Minimal zero entropy subshifts can be unrestricted along any sparse set

Abstract

We present a streamlined proof of a result essentially presented by the author in [Some counterexamples in topological dynamics. Ergod. Th. & Dynam. Sys. 28(4) (2008), 1291–1322], namely that for every set $S = \{s_1, s_2, \ldots \} \subset \mathbb {N}$ of zero Banach density and finite set A, there exists a minimal zero-entropy subshift $(X, \sigma )$ so that for every sequence $u \in A^{\mathbb {Z}}$, there is $x_u \in X$ with $x_u(s_n) = u(n)$ for all $n \in \mathbb {N}$. Informally, minimal deterministic sequences can achieve completely arbitrary behavior upon restriction to a set of zero Banach density. As a corollary, this provides counterexamples to the polynomial Sarnak conjecture reported by Eisner [A polynomial version of Sarnak’s conjecture. C. R. Math. Acad. Sci. Paris 353(7) (2015), 569–572] which are significantly more general than some recently provided by Kanigowski, Lemańczyk and Radziwiłł [Prime number theorem for analytic skew products. Ann. of Math. (2) 199 (2024), 591–705] and by Lian and Shi [A counter-example for polynomial version of Sarnak’s conjecture. Adv. Math. 384 (2021), Paper no. 107765] and shows that no similar result can hold under only the assumptions of minimality and zero entropy.

1. Introduction

The well-known Sarnak conjecture states that the Möbius function $\mu $ is uncorrelated with all deterministic sequences. A sequence is called deterministic if it is the image under a continuous function of the trajectory of a point in a topological dynamical system with zero entropy (see §2 for definitions of this and other concepts not defined in this introduction). More formally, we state the Sarnak conjecture.

Conjecture 1.1. (Sarnak conjecture)

If $(X,T)$ is a topological dynamical system with zero entropy, $x_0 \in X$ , and $f \in C(X)$ , then

$$ \begin{align*} \frac{1}{N} \sum_{n=1}^N \mu(n) f(T^n x_0) \rightarrow 0. \end{align*} $$

Although this problem is still open, there are many recent works on the topic, which have made significant progress and resolved it for some classes of dynamical systems. In [1], a potential stronger ‘polynomial’ (meaning that only polynomial iterates of $x_0$ are taken rather than all) version of the Sarnak conjecture was conjectured. To rule out some degenerate examples, the assumption of minimality was added on $(X, T)$ , meaning that for every $x \in X$ , the set $\{T^n x\}$ is dense.

Conjecture 1.2. (Polynomial Sarnak conjecture [1, Conjecture 2.3])

If $(X,T)$ is a minimal topological dynamical system with zero entropy, $x_0 \in X$ , $f \in C(X)$ , and $p: \mathbb {N} \rightarrow \mathbb {N}_0$ is a polynomial, then

$$ \begin{align*} \frac{1}{N} \sum_{n=1}^N \mu(n) f(T^{p(n)} x_0) \rightarrow 0. \end{align*} $$

This conjecture is now known to be false; recently, Kanigowski, Lemańczyk, and Radziwiłł[4] and Lian and Shi [6] have separately provided counterexamples. The counterexamples from [6] are symbolically defined dynamical systems called Toeplitz subshifts, are specific to the case $p(n) = n^2$ (though they could perhaps be generalized), and attain correlation with $\mu $ arbitrarily close (but not equal to) $1$ . The counterexamples from [4] are skew products on manifolds, and although stated explicitly only for $p(n) = n^2$ , they can be applied to a much larger class called ‘almost sparse sequences’ (which includes the primes). Both constructions make strong usage of arithmetic properties of the Möbius function, but can be generalized to other arithmetic functions.

The purpose of this note is to show that even much weaker versions of Conjecture 1.2 are false, because minimal zero entropy systems can achieve any possible behavior (that is, not just correlation with $\mu $ ) along any prescribed set $S \subset \mathbb {N}$ of zero Banach density (that is, not just the image of a polynomial). One such result had already been proved by the author in [7], which already immediately refutes the polynomial Sarnak conjecture.

Theorem 1.3. [7, Corollary 5.1]

Assume that $d \in \mathbb {N}$ , $(w_n)$ is an increasing sequence of positive integers where $w_{n+1} < (w_{n+1} - w_n)^{d+1}$ for large enough n, and $(z_n)$ is any sequence in $\mathbb {T} := \mathbb {Z}/\mathbb {N}$ . Then there exists a totally minimal, totally uniquely ergodic, topologically mixing zero entropy map S on $\mathbb {T}^{2d+4}$ so that if $\pi $ is projection onto the final coordinate, $\pi (S^{w_n} \mathbf {0}) = z_n$ for sufficiently large n.

(We do not further work with the properties of unique ergodicity and topological mixing, and so do not provide definitions here. However, we do note that Theorem 1.3 shows that even adding these hypotheses to Conjecture 1.2 would not make it true.) We note that the entropy of the transformation S was never mentioned in [7]. However, S is defined as a suspension flow of a product of a toral rotation and a skew product T under a roof function $1 < g < 3$ . The skew product T is of the form $(x_1, x_2, x_3, \ldots , x_m) \mapsto (x_1 + \alpha , x_2 + f(x_1), x_3 + x_2, \ldots , x_m + x_{m-1})$ for a continuous self-map f of $\mathbb {T}$ . Since its first coordinate is an irrational rotation, known to have zero entropy, the map T also has zero entropy by Abramov’s skew product entropy formula. Then S has zero entropy as well, by Abramov’s suspension flow entropy formula.

Remark 1.4. Here are a few more relevant facts about the construction from [7].

1. (1) The map S is distal, meaning that for all $x \neq y$ , $\{d(T^n x, T^n y)\}_n$ is bounded away from $0$ .

2. (2) The roof function g is $C^{\infty }$ . The function f is $C^1$ , and for any desired k, it can be made $C^k$ by increasing the dimension m; this means that the same is true of the map S.

The second fact may be of interest since the authors of [4] prove a positive result for convergence along prime iterates of similar skew products $(x, y) \mapsto (x+\alpha , y+f(x))$ under the assumption that the function f is real analytic and provide some counterexamples with continuous f. Though the constructions are not exactly the same, and though the primes absolutely do not satisfy the assumption of Theorem 1.3, in some sense, fact (2) suggests that no $C^k$ condition is sufficient for good averaging of skew products along sparse sequences.

We note that Theorem 1.3 clearly applies to any sequence $w_n = p(n)$ for a non-constant polynomial $p: \mathbb {N} \rightarrow \mathbb {N}_0$ (possibly omitting finitely many terms), and so, by simply defining $z_n$ to be $\tfrac 12$ when $\mu (n) = 1$ and $0$ otherwise, one achieves

$$ \begin{align*} \frac{1}{N} \sum_{n=1}^N \mu(n) \pi(S^{p(n)} \textbf{0}) = \frac{0.5 |\mu^{-1}(\{1\}) \cap \{1, \ldots, N\}|}{N}, \end{align*} $$

which does not approach $0$ as $N \rightarrow \infty $ , disproving the polynomial Sarnak conjecture for every non-constant p. The same is true of any function p with polynomial growth, even for degree less than $2$ , e.g. $p(n) = \lfloor n^{1.01} \rfloor $ . Theorem 1.3 cannot be applied to more slowly growing p such as $\lfloor n \ln n \rfloor $ . However, the author proved a different result in [7] using subshifts, which applies to all sequences of zero Banach density. (A subshift is a closed shift-invariant subset of $A^{\mathbb {Z}}$ (for some finite alphabet A) endowed with the left-shift transformation.)

Specifically, [7, Corollary 3.1] states that for any sequence of zero Banach density (regardless of growth rate), there exists a minimal subshift whose points can achieve arbitrary behavior along that sequence. However, entropy was not mentioned there, and although the proof there can indeed yield a zero entropy subshift, it is not easy to verify; the construction is quite complicated to achieve $(X,T)$ which is totally minimal, totally uniquely ergodic, and topologically mixing.

In this note, we present a streamlined self-contained proof of the following result, which shows that minimal zero entropy subshifts can realize arbitrary behavior along any sequence of zero Banach density.

Theorem 1.5. For any $S = \{s_1, s_2, \ldots \} \subset \mathbb {N}$ with $d^*(S) = 0$ and any finite alphabet A, there exists a minimal zero entropy subshift $X \subset A^{\mathbb {Z}}$ so that for every $u \in A^{\mathbb {N}}$ , there is $x_u \in X$ where $x_u(s_n) = u(n)$ for all $s \in S$ .

We note that this proves that even with substantially weaker hypotheses, nothing in the spirit of the polynomial Sarnak conjecture can hold under only the assumptions of minimality and zero entropy. Even if p is only assumed to have a range of zero Banach density and $\rho : \mathbb {N} \rightarrow \mathbb {Z}$ is only assumed to have $\limsup\ ({1}/{N}) \sum _{n=1}^N |\rho (n)|> 0$ (equivalently, $\rho $ takes non-zero values on a set of positive upper density), one can define a subshift X on $\{-1,0,1\}$ and $x_u \in X$ as in Theorem 1.5 for $u(n) = \textrm {sgn}(\rho (n))$ . Then, for $f \in C(X)$ defined by $x \mapsto x(0)$ , the limit supremum of the averages

$$ \begin{align*} \frac{1}{N} \sum_{n=1}^N \rho(n) f(\sigma^{p(n)} x_u) &=\frac{1}{N} \sum_{n=1}^N \rho(n) x_u(p(n)) = \frac{1}{N} \sum_{n=1}^N \rho(n) u(n)\\ &= \frac{1}{N} \sum_{n=1}^N \rho(n) \textrm{sgn}(\rho(n)) =\frac{1}{N} \sum_{n=1}^N |\rho(n)| \end{align*} $$

is positive by assumption.

We note that when $\rho = \mu $ is the Möbius function, this means that

$$ \begin{align*} \frac{1}{N} \sum_{n=1}^N \mu(n) f(\sigma^{p(n)} x_u) \end{align*} $$

can be made to approach ${6}/{\pi ^2}$ (for $x_u$ in a minimal zero-entropy subshift), a slight improvement of [6] which showed that it could attain values arbitrarily close to ${6}/{\pi ^2}$ .

Remark 1.6. The phenomenon of a subshift achieving arbitrary values along a sequence is closely related to the notion of interpolation sets defined in [2]; see [5] for a recent survey.

2. Definitions

A topological dynamical system $(X, T)$ is defined by a compact metric space X and homeomorphism $T: X \rightarrow X$ . A subshift is a topological dynamical system defined by some finite set A (called the alphabet) and the restriction of the left shift map $\sigma : A^{\mathbb {Z}} \rightarrow A^{\mathbb {Z}}$ defined by $(\sigma x)(n) = x(n+1)$ to some closed and $\sigma $ -invariant $X \subset A^{\mathbb {Z}}$ (with the induced product topology). A subshift $(X, \sigma )$ is minimal if for every $x \in X$ , $\{\sigma ^n x\}_{n \in \mathbb {Z}}$ is dense in X.

A word over A is any finite string of symbols from A; a word $w = w(1) \cdots w(n)$ is said to be a subword of a word or infinite sequence x if there exists i so that $w(1) \cdots w(n) = x(i+1) \cdots x(i+n)$ . The language $L(X)$ of a subshift $(X, \sigma )$ is the set of all subwords of sequences in X, and for any $n \in \mathbb {N}$ , we denote $L_n(X) = L(X) \cap A^n$ . For two words $u = u(1) \cdots u(m)$ and $v = v(1) \cdots v(n)$ , denote by $uv$ their concatenation $u(1) \cdots u(m) v(1) \cdots v(n)$ .

We do not give a full definition of topological entropy here, but note that it is a number $h(X,T) \in [0, \infty ]$ associated to any TDS $(X, T)$ which is conjugacy-invariant. We will only need the following definition for subshifts: for any $(X, \sigma )$ ,

$$ \begin{align*} h(X, \sigma) = \lim \frac{\ln |L_n(X)|}{n}. \end{align*} $$

The Banach density of a set $S \subset \mathbb {N}$ is

$$ \begin{align*} d^*(S) := \lim_{n \rightarrow \infty} \sup_{k \in \mathbb{N}} \frac{|S \cap \{k, \ldots, k+n-1\}|}{n}. \end{align*} $$

3. Proof of Theorem 1.5

Proof. As in [7], we adapt the block-concatenation construction of Hahn and Katznelson [3].

We construct X iteratively via auxiliary sequences $m_k$ of odd positive integers, $A_k \subset A^{m_k}$ , and $w_k \in A_k$ . Define $m_0 = 1$ , $A_0 = A$ , and $w_0 = 0$ (which we assume without loss of generality to be in A). Now, suppose that $m_k$ , $A_k$ , and $w_k$ are defined. Define $m_{k+1}> \max (3m_k |A_k|, 12(\ln 2)(4/3)^{k+1})$ to be an odd multiple of $3m_k$ large enough that $|S \cap I|/|I| < (3m_k)^{-1}$ for all intervals I of length $m_{k+1}$ (using the fact that $d^*(S)\! = 0$ ). Define $A_{k+1}$ to be the set of all concatenations of ${m_{k+1}}/{m_k}$ words in $A_k$ in which every word in $A_k$ is used at least once and in which at least one-third of the concatenated words are equal to $w_k$ . Define $Y_k$ to be the set of shifts of biinfinite (unrestricted) concatenations of words in $A_k$ , define $Y = \bigcap _k Y_k$ , and define X to be the subshift of Y consisting of sequences in which every subword is a subword of some $w_k$ .

We claim that $(X, \sigma )$ is minimal. Indeed, consider any $x \in X$ and $w \in L(X)$ . By definition, w is a subword of $w_k$ for some k. By definition, $w_k$ is a subword of every word in $A_{k+1}$ . Finally, x is a shift of a concatenation of words in $A_{k+1}$ , each of which contains $w_k$ and therefore w. So, x contains w, and since $w \in L(X)$ was arbitrary, the orbit of x is dense. Since $x \in X$ was arbitrary, $(X, \sigma )$ is minimal.

We also claim that $(X, \sigma )$ has zero entropy. We see this by bounding $|A_k|$ from above. For every k, each word in $A_{k+1}$ is defined by an ordered $(m_{k+1}/m_k)$ -tuple of words in $A_k$ , where at least one-third are $w_k$ . The number of such tuples can be bounded from above by

$$ \begin{align*} {m_{k+1}/m_k \choose m_{k+1}/3m_k} |A_k|^{2m_{k+1}/3m_k} \leq 2^{m_{k+1}/m_k} |A_k|^{2m_{k+1}/3m_k}. \end{align*} $$

Therefore,

$$ \begin{align*} \frac{\ln |A_{k+1}|}{m_{k+1}} \leq \frac{\ln 2}{m_k} + \frac{2}{3} \frac{\ln |A_k|}{m_k}. \end{align*} $$

Now, it is easily checked that $({\ln |A_k|})/{m_k} \leq \ln |A| (3/4)^k$ for all k by induction. The base case $k = 0$ is immediate. For the inductive step, if we assume that $({\ln |A_k|})/{m_k} \leq \ln |A| (3/4)^k$ , then recalling that $m_k> 12(\ln 2) (4/3)^k$ ,

$$ \begin{align*} \frac{\ln |A_{k+1}|}{m_{k+1}}\! < \!\frac{1}{12} (3/4)^{k} + \frac{2}{3} \ln |A| (3/4)^k\! \leq\! \frac{\ln |A|}{12} (3/4)^{k} + \frac{2}{3} \ln |A| (3/4)^k = \ln |A| (3/4)^{k+1}. \end{align*} $$

Therefore, for all k, $|A_k| \leq e^{\ln |A| (3/4)^k m_k}$ . Finally, we note that every word in $L_{m_k}(X)$ is a subword of a concatenation of a pair of words in $A_k$ , so determined by such a pair and by the location of the first letter. Therefore, $|L_{m_k}(X)| \leq m_k |A_k|^2 < m_k e^{2\ln |A|(3/4)^k m_k}$ . This clearly implies that

$$ \begin{align*} h(X) = \lim_{k \rightarrow \infty} \frac{\ln |L_{m_k}(X)|}{m_k} \leq \limsup_{k \rightarrow \infty} \frac{\ln m_k}{m_k} + 2\ln |A|(3/4)^k = 0, \end{align*} $$

that is, $\mathbf {X}$ has zero entropy.

It remains, for $u \in A^{\mathbb {N}}$ , to construct $x_u \in X$ with $x_u(s_n) = u(n)$ for all $s_n \in S$ . The construction of $x_u$ proceeds in steps, where it is continually assigned letters from A on portions of $\mathbb {Z}$ , with undefined portions labeled by $*$ . Formally, define $x^{(0)} \in A \sqcup \{*\}^{\mathbb {Z}}$ by $x^{(0)}(s_n) = u(n)$ for $s \in S$ and $*$ for all other locations.

Now partition $\mathbb {Z}$ into the intervals $((i-0.5)m_1, (i+0.5)m_1)$ (herein, all intervals are assumed to be intersected with $\mathbb {Z}$ ). For every i for which $S \cap ((i-0.5)m_1, (i+0.5)m_1) \neq \varnothing $ , consider the $m_1$ -letter word $x^{(0)}(((i-0.5)m_1, (i+0.5)m_1))$ . By definition of $m_1$ , $|S \cap ((i-0.5)m_1, \ldots , (i+0.5)m_1)| < m_1/3m_0 = m_1/3$ , and so at most one-third of the letters in this word are non- $*$ . Fill the remaining locations by assigning the first $m_1/3$ as $w_0 = 0$ . At least $m_1/3$ letters remain, which is larger than $|A_0| = |A|$ by definition of $m_1$ . Fill those in an arbitrary way which uses all letters from A at least once. The resulting $m_1$ -letter word is in $A_1$ by definition, call it $w^{(1)}_i$ . Now, define $x^{(1)}$ by setting $x^{(1)}(((i-0.5)m_1, (i+0.5)m_1)) = w^{(1)}_i$ for all i as above (that is, those for which $S \cap ((i-0.5)m_1, (i+0.5)m_1) \neq \varnothing $ ) and $*$ elsewhere. Note that $x^{(1)}$ is an infinite concatenation of words in $A_1$ and blocks of $*$ of length $m_1$ and that $x^{(1)}$ contains $*$ on any interval $((i-0.5)m_1, (i+0.5)m_1)$ which is disjoint from S.

Now, suppose that $x^{(k)}$ has been defined as an infinite concatenation of words in $A_k$ and blocks of $*$ of length $m_k$ which contains $*$ on any interval $((i-0.5)m_k, (i+0.5)m_k)$ which is disjoint from S. We wish to extend $x^{(k)}$ to $x^{(k+1)}$ by changing some $*$ symbols to letters in A. Consider any i for which $S \cap ((i-0.5)m_{k+1}, \ldots , (i+0.5)m_{k+1}) \neq \varnothing $ . The portion of $x^{(k)}$ occupying that interval is a concatenation of words in $A_k$ and blocks of $*$ of length $m_k$ (we use here the fact that $m_{k+1}$ is odd), and the number which are words in $A_k$ is bounded from above by the number of $j \in ((i-0.5)m_{k+1}/m_k, (i+0.5)m_{k+1}/m_k)$ for which $((j-0.5)m_k, (j+0.5)m_k)$ is not disjoint from S, which in turn is bounded from above by $|S \cap ((i-0.5)m_{k+1}, (i+0.5)m_{k+1})|$ , which by definition of $m_{k+1}$ is less than $m_{k+1}/3m_k$ . Therefore, at least two-thirds of the concatenated $m_k$ -blocks comprising $x^{(k)}(((i-0.5)m_{k+1}, (i+0.5)m_{k+1}))$ are blocks of $*$ . Fill the first $m_{k+1}/3m_k$ of these with $w_k$ . Then at least $m_{k+1}/3m_k$ blocks remain, which is more than $|A_k|$ by definition of $m_{k+1}$ . Fill these in an arbitrary way which uses each word in $|A_k|$ at least once. By definition, this creates a word in $A_{k+1}$ , which we denote by $w^{(k+1)}_i$ . Define $x^{(k+1)}(((i-0.5)m_{k+1}, (i+0.5)m_{k+1})) = w^{(k+1)}_i$ for any i as above (that is, those for which $S \cap ((i-0.5)m_{k+1}, (i+0.5)m_{k+1}) \neq \varnothing $ ) and as $*$ elsewhere. Note that $x^{(k+1)}$ is an infinite concatenation of words in $A_{k+1}$ and blocks of $*$ of length $m_{k+1}$ which contains $*$ on any interval $((i-0.5)m_{k+1}, (i+0.5)m_{k+1})$ which is disjoint from S.

We now have defined $x^{(k)} \in (A \sqcup \{*\})^{\mathbb {Z}}$ for all $k \in \mathbb {N}$ . Since each is obtained from the previous by changing some $*$ symbols to letters from A, they approach a limit $x_u$ which agrees with $x^{(0)}$ on all locations where $x^{(0)}$ had letters from A, that is, $x_u(s_n) = u(n)$ for all $n \in \mathbb {N}$ . Since $S \neq \varnothing $ , $S \cap (-0.5m_k, 0.5m_k) \neq \varnothing $ for all large enough k, and so $x^{(k)}((-0.5m_k, 0.5m_k))$ has no $*$ , meaning that $x_u \in A^{\mathbb {Z}}$ .

It remains only to show that $x_u \in X$ . By definition, $x_u$ is a concatenation of words in $A_k$ for every k, so $x_u \in Y = \bigcap _k Y_k$ as in the definition of X. Finally, every subword w of $x_u$ is contained in $x_u((-0.5m_k, 0.5m_k))$ for large enough k, and this word is in $A_k$ by definition. Since all words in $A_k$ are subwords of $w_{k+1}$ , w is also. Therefore, by definition, ${x_u \in X}$ and ${x_u(s_n) = u(n)}$ for all n, completing the proof.

Remark 3.1. We observe that the assumption of zero Banach density cannot be weakened in Theorem 1.5. Assume for a contradiction that $S \subset \mathbb {N}$ has $d^*(S) = \alpha> 0$ , and that every $u \in A^{\mathbb {N}}$ could be assigned $x_u$ as in Theorem 1.5. By definition of Banach density, there exist intervals $I_n$ with lengths approaching infinity so that $|S \cap I_n|/|I_n|> \alpha /2$ for all n. For every n, since all possible assignments of letters from A to locations in $S \cap I_n$ give rise to sequences in X, $|L_{|I_n|}(X)| \geq 2^{|S \cap I_n|}> |A|^{\alpha |I_n|/2}$ . Then,

$$ \begin{align*} h(X) = \lim_n \frac{\ln |L_{|I_n|}(X)|}{|I_n|} \geq \limsup \frac{\ln |A|^{\alpha|I_n|/2}}{|I_n|} = \alpha (\ln |A|)/2> 0. \end{align*} $$

Therefore, no such X, minimal or otherwise, can have zero entropy.