On a group theoretic generalization of the Morse-Hedlund theorem

TL;DR

This paper generalizes the Morse-Hedlund theorem using group actions, establishing bounds on factor complexity of infinite words based on subgroup orbits, and characterizes Sturmian words through these bounds.

Contribution

It introduces a group-theoretic framework to extend the Morse-Hedlund theorem, connecting factor complexity with subgroup orbits and characterizing Sturmian words.

Findings

For aperiodic words, complexity is at least the number of subgroup orbits plus one.

Equality in complexity characterizes Sturmian words.

The classical Morse-Hedlund theorem is recovered as a special case.

Abstract

In their 1938 seminal paper on symbolic dynamics, Morse and Hedlund proved that every aperiodic infinite word $x\in A^N,$ over a non empty finite alphabet $A,$ contains at least $n+1$ distinct factors of each length $n.$ They further showed that an infinite word $x$ has exactly $n+1$ distinct factors of each length $n$ if and only if $x$ is binary, aperiodic and balanced, i.e., $x$ is a Sturmian word. In this paper we obtain a broad generalization of the Morse-Hedlund theorem via group actions. Given a subgroup $G$ of the symmetric group $S_n, $ let $1\leq \epsilon(G)\leq n$ denote the number of distinct $G$-orbits of $\{1,2,\ldots ,n\}.$ Since $G$ is a subgroup of $S_n,$ it acts on $A^n=\{a_1a_2\cdots a_n\,|\,a_i\in A\}$ by permutation. Thus, given an infinite word $x\in A^N$ and an infinite sequence $\omega=(G_n)_{n\geq 1}$ of subgroups $G_n \subseteq S_n,$ we consider the complexity function $p_{\omega ,x}:N \rightarrow N$ which counts for each length $n$ the number of equivalence classes of factors of $x$ of length $n$ under the action of $G_n.$ We show that if $x$ is aperiodic, then $p_{\omega, x}(n)\geq\epsilon(G_n)+1$ for each $n\geq 1,$ and moreover, if equality holds for each $n,$ then $x$ is Sturmian. Conversely, let $x$ be a Sturmian word. Then for every infinite sequence $\omega=(G_n)_{n\geq 1}$ of Abelian subgroups $G_n \subseteq S_n,$ there exists $\omega '=(G_n')_{n\geq 1}$ such that for each $n\geq 1:$ $G_n'\subseteq S_n$ is isomorphic to $G_n$ and $p_{\omega',x}(n)=\epsilon(G'_n)+1.$ Applying the above results to the sequence $(Id_n)_{n\geq 1},$ where $Id_n$ is the trivial subgroup of $S_n$ consisting only of the identity, we recover both directions of the Morse-Hedland theorem.