On The Structure of Dyck Languages

TL;DR

This paper proves a fundamental property of Dyck languages, showing that the closure of the one-sided Dyck language in a free monoid results in a two-sided Dyck language, revealing structural insights.

Contribution

It establishes a novel theoretical connection between one-sided and two-sided Dyck languages within free monoids.

Findings

Closure of one-sided Dyck language is a two-sided Dyck language

Provides a new structural understanding of Dyck languages

Advances formal language theory

Abstract

We prove that the closure of the one-sided Dyck language in a free monoid is a two-sided Dyck language.

Full text

Keywords: Dyck Languages, Monoid, Homomorphism, Closed set.

1. Introduction

A Dyck language (cf. [1], p.27) consists of ”well-formed” words over a finite number of pairs of parentheses. The restricted or one-sided Dyck languages $D_{n}^{{}^{\prime}*},n\geqslant 1$ are formed of the words over $n$ pairs of parentheses which are ”correct” in the usual sense, i.e. in each pair of canceling brackets an opening bracket precedes the closing bracket. Thus $([()()]\{\}[])()$ is a word in $D_{3}^{{}^{\prime}*}$.

For the unrestricted or two-sided Dyck languages $D^{*}_{n},n\geqslant 1$, the interpretation of the parentheses is different. Two parentheses of the same type are considered as formal inverses for each other. A word is considered as ”correct” if and only if successive deletion of pairs of associated parentheses of the form $()$ and of the form $)($ yields the empty word. Thus $)()(][)($ is a word in $D_{2}^{*}$.

Note that the word $)($ is a word in the two-sided Dyck language, but not in the one-sided Dyck language.

Let $D^{*}_{2}$ be a two-sided (unrestricted) Dyck language on two pairs of letters $\{a,\bar{a}\}$ and $\{b,\bar{b}\}$. Denote by $D^{{}^{\prime}*}_{2}$ the corresponding one-sided (restricted) Dyck language.

Let $M=\{a,\bar{a},b,\bar{b}\}^{*}$ be the free monoid on $\{a,\bar{a},b,\bar{b}\}$ and let $F$ be the free group on the free generators $\{a,\bar{a},b,\bar{b}\}$. We endow $F$ with the profinite topology in which all the subgroups of finite index in $F$ are the open neighborhoods of $1_{F}$. We consider the topology on $M$ induced by the embedding $M\rightarrow F$.

The main result of this paper is the following theorem.

Theorem 1**.**

The closure of $D^{{}^{\prime}*}_{2}$ in $M$ is $D^{*}_{2}$.

2. Proof of Theorem 1

Let $F_{2}$ be the free group on the generators $x$ and $y$. Let $\phi:F\rightarrow F_{2}$ be the homomorphism given by $\phi(a)=x,\phi(\bar{a})=x^{-1},\phi(b)=y$, and $\phi(\bar{b})=y^{-1}$. Then $Ker(\phi)\cap M=D^{*}_{2}$ because $Ker(\phi)$ consists of words in $F$ which reduce to $1_{F}$ when we delete the subwords $a\bar{a},\bar{a}a,b\bar{b},\bar{b}b,a^{-1}(\bar{a})^{-1},(\bar{a})^{-1}a^{-1},b^{-1}(\bar{b})^{-1}$, and $(\bar{b})^{-1}b^{-1}$. All such words constitute the two-sided Dyck language $D^{*}_{2}$.

The homomorphism $\phi$ is continuous in the profinite topologies on $F$ and $F_{2}$ because a preimage of a subgroup of finite index in $F_{2}$ is a subgroup of finite index in $F$. Therefore, $Ker(\phi)$ is closed in $F$ and $Ker(\phi)\cap M=D^{*}_{2}$ is closed in the induced topology on $M$. Thus, the closure of $D^{{}^{\prime}*}_{2}$ in $M$ is a subset of $D^{*}_{2}$.

On the other hand, let $U$ be a normal subgroup of finite index in $F$. There exists $n$ such that $(aU)^{n}=(\bar{a}U)^{n}=(bU)^{n}=(\bar{b}U)^{n}=U$. It follows that $a^{n-1}\equiv a^{-1}(modU),(\bar{a})^{n-1}\equiv(\bar{a})^{-1}(modU),b^{n-1}\equiv b^{-1}(modU),(\bar{b})^{n-1}\equiv(\bar{b})^{-1}(modU)$. Hence $\bar{a}a=({(\bar{a}a)^{-1})}^{-1}={(a^{-1}(\bar{a})^{-1})}^{-1}\equiv{(a^{n-1}(\bar{a})^{n-1})}^{-1}(modU)$ and $\bar{b}b=({(\bar{b}b)^{-1})}^{-1}={(b^{-1}(\bar{b})^{-1})}^{-1}\equiv{(b^{n-1}(\bar{b})^{n-1})}^{-1}(modU)$.

Therefore $D^{*}_{2}\subseteq\langle D^{{}^{\prime}*}_{2}\rangle U$, where $\langle D^{{}^{\prime}*}_{2}\rangle$ is the subgroup of $F$ generated by $D^{{}^{\prime}*}_{2}$ in $F$. It follows that the closure of $D^{*}_{2}$ in $F$ is a subset of the closure of $\langle D^{{}^{\prime}*}_{2}\rangle$ in $F$.

So ${\overline{D^{*}_{2}}}^{M}={\overline{D^{{}^{\prime}*}_{2}}}^{F}\cap M=\overline{D^{{}^{\prime}*}_{2}}^{M}=D^{{}^{\prime}*}_{2}$. Hence, ${\overline{D^{{}^{\prime}*}_{2}}}^{M}={\overline{D^{*}_{2}}}^{M}=D^{*}_{2}$, as required.

3. Acknowledgment

The first author would like to thank the Albert Einstein Institute of Mathematics of the Hebrew University for generous support.