# Markoff–Lagrange spectrum of one-sided shifts

Hajime Kaneko, Wolfgang Steiner

► To cite this version:

Hajime Kaneko, Wolfgang Steiner. Markoff–Lagrange spectrum of one-sided shifts. Mathematika, 2024, 70 (3), pp.e12250. 10.1112/mtk.12250. hal-04598637

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# Markoff–Lagrange spectrum of one-sided shifts

Hajime Kaneko, Wolfgang Steiner

► To cite this version:

Hajime Kaneko, Wolfgang Steiner. Markoff–Lagrange spectrum of one-sided shifts. Mathematika, 2024, 70 (3), pp.e12250. 10.1112/mtk.12250. hal-04598637

HAL Id: hal-04598637

https://u-paris.hal.science/hal-04598637v1

Submitted on 3 Jun 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
# Markoff-Lagrange spectrum of one-sided shifts

Hajime Kaneko¹ | Wolfgang Steiner²

¹Institute of Mathematics / Research Core for Mathematical Sciences, University of Tsukuba, Tsukuba, Japan

²Université Paris Cité, CNRS, IRIF, Paris, France

## Correspondence

Wolfgang Steiner, Université Paris Cité, CNRS, IRIF, F-75006 Paris, France.

Email: steiner@irif.fr

## Funding information

Agence Nationale de la Recherche, Grant/Award Numbers:

ANR-18-CE40-0007, ANR-22-CE40-0011, ANR-23-CE40-0024-01; JSPS KAKENHI, Grant/Award Number: 19K03439

## Abstract

For the Lagrange spectrum and other applications, we determine the smallest accumulation point of binary sequences that are maximal in their shift orbits. This problem is trivial for the lexicographic order, and its solution is the fixed point of a substitution for the alternating lexicographic order. For orders defined by cylinders, we show that the solutions are S-adic sequences, where S is a certain infinite set of substitutions that contains Sturmian morphisms. We also consider a similar problem for symmetric ternary shifts, which is applicable to the multiplicative version of the Markoff-Lagrange spectrum.

## MSC 2020

11J06, 37B10 (primary), 11A63, 11K16 (secondary)

# 1 | INTRODUCTION

Let $A^\infty$ be the set of infinite words on a finite alphabet $A$, equipped with a total order $\le$ and the ultrametric $d$ given by $d(a_1a_2\cdots, b_1b_2\cdots) = 2^{-\min\{n\ge1 : a_n\ne b_n\}}$ for $a_1a_2\cdots \ne b_1b_2\cdots$. We study properties of the set of sup-words

$$ \mathcal{M}_{\le} := \{s_{\le}(\mathbf{a}) : \mathbf{a} \in A^{\infty}\}, \quad \text{with} \quad s_{\le}(a_1a_2\cdots) := \sup_{n\ge1} a_n a_{n+1}\cdots, $$

for a large class of orders on $A^\infty$. In particular, we are interested in the smallest accumulation point $\mathbf{m}_{\le} \in \mathcal{M}_{\le}$. For the lexicographic order $\le_{lex}$, words in $\mathcal{M}_{\le_{lex}}$ occur as (quasi-greedy) $\beta$-expansions of 1 for real bases $\beta > 1$ (see [20]), with $\mathbf{m}_{\le_{lex}} = 1000\cdots$ being the limit of these expansions as $\beta \to 1$. For the alternating lexicographic order $\le_{alt}$, most elements of $\mathcal{M}_{\le_{alt}}$ are $(-\beta)$-expansions
# Markoff-Lagrange spectrum of one-sided shifts

Hajime Kaneko¹ | Wolfgang Steiner²

¹Institute of Mathematics / Research Core for Mathematical Sciences, University of Tsukuba, Tsukuba, Japan

²Université Paris Cité, CNRS, IRIF, Paris, France

## Correspondence

Wolfgang Steiner, Université Paris Cité, CNRS, IRIF, F-75006 Paris, France.

Email: steiner@irif.fr

## Funding information

Agence Nationale de la Recherche, Grant/Award Numbers:

ANR-18-CE40-0007, ANR-22-CE40-0011, ANR-23-CE40-0024-01; JSPS KAKENHI, Grant/Award Number: 19K03439

## Abstract

For the Lagrange spectrum and other applications, we determine the smallest accumulation point of binary sequences that are maximal in their shift orbits. This problem is trivial for the lexicographic order, and its solution is the fixed point of a substitution for the alternating lexicographic order. For orders defined by cylinders, we show that the solutions are S-adic sequences, where S is a certain infinite set of substitutions that contains Sturmian morphisms. We also consider a similar problem for symmetric ternary shifts, which is applicable to the multiplicative version of the Markoff-Lagrange spectrum.

## MSC 2020

11J06, 37B10 (primary), 11A63, 11K16 (secondary)

# 1 | INTRODUCTION

Let $A^\infty$ be the set of infinite words on a finite alphabet $A$, equipped with a total order $\le$ and the ultrametric $d$ given by $d(a_1a_2\cdots, b_1b_2\cdots) = 2^{-\min\{n\ge1 : a_n\ne b_n\}}$ for $a_1a_2\cdots \ne b_1b_2\cdots$. We study properties of the set of sup-words

$$ \mathcal{M}_{\le} := \{s_{\le}(\mathbf{a}) : \mathbf{a} \in A^{\infty}\}, \quad \text{with} \quad s_{\le}(a_1a_2\cdots) := \sup_{n\ge1} a_n a_{n+1}\cdots, $$

for a large class of orders on $A^\infty$. In particular, we are interested in the smallest accumulation point $\mathbf{m}_{\le} \in \mathcal{M}_{\le}$. For the lexicographic order $\le_{lex}$, words in $\mathcal{M}_{\le_{lex}}$ occur as (quasi-greedy) $\beta$-expansions of 1 for real bases $\beta > 1$ (see [20]), with $\mathbf{m}_{\le_{lex}} = 1000\cdots$ being the limit of these expansions as $\beta \to 1$. For the alternating lexicographic order $\le_{alt}$, most elements of $\mathcal{M}_{\le_{alt}}$ are $(-\beta)$-expansions

© 2024 The Authors. *Mathematika* is copyright © University College London and published by the London Mathematical Society on behalf of University College London. This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited.
of $\frac{-\beta}{\beta+1}$ in the sense of [16, 24], and again, $\mathbf{m}_{\leq \text{alt}}$ is the limit of these expansions as $\beta \to 1$. An image of $\mathcal{M}_{\leq \text{alt}}$ occurs in a multiplicative version of the (Markoff-)Lagrange spectrum w.r.t. an integer base, which is defined in terms of well approximable numbers [2, 12]; see Proposition 10. Below the image of $\mathbf{m}_{\leq \text{alt}}$, which is the fixed point of a substitution [3, 5, 13], we find the discrete part of this spectrum. The classical Markoff and Lagrange spectra are given by two-sided versions of $\mathcal{M}_{\leq \text{alt}}$ (and the Lagrange spectrum is defined by $\limsup$ instead of $\sup$). The unimodal order $\leq_{\text{uni}}$ yields kneading sequences of unimodal maps [19], and $\mathbf{m}_{\leq_{\text{uni}}}$ is the fixed point of the period-doubling (or Feigenbaum) substitution. Sup-words are also closely related to infinite Lyndon words, which are defined by $a_1 a_2 \cdots < a_n a_{n+1} \cdots$ for all $n \geq 2$; see [23].

We consider orders satisfying that

$$ \mathbf{a} \leq \mathbf{b} \leq \mathbf{c} \quad \text{implies} \quad d(\mathbf{a}, \mathbf{b}) \leq d(\mathbf{a}, \mathbf{c}) \quad \text{for all } \mathbf{a}, \mathbf{b}, \mathbf{c} \in A^{\infty} \quad (1) $$

(note that $d(\mathbf{a}, \mathbf{b}) \leq d(\mathbf{a}, \mathbf{c})$ is equivalent to $d(\mathbf{b}, \mathbf{c}) \leq d(\mathbf{a}, \mathbf{c})$ by the strong triangle inequality), and we call them cylinder orders because the elements of each cylinder of words are contiguous. Here, the cylinder (of length $n$) given by $a_1 \cdots a_n \in A^n$ is

$$ [a_1 \cdots a_n] := \{a'_1 a'_2 \cdots \in A^{\infty} : a'_1 \cdots a'_n = a_1 \cdots a_n\}, $$

and we write $[a_1 \cdots a_n] < [b_1 \cdots b_n]$ if $(a_1 \cdots a_n)^{\infty} < (b_1 \cdots b_n)^{\infty}$ (or, equivalently, $\mathbf{a} < \mathbf{b}$ for all $\mathbf{a} \in [a_1 \cdots a_n], \mathbf{b} \in [b_1 \cdots b_n]$), where $w^{\infty}$ denotes the infinite repetition of a word $w$. Note that (1) is equivalent to the condition (*) of [21], which states that $ux < vy$ for all $x, y \in A^{\infty}$, $u, v \in A^n$, $n \geq 1$, such that $u^{\infty} < v^{\infty}$. Generalized lexicographic orders as considered in [22] satisfy this condition.

In Section 2, we give basic properties of $\mathcal{M}_{\leq}$. Section 3 contains our main result, an algorithm for determining the smallest accumulation point $\mathbf{m}_{\leq}$ of $\mathcal{M}_{\leq}$, for any cylinder order $(A^{\infty}, \leq)$. We also give a complete description, in terms of S-adic sequences, of all words $\mathbf{m}_{\leq}$ obtained by cylinder orders and of the discrete part of $\mathcal{M}_{\leq}$, that is, all its elements below $\mathbf{m}_{\leq}$. Since the words $\mathbf{m}_{\leq}$ have linear factor complexity, real numbers having such $\beta$-expansions are in $\mathbb{Q}(\beta)$ or transcendental, for all Pisot or Salem bases $\beta \geq 2$. In Section 4, we determine $\mathbf{m}_{\leq}$ for some classical examples of cylinder orders and show that all Sturmian sequences that are sup-words can occur. We consider cylinder orders on symmetric alphabets in Section 5 and apply our results to the multiplicative Lagrange spectrum and other problems in Section 6,

## 2 | MARKOFF-LAGRANGE SPECTRUM

We first show that the set of periodic words in $\mathcal{M}_{\leq}$ is dense, and that $\mathcal{M}_{\leq}$ is equal to

$$ \mathcal{L}_{\leq} := \{\ell_{\leq}(\mathbf{a}) : \mathbf{a} \in A^{\infty}\}, \quad \text{with} \quad \ell_{\leq}(a_1 a_2 \cdots) := \limsup_{n \to \infty} a_n a_{n+1} \cdots. $$

Note that $\mathcal{M}_{\leq}$ and $\mathcal{L}_{\leq}$ can be seen as generalizations of the *Markoff and Lagrange spectrum*, respectively. In the classical case, these spectra are defined by two-sided sequences, and the Lagrange spectrum is a strict subset of the Markoff spectrum [11, 14].

**Theorem 1.** Let $\leq$ be a cylinder order on $A^{\infty}$. Then

$$ \mathcal{L}_{\leq} = \mathcal{M}_{\leq} = \text{closure}(\{s_{\leq}(\mathbf{a}) : \mathbf{a} \in A^{\infty} \text{ purely periodic}\}). $$

In particular, the set $\mathcal{M}_{\leq}$ is closed.
of $\frac{-\beta}{\beta+1}$ in the sense of [16, 24], and again, $\mathbf{m}_{\leqslant \text{alt}}$ is the limit of these expansions as $\beta \to 1$. An image of $\mathcal{M}_{\leqslant \text{alt}}$ occurs in a multiplicative version of the (Markoff-)Lagrange spectrum w.r.t. an integer base, which is defined in terms of well approximable numbers [2, 12]; see Proposition 10. Below the image of $\mathbf{m}_{\leqslant \text{alt}}$, which is the fixed point of a substitution [3, 5, 13], we find the discrete part of this spectrum. The classical Markoff and Lagrange spectra are given by two-sided versions of $\mathcal{M}_{\leqslant \text{alt}}$ (and the Lagrange spectrum is defined by $\limsup$ instead of $\sup$). The unimodal order $\leqslant_{\text{uni}}$ yields kneading sequences of unimodal maps [19], and $\mathbf{m}_{\leqslant_{\text{uni}}}$ is the fixed point of the period-doubling (or Feigenbaum) substitution. Sup-words are also closely related to infinite Lyndon words, which are defined by $a_1 a_2 \cdots < a_n a_{n+1} \cdots$ for all $n \geq 2$; see [23].

We consider orders satisfying that

$$ \mathbf{a} \leq \mathbf{b} \leq \mathbf{c} \quad \text{implies} \quad d(\mathbf{a}, \mathbf{b}) \leq d(\mathbf{a}, \mathbf{c}) \quad \text{for all } \mathbf{a}, \mathbf{b}, \mathbf{c} \in A^{\infty} \quad (1) $$

(note that $d(\mathbf{a}, \mathbf{b}) \leq d(\mathbf{a}, \mathbf{c})$ is equivalent to $d(\mathbf{b}, \mathbf{c}) \leq d(\mathbf{a}, \mathbf{c})$ by the strong triangle inequality), and we call them cylinder orders because the elements of each cylinder of words are contiguous. Here, the cylinder (of length $n$) given by $a_1 \cdots a_n \in A^n$ is

$$ [a_1 \cdots a_n] := \{a'_1 a'_2 \cdots \in A^{\infty} : a'_1 \cdots a'_n = a_1 \cdots a_n\}, $$

and we write $[a_1 \cdots a_n] < [b_1 \cdots b_n]$ if $(a_1 \cdots a_n)^{\infty} < (b_1 \cdots b_n)^{\infty}$ (or, equivalently, $\mathbf{a} < \mathbf{b}$ for all $\mathbf{a} \in [a_1 \cdots a_n], \mathbf{b} \in [b_1 \cdots b_n]$), where $w^{\infty}$ denotes the infinite repetition of a word $w$. Note that (1) is equivalent to the condition (*) of [21], which states that $ux < vy$ for all $x, y \in A^{\infty}, u, v \in A^n, n \geq 1$, such that $u^{\infty} < v^{\infty}$. Generalized lexicographic orders as considered in [22] satisfy this condition.

In Section 2, we give basic properties of $\mathcal{M}_{\leqslant}$. Section 3 contains our main result, an algorithm for determining the smallest accumulation point $\mathbf{m}_{\leqslant}$ of $\mathcal{M}_{\leqslant}$, for any cylinder order $(A^{\infty}, \leqslant)$. We also give a complete description, in terms of S-adic sequences, of all words $\mathbf{m}_{\leqslant}$ obtained by cylinder orders and of the discrete part of $\mathcal{M}_{\leqslant}$, that is, all its elements below $\mathbf{m}_{\leqslant}$. Since the words $\mathbf{m}_{\leqslant}$ have linear factor complexity, real numbers having such $\beta$-expansions are in $\mathbb{Q}(\beta)$ or transcendental, for all Pisot or Salem bases $\beta \geq 2$. In Section 4, we determine $\mathbf{m}_{\leqslant}$ for some classical examples of cylinder orders and show that all Sturmian sequences that are sup-words can occur. We consider cylinder orders on symmetric alphabets in Section 5 and apply our results to the multiplicative Lagrange spectrum and other problems in Section 6,

## 2 | MARKOFF-LAGRANGE SPECTRUM

We first show that the set of periodic words in $\mathcal{M}_{\leqslant}$ is dense, and that $\mathcal{M}_{\leqslant}$ is equal to

$$ \mathcal{L}_{\leqslant} := \{\ell_{\leqslant}(\mathbf{a}) : \mathbf{a} \in A^{\infty}\}, \quad \text{with} \quad \ell_{\leqslant}(a_1 a_2 \cdots) := \limsup_{n \to \infty} a_n a_{n+1} \cdots. $$

Note that $\mathcal{M}_{\leqslant}$ and $\mathcal{L}_{\leqslant}$ can be seen as generalizations of the *Markoff and Lagrange spectrum*, respectively. In the classical case, these spectra are defined by two-sided sequences, and the Lagrange spectrum is a strict subset of the Markoff spectrum [11, 14].

**Theorem 1.** Let $\leqslant$ be a cylinder order on $A^{\infty}$. Then

$$ \mathcal{L}_{\leqslant} = \mathcal{M}_{\leqslant} = \text{closure}(\{s_{\leqslant}(\mathbf{a}) : \mathbf{a} \in A^{\infty} \text{ purely periodic}\}). $$

In particular, the set $\mathcal{M}_{\leqslant}$ is closed.
In the proof of the theorem, we use the following characterization of cylinder orders.

**Lemma 2.** An order $\le$ on $A^\infty$ is a cylinder order if and only if

$$ \mathbf{a} \le \mathbf{b} \text{ implies } \mathbf{a}' \le \mathbf{b}' \text{ for all } \mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}' \in A^\infty \text{ with } \max(d(\mathbf{a}, \mathbf{a}'), d(\mathbf{b}, \mathbf{b}')) < d(\mathbf{a}, \mathbf{b}). \quad (2) $$

*Proof.* Let $\le$ be a cylinder order, and $\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}' \in A^\infty$ such that $\mathbf{a} \le \mathbf{b}, d(\mathbf{a}, \mathbf{a}') < d(\mathbf{a}, \mathbf{b}), d(\mathbf{b}, \mathbf{b}') < d(\mathbf{a}, \mathbf{b})$. Then both $\mathbf{a} \le \mathbf{b} \le \mathbf{a}'$ and $\mathbf{b}' \le \mathbf{a}' \le \mathbf{b}$ are impossible by (1), using that $d(\mathbf{a}', \mathbf{b}') = d(\mathbf{a}, \mathbf{b})$ by the strong triangle inequality. This implies that $\mathbf{b} > \mathbf{a}'$ and thus $\mathbf{b}' > \mathbf{a}'$, that is, (2) holds. Let now $\le$ be an order satisfying (2). Then, $\mathbf{a} \le \mathbf{b}$ and $d(\mathbf{a}, \mathbf{c}) < d(\mathbf{a}, \mathbf{b})$ imply that $\mathbf{c} < \mathbf{b}$, thus $\mathbf{a} \le \mathbf{b} \le \mathbf{c}$ with $d(\mathbf{a}, \mathbf{c}) < d(\mathbf{a}, \mathbf{b})$ is impossible, that is, (1) holds. $\square$

*Proof of Theorem 1.* We show first that $M_\le$ is the closure of $s_\le(\mathbf{a}')$ with purely periodic $\mathbf{a}' \in A^\infty$. Since $s_\le(s_\le(\mathbf{a})) = s_\le(\mathbf{a})$, it suffices to consider $\mathbf{a} \in A^\infty$ with $s_\le(\mathbf{a}) = \mathbf{a}$. We have $(a_1 \cdots a_n)^\infty \in M_\le$ whenever

$$ 2^{-n} d(\mathbf{a}, a_{n+1} a_{n+2} \cdots) < 2^{-i} d(\mathbf{a}, a_{i+1} a_{i+2} \cdots) \quad \text{for all } i \in [1, n). \quad (3) $$

Indeed, we have, for all $i \in [1, n)$, that $a_{i+1} a_{i+2} \cdots \le \mathbf{a}$ (because $s_\le(\mathbf{a}) = \mathbf{a}$) and

$$ \begin{aligned} d(\mathbf{a}, a_{i+1} a_{i+2} \cdots) &> 2^{i-n} d(\mathbf{a}, a_{n+1} a_{n+2} \cdots) = 2^{i-n} d((a_1 \cdots a_n)^\infty, a_{n+1} a_{n+2} \cdots) \\ & = 2^i d((a_1 \cdots a_n)^\infty, \mathbf{a}) = d(a_{i+1} \cdots a_n (a_1 \cdots a_n)^\infty, a_{i+1} a_{i+2} \cdots), \end{aligned} $$

hence (2) gives that $a_{i+1} \cdots a_n (a_1 \cdots a_n)^\infty \le (a_1 \cdots a_n)^\infty$, thus $(a_1 \cdots a_n)^\infty \in M_\le$. Since $\lim_{n \to \infty} 2^{-n} d(\mathbf{a}, a_{n+1} a_{n+2} \cdots) = 0$, we have either $d(\mathbf{a}, a_{n+1} a_{n+2} \cdots) = 0$ for some $n \ge 1$, that is, $\mathbf{a}$ is purely periodic, or there are infinitely many $n$ such that (3) holds and thus $(a_1 \cdots a_n)^\infty \in M_\le$ infinitely often. Therefore, $M_\le$ is contained in the closure of $\{s_\le(\mathbf{a}) : \mathbf{a} \in A^\infty \text{ purely periodic}\}$. For the opposite inclusion, we have to show that $M_\le$ is closed. Consider $\mathbf{a} = \lim_{k \to \infty} \mathbf{a}^{(k)}$ with $\mathbf{a}^{(k)} \in M_\le$. If $a_{n+1} a_{n+2} \cdots > \mathbf{a}$ for some $n \ge 1$, then we would also have $a_{n+1}^{(k)} a_{n+2}^{(k)} > \mathbf{a}^{(k)}$ for all large enough $k$, contradicting that $\mathbf{a}^{(k)} \in M_\le$. This implies that $s_\le(\mathbf{a}) = \mathbf{a}$, that is, $\mathbf{a} \in M_\le$.

Since $s_\le(\ell_\le(\mathbf{a})) = \ell_\le(\mathbf{a})$ for all $\mathbf{a} \in A^\infty$, we have $\mathcal{L}_\le \subseteq M_\le$. For the opposite inclusion, let $\mathbf{a} = \lim_{k \to \infty} \mathbf{a}^{(k)}$ for some purely periodic words $\mathbf{a}^{(k)} \in M_\le$, let $(p_k)$ be an increasing sequence satisfying $\mathbf{a}^{(k)} = (a_1^{(k)} \cdots a_{p_k}^{(k)})^\infty$, and let $\mathbf{b} = a_1^{(1)} \cdots a_{p_1}^{(1)} a_1^{(2)} \cdots a_{p_2}^{(2)} \cdots$. Then, $\ell_\le(\mathbf{b}) \ge \mathbf{a}$. If $a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{p_{k+1}}^{(k+1)} > \mathbf{a}$ for some $k \ge 1$, $0 \le i < p_k$, then

$$ \delta_{i,k} := d(a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{p_{k+1}}^{(k+1)} \cdots, \mathbf{a}) \le \max(d(\mathbf{a}^{(k)}, \mathbf{a}), d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)}), 2^{-p_k}). \quad (4) $$

Indeed, for $1 \le j \le p_k - i$, we cannot have $[a_{i+1}^{(k)} \cdots a_{i+j}^{(k)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}]$ because this contradicts $\mathbf{a}^{(k)} \in M_\le$, thus $\delta_{i,k} = 2^{-j}$ implies $d(\mathbf{a}^{(k)}, \mathbf{a}) \ge 2^{-j}$; similarly, for $p_k - i < j \le 2p_k - i$, $[a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k)} \cdots a_{i+j-p_k}^{(k)}] = [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{i+j-p_k}^{(k+1)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}]$

$$ [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k)} \cdots a_{i+j-p_k}^{(k)}] = [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{i+j-p_k}^{(k+1)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}] $$

is impossible, thus $\delta_{i,k} = 2^{-j}$ implies $d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)}) \ge 2^{p_k - i - j}$ or $d(\mathbf{a}^{(k)}, \mathbf{a}) \ge 2^{-j}$; hence, we have $\delta_{i,k} < 2^{i-2p_k}$ or $\delta_{i,k} \le 2^{i-p_k} d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)})$ or $\delta_{i,k} \le d(\mathbf{a}^{(k)}, \mathbf{a})$, which implies (4). Since the right-hand side of (4) tends to 0 as $k \to \infty$, we have $\ell_\le(\mathbf{b}) \le \mathbf{a}$, thus $\mathbf{a} = \ell_\le(\mathbf{b}) \in \mathcal{L}_\le$. $\square$
In the proof of the theorem, we use the following characterization of cylinder orders.

**Lemma 2.** An order $\le$ on $A^\infty$ is a cylinder order if and only if

$$ \mathbf{a} \le \mathbf{b} \text{ implies } \mathbf{a}' \le \mathbf{b}' \quad \text{for all } \mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}' \in A^\infty \text{ with } \max(d(\mathbf{a}, \mathbf{a}'), d(\mathbf{b}, \mathbf{b}')) < d(\mathbf{a}, \mathbf{b}). \quad (2) $$

*Proof.* Let $\le$ be a cylinder order, and $\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}' \in A^\infty$ such that $\mathbf{a} \le \mathbf{b}, d(\mathbf{a}, \mathbf{a}') < d(\mathbf{a}, \mathbf{b}), d(\mathbf{b}, \mathbf{b}') < d(\mathbf{a}, \mathbf{b})$. Then both $\mathbf{a} \le \mathbf{b} \le \mathbf{a}'$ and $\mathbf{b}' \le \mathbf{a}' \le \mathbf{b}$ are impossible by (1), using that $d(\mathbf{a}', \mathbf{b}') = d(\mathbf{a}, \mathbf{b})$ by the strong triangle inequality. This implies that $\mathbf{b} > \mathbf{a}'$ and thus $\mathbf{b}' > \mathbf{a}'$, that is, (2) holds. Let now $\le$ be an order satisfying (2). Then, $\mathbf{a} \le \mathbf{b}$ and $d(\mathbf{a}, \mathbf{c}) < d(\mathbf{a}, \mathbf{b})$ imply that $\mathbf{c} < \mathbf{b}$, thus $\mathbf{a} \le \mathbf{b} \le \mathbf{c}$ with $d(\mathbf{a}, \mathbf{c}) < d(\mathbf{a}, \mathbf{b})$ is impossible, that is, (1) holds. $\square$

*Proof of Theorem 1.* We show first that $M_\le$ is the closure of $s_\le(\mathbf{a}')$ with purely periodic $\mathbf{a}' \in A^\infty$. Since $s_\le(s_\le(\mathbf{a})) = s_\le(\mathbf{a})$, it suffices to consider $\mathbf{a} \in A^\infty$ with $s_\le(\mathbf{a}) = \mathbf{a}$. We have $(a_1 \cdots a_n)^\infty \in M_\le$ whenever

$$ 2^{-n}d(\mathbf{a}, a_{n+1}a_{n+2}\cdots) < 2^{-i}d(\mathbf{a}, a_{i+1}a_{i+2}\cdots) \quad \text{for all } i \in [1, n). \quad (3) $$

Indeed, we have, for all $i \in [1, n)$, that $a_{i+1}a_{i+2}\cdots \le \mathbf{a}$ (because $s_\le(\mathbf{a}) = \mathbf{a}$) and

$$ \begin{aligned} d(\mathbf{a}, a_{i+1}a_{i+2}\cdots) &> 2^{i-n}d(\mathbf{a}, a_{n+1}a_{n+2}\cdots) = 2^{i-n}d((a_1\cdots a_n)^\infty, a_{n+1}a_{n+2}\cdots) \\ & = 2^i d((a_1\cdots a_n)^\infty, \mathbf{a}) = d(a_{i+1}\cdots a_n(a_1\cdots a_n)^\infty, a_{i+1}a_{i+2}\cdots), \end{aligned} $$

hence (2) gives that $a_{i+1}\cdots a_n(a_1\cdots a_n)^\infty \le (a_1\cdots a_n)^\infty$, thus $(a_1\cdots a_n)^\infty \in M_\le$. Since $\lim_{n\to\infty} 2^{-n}d(\mathbf{a}, a_{n+1}a_{n+2}\cdots) = 0$, we have either $d(\mathbf{a}, a_{n+1}a_{n+2}\cdots) = 0$ for some $n \ge 1$, that is, $\mathbf{a}$ is purely periodic, or there are infinitely many $n$ such that (3) holds and thus $(a_1\cdots a_n)^\infty \in M_\le$ infinitely often. Therefore, $M_\le$ is contained in the closure of $\{s_\le(\mathbf{a}) : \mathbf{a} \in A^\infty \text{ purely periodic}\}$. For the opposite inclusion, we have to show that $M_\le$ is closed. Consider $\mathbf{a} = \lim_{k\to\infty} \mathbf{a}^{(k)}$ with $\mathbf{a}^{(k)} \in M_\le$. If $a_{n+1}a_{n+2}\cdots > \mathbf{a}$ for some $n \ge 1$, then we would also have $a_{n+1}^{(k)}a_{n+2}^{(k)} > \mathbf{a}^{(k)}$ for all large enough $k$, contradicting that $\mathbf{a}^{(k)} \in M_\le$. This implies that $s_\le(\mathbf{a}) = \mathbf{a}$, that is, $\mathbf{a} \in M_\le$.

Since $s_\le(\ell_\le(\mathbf{a})) = \ell_\le(\mathbf{a})$ for all $\mathbf{a} \in A^\infty$, we have $\mathcal{L}_\le \subseteq M_\le$. For the opposite inclusion, let $\mathbf{a} = \lim_{k\to\infty} \mathbf{a}^{(k)}$ for some purely periodic words $\mathbf{a}^{(k)} \in M_\le$, let $(p_k)$ be an increasing sequence satisfying $\mathbf{a}^{(k)} = (a_1^{(k)}\cdots a_{p_k}^{(k)})^\infty$, and let $\mathbf{b} = a_1^{(1)}\cdots a_{p_1}^{(1)}a_1^{(2)}\cdots a_{p_2}^{(2)}\cdots$. Then, $\ell_\le(\mathbf{b}) \ge \mathbf{a}$. If $a_{i+1}^{(k)}\cdots a_{p_k}^{(k)}a_1^{(k+1)}\cdots a_{p_{k+1}}^{(k+1)} > \mathbf{a}$ for some $k \ge 1$, $0 \le i < p_k$, then

$$ \delta_{i,k} := d(a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{p_{k+1}}^{(k+1)} \cdots, \mathbf{a}) \le \max(d(\mathbf{a}^{(k)}, \mathbf{a}), d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)}), 2^{-p_k}). \quad (4) $$

Indeed, for $1 \le j \le p_k - i$, we cannot have $[a_{i+1}^{(k)} \cdots a_{i+j}^{(k)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}]$ because this contradicts $\mathbf{a}^{(k)} \in M_\le$, thus $\delta_{i,k} = 2^{-j}$ implies $d(\mathbf{a}^{(k)}, \mathbf{a}) \ge 2^{-j}$; similarly, for $p_k - i < j \le 2p_k - i$, $[a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k)} \cdots a_{i+j-p_k}^{(k)}] = [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{i+j-p_k}^{(k+1)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}]$

$$ [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k)} \cdots a_{i+j-p_k}^{(k)}] = [a_{i+1}^{(k)} \cdots a_{p_k}^{(k)} a_1^{(k+1)} \cdots a_{i+j-p_k}^{(k+1)}] > [a_1 \cdots a_j] = [a_1^{(k)} \cdots a_j^{(k)}] $$

is impossible, thus $\delta_{i,k} = 2^{-j}$ implies $d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)}) \ge 2^{p_k - i - j}$ or $d(\mathbf{a}^{(k)}, \mathbf{a}) \ge 2^{-j}$; hence, we have $\delta_{i,k} < 2^{i-2p_k}$ or $\delta_{i,k} \le 2^{i-p_k}d(\mathbf{a}^{(k)}, \mathbf{a}^{(k+1)})$ or $\delta_{i,k} \le d(\mathbf{a}^{(k)}, \mathbf{a})$, which implies (4). Since the right-hand side of (4) tends to 0 as $k \to \infty$, we have $\ell_\le(\mathbf{b}) \le \mathbf{a}$, thus $\mathbf{a} = \ell_\le(\mathbf{b}) \in \mathcal{L}_\le$. $\square$
3 | SMALLEST ACCUMULATION POINT OF M_≤

For determining the smallest accumulation point $\mathbf{m}_{\leq}$ of $\mathcal{M}_{\leq}$ for a cylinder order $(A^{\infty}, \leq)$, we can restrict to two-letter alphabets, w.l.o.g., $A = \{0, 1\}$. Indeed, if $\{0, 1\} \subseteq A$, $[0] < [1]$, then $\mathcal{M}_{\leq}$ has the accumulation point $10^{\infty}$ (because $(10^n)^{\infty} \in \mathcal{M}_{\leq}$ for all $n \geq 1$), and we clearly have $s_{\leq}(\mathbf{a}) > 10^{\infty}$ if $\mathbf{a} \in A^{\infty}$ contains a letter $a_n \in A$ with $[a_n] > [1]$.

We use substitutions (also called word morphisms) and limit words (or S-adic sequences). Let $A^*$ be the monoid of finite words over the alphabet $A$, with concatenation as operation. A substitution $\sigma : A^* \to A^*$ satisfies $\sigma(vw) = \sigma(v)\sigma(w)$ for all $v, w \in A^*$ and is extended naturally to $A^{\infty}$; it suffices to give $\sigma(\mathbf{a})$ for $\mathbf{a} \in A$ to define $\sigma$. For a sequence $\sigma = (\sigma_n)_{n \geq 1}$ of substitutions on the alphabet $A$ and an infinite word $\mathbf{a} \in A^{\infty}$, the limit word is

$$ \sigma(\mathbf{a}) := \lim_{n \to \infty} \sigma_{[1,n]}(\mathbf{a}), $$

if this limit exists; we use the notation $\sigma_{[1,n]} := \sigma_1 \circ \sigma_2 \circ \dots \circ \sigma_n$ for $n \geq 0$, with $\sigma_{[1,0]}$ being the identity map. For a set of substitutions $S$, we denote the monoid generated by the composition of substitutions in $S$ by $S^*$. We use the set of substitutions

$$ S = \{\tau_{j,k} : 0 \le j < k\}, \quad \text{with} \quad \tau_{j,k} : \begin{cases} 0 \mapsto 10^j, \\ 1 \mapsto 10^k. \end{cases} $$

Our main result is the following characterization of the smallest accumulation point $\mathbf{m}_{\leq}$ and the discrete part of $\mathcal{M}_{\leq}$ for cylinder orders $\leq$.

**Theorem 3.** Let $\mathbf{m} \in \{0, 1\}^{\infty}$. Then, $\mathbf{m} = \mathbf{m}_{\leq}$ for some cylinder order $\leq$ on $\{0, 1\}^{\infty}$ with $0^{\infty} < 1^{\infty}$ if and only if $\mathbf{m} = \sigma(10^{\infty})$ for some $\sigma \in S^*$ or $\mathbf{m} = \sigma(1^{\infty})$ for some $\sigma \in S^{\infty}$.

$$ \text{If } \mathbf{m}_{\leq} = \sigma(1^{\infty}), \sigma = (\sigma_n)_{n \geq 1} \in S^{\infty}, \text{ then} $$

$$ \{\mathbf{a} \in \mathcal{M}_{\leq} : \mathbf{a} < \mathbf{m}_{\leq}\} = \{(\sigma_{[1,n]}(0))^{\infty} : n \geq 0\}. \quad (5) $$

$$ \text{If } \mathbf{m}_{\leq} = \sigma_{[1,h]}(10^{\infty}), \sigma_1, \dots, \sigma_h \in S, h \ge 0, \text{ then} $$

$$ \{\mathbf{a} \in \mathcal{M}_{\leq} : \mathbf{a} < \mathbf{m}_{\leq}\} = \{(\sigma_{[1,n]}(0))^{\infty} : 0 \le n \le h\} \cup \{\sigma_{[1,h]}((10^j)^{\infty}) : j \ge 0, \sigma_{[1,h]}((10^j)^{\infty}) < \mathbf{m}_{\leq}\}, \quad (6) $$

and there is at most one $j \geq 0$ such that $\sigma_{[1,h]}((10^j)^{\infty}) < \mathbf{m}_{\leq}$.

The following proposition constitutes the core of the proof of Theorem 3 and provides an algorithm for calculating $\mathbf{m}_{\leq}$.

**Proposition 4.** Let $\leq$ be a cylinder order on $\{0, 1\}^{\infty}$ with $[0] < [1]$.

If $[10^j1] < [10^j0]$ holds for at most one $j \geq 0$, then $\mathbf{m}_{\leq} = 10^{\infty}$, and $\mathbf{a} < \mathbf{m}_{\leq}$ implies that $\mathbf{a} = 0^{\infty}$ or $\mathbf{a} = (10^j)^{\infty}$ (in case $[10^j1] < [10^j0]$).

Otherwise, we have $\mathbf{m}_{\leq} = \tau_{j,k}(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ on $\{0, 1\}^{\infty}$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\tau_{j,k}(\mathbf{a}) \leq \tau_{j,k}(\mathbf{b})$, where $j, k$ are minimal such that $0 \le j < k$, $[10^j1] < [10^j0]$, $[10^k1] < [10^k0]$; we have $[0] < [1]$ and $\{\mathbf{a} \in \mathcal{M}_{\leq} : \mathbf{a} < \mathbf{m}_{\leq}\} = \{0^{\infty}\} \cup \{\tau_{j,k}(\mathbf{a}) : \mathbf{a} \in \mathcal{M}_{\leq}, \mathbf{a} < \mathbf{m}_{\leq}\}$.
3 | SMALLEST ACCUMULATION POINT OF $\mathcal{M}_{\le}$

For determining the smallest accumulation point $\mathbf{m}_{\le}$ of $\mathcal{M}_{\le}$ for a cylinder order $(A^{\infty}, \le)$, we can restrict to two-letter alphabets, w.l.o.g., $A = \{0, 1\}$. Indeed, if $\{0, 1\} \subseteq A$, $[0] < [1]$, then $\mathcal{M}_{\le}$ has the accumulation point $10^{\infty}$ (because $(10^n)^{\infty} \in \mathcal{M}_{\le}$ for all $n \ge 1$), and we clearly have $s_{\le}(\mathbf{a}) > 10^{\infty}$ if $\mathbf{a} \in A^{\infty}$ contains a letter $a_n \in A$ with $[a_n] > [1]$.

We use substitutions (also called word morphisms) and limit words (or S-adic sequences). Let $A^*$ be the monoid of finite words over the alphabet $A$, with concatenation as operation. A substitution $\sigma : A^* \to A^*$ satisfies $\sigma(vw) = \sigma(v)\sigma(w)$ for all $v, w \in A^*$ and is extended naturally to $A^{\infty}$; it suffices to give $\sigma(\mathbf{a})$ for $\mathbf{a} \in A$ to define $\sigma$. For a sequence $\sigma = (\sigma_n)_{n \ge 1}$ of substitutions on the alphabet $A$ and an infinite word $\mathbf{a} \in A^{\infty}$, the limit word is

$$ \sigma(\mathbf{a}) := \lim_{n \to \infty} \sigma_{[1,n]}(\mathbf{a}), $$

if this limit exists; we use the notation $\sigma_{[1,n]} := \sigma_1 \circ \sigma_2 \circ \dots \circ \sigma_n$ for $n \ge 0$, with $\sigma_{[1,0]}$ being the identity map. For a set of substitutions $S$, we denote the monoid generated by the composition of substitutions in $S$ by $S^*$. We use the set of substitutions

$$ S = \{\tau_{j,k} : 0 \le j < k\}, \quad \text{with} \quad \tau_{j,k} : \begin{cases} 0 \mapsto 10^j, \\ 1 \mapsto 10^k. \end{cases} $$

Our main result is the following characterization of the smallest accumulation point $\mathbf{m}_{\le}$ and the discrete part of $\mathcal{M}_{\le}$ for cylinder orders $\le$.

**Theorem 3.** Let $\mathbf{m} \in \{0, 1\}^{\infty}$. Then, $\mathbf{m} = \mathbf{m}_{\le}$ for some cylinder order $\le$ on $\{0, 1\}^{\infty}$ with $0^{\infty} < 1^{\infty}$ if and only if $\mathbf{m} = \sigma(10^{\infty})$ for some $\sigma \in S^*$ or $\mathbf{m} = \sigma(1^{\infty})$ for some $\sigma \in S^{\infty}$.

$$ \text{If } \mathbf{m}_{\le} = \sigma(1^{\infty}), \sigma = (\sigma_n)_{n \ge 1} \in S^{\infty}, \text{ then} $$

$$ \{\mathbf{a} \in \mathcal{M}_{\le} : \mathbf{a} < \mathbf{m}_{\le}\} = \{(\sigma_{[1,n]}(0))^{\infty} : n \ge 0\}. \quad (5) $$

$$ \text{If } \mathbf{m}_{\le} = \sigma_{[1,h]}(10^{\infty}), \sigma_1, \dots, \sigma_h \in S, h \ge 0, \text{ then} $$

$$ \{\mathbf{a} \in \mathcal{M}_{\le} : \mathbf{a} < \mathbf{m}_{\le}\} = \{(\sigma_{[1,n]}(0))^{\infty} : 0 \le n \le h\} \cup \{\sigma_{[1,h]}((10^j)^{\infty}) : j \ge 0, \sigma_{[1,h]}((10^j)^{\infty}) < \mathbf{m}_{\le}\}, \quad (6) $$

and there is at most one $j \ge 0$ such that $\sigma_{[1,h]}((10^j)^{\infty}) < \mathbf{m}_{\le}$.

The following proposition constitutes the core of the proof of Theorem 3 and provides an algorithm for calculating $\mathbf{m}_{\le}$.

**Proposition 4.** Let $\le$ be a cylinder order on $\{0, 1\}^{\infty}$ with $[0] < [1]$.

If $[10^j1] < [10^j0]$ holds for at most one $j \ge 0$, then $\mathbf{m}_{\le} = 10^{\infty}$, and $\mathbf{a} < \mathbf{m}_{\le}$ implies that $\mathbf{a} = 0^{\infty}$ or $\mathbf{a} = (10^j)^{\infty}$ (in case $[10^j1] < [10^j0]$).

Otherwise, we have $\mathbf{m}_{\le} = \tau_{j,k}(\mathbf{m}_{\le})$ for the cylinder order $\le$ on $\{0, 1\}^{\infty}$ defined by $\mathbf{a} \le \mathbf{b}$ if $\tau_{j,k}(\mathbf{a}) \le \tau_{j,k}(\mathbf{b})$, where $j, k$ are minimal such that $0 \le j < k$, $[10^j1] < [10^j0]$, $[10^k1] < [10^k0]$; we have $[0] < [1]$ and $\{\mathbf{a} \in \mathcal{M}_{\le} : \mathbf{a} < \mathbf{m}_{\le}\} = \{0^{\infty}\} \cup \{\tau_{j,k}(\mathbf{a}) : \mathbf{a} \in \mathcal{M}_{\le}, \mathbf{a} < \mathbf{m}_{\le}\}$.
**Proof.** We have $\mathbf{m}_{\leq} \in [1]$ because $\mathcal{M}_{\leq} \cap [0] = \{0^{\infty}\}$, and $\mathbf{m}_{\leq} \leq 10^{\infty}$ because $(10^n)^{\infty} \in \mathcal{M}_{\leq}$ for all $n \geq 0$. If $[10^j0] < [10^j1]$ for all $j \geq 0$, that is, $\min[1] = 10^{\infty}$, then $\mathbf{m}_{\leq} = 10^{\infty}$. Otherwise, let $j$ be minimal such that $[10^j1] < [10^j0]$, that is, $\min[1] \in [10^j]$. Then $\mathcal{M}_{\leq} \cap [10^j1] = \{(10^j)^{\infty}\}$ implies that $\mathbf{m}_{\leq} \in [10^j0]$. If $[10^k0] < [10^k1]$ for all $k > j$, that is, $\min[10^j0] = 10^{\infty}$, then $\mathbf{m}_{\leq} = 10^{\infty}$. Otherwise, let $k > j$ be minimal such that $[10^k1] < [10^k0]$, that is, $\min[10^j0] \in [10^k1]$. We have $\mathbf{m}_{\leq} \leq \tau_{j,k}(10^{\infty})$ because $\tau_{j,k}((10^n)^{\infty}) \in \mathcal{M}_{\leq}$ for all $n \geq 0$, thus $\mathbf{m}_{\leq} \in [10^k1]$. Each word in $\mathcal{M}_{\leq} \cap [10^k1]$ is a concatenation of blocks $10^j$ and $10^k$, thus $\mathcal{M}_{\leq} \cap [10^k1] \subseteq \tau_{j,k}([1])$. This proves that $\mathbf{m}_{\leq} = \tau_{j,k}(\mathbf{m}_{\leq})$, where $\mathbf{a} \leq \mathbf{b}$ if $\tau_{j,k}(\mathbf{a}) \leq \tau_{j,k}(\mathbf{b})$; note that $\mathbf{m}_{\leq} \in [1]$ because $[0] < [1]$. Since $\tau_{j,k}([w0]) \subseteq [\tau_{j,k}(w0)1]$ and $\tau_{j,k}([w1]) \subseteq [\tau_{j,k}(w0)0]$ for all $w \in \{0,1\}^*$, $\leq$ is a cylinder order. If $\mathbf{a} \in \mathcal{M}_{\leq}$ with $\mathbf{a} < \mathbf{m}_{\leq}$, then $\mathbf{a} = 0^{\infty}$ or $\mathbf{a} = (10^j)^{\infty} = \tau_{j,k}(0^{\infty})$ or $\mathbf{a} = \tau_{j,k}(\mathbf{a}')$ with $\mathbf{a}' \in \mathcal{M}_{\leq} \cap [1]$, $\mathbf{a}' < \mathbf{m}_{\leq}$. $\square$

The following lemma is used in the construction of a cylinder order $\leq$ such that $\mathbf{m}_{\leq} = \mathbf{m}$ for a given word $\mathbf{m}$. Here, $\varepsilon$ denotes the empty word.

**Lemma 5.** Let $\sigma = (\sigma_n)_{n \geq 1} \in S^\infty$, $w_n = \sigma_{[1,n]}(0) \cdots \sigma_{[1,1]}(0)$ for $n \geq 0$, with $w_0 = \varepsilon$.

For all even $n \geq 0$, we have $\sigma_{[1,n]}([0]) \subseteq [w_n0]$, $\sigma_{[1,n]}([1]) \subseteq [w_n1]$.

For all odd $n \geq 1$, we have $\sigma_{[1,n]}([0]) \subseteq [w_n1]$, $\sigma_{[1,n]}([1]) \subseteq [w_n0]$.

**Proof.** Since $\sigma_{[1,0]}$ is the identity, the statement is trivial for $n = 0$. Suppose that it is true for all $\sigma \in S^\infty$ for some even $n \geq 0$. Then,

$$ \sigma_{[1,n+1]}([0]) = \sigma_1 \circ \sigma_{[2,n+1]}([0]) \subseteq \sigma_1([\sigma_{[2,n+1]}(0) \cdots \sigma_{[2,2]}(0)0]) \subseteq [\sigma_{[1,n+1]}(0) \cdots \sigma_{[1,2]}(0)\sigma_1(0)1], $$

$$ \sigma_{[1,n+1]}([1]) = \sigma_1 \circ \sigma_{[2,n+1]}([1]) \subseteq \sigma_1([\sigma_{[2,n+1]}(0) \cdots \sigma_{[2,2]}(0)1]) \subseteq [\sigma_{[1,n+1]}(0) \cdots \sigma_{[1,2]}(0)\sigma_1(0)0], $$

thus the statement is true for all $\sigma \in S^\infty$ for $n+1$. The case of odd $n$ is similar. $\square$

**Proof of Theorem 3.** Let first $\mathbf{m}_{\leq}$ be a cylinder order with $[0] < [1]$. By iterating Proposition 4, we obtain a finite sequence $\sigma_1, \dots, \sigma_h \in S$ such that $\mathbf{m}_{\leq} = \sigma_{[1,h]}(10^{\infty})$ or an infinite sequence $\sigma = (\sigma_n)_{n \geq 1} \in S^\infty$ such that $\mathbf{m}_{\leq} \in \sigma_{[1,n]}([1])$ for all $n \geq 1$. Since $\sigma_{[1,n+1]}(1)$ starts with $\sigma_{[1,n]}(1)$ and is longer than $\sigma_{[1,n]}(1)$ for all $n \geq 1$, we have $\bigcap_{n \geq 1} \sigma_{[1,n]}([1]) = \{\sigma(1^\infty)\}$. Equations (5) and (6), respectively, follow from Proposition 4.

Let now $\sigma = (\sigma_n)_{n \geq 1} = (\tau_{j_n,k_n})_{n \geq 1} \in S^\infty$. By Proposition 4 and Lemma 5, we have $\sigma(1^\infty) = \mathbf{m}_{\leq}$ for all cylinder orders $\leq$ satisfying

$$ \begin{align} [\sigma_{[1,n]}(10^i)w_n0] < [\sigma_{[1,n]}(10^i)w_n1] \quad &\text{for all even } n \geq 0, j_{n+1} \neq i < k_{n+1}, \nonumber \\ &\text{and for all odd } n \geq 1, i \in \{j_{n+1}, k_{n+1}\}, \tag{7} \\ [\sigma_{[1,n]}(10^i)w_n1] < [\sigma_{[1,n]}(10^i)w_n0] \quad &\text{for all odd } n \geq 1, j_{n+1} \neq i < k_{n+1}, \nonumber \\ &\text{and for all even } n \geq 0, i \in \{j_{n+1}, k_{n+1}\}. \nonumber \end{align} $$

Such cylinder orders exist since $\sigma_{[1,n+1]}(1)w_{n+1}$ is longer than $\sigma_{[1,n]}(10^{k_{n+1}})w_n = \sigma_{[1,n+1]}(1)w_n$ for all $n \geq 0$. To obtain $\mathbf{m}_{\leq} = \sigma_{[1,h]}(10^{\infty})$, $h \geq 0$, we use cylinder orders $\leq$ such that (7) holds only for $n < h$ and such that, for all $i \geq 0$, $[\sigma_{[1,h]}(10^i)w_h0] < [\sigma_{[1,h]}(10^i)w_h1]$ if $h$ is even, $[\sigma_{[1,h]}(10^i)w_h1] < [\sigma_{[1,h]}(10^i)w_h0]$ if $h$ is odd. $\square$
**Proof.** We have $\mathbf{m}_{\leq} \in [1]$ because $\mathcal{M}_{\leq} \cap [0] = \{0^{\infty}\}$, and $\mathbf{m}_{\leq} \leq 10^{\infty}$ because $(10^n)^{\infty} \in \mathcal{M}_{\leq}$ for all $n \geq 0$. If $[10^j0] < [10^j1]$ for all $j \geq 0$, that is, $\min[1] = 10^{\infty}$, then $\mathbf{m}_{\leq} = 10^{\infty}$. Otherwise, let $j$ be minimal such that $[10^j1] < [10^j0]$, that is, $\min[1] \in [10^j]$. Then $\mathcal{M}_{\leq} \cap [10^j1] = \{(10^j)^{\infty}\}$ implies that $\mathbf{m}_{\leq} \in [10^j0]$. If $[10^k0] < [10^k1]$ for all $k > j$, that is, $\min[10^j0] = 10^{\infty}$, then $\mathbf{m}_{\leq} = 10^{\infty}$. Otherwise, let $k > j$ be minimal such that $[10^k1] < [10^k0]$, that is, $\min[10^j0] \in [10^k1]$. We have $\mathbf{m}_{\leq} \leq \tau_{j,k}(10^{\infty})$ because $\tau_{j,k}((10^n)^{\infty}) \in \mathcal{M}_{\leq}$ for all $n \geq 0$, thus $\mathbf{m}_{\leq} \in [10^k1]$. Each word in $\mathcal{M}_{\leq} \cap [10^k1]$ is a concatenation of blocks $10^j$ and $10^k$, thus $\mathcal{M}_{\leq} \cap [10^k1] \subseteq \tau_{j,k}([1])$. This proves that $\mathbf{m}_{\leq} = \tau_{j,k}(\mathbf{m}_{\leq})$, where $\mathbf{a} \leq \mathbf{b}$ if $\tau_{j,k}(\mathbf{a}) \leq \tau_{j,k}(\mathbf{b})$; note that $\mathbf{m}_{\leq} \in [1]$ because $[0] < [1]$. Since $\tau_{j,k}([w0]) \subseteq [\tau_{j,k}(w0)1]$ and $\tau_{j,k}([w1]) \subseteq [\tau_{j,k}(w0)0]$ for all $w \in \{0,1\}^*$, $\leq$ is a cylinder order. If $\mathbf{a} \in \mathcal{M}_{\leq}$ with $\mathbf{a} < \mathbf{m}_{\leq}$, then $\mathbf{a} = 0^{\infty}$ or $\mathbf{a} = (10^j)^{\infty} = \tau_{j,k}(0^{\infty})$ or $\mathbf{a} = \tau_{j,k}(\mathbf{a}')$ with $\mathbf{a}' \in \mathcal{M}_{\leq} \cap [1]$, $\mathbf{a}' < \mathbf{m}_{\leq}$. $\square$

The following lemma is used in the construction of a cylinder order $\leq$ such that $\mathbf{m}_{\leq} = \mathbf{m}$ for a given word $\mathbf{m}$. Here, $\varepsilon$ denotes the empty word.

**Lemma 5.** Let $\sigma = (\sigma_n)_{n \geq 1} \in S^{\infty}$, $w_n = \sigma_{[1,n]}(0) \cdots \sigma_{[1,1]}(0)$ for $n \geq 0$, with $w_0 = \varepsilon$.

For all even $n \geq 0$, we have $\sigma_{[1,n]}([0]) \subseteq [w_n0]$, $\sigma_{[1,n]}([1]) \subseteq [w_n1]$.

For all odd $n \geq 1$, we have $\sigma_{[1,n]}([0]) \subseteq [w_n1]$, $\sigma_{[1,n]}([1]) \subseteq [w_n0]$.

**Proof.** Since $\sigma_{[1,0]}$ is the identity, the statement is trivial for $n = 0$. Suppose that it is true for all $\sigma \in S^{\infty}$ for some even $n \geq 0$. Then,

$$\sigma_{[1,n+1]}([0]) = \sigma_1 \circ \sigma_{[2,n+1]}([0]) \subseteq \sigma_1([\sigma_{[2,n+1]}(0) \cdots \sigma_{[2,2]}(0)0]) \subseteq [\sigma_{[1,n+1]}(0) \cdots \sigma_{[1,2]}(0)\sigma_1(0)1],$$

$$\sigma_{[1,n+1]}([1]) = \sigma_1 \circ \sigma_{[2,n+1]}([1]) \subseteq \sigma_1([\sigma_{[2,n+1]}(0) \cdots \sigma_{[2,2]}(0)1]) \subseteq [\sigma_{[1,n+1]}(0) \cdots \sigma_{[1,2]}(0)\sigma_1(0)0],$$

thus the statement is true for all $\sigma \in S^{\infty}$ for $n+1$. The case of odd $n$ is similar. $\square$

**Proof of Theorem 3.** Let first $\mathbf{m}_{\leq}$ be a cylinder order with $[0] < [1]$. By iterating Proposition 4, we obtain a finite sequence $\sigma_1, \dots, \sigma_h \in S$ such that $\mathbf{m}_{\leq} = \sigma_{[1,h]}(10^{\infty})$ or an infinite sequence $\sigma = (\sigma_n)_{n \geq 1} \in S^{\infty}$ such that $\mathbf{m}_{\leq} \in \sigma_{[1,n]}([1])$ for all $n \geq 1$. Since $\sigma_{[1,n+1]}(1)$ starts with $\sigma_{[1,n]}(1)$ and is longer than $\sigma_{[1,n]}(1)$ for all $n \geq 1$, we have $\bigcap_{n \geq 1} \sigma_{[1,n]}([1]) = \{\sigma(1^{\infty})\}$. Equations (5) and (6), respectively, follow from Proposition 4.

Let now $\sigma = (\sigma_n)_{n \geq 1} = (\tau_{j_n,k_n})_{n \geq 1} \in S^{\infty}$. By Proposition 4 and Lemma 5, we have $\sigma(1^{\infty}) = \mathbf{m}_{\leq}$ for all cylinder orders $\leq$ satisfying

$$\begin{align}
[\sigma_{[1,n]}(10^i)w_n0] < [\sigma_{[1,n]}(10^i)w_n1] \quad &\text{for all even } n \geq 0, j_{n+1} \neq i < k_{n+1}, \nonumber \\
&\text{and for all odd } n \geq 1, i \in \{j_{n+1}, k_{n+1}\}, \tag{7} \\
[\sigma_{[1,n]}(10^i)w_n1] < [\sigma_{[1,n]}(10^i)w_n0] \quad &\text{for all odd } n \geq 1, j_{n+1} \neq i < k_{n+1}, \nonumber \\
&\text{and for all even } n \geq 0, i \in \{j_{n+1}, k_{n+1}\}. \nonumber
\end{align}$$

Such cylinder orders exist since $\sigma_{[1,n+1]}(1)w_{n+1}$ is longer than $\sigma_{[1,n]}(10^{k_{n+1}})w_n = \sigma_{[1,n+1]}(1)w_n$ for all $n \geq 0$. To obtain $\mathbf{m}_{\leq} = \sigma_{[1,h]}(10^{\infty})$, $h \geq 0$, we use cylinder orders $\leq$ such that (7) holds only for $n < h$ and such that, for all $i \geq 0$, $[\sigma_{[1,h]}(10^i)w_h0] < [\sigma_{[1,h]}(10^i)w_h1]$ if $h$ is even, $[\sigma_{[1,h]}(10^i)w_h1] < [\sigma_{[1,h]}(10^i)w_h0]$ if $h$ is odd. $\square$
The sequence $\mathbf{m}_{\le}$ is thus either eventually periodic or an S-adic sequence. A word $\mathbf{a} = a_1 a_2 \cdots$ is eventually periodic if and only if the factor complexity

$$p_{a_1 a_2 \cdots}(n) := \#\{a_{k+1} a_{k+2} \cdots a_{k+n} : k \ge 0\}$$

is bounded; see, for example, [18, Theorem 1.3.13]. The smallest complexity for an aperiodic sequence is $p_a(n) = n+1$, which is attained precisely by Sturmian sequences; see, for example, [18, Theorem 2.1.5]. By [10, Proposition 2.1], all aperiodic words with $\limsup \frac{p_a(n)}{n} < \frac{4}{3}$ are essentially equal to $\sigma(1^{\infty})$ with $\sigma = (\tau_{j_n, k_n})_{n \ge 1} \in S^{\infty}$ (and $k_n \le 2j_n+1$ or $(j_n, k_n) = (0, 2)$). Without conditions on $j_n, k_n$, we get the following upper bound for $p_{\sigma(1^{\infty})}(n)$, which is optimal since $p_{\tau_{\sigma(1^{\infty})}(n)} = 3n-2$ for all $k \ge 2, 2 \le n \le k$.

**Proposition 6.** Let $\le$ be a cylinder order. Then $p_{\mathbf{m}_{\le}}(n) \le 3n-2$ for all $n \ge 2$.

*Proof.* The proof is similar to that of [6, Theorem 17]; see also [10, Proposition 4.1]. Recall that the set of factors of a word $\mathbf{a} = a_1 a_2 \cdots \in \{0, 1\}^{\infty}$ is $\{a_{k+1} a_{k+2} \cdots a_{k+n} : k, n \ge 0\}$, a factor $v$ of $\mathbf{a}$ is strong bispecial if all four words $0v0, 0v1, 1v0, 1v1$ are factors of $\mathbf{a}$, weak bispecial if $0v1, 1v0$ are factors and $0v0, 1v1$ are not factors of $\mathbf{a}$. Then, $v$ is a strong/weak bispecial factor of $\sigma(1^{\infty})$, $\sigma = (\sigma_n)_{n \ge 1} = (\tau_{j_n, k_n})_{n \ge 1} \in S^{\infty}$, if and only if $v = 0^{j_1}/0^{k_1-1}$, $k_1 \ge j_1+2$, or $v = 0^{j_1}\sigma_1(v'0)$, where $v'$ is a strong/weak bispecial factor of $\lim_{n \to \infty} \sigma_{[2,n]}(1^{\infty})$. By iterating, we obtain that all strong/weak bispecial factors of $\sigma(1^{\infty})$ are of the form

$$w_{\ell,h} = \sigma_{[1,0]}(0^{j_1}) \cdots \sigma_{[1,h-1]}(0^{j_h}) \sigma_{[1,h]}(0^{\ell}) \sigma_{[1,h]}(0) \cdots \sigma_{[1,1]}(0),$$

with $h \ge 0$ such that $k_{h+1} \ge j_{h+1}+2$, where $\ell = j_{h+1}$ for a strong bispecial factor, $\ell = k_{h+1}-1$ for a weak one; here, $w_{\ell,0} = 0^{\ell}$. For any recurrent word $\mathbf{a} \in \{0, 1\}^{\infty}$ (where recurrence means that each prefix of $\mathbf{a}$ occurs infinitely often in $\mathbf{a}$), the difference of $p_a(n+2) - p_a(n+1)$ and $p_a(n+1) - p_a(n)$ equals the difference of the number of strong and weak bispecial factors of $\mathbf{a}$ of length $n$; see [7, Proposition 3.2]. By telescoping and since $p_a(1) - p_a(0) = 1$, $p_a(n+2) - p_a(n+1) - 1$ is equal to the difference of the number of strong and weak bispecial factors of $\mathbf{a}$ up to length $n$. Since $|w_{k_{h+1}-1,h}| < |\sigma_{[1,h+1]}(1)| \le |\sigma_{[1,h+2]}(0)| < |w_{j_{h+3},h+2}|$ for all $h \ge 0$, this difference for $\mathbf{a} = \sigma(1^{\infty})$ is at most 2 for all $n \ge 0$; note that $\sigma(1^{\infty})$ is recurrent since $\sigma_{[n,n+1]}(1)$ starts with $10^{k_n}1$ for all $n \ge 1$. Since $p_a(2) \le 4$, we have thus $p_{\sigma(1^{\infty})} \le 3n-2$ for all $n \ge 2$. Since $p_{\sigma_{[1,h]}(10^{\infty})}(n) \le p_{\sigma(1^{\infty})}(n)$ if $k_{h+1}$ is sufficiently large, we also have $p_{\sigma_{[1,h]}(10^{\infty})}(n) \le 3n-2$ for all $n \ge 2$, which proves the proposition by Theorem 3. $\square$

Since the factor complexity is bounded by a linear function, we can apply the results of [1]. For $\beta > 1$, let

$$\pi_{\beta}(a_1 a_2 \cdots) := \sum_{n=1}^{\infty} \frac{a_n}{\beta^n}.$$

Recall that Pisot and Salem numbers are algebraic integers $\beta > 1$ with all Galois conjugates (except $\beta$ itself) having absolute value $\le 1$; $\beta$ is a Salem number if a conjugate lies on the unit circle, a Pisot number otherwise. In particular, all integers $\beta \ge 2$ are Pisot numbers.
The sequence $\mathbf{m}_{\leq}$ is thus either eventually periodic or an S-adic sequence. A word $\mathbf{a} = a_1 a_2 \cdots$ is eventually periodic if and only if the factor complexity

$$p_{a_1 a_2 \cdots}(n) := \#\{a_{k+1} a_{k+2} \cdots a_{k+n} : k \ge 0\}$$

is bounded; see, for example, [18, Theorem 1.3.13]. The smallest complexity for an aperiodic sequence is $p_{\mathbf{a}}(n) = n+1$, which is attained precisely by Sturmian sequences; see, for example, [18, Theorem 2.1.5]. By [10, Proposition 2.1], all aperiodic words with $\limsup \frac{p_{\mathbf{a}}(n)}{n} < \frac{4}{3}$ are essentially equal to $\sigma(1^{\infty})$ with $\sigma = (\tau_{j_n, k_n})_{n \ge 1} \in S^{\infty}$ (and $k_n \le 2j_n+1$ or $(j_n, k_n) = (0, 2)$). Without conditions on $j_n, k_n$, we get the following upper bound for $p_{\sigma(1^{\infty})}(n)$, which is optimal since $p_{\tau_{\sigma(1^{\infty})}(n)} = 3n-2$ for all $k \ge 2$, $2 \le n \le k$.

**Proposition 6.** Let $\leq$ be a cylinder order. Then $p_{\mathbf{m}_{\leq}}(n) \leq 3n-2$ for all $n \geq 2$.

*Proof.* The proof is similar to that of [6, Theorem 17]; see also [10, Proposition 4.1]. Recall that the set of factors of a word $\mathbf{a} = a_1 a_2 \cdots \in \{0, 1\}^{\infty}$ is $\{a_{k+1} a_{k+2} \cdots a_{k+n} : k, n \ge 0\}$, a factor $v$ of $\mathbf{a}$ is strong bispecial if all four words $0v0, 0v1, 1v0, 1v1$ are factors of $\mathbf{a}$, weak bispecial if $0v1, 1v0$ are factors and $0v0, 1v1$ are not factors of $\mathbf{a}$. Then, $v$ is a strong/weak bispecial factor of $\sigma(1^{\infty})$, $\sigma = (\sigma_n)_{n \ge 1} = (\tau_{j_n, k_n})_{n \ge 1} \in S^{\infty}$, if and only if $v = 0^{j_1}/0^{k_1-1}$, $k_1 \ge j_1+2$, or $v = 0^{j_1}\sigma_1(v'0)$, where $v'$ is a strong/weak bispecial factor of $\lim_{n \to \infty} \sigma_{[2,n]}(1^{\infty})$. By iterating, we obtain that all strong/weak bispecial factors of $\sigma(1^{\infty})$ are of the form

$$w_{\ell,h} = \sigma_{[1,0]}(0^{j_1}) \cdots \sigma_{[1,h-1]}(0^{j_h}) \sigma_{[1,h]}(0^{\ell}) \sigma_{[1,h]}(0) \cdots \sigma_{[1,1]}(0),$$

with $h \ge 0$ such that $k_{h+1} \ge j_{h+1}+2$, where $\ell = j_{h+1}$ for a strong bispecial factor, $\ell = k_{h+1}-1$ for a weak one; here, $w_{\ell,0} = 0^{\ell}$. For any recurrent word $\mathbf{a} \in \{0, 1\}^{\infty}$ (where recurrence means that each prefix of $\mathbf{a}$ occurs infinitely often in $\mathbf{a}$), the difference of $p_{\mathbf{a}}(n+2) - p_{\mathbf{a}}(n+1)$ and $p_{\mathbf{a}}(n+1) - p_{\mathbf{a}}(n)$ equals the difference of the number of strong and weak bispecial factors of $\mathbf{a}$ of length $n$; see [7, Proposition 3.2]. By telescoping and since $p_{\mathbf{a}}(1) - p_{\mathbf{a}}(0) = 1$, $p_{\mathbf{a}}(n+2) - p_{\mathbf{a}}(n+1) - 1$ is equal to the difference of the number of strong and weak bispecial factors of $\mathbf{a}$ up to length $n$. Since $|w_{k_{h+1}-1,h}| < |\sigma_{[1,h+1]}(1)| \le |\sigma_{[1,h+2]}(0)| < |w_{j_{h+3},h+2}|$ for all $h \ge 0$, this difference for $\mathbf{a} = \sigma(1^{\infty})$ is at most 2 for all $n \ge 0$; note that $\sigma(1^{\infty})$ is recurrent since $\sigma_{[n,n+1]}(1)$ starts with $10^{k_n}1$ for all $n \ge 1$. Since $p_{\mathbf{a}}(2) \le 4$, we have thus $p_{\sigma(1^{\infty})} \le 3n-2$ for all $n \ge 2$. Since $p_{\sigma_{[1,h]}(10^{\infty})}(n) \le p_{\sigma(1^{\infty})}(n)$ if $k_{h+1}$ is sufficiently large, we also have $p_{\sigma_{[1,h]}(10^{\infty})}(n) \le 3n-2$ for all $n \ge 2$, which proves the proposition by Theorem 3. $\square$

Since the factor complexity is bounded by a linear function, we can apply the results of [1]. For $\beta > 1$, let

$$\pi_{\beta}(a_1 a_2 \cdots) := \sum_{n=1}^{\infty} \frac{a_n}{\beta^n}.$$

Recall that Pisot and Salem numbers are algebraic integers $\beta > 1$ with all Galois conjugates (except $\beta$ itself) having absolute value $\le 1$; $\beta$ is a Salem number if a conjugate lies on the unit circle, a Pisot number otherwise. In particular, all integers $\beta \ge 2$ are Pisot numbers.
**Proposition 7.** Let $\beta \ge 2$ be a Pisot or Salem number, and let $\le$ be a cylinder order on $\{0, 1\}^\infty$. Then, $\pi_\beta(\mathbf{m}_\le)$ is in $\mathbb{Q}(\beta)$ or transcendental.

*Proof.* This is a direct consequence of Proposition 6 and [1, Theorem 1A]. □

# 4 | EXAMPLES

## 4.1 | Lexicographic order

The classical order on the set of infinite words $A^\infty$ with an ordered alphabet $(A, \le)$ is the lexicographic order, defined by $[wa] <_\text{lex} [wb]$ for all $w \in A^*$, $a, b \in A$ with $a < b$. For $A = \{0, 1\}$, we have $\mathbf{m}_\le_\text{lex} = 10^\infty = \min \mathcal{M}_\le_\text{lex} \setminus \{0^\infty\}$.

By [20, §2], a sequence $\mathbf{a} \in \{0, 1\}^\infty$ is the greedy $\beta$-expansion of 1 for some $\beta \in (1, 2)$ if and only if $\mathbf{a} \in \mathcal{M}_\le_\text{lex} \setminus \{10^\infty\}$ and $\mathbf{a}$ is not purely periodic; it is the quasi-greedy $\beta$-expansion of 1 for some $\beta \in (1, 2]$ if and only if $\mathbf{a} \in \mathcal{M}_\le_\text{lex}$ and $\mathbf{a}$ does not end with $0^\infty$. Here, the greedy $\beta$-expansion of 1 is the lexicographically largest sequence $\mathbf{a} \in \{0, 1\}^\infty$ with $\pi_\beta(\mathbf{a}) = 1$, the quasi-greedy $\beta$-expansion of 1 is the largest such sequence that does not end with $0^\infty$. By [15, Theorem 1], we also have that $(0^\infty, \mathbf{a})$ is the pair of kneading sequences of a Lorenz map if and only if $\mathbf{a} \in \mathcal{M}_\le_\text{lex}$ does not end with $0^\infty$.

## 4.2 | Alternating lexicographic order

The alternating lexicographic order is defined by $[wa] <_\text{alt} [wb]$ if $a < b$ and $|w|$ is even, or $a > b$ and $|w|$ is odd, where $|w|$ denotes the length of a word $w \in A^*$. For $A = \{0, 1\}$ with $0 < 1$, we have $[11] <_\text{alt} [10]$, $[101] >_\text{alt} [100]$, $[1001] <_\text{alt} [1000]$, thus $\mathbf{m}_\le_\text{alt} = \tau_{0,2}(\mathbf{m}_\le)$ by Proposition 4, with $\mathbf{a} \le \mathbf{b}$ if $\tau_{0,2}(\mathbf{a}) \le_\text{alt} \tau_{0,2}(\mathbf{b})$. Since $\le$ is equal to $\le_\text{alt}$, we obtain that $\mathbf{m}_\le_\text{alt}$ is the fixed point of $\tau_{0,2}$, that is,

$$ \mathbf{m}_\le_\text{alt} = \tau_{0,2}(\mathbf{m}_\le_\text{alt}) = 100111001001001110011 \cdots; $$

see also [3, 5, 13]. By Theorem 3, we have

$$ \{\mathbf{a} \in \mathcal{M}_{\leq \text{alt}} : \mathbf{a} <_{\text{alt}} \mathbf{m}_{\leq \text{alt}}\} = \{(\tau_{0,2}^{n}(0))^{\infty} : n \geq 0\} = \{0^{\infty}, 1^{\infty}, (100)^{\infty}, (10011)^{\infty}, \dots\}. \quad (8) $$

According to [16], the $(-\beta)$-expansion of $x \in [\frac{-\beta}{\beta+1}, \frac{1}{\beta+1})$, $\beta > 1$, is the sequence $a_1 a_2 \cdots$ with $a_n = \lfloor \frac{\beta}{\beta+1} - \beta T_{-\beta}^{n-1}(x) \rfloor$, given by the $(-\beta)$-transformation $T_{-\beta}(y) := -\beta y - \lfloor \frac{\beta}{\beta+1} - \beta y \rfloor$, and the set of $(-\beta)$-expansions is characterized by that of $\frac{-\beta}{\beta+1}$. By [24, Theorem 2], a sequence $\mathbf{a}$ is the $(-\beta)$-expansion of $\frac{-\beta}{\beta+1}$ for some $\beta \in (1, 2)$ if and only if $\mathbf{a} \in \mathcal{M}_\le_\text{alt} \setminus \{(10)^\infty\}$, $\mathbf{a} >_\text{alt} \mathbf{m}_\le_\text{alt}$, and $\mathbf{a} \notin \{w1, w00\}^\infty \setminus \{(w1)^\infty\}$ for all $w \in \{0, 1\}^*$ such that $(w1)^\infty >_\text{alt} \mathbf{m}_\le_\text{alt}$. Note that continued fractions are also ordered by the alternating lexicographic order on the sequences of partial quotients, and $\mathbf{m}_\le_\text{alt}$ occurs, for example, in [8, Theorem 2] and [17, Remark 11.1].
**Proposition 7.** Let $\beta \ge 2$ be a Pisot or Salem number, and let $\le$ be a cylinder order on $\{0, 1\}^{\infty}$. Then, $\pi_{\beta}(\mathbf{m}_{\le})$ is in $\mathbb{Q}(\beta)$ or transcendental.

*Proof.* This is a direct consequence of Proposition 6 and [1, Theorem 1A]. □

# 4 | EXAMPLES

## 4.1 | Lexicographic order

The classical order on the set of infinite words $A^{\infty}$ with an ordered alphabet $(A, \le)$ is the lexicographic order, defined by $[wa] <_{lex} [wb]$ for all $w \in A^*$, $a, b \in A$ with $a < b$. For $A = \{0, 1\}$, we have $\mathbf{m}_{\le_{lex}} = 10^{\infty} = \min \mathcal{M}_{\le_{lex}} \setminus \{0^{\infty}\}$.

By [20, §2], a sequence $\mathbf{a} \in \{0, 1\}^{\infty}$ is the greedy $\beta$-expansion of 1 for some $\beta \in (1, 2)$ if and only if $\mathbf{a} \in \mathcal{M}_{\le_{lex}} \setminus \{10^{\infty}\}$ and $\mathbf{a}$ is not purely periodic; it is the quasi-greedy $\beta$-expansion of 1 for some $\beta \in (1, 2]$ if and only if $\mathbf{a} \in \mathcal{M}_{\le_{lex}}$ and $\mathbf{a}$ does not end with $0^{\infty}$. Here, the greedy $\beta$-expansion of 1 is the lexicographically largest sequence $\mathbf{a} \in \{0, 1\}^{\infty}$ with $\pi_{\beta}(\mathbf{a}) = 1$, the quasi-greedy $\beta$-expansion of 1 is the largest such sequence that does not end with $0^{\infty}$. By [15, Theorem 1], we also have that $(0^{\infty}, \mathbf{a})$ is the pair of kneading sequences of a Lorenz map if and only if $\mathbf{a} \in \mathcal{M}_{\le_{lex}}$ does not end with $0^{\infty}$.

## 4.2 | Alternating lexicographic order

The alternating lexicographic order is defined by $[wa] <_{alt} [wb]$ if $a < b$ and $|w|$ is even, or $a > b$ and $|w|$ is odd, where $|w|$ denotes the length of a word $w \in A^*$. For $A = \{0, 1\}$ with $0 < 1$, we have $[11] <_{alt} [10]$, $[101] >_{alt} [100]$, $[1001] <_{alt} [1000]$, thus $\mathbf{m}_{\le_{alt}} = \tau_{0,2}(\mathbf{m}_{\le})$ by Proposition 4, with $\mathbf{a} \le \mathbf{b}$ if $\tau_{0,2}(\mathbf{a}) \le_{alt} \tau_{0,2}(\mathbf{b})$. Since $\le$ is equal to $\le_{alt}$, we obtain that $\mathbf{m}_{\le}$ is the fixed point of $\tau_{0,2}$, that is,

$$ \mathbf{m}_{\le_{alt}} = \tau_{0,2}(\mathbf{m}_{\le_{alt}}) = 100111001001001110011 \cdots; $$

see also [3, 5, 13]. By Theorem 3, we have

$$ \{\mathbf{a} \in \mathcal{M}_{\le_{alt}} : \mathbf{a} <_{alt} \mathbf{m}_{\le \le_{alt}}\} = \{(\tau_{0,2}^{n}(0))^{\infty} : n \ge 0\} = \{0^{\infty}, 1^{\infty}, (100)^{\infty}, (10011)^{\infty}, \dots\}. \quad (8) $$

According to [16], the $(-\beta)$-expansion of $x \in [\frac{-\beta}{\beta+1}, \frac{1}{\beta+1})$, $\beta > 1$, is the sequence $a_1 a_2 \cdots$ with $a_n = \lfloor \frac{\beta}{\beta+1} - \beta T_{-\beta}^{n-1}(x) \rfloor$, given by the $(-\beta)$-transformation $T_{-\beta}(y) := -\beta y - \lfloor \frac{\beta}{\beta+1} - \beta y \rfloor$, and the set of $(-\beta)$-expansions is characterized by that of $\frac{-\beta}{\beta+1}$. By [24, Theorem 2], a sequence $\mathbf{a}$ is the $(-\beta)$-expansion of $\frac{-\beta}{\beta+1}$ for some $\beta \in (1, 2)$ if and only if $\mathbf{a} \in \mathcal{M}_{\le_{alt}} \setminus \{(10)^{\infty}\}$, $\mathbf{a} >_{alt} \mathbf{m}_{\le_{alt}}$, and $\mathbf{a} \notin \{w1, w00\}^{\infty} \setminus \{(w1)^{\infty}\}$ for all $w \in \{0, 1\}^*$ such that $(w1)^{\infty} >_{alt} \mathbf{m}_{\le_{alt}}$. Note that continued fractions are also ordered by the alternating lexicographic order on the sequences of partial quotients, and $\mathbf{m}_{\le_{alt}}$ occurs, for example, in [8, Theorem 2] and [17, Remark 11.1].
## 4.3 | Unimodal maps

Let $A = \{0, 1\}$ and define the *unimodal order* by $[w0] <_{uni} [w1]$ if $|w|_1$ is even, $[w0] >_{uni} [w1]$ if $|w|_1$ is odd, where $|w|_1$ denotes the number of occurrences of 1 in $w \in \{0, 1\}^*$. Then we have $[11] <_{uni} [10]$, $[101] <_{uni} [100]$, and

$$ \mathbf{m}_{\leq \text{uni}} = \tau_{0,1}(\mathbf{m}_{\leq \text{alt}}) = 1011101010111011110111101010111010 \dots $$

This is the fixed point of the period-doubling (or Feigenbaum) substitution $0 \mapsto 11$, $1 \mapsto 10$. The set $\mathcal{M}_{\leq \text{uni}}$ is the set of kneading sequences of unimodal maps [9, 19].

We define the *flipped unimodal order* by $[w0] <_{flip} [w1]$ if $|w|_0$ is even, $[w0] >_{flip} [w1]$ if $|w|_0$ is odd, where $|w|_0$ denotes the number of occurrences of 0 in $w \in \{0, 1\}^*$. Then we have $[11] >_{flip} [10]$, $[101] <_{flip} [100]$, $[1001] >_{flip} [1000]$, $[10001] >_{flip} [10000]$, and

$$ \mathbf{m}_{\leq \text{flip}} = \tau_{1,3}(\mathbf{m}_{\leq \text{alt}}) = 1000101010001000100010101000101010001000 \dots $$

Note that $\mathbf{m}_{\leq \text{flip}} = F(\mathbf{m}_{\leq \text{uni}})$, where $F(a_1 a_2 \dots) := (1-a_1)(1-a_2)\dots$, and we have

$$ \mathcal{M}_{\leq \text{uni}} = 1F(\mathcal{M}_{\leq \text{flip}}) \cup \{0^{\infty}\}. $$

## 4.4 | Sturmian sequences

The set of substitutions $\{\theta_k : k \geq 1\}$ defined by $\theta_k(0) = 0^{k-1}1$, $\theta_k(1) = 0^{k-1}10$, generates the standard Sturmian words; see [18, Corollary 2.2.22]. Since $\tau_{k-1,k}$ is rotationally conjugate to $\theta_k$, more precisely $\theta_k(w)0^{k-1} = 0^{k-1}\tau_{k-1,k}(w)$ for all $w \in \{0, 1\}^*$, the set of substitutions $\{\tau_{k-1,k} : k \geq 1\}$ generates the same shifts as $\{\theta_k : k \geq 1\}$. Therefore, the limit words of sequences in $\{\tau_{k-1,k} : k \geq 1\}^{\infty}$ provide elements of all Sturmian shifts. For example, the limit word of the sequence $(\tau_{0,1})^{\infty}$ is the Fibonacci word. Note that Sturmian sequences that are sup-words are also considered in [4].

## 5 | SYMMETRIC ALPHABETS

For a real number $q > 1$, the set

$$ \tilde{\mathcal{L}}_q := \{ \limsup_{n \to \infty} \|xq^n\| : x \in \mathbb{R} \}, \qquad (9) $$

where $\|\cdot\|$ denotes the distance to the nearest integer, is a multiplicative version of the Lagrange spectrum, and was studied in [2, 12]. If $q$ is an integer, then representing $x = \sum_{k=-\infty}^{\infty} a_k q^{-k}$ with $a_k \in \mathbb{Z}$, $a_k \neq 0$ for finitely many $k \leq 0$, $|\sum_{k=n+1}^{\infty} a_k q^{n-k}| \leq 1/2$, gives that $\|xq^n\| = |\sum_{k=n+1}^{\infty} a_k q^{n-k}|$; see also Proposition 10 below. This leads us to consider

$$ \mathcal{M}_{\leq}^{\text{abs}} = \{s_{\leq}^{\text{abs}}(\mathbf{a}) : \mathbf{a} \in \{0, \pm 1\}^{\infty}\} \quad \text{with} \quad s_{\leq}^{\text{abs}}(a_1 a_2 \dots) = \sup_{n \geq 1} \text{abs}(a_n a_{n+1} \dots), $$
## 4.3 | Unimodal maps

Let $A = \{0, 1\}$ and define the *unimodal order* by $[w0] <_{uni} [w1]$ if $|w|_1$ is even, $[w0] >_{uni} [w1]$ if $|w|_1$ is odd, where $|w|_1$ denotes the number of occurrences of 1 in $w \in \{0, 1\}^*$. Then we have $[11] <_{uni} [10]$, $[101] <_{uni} [100]$, and

$$ \mathbf{m}_{\leq \text{uni}} = \tau_{0,1}(\mathbf{m}_{\leq \text{alt}}) = 1011101010111011110111101010111010 \dots $$

This is the fixed point of the period-doubling (or Feigenbaum) substitution $0 \mapsto 11$, $1 \mapsto 10$. The set $\mathcal{M}_{\leq \text{uni}}$ is the set of kneading sequences of unimodal maps [9, 19].

We define the *flipped unimodal order* by $[w0] <_{flip} [w1]$ if $|w|_0$ is even, $[w0] >_{flip} [w1]$ if $|w|_0$ is odd, where $|w|_0$ denotes the number of occurrences of 0 in $w \in \{0, 1\}^*$. Then we have $[11] >_{flip} [10]$, $[101] <_{flip} [100]$, $[1001] >_{flip} [1000]$, $[10001] >_{flip} [10000]$, and

$$ \mathbf{m}_{\leq \text{flip}} = \tau_{1,3}(\mathbf{m}_{\leq \text{alt}}) = 1000101010001000100010101000101010001000 \dots $$

Note that $\mathbf{m}_{\leq \text{flip}} = F(\mathbf{m}_{\leq \text{uni}})$, where $F(a_1 a_2 \dots) := (1-a_1)(1-a_2)\dots$, and we have

$$ \mathcal{M}_{\leq \text{uni}} = 1F(\mathcal{M}_{\leq \text{flip}}) \cup \{0^{\infty}\}. $$

## 4.4 | Sturmian sequences

The set of substitutions $\{\theta_k : k \geq 1\}$ defined by $\theta_k(0) = 0^{k-1}1$, $\theta_k(1) = 0^{k-1}10$, generates the standard Sturmian words; see [18, Corollary 2.2.22]. Since $\tau_{k-1,k}$ is rotationally conjugate to $\theta_k$, more precisely $\theta_k(w)0^{k-1} = 0^{k-1}\tau_{k-1,k}(w)$ for all $w \in \{0, 1\}^*$, the set of substitutions $\{\tau_{k-1,k} : k \geq 1\}$ generates the same shifts as $\{\theta_k : k \geq 1\}$. Therefore, the limit words of sequences in $\{\tau_{k-1,k} : k \geq 1\}^{\infty}$ provide elements of all Sturmian shifts. For example, the limit word of the sequence $(\tau_{0,1})^{\infty}$ is the Fibonacci word. Note that Sturmian sequences that are sup-words are also considered in [4].

# 5 | SYMMETRIC ALPHABETS

For a real number $q > 1$, the set

$$ \tilde{\mathcal{L}}_q := \{ \limsup_{n \to \infty} \|xq^n\| : x \in \mathbb{R} \}, \qquad (9) $$

where $\|\cdot\|$ denotes the distance to the nearest integer, is a multiplicative version of the Lagrange spectrum, and was studied in [2, 12]. If $q$ is an integer, then representing $x = \sum_{k=-\infty}^{\infty} a_k q^{-k}$ with $a_k \in \mathbb{Z}$, $a_k \neq 0$ for finitely many $k \leq 0$, $|\sum_{k=n+1}^{\infty} a_k q^{n-k}| \leq 1/2$, gives that $\|xq^n\| = |\sum_{k=n+1}^{\infty} a_k q^{n-k}|$; see also Proposition 10 below. This leads us to consider

$$ \mathcal{M}_{\leq}^{\text{abs}} = \{s_{\leq}^{\text{abs}}(\mathbf{a}) : \mathbf{a} \in \{0, \pm 1\}^{\infty}\} \text{ with } s_{\leq}^{\text{abs}}(a_1 a_2 \dots) = \sup_{n \geq 1} \text{abs}(a_n a_{n+1} \dots), $$
where

$$
\mathrm{abs}(\mathbf{a}) = \begin{cases} \mathbf{a} & \text{if } \mathbf{a} \ge \mathrm{lex} \ 0^{\infty}, \\ -\mathbf{a} & \text{if } \mathbf{a} \le \mathrm{lex} \ 0^{\infty}, \end{cases} \quad \text{with } -(a_1 a_2 \cdots) = (-a_1)(-a_2) \cdots.
$$

We denote the smallest accumulation point of $\mathcal{M}_{\leq}^{\text{abs}}$ by $\mathbf{m}_{\leq}^{\text{abs}}$.

The same proof as for Theorem 1 shows for all cylinder orders $\leq$ on $\{0, \pm 1\}^{\infty}$ that

$$
\mathcal{L}_{\leq}^{\text{abs}} = \mathcal{M}_{\leq}^{\text{abs}} = \text{closure}(\{s_{\leq}^{\text{abs}}(\mathbf{a}) : \mathbf{a} \in \{0, \pm 1\}^{\infty} \text{ purely periodic}\}),
$$

$$
\text{where } \mathcal{L}_{\leq}^{\text{abs}} := \{\limsup_{n \ge 1} \text{abs}(a_n a_{n+1} \cdots) : a_1 a_2 \cdots \in \{0, \pm 1\}^{\infty}\}.
$$

In the following, we assume that a cylinder order on {0, ±1}∞ is consistent (with the natural order on {0, ±1}), which means that, for each w ∈ {0, ±1}* , we have [w(−1)] < [w0] < [w1] or [w(−1)] > [w0] > [w1]. In order to describe m ∞ ≤ , we define maps ϑ₀, ϑ₁, ϑ₂ from {0, 1}* to {0, ±1}* by ϑ₀(ε) = ϑ₁(ε) = ϑ₂(ε) = ε for the empty word ε, and

$$
\vartheta_0(w0) = \begin{cases} \vartheta_0(w)1 & \text{if } |w|_0 \text{ is even,} \\ \vartheta_0(w)(-1) & \text{if } |w|_0 \text{ is odd,} \end{cases}
$$

$$
\vartheta_0(w1) = \begin{cases} \vartheta_0(w)10 & \text{if } |w|_0 \text{ is even,} \\ \vartheta_0(w)(-1)0 & \text{if } |w|_0 \text{ is odd,} \end{cases}
$$

$$
\vartheta_1(w0) = \begin{cases} \vartheta_1(w)1 & \text{if } |w|_1 \text{ is even,} \\ \vartheta_1(w)(-1) & \text{if } |w|_1 \text{ is odd,} \end{cases}
$$

$$
\vartheta_1(w1) = \begin{cases} \vartheta_1(w)10 & \text{if } |w|_1 \text{ is even,} \\ \vartheta_1(w)(-1)0 & \text{if } |w|_1 \text{ is odd,} \end{cases}
$$

$$
\vartheta_2(w0) = \begin{cases} \vartheta_2(w)1 & \text{if } |w| \text{ is even,} \\ \vartheta_2(w)(-1) & \text{if } |w| \text{ is odd,} \end{cases}
$$

$$
\vartheta_2(w1) = \begin{cases} \vartheta_2(w)10 & \text{if } |w| \text{ is even,} \\ \vartheta_2(w)(-1)0 & \text{if } |w| \text{ is odd,} \end{cases}
$$

for all $w \in \{0, 1\}^*$, where $|w|$ is the length of a word $w$ and $|w|_i$ the number of occurrences of the letter $i$ in $w$. As for substitutions, the maps $\vartheta_i$ are extended naturally to $\{0, 1\}^\infty$.

**Theorem 8.** Let $\mathbf{m} \in \{0, \pm 1\}^\infty$. Then $\mathbf{m} = \mathbf{m}_{\leq}^{\text{abs}}$ for some consistent cylinder order $\leq$ on $\{0, \pm 1\}^\infty$ with $0^\infty < 1^\infty$ if and only if $\mathbf{m} = \sigma(\mathbf{m}_{\leq})$ for some $\sigma \in \{\vartheta_0, \vartheta_1, \vartheta_2, \tau_{0,1}\}$ and some cylinder order $\leq$ on $\{0, 1\}^\infty$ with $0^\infty < 1^\infty$.

If $\mathbf{m}_{\leq}^{\text{abs}} = \sigma(\mathbf{m}_{\leq})$, then we can assume that $\mathbf{a} \leq \mathbf{b}$ if and only if $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$, and we have

$$
\{\mathbf{a} \in \mathcal{M}_{\leq}^{\text{abs}} : \mathbf{a} < \mathbf{m}_{\leq}^{\text{abs}}\} = \{0^{\infty}\} \cup \{\sigma(\mathbf{a}) : \mathbf{a} \in \mathcal{M}_{\leq}, \mathbf{a} < \mathbf{m}_{\leq}\}. \quad (10)
$$

*Proof.* Let first $\leq$ be a consistent cylinder order on $\{0, \pm 1\}^{\infty}$. Then, the order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$ is a cylinder order for all $\sigma \in \{\vartheta_0, \vartheta_1, \vartheta_2, \tau_{0,1}\}$. Indeed, for any $w \in \{0, 1\}^*$, we have $\sigma([w0]) \subset [\sigma(w)xy]$, $\sigma([w1]) \subset [\sigma(w)x0]$, where $x, y \in \{\pm 1\}$.

Assume first that $[1\bar{1}] < [10]$; here and in the following, we use the notation $\bar{1} = -1$. Then, $\mathcal{M}_{\leq}^{\text{abs}} \cap [1\bar{1}] = \{(1\bar{1})^{\infty}\}$, thus $\mathbf{m}_{\leq}^{\text{abs}} \in [10]$. If $[10\bar{1}] < [100]$, then each 1 in a word in $\mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}]$ is followed by $\bar{1}$ or $0\bar{1}$, and each $\bar{1}$ is followed by 1 or 01, that is,

$$
\mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}] \subseteq 10(\{\bar{1}, \bar{0}\}\{1, 10\})^{\infty} = \vartheta_2([1]).
$$

Therefore, $\mathbf{m}_{\leq}^{\text{abs}} = \vartheta_2(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\vartheta_2(\mathbf{a}) \leq \vartheta_2(\mathbf{b})$. If $[101] < [100]$, then each 1 in a word in $\mathcal{M}_{\leq}^{\text{abs}} \cap [101]$ is followed by $\bar{1}$ or 01, and each $\bar{1}$ is followed by 1 or
where

$$ \mathrm{abs}(\mathbf{a}) = \begin{cases} \mathbf{a} & \text{if } \mathbf{a} \ge \mathrm{lex} \ 0^{\infty}, \\ -\mathbf{a} & \text{if } \mathbf{a} \le \mathrm{lex} \ 0^{\infty}, \end{cases} \quad \text{with } -(a_1 a_2 \cdots) = (-a_1)(-a_2) \cdots. $$

We denote the smallest accumulation point of $\mathcal{M}_{\leq}^{\text{abs}}$ by $\mathbf{m}_{\leq}^{\text{abs}}$.

The same proof as for Theorem 1 shows for all cylinder orders $\leq$ on $\{0, \pm 1\}^{\infty}$ that

$$ \mathcal{L}_{\leq}^{\text{abs}} = \mathcal{M}_{\leq}^{\text{abs}} = \text{closure}(\{s_{\leq}^{\text{abs}}(\mathbf{a}) : \mathbf{a} \in \{0, \pm 1\}^{\infty} \text{ purely periodic}\}), $$

$$ \text{where } \mathcal{L}_{\leq}^{\text{abs}} := \{\limsup_{n \ge 1} \text{abs}(a_n a_{n+1} \cdots) : a_1 a_2 \cdots \in \{0, \pm 1\}^{\infty}\}. $$

In the following, we assume that a cylinder order on $\{0, \pm 1\}^{\infty}$ is consistent (with the natural order on $\{0, \pm 1\}$), which means that, for each $w \in \{0, \pm 1\}^{*}$, we have $[w(-1)] < [w0] < [w1]$ or $[w(-1)] > [w0] > [w1]$. In order to describe $\mathbf{m}_{\leq}^{\text{abs}}$, we define maps $\varrho_0, \varrho_1, \varrho_2$ from $\{0, 1\}^{*}$ to $\{0, \pm 1\}^{*}$ by $\varrho_0(\varepsilon) = \varrho_1(\varepsilon) = \varrho_2(\varepsilon) = \varepsilon$ for the empty word $\varepsilon$, and

$$ \varrho_0(w0) = \begin{cases} \varrho_0(w)1 & \text{if } |w|_0 \text{ is even,} \\ \varrho_0(w)(-1) & \text{if } |w|_0 \text{ is odd,} \end{cases} $$

$$ \varrho_0(w1) = \begin{cases} \varrho_0(w)10 & \text{if } |w|_0 \text{ is even,} \\ \varrho_0(w)(-1)0 & \text{if } |w|_0 \text{ is odd,} \end{cases} $$

$$ \varrho_1(w0) = \begin{cases} \varrho_1(w)1 & \text{if } |w|_1 \text{ is even,} \\ \varrho_1(w)(-1) & \text{if } |w|_1 \text{ is odd,} \end{cases} $$

$$ \varrho_1(w1) = \begin{cases} \varrho_1(w)10 & \text{if } |w|_1 \text{ is even,} \\ \varrho_1(w)(-1)0 & \text{if } |w|_1 \text{ is odd,} \end{cases} $$

$$ \varrho_2(w0) = \begin{cases} \varrho_2(w)1 & \text{if } |w| \text{ is even,} \\ \varrho_2(w)(-1) & \text{if } |w| \text{ is odd,} \end{cases} $$

$$ \varrho_2(w1) = \begin{cases} \varrho_2(w)10 & \text{if } |w| \text{ is even,} \\ \varrho_2(w)(-1)0 & \text{if } |w| \text{ is odd,} \end{cases} $$

for all $w \in \{0, 1\}^{*}$, where $|w|$ is the length of a word $w$ and $|w|_i$ the number of occurrences of the letter $i$ in $w$. As for substitutions, the maps $\varrho_i$ are extended naturally to $\{0, 1\}^{\infty}$.

**Theorem 8.** Let $\mathbf{m} \in \{0, \pm 1\}^{\infty}$. Then $\mathbf{m} = \mathbf{m}_{\leq}^{\text{abs}}$ for some consistent cylinder order $\leq$ on $\{0, \pm 1\}^{\infty}$ with $0^{\infty} < 1^{\infty}$ if and only if $\mathbf{m} = \sigma(\mathbf{m}_{\leq})$ for some $\sigma \in \{\varrho_0, \varrho_1, \varrho_2, \tau_{0,1}\}$ and some cylinder order $\leq$ on $\{0, 1\}^{\infty}$ with $0^{\infty} < 1^{\infty}$.

If $\mathbf{m}_{\leq}^{\text{abs}} = \sigma(\mathbf{m}_{\leq})$, then we can assume that $\mathbf{a} \leq \mathbf{b}$ if and only if $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$, and we have

$$ \{\mathbf{a} \in \mathcal{M}_{\leq}^{\text{abs}} : \mathbf{a} < \mathbf{m}_{\leq}^{\text{abs}}\} = \{0^{\infty}\} \cup \{\sigma(\mathbf{a}) : \mathbf{a} \in \mathcal{M}_{\leq}, \mathbf{a} < \mathbf{m}_{\leq}\}. \quad (10) $$

*Proof.* Let first $\leq$ be a consistent cylinder order on $\{0, \pm 1\}^{\infty}$. Then, the order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$ is a cylinder order for all $\sigma \in \{\varrho_0, \varrho_1, \varrho_2, \tau_{0,1}\}$. Indeed, for any $w \in \{0, 1\}^{*}$, we have $\sigma([w0]) \subset [\sigma(w)xy]$, $\sigma([w1]) \subset [\sigma(w)x0]$, where $x, y \in \{\pm 1\}$.

Assume first that $[1\bar{1}] < [10]$; here and in the following, we use the notation $\bar{1} = -1$. Then, $\mathcal{M}_{\leq}^{\text{abs}} \cap [1\bar{1}] = \{(1\bar{1})^{\infty}\}$, thus $\mathbf{m}_{\leq}^{\text{abs}} \in [10]$. If $[10\bar{1}] < [100]$, then each 1 in a word in $\mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}]$ is followed by $\bar{1}$ or $0\bar{1}$, and each $\bar{1}$ is followed by 1 or 01, that is,

$$ \mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}] \subseteq 10(\{\bar{1}, \bar{0}\}\{1, 10\})^{\infty} = \varrho_2([1]). $$

Therefore, $\mathbf{m}_{\leq}^{\text{abs}} = \varrho_2(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\varrho_2(\mathbf{a}) \leq \varrho_2(\mathbf{b})$. If $[101] < [100]$, then each 1 in a word in $\mathcal{M}_{\leq}^{\text{abs}} \cap [101]$ is followed by $\bar{1}$ or 01, and each $\bar{1}$ is followed by 1 or
0$\bar{1}$, that is,

$$
\begin{align*}
\mathcal{M}_{\leq}^{\text{abs}} \cap [101] &\subseteq 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \cup 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \cup 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \\
&= \varrho_0(1(1^{*01}0^*)^{\infty}) \cup \varrho_0(1(1^{*01}0^*)^{*1\infty}) \cup \varrho_0(1(1^{*01}0^*)^{*1^{*01}}) = \varrho_0([1]).
\end{align*}
$$

Therefore, we have $\mathbf{m}_{\leq}^{\text{abs}} = \varrho_0(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\varrho_0(\mathbf{a}) \leq \varrho_0(\mathbf{b})$. Assume now $[11] < [10]$. Then $\mathcal{M}_{\leq}^{\text{abs}} \cap [11] = \{1^{\infty}\}$, thus $\mathbf{m}_{\leq} \in [10]$. If $[10\bar{1}] < [100]$, then

$$
\mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}] \subseteq 10(\bar{1}^*\bar{1}01^*10)^{\infty} \cup 10(\bar{1}^*\bar{1}01^*10)^*\bar{1}^{\infty} \cup 10(\bar{1}^*\bar{1}01^*10)^*\bar{1}^*\bar{1}01^{\infty} = \varrho_1([1]),
$$

thus $\mathbf{m}_{\leq}^{\text{abs}} = \varrho_1(\mathbf{m}_{\leq})$, with $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\varrho_1(\mathbf{a}) \leq \varrho_1(\mathbf{b})$. If $[101] < [100]$, then

$$
\mathcal{M}_{\leq}^{\text{abs}} \cap [101] \subseteq 10\{1, 10\}^{\infty} = \tau_{0,1}([1]),
$$

thus $\mathbf{m}_{\leq}^{\text{abs}} = \tau_{0,1}(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\tau_{0,1}(\mathbf{a}) \leq \tau_{0,1}(\mathbf{b})$. Since $\varrho_0(0^\infty) = \varrho_2(0^\infty) = (\bar{1}\bar{1})^\infty$ and $\varrho_1(0^\infty) = \tau_{0,1}(0^\infty) = 1^\infty$, Equation (10) holds.

Let now $\leq$ be a cylinder order on $\{0, 1\}^\infty$ with $0^\infty < 1^\infty$ and $\sigma \in \{\varrho_0, \varrho_1, \varrho_2, \tau_{0,1}\}$. Then, there exists a consistent cylinder order $\leq$ on $\{0, \pm 1\}^\infty$ satisfying $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$ if $\mathbf{a} \leq \mathbf{b}$ and $0^\infty < 1^\infty$. Indeed, for $w \in \{0, \pm 1\}^*$ and distinct $a, b \in \{0, \pm 1\}$, we set $[wa] < [wb]$ if $\mathbf{a}' < \mathbf{b}'$ for some $\mathbf{a}', \mathbf{b}' \in \{0, 1\}^\infty$ with $\sigma(\mathbf{a}') \in [wa]$, $\sigma(\mathbf{b}') \in [wb]$; by Lemma 2, this does not depend on the choice of $\mathbf{a}', \mathbf{b}'$. Moreover, since $\sigma(\mathbf{a}) \in [w\bar{1}]$ and $\sigma(\mathbf{b}) \in [w1]$ is impossible, we have no obstruction to a consistent cylinder order. Since $0^\infty < 1^\infty$, we have $[\bar{1}] < [10]$ in case $\sigma \in \{\varrho_0, \varrho_2\}$, $[11] < [10]$ in case $\sigma \in \{\varrho_1, \tau_{0,1}\}$. We can set $[10\bar{1}] < [100]$ in case $\sigma \in \{\varrho_1, \varrho_2\}$ because $([100] \cup [101]) \cap \sigma(\{0, 1\})^\infty = \emptyset$; similarly, we can set $[101] < [100]$ in case $\sigma \in \{\varrho_0, \tau_{0,1}\}$. Then, we have $\mathbf{m}_{\leq}^{\text{abs}} = \sigma(\mathbf{m}_{\leq})$. $\square$

**Proposition 9.** Let $\beta \ge 3$ be a Pisot or Salem number, and let $\le$ be a consistent cylinder order on $\{0, \pm 1\}^\infty$. Then $\pi_\beta(\mathbf{m}_{\le}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental.

*Proof.* Let $G(a_1a_2\cdots) = |a_1| |a_2| \cdots$. Then $G \circ \varrho_l = \tau_{0,1}$ for all $l \in \{0, 1, 2\}$, thus $p_{G(\mathbf{m}_{\le}^{\text{abs}})}(n) \le 3n-2$ by Proposition 6, Theorems 3 and 8. Moreover, the map $G$ is 2-to-1 from the set of factors of $\mathbf{m}_{\le}^{\text{abs}}$ to the set of factors of $G(\mathbf{m}_{\le}^{\text{abs}})$, thus $p_{\mathbf{m}_{\le}^{\text{abs}}}(n) \le 6n-4$. By [1, Theorem 1A] and by adding 1 to each digit of $\mathbf{m}_{\le}^{\text{abs}}$, we obtain that $\pi_\beta(\mathbf{m}_{\le}^{\text{abs}}) + \frac{1}{\beta-1}$ is in $\mathbb{Q}(\beta)$ or transcendental; thus, also $\pi_\beta(\mathbf{m}_{\le}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental. $\square$

**6** | EXAMPLES OF ORDERS ON SYMMETRIC SHIFT SPACES

**6.1** | Lexicographic order

For the lexicographic order on $\{0, \pm 1\}^\infty$ (with $-1 < 0 < 1$), we have $[1\bar{1}] < [10]$, $[10\bar{1}] < [100]$, and we obtain that

$$
\mathbf{m}_{\leq \text{lex}}^{\text{abs}} = \varrho_2(\mathbf{m}_{\leq \text{alt}}) = 10\bar{1}1\bar{1}010\bar{1}01\bar{1}0\bar{1}1\bar{1}01\bar{1}0\bar{1}010\bar{1}1\bar{1}010\cdots.
$$
0$\bar{1}$, that is,

$$
\begin{align*}
\mathcal{M}_{\leq}^{\text{abs}} \cap [101] &\subseteq 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \cup 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \cup 10((10)^*1(\bar{1}0)^*\bar{1})^{\infty} \\
&= \varrho_0(1(1^{*01}0^*0)^{\infty}) \cup \varrho_0(1(1^{*01}0^*0)^{*1\infty}) \cup \varrho_0(1(1^{*01}0^*0)^{*101}^{\infty}) = \varrho_0([1]).
\end{align*}
$$

Therefore, we have $\mathbf{m}_{\leq}^{\text{abs}} = \varrho_0(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\varrho_0(\mathbf{a}) \leq \varrho_0(\mathbf{b})$. Assume now $[11] < [10]$. Then $\mathcal{M}_{\leq}^{\text{abs}} \cap [11] = \{1^{\infty}\}$, thus $\mathbf{m}_{\leq} \in [10]$. If $[10\bar{1}] < [100]$, then

$$
\mathcal{M}_{\leq}^{\text{abs}} \cap [10\bar{1}] \subseteq 10(\bar{1}^*\bar{1}01^*10)^{\infty} \cup 10(\bar{1}^*\bar{1}01^*10)^*\bar{1}^{\infty} \cup 10(\bar{1}^*\bar{1}01^*10)^*\bar{1}^*\bar{1}01^{\infty} = \varrho_1([1]),
$$

thus $\mathbf{m}_{\leq}^{\text{abs}} = \varrho_1(\mathbf{m}_{\leq})$, with $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\varrho_1(\mathbf{a}) \leq \varrho_1(\mathbf{b})$. If $[101] < [100]$, then

$$
\mathcal{M}_{\leq}^{\text{abs}} \cap [101] \subseteq 10\{1, 10\}^{\infty} = \tau_{0,1}([1]),
$$

thus $\mathbf{m}_{\leq}^{\text{abs}} = \tau_{0,1}(\mathbf{m}_{\leq})$ for the cylinder order $\leq$ defined by $\mathbf{a} \leq \mathbf{b}$ if $\tau_{0,1}(\mathbf{a}) \leq \tau_{0,1}(\mathbf{b})$. Since $\varrho_0(0^\infty) = \varrho_2(0^\infty) = (\bar{1}\bar{1})^\infty$ and $\varrho_1(0^\infty) = \tau_{0,1}(0^\infty) = 1^\infty$, Equation (10) holds.

Let now $\leq$ be a cylinder order on $\{0, 1\}^\infty$ with $0^\infty < 1^\infty$ and $\sigma \in \{\varrho_0, \varrho_1, \varrho_2, \tau_{0,1}\}$. Then, there exists a consistent cylinder order $\leq$ on $\{0, \pm 1\}^\infty$ satisfying $\sigma(\mathbf{a}) \leq \sigma(\mathbf{b})$ if $\mathbf{a} \leq \mathbf{b}$ and $0^\infty < 1^\infty$. Indeed, for $w \in \{0, \pm 1\}^*$ and distinct $a, b \in \{0, \pm 1\}$, we set $[wa] < [wb]$ if $\mathbf{a}' < \mathbf{b}'$ for some $\mathbf{a}', \mathbf{b}' \in \{0, 1\}^\infty$ with $\sigma(\mathbf{a}') \in [wa]$, $\sigma(\mathbf{b}') \in [wb]$; by Lemma 2, this does not depend on the choice of $\mathbf{a}', \mathbf{b}'$. Moreover, since $\sigma(\mathbf{a}) \in [w\bar{1}]$ and $\sigma(\mathbf{b}) \in [w1]$ is impossible, we have no obstruction to a consistent cylinder order. Since $0^\infty < 1^\infty$, we have $[\bar{1}] < [10]$ in case $\sigma \in \{\varrho_0, \varrho_2\}$, $[11] < [10]$ in case $\sigma \in \{\varrho_1, \tau_{0,1}\}$. We can set $[10\bar{1}] < [100]$ in case $\sigma \in \{\varrho_1, \varrho_2\}$ because $([100] \cup [101]) \cap \sigma(\{0, 1\})^\infty = \emptyset$; similarly, we can set $[101] < [100]$ in case $\sigma \in \{\varrho_0, \tau_{0,1}\}$. Then, we have $\mathbf{m}_{\leq}^{\text{abs}} = \sigma(\mathbf{m}_{\leq})$. $\square$

**Proposition 9.** Let $\beta \geq 3$ be a Pisot or Salem number, and let $\leq$ be a consistent cylinder order on $\{0, \pm 1\}^\infty$. Then $\pi_\beta(\mathbf{m}_{\leq}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental.

*Proof.* Let $G(a_1a_2\cdots) = |a_1| |a_2| \cdots$. Then $G \circ \varrho_l = \tau_{0,1}$ for all $l \in \{0, 1, 2\}$, thus $p_{G(\mathbf{m}_{\leq}^{\text{abs}})}(n) \le 3n-2$ by Proposition 6, Theorems 3 and 8. Moreover, the map $G$ is 2-to-1 from the set of factors of $\mathbf{m}_{\leq}^{\text{abs}}$ to the set of factors of $G(\mathbf{m}_{\leq}^{\text{abs}})$, thus $p_{\mathbf{m}_{\leq}^{\text{abs}}}(n) \le 6n-4$. By [1, Theorem 1A] and by adding 1 to each digit of $\mathbf{m}_{\leq}^{\text{abs}}$, we obtain that $\pi_\beta(\mathbf{m}_{\leq}^{\text{abs}}) + \frac{1}{\beta-1}$ is in $\mathbb{Q}(\beta)$ or transcendental; thus, also $\pi_\beta(\mathbf{m}_{\leq}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental. $\square$

**6** | EXAMPLES OF ORDERS ON SYMMETRIC SHIFT SPACES

**6.1** | Lexicographic order

For the lexicographic order on $\{0, \pm 1\}^\infty$ (with $-1 < 0 < 1$), we have $[1\bar{1}] < [10]$, $[10\bar{1}] < [100]$, and we obtain that

$$
\mathbf{m}_{\leq \text{lex}}^{\text{abs}} = \varrho_2(\mathbf{m}_{\leq \text{alt}}) = 10\bar{1}1\bar{1}010\bar{1}01\bar{1}0\bar{1}1\bar{1}01\bar{1}0\bar{1}010\bar{1}1\bar{1}010\cdots.
$$
The following proposition relates the Lagrange spectrum $\tilde{\mathcal{L}}_q$, defined in (9), and its smallest accumulation point $\tilde{\mathbf{m}}_q$ to $\mathcal{M}_{\le lex}^{\text{abs}}$ and $\mathbf{m}_{\le lex}^{\text{abs}}$; it slightly improves results of [2, 12].

**Proposition 10.** We have

$$ \tilde{\mathcal{L}}_2 = \pi_2(\mathcal{M}_{\le lex}^{\text{abs}}) \cap [0, \frac{1}{2}] \neq \pi_2(\mathcal{M}_{\le lex}^{\text{abs}}), \quad \tilde{\mathcal{L}}_3 = \pi_3(\mathcal{M}_{\le lex}^{\text{abs}}), \qquad (11) $$

$$ \pi_q(\mathcal{M}_{\le lex}^{\text{abs}}) = \tilde{\mathcal{L}}_q \cap [0, \frac{1}{q-1}] \neq \tilde{\mathcal{L}}_q \quad \text{for all integers } q \ge 4. \qquad (12) $$

For all integers $q \ge 2$, we have $\tilde{\mathbf{m}}_q = \pi_q(\mathbf{m}_{\le lex}^{\text{abs}}) = \pi_q(\varrho_2(\mathbf{m}_{\le alt}))$ and

$$ \tilde{\mathcal{L}}_q \cap [0, \tilde{\mathbf{m}}_q) = \pi_q(\mathcal{M}_{\le lex}^{\text{abs}}) \cap [0, \tilde{\mathbf{m}}_q) = \{0\} \cup \{\pi_q(\varrho_2(\tau_{0,2}^n(0^\infty))) : n \ge 0\}. $$

*Proof.* As mentioned at the beginning of Section 5, for integer $q \ge 2$, $\|xq^n\|$ can be determined by a symmetric $q$-expansion of $x$. We can assume w.l.o.g. $|x| \le \frac{1}{2}$. Let

$$ \mathcal{A}_q := \{a_1 a_2 \cdots \in A_q^\infty : |\pi_q(a_k a_{k+1} \cdots)| \le \frac{1}{2} \text{ for all } k \ge 1\}, \text{ with } \mathcal{A}_q := \{0, \pm 1, \dots, \pm \lfloor q/2 \rfloor\}. $$

For each $x \in [-\frac{1}{2}, \frac{1}{2}]$, we obtain a sequence $\mathbf{a} = a_1 a_2 \cdots \in \mathcal{A}_q$ satisfying $x = \pi_q(\mathbf{a})$ by taking $a_k = \lfloor q \tilde{T}_q^{k-1}(x) + \frac{1}{2} \rfloor$ where $\tilde{T}_q(y) := qy - \lfloor qy + \frac{1}{2} \rfloor$. Then,

$$ \|xq^n\| = |\pi_q(a_{n+1}a_{n+2}\cdots)| = \pi_q(\text{abs}(a_{n+1}a_{n+2}\cdots)). $$

Note that $\mathbf{a} \le lex \mathbf{b}$ implies $\pi_q(\mathbf{a}) \le \pi_q(\mathbf{b})$ for all $\mathbf{a}, \mathbf{b} \in \mathcal{A}_q$. Since $\mathcal{L}_{\le lex}^{\text{abs}} = \mathcal{M}_{\le lex}^{\text{abs}}$ (and a similar relation holds for larger alphabets), we obtain that

$$ \tilde{\mathcal{L}}_q = \{\pi_q(s_{\le lex}^{\text{abs}}(\mathbf{a})) : \mathbf{a} \in \mathcal{A}_q\}. $$

For $q \in \{2, 3\}$, we have $\mathcal{A}_q = \{0, \pm 1\}$, thus $\tilde{\mathcal{L}}_q \subseteq \pi_q(\mathcal{M}_{\le lex}^{\text{abs}})$. For $q \ge 3$, we have $\{0, \pm 1\}^\infty \subseteq \mathcal{A}_q$, thus $\pi_q(\mathcal{M}_{\le lex}^{\text{abs}}) \subseteq \tilde{\mathcal{L}}_q$. Since $\pi_2(\mathcal{M}_{\le lex}^{\text{abs}}) \cap [0, \frac{1}{2}] \subseteq \tilde{\mathcal{L}}_2$ and $1 = \pi_2(1^\infty) \in \pi_2(\mathcal{M}_{\le lex}^{\text{abs}}) \setminus \tilde{\mathcal{L}}_2$, this proves (11). For $q \ge 4$, we have $\pi_q(s_{\le lex}^{\text{abs}}(\mathbf{a})) \ge \pi_q((\bar{2})^\infty) = \frac{2}{q+1} > \frac{1}{q-1}$ for all $\mathbf{a} \in A_q^\infty \setminus \{0, \pm 1\}^\infty$, thus $\tilde{\mathcal{L}}_q \cap [0, \frac{1}{q-1}] \subseteq \pi_q(\mathcal{M}_{\le lex}^{\text{abs}})$. Together with $\frac{2}{q+1} \in \tilde{\mathcal{L}}_q \setminus \pi_q(\mathcal{M}_{\le lex}^{\text{abs}})$, we obtain (12). Since $\pi_q$ is order-preserving on $\mathcal{A}_q$, we obtain that $\tilde{\mathbf{m}}_q = \pi_q(\mathbf{m}_{\le lex}^{\text{abs}})$ and that $\tilde{\mathcal{L}}_q$ and $\pi_q(\mathcal{M}_{\le lex}^{\text{abs}})$ agree on $[0, \tilde{\mathbf{m}}_q)$. Since $\{\mathbf{a} \in \mathcal{M}_{\le lex}^{\text{abs}} : \mathbf{a} < \mathbf{m}_{\le lex}^{\text{abs}}\}$ is equal to $\{0^\infty\} \cup \{\varrho_2(\tau_{0,2}^n(0^\infty)) : n \ge 0\}$ by Theorem 8 and (8), this completes the proof of the proposition. $\square$

## 6.2 | Alternating lexicographic order

For the alternating lexicographic order on $\{0, \pm 1\}^\infty$ (with $-1 < 0 < 1$), we have $[11] <_{\text{alt}} [10]$ and $[10\bar{1}] <_{\text{alt}} [100]$,

$$ \mathbf{m}_{\le \text{alt}}^{\text{abs}} = \varrho_1(\mathbf{m}_{\le \text{alt}}) = 10\bar{1}\bar{1}\bar{1}010\bar{1}01110\bar{1}\bar{1}\bar{1}01110\bar{1}010\bar{1}\bar{1}\bar{1}010\cdots. $$
The following proposition relates the Lagrange spectrum $\tilde{\mathcal{L}}_q$, defined in (9), and its smallest accumulation point $\tilde{\mathbf{m}}_q$ to $\mathcal{M}_{\leq lex}^{\text{abs}}$ and $\mathbf{m}_{\leq lex}^{\text{abs}}$; it slightly improves results of [2, 12].

**Proposition 10.** We have

$$ \tilde{\mathcal{L}}_2 = \pi_2(\mathcal{M}_{\leq lex}^{\text{abs}}) \cap [0, \frac{1}{2}] \neq \pi_2(\mathcal{M}_{\leq lex}^{\text{abs}}), \quad \tilde{\mathcal{L}}_3 = \pi_3(\mathcal{M}_{\leq lex}^{\text{abs}}), \qquad (11) $$

$$ \pi_q(\mathcal{M}_{\leq lex}^{\text{abs}}) = \tilde{\mathcal{L}}_q \cap [0, \frac{1}{q-1}] \neq \tilde{\mathcal{L}}_q \quad \text{for all integers } q \geq 4. \qquad (12) $$

For all integers $q \geq 2$, we have $\tilde{\mathbf{m}}_q = \pi_q(\mathbf{m}_{\leq lex}^{\text{abs}}) = \pi_q(\varrho_2(\mathbf{m}_{\leq alt}))$ and

$$ \tilde{\mathcal{L}}_q \cap [0, \tilde{\mathbf{m}}_q) = \pi_q(\mathcal{M}_{\leq lex}^{\text{abs}}) \cap [0, \tilde{\mathbf{m}}_q) = \{0\} \cup \{\pi_q(\varrho_2(\tau_{0,2}^n(0^\infty))) : n \geq 0\}. $$

*Proof.* As mentioned at the beginning of Section 5, for integer $q \geq 2$, $\|xq^n\|$ can be determined by a symmetric $q$-expansion of $x$. We can assume w.l.o.g. $|x| \leq \frac{1}{2}$. Let

$$ \mathcal{A}_q := \{a_1 a_2 \cdots \in A_q^\infty : |\pi_q(a_k a_{k+1} \cdots)| \leq \frac{1}{2} \text{ for all } k \geq 1\}, \text{ with } \mathcal{A}_q := \{0, \pm 1, \dots, \pm \lfloor q/2 \rfloor\}. $$

For each $x \in [-\frac{1}{2}, \frac{1}{2}]$, we obtain a sequence $\mathbf{a} = a_1 a_2 \cdots \in \mathcal{A}_q$ satisfying $x = \pi_q(\mathbf{a})$ by taking $a_k = \lfloor q \tilde{T}_q^{k-1}(x) + \frac{1}{2} \rfloor$ where $\tilde{T}_q(y) := qy - \lfloor qy + \frac{1}{2} \rfloor$. Then,

$$ \|xq^n\| = |\pi_q(a_{n+1}a_{n+2}\cdots)| = \pi_q(\text{abs}(a_{n+1}a_{n+2}\cdots)). $$

Note that $\mathbf{a} \leq_{lex} \mathbf{b}$ implies $\pi_q(\mathbf{a}) \leq \pi_q(\mathbf{b})$ for all $\mathbf{a}, \mathbf{b} \in \mathcal{A}_q$. Since $\mathcal{L}_{\leq lex}^{\text{abs}} = \mathcal{M}_{\leq lex}^{\text{abs}}$ (and a similar relation holds for larger alphabets), we obtain that

$$ \tilde{\mathcal{L}}_q = \{\pi_q(s_{\leq lex}^{\text{abs}}(\mathbf{a})) : \mathbf{a} \in \mathcal{A}_q\}. $$

For $q \in \{2, 3\}$, we have $\mathcal{A}_q = \{0, \pm 1\}$, thus $\tilde{\mathcal{L}}_q \subseteq \pi_q(\mathcal{M}_{\leq lex}^{\text{abs}})$. For $q \geq 3$, we have $\{0, \pm 1\}^\infty \subseteq \mathcal{A}_q$, thus $\pi_q(\mathcal{M}_{\leq lex}^{\text{abs}}) \subseteq \tilde{\mathcal{L}}_q$. Since $\pi_2(\mathcal{M}_{\leq lex}^{\text{abs}}) \cap [0, \frac{1}{2}] \subseteq \tilde{\mathcal{L}}_2$ and $1 = \pi_2(1^\infty) \in \pi_2(\mathcal{M}_{\leq lex}^{\text{abs}}) \setminus \tilde{\mathcal{L}}_2$, this proves (11). For $q \geq 4$, we have $\pi_q(s_{\leq lex}^{\text{abs}}(\mathbf{a})) \geq \pi_q((\bar{2})^\infty) = \frac{2}{q+1} > \frac{1}{q-1}$ for all $\mathbf{a} \in A_q^\infty \setminus \{0, \pm 1\}^\infty$, thus $\tilde{\mathcal{L}}_q \cap [0, \frac{1}{q-1}] \subseteq \pi_q(\mathcal{M}_{\leq lex}^{\text{abs}})$. Together with $\frac{2}{q+1} \in \tilde{\mathcal{L}}_q \setminus \pi_q(\mathcal{M}_{\leq lex}^{\text{abs}})$, we obtain (12). Since $\pi_q$ is order-preserving on $\mathcal{A}_q$, we obtain that $\tilde{\mathbf{m}}_q = \pi_q(\mathbf{m}_{\leq lex}^{\text{abs}})$ and that $\tilde{\mathcal{L}}_q$ and $\pi_q(\mathcal{M}_{\leq lex}^{\text{abs}})$ agree on $[0, \tilde{\mathbf{m}}_q)$. Since $\{\mathbf{a} \in \mathcal{M}_{\leq lex}^{\text{abs}} : \mathbf{a} < \mathbf{m}_{\leq lex}^{\text{abs}}\}$ is equal to $\{0^\infty\} \cup \{\varrho_2(\tau_{0,2}^n(0^\infty)) : n \geq 0\}$ by Theorem 8 and (8), this completes the proof of the proposition. $\square$

## 6.2 | Alternating lexicographic order

For the alternating lexicographic order on $\{0, \pm 1\}^\infty$ (with $-1 < 0 < 1$), we have $[11] <_{\text{alt}} [10]$ and $[10\bar{1}] <_{\text{alt}} [100]$,

$$ \mathbf{m}_{\leq \text{alt}}^{\text{abs}} = \varrho_1(\mathbf{m}_{\leq \text{alt}}) = 10\bar{1}\bar{1}\bar{1}010\bar{1}01110\bar{1}\bar{1}\bar{1}01110\bar{1}010\bar{1}\bar{1}\bar{1}010\cdots. $$
## 6.3 | Bimodal order

Similarly to the unimodal order, we define the bimodal order on {$0, \pm 1\}^\infty$ by $[wa] <_{bi} [wb]$ if $a < b$ (with $-1 < 0 < 1$) and $|w|_1 + |w|_{-1}$ is even, or $a > b$ and $|w|_1 + |w|_{-1}$ is odd. Then $\mathbf{m}_{\leq_{bi}}^{\text{abs}} = \tau_{0,1}(\mathbf{m}_{\leq_{alt}}) = \mathbf{m}_{\leq_{uni}}$. We get the same result for the order defined by $[wa] < [wb]$ if $a < b$ and $|w|_1$ is even, or $a > b$ and $|w|_1$ is odd.

We also define the flipped bimodal order on {$0, \pm 1\}^\infty$ by $[wa] <_{\text{biflip}} [wb]$ if $a < b$ and $|w|_0$ is even, or $a > b$ and $|w|_0$ is odd. Then

$$ \mathbf{m}_{\leq_{\text{biflip}}}^{\text{abs}} = \rho_0(\mathbf{m}_{\leq_{\text{alt}}}) = 101\bar{1}1010101\bar{1}101\bar{1}101\bar{1}1010101\bar{1}1010\dots $$

## 6.4 | Other orders

For $e \in \{\pm 1\}^\infty$, we define a cylinder order $\leq_e$ on $\{0, \pm 1\}^\infty$ by

$$ a \leq_e b \quad \text{if } e \cdot a \leq_{\text{lex}} e \cdot b, $$

where $(e_1e_2\dots) \cdot (a_1a_2\dots) = (e_1a_1)(e_2a_2)\dots$. We know from Proposition 7 that $\pi_\beta(\mathbf{m}_{\leq_e}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental for all Pisot or Salem numbers $\beta$. However, here the value of $\pi_\beta(\mathbf{e} \cdot \mathbf{m}_{\leq_e}^{\text{abs}})$, which is the smallest accumulation point of $\{\limsup_{n\to\infty} |\sum_{k=n+1}^\infty \frac{e_k a_k}{\beta^{k-n}}| : a_1a_2\dots \in \{0, \pm 1\}^\infty\}$ when $e_1 = 1$ and $\beta \geq 3$, is more relevant. If $e$ is periodic with period $k$, then $p_{e \cdot a}(n) \leq k p_a(n)$, and hence, $\pi_\beta(\mathbf{e} \cdot \mathbf{m}_{\leq_e}^{\text{abs}})$ is also in $\mathbb{Q}(\beta)$ or transcendental for all Pisot or Salem numbers $\beta$. We do not know whether the same result holds when $e$ is aperiodic.

## ACKNOWLEDGMENTS

This work was supported by the Agence Nationale de la Recherche through the projects CODYS (ANR-18-CE40-0007), IZES (ANR-22-CE40-0011), and SymDynAr (ANR-23-CE40-0024-01), as well as by the JSPS KAKENHI Grant Number 19K03439.

## JOURNAL INFORMATION

Mathematika is owned by University College London and published by the London Mathematical Society. All surplus income from the publication of Mathematika is returned to mathematicians and mathematics research via the Society's research grants, conference grants, prizes, initiatives for early career researchers and the promotion of mathematics.

## REFERENCES

1. B. Adamczewski and Y. Bugeaud, *On the complexity of algebraic numbers. I. Expansions in integer bases*, Ann. of Math. (2) **165** (2007), no. 2, 547–565.

2. S. Akiyama and H. Kaneko, *Multiplicative analogue of Markoff-Lagrange spectrum and Pisot numbers*, Adv. Math. **380** (2021), Paper No. 107547, 39, Corrigendum, **394** (2022), 107996.

3. J.-P. Allouche, *Théorie des nombres et automates*, Thèse d'État, Bordeaux, 1983.

4. J.-P. Allouche, *A note on univoque self-Sturmian numbers*, Theor. Inform. Appl. **42** (2008), no. 4, 659–662.

5. J.-P. Allouche and M. Cosnard, *Itérations de fonctions unimodales et suites engendrées par automates*, C. R. Acad. Sci. Paris Sér. I Math. **296** (1983), no. 3, 159–162.

6. L. Balková, *Complexity for infinite words associated with quadratic non-simple Parry numbers*, J. Geom. Symmetry Phys. **7** (2006), 1–11.
## 6.3 Bimodal order

Similarly to the unimodal order, we define the bimodal order on {$0, \pm 1\}^\infty$ by $[wa] <_{bi} [wb]$ if $a < b$ (with $-1 < 0 < 1$) and $|w|_1 + |w|_{-1}$ is even, or $a > b$ and $|w|_1 + |w|_{-1}$ is odd. Then $\mathbf{m}_{\leq_{bi}}^{\text{abs}} = \tau_{0,1}(\mathbf{m}_{\leq_{alt}}) = \mathbf{m}_{\leq_{uni}}$. We get the same result for the order defined by $[wa] < [wb]$ if $a < b$ and $|w|_1$ is even, or $a > b$ and $|w|_1$ is odd.

We also define the flipped bimodal order on {$0, \pm 1\}^\infty$ by $[wa] <_{\text{biflip}} [wb]$ if $a < b$ and $|w|_0$ is even, or $a > b$ and $|w|_0$ is odd. Then

$$ \mathbf{m}_{\leq_{\text{biflip}}}^{\text{abs}} = \rho_0(\mathbf{m}_{\leq_{\text{alt}}}) = 101\bar{1}1010101\bar{1}101\bar{1}101\bar{1}1010101\bar{1}1010\dots $$

## 6.4 Other orders

For $e \in \{\pm 1\}^\infty$, we define a cylinder order $\leq_e$ on $\{0, \pm 1\}^\infty$ by

$$ a \leq_e b \quad \text{if } e \cdot a \leq_{\text{lex}} e \cdot b, $$

where $(e_1e_2\dots) \cdot (a_1a_2\dots) = (e_1a_1)(e_2a_2)\dots$. We know from Proposition 7 that $\pi_\beta(\mathbf{m}_{\leq_e}^{\text{abs}})$ is in $\mathbb{Q}(\beta)$ or transcendental for all Pisot or Salem numbers $\beta$. However, here the value of $\pi_\beta(\mathbf{e} \cdot \mathbf{m}_{\leq_e}^{\text{abs}})$, which is the smallest accumulation point of $\{\limsup_{n \to \infty} |\sum_{k=n+1}^\infty \frac{e_k a_k}{\beta^{k-n}}| : a_1 a_2 \dots \in \{0, \pm 1\}^\infty\}$ when $e_1 = 1$ and $\beta \geq 3$, is more relevant. If $e$ is periodic with period $k$, then $p_{e \cdot a}(n) \leq k p_a(n)$, and hence, $\pi_\beta(\mathbf{e} \cdot \mathbf{m}_{\leq_e}^{\text{abs}})$ is also in $\mathbb{Q}(\beta)$ or transcendental for all Pisot or Salem numbers $\beta$. We do not know whether the same result holds when $e$ is aperiodic.

## ACKNOWLEDGMENTS

This work was supported by the Agence Nationale de la Recherche through the projects CODYS (ANR-18-CE40-0007), IZES (ANR-22-CE40-0011), and SymDynAr (ANR-23-CE40-0024-01), as well as by the JSPS KAKENHI Grant Number 19K03439.

## JOURNAL INFORMATION

Mathematika is owned by University College London and published by the London Mathematical Society. All surplus income from the publication of Mathematika is returned to mathematicians and mathematics research via the Society's research grants, conference grants, prizes, initiatives for early career researchers and the promotion of mathematics.

## REFERENCES

1. B. Adamczewski and Y. Bugeaud, *On the complexity of algebraic numbers. I. Expansions in integer bases*, Ann. of Math. (2) **165** (2007), no. 2, 547–565.

2. S. Akiyama and H. Kaneko, *Multiplicative analogue of Markoff-Lagrange spectrum and Pisot numbers*, Adv. Math. **380** (2021), Paper No. 107547, 39, Corrigendum, **394** (2022), 107996.

3. J.-P. Allouche, *Théorie des nombres et automates*, Thèse d'État, Bordeaux, 1983.

4. J.-P. Allouche, *A note on univoque self-Sturmian numbers*, Theor. Inform. Appl. **42** (2008), no. 4, 659–662.

5. J.-P. Allouche and M. Cosnard, *Itérations de fonctions unimodales et suites engendrées par automates*, C. R. Acad. Sci. Paris Sér. I Math. **296** (1983), no. 3, 159–162.

6. L. Balková, *Complexity for infinite words associated with quadratic non-simple Parry numbers*, J. Geom. Symmetry Phys. **7** (2006), 1–11.
7. J. Cassaigne, *Complexité et facteurs spéciaux*, Bull. Belg. Math. Soc. Simon Stevin **4** (1997), no. 1, 67-88, Journées Montoises (Mons, 1994).

8. J. Cassaigne, *Limit values of the recurrence quotient of Sturmian sequences*, Theoret. Comput. Sci. **218** (1999), no. 1, 3-12.

9. P. Collet and J.-P. Eckmann, *Iterated maps on the interval as dynamical systems*, Progress in Physics, vol. 1, Birkhäuser, Boston, MA, 1980.

10. D. Creutz and R. Pavlov, *Low complexity subshifts have discrete spectrum*, Forum Math. Sigma **11** (2023), Paper No. e96, 22.

11. T. W. Cusick and M. E. Flahive, *The Markoff and Lagrange spectra*, Mathematical Surveys and Monographs, vol. 30, American Mathematical Society, Providence, RI, 1989.

12. A. Dubickas, *On the distance from a rational power to the nearest integer*, J. Number Theory **117** (2006), no. 1, 222-239.

13. A. Dubickas, *On a sequence related to that of Thue-Morse and its applications*, Discrete Math. **307** (2007), no. 9-10, 1082-1093.

14. G. A. FreiMAN, *Non-coincidence of the spectra of Markov and of Lagrange*, Mat. Zametki **3** (1968), 195-200.

15. J. H. Hubbard and C. T. Sparrow, *The classification of topologically expansive Lorenz maps*, Comm. Pure Appl. Math. **43** (1990), no. 4, 431-443.

16. S. Ito and T. Sadahiro, *Beta-expansions with negative bases*, Integers **9** (2009), A22, 239-259.

17. C. Kraaikamp, T. A. Schmidt, and W. Steiner, *Natural extensions and entropy of α-continued fractions*, Nonlinearity **25** (2012), no. 8, 2207-2243.

18. M. Lothaire, *Algebraic combinatorics on words*, Encyclopedia of Mathematics and its Applications, vol. 90, Cambridge University Press, Cambridge, 2002.

19. J. Milnor and W. Thurston, *On iterated maps of the interval*, Dynamical systems (College Park, MD, 1986-87), Lecture Notes in Math., vol. 1342, Springer, Berlin, 1988, pp. 465-563.

20. W. Parry, *On the β-expansions of real numbers*, Acta Math. Acad. Sci. Hungar. **11** (1960), 401-416.

21. M. Postic and L. Q. Zamboni, *ω-Lyndon words*, Theoret. Comput. Sci. **809** (2020), 39-44.

22. C. Reutenauer, *Mots de Lyndon généralisés*, Sém. Lothar. Combin. **54** (2006), Art. B54h, 16.

23. R. Siromoney, L. Mathew, V. R. Dare, and K. G. Subramanian, *Infinite Lyndon words*, Inform. Process. Lett. **50** (1994), no. 2, 101-104.

24. W. Steiner, *Digital expansions with negative real bases*, Acta Math. Hungar. **139** (2013), no. 1-2, 106-119.
7. J. Cassaigne, *Complexité et facteurs spéciaux*, Bull. Belg. Math. Soc. Simon Stevin **4** (1997), no. 1, 67-88, Journées Montoises (Mons, 1994).

8. J. Cassaigne, *Limit values of the recurrence quotient of Sturmian sequences*, Theoret. Comput. Sci. **218** (1999), no. 1, 3-12.

9. P. Collet and J.-P. Eckmann, *Iterated maps on the interval as dynamical systems*, Progress in Physics, vol. 1, Birkhäuser, Boston, MA, 1980.

10. D. Creutz and R. Pavlov, *Low complexity subshifts have discrete spectrum*, Forum Math. Sigma **11** (2023), Paper No. e96, 22.

11. T. W. Cusick and M. E. Flahive, *The Markoff and Lagrange spectra*, Mathematical Surveys and Monographs, vol. 30, American Mathematical Society, Providence, RI, 1989.

12. A. Dubickas, *On the distance from a rational power to the nearest integer*, J. Number Theory **117** (2006), no. 1, 222-239.

13. A. Dubickas, *On a sequence related to that of Thue-Morse and its applications*, Discrete Math. **307** (2007), no. 9-10, 1082-1093.

14. G. A. Freiman, *Non-coincidence of the spectra of Markov and of Lagrange*, Mat. Zametki **3** (1968), 195-200.

15. J. H. Hubbard and C. T. Sparrow, *The classification of topologically expansive Lorenz maps*, Comm. Pure Appl. Math. **43** (1990), no. 4, 431-443.

16. S. Ito and T. Sadahiro, *Beta-expansions with negative bases*, Integers **9** (2009), A22, 239-259.

17. C. Kraaikamp, T. A. Schmidt, and W. Steiner, *Natural extensions and entropy of α-continued fractions*, Nonlinearity **25** (2012), no. 8, 2207-2243.

18. M. Lothaire, *Algebraic combinatorics on words*, Encyclopedia of Mathematics and its Applications, vol. 90, Cambridge University Press, Cambridge, 2002.

19. J. Milnor and W. Thurston, *On iterated maps of the interval*, Dynamical systems (College Park, MD, 1986-87), Lecture Notes in Math., vol. 1342, Springer, Berlin, 1988, pp. 465-563.

20. W. Parry, *On the β-expansions of real numbers*, Acta Math. Acad. Sci. Hungar. **11** (1960), 401-416.

21. M. Postic and L. Q. Zamboni, *ω-Lyndon words*, Theoret. Comput. Sci. **809** (2020), 39-44.

22. C. Reutenauer, *Mots de Lyndon généralisés*, Sém. Lothar. Combin. **54** (2006), Art. B54h, 16.

23. R. Siromoney, L. Mathew, V. R. Dare, and K. G. Subramanian, *Infinite Lyndon words*, Inform. Process. Lett. **50** (1994), no. 2, 101-104.

24. W. Steiner, *Digital expansions with negative real bases*, Acta Math. Hungar. **139** (2013), no. 1-2, 106-119.