beta$ intersects $\alpha$ transversely,
then $\alpha$ is realised in $\cal{T}$ (cf. \cite{meister2}).
If $\beta$ and $\alpha$ are
disjoint, then $\alpha$ is collapsed in $\cal{T}$.
It remains to deal with the case when $\alpha$ is a
connected component of $\beta$.
Let $(\eta_i)$ be a sequence of positive numbers converging to $0$.
Let $\kappa_i\subset F$ be a sequence of segments intersecting $\alpha$ transversely such
that  $\kappa_{i+1}\subset\kappa_i$ and $\int_{\kappa_i}
d\beta\leq\frac{1}{2}\eta_i$.
Let $\theta_i$ be a train track minimally carrying $\alpha$ and having only one switch
which is $\kappa_i$ (refer to \cite[$\S$ 3.2]{meister1} for the construction of such a
train track).
Let $p'$ be a point of $\bigcap_i\kappa_i$, and $\hat p'\in {\mathbb H}^2$ a lift
of $p'$.
This point $\hat p'$ corresponds to a point of $\cal{T}_\beta$ which we shall also
denote by $\hat p'$. Let $p\in \cal{T}$ be the image of $\hat p'$ under $\phi$.
Let $\hat \kappa_i\subset{\mathbb H}^2$ be the lift of $\kappa_i$ that contains $\hat p'$, and
$\hat\theta_i$ the lift of $\theta_i$ that contains $\hat\kappa_i$.
Define $\phi_i$ on
$\hat\kappa_i$ by $\phi_i(\hat\kappa_i)=p$.
Extend $\phi_i$ to an equivariant map from
the union of switches of $\hat\theta_i$ to $\cal{T}$.
If $\hat b$ is a branch of
$\hat\theta_i$, we define $\phi_i(\hat b)$ to be the segment of $\cal{T}$ which connects
the images of the vertical sides of $\hat b$. Finally, extend $\phi_i$ to a
$\pi_1(S)$-equivariant map $\phi_i:{\mathbb H}^2\rightarrow\cal{T}$. Let $\hat b$ be a branch of
$\hat\theta_i$.
Translating it by an element of $\pi_1(S)$, we can assume that
$\hat\kappa_i$ contains a vertical side of $\hat b$. Since $\theta_i$ has only one switch,
there is some $g\in\pi_1(S)$ such that $g(\hat\kappa_i)$ contains the other vertical side
of $\hat b$. Let $\hat k_1$ be an arc joining $\hat\kappa_i$ to $g(\hat\kappa_i)$ whose
projection $k_1$ on $S$ lies in $b-|\beta|$. Then we have $\int_{k_1} d\beta=0$.
Let $\hat k_2\subset\hat\kappa_i$ be an arc joining $\hat p'$ to $\hat k_1$ and let $\hat
k_3\subset
g(\hat\kappa_i)$ be an arc joining $g(\hat p')$ to $\hat k_1$. The $\beta$-measures
of
$k_2$ and $k_3$ are less than $\int_{\kappa_i} d\beta\leq\frac{1}{2}\eta_i$.
Therefore we have
$\int_{k_1\cup k_2\cup k_3} d\beta\leq\eta_i$.
This implies that the distance between $\hat
p'$ and $g(\hat p')$ in ${\cal T}_\beta$ is less than $\eta_i$. It follows then from the
construction of $\phi_i$ that the length of $\phi_i(\hat b)$ is less than $\eta_i$.
Thus we
conclude that when $F$ is incompressible, one of the situations of our
lemma holds.


\underline{Second case} : $F$ is compressible in $N$. We shall use the assumption
that
$\lambda$ lies in $\cal{D}(M)$ and some results of \cite{boches} to reduce this case to
the previous case when $F$ is incompressible.
In the following, we shall be lead to consider the case when
we can construct some homoclinic leaves using the results of \cite{boches}.
Let us
first consider what will happen if $F$ contains a homoclinic leaf $h$.

\begin{cl}   \label{homoclinic} If $F$ contains a homoclinic leaf $h$ which does not
intersect $|\alpha|$ transversely, then $S(\alpha)$ is incompressible, the closure of $h$
contains two geodesic laminations $|\alpha|$ and $A$, and there is an essential $I$-bundle
$S\times I$ in $N$  over a surface $S$ homeomorphic to $S(\alpha)$ such that
$S\times I \cap \partial N= S \times \{0,1\}$,  $S\times\{0\} =S(\alpha)$, and
 $A$ is arational in $S\times\{1\}$.
\end{cl}

\begin{proof} As we saw in the section 2.6, an incompressible surface can not
contain a
homoclinic leaf. By construction, $F-S(\alpha)$ is incompressible, therefore the geodesic
$h$ does not lie in $F- \bar S(\alpha)$. Since $\alpha$ is arational in $S(\alpha)$, this
implies that a half-leaf $h_+$ of $h$ is asymptotic to a half-leaf of $\alpha$ in $\partial N$. Let $h_-$
be a half-leaf of $h$ which is disjoint from $h_+$.\\
\indent
Let us first assume that $h_-$ is asymptotic to a half-leaf of $\alpha$. By
shortening
$h_-$ and $h_+$ if necessary, we can assume that they both lie in $S(\alpha)$ and
are disjoint from $\lambda$.
Since $h$ is homoclinic, it follows then from Lemma
\ref{saintptot} that $h_-$ is asymptotic to $h_+$ in $\partial N$.
In particular, since $h_-$ is asymptotic to
a half-leaf of $\alpha$, the two half-leaves $h_-$ and $h_+$ are
asymptotic to the same half-leaf of $\alpha$.
Take a short arc $k$ connecting $h_-$ and $h_+$ so that they are lifted to  a triangle with one vertex at infinity
in $\partial \tilde{N}$.
Then any lift of a half-leaf of $\alpha$ going into the inside of this triangle
must have the same endpoint as the lifts of $h_-, h_+$.
This means that such a half-half leaf captured between $h_-$ and $h_+$ intersects
a short arc $k$ however short we take $k$ to be.
In other words $\int_k d\alpha$ is independent of our choice of $k$.
Since
$\alpha$ is not a closed curve and the length of $k$ can be made to tend to $0$, we
have $\int_{k} d\alpha
=0$.
Therefore, the union of $k$ and the
bounded component of $h-\partial k$ is homotopic to a meridian $m\subset F$ with
$i(m,\alpha) = 0$. It follows that $m\subset F-S(\alpha)$,
contradicting the fact that $ F-S(\alpha)$ is incompressible.


If $h_-$ is not asymptotic to a half-leaf of $\alpha$, by shortening $h_-$, we can assume
that $h_-$ lies in $F-  S(\alpha)$. If $S(\alpha)$ is compressible then by Lemma
\ref{tightposition}, $S(\alpha)$ contains a meridian $m$ with respect to which $\alpha$ is
in tight position. This implies that $h_+$ is also in tight position with respect to $m$.
Since $h$ is homoclinic, $h_-$ also intersects $m$ contradicting the assumption $h_-\subset
F-S(\alpha)$.\\
\indent
If $S(\alpha)$ is incompressible, let $\cal{V}\subset F$ be a small annular neighbourhood of $\partial'\bar S(\alpha)$. By Proposition \ref{carac}, $h_-\cup h_+$ lies
in the boundary of a component of the characteristic submanifold of $(N,F-\cal V)$.
More precisely, there is an essential $I$-bundle in\linebreak $(N,F-\cal V)$ whose boundary
contains $h_-\cup h_+$. Since $h_-$ is not asymptotic to a half leaf of $\alpha$, this
$I$-bundle is not twisted. Hence it is homeomorphic to $S\times I $ with $S\times\{0\}
=S(\alpha)$. Let $q:S\times I\rightarrow S$ be the projection along the fibres.
Since $h$ is homoclinic, the half-geodesics $q(h_+)$ and $q(h_-)$ are asymptotic. Furthermore, $q(h_+)$ is asymptotic to $q(\alpha)$. So $q(h_-)$ is asymptotic to $q(\alpha)$. It follows that $A=S\times\{1\}\cap q^{-1}(q(\alpha))$ is an arational lamination (in $ S\times\{1\}$) that lies in the closure of $h_-$. This concludes the proof of Claim \ref{homoclinic}.
\end{proof}

Let $(L_n)$ be a sequence of multi-curves in $S(\alpha)$ converging to the support of
$\alpha$ in the
Hausdorff topology.
If
$N$ contains an incompressible $I$-bundle $S\times I$ with $S\times\{0\}=S(\alpha)$,
then let
$c\subset F-\bar S(\alpha)$ be a simple closed curve such that
$c\subset S \times\{1\}$ and $c$ cannot be homotoped into a component of
$\partial S\times\{1\}$.
Otherwise we set $c$ to be empty.
By \cite{chefm}, there exist a measured
geodesic lamination
$\beta_n\in\cal{ML}(F)$ and a morphism $\phi_n:\cal{T}_{\beta_n}\rightarrow
\cal{T}_N$ from the dual tree of $\beta_n$ to $\cal{T}_N$ such that for any closed curve
$l_n$ which lies in $L_n\cup\partial'\bar S(\alpha)\cup c$, either $\delta_\cal{T}(l_n)>0$
and the restriction of $\phi_n$ to the axis of $l_n$ is an isometry or
$\delta_\cal{T}(l_n)=0$ and $i(l_n,\beta_n)=0$ (see \cite{boches} or \cite{espoir}).
We extract a subsequence so that $|\beta_n|$ converges in the Hausdorff topology to some
geodesic lamination $B$.

If $\alpha$ does not intersect $B$, then $\alpha$ does not
intersect $\beta_n$ for sufficiently large $n$.
This implies that $\alpha$ is collapsed in
$\cal{T}_N$ and therefore
in $\cal{T}$. If $\alpha$ intersects $B$ transversely, then
$\alpha$ intersects $\beta_n$ transversely for sufficiently large $n$.
In this
case $\alpha$ is realised in
$\cal{T}_N$ (cf. \cite[Lemma 11]{boches}) and therefore $\alpha$ is realised in $\cal{T}$.
It remains to deal with the case when $|\alpha|$ is contained in $B$.

For $n\in{\mathbb N}$, let $\beta_n$ be the measured geodesic lamination which we have constructed
above. Since we are considering the case when $F$ is compressible, by Claim \ref{limit} either there is a meridian $m\subset F$ such that $\beta_n$ contains
no $m$-waves or there is
a sequence of meridians converging in the Hausdorff topology to a geodesic
lamination which does not intersect $\beta_n$ transversely.
By the proof of \cite[Proposition 2]{boches}, if $\beta_n$
intersects a meridian $m$ and contains no $m$-waves, then $|\beta_n|$ can be extended to
a geodesic lamination with a homoclinic leaf $h_n$.
Furthermore by the proof of
\cite[Proposition 1]{boches}, any neighbourhood of $h_n$ contains a meridian.
Thus we have found in any case a sequence of meridians converging in the Hausdorff topology to a geodesic lamination $H$
which does not intersect $B$ transversely. By Casson's criterion (Lemma \ref{casson}),
the lamination $H$ contains a homoclinic leaf.

Recall that we are considering the case when $|\alpha|\subset B$. Therefore by
Claim \ref{homoclinic}, $S(\alpha)$ is incompressible, $H$ contains two geodesic
laminations $|\alpha|$ and $A$ and there is an essential $I$-bundle $S\times I$ over a surface $S$ homeomorphic to $S(\alpha)$ such that $S\times \{0\}=S(\alpha)$ and
$A$ is arational in $S\times\{1\}$.

If $F-S(A)$ contains a meridian, then by Claim \ref{limit}, either there is a meridian
$m\subset F-S(A)$ such that $\alpha$ contains no $m$-waves or $F-S(A)$ contains a
homoclinic leaf $h$ which does not intersect $\alpha$ transversely.

Since $F-S(\alpha)$ is incompressible, in the former case $m$ intersects $\alpha$
transversely. By construction the lifts of $\alpha$ and $A$ to the universal cover of
$S\times I$ have the same endpoints. Since $\alpha$ contains no $m$-waves and
intersects $m$ transversely, $A$ also intersects $m$ transversely (this is derived from the
proof of
\cite[Affirmation 3.4]{espoir}). This contradicts the fact that $m$ lies in $F-S(A)$.

In the latter case, $F-S(A)$ contains a homoclinic leaf $h$ which does not
intersect $\alpha$ transversely. Applying Claim \ref{homoclinic}, we get an essential
$I$-bundle $S\times I$ such that $S\times \{0\}=S(\alpha)$ and that $S\times \{1\}\subset
F-S(A)$.
By the uniqueness of the characteristic submanifold however (see \cite{jacs} or
\cite{jojo}), this is impossible.
Thus we have proved that $F-S(A)$ is an incompressible surface.

Any leaf of $B$
intersecting $\partial(F-S(A))$ contains a half-leaf asymptotic to $A$. Such a half-leaf
intersects infinitely many times the simple closed curve\linebreak $c\subset (F-S(\alpha))$ (the one
which we have chosen before constructing $\beta_n$). This implies that
$i(\beta_n,\partial(F-S(A)))$ is $o(i(\beta_n,c))$.
On the other hand, by construction, we have
$i(\beta_n,c)=\delta_\cal{T}(c)$ for any $n$. Thus we get $i(\beta_n,\partial
(F-S(A)))\longrightarrow 0$, which implies that the conjugacy class of $\pi_1(F)$
represented by each
component of $
\partial (F-S(A))$ has a fixed point in $\cal{T}$. This enables us to use Skora's Theorem
on the minimal subtree of $\cal{T}$ which is invariant under the action of $\pi_1(F-S(A))$
and we can argue in the same way as when $F$ is incompressible.\\

Thus we have proved the first part of Lemma \ref{quivabien}. It only remains to
show that at least one component of $\lambda$ is realised in $\cal{T}$. This is
proved in \cite[Proposition 6.1]{masurext} (see also \cite[Proposition 6]{espoir}).
Let us
briefly review the proof. Let $(L_n)$ be a sequence of multi-curves
converging to $\lambda$
in the Hausdorff topology. As we have already seen, there exist a measured geodesic
lamination $\beta_n\in\cal{ML}(\partial M)$ and a morphism
$\phi_n:\cal{T}_{\beta_n}\rightarrow \cal{T}$ from the dual tree of $\beta$ to $\cal{T}$
such that for any closed curve $l_n$ which lies in $L_n$,
either $\delta_\cal{T}(l_n)>0$ and the restriction of $\phi_n$ to the axis of $l_n$ is an
isometry or $\delta_\cal{T}(l_n)=0$ and $i(l_n,\beta_n)=0$. Extract a
subsequence such that $(\beta_n)$ converges in the Hausdorff topology to a geodesic
lamination $B$. As we have seen above, any connected component of $\lambda$ that
intersects $B$ transversely is realised in $\cal{T}$.


If $S(B)$ is compressible, then, by the proof of \cite[Proposition 2]{boches}, $S(B)$
contains a homoclinic leaf $h$ which does not intersect $B$ transversely. Such a homoclinic
leaf intersects $\lambda$ transversely (\cite{masurext}, see also the proof of Claim
\ref{homoclinic} above). Thus, if $S(B)$ is compressible, $\lambda$ intersects $B$
transversely.

If $S(B)$ is incompressible, then we can apply Skora's theorem to each component of
$S(B)$. It follows that $\beta_n$ does not depend on $n$ for sufficiently large $n$.
Denote by $\beta$ this constant geodesic measured lamination $\beta_n$. We deduce then from
\cite{mors3} that $\beta$ is an annular lamination (see \cite[Lemma 14]{meister1} or
\cite[Lemme 4.4]{espoir}).
Thus we see that $\lambda$ intersects transversely the support $B$ of
$\beta$ also in this case.
\end{proof}

\section{Mapping the train tracks to $\epsilon_n^{-1}{\mathbb H}^3$}
In this section, we  begin the proof Theorem \ref{th&}. The first step in this proof
is to construct a $\rho_n$-equivariant map from the universal cover of the train tracks
constructed in Lemma \ref{quivabien} to ${\mathbb H}^3$ which has some nice properties.
Recall that when $\hat{\tau}$ is a universal cover of a train track $\tau$ on a
component $S$ of $\partial M$, we
say that a map $\hat{h}:
\hat{\tau} {\rightarrow} {\mathbb H}^3$ is $\rho_n$-equivariant if and only if for every $g \in \pi_1(S)$
and $x \in \hat{\tau}$,
we have $\hat{h}(gx)= \rho_n(i_*(g))\hat{h}(x)$, where $i$ denotes the inclusion from $S$ to $M$.

Now let us recall the statement of Theorem \ref{th&}.

\setcounter{thm}{1}
\begin{thm} Let $\rho_n:\pi_1(M)\rightarrow \mathrm{Isom}({\mathbb H}^3)$ be a sequence of discrete
faithful representations which uniformise $M$ and let $(\lambda_n)\subset\cal{ML}(
\partial M)$ be a sequence of measured geodesic laminations such that $(l_{\rho_n}(\lambda_n))$
is a bounded sequence and $(\lambda_n)$ converges in $\cal{ML}(
\partial M)$ to a measured geodesic lamination $\lambda\in\cal{D}(M)$. Then the sequence
$(\rho_n)$ has a compact closure in $AH(M)$, namely any subsequence contains an
algebraically converging subsequence.
\end{thm}

\begin{proof} Assume, seeking a contradiction, that no subsequences of
$(\rho_n)$ converge algebraically. Let us choose a non-principal ultrafilter $\omega$. In the section 2 :
Preliminaries, we have seen that $(\rho_n)$, regarded as a sequence of actions on
$\epsilon_n {\mathbb H}^3$,
tends (with
respect to $\omega$) to a small minimal action of $\pi_1(M)$ on an ${\mathbb R}$-tree $\cal{T}$.


By Thurston's result (cf. \cite{flp} and \cite{penner}), weighted simple closed curves are
dense in $\cal{ML}(S)$ for every component $S$ of $\partial M$. By approximating each
$(\lambda_n)$ by a sequence of such unions of weighted simple closed curves, and
by a diagonal extraction, we get a sequence of unions of weighted simple closed
curves on components of $\partial M$ satisfying the hypothesis of Theorem \ref{th&}.
Therefore we can assume that
each $\lambda_n \cap S$ is a weighted simple closed curve if it is not empty
for every component $S$ of $\partial M$.
We further
extract a subsequence in such a way that
$(|\lambda_n|)$ converges in the Hausdorff topology to a geodesic lamination $L_\infty$.
Let $L_{rec}$ be the union of the recurrent leaves of $L_\infty$. We have
$|\lambda|\subset L_{rec}$. Note that $L_{rec}$ admits a transverse measure
with full support whose restriction to $|\lambda|$ coincides with the
transverse measure of $\lambda$, and
endowed with this measure,
$L_{rec}$  is contained in $\cal{D}(M)$ since $\lambda$ is contained in
$\cal{D}(M)$.
\begin{lemma}  \label{rtt} After taking a subsequence of $(\rho_n)$, there are train
tracks $\tau_n$, $\tau^1$ and $\tau^2_n$
such that
\begin{itemize}
\item each minimal sublamination of $L_\infty$ is carried by $\tau^1$ or $\tau^2_n$;
\item $\lambda_n$ is minimally carried by $\tau_n$;
\item $\tau^1$ and $\tau^2_n$ are disjoint subtracks of $\tau_n$;
\item the switches of $\tau_n$ lie in $\tau^1\cup\tau^2_n$;
\item the sum of the weights with which $\tau_n$ carries $\lambda_n$ is bounded
independently of $n$;
\end{itemize}
and there is a $\rho_n$-equivariant map $\tilde h_n : \hat\tau_n\rightarrow{\mathbb H}^3$ from
the preimage of $\tau_n$ in the universal cover of $
\partial M$ to ${\mathbb H}^3$ such that :
\begin{enumerate}[\indent\indent a)]
\item for any branch $\hat b$ of $\hat\tau_n$, its image $\tilde h_n(\hat b)$ is either
a geodesic segment or a point;
\item there are $R>0$ and $n(R) \in {\mathbb N}$ such that, for $n\geq n(R)$, if
$\hat b$ is a branch
of $\hat\tau_n$ which projects into $\tau^1$, then $l(\tilde h_n(\hat b))\geq R
\epsilon_n^{-1}$;
\item there is a sequence of positive numbers $\delta_n\longrightarrow 0$ such that,
for any $n\in{\mathbb N}$,
if $\hat b_1,\hat b_2$ are two adjacent branches of $\hat\tau_n$ which are separated by a switch and which both project
into
$\tau^1$, then the exterior angle between $\tilde h_n(\hat b_1)$ and $\tilde h_n(\hat
b_2)$ is less than $\delta_n$;
\item there is a sequence of positive numbers $\eta_n\longrightarrow 0$ such that, for
any $n\in{\mathbb N}$, if
$\hat b$ is a branch of $\hat\tau_n$ which projects into $\tau^2_n$, then we have $\epsilon_n
l(\tilde h_n(b))\leq\eta_n$, where $\epsilon_n$ is the rescaling factor of ${\mathbb H}^3$ which
appeared above;
\item for any $n\in{\mathbb N}$, if $\hat b$ is a branch of $\hat\tau_n$ which projects into
$\tau_n-(\tau^1\cup\tau^2_n)$, then $\lambda_n(b)(\epsilon_n l(\tilde h_n(\hat b)))$ is less
than $\eta_n$ for $(\eta_n)$ given in (d).
\end{enumerate}
\end{lemma}

\begin{proof}
Since we have only to construct train tracks on each component of $\partial M$ with
non-empty intersection with $L_\infty$, we can assume that $L_\infty$ is
contained in a component $S$ of $
\partial M$.
 Let $L$ be a minimal sublamination of $L_{rec}$.

Let us first consider the case when there is a train track $\theta$ minimally
carrying $L$ which is realised in $\cal{T}_S$ (recall that $\cal{T}_S$ is the minimal subtree of $\cal{T}$ invariant under $i_*\pi_1(S)$). Let
$\hat\theta$ be a lift of $\theta$
to the universal cover ${\mathbb H}^2$ of $S$. There is a continuous $\pi_1(S)$-equivariant map
$\phi_n:{\mathbb H}^2\rightarrow\cal{T}_S$ such that $\phi_n$ is constant on every tie
of $\hat\theta$ and the restriction of $\phi_n$ to any train route is injective.
Following \cite{conti}, we shall construct a $\rho_n$-equivariant map $\tilde
h_n:\hat\theta\rightarrow{\mathbb H}^3$. Let $\kappa_1,...,\kappa_p$ be the switches of $\theta$
and $\hat\kappa_1,...,\hat\kappa_p\subset {\mathbb H}^2$ lifts of $\kappa_1,...,\kappa_p$.
Denote by $(\tilde x_{i,n})\in\cal{T}_S\subset(X_\omega,x)$ the point
$\phi_n(\hat\kappa_i)$. We first define $\tilde h_n$ on
$\{\hat\kappa_1,...,\hat\kappa_p\}$ by setting $\tilde h_n(\hat\kappa_i)=\tilde x_{i,n}$.
We extend this map to $\pi_1(S)(\{\hat\kappa_1,...,\hat\kappa_p\})$ by $\tilde
h_n(g(\kappa_i))=\rho_n(g)\circ\tilde h_n(\hat\kappa_i)$ for any $g\in\pi_1(S)$ and any
$1\leq i\leq p$. Let $\hat b$ be a branch of $\hat \theta$. The vertical sides of $\hat b$
lie in two switches $\hat\kappa$ and $\hat\kappa'$ whose images by $\tilde h_n$ have
already been defined. On $\hat b$, we let $\tilde h_n$ be the map which is constant on
each
tie of $\hat b$, and which induces a parametrisation of
the geodesic segment joining $\tilde h_n(\hat \kappa)$ to $\tilde h_n(\hat\kappa')$ with
constant speed on a horizontal side of $\hat b$. Then for any branch $\hat b$ of
$\hat\theta$,
we have $\epsilon_n l(\tilde h_n(\hat b))\longrightarrow l_{\cal{T}_S}(\phi_n(\hat b))>0$.

Let $\theta'$ be the first subdivision of $\theta$ as defined in \cite[Chapitre
4, \S4.1]{meister2}, and $\hat\theta'\subset\hat\theta$ a lift of $\theta'$. We shall
deform the map $\tilde h_n$ to one which is adapted to $\hat\theta'$.
For a branch $\hat b$ of $\hat\theta'$, its image by $\tilde h_n$ is a broken geodesic segment, more
precisely, it is the union of two geodesic segments. We deform $\tilde h_n$ by a
homotopy which is constant on the vertical sides of $\hat b$ to a map which is constant on
each tie of $\hat b$ and takes $\hat b$ to the geodesic segment joining the images
under $\tilde h_n$ of the vertical sides of $\hat b$.

Since $\theta$ is realised in $\cal{T}_S$, it follows from the argument of
\cite[Chapitre 4]{meister2} that $\tilde h_n$ has the following properties :
\begin{enumerate}[\indent\indent a)]
\setcounter{enumi}{1}
\item there are $R>0$ and $n(R)$ such that, for $n\geq n(R)$, if $\hat b$ is a branch
of $\hat\theta'$, we have $l(\tilde h_n(\hat b))\geq R\epsilon_n^{-1}$;
\item there is a sequence of positive numbers $\delta_n\longrightarrow 0$ such that if
$\hat b_1,\hat
b_2$ are two adjacent branches of $\hat\theta$ which are separated by a switch, then the
external angle between $\tilde h_n(\hat b_1)$ and $\tilde h_n(\hat b_2)$ is smaller than
$\delta_n$.
\end{enumerate}
We perform the same for all the components of $L_{rec}$ that are
realised in $\cal{T}_S$. Denote by $\tau^1$ the union of the train tracks $\theta'$
thus obtained and by $\tilde h_n:\hat\tau^1\rightarrow{\mathbb H}^3$ the maps which agree with the
map defined above on each connected component of $\hat\tau^1$. By Lemma \ref{quivabien},
$\tau^1$ is not empty.
We also see that $\lambda_i$ passes through every branch of $\tau^1$ for every $i$ after taking
a subsequence.

When a component $L$ of $L_{rec}$ is not realised in $\cal{T}$, Lemma
\ref{quivabien} gives rise to a sequence of train tracks $\theta_i$ each
carrying
$L$ minimally.
We can assume that $\lambda_i$ passes through every branch of $\theta_i$.
Let us denote by $\tau_i^2$ the union of the train tracks thus obtained from
the
components of $L_{rec}$ which are not realised in $\cal{T}$. Finally we add branches to
$\tau^1\cup\tau^2_i$ to get a train track $\tau_i$ which minimally carries $L_\infty$
and $\lambda_i$.

We shall now extend the map $\tilde h_n$ to the preimage of $\tau_i^2$. Consider a
connected component $L$ of $L_{rec}$ which is not realised in $\cal{T}$. Consider the
subtrack $\theta_i$ of $\tau_i$ which minimally carries $L$. We get from Lemma
\ref{quivabien} that there are a point $x\in\cal{T}$, a sequence $\eta_i\rightarrow
0$, and
a sequence of $\pi_1(S)$-equivariant maps $\phi_i:{\mathbb H}^2\rightarrow\cal{T}$ such that
$\phi_i$ maps each branch of the preimage of $\theta_i$ to a geodesic segment (which may be
a point) with length smaller than $\eta_i$ and a lift of $\kappa_i$ is mapped to $x$
under $\phi_i$.
Set $x=(\tilde x_n)\in\Pi_n(\epsilon_n{\mathbb H}^3)$ and fix some $i\in{\mathbb N}$.
Let $\hat\theta_i\subset{\mathbb H}^2$ be a lift of $\theta_i$, and $\hat\kappa_i\subset\hat\theta_i$ the lift
of $\kappa_i$ that is mapped to $x$ by $\phi_i$. Let $G_i\subset\pi_1(
\partial M)$ be a finite set consisting of all $g \in \pi_1(\partial M)$ such
that if $\hat b$ is a branch of $\hat\theta_i$ and
$\hat\kappa_i$ contains a vertical side of $\hat b$, then the other vertical side of $\hat
b$ lies in $g(\hat\kappa_i)$.
Recall that for each branch $\hat b$, the length
of $\phi_i(\hat b)$ is less than $\eta_i$.
Therefore, we have $d(x,gx)\leq\eta_i$ for any $g\in
G_i$.
Since $\cal{T}$ is the $\omega$-ultralimit of $\epsilon_n {\mathbb H}^3$,
we have $\epsilon_n d(\tilde x_n,\rho_n(g)(\tilde x_n))\leq 2\eta_i$ for any $g\in G_i$ for $n$
large enough.
For any $g\in G_i\cup \{id\}$, we define $\tilde h_n(g(\hat\kappa_i))$ by $\tilde
h_n(g(\hat\kappa_i))=\rho_n(g)(\tilde x_n)$.
Let $\hat b$ be a branch of $\hat\theta_i$
with vertical sides lying in two switches $\hat\kappa_i$ and $g(\hat\kappa_i)$ for some $g\in
G_i$.
If $\tilde h_n(\hat\kappa_i)=\tilde h_n(\rho_n(g)(\hat\kappa_i))$, then we set
$\tilde h_n(\hat b)=\tilde h_n(\hat\kappa_i)$.
Otherwise, we set $\tilde h_n$ to be the
map which is constant on each tie of $\hat b$ and takes $\hat b$ to the geodesic
segment joining $\tilde h_n(\hat \kappa_i)$ to $\tilde h_n(\rho_n(g)(\hat\kappa_i))$.
Extend $\tilde h_n$ to an equivariant map from $\hat\theta_i$ to ${\mathbb H}^3$.
For sufficiently large $n$ and any branch $\hat b$ of $\hat\theta_i$, we have $\epsilon_n
l(\tilde h_n(b))\leq
2\eta_i$. Furthermore the sum of the weights with which $\theta_i$ carries $\lambda_n$ is
less than $\int_{\kappa_i}d\lambda_n\leq\int_{\kappa_1}d\lambda_n\longrightarrow
\int_{\kappa_1} d\lambda$.

We do the same construction for all the components of $L_{rec}$ that are not realised in
$\cal{T}_S$, and denote by $\tilde h_n:\hat\tau^1\cup\hat\tau^2_i\rightarrow{\mathbb H}^3$ the
maps whose restriction to each connected component of
$\hat\tau^1 \cup \hat\tau_2$ is the maps thus defined. It follows from the
construction that there is $N(i)$ such
that for $n\geq N(i)$, if $\tilde b$ is a lift of a branch of $\tau^2_i$, we have $\epsilon_n
l(\tilde h_n(\tilde b))\leq 2\eta_i$.

It remains to define $\tilde h_n$ on the lifts of the branches of
$\tau_i-(\tau^1\cup\tau^2_i)$. Let $\hat b$ be such a lift. Let $\hat \kappa$ and $\hat
\kappa'$ be the two vertical sides of $\hat b$.
Their projections $\kappa$ and $\kappa'$
lie in $\tau^1\cup\tau^2_i$.
Hence their images by $\tilde h_n$ are already defined, and
there are two points $x=(\tilde x_n), x'=(\tilde x'_n)$ in $\cal{T}$ such that $\tilde
h_n(\bar \kappa)=\tilde x_n$ and $\tilde
h_n(\bar\kappa')=\tilde x'_n$. We set $\tilde h_n$ to be the map which is
constant on each tie of $\hat b$ and takes $\hat b$ to the geodesic segment joining
$\tilde x_n$ to $\tilde x'_n$. We then have $\epsilon_n d(\tilde
x_n,\tilde x'_n)\longrightarrow d(x,x')$. Furthermore, since $\lambda$
is carried by $\tau^1\cup\tau^2_i$, we have $\lambda_n(b)\longrightarrow 0$. So, for $n$
large enough, we have $\lambda_n(b)(\epsilon_n l(\tilde h_n(\tilde b)))\leq 2\eta_i$.

Thus we have proved that there is $N(i)$ such that for $n\geq N(i)$, for a
branch $b$ of $\tau_i-\tau^1$, either $b$ is a branch of $\tau^2_i$ and we have
$\epsilon_n
l(\tilde h_n(\hat b))\leq 2\eta_i$ or $b$ is a branch of $\tau_i - (\tau^1 \cup
\tau^2_i)$ and we
have $\lambda_n(b)(\epsilon_n l(\tilde h_n(\hat
b)))\leq 2\eta_i$.
Now by choosing $N(i)$ such that $N(i)< N(i+1)$,
and taking a subsequence $\lambda_{N(n)}$ so that the $n$-th term is the
original $N(n)$-th term, we obtain the expected train track.
\end{proof}

\indent Since the $\lambda_n \cap S$ are weighted simple closed curves, $\tau_n\cap S$ is
connected.
Assume that $\tau_n \cap S \subset \tau^1$ for some component $S$ of $\partial M$.
For large $n$, $\lambda_n$ is carried
by $\tau^1$. By \cite[Chapter 4]{conti}, there is $\xi>0$ such that for $n$ large enough
$l_{\rho_n}(\lambda_n)\geq\xi l_{\rho_n}(h_n(\lambda_n))$. It then follows
from the
condition
(b) that $l_{\rho_n}(h_n(\lambda_n))$ tends to $\infty$. Therefore we have
$l_{\rho_n}(\lambda_n)\geq\xi l_{\rho_n}(h_n(\lambda_n)) \longrightarrow\infty$,
contradicting the assumption that $(l_{\rho_n}(\lambda_n))$ is a bounded sequence. From
now on, we continue our argument under the assumption that $\tau^1 \cap S \subsetneq\tau_n
\cap S$ for every component $S$ of $
\partial M$.






\section{Constructing annuli} \label{asum}
In this section, we shall work under the following assumptions to which we shall
refer as the assumptions of the section \ref{asum}:

Let $(\rho_n:\pi_1(M)\rightarrow \mathrm{Isom}({\mathbb H}^3))$ be a sequence of discrete faithful representations that
uniformise  $M$ such that the action of $\rho_n(\pi_1(M))$ on $\epsilon_n{\mathbb H}^3$ tends (with respect to $\omega$)
to  a small minimal action of $\pi_1(M)$ on an ${\mathbb R}$-tree ${\cal T}$.
Let $\lambda_n\subset\cal{ML}(\partial M)$ be a sequence of measured geodesic
laminations such that $l_{\rho_n}(\lambda_n)$ is a bounded sequence and that $\lambda_n$ converges in $\cal{ML}(\partial M)$ to some measured geodesic lamination $\lambda$.\\



At the end of this section, we shall reach a contradiction by showing that, under
these
assumptions, $\lambda$ cannot lie in ${\cal D}(M)$.
This will conclude the proof
of Theorem \ref{th&}.

Take a component $S$ of $\partial M$ such that $S \cap \tau^1 \neq \emptyset$.
In this section, we have only to pay attention to the behaviour of $\lambda_n$ on $S$.
Therefore, for simplicity, we denote $\lambda_n \cap S$ by $\lambda_n$, and $\tau_n \cap
S$ by $\tau_n$, etc.
In particular, $\lambda_n$ is assumed to be a weighted simple closed curve.

Let us denote by $c_n\subset S$ the support of
$\lambda_n$ and by $w_n$ the weight of $\lambda_n$ on $c_n$.
Let $c^*_n$ be the closed geodesic representative of $c_n$ in
${\mathbb H}^3/\rho_n(\pi_1(M))$ and let\linebreak
$h_n:\tau_n\rightarrow{\mathbb H}^3/\rho_n(\pi_1(M))$ be the projection of $\tilde h_n$.
By construction, $\tau_n$ carries $c_n$.
We set $c^1_n=c_n\cap\tau^1$,
$c^2_n=c_n\cap\tau^2_n$ and $c^3_n=c_n\cap(\tau_n-(\tau^1\cup\tau^2_n))$.
Then we
have $w_nl_{\rho_n}(h_n(c_n^2))=\sum_{b\text{ : the branches of
}\tau_n^2}\lambda_n(b)l_{\rho_n}(h_n(b))$ and we have \linebreak
$w_nl_{\rho_n}(h_n(c_n^3))=\sum_{b\text{: the branches of
}\tau-(\tau_1\cup\tau_n^2)}\lambda_n(b)l_{\rho_n}(h_n(b))$.
Therefore, by the property (d), we have $w_nl_{\rho_n}(h_n(c_n^2))=o(\epsilon_n^{-1})$
and by the property
(e), we have\linebreak $w_nl_{\rho_n}(h_n(c_n^3))=o(\epsilon_n^{-1})$.

The curve $h_n(c_n)$ is a piecewise geodesic, whose geodesic segments we call  edges and
whose points where it failed to be a geodesic we call vertices. The
vertices are the images
of the intersection of $c_n$ and the switches of $\tau_n$.
Let
$x_{n,1},...,x_{n,p_n}$ be the vertices of $h_n(c_n)$, and choose $p_n$
points
$y_{n,1},...,y_{n,p_n}$ on $c^*_n$. For $1\leq i\leq p_n$, we consider the
geodesic
triangle with vertices $y_{n,i},x_{n,i},x_{n,i+1}$ (with $x_{n,p_{n}+1}=x_{n,1}$ and
$y_{n,p_{n}+1}=y_{n,1}$) and the geodesic triangle with vertices
$x_{n,i+1},y_{n,i},y_{n,i+1}$. The union of these triangles for $i=1, \dots ,
p_n$
is a
simplicial annulus $A_n=S^1\times [0,1]$ cobounded by $c^*_n$ and $h_n(c_n)$. The
metric
$\nu_n$ induced on this annulus by the lengths of paths is a hyperbolic metric
with piecewise geodesic boundary.
By
the Gauss-Bonnet formula, the area of $A_n$ is less than $2p_n\pi$. By Lemma
\ref{rtt}, the
sequence $(w_np_n)=(\sum_{b\text{: the branches of }\tau_n} \lambda_n(b))$ is
bounded. We
parametrise $A_n$ by $S^1\times[0,1]$ so that the projection of $S^1\times\{1\}$ to
${\mathbb H}^3/\rho_n(\pi_1(M))$ is $c_n^*$. We denote by $\bar c_n^j$ the part of
$S^1\times\{0\}$ corresponding to $h_n(c_n^j)$ for $j=1,2,3$, and by $\bar
c_n$ the union of the three.

For a positive number $\epsilon$, which we shall specify later, and each point
$x\in\bar c_n^1$, we consider a geodesic arc $a_x$ on $(S^1 \times I, \nu_n)$
perpendicular to $S^1 \times \{0\}$ at $x$ having length $\epsilon$ with respect to
$\nu_n$. If the perpendicular reaches $ S^1 \times \partial I$ before the length
$\epsilon$ is attained, we define $a_x$ to be the geodesic arc having both endpoints on $
S^1 \times \partial I$.

We are going to estimate from below the length of the set of points $x$ for which
the $a_x$ reach $S^1 \times \{1\}$ without intersecting themselves or each
other.
Since the length of this set of points is bounded above
by $\mathrm{length}(c_n^*)$ (with respect to $\nu_n)$, we get an inequality, which will
appear as the inequality (i) below. For that, we need to subtract from the length of $\bar
c_n^1$ the lengths of (I) the set of points $x$ for which $a_x$ has self-intersection,
(II) the set of points $x$ for which $a_x$ intersects $a_y$ with $x \neq y$, (III) the set
of points $x$ for which $a_x$ has an endpoint on either $\bar c_n^2$ or $\bar c_n^3$,
and (IV) the set of points $x$ which are neither of type (I) nor of type (II) and for
which $a_x$ has an endpoint in the interior of $S^1\times I$.

We first consider the contribution of the points of type (I) to the length,
i.e., $x$ for which
$a_x$ intersects itself transversely. By the Gauss-Bonnet formula, a geodesic loop
formed by a
subarc of $a_x$ cannot be null-homotopic. Hence, there must be a loop formed by a subarc
of $a_x$ freely homotopic to $S^1 \times \{1\}$. It follows that if both of two
perpendiculars $a_{x_1}, a_{x_2}$ with $x_1 \neq x_2$ have self-intersection, then
$a_{x_1} \cap a_{x_2} \neq \emptyset$. Thus, the contribution of
the set of
$x$ with self-intersecting $a_x$ to the length is absorbed in the contribution of $x$ with $a_x$
intersecting another $a_y$, that is, of type (II), which will be dealt with below.

We next consider the points $x$ of type (II) namely $a_x$ intersects $a_y$ for
some $y\in\bar c_n^1$. Let $m$ be a point in the intersection $a_x \cap a_y$, and let
$a_x', a_y'$ be subarcs of $a_x, a_y$ between $x$ and $m$ and $y$ and $m$ respectively.
Let $\beta$ be an arc on $\bar c_n$ to which $a_x' \cup a_y'$ is homotopic fixing the
endpoints. Suppose that $\beta$ is also contained in $\bar c_n^1$. We then say that $x$ is
an {\em inessential} point of type (II). It was shown in \cite[Lemme
5.11]{Bouts} that,
in this situation, there is a constant $\xi_n$ depending only on $\epsilon$ and the
maximal exterior angle of the vertices on $\bar c_n^1$, which is less than
$\delta_n$ in our
case, such that $x$ is within the distance $\xi_n$ with respect to $\nu_n$ from a vertex
of $\bar c_n^1$. It was also shown in \cite{Bouts} that the constant $\xi_n$ tends to $0$
as either $\delta_n$ or $\epsilon$ goes to $0$. If $\beta$ does not lie on $\bar c_n^1$,
then we say that $x$ is an {\em essential} point of type (II).
We also call an arc such as $a_x' \cup a_y'$, which is not homotopic to an arc
in $\bar c_n^1$, an {\em essential arc}.
We should note that the length of such an essential arc is less than or equal
to $2\epsilon$.
Let
$\bar c_n^+$ be the
union of the essential point of type (II). We shall use the essential points of
type (II)
to construct a sequence of annuli in Lemma \ref{an} and then these annuli
will lead us to a contradiction.

Now we consider the points of type (III). The total length with respect to $\nu_n$
of the set of points $x$ on $\bar c_n^1$ for which $a_x$ reaches a point on $\bar c_n^2$
is bounded above by the length of $\bar c_n^2$. Similarly, the total length of the set of
points $x$ for which $a_x$ reaches a point on $\bar c_n^3$ is bounded above by
the length of $\bar c_n^3$.

Finally we consider the points of type (IV); the points $x$ such that $a_x
\setminus \{x\}$ is contained in the interior of $S^1\times I$ while $a_x$ has
neither
self-intersection nor intersection with another $a_y$. Since the union of $a_x$ for $x$
of type (IV) has area bounded below by the length of the set of points $x$ of type (IV)
multiplied by $\mbox{sh}(\epsilon)$, we can bound the length from above by
$\mathrm{Area}(A_n)/\mbox{sh}(\epsilon)$.

Putting all of these considerations together, we get an inequality:
\begin{eqnarray*}
\text{(i)} \hspace{3mm} \l(\bar{c}_n^1) -2p_n\xi_n -
\l(\bar{c}_n^2) -\l(\bar{c}_n^3) -\mathrm{Area}(A_n)/\mbox{sh}(\epsilon) -\l(\bar{c}_n^+) \leq
\l(c_n^*).
\end{eqnarray*}

\smallskip
As we saw at the beginning of this section, from the properties
(d) and
(e), we get $w_n\epsilon_n l_{\rho_n}(\bar c_n^2)\longrightarrow 0$ and $w_n\epsilon_n
l_{\rho_n}(\bar c_n^3)\longrightarrow 0$.
We also saw that $(w_np_n)$ is a
bounded sequence. It follows that we have
$2w_n p_n\xi_n\longrightarrow 0$. This implies also that $w_n\mathrm{Area}(A_n)\leq 2 w_n
p_n\pi$ is a bounded sequence and that we have\linebreak $\epsilon_n
w_n\mathrm{Area}(A_n)\longrightarrow 0$.
By assumption $(w_n l_{\rho_n}(c^*_n))=(l_n(\lambda_n))$ is a bounded sequence;
hence $\epsilon_n w_n l_{\rho_n}(c^*_n)$ tends to $0$.
Thus we get
$w_n\epsilon_n(l_{\nu_n}(\bar{c}_n^1)-l_{\nu_n}(\bar{c}_n^+))\longrightarrow 0$.
We
put it into a form of a claim for later use.
\begin{cl} \label{zero} We have
$w_n\epsilon_n(l_{\nu_n}(\bar{c}_n^1)-l_{\nu_n}(\bar{c}_n^+))\longrightarrow 0$.
\end{cl}

From this claim, we shall deduce the following lemma which will prevent $\lambda$ from lying in ${\cal D}(M)$.

\begin{lemma} \label{an}
Under the assumptions of the section \ref{asum}, there is a homoclinic leaf or an
annular lamination that does not intersect $|\lambda|$ transversely.
\end{lemma}


\begin{proof}
We shall prove this lemma in several steps.

Fix an orientation for $c_n$. Let $s$ be a segment lying in $c_n^1$, and
put on
it the orientation induced by that on $c_n$. The {\em train route}
$b(1),...,b(t)$ of $s$ is the ordered finite sequence of branches of $\tau^1$ through
which $s$ goes in this order : $b(i)$ is an element of the set $\{b_1,..., b_p\}$
of branches of
$\tau^1$. We fix an orientation for each branch of $\tau^1$. A branch $b(i)$ in
the
train route of $s$ is said to be {\em positively oriented} if its orientation
coincides with the
orientation of $s$ and {\em negatively oriented} otherwise. The {\em oriented train
route} $bo(1),...,bo(t)$ of $s$ is the ordered finite sequence of oriented branches
of
$\tau^1$ through which $s$ goes in this order with the assigned orientations:
$bo(i)$ is an element of
$\{b_1,..., b_p\}\times\{+,-\}$. When $(bo(i))_{i\in I}$ is an oriented train route, we
shall denote by $(b(i))_{i\in I}$ the corresponding non-oriented train route.

Now we are going to construct some long and thin bands connecting two segments
of $h_n(c_n)$.

\begin{lemma}
\label{band}
 There are two infinite oriented train routes
$bo,bo':{\mathbb N}\rightarrow \{b_1,..., b_p\} \times \{+, -\}$ in $\tau^1$, a map
$V:{\mathbb N}\rightarrow{\mathbb N}$,
and a map $F:{\mathbb N}\times
{\mathbb N}\rightarrow{\mathbb N}$ such that for any $n,t\in{\mathbb N}$ there are $F(n,t)$ disjoint segments
$s(n,j)\subset c_n^1$ $(1\leq j\leq F(n,t))$ satisfying the following :
\begin{itemize}
\item the oriented train route of $s(n,j)$ is $(bo(i))_{i\leq t}$;
\item there is a segment $s'(n,j)\subset c_n^1$ whose oriented train route is
$(bo'(i))_{0\leq i\leq V(t)}$;
\item there is a homeomorphism $g_{n,j}: h_n(s(n,j))\rightarrow h_n(s'(n,j))$;
\item $g_{n,j}(h_n(v(i)))\in h_n(b'(V(i)))$ for any $i\leq t$ and  $n \in {\mathbb N}$ where
$v(i)=b(i)\cap b(i+1)$;
\item any point $x\in h_n(s(n,j))$ is connected to $g_{n,j}(x)$ by an essential arc
$\zeta_n(x)$ on $A_n$ with length less than $6\epsilon$;
\item the simple closed curve $s(n,j)\cup \zeta_{n}(
\partial s(n,j))\cup s'(n,j)$ bounds a disc $D_{n,j}$ in $A_n$;
\item for any $t\in{\mathbb N}$, $F(n,t)$ tends to $\infty$ as $n\longrightarrow \infty$.
\end{itemize}
\end{lemma}


Notice that we have adopted the convention $0\in{\mathbb N}$, which is necessary since
the map $F$ may vanish for many $(n,t)$.
\begin{proof}
Let $\{\sigma_{1,n},\sigma_{2,n},\dots \}\subset\bar c_n$ be a maximal family of
disjoint
segments with diameter $6\epsilon$ such that the middle point $x_{i,n}$ of $\sigma_{i,n}$ lies
in $\bar c_n^+$. Consider a segment $\sigma_{i,n}$ and its middle point $x_{i,n}$. In the
family of essential arcs joining $x_{i,n}$ to $\bar c_n^1$, we take an arc
$a_{i,n}$
to be the shortest (with respect to $\nu_n$). Since $x_{i,n}$ lies in
$\bar c_n^+$, the length of $a_{i,n}$ is less than $2\epsilon$.

 If $
\partial a_{i,n}-x_{i,n}$ lies in some $\sigma_{j,n}$, we denote by $\zeta_{i,n}$
the
geodesic segment joining $x_{i,n}$ to $x_{j,n}$ which is homotopic to $a_{i,n}$ relative
to $x_{i,n}\cup\sigma_{j,n}$. If $
\partial a_{i,n}-x_{i,n}$ is disjoint from $\bigcup\sigma_{j,n}$, we define
$\zeta_{i,n}$
to be $a_{i,n}$. The length of each segment $\zeta_{i,n}$ thus obtained is less
than $5\epsilon$.
We shall next show that the segments $\zeta_{i,n}$ have mutually disjoint
interiors.

Consider two different segments $\zeta_{1,n}$ and $\zeta_{2,n}$ and assume that
their interiors intersect. Then the interiors of $a_{1,n}$ and $a_{2,n}$
also intersect.
Let  $y$ be a point in the intersection. Let $[x_{i,n},y[$ be the connected
component of $a_{i,n}-\{y\}$ containing
$x_{i,n}$. Let $\kappa$ be the shortest segment among $a_{1,n}-[x_{1,n},y[$ and
$a_{2,n}-[x_{2,n},y[$. Then, for $i=1,2$, the length of the arc
$[x_{i,n},y]\cup\kappa$ is less
than or equal to the length of $a_{i,n}$. Furthermore one of the two arcs
$[x_{i,n},y]\cup\kappa$, say $[x_{1,n},y]\cup\kappa$ is not the shortest in its
homotopy class relative to the endpoints.
Let $a'_1$ be the shortest arc homotopic to $[x_{1,n},y]\cup\kappa$ relative
to the endpoints.
Then
the length of the segment $a'_1$ (with respect to $\nu_n$) is less than the length of $a_{1,n}$.
Recall that we
chose $a_{1,n}$ which is shortest among all essential arcs joining
$x_{1,n}$ to $\bar c_n^1$. It follows that $a'_1$ is not essential, namely there is a
segment $\beta\subset \bar c_n^1$ homotopic to $a'_1$ relative to the endpoints.
The endpoints of $\beta$ are
$x_{1,n}$ and another point which we call $y_1$. The distance (with respect to $\nu_n$)
between $x_{1,n}$ and $y_1$ is less than the length of $a'_1$ which is less than $2\epsilon$.
By the properties (b) and (c), each component of $\bar c_n^1$ (in particular
the one
containing $\beta$) is a union of long geodesic segments such that the exterior angle
between two consecutive segments is small. By \cite[Lemma 4.2.10]{ceg} such a component of
$\bar c_n^1$ is a $(K, \eta)$-quasi-geodesic with $K {\rightarrow} 1, \eta {\rightarrow} 0$ as $n
{\rightarrow} \infty$.
It follows that there is $N$ (independent of $\beta$)
such that for $n\geq N$, the length of $\beta$ is less than $3\epsilon$. This implies that
$y_1$ lies in $\sigma_{1,n}$. By our definition of $\zeta_{1,n}$, it has an
endpoint on $x_{1,n}$, not on $y_1$ (see figure
\ref{dis}). This contradicts our assumption that the interiors of $\zeta_{1,n}$ and $\zeta_{2,n}$ intersect.


\begin{figure}[hbtp]
\psfrag{a}{$x_{1,n}$}
\psfrag{b}{$y_1$}
\psfrag{c}{$y$}
\psfrag{d}{$x_{2,n}$}
\psfrag{f}{$\bar c_n$}
\psfrag{g}{$\sigma_{1,n}$}
\psfrag{h}{$\sigma_{2,n}$}
\psfrag{i}{$c^*_n$}
\psfrag{j}{$a_{2,n}$}
\psfrag{k}{$a_{1,n}$}
\psfrag{l}{$\zeta_{2,n}$}
\psfrag{m}{$\zeta_{1,n}$}
\psfrag{n}{$\kappa$}
\centerline{\includegraphics{dis.eps}}
\caption{From $a_{i,n}$ to $\zeta_{i,n}$}
\label{dis}
\end{figure}

\indent
Even if some segment $a_{i,n}$ has a self-intersection, the same argument shows
that $\zeta_{i,n}$ does not have any self-intersection.

Thus we have proved the following claim.

\begin{cl} There is a map $f_n:\{x_{1,n},x_{2,n},...\}\rightarrow \bar c^1_n$ and
a family $(\zeta_i)$ of essential segments with disjoint interiors such that the
length of $\zeta_i$ is less than $5\epsilon$, \linebreak $\partial\zeta_i=\{x_i,f_n(x_i)\}$, and
$\{f_n(x_{1,n}),f_n(x_{2,n}),...\}\cap\bigcup_i \sigma_{i, n
}\subset\{x_{1,n},x_{2,n},...\}$.
\end{cl}

Let $j$ be an integer, and cut $\bar c^1_n$ into disjoint segments
$s_{1,n},s_{2,n},...$, each containing $j$ edges (if the number of edges of some
component of $\bar c^1_n$ is not a multiple of $j$, then there are some edges of
$\bar
c^1_n$ not belonging to any one of these segments). The following claim will
help us to evaluate the number of segments thus obtained.
\begin{cl}
\label{order of number}
Let $r_n$ be the number of the components of $\bar c_n^1$. Then, $w_nr_n
\longrightarrow 0$ as $n \rightarrow \infty$.
\end{cl}

\begin{proof}
 Note that any train route on $\tau_n$ connecting a point in $\tau^1$ and a
point in $\tau^2_n$ must pass through a point in $\tau_n - (\tau^1 \cup
\tau_n^2))$. Therefore between any two components of $c_n^1$, there is a
component of $c_n\cap(\tau_n- (\tau^1 \cup \tau_n^2))$. Hence $(w_nr_n)$ is
bounded above by $\sum_{b \subset \tau_n - (\tau^1 \cup\tau^2_n)}\lambda_n(b)$.
Since $|\lambda|\subset L_{rec}$ is carried by $\tau^1\cup\tau^2_n$, the sum $\sum_{b \subset
\tau_n - (\tau^1 \cup\tau^2_n)}\lambda_n(b)$ tends to $0$ as $n
\rightarrow \infty$. It follows that we have $w_nr_n \longrightarrow 0$.
\end{proof}

Since $w_n (\text{the number of edges of}\ \bar c^1_n)=w_n(\sum_{b\subset \tau^1}
c_n(b))
\longrightarrow \sum_{b\subset\tau^1}\lambda(b)$, the number of edges of $\bar
c^1_n$ is $\Theta(w_n^{-1})=\Theta(p_n)$.
The
number of edges not lying in one of the $s_{i,n}$ is less than $j r_n=o(p_n)$.
It follows that the number of segments $s_{i,n}$ is $\Theta(p_n)$.

Let $t_n$ be the number of edges of $\bar c^1_n$ containing no segment
among the $\sigma_{i,n}$. If an edge $e$
contains no
segment among the $\sigma_{i,n}$, then there is no point of $\bar c_n^+$ in $e$
outside the $6\epsilon$-neighbourhood of $
\partial e$. By the property (b), the total length of
these edges is greater than $t_n R\epsilon_n^{-1}$ and smaller than $l_{\nu_n}(\bar
c_n^1)-l_{\nu_n}(\bar c_n^+)+12t_n\epsilon$. By Claim \ref{zero}, we have $w_n\epsilon_n t_n
R\epsilon_n^{-1}\longrightarrow 0$. Therefore $t_n$ is  $o(w_n^{-1})=o(p_n)$.

Thus we have seen that among the $s_{i,n}$, there are $\Theta(p_n)$ disjoint
segments lying in
$\bar c^1_n$ and containing $j$ edges each of which contains a segment
among the $\sigma_{i,n}$.
We shall denote these $\Theta(p_n)$-many segments again by $s_{1,n},
s_{2,n}, \dots $.

Let $s\in\{s_{1,n},s_{2,n},\dots \}$ be a segment with the following
property : there are at least two distinct components of
$\bar c^1_n$ containing an endpoint of $\zeta_{i,n}$ for some $x_{i,n}\in s$.
Let $t'_n$ be the
number of those with this property among the $s_{i,n}$. In each such a segment
$s$, we choose
two points in $\{x_{1,n},x_{2,n},...\}\cap s$, say $x_{1,n},x_{2,n}$, such
that $\zeta_{1,n}$ and $\zeta_{2,n}$ connect $s$ to distinct components of
$\bar c^1_n$ and
the segment $]x_{1,n},x_{2,n}[\subset s$ does not contain any $x_{k,n}$. There
may be several $\zeta_{i,n}$ which have $x_{1,n}$ or $x_{2,n}$ as an endpoint.
There are two points $y_{1,n}$ and $y_{2,n}$ which lie in two distinct
components of $\bar c_n^1$ such that $y_{1,n}$ (resp. $y_{2,n}$) is connected to
$x_{1,n}$ or $x_{2,n}$ by some $\zeta_{i,n}$ and that if $[y_{1,n},y_{2,n}]$ is the
segment of $\bar c_n$ joining $y_{1,n}$ to $y_{2,n}$ whose interior does not
contain $x_{1,n}$, then there is no $\zeta_{i,n}$ connecting $]y_{1,n},y_{2,n}[$
to $\{x_{1,n},x_{2,n}\}$. Let us embed $A_n$ in a round disc such that $\bar
c_n$ is the boundary and connect $y_{1,n}$ to $y_{2,n}$ by a geodesic segment
with respect to the ordinary Euclidean metric on the disc.

Assume that for another segment $s' \in \{s_{1,n}, \dots\}$ with the same
property, the resulting geodesic segment intersects
transversely the geodesic segment produced from $s$ above (i.e. the geodesic segment
connecting $y_{1,n}$ to $y_{2,n}$). Suppose that $x_{3,n}$ and $x_{4,n}$ are the
points on $s'$ chosen in the same way as $x_{1,n}, x_{2,n}$ for $s$. Then we have
$[x_{3,n},x_{4,n}]\subset
[y_{1,n},y_{2,n}]$ and either $y_{1,n}\in\{x_{3,n},x_{4,n}\}$ or
$y_{2,n}\in\{x_{3,n},x_{4,n}\}$. There is only one segment $s_{i,n}$
containing $f(x_{1,n})$ and only one containing $f(x_{2,n})$. Since there is no
$\zeta_i$ connecting $]y_{1,n},y_{2,n}[$ to ${x_{1,n},x_{2,n}}$, if we have
$f(x_{3,n})\in\{x_{1,n},x_{2,n}\}$ then $x_{3,n}$ is equal to $f(x_{1,n})$
or $f(x_{2,n})$. It follows that there are only two possibilities for the
configuration of $s,s'$ and the geodesic segments constructed above (see Figure \ref{poly}).


\begin{figure}[hbtp]
\psfrag{a}{$x_{1,n}$}
\psfrag{b}{$x_{2,n}$}
\psfrag{c}{$y_{1,n}=x_{3,n}$}
\psfrag{d}{$x_{4,n}$}
\psfrag{e}{$y_{4,n}$}
\psfrag{f}{$y_{2,n}$}
\psfrag{g}{$\bar c_n$}
\psfrag{h}{$c^*_n$}
\centerline{\includegraphics{poly.eps}}
\caption{A segment intersects at most two other segments}
\label{poly}
\end{figure}

From that we deduce that there are at least $\frac{1}{3} t'_n$ disjoint
geodesic segments in the round disc, each connecting two distinct components
of $\bar c_n^1$.
Furthermore, since the $\zeta_i$ have disjoint interiors, any pair of connected
components of $\bar c_n^1$ is connected by at most two of these disjoint
segments.
Consider a map from $\bar c_n$ to a round circle that preserves the order and maps
each connected component $C_i$ of $\bar c_n^1$ to a point $Q_i$. Join two points
$Q_i$ and $Q_j$ by a segment if and only if there is one of the segments constructed above which
joins $C_i$ and $C_j$. Thus we have constructed at least $\frac{1}{6} t'_n$
segments with
disjoint interiors. We can now add some geodesic segments in the disc to get a
triangulation of
the polygon with vertices $Q_1,...,Q_{r_n}$. Such a triangulation has $2r_n-3$
edges (this can easily be computed with the Euler formula). Therefore we have
$\frac{1}{6} t'_n\leq 2r_n-3=o(p_n)$.

Since we initially had $\Theta(p_n)$ segments $\{s_{1,n},s_{2,n},...\}$,
after excluding $o(p_n)$ segments as above from them,
there remains $\Theta(p_n)$ disjoint segments $\{s_{1,n},
s_{2,n},...\}$ in $\bar c_n^1$, (for which we use the same symbols) such that
if $s$ is one of those segments, then we have :
\begin{itemize}
\item $s$ contains $j$ edges;
\item i) each edge of $s$ contains some $x_{i,n}$;
\item ii) there is a unique component $C$ (depending on $s$) of $\bar c_n^1$ such
that for any $x_{i,n}\in s$, we have $f_n(x_{i,n})\in C$.
\end{itemize}

Let $s$ be one of these segments, $C$ the associated component of
$\bar c_n^1$, and $x$ a point of $s\cap\{x_{1,n},...\}$.
Denote by
$\zeta_x$ the corresponding $\zeta_{i,n}$. Since $A_n$ is an annulus and
$\zeta_x$ is embedded, there are only two possibilities for the homotopy class
of $\zeta_x$ relative to $s\cup C$. Therefore, taking $2j$ instead of $j$ at the
beginning and cutting each segment into halves, we get
$\Theta(p_n)$ disjoint segments $\{s_{1,n},...\}$ in $\bar c_n^1$ each one
containing $j$ edges and satisfying (i), (ii) above and :\smallskip\\
\indent iii) for any $x,y\in s\cap \bar c_n^+$, $\zeta_{x}$ and $\zeta_{y}$ are
homotopic relative to $s\cup C$.\medskip\\
\indent
Let $s$ be one of the segments produced above, and $C$ the
corresponding component of $\bar c_n^1$. Let $x$ and $y$ be the extremal points
of $s\cap\{x_{1,n},...\}$, and  $[x,y]$ the segment in $s$ joining $x$ to $y$.
The segment $[x,y]$ contains at least $(j-2)$ edges. We have the following :

\begin{lemma}  \label{anne} There is $N\in{\mathbb N}$ which does not depend on $s$ such
that for any $n\geq N$ we have the following by homotoping $A_n$ keeping
$\partial A_n$ and $\bar c_n^i$ unchanged:\\
\indent - there is a homeomorphism $g_n:[x,y]\rightarrow [f_n(x),f_n(y)]$ such
that for any $n\geq N$ and any $z\in[x,y]$, the two points $z$ and $g_n(z)$ are
connected by an
essential arc whose length (with respect to $\nu_n$) is smaller than $6\epsilon$.
\end{lemma}
\begin{proof} By the property (iii), the simple closed curve $\zeta_x\cup
[x,y]\cup
\zeta_y\cup [f_n(x),f_n(y)]$ bounds a disc in $A_n$.
Since both $[x,y]$ and $[f_n(x),f_n(y)]$ lie in $\bar c_n^1$, by the
properties (b) and (c), they
consist of long geodesic segments such that the external angles formed by two
adjacent segments are less than $\delta_n$. Let $k$ be the geodesic segment in
${\mathbb H}^3/\rho_n(\pi_1(M))$ joining $x$ to $y$ that is homotopic to $[x,y]$, and let
$k'$ be the one in ${\mathbb H}^3/\rho_n(\pi_1(M))$ joining $f_n(x)$ to
$f_n(y)$ that is homotopic to $[f_n(x),f_n(y)]$. We parametrise the arcs
$[x,y]$, $[f_n(x),f_n(y)]$, $k$ and $k'$ by the arc-length.

By \cite[Lemma
4.2.10]{ceg}, for sufficiently large $n$, we have\linebreak
$d([x,y](t),k(t))\leq\epsilon'_n$
and $d([f_n(x),f_n(y)](t),k'(t))\leq\epsilon'_n$ for any $t$ with $\epsilon'_n\longrightarrow
0$.
 It follows that
$1\leq\frac{l([x,y])}{l(k)}\leq\frac{l(k)+\epsilon'_n}{l(k)}\leq 1+\epsilon_n'$ and
that
$1\leq\frac{l([f_n(x),f_n(y))}{l(k')}\leq 1+\epsilon_n'$ for sufficiently large
$n$, where $l(.)$ denotes the length in ${\mathbb H}^3/\rho_n(\pi_1(M))$.
Therefore, we have
$d([x,y](\frac{l([x,y])}{l(k)}t),k(t))\leq 2\epsilon'_n$ for sufficiently large
$n$. For the same reason, we have also
$d([f_n(x),f_n(y)](\frac{l([f_n(x),f_n(y)])}{l(k')} t),k'(t))\leq 2\epsilon'_n$.

By the property (iii), the simple closed curve $\zeta_x\cup k\cup
\zeta_y\cup k'$ bounds a disc. Since $k$ and $k'$ are geodesic segments, the
function $d(k(t),k'(\frac{l(k')}{l(k)} t))$ is convex. Therefore we have
$d(k(t),k'(\frac{l(k')}{l(k)} t))\leq 5\epsilon$ for any $t$.

We define $g_n:[x,y]\rightarrow [f_n(x),f_n(y)]$ by the following formula \linebreak
$g_n([x,y](\frac{l([x,y])}{l(k)}t))=[f_n(x),f_n(y)](\frac{l([f_n(x),f_n(y))}{l(k)}t)$.
Setting $z=[x,y](\frac{l([x,y])}{l(k)}t)$, we have that
the distance
$d(z,g_n(z))$ is less than the following quantity\linebreak $d([x,y](\frac{l([x,y])}{l(k)}t),k(t))+d(k(t),k'(\frac{l(k')}{l(k)}
t))+d(k'(\frac{l(k')}{l(k)}
t,[f_n(x),f_n(y)](\frac{l([f_n(x),f_n(y))}{l(k)}t))$. Then we get from the above
$d(z,g_n(z))\leq 5\epsilon+4\epsilon'_n$. Now we conclude by taking $N$ such that $4\epsilon'_n\leq \epsilon$
for $n\geq N$
and by changing $A_n$ by a homotopy so that
the
geodesic segment $\zeta_n(x)$ connecting $z$ to $g_n(z)$ lies in $A_n$ for any
$z$ in $s$.
\end{proof}


We remove from $s$ its two extremal edges so that $g_n$ is defined on the
entire
$s$.

By construction, there is a constant $R'$ such that for any branch $\hat b$ of
$\hat\tau^1$, we have $l(\tilde h_n(\hat b))\leq R'\epsilon_n^{-1}$. By Lemma
\ref{anne} and  since $\delta_n \longrightarrow 0$ as $n \rightarrow
\infty$, we have $l_{\rho_n}(g_n(s))\leq l_{\rho_n}(s)+13\epsilon\leq(j-2)
R'\epsilon_n^{-1}+13\epsilon$.
Therefore if $j'_n$ is the number of edges which $g_n(s)$ contains, $j'_n\leq (j-2)
(\frac{R'}{R}+1)$ for large $n$.

Since $\tau^1$ has only finitely many branches, there are only finitely
many possibilities for the oriented train routes of $s$ and of $g_n(s)$. Thus we
can find $\Theta(p_n)$-many disjoint segments $\{s_{1,n},...\}$ with the properties
(i), (ii) and (iii), having the same oriented train route such that the $g_n(s)$ also
have the same oriented train route for all $s \in \{s_{1,n}, \dots \}$.

Now we fix $j$ to be $3$ and extract a subsequence so that the oriented
train routes of $s_{i,n}$ and $g_n(s_{i,n})$ do not depend on $n$.


To each $s_{i,n}$, we add the edge of $\bar c_n$ which is adjacent to the last (with respect
to the orientation of $s_{i,n}$) edge of $s_{i,n}$. In this family of segments, take a
maximal family of disjoint segments lying in $\bar c_n^1$, which we denote by
$\{s'_{1,n},s'_{2,n},...\}$.
By Claim \ref{order of number}, the number of elements in this family is $\Theta(p_n)$.
We shall find in this family $\Theta(p_n)$ segments which have the same properties as
the segments $\{s_{1,n}, \dots \}$ above.

 We have already seen that the number of edges of $\bar c_n^1$ not containing a
segment $\sigma_{i,n}$ is $o(p_n)$.
Therefore we can find $\Theta(p_n)$ segments
among the $s'_{i,n}$ each of whose edges contains an element of
$\{\sigma_{1,n},...\}$.
We denote this family of segments again by $s'_{i,n}$ changing the indices.

By the same argument as above (see the paragraph before the statement of property
(ii)) the number of segments $s'_{i,n}$ such that there are two distinct
components of
$\bar c^1_n$ containing a point $f_n(x)$ for some $x\in s'_{i,n}\cap \{x_1,...\}$ is
$o(p_n)$. Therefore we have $\Theta(p_n)$ segments among the $s'_{i,n}$ satisfying
(i) and (ii).
We denote this subfamily of segments again by $s'_{i,n}$.

Let $s'_{1,n}$ be one of these segments, and $C_1$ the connected component
of $\bar c_n^1$ such that we have $f_n(x)\in C_1$ for any $x\in s'_{1,n}\cap \{x_{1,n},...\}$.
Assume that there are two arcs $\zeta_x$ and $\zeta_y$ with $x,y\in s'_{1,n}\cap
\{x_{1,n},...\}$ which are not homotopic relative to $s'_1\cup C_1$. Let $s'_{2,n}$ be
another one of the $s'_{i,n}$ and $C_2$ the connected component of $\bar c_n^1$ such that
$f_n(x)\in C_2$ for any $x\in s'_{2,n}\cap \{x_{1,n},...\}$. Since the $\zeta_{i,n}$ have
disjoint interiors, if $C_1=C_2$, then $s'_{2,n}$ has property (iii). It follows that
there are at most $r_n=o(p_n)$ segments $s'_{i,n}$ that do not have property (iii).

Thus we have seen that there are $\Theta(p_n)$ disjoint segments $\{s'_{1,n},...\}$ with the
 properties (i), (ii)
and (iii) such that each $s'_{i,n}$ is obtained by adding to one of the $s_{i,n}$ the edge
adjacent to the last edge. The proof of Lemma \ref{anne} applies to these segments,
yielding an homeomorphism $g_n$ which can be chosen to coincide with the one
defined on the segments $s_{i,n}$ if restricted to them.

From this last family, we take $\Theta(p_n)$ segments $s'_{i,n}$ such that the oriented
train routes of $s'_{i,n}$ and $g_n(s'_{i,n})$ do not depend on $i$. Then we extract a
subsequence (with respect to $n$) such that for sufficiently large $n$, the oriented train
routes of $s'_{i,n}$ and of $g_n(s'_{i,n})$ do not depend on $n$.

By doing this argument recursively, increasing $j$ one by one, we complete the proof of
Lemma \ref{band}.
\end{proof}




When $(bo(i))_{i\in{\mathbb N}}$ is an oriented train route, we denote by $(b(i))_{i\in{\mathbb N}}$ the
corresponding non-oriented train route. Since for any $t$, the simple closed
curve $c_n$ goes through the oriented train
routes $(bo(i))_{i\leq t}$ and $(bo'(i))_{i\leq t}$ for $n$ large enough, there are two
half-leaves $l_+$ and $l'_+$ of $L_{rec}$ whose oriented train routes are $(bo(i))_{i\in{\mathbb N}}$ and
$(bo'(i))_{i\in {\mathbb N}}$ respectively.

Let $e\in{\mathbb N}$ be an integer which we shall specify later, and let
$\varphi:{\mathbb N}\rightarrow{\mathbb N}$ be a strictly increasing map such that : (*)
\begin{enumerate}
\item $bo(\varphi(i)+j)=bo(\varphi(0)+j)$ for any $i\in{\mathbb N}$ and any $0\leq j\leq e$.
\item
$bo'(V(\varphi(i))+j)=bo'(V(\varphi(0))+j)$ for any $i\in{\mathbb N}$ and any $0\leq j\leq
\frac{R'}{R}e +1$.
\item Suppose that $l_+$ (resp. $l'_+$) is not a closed curve, and let $\alpha_i$ be a sub-arc
of $l_+$ with train route $(b(j))_{\varphi(0)\leq j<\varphi(i)}$ (resp. $\alpha'_i$ the arc
of $l'_+$ with train route $(b'(i))_{\varphi(0)\leq i<\varphi(i)}$). We can assume that the
two endpoints
of $\alpha_i$ (resp $\alpha'_i$) lie in the same switch of $\tau^1$. Then the sequences
$(\partial
\alpha_i)$ (resp. $\partial \alpha'_i$) converges to a single
point with respect to the Hausdorff topology.
\end{enumerate}
\indent The existence of such a map $\varphi$ follows from the fact that the number of
branches of $\tau^1$ is bounded.

For $t\in{\mathbb N}$, let $N(t)$ be an integer such that $F(n,t)\geq 1$ for any
$n\geq N(t)$ where $F(.,.)$ is the map defined in Lemma \ref{band}. The map
$N:{\mathbb N}\rightarrow{\mathbb N}$ can be assumed to be increasing and non-constant since we can
assume that $F(n,t)$ is non-constant with respect to $t$.
For $n\geq N(e)$,
let $k_n$ be the maximal number such that $N(\varphi(k_n)+e)\leq n< N(\varphi(k_n+1)+e)$.
Since
$\tau^1\subsetneq\tau$, the train route of $c_n$ never contains the entire $(b(i))_{i\in{\mathbb N}}$,
which lies in $\tau^1$.
This implies that $F(n,t)=0$ for sufficiently large $t$; hence we
have $k_n<\infty$. Let $\psi:{\mathbb N}\rightarrow{\mathbb N}$ be the map defined by $\psi(n)=k_n$. Since
$\varphi$ is strictly increasing and $N$ is increasing and non-constant, $\psi(n)$ tends
to $\infty$ as $n$ tends to $\infty$. To get a better idea on this map $\psi$, the reader
should notice that if we forget the condition (3), $\psi(n)$ denotes the number of times
that
$s(n,j)$ comes back to the oriented train route $(bo(k))_{\varphi(0)\leq k \leq e+\varphi(0)}$ at the same times as
$s'(n,j)$ comes back to the oriented train route $(bo'(k))_{\varphi(0)\leq k \leq \frac{R'}{R}e +1+\varphi(0)}$.

Let us denote by $b'$ the branch $b'(V(\varphi(0)))$. For each $n\geq N(e)$, cut
the geodesic segment $h_n(b')$ into $\psi(n)$ isometric segments $\varsigma(j,n)$. The
length $l_{\rho_n}(\varsigma(j,n))$ of such a segment is
$\frac{l_{\rho_n}(h_n(b'))}{\psi(n)}\leq
R'\frac{\epsilon_n^{-1}}{\psi(n)}=o(\epsilon_n^{-1})$.

Let $s(n,\varphi(\psi(n))+e)\subset c_n^1$ be a segment which we obtained in Lemma
\ref{band}.
For simplicity, we shall denote it by $s_n$ and the
corresponding segment $s'(n,\varphi(\psi(n))+e)$ by $s'_n$.
Note that $s_n$ goes through
the train
route $(bo(k))_{0\leq k\leq \varphi(\psi(n))+e}$, and hence it travels through
$(bo(k))_{\varphi(0)\leq
k \leq e+\varphi(0)}$ at least $\psi(n)+1$ times. The same thing holds for $s_n'$ by the property (2)
of (*). By Lemma \ref{band}, we see that for any $0\leq i\leq \psi(n)$,\linebreak $g_n\circ
h_n \circ
v(\varphi(i))$ lies in $h_n(b'(V(\varphi(i))))=h_n(b')$ (where $g_n$ stands for
$g_{n,\varphi(\psi(n))+e}$).
Recall that we have $\psi(n)$ segments $\varsigma(j,n)$. Among the
$\psi(n)+1$ points\linebreak $ g_n\circ h_n\circ v(\varphi(i))$, $0\leq i\leq \psi(n)$, we can find two
points $g_n\circ h_n \circ v(\varphi(i_n))$ and $g_n\circ h_n \circ v(\varphi(j_n))$ which
lie in the
same segment among the $\varsigma(j,n)$, which we denote by $\varsigma_n$. Let
$I_n\subset s_n$ be the  segment between $v(\varphi(i_n))$ and $v(\varphi(j_n))$. Assume that $j_n>i_n$
and let $J_n$ be the sub-segment of $s_n$ containing the $e$ vertices following
$v(\varphi(j_n))$.

Let $\tilde s_n\subset{\mathbb H}^3$ be a lift of $h_n(s_n)$, and let $\tilde v(\varphi(i_n))$,
$\tilde v(\varphi(j_n))$, $\tilde I_n$ and $\tilde J_n$ be lifts of $h_n \circ v(\varphi(i_n))$,
$h_n \circ v(\varphi(j_n))$, $h_n(I_n)$ and $h_n(J_n)$ respectively, lying in $\tilde s_n$.
We lift the
map $g_n$ to a map $\tilde g_n$ from $\tilde s_n$ to a lift $\tilde s'_n$ of $s'_n$. Let
$\rho_n(a_n)\in\rho_n(\pi_1(M))$ be the covering translation that takes $\tilde
v(\varphi(i_n))$
to $\tilde v(\varphi(j_n))$. Since all the branches $bo(\varphi(i))$ have the same
orientation, $\rho_n(a_n)$ acts on $\tilde I_n$ as a translation. Since\linebreak
$bo(\varphi(i_n)+j)=bo(\varphi(j_n)+j)$ for all $j\leq e$, the isometry $\rho_n(a_n)$ also
acts
as a translation on $\tilde I_n\cup\tilde J_n$. Let $\tilde \varsigma_n$ be the lift of
$\varsigma_n$ that contains $\tilde g_n\circ\tilde v(\varphi(i_n))$, let $\tilde
\varsigma'_n$ be the lift that contains $\tilde g_n\circ\tilde v(\varphi(j_n))$, and let
$\rho_n(a'_n)$ be the isometry that maps $\tilde \varsigma_n$ to $\tilde \varsigma'_n$.
Let $\tilde I'_n\subset\tilde s'_n$ be the piecewise geodesic segment between $\tilde
g_n\circ\tilde v(\varphi(i_n))$ and $\rho_n(a'_n)\circ\tilde g_n\circ\tilde
v(\varphi(i_n))$, and let $\tilde J'_n\subset\tilde s'_n$ be the segment between
$\rho_n(a'_n)\circ\tilde g_n\circ\tilde v(\varphi(i_n))$ and $\tilde g_n\circ\tilde
v(\varphi(j_n)+e)$. From the assumption that
$bo'(V(\varphi(i_n))+j)=bo'(V(\varphi(j_n))+j)$ for any $0 \leq j \leq \frac{R'}{R}e
+1$, it follows
that $\rho_n(a'_n)$ acts as a translation on $\tilde I'_n\cup\tilde J'_n$.

As we saw before, the length of $\tilde \varsigma_n$ is
$o(\epsilon_n^{-1})$. It follows that the distance\linebreak $d(\tilde g_n\circ\tilde
v(\varphi(j_n)),\rho_n(a'_n)\circ\tilde g_n\circ\tilde v(\varphi(i_n))$ is
$o(\epsilon_n^{-1})$. From this and the facts that $\rho_n(a_n)$ acts as a translation on
$\tilde I_n\cup \tilde J_n$ and that $\rho_n(a'_n)$ acts as a translation on $\tilde
I'_n\cup\tilde J'_n$, we shall deduce the following claim :

\begin{cl}   \label{piti} For $R>0$, let $\cal{V}_R(\tilde J_n)$ be the
$R$-neighbourhood of $\tilde J_n$, then for any sequence of points $\tilde z_n\in
\cal{V}_R(\tilde
J_n)$, we have $d(\tilde z_n,\rho_n(a^{-1}_n a'_n)(\tilde z_n))$ is $o(\epsilon_n^{-1})$.
\end{cl}
\begin{proof} It is sufficient to prove this claim for any sequence $\tilde z_n\in \tilde
J_n$.
Since $\rho_n(a_n^{-1})$ acts as a translation
on $\tilde I_n\cup\tilde J_n$, the point $\rho_n(a_n^{-1})(\tilde z_n)$ is the point
$\tilde z'_n$ on $\tilde I_n\cup\tilde J_n$ that is at a distance (measured on $\tilde
I_n\cup\tilde J_n$) equal to $d(\tilde z_n,\tilde v(\varphi(j_n))$ from $\tilde v(\varphi(i_n))$.
Hence we have $d(\tilde z_n,\tilde v(\varphi(j_n))=d(\tilde z'_n,\tilde
v(\varphi(i_n)))$.

The point $\tilde z''_n=\rho_n(a'_n)\circ\tilde g_n(\tilde z'_n)$ is the point of
$\tilde J'_n$ that is at a distance (measured on $\tilde I'_n\cup\tilde J'_n$) of $d(\tilde
g_n(z'_n),\tilde g_n\circ\tilde v(\varphi(i_n))$ from $\rho_n(a'_n)\circ\tilde
g_n\circ\tilde v(\varphi(i_n))$. The points $g_n\circ h_n \circ v(\varphi(i_n))$ and $g_n\circ h_n\circ
v(\varphi(j_n))$ both lie in $\varsigma_n$ whose length is $o(\epsilon_n^{-1})$. It follows
that the distance $d(\tilde g_n\circ\tilde v(\varphi(j_n)),\rho_n(a'_n)\circ\tilde
g_n\circ\tilde v(\varphi(i_n)))$ is $o(\epsilon_n^{-1})$. Therefore we have $d(\tilde
z''_n, \tilde g_n\circ\tilde v(\varphi(j_n)))= d(\tilde g_n(\tilde z'_n),\tilde
g_n\circ\tilde v(\varphi(i_n)))+o(\epsilon_n^{-1})$. From Lemma \ref{band}, we get then
the equality $d(\tilde z''_n, \tilde g_n\circ\tilde v(\varphi(j_n)))=d(\tilde z'_n,\tilde
v(\varphi(i_n)))+o(\epsilon_n^{-1})$. Using the equality of the paragraph above, we get
$d(\tilde z''_n, \tilde g_n\circ\tilde v(\varphi(j_n)))=d(\tilde z_n,\tilde
v(\varphi(j_n))+o(\epsilon_n^{-1})$. Since almost geodesics $\tilde J_n,\tilde J_n'$ lie
$6\epsilon$-neighborhood each other, it follows that we have $d(\tilde z''_n,\tilde
z_n)=o(\epsilon_n^{-1})$.

By definition, we have $\tilde z''_n=\rho_n(a'_n)\circ\tilde
g_n\circ\rho_n(a_n^{-1})(\tilde z_n)$.
Therefore we have
$$d(\tilde z_n, \tilde z''_n)-d(\tilde z''_n,\rho_n( a'_n a^{-1}_n)(\tilde z_n))\leq d(\rho_n( a'_n a^{-1}_n)(\tilde z_n),\tilde z_n)\mbox{   and}$$
$$d(\rho_n( a'_n a^{-1}_n)(\tilde z_n),\tilde z_n)\leq d(\tilde z_n, \tilde z''_n)+ d(\tilde z''_n,\rho_n( a'_n a^{-1}_n)(\tilde z_n)).$$
We have also
 $$d(\rho_n((a'_n)^{-1})\tilde z''_n, \rho_n(a_n^{-1})\tilde z_n)=d(\tilde g_n\circ\rho_n(a_n^{-1})(\tilde z_n),\rho_n(a_n^{-1})(\tilde z_n))=o(\epsilon_n^{-1}).$$
Thus we finally get the
equality $d(\rho_n(a'_n a^{-1}_n)(\tilde z_n),\tilde z_n)=o(\epsilon_n^{-1})$.

We should notice that the $o(\epsilon_n^{-1})$ is ``uniform'', namely there is a
sequence $\delta_n\rightarrow 0$ independent of $(\tilde z_n)$ such that $d(\rho_n(a'_n a^{-1}_n)(\tilde z_n),\tilde
z_n) \leq \delta_n \epsilon_n^{-1}$.
\end{proof}


We shall use this claim to prove the following lemma.

\begin{lemma}  \label{ano} There is $N$ such that for $n\geq N$, $\rho_n(a^{-1}_n
a'_n)=id$.
\end{lemma}

We will use the following lemma
\begin{lemma}  \label{pitito} Let $[A_n,B_n]\subset{\mathbb H}^3$ be a sequence of geodesic
segments between $A_n$ and $B_n$ such that $l([A_n,B_n])$ is $\Theta(\epsilon_n^{-1})$ and
let $\delta_n,\delta'_n\in\pi_1(M)$ be two sequences such that the distances
$d(A_n,\rho_n(\delta_n)(A_n))$, $d(A_n,\rho_n(\delta'_n)(A_n))$,
$d(B_n,\rho_n(\delta_n)(B_n))$ and $d(B_n,\rho_n(\delta'_n)(B_n))$ are
all $o(\epsilon_n^{-1})$. Then there is $N$ such that for $n\geq N$,
$[\rho_n(\delta_n),\rho_n(\delta'_n)]=Id$.
\end{lemma}
\begin{proof} This comes directly from the arguments that M. Kapovich used in
\cite[Theorem 10.24]{kapo} to prove that the action is small (cf. \cite[p. 239]{kapo}).
\end{proof}
\begin{proof}[Proof of Lemma \ref{ano}]
Set $e=2p+1$ where $p$ is the number of the branches of $\tau^1$. If we fix some
$n\in{\mathbb N}$, we can find two different integers $i_1$ and $i_2$ between $\psi(n)$ and $\psi(n)+e$ such
that $bo(i_1)=bo(i_2)$. Let $K_n\subset J_n$ be the segment of $s_n$ with train route
$(b(k))_{i_1\leq k\leq i_2}$, and let $\delta_n\in\pi_1(M)$ be the element such that
$\rho_n(\delta_n)$
takes
$\tilde v(i_1)$ to $\tilde v(i_2)$. The isometry $\rho_n(\delta_n)$ acts as a translation
on the lift $\tilde K_n$ that lies in $\tilde s_n$. Since $|i_2-i_1|\leq e$,we can extract a subsequence such that
 $\delta_n$ does not depend on $n$. Let us denote it by $g$.

Let $\tilde K_{n-}$ and $\tilde K_{n+}$ be the two extremal edges of $\tilde K_n$,
such that $\rho_n(g)$ maps $\tilde K_{n-}$ to $\tilde K_{n+}$. By Claim \ref{piti},
up to $o(\epsilon_n^{-1})$, $\rho_n(a^{-1}_n a'_n)$ is the identity on $\tilde I_n\cup\tilde
J_n$. By the same arguments, $\rho_n(a^{\prime-1}_n a_n)$ is the
identity on $\tilde I_n\cup\tilde J_n$ up to $o(\epsilon_n^{-1})$. It follows that we have :\\
\indent - for any sequence $\tilde z_n\in \cal{V}_R(\tilde K_{n+})$, $d(\tilde
z_n,[\rho_n(a^{-1}_n a'_n),\rho_n(g)](\tilde z_n))=o(\epsilon_n^{-1})$;\\
\indent - for any sequence $\tilde z_n\in \cal{V}_R(\tilde K_{n-})$, $d(\tilde
z_n,[\rho^{-1}_n(g),\rho_n(a^{-1}_n a'_n)](\tilde z_n))=o(\epsilon_n^{-1})$;\\
\indent - for any sequence $\tilde z_n\in \cal{V}_R(\tilde K_{n+})$, $d(\tilde
z_n,[\rho_n(g),\rho_n(a^{\prime-1}_n a_n)](\tilde z_n))=o(\epsilon_n^{-1})$.\\
\indent Since $\tilde K_{n+}$ and $\tilde K_{n-}$ are edges of $\tilde h_n(\hat c^1_n)$,
their lengths are $\Theta(\epsilon_n^{-1})$. Applying Lemma \ref{pitito} to the segments
$\tilde K_{n-}$ and $\tilde K_{n+}$, we see that for sufficiently large $n$, $[\rho_n(a^{-1}_n
a'_n),\rho_n(g)]$, $[\rho^{-1}_n(g),\rho_n(a^{-1}_n a'_n)]$ and
$[\rho_n(g),\rho_n(a^{\prime-1}_n a_n)]$ commute with $\rho_n(a^{-1}_n a'_n)$.
Therefore they belong to an elementary subgroups of $\rho_n(\pi_1(M))$. By \cite[p.
239]{kapo}, it follows that the group generated by $\rho_n(a^{-1}_n a'_n)$ and
$\rho_n(g)$ is elementary.

Since $i_1\neq i_2$, the distance of translation of $\rho_n(g)$ is $\Theta(\epsilon^{-1}_n)$.In particular $\rho_n(g)$ is not a parabolic isometry. Since the group generated by
$\rho_n(a^{-1}_n a'_n)$ and $\rho_n(g)$ is elementary, there are $f\in\pi_1(M)$, $t,t_n\in{\mathbb N}$
such that $g=f^t$ and $a^{-1}_n a'_n=f^{t_n}$. Since the distance of translation of $\rho_n(g)$ is $\Theta(\epsilon^{-1}_n)$,
the distance of translation of $\rho_n(f)$ is $\Theta(\epsilon^{-1}_n)$. But by Lemma \ref{piti}, the distance of translation of
$\rho_n(a^{-1}_n a'_n)$ is $o(\epsilon^{-1}_n)$. Therefore we have $\rho_n(a^{-1}_n
a'_n)=id$ for sufficiently large $n$.
\end{proof}

In the construction of $a_n$ and $a'_n$, we chose $i_n$ and $j_n$
so that $g_n\circ h_n(v(\varphi(i_n)))$ and $g_n\circ h_n(v(\varphi(j_n)))$ lie in the same
segment $\varsigma_n$. The consequence of this choice is that the distance
measured on $h_n(s_n)$ from $g_n\circ h_n(v(\varphi(i_n)))$ to $g_n\circ h_n(v(\varphi(j_n)))$ is
smaller than $R'\frac{\epsilon_n^{-1}}{\psi(n)}$. In fact we only used the fact that
the distance measured on $h_n(s_n)$ from $g_n\circ h_n(v(\varphi(i_n)))$ to
$g_n\circ h_n(v(\varphi(j_n)))$ is $o(\epsilon_n^{-1})$.
Therefore we could have imposed a weaker hypothesis on $i_n$ and $j_n$, namely
that $g_n\circ h_n(v(\varphi(i_n)))$
and $g_n\circ h_n(v(\varphi(j_n)))$ are separated by $o(\psi(n))$-many segments
$\varsigma(j,n)$. This allows us to assume that $i_n$
tends to $\infty$ while the result of Lemma \ref{ano} that
$\rho_n(a^{-1}_n a'_n)=id$ still holds for sufficiently large $n$.\\

\indent
The exhausted reader will probably be glad to know that we shall now construct the homoclinic leaf or the annular lamination of Lemma \ref{an}.\\
\indent
Let us fix a reference hyperbolic metric on $\partial M$. Recall that $l_+$ and $l'_+$ are two half-leaves $l_+$ and $l'_+$ of $L_{rec}$ whose train routes are $(bo(i))_{i\in{\mathbb N}}$ and $(bo'(i))_{i\in {\mathbb N}}$ respectively. Let $k_n\subset l_+$ and $k'_n\subset l'_+$ be the geodesic arcs with train routes $(bo(i))_{\varphi(i_n)\leq i<\varphi(j_n)}$ and $(bo'(i))_{\varphi(i_n)\leq i<\varphi(j_n)}$ respectively. The endpoints of $k_n$ (resp. $k'_n$) are connected by an arc $\kappa_n$ (resp. $\kappa'_n$) which lies in a switch of $\tau^1$. Let $e_n$ (resp. $e'_n$) be the closed geodesic in the free homotopy class of $k_n\cup\kappa_n$ (resp. $k'_n\cup\kappa'_n$). These curves may have some self-intersections. Note that $e_n$ (resp. $e_n'$) represents $a_n$ (resp. $a_n'$).\\
\indent
Since $i_n$ tends to $\infty$, it follows from the choice of
$\varphi$ (by property (3) of (*)) that the length of $\kappa_n$
tends to $0$ (with respect to our reference metric on
$\partial M$). Moreover, the orientation of $b(\varphi(i_n))$
coincides with the orientation of $b(\varphi(j_n))$, so the curve
$k_n\cup\kappa_n$ is getting closer to a geodesic. Let
$\cal{V}(L_{rec})$ be a neighbourhood of $L_{rec}$. Since the lengths
of $\kappa_n$ tend to $0$ and since $k_n\subset L_{rec}$, the curve
$k_n\cup\kappa_n$ lies in $\cal{V}(L_{rec})$ for $n$ large enough.
So we get that $e_n$ lies in $\cal{V}(L_{rec})$ for $n$ large
enough. It follows that the sequence $(e_n)$ converges to a
connected component of $L_{rec}$ in the Hausdorff topology. The same
arguments apply to the sequence $(e'_n)$ and shows that it also
converges to a connected component of $L_{rec}$ in the Hausdorff topology. Furthermore it follows also from the choice of $k_n$ and $\kappa_n$ that we have $i(\lambda,e_n)\longrightarrow 0$ and $i(\lambda,e'_n)\longrightarrow 0$.\\
\indent
By Lemma \ref{ano},there is an annulus $E_n$ (not necessarily embedded) connecting $e_n$ to $e'_n$. Notice that since $M$ is irreducible and atoroidal, such an annulus $E_n$ is unique up to homotopy. Since $h_n$ is homotopic to the inclusion, there is an annulus $F_n$ (not necessarily embedded) in $M_n={\mathbb H}^3/\rho_n(\pi_1(M))$ connecting $h_n(e_n)$ to $h_n(e'_n)$. Furthermore, we can change $A_n$ by a homotopy so that $F_n$ is a subset of $A_n$.\\

\indent
Assume first that $E_n$ can be homotoped in $\partial M$. Then $e_n$ is homotopic to $e'_n$ in $\partial M$. Since $e_n$ and $e'_n$ are closed geodesics with respect to the reference metric on $\partial M$, they are equal. So $E_n$ is homotopic to $e_n$. It follows that $F_n$ is homotopic to $h_n(e_n)$. Let $\zeta_n\subset F_n\subset A_n$ be an arc produced in Lemma \ref{band}. The homotopy between $F_n$ and $h_n(e_n)$ sends $\zeta_n$ to an arc $\beta_n\subset h_n(e_n)$. On the other hand $\zeta_n$ is an essential arc. Therefore $\zeta_n$ is homotopic in $A_n$ to an arc $\zeta'_n\subset\bar c_n$ such that $\zeta'_n\not\subset\bar c_n^1$. Let us now consider $\zeta'_n\subset\bar c_n$ as lying in $c_n$. By the definition of $\beta_n$, there is an arc $\beta'_n\subset h_n^{-1}(\beta_n)$ joining the endpoints of $\zeta'_n$ such that $\beta'_n\cup\zeta'_n$ bounds a disc $D'_n$. Since $\zeta'_n\not\subset\bar c_n^1$ and since $\beta_n\subset h_n(\tau_1)$, $D'_n$ can not be homotoped in $\partial M$. Since $\zeta'_n$ lies in $c_n=|\lambda_n|$, we have $i(\partial D'_n,\lambda_n)\leq \int_{\beta'_n} d\lambda_n$. By the Loop Theorem, there is an essential disc $D_n$ such that $i(\partial D_n,\lambda_n)\leq \int_{\beta'_n} d\lambda_n$. In order to get a homoclinic leaf which does not intersect $|\lambda|$ transversely, we are going to show that $i(\lambda_n,\partial D_n)\longrightarrow 0$.\\
\indent
As we have seen above, $\kappa_n\cup k_n$ and $\kappa'_n\cup k'_n$ are close to $e_n=e'_n$. Especially, there is a small arc $\beta''_n$ joining $\kappa_n$ to $\kappa'_n$ with $\int_{\beta''_n} d\lambda_n\longrightarrow 0$. By construction, $\beta'_n$ and $\beta''_n$ are homotopic relative to $\kappa_n\cup k_n\cup\kappa'_n\cup k'_n$. Therefore we have $\int_{\beta'_n}d\lambda_n \leq \int_{\beta''_n}d\lambda_n +q_n i(e_n,\lambda_n)$, where $q_n$ is an integer which depends on the number of times that $\beta'_n$ spirals toward $e_n$.  Let us show that $\beta_n\subsetneq h_n(e_n)$, it will follow that $q_n\leq 1$. By Lemma \ref{band}, the length of $\zeta_n$ in $M_n={\mathbb H}^3/\rho_n(\pi_1(M))$ is less than $6\epsilon$. Since $\beta_n\subset h_n(\tau_1)$, it follows from properties $b)$ and $c)$ of Lemma \ref{rtt} that $\beta_n$ is a quasi-geodesic segment. Therefore, for sufficiently large $n$, the length of $\beta_n$ in $M_n$ is less than $7\epsilon$. By property $b)$ of $h_n(\tau_n)$, the length of $h_n(e_n)$ goes to $\infty$. Thus we get $\beta_n\subsetneq h_n(e_n)$ and it follows that $q_n\leq 1$. Since we have $\int_{\beta''_n} d\lambda_n\longrightarrow 0$ and $i(\lambda_n,e_n)\longrightarrow 0$, we can conclude that $\int_{\beta'_n} d\lambda_n\longrightarrow 0$. It follows then from the inequality $i(\partial D_n,\lambda_n)\leq \int_{\beta'_n} d\lambda_n$ that $i(\lambda_n,\partial D_n)\longrightarrow 0$. Extract a subsequence such that $(\partial D_n)$ converges in the Hausdorff topology to some geodesic lamination $H$. By Casson's criterion (Lemma \ref{casson}), $H$ contains a homoclinic leaf. Since $i(\lambda_n,\partial D_n)\longrightarrow 0$, $H$ does not intersect $|\lambda|$ transversely.\\

\indent
Assume now that $E_n$ can not be homotoped in $\partial M$. As we have seen above, $\partial E_n=e_n\cup e'_n$ converges in the Hausdorff topology to a sublamination $E$ of $L_{rec}$. Let $\mu$ be a measured lamination with $|\mu|=E$. By construction $e_n$ is homotopic to the union of a segment lying in $L_{rec}$ and of a small segment $\kappa_n$ (whose length with respect to the reference metric tends to $0$). It follows that $i(\mu,e_n)\longrightarrow 0$. The same is true for $e'_n$. By Lemma \ref{annulus} either the lamination $E$ is annular or $S(E)$ contains a homoclinic leaf which does not intersect $E$ transversely. On the other hand, $E$ is a sublamination of $L_{rec}$. So we have constructed a homoclinic leaf or an annular lamination which does not intersect $L_{rec}$ transversely. Since $|\lambda|$ is a sublamination of $L_{rec}$, this concludes the proof of Lemma \ref{an}.
\end{proof}













Now, we can complete the proof of Theorem \ref{th&}.
Let $\rho_n$, $\lambda_n$
and $\lambda$ be as in Theorem \ref{th&}.
If no subsequence of $\rho_n$
converges algebraically, it follows from Lemma \ref{rtt} that $\rho_n$ and $\lambda_n$ satisfy the assumptions of
\S\ref{asum}.
By Lemma \ref{an}, there is a homoclinic leaf or an annular lamination which does not intersect $|\lambda|$ transversely.
By Lemma \ref{anho}, this contradicts the assumption that $\lambda\in{\cal D}(M)$. This
completes the proof of Theorem \ref{th&}.
\end{proof}


\section{Conclusion}

We shall now deduce Theorem \ref{main} from Theorem \ref{th&}.
\setcounter{thm}{0}

\begin{thm} Let $M$ be a compact irreducible atoroidal $3$-manifold with
boundary.
Let $(m_n)$ be a sequence in the Teichm{\"u}ller space $\mathcal{T}(
\partial M)$ which converges in the Thurston compactification to a projective
lamination $[\lambda]$ contained in $\cal{PD}(M)$. Let $q : \mathcal{T}(
\partial M)\rightarrow CC_0(M)$ be the Ahlfors-Bers map, and suppose that
$(\rho_n:\pi_1(M){\rightarrow} G_n \subset PSL(2,{\mathbb C}))$ is a sequence of discrete
faithful representations corresponding to $(q(m_n))$. Then passing to a
subsequence, $(\rho_n)$ converges in $AH(N)$.
\end{thm}

\begin{proof} For a simple closed curve $c\subset
\partial M$, we denote by $l_{m_n}(c)$ the length of $c$ with respect to the
metric $m_n$ and by $l_{\rho_n}(c)$ the length of the closed geodesic of
${\mathbb H}^3/\rho_n(\pi_1(M))$ in the free homotopy class of $c$. By \cite{thuii} (see
also \cite{flp}), there is a sequence of simple closed curves
$c_n\subset\partial M$ whose projective classes converge to $[\lambda]$ in
$\cal{PML}(\partial
M)$ such that $\frac{l_{m_n}(c_n)}{l_{m_0}(c_n)}$ tends to $0$ as $n$ goes to
infinity.

Using the following result of \cite{BC}, we shall get an upper bound for
the sequence $(l_{\rho_n}(\lambda_n))$.
\begin{thm}[Bridgeman-Canary]
\label{B-C} For any $Q >0$, there is a constant $K>0$ depending only on $Q$ with
the following conditions. Let $\Gamma$ be a finitely generated Kleinian group
without torsion such that the shortest meridian length is greater than $Q$. Let
$C(\Gamma)$ be the convex core of ${\mathbb H}^3/\Gamma$, and consider the nearest point
retraction $r: \Omega_\Gamma/\Gamma \rightarrow
\partial C(\Gamma)$. Then $r$ is $K$-Lipschitz and has a homotopically inverse
$K$-Lipschitz map.
\end{thm}


Let us first verify that we are considering a situation where the
hypothesis of this theorem is fulfilled.
\begin{lemma}
\label{length grows}
There is a positive number $Q$ such that $l_{m_n}(d) \geq Q$ for any meridian $d$ of  $\partial M$.
\end{lemma}

\begin{proof} Assuming the contrary, we have a sequence of meridians $(d_n)$ such
that $(l_{m_n}(d_n))$ tends to $0$. Let us extract a subsequence so that
$(d_n)$ converges with respect to the Hausdorff topology to a geodesic lamination
$D\subset M$. By Casson's criterion, $D$ contains a homoclinic leaf. Since
$[\lambda]\in\cal{PD}(M)$, the lamination $D$ intersects the support of
$[\lambda]$ transversely (\cite[Lemma 3.5]{masurext}, see also \cite[Lemme
3.3]{espoir}). It follows that the sequence $i(\lambda,d_n)$ is bounded away
from $0$. This implies that the sequence $l_{m_n}(d_n)$ tends to $\infty$. Thus
we get a contradiction.
\end{proof}


\indent The length $l_{\rho_n}(c_n)$ is clearly less than the length of any
curve in $
\partial C(\rho_n(\pi_1(M)))$ which is freely homotopic to $c_n$. Thus, applying
Theorem \ref{B-C}, we get that $\frac{l_{\rho_n}(c_n)}{l_{m_0}(c_n)}$ tends to
$0$.

Let us denote by $\lambda_n$ the measured geodesic lamination obtained by
endowing $c_n$ with a Dirac measure whose weight is equal to
$l_{m_0}(c_n)^{-1}$. The sequence $\lambda_n$ converges in $\cal{ML}(\partial
M)$ to a measured geodesic lamination $\lambda$ which lies in the projective
class $[\lambda]$. Since $\frac{l_{\rho_n}(c_n)}{l_{m_0}(c_n)}$ tends to
$0$, we have $l_{\rho_n}(\lambda_n)\longrightarrow
0$. Since $\lambda$ lies in the projective class $[\lambda]\in{\cal PD}(M)$, the
measured geodesic lamination $\lambda$ lies in ${\cal D}(M)$. Applying Theorem
\ref{th&}, we see that a subsequence of $(\rho_n)$ converges algebraically.
\end{proof}

\addcontentsline{toc}{chapter}{Bibliographie}
