Proof. Let $P_1,P_2,P_3$ be the real polynomials appearing in Lemma 2.3 and Q be the real polynomial appearing in equation (2.1). Since K is compact, $w=\max \{|\psi (g)|:g\in K\}$ is a positive real number. Put $\widetilde Q(n)=Q(n,w,w,\ldots ,w)$ and $\widetilde P_2(n)=w^{n_G}P_2(n)$ , where $n_G$ is the constant appearing in Lemma 2.3(2).

Let ${\mathbf {g}}(n)=g_0^{\tbinom {n}{0}}g_1^{\tbinom {n}{1}}\ldots g_s^{\tbinom {n}{s}}$ and $\widetilde {\mathbf {g}}(n)=\widetilde g_0^{\tbinom {n}{0}} \widetilde g_1^{\tbinom {n}{1}}\ldots \widetilde g_s^{\tbinom {n}{s}}$ be two polynomials adapted to the lower central series filtration with coefficients $g_0,\ldots ,g_s,\widetilde g_0,\ldots ,\widetilde g_s\in K$ . A simple computation yields

(2.2) $$ \begin{align} d({\mathbf{g}}(n),\widetilde {\mathbf{g}}(n))&\le \sum_{i=0}^{s-1}d\big(g_0^{\tbinom{n}{0}} \ldots g_{i-1}^{\tbinom{n}{i-1}}\widetilde g_{i}^{\tbinom{n}{i}}\ldots \widetilde g_s^{\tbinom{n}{s}},g_0^{\tbinom{n}{0}}\ldots g_{i}^{\tbinom{n}{i}} \widetilde g_{i+1}^{\tbinom{n}{i+1}}\ldots \widetilde g_s^{\tbinom{n}{s}}\big)\nonumber\\ &=\sum_{i=0}^{s-1}d\big(g_0^{\tbinom{n}{0}}\ldots g_{i-1}^{\tbinom{n}{i-1}} \widetilde g_{i}^{\tbinom{n}{i}},g_0^{\tbinom{n}{0}}\ldots g_{i}^{\tbinom{n}{i}}\big)\nonumber\\ &\le \sum_{i=0}^{s-1}P_1(|\psi(g_0^{\tbinom{n}{0}}\ldots g_{i-1}^{\tbinom{n}{i-1}})|, |\psi(\widetilde g_{i}^{\tbinom{n}{i}})|,|\psi(g_{i}^{\tbinom{n}{i}})|)d \big(\widetilde g_{i}^{\tbinom{n}{i}},g_{i}^{\tbinom{n}{i}}\big)\nonumber\\ &\le \sum_{i=0}^{s-1}P_1(\widetilde Q(n),\widetilde P_2(\tbinom{n}{i}), \widetilde P_2(\tbinom{n}{i}))d\big(\widetilde g_{i}^{\tbinom{n}{i}},g_{i}^{\tbinom{n}{i}}\big)\nonumber\\ &\le \widetilde P(n)\sum_{i=0}^{s-1}d\big(\widetilde g_{i}^{\tbinom{n}{i}},g_{i}^{\tbinom{n}{i}}\big) \end{align} $$

for all $n\in \mathbb {N}$ , where $\widetilde P(n)=\sum _{i=0}^{s-1}P_1(\widetilde Q(n),\widetilde P_2(\tbinom {n}{i}),\widetilde P_2(\tbinom {n}{i}))$ is a polynomial of n.

Now we are going to show that there is a real polynomial $P_4: \mathbb {R}\rightarrow \mathbb {R}$ such that $d(\widetilde g^{n},g^{n})\le P_4(n)d(\widetilde g,g)$ for all $g,\widetilde g\in K$ . In fact, it follows from the fact

(2.3) $$ \begin{align}\begin{split} d(\widetilde g^{n},g^{n})&\le \sum_{i=0}^{n-1}d(\widetilde g^ig^{n-i},\widetilde g^{i+1}g^{n-i-1})=\sum_{i=0}^{n-1}d(\widetilde g^ig,\widetilde g^{i+1})\\ &\le\sum_{i=0}^{n-1}P_1(|\psi(\widetilde g^i)|,|\psi(\widetilde g)|,|\psi(g)|)d(\widetilde g,g)\\ &\le\sum_{i=0}^{n-1}P_1(\widetilde P_2(i),w,w)d(\widetilde g,g)\\ &\le P_4(n)d(\widetilde g,g) \end{split} \end{align} $$

for all $n\in \mathbb {N}$ , where $P_4(n)=\sum _{i=0}^{n-1}P_1(\widetilde P_2(i),w,w)$ is a real polynomial of n. Summing up, we obtain

$$ \begin{align*} d({\mathbf{g}}(n),\widetilde {\mathbf{g}}(n))&\overset{(2.2)}\le \widetilde P(n)\sum_{i=0}^{s-1}d\big(\widetilde g_{i}^{\tbinom{n}{i}},g_{i}^{\tbinom{n}{i}}\big)\\ &\overset{(2.3)}\le \widetilde P(n)\sum_{i=0}^{s-1}P_4(\tbinom{n}{i})d(\widetilde g_i,g_i)\\ &\le P(n)\max\{d(g_i,\widetilde g_i):0\le i\le s\} \end{align*} $$

for all $n\in \mathbb {N}$ , where $P(n)=\widetilde P(n)\sum _{i=0}^{s-1}P_4(\tbinom {n}{i})$ is a real polynomial of n. Then $P(n)$ is the real polynomial as required. This ends the proof of Lemma 2.6.

Proof of Proposition 2.5

By Lemma 2.4, there exists a compact subset K of G such that any polynomial sequence ${\mathbf {g}}$ adapted to the lower central series filtration can be factorized as ${\mathbf {g}}={\mathbf {g}}'\mathbf {\gamma }$ , where ${\mathbf {g}}'$ is a polynomial sequence adapted to the lower central series filtration with coefficients in K and $\mathbf {\gamma }$ is a polynomial sequence with coefficients in $\Gamma $ . Since K is compact, by Lemma 2.6, there is a real polynomial $P: \mathbb {R}\rightarrow \mathbb {R}$ such that

(2.4) $$ \begin{align} d({\mathbf{g}}(j),\widetilde {\mathbf{g}}(j))\le P(j)\max\{d(g_i,\widetilde g_i):0\le i\le s\} \quad\text{for all }j\in \mathbb{N},\end{align} $$

and any polynomials ${\mathbf {g}}$ , $\widetilde {\mathbf {g}}\in \mathrm{Poly}(G)$ with coefficients $g_0,\ldots ,g_s,\widetilde g_0,\ldots ,\widetilde g_s\in K$ . It is not hard to see that there exists $k_0\in \mathbb {N}$ and $C>1$ such that

(2.5) $$ \begin{align} P(n)<Cn^{k_0}\quad\text{for all }n\in\mathbb{N}. \end{align} $$

Since K is compact, for $\epsilon> 0$ , we let $N_\epsilon (K)$ be the smallest number of open balls of ratio $\epsilon $ needed to cover K. The upper Minkowski dimension or box dimension (see [Reference Mattila17]) is defined by

$$ \begin{align*}\limsup_{\epsilon\to0}\frac{-\log N_\epsilon(K)}{\log\epsilon}.\end{align*} $$

This dimension of K is not larger than the usual dimension of G since K is a subset of G. Hence, there exists a positive constant L such that

(2.6) $$ \begin{align} N_\epsilon(K)\le L\bigg(\frac{1}{\min\{ \epsilon,1\}}\bigg)^{\mathrm{dim}(G)+1}. \end{align} $$

Set

$$ \begin{align*}k=k_0(s+1)(\mathrm{dim}(G)+1)\quad\text{and}\quad C(\epsilon)=\bigg(L\bigg(\frac{2C}{\min\{ \epsilon,1\}}\bigg)^{\mathrm{dim}(G)+1}\bigg)^{s+1}\quad\text{for }\epsilon>0.\end{align*} $$

We are going to show that

$$ \begin{align*}s_n(G/\Gamma,\epsilon)\le C(\epsilon) n^k\end{align*} $$

for $n\in \mathbb {N}$ and $\epsilon>0$ . To do this, let $\pi $ be the projection from $\mathrm{Poly}(K)$ to $\mathrm{Poly}(G/\Gamma )$ defined by $\pi ({\mathbf {g}})(n)={\mathbf {g}}(n)\Gamma $ for all $n\in \mathbb {Z}$ . By Lemma 2.4, $\pi $ is surjective and

$$ \begin{align*}d({\mathbf{g}}(j),\mathbf{\widetilde g}(j))\ge d(\pi(\mathbf{ g})(j),\pi(\mathbf{\widetilde g})(j))\quad\text{for all } j\in\mathbb{Z}.\end{align*} $$

Hence,

(2.7) $$ \begin{align}s_n(\text{Poly}(G/\Gamma),\epsilon)\le s_n(\text{Poly}(K),\epsilon)\quad\text{for all } n\in\mathbb{N} \quad\text{and}\quad\epsilon>0. \end{align} $$

For $\tau>0$ , we let $E_{\tau }$ be a finite subset of K such that

$$ \begin{align*} \sharp E_\tau\le N_{\tau}(K)\quad\text{and}\quad K\subset \bigcup_{g\in E_\tau}B(g,\tau). \end{align*} $$

For $0\le i\le s$ , we let $E_{\tau }^{(i)}$ be a subset of $K\cap G_i$ such that

(2.8) $$ \begin{align} \sharp E_\tau^{(i)}\le N_{\tau}(K)\quad\text{and}\quad K\cap G_i\subset \bigcup_{g\in E^i_\tau}B(g,2\tau). \end{align} $$

Put $P_\tau $ to be the collection of all polynomial sequences ${\mathbf {g}}$ adapted to the lower central series filtration with coefficients $g_i\in E_\tau ^{(i)}$ , $i=0,1,\ldots ,s$ . Then for $n\in \mathbb {N}$ and $\epsilon>0$ ,

(2.9) $$ \begin{align}\sharp P_{{\epsilon}/{2Cn^{k_0}}}= \prod_{i=0}^s\sharp E^{(i)}_{{\epsilon}/{2Cn^{k_0}}}\overset{(2.8),(2.6)} \le \left(L\left(\frac{2Cn^{k_0}}{\min\{ \epsilon,1\}}\right)^{\mathrm{dim}(G)+1}\right)^{s+1}=C(\epsilon)n^k. \end{align} $$

Now we fix $n\in \mathbb {N}$ and $\epsilon>0$ . By equation (2.8), for any polynomial sequence ${\mathbf {g}}\in \mathrm{Poly}(K)$ with coefficients $g_0,\ldots ,g_s\in K$ , we have that $g_i\in K\cap G_i$ . Thus, there exists ${\bar {\mathbf g}}\in P_{{\epsilon }/{2Cn^{k_0}}}$ with coefficients $\bar g_0\in E^{(0)}_{{\epsilon }/{2Cn^{k_0}}} ,\ldots ,\bar g_s\in E^{(s)}_{{\epsilon }/{2Cn^{k_0}}}$ such that

$$ \begin{align*}d(g_i,\bar g_i)\overset{(2.8)}<\frac{\epsilon}{Cn^{k_0}}\quad\text{for all } 0\le i\le s.\end{align*} $$

Therefore,

$$ \begin{align*}d({\mathbf{g}}(0),\bar{\mathbf{ g}}(0))=d(g_0,\bar{g}_0)<\frac{\epsilon}{Cn^{k_0}}<\epsilon\end{align*} $$

and for $1\le j\le n-1$ , one has

$$ \begin{align*}d({\mathbf{g}}(j),\bar{\mathbf{g}}(j))\overset{(2.4)}\le P(j) \max\{d(g_i,\bar g_i):0\le i\le s\}\overset{(2.5)}< Cj^{k_0}\times\frac{\epsilon}{Cn^{k_0}}\le \epsilon.\end{align*} $$

Hence,

$$ \begin{align*}s_n(\text{Poly}(G/\Gamma),\epsilon)\overset{(2.7)}\le s_n(\text{Poly}(K),\epsilon) \le \sharp P_{{\epsilon}/{2Cn^{k_0}}}\overset{(2.9)}\le C(\epsilon)n^k.\end{align*} $$

Since the above inequality holds for all $n\in \mathbb {N}$ and $\epsilon>0$ , we end the proof of Proposition 2.5.

Proof of (2) $\Longrightarrow $ (1) in Theorem 1.1

Assume that Theorem 1.1(2) holds, that is, the logarithmic Sarnak conjecture holds for any t.d.s. with polynomial mean complexity. In what follows, we aim to show that the logarithmic Sarnak conjecture holds.

Assume the contrary that this is not the case, then by Tao’s result [Reference Tao22], the Conjecture 2.1 does not hold. This means that there exist an $s\in \mathbb {N}$ , an s-step nilmanifold $G/\Gamma $ , a $\mathrm{Lip}$ -continuous function $F:G/\Gamma \to \mathbb {C}$ , and an $x_0\in G/\Gamma $ such that

(2.10) $$ \begin{align} \limsup_{H\to+\infty}\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log}\sup_{g\in G}| \mathbb{E}_{h\le H}\mu(n+h)F(g^hx_0)|>0. \end{align} $$

It is clear that $\|F\|_{\infty }:=\max _{x\in G/\Gamma }|F(x)|>0$ . Without loss of generality, we assume that

(2.11) $$ \begin{align} \|F\|_\infty=1. \end{align} $$

Now we add an extra point p to the compact metric space $G/\Gamma $ . We then extend the metric d on $G/\Gamma $ to the space $G/\Gamma \cup \{p\}$ by letting $d(p,x)=1$ for all $x\in G/\Gamma $ . So, $(G/\Gamma \cup \{p\},d)$ is also a compact metric space. Let $\widetilde F: (G/\Gamma \cup \{p\})^{\mathbb {Z}}\to \mathbb {C}$ be defined by $\widetilde F(z)=F(z(0))$ if $z(0)\in G/\Gamma $ and $0$ if $z(0)=p$ . It is clear that $\widetilde F$ is a continuous function and

(2.12) $$ \begin{align}\|\widetilde F\|_\infty =1 \end{align} $$

by equation (2.11).

In what follows, we will find a point $y\in (G/\Gamma \cup \{p\})^{\mathbb {Z}}$ such that

(2.13) $$ \begin{align}\limsup_{N\to\infty}|\mathbb{E}_{n\le N}^{\log} \mu (n)\widetilde F(\sigma^ny)|>0,\end{align} $$

and the t.d.s. $(X_y,\sigma )$ has polynomial mean complexity, where $\sigma : (G/\Gamma \cup \{p\})^{\mathbb {Z}}\rightarrow (G/\Gamma \cup \{p\})^{\mathbb {Z}}$ is the left shift and $X_y=\overline {\{\sigma ^ny:n\in \mathbb {Z}\}}$ is a $\sigma $ -invariant compact subset of $(G/\Gamma \cup \{p\})^{\mathbb {Z}}$ . Clearly, this is a contradiction to our assumption and thus proves that statement (2) implies statement (1) in Theorem 1.1.

We divide the remaining proof into two steps.

Step 1. The construction of the point y. First, we note that

$$ \begin{align*}|z|\le \sum_{j=0}^{3} \max\bigg\{ \mathrm{Re}\bigg(e\bigg(\frac{j}{4}\bigg)z\bigg),0\bigg\}\end{align*} $$

for $z\in \mathbb {C}$ . Thus, by equation (2.10), there is $\beta \in \{0,\tfrac 14,\tfrac 24,\tfrac 34\}$ such that

$$ \begin{align*}\limsup_{H\to+\infty}\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log}\max\Big\{\! \sup_{g\in G} \mathrm{Re}( e(\beta)\mathbb{E}_{h\le H}\mu(n+h)F(g^hx_0)),0\Big\}>0.\end{align*} $$

Thus, we can find $\tau \in (0,1)$ with

$$ \begin{align*} E:=\Big\{H\in \mathbb{N}: \limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log}\max\!\Big\{\! \sup_{g\in G} \mathrm{Re}( e(\beta)\mathbb{E}_{h\le H}\mu(n+h)F(g^hx_0)),0\Big\}>\tau\Big\} \end{align*} $$

is an infinite set. Moreover, putting and by induction, we can find strictly increasing sequences $\{H_i\}_{i=1}^{\infty }$ of E and $\{N_i\}_{i=1}^\infty $ of natural numbers such that for each $i\in \mathbb {N}$ , one has

(2.14) $$ \begin{align} H_i<\sigma N_i^\sigma<\frac{\sigma}{10}H_{i+1}^\sigma, \end{align} $$

and there exist $g_{n,i}\in G$ for $ 1\le n\le N_i$ satisfying

(2.15) $$ \begin{align} \mathbb{E}_{n\le N_i}^{\log}\max\{\mathrm{Re}(e(\beta)\mathbb{E}_{h\le H_i}\mu(n+h)F(g_{n,i}^hx_0)),0\}>\tau. \end{align} $$

For $i\in \mathbb {N}$ , let $M_i=\sum _{n=1}^{N_i}({1}/{n})$ and

(2.16) $$ \begin{align} S_i=\bigg\{n\in [1,N_i]\cap \mathbb{Z}: \mathrm{Re}(e(\beta)\mathbb{E}_{h\le H_i}\mu(n+h)F(g_{n,i}^hx_0))>\frac{\tau}{2}\bigg\}. \end{align} $$

Then by equations (2.12) and (2.15), we have

(2.17) $$ \begin{align} \sum_{n\in S_i}\frac{1}{n}>\frac{\tau}{2}M_i. \end{align} $$

Notice that $\lim _{N\to +\infty }({\sum _{n\le N^\sigma }({1}/{n})}/{\sum _{n\le N}({1}/{n})})=\sigma .$ So, when $i\in \mathbb {N}$ is large enough, we have

$$ \begin{align*}\sum_{n\in S_i\setminus[1, N_{i}^\sigma]}\frac{1}{n}\overset{(2.17)}>\frac{\tau}{2}M_i-\sum_{n\le N_{i}^\sigma}\frac{1}{n}>\frac{\tau}{2}M_i-2\sigma M_{i}>\frac{\tau}{4}M_i. \end{align*} $$

Hence, we can select $S_i'\subset S_i\setminus [1,N_{i}^\sigma ]$ with each gap not less than $2H_i$ and

(2.18) $$ \begin{align} \sum_{n\in S^{\prime}_i}\frac{1}{n}> \frac{\tau M_i}{8H_i} \end{align} $$

for $i\in \mathbb {N}$ large enough.

Define $y: \mathbb {Z}\to G/\Gamma \cup \{p\}$ such that

$$ \begin{align*}y(n+h):=g_{n,i}^hx_0\quad\text{for }n\in S_i', h=1,2,\ldots,H_i,i\in\mathbb{N}\end{align*} $$

and $y(m)=p$ for $m\in \mathbb {Z}\setminus \bigcup _{i=1}^{\infty } \bigcup _{n\in S_i'}\{n+1,n+2,\ldots , n+H_i\}$ .

Clearly, y is well defined since $N_{i+1}>N_i+H_i$ by equation (2.14). Then one has by equations (2.16) and (2.18) that

$$ \begin{align*} \mathrm{Re}\bigg(e(\beta)\sum_{n\in S_i'}\frac{1}{n}\sum_{h\le H_i}\mu(n+h)\widetilde F(\sigma^{n+h}y)\bigg)>\frac{\tau^2}{16}M_i \end{align*} $$

for $i\in \mathbb {N}$ large enough. This implies

(2.19) $$ \begin{align} \bigg|\sum_{n\in S_i'}\frac{1}{n}\sum_{h\le H_i}\mu(n+h)\widetilde F(\sigma^{n+h}y)\bigg|>\frac{\tau^2}{16}M_i \end{align} $$

for $i\in \mathbb {N}$ large enough. Moreover, for $i\in \mathbb {N}$ large enough,

$$ \begin{align*} \bigg|&\sum_{n\in S_i'}\frac{1}{n}\sum_{h\le H_i}\mu(n+h)\widetilde F(\sigma^{n+h}y)-\sum_{n\in S_i'} \sum_{h\le H_i}\frac{\mu(n+h)\widetilde F(\sigma^{n+h}y)}{n+h}\bigg|\\ &\overset{(2.12)}\le \sum_{n\in S_i'} \sum_{h\le H_i}\bigg(\frac{1}{n}-\frac{1}{n+h}\bigg)\le \sum_{n\in S_i'} \sum_{h\le H_i}\frac{H_i}{n(n+H_i)}\\ &\,\le \sum_{n\in S_i'} \frac{H_i}{n(N_i^\sigma+H_i)}\le \sum_{n\in S_i'} \frac{H_i}{nN_i^\sigma}\\ &\overset{(2.14)}\le \sigma\sum_{n\in S_i'}\frac{1}{n}\le \frac{\tau^2}{32}M_i. \end{align*} $$

Combining this inequality with equation (2.19), one has

(2.20) $$ \begin{align} \bigg|\sum_{N_{i}^\sigma<n \le N_i+H_i}\frac{\mu (n )\widetilde F(\sigma^{n}y)}{n}\bigg|=\bigg|\sum_{n\in S_i'} \sum_{h\le H_i}\frac{\mu(n+h)\widetilde F(\sigma^{n+h}y)}{n+h}\bigg|\ge \frac{\tau^2M_i}{32} \end{align} $$

for $i\in \mathbb {N}$ large enough. Thus,

$$ \begin{align*} \bigg|&\frac{1}{M_i}\sum_{n\le N_i}\frac{\mu (n )\widetilde F(\sigma^{n}y)}{n}\bigg|\\&\,\ge\frac{1}{M_i} \bigg|\sum_{N_{i}^\sigma<n\le N_i+H_i}\frac{\mu (n )\widetilde F(\sigma^{n}y)}{n}\bigg| -\frac{1}{M_i}\sum_{n\le N_{i}^\sigma\text{ or}\atop N_i<n\le N_i+H_i}\bigg|\frac{\mu (n )\widetilde F(\sigma^{n}y)}{n}\bigg|\\&\overset{(2.20)}\ge\frac{\tau^2}{32}-\frac{\|\widetilde F\|_\infty}{M_i} \sum_{ N_i<n\le N_i+H_i}\frac{1}{n}-\frac{\|\widetilde F\|_\infty}{M_i}\sum_{ n\le N_{i}^\sigma}\frac{1}{n}\\&\overset{(2.12)}\ge\frac{\tau^2}{32}-\frac{H_i}{N_i}-2\sigma\overset{(2.14)}\ge \frac{\tau^2}{32}-3\sigma \overset{(2.14)}\ge \frac{\tau^2}{100} \end{align*} $$

for $i\in \mathbb {N}$ large enough. This deduces that

$$ \begin{align*}\limsup_{N\to\infty}|\mathbb{E}_{n\le N}^{\log} \mu (n)\widetilde F(\sigma^ny)|\ge \frac{\tau^2}{100}>0. \end{align*} $$

Therefore, y is the point as required.

Step 2. $(X_y,\sigma )$ has polynomial mean complexity. Recall that $X_y =\overline {\{\sigma ^ny:n\in \mathbb {Z}\}}$ is a compact $\sigma $ -invariant subset of $(G/\Gamma \cup \{p\})^{\mathbb {Z}}$ . The metric on $(G/\Gamma \cup \{p\})^{\mathbb {Z}}$ is defined by

(2.21) $$ \begin{align}D(x,x')=\sum_{n\in\mathbb{Z}}\frac{d(x(n),x'(n))}{2^{|n|+2}}\end{align} $$

for $x=(x(n))_{n\in \mathbb {Z}}, x'=(x'(n))_{n\in \mathbb {Z}}\in (G/\Gamma \cup \{p\})^{\mathbb {Z}}$ . By Proposition 2.5, we can find $k>1$ such that

(2.22) $$ \begin{align} \lim_{n\to+\infty}\frac{s_{n}(\text{Poly}(G/\Gamma),\epsilon)}{n^{k}}=0\quad\text{for all } \epsilon>0. \end{align} $$

Now we are going to show that

$$ \begin{align*}\liminf_{n\to+\infty}\frac{S_{n}(D,\sigma,X_y,\epsilon)}{n^{k+1}}=0\quad\text{for all }\epsilon>0.\end{align*} $$

For $n\in \mathbb {Z}_+$ and $-n\le q\le n$ , let $X_{n,q}$ be the collection of all points $z\in (G/\Gamma \cup \{p\})^{\mathbb {Z}}$ with

$$ \begin{align*}z(j)=\begin{cases} p & \text{if }-n\le j< q, \\ {\mathbf{g}}(j)\Gamma & \text{if } q\le j\le n, \end{cases}\!\! \end{align*} $$

where ${\mathbf {g}}$ is some polynomial sequence of G adapted to the lower central series filtration; and let $X_{n,q}^*$ be the collection of all points $z\in (G/\Gamma \cup \{p\})^{\mathbb {Z}}$ with

$$ \begin{align*}z(j)=\begin{cases} {\mathbf{g}}(j)\Gamma & \text{if }-n\le j< q, \\ p & \text{if } q\le j\le n, \end{cases}\!\! \end{align*} $$

where ${\mathbf {g}}$ is some polynomial sequence of G adapted to the lower central series filtration.

For $i\in \mathbb {N}$ , put $t_i=[H_i/2]$ , where $[u]$ is the integer part of the real number u. Then,

(2.23) $$ \begin{align} X_y\subset \bigcup\limits_{-t_i\le q\le t_i} X_{t_i,q}\cup\bigcup\limits_{-t_i\le q\le t_i} X_{t_i,q}^*\cup\{\sigma^jy:-H_i\le j\le H_i\}. \end{align} $$

In fact, since $H_{j+1}>N_j+H_j$ for all $j\in \mathbb {N}$ , one has

(2.24) $$ \begin{align} \sigma^ny\in \bigcup\limits_{-t_i\le q\le t_i} X_{t_i,q}\cup\bigcup\limits_{-t_i\le q\le t_i} X_{t_i,q}^* \cup\{\sigma^jy:-H_i\le j\le H_i\} \quad\text{for all } n\in \mathbb{Z} \end{align} $$

by the construction of y. It is not hard to see that $X_{t_i,q},X_{t_i,q}^*$ are all compact subsets of $(G/\Gamma \cup \{p\})^{\mathbb {Z}}$ for each $-t_i\le q \le t_i$ and $i\in \mathbb {N}$ by Lemma 2.4. Hence, the set in right part of equation (2.23) is also a compact subset of $(G/\Gamma \cup \{p\})^{\mathbb {Z}}$ . Now equation (2.23) follows from equation (2.24).

Now we fix $\epsilon>0$ . We have the following claim.

Proof of the Claim

We prove part (1) first. For $i\in \mathbb {N}$ and $-t_i\le q\le t_i$ , we let $\pi _{i,q}: \mathrm{Poly}(G/\Gamma )\to X_{t_i,q}$ be defined by

$$ \begin{align*}\pi_{i,q}(z)(j)=\begin{cases} p & \text{ if }-t_i\le j<q,\\ z(j) & \text{ otherwise},\end{cases}\end{align*} $$

for $z\in \mathrm{Poly}(G/\Gamma )$ . For $i\in \mathbb {N}$ large enough, if $z,\widetilde z\in \mathrm{Poly}(G/\Gamma )$ with $d(z(j),\widetilde z(j))<{\epsilon }/{2}$ for all $-t_i\le j\le t_i$ , then for $q\in [-t_i,t_i]\cap \mathbb {Z}$ ,

(2.25) $$ \begin{align} &\!\!\bar{D}_{[t_i/2]}(\pi_{i,q}(z),\pi_{i,q}(\widetilde z))=\frac{1}{[t_i/2]}\sum_{l=0}^{[t_i/2]-1}D(\sigma^lz,\sigma^l\widetilde z)\nonumber\\ &\ \ \overset{(2.21)}= \frac{1}{[t_i/2]}\sum_{l=0}^{[t_i/2]-1}\sum_{n\in\mathbb{Z}}\frac{d(z(n+l),\widetilde z(n+l))}{2^{|n|+2}}\nonumber\\ &\quad\le \frac{1}{[t_i/2]}\sum_{l=0}^{[t_i/2]-1}\bigg(\sum_{|n|\le [t_i/2] }\frac{d(z(n+l), \widetilde z(n+l))}{2^{|n|+2}}+\sum_{|n|> [t_i/2] }\frac{d(z(n+l),\widetilde z(n+l))}{2^{|n|+2}}\bigg)\nonumber\\ &\quad\le \frac{1}{[t_i/2]}\sum_{l=0}^{[t_i/2]-1}\bigg( \frac{\epsilon}{2}+\frac{\text{diam} (G/\Gamma)}{2^{t_i/2}}\bigg) <\epsilon, \end{align} $$

where we use the fact $t_i\to +\infty $ as $i\to +\infty $ in the last inequality. Notice that the map $\pi _{i,q}$ is surjective for all $i\in \mathbb {N}$ and $-t_i\le q\le t_i$ . By equation (2.25), for $i\in \mathbb {N}$ large enough, one has

$$ \begin{align*}S_{[t_i/2]}(D,\sigma,X_{t_i,q},\epsilon)\le s_{t_i}\bigg(\text{Poly}(G/\Gamma),\frac{\epsilon}{2}\bigg)\quad\text{for all }q\in [-t_i,t_i]\cap\mathbb{Z}.\end{align*} $$

By the similar arguments, one has part (2). This ends the proof of the Claim.

Hence, by the above Claim and equation (2.23), one has

$$ \begin{align*} S_{[t_i/2]}(D,\sigma,X_y,\epsilon)&\le (2H_i+1)+ \sum_{q=-t_i}^{t_i}(S_{[t_i/2]}(D,\sigma,X_{t_i,q},\epsilon)+ S_{[t_i/2]}(D,\sigma,X_{t_i,q}^*,\epsilon))\\ \nonumber &\le (2H_i+1)+(2t_i+1)s_{t_i}\bigg(\text{Poly}(G/\Gamma),\frac{\epsilon}{2}\bigg)\end{align*} $$

for $i\in \mathbb {N}$ large enough. Combining this with equation (2.22),

$$ \begin{align*} \liminf_{n\to+\infty}\frac{S_{n}(D,\sigma,X_y,\epsilon)}{n^{k+1}} &\le \liminf_{i\to+\infty}\frac{S_{[t_i/2]}(D,\sigma,X_y,\epsilon)}{[t_i/2]^{k+1}}\\ &\le \liminf_{i\to+\infty}\frac{ (2H_i+1)+(2t_i+1)s_{t_i}(\text{Poly}(G/\Gamma),{\epsilon}/{2})}{[t_i/2]^{k+1}}\\ &=0, \end{align*} $$

where we used the assumption $t_i=[H_i/2]$ . This implies that $(X_y,\sigma )$ has polynomial mean complexity, since the above inequality is true for all $\epsilon>0$ . This ends the proof of Step 2.

Proof. First, we assume that equation (2.26) holds for any subset U of X with small boundary and $x\in X$ . For a given $f\in C(X)$ and fixed $\delta>0$ , let

$$ \begin{align*}\epsilon=\epsilon(\delta) =\sup_{x,y\in X,d(x,y)<\delta}|f(x)-f(y)|.\end{align*} $$

Let $\mathcal {P}=\{P_1,P_2,\ldots ,P_k\}$ be a partition of X with diameter smaller than $\delta $ and each element of $\mathcal {P}$ has a small boundary. For $1\le i\le k$ , we fix points $x_i\in P_i$ and define $\bar f(x)=f(x_i)$ if $x\in P_i$ . Then, $\bar f(x)=\sum _{i=1}^k f(x_i)\mathbf {1}_{P_i}(x)$ and by equation (2.26),

$$ \begin{align*}\lim_{N\to+\infty}\mathbb{E}_{n\le N}^{\log}\bar f(T^nx)\mu(n)=0\end{align*} $$

for all $x\in X$ . Since $\|\bar f-f\|_{\infty }\le \epsilon $ , we have

$$ \begin{align*} &\limsup_{N\to+\infty}|\mathbb{E}_{n\le N}^{\log} f(T^nx)\mu(n)|\\ &\quad\le \limsup_{N\to+\infty}|\mathbb{E}_{n\le N}^{\log}\bar f(T^nx)\mu(n)|+\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} \|\bar f-f\|_{\infty}\cdot|\mu(n)|\\ &\quad\le \epsilon \end{align*} $$

for all $x\in X$ . By taking $\delta \to 0$ and then $\epsilon \to 0$ , one has

$$ \begin{align*}\lim_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} f(T^nx)\mu(n)=0\end{align*} $$

for all $x\in X$ . This implies the logarithmic Sarnak conjecture holds for $(X,T)$ since f is arbitrary.

Conversely, we assume that the logarithmic Sarnak conjecture holds for $(X,T)$ . Let U be a subset of X with small boundary. Fix $\delta>0$ . By a result of Shub and Weiss (see [Reference Shub and Weiss20, p. 537]), we can find $\epsilon>0$ such that for N large enough,

$$ \begin{align*} \frac{1}{N}\sum_{n=1}^{N}\mathbf{1}_{B(\partial U,\epsilon)}(T^nx)\le \frac{\delta}{2} \end{align*} $$

for all $x\in X$ , where $B(\partial U,\epsilon )=\{y\in X: d(y,\partial U)<\epsilon \}$ . Moreover, for N large enough,

(2.27) $$ \begin{align} \begin{split} \mathbb{E}_{n\le N}^{\log}\mathbf{1}_{B(\partial U,\epsilon)}(T^nx)&= \frac{1}{M_N}\sum_{n=1}^{N}\frac{\mathbf{1}_{B(\partial U,\epsilon)}(T^nx)}{n}\\ &=\frac{1}{M_N}\bigg(\frac{S_N(x)}{N}+\sum_{j=1}^{N-1}\frac{S_j(x)}{j} \frac{1}{j+1}\bigg)\\ &\le \delta \end{split} \end{align} $$

for all $x\in X$ , where we simply write $M_N=\kern-1pt\sum _{n=1}^N({1}/{n})$ and $S_j(x)=\kern-1pt\sum _{n=1}^j \kern-1pt\mathbf {1}_{B(\partial U,\epsilon )}(T^nx)$ for $j\in \mathbb {N}$ .

Using Urysohn’s lemma, there exists a continuous function $h{\kern-1.2pt}:{\kern-1.2pt}X{\kern-1.3pt}\to{\kern-1.2pt} \mathbb {R}$ with $0{\kern-1.2pt}\le{\kern-1.2pt} h{\kern-1.2pt}\le{\kern-1.2pt} 1$ such that $h(x)=1$ for $x\in U\setminus B(\partial U,\epsilon )$ and $h(x)=0$ for $x\in X\setminus \big (U\cup B(\partial U,\epsilon )\big )$ . Since the logarithmic Sarnak conjecture holds for $(X,T)$ , one has

$$ \begin{align*}\lim_{N\to+\infty} \mathbb{E}_{n\le N}^{\log} h(T^nx)\mu(n)=0\end{align*} $$

for all $x\in X$ . Combining this equality with equation (2.27), we obtain

$$ \begin{align*} &\limsup_{N\to+\infty}|\mathbb{E}_{n\le N}^{\log} 1_U(T^nx)\mu (n)|\\ &\quad\le \limsup_{N\to+\infty}|\mathbb{E}_{n\le N}^{\log} h(T^nx)\mu (n)| +\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} |h(T^nx)-1_U(T^nx)|\\ &\quad\le \limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} 1_{B(\partial U,\epsilon)}(T^nx)\le \delta \end{align*} $$

for all $x\in X$ . By taking $\delta \to 0$ , we have

$$ \begin{align*}\lim_{N\to+\infty}\mathbb{E}_{n\le N}^{\log}1_U(T^nx)\mu(n)=0\end{align*} $$

for all $x\in X$ . This ends the proof of Lemma 2.8.

Proof. We fix an $\delta \in (0,+\infty )$ and a non-empty subset U of X with small boundary. By a result of Shub and Weiss (see [Reference Shub and Weiss20, p. 537]), there exist $N_\delta \in \mathbb {N}$ and $\epsilon _0\in (0,+\infty )$ such that

(2.28) $$ \begin{align}\sup_{x\in X,N\ge N_\delta}\frac{1}{N}\sum_{n=0}^{N-1}\mathbf{1}_{B(\partial U,\epsilon_0)}(T^nx)<\delta, \end{align} $$

where $B(\partial U,\epsilon _0)=\{z\in X:d(z,z')<\epsilon _0\text { for all }z'\in \partial U \}$ if $\partial U $ is not empty and $B(\partial U,\epsilon _0)=\emptyset $ if $\partial U $ is empty.

We notice that $\overline {U\setminus B(\partial U,\epsilon _0)}\cap \overline {X\setminus U}=\emptyset $ and $\overline {(X\setminus U)\setminus B(\partial U,\epsilon _0)}\cap \overline {U}=\emptyset $ . Thus, we can find $\epsilon \in (0,\delta ^2)$ such that when $x,y\in X$ with $d(x,y)<\sqrt {\epsilon }$ , if $x\in U\setminus B(\partial U,\epsilon _0)$ (respectively $x\in (X\setminus U)\setminus B(\partial U,\epsilon _0)$ ), then $y\in U$ (respectively $y\in X\setminus U$ ). We are to show that $\epsilon $ is the constant as required. We fix $N\ge N_\delta $ and $x_1,x_2\in X$ with $\bar d_N(x_1,x_2)< \epsilon $ . Set

$$ \begin{align*}C=\{0\le n\le N-1:T^nx_1\in B(\partial U,\epsilon_0)\}.\end{align*} $$

By equation (2.28), $\sharp C\le \delta N$ . Put

$$ \begin{align*}\mathcal{A}=\{0\le n\le N-1: d(T^nx_1,T^nx_2)<\sqrt{\epsilon} \}.\end{align*} $$

One has $\sharp \mathcal {A}\ge (1-\sqrt {\epsilon })N$ and $\hat {x_1}(n)=\hat {x_2}(n) \text { for all }n\in \mathcal {A}\setminus C$ . Therefore,

$$ \begin{align*}\sharp\{0\le n\le N-1:\hat {x_1}(n)=\hat{x_2}(n)\}\ge \sharp \mathcal{A}-\sharp C\ge(1-\sqrt{\epsilon}-\delta)N.\end{align*} $$

Since $\delta>\sqrt {\epsilon }$ , one has

$$ \begin{align*}\sharp\{0\le n\le N-1:\hat {x_1}(n)\neq\hat{x_2}(n)\}\le2\delta N.\end{align*} $$

This ends the proof of Lemma 2.9.

Proof. We fix a $\delta>0$ and a non-empty subset U of X with small boundary. We are to find $\epsilon \in (0,+\infty )$ such that $S_N(d,\sigma ,\hat {X},\delta )\le S_N(d, T,X,\epsilon )$ for N large enough. To do this, we choose $L\in \mathbb {N}$ and $\delta '>0$ such that

(2.30) $$ \begin{align} 4\delta'L+\frac{2}{2^L}<\delta. \end{align} $$

By Lemma 2.9, there exists $\epsilon :=\epsilon (\delta ')>0$ such that for $N\in \mathbb {N}$ large enough and $x_1,x_2\in X$ with $\bar d_N(x_1,x_2)< \epsilon $ , one has

(2.31) $$ \begin{align} \sharp\{0\le n\le N-1:\hat {x_1}(n)\neq\hat{x_2}(n)\}\le2\delta' N. \end{align} $$

Fix $x_1,x_2\in X$ with $\bar d_N(x_1,x_2)< \epsilon $ and put

$$ \begin{align*} \mathcal{C}_N=\{0\le n\le N-1:\hat {x_1}(n+l)\neq\hat{x_2}(n+l)\text{ for some }-L+1\le l\le L-1\}. \end{align*} $$

By equation (2.31), we have for $N\in \mathbb {N}$ large enough,

$$ \begin{align*}\sharp\mathcal{C}_N\le 4\delta'LN.\end{align*} $$

Notice that $d(\sigma ^n\hat {x_1},\sigma ^n\hat {x_2})\le 1$ for $n\in \mathcal {C}_N$ . One has

$$ \begin{align*} \bar d_N(\hat{x_1},\hat{x_2})&=\frac{1}{N}\bigg(\sum_{n\in \mathcal{C}_N}d(\sigma^n\hat{x_1},\sigma^n\hat{x_2})+\sum_{n\in[0,N-1]\setminus\mathcal{C}_N}d(\sigma^n\hat{x_1},\sigma^n\hat{x_2})\bigg) \\ &\overset{(2.29)}\le \frac{1}{N}\bigg(\sum_{n\in \mathcal{C}_N}1+\sum_{n\in[0,N-1]\setminus\mathcal{C}_N}\frac{2}{2^L}\bigg) \\ &=\frac{1}{N}\bigg(\sharp\mathcal{C}_N+\frac{2}{2^L}(N-\sharp\mathcal{C}_N)\bigg) \\ &\overset{(2.30)} \le4\delta'L+\frac{2}{2^L}<\delta. \end{align*} $$

Therefore, $S_N(d,\sigma ,\hat {X},\delta )\le S_N(d, T,X,\epsilon )$ for $N\in \mathbb {N}$ large enough and $\epsilon $ is the constant as required. This ends the proof of Proposition 2.10.

Proof of Theorem 1.1: (3) $\Longrightarrow $ (2)

Assume that Theorem 1.1(3) holds. Now we are going to show that Theorem 1.1(2) holds. Assume the contrary that Theorem 1.1(2) does not hold, then there exists a t.d.s. $(X,T)$ with polynomial mean complexity such that the logarithmic Sarnak conjecture does not hold for $(X,T)$ .

Let $(Y,S)$ be an irrational rotation on the circle. Then $(X\times Y,T\times S)$ has polynomial mean complexity as well as zero mean dimension and admits a non-periodic minimal factor $(Y,S)$ . Hence, $(X \times Y,T\times S)$ has small boundary property by [Reference Lindenstrauss14, Theorem 6.2]. Since the logarithmic Sarnak conjecture does not hold for $(X,T)$ , neither does $(X \times Y,T\times S)$ . By Lemma 2.8, there is a subset U of $X\times Y$ with small boundary and $w\in X\times Y$ such that

$$ \begin{align*}\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} \mathbf{1}_{U}\big((T\times S)^nw\big)\mu(n)>0.\end{align*} $$

Combining this with Proposition 2.10, the $\{0,1\}$ -symbolic system $(\overline {\{\hat z:z\in X\times Y\}},\sigma )$ has polynomial mean complexity and

$$ \begin{align*}\limsup_{N\to+\infty}\mathbb{E}_{n\le N}^{\log} F_0(\sigma^n\hat w)\mu(n)>0,\end{align*} $$

where $F_0(\hat z)=\hat z(0)$ for $z\in X\times Y$ , which contradicts the assumption that Theorem 1.1(3) holds. This ends the proof of (3) $\Longrightarrow $ (2) in Theorem 1.1, and hence the proof of Theorem 1.1.