Balanced words in higher dimensions

TL;DR

This paper investigates higher-dimensional balanced words, demonstrating the existence of two-dimensional balanced words with irrational densities, thus answering a previously open question in the field.

Contribution

It proves that balanced words in two dimensions can have irrational densities, extending the understanding of their properties beyond rational values.

Findings

Existence of two-dimensional balanced words with irrational densities

Answer to Berthé and Tijdeman's open question

Extension of balanced word theory to higher dimensions

Abstract

For $d\ge 1$, a word $w\in \{ 0,1\}^{\Z^d}$ is called balanced if there exists $M > 0$ such that for any two rectangles $R, R^{'}\subset\Z^d$ that are translates of each other, the number of occurrences of the symbol $1$ in $R$ and $R^{'}$ differ by at most $M$. It is known that for every balanced word $w$, the asymptotic frequency of the symbol $1$ ( called the density of $w$ ) exists. In this paper we show that there exist two dimensional balanced words with irrational densities, answering a question raised by Berth\'e and Tijdeman.

Equations

$$S_w=\{g\in {\mathbb{Z}}^d:w(g)=1\}.$$

$$|\sum _{g\in R}f(g)|\le M.$$

$$\sum _{g\in R}f(g)=a|R|-|S_{w(f)}\cap R|.$$

$$||S_{w(f)}\cap R|-|S_{w(f)}\cap R^{{}^{\prime }}||=|\sum _{g\in R}f(g)-\sum _{g\in R^{{}^{\prime }}}f(g)|\le 2M.$$

$$p\cdot f(x)=\sum _g{c}_{g}f(x+g).$$

$$\theta ((n_1,\dots ,n_d)=\sum _1^d{n}_i,$$

$$u^g\cdot (f\circ \theta )(g^{{}^{\prime }})=f\circ \theta (g+g^{{}^{\prime }})=u^{\theta (g)}f(\theta (g^{{}^{\prime }})).$$

$$p_0=\prod _{i=1}^d(u_i-1),h=p_0\cdot (f\circ \theta ).$$

$$\sum _{g\in R}h(g)=\sum _{g\in R}(u^g\cdot h)(0)=(q\cdot h)(0).$$

$$q\cdot h=(\prod _{i=1}^d(u_i^{b_i}-u_i^{a_i}))\cdot (f\circ \theta ).$$

$$T_h(x,y)=(x+y,y+h(x,y)).$$

$$((u-1{)}^2\cdot x{)}_n=y_{n+1}-y_n=h(x_n,y_n)\in \{a,a-1\}.$$

$$y_{l-1}+a=y_l+1<0,$$

$$l=\max \{i\le k:x_i\le 0\}.$$

$$x_{j+1}\ge x_{j+2}\ge \cdots \ge x_k>\alpha .$$

$$\alpha <x_{j+1}\le j+1-l\le 2+\frac{1}{1-a}$$

$$l=\max \{i\le k:x_i\ge 0\}.$$

$$x_{j+1}\le x_{j+2}\le \cdots \le x_k<-\beta .$$

$$-\beta >x_{j+1}\ge l-1-j\ge -(2+\frac{1}{a}).$$

Full text

1. Introduction

A word $w\in\{0,1\}^{\mathbb{Z}}$ is called $k$-balanced if for any two blocks $B,B^{{}^{\prime}}\subset\mathbb{Z}$ of equal length, the number of occurrences of the symbol $1$ in $B$ and $B^{{}^{\prime}}$ differ by at most $k$. We will call a word $w$ balanced if it is $k$-balanced for some $k\geq 1$. In the literature $1$-balanced words are often called balanced. However, following [1], we will reserve this term for the weaker property.

Balanced words occur naturally in many different areas, including ergodic theory, number theory, and theoretical computer science ([2], [3],[6]). The study of balanced words was initiated by Morse and Hedlund ([4], [5]). They obtained a classification theorem for $1$-balanced words involving the density (asymptotic frequency of the symbol 1) of the underlying word. In particular, they proved that any $1$-balanced word $w\in\{0,1\}^{\mathbb{Z}}$ admits a density, and $1$-balanced words in $\{0,1\}^{\mathbb{Z}}$ with irrational densities are Sturmian words corresponding to certain codings of irrational rotations of the circle.

In [1], Berthé and Tijdeman studied balanced words in $\{0,1\}^{\mathbb{Z}^{d}}$ for $d>1$. They showed that higher dimensional balanced words also admit densities, but for $d\geq 2$, the density of a $1$-balanced word is always rational ([1, the main corollary]). The following question has been raised by several authors ([1],[7, Conjecture 2]) :

Question : *For $d\geq 2$, does there exist a balanced word in $\{0,1\}^{\mathbb{Z}^{d}}$ with irrational density ? *

In this paper we give an affirmative answer to this question.

Theorem 1.1**.**

For any $a\in[0,1]$ there exists a balanced word $w$ in $\{0,1\}^{\mathbb{Z}^{2}}$ with density $a$.

If $a$ is an irrational number sufficiently close to ${\frac{1}{2}}$ then our proof shows that there exists a $k$-balanced word $w$ in $\{0,1\}^{\mathbb{Z}^{2}}$ with $k=33$ that has density $a$.

2. Density of balanced words

For a set $B$, $|B|$ will denote the cardinality of $B$. For $a,b\in\mathbb{Z}$ with $a\leq b$ we will denote the set $\{a,a+1,\ldots,b-1\}$ by $[a,b)$. A rectangle $R$ in $\mathbb{Z}^{d}$ is a set of the form $[a_{1},b_{1})\times\cdots\times[a_{d},b_{d})$. For $d\geq 1$, $W_{d}$ will denote the set $\{0,1\}^{\mathbb{Z}^{d}}$, the collection of all $d$-dimensional words with $\{0,1\}$ as the alphabet set. If $w\in W_{d}$, we define $S_{w}\subset\mathbb{Z}^{d}$ by

[TABLE]

A word $w$ is said to have a *density * if there exists $a\in[0,1]$ such that $|S_{w}\cap R_{n}|/|R_{n}|\mapsto a$ as $n\mapsto\infty$ for every sequence of rectangles $\{R_{n}\}$ with $|R_{n}|\mapsto\infty$.

Suppose $d\geq 1$, $a\in[0,1]$ and $f$ is a function from $\mathbb{Z}^{d}$ to the two element set $\{a,a-1\}$. We define $w(f)\in W_{d}$ by $w(f)(g)=a-f(g)$.

Lemma 2.1**.**

Let $a$,$f$ and $w(f)$ be as above. Suppose there exists $M>0$ such that for every rectangle $R$,

[TABLE]

Then $w(f)$ is a balanced word with density $a$.

Proof. Since $f(g)=a-w(f)(g)$ for all $g\in\mathbb{Z}^{d}$, it follows that for any rectangle $R$,

[TABLE]

Hence $|{\frac{|S_{w(f)}\cap R|}{|R|}}-a|\leq{\frac{M}{|R|}}$. This implies that $|S_{w(f)}\cap R|/|R|\mapsto a$ as $|R|\mapsto\infty$, i.e., $w(f)$ has density $a$. If $R^{{}^{\prime}}$ is a translate of $R$ then the above equation also shows that

[TABLE]

Hence $w(f)$ is a $2M$-balanced word. $\Box$

For $d\geq 1$, let $Q_{d}$ denote the ring ${\mathbb{Z}}[u_{1},\ldots,u_{d}]$, the polynomial ring in $d$ commuting variables with integer coefficients. For $g=(n_{1},\ldots,n_{d})\in{\mathbb{Z}}^{d}$, we will denote the monomial $u_{1}^{n_{1}}\cdots u_{d}^{n_{d}}$ by $u^{g}$. Every element of $Q_{d}$ can be expressed as $\sum c_{g}u^{g}$, with $c_{g}\in\mathbb{Z}$ and $c_{g}=0$ for all but finitely many $g$. For $d\geq 1$, $l^{\infty}(\mathbb{Z}^{d})$ will denote the space of all bounded functions from $\mathbb{Z}^{d}$ to $\mathbb{R}$. For $p=\sum_{g}c_{g}u^{g}$ and $f\in l^{\infty}(\mathbb{Z}^{d})$ we define $p\cdot f\in l^{\infty}(\mathbb{Z}^{d})$ by

[TABLE]

Theorem 2.2**.**

Suppose $d\geq 1$ and $a\in[0,1]$. If there exists $f\in l^{\infty}(\mathbb{Z})$ such that $((u-1)^{d}\cdot f)(m)\in\{a,a-1\}$ for all $m\in\mathbb{Z}$, then there exists a balanced word $w\in W_{d}$ with density $a$.

*Proof. * Let $\theta:\mathbb{Z}^{d}\rightarrow\mathbb{Z}$ denote the homomorphism defined by

[TABLE]

and let $\theta^{*}$ denote the ring homomorphism from $Q_{d}$ to $Q_{1}$ induced by $\theta$, i.e., $\theta^{*}(\sum c_{g}u^{g})=\sum c_{g}u^{\theta(g)}$. We note that $f\circ\theta$ is an element of $l^{\infty}(\mathbb{Z}^{d})$, and for any $g,g^{{}^{\prime}}\in\mathbb{Z}^{d}$,

[TABLE]

Hence $p\cdot(f\circ\theta)=(\theta^{*}(p)\cdot f)\circ\theta$ for all $p\in Q_{d}$. We define $h\in l^{\infty}(\mathbb{Z}^{d})$ and $p_{0}\in Q_{d}$ by

[TABLE]

It is easy to see that $\theta^{*}(p_{0})=(u-1)^{d}$. Since the image of $\theta^{*}(p_{0})\cdot f$ is contained in $\{a,a-1\}$, we deduce that the image of $h=(\theta^{*}(p_{0})\cdot f)\circ\theta$ is also contained in $\{a,a-1\}$. Let $R=\prod_{i}[a_{i},b_{i})$ be an arbitrary rectangle in $\mathbb{Z}^{d}$. Let $q\in Q_{d}$ denote the polynomial $\sum_{g\in R}u^{g}$. Then

[TABLE]

We note that $q=q_{1}\cdots q_{d}$, where $q_{i}=\sum_{k\in[a_{i},b_{i})}u_{i}^{k}$. As $q\cdot h=qp_{0}\cdot(f\circ\theta)$, this implies that

[TABLE]

We note that $||s||_{\infty}=||u^{g}s||_{\infty}$ for any $s\in l^{\infty}(\mathbb{Z}^{d})$ and $g\in\mathbb{Z}^{d}$. Since the right hand side of the above equality has $2^{d}$ terms of the form $\pm u^{g}\cdot(f\circ\theta)$, we conclude that $||q\cdot h||_{\infty}\leq 2^{d}||f||_{\infty}$. In particular, $\sum_{g\in R}h(g)=(q\cdot h)(0)\leq 2^{d}||f||_{\infty}$. As $R\subset\mathbb{Z}^{d}$ is arbitrary, the given assertion follows from the previous lemma. $\Box$

It is easy to see that for any $f\in l^{\infty}(\mathbb{Z})$ the value of $(u-1)^{d}\cdot f$ at $m$ depends only on $\{f(m),f(m+1),\ldots\}$. Hence if ${\mathcal{S}}$ denotes the space of all sequences taking values in $\mathbb{R}$ then $x\mapsto(u-1)^{d}\cdot x$ is a well defined map from ${\mathcal{S}}$ to ${\mathcal{S}}$. We note the following consequence of the previous result :

Corollary 2.3**.**

Suppose $d\geq 1$ and $a\in[0,1]$. If there exists a bounded sequence $x$ such that $x_{1}=\cdots=x_{d}=0$ and $((u-1)^{d}\cdot x)_{n}$ lies in $\{a,a-1\}$ for all $n\geq 1$, then there exists a balanced word $w\in W_{d}$ with density $a$.

Proof. In view of the previous theorem it is enough to construct $f\in l^{\infty}(\mathbb{Z})$ such that $((u-1)^{d}\cdot f)(m)\in\{a,a-1\}$ for all $m\in\mathbb{Z}$. We note that $(u-1)^{d}=\sum c_{i}u^{i}$, where $c_{i}=(-1)^{d-i}{\binom{d}{i}}$. Suppose $d$ is even. We define $f:\mathbb{Z}\rightarrow\mathbb{R}$ by $f(m)=x_{m}$ if $m\geq 1$, and $f(m)=x_{d+1-m}$ if $m\leq 0$. Clearly $f$ is an element of $l^{\infty}(\mathbb{Z})$. Since $((u-1)^{d}\cdot x)_{n}$ lies in $\{a,a-1\}$ for all $n\geq 1$, and $c_{i}=c_{d-i}$ for all $i$, it follows that $((u-1)^{d}\cdot f)(m)$ lies in $\{a,a-1\}$ for all $m\in\mathbb{Z}$. If $d$ is odd, we define $f(m)=x_{m}$ if $m\geq 1$, and $f(m)=-x_{d+1-m}$ if $m\leq 0$. Since $c_{i}=-c_{d-i}$ for all $i$, we deduce that $((u-1)^{d}\cdot f)(m)$ lies in $\{a,a-1\}$ for all $m\in\mathbb{Z}$. $\Box$

3. Two dimensional balanced words

Now we consider the case when $d=2$. Let $h$ be an arbitrary function from $\mathbb{R}^{2}$ to $\{a,a-1\}$. We define a function $T_{h}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ by

[TABLE]

We define $(x_{n},y_{n})=T_{h}^{n}(0,0)$, $x=\{x_{i}\}$ and $y=\{y_{i}\}$. Since $x_{n+1}=x_{n}+y_{n}$, it follows that $(u-1)\cdot x=y$. Hence for all $n\geq 1$,

[TABLE]

As $x_{1}=x_{2}=0$, Theorem 1.1 follows from Corollary 2.3 and the following result :

Theorem 3.1**.**

For any $a\in[0,1]$, there exists $h:\mathbb{R}^{2}\rightarrow\{a,a-1\}$ such that the set $\{T_{h}^{n}(0,0):n\geq 1\}\subset\mathbb{R}^{2}$ is bounded.

*Proof. * We define $h:\mathbb{R}^{2}\rightarrow\{a,a-1\}$ as follows : $h(x,y)=a-1$ if $y+a>1$, or if both $x$ and $y+a$ are positive. Otherwise, $h(x,y)=a$. We define $(x_{n},y_{n})=T_{h}^{n}(0,0)$ and claim that $-1\leq y_{i}\leq 1$ for all $i$. Suppose this is not the case. If there exists $m$ such that $y_{m}>1$, we set $l=\min\{i:y_{i}>1\}$. Then $y_{l}=y_{l-1}+h(x_{l-1},y_{l-1})>1$. From the minimality of $l$ we deduce that $h(x_{l-1},y_{l-1})>0$. As the range of $h$ is $\{a,a-1\}$, this shows that $h(x_{l-1},y_{l-1})=a$. But this can happen only if $y_{l-1}+a\leq 1$, i.e., $y_{l}\leq 1$. Since this contradicts $y_{l}>1$, we conclude that $y_{i}\leq 1$ for all $i$. Now suppose $y_{i}<-1$ for some $i$. We define $l=\min\{i:y_{i}<-1\}$. Clearly $y_{l-1}>y_{l}$, i.e., $h(x_{l-1},y_{l-1})=y_{l}-y_{l-1}$ is negative. Hence $h(x_{l-1},y_{l-1})=a-1$ and $y_{l-1}=y_{l}+1-a$. But this implies that

[TABLE]

which is not possible since by construction $h(x_{l-1},y_{l-1})=a$ whenever $y_{l-1}+a\leq 0$. This proves the claim.

Now we will show that $x_{i}\leq\alpha={\frac{1}{1-a}}+2$ for all $i$. Suppose this is not the case. We pick any $k$ such that $x_{k}>\alpha$ and define

[TABLE]

Clearly $x_{i}>0$ for $i\in[l+1,k]$. We look at the finite sequence $y_{l},y_{l+1},\ldots,y_{k-1}$. If $y_{i}>0$ for some $i\in[l,k-1]$, then $y_{i}+a$ is also positive. Since $x_{i}>0$, this implies that $h(x_{i},y_{i})=a-1$. Hence $y_{i+1}=y_{i}+a-1$. If $y_{i}\leq 0$ then there are two possibilities. In the first case, when $y_{i}+a>0$, from the definition of $h$ it follows that $h(x_{i},y_{i})=a-1$. In particular, $y_{i+1}\leq y_{i}\leq 0$. In the second case, when $y_{i}+a\leq 0$, it is easy to see that $y_{i+1}=y_{i}+h(x_{i},y_{i})\leq 0$ irrespective of whether $h(x_{i},y_{i})=a$ or $a-1$. Combining these two cases, we observe that $y_{i+1}\leq 0$ whenever $y_{i}\leq 0$. Therefore the sequence $y_{l},y_{l+1},\ldots,y_{k-1}$ decreases by $1-a$ at each step as long as $y_{i}$ is positive, and once it becomes non-positive it remains non-positive. So there exists a unique $j\in[l,k-1]$ such that $y_{i}>0$ for $l\leq i<j$ and $y_{i}\leq 0$ for $i\geq j$. Since $x_{i+1}-x_{i}=y_{i}$, it follows that

[TABLE]

We note that $y_{l}\leq 1$, $y_{j-1}>0$, and $y_{i+1}=y_{i}-(1-a)$ for $i<j$. This shows that $(j-1-l)\leq{\frac{1}{1-a}}$. From the previous claim we see that $x_{i+1}=x_{i}+y_{i}\leq x_{i}+1$ for all $i$. As $x_{l}\leq 0$, we obtain that

[TABLE]

. This contradiction shows that $x_{i}\leq\alpha$ for all $i$.

To complete the proof of the given assertion we need to show that the sequence $\{x_{i}\}$ is also bounded from below. We set $\beta=2+{\frac{1}{a}}$ and claim that $x_{i}\geq-\beta$ for all $i$. Suppose this is not the case. We pick $k$ such that $x_{k}<-\beta$ and define

[TABLE]

Clearly $x_{i}<0$ for $i\in[l+1,k]$. As before, we look at the finite sequence $y_{l},y_{l+1},\ldots,y_{k-1}$. If $y_{i}<0$ for some $i\in[l,k-1]$, then $y_{i}+a<1$. Since $x_{i}\leq 0$, this implies that $h(x_{i},y_{i})=a$. Hence $y_{i+1}=y_{i}+a$. If $y_{i}\geq 0$ then there are two possibilities. In the first case, when $y_{i}+a\leq 1$, from the definition of $h$ it follows that $h(x_{i},y_{i})=a$. In particular, $y_{i+1}>y_{i}\geq 0$. In the second case, when $y_{i}+a>1$, it is easy to see that $y_{i+1}=y_{i}+h(x_{i},y_{i})>0$ irrespective of whether $h(x_{i},y_{i})=a$ or $a-1$. Combining these two cases, we observe that $y_{i+1}\geq 0$ whenever $y_{i}\geq 0$. Therefore the sequence $y_{l},y_{l+1},\ldots,y_{k-1}$ increases by $a$ at each step as long as $y_{i}$ is negative, and once it becomes non-negative it remains non-negative. So there exists a unique $j\in[l,k-1]$ such that $y_{i}<0$ for $l\leq i<j$ and $y_{i}\geq 0$ for $i\geq j$. Since $x_{i+1}-x_{i}=y_{i}$, it follows that

[TABLE]

We note that $y_{l}\geq-1$, $y_{j-1}<0$, and $y_{i+1}=y_{i}+a$ for $i<j$. This shows that $(j-1-l)\leq{\frac{1}{a}}$. From the previous claim we see that $x_{i+1}=x_{i}+y_{i}\geq x_{i}-1$ for all $i$. As $x_{l}\geq 0$, we obtain that

[TABLE]

This contradiction shows that $x_{i}\geq-\beta$ for all $i$. $\Box$