Finite beta-expansions with negative bases

TL;DR

This paper investigates the negative finiteness property of beta-expansions for certain Pisot numbers, including the Tribonacci number, and analyzes the length of fractional parts in their arithmetic operations.

Contribution

It identifies classes of Pisot numbers with the negative finiteness property and computes maximal fractional lengths for specific cases like the Tribonacci number.

Findings

Identified classes of Pisot numbers with the negative finiteness property

Computed maximal fractional lengths for addition and subtraction of $(-eta)$-integers

Provided conditions that exclude the negative finiteness property

Abstract

The finiteness property is an important arithmetical property of beta-expansions. We exhibit classes of Pisot numbers $\beta$ having the negative finiteness property, that is the set of finite $(-\beta)$-expansions is equal to $\mathbb{Z}[\beta^{-1}]$. For a class of numbers including the Tribonacci number, we compute the maximal length of the fractional parts arising in the addition and subtraction of $(-\beta)$-integers. We also give conditions excluding the negative finiteness property.

Equations

$$T_{-\beta }:[{\ell }_{\beta },{\ell }_{\beta }+1)\to [{\ell }_{\beta },{\ell }_{\beta }+1),x\mapsto -\beta x-\lfloor -\beta x-{\ell }_{\beta }\rfloor ,\text{mbox}with{\ell }_{\beta }=\frac{-\beta }{\beta +1};$$

$$x=\sum _{i=1}^{\infty }\frac{x_i}{(-\beta {)}^i}\text{mbox}with{x}_i=\lfloor -\beta T_{-\beta }^{i-1}(x)-{\ell }_{\beta }\rfloor \text{mbox}foralli\ge 1.$$

$$⟨x{⟩}_{-\beta }=\{\begin{array}{ll}{x}_1\cdots x_{k-1}{x}_k\bullet x_{k+1}{x}_{k+2}\cdots & \text{ if k≥1,}\\ 0\bullet x_1{x}_2{x}_3\cdots & \text{ if k=0.}\end{array}$$

$${\mathbb{Z}}_{-\beta }=\{x\in \mathbb{R}:⟨x{⟩}_{-\beta }=x_1\cdots x_{k-1}{x}_k\bullet 0^{\omega }\}=\bigcup _{k\ge 0}(-\beta {)}^k{T}_{-\beta }^{-k}(0),$$

$$\mathrm{Fin}(-\beta )=\{x\in \mathbb{R}:⟨x{⟩}_{-\beta }=x_1\cdots x_{k-1}{x}_k\bullet x_{k+1}\cdots x_{k+n}{0}^{\omega }\}=\bigcup _{n\ge 0}\frac{{\mathbb{Z}}_{-\beta }}{(-\beta {)}^n}.$$

$$\frac{-\beta }{\beta +1}=\frac{d_1}{-\beta }+\frac{d_2}{(-\beta {)}^2}+\cdots +\frac{d_k}{(-\beta {)}^k}$$

$$p(x-1)=xf(x)+p(-1),$$

$${\tau }_{\mathbf{r},\alpha }(z_0,z_1,\dots ,z_{d-1})=(z_1,\dots ,z_{d-1},z_d),$$

$$0\le r_0{z}_0+r_1{z}_1+\cdots +r_{d-1}{z}_{d-1}+z_d+\alpha <1.$$

$${\tau }_{\mathbf{r},\alpha }(z_0,z_1,\dots ,z_{d-1})=(z_1,\dots ,z_{d-1},-\lfloor \mathbf{rz}+\alpha \rfloor ),$$

$${\mathcal{D}}_{d,\alpha }^0=\{\mathbf{r}\in {\mathbb{R}}^d:\forall \mathbf{z}\in {\mathbb{Z}}^d,\exists k,{\tau }_{\mathbf{r},\alpha }^k(\mathbf{z})=0\}.$$

$$x^d+(-1)a_1{x}^{d-1}+\cdots +(-1{)}^d{a}_d=(x+\beta )(x^{d-1}+r_{d-2}{x}^{d-2}+\cdots +r_{1}x+r_0),$$

$$\displaystyle\mbox{i.e.,}\quad r_{i}=(-1)^{d-i}\bigg{(}\frac{a_{d-i}}{\beta}+\dots+\frac{a_{d}}{\beta^{i+1}}\bigg{)}\quad\mbox{for}\ i=0,1,\dots,d-2.$$

$$\begin{array}{ccc}{\mathbb{Z}}^{d-1}& \stackrel{{\tau }_{r,\alpha }}{\to }& {\mathbb{Z}}^{d-1}\\ \downarrow \varphi & & \downarrow \varphi \\ \mathbb{Z}[\beta ]\cap [{\ell }_{\beta },{\ell }_{\beta }+1)& \stackrel{T_{-\beta }}{\to }& \mathbb{Z}[\beta ]\cap [{\ell }_{\beta },{\ell }_{\beta }+1)\end{array}$$

$$T_{-\beta }(\varphi (\mathbf{z}))=-\beta \varphi (\mathbf{z})+n=\sum _{i=1}^{d-1}{r}_{i-1}{z}_i+n^{\prime }=\varphi (z_1,\dots ,z_{d-2},z_{d-1})=\varphi ({\tau }_{\mathbf{r},\alpha }(\mathbf{z})),$$

$$\sum _{1\le j\le d/k}{r}_{d-jk}>\alpha -1.$$

$$-\sum _{r_i<0}{r}_i\le \frac{a_1-2}{{\beta }^2}+\frac{a_1-2}{{\beta }^4}+\cdots +\frac{a_1-2}{{\beta }^{2\lceil d/2\rceil -2}}\le \frac{a_1-2}{{\beta }^2-1}<\frac{1}{\beta +1}$$

$$\displaystyle\frac{\beta+1}{\beta}\sum_{i=0}^{d-1}|r_{i}|\leq\frac{\beta+1}{\beta}\bigg{(}\frac{a_{2}+\dots+a_{d}}{\beta}+\frac{a_{3}+\dots+a_{d}}{\beta^{2}}+\dots+\frac{a_{d}}{\beta^{d-1}}\bigg{)}$$

$$=\frac{a_2+\cdots +a_d}{\beta }+\frac{a_2+2a_3+\cdots +2a_d}{{\beta }^2}+\frac{a_3+2a_4+\cdots +2a_d}{{\beta }^3}+\cdots +\frac{a_{d-1}+2a_d}{{\beta }^{d-1}}+\frac{a_d}{{\beta }^d}$$

$$\le \frac{a_1-2}{\beta }+\frac{2(a_1-2)-a_2}{{\beta }^2}+\frac{2(a_1-a_2-2)-a_3}{{\beta }^3}+\cdots +\frac{2(a_1-a_2-\cdots -a_{d-1}-2)-a_d}{{\beta }^d}$$

$$\displaystyle\leq 1-2\bigg{(}\frac{1}{\beta}-\frac{a_{1}-2}{\beta^{2}}-\frac{a_{1}-a_{2}-2}{\beta^{3}}-\dots-\frac{a_{1}-a_{2}-\dots-a_{d-1}-2}{\beta^{d}}\bigg{)}$$

$$\displaystyle\leq 1-\frac{2}{\beta}\bigg{(}1-\frac{a_{1}-2}{\beta-1}\bigg{)}<1.$$

$$|b+1|<a+c\text{ and }b+c^2<\mathrm{sgn}(c)(1+ac).$$

$$(-1,1)\mapsto (1,1)\mapsto (1,0)\mapsto (0,-1)\mapsto (-1,-1)\mapsto (-1,0)\mapsto (0,0),$$

$$(0,1)\mapsto (1,0),(1,-1)\mapsto (-1,-1).$$

$$(1,0)\mapsto (0,-1)\mapsto (-1,1)\mapsto (1,-1)\mapsto (-1,0)\mapsto (0,0),(0,1)\mapsto (1,-1),$$

$$(-1,-1)\mapsto (-1,1)\mapsto (1,0)\mapsto (0,-1)\mapsto (-1,0)\mapsto (0,0),$$

$$(0,1)\mapsto (1,0),(1,1)\mapsto (1,-1)\mapsto (-1,0).$$

$$(0,1)\mapsto (1,1)\mapsto (1,0)\mapsto (0,-1)\mapsto (-1,-1)\mapsto (-1,0)\mapsto (0,0),$$

$$(0,1,0,0)\mapsto (1,0,0,1)\mapsto (0,0,1,0)\mapsto (0,1,0,-1)\mapsto (1,0,-1,0)\mapsto (0,-1,0,0)\mapsto$$

$$(-1,0,0,-1)\mapsto (0,0,-1,-1)\mapsto (0,-1,-1,0)\mapsto (-1,-1,0,0)\mapsto (-1,0,0,0)\mapsto (0,0,0,0),$$

$$(0,0,-1,0)\mapsto (0,-1,0,1)\mapsto (-1,0,1,0)\mapsto (0,1,0,-1),$$

$$(0,1,1,1)\mapsto (1,1,1,1)\mapsto (1,1,1,0)\mapsto (1,1,0,-1)\mapsto (1,0,-1,-1)\mapsto$$

$$(0,-1,-1,-1)\mapsto (-1,-1,-1,-1)\mapsto (-1,-1,-1,0)\mapsto (-1,-1,0,0),$$

$$(0,0,0,1)\mapsto (0,0,1,1)\mapsto (0,1,1,0)\mapsto (1,1,0,0)\mapsto (1,0,0,0)\mapsto (0,0,0,-1)\mapsto (0,0,-1,-1),$$

$$(-1,-1,0,1)\mapsto (-1,0,1,1)\mapsto (0,1,1,0).$$

$$\max\{\operatorname{fr}(x\pm y):\,x,y\in\mathbb{Z}_{-\beta}\}=3m+\begin{cases}3&\mbox{if $m=1$ or $m$ is even},\\ 4&\mbox{if $m\geq 3$ is odd}.\end{cases}$$

Full text

1 Introduction

Digital expansions in real bases $\beta>1$ were introduced by Rényi [23]. Of particular interest are bases $\beta$ satisfying the finiteness property, or Property (F), which means that each element of $\mathbb{Z}[\beta^{-1}]\cap[0,\infty)$ has a finite (greedy) $\beta$-expansion. We know from Frougny and Solomyak [13] that each base with Property (F) is a Pisot number, but the converse is not true. Partial characterizations are due to [13, 16, 1]. In [2], Akiyama et al. exhibited an intimate connection to shift radix systems (SRS), following ideas of Hollander [16]. For results on shift radix systems (with the finiteness property), we refer to the survey [18].

Numeration systems with negative base $-\beta<-1$, or $(-\beta)$-expansions, received considerable attention since the paper [17] of Ito and Sadahiro in 2009. They are given by the $(-\beta)$-transformation

[TABLE]

see Section 2 for details. Certain arithmetic aspects seem to be analogous to those for positive base systems [12, 20], others are different, e.g., both negative and positive numbers have $(-\beta)$-expansions; for $\beta<\frac{1+\sqrt{5}}{2}$, the only number with finite $(-\beta)$-expansion is [math]. We say that $\beta>1$ has the negative finiteness property, or Property ($-$F), if each element of $\mathbb{Z}[\beta^{-1}]$ has a finite $(-\beta)$-expansion. By Dammak and Hbaib [10], we know that $\beta$ must be a Pisot number, as in the positive case. It was shown in [20] that the Pisot roots of $x^{2}-mx+n$, with positive integers $m,n$, $m\geq n+2$, satisfy the Property ($-$F). This gives a complete characterization for quadratic numbers, as $\beta$ does not possess Property ($-$F) if $\beta$ has a negative Galois conjugate, by [20].

First, we give other simple criteria when $\beta$ does not satisfy Property ($-$F). Surprisingly, this happens when $\ell_{\beta}$ has a finite $(-\beta)$-expansion, which is somewhat opposite to the positive case, where Property (F) implies that $\beta$ is a simple Parry number.

Theorem 1**.**

If $T_{-\beta}^{k}(\ell_{\beta})=0$ for some $k\geq 1$, or if $\beta$ is the root of a polynomial $p(x)\in\mathbb{Z}[x]$ with $|p(-1)|=1$, then $\beta$ does not possess Property ($-$F).

The main tool we use is a generalization of shift radix systems. We show that the $(-\beta)$-transformation is conjugated to a certain $\alpha$-SRS. Then we study properties of this dynamical system. We obtain a complete characterization for cubic Pisot units.

Theorem 2**.**

Let $\beta>1$ be a cubic Pisot unit with minimal polynomial $x^{3}-ax^{2}+bx-c$. Then $\beta$ has Property ($-$F) if and only if $c=1$ and $-1\leq b<a$, $|a|+|b|\geq 2$.

Considering Pisot numbers of arbitrary degree, we have the following results.

Theorem 3**.**

Let $\beta>1$ be a root of $x^{d}-mx^{d-1}-\dots-mx-m$ for some positive integers $d,m$. Then $\beta$ has Property ($-$F) if and only if $d\in\{1,3,5\}$.

Theorem 4**.**

Let $\beta>1$ be a root of $x^{d}-a_{1}x^{d-1}+a_{2}x^{d-2}+\dots+(-1)^{d}a_{d}\in\mathbb{Z}[x]$ with $a_{i}\geq 0$ for $i=1,\dots,d$, and $a_{1}\geq 2+\sum_{i=2}^{d}a_{i}.$ Then $\beta$ has Property ($-$F).

These theorems are proved in Section 3. In Section 4, we give a precise bound on the number of fractional digits arising from addition and subtraction of $(-\beta)$-integers in case $\beta>1$ is a root of $x^{3}-mx^{2}-mx-m$ for $m\geq 1$. This is based on an extension of shift radix systems. The corresponding numbers for $\beta$-integers have not been calculated yet, although they can be determined in a similar way.

2 ($-\beta$)-expansions

For $\beta>1$, any $x\in[\ell_{\beta},\ell_{\beta}+1)$ has an expansion of the form

[TABLE]

This gives the infinite word $d_{-\beta}(x)=x_{1}x_{2}x_{3}\cdots\in\mathcal{A}^{\mathbb{N}}$ with $\mathcal{A}=\{0,1,\ldots,\lfloor\beta\rfloor\}$. Since the base is negative, we can represent any $x\in\mathbb{R}$ without the need of a minus sign. Indeed, let $k\in\mathbb{N}$ be minimal such that $\frac{x}{(-\beta)^{k}}\in(\ell_{\beta},\ell_{\beta}+1)$ and $d_{-\beta}(\frac{x}{(-\beta)^{k}})=x_{1}x_{2}x_{3}\cdots$. Then the $(-\beta)$-expansion of $x$ is defined as

[TABLE]

Similarly to positive base numeration systems, the set of $(-\beta)$-integers can be defined using the notion of $\langle x\rangle_{-\beta}$, by

[TABLE]

where $0^{\omega}$ is the infinite repetition of zeros. The set of numbers with finite $(-\beta)$-expansion is

[TABLE]

If $\langle x\rangle_{-\beta}=x_{1}\cdots x_{k-1}x_{k}\bullet x_{k+1}\cdots x_{k+n}0^{\omega}$ with $x_{k+n}\neq 0$, then $\operatorname{fr}(x)=n$ denotes the length of the fractional part of $x$; if $x\in\mathbb{Z}_{-\beta}$, then $\operatorname{fr}(x)=0$.

3 Finiteness

In this section, we discuss the Property ($-$F) for several classes of Pisot numbers $\beta$. Note that $\operatorname{Fin}(-\beta)$ is a subset of $\mathbb{Z}[\beta^{-1}]$ since $\beta$ is an algebraic integer, hence Property ($-$F) means that $\operatorname{Fin}(-\beta)=\mathbb{Z}[\beta^{-1}]$, i.e., $\operatorname{Fin}(-\beta)$ is a ring. We start by showing that bases $\beta$ satisfying $d_{-\beta}(\ell_{\beta})=d_{1}d_{2}\dots d_{k}0^{\omega}$, which can be considered as analogs to simple Parry numbers, do not possess Property ($-$F). This was conjectured in [19] and supported by the fact that $d_{-\beta}(\ell_{\beta})=d_{1}d_{2}\dots d_{k}0^{\omega}$ with $d_{1}\geq d_{j}+2$ for all $2\leq j\leq k$ implies that $d_{-\beta}(\beta{-}1{-}d_{1})=(d_{2}{+}1)(d_{3}{+}1)\cdots(d_{k}{+}1)1^{\omega}$. However, the assumption $d_{1}\geq d_{j}+2$ is not necessary for showing that Property ($-$F) does not hold.

We also prove that a base with Property ($-$F) cannot be the root of a polynomial of the form $a_{0}x^{d}+a_{1}x^{d-1}+\dots+a_{d}$ with $|\sum_{i=0}^{d}(-1)^{i}a_{i}|=1$.

Proof of Theorem 1.

If $T_{-\beta}^{k}(\ell_{\beta})=0$, i.e., $d_{-\beta}(\ell_{\beta})=d_{1}d_{2}\dots d_{k}0^{\omega}$, then we have

[TABLE]

and thus $\frac{-1}{\beta+1}\in\mathbb{Z}[\beta^{-1}]$. However, we have $\frac{-1}{\beta+1}\notin\operatorname{Fin}(-\beta)$ since $T_{-\beta}(\frac{-1}{\beta+1})=\frac{-1}{\beta+1}$, i.e., $d_{-\beta}(\frac{-1}{\beta+1})=1^{\omega}$. Hence $\beta$ does not possess Property ($-$F).

If $p(\beta)=0$ with $|p(-1)|=1$, then write

[TABLE]

with $f(x)\in\mathbb{Z}[x]$. Then we have $\frac{1}{\beta+1}=|f(\beta+1)|\in\mathbb{Z}[\beta]$ and thus $\frac{-(-\beta)^{-j}}{\beta+1}\in\mathbb{Z}[\beta^{-1}]$ for some $j\geq 0$. Now, $d_{-\beta}(\frac{-(\beta)^{-j}}{\beta+1})=0^{j}\,1^{\omega}$ implies that $\beta$ does not have the Property ($-$F). ∎

The main tool we will be using in the rest of the paper are $\alpha$-shift radix systems. An $\alpha$-SRS is a dynamical system acting on $\mathbb{Z}^{d}$ in the following way. For $\alpha\in\mathbb{R}$, $\mathbf{r}=(r_{0},r_{1},\dots,r_{d-1})\in\mathbb{R}^{d}$, and $\mathbf{z}=(z_{0},z_{1},\dots,z_{d-1})\in\mathbb{Z}^{d}$, let $\tau_{\mathbf{r},\alpha}$ be defined as

[TABLE]

where $z_{d}$ is the unique integer satisfying

[TABLE]

Alternatively, we can say that

[TABLE]

where $\mathbf{rz}$ stands for the scalar product.

The usefulness of $\alpha$-SRS with $\alpha=0$ for the study of finiteness of $\beta$-expansions was first shown by Hollander in his thesis [16]. His approach was later formalized in [2] where the case $\alpha=0$ was extensively studied. The symmetric case with $\alpha=\frac{1}{2}$ was then studied in [4]. Finally, general $\alpha$-SRS were considered by Surer [24].

We say that $\tau_{\mathbf{r},\alpha}$ has the finiteness property if for each $\mathbf{z}\in\mathbb{Z}^{d}$ there exists $k\in\mathbb{N}$ such that $\tau^{k}_{\mathbf{r},\alpha}(\mathbf{z})=\boldsymbol{0}.$ The finiteness property of $\tau_{\mathbf{r},\alpha}$ is closely related to the Property ($-$F), thus it is desirable to study the set

[TABLE]

The following proposition shows the link between $(-\beta)$-expansions and $\alpha$-SRS.

Proposition 5**.**

Let $\beta>1$ be an algebraic integer with minimal polynomial $x^{d}+a_{1}x^{d-1}+\dots+a_{d-1}x+a_{d}$. Set $\alpha=\frac{\beta}{\beta+1}$ and let $(r_{0},r_{1},\dots,r_{d-2})\in\mathbb{R}^{d-1}$ be such that

[TABLE]

Then $\beta$ has Property ($-$F) if and only if $(r_{0},r_{1},\dots,r_{d-2})\in\mathcal{D}_{d-1,\alpha}^{0}$.

Proof.

Let $\mathbf{r}=(r_{0},r_{1},\dots,r_{d-2})$. First we show that for $\phi:\mathbf{z}\mapsto\mathbf{rz}-\lfloor\mathbf{rz}+\alpha\rfloor$ the following commutation diagram holds, i.e., the systems $(\tau_{\mathbf{r},\alpha},\mathbb{Z}^{d-1})$ and $(T_{-\beta},\mathbb{Z}[\beta]\cap[\ell_{\beta},\ell_{\beta}+1))$ are conjugated.

[TABLE]

Since $r_{i}=(-1)^{d-i-1}(\beta^{d-i-1}+a_{1}\beta^{d-i-2}+\cdots+a_{d-i-1})$ for $0\leq i\leq d-2$, the set $\{r_{i}:\,0\leq i<d\}$ with $r_{d-1}=1$ forms a basis of $\mathbb{Z}[\beta]$, hence $\phi$ is a bijection. Moreover, we have $-\beta r_{i}=r_{i-1}+c_{i}$ with $c_{i}\in\mathbb{Z}$ and $r_{-1}=0.$ For $\mathbf{z}=(z_{0},z_{1},\dots,z_{d-2})$, we have $\phi(\mathbf{z})=\sum_{i=0}^{d-1}r_{i}z_{i}$ with $z_{d-1}=-\lfloor\mathbf{rz}+\alpha\rfloor$, thus

[TABLE]

where $n$ and $n^{\prime}$ are integers; for the third equality, we have used that $T_{-\beta}(\phi(\mathbf{z}))\in[\ell_{\beta},\ell_{\beta}+1)$.

Therefore, we have $\mathbf{r}\in\mathcal{D}_{d-1,\alpha}^{0}$ if and only if for each $x\in\mathbb{Z}[\beta]\cap[\ell_{\beta},\ell_{\beta}+1)$ there exists $k\geq 0$ such that $T_{-\beta}^{k}(x)=0$. Since for each $x\in\mathbb{Z}[\beta^{-1}]\cap[\ell_{\beta},\ell_{\beta}+1)$ we have $T^{n}_{-\beta}(x)\in\mathbb{Z}[\beta]$ for some $n\in\mathbb{N}$, Property ($-$F) is equivalent to $\mathbf{r}\in\mathcal{D}_{d-1,\alpha}^{0}$. ∎

Thus the problem of finiteness of $(-\beta)$-expansions can be interpreted as the problem of finiteness of the corresponding $\alpha$-SRS. This problem is often decidable by checking the finiteness of $\alpha$-SRS expansions of a certain subset of $\mathbb{Z}^{d}$. A set of witnesses of $\mathbf{r}\in\mathbb{R}^{d}$ is a set $\mathcal{V}\subset\mathbb{Z}^{d}$ that satisfies

$\pm\mathbf{e}_{i}\in\mathcal{V}$ where $\mathbf{e}_{i}$ denotes the standard basis of $\mathbb{R}^{d},$ 2. 2.

if $\mathbf{z}\in\mathcal{V}$, then $\tau_{\mathbf{r},0}(\mathbf{z}),-\tau_{\mathbf{r},0}(-\mathbf{z})\in\mathcal{V}$.

The following proposition is due to Surer [24] and Brunotte [7].

Proposition 6**.**

Let $\alpha\in[0,1)$ and $\mathbf{r}\in\mathbb{R}^{d}$. Then $\mathbf{r}\in\mathcal{D}_{d,\alpha}^{0}$ if and only if there exists a set of witnesses that does not contain nonzero periodic elements of $\tau_{\mathbf{r},\alpha}$.

Sets of witnesses for several classes of $\mathbf{r}\in\mathbb{R}^{d}$ were derived in [3]. Exploiting their explicit form, several regions of finiteness can be determined; see in particular [3, Theorems 3.3–3.5]. An $\alpha$-SRS analogy of some of those regions was given by Brunotte [7]. Brunotte’s result, however, is unsuitable for our purposes. The next proposition gives several regions of finiteness of $\alpha$-SRS.

Proposition 7**.**

Let $\mathbf{r}=(r_{0},r_{1},\dots,r_{d-1})\in\mathbb{R}^{d}$ and $\alpha\in[0,1).$

If $\sum_{i=0}^{d-1}|r_{i}|\leq\alpha$ and $\sum_{r_{i}<0}r_{i}>\alpha-1$, then $\mathbf{r}\in\mathcal{D}_{d,\alpha}^{0}.$ 2. 2.

If $0\leq r_{0}\leq r_{1}\leq\dots\leq r_{d-1}\leq\alpha$, then $\mathbf{r}\in\mathcal{D}_{d,\alpha}^{0}$. 3. 3.

If $\sum_{i=0}^{d-1}|r_{i}|\leq\alpha$ and $r_{i}<0$ for exactly one index $i=d-k$, then $\mathbf{r}\in\mathcal{D}_{d,\alpha}^{0}$ if and only if

[TABLE]

Proof.

The set $\mathcal{V}=\{-1,0,1\}^{d}$ is closed under $\tau_{\mathbf{r},0}(\mathbf{z})$ and $-\tau_{\mathbf{r},0}(-\mathbf{z})$, hence it is a set of witnesses. For any $\mathbf{z}\in\mathcal{V}$ we have $|\mathbf{rz}|\leq\alpha$, thus $\lfloor\mathbf{rz}+\alpha\rfloor\in\{0,1\}$. Hence any periodic point of $\tau_{\mathbf{r},\alpha}$ is in $\{0,-1\}^{d}$. For $\mathbf{z}\in\{0,-1\}^{d}$ we have $\mathbf{rz}+\alpha\leq-\sum_{r_{j}<0}r_{j}+\alpha<1$. Therefore $\lfloor\mathbf{rz}+\alpha\rfloor=0$, so the only period is the trivial one. 2. 2.

In this case we take as a set of witnesses the elements of $\{-1,0,1\}^{d}$ with alternating signs, i.e., $z_{i}z_{j}\leq 0$ for any pair of indices $i<j$ such that $z_{k}=0$ for each $i<k<j$. For any $\mathbf{z}\in\mathcal{V}$ we have again $|\mathbf{rz}|\leq\alpha$, thus $\lfloor\mathbf{rz}+\alpha\rfloor\in\{0,1\}$ and $\tau_{\mathbf{r},\alpha}(\mathbf{z})\in\mathcal{V}$. Therefore, we have $\tau_{\mathbf{r},\alpha}^{n}(\mathbf{z})=(-1,0,\dots,0)$ for some $n\geq 0$, hence $\tau_{\mathbf{r},\alpha}^{n+1}(\mathbf{z})=\boldsymbol{0}$. 3. 3.

In this case we have $\mathcal{V}=\{-1,0,1\}^{d}$. As above, all periodic points of $\tau_{\mathbf{r},\alpha}$ are in $\{0,-1\}^{d}$. If $\mathbf{z}=(z_{0},z_{1},\dots,z_{d-1})$ is a periodic point with $z_{d}=-\lfloor\mathbf{rz}+\alpha\rfloor=-1$, then we must have $z_{d-k}=-1$, and consequently $z_{d-jk}=-1$ for all $1\leq j\leq d/k$. Then $z_{d}=-1$ also implies that $-\sum_{1\leq j\leq d/k}r_{d-jk}+\alpha\geq 1$, i.e., (2) does not hold. On the other hand, if (2) holds, then the vector $(z_{0},z_{1},\ldots,z_{d-1})$ with $z_{d-jk}=-1$ for $1\leq j\leq d/k$, $z_{i}=0$ otherwise, is a periodic point of $\tau_{\mathbf{r},\alpha}$. ∎

Next we prove Property ($-$F) when $\beta$ is a root of a polynomial with alternating coefficients, where the second highest coefficient is dominant.

Proof of Theorem 4.

Let $\beta>1$ be a root of $p(x)=x^{d}-a_{1}x^{d-1}+a_{2}x^{d-2}+\dots+(-1)^{d}a_{d}\in\mathbb{Z}[x]$ with $a_{i}\geq 0$ for $i=1,\dots,d$, and $a_{1}\geq 2+\sum_{i=2}^{d}a_{i}$. As $\frac{\mathrm{d}}{\mathrm{d}x}(p(x)x^{-d})\geq\frac{a_{1}}{x^{2}}-\frac{a_{1}-2}{x^{3}}>0$ for $x>1$, the polynomial $p(x)$ has a unique root $\beta>1$, and we have $\beta>a_{1}-1$ since $p(a_{1}-1)\leq-(a_{1}-1)^{d-1}+(a_{1}-2)(a_{1}-1)^{d-2}<0$. By Proposition 5, Property ($-$F) holds if and only if $(r_{0},r_{1},\dots,r_{d-2})\in\mathcal{D}_{d-1,\alpha}^{0}$, with $r_{i}=a_{d-i}\beta^{-1}-a_{d-i+1}\beta^{-2}+a_{d-i+2}\beta^{-3}-\dots+(-1)^{d-i}a_{d}\beta^{-d+i-1}$. We have

[TABLE]

and

[TABLE]

Therefore, item 1 of Proposition 7 gives that Property ($-$F) holds. ∎

Now we can classify the cubic Pisot units with Property ($-$F). The following description of cubic Pisot numbers in terms of the coefficients of the minimal polynomial is due to Akiyama [1, Lemma 1].

Lemma 8**.**

A number $\beta>1$ with minimal polynomial $x^{3}-ax^{2}+bx-c$ is Pisot if and only if

[TABLE]

Proof of Theorem 2.

Let $\beta>1$ be a cubic Pisot unit with minimal polynomial $x^{3}-ax^{2}+bx-c$. If $c=-1$, then $\beta$ has a negative conjugate, which contradicts Property ($-$F) by [20]. Therefore, we assume in the following that $c=1$. Then from Lemma 8 we have that $-a-1\leq b<a$. By Proposition 5, Property ($-$F) holds if and only if $(r_{0},r_{1})\in\mathcal{D}_{2,\alpha}^{0}$, with $(r_{0},r_{1})=(\frac{1}{\beta},\frac{b}{\beta}-\frac{1}{\beta^{2}})$ and $\alpha=\frac{\beta}{\beta+1}$. We distinguish five cases for the value of $b$.

$b=0$: If $a\geq 2$, then we have $|r_{0}|+|r_{1}|=\frac{1}{\beta}+\frac{1}{\beta^{2}}<\alpha$ and $r_{0}+r_{1}>0>\alpha-1$, so we apply item 3 of Proposition 7. If $a=1$, then we have $T^{-1}_{-\beta}(0)=\{0\}$ as $\beta<\frac{1+\sqrt{5}}{2}$, thus $\operatorname{Fin}(-\beta)=\{0\}$. 2. 2.

$b=-1$: If $a\geq 1$, then $r_{0}+r_{1}=-\frac{1}{\beta^{2}}>\frac{-1}{\beta+1}=\alpha-1$. If $a\geq 3$, then we also have $|r_{0}|+|r_{1}|<\alpha$ and use item 3 of Proposition 7. If $a=2$, then $r_{0}\approx 0.39$, $r_{1}\approx-0.55$, $\alpha\approx 0.72$, $\{-1,0,1\}^{2}$ is a set of witnesses, and Property ($-$F) holds because $\tau_{\mathbf{r},\alpha}$ acts on this set in the following way:

[TABLE]

For $a=1$, we refer to Theorem 3, which is proved below. If $a=0$, then $\beta<\frac{1+\sqrt{5}}{2}$ and thus $\operatorname{Fin}(-\beta)=\{0\}$. 3. 3.

$1\leq b\leq a-2$: For $b\geq 2$, we have $0<r_{0}<r_{1}<\alpha$ and thus $(r_{0},r_{1})\in\mathcal{D}_{2,\alpha}^{0}$ by item 2 of Proposition 7. If $b=1$, then we can use item 1 of Proposition 7 because $r_{0},r_{1}>0$ and $r_{0}+r_{1}<\alpha$. 4. 4.

$1\leq b=a-1$: We have $\beta=b+\frac{1}{\beta(\beta-1)}$. For $b\geq 3$, we have $0<r_{0}<\alpha<r_{1}<1$, the set $\{-1,0,1\}^{2}\setminus\{(1,1),(-1,-1)\}$ is a set of witnesses, and $\tau_{\mathbf{r},\alpha}$ acts on this set by

[TABLE]

thus Property ($-$F) holds. If $b=2$, then $0<r_{0}<r_{1}<\alpha$ and we can use item 2 of Proposition 7. If $b=1$, then $r_{0}\approx 0.57$, $r_{1}\approx 0.25$, $\alpha\approx 0.64$, thus $\{-1,0,1\}^{2}$ is a set of witnesses, with

[TABLE] 5. 5.

$-a-1\leq b\leq-2$: We have $-r_{0}-r_{1}+\alpha=\frac{-b-1}{\beta}+\frac{1}{\beta^{2}}+\frac{\beta}{\beta+1}>1$, thus $\tau_{\mathbf{r},\alpha}(-1,-1)=(-1,-1)$, hence $(r_{0},r_{1})\notin\mathcal{D}_{2,\alpha}^{0}$.

Therefore, $\beta$ has Property ($-$F) if and only if $-1\leq b<a$, $|a|+|b|\geq 2$. ∎

Finally, we study generalized $d$-bonacci numbers.

Proof of Theorem 3.

Let $\beta>1$ be a root of $x^{d}-mx^{d-1}-\dots-mx-m$ with $d,m\in\mathbb{N}$.

If $d=1$ (and $m\geq 2$), then $\beta$ is an integer, and Property ($-$F) follows from $\mathbb{Z}_{-\beta}=\mathbb{Z}$; see e.g. [20].

If $d=3$, then $\mathbf{r}=(\frac{m}{\beta},-\frac{m}{\beta}-\frac{m}{\beta^{2}})$, $0<r_{0}<\alpha<-r_{1}<1$, with $\alpha=\frac{\beta}{\beta+1}$, and $\tau_{\mathbf{r},\alpha}$ satisfies

[TABLE]

with $\{-1,0,1\}^{2}\setminus\{(1,-1),(-1,1)\}$ being a set of witnesses.

If $d=5$, then $\mathbf{r}=(\frac{m}{\beta},-\frac{m}{\beta}-\frac{m}{\beta^{2}},\frac{m}{\beta}+\frac{m}{\beta^{2}}+\frac{m}{\beta^{3}},-\frac{m}{\beta}-\frac{m}{\beta^{2}}-\frac{m}{\beta^{3}}-\frac{m}{\beta^{4}})$, which gives $0<r_{0}<\alpha<-r_{1}<r_{2}<-r_{3}<1$ and the $\tau_{\mathbf{r},\alpha}$-transitions

[TABLE]

Let $\mathcal{V}$ be the set of these states. We have $\pm\mathbf{e}_{i}\in\mathcal{V}$, $\mathbf{z}\in\mathcal{V}$ if and only if $-\mathbf{z}\in\mathcal{V}$ and $\tau_{\mathbf{r},0}(\mathbf{z})\in\mathcal{V}$ for all $\mathbf{z}\in\mathcal{V}$, thus $\mathcal{V}$ is a set of witnesses. As $\tau_{\mathbf{r},\alpha}^{11}(\mathbf{z})=(0,0,0,0)$ for all $\mathbf{z}\in\mathcal{V}$, $\beta$ has Property ($-$F).

For odd $d\geq 7$, Property ($-$F) does not hold since $T_{-\beta}^{d-1}(\frac{m}{\beta^{2}}+\frac{m}{\beta^{3}}+\frac{m}{\beta^{4}}-1)=\frac{m}{\beta^{2}}+\frac{m}{\beta^{3}}+\frac{m}{\beta^{4}}-1$, i.e., $\tau_{\mathbf{r},\alpha}^{d-1}(-1,0,0,-1,0,0,\dots,0)=(-1,0,0,-1,0,0,\dots,0)$. For even $d\geq 2$, we use the second condition of Theorem 1, or that $\tau_{\mathbf{r},\alpha}(-1,\dots,-1)=(-1,\dots,-1)$.

Therefore, $\beta$ has Property ($-$F) if and only if $d\in\{1,3,5\}$. ∎

4 Addition and subtraction

In this section, we consider the lengths of fractional parts arising in the addition and subtraction of $(-\beta)$-integers; we prove the following theorem.

Theorem 9**.**

Let $\beta>1$ be a root of $x^{3}-m\beta^{2}-m\beta-m$, $m\geq 1$. We have

[TABLE]

Throughout the section, let $\beta$ be as in Theorem 9, $\mathbf{r}=(r_{0},r_{1})=(\frac{m}{\beta},-\frac{m}{\beta}-\frac{m}{\beta^{2}})$ and $\alpha=\frac{\beta}{\beta+1}$. Recall that $x,y\in\mathbb{Z}_{-\beta}$ means that $T_{-\beta}^{k}(\frac{x}{(-\beta)^{k}})=0=T_{-\beta}^{k}(\frac{y}{(-\beta)^{k}})$, and $\operatorname{fr}(x\pm y)=n$ is the minimal $n\geq 0$ such that $T_{-\beta}^{k+n}(\frac{x\pm y}{(-\beta)^{k}})=0$, with $k\geq 0$ such that $\frac{x}{(-\beta)^{k}},\frac{y}{(-\beta)^{k}},\frac{x\pm y}{(-\beta)^{k}}\in(\ell_{\beta},\ell_{\beta}+1)$. To determine $\operatorname{fr}(x-y)$, set

[TABLE]

for $j\geq 0$. Then we have $s_{j}=T_{-\beta}^{j}(\frac{x-y}{(-\beta)^{k}})$ for $j\geq k$, and, for all $j\geq 0$,

[TABLE]

As $s_{0}=0$, we have $s_{j}\in\mathbb{Z}[\beta]$ for $j\geq 0$. Therefore, we extend the bijection $\phi:\,\mathbb{Z}^{2}\to\mathbb{Z}[\beta]\cap[\ell_{\beta},\ell_{\beta}+1)$ to

[TABLE]

Note that $\Phi(\mathbf{z},0)=\phi(\mathbf{z})$.

Lemma 10**.**

Let $\mathbf{z}=(z_{0},z_{1})\in\mathbb{Z}^{2}$, $h\in\{-1,0,1\}$ and $b\in\mathcal{B}$. Then

[TABLE]

Proof.

We have

[TABLE]

∎

Hence, we have $s_{j}\in\Phi(\tilde{\tau}_{\mathbf{r},\alpha}^{j}(\mathbf{0},0))$, where $\tilde{\tau}_{\mathbf{r},\alpha}$ extends $\tau_{\mathbf{r},\alpha}$ to a set-valued function by

[TABLE]

To give a bound for the sets $\tilde{\tau}_{\mathbf{r},\alpha}^{j}(\mathbf{0},0)$, let

[TABLE]

Then $\bigcup_{k\geq 0}\{A_{k},B_{k},C_{k},D_{k},E_{k},F_{k}\}$ forms a partition of $\mathbb{Z}^{2}\setminus\{(0,0),(-1,-1)\}$, with the sets $B_{0}$, $C_{0}$, $D_{0}$, $E_{0}$, $F_{0}$, and $E_{1}$ being empty, see Figure 1. If $m\geq 2$, then let

[TABLE]

If $m=1$, then we add the point $(-2,0,-1)$ to this set, i.e.,

[TABLE]

We call a point $\mathbf{z}\in\mathbb{Z}^{2}$ full if $\{\mathbf{z}\}\times\{-1,0,1\}\subset V$.

The following result is the key lemma of this section.

Lemma 11**.**

Let $x,y\in[\ell_{\beta},\ell_{\beta}+1)$ such that $x-y\in[\ell_{\beta},\ell_{\beta}+1)$. Then $T_{-\beta}^{j}(x-y)+T_{-\beta}^{j}(y)-T_{-\beta}^{j}(x)\in\Phi(V)$ for all $j\geq 0$.

To prove Lemma 11, we first determine the value of $\lfloor\mathbf{rz}+\alpha\rfloor$ for $(\mathbf{z},h)\in V$.

Lemma 12**.**

Let $\mathbf{z}=(z_{0},z_{1})\in\mathbb{Z}^{2}$ with $-m-1\leq z_{0}\leq m$, $|z_{1}|\leq m+1$ and $|z_{0}-z_{1}|\leq m+1$. Then

[TABLE]

Proof.

We have $z_{0}r_{0}+z_{1}r_{1}=z_{0}-z_{1}-z_{0}\frac{m}{\beta^{2}}+(z_{1}-z_{0})\frac{m}{\beta^{3}}$ and

[TABLE]

thus $\lfloor\mathbf{rz}+\alpha\rfloor\in z_{0}-z_{1}+\{0,1\}$.

If $z_{0}\geq 0$, then we have $-z_{0}\frac{m}{\beta^{2}}+(z_{1}-z_{0})\frac{m}{\beta^{3}}\leq(m+1)\frac{m}{\beta^{3}}<\frac{1}{\beta+1}$. If $z_{1}\leq z_{0}=-1$, then $-z_{0}\frac{m}{\beta^{2}}+(z_{1}-z_{0})\frac{m}{\beta^{3}}\leq\frac{m}{\beta^{2}}<\frac{1}{\beta+1}$. This shows that $\lfloor\mathbf{rz}+\alpha\rfloor=z_{0}-z_{1}$ in these two cases.

If $z_{1}>z_{0}=-1$, then we have $-z_{0}\frac{m}{\beta^{2}}+(z_{1}-z_{0})\frac{m}{\beta^{3}}\geq\frac{m}{\beta^{2}}+\frac{m}{\beta^{3}}>\frac{1}{\beta+1}$. Finally, if $z_{0}\leq-2$, then $-z_{0}\frac{m}{\beta^{2}}+(z_{1}-z_{0})\frac{m}{\beta^{3}}\geq 2\frac{m}{\beta^{2}}-(m-1)\frac{m}{\beta^{3}}>\frac{1}{\beta+1}$, thus $\lfloor\mathbf{rz}+\alpha\rfloor=z_{0}-z_{1}+1$ in the latter cases. ∎

Proof of Lemma 11.

We have already seen above that $T_{-\beta}^{j}(x-y)+T_{-\beta}^{j}(y)-T_{-\beta}^{j}(x)\in\Phi(\tilde{\tau}_{\mathbf{r},\alpha}^{j}(\mathbf{0},0))$. As $(\mathbf{0},0)\in V$, it suffices to show that $\tilde{\tau}_{\mathbf{r},\alpha}(V)\subseteq V$.

Let $(\mathbf{z},h)\in V$ and $b\in B$ such that

[TABLE]

i.e., $(z_{1},h-\lfloor\mathbf{rz}+\alpha\rfloor,h^{\prime})\in\tilde{\tau}_{\mathbf{r},\alpha}(\mathbf{z},h)$. If $(z_{1},h-\lfloor\mathbf{rz}+\alpha\rfloor)$ is full, then we clearly have $\tilde{\tau}_{\mathbf{r},\alpha}(\mathbf{z},h)\subset V$. Otherwise, we have to consider the possible values of $h^{\prime}$. We distinguish seven cases.

$\mathbf{z}\in\{(0,0),(-1,-1)\}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=0$.

If $m\geq 2$, then $(0,h)$ and $(-1,h)$ are full since $(0,-1)\in D_{1}$, $(0,1),(-1,1)\in A_{1}$, $(-1,0)\in A_{0}$, and $(0,0)$, $(-1,-1)$ are also full.

If $m=1$, then $(0,-1)$, $(0,0)$, $(-1,-1)$, $(-1,0)$ are full. For $h=1$, we have $h^{\prime}=-1+{\lfloor r_{0}z_{1}+r_{1}+\alpha\rfloor}+b=b-2$, thus $h^{\prime}=1$. The points $(z_{1},1,-1)$ for $z_{1}\in\{-1,0\}$ are in $V$. 2. 2.

$\mathbf{z}=(j,k)\in A_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=-k$ for $j=-1$ and $\lfloor\mathbf{rz}+\alpha\rfloor=j-k$ for $0\leq j<k$.

If $k=0$, then $\mathbf{z}=(-1,0)$, and $(0,h)$ is full for $m\geq 2$. If $m=1$, then $(0,0)$ and $(0,-1)$ are full, and $h=1$ gives that $h^{\prime}=\lfloor r_{1}+\alpha\rfloor+b=b-1\in\{-1,0\}$, thus $\tilde{\tau}_{\mathbf{r},\alpha}(\mathbf{z},1)\subset V$.

If $1\leq k<m$, then $(k,h+k)$, $(k,h-j+k)$ lie in $B_{k}\cup C_{k}\cup\{(k,k+1)\}$ and are full.

If $k=m$, then we have either $h\in\{-1,0\}$, thus $(m,h+m)$ and $(m,h-j+m)$ lie in the set of full points $B_{m}\cup C_{m}$, or $h=1$ and $1\leq j<m$, in which case $(m,1-j+m)\in B_{m}$ is also full. (Note that $(-1,m,1),(0,m,1)\notin V$.) 3. 3.

$\mathbf{z}=(k,j)\in B_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=k-j$, $1\leq j\leq k$.

For $h\in\{0,1\}$, the point $(j,h+j-k)$ is in $C_{k}$ and $C_{k-1}\cup B_{k}$ respectively, hence full. The point $(j,j-k-1)\in C_{k+1}$ is full if $k<m$. Finally, if $k=m$ and $h=-1$, then $h^{\prime}=m+\lfloor r_{0}j+r_{1}(j-m-1)+\alpha\rfloor+b=2m+1+b=1$, and $(j,j-m-1,1)\in V$. 4. 4.

$\mathbf{z}=(j,j-k)\in C_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=k$, $1\leq j\leq k$.

The point $(j-k,h-k)$, $h\in\{-1,0,1\}$, is in $D_{k+1}$, $D_{k}$, and $D_{k-1}\cup E_{k-1}\cup\{(0,0),(-1,-1)\}$ respectively, hence full for all $k<m$, $k\leq m$, and $k\leq m+1$ respectively. It remains to consider $h=-1$, $k=m$. For $1\leq j\leq m-3$, the point $(j-m,-m-1)$ is full; we have

[TABLE]

$(0,-m-1,1)\in V$, and $\{(j-m,-m-1)\}\times\{0,1\}\subset V$ for $\max(1,m-2)\leq j<m$. 5. 5.

$\mathbf{z}=(-j,-k)\in D_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=k-j$ if $j\in\{0,1\}$, $\lfloor\mathbf{rz}+\alpha\rfloor=k-j+1$ if $2\leq j<k$.

Let first $k=1$, i.e., $\mathbf{z}=(0,-1)$. The point $(-1,h-1)$ lies in $D_{2}\cup\{(-1,-1)\}\cup A_{0}$ and is full, except if $m=1$, $h=-1$; in the latter case, we have $h^{\prime}=1+\lfloor-r_{0}-2r_{1}+\alpha\rfloor+b=b+2\in\{0,1\}$, and $\{(-1,-2)\}\times\{0,1\}\in V$.

For $2\leq k\leq m$, the points $(-k,h+j-k)$, $j\in\{0,1\}$, and $(-k,h+j-k-1)$, $2\leq j<k$, lie in $\{(-k,-i):\,0\leq i\leq k+1\}$, and are full, except for $k=m=2$, $h=-1$, $j=0$; in the latter case, we have $h^{\prime}=2+\lfloor-2r_{0}-3r_{1}+\alpha\rfloor+b=b+4\in\{0,1\}$, and $\{(-2,-3)\}\times\{0,1\}\in V$.

Finally, for $k=m+1$, we have $j=0$, $h=1$, or $1\leq j\leq\min(m,2)$, $h\in\{0,1\}$, or $3\leq j\leq m$, $h\in\{-1,0,1\}$, thus the points $(-m-1,h+j-m-1)$, $j\in\{0,1\}$, and $(-m-1,h+j-m-2)$, $2\leq j\leq m$, lie in $\{(-m-1,-i):\,\min(m-1,1)\leq i\leq m\}$ and are full, except for $m=j=h=1$; in the latter case, we have $h^{\prime}=-1+\lfloor-2r_{0}+\alpha\rfloor+b=b-2=-1$, and $(-2,0,-1)\in V$. 6. 6.

$\mathbf{z}=(-k,-j)\in E_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=j-k+1$, $2\leq j\leq k$.

The point $(-j,h-j+k-1)\in F_{k-2}\cup F_{k-1}\cup F_{k}\cup\{(-k,-2)\}$ is full, except for $k=m+1$, $h=1$; in the latter case, we have $2\leq j\leq m$, $h^{\prime}=\lfloor-r_{0}j+r_{1}(m-j+1)+\alpha\rfloor+b=b-m\in\{-1,0\}$, and $\{(-j,m-j+1)\}\times\{-1,0\}\subset V$. 7. 7.

$\mathbf{z}=(-j,k-j)\in F_{k}$: We have $\lfloor\mathbf{rz}+\alpha\rfloor=1-k$, $2\leq j\leq k+1$.

If $1\leq k\leq m$, then the point $(k-j,h+k-1)\in A_{k-2}\cup A_{k-1}\cup A_{k}\cup\{(k-2,k-2)\}$ is full, except for $k=m$, $j\in\{m,m+1\}$, $h=1$; in the latter case, we have $h^{\prime}=\lfloor r_{0}(m-j)+r_{1}m+\alpha\rfloor+b=b-m\in\{-1,0\}$, and $\{(m-j,m)\}\times\{-1,0\}\subset V$.

If $k=m+1$, then $2\leq j\leq m$, $h\in\{-1,0\}$, or $m=1$, $j=2$, $h=-1$, and ${(m+1-j,h+m)}\in A_{m-1}\cup A_{m}\cup\{(m-1,m-1)\}$ is full. ∎

Lemma 13**.**

For the following chains of sets, $\tau_{\mathbf{r},\alpha}$ maps elements of a set into its successor:

[TABLE]

On the remaining $\mathbf{z}=(z_{0},z_{1})\in\mathbb{Z}^{2}$ with $-m-1\leq z_{0}\leq m$, $-m-1\leq z_{1}\leq m$ and $|z_{0}-z_{1}|\leq m+1$, $\tau_{\mathbf{r},\alpha}$ acts by

[TABLE]

Proof.

This is a direct consequence of Lemma 12, except for $(-m-2,-1)\in F_{m+1}$; see also the proof of Lemma 11. As $\frac{1}{\beta+1}<\frac{(m+2)m}{\beta^{2}}+\frac{(m+1)m}{\beta^{3}}<1+\frac{m}{\beta^{2}}<1+\frac{1}{\beta+1}$, the proof of Lemma 12 shows that $\tau_{\mathbf{r},\alpha}(-m-2,-1)=(-1,m)\in A_{m}$. ∎

Proposition 14**.**

We have

[TABLE]

Proof.

Let $k\geq 0$ be such that $\frac{x}{(-\beta)^{k}},\frac{y}{(-\beta)^{k}},\frac{x-y}{(-\beta)^{k}}\in(\ell_{\beta},\ell_{\beta}+1)$. Then $\operatorname{fr}(x-y)=n$ is the minimal $n\geq 0$ such that $T_{-\beta}^{k+n}(\frac{x-y}{(-\beta)^{k}})=0$. Let $\mathbf{z}\in\mathbb{Z}^{2}$ be such that $T_{-\beta}^{k}(\frac{x-y}{(-\beta)^{k}})=\phi(\mathbf{z})$. Then $\operatorname{fr}(x-y)$ is the minimal $n\geq 0$ such that $\tau_{\mathbf{r},\alpha}^{n}(\mathbf{z})=0$, and we have $(\mathbf{z},0)\in V$, i.e.,

[TABLE]

Therefore, $\operatorname{fr}(x-y)$ is bounded by the maximal length of the path from $\mathbf{z}$ to $(0,0)$ given by Lemma 13.

For $1\leq k\leq m/2$, the sets $F_{2k+1}$, $A_{2k}$, $B_{2k}$, and $C_{2k}$ are mapped to $D_{2}$ in $6k-2$, $6k-3$, $6k-4$, and $6k-5$ steps respectively. For $2\leq k\leq(m+1)/2$, the sets $D_{2k}$ and $E_{2k}$ are mapped to $D_{2}$ in $6k-6$ and $6k-7$ steps respectively. The points $(0,-2)$ and $(-1,-2)$ in $D_{2}$ are mapped to $(0,0)$ in 4 and 3 steps respectively.

Similarly, for $1\leq k\leq(m+1)/2$, the sets $F_{2k}$, $A_{2k-1}$, $B_{2k-1}$, and $C_{2k-1}$ are mapped to $D_{1}=\{(0,-1)\}$ in $6k-2$, $6k-3$, $6k-4$, and $6k-5$ steps respectively. For $1\leq k\leq m/2$, the sets $D_{2k+1}$ and $E_{2k+1}$ are mapped to $D_{1}$ in $6k$ and $6k-1$ steps respectively. Finally, the point $(0,-1)\in D_{1}$ is mapped to $(0,0)$ in 3 steps.

For even $m$, the longest path comes thus from $D_{m+1}$ and has length $3m+3$. For odd $m\geq 3$, the longest path comes from $F_{m+1}$ and has length $3m+4$. For $m=1$, the longest path comes from $A_{1}$ (since $F_{2}\times\{0\}\cap V=\emptyset$ in this case) and has length 6. This proves the upper bound for $\operatorname{fr}(x-y)$.

For $m=1$, this bound is attained by $x=1-\beta$, $y=\beta^{4}-\beta^{3}$, since $\operatorname{fr}(x-y)=\operatorname{fr}(\frac{1}{\beta^{3}}-\beta^{3})=\operatorname{fr}(\frac{1}{\beta^{3}})=6$. Assume in the following that $m\geq 2$. Then the points $(-m,-m-1,0)\in D_{m+1}\times\{0\}$ and ${(-2,m-1,0)}\in F_{m+1}\times\{0\}$ are in $\tilde{\tau}_{\mathbf{r},\alpha}^{j}(0,0,0)$ for sufficiently large $j$ because they can be attained from $(0,0,0)$ by transitions

[TABLE]

with $b\in B$ by the following paths (cf. the proof of Lemma 11):

[TABLE]

For even $m\geq 2$, these paths correspond to

[TABLE]

for the second equality, we have used that $(1mmm00)^{m/2}\,0mmm\bullet{d_{-\beta}(\phi(-m,-m-1))}$ is a $(-\beta)$-expansion. Indeed, this follows from the lexicographic conditions given in [17] since $d_{-\beta}(\ell_{\beta})=m0m^{\omega}$ and $d_{-\beta}(\phi(-m,-m-1))$ starts with $2$ (as $-\beta\phi(-m,-m-1)-2=\phi(-m-1,-2)$). For odd $m\geq 3$, we have

[TABLE]

This concludes the proof of the proposition. ∎

Proposition 15**.**

We have

[TABLE]

Proof.

Let $\mu=\max\{\operatorname{fr}(x-y):\,x,y\in\mathbb{Z}_{-\beta}\}$. For $x,y\in\mathbb{Z}_{-\beta}$, $\operatorname{fr}(x+y)$ is the minimal $n\geq 0$ such that $T_{-\beta}^{k+n}(\frac{x+y}{(-\beta)^{k}})=0$, with $k\geq 0$ such that $\frac{x}{(-\beta)^{k}},\frac{y}{(-\beta)^{k}},\frac{x+y}{(-\beta)^{k}}\in(\ell_{\beta},\ell_{\beta}+1)$. By Lemma 11, we have

[TABLE]

for all $j\geq 0$, thus $T_{-\beta}^{k}(\frac{x+y}{(-\beta)^{k}})\in-\Phi(V)$. Therefore, we have $T_{-\beta}^{k}(\frac{x+y}{(-\beta)^{k}})=\phi(\mathbf{z})=-\Phi(-\mathbf{z},h)$ for some $\mathbf{z}=(z_{0},z_{1})\in\mathbb{Z}^{2}$ and $h\in\{0,1\}$ with $(-\mathbf{z},h)\in V$.

If $(\mathbf{z},0)\in V$, then the proof of Proposition 14 shows that $\tau_{\mathbf{r},\alpha}^{\mu}(\mathbf{z})=\mathbf{0}$, thus $\operatorname{fr}(x+y)\leq\mu$.

Assume now that $(\mathbf{z},0)\notin V$. Then

[TABLE]

We can exclude $-\mathbf{z}=(-j,m-j+1)$, $1\leq j\leq m$, because this would imply $h=0$ and

[TABLE]

This means that $\mathbf{z}\in(A_{m+1}\cup B_{m+1})\setminus\{(-1,m+1),(m+1,m+1)\}$. With the notation of Lemma 13, we have

[TABLE]

where we have used Lemma 12 and that $\lfloor r_{0}(m+1)+r_{1}j+\alpha\rfloor=m-j$ for $1\leq j\leq m$, as

[TABLE]

Hence, the points in $A_{m+1}\setminus\{(-1,m+1)\}$ and $B_{m+1}\setminus\{(m+1,m+1)\}$ are mapped to $(0,0)$ in the same number of steps as those in $A_{m}$ and $B_{m}$ respectively, thus $\operatorname{fr}(x+y)\leq\mu$. ∎

Now, Theorem 9 is an immediate consequence of Propositions 14 and 15.

Remark 16**.**

It is also possible to determine the exact value of $\max\{\operatorname{fr}(x+y):\,x,y\in\mathbb{Z}_{-\beta}\}$ in the same fashion as in the proof of Proposition 14; we have

[TABLE]