1$. Suppose that $\rho(f)= \left\{ \sfrac{p}{q} \right\}$. Then there exists a family of pairwise disjoint simple curves $(\gamma_{i})_{i \in \mathbb{Z} / nq \mathbb{Z}}$ which join the two boundary components of the annulus such that, for any index $i$: \begin{enumerate} \item The curve $\gamma_{i}$ lies strictly between the curves $\gamma_{i-1}$ and $\gamma_{i+1}$. \item The curve $f(\gamma_{i})$ lies strictly between the curves $\gamma_{i+np-1}$ and $\gamma_{i+np+1}$. \end{enumerate} \end{proposition} \begin{rema} In the case of the rotation $R_{\sfrac{p}{q}}$, note that it suffices to take $\gamma_{i}(t)=(\sfrac{i}{nq},t)$. \end{rema} \begin{figure}[ht] \begin{center} \vspace*{-\baselineskip} \includegraphics[scale=0.45]{militon_figures/courbesgamma} \end{center} \caption{Illustration of Proposition \ref{pavageverticalanneau} in the case $p=0$, $q=1$ and $n=4$} \label{fig} \end{figure} For technical reasons, it is more convenient to prove the following stronger proposition. \begin{proposition} Let $f$ be a homeomorphism in $\Homeo_{0}(\mathbb{A})$ and $p$ and $q$ be integers such that either $q>0$, $0

0$ and $N>0$. Then there exists a family of pairwise disjoint simple curves $(\gamma'_{i})_{ i \in \mathbb{Z} / nq \mathbb{Z}}$ which join the two boundary components of the annulus such that, for any index $i$ and any integer $0\leq k \leq N$:
\begin{enumerate}
\item If $n \neq 1$ or $q \neq 1$, the curve $f^{k}(\gamma'_{i})$ lies strictly between the curves $\gamma'_{i+knp-1}$ and $\gamma'_{i+knp+1}$.
\item If $n\!=\!q\!=\!1$, any lift of the curve $f^{k}(\gamma'_{1})$ meets at most one lift of the curve~$\gamma'_{1}$.
\end{enumerate}
\end{proposition}
The case $N=1$ of this proposition yields directly Proposition \ref{pavageverticalanneau}.
\begin{proof}
We say that a finite sequence of curves $(\gamma'_{i})_{ i \in \mathbb{Z} / nq \mathbb{Z}}$ satisfies property $P(N,n)$ if it satisfies the conclusion of the proposition. We prove by induction on $n$ that, for any $N$, there exist curves $(\gamma'_{i})_{i \in \mathbb{Z}/qn \mathbb{Z}}$ which satisfy property $P(N,n)$ and such that, for any index $n \leq i 0$. Apply Proposition \ref{pavageverticalanneau} to the homeomorphism $f$ with $n=N$: this proposition provides curves $(\gamma_{i})_{ i \in \mathbb{Z}/Nq \mathbb{Z}}$. Consider a finite sequence $(\alpha_{j})_{j \in \llbracket 0,N' \rrbracket}$ of pairwise disjoint loops $\mathbb{S}^{1} \to \mathbb{A}$ such that:
\begin{enumerate}
\item For any $t \in \mathbb{S}^{1}$, $\alpha_{0}(t)=(t,0)$ and $\alpha_{N'}(t)=(t,1)$.
\item The loops $\alpha_{j}$ are homotopic to $\alpha_{0}$.
\item For any index $1 \leq j *0$ such that, for any $n$,
$$ \diam (\tilde{f}^{k_{n}}(D_{b})) \geq C \newell_{\mathcal{G}}(\gamma_{n})-C'.$$
Therefore
$$ \frac{\diam (\tilde{f}^{k_{n}}(D_{b}))}{k_{n}} \geq \frac{C \newell_{\mathcal{G}}(\gamma_{n})}{N (\newell_{\mathcal{G}}(\gamma_{n})+1)}- \frac{C'}{N(\newell_{\mathcal{G}}(\gamma_{n})+1)}.$$
the right hand side of this inequality has a positive limit as $n \to + \infty$. Moreover, the sequence $(|k_{n}|)_{n}$ has to tend to $+\infty$: otherwise, one of the sets of the form $\tilde{f}^{l}(D_{b})$, with $l$ in $\mathbb{Z}$, would have infinite diameter as it would cross infinitely many sets of the form $\gamma_{n}(D_{b})$. This contradicts the hypothesis $\lim_{n \to +\infty} \sfrac{\diam (\tilde{f}^{n}(D_{b}))}{n}=0$ (recall that this hypothesis is independent of the chosen fundamental domain).
\end{proof}
We now want to apply Lemma \ref{inf} to complete this first step. We have first to check that the family of essential arcs we will consider satisfies the hypothesis of this lemma. For any essential arc $c$ of $\tilde{S}$ we denote by $c^{0}$ the arc with the opposite orientation. We denote by $\mathcal{C}$ the set of connected components of $\partial \tilde{S}$ which lie on the left of $\tilde{\beta}_{0}$.
Recall first that the essential arcs among the lifts of the arc $\beta_{0}$ which lie strictly on the right of $\tilde{\beta}_{0}$ are the curves of the form $\gamma(\tilde{\beta}_{0})$, where $\gamma$ belongs to $a \pi_{1}(S)$. Therefore the arc $\tilde{\beta}_{0}$ lies strictly on the left of the curves of the form $\gamma(\tilde{\beta}_{0})$, with $\gamma \in a \pi_{1}(S)-a\pi_{1}(S)a^{-1}$, as the curve $\gamma^{-1}(\tilde{\beta}_{0})$ is strictly on the left of the curve~$\tilde{\beta}_{0}$. Moreover, the arc $\tilde{\beta}_{0}$ lies strictly on the right of the curves of the form $\gamma(\tilde{\beta}_{0})$, with $\gamma \in a\pi_{1}(S) a^{-1}$. By this discussion, the components in $\mathcal{C}$ lie strictly on the left of the curves of the form $\gamma(\tilde{\beta}_{0})$, hence also of the form $\gamma \tilde{f}^{k}(\tilde{\beta}_{0})$, with $k \in \mathbb{Z}$ and $\gamma \in\nobreak a \pi_{1}(S)-a\pi_{1}(S)a^{-1}$. They also lie strictly on the left of the curves of the form $\gamma \tilde{f}^{k}(\tilde{\beta}_{0}^{0})$, with $k \in \mathbb{Z}$ and $\gamma \in a \pi_{1}(S)a^{-1}$.
For any $1 \leq i \leq b$, denote by $A_{i}$ the subset of $a\pi_{1}(S)$ consisting of deck transformations $\gamma$ such that $D_{b}$ lies strictly on the left of the arc $\gamma(\tilde{\alpha}_{i})$. Denote by $A_{i}^{c}$ the complement of this set in $a \pi_{1}(S)$.
Consider the family $\mathcal{F}$ of essential arcs consisting of the arcs which meet the arc $\tilde{\beta}_{0}$ of one of the following forms.
\begin{enumerate}
\item $\gamma \tilde{f}^{k}(\tilde{\beta}_{0}), \gamma \in a \pi_{1}(S)-a\pi_{1}(S)a^{-1}, \left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$.
\item $\gamma \tilde{f}^{k}(\tilde{\beta}_{0}^{0}), \gamma \in a\pi_{1}(S)a^{-1}, \left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$.
\item $\gamma\tilde{f}^{k}(\tilde{\alpha_{i}}), 1 \leq i \leq b, \gamma \in A_{i}, \left| k \right| \leq N(\newell_{\mathcal{G}}(\gamma)+1)$
\item $\gamma \tilde{f}^{k}(\tilde{\alpha}_{i}^{0}), 1 \leq i \leq b, \gamma \in A_{i}^{c}, \left| k \right| \leq N(\newell_{\mathcal{G}}(\gamma)+1)$
\end{enumerate}
By Lemma \ref{finitenumber}, the family $\mathcal{F}$ is finite. Take $\tilde{\beta}_{1}=\inf_{\tilde{\beta}_{0}}( \mathcal{F}).$ Let us check that this curve satisfies the wanted properties.
By construction of the curve $\tilde{\beta}_{1}$, for any deck transformation $\gamma$ in $a \pi_{1}(S)$ and any integer $\left| k \right| \leq N (\newell_{\mathcal{G}}(\gamma)+1)$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\alpha}_{i})$ lie on the right of the curve~$\tilde{\beta}_{1}$.
Now, take a deck transformation $\gamma \in a \pi_{1}(S)- a \pi_{1}(S)a^{-1}$ and $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$. Observe that, by the uniqueness part of Lemma \ref{inf}, $\gamma\tilde{f}^{k}(\tilde{\beta}_{1})=\inf_{\gamma \tilde{f}^{k}(\tilde{\beta}_{0})}(\gamma\tilde{f}^{k}(\mathcal{F}))$, where $\gamma\tilde{f}^{k}(\mathcal{F})= \bigl\{ \gamma \tilde{f}^{k}(\tilde{\beta})\sep \tilde{\beta} \in \mathcal{F} \bigr\}$. It is easy to check that any curve in $\gamma\tilde{f}^{k}(\mathcal{F})$ is either equal to one of the curves in $\mathcal{F}$ or lies strictly on the right of $\tilde{\beta}_{0}$. Hence any of these curves lie on the right of $\tilde{\beta}_{1}$. As the curve $\tilde{f}(\tilde{\beta}_{0})$ also lies on the right of the curve $\tilde{\beta}_{1}$ by construction, we deduce that the curve $\gamma\tilde{f}^{k}(\tilde{\beta}_{1})$ lies on the right of the curve $\tilde{\beta}_{1}$, by Lemma \ref{inf}.
Finally, take a deck transformation $\gamma \in a \pi_{1}(S)a^{-1}$ and $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$. As the arc~$\tilde{\beta}_{1}$ lies on the left of the arc $\tilde{\beta}_{0}$, the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{1})$ lies on the left of the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{0})$. As the arc $\tilde{\beta}_{1}$ lies on the right of the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{0})$ by construction and as the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{0})$ is on the right of the arc $\tilde{\beta}_{1}$ and of the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{1})$, the arc $\gamma \tilde{f}^{k}(\tilde{\beta}_{1})$ lies on the right of the arc $\tilde{\beta}_{1}$.
\end{proof}
\begin{proof}[Proof of Lemma \ref{goodcurve}: second step]
\emph{We prove that we can perturb the arc $\tilde{\beta}_{1}$ to obtain an arc $\tilde{\beta}'_{1}$ such that
\begin{enumerate}
\item For any deck transformation $\gamma$ in $a \pi_{1}(S)$ and any integer $\left| k \right| \leq N (\newell_{\mathcal{G}}(\gamma)+1)$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\alpha}_{i})$ lie \emph{strictly} on the right of the curve $\tilde{\beta}'_{1}$.
\item For any deck transformation $\gamma$ in $a \pi_{1}(S)$ and any integer $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\beta}'_{1})$ lie \emph{strictly} on the right of the curve $\tilde{\beta}'_{1}$.
\end{enumerate}}
Let $M$ be the maximal length with respect to $\mathcal{G}$ of an element $\gamma$ in $a \pi_{1}(S)-a\pi_{1}(S)a^{-1}$ such that there exists $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$ with $\gamma \tilde{f}^{k}(\tilde{\beta}_{1}) \cap \tilde{\beta}_{1} \neq \emptyset$. As in the proof of Lemma \ref{finitenumber}, one can prove that $M$ is well-defined. Denote by $K$ the compact set consisting of points on $\tilde{\beta}_{1}$ which belong to an arc of the form $\gamma \tilde{f}^{k}(\tilde{\beta}_{1})$, where $\newell_{\mathcal{G}}(\gamma)=M$ and $\left| k \right| \leq N M$. By maximality of $M$, for any deck transformation $\gamma \in a \pi_{1}(S)-a\pi_{1}(S)a^{-1}$ and any $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$, the image under $\gamma \tilde{f}^{k}$ of $K$ is sent strictly on the right of $\tilde{\beta}_{1}$. Take a disjoint union $(U_{i})_{i}$ of open disks which cover $K$, such that, for any $i$, $\tilde{\beta}_{1} \cap U_{i}$ is connected and $U_{i}$ is sent strictly to the right of $\tilde{\beta}_{1}$ under a homeomorphism of the form $\gamma \tilde{f}^{k}$, with $\gamma \in a \pi_{1}(S)-a\pi_{1}(S)a^{-1}$ and $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$. Fix a parameterization of $\tilde{\beta}_{1}$. For any $i$, choose parameters $t_{1,i} j$. In both cases, the claim is true as the arc $\gamma^{-1}\gamma_{0} \tilde{f}^{k_{0}-k}(\tilde{\alpha}_{i_{0}})$ is disjoint from $\tilde{\delta}_{j+1}$ by induction hypothesis.
Let us see why, for any element $\gamma$ in $a \pi_{1}(S)$ and any integer $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\delta}'_{j})$ lie strictly on the right of the curve $\tilde{\delta}'_{j}$. Fix such an element $\gamma_{0}$ in $a \pi_{1}(S)$ and such an integer $k_{0}$. Here we distinguish the cases $\gamma_{0} \in a \pi_{1}(S)-a \pi_{1}(S)a^{-1}$ and $\gamma_{0} \in a \pi_{1}(S)a^{-1}$. In the first case, notice that the curve $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}'_{j})$ is on the right of the curve $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j+1})$ by definition of $\tilde{\delta}'_{j}$. Hence it suffices to prove that this last arc is strictly on the right of any arc in $\mathcal{F}_{j}\cup\{ \tilde{\delta}_{j+1}\}$. To do this, it suffices to prove that any arc in $\mathcal{F}_{j} \cup\{ \tilde{\delta}_{j+1}\}$ is on the left of the arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j+1})$, which is easily done by using the induction hypothesis. Now suppose that $\gamma_{0} \in a \pi_{1}(S)a^{-1}$. As usual, as one of the endpoints of the arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j})$ is strictly on the right of the curve $\tilde{\delta}_{j}$, it suffices to prove that $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j}) \cap \tilde{\delta}_{j}= \emptyset$. To do this, it suffices to check that the image under $\gamma_{0}\tilde{f}^{k_{0}}$ of any essential arc in $\mathcal{F}_{j}\cup \{ \tilde{\delta}_{j+1}\}$ is disjoint from any arc in $\mathcal{F}_{j} \cup\{ \tilde{\delta}_{j+1}\}$ which can be done without serious difficulty by using the induction hypothesis and the properties of the curves $\tilde{\alpha}_{i}$.
By construction, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\alpha}_{i})$ with $1 \leq i \leq b$, $\gamma \in \Gamma \cup \pi_{1}(S)a^{-1}- a \pi_{1}(S)a^{-1}$, $\newell_{\mathcal{G}}(\gamma)=j$, lie on the left of the arc $\tilde{\delta}'_{j}$. By induction hypothesis, as the arc $\tilde{\delta}_{j+1}$ lies on the left of the arc $\tilde{\delta}'_{j}$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\alpha}_{i})$ with $1 \leq i \leq b$, $\gamma \in \Gamma \cup \pi_{1}(S)a^{-1}- a \pi_{1}(S)a^{-1}$, $\newell_{\mathcal{G}}(\gamma)>j$ lie strictly on the left of the arc $\delta'_{j}$.
Finally, let us check that, for any non-trivial element $\gamma$ in $\Gamma$ with $\newell_{\mathcal{G}}(\gamma) \geq j$ and any integer $\left| k \right| \leq N \newell_{\mathcal{G}}(\gamma)$, the curves of the form $\gamma \tilde{f}^{k}(\tilde{\delta}'_{j})$ lie on the left of the curve~$\tilde{\delta}'_{j}$. Fix such an element $\gamma_{0}\in \Gamma$ and such an integer $k_{0}$. This results from the following facts.
\begin{enumerate}
\item The arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}'_{j})$ lies on the right of the arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j+1})$.
\item The arc $\tilde{\delta}'_{j}$ lies on the left of the arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j+1})$.
\item The arc $\gamma_{0} \tilde{f}^{k_{0}}(\tilde{\delta}_{j+1})$ is on the left of the arc $\tilde{\delta}'_{j}$.
\end{enumerate}
With arguments similar to those used during Step 2, we then perturb the arc $\tilde{\delta}'_{j}$ to obtain an arc $\tilde{\delta}_{j}$ which satisfies the required properties.
\end{proof}
\backmatter
\bibliographystyle{jepplain}
\bibliography{militon}
\egroup
\end{document}
*