Low complexity subshifts have discrete spectrum

Abstract

We prove results about subshifts with linear (word) complexity, meaning that $\limsup \frac {p(n)}{n} < \infty $ , where for every n, $p(n)$ is the number of n-letter words appearing in sequences in the subshift. Denoting this limsup by C, we show that when $C < \frac {4}{3}$ , the subshift has discrete spectrum, that is, is measurably isomorphic to a rotation of a compact abelian group with Haar measure. We also give an example with $C = \frac {3}{2}$ which has a weak mixing measure. This partially answers an open question of Ferenczi, who asked whether $C = \frac {5}{3}$ was the minimum possible among such subshifts; our results show that the infimum in fact lies in $[\frac {4}{3}, \frac {3}{2}]$ . All results are consequences of a general S-adic/substitutive structure proved when $C < \frac {4}{3}$ .

Markoff–Lagrange spectrum of one‐sided shifts

“…Moreover, since 𝜎(𝐚) ∈ [𝑤 1] and 𝜎(𝐛) ∈ [𝑤1] is impossible, we have no obstruction to a consistent cylinder order. Since 0 ∞ ≺ 1 ∞ , we have [11] < [10] in case 𝜎 ∈ {𝜚 0 , 𝜚 2 }, [11] < [10] ; it slightly improves results of [2,12].…”

“…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 ∈ 𝐴 * . For 𝐴 = {0, 1} with 0 < 1, we have [11] < alt [10]…”

“…if |𝑤| 1 is odd, where |𝑤| 1 denotes the number of occurrences of 1 in 𝑤 ∈ {0, 1} * . Then we have [11] < uni [10], [101] < uni [100], and…”

“…where 𝑥, 𝑦 ∈ {±1}. Assume first that [1 1] < [10]; here and in the following, we use the notation 1 = −1. Then,…”

“…If Therefore, we have 𝐦 abs ⩽ = 𝜚 0 (𝐦 ⪯ ) for the cylinder order ⪯ defined by 𝐚 ⪯ 𝐛 if 𝜚 0 (𝐚) ⩽ 𝜚 0 (𝐛). Assume now [11] < [10]. Then  abs ⩽ ∩ [11] = {1 ∞ }, thus 𝐦 ⩽ ∈ [10].…”