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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Properties of Abelian Return Words
}
\vskip 1cm
\large
M. Rigo, 
P. Salimov,\footnote{The second author is supported by the Russian President's grant no. MK-4075.2012.1, the Russian Foundation for Basic Research grant no. 11-01-00997 and by a University of Li{\`e}ge post-doctoral grant.}
and E. Vandomme\\
Department of Mathematics \\
University of Li{\`e}ge \\
Grande traverse 12 (B37)\\
B-4000 Li{\`ege} \\
Belgium \\
\href{mailto:M.Rigo@ulg.ac.be}{\tt M.Rigo@ulg.ac.be}
\href{mailto:E.Vandomme@ulg.ac.be}{\tt E.Vandomme@ulg.ac.be} \\
\end{center}

\vskip .2 in



\begin{abstract}
    We investigate some properties of abelian return words as recently
    introduced by Puzynina and Zamboni. In
    particular, we obtain a characterization of Sturmian words with
    nonzero intercept in terms of the finiteness of the set of
    abelian return words to all prefixes. We describe this set of
    abelian returns for the Fibonacci word but also for the
    $2$-automatic Thue--Morse word. We also investigate the
    relationship existing between abelian complexity and finiteness of
    the set of abelian returns to all prefixes. 
\end{abstract}

\section{Introduction}
Many notions occurring in combinatorics on words have been recently
and fruitfully extended to an abelian context. Two words $u$ and $v$
are said to be {\em abelian equivalent} if $u$ is a permutation of the letters in $v$
and usually, the corresponding concepts are defined up to such an
equivalence. This framework gives rise to many challenging questions
in combinatorics on words: what kind of information is preserved in the abelian context? 
To what extent can the
classical results be applied? What kind of characterization can we obtain? For instance, consider the classical notion
of {\em factor complexity} $p_{\mathbf{x}}$ which maps an integer $n\ge 0$ to the number of distinct
factors of length $n$ occurring in an infinite word $\mathbf{x}$. 
The well-known theorem of Morse--Hedlund  gives a characterization of the ultimately periodic
words. See for instance
\cite{Lot1}. Sturmian words are defined by the property $p_{\mathbf{x}}(n)=n+1$ for all $n\ge 0$. The analogue to factor complexity is the {\em abelian complexity} of $\mathbf{x}$
which maps $n\ge 0$ to the number of distinct abelian classes partitioning the set
of factors of length $n$ occurring in $\mathbf{x}$. This latter notion was already introduced in the 1970's \cite{Cov}. Some other important questions in combinatorics on words such as avoiding abelian repetitions, were initiated at the same period. See for instance \cite{Dek}. See also the reference \cite{Rich} on abelian complexity which contains many relevant bibliographic pointers.

The return word is a classical notion in combinatorics on words and 
symbolic dynamical systems \cite{r1,r2,r3}. For instance, Durand obtained a
characterization of primitive substitutive sequences in terms of
return words and derived sequences \cite{Dur}. Let $u$ be a recurrent
factor of $\mathbf{x}$, i.e., a factor occurring infinitely
many times in $\mathbf{x}$. A {\em return word} to $u$ is a factor
separating two consecutive occurrences of $u$.  In this paper, we
consider the abelian analogue of this notion of return word. Such a
study has been recently presented by Puzynina and Zamboni during
the WORDS 2011 conference. Here we focus on different aspects of
abelian returns and we hope that our results can be seen as
complementary to those found by Puzynina and Zamboni \cite{PZ}. 

The main difference is that we usually consider the set of abelian
returns with respect to all the factors of an infinite word
$\mathbf{x}$, while Puzynina and Zamboni \cite{PZ} study the set of abelian
returns with respect to each factor taken separately. In particular,
their main contribution is a characterization of
Sturmian words: {\em a recurrent infinite word is Sturmian if and only
  if each of its factors has two or three abelian returns}. Sturmian
words have been extensively studied and, in particular, every Sturmian
word can be obtained by coding the orbit $(R_\alpha^n(\rho))_{n\ge 0}$
of a point $\rho$ on a circle under a rotation $R_\alpha$ by an
irrational angle $\alpha$ when the circle is partitioned in a suitable
way into two complementary intervals. The irrational $\alpha$ is
called the {\em slope} of the Sturmian word and the initial point
$\rho$ is its {\em intercept}. See for instance \cite{Lot,MH}. Many of
our results on Sturmian words rely on this definition of Sturmian
words.

This paper is organized as follows.  In Section~\ref{sec:1}, we
present the main definitions and notation used in this paper. In
Section~\ref{sec:2}, we discuss the relationship with periodicity and
we prove that a recurrent word is periodic if and only if its set of
abelian returns is finite. We also construct an abelian uniformly
recurrent word which is not eventually recurrent. In
Section~\ref{sec:3}, we restrict ourselves to the set
$\mathcal{APR}_\mathbf{x}$ of abelian returns to all prefixes. In
particular, this set is finite for any uniformly recurrent and abelian
periodic word. We study the special case of the Thue--Morse word
$\mathbf{t}$ \cite{astm} and show that the set of abelian returns to
all prefixes of $\mathbf{t}$ contains 16 elements. Next, we obtain a
characterization of Sturmian words with (non)zero intercept as
follows. Let $\mathbf{x}$ be a Sturmian word coding an orbit
$(R_\alpha^n(\rho))_{n\ge 0}$. The set $\mathcal{APR}_\mathbf{x}$ of
abelian returns to the prefixes of $\mathbf{x}$ is finite if and only
if $\mathbf{x}$ does not have a null intercept (see
Theorem~\ref{thmSturmianImplF}). The celebrated Fibonacci word
$\mathbf{f}$ can be defined with a slope and an intercept both equal
to $1/\tau^2$ where $\tau$ is the Golden mean.  Therefore our result
implies that $\mathcal{APR}_\mathbf{f}$ is finite. We show that this
set contains exactly $5$ elements. Interestingly, our developments can
be related to the lengths of the palindromic prefixes of $\mathbf{f}$. See for instance  
\cite{luca,Fis}. By contrast the set of abelian returns to all
prefixes for the word ${\tt 0}\mathbf{f}$ is infinite.  Then we show
that if $\mathbf{x}$ is an abelian recurrent word such that
$\mathcal{APR}_{\mathbf{x}}$ is finite, then $\mathbf{x}$ has bounded
abelian complexity.  In the last section of this paper, we introduce
the notion of abelian derived sequence.  If a word $\mathbf{x}$ is uniformly recurrent, then
$\mathbf{x}$ can be factored in terms of abelian returns to a given
prefix of $\mathbf{x}$. This gives rise to a coding that allows one to
define a new sequence. Contrary to the non-abelian case and the
characterization obtained by Durand, the
Thue--Morse word is an example of word having infinitely many derived sequences.


\section{Preliminaries}\label{sec:1}
An infinite word $\mathbf{x}$ is said to be
{\em recurrent} if every factor $u$ of $\mathbf{x}$ appears
infinitely often in $\mathbf{x}$. An infinite word $\mathbf{x}$ is said to be
{\em uniformly recurrent}, if it is recurrent and the distance between any two consecutive
occurrences of any of its factors $u$ is bounded by a constant depending only on $u$. The language of all the finite factors (resp. prefixes) of an infinite word $\mathbf{x}$ is denoted by $\Fac(\mathbf{x})$ (resp. $\Pref(\mathbf{x})$). Let $i,j$ be such that $i\le j$. The factor $x_ix_{i+1}\cdots x_j$ of $\mathbf{x}=x_0x_1\cdots$ is denoted by $\mathbf{x}[i,j]$. The notation $\mathbf{x}[i,i]$ is shortened to $\mathbf{x}_i$.


\subsection{Return words}\label{sec:21}
 The classical notion of return word
has been used by Durand \cite{Dur} but was previously introduced by Boshernitzan
\cite{Bos} (see also \cite{Gri} for the notion of induced
transformation in a dynamical context). Let $u$ be a prefix of a
uniformly recurrent word $\mathbf{x}$. A nonempty factor $w$ of
$\mathbf{x}$ is a {\em return word} to $u$, if there exists some $i\ge
0$ such that
\begin{itemize}
  \item $\mathbf{x}[i,i+|w|-1]=w$,
  \item $\mathbf{x}[i,i+|u|-1]=u=\mathbf{x}[i+|w|,i+|w|+|u|-1]$,
  \item $\mathbf{x}[i+j,i+j+|u|-1]\neq u$ for all $j\in\{1,\ldots,|w|-1\}$.
\end{itemize}
We denote by
$\mathcal{R}_{\mathbf{x},u}$ the set of return words to $u$. Since $\mathbf{x}$ is uniformly recurrent, this set is finite because the length of the longest return word to $u$ is bounded by the maximal distance between two successive occurrences of $u$. If we order the return words to $u$ with respect to their first occurrence in $x$, then the corresponding map is $\Lambda_{\mathbf{x},u}:\mathcal{R}_{\mathbf{x},u}\to\{{\tt 1},\ldots,\#(\mathcal{R}_{\mathbf{x},u})\}=:{R}_{\mathbf{x},u}$. Since $\mathcal{R}_{\mathbf{x},u}$ is a code \cite{Dur}, i.e., any element in $\mathcal{R}_{\mathbf{x},u}^*$ has a unique factorization as return words to $u$, $\mathbf{x}$ can be written in a unique way as $\mathbf{x}=m_0m_1m_2\cdots$ and the {\em derived sequence} $\mathcal{D}_u(\mathbf{x})$ is an infinite word $\Lambda_{\mathbf{x},u}(m_0)\Lambda_{\mathbf{x},u}(m_1)\Lambda_{\mathbf{x},u}(m_2)\cdots$ over ${R}_{\mathbf{x},u}$.

\subsection{Abelian returns}

Recently, the notion of return words has been generalized to an
abelian framework \cite{PZ}.  In this paper, we will distinguish two cases: abelian
return to a prefix and abelian return to a factor.  We make such a
distinction to be able to define in the first case the abelian derived
sequence. Let us start with a few definitions.

Let $A=\{a_1,\ldots,a_k\}$ be a $k$-letter alphabet. 
We denote by $|w|_{a_i}$ the number of occurrences of the letter $a_i$ in a word $w\in A^*$. 
The Parikh mapping $\Psi:A^*\to\mathbb{N}^k$ is defined by $\Psi(w)=(|w|_{a_1},\ldots,|w|_{a_k})$. Let $u,v$ be two
finite words of the same length. We say that $u$ and $v$ are {\em
  abelian equivalent} and we write $u\sim_{ab} v$ if $\Psi(u)=\Psi(v)$. 

An infinite word $\mathbf{x}$ is {\em abelian periodic (of period
  $m$)}, if it can be factored as $\mathbf{x}=u_1u_2u_3\cdots$
where, for all $i,j$, the finite words $u_i$ and $u_j$ have the same length $m$ and 
are abelian equivalent.  The smallest $m$ for which such a
factorization exists is called the {\em abelian period} of $\mathbf{x}$. For instance, the Thue--Morse word is an
infinite concatenation of ${\tt 01}$ and ${\tt 10}$ and is thus
abelian periodic of period $2$.

 Let $\mathbf{x}$ be an infinite word. If, for each factor $u$ of
    $\mathbf{x}$, there exist infinitely many $i$ such that
    $\mathbf{x}[i,i+|u|-1]\sim_{ab} u$, then $\mathbf{x}$ is said to be
    {\em abelian recurrent}.  

    If $\mathbf{x}$ is abelian recurrent and if, for each factor $u$
    of $\mathbf{x}$, the distance between any two consecutive
    occurrences of factors abelian equivalent to $u$ is bounded by a
    constant depending only on $u$, then $\mathbf{x}$ is said to be
    {\em abelian uniformly recurrent}.

    \begin{remark}
        Note that uniform recurrence implies obviously abelian uniform
        recurrence. We will show in Proposition~\ref{pro:aur} that the
        converse does not hold.
    \end{remark}

\begin{definition}
    Let $u$ be a prefix of an abelian 
    uniformly recurrent word $\mathbf{x}$. We say that a nonempty
    factor $w$ of $\mathbf{x}$ is an {\em abelian return} to $u$, if
    there exists some $i\ge 0$ such that
\begin{itemize}
  \item $\mathbf{x}[i,i+|w|-1]=w$,
  \item $\mathbf{x}[i,i+|u|-1]\sim_{ab} u \sim_{ab} \mathbf{x}[i+|w|,i+|w|+|u|-1]$,
  \item $\mathbf{x}[i+j,i+j+|u|-1]\not\sim_{ab} u$, for all $j\in\{1,\ldots,|w|-1\}$.
\end{itemize}
We denote by $\mathcal{APR}_{\mathbf{x},u}$ the set of abelian returns
to the prefix $u$. Since $\mathbf{x}$ is abelian uniformly recurrent,
then the set $\mathcal{APR}_{\mathbf{x},u}$ is finite. We define the
set of abelian returns to prefixes as
$$\mathcal{APR}_{\mathbf{x}}:=\bigcup_{u\in\Pref(\mathbf{x})}
\mathcal{APR}_{\mathbf{x},u}.$$ Observe that if $\mathbf{x}$ is
uniformly recurrent, then the length of the longest element in
$\mathcal{APR}_{\mathbf{x},u}$ is bounded by the length of the
longest element in $\mathcal{R}_{\mathbf{x},u}$.
\end{definition}

We will also consider a more general situation where $u$ is
not restricted to be a prefix of $\mathbf{x}$. Puzynina and Zamboni \cite{PZ} 
called this notion a {\em semi-abelian return} to the abelian class of $u$ and
the number of abelian returns is the number of distinct abelian
classes of semi-abelian returns.

\begin{definition}
   If $\mathbf{x}$ is abelian recurrent and
    if $u$ is a factor of $\mathbf{x}$, we can define as above the
    notion of abelian return to $u$. The corresponding set
    $\mathcal{AR}_{\mathbf{x},u}$ of abelian returns to $u$ is well
    defined.  We define the set of abelian returns as
    $$\mathcal{AR}_{\mathbf{x}}:=\bigcup_{u\in \Fac(\mathbf{x})}
    \mathcal{AR}_{\mathbf{x},u}.$$
\end{definition}

\begin{remark}
    Let $\mathbf{x}$ be an abelian recurrent word. The set
    $\mathcal{AR}_{\mathbf{x},u}$ is finite, for each factor $u$ of
    $\mathbf{x}$, if and only if $\mathbf{x}$ is abelian uniformly
    recurrent.
\end{remark}

\section{Finiteness of the set of abelian returns}\label{sec:2}

Puzynina and Zamboni \cite{PZ} provided a discussion between periodicity and the number of abelian returns. Here we take the finiteness of the set
of abelian returns to characterize periodicity.

\begin{theorem} Let $\mathbf{x}$ be a recurrent word. The set
    $\mathcal{AR}_{\mathbf{x}}$ is finite if and only if $\mathbf{x}$
    is periodic.
\end{theorem}
\begin{proof} The ``if'' part is obvious. We prove the ``only if'' part.

    Suppose that $\mathcal{AR}_{\mathbf{x}}$ is finite and that
    $\mathbf{x}$ is recurrent but not periodic.  In this case, for
    each $k$, there exists a word $u$ satisfying $|u|&gt;k$ such that
    $au,bu \in \Fac(\mathbf{x})$ for some letters $a\not=b$.  Hence
    there exist $i,j$ such that $i&lt;j$, $\mathbf{x}[i,i+|u|]=au$ and
    $\mathbf{x}[j,j+|u|]=bu$.  Define $v=\mathbf{x}[i,j-1]$.  Since
    $\mathbf{x}[i+d,j-1+d]\not\sim_{ab} v$ for all $d\in\{1,\ldots,|u|\}$,
    there is an abelian return to $v$ in $\mathbf{x}$ of length at
    least $k$. As we can do the same for arbitrarily large $k$, the
    set $\mathcal{AR}_{\mathbf{x}}$ is infinite.
\end{proof}

Obviously, uniform recurrence implies abelian uniform recurrence, but
the converse is not true. Recall that an {\em eventually recurrent word} is an infinite 
word having a recurrent suffix.

\begin{proposition}\label{pro:aur} There exists an abelian uniformly recurrent word
    which is not eventually recurrent.
\end{proposition}
\begin{proof} Let $\mathbf{t}=t_0t_1\cdots={\tt 01101001}\cdots$ be
    the Thue--Morse word and $\varphi_{TM}$ be the Thue--Morse
    morphism, $\varphi_{TM}(\mathbf{t})=\mathbf{t}$. Define the set $I=\{i_0&lt;i_1&lt;\ldots\}$ of all positions
    where an isolated ${\tt 1}$ occurs. That is, for all $n$, we have
    $t_{i_n}={\tt 1}$ and $t_{i_n-1}=t_{i_n+1}={\tt 0}$. Moreover we set $J=\{
    i_{2^k}\mid k&gt;0\}$.

    Let $\mathbf{y}=y_0y_1\cdots$ be the word defined by $y_j={\tt
      0}$, if $j\in J$, and $y_j=t_j$ otherwise. Define
    $\mathbf{x}=\varphi_{TM}(\mathbf{y})$.

    The word $\mathbf{x}$ coincides with $\mathbf{y}$ almost
    everywhere, except for positions from the set $2J\cup(2J+1)$.
    Hence, each factor of the Thue--Morse word occurs in $\mathbf{x}$
    uniformly, i.e., with bounded gaps. At the same time, the
    factor $\varphi_{TM}({\tt 000})$ occurs in $\mathbf{x}$ with
    strictly growing gaps. Hence $\mathbf{x}$ is not eventually
    recurrent.

    Let us now prove that $\mathbf{x}$ is abelian uniformly recurrent.
    First we point out a property of the Thue--Morse word: for all
    $d&gt;0$ and all $a\in\{{\tt 0},{\tt 1}\}$, there exists $k$ such
    that $t_k=a \neq t_{k+d}$. This property follows from the
    well-known fact that the Thue--Morse word does not contain any
    constant infinite arithmetical subsequence \cite{Sh}.

    As $\mathbf{x}$ is abelian periodic (of period $2$), the weight
    (i.e., the sum of digits) of each factor $u$ of
    $\mathbf{x}$ of odd length is either $\frac{|u|+1}{2}$ or
    $\frac{|u|-1}{2}$. Note that $y_i={\tt 0}$ implies
    $|\mathbf{x}[2i+1,2i+|u|]|_{\tt 1}=\frac{|u|+1}{2}$ and $y_i={\tt
      1}$ implies $|\mathbf{x}[2i+1,2i+|u|]|_{\tt 1}=\frac{|u|-1}{2}$.
    Since ${\tt 0}$ (resp. ${\tt 1}$) occurs with bounded gaps in
    $\mathbf{y}$, gaps between abelian occurrences in
    $\mathbf{x}$ of a factor of odd length are bounded.

    The weight of a factor $u$ of even length of $\mathbf{x}$ can take
    values $\frac{|u|}{2}$, $\frac{|u|}{2}+1$ and $\frac{|u|}{2}-1$.
    The first case takes place when $u$ occurs at an even
    position in $\mathbf{x}$, meaning that the gaps between abelian
    occurrences of $u$ of weight $\frac{|u|}{2}$ in $\mathbf{x}$ are
    bounded.  The last two cases take place if $u$ occurs in
    $\mathbf{x}$ at an odd position $i$ and if $y_{\frac{i-1}{2}}={\tt
      1}$ and $y_{\frac{i-1+|u|}{2}}={\tt 0}$ or,
    $y_{\frac{i-1}{2}}={\tt 0}$ and $y_{\frac{i-1+|u|}{2}}={\tt 1}$.
    Due to the mentioned property of the Thue--Morse word, there
    exists $k$ such that $t_k={\tt 1}\neq t_{k+\frac{|u|}{2}}$ (resp.
    $t_k={\tt 0}\neq t_{k+\frac{|u|}{2}}$) and since $\mathbf{t}$ is
    uniformly recurrent, the factor $\mathbf{t}[k,k+\frac{|u|}{2}]$
    occurs infinitely often with bounded gaps in $\mathbf{t}$.  Hence
    abelian occurrences of $u$ in $\mathbf{x}$ appear infinitely often
    with bounded gaps.

\end{proof}

\section{Finiteness of the set of abelian returns to prefixes}\label{sec:3}
Contrary to the finiteness of $\mathcal{AR}_{\mathbf{x}}$, the
finiteness of $\mathcal{APR}_{\mathbf{x}}$ does not imply periodicity
nor abelian periodicity of $\mathbf{x}$. Moreover, if $\mathbf{x}$
is uniformly recurrent, it is well-known that
$$\min_{v\in\mathcal{R}_{\mathbf{x},u}} |v|\to\infty, \text{ if }
|u|\to\infty,$$ meaning that taking longer prefixes eventually leads
to longer return words. Here we show that such a result does not hold
for abelian returns to prefixes. Indeed, for the Thue--Morse word the
corresponding set $\mathcal{APR}_{\mathbf{t}}$ is finite and can be
described precisely. Such a result also holds for the Fibonacci word.
In particular, amongst the set of Sturmian words, the finiteness of
$\mathcal{APR}_{\mathbf{x}}$ characterizes Sturmian
words with nonzero intercept.

\begin{lemma}\label{lemmaAPandURImplF} If $\mathbf{x}$ is a uniformly
    recurrent and abelian periodic word, then the set
    $\mathcal{APR}_{\mathbf{x}}$ is finite.
\end{lemma}
\begin{proof} Let $m$ be the (minimal) abelian period of $\mathbf{x}$.
    Let us find an upper bound for the length of an abelian return $u$
    to a prefix $p$ of $\mathbf{x}$.

    Suppose first that $|p|=mk$. In this case, due to abelian
    periodicity, for all $i$, we have $\mathbf{x}[mi,m(i+k)-1]\sim_{ab} p$. Hence we get $|u|\leqslant m$.

    Suppose now that $|p|=mk+\ell$, where $0&lt;\ell&lt;m$. Let us denote the
    word $\mathbf{x}[mk,m(k+1)-1]$ by $s$. As the word $\mathbf{x}$ is abelian periodic,
    if there exists $i$ such that the equality $\mathbf{x}[mi,m(i+1)-1]=s$ holds, then 
    $\mathbf{x}[m(i-k),mi+\ell-1]\sim_{ab} p$. Hence, it is sufficient to prove that the
    set $$\{i\ge 0 \mid \mathbf{x}[mi,m(i+1)-1]=s\}$$ has bounded gaps.

Let us consider the word $\mathbf{x}'$ on the alphabet of factors of $\mathbf{x}$ of length $m$, such that $\mathbf{x}'_i = \mathbf{x}[mi,m(i+1)-1]$. It is well-known that the uniform recurrence of $\mathbf{x}$ implies uniform recurrence of $\mathbf{x}'$ (see for instance \cite{PS}). Hence, for each letter of $\mathbf{x}'$ there is an upper bound on the gap between two consecutive occurrences of it in $\mathbf{x}'$. Denoting the maximum of such constants by $D$, we get $|u|\leqslant mD$. \end{proof}

\begin{remark} In Lemma \ref{lemmaAPandURImplF}, the condition on a
    word $\mathbf{x}$ to be uniformly recurrent is essential: there exists an abelian periodic word $\mathbf{x}$ which is not uniformly recurrent and such that $\mathcal{APR}_{\mathbf{x},u}$ is infinite for some prefix $u$ of $\mathbf{x}$.  Consider the
    abelian periodic word of period $4$ given by $\mathbf{x}= \phi \varphi^{\omega}({\tt 0})$
    where $\varphi: {\tt 0} \mapsto {\tt 010}, {\tt 1} \mapsto {\tt 111}$ and $\phi: {\tt 0} \mapsto
    {\tt 01230123}, {\tt 1} \mapsto {\tt 0213}$:
\begin{equation*}
    \mathbf{x} = {\tt 01230123 \;\; 0213 \;\; 01230123 \;\; 0213 \;\; 0213 \;\; 0213} \cdots
\end{equation*} In $\mathbf{x}$ there are unbounded gaps between consecutive abelian occurrences of its prefix ${\tt 012301}$ that correspond to the occurrences of $\phi({\tt 1}^m)$.
\end{remark}

\begin{remark} In Lemma \ref{lemmaAPandURImplF}, the condition on
    abelian periodicity of $\mathbf{x}$ is not necessary to get
    finiteness of $\mathcal{APR}_{\mathbf{x}}$. We shall give an
    example below when discussing the case of Sturmian words. Indeed,
    Sturmian words are not abelian periodic (see
    Lemma~\ref{lem:sturmper}) but for instance, the Fibonacci word
    $\mathbf{f}$ is uniformly recurrent and the corresponding set
    $\mathcal{APR}_{\mathbf{f}}$ is finite.
\end{remark}

\begin{proposition}
    A word $\mathbf{x}$ is periodic if and only there exists some
    prefix $u$ such that infinitely many factors of $\mathbf{x}$ are
    abelian equivalent to $u$ and all the abelian returns in
    $\mathcal{APR}_{\mathbf{x},u}$ have length $1$.
\end{proposition}

\begin{proof}
    If $\mathbf{x}=u^\omega$, then $\mathbf{x}[i,i+|u|-1]\sim_{ab} u$
    for all $i\ge 0$. Conversely, if all the abelian returns to some
    prefix $u$ in $\mathcal{APR}_{\mathbf{x},u}$ have length $1$, then $\mathbf{x}[i,i+|u|-1]\sim_{ab}
    u\sim_{ab}\mathbf{x}[i+1,i+|u|]$ for all $i\ge 0$. There is an abelian return $a$ of length $1$
    at position $i$ in $\mathbf{x}$ and it also occurs in
    position $i+|u|$. It follows that $|u|$ is a period of $\mathbf{x}$.
\end{proof}

\subsection{Finiteness of $\mathcal{APR}_{\mathbf{t}}$ for the Thue--Morse word}

We already know from Lemma~\ref{lemmaAPandURImplF} that the
Thue--Morse word has a finite set of abelian returns to all its
prefixes. Here we describe precisely this set.

\begin{lemma}\label{lem:2}
    Let $\mathbf{x}$ be a uniformly recurrent word.  Let $n\ge 1$ and
    $i,j$ be such that $i&lt;j$. Assume that $\mathbf{x}[i,i+n-1]\sim_{ab} \mathbf{x}[j,j+n-1]$ and 
    there exists a prefix $u$ of length $j-i$ of $\mathbf{x}$ such
    that $u\sim_{ab} \mathbf{x}[i,j-1]$. The word $\mathbf{x}[i,i+n-1]$is an occurrence of an 
    abelian return to the prefix $u$ if and only if, for all
    $\ell\in\{0,\ldots,n-2\}$, $\mathbf{x}[i,i+\ell]\not\sim_{ab}
    \mathbf{x}[j,j+\ell]$. 
\end{lemma}

\begin{proof}
    Since $|u|=j-i$, by assumption we have $\mathbf{x}[i,i+|u|-1]\sim_{ab}
    u$. Observe first that there exists $\ell\in\{0,\ldots,n-1\}$ such
    that $\mathbf{x}[i,i+\ell]\sim_{ab} \mathbf{x}[j,j+\ell]$ if and only
    if $\mathbf{x}[i+\ell+1,i+\ell+|u|]\sim_{ab} u$. In particular, since
    $\mathbf{x}[i,i+n-1]\sim_{ab} \mathbf{x}[j,j+n-1]$, we get
    $\mathbf{x}[i+n,i+n+|u|-1]=\mathbf{x}[i+n,j-1+n]\sim_{ab} u$. Moreover,
    $\ell\in\{0,\ldots,n-2\}$ is such that
    $\mathbf{x}[i+\ell+1,i+\ell+|u|]\not\sim_{ab} u$ if and only if
    $\mathbf{x}[i,i+\ell]\not\sim_{ab} \mathbf{x}[j,j+\ell]$.
\end{proof}

\begin{remark}
From this lemma, we can derive a necessary condition for a word to be
an abelian return to a prefix. If a word $w=w_1\cdots w_n$ of length $n$ is an abelian
return to a prefix, then there exists some factor $y=y_1\cdots y_n$ of
$\mathbf{x}$ such that 
\begin{equation}
    \label{eq:cond}
    w\sim_{ab} y \text{ and, for all }\ell\in\{1,\ldots,n-1\}, w_1\cdots w_\ell\not\sim_{ab} y_1\cdots y_\ell.
\end{equation}
This condition is not sufficient. For instance, $w={\tt 001011}$ and
$y= {\tt 110010}$ are two factors of length $6$ satisfying\eqref{eq:cond} and occurring in the Thue--Morse word
$\mathbf{t}$. But, as shown in the following proposition, $w$ is not an abelian return to any prefix.
\end{remark}

\begin{theorem}
    The set $\mathcal{APR}_{\mathbf{t}}$ of abelian returns to prefixes for the Thue--Morse word $\mathbf{t}$ is
$$\{{\tt 0},\ {\tt 1},\  {\tt 01},\  {\tt 10},\  {\tt 001},\  {\tt 011},\  {\tt 100},\  {\tt 110},\  {\tt 0011},\  {\tt 0101},\  {\tt 1010},\  {\tt 1100},\  {\tt 00101},\  {\tt 01011},\  {\tt 10100},\  {\tt 11010}\}.$$
\end{theorem}

\begin{proof}
    One can check with some computer experiments that the factors
    given above appear as abelian returns to some prefix of
    $\mathbf{t}$. Moreover, one can also check that these are the only
    factors of length $2,\ldots,5$ in $\mathbf{t}$ satisfying
    condition~\eqref{eq:cond}.

    Assume that there exists some abelian return
    $w=w_1\cdots w_n=\mathbf{t}[i,i+n-1]$ of length $n\ge 2$ to a prefix of $\mathbf{t}$ occurring at position $i$. 
    In particular, we may assume that $w$ is an abelian return to the prefix $u$ of length $j-i&gt;0$ and $y=y_1\cdots y_n=\mathbf{t}[j,j+n-1]$ satisfies ~\eqref{eq:cond}. 
    We will show that the length of $w$ is at most $5$.
    Recall that $\mathbf{t}[2k,2k+1]\in\{{\tt 01},{\tt 10}\}$ for all $k\ge 0$.

    Assume first that $i,j$ are even. Since $\mathbf{t}[i,i+1]$ and
    $\mathbf{t}[j,j+1]$ belong to $\{{\tt 01},{\tt 10}\}$, we conclude
    that $\mathbf{t}[i,i+1]\sim_{ab}\mathbf{t}[j,j+1]$ and, in that
    situation, we can only have an abelian return of length at most $2$.

    Assume now that $i$ is odd and $j$ is even and that
    $\mathbf{t}_i={\tt 0}$ (symmetric cases can be treated in the
    same way). Our aim is to build the longest possible abelian return. Since
    $\mathbf{t}_i={\tt 0}$ and $j$ is even, we consider
    $\mathbf{t}[j,j+1]={\tt 10}$ because otherwise,
    $\mathbf{t}[j,j+1]={\tt 01}$ and $w_1=y_1$ (i.e., $\mathbf{t}[i+1,j+1]\sim_{ab} \mathbf{t}[i,j]\sim_{ab} u$ and we get directly an abelian return of length $n=1$). Now
    $\mathbf{t}[i,i+2]={\tt 001}$ because otherwise,
    $\mathbf{t}[i,i+2]={\tt 010}$ and $w_1w_2\sim_{ab} y_1y_2$. Continuing
    this way, we have $\mathbf{t}[j,j+3]={\tt 1010}$ and
    $\mathbf{t}[i,i+4]={\tt 00101}$. Since $({\tt 10})^3$ is not a
    factor of $\mathbf{t}$, we have $\mathbf{t}[j,j+5]={\tt 101001}$
    and $\mathbf{t}[i,i+4]\sim_{ab} \mathbf{t}[j,j+4]$. In that situation,
    we can only have an abelian return of length at most $5$.

    The last case is when $i$ and $j$ are odd. Assume
    $\mathbf{t}_i={\tt 0}$ and $\mathbf{t}_j={\tt 1}$. 
    We have $\mathbf{t}_i={\tt 0}$ and $\mathbf{t}_{j-1}={\tt 0}$
    because $\mathbf{t}[j-1,j]={\tt 01}$. Moreover,
    $z=\mathbf{t}[i+1,j-2]\in\{{\tt 01},{\tt 10}\}^*$ and thus
    $v=\mathbf{t}[i,j-1]={\tt 0}z{\tt 0}$ is a word of even length such
    that $|v|_{\tt 0}=2+|v|_{\tt 1}$. Therefore $v$ cannot be abelian
    equivalent to a prefix $u$ of $\mathbf{t}$.
    So in such a situation, we cannot have an abelian return
    to some prefix of $\mathbf{t}$.
\end{proof}

\begin{proposition}
If a factor of length $n\ge 6$ of the Thue--Morse word 
satisfies \eqref{eq:cond}, then $n$ is even.    
\end{proposition}

\begin{proof}
    Let $w=\mathbf{t}[i,i+n-1]$ and $y=\mathbf{t}[j,j+n-1]$ be factors 
    of $\mathbf{t}$ of length $n\ge 6$ satisfying ~\eqref{eq:cond}.
    As $n\ge 6$, $i$ and $j$ are odd. 
    Hence, to satisfy the condition~\eqref{eq:cond}, we must have
    $$
    \begin{pmatrix}
	\mathbf{t}[i,i+n-1]\\
	\mathbf{t}[j,j+n-1]\\
    \end{pmatrix}\in \begin{pmatrix}
	    {\tt 0}\\{\tt 1}\\
	\end{pmatrix}
    \left\{
	\begin{pmatrix}
	    {\tt 01}\\{\tt 01}\\
	\end{pmatrix},
      \begin{pmatrix}
	    {\tt 10}\\{\tt 10}\\
	\end{pmatrix},  \begin{pmatrix}
	    {\tt 01}\\{\tt 10}\\
	\end{pmatrix}
    \right\}^* \begin{pmatrix}
	    {\tt 1}\\{\tt 0}\\
	\end{pmatrix}
    \cup
    \begin{pmatrix}
	    {\tt 1}\\{\tt 0}\\
	\end{pmatrix}
    \left\{
	\begin{pmatrix}
	    {\tt 01}\\{\tt 01}\\
	\end{pmatrix},
      \begin{pmatrix}
	    {\tt 10}\\{\tt 10}\\
	\end{pmatrix},  \begin{pmatrix}
	    {\tt 10}\\{\tt 01}\\
	\end{pmatrix}
    \right\}^* \begin{pmatrix}
	    {\tt 0}\\{\tt 1}\\
	\end{pmatrix}.
    $$
    So $n$ must be even.
\end{proof}
For $n=6,8,10,\ldots,104$, with a computer search, we get the following number of factors of length $n$ satisfying  \eqref{eq:cond}: $6$, $4$, $8$, $12$, $12$, $4$, $8$, $8$, $4$, $0$, $0$, $8$, $0$, $0$, $4$, $8$, $4$, $0$, $0$, $0$, $0$, $0$, $4$, $0$, $4$, $0$, $0$, $0$, $0$, $0$, $4$, $8$, $4$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $0$, $8$, $0$, $0$.

\subsection{Sturmian words}


Sturmian words form one of the most studied classes of infinite words.
It can be defined in several ways. For our purpose, the definition
in terms of rotations is convenient.

Let $C$ be the one-dimensional torus $\mathbb{R}/\mathbb{Z}$
identified with the interval $[0,1)$. As usual, we
denote by $\{x\}$ the fractional part of $x$. The \emph{rotation}
$R_{\alpha}$ defined for a real number $\alpha$ is a mapping
$C\rightarrow C$ that maps an $x$ to $\{x+\alpha\}$.
\begin{theorem}[Kronecker \cite{AS}]\label{theoremKronecker} If a
    number $\alpha$ is irrational, then the set of points
    $\{R^i_\alpha(\rho) \mid i\in \mathbb{N} \}$ is dense in $C$
    for all initial points $\rho\in C$.
\end{theorem}

By an \emph{interval} (resp. \emph{half-interval}) of $C$ we mean a set
of points that is an image of an interval (resp. half-interval) of
$\mathbb{R}$ under operation $\{\cdot\}$. For instance, if $0\le b&lt;a&lt;1$, then $[a,1)\cup [0,b)$ is denoted by $[a,b)$. 

Let $\alpha$ be irrational and $\rho$ be real. Without loss of generality we can assume $0\le \alpha,\rho&lt;1$. A \emph{Sturmian word}
$\mathbf{x}=St(\alpha, \rho)$ (resp. $\mathbf{x}=St'(\alpha, \rho)$) can be defined
 as
\begin{equation}
    x_i=\begin{cases}
            {\tt 0}, &amp;  \text{ if } R^i_{\alpha}(\rho) \in I_0;\\
  {\tt 1}, &amp;  \text{ if } R^i_{\alpha}(\rho) \in I_1 , 
          \end{cases}
    \end{equation} where $I_0 = [0, 1 - \alpha)$ and $I_1 = [1 - \alpha,1)$ (resp. $I_0 = (0, 1 - \alpha]$ and $I_1 = (1 - \alpha,1]$). See \cite{AS,Lot}.

If $\rho=0$, then $$St(\alpha, 0)={\tt 0} \mathbf{c}_\alpha \text{ and } St'(\alpha, 0)={\tt 1} \mathbf{c}_\alpha$$ and $\mathbf{c}_\alpha$ is said to be the {\em characteristic Sturmian word of slope} $\alpha$ \cite{Lot}. If $\mathbf{x}=St(\alpha, 0)$, we say that $\mathbf{x}$ is a Sturmian word {\em with null intercept}.
\begin{example}
    If $\tau=(1+\sqrt{5})/2$ is the Golden mean, then $St(1/\tau^2, 0)={\tt 0}\mathbf{f}$ where $\mathbf{f}$ is the Fibonacci word ${\tt 0100101001}\cdots$ which is the unique fixed point of the morphism $\varphi:{\tt 0}\mapsto {\tt 01}, {\tt 1}\mapsto {\tt 0}$. In particular, we have $\mathbf{f}=St(1/\tau^2,1/\tau^2)$.
\end{example}

    \begin{theorem}\cite[Theorem 2.1.5]{Lot}
    An infinite word $\mathbf{x}\in\{{\tt 0},{\tt
      1}\}^\omega$ is Sturmian if and only if it is aperiodic and
    balanced, i.e., for all $u,v\in Fac(\mathbf{x})$ of same length, we have $\left| |u|_{\tt 1}-|v|_{\tt 1} \right|\le 1$. 
    \end{theorem}

Let $\mathbf{x}=St(\alpha, \rho)$ be a Sturmian word. For a binary word $v=v_0v_1\cdots v_m$, we define a half-interval $I_v$ of $C$ as 
\begin{equation}
    \label{eq:Iv}
I_v:=I_{v_0}\cap R^{-1}_{\alpha}(I_{v_1})\cap\dots\cap R^{-m}_{\alpha}(I_{v_m}).    
\end{equation}
Hence $\mathbf{x}[i,i+m]=v$ if and only if $R^i_{\alpha}(\rho) \in I_v$. See \cite[Section 2.1.2]{Lot}.

    \begin{definition}
        Let $\mathbf{x}=St(\alpha, \rho)$ be a Sturmian word.  For each $k$ the number
        of ${\tt 1}$'s in a factor of length $k$ in $\mathbf{x}$ can
        only take the values $\lceil k\alpha \rceil$ or $\lceil
        k\alpha \rceil-1$. The corresponding factors will be called
        respectively {\em heavy} and {\em light}. If $\mathbf{x}$ is
        understood from the context, $H(k)$ (resp. $L(k)$) will denote the set of
        heavy (resp. light) factors of length $k$ in $\mathbf{x}$. Define 
	$$I_H(k) := \bigcup_{v \in H(k)} I_v \text{ and }
        I_L(k) := \bigcup_{v \in L(k)} I_v.$$ So, the word
        $\mathbf{x}[i,i+k-1]$ is heavy if and only if $R^i_{\alpha}(\rho) \in
        I_H(k)$.
    \end{definition}


\begin{theorem}\label{thmSturmianImplF} 
Let $\mathbf{x}$ be a Sturmian word. 
The set $\mathcal{APR}_{\mathbf{x}}$ is finite if and only if $\mathbf{x}$ does not have a null intercept.
\end{theorem}

\begin{proof} For the sake of convenience, let $\mathbf{x}$ be defined as $St(\alpha, \rho)$ for half-intervals $I_0 = [0, 1 - \alpha)$ and $I_1 = [1 - \alpha,1)$. Let us prove by induction on $k\ge 1$ that 
    \begin{equation}
        \label{eq:IHIL}
I_H(k) = [1-\{k\alpha\},1)\text{ and }I_L(k) = [0,1-\{k\alpha\}).        
    \end{equation}
It holds true for $k=1$. Suppose now that the statement holds true for some $k\ge 1$. We consider two cases.

\begin{itemize}
  \item Assume that $0 \not\in R^{-k}_\alpha (I_1)$. Therefore we get
    $R^{-k}_\alpha (I_1) = [1-\{(k+1)\alpha\},1-\{k\alpha\})$ with $1-\{(k+1)\alpha\}&lt;1-\{k\alpha\}$. By the
    induction hypothesis, we have $I_H(k) = [1-\{k\alpha\},1)$ and
    consequently, $$R^{-k}_\alpha (I_1) \cap I_H(k) = \emptyset.$$ This
    means that all the heavy factors of length $k$ of $\mathbf{x}$ can
    only be extended with ${\tt 0}$ to factors of length $k+1$ of
    $\mathbf{x}$. In particular, the weights of heavy factors of
    length $k$ and $k+1$ are the same. At the same time, we have
    $R^{-k}_\alpha (I_1) \cap I_L(k) = R^{-k}_\alpha (I_1)$, which means
    that the factors corresponding to elements belonging to this
    latter set are the light factors of length $k$ that are extended
    with ${\tt 1}$ to heavy factors of length $k+1$.  We conclude that 
$$I_H(k+1) =
I_H(k) \cup R^{-k}_\alpha (I_1) = [1-\{(k+1)\alpha\},1)$$  and $I_L(k+1) =
I_L(k)\backslash R^{-k}_\alpha (I_1)=[0,1-\{(k+1)\alpha\})$.

\item Assume now that $0 \in R^{-k}_\alpha (I_1)$, i.e., $1-\{(k+1)\alpha\}&gt;1-\{k\alpha\}$. In this case, using again the induction hypothesis, 
  $R^{-k}_\alpha (I_1) \cap I_H(k) = [1-\{(k+1)\alpha\},1)$ is
  nonempty.  This interval corresponds to the heavy factors of length
  $k$ having an extension with ${\tt 1}$ making them the only heavy
  factors of length $k+1$ in $\mathbf{x}$.
\end{itemize}

Now we are ready to prove the main part of the statement. First of
all, let us prove that, if $\mathbf{x}$ has a null intercept, then
$\mathcal{APR}_{\mathbf{x}}$ is infinite. Let $k\ge 1$ and $p$ be the
prefix of length $k$ of $\mathbf{x}$. As $\rho=0$, we have $0\in
I_p$. Since the interval $I_p$ corresponds exactly to one word $p$ which 
is either light or heavy, we have $I_p \subseteq I_L(k)$ or $I_p \subseteq I_H(k)$.
As $0\in I_p$, we conclude that $I_p \subseteq I_L(k)$ using \eqref{eq:IHIL}.
In other words, we have just shown that each prefix of $\mathbf{x}$ is a light factor. 

Now we show that, for all $n$, there exists a length $\ell$ such that
gaps between consecutive occurrences of light factors of length $\ell$
in $\mathbf{x}$ can be larger than $n$. Let $n\ge 1$. Define the set of points 
$$S_n := \{R^i_\alpha(0) \mid
0\leqslant i \leqslant n \}$$ and denote by $d$ the minimal length of
intervals having endpoints in $S_n$. Due to Kronecker's theorem, we
can find some $\ell$ such that $|I_L(\ell)|&lt;d$ and it follows that $I_L(\ell)
\cap S_n = \{0\}$. With our definitions, it means that the light prefix
of $\mathbf{x}$ of length $\ell$ is followed by at least $n$ heavy
consecutive factors of length $\ell$. Since this can be done for any $n$,
the set $\mathcal{APR}_{\mathbf{x}}$ is infinite.
\smallskip

Let us prove that, if $\mathbf{x}$ does not have a null intercept,
then $\mathcal{APR}_{\mathbf{x}}$ is finite. The main difference with
the previous situation is that a prefix can now be a heavy or a light
factor depending on its length: $\rho\ge 1-\{k\alpha\}$ if and only
the prefix of length $k$ is heavy. We will show that there exists a
constant $c$ such that, for all prefixes $p$ of $\mathbf{x}$, the gap
between consecutive occurrences of factors abelian equivalent to $p$
is bounded by $c$.

Let $n\ge 1$. Consider as before the set $S_n$ and order its elements
$0=s_0&lt;s_1&lt;\cdots &lt;s_n$. Denote by $D(n)$ the maximal length of the
intervals $[s_0,s_1)$, \ldots, $[s_{n-1},s_n)$, $[s_n,s_0)$ whose endpoints are consecutive points in $S_n$. Due to Kronecker's theorem,
there exists some $c$ such that $2D(c)&lt;\min\{\rho, 1-\rho\}$.

Suppose that the prefix of length $k$ of $\mathbf{x}$ is a light word.
Then we have $\rho \in I_L(k)$ and, consequently, $|I_L(k)|&gt;\rho$.
Assume that there is a light factor of length $k$ occurring at
position $i$ in $\mathbf{x}$, i.e., $R^{i}_\alpha (\rho)\in
I_L(k)$. We consider two cases. If $R^{i}_\alpha (\rho)\ge D(c)$, there exists $j \in\{1,\ldots,
c\}$ such that $R^{j}_\alpha(0)=s_c$ and $\theta=1-R^{j}_\alpha(0)\in(0,D(c)]$. Hence the
point $R^{j}_\alpha(R^{i}_\alpha (\rho))=R^{i+j}_\alpha (\rho)=R^{i}_\alpha (\rho)-\theta$ belongs to
$I_L(k)$, i.e., the factor of length $k$ at position $i+j$ in
$\mathbf{x}$ is light again. If $R^{i}_\alpha (\rho)&lt; D(c)$, there exists $j \in\{1,\ldots,
c\}$ such that $R^{j}_\alpha(0)=s_1\le D(c)&lt;\rho/2$. Hence the
point $R^{j}_\alpha(R^{i}_\alpha (\rho))=R^{i+j}_\alpha (\rho)=R^{i}_\alpha (\rho)+s_1$ is less than $\rho$ and belongs to
$I_L(k)$.

A similar proof can be done for the case of a heavy prefix of length
$k$. Assume that $\rho\in I_H(k)$ and that, for some $i\ge 0$,
$R^{i}_\alpha (\rho)\in I_H(k)$. If $R_\alpha^i(\rho)&lt;1-D(c)$, then
$R_\alpha^i(\rho)+s_1&lt;1$. If $R_\alpha^i(\rho)\ge 1-D(c)$, then
$R_\alpha^i(\rho)-(1-s_c) \ge 1-2D(c)&gt;\rho$. We can derive the same conclusion as above.

Hence the number $c$ is an upper bound
on the length of abelian returns to any prefix and therefore
$\mathcal{APR}_{\mathbf{x}}$ is finite.
\end{proof}

\begin{lemma}\label{lem:sturmper}
    No Sturmian word is abelian periodic.
\end{lemma}

\begin{proof}
    Proceed by contradiction and assume that
    $\mathbf{x}=St(\alpha,\rho)$ is abelian periodic of period $m$
    with $\alpha$ irrational.  Then all factors of the kind
    $\mathbf{x}[tm,(t+1)m-1]$, $t\in\mathbb{N}$, are abelian
    equivalent, i.e., have the same weight. Assume that, for
    all $t$, $R_\alpha^{tm}(\rho)= R_{m\alpha}^{t}(\rho)\in I_L(m)$.
    But since $\alpha$ is irrational, $m\alpha$ is also irrational and
    thanks to Kronecker's theorem, $\{R_{m\alpha}^{t}(\rho)\mid t\ge
    0\}$ is dense in $C$ contradicting the fact that
    $\{R_{m\alpha}^{t}(\rho)\mid t\ge 0\}\cap I_H(m)$ should be empty.
\end{proof}

\subsection{Finiteness of $\mathcal{APR}_{\mathbf{f}}$ for the Fibonacci word}

From Theorem~\ref{thmSturmianImplF}, since the Fibonacci word
$\mathbf{f}$ is given by $St(1/\tau^2,1/\tau^2)$, we already know that
$\mathcal{APR}_{\mathbf{f}}$ is finite. Here we exhibit exactly the
elements of this set in Theorem~\ref{the:fib}. As a first attempt,
\eqref{eq:cond} gives a necessary condition that allows one to exclude
some words as abelian returns. This condition will not be used in the
proof of Theorem~\ref{the:fib} but,  interestingly, our developments can
be related to the lengths of the palindromic prefixes of $\mathbf{f}$,
\cite{luca,Fis}.

\begin{proposition}\label{propfibofinite}
For the Fibonacci word, there exist exactly two factors of length $n$
satisfying \eqref{eq:cond} if $n$ is a Fibonacci number. Otherwise, no
factor of length $n$ satisfies such a condition.
\end{proposition}

\begin{proof}
    Consider two factors $x,y$ of length $n$ satisfying
    \eqref{eq:cond} and occurring in 
    $$\mathbf{f}={\tt 010010100100}\cdots .$$ 
    Assume that $x$ starts with ${\tt 0}$.
    Then to fulfill \eqref{eq:cond}, $y$ starts with ${\tt 1}$. Since
    $\mathbf{f}$ is Sturmian, for any two words of
    the same length $x'$ and $y'$ which are prefixes of $x$ and $y$ respectively, we have $\left|
        |x'|_{\tt 1}-|y'|_{\tt 1} \right|\le 1$. Therefore, we deduce
    that $x$ and $y$ must be of the form $x={\tt 0}u{\tt 1}$ and
    $y={\tt 1}u{\tt 0}$ for some $u\in\{{\tt 0},{\tt 1}\}^*$. This
    means that $u$ is a bispecial factor of the Fibonacci word.

    Recall that the left special factors in $\mathbf{f}$ are its
    prefixes and its right special factors are the mirror images of
    its prefixes \cite[Prop. 4.10.3]{CANT}. So bispecial factors of
    $\mathbf{f}$ are its palindromic prefixes. If $(\ell_i)_{i\ge 1}$
    denotes the increasing sequence of all lengths of palindromic
    prefixes in $\mathbf{f}$, it is well-known that $(\ell_i)_{i\ge
      1}=(0,1,3,6,11,\ldots)$ is given by $\ell_i=F_{i+1}-2$ where
    $F_i$ is the $i$th Fibonacci number. See \cite[Thm.  5]{luca} and  \cite{Fis}.
    Hence $n$ must be a Fibonacci number.

    Conversely, for any bispecial factor $u$ of $\mathbf{f}$, it is easy to
    show that either ${\tt 0}u{\tt 0}$ or ${\tt 1}u{\tt 1}$ is not a factor
    occurring in $\mathbf{f}$ (see for instance \cite[p. 47]{Lot}).
    Therefore, amongst the four words ${\tt 0}u{\tt 0}$, ${\tt 1}u{\tt
      1}$, ${\tt 0}u{\tt 1}$ and ${\tt 1}u{\tt 0}$, the last two must
    occur in $\mathbf{f}$ and we get exactly two factors of length $|u|+2$
satisfying \eqref{eq:cond}. Indeed, assume that ${\tt 0}u{\tt 0}$ does
    not occur in $\mathbf{f}$. Then for $u$ to be left (resp. right)
    special, ${\tt 0}u{\tt 1}$ (resp. ${\tt 1}u{\tt 0}$) must occur in
    $\mathbf{f}$.
\end{proof}

The reader may notice that the computations carried out in the proofs of the next two results could also be adapted to other Sturmian words.

\begin{theorem}\label{the:fib}
    The set $\mathcal{APR}_{\mathbf{f}}$ of abelian returns to prefixes for the Fibonacci word $\mathbf{f}$ contains exactly the words ${\tt 0},\ {\tt 1},\  {\tt 01},\  {\tt 10},\  {\tt 001}$.
\end{theorem}

\begin{proof}
Using the same notation as in Theorem~\ref{thmSturmianImplF}, for
$c=7$, we have $D(7)\approx 0.145898$ which is such that
$2D(7)&lt;\min\{1/\tau^2, 1-1/\tau^2\}\approx 0.381$. Hence, all abelian
returns to prefixes of the Fibonacci word have length at most $7$.
Actually, this value can be reduced:

\begin{lemma}\label{lemFibRedToThree}
    There is no abelian return of length greater than $3$ to prefixes in the Fibonacci word.
\end{lemma}

\begin{proof} 
    With the notation of the proof of Theorem~\ref{thmSturmianImplF},
    we set $\rho=\alpha = 1/\tau^2\approx 0.381$. Let $i$ be a natural number. Define the four points
    $\rho_{i,t} = R^{i+t}_\alpha(\rho)$ for $t=0,1,2,3$.

    Recall that, for all $k\ge 1$, the unit circle $[0,1)$ is split
    into two half-intervals $I_H(k) = [1-\{k\alpha\},1)$ and
    $I_L(k)=[0,1-\{k\alpha\})$ such that two factors $\mathbf{f}[i,i+k-1]$
    and $\mathbf{f}[j,j+k-1]$ are abelian equivalent if and only if the 
    points $R^i_\alpha(\rho)$ and $R^j_\alpha(\rho)$ belong to the
    same interval $I_H(k)$ or $I_L(k)$. 

    Let $I$ be any of the two intervals $I_H(k)$ or
    $I_L(k)$. What we are going to prove is that if $\rho$ and
    $\rho_{i,0}$ belong to $I$, then either $\rho_{i,2}$ or
    $\rho_{i,3}$ belongs to $I$.  In other words, if
    $\mathbf{f}[i,i+k-1]\sim_{ab}\mathbf{f}[0,k-1]$, then either
    $\mathbf{f}[i+2,i+k+1]$ or $\mathbf{f}[i+3,i+k+2]$ is abelian
    equivalent to $\mathbf{f}[i,i+k-1]$ which gives the upper bound on
    the length of any abelian return to a prefix of $\mathbf{f}$.
 
    Note, that $\rho_{i,0}=R_\delta(\rho_{i,2})$, $\rho_{i,2}=R_{-\delta}(\rho_{i,0})$ and
    $\rho_{i,3}=R_{\alpha-\delta}(\rho_{i,0})$, where
    $\delta=1-2\alpha\approx 0.2361$. Assume that the factor of length $k$ starting in position $i$ is abelian equivalent to the prefix of $\mathbf{f}$ of length $k$, i.e., $\rho$ and $\rho_{i,0}$ are both light or heavy words. We consider two cases.
    Suppose first that $\rho,\rho_{i,0}\in I_L(k)$. In this case, we have $[0,\rho] \subseteq I_L(k)$.
\begin{itemize}
    \item If $\rho\le \rho_{i,0}&lt;1$, then  $[0,\rho_{i,0}] \subseteq I_L(k)$ and $0&lt;\rho_{i,2}=R_{-\delta}(\rho_{i,0})&lt;\rho_{i,0}$. Thus $\rho_{i,2}$  belongs also to $I_L(k)$.
    \item If $\rho&gt;\rho_{i,0}&gt;0$, either $\rho_{i,0}\ge \delta$ and
      then $\rho_{i,2}=R_{-\delta}(\rho_{i,0})\in[0,\rho)$ meaning
      that $\rho_{i,2}\in I_L(k)$ or, $0&lt;\rho_{i,0}&lt; \delta$, i.e.,
      $-\delta&lt;\rho_{i,2}-1&lt;0$ and then
      $0&lt;\alpha-\delta&lt;\rho_{i,3}=R_{\alpha}(\rho_{i,2})&lt;\rho$ meaning that
      $\rho_{i,3}\in I_L(k)$.
\end{itemize}
Suppose now that $\rho,\rho_{i,0}\in I_H(k)$. In this case, as $\rho\in I_H(k)$, we have $[\rho,1)\subseteq I_H(k)$.
\begin{itemize}
  \item If $\rho&gt;\rho_{i,0}&gt;0$, then $[\rho_{i,0},1)\subseteq I_H(k)$ and $\rho_{i,3}=R_{\alpha-\delta}(\rho_{i,0})$ belongs to $I_H(k)$.
  \item If $\rho\le \rho_{i,0}&lt;1$,  either $\rho_{i,0}\ge \rho+\delta$ and
      then $\rho_{i,2}=R_{-\delta}(\rho_{i,0})\in I_H(k)$ or, $\rho\le \rho_{i,0}&lt; \rho+\delta$, i.e.,
      $\rho-\delta\le\rho_{i,2}&lt;\rho$ and then
      $\rho&lt;\rho-\delta+\alpha\le \rho_{i,3}=R_{\alpha}(\rho_{i,2})&lt;\rho+\alpha&lt;1$ meaning that
      $\rho_{i,3}\in I_H(k)$.
\end{itemize}
\end{proof}

The factors of length at most $3$ occurring in $\mathbf{f}$ are
$\varepsilon$, ${\tt 0}$, ${\tt 1}$, ${\tt 00}$, ${\tt 01}$, ${\tt
  10}$, ${\tt 001}$, ${\tt 010}$, ${\tt 100}$ and ${\tt 101}$.
Clearly, ${\tt 00}$, ${\tt 010}$ and ${\tt 101}$ do not satisfy
\eqref{eq:cond} and cannot be abelian returns. To conclude the proof,
we just have to show that ${\tt 100}$ is also forbidden.

\begin{lemma}
    The set $\mathcal{APR}_{\mathbf{f}}$ of abelian returns to
    prefixes for the Fibonacci word $\mathbf{f}$ does not contain
    ${\tt 100}$.
\end{lemma}

\begin{proof}
    We continue with notation of Lemma~\ref{lemFibRedToThree}.
    Suppose that ${\tt 100}\in \mathcal{APR}_{\mathbf{f}}$. There exists a prefix $p$ of $\mathbf{f}$ of length $k$ and a
    position $i\ge 0$ such that
    \begin{enumerate}
      \item $\mathbf{f}[i,i+2]={\tt 100}$,
      \item $\mathbf{f}[i,i+k-1]\sim_{ab}p$, i.e., $\rho$ and
    $\rho_{i,0}$ belong to the same interval $I\in\{I_L(k),I_H(k)\}$,
  \item for $t=1,2$, $\mathbf{f}[i+t,i+t+k-1]\not\sim_{ab}p$, i.e., $\rho_{i,1}$ and
    $\rho_{i,2}$ do not belong to $I$,
\item $\mathbf{f}[i+3,i+2+k]\sim_{ab}p$.
    \end{enumerate}
    To get a contradiction, let us prove that either $\rho_{i,1}$ or
    $\rho_{i,2}$ belongs to $I$. Since $\mathbf{f}_i={\tt 1}$, $\rho_{i,0}$ belongs to $I_1=[1-\alpha,1)$.
If $I=I_L(k)$, then we have $\rho_{i,1}\in [0,\rho)\subseteq I_L(k)$. If $I=I_H(k)$, then  
 we have $\rho_{i,2}\in [\rho,\rho+\alpha)\subseteq I_H(k)$.
\end{proof}

That concludes the proof of Theorem~\ref{the:fib}.
\end{proof}

\subsection{Link with abelian complexity}

The {\em abelian complexity} of an infinite word $\mathbf{x}$ is the
function $p^{ab}_\mathbf{x}:\mathbb{N}\to\mathbb{N}$ that maps $n\ge
0$ to the number of distinct abelian equivalence classes of
factors of length $n$ in $\mathbf{x}$. Let $C&gt;0$. Recall that an
infinite word $\mathbf{x}\in A^\omega$ is {\em $C$-balanced}, if for
all $u,v\in\Fac(\mathbf{x})$ such that $|u|=|v|$, we have $\left|
    |u|_a-|v|_a\right|\le C$ for all $a\in A$.

\begin{lemma}\cite{Rich}
    An infinite word has bounded abelian complexity if and only if it is $C$-balanced for some $C&gt;0$.
\end{lemma}

\begin{proposition} If $\mathbf{x}$ is an abelian recurrent word such that 
    $\mathcal{APR}_{\mathbf{x}}$ is finite, then $\mathbf{x}$ has bounded abelian complexity.
\end{proposition}
\begin{proof} Suppose $\mathbf{x}$ satisfies the assumptions of the
    proposition but that $\mathbf{x}$ has unbounded abelian
    complexity. From the previous lemma, we deduce that there exists a
    symbol $a$ such that the maximum of differences $|u|_a-|v|_a$ for
    factors $u,v$ in $\mathbf{x}$ having equal length can be
    arbitrarily large.

    Let $\delta&gt;0$. There exist $u,v\in \Fac(\mathbf{x})$ of
    equal length $n$ such that $|u|_a-|v|_a\ge\delta$. Let
    $p=x_0x_1\ldots x_{n-1}$ be the prefix of length $n$ of
    $\mathbf{x}$. Without loss of generality, we may assume that
    $$||u|_a-|p|_a|\geqslant\frac{\delta}{2}.$$ Indeed, if $||u|_a-|p|_a|&lt;\delta/2$ and $||v|_a-|p|_a|&lt;\delta/2$, then one would deduce that $||u|_a-|v|_a|&lt;\delta$.

As $\mathbf{x}$ is abelian recurrent, factors abelian equivalent to $p$ (resp. to $u$) occur infinitely often in $\mathbf{x}$. Therefore there exist $i&lt;j&lt;k$ such that
\begin{enumerate}
  \item $\mathbf{x}[i,i+n-1]\sim_{ab} p$, $\mathbf{x}[k,k+n-1]\sim_{ab} p$,
  \item for all $t$ such that $i&lt;t&lt;k$, we have $\mathbf{x}[t,t+n-1]\not\sim_{ab} p$,
  \item $\mathbf{x}[j,j+n-1]\sim_{ab} u$.
\end{enumerate}
This just means that we can consider two consecutive factors abelian
equivalent to $p$ separated by a factor abelian equivalent to $u$.
Note that, for all $t$, $$||x[t+c,t+n-1+c]|_a-|x[t,t+n-1]|_a|\leqslant c,\quad
\forall c\le n.$$ Hence, $j-i\ge\delta/2$ and $k-j\geqslant \delta/2$.
Therefore we get $k-i\ge\delta$ which means that the abelian return
$\mathbf{x}[i,k-1]$ to the prefix $p$ has length at least $\delta$. As
$\delta$ can be chosen arbitrarily large, the set
$\mathcal{APR}_{\mathbf{x}}$ is infinite and that is a contradiction.
\end{proof}

Note that any Sturmian word $\mathbf{x}$ satisfies $p^{ab}_\mathbf{x}(n)=2$ for all $n\ge 1$ : there are exactly two kinds of factors of length $n$, the light ones and the heavy ones. But thanks to Theorem~\ref{thmSturmianImplF}, if $\mathbf{x}$ is a Sturmian word with null intercept, then $\mathcal{APR}_{\mathbf{x}}$ is infinite. In other words, bounded abelian complexity does not imply the finiteness of $\mathcal{APR}_{\mathbf{x}}$.

\section{Abelian derived sequences}

We refer the reader to definitions and notation introduced in Section~\ref{sec:21}. 
As was studied by Durand \cite{Dur} for classical return words, we
introduce the notion of abelian derived sequence which is the
factorization of an infinite word with respect to its abelian returns
to prefixes in their order of occurrence. The next result allows us to define such a sequence.

\begin{lemma}\label{lem:algo}
    Let $u$ be a prefix of a uniformly recurrent word $\mathbf{x}$.
    The word $\mathbf{x}$ has a factorization as a sequence
    $m_0m_1m_2\cdots$ of elements in $\mathcal{APR}_{\mathbf{x},u}$
    computed as follows. Consider the sequence of indices $(i_n)_{n\ge
      0}$ such that, for all $j\ge 0$, $\mathbf{x}[i_j,i_j+|u|-1]\sim_{ab}
    u$ and, for all $i\not\in\{i_n\mid n\ge 0\}$, we have
    $\mathbf{x}[i,i+|u|-1]\not\sim_{ab} u$. Set
    $m_n:=\mathbf{x}[i_n,i_{n+1}-1]$.
\end{lemma}

As shown in Example~\ref{exa1}, the factorization of $\mathbf{x}$ with elements in $\mathcal{APR}_{\mathbf{x},u}$ is not necessarily unique.

\begin{definition}
    We define a map
    $\mu_{\mathbf{x},u}:\mathcal{APR}_{\mathbf{x},u}\to\{{\tt
      1},\ldots,\#(\mathcal{APR}_{\mathbf{x},u})\}=:{A}_{\mathbf{x},u}$
    analogous to $\Lambda_{\mathbf{x},u}$.  The {\em abelian derived
      sequence} $\mathcal{E}_u(\mathbf{x})$ is the corresponding
    infinite word
$$\mu_{\mathbf{x},u}(m_0)\mu_{\mathbf{x},u}(m_1)\mu_{\mathbf{x},u}(m_2)\cdots$$
    over ${A}_{\mathbf{x},u}$ where the sequence $m_0m_1m_2\cdots\in \mathcal{APR}_{\mathbf{x},u}^\omega$ is the one computed in the previous lemma. 
    The inverse map $\mu_{\mathbf{x},u}^{-1}$ defines a morphism $\theta_{\mathbf{x},u}$ from ${A}^*_{\mathbf{x},u}$ to $\mathcal{APR}_{\mathbf{x},u}^*$
\end{definition}

Observe that $\mathcal{E}_u(\mathbf{x})$ is uniformly recurrent.
Indeed, if $a_1\cdots a_n$ is a factor occurring in
$\mathcal{E}_u(\mathbf{x})$, it comes from a factor $m_1\cdots
m_n\in\mathcal{APR}_{\mathbf{x},u}^*$ such that $m_1\cdots m_nv$ occurs
in $\mathbf{x}$ for some $v\sim_{ab} u$ and $\mu_{\mathbf{x},u}(m_1)\cdots
\mu_{\mathbf{x},u}(m_n)=a_1\cdots a_n$. Since $\mathbf{x}$ is
uniformly recurrent, $m_1\cdots m_nv$ occurs infinitely often with
bounded gaps in $\mathbf{x}$.

\begin{example}\label{exa1}
Consider the Thue--Morse word where the first few symbols are 
$$\mathbf{t}={\tt 01101001100101101001011001101001100101100110100101}\cdots.$$
Take the prefix $u={\tt 011}$. We get the derived sequence over ${R}_{\mathbf{t},u}=\{{\tt 1},\ldots,{\tt 4}\}$
$$\mathcal{D}_u(\mathbf{t})={\tt 12341243123431241234124312412343123412431234312412}\cdots$$
where the set of return words to $u$  in order of occurrence in $\mathbf{t}$ is given by
$$\mathcal{R}_{\mathbf{t},u}=\{{\tt 011010},\ {\tt 011001},\ {\tt 01101001},\ {\tt 0110}\}.$$
The abelian derived sequence over ${A}_{\mathbf{t},u}=\{{\tt 1},\ldots,{\tt 6}\}$ is
$$\mathcal{E}_u(\mathbf{t})={\tt 12314212521612314216125212314212521612521231421612}\cdots$$
where the set of abelian returns to $u$ in order of occurrence in $\mathbf{t}$ is given by
$$\mathcal{APR}_{\mathbf{t},u}=\{{\tt 0},\ {\tt 1},\ {\tt 1010},\ {\tt 1100},\ {\tt 10100},\ {\tt 110}\}.$$
Note that, since ${\tt 0}, {\tt 1}\in\mathcal{APR}_{\mathbf{t},u}$, there are infinitely many factorizations of $\mathbf{t}$ in terms of elements belonging to $\mathcal{APR}_{\mathbf{t},u}$.
\end{example}

\begin{proposition}
    Let $u$ be a prefix of a uniformly recurrent word $\mathbf{x}$.
    There exists a morphism $h_u$ from ${R}_{\mathbf{x},u}$ to
    ${A}_{\mathbf{x},u}^*$ such that
    $h_u(\mathcal{D}_u(\mathbf{x}))=\mathcal{E}_u(\mathbf{x})$.
\end{proposition}

\begin{proof}
    Each return word $m$ occurring in $\mathbf{x}$ is followed by $u$.
    Consider the procedure of Lemma~\ref{lem:algo} applied to $mu$. It will define
     the image by $h_u$ of $\Lambda_{\mathbf{x},u}(m)$. Indeed,
    one has to take into account a factor $u$ appended to $m$ because some suffix of $m$
    and a prefix of $u$ can give a word $v\sim_{ab} u$ leading to some
    abelian return in the decomposition of $m$. More precisely, we consider all the occurrences 
    $0=i_1&lt;\cdots&lt;i_t=|m|$ of factors abelian equivalent to $u$ in $w=mu$. Then 
    $$h_u(\Lambda_{\mathbf{x},u}(m)):=\mu_{\mathbf{x},u}(w[i_1,i_2-1])\cdots\mu_{\mathbf{x},u}(w[i_{t-1},i_t-1]).$$
\end{proof}

\begin{example}[Example~\ref{exa1} continued]
There exists a morphism $h_u$ from ${R}_{\mathbf{t},u}$ to ${A}_{\mathbf{t},u}^*$ such that 
$h_u(\mathcal{D}_u(\mathbf{t}))=\mathcal{E}_u(\mathbf{t})$. Take
$$h_u({\tt 1})={\tt 123},\ h_u({\tt 2})={\tt 142},\ h_u({\tt 3})={\tt 1252},\ h_u({\tt 4})={\tt 16}.$$
Let us explain how to get $h_u({\tt 2})$. We have the following factorization where the vertical bars
 indicate the occurrence of a factor abelian equivalent to $u$:
$$\Lambda^{-1}_{\mathbf{t},u}({\tt 2})\, u={\tt (|0|1100|1)\, 011}.$$
\end{example}

\begin{definition}
A map $h:A^\omega\to B^\omega$ is a {\em $t$-block morphism}, if there exists some map $f:A^t\to B^*$ 
such that, for all $\mathbf{w}\in A^\omega$,  
$$h(\mathbf{w})=f(\mathbf{w}[0,t-1])f(\mathbf{w}[1,t])f(\mathbf{w}[2,t+1])\cdots.$$
By abuse of notation, the second map $f$ will also be denoted by $h$.
\end{definition}

\begin{proposition}
    Let $u$ be a prefix of a uniformly recurrent word $\mathbf{x}$.
    Let $v$ be a prefix of $\mathbf{y}=\mathcal{E}_u(\mathbf{x})$. There exist 
    $t\le |u|-1$ and a $t$-block morphism $h_{u,v}:({A}_{\mathbf{y},v})^t\to {A}_{\mathbf{x},u}^*$ such that
    $$h_{u,v}(\mathcal{E}_v(\mathcal{E}_u(\mathbf{x})))=\mathcal{E}_u(\mathbf{x}).$$
\end{proposition}

\begin{proof}
    Note that any element $\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a))$
    with $a\in{A}_{\mathbf{y},v}$ is a concatenation of abelian
    returns to $u$.  Now consider a factor $a_0a_1\cdots a_{t-1}$
    occurring in $\mathcal{E}_v(\mathcal{E}_u(\mathbf{x}))$. We have
    to determine the unique factorization of
    $\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_0))$ with abelian
    returns to $u$ given by Lemma~\ref{lem:algo}. This one is
    completely determined when one knows the $|u|-1$ symbols occurring
    next. Without that extra knowledge we cannot uniquely determine
    the factorization for the last $|u|-1$ symbols possibly occurring in $\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_0))$.  This is the
    reason to consider the suffix $a_1\cdots a_{t-1}$ in such a way that
    $\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_1))\cdots
    \theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_{t-1}))$ has length at
    least $|u|-1$. One takes $t$ large enough to ensure this property for any initial symbol $a_0\in{A}_{\mathbf{y},v}$. More precisely, consider all the occurrences $0=i_1&lt;\cdots&lt;i_s$ 
    of factors abelian equivalent to $u$ in $w=\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_0))\cdots
    \theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_{t-1}))$. Let $r$ be the largest integer 
    such that $i_r&lt;|\theta_{\mathbf{x},u}(\theta_{\mathbf{y},v}(a_0))|$. Then 
    $$h_{u,v}(a_0\cdots a_{t-1}):=\mu_{\mathbf{x},u}(w[i_1,i_2-1])\cdots\mu_{\mathbf{x},u}(w[i_r,i_{r+1}-1]).$$
Note that the above definition is only meaningful if $a_0\cdots a_{t-1}$ is a factor of $\mathcal{E}_v(\mathcal{E}_u(\mathbf{x}))$. Since this is the only relevant situation, in any other case, the image of 
$h_{u,v}$ is set to $\varepsilon$.
\end{proof}

Observe that if we iterate the process, since the composition of a
$t$-block morphism and an $s$-block morphism is an $(st)$-block
morphism, then there exists an $r$-block morphism $h$ such that
$h(\mathcal{E}_{u_k}(\cdots (\mathcal{E}_{u_2}(
\mathcal{E}_{u_1}(\mathbf{x})))\cdots
))=\mathcal{E}_{u_1}(\mathbf{x})$ where prefixes $u_1,\ldots,u_k$ are
considered accordingly.

\begin{example}[Example~\ref{exa1} continued]
We can iterate the process of computing the abelian derived
sequence, for instance by taking each time the corresponding prefix of length $3$:
$$\mathcal{E}_{\tt 123}(\mathcal{E}_{\tt 011}(\mathbf{t}))={\tt 12131415121315141213141514121315121314151213151412}\cdots,$$
$$\mathcal{E}_{\tt 121}(\mathcal{E}_{\tt 123}(\mathcal{E}_{\tt 011}(\mathbf{t})))={\tt 12341243123431241234124312412343123412431234312412}\cdots,$$
$$\mathcal{E}_{\tt 123}(\mathcal{E}_{\tt 121}(\mathcal{E}_{\tt 123}(\mathcal{E}_{\tt 011}(\mathbf{t}))))={\tt 12341432123432141234143214123432123414321234321412}\cdots.$$
    Let us illustrate the previous result. Take again $u={\tt 011}$, $\mathbf{y}=\mathcal{E}_u(\mathbf{t})$, $v={\tt 123}$.
We have $$\mathcal{APR}_{\mathbf{y},v}=\{{\tt 1},\ {\tt 23142125216},\ {\tt 23142161252},\ {\tt 231421252161252},\ {\tt 2314216}\}.$$
Observe that $\theta_{\mathbf{t},u}(\theta_{\mathbf{y},v}({\tt 1}))= \theta_{\mathbf{t},u}({\tt 1})={\tt 0}$ and, 
for all $a\in\{{\tt 2},\ldots,{\tt 5}\}$, $\theta_{\mathbf{y},v}(a)$ has a prefix ${\tt 23}$, 
so $\theta_{\mathbf{t},u}(\theta_{\mathbf{y},v}(a))$ has prefix ${\tt 110}\sim_{ab} u$.
Let us assume that $h_{u,v}$ is a $3$-block morphism. We define $h_{u,v}({\tt 1}ab)=1$, 
for all $a,b\in A_{\mathbf{y},v}$ and $a\neq{\tt 1}$. We get $h_{u,v}({\tt 213})={\tt 23142125216}$ because, 
if vertical bars denote occurrences of a factor abelian equivalent to $u$, we get the following factorization:
$$\theta_{\mathbf{t},u}(\theta_{\mathbf{y},v}({\tt 2}))\, \theta_{\mathbf{t},u}(\theta_{\mathbf{y},v}({\tt 13}))= $$
$$
({\tt |1|1010|0|1100|1|0|1|10100|1|0|110})\, {\tt |011010011001011001101001}.$$
\end{example}

\begin{proposition}
    Let $\sigma$ be a primitive substitution and $u$ be a prefix of its fixed point $\mathbf{x}=\sigma(\mathbf{x})\in A^\omega$. 
    There exists a $2$-block morphism $\sigma_u:{A}_{\mathbf{x},u}^*\to{A}_{\mathbf{x},u}^*$ such that 
    $\mathcal{E}_u(\mathbf{x})$ is fixed point of $\sigma_u$ and 
    $$\theta_{\mathbf{x},u}(\sigma_u (\mathcal{E}_u(\mathbf{x})))=\sigma(\theta_{\mathbf{x},u}(\mathcal{E}_u(\mathbf{x}))).$$
\end{proposition}

\begin{proof}
    We may replace $\sigma$ by a convenient power of $\sigma$ in such
    a way that, for all $a\in A$, $\sigma(a)$ contains an occurrence
    of a factor abelian equivalent to $u$. For all $a,b\in{A}_{\mathbf{x},u}$, consider
    all the occurrences $i_1&lt;\cdots&lt;i_t$ of a factor abelian equivalent to $u$
    occurring in $w=\sigma(\theta_{\mathbf{x},u}(ab))$.  With our choice of
    $\sigma$, at least one of these $i_j$ belongs to
    $[0,|\sigma(\theta_{\mathbf{x},u}(a))|-1]$ (resp.
    $[|\sigma(\theta_{\mathbf{x},u}(a))|,|w|-1]$). Let $r$ be the largest
    integer such that $i_r&lt; |\sigma(\theta_{\mathbf{x},u}(a))|$. We define
    $$\sigma_u(ab)= \mu_{\mathbf{x},u}(w[i_1,i_2-1])\cdots
    \mu_{\mathbf{x},u}(w[i_r,i_{r+1}-1]).$$
\end{proof}

\begin{corollary}
    Let $\sigma$ be a primitive substitution
    and $u$ be a prefix of its fixed point $\mathbf{x}=\sigma(\mathbf{x})\in A^\omega$. The sequence
    $\mathcal{E}_u(\mathbf{x})$ is primitive substitutive, i.e., there exists a
    primitive morphism $\tau_u:B\to B^*$ and a coding $\phi:B\to
    {A}_{\mathbf{x},u}$ such that
    $\mathcal{E}_u(\mathbf{x})=\phi(\tau_u^\omega(b))$ for some $b\in
    B$.
\end{corollary}

\begin{proof}
    We may replace $\sigma$ by a convenient power of $\sigma$ in such
    a way that, for all $a\in A$, $\sigma(a)$ contains occurrences of
    two factors abelian equivalent to $u$.  Consider the alphabet $$B=\{(a,b)\mid
    a,b\in {A}_{\mathbf{x},u} \wedge ab \text{ is a factor of }\mathcal{E}_u(\mathbf{x})\}.$$ 

    For all $a,b\in{A}_{\mathbf{x},u}$ such that $(a,b)\in B$, consider
    all the occurrences $i_1&lt;\cdots&lt;i_t$ of a factor abelian equivalent to $u$
    occurring in $w=\theta_{\mathbf{x},u}(ab)$.  Let $r$ be the smallest
    integer such that $i_r\ge |\theta_{\mathbf{x},u}(a)|$. Note that $r\ge 3$.
    We define
    $$\tau_u((a,b))= (\mu_{\mathbf{x},u}(w[i_1,i_2-1]),\mu_{\mathbf{x},u}(w[i_2,i_3-1]))\cdots
    (\mu_{\mathbf{x},u}(w[i_{r-1},i_r-1]),\mu_{\mathbf{x},u}(w[i_{r},i_{r+1}-1])).$$
    Let $e_0e_1$ be the prefix of length $2$ of $\mathcal{E}_u(\mathbf{x})$. We have
    $$\mathcal{E}_u(\mathbf{x})=\phi(\tau_u^\omega((e_0,e_1)))$$
    where $\phi:B\to{A}_{\mathbf{x},u}$ is the coding that maps $(a,b)\in B$ to $a$.

Observe that, for all $(a,b)\in B$, $|\tau_u^n((a,b))|\ge 2^n$. Let us
show that $\tau_u$ is primitive. Since $\mathcal{E}_u(\mathbf{x})$ is
uniformly recurrent, there exists $K$ such that any factor of length
$K$ of $\mathcal{E}_u(\mathbf{x})$ contains all elements in $\{cd\mid
(c,d)\in B\}$. Therefore any factor of length $K$ of
$\tau_u^\omega((e_0,e_1))$ contains all the elements of $B$. Take $N$
such that $2^N\ge K$. Then, for all $(a,b),(c,d)\in B$,
$\tau_u^N((a,b))$ contains $(c,d)$ which means that $\tau_u$ is
primitive.
\end{proof}

\begin{example}[Example~\ref{exa1} continued]
    Take again $u={\tt 011}$ and the morphism 
    $\sigma:{\tt 0}\mapsto {\tt 01101001}, {\tt 1}\mapsto {\tt 10010110}$ generating $\mathbf{t}$. We have 
$$\sigma(\theta_{\mathbf{t},u}({\tt 12}))=({\tt |0|1|1010|0|1}) {\tt 100|1|0|110} \text{ and } \sigma_u({\tt 12})={\tt 12314}$$
$$\sigma(\theta_{\mathbf{t},u}({\tt 23}))=({\tt 100|1|0|1|10}) {\tt 100|1|0|110}\cdots \text{ and } \sigma_u({\tt 23})={\tt 2125}$$
$$\sigma(\theta_{\mathbf{t},u}({\tt 31}))=({\tt 100|1|0|110|0|1|1010|0|1100|1|0|110|0|1|10100|1}) {\tt |0|1|101001}$$
$$ \text{ and } \sigma_u({\tt 21})={\tt 216123142161252}.$$
Using the above corollary, we get
$$\tau_u(1,2)= (1,2)(2,3)(3,1)(1,4)(4,2),\ \tau_u(2,3)=(2,1)(1,2)(2,5)(5,2),\ldots.$$
\end{example}

\subsection{Abelian derivatives of the Thue--Morse word}

\begin{proposition}
    For the Thue--Morse word $\mathbf{t}$, the set $\{\mathcal{E}_u(\mathbf{t})\mid u \in\Pref(\mathbf{t})\}$ is infinite.
\end{proposition}

\begin{proof} 
	It is sufficient to show that the set $\{\mathcal{E}_u(\mathbf{t})\mid u \in\Pref(\mathbf{t}) : |u|\equiv 1\pmod{2}\}$
	is infinite. 
	Proceed by contradiction and suppose that the set $\{\mathcal{E}_u(\mathbf{t})\mid u \in\Pref(\mathbf{t}): |u|\equiv 1\pmod{2}\}$ is finite. 
	Then there exist $u$ and $v$ distinct prefixes of odd length of the Thue--Morse word $\mathbf{t}$ such that 
	$\mathcal{E}_u(\mathbf{t})=\mathcal{E}_v(\mathbf{t})$. Since $\mathcal{APR}_{\mathbf{t}}$ is finite, we can moreover assume that $\theta_{\mathbf{t},u}=\theta_{\mathbf{t},v}$. Indeed, infinitely many sequences of the kind $\mathcal{E}_u(\mathbf{t})$ are equal and thus defined on the same alphabet $A_{\mathbf{t},u}$. For all such sequences, there are finitely many morphisms of the kind $\theta_{\mathbf{t},u}$ associating with each element of $A_{\mathbf{t},u}$ an element of the finite set $\mathcal{APR}_{\mathbf{t}}$. So we can impose the extra condition on $\theta_{\mathbf{t},u}$.
Let $$I(w):=\{i\in \mathbb{N}\mid \mathbf{t}[i,i+|w|-1]\sim_{ab} w\}$$
    	denote the set of occurrences of factors of $\mathbf{t}$ abelian equivalent to a word $w$. We have $I(u)=I(v)$ as $\theta_{\mathbf{t},u}=\theta_{\mathbf{t},v}$. 
	Without loss of generality, we may suppose that $|u|=2k+1$, 
	$|v|=2\ell+1$ with $k&lt;\ell$. We have $\Psi(u)=(k,k+1)$ or $\Psi(u)=(k+1,k)$ and $\Psi(v)=(\ell,\ell+1)$ 
	or $\Psi(v)=(\ell+1,\ell)$. Let $a_u$ (resp. $a_v$) denote the letter having $k+1$ (resp. 
	$\ell+1$) occurrences in the prefix $u$ (resp. $v$). Note that $a_u$ and $a_v$ are 
	respectively the last letters of $u$ and $v$. 

For any odd position $j$, recalling that $\mathbf{t}[2m,2m+1]\in\{{\tt 10},{\tt 01}\}$, we have
	$$\mathbf{t}_j=a_u \Leftrightarrow \mathbf{t}[j,j+|u|-1]\sim_{ab} u
	\Leftrightarrow\mathbf{t}[j,j+|v|-1]\sim_{ab} v \Leftrightarrow \mathbf{t}_j=a_v$$
where the central equivalence comes from the fact that $I(u)=I(v)$. 
	As there exists at least one such $j$, we have $a_u=a_v=:a$. 

For any even position $j$, we have 
$$\mathbf{t}_{j+|u|-1}=a\Leftrightarrow j\in I(u) \Leftrightarrow j\in I(v)\Leftrightarrow \mathbf{t}_{j+|v|-1}=a$$
since $I(u)=I(v)$.
	Using this observation, we can show by induction that $$\mathbf{t}_{|u|-1+n(|v|-|u|)}=a$$ 
	for all $n\in\mathbb{N}$. In other words, there exists a constant infinite arithmetical subsequence in $\mathbf{t}$, 
	which is a contradiction, since it is well-known that the Thue-Morse word does not contain any such subsequence.
	Indeed, for $n=0$, it is clear that the last letter of $u$ is $a$. Suppose now 
	that the result holds true for $n\ge 0$. We have $\mathbf{t}_{|u|-1+n(|v|-|u|)}=a$. Since $|u|,|v|$ are odd,  
	$n(|v|-|u|)$ is an even number and belongs to $I(u)=I(v)$. Therefore $\mathbf{t}_{|v|-1+n(|v|-|u|)}=a$ and 
	$|v|-1+n(|v|-|u|)=|u|-1+(n+1)(|v|-|u|)$. 
\end{proof}

\begin{remark}
    Using the same notation as in the previous proof, we show that the set $\{\mathcal{E}_u(\mathbf{t})\mid u \in\Pref(\mathbf{t}) : |u|\equiv 0\pmod{2}\}$ is infinite. Proceed by contradiction. Then there exist $u$ and $v$ distinct prefixes of even length of $\mathbf{t}$ such that $\mathcal{E}_u(\mathbf{t})=\mathcal{E}_v(\mathbf{t})$ and $\theta_{\mathbf{t},u}=\theta_{\mathbf{t},v}$. Hence $I(u)=I(v)$. Note that $ 2\mathbb{N}\subseteq I(u)=I(v)$.
    Suppose that $|u|=2k$, $|v|=2\ell$, with $k &lt; \ell$. Since a prefix of even length has a Parikh vector of the kind $(r,r)$, $2i+1$ is in $I(u)$ if and only if $\mathbf{t}_i=\mathbf{t}_{i+k}$. Similarly, $2i+1$ is in $I(v)$ if and only if $\mathbf{t}_i=\mathbf{t}_{i+\ell}$.
    From $I(u)\setminus 2\mathbb{N} = I(v)\setminus 2\mathbb{N}$, we deduce that, for all $i\in\mathbb{N}$, $\mathbf{t}_i=\mathbf{t}_{i+k}$ implies $ \mathbf{t}_i=\mathbf{t}_{i+\ell}$ and conversely.
     This leads to the contradiction that $\mathbf{t}$ is ultimately periodic of period $\ell-k$. Indeed, suppose to the contrary that for some $i$, $\mathbf{t}_{i+k}\neq\mathbf{t}_{i+\ell}$. In this case, either $\mathbf{t}_{i+k}$ or $\mathbf{t}_{i+\ell}$ is equal to $\mathbf{t}_i$. From our last deduction, we get that all three letters $\mathbf{t}_i$, $\mathbf{t}_{i+k}$, $\mathbf{t}_{i+\ell}$ are equal. 
\end{remark}

\begin{remark} 
    Using the same notation as in the previous remark, there exist no prefixes $u,v$ of $\mathbf{t}$ such that $|u|$
    is even, $|v|$ is odd and $I(u)=I(v)$. (The symmetric
    case can be treated in the same way.) Assume that $|u|=2k$,
    $|v|=2\ell+1$ for some positive integers $k\ne\ell$. We get
    $\Psi(u)=(k,k)$ and $\Psi(v)=(\ell,\ell+1)$ or
    $\Psi(v)=(\ell+1,\ell)$. Let $a$ denote the letter that has
    $\ell+1$ occurrences in $v$. As $v\in\Pref(\mathbf{t})$,
    $\mathbf{t}_{|v|-1}=a$. Note that, for all even positions $j$,
    if $j$ is in $I(v)$, then $\mathbf{t}_{j+|v|-1}=a$.
Moreover, for all even $j$, we have $j\in I(u)$. 
Since $I(u)=I(v)$, we get $2\mathbb{N}\subseteq I(v)$ and thus  $\mathbf{t}_{j+|v|-1}= a$ for all even $j$.
    Therefore, for all even $j\ge |v|-1$, we have 
    $\mathbf{t}_j=a$ and also $\mathbf{t}_{j+1}={\tt 1}-a$ since $\mathbf{t}$ is made up of blocks ${\tt 01}$ or ${\tt 10}$. This 
    means that the Thue--Morse word is ultimately periodic of period $2$ which is a contradiction.
\end{remark}

\section{Acknowledgment}

The authors are very pleased to contribute to this special issue in
honor of Jean-Paul Allouche. This is a nice way to wish him 
a happy birthday with some combinatorial flavor.
In particular, the first author
is very grateful to Jean-Paul Allouche who is always  enthusiastic,
positive and ready to usefully comment on his research work.

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 68R15; Secondary 68Q68.

\noindent \emph{Keywords: } 
return word, recurrence, abelian equivalence, Sturmian word.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A001911},
\seqnum{A003622},
\seqnum{A003849},
\seqnum{A010060}, and
\seqnum{A022342}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received  June 2 2012;
revised version received   September 24 2012.
Published in {\it Journal of Integer Sequences}, March 2 2013.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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