%~Mouliné par MaN_auto v.0.38.0 (124554dc) 2026-02-04 11:21:38 \documentclass[AHL,Unicode,longabstracts,published]{cedram} %\usepackage{enumitem} \usepackage{tikz} \usepackage{aligned-overset} \DeclareMathOperator{\symdiff}{\triangle} \DeclareMathOperator{\lst}{\mathsf{e}} \DeclareMathOperator{\bad}{\mathsf{V}} \DeclareMathOperator{\p}{\mathsf{p}} \DeclareMathOperator{\q}{\mathsf{q}} \DeclareMathOperator{\Q}{\mathsf{Q}} \usetikzlibrary{decorations.pathreplacing,calligraphy} \newcommand{\noopsort}[1]{} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\diam}{diam} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\bbC}{\mathbb{C}} \newcommand{\metric}{\mathsf{d}} \newcommand{\intd}{\;\mathsf{d}} \newcommand{\Lip}{\mathsf{Lip}} \newcommand{\lp}{\mathsf{L}} \newcommand{\ltwo}{\lp^{\!\mathsf{2}}} \newcommand{\KR}{\mathsf{KR}} \newcommand{\id}{\mathsf{Id}} \newcommand{\lft}{\mathsf{L}} \newcommand{\rht}{\mathsf{R}} \renewcommand{\P}{\mathsf{P}} \newcommand{\scl}{\mathsf{J}} \newcommand{\meas}{\mathcal{M}} \newcommand{\clT}{\mathcal{T}} \def\mathbi#1{\textbf{\em #1}} \newcommand{\bfr}{\mathbf{r}} \newcommand{\bir}{\mathbi{r}} %\newcommand{\define}[1]{\textbf{#1}} \renewcommand{\emptyset}{\varnothing} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %~ This fixes spacing issues around \left( and \right) and other similar symbols \let\originalleft\left \let\originalright\right \renewcommand{\left}{\mathopen{}\mathclose\bgroup\originalleft} \renewcommand{\right}{\aftergroup\egroup\originalright} \renewcommand*{\to}{\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}} \let\oldmapsto\mapsto \renewcommand*{\mapsto}{\mathchoice{\longmapsto}{\oldmapsto}{\oldmapsto}{\oldmapsto}} \renewcommand*{\implies}{\mathchoice{\Longrightarrow}{\Rightarrow}{\Rightarrow}{\Rightarrow}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \graphicspath{{./figures/}} \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} %\hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\relabel}{\renewcommand{\labelenumi}{(\theenumi)}} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}\relabel} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}\relabel} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}\relabel} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}\relabel} \newcommand*{\Eenumi}{\renewcommand*{\theenumi}{E\arabic{enumi}}\relabel} \newcommand*{\Penumi}{\renewcommand*{\theenumi}{P\arabic{enumi}}\relabel} \newcommand*{\PJenumi}{\renewcommand*{\theenumi}{PJ\arabic{enumi}}\relabel} \newcommand*{\Jenumi}{\renewcommand*{\theenumi}{J\arabic{enumi}}\relabel} \newcommand*{\Fenumi}{\renewcommand*{\theenumi}{F\arabic{enumi}}\relabel} \let\oldtilde\tilde \renewcommand*{\tilde}[1]{\mathchoice{\widetilde{#1}}{\widetilde{#1}}{\oldtilde{#1}}{\oldtilde{#1}}} \let\oldexists\exists \renewcommand*{\exists}{\mathrel{\oldexists}} \let\oldforall\forall \renewcommand*{\forall}{\mathrel{\oldforall}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{A rank one mild mixing system without minimal self joinings} \alttitle{Un système de rang un modérément mélangeant sans autocouplages minimaux} \subjclass{37A05} \keywords{Mild mixing, Rank one, Self-joinings} \author[\initial{J.} \lastname{Chaika}]{\firstname{Jon} \lastname{Chaika}} \address{Office 217 Fuld Hall\\ Einstein Dr, Princeton\\ NJ 08540 (USA)} %demander si une adresse email pro est présente. \email{jonchaika@gmail.com} \thanks{We thank Mariusz Lemanczyk for bringing this question to our attention and a helpful conversation. We thank the anonymous referee for suggestions that improved the clarity of the arguments. JC was partially supported by NSF grants DMS-2055354 and DMS-2350393.} \CDRGrant[NSF]{DMS-2055354} \CDRGrant[NSF]{DMS-2350393} \author[\initial{D.} \lastname{Robertson}]{\firstname{Donald} \lastname{Robertson}} \address{Department of Mathematics\\ University of Manchester\\ Oxford Road, Manchester\\ M13 9PL (United Kingdom)} \email{donald.robertson@manchester.ac.uk} \begin{abstract} We show that there is a rank $1$ transformation that is mildly mixing but does not have minimal self-joinings, answering a question of Thouvenot. \end{abstract} \begin{altabstract} Nous montrons l'existence d'une transformation de rang $1$ qui est modérément mélangeante mais admet des autocouplages non minimaux, ce qui répond à une question de Thouvenot. \end{altabstract} \datereceived{2024-11-26} \daterevised{2025-09-05} \dateaccepted{2025-10-03} \editors{S.~Gou\"ezel and Y.~Coudène} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \dateposted{2026-03-17} \begin{document} \maketitle \section{Introduction} A \emph{joining} of two measure-preserving system $(X,\mu,T)$ and $(Y, \nu,S)$ is any $T \times S$ invariant measure on $X \times Y$ that is mapped to $\mu$ by the first coordinate projection and to $\nu$ by the second. The product measure $\mu \otimes \nu$ is always a joining. The existence of other joinings of $(X\,\mu,T)$ and $(Y,\nu,S)$ detects whether the systems have properties in common. In particular, if $(X,\mu,T)$ and $(Y,\nu,S)$ have no other joinings they tend to have orthogonal dynamical properties. When $\mu \otimes \nu$ is the only joining the systems $(X,\mu,T)$ and $(Y,\nu,S)$ are called \emph{disjoint}. Joinings and disjointness were first studied by Furstenberg~\cite{MR0213508}. The survey of de la Rue~\cite{MR4647076} is a good introduction. A non-trivial system $(X,\mu,T)$ is never disjoint from itself: there are always \emph{off-diagonal} joinings obtained by embedding $X$ in $X \times X$ using the map $x \mapsto (x,T^n x)$ for any $n \in \Z$. When these and the product measure are the only ergodic joinings of $(X,\mu,T)$ with itself the system $(X,\mu,T)$ is said to have \emph{minimal self-joinings}. Rudolph~\cite{MR0555301} gave the first example of a system with minimal self-joinings. King~\cite{MR0963154} showed that all mixing rank one systems have minimal self-joinings. The survey of Ferenczi~\cite{MR1436950} collects many equivalent definitions of rank one and we record here the definition most useful for our purposes. \begin{defi}[{\cite[Definition~6]{MR1436950}}] A system $(X,\mu,T)$ is \emph{rank one} if for every measurable set $A \subset X$ and every $\epsilon > 0$ there is a measurable set $F \subset X$ and $h \in \N$ and a measurable set $A' \subset X$ such that: \begin{itemize} \item the sets $F, TF,\dots,T^{h-1} F$ are pairwise disjoint, \item $\mu(A' \symdiff A) < \epsilon$, \item $\mu(F \cup TF \cup \dots \cup T^{h-1} F) > 1 - \epsilon$, \item $A'$ belongs to the $\sigma$-algebra generated by $\{ F, TF,\dots, T^{h-1} F\}$, \end{itemize} all hold. \end{defi} Thouvenot asked the following question. \begin{ques}[{\cite[\S~1.2 in Section~12]{MR4232241}}]\label{q:thouvenot} Does a non-rigid rank one transformation without factors have the minimal self-joinings property? \end{ques} To make sense of this question we recall the following definitions. \begin{defi} Fix a system $(X,\mu,T)$. Say that $f \in \ltwo(X,\mu)$ is \emph{rigid} for $T$ if there is a sequence $ \bir$ in $\Z$ with $|r(n)| \to \infty$ and $T^{r(n)} f \to f$ in $\ltwo(X,\mu)$. In this situation we say that $f$ is \emph{rigid} along $\bir$. Any such sequence is called an $f$- \emph{rigidity sequence}. One says that $(X,\mu,T)$ is \emph{mild mixing} if the only rigid functions are the constant functions. \end{defi} King~\cite[p.~377]{MR0863200} proved that all non-trivial factors of rank one systems are rigid. As mild mixing systems cannot have non-trivial rigid factors, Question~\ref{q:thouvenot} is equivalent to asking whether mildly mixing rank-one systems have minimal self-joinings. We construct a mild-mixing rank-one system that does not have minimal self-joinings, thereby answering negatively Question~\ref{q:thouvenot}. \begin{theo}\label{thm:main} There is a mild mixing rank one system $(X,\mu,T)$ without minimal self-joinings. \end{theo} Ryzhikov has results connected to this theorem. On the one hand he showed that such an example can not be ``bounded'' rank 1. That is, bounded rank 1 mildly mixing systems have minimal self-joinings~\cite{MR3235793}. On the other hand, he showed that for any $\epsilon>0$ there is a mildly mixing transformation without minimal self-joinings that has a ``local rank'' of $1-\epsilon$~\cite{Ryzarx}. \subsection{Outline of proof} We define in Section~\ref{sec:system} a measure-preserving system $(X,\mu,T)$. The following properties of the system are established in Sections~\ref{sec:rank one}, \ref{sec:ergodic joining} and~\ref{sec:mild mixing} respectively. \begin{theo}\label{thm:rank one} The system constructed in Section~\ref{sec:system} is rank one. \end{theo} \begin{theo}\label{thm:ergodic joining} The system constructed in Section~\ref{sec:system} does not have the minimal self-joinings property. \end{theo} \begin{theo}\label{thm:mild mixing} The system constructed in Section~\ref{sec:system} is mild mixing. \end{theo} Our system will be the induced transformation of an odometer on an explicit subset obtained by removing a countable number of cylinders. First consider the odometer \[ \Omega = \prod_{i \in \N} \{0, \dots, 9 \} \] on which $S : \Omega \to \Omega$ is the usual ``addition with carry''. The product of the uniform measures on $\{0,\dots,9\}$, denoted $\nu$, is $S$-invariant and the system $(\Omega, \nu, S)$ is ergodic. Put \[ W_i = \{ x \in \Omega : x(1) = 7,\dots, x(i-1) = 7, x(i) = 8 \} \] for all $i \ge 2$ and $W_1 = \{ x \in \Omega : x(1) = 8 \}$. The closed subset \[ X = \Omega \big\backslash \bigcup_{i \in \N} W_i \] has positive measure and we can therefore consider the induced system $(X,\mu,T)$ where $\mu$ is the normalization of the restriction of $\nu$ to $X$ and $T$ is the first return map defined by \[ T(x) = S^{\min \left\{ n \in \N\, :\, S^n(x) \in X\right\}}(x) \] for (in this case) all $x \in X$. The removal of the cylinder sets $W_i$ dramatically alters the dynamical properties of the system. While $(\Omega,\nu,S)$ is rigid, the system $(X,\mu,T)$ is mild mixing. This is due to the abundance for any large $n \in \N$ of pairs of nearby points $x,y$ with the property that $T^n(x)$ and $T^{n+1}(y)$ are also nearby. For example, if $n = 10^i$ and $x(i) = 8$ and $y(i) = 9$ and $x(j) = y(j)$ for all $j < i$ and $(x(1),\dots,x(i-1))$ precedes $(7,\dots,7)$ lexicographically in the collection of cylinders that intersect $X$ then: \begin{itemize} \item $x$ and $y$ are close, \item there will be $1 \le k \le n$ with $S^k(x) \in W_i$, \item for no $1 \le k \le n$ will $S^k(y)$ belong to any $W_j$, \end{itemize} so $T^n(x)$ will be close to $T^{n+1}(y)$. One can use such points to show that every rigid function is approximately $T$-invariant and therefore constant by ergodicity. For more general $n$ the decimal expansion of $n$ dictates the scale $i$ for which the removed cylinder $W_i$ can be used to ``pick up invariance'' in the above sense. It is likely, using an argument analogous to that of~\cite{MR2129107} for showing that linear recurrent 3-IETs have minimal self-joinings, that pairs of points with the above behaviour can be used to prove the system $(X,\mu,T)$ has the much stronger property of minimal self-joinings. To produce a system that does not have minimal self-joinings we build on a method from~\cite{MR4269425} for producing ergodic joinings that are neither one of the off-diagonal joinings nor the product measure. To do so we modify $\Omega$ to be of the form \[ \Omega = \prod_{i \in \N} \{ 0,\dots,c(i) \} \] where $c(i) = 9$ unless $i = 2^{j+5}$ in which case $c(i) = 2^{j+6} - 1$; and by modifying $W_i$ when $i = 2^{j+5}$ to consist of many more cylinders. The formal definition is in Section~\ref{sec:system}. In modifying the odometer to allow the removal of many cylinders at certain scales we are able to find for each $k \in \N$ sets $A_k$ and $B_k$ of measure about 0.5 and times $\p(k)$, $\q(k)$ with \begin{equation}\label{intro:reduction} T^{\p(k)} \approx I \;\text{on}\; A_k \quad\text{and}\quad T^{\p(k)} \approx T^{\q(k)} \;\text{on}\; B_k \end{equation} and the relation $q_{k+1} = 2p_k - q_k$. We use these properties to build a non-trivial joining in Section~\ref{sec:ergodic joining}. See in particular the paragraph after Proposition~\ref{prop:ce}, for how these properties facilitate building the joining. Having modified the sets $W_i$ it is no longer clear that the resulting system $(X,\mu,T)$ is mild mixing; the bulk of our effort is in verifying that property. We wish to pursue the same strategy as described above, but the decimal expansion of a time $n$ may no longer be relevant. Instead one must expand \begin{equation}\label{intro:expansion} n = e(\scl(n)) \p(\scl(n)) + e(\scl(n) - 1) \p(\scl(n) - 1) + \dots + e(1) \p(1) \end{equation} where $k \mapsto \p(k)$ is a specific sequence of quickly growing return times for our system $(X,\mu,T)$. When the scale $\scl(n)$ of the expansion~\eqref{intro:expansion} is not of the form $2^{j+5}$ we can again ``pick up invariance'' using the single removed cylinder $W_{\scl(n)}$. However, if $\scl(n)$ is of the form $2^{j+5}$ we cannot take this approach: the set $W_{\scl(n)}$ that we remove consists in this case of too many cylinders to run that argument. To overcome this difficulty we appeal to the relations~\eqref{intro:reduction} which will enable us to ``reduce'' our attention to $n - e(\scl(n)) \p(\scl(n))$ and $n - (\p(\scl(n)) - \q(\scl(n))) e(\scl(n))$ on $A_{\scl(n)}$ and $B_{\scl(n)}$ respectively. A single application of this reduction procedure may not suffice, as these reduced times may themselves not have expansions at scales where one can pick up invariance. Moreover, even after several iterations of this reduction procedure we might not be in a position to pick up invariance. To overcome this difficulty we consider also the reductions of a multiple $sn$ of $n$. Our main technical result (Proposition~\ref{prop:reduction}) keeps track of what happens as we apply the reduction procedure to $n$ and $sn$ simultaneously. The upshot is that the reduction of $sn$ is roughly equal to $s$ multiplied by the reduction of $n$. Fixing in advance a relatively small scale compared to that of $n$ we choose $s$ large enough that $s$ multiplied by the reduction of $n$ must ``overflow'' and have an expansion at which one can again pick up invariance. This approach is quite technical, and it is natural to wonder whether the careful bookkeeping is necessary. Two facts suggest this care is important. The first fact is the real possibility of ``secret'' rigidity times. Such ``secret'' times are present in exchanges of three intervals (see for instance Case~2 in the proof of~\cite[Theorem~5.1]{MR2129107}) and transformations such as Katok's map (see~\cite[Section~1.4.3]{MR1436950}). The second fact is that there are some natural candidates for rigidity sequences for our transformation: we build our joining by showing that there is a sequence $n_1,n_2,\ldots\in \Z$ so that the corresponding sequence of off-diagonal joinings is Cauchy in the weak$^*$ topology; it is then natural to wonder if say, ${n_{i+1}-n_i}$ is a rigidity sequence. It is an open question of King~\cite[p.~382]{MR0863200} whether rank one transformations always have that $\{(id \times T^n)_*\mu\}_{n \in \Z}$ is weakly dense in the set of ergodic self-joinings. We hope that our strategy will help to further differentiate mildly mixing rank one transformations from mixing rank one transformations. We comment further on this in Remark~\ref{rem:rank two}. \subsection{Outline of paper} In Section~\ref{sec:system} we define our system $(X,\mu,T)$ and introduce the numeration system relevant to it. Additionally, we prove some basic properties of the system including that it is partially rigid and weakly mixing. We prove in Sections~\ref{sec:rank one} and~\ref{sec:ergodic joining} that our system is rank one and has a non-trivial ergodic joining, respectively. The most involved argument (that $(X,\mu,T)$ is mild mixing) is in Section~\ref{sec:mild mixing}. We include two appendices: one proves a minor modification of work in~\cite{MR4269425} needed in Section~\ref{sec:ergodic joining} and the other lists notation used throughout the paper. \section{Building the system}\label{sec:system} In this section we define our measure-preserving system $(X,\mu,T)$ from an odometer $(\Omega,\nu,S)$ by taking $X$ to be a subset of $\Omega$ and $T$ to be the induced transformation and $\mu$ to be the normalized restriction of $\nu$ to $X$. In fact $X$ will be $\Omega$ less a countable union of odometer cylinders and thus a closed subset of $\Omega$. \subsection{An odometer} Define \begin{align*} d(j) &{} = 2^{j+5} \\ a(j) &{} = 2^{j+5} \end{align*} for all $n \in \N$. Define \[ c(n) = \begin{cases} 9 & n \notin \{ a(j) : j \in \N \} \\ 2 d(j) + 1 & n = a(j) \end{cases} \] for all $n \in \N$ and put \[ \Omega = \prod_{n \in \N} \{0,\dots,c(n) \} \] with the product topology. Given $x,y \in \Omega$ and $k \in \N$ write \[ x \overset{k}{=} y \] when $x(i) = y(i)$ for all $1 \le i \le k$. For concreteness, define a metric on $\Omega$ by \[ \metric(x,y) = \begin{cases} 0 &\text{if } x=y \\ 2^{-j} &\text{where } j = \min\{i:x(i)\neq y(i)\} \end{cases} \] for all $x,y \in \Omega$. Define $S : \Omega \to \Omega$ as follows: put $S(c) = 0$ and, for every $x \ne c$ define \[ S(x(1),x(2),\dots) = (0,\dots,0,\,x(j) + 1,\,x(j+1),\dots) \] where $j = \min \{ k \in \N : x(k) < c(k) \}$. The map $S$ is a homeomorphism of $\Omega$ and preserves the product Borel measure \[ \nu = \bigotimes_{k \in \N} \frac{1}{c(k) + 1} \left(\delta_0 + \dots + \delta_{c(k)}\right) \] on $\Omega$. By a \emph{cylinder} we mean any subset of $\Omega$ of the form \[ \left\{ x \in \Omega : x(i(1)) = b(1),\dots,x(i(k)) = b(k) \right\} \] for some $k \in \N$ and values $0 \le b(j) \le c(i(j))$ for each $1 \le j \le k$. We call $i(1),\dots,i(k)$ the \emph{defining indices} of the cylinder. The \emph{length} of a cylinder is the numbers of coordinates it fixes. We sometimes write $|C|$ for the length of the cylinder $C$. We use the notation % \begin{multline}\label{eqn:cylinder} [b(1)\ b(2) \dots b(k)]\\ = \{ x \in \Omega : x(1) = b(1) \bmod c(1),\dots,x(k) = b(k) \bmod c(k) \} \end{multline} given prescribed values $b(1), \dots, b(k) \in \N \cup \{0\}$ for cylinders defined by the first $k$ coordinates. We for example write $[0^k]$ for the cylinder $[00\dots 0]$ of length $k$ and $[7^{k-1} 8]$ for the cylinder $[77\dots 78]$ of length $k$. We sometimes refer to the index $n \in \N$ defining an odometer digit as a \emph{scale}. When $n$ belongs to $\{ a(j) : j \in \N \}$ we call $n$ an \emph{unbounded} scale; otherwise $n$ is a \emph{bounded} scale. For any index $n \ge 2$ we picture the odometer as a collection of $t(n)$ cylinders arranged as in Figure~\ref{fig:odometerTower} mapped to one another by the transformation $S$. For each $s \in \N$ define $\lambda(s)$ to be the largest unbounded scale strictly smaller than $s$. Thus $\lambda(a(j)+1) = a(j)$ and $\lambda(a(j)) = a(j-1)$. For example $\lambda(2^{j+5}) = 2^{j+4}$. We will sometimes iterate $\lambda$ so that $\lambda^2(a(j)+1) = a(j-1)$. For example $\lambda^3(2^{j+5} + 6) = 2^{j+3}$. Put \[ t(n) = \prod_{j=1}^n 1 + c(j) \] for each $n \in \N$. Note that $t(n)$ is the minimal $m \in \N$ with $S^m[0^{n+1}]=[0^n 1]$. \begin{figure}[!h] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-2,0)}] \node at (0,4) {$c(1) c(2) \dots c(n-1)$}; \node at (0,3.2) {$\vdots$}; \node at (0,2) {$20 \dots 0$}; \node at (0,1) {$10 \dots 0$}; \node at (0,0) {$00 \dots 0$}; \end{scope} \begin{scope} \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$0$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$1$}; \end{scope} \begin{scope}[shift={(3,0)}] \node at (0.5,3) {$\dots$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$c(n)$}; \end{scope} \end{tikzpicture} \caption{A partition of an $\Omega$ into cylinders (drawn schematically as intervals) of length $n$. Rows are labeled by the first $n - 1$ coordinates and columns are labeled by the $n^{\rm th}$ coordinate. The transformation $S$ maps a cylinder to the one above, with the cylinder at the top of each column mapped to the bottom cylinder of the tower to the right, and the top right cylinder mapped to the bottom left cylinder.}\label{fig:odometerTower} \end{figure} \subsection{Inductive removals to define the system}\label{subsec:removals} For each $k \in \N$ we define a union $W_k$ of cylinders of length $k$ that will be removed from $\Omega$ to form our space $X$. When $k \notin \{ a(n) : n \in \N \}$ the set $W_k$ will consist of the single cylinder $[7^{k-1}8]$ of length $k$. When $k$ is equal to some $a(n)$ the set $W_k$ will consist of a large number of cylinders of length $k$ defined by an inductive procedure. The formal construction follows. Put $Y_0 = \Omega$. Put $W_1 = [8]$ and for each $1 < k < a(1)$ define \[ W_k = \left\{ x \in \Omega\,:\, x(i) = 7, \quad\forall 1 \le i < k \quad\text{and}\quad x(k) = 8 \right\} = \left[7^{k-1}8\right] \] so that $\nu(W_k) = 1/t(k)$ for all $1 \le k < a(1)$. Write \[ Y_k = \Omega \setminus (W_1 \cup \dots \cup W_k) \] and let $S|Y_k$ be the induced transformation on $Y_k$ for all $1 \le k < a(1)$. See Figure~\ref{fig:boundedRemoval} for a schematic. For each $1 \le k < a(1)$ define $\q(k) = 0$ and $\p(k)$ to be the minimal $m \in \N$ with $(S|Y_k)^m [0^{k-1}0] = [0^{k-1}1]$. For example $\p(1) = 1$ and $\p(2) = 9$ and $\p(3) = 89$. \begin{figure}[!h] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-1.5,0)}] \node at (0,9) {$99 \dots 99$}; \node at (0,8) {$79 \dots 99$}; \node at (0,7.2) {$\vdots$}; \node at (0,6) {$97 \dots 77$}; \node at (0,5) {$77 \dots 77$}; \node at (0,4) {$67 \dots 77$}; \node at (0,3.2) {$\vdots$}; \node at (0,2) {$20 \dots 00$}; \node at (0,1) {$10 \dots 00$}; \node at (0,0) {$00 \dots 00$}; \end{scope} \begin{scope} \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$0$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \node at (0.5,7) {$\dots$}; \node at (0.5,3) {$\dots$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$7$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$8$}; \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$9$}; \end{scope} \begin{scope}[shift={(7.5,0)}] \draw [decorate, decoration = {brace}] (0,9.25) -- (0,-0.25); \node[align=left] at (1.2,4.5) {$\p(k)$\\cylinders}; \end{scope} \end{tikzpicture} \caption{The picture of $Y_{k-1}$ at scale $k$ for some $1 < k < a(1)$. The marked cylinder is $W_k$. Note that the rows labeled $89 \dots 99$ and $87 \dots 77$ are not shown as they were removed via $W_1$.}\label{fig:boundedRemoval} \end{figure} Put $\Q(1) = \q(a(1)) = 1$ and define \begin{align*} W_{a(1)} &= \bigcup_{j=1}^{\Q(1)} \left\{ x \in Y_{a(1)-1} : x(a(1)) \ge d(1) + 1 \quad\text{and}\quad \left(S|Y_{a(1)-1}\right)^j(x) \in \left[0^{a(1) -1}\right]\right\}\\ &= \bigcup_{i=d(1)+1}^{2d(1)+1} \left[9^{a(1)-1} i\right] \end{align*} cf.\ Figure~\ref{fig:firstBigRemoval}. We have $\nu(W_{a(1)}) = (d(1) + 1) \Q(1) /t(a(1))$. Put \[ Y_{a(1)} = \Omega \setminus \left(W_1 \cup \dots \cup W_{a(1)}\right) \] and let $S|Y_{a(1)}$ be the transformation induced on $Y_{a(1)}$ by $S$. Analogous to what we did before, define $\p(a(n))$ to be the minimal $m \in \N$ with $(S|Y_{a(1)})^m [0^{a(1)-1}0] = [0^{a(1) - 1} 1]$. Write $\P(1) = \p(a(n))$. \begin{figure}[t] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-1.5,0)}] \node at (0,5) {$99 \dots 99$}; \node at (0,4) {$79 \dots 99$}; \node at (0,3.2) {$\vdots$}; \node at (0,2) {$20 \dots 00$}; \node at (0,1) {$10 \dots 00$}; \node at (0,0) {$00 \dots 00$}; \end{scope} \begin{scope} \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$0$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \node at (0.5,3) {$\dots$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$d(1)$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$d(1)+1$}; \end{scope} \begin{scope}[shift={(6,0)}] \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$d(1)+2$}; \end{scope} \begin{scope}[shift={(7.5,0)}] \node at (0.5,3) {$\dots$}; \end{scope} \begin{scope}[shift={(9,0)}] \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$2d(1)+1$}; \end{scope} \end{tikzpicture} \caption{The picture of $Y_{a(1)-1}$ at scale $a(1)$. The marked cylinders constitute $W_{a(1)}$. Note that the row labeled $89\dots 99$ is not shown as it was removed via~$W_1$.}\label{fig:firstBigRemoval} \end{figure} \pagebreak We now proceed inductively, supposing that subsets $W_1,\dots,W_{a(n)}$ of $\Omega$ and $\p(1),\dots,\p(a(n))$ in $\N$ and $\q(1),\dots,\q(a(n))$ in $\N$ have been defined, with $Y_i = \Omega \setminus (W_1 \cup \dots \cup W_i)$ for all $1 < i \le a(n)$ and $\P(i) = \p(a(i))$ and $\Q(i) = \q(a(i))$ for all $1 \le i \le n$. Define \[ W_k = \left\{ x \in \Omega : x(k) = 7 \quad\forall\; 1 \le i < k \quad\text{and}\quad x(k) = 8 \right\} = \left[7^{k-1}8\right] \] for each $a(n) < k < a(n+1)$. Note that $W_k$ was not removed at any earlier stage. We have \begin{equation}\label{eqn:meas_bounded_removal} \nu(W_k) = \frac{1}{t(k)} \end{equation} for all $a(n) < k < a(n+1)$. Let \[ Y_k = \Omega \setminus (W_1 \cup \dots \cup W_k) = Y_{a(n)} \setminus \left(W_{a(n)+1} \cup \dots \cup W_k\right) \] for all $a(n) < k < a(n+1)$. Define $\p(k)$ to be the minimal $m \in \N$ with \[ (S|Y_k)^m \left[0^{k-1}0\right] = \left[0^{k-1}1\right] \] and define $\q(k) = 0$ for all $a(n) < k < a(n+1)$. Next, let $\Q(n+1) = \q(a(n+1)) = 2 \P(n) - \Q(n)$ and put \[ W_{a(n+1)} = \bigcup_{j=1}^{\Q(n+1)} \left\{ \begin{aligned} &x \in Y_{a(n+1)-1} : x(a(n+1)) \ge d(n+1) + 1\\ &\mkern150mu\text{and}\quad \left(S|Y_{a(n+1)-1}\right)^j(x) \in \left[0^{a(n+1)-1}\right] \end{aligned} \right\} \] which is shown in Figure~\ref{fig:unbounded_removal}. We have \begin{equation}\label{eqn:meas_unbounded_removal} \nu\left(W_{a(n+1)}\right) = \frac{(d(n+1)+1) \Q(n+1)}{t(a(n+1))} \end{equation} Let \[ Y_{a(n+1)} = \Omega \setminus \left(W_1 \cup \dots \cup W_{a(n+1)}\right) \] and define $\p(a(n)+1) \in \N$ to be minimal satisfying \[ \left(S|Y_{a(n+1)}\right)^{\p(a(n+1))}\left[0^{a(n+1)-1} 0\right] = \left[0^{a(n+1) - 1}1\right] \] and put $\P(n+1) = \p(a(n+1))$ to complete the inductive construction. It is immediate from Figures~\ref{fig:boundedRemoval} and~\ref{fig:unbounded_removal} that \begin{equation}\label{eqn:pn_bound} (c(n) + 1) \p(n) \le t(n) \end{equation} for all $n \in \N$ and in particular, $t(n)$ is the number of levels of $Y$ at stage $n$ and $p(n)$ is morally the number of levels of $X$ at $p(n-1)$. \begin{lemm}\label{lem:removed_mass} $ \sum_{k \in \N} \nu(W_k) < \frac{1}{8}. $ \end{lemm} \begin{proof} First estimate \begin{equation}\label{eq:unbounded removal est} \begin{aligned} \frac{(d(n)+1) \Q(n)}{t(a(n))} &= \frac{d(n)+1}{c(a(n))+1} \cdot \frac{\Q(n)}{t(a(n)-1)} \le \frac{2\P(n-1)}{t(a(n)-1)} \le \frac{t(a(n-1))}{t(a(n)-1)}\\ &= \frac{1}{10^{a(n) - a(n-1)-1}} \end{aligned} \end{equation} using~\eqref{eqn:pn_bound}. Then use~\eqref{eqn:meas_bounded_removal} and~\eqref{eqn:meas_unbounded_removal} to get \[ \sum_{k \in \N} \nu(W_k) \le \sum_{k \in \N} \frac{1}{t(k)} + \sum_{n \in \N} \frac{(d(n)+1) \Q(n)}{t(a(n))} \le \sum_{n \in \N} \frac{1}{10^n} + \frac{1}{10^{a(n+1) - a(n)-1}} \] which is less than $1/8$. \end{proof} By Lemma~\ref{lem:removed_mass} the set \[ X = \Omega \big\backslash \bigcup_{k \in \N} W_k \] has positive measure. It is a closed subset of $\Omega$ as each $W_k$ is open. The set \[ G = \left\{ \begin{aligned} x \in X : S^n(x) \in X_0\quad &\text{for infinitely many } n \in \N\\ \qquad\text{and } S^{-n}(x) \in X_0\quad &\text{for infinitely many } n \in \N \end{aligned} \right\} \] has full measure in $X$ by Poincaré{} recurrence. We define $T : X \to X$ by taking $T$ to be the first return map on $G$ and the identity map on $X \setminus G$. Define $\mu(B) = \nu(B \cap X) / \nu(X)$ for every set $B \subset X$ of the form $C \cap X$ with $C$ a Borel subset of $\Omega$. Our system is $(X,\mu,T)$. As $(\Omega,\nu,S)$ is ergodic $(X,\mu,T)$ is as well. \begin{figure}[!h] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-0.5,0)}] \draw [decorate, decoration = {brace}] (0,-1) -- (0,8); \node[align=right] at (-1.2,3.5) {$\P(n)$\\cylinders}; \end{scope} \begin{scope} \draw (0,8) -- (1,8); \draw (0,7) -- (1,7); \node at (0.5,6.2) {$\vdots$}; \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$0$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \node at (0.5,6) {$\dots$}; \node at (0.5,1) {$\dots$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,8) -- (1,8); \draw (0,7) -- (1,7); \node at (0.5,6.2) {$\vdots$}; \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$d(n)$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw[color=red] (0,8) -- (1,8); \draw[color=red] (0.4,7.7) -- (0.6,8.3); \draw[color=red] (0,7) -- (1,7); \draw[color=red] (0.4,6.7) -- (0.6,7.3); \node at (0.5,6.2) {$\vdots$}; \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$d(n)+1$}; \end{scope} \begin{scope}[shift={(6,0)}] \node at (0.5,6) {$\dots$}; \node at (0.5,1) {$\dots$}; \end{scope} \begin{scope}[shift={(7.5,0)}] \draw[color=red] (0,8) -- (1,8); \draw[color=red] (0.4,7.7) -- (0.6,8.3); \draw[color=red] (0,7) -- (1,7); \draw[color=red] (0.4,6.7) -- (0.6,7.3); \node at (0.5,6.2) {$\vdots$}; \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$2d(n)+1$}; \end{scope} \begin{scope}[shift={(9,0)}] \draw [decorate, decoration = {brace}] (0,8) -- (0,5); \node[align=left] at (1.2,6.5) {$\Q(n)$\\cylinders}; \end{scope} \end{tikzpicture} \caption{The picture of $Y_{a(n)-1}$ at scale $a(n)$. The marked cylinders constitute~$W_{a(n)}$.}\label{fig:unbounded_removal} \end{figure} \subsection{Measures of cylinders} We will note that our choices for $d(n)$ and $a(n)$ give the following properties. \begin{align} \tag{$\rm P1$} \P(n) \le t(a(n)-1)\label{p:tower_height}\\ \tag{$\rm P2$}\label{p:summing_gaps} \sum_{n=1}^\infty \frac{\P(n)}{\P(n+1)} < \frac{1}{10^5} \end{align} We will want often to study partial $T$ orbits as orbits of some $S|Y_n$. For convenience we define for any $r \in \Z$ and any $b \in \N$ \[ \bad(b,r) = \bigcup_{a(m) > b} \left\{x \in X : (S|Y_b)^i(x) \in W_{a(m)} \quad\text{for some }\; -|r| \le i \le |r| \right\} \] to consist of those points whose $S|Y_b$ orbits visit some finer $W_{a(m)}$ between times $-|r|$ and $|r|$. When $x$ does not belong to $\bad(b,r)$ we have \[ T^i(x) = (S|Y_b)^i(x)\quad \forall -|r| \le i \le |r| \] unless some $(S|Y_b)^i(x)$ belongs to some $[7^i 8]$ with $i \ge b$. If a cylinder set $C$ of length $k$ does not belong to $W_1 \cup \dots \cup W_k$ then $\mu$ assigns it positive measure. In other words, any cylinder set $C$ not removed by scale $|C|$ has positive measure in $X$. Slightly more is true: the following lemma will be used to prove $\mu(C) t(|C|)$ is bounded away from zero over such cylinder sets. \begin{lemm}\label{lem:cylinder_decay} If $C \ne [7^{|C|}]$ is a cylinder not contained in $W_1 \cup \dots \cup W_{|C|}$ then \[ \nu(C \setminus X) \le \frac{4}{t(|C|)} \sum_{a(n) > |C|} \frac{\P(n-1)}{\P(n)} \] holds. \end{lemm} \pagebreak{} \begin{proof} Fix a cylinder set $C \ne [7^{|C|}]$ disjoint from $W_1 \cup \dots \cup W_{|C|}$. We can only possibly have $C \cap W_j \ne \emptyset$ when $j = a(n) > |C|$ for some $n \in \N$. For any such $n$, note that $C$ consists of $t(a(n)) / t(|C|)$ cylinders of length $a(n)$ and that at most \[ \frac{2 (d(n) + 1) \Q(n)}{(2d(n)+1) \P(n)} \le \frac{2 \Q(n)}{\P(n)} \le \frac{4 \P(n-1)}{\P(n)} \] of them belong to $W_{a(n)}$. Therefore \[ \nu\left(C \cap W_{a(n)}\right) \le \frac{4}{t(|C|)} \frac{\P(n-1)}{\P(n)} \] from which the conclusion follows. \end{proof} \begin{lemm}\label{lem:cylinder_size} There is $\xi > 0$ such that \[ \mu(C) \ge \frac{\xi}{t(|C|)} \] whenever $C$ is a cylinder set not contained in $W_1 \cup \dots \cup W_{|C|}$. \end{lemm} \begin{proof} If $C = [7^{|C|}]$ then \[ \nu(C \cap W_n) \le \frac{1}{t(n)} \] whenever $n > |C|$ is a bounded scale. Thus for every $C$ not contained in $W_1 \cup \dots \cup W_{|C|}$ we have \[ \nu(C \setminus X) \le \sum_{n > |C|} \frac{1}{t(n)} + \frac{4}{t(|C|)} \sum_{a(n) > |C|} \frac{\P(n-1)}{\P(n)} \le \frac{1}{t(|C|)} \left(\frac{1}{9} + \frac{4}{10^5} \right) \] after Lemma~\ref{lem:cylinder_decay} and~\ref{p:summing_gaps}. Thus \[ \xi = \frac{55}{63} \frac{1}{\nu(X)} \] suffices. \end{proof} \begin{lemm}\label{lem:indep} Let $D \subset \Omega$ be a cylinder whose largest defining index is $r$ and let $C \subset \Omega$ be a cylinder whose smallest defining index is $r+k$ for some $k \ge 1$. Further assume $D \cap X \neq \emptyset$ and $D \cap C \cap X \neq \emptyset$. Then \[ \frac{\mu(D\cap C \cap X)}{\mu(D)\mu(C)}\geq 1-\frac{1}{2^k} \] holds. \end{lemm} \begin{proof} We first consider the case when $D$ has every index defined up to $r$ and the indices defining $C$ are from $r+1$ to $b$. (This is a special version of the case $k=1$.) Thus $C \cap D$ is a single cylinder defined at entries $1$ to $b$ that is non-empty. Thus $ C \cap D \subset Y_b$ and so $\mu(C \cap D)=\nu(X)^{-1}(\frac 1 {t(b)} -\nu(C \cap D \cap X^c))$. Thus we compute \begin{align*} \nu(Y_b \setminus X) &\leq \sum_{j>b} \frac{1}{t(j)} + \sum_{a(j)>b} \frac{4\P(j-1)}{\P(j)}<\frac{1}{8} \nu(C \cap D) = \frac{1}{8} \frac {1}{t(b)} \\ \intertext{so} \mu(C\cap D) &\geq \left(\frac{7}{8} \right)^2 \frac{1}{t(r)} \frac{1}{\nu(X)} \frac{t(r)}{t(b)} \frac{1}{\nu(X)} \geq \left(\frac{7}{8}\right)^2\mu(C) \mu(D) \end{align*} establishing the first case. We now treat $D$ defined at all indices at most $r$ and $C$ is defined at all indices from $r+k$ to $b$. We consider $D \cap Y_{r+k-1},$ a union of $N\geq 8^k $ cylinders. Now \begin{equation}\label{eq:D bound} \left| \mu(D)-\frac{N}{\p(r+k+1)} \right| \leq \max \left\{ \nu\left(Y_{r+k-1}\setminus X\right)\nu(X)^{-1},\; \frac{\nu(X)^{-1}}{\nu(Y_{r+k-1})^{-1}} \right\}. \end{equation} Indeed we estimate the maximum amount that could be removed by going from $Y_{r+k-1}$ to $X$ and assume it is all removed from $D$ and on the other hand, if nothing is removed from $D$, how much we have to scale it to obtain its proportion of $X$ from its proportion of $Y_{r+k-1}$. Similarly \begin{equation}\label{eq:C bound} \left| \mu(C)-\frac{\p(r+k+1)}{\p(b+1)} \right| \leq \max \left\{ \nu(Y_{b}\setminus X)\nu(X)^{-1},\; \frac{\nu(X)^{-1}}{\nu(Y_{b})^{-1}} \right\}. \end{equation} At most one of the $N$ cylinders defined at entries 1 to $r+k-1$ intersects $\bigcup_{j \notin \{a(k)\}}W(j)$. For each of $N-1$ other cylinders, we intersect them with $C$ and obtain cylinders defined at entries from $1$ to $b$ which do not intersect $W_j$ for $j \notin \{a(i)\}_{i \in \N}.$ Name these cylinders $E_1,\dots,E_{N-1}$ and note that \[ \nu \left(\bigcup_{i=1}^{N-1}E_i \cap C \cap Y_b^c \right) < \sum_{a(j)>b}\frac 4 {t(b)}\frac{\P(j-1)}{\P(j)}< \frac 1 {2^{b+5}}\frac 1 {\p(b+1)} \] holds. We obtain that \[ \mu(C \cap D) \geq \frac{N-1}{\p(b+1)}-\frac{1}{2^{b+5}}\frac{1}{\p(b+1)}. \] Since $N\geq 8^k$, combining this with~\eqref{eq:D bound} and~\eqref{eq:C bound} we have the claim. The general case is similar: treat $D$ as a union of $A$ cylinders defined on every index from 1 to $r+k-1$ and $C$ as a union of $B$ cylinders defined on every index from $r+k$ to $b$. \end{proof} \subsection{Expansion and reduction}\label{subsec:expansion_reduction} Given $r \in \Z$ we will analyze $T^r$ by expanding $r$ using a numeration system closely tied to the dynamics of $T$. The numeration system is based on the heights $\p(j)$ of the left-most towers in the scale $j$ picture. At unbounded scales we might instead have chosen to use $\p(j) - \q(j)$ in our numeration system as each height is witnessed by about half of the points in the system. Although we can only use one in our numeration system, we will in effect keep track of both in the technical parts of our arguments using the reductions in Definition~\ref{def:reduction}. To define this expansion we need the following lemma. \begin{lemm}\label{lem:expansion_defined} We have \[ \left(\frac{c(n) + 1}{2} + 1 \right) \p(n) \le \p(n+1) \] for all $n \in \N$. \end{lemm} \begin{proof} We have \[ \p(n+1) = \begin{cases} (c(n) + 1) \p(n) - 1 & n \text{ a bounded scale} \\ (c(n) + 1) \p(n) - \q(n) (c(n)+1)/2 & n \text{ an unbounded scale} \\ \end{cases} \] from which the statement follows. \end{proof} \begin{defi} Fix $r$ in $\Z$. By the \emph{tower expansion} of $r$ we mean the expansion \begin{equation}\label{eqn:rigidity_expan} r = \sum_{j=1}^{\scl(r)} e(j) \p(j) \end{equation} with $e(\scl(r)) \ne 0$ and $2|e(j)| \le c(j)+1$ chosen greedily. Formally, this means the following: take \[ \scl(r) = \max \left\{ j \in \N : e \mapsto |r - e \p(j)|\quad \text{is not minimized by } e = 0 \right\} \] with the convention that $\scl(0) = 0$. Define $\lst(r) = e(\scl(r))$ to be the associated minimizing value, choosing $|\!\lst(r)|$ as large as possible whenever there is a choice. Since $\p(1) = 1$ both $1$ and $-1$ have tower expansions, and since $|r - \lst(r) \p(\scl(r))| < |r|$ one can define the tower expansion of any integer by induction on $|r|$. Indeed letting $r'=r-e(r)J(r)$ we can inductively (using $|r'|<|r|$) assume $r'=\sum_{i=0}^{J(r')}e(i)p(i)$ and represent $r=e(r)p(J(r))+\sum_{i=0}^{J(r')}e(i)p(i)$. Since $J(r') c(\scl(r))/2$ satisfy $T^{\p(\scl(r)) - \q(\scl(r))}(x) \approx x$ (see Corollaries~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T right} and~\ref{cor:neg reduc conseq}$\MK$\eqref{cor:neg red T right}). Motivated by this, the next definition will be used to relate $T^r(x)$ for $r$ at unbounded scales to $T^{r'}(x)$ for $r'$ at smaller scales. \begin{defi}\label{def:reduction} Define the \emph{left} and \emph{right reductions} by \begin{align*} \lft(r) &= r - \lst(r) \p(\scl(r)) \\ \rht(r) &= r - \lst(r) \p(\scl(r)) + \lst(r) \q(\scl(r)) \end{align*} respectively whenever $\scl(r)$ is an unbounded scale. \end{defi} A definite proportion of points $x$ satisfy $T^r(x) \approx T^{\lft(r)}(x)$ and a definite proportion of points $x$ satisfy $T^r(x) \approx T^{\rht(r)}(x)$ whenever $\scl(r)$ is unbounded. \begin{lemm}\label{lem:need it later} Fix $r \in \Z$ with $\scl(r)$ unbounded: \begin{enumerate} \item\label{lemm2.12.1} $|\lft(r)| \le \p(\scl(r))$, \item\label{lemm2.12.2} $|\rht(r)| \le \p(\scl(r)) - \q(\scl(r))$. \end{enumerate} \end{lemm} \begin{proof} For~\eqref{lemm2.12.1}, apply Lemma~\ref{lem:expansion_bound} with $J = \scl(r) - 1$. For~\eqref{lemm2.12.2}, apply Lemma~\ref{lem:expansion_bound} with $J = \scl(r) - 1$ and Lemma~\ref{lem:lot_red_bound}. \end{proof} The following lemmas describe how reductions and scales interact. \begin{lemm}\label{lem: reduction not both small} Fix $m \in \Z$ with $\scl(m)$ an unbounded scale. We have \[ \max\{\scl(\lft(m)), \scl(\rht(m))\} \ge \lambda(\scl(m)) \] and if $\lft(m)$ and $\rht(m)$ are both at an unbounded scale then neither is at scale $\scl(m)$ and at least one of $\scl(\lft(m)) = \lambda(\scl(m))$ and $\scl(\rht(m)) = \lambda(\scl(m))$ holds. \end{lemm} \begin{proof} Since \begin{equation}\label{eq:differnce in reduction} \rht(m) - \lft(m) = \lst(m) \q(\scl(m)) = 2 \lst(m) \p(\lambda(\scl(m))) - \lst(m) \q(\lambda(\scl(m))) \end{equation} we have in general \begin{equation}\label{eq:one correct} \max\{|\rht(m)|,\;|\lft(m)|\}\geq \frac{3}{4} \p(\lambda(m)) \end{equation} and so the larger of $\rht(m)$, $\lft(m)$ in absolute value cannot be at a scale smaller than $\lambda(\scl(m))$, establishing the first claim. We now assume that both $\lft(m),\rht(m)$ are at unbounded scales. It is immediate that (in general) $\scl(\lft(m)) < \scl(m)$. As $\lft(m)$ is assumed to be at an unbounded scale we must have $\scl(\lft(m)) \le \lambda(\scl(m))$. That implies \begin{align*} |\lft(m)| &\le \frac{\p(\lambda(\scl(m))+1)}{2} \\ \intertext{and, as we have} |\lst(m) \q(\scl(m))| &\le \p(\scl(m)-1) \end{align*} from Lemma~\ref{lem:lot_red_bound}, the triangle inequality gives \[ 2|\rht(m)| \le \p(\lambda(\scl(m))+1) + 2\p(\scl(m)-1) < \p(\scl(m)) \] which means $\scl(\rht(m)) < \scl(m)$. As $\rht(m)$ is also assumed to be at an unbounded scale we have $\scl(\rht(m)) \le \lambda(\scl(m))$ as well. Thus when they are both at an unbounded scale, it must be that one of $\rht(m)$ and $\lft(m)$ is at scale $\lambda(\scl(m))$. \end{proof} \begin{lemm}\label{lem:reduction LR} Let $m \in \Z$ be at an unbounded scale $a(b)$. At least one of $\scl(\rht(m)) < \scl(m)$ and $|\rht(m)|<\frac{3}{4}\p(\scl(m))$ is true. Also $\scl(\lft(m)) < \scl(m)$. \end{lemm} \begin{proof} This follows by the greedy algorithm to define $\p$ and the fact that $(2d(b)+2)\q(\scl(m))$ is much smaller than $\frac{1}{8} \p(\scl(m))$. \end{proof} \subsection{Some dynamical properties} We mention here briefly some dynamical properties of $(X,\mu,T)$ that follow quickly from the construction. \subsubsection{Partial rigidity} The task of proving $(X,\mu,T)$ is mild mixing is made difficult by partial rigidity in the system, which we remark upon here. Recall that $(X,\mu,T)$ is \emph{partially rigid} if there is a constant $\delta > 0$ such that, for every measurable $A \subset X$ one has $\mu(A \cap T^n A) \ge \delta \mu(A)$ infinitely often. Using Lusin's theorem, one can deduce partial rigidity from the following proposition. \begin{prop} For all $\epsilon>0$ \[ \lim_{n \to \infty} \mu\left(\left\{x\in X: d\left(T^{\p(n)}x,x\right)<\epsilon\right\}\right)\geq \frac{1}{9} \] holds. \end{prop} Let $k_x:\Z \to \Z$ by $k_x(n)=m$ where $S^m(x) = T^{n}(x)$. It is not hard to check that \[ \nu\left(\left\{x \in X\subset \Omega:k_x(\p(n))=t(n-1)\right\}\right)\geq \frac{1}{9} \] for all large enough $n$. Indeed, this follows from estimating $1-t(n-1)\nu(\bigcup_{k\geq n}W_k).$ Note that there are two cases, one where $n$ is at an unbounded scale and one where $n$ is at a bounded scale. \subsubsection{Weak mixing}\label{subsec:weak_mixing} We prove here that $(X,\mu,T)$ is weak mixing. We follow roughly a well-known argument but give the proof from scratch both for completeness and because the argument proceeds along lines similar to, but much simpler than, those in the proof of our main result. More precisely, in both arguments we study $T$ by relating it to $S|Y_m$ for appropriate $m$. It is particularly important to examine the effect of $W_m$. \begin{prop} $(X,\mu,T)$ is weakly mixing. \end{prop} \begin{proof} Fix an eigenfunction $f$ of $T$ with eigenvalue $\lambda$. Recall that ergodicity of $T$ gives $|f(x)| = 1$ almost surely. We must prove that $\lambda = 1$. Fix $\epsilon > 0$ and $s > 0$. There is a compact set $G \subset X$ with $\nu(G) \ge 1 - s$ on which $f$ is uniformly continuous. Fix $\delta > 0$ so that $\metric(x,y) < \delta$ guarantees $|f(x) - f(y)| < \epsilon$ whenever $\{x,y\} \subset G$. Fix a bounded scale $n \in \N$ with $2^{-(n-1)} < \delta$ and $n-1$ an unbounded scale. We wish to choose points $x,y$ in $X$ with all the following properties. %\begin{enumerate} %\item\label{property1} $\{ x,y, T^{2\p(n)+1}(x), T^{2\p(n)}(y) \} \subset G$ %\item\label{property2} $x(i) = y(i)$ for all $1 \le i \le n-1$ %\item\label{property3} $x(n) = 4$ and $x(n) = 7$ %\item\label{property4} $T^i(x) = (S|Y_n)^i(x)$ and $T^i(y) = (S|Y_n)^i(y)$ for all $1 \le i \le 2 \p(n)+1$ %\item\label{property5} $f(T^{2\p(n)+1}(x)) = \lambda^{2 \p(n) + 1} f(x)$ and $f(T^{2\p(n)}(y)) = \lambda^{2 \p(n)} f(y)$ %\end{enumerate} \begin{gather} \tag{1}\label{property1} \left\{ x,y, T^{2\p(n)+1}(x),\; T^{2\p(n)}(y) \right\} \subset G\\ \tag{2}\label{property2} x(i) = y(i),\quad\forall 1 \le i \le n-1\\ \tag{3}\label{property3} x(n) = 4\quad\text{and} \quad x(n) = 7\\ \tag{4}\label{property4} T^i(x) = (S|Y_n)^i(x)\quad\text{and} \quad T^i(y) = (S|Y_n)^i(y) \quad \forall 1 \le i \le 2 \p(n)+1\\ \tag{5}\label{property5} f\left(T^{2\p(n)+1}(x)\right) = \lambda^{2 \p(n) + 1} f(x)\quad\text{and} \quad f\left(T^{2\p(n)}(y)\right) = \lambda^{2 \p(n)} f(y) \end{gather} Indeed, if we are able to do this then we have \begin{itemize} \item $|f(x) - f(y)| < \epsilon$ by properties~\eqref{property1} and~\eqref{property2}. \item $T^{2 \p(n) + 1}(x)$ and $T^{2 \p(n)}(y)$ agree in their first $n-1$ coordinates by~\eqref{property2}, \eqref{property3}, \eqref{property4} and property~\eqref{property5}. Indeed our third condition and the definition of $W_n$ gives that $(S|Y_n)^{2\p(n)+1}x\overset{n}{=}(S|Y_n)x$ and $(S|Y_n)^{2\p(n)}y\overset{n}{=}(S|Y_n)y$. \item $|f(T^{2\p(n) +1}(x)) - f(T^{2\p(n)}(y))| < \epsilon$ by property~\eqref{property1} and the preceding item from which we conclude that $|1 - \lambda| < 2\epsilon$. \end{itemize} To obtain the five properties we note that $G \cap T^{-2\p(n)} G$ and $G \cap T^{-2\p(n)-1} G$ are both very large if $s$ is small enough, and that $\bad(n,2\p(n)+1)$ will have small measure if $n$ is large, so that we can guarantee the first four properties by shrinking $s$. \eqref{property5} holds almost surely. \end{proof} \pagebreak{} \section{The rank one property}\label{sec:rank one} We prove here that our system is rank one. \begin{proof}[Proof of Theorem~\ref{thm:rank one}] Fix $A \subset X$ measurable and $\epsilon > 0$. For each ${n \in \N}$ let $h(n) \in \N$ be minimal with the property that $(S|Y_{a(n)})^{h(n)} [8 7^{a(n)-1}]$ is the lexicographically largest cylinder of length $a(n)$ in $Y_{a(n)}$. Certainly $h(n) < t(a(n))$. All subsequent removals $W_k$ with $k \not\in \{a(m) : m > n \}$ are contained within $[7^{a(n)}]$. Let \[ F_n=\left[87^{a(n)-1}\right]\setminus \bad \bigl(a(n),h(n)\bigr) \subset X \] which is shown in Figure~\ref{fig:rank_one}. We now bound the measure removed by removing $\bad(a(n),h(n))$. For each $m > n$ we remove from $[8 7^{a(n)-1}]$ at most $d(m) + 1$ cylinders of length $a(m)$. Thus \begin{multline*} \mu\left(X \setminus F_n \cup T^1 F_n \cup \dots \cup T^{h(n)-1} F_n\right) \\ \begin{aligned} &\le \frac{1}{\nu(X)} \left(8 \P(n) \nu\left(\left[0^{a(n)}\right]\right) + \sum_{m > n} h(n) \frac{d(m)+1}{t(a(m))} \right) \\ &\le \frac{8}{\nu(X)} \frac{t(a(n)-1)}{t(a(n))} + \frac{2}{\nu(X)} \sum_{m > n} \frac{t(a(n)) d(m)}{t(a(m))} \\ &\le \frac{8}{\nu(X)} \frac{1}{2d(n)+2} + \frac{2}{\nu(X)} \sum_{m > n} \frac{1}{10^{a(m) - a(n)-1}} \end{aligned} \end{multline*} where the term $8\P(n) \nu([0^{a(n)}])$ accounts for the cylinders of length $a(n)$ that precede $[87^{a(n)-1}]$ lexicographically and we have used $\P(n) \le t(a(n)-1)$. We thus have \[ \mu\left(F_n \cup T F_n \cup \dots \cup T^{h(n)} F_n\right) \to 1 \] as $n \to \infty$. \begin{figure}[!h] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-4.5,0)}] \node at (0,4) {$87 \dots 77$}; \node at (0,3) {$77 \dots 77$}; \node at (0,2) {$67 \dots 77$}; \node at (0,0) {$10 \dots 00$}; \node at (0,-1) {$00 \dots 00$}; \end{scope} \begin{scope}[shift={(-3,0)}] \draw (0,9) -- (1,9); \node at (0.5,8.2) {$\vdots$}; \draw (0,7) -- (1,7); \draw (0,6) -- (1,6); \node at (0.5,5.2) {$\vdots$}; \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$0$}; \end{scope} \begin{scope}[shift={(-1.5,0)}] \node at (0.5,8) {$\dots$}; \node at (0.5,5) {$\dots$}; \node at (0.5,1) {$\dots$}; \end{scope} \begin{scope}[shift={(0,0)}] \draw (0,9) -- (1,9); \node at (0.5,8.2) {$\vdots$}; \draw (0,7) -- (1,7); \draw (0,6) -- (1,6); \node at (0.5,5.2) {$\vdots$}; \draw (0,4) -- (1,4); \draw[dashed] (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$7$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \node at (0.5,8) {$\dots$}; \node at (0.5,5) {$\dots$}; \node at (0.5,1) {$\dots$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,9) -- (1,9); \node at (0.5,8.2) {$\vdots$}; \draw (0,7) -- (1,7); \draw (0,6) -- (1,6); \node at (0.5,5.2) {$\vdots$}; \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$d(n)$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw[color=red] (0,9) -- (1,9); \draw[color=red] (0.4,8.7) -- (0.6,9.3); \node at (0.5,8.2) {$\vdots$}; \draw[color=red] (0,7) -- (1,7); \draw[color=red] (0.4,6.7) -- (0.6,7.3); \draw (0,6) -- (1,6); \node at (0.5,5.2) {$\vdots$}; \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$d(n)+1$}; \end{scope} \begin{scope}[shift={(6,0)}] \node at (0.5,8) {$\dots$}; \node at (0.5,5) {$\dots$}; \node at (0.5,1) {$\dots$}; \end{scope} \begin{scope}[shift={(7.5,0)}] \draw[color=red] (0,9) -- (1,9); \draw[color=red] (0.4,8.7) -- (0.6,9.3); \node at (0.5,8.2) {$\vdots$}; \draw[color=red] (0,7) -- (1,7); \draw[color=red] (0.4,6.7) -- (0.6,7.3); \draw (0,6) -- (1,6); \node at (0.5,5.2) {$\vdots$}; \draw (0,4) -- (1,4); \draw (0,3) -- (1,3); \draw (0,2) -- (1,2); \node at (0.5,1.2) {$\vdots$}; \draw (0,0) -- (1,0); \draw (0,-1) -- (1,-1); \node at (0.5,-2) {$2d(n)+1$}; \end{scope} \end{tikzpicture} \caption{The picture of $Y_{a(n)-1}$ at scale $a(n)$. The marked cylinders make up $W_{a(n)}$. The dashed cylinder contains all $W_j$ with $j > a(n)$ and $j$ a bounded scale. A subset of $[87^{a(n)-1}]$ will be used as the set $F$ exhibiting $T$ as rank one.}\label{fig:rank_one} \end{figure} \pagebreak Choose $n$ so large that there exists cylinders $C_1,\dots,C_t\subset Y$ with defining coordinates $1,\dots.,a(n)$ and \begin{itemize} \item $\mu (A \triangle (X \cap C_1 \cup \dots \cup X \cap C_t) ) < \epsilon/100 \nu(X)$, \item $8 \P(n) \nu([0^{a(n)}]) < \epsilon/10$, \item $\mu(X \setminus F_n \cup T^1 F_n \cup \dots \cup T^{h(n)-1} F_n) < \epsilon/2$, \end{itemize} all hold. Then the $\mu$ measure of the intersection of $A$ with the union of all cylinders that precede $[87^{a(n) - 1}]$ lexicographically is at most $\epsilon/10 \nu(X)$. For $0 \le i < h(n)$ put $A_i = T^i F_n$ if \[ \left(C_1 \cup \dots \cup C_t\right) \cap T^i F_n \ne \emptyset \] and $A_i = \emptyset$ otherwise. The set $A' = A_0 \cup \dots \cup A_{h(n)-1}$ is measurable with respect to the $\sigma$-algebra generated by $\{ F_n, T F_n, \dots, T^{h(n)-1} F_n \}$ and $\mu(A' \triangle A) < \epsilon$. \end{proof} \begin{rema}\label{rem:rank two} It is natural to try to determine whether our procedure can be modified to construct a mildly mixing rank 1 transformation $T$ with the property that $T^n$ has rank two for all $n\geq 2$. (Note that $T$ being mild mixing precludes $T^n$ from being rank one: indeed $T$ is in its centralizer and were $T^n$ rank one it could be concluded that $T$ is rigid by the weak closure theorem. We thank Mariusz~Lemanczyk for pointing this out.) Indeed, one could try to modify our construction (perhaps by removing some extra levels at certain bounded scales and changing the definition of $\Q(n)$) so that for every $n$ there are infinitely many $k$ so that at the unbounded scales $a(k)$ we have $\P(k)$ and $\P(k)-\Q(k)$ both relatively prime to $n$. One could use the following definitions. \begin{align*} F_k &=\bigcup_{i=0}^{n-1}(S|Y_{a(k)})^i\left[87^{a(k)-1}\right]\setminus \bad(a(k),\p(k)+2) \\ F_k &=\bigcup_{i=0}^{n-1}(S|Y_{a(k)})^i\left[87^{a(k)-2}d(k)+2\right]\setminus \bad(a(k),\,\p(k)+2). \end{align*} Letting \begin{align*} h &=\min\left\{i>0:(S|Y_{a(k)})^{in}F \cap \left[0^{a(k)-1}d(k)\right] \neq \emptyset\right\}-1 \\ \intertext{and} h' &=\min\left\{i>0:(S|Y_{a(k)})^{in}F' \cap \left[0^{a(k)}\right] \neq \emptyset\right\}-1 \end{align*} we see our transformation is rank 2. Having done the modification to the construction, one still needs to check that the proof of mild mixing goes through. We do not pursue this here. \end{rema} \section{An ergodic joining}\label{sec:ergodic joining} The purpose of this section is to construct an ergodic self-joining of $(X,\mu,T)$ that is neither the product measure nor an off-diagonal joining. The construction is an application of~\cite[Proposition~3.1]{MR4269425}; to prepare for its application in Section~\ref{subsec:apply_CE} we give in Section~\ref{subsec:two_toweres} a technical description of towers for $(X,\mu,T)$. We briefly describe the proof. We find a pair of sequences of integers $\alpha(i)$ and $\gamma(i)$ and measurable sets $A(i)$, $B(i)$ so that \begin{itemize} \item $T^{\alpha(i)}|_{A(i)}\approx T^{\gamma(i-1)}$ and $T^{\alpha(i)}|_{B(i)} \approx T^{\alpha(i-1)} $ \item $T^{\gamma(i)}|_{A(i)}\approx T^{\alpha(i-1)}$ and $T^{\gamma(i)}|_{B(i)} \approx T^{\gamma(i-1)} $. \end{itemize} We use this to show \begin{itemize} \item $\nu_{\alpha(i)}$ and $\mu_{\gamma(i)}$ are Cauchy with the same limit. \item This limit is close to $c(i) \nu_{\alpha(k)}+(1-c(i))\nu_{\gamma(k)}$ for all $k$. \item This common limit is neither one-to-one on almost every fiber nor the product measure. \item The limit is ergodic. \end{itemize} This strategy is from~\cite{MR4269425} where it was used to build joinings of interval exchange transformations. Proposition~\ref{prop:ce} is most of the technical argument from that paper adapted to our setting. Proposition~\ref{prop:joining_prep} proves the required facts about our system to apply Proposition~\ref{prop:ce}. We describe the choices of the $\alpha(i)$ and $\gamma(i)$, between the statement of Proposition~\ref{prop:ce} and the proof of Theorem~\ref{thm:ergodic joining}. The sets $A(i)$, $B(i)$ are given in the statements of Proposition~\ref{prop:joining_prep}. \subsection{A tower}\label{subsec:two_toweres} In the picture at scale $a(k)$ we have a tower \[ \clT = \bigcup_{n=0}^{(d(k) + 1)\P(k)-1} (S|Y_{a(k)})^n \left[0^{a(k)}\right] \subset Y_{a(k)} \] of height $(d(k)+1) \P(k)$ for the transformation $S|Y_{a(k)}$. We will wish often to think of $\clT \cap X$ as a tower for $T$ on which $T^n$ and $S|Y_{a(j)}^n$ agree for all $-2 \P(k) \le n \le 2 \P(k)$ for most points $x$ in $\clT \cap X$. In this subsection we produce subsets $J(k)$ of $[0^{a(k)}]$ that are the bases of towers for $T$ with such properties. We will do so by removing ``columns'' from the tower. \begin{defi} By a \emph{column} at scale $i$ we mean a subset of $\Omega$ defined by fixing values for all coordinates $i \le n \le I$. We call $I \ge i$ the \emph{rank} of the column. The \emph{base} of a column is its intersection with $[0^{i-1}]$. \end{defi} For example $\{ x \in \Omega : x(4) = 2, x(5) = 3 \}$ is a column at scale 4 and rank 5 while \[ \left\{ x \in \Omega : 5 \le x(a(j)) \le 10 \right\} \] is a union of six columns at scale $a(j)$ and rank $a(j)$. \begin{prop}\label{prop:joining_prep} For every $k \in \N$ there are sets $J(k), U(k) \subset X$ and a~constant $c>0$ with the following properties. \begin{enumerate}\PJenumi \item\label{jt:return} $\min \{ n \in \N : T^n(x) \in J(k) \} \ge (d(k)+1)(2\P(k) - \Q(k))-3$ for every $x \in J(k)$, \item\label{jt:shrinking} $(d(k)+1) \P(k) \sum_{j > k} \mu(J(j)) \to 0$. \end{enumerate} Writing \[ A(k) = \left(\bigcup_{n=0}^{(d(k)+1)\P(k)-1} T^n J(k)\right) \big\backslash U(k) \quad \text{and} \quad B(k) = \bigl(X \setminus A(k)\bigr) \setminus U(k) \] for every $k \in \N$ we have all the following properties. \begin{enumerate}\PJenumi \setcounter{enumi}{2} \item\label{jt:left} $T^n(x) = (S|Y_{a(k)})^n(x)$ for every $x \in A(k)$ and every $-2 \P(k) \le n \le 2 \P(k).$ \item \label{conc:same left} For every $x \in A(k)$ we have $(S|Y_{a(k)})^{\P(k)}(x) \overset{a(k)-1}{=}x$. \item\label{jt:right} $T^n(x) = (S|Y_{a(k)})^n(x)$ for every $x \in B(k)$ and every \[ -2\P(k) + 2\Q(k) \le n \le 2\P(k) - 2\Q(k). \] \item \label{conc:same right} For every $x \in B(k)$ we have $(S|Y_{a(k)})^{\P(k)-\Q(k)}(x) \overset{a(k)-1}{=}x$. \item\label{jt:discard} $\mu(U(k)) \le \frac {15} {2^k}$, \item\label{jt:proportion} $\mu(A(k)) \ge c$ and $\mu(B(k)) \ge c$. \end{enumerate} \end{prop} \begin{proof} For each $k \in \N$ define \[ V(k) = \bad \bigl(a(k),(d(k)+1)(2\P(k)-\Q(k))-1\bigr) \] and put $J(k) = [0^{a(k)}] \setminus V(k)$. Note that $J(k) \subset X$ for all $k$. For every $x \in J(k)$ and every $0 \le n \le (d(k)+1)(2\P(k)-\Q(k))-3$ we have \begin{equation}\label{eqn:tower_discrep} T^n(x) = (S|Y_{a(k)})^n(x) \quad\text{or}\quad T^n(x) = (S|Y_{a(k)})^{n+1}(x) \end{equation} and \begin{equation}\label{eq:equal} (S|Y_{a(k)})^j(x) \notin \bigcup_{\substack{j>a(k)\\j \notin\{a(i)\}}}W_j \implies T^n(x)=(S|Y_{a(k)})^n(x). \end{equation} Indeed, if $T^n(x) \ne (S|Y_{a(k)})^n(x)$ for some $n$ in the above range then \[ (S|Y_{a(k)})^j(x) \in \bigcup_{i>a(k)}W_i \] for some $0 < j \le n$. But as $x \notin V(k)$ this restriction becomes \[ (S|Y_{a(k)})^j(x) \in \bigcup_{\substack{i>a(k)\\i \text{ bounded}}} W_i \] and \begin{multline*} \min \left\{ \ell>0 \,\middle|\, \begin{aligned} &(S|Y_{a(k)})^\ell x \in W_j\quad \text{for some } x \in W_i \setminus V(k)\\ &\qquad\text{with }i\neq j, \, i,j\in \{a(k)+1,a(k)+2,\dots\}\setminus \{a(r)\} \end{aligned} \right\}\\ \ge \p(a(k)+1)-1 \end{multline*} so for each $x \in J(k)$ \begin{align*} \left| \left\{ 0\leq j\leq n:(S|Y_{(a(k)})x \in \bigcup_{\ell>a(k)}W_\ell \right\} \right| &\leq 1\\ \intertext{and} \left| \left\{ 0\geq j\geq -n:(S|Y_{(a(k)})x \in \bigcup_{\ell>a(k)}W_\ell \right\} \right| &\leq 1 \end{align*} which establishes~\eqref{jt:return} because $(d(k)+1)(2\P(k)-\Q(k))$ is the first return time to $[0^{a(k)}]$ for $S|Y_{a(k)}$. For~\eqref{conc:same left} and~\eqref{conc:same right} note that the definition of $A(k)$ gives that $x \in A(k)$ implies $0\leq x(\ell)\leq d(k)+1$ and $y \in A(k)^c\supset B(k)$ implies $0\leq y(\ell)\leq 2d(k)+1$, consulting Figure~\ref{fig:unbounded_removal} gives these claims. For~\eqref{jt:shrinking} note that $\mu(J(k)) \le 1/\nu(X) t(a(k)+1)$ giving \begin{equation}\label{eq:est JT} (d(k)+1)\P(k) \sum_{j > k} \mu(J(j)) \le \frac{1}{\nu(X)} \sum_{j > k} \frac{(d(k)+1)\P(k)}{t(a(j)+1)} \le \frac{1}{\nu(X)} \sum_{j > k} \frac{1}{10^{a(j) - a(k)}} \end{equation} because $(d(k)+1)\P(k) \le t(a(k))$ for all $j > k$. To get~\eqref{jt:left} and~\eqref{jt:right} we define the sets $U(k)$ in terms of columns. First define \begin{multline}\label{eq:U} U_0(k)\\ =\bigl\{x\in \Omega:x(a(k)\notin \{0,1,5,6,7,8,9,d(k)-1,\dots,d(k)+2,2d(k),2d(k)+1\}\bigr\} \end{multline} which is a union of columns at scale $a(k)$. We have \[ \nu(U_0(k)) = \frac{13 \P(k)}{t(a(k))} \le \frac{13}{d(k)} \] and therefore \begin{equation}\label{eq:JT est 2} \mu(U_0(k) \cap X) \le \frac{13}{d(k) \nu(X)} \le \frac{15}{2^k} \end{equation} which goes to zero as $k \to \infty$. Now, if $ x \in X \setminus \bigl(U_0(k) \cup V(k)\bigr) $ then for all $0\leq |n|\leq \P(k)$ because $x \notin V(k)$ and $(T^nx)(a(k))\neq 7$, as in~\eqref{eq:equal}, $(S|Y_{a(k)})^nx=T^nx$. Setting $U(k)=U_0(k) \cup V(k)$ we have~\eqref{jt:left} and~\eqref{jt:right}. Using~\eqref{eq:JT est 2} we have \[ \mu(U(k)) \leq \frac{14}{2^k} + \frac 1 {\nu(X)} (2( (d(k) + 1) (2P(k) - Q(k)) - 1) + 1) \nu\left(\bigcup_{j>k}W_{a(j)}\right)\leq \frac {15} {2^k} \] which is~\eqref{jt:discard}. Lastly, \eqref{jt:proportion} is a consequence of $\{ x \in Y_{a(k)} : x(k) \le d(k) \}$ and $\{x \in Y_{a(k)} : x(k) > d(k) \}$ both having measure at least $\frac{1}{4}$ and~\eqref{jt:discard}. \end{proof} \subsection{Building the joining}\label{subsec:apply_CE} In this section we prove Theorem~\ref{thm:ergodic joining}. We begin by recalling that the Kantorovich--Rubinstein metric on the space $\meas_1(V)$ of Borel probability measures on a compact metric space $V$ is defined by \[ \metric_\KR(\eta,\xi) = \sup \left\{ | \eta(f) - \xi(f) | \,:\, f \quad\text{is 1-Lipschitz on $X$ with respect to } \metric|_{X\times X} \right\} \] for all $\eta, \xi$ in $\meas_1(V)$. The joining $\lambda$ will be built using a minor modification of~\cite[Proposition~3.1]{MR4269425} (in the case $d=2$). We highlight the differences in bold and in the appendix we prove the version of~\cite[Corollary~3.3]{MR4269425}, which is the only part of the proof that requires significant modification. Note that~\eqref{ce:kr} is a strengthening of~\cite[Proposition~3.1$\MK$(B)]{MR4269425}, the corresponding condition in~\cite{MR4269425}, and this is used for~\eqref{eq:CEchanged} in the appendix. \pagebreak{} \begin{prop}\label{prop:ce} Fix sequences \begin{itemize} \item $k \mapsto J(k)$ of \emph{subsets of} $X$, \item $k \mapsto U(k)$ of measurable subsets of $X$, \item $k \mapsto r(k)$ of natural numbers, \item $k \mapsto \alpha(k)$ and $k \mapsto \gamma(k)$ of \emph{integers}, \item $k \mapsto \epsilon(k)$ of positive real numbers, \end{itemize} and set \begin{align}\label{eq:U only bad} &A(k) = \bigcup_{i=0}^{r(k)-1} T^i (J(k)) \backslash U(k) &&B(k) = (X \setminus A(k)) \setminus U(k) \end{align} and let $\nu_i$ on $X \times X$ be the push-forward of $\mu$ under the map $x \mapsto (x, T^i(x))$. Assume all of the following. \begin{enumerate}\Jenumi \item\label{ce:proportion} There is $c > 0$ such that for all $k \in \N$ one has $\mu(A(k)) \ge c$ and $\mu(B(k)) \ge c$. \item\label{ce:return} For all $k \in \N $ the minimal return time of $T$ on $J(k)$ is at least $\frac{3}{2} r(k)$. \item\label{ce:discard} For all $k \in \N$ we have $\mu(U(k)) < \epsilon(k)$. \item\label{ce:shrinking} $\lim_{k \to \infty} r(k) \sum{j > k} \mu(J(j)) = 0$. \item\label{ce:error} The sequence $k \mapsto \epsilon(k)$ is non-increasing and summable. \item\label{ce:nearness} For every $k \in \N$ and every $x \in A(k)$ one has \begin{equation}\label{ce:nearness_1} \metric\left(T^{\alpha(k)} x, T^{\gamma(k-1)} x\right) < \epsilon(k) \quad \text{and} \quad \metric\left(T^{\gamma(k)} x, T^{\alpha(k-1)} x\right) < \epsilon(k) \end{equation} and for every $k \in \N$ and every $x \in B(k)$ one has \begin{equation}\label{ce:nearness_2} \metric\left(T^{\alpha(k)} x, T^{\alpha(k-1)} x\right) < \epsilon(k) \quad\text{and}\quad \metric\left(T^{\gamma(k)} x, T^{\gamma(k-1)} x\right) < \epsilon(k). \end{equation} \item\label{ce:kr} For every $k \in \N$ the estimates \begin{multline}\label{eq:ce changed} \mu \left(\, \bigcup_{9L \ge r(k+1)} \left\{ x \in X : \metric_\KR \left(\nu_{\alpha(k)}, \frac{1}{L} \sum_{i=1}^L (T \times T)^i \delta\left(x,T^{\alpha(k)} x\right) \right) > \epsilon(k) \right\} \right) \\ < \epsilon(k) \end{multline} \begin{multline}\label{eq:ce changed 2} \mu \left(\bigcup_{9L \ge r(k+1)} \left\{ x \in X : \metric_\KR \left(\nu_{\gamma(k)}, \frac{1}{L} \sum_{i=1}^L (T \times T)^i \delta\left(x,T^{\gamma(k)} x\right) \right) > \epsilon(k) \right\} \right)\\ < \epsilon(k) \end{multline} both hold. \end{enumerate} Then \[ \lim_{j \to \infty} \nu_{\alpha(j)} = \lim_{j \to \infty} \frac{\nu_{\alpha(j)} + \nu_{\gamma(j)}}{2} = \lim_{j \to \infty} \nu_{\gamma(j)} \] and in particular the above limits exist. Moreover the common limiting value $\lambda$ is ergodic for $T \times T$ and there is $C > 0$ such that \[ \metric_\KR \left(\lambda, \frac{\nu_{\alpha(k)} + \nu_{\gamma(k)}}{2} \right) \le C \sum_{i > k} \epsilon(i) \] for all $k \in \N$. \end{prop} In order to apply Proposition~\ref{prop:ce} to establish Theorem~\ref{thm:ergodic joining} the central assumption to establish is~\eqref{ce:nearness}. The times $\alpha(k)$ and $\gamma(k)$ will be chosen based on the picture at the unbounded scale $a(k)$. The set $A(k)$ will consist of points in the left tower at scale $a(k)$ while the set $B(k)$ will consist of points in the right tower at scale $a(k)$. Indeed, if $x \in A(x)$, then by~\eqref{eq:U} we have $x(a(k))\in [2,d(k)-2]$ and so $T^n(x)$ is in the left tower at scale $a(k)$ for all $0\leq |n|\leq \P(k)$. Analogously for $x \in B(k)$ and the right tower. Recall that $\P(k)$ is the height of the left tower and $\P(k) - \Q(k)$ is the height of the right tower at scale $a(k)$. Taking \begin{align*} &\alpha(0) = 1, &&\alpha(1) = \P(1), \\ &\alpha(2) = \P(2) - \P(1) + 1, &&\alpha(3) = \P(3) - \P(2) + \P(1), \end{align*} and \begin{align*} &\gamma(0) = 0, &&\gamma(1) = - \P(1) + 1, \\ &\gamma(2) = - \P(2) + \P(1), &&\gamma(3) = -\P(3) + \P(2) - \P(1) + 1 \end{align*} and using the inductive definition $\Q(n+1) = 2\P(n) - \Q(n)$ we have for example that $T^{\alpha(3)} \approx T^{\gamma(2)}$ and $T^{\gamma(3)} \approx T^{\alpha(2)}$ for many points in $A(3)$ and $T^{\alpha(3)} \approx T^{\alpha(2)}$ and $T^{\gamma(3)} \approx T^{\gamma(2)}$ for many points in $B(3)$. \begin{proof}[Proof of Theorem~\ref{thm:ergodic joining}] We will apply Proposition~\ref{prop:ce} to produce an ergodic joining that is neither the product measure nor an off diagonal joining. Let the constant $c > 0$ and the sequences $k \mapsto J(k)$ and $k \mapsto U(k)$ be those from Proposition~\ref{prop:joining_prep}. Define $r(k) = (d(k)+1) \P(k) -1$ and put $\epsilon(k) = 15 \cdot 2^{-k}$ With these choices the sets $A(k)$ and $B(k)$ in the hypothesis of Proposition~\ref{prop:ce} are the same as the sets $A(k)$ and $B(k)$ in the statement of Proposition~\ref{prop:joining_prep}. In particular we get~\eqref{ce:proportion} from~\eqref{jt:proportion} and~\eqref{ce:return} from \[ 3(d(k)+1) \P(k) \le 4(d(k)+1) (\P(k) - \Q(k)) \] and~\eqref{jt:return}. Our choice of $\epsilon(k)$ gives~\eqref{ce:discard} immediately and~\eqref{ce:shrinking} is just~\eqref{jt:shrinking}. We get \eqref{ce:error} from~\eqref{jt:discard}. To establish~\eqref{ce:nearness} and~\eqref{ce:kr} we need to define the sequences $\alpha(k)$ and $\gamma(k)$. Put $\alpha(0) = 1$ and $\gamma(0) = 0$ and inductively define \[ \alpha(k+1) = \P(k+1) + \gamma(k) \quad\text{and}\quad \gamma(k+1) = - \P(k+1) + \alpha(k) \] for all $k \in \N$. One can check by induction that \begin{equation}\label{eqn:joining_tower_relations} \alpha(k) + \Q(k) = \alpha(k-1) + \P(k) \quad\text{and}\quad \gamma(k) + \P(k) = \Q(k) + \gamma(k-1) \end{equation} for all $k \in \N$. We verify that~\eqref{ce:nearness} holds. We calculate \begin{align*} T^{\alpha(k)}(x) = T^{\P(k) + \gamma(k-1)}(x) \overset{a(k)-1}{=} T^{\gamma(k-1)}(x) &\implies \metric\left(T^{\alpha(k)}(x), T^{\gamma(k-1)}(x)\right) \le \epsilon(k) \\ \intertext{and} T^{\gamma(k)}(x) = T^{-\P(k) + \alpha(k-1)}(x) \overset{a(k)-1}{=} T^{\alpha(k-1)}(x) &\implies \metric\left(T^{\gamma(k)}(x), T^{\alpha(k-1)}(x)\right) \le \epsilon(k) \end{align*} using~\eqref{jt:left} and~\eqref{conc:same left}, which gives~\eqref{ce:nearness_1}. Similarly, for all $x \in B(k)$ we have \begin{align*} T^{\alpha(k)}(x) \overset{a(k)-1}{=} T^{\alpha(k) - \P(k) + \Q(k)}(x) = T^{\alpha(k-1)}(x) &\implies \metric\left(T^{\alpha(k)}(x), T^{\alpha(k-1)}(x)\right) < \epsilon(k) \\ \intertext{and} T^{\gamma(k)}(x) \overset{a(k)-1}{=} T^{\gamma(k) + \P(k) - \Q(k)}(x) = T^{\gamma(k-1)}(x) &\implies \metric\left(T^{\gamma(k)}(x), T^{\gamma(k-1)}(x)\right) < \epsilon(k) \end{align*} using~\eqref{jt:right} and~\eqref{conc:same right}. This establishes~\eqref{ce:nearness_2} and thus~\eqref{ce:nearness}.%essayer de faire revenir cette ligne p. d'avant Finally, we establish~\eqref{eq:ce changed} to give half of~\eqref{ce:kr}, the other half, \eqref{eq:ce changed 2}, is analogous and omitted. Observe that $\p(n+1)$ is the number of cylinders of length $n$ that non-trivially intersect $X$. Also, observe that if $x \notin \bad(n,\p(n+1))$ then either \begin{itemize} \item $\{T^jx\}_{0\leq j \leq \p(n+1)}$ intersects every cylinder of length $n$ in $Y_n$ exactly once \item or it does not intersect $[7^n]$ and it intersects one such cylinder twice. \end{itemize} Combining these two facts, we have that if $x,y \notin \bad(\p(n+1)+a(k),n) $ then \begin{multline}\label{eq:pair sum close} \left|\frac{1}{\p(n+1)}\sum_{i=0}^{\p(n+1)-1} f\left(T^ix,T^{a(k)}T^ix\right)-f\left(T^iy,T^{a(k)}T^iy\right) \right|\\ \leq 2^{-n}\Lip(f)+\frac {a(k)+2} {\p(n+1)}\cdot \diam(X) \Lip(f). \end{multline} Indeed, if $T^ix,T^jy$ are in the same cylinder of length $n$ and $T^{a(k)+i}x,T^{a(k)+j}y$ are in the same cylinder of length $n$ then \[ \left|f\left(T^ix,T^{a(k)+i}x\right)-f\left(T^jy,T^{a(k)+j}y\right)\right|\leq 2^{-n-1}\Lip(f). \] This allows us to treat at least $\p(n+1)-(a(k)+2)$ terms in each summand and we estimate trivially on the rest. Thus, if $x \notin V(n,\p(n+1)+a(k))$, $f$ is $1$-Lipschitz and $\|f\|_{\sup}=1$ we have \begin{multline}\label{eq:for kr} \left|\int_X f d\mu-\frac 1 {\p(n+1)}\sum_{i=0}^{\p(n+1)}f(T^ix) \right|\\ \leq 2\mu\bigg(\bad\bigl(n,\p(n+1)+a(k)\bigr)\bigg)+2^{-n}\frac {a(k)} {\p(n+1)}. \end{multline} We now let $n=\frac 3 2 \cdot a(k)$ and we have \begin{equation}\label{eq:bad set bound} \begin{aligned} \mu\bigl(\bad\left(n,\p(n+1)+a(k)\right)\bigr) &\leq 2 \cdot \bigl(\p(n+1)+a(k)\bigr) \nu(X)^{-1}\nu\left(\bigcup_{a(j)>n}W_{a(j)}\right)\\ &\leq 8 \cdot \frac{10^{1.6a(k)}}{10^{a(k+1)-a(k)-1}}. \end{aligned} \end{equation} This uses~\eqref{eq:unbounded removal est}, the fact that $|\p(n+1)|\leq 10^n \cdot 2^{k+6}.$ To complete the proof we need straightforward estimate: Assume \[ \mu \left(\left\{ x: \left| \frac 1 r \sum_{i=0}^{r-1}f\left(T^ix,T^{i+m}x\right) \right| >\epsilon \right\} \right)<\delta \] then for $Lr\leq N<(L+1)N$ \begin{equation}\label{eq:abstract estimate} \mu \left(\left\{ x : \left| \frac 1 N \sum_{i=0}^{N-1}f\left(T^ix,T^{i+m}x\right) \right| >\epsilon+\sqrt{\delta}\|f\|_{\sup}+\frac 2 {L} \right\} \right) < \sqrt{\delta}. \end{equation} Indeed, we say $j$ is bad for $x$ if \[ \left| \frac 1 r \sum_{i=jr}^{(j+1)r-1}f\left(T^ix,T^{i+m}x\right) \right| >\epsilon. \] Now \[ \mu\left( \{x: |\{0\leq j\leq L-1:j \quad\text{is bad for }x\}|\geq \left\lceil \sqrt{\delta}L\right\rceil \right)<\sqrt{\delta}. \] On this set the estimate in~\eqref{eq:abstract estimate} holds. Observing that $\frac 1 9 r(k+1)\geq 9^{n-a(k+1)-2} \p(n)$, plugging in~\eqref{eq:for kr} and~\eqref{eq:bad set bound} into \eqref{eq:abstract estimate} completes the proof that the assumptions of Proposition~\ref{prop:ce} hold. We now have to prove that our limiting joining, $\sigma$, is neither one-to-one nor the product measure. To do this we require some additional notation. Let $\tau$ be a coupling of $\mu$ on $X \times X$. Applying disintegration of measures to projection onto the first coordinate, one defines a $\mu$-a.e. uniquely defined family of measure $\tau_x$ so that \begin{itemize} \item $\tau_x$ is supported on $\{x\} \times X$, \item $\int \tau_xd\mu(x)=\tau$. \end{itemize} Observe that \begin{multline}\label{eq:disintegration understood} \left\{x:d_{KR}\left(\sigma_x, \frac 1 2 \left[\delta\left(x,T^{\gamma(k)}x\right)+\delta\left(x,T^{\alpha(x)}x\right)\right]\right)\leq 2^{-a(k)} \right\}\\ \subset \bigcup_{j>k}A(k) \cup B(k)= \left(\bigcup_{j>k}U(k)\right)^c. \end{multline} The containment uses~\eqref{ce:nearness} and the equality uses~\eqref{eq:U only bad}. Following an analysis as in the proof of~\eqref{ce:kr}, applying this with $k=1$ shows that $\sigma$ is not the product measure. To prove that the joining is not one-to-one on a.e. fiber we use the following: \begin{lemm}[{\cite[Lemma~4.7]{MR4269425}}] If $\sigma$ is a measure that is one-to-one on almost every fiber then $\sigma$ cannot be the weak-* limit of a sequence of measures $\tau_i$ that are two-to-one on almost every fiber and such that there exists a $\delta$ satisfying \[ \mu(\{x : \diam(\supp(\tau_i|_x) > \delta\}) > 3/4 \] for infinitely many $i$ where $\tau_i|_x$ is the a.e. defined measure given by applying disintegration of measures to the projection of $\tau$ onto the first coordinate. \end{lemm} Using~\eqref{ce:discard}, \eqref{ce:error} and~\eqref{eq:disintegration understood} provides the assumption of the above lemma. \end{proof} \begin{rema} One can use~\eqref{eq:disintegration understood} a little more and show that $\sigma$ is two-to-one on a.e. fiber. Indeed, this equation gives the atoms in $\sigma|_x$. This fact is not needed for any statements of this paper and we omit the details. \end{rema} \section{Mild mixing}\label{sec:mild mixing} In this section we will prove Theorem~\ref{thm:mild mixing}. Given a non-constant function $f$ in $\ltwo(X,\mu)$ we wish to prove that no sequence $\bir$ in $\Z$ with $|r(n)| \to \infty$ satisfies\linebreak $\| f - T^{r(n)} f \| \to 0$. In Section~\ref{subsec:buddies} we define the essential notion of ``buddies'' and give in Theorem~\ref{thm:getting_constant weak} a criterion for a non-constant $f \in \ltwo(X,\mu)$ not to have $\bir$ as a rigidity sequence. The thrust of the criterion is that the existence of a uniformly positive proportion of $(r(\tau),\tau)$ buddies as $\tau \to \infty$ precludes $f$ from being rigid along $\bir$. Thus, given $\bir$ we need to be able to produce buddies. There are certain situations where producing buddies is relatively straightforward. Those are the situations where we do not need to carry out the reduction procedure mentioned in the introduction. In those situations are able to instead produce ``friendly pairs'' which are defined in Section~\ref{subsec:buddies from pairs}. In fact, we would prefer to work with friendly pairs throughout and avoid altogether the notion of buddies. However, when using the reduction procedure (see Section~\ref{subsec:expansion_reduction}) to prove Proposition~\ref{prop:reduction} we are unable to work directly with friendly pairs. We give in Section~\ref{subsec:reduction summary} a technical summary of the reduction procedure. In Section~\ref{subsec:reduction once} we given some preparatory results describing what happens when we apply the reduction procedure once. The main technical result is proved in Section~\ref{subsec:reduction main}. Finally, the proof of Theorem~\ref{thm:mild mixing} is given in Section~\ref{subsec:mild mixing}. \subsection{Buddies}\label{subsec:buddies} Essential to our proof is the following definition. \begin{defi}[Buddies]\label{def:weak_friends} Fix $r,j \in \N$. We will say that $(y,z)$ in $X \times X$ are \emph{$(r,j)$ buddies} if the following properties all hold. \begin{enumerate} \item\label{cond:start same 2} $y \overset{j-1}{=} z$. \item\label{cond:differ by 1} There exists $0 < |i| <9$ so that $(S|Y_{j-1}^i \circ T^r)(y) \overset{j-1}{=} (T^r)(z)$ \end{enumerate} \end{defi} \begin{rema} In our argument, we will have $|i|=1$. However, in related arguments, $|i|$ will often be in a larger range (see e.g.~\cite{MR4251942}). As such we phrase things more generally to have Theorem~\ref{thm:getting_constant weak} in a more general situation. \end{rema} The points $y$ and $z$ are $(r,j)$ buddies if they are initially close (in that they agree in the first $j-1$ coordinates) and through $r$ iterations of $T$ differ by at least one and at most eight iterations of $S|Y_{j-1}$. \begin{exam} Fix $k \in \N$ a bounded scale. When $y \in X \cap [6 7^{k-2} 7]$ with $S(y) = T(y)$ and $z \in X \cap [6 7^{k-2}8]$ with $S^2(y) = T(y)$ the points $y$ and $z$ are $(1,k)$ buddies. \end{exam} We record here a lemma about buddies that will be needed later. \begin{lemm}\label{lem:buddies through reduc} Assume $r,u\in \N$, $j0$ so that for all $\tau \in \N$ there exists $ i_0 \in \N$ so that for all $i\geq i_0$ there exists a Borel set $\Gamma_i \subset X$ and a~measure preserving function $\phi_i : X \to X$ with \begin{itemize} \item $\mu(\Gamma_i) > \rho$, \item for all $z \in \Gamma_i$ we have that $(\phi_i(z),z)$ are $(r(i),\tau)$-buddies \end{itemize} all hold. Then $\bfr$ is not a rigidity sequence for $f$. \end{theo} \begin{lemm}\label{lem:same power} For all $\epsilon>0$ and $B\in \N$ there exists $\tau$ so that if $r \in \Z$, $\Gamma$ is a~Borel set and $\phi:\Gamma \to X$ is a measure preserving function satisfying $(\phi(z),z)$ are $(r,\tau)$-buddies for then \[ \mu \left( \left\{ y\in \Gamma\,\middle|\, \begin{aligned} &a\exists \ell \text{ with }0<|\ell|<10 \text{ so that for all }-B\leq a\leq B\\ &\mkern150mu\text{we have } \metric \Big(\left(T^{r+a} \circ \phi_i\right)(y),\; T^{r+a+\ell}(y) \Big) > \epsilon \end{aligned} \right\} \right) <\epsilon. \] \end{lemm} \begin{proof} The definition of $(y,z)$-buddies gives that for any $\delta>0$ there exists $j_0$ so that for all $j'\geq j_0$ we have that for every $(y,z)$ which are $(r,j')$-buddies we have that there exists $0<|k|<10$ so that \begin{equation}\label{eq:buddies close} \min_{0<|k|<10}\metric\left((T^{r} \circ \phi_i)(y),\quad\left((S|Y_{j'-1})^k \circ T^{r}\right)(y)\right)<\delta. \end{equation} Because $T^k:X \to X$ is measurable for all $k \in \Z$, by Lusin's theorem and the fact that continuous functions on compact sets are uniformly continuous, for any $B \in \N$ there exists $\delta>0$ and $K \subset X$, compact with $\mu(K)>1-\epsilon$, so that for all $x,y \in K$ with $d(x,y)<\delta$ and $-B\leq a\leq B $ we have \[ \metric(T^ax,T^ay)<\epsilon. \] Combining this with~\eqref{eq:buddies close} gives the lemma. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:getting_constant weak}] Let $\rho>0$ be given and $f$ be a non-constant $\ltwo(\mu)$ function. Because $T$ is weakly mixing and thus every power of $T$ is ergodic, for all $0<|\ell|<9$ there exists $c_\ell>0$ so that \[ \int \left|f-f\circ T^\ell\right| \intd \mu>c_\ell \] holds. Let $\tilde{c}=\min\{1,c_\ell:0<|\ell|<9\}$ and $0<\varepsilon<\frac{\tilde{c}}9 \frac{\rho}{400}.$ By the ergodic theorem (either Birkhoff or Von Neumann), for each $0<|\ell|<9$ there exists $B_\ell \in \N$ so that for all $B\geq B_\ell$ we have \begin{equation}\label{eq:erg} \mu \left(\left\{ y : \left|\frac 1 {B+1} \sum_{k=0}^{B} \left|f(T^{k}y)-f(T^{\ell+k}y)\right|-\int\left|f-f \circ T^{\ell}\right| \intd\mu\right|>\varepsilon\right\} \right) < \varepsilon. \end{equation} Denote these sets $A_{\ell,B}$. Let $\tilde{B}=\max\{B_\ell:0<|\ell|<9\}$. By Lusin's theorem, there exists $K:=K_{\varepsilon,\tilde{B}}\subset X$ compact so that $\mu(K)>1-\varepsilon$ and $f\circ T^i$ restricted to $K$ is continuous for all $-9\frac{1}{20} \rho$ and \begin{equation}\label{eq:specific ell} \left(T^{n_i+a} \circ \phi_i\right)(y) \overset{\tau}{=} \left(T^{n_i+a+\ell}\right)(z)\quad \text{for all } -\tilde{B}\leq a \leq \tilde{B} \quad\text{and all }\; z \in G_i \end{equation} both hold. Now let $D$ denote the set of $z$ so that: \begin{itemize} \item $z \in K$, \item $\phi_i(z) \in K$, \item $T^{n_i}(z) \in K$, \item $(T^{n_i} \circ \phi_i)(z)\in K$, \item $z \in G_i$, \item $T^{n_i}(z) \in A_{\ell,\tilde{B}}.$ \end{itemize} Observe that $\mu(D)=\mu(G_i)-\mu(A_{\ell,\tilde{B}}^c)-4 \mu(K^c)>\frac 1 {40}\rho.$ For all $z\in D$ we have by~\eqref{eq:erg}, \eqref{eq:all close} and~\eqref{eq:specific ell} that \[ \left| \frac{1}{\tilde{B}+1} \sum_{j=0}^{\tilde{B}} \left|f\left(T^{n_i+j}(z)\right)-f\left(\left(T^{n_i+j} \circ \phi_i\right)(z)\right)\right| - \int \left|f-f \circ T^\ell\right| \intd\mu\right| < 2\varepsilon \] and so \[ \frac{1}{\tilde{B}+1} \sum_{j=0}^{\tilde{B}} \Bigl| f\left(T^{n_i+j}(z)\right)-f\left((T^{n_i+j} \circ \phi_i)(z)\right) \Bigr| > \tilde{c} - 2\varepsilon \] as well. Because the triangle inequality gives \begin{multline*} \left|f(T^{n_i+j}(z))-f\left((T^{n_i+j} \circ \phi_i)(z)\right)\right|- \left|f(T^j(z))-f\left((T^j \circ \phi_i)(z)\right)\right|\\ \leq \left|f(T^{n_i+j}(z))-f(T^j(z))\right|+\left|f\left((T^{n_i+j} \circ \phi_i)(z)\right)-f\left((T^j\circ \phi_i)(z)\right)\right| \end{multline*} we can estimate \begin{multline*} \frac{1}{\tilde{B}+1} \sum_{j=0}^{\tilde{B}}\int_{D} \left| f(T^{n_i+j}(z))-f(T^j(z)) \right| + \left| f\left((T^{n_i+j} \circ \phi_i)(z)\right)-f\left((T^j \circ \phi_i)(z)\right) \right| \intd\mu \\ \ge\mu(D) \tilde{c} - 2\varepsilon - \frac{1}{\tilde{B}+1} \sum_{j=0}^{\tilde{B}}\int_{D} \left|f(T^j(z))-f\left((T^j \circ \phi_i)(z)\right)\right| \intd \mu\geq \frac{1}{80} \rho \tilde{c} \end{multline*} \pagebreak{} \noindent where the last inequality uses~\eqref{eq:all close}. Thus \begin{equation}\label{eq:not rigid} \int_X \left|f(T^{n_i}x)-f(x)\right|\geq \frac 1 {160} \rho \tilde{c} \end{equation} proving the Theorem~\ref{thm:getting_constant weak}. \end{proof} In fact the proof gives the following. \begin{coro}\label{cor:end} For any non-constant $f \in \ltwo(X,\mu)$ and any $\rho > 0$ there exists $\tau':=\tau'(\rho,f) \in \N$ and $\delta:=\delta(\rho,f)>0$ so that if \begin{itemize} \item $m \in \Z$, \item $\Gamma$ is a Borel set, \item $\phi:\Gamma \to X$ is $\mu$-measure preserving \end{itemize} %revoir le placement de satisfy \begin{itemize} \item $\mu(\Gamma)\geq \rho$ and \item for all $z \in \Gamma$ we have that $(z,\phi(z))$ are $(m,\tau)$-buddies \end{itemize} then \[ \int_X \left|f -f\circ T^m\right| \intd\mu>\delta \] holds. \end{coro} \subsection{Buddies from friendly pairs}\label{subsec:buddies from pairs} We define friendly pairs and prove that friendly pairs are always buddies. We then show how to produce friendly pairs at bounded scales, establishing that if $f\in \ltwo$, non-constant and $\bir$ is a sequence such that $\scl(r(n))$ is at a bounded scale for infinitely many $n$ then $\bir$ is not an $f$-rigidity sequence (see Corollary~\ref{cor:bounded mmix}). \begin{defi} [Friendly pairs]\label{def:friends} Fix $r \in \Z$ and $j \in \N$ with $j$ at a bounded scale. We will say that $(y,z)$ in $X \times X$ is an \emph{$(r,j)$ friendly pair} if the following properties all hold. \begin{enumerate}\Fenumi \item \label{cond:start same} $y(n) \overset{j-1}{=} z(n)$ for all $1 \le n \le j-1$, \item \label{cond:T defined 1} $T^n(y)= (S|Y_j)^n(y)$ for all $1 \le |n| \le |r|+1$ with $sgn(n)=sgn(r)$, \item \label{cond:T defined 2} $T^n(z) = (S|Y_j)^n(z)$ for all $1 \le |n| \le |r|+1$ with $sgn(h)=sgn(r)$, \item \label{cond:friend miss} $(S|Y_{j-1})^n(y) \not\in W_j$ for all $1 \le |n| \le |r|+1$ with $sgn(h)=sgn(r)$, \item \label{cond:friend hit} $(S|Y_{j-1})^h(z) \in W_j$ for a unique $1 \le |h| \le |r|$ with $sgn(h)=sgn(r)$. \end{enumerate} \end{defi} The following lemma says that friendly pairs are always buddies. \begin{lemm}\label{lem:friend to weak} If $(y,z)$ is an $(r,j)$ friendly pair then $(y,z)$ are $(r,j)$-buddies. \end{lemm} \begin{proof} By conditions~\eqref{cond:T defined 2} and~\eqref{cond:friend hit} we have that $T^r(z)=(S|Y_{j-1})^{r+1}(z)$ and by conditions~\eqref{cond:T defined 1} and~\eqref{cond:friend miss} we have that $T^r(y)=(S|Y_{j-1})^r(y)$ and $T^{r+1}(y)=(S|Y_{j-1})^{r+1}(y)$. By condition~\eqref{cond:start same} we have that the first $j-1$ entries of $(S|Y_{j-1})^k(y)$ are the same as the first $j-1$ entries of $(S|Y_{j-1})^k(z)$ for all $k$ and in particular for $k \in \{r,r+1\}$ proving the second condition in the definition of buddies. This proves Lemma~\ref{lem:friend to weak}. \end{proof} There are some situations where it is relatively straightforward to produce friendly pairs. The first, formalized by the next result, is when $r \in \Z$ is at a bounded scale. The second, given by Proposition~\ref{prop:small friend}, is when $r$ is small for its scale and $\scl(r) - 1$ is a bounded scale. The second applies in particular when $r$ is at an unbounded scale, but small at that scale. The salient point is that neither of these situations require us to reduce to smaller scales. \begin{figure}[t] \centering \begin{tikzpicture}[yscale=0.5, line width=0.5mm] \begin{scope}[shift={(-1.5,0)}] \node at (0,5) {$77 \dots 77$}; \end{scope} \begin{scope} \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$0$}; \end{scope} \begin{scope}[shift={(1.5,0)}] \node at (0.5,7) {$\dots$}; \node at (0.5,3) {$\dots$}; \end{scope} \begin{scope}[shift={(3,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \fill (0.4,1) ellipse (0.1 and 0.2); \draw (0,0) -- (1,0); \node at (0.5,-1) {$7$}; \end{scope} \begin{scope}[shift={(4.5,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw[color=red] (0,5) -- (1,5); \draw[color=red] (0.4,4.7) -- (0.6,5.3); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \fill (0.7,1) ellipse (0.1 and 0.2); \draw (0,0) -- (1,0); \node at (0.5,-1) {$8$}; \end{scope} \begin{scope}[shift={(6,0)}] \draw (0,9) -- (1,9); \draw (0,8) -- (1,8); \node at (0.5,7.2) {$\vdots$}; \draw (0,6) -- (1,6); \draw (0,5) -- (1,5); \draw (0,4) -- (1,4); \node at (0.5,3.2) {$\vdots$}; \draw (0,2) -- (1,2); \draw (0,1) -- (1,1); \draw (0,0) -- (1,0); \node at (0.5,-1) {$9$}; \end{scope} \begin{scope}[shift={(7.5,0)}] \draw [decorate, decoration = {brace}] (0,9.25) -- (0,-0.25); \node[align=left] at (1.2,4.5) {$\p(j)$\\cylinders}; \end{scope} \end{tikzpicture} \caption{The picture of $Y_{j-1}$ at the bounded scale $j$. The marked interval is~$W_j$. The marked points form a $(\p(j), j)$ friendly pair if both points are outside $\bad(j,\p(j))$.}\label{fig:acquainted_pair} \end{figure} \begin{theo}\label{thm:bounded wfriends} For all $r \in \Z$ with $\scl(r)$ bounded there is a set $B \subset X$ with the following properties. \begin{itemize} \item $\mu(B) \ge 10^{-5}$. \item There is $\Phi : B \setminus \bad(\scl(r),8r) \to X$ a measure-preserving map that is a bijection onto its image such that $(\Phi(z), z)$ is are $(r,\scl(r))$-friends for all $z \in B \setminus \bad(\scl(r),8r)$. \end{itemize} \end{theo} \begin{proof} Fix $r \in \Z$ with $\scl(r)$ bounded. We prove the case $r < 0$. The case $r > 0$ is similar and therefore omitted. Note that \begin{equation}\label{eqn:wee bound} \p(\scl(r)+1)/2\ge|r| \ge \p(\scl(r)) / 2 \end{equation} holds from the definition of the expansion. Define \begin{multline}\label{eqn:Bset} B = \biggl\{ x \in X : x(\scl(r)) = 8,\\ x(\scl(r)+1)\in \left[\frac{(c(\scl(r)+1)+1)}{5},\;\frac{(c(\scl(r)+1)+1)}{2} \right] \\ \text{and } \left(S|Y_{\scl(r)-1}\right)^i(x) \in W_{\scl(r)} \text{ for some } -|r| < i \le -1 \biggr\} \end{multline} and note that $\nu(B) \ge 10^{-3}$ by independence so $\mu(B) \ge 10^{-3}$ by definition of $\mu$. Given $y \in B \setminus \bad(\scl(r),8r)$ define $\Phi(y)$ by $(\Phi(y))(\scl(r)) = 6$ and $(\Phi(y))(n) = y(n)$ otherwise. Fix $z \in B \setminus \bad(\scl(r),8r)$. The pair $(\Phi(z),z)$ satisfies~\eqref{cond:start same}. Observe that $(S|Y_{\scl(r)-1})^h(z) \in W_{\scl(r)}$ for exactly one $r< h\leq-1$ because of~\eqref{eqn:wee bound} giving~\eqref{cond:friend hit}. We also have~\eqref{cond:T defined 2} because \[ (S|Y_{\scl(r)})^h(z) \notin \bigcup_{i>\scl(r)} W_i \] for all $-r \le h \le -1$ follows upon combining $z \notin \bad(\scl(r),8r)$ with the condition on $z(\scl(r) + 1)$. As $(\Phi(z))(\scl(r)) = 6$ we have $(S|Y_{\scl(r)-1})^h(\Phi(z)) \notin W_{\scl(r)}$ for all $-r-1 \le h \le -1$. That gives~\eqref{cond:friend miss}. Next note that $\Phi(z) = T^n(z)$ for some $4r\scl(r)}W_{a(k)} \] which, together with the fact that $(\Phi(z))(\scl(r)) = 6$ implies \[ (S|Y_{\scl(r)-1})^h(\Phi(z)) \notin \bigcup_{j>\scl(r)} W_j \] for all $r \le h \le 0$, gives~\eqref{cond:T defined 1}. Thus $(\Phi(z),z)$ is an $(r,\scl(r))$ friendly pair. \end{proof} \begin{coro}\label{cor:bounded mmix} If $\bfr$ is a sequence in $\Z$ with $|r(i)| \to \infty$ and $r(i)$ at a~bounded scale infinitely often then $\bfr$ is not a rigidity sequence for any non-constant $\lp^2(X,\mu)$ function. \end{coro} \begin{proof} Pass to a subsequence with the property that $r(i)$ is at a bounded scale for all $i \in \N$. Let $B_i$ and $\Phi_i$ result from applying Theorem~\ref{thm:bounded wfriends} to each $r(i)$. By Lemma~\ref{lem:friend to weak} and Theorem~\ref{thm:getting_constant weak} it suffices to show \[ \mu\Bigl(B_i \cap \bad\bigl(\scl(r(i)),\;8r(i)\bigr) \Bigr) < \frac{8}{9} \cdot \frac{1}{10^5} \] for all large enough $i$. Fix $i$ and write $r$ for $r(i)$ and $B$ for $B_i$. For convenience of exposition we assume $r < 0$. Because of the fast decay of $j \mapsto \mu(W_{a(j)})$, it suffices to show that if $a(k)$ is the smallest unbounded scale strictly greater than $\scl(r)$ we have \[ \mu\left(\left\{x\in B:(S|Y_{\scl(r)})^i(x)\in W_{a(k)}\quad \text{for some}\; -8|r|10^{-5}$ and $\Phi:B \setminus \bad(\scl(r),2r) \to X$ measure preserving bijection onto its image, so that for all $x \in B$, $(x,\Phi(x))$ are $(r,\scl(r)-1)$ friends. \end{prop} \begin{proof} By Theorem~\ref{thm:bounded wfriends} may assume $r$ is at an unbounded scale. Thus $\scl(r) \in \{a(\ell)\}$. For readability, in this proof we write $b$ for $\scl(r)$. It suffices to construct a set of $(x,x')$ that form an $(r,b-1)$-friendly pair. We present this in the case that $r>0$. The case $r<0$ is similar. The pairs are formed by $x$ so that $x(b-1)=9$, \begin{equation}\label{eq:index good} x(b) < d(b)-2, \quad\text{and}\quad x(b+1)\notin \{6, 7,8\}. \end{equation} and $x'$ so that \begin{equation}\label{eq:index same} x'(b-1) = 6 \quad\text{and}\quad x'(\ell)=x(\ell)\quad \forall\; \ell \neq b-1. \end{equation} Observe that $x \overset{b-2}{=} y$ gives~\eqref{cond:start same}. Also, from~\eqref{eq:index good} and~\eqref{eq:index same}, $(S|Y_{b-1})^k(x') \in W_{b-1}$ for some $0 \scl(r)$. Thus $(S|Y_{\scl(r)})^i(x) = T^i(x)$ for all $-|\lft(r)| \le i\le r$. As $x$ satisfies in particular the hypothesis of~\eqref{lem:pos red S left} we get the desired agreement in the first $\scl(r)-1$ coordinates. \let\qed\relax \end{proof} \begin{proof}[{\eqref{cor:pos red T right}}] Note there is no possibility of $(S|Y_{\scl(r)})^i(x)$ belonging to some $W_k$ with $k > \scl(r)$ for some $-|\rht(r)| \le i \le r$ and we have the desired conclusion by appealing to~\eqref{lem:pos red S right}. \let\qed\relax \end{proof} \begin{proof}[{\eqref{cor:pos red T friends}}] We need to verify the properties in Definition~\ref{def:friends}. Condition~\eqref{cond:start same} is one of our assumptions. To prove~\eqref{cond:T defined 1} and~\eqref{cond:friend miss} combine $y \not\in \bad(\scl(r),r)$ with $y(\scl(r) + 1) = 3$ to get $(S|Y_{\scl(r)})^i(y) \not\in W_k$ for all $k \ge \scl(r)+1$ and all $1 \le i \le r$. For~\eqref{cond:T defined 2} and~\eqref{cond:friend hit} combine $x \not\in \bad(\scl(r),r+1)$ with the facts that there is a unique $1 \le i \le r$ for which $(S|Y_{\scl(r)})^i(x)$ belongs to $[7^{\scl(r)}]$ and that $x(\scl(r) + 1) = 7$ allows~\eqref{lem:pos red S bdd} to guarantee there is $1 \le i < r$ with $(S|Y_{\scl(r)})^i(x)$ in $W_{\scl(r)+1}$. \end{proof}\let\qed\relax \end{proofc} We have additionally the following reduction statements when the time $r$ is negative. The proofs are entirely analogous. \begin{lemm}\label{lem:reduction negative} Fix $r \in -\N$ with $\scl(r)$ an unbounded scale $a(b)$ and fix $x \in X$. \begin{enumerate} \item\label{lem:neg red S left} If $-\lst(r)+1 \le x(\scl(r)) \le d(b) -1$ then $(S|Y_{\scl(r)})^r(x)$ and $(S|Y_{\scl(r)})^{\lft(r)}(x)$ agree in their first $\scl(r) - 1$ coordinates. \item\label{lem:neg red S right} If $d(b) + 2 - \lst(r) \le x(\scl(r)) \le c(b)-2$ then $(S|Y_{\scl(r)})^r(x)$ and $(S|Y_{\scl(r)})^{\rht(r)}(x)$ agree in their first $\scl(r)-1$ coordinates. \item\label{lem: neg red S bdd} If $\lst(r) \leq -11$ and $10 \le x(\scl(r)) \le -\lst(r) + 1$ then $((S|Y_{\scl(r)})^i(x))(\scl(r)) = 0$ for some $0 \ge i \ge r$. If $x(\scl(r) + 1) = 8$ as well there is $0 \ge i \ge r$ with $(S|Y_{\scl(r)})^i(x) \in W_{\scl(r) + 1}$. \end{enumerate} \end{lemm} \begin{coro}\label{cor:neg reduc conseq} Fix $r \in -\N$ with $\scl(r)$ an unbounded scale $a(b)$ and fix $x \in X$. \begin{enumerate} \item\label{cor:neg red T left} If $- \lst(r) + 10 \le x(\scl(r)) \le d(b)-1$ and $x \notin \bad(\scl(r),r)$ then \begin{align*} &(S|Y_{\scl(r)})^r(x) = T^r(x) &&(S|Y_{\scl(r)})^{\lft(r)}(x) = T^{\lft(r)}(x) \end{align*} and these two points agree in the first $\scl(r)-1$ coordinates. \item\label{cor:neg red T right} If $d(b) + 2 - \lst(r) \le x(\scl(r)) \le c(b)-2$ and $x \notin \bad(\scl(r),r)$ then \begin{align*} &(S|Y_{\scl(r)})^r(x)=T^r(x) &&(S|Y_{\scl(r)})^{\rht(r)}(x)=T^{\rht(r)}(x) \end{align*} and these two points agree in their first $\scl(r)-1$ coordinates. \item\label{conc:neg reduc friends} If $\lst(r) \leq -11$ and $10 \le x(\scl(r)) \le -\lst(r) + 1$ and $x(\scl(r)+1) = 8$ and $x \notin \bad(\scl(r),r+1)$ then $(y,x)$ are $(r,\scl(r) + 1)$-friends whenever $y$ satisfies $x(k) = y(k)$ for all $k \le \scl(r)$ and $y(\scl(r) + 1) = 3$ and $y \notin \bad(\scl(r),r)$. \end{enumerate} \end{coro} %\pagebreak{} \begin{coro}\label{cor:small e(J)}% \!\! Fix $r \in \Z$ with $\scl(r)$ an unbounded scale $a(b)$ and ${|\lst(r)| < 10}$. There exists $B \in X$ measurable and $\Phi:B \to X$ bijective and measurable on its image so that $(x,\phi(x))$ are $(r,\scl(r)+1)$-friends and \[ \mu(B)\geq \frac{1}{2} \frac {1}{20} \frac{1}{20} \frac{1}{d(b)}. \] \end{coro} \begin{proof} We prove the case $r>0$. The case $r<0$ is analogous. Let $J = \scl(r)$ and let \[ B=\left\{ \begin{multlined} y\in \Omega :(S|Y_{J(r)})^i(x) \in W_{\scl(r) + 1}\\ \text{for some } 0 \leq i\frac r {t(J+1)}>\frac 1 {20d(J)}. \] By our independence result, Lemma~\ref{lem:indep} applied with $k=1$, and the fact that there are 10 symbols we have that adding the condition that $x(J+2)=5$ reduces the measure by less than a factor of $2 \cdot 10$. Lastly, the condition not being in $V(b,r)$ removes less than \[ 2 \cdot r \sum_{a(j)>J(r)}\frac{Q(j)}{P(j)}< 2 \cdot 10 p(J(r)) \sum_{a(j)>J(r)}\frac{Q(j)}{P(j)}<\frac 1 2 \cdot \frac 1 {20}\frac 1 {20} \frac 1 {d(J)} \] giving the claim. \end{proof} By similar estimates to what we have done earlier Corollary~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T friends}, Corollary~\ref{cor:neg reduc conseq}$\MK$\eqref{conc:neg reduc friends} and Corollary~\ref{cor:small e(J)} give the following as well. \begin{lemm}\label{lem:friends} Fix $r \in \N$ with $\scl(r)$ unbounded. With \[ A_0 = \{ x \in X : x \text{ does not satisfy first assumption of Corollaries~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T left} or~\eqref{cor:pos red T right}} \} \] there is $B\subset A_0$, and $\Phi:B \to X$, a measure preserving injection so that\linebreak ${\mu(B)>\mu(A_0)10^{-6}}$ and $(x,\Phi(x))$ are $(r,\scl(r)+1)$-friends. The same is true with \[ A_0 = \{ x \in X : x \text{ does not satisfy first assumption of Corollaries~\ref{cor:neg reduc conseq}$\MK$\eqref{cor:neg red T left} or~\eqref{cor:neg red T right}} \} \] instead. \end{lemm} \subsection{Key result for unbounded scales}\label{subsec:reduction main} In this section we state and prove the key result needed for the proof of Theorem~\ref{thm:mild mixing}. Fix an unbounded scale $\tau \in \N$. Given $r \in \Z$ with $\scl(r) > \tau$ the proposition keeps track -- using Corollary~\ref{cor:pos reduc conseq} and Corollary~\ref{cor:neg reduc conseq} -- of the behavior of points under first $T^{\lft(r)}$ and $T^{\rht(r)}$ and then under $T^{\lft(\lft(r))}, T^{\rht(\lft(r))}, T^{\lft(\rht(r))}$ and $T^{\rht(\rht(r))}$ and so on until we have applied compositions of $\lft$ and $\rht$ enough times to arrive at the scale~$\tau$. There are various intricacies to keep track of. For example, it may be the case that $\lft(\rht(\lft(r)))$ is at a bounded scale, it may be that $\lst(\lft(\lft(r)))$ is either very small or very large, and it may be that some compositions of $\lft$ and $\rht$ reduce $r$ to a scale smaller than $\tau$. We begin with a preparatory lemma. \begin{lemm}\label{lem:for pairing} If $0<|k|\leq 2d(r)+3$ then either $\scl(k\Q(r))=a(r-1)$ or $k\Q(r)$ is at a bounded scale. Moreover, either \[ \scl \bigl(\lft(k\Q(r))\bigr) = \scl \bigl(\rht(k\Q(r))\bigr) = a(r-2) \] or at least one of $\scl(\lft(k\Q(r)))$, $\scl(\rht(k\Q(r)))$ is at a bounded scale. Moreover, in the first case, both $\scl(\lft(k\Q(r)))$ and $\scl(\rht(k\Q(r)))$ have the form $k'\Q(r-1)$ for $|k'|\leq 2d(r-2)+3$. If $\scl(b)=a(r-\ell)$ for $\ell \geq 2$ and $\scl(k \Q(r)+b) \tau$. Fix $I \in \N$ with $\lambda^I(\scl(m)) \ge \tau$. For every $1 \le i \le I$ there is a tuple \begin{equation}\label{eq:tuple} \left(A_0^{(i)}, A_1^{(i)},n_1^{(i)}, \dots, A_{k(i)}^{(i)},n^{(i)}_{k(i)} \right) \end{equation} with all the following properties. \begin{enumerate}\Eenumi \item\label{cond:setup} $A_j^{(i)} \subset X$ for all $0 \le j \le k(i)$ and $n^{(i)}_j \in \Z$ for all $1 \le j \le k(i)$ and $k(i) \in 2\N \cup \{0\}$. \item\label{cond:partition} The sets $A^{(i)}_0,\dots,A^{(i)}_{k(i)}$ form a measurable partition of $X$ in which each $A^{(i)}_j$ is the intersection with $X$ of a finite union of cylinders, each of which is defined by fixing values at unbounded scales between $\scl(m)$ and $\lambda^{i-1}(\scl(m))$. \item\label{cond:agree} Every $1 \le j \le k(i)$ has an ancestor $1 \le \alpha(j) \le k(i-1)$. If $j \ne 0$ and $x \in A_j^{(i)}$ and \begin{equation}\label{eqn:bad ancestors} x \notin \bigcup_{s=0}^i \bad \left(\scl \bigl(n^{(i-s)}_{\alpha^s(j)} \bigr), n^{(i-s)}_{\alpha^s(j)} \right) \end{equation} then \begin{equation}\label{eq:same coords} (T^m(x))(\ell) = \left(T^{n_j^{(i)}}(x)\right)(\ell),\quad \forall\ell\leq \lambda^{i-1}(\scl(m))-1. \end{equation} \item\label{cond:Azero friends} There exists $F\subset A^{(i)}_0$ with $\mu(F)>10^{-9}\mu(A^{(i)}_0) $ and $\psi:F \to X$ measure preserving and $j \ge \tau$ so that $(x,\psi(x))$ are $(m,j)$ buddies and in particular $(m,\tau)$ buddies. \item\label{cond:differ} The numbers $1,\dots,k(i)$ are paired so that each $1 \le j \le k(i)$ is in exactly one pair. Moreover, whenever $j$ and $h$ are paired then at least one of $n_j^{(i)}$ and $n_h^{(i)}$ is at scale $\lambda^i(\scl(m))$. \item\label{cond:paired numbers} If $j$ and $h$ are paired with $\lambda^i(\scl(m)) = \scl(n_j^{(i)}) \ne \scl(n_h^{(i)})$ then \[ n_j^{(i)}=k\q\left(\lambda^{i-1}(\scl(m))\right)+n_h^{(i)}\quad\text{with}\quad 2|k| \le c\left(\lambda^{i-1}\scl(m)\right) + 1. \] \item\label{cond:paired cyl} If $j$ and $h$ are paired the cylinders defining $A_j^{(i)}$ and $A_h^{(i)}$ differ in at most one unbounded scale. In that index one of them satisfies the first hypothesis of either Lemma~\ref{lem:reduction positive}$\MK$\eqref{lem:pos red S left} or Lemma~\ref{lem:reduction negative}$\MK$\eqref{lem:neg red S left} and the other satisfies the first hypothesis of either Lemma~\ref{lem:reduction positive}$\MK$\eqref{lem:pos red S right} or Lemma~\ref{lem:reduction negative}$\MK$\eqref{lem:neg red S right} respectively with $\max\{|n_j|,|n_h|\}$ in place of $r$ therein. \end{enumerate} \end{prop} \begin{proof} Let $\tau$, $m$ and $I$ be as in the statement. The proof is recursive. \begin{proof}[{Initial step}] First suppose both $\scl(\lft(m))$ and $\scl(\rht(m))$ are unbounded scales. By Lemma~\ref{lem: reduction not both small} neither can be at scale $\scl(m)$ and at least one is at scale $\lambda(\scl(m))$. If $m$ is positive let $A^{(1)}_1$ and $A^{(1)}_2$ be the sets of points in $X$ satisfying the hypothesis $9 \le x(\scl(m)) \le d(b)-\lst(r)$ of Corollary~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T left} and the hypothesis $d(b) + 2 \le x(\scl(m)) \le c(b) - \lst(r)$ of Corollary~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T right} respectively. When $m$ is negative use corresponding hypotheses of Corollaries~\ref{cor:neg reduc conseq}$\MK$\eqref{cor:neg red T left} and~\eqref{cor:neg red T right} respectively instead. In either case define $A^{(1)}_0$ to be the intersection with $X$ of the cylinders defined by the $\scl(m)$ coordinate that contribute neither to $A^{(1)}_1$ nor to $A^{(1)}_2$. Thus $A^{(1)}_0,A^{(1)}_1,A^{(1)}_2$ is a measurable partition of $X$ and~\eqref{cond:partition} holds. Set $k(1) = 2$ and $n^{(1)}_1 = \lft(m)$ and $n^{(1)}_2 = \rht(m)$. Thus~\eqref{cond:setup} holds. We get~\eqref{cond:agree} from the conclusions of Corollary~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T left} and~\eqref{cor:pos red T right} respectively. (When $m$ is negative use Corollary~\ref{cor:neg reduc conseq} instead.) As $k(1) = 2$ we are forced to pair $n^{(1)}_1$ and $n^{(1)}_2$. By Lemma~\ref{lem: reduction not both small} at least one of $n^{(1)}_1$ and $n^{(2)}_2$ is at scale $\lambda(\scl(m))$ giving~\eqref{cond:differ}. As $|\lft(m) - \rht(m)| = |\lst(m)| \q(\scl(m))$ and $2|\lst(m)| \le c(\scl(m)) + 1$ we have~\eqref{cond:paired numbers}. Property~\eqref{cond:paired cyl} holds as the cylinders defining $A^{(1)}_1$ and $A^{(1)}_2$ are defined in the $\scl(m)$ coordinate alone to satisfy the appropriate hypotheses. Secondly, suppose instead that at least one of $\lft(m)$ and $\rht(m)$ is at a bounded scale greater than or equal to $\tau$. In this case set $k(1) = 0$ and $A^{(1)}_0 = X$. All properties except~\eqref{cond:Azero friends} are immediate or vacuous. We now treat~\eqref{cond:Azero friends}, assuming without loss of generality that $\scl(\lft(m))=k>\tau$, a bounded scale. By Theorem~\ref{thm:bounded wfriends} we get sufficiently many $(\lft(m),\scl(\lft(m)))$-friends. By Corollary~\ref{cor:pos reduc conseq}$\MK$\eqref{cor:pos red T left} if $m>0$ or Corollary~\ref{cor:neg reduc conseq}$\MK$\eqref{cor:neg red T left} if $m<0$, Lemmas~\ref{lem:friend to weak} and~\ref{lem:buddies through reduc} we have~\eqref{cond:Azero friends}. Finally, suppose one of $\lft(m)$, $\rht(m)$ is at an unbounded scale and the other is at a~bounded scale less than $\tau$. (We will not find both $\lft(m)$ and $\rht(m)$ at bounded scales less than $\tau$ because $\lft(m) - \rht(m) = \lst(m) \q(\scl(m))$ forces at least one of $\lft(m)$ and $\rht(m)$ to be at a scale at least as large as $\lambda(\scl(m)) - 1$.) Then we proceed as in the first case, except that for~\eqref{cond:differ} we instead note here that if $\lft(m)$ is not at scale $\lambda(\scl(m))$ then $\scl(\lft(m)) \le \lambda^2(\scl(m))$ and so $\scl(\rht(m)) = \lambda(\scl(m))$ because $\rht(m) = \lft(m) + \lst(m) \q(\scl(m))$ gives $\scl(\rht(m)) \ge \lambda(\scl(m))$ and it cannot be larger than $\lambda(\scl(m))$. \let\qed\relax \end{proof} \begin{proof}[{The recursion}] For the inductive step, suppose for some $1 \le i < I$ that a tuple \[ \left(A^{(i)}_0, A^{(i)}_1, n^{(i)}_1,\dots, A^{(i)}_{k(i)}, n^{(i)}_{k(i)}\right) \] satisfying all the properties has been produced. If $k(i) = 0$ we set $k(i+1) = 0$ and $A^{(i+1)}_0 = A^{(i)}_0$. All properties continue to hold. Assuming, then, that $k(i) > 0$, we proceed by reducing further each $n^{(i)}_j$ unless one of the following holds. Say that the pair $h \ne j$ in $\{1,\dots,k(i) \}$ is \emph{bad} if at least one of the following is true. \begin{itemize} \item $\scl(n_j^{(i)}) = \lambda^i(\scl(m))$ and $n_j^{(i)} < \frac{3}{4} \p(\lambda^i(\scl(m)))$, \item $\scl(n_j^{(i)}) = \lambda^i(\scl(m))$ and $\lft(n_j^{(i)})$ is at a bounded scale not smaller than $\tau$, \item $\scl(n_j^{(i)}) = \lambda^i(\scl(m))$ and $\rht(n_j^{(i)})$ is at a bounded scale not smaller than $\tau$, \item $\scl(n_h^{(i)}) = \lambda^i(\scl(m))$ and $n_h^{(i)}<\frac{3}{4} \p(\lambda^i(\scl(m)))$, \item $\scl(n_h^{(i)}) = \lambda^i(\scl(m))$ and $\lft(n_h^{(i)})$ is at a bounded scale not smaller than $\tau$, \item $\scl(n_h^{(i)}) = \lambda^i(\scl(m))$ and $\rht(n_h^{(i)})$ is at a bounded scale not smaller than $\tau$. \end{itemize} When the pair to which $1 \le j \le k(i)$ belongs is not bad we call $j$ \emph{reducible} if $\scl(n_j^{(i)}) = \lambda^{i}(\scl(m))$. Note that~\eqref{cond:differ} says that if neither element of a pair is bad then at least one element of each pair is reducible. As a reducible $j$ does not belong to a bad pair it must be the case that $\lft(n^{(i)}_j)$ and $\rht(n^{(i)}_j)$ are at unbounded scales. For each $1 \le j \le k(i)$ that is reducible set \[ F(j) = \begin{cases} \left\{ x \in A^{(i)}_j : x\left(\scl\left(n^{(i)}_j\right)\right) \in [0,8] \cup [d-e+1, d+1] \cup [c-e+1, c] \right\}\\ & \mkern-20mu\text{if } n^{(i)}_j>0 \\ \left\{ x \in A^{(i)}_j : x\left(\scl\left(n^{(i)}_j\right)\right) \in [0,9-e] \cup [d,d+1-e] \cup [c-1,c] \right\}\\ & \mkern-20mu\text{if } n^{(i)}_j < 0 \end{cases} \] \pagebreak{} \noindent where $c = c(\lambda^i(\scl(m)))$ and $d = d(\lambda^i(\scl(m)))$ and $e = \lst(n^{(i)}_j)$. For each $1 \le h \le k(i)$ that is neither in a bad pair nor reducible set \[ F(h) = \begin{cases} \left\{ x \in A^{(i)}_h : x\left(\scl\left(n^{(i)}_j\right)\right) \in [0,8] \cup [d-e+1, d+1] \cup [c-e+1, c] \right\} \\ & \mkern-20mu\text{if } n^{(i)}_k > 0 \\ \left\{ x \in A^{(i)}_h : x\left(\scl\left(n^{(i)}_j\right)\right) \in [0,9-e] \cup [d,d+1-e] \cup [c-1,c] \right\}\\ & \mkern-20mu\text{if } n^{(i)}_k < 0 \end{cases} \] where $c = c(\lambda^i(\scl(m)))$ and $d = d(\lambda^i(\scl(m)))$ and $e = \lst(n^{(i)}_k)$ and $j$ is the (necessarily reducible) pair of $h$ as in~\eqref{cond:paired cyl}. Note all $1 \le h \le k(i)$ that are neither bad nor reducible are paired with an index that is reducible and not bad. We define \begin{equation}\label{eqn:A0 recursive def} A_0^{(i+1)} = A_0^{(i)} \cup \bigcup_{j \text{ in a bad pair}} A_j^{(i)} \cup \bigcup_{j \text{ reducible }} F(j) \cup \bigcup_{\substack{h\text{ neither bad}\\\text{nor reducible}}} F(h) \end{equation} and note that the sets making up~\eqref{eqn:A0 recursive def} are pairwise disjoint as $A^{(i)}_0,\dots,A^{(i)}_{k(i)}$ is a~partition of $X$. We now verify~\eqref{cond:Azero friends} for $A^{(i+1)}_0$ by checking~\eqref{cond:Azero friends} holds for each of its constituents. \begin{itemize} \item It follow by the inductive assumption for $A_0^{(i)}$. \item Fix a bad pair $h \ne j$. We first check that \begin{equation}\label{eqn:bad pair sets comparable} \frac{1}{2} < \frac{\mu\left(A^{(i)}_j\right)}{\mu\left(A^{(i)}_k\right)} < 2 \end{equation} holds. By~\eqref{cond:partition} and~\eqref{cond:paired cyl} there exist 3 unions of cylinders, $B,C,D$ all defined only by fixing indices at unbounded scales greater than $\tau$ so that $A_j^{(i)}=B \cap C$ and $A_k^{(i)}=B \cap D$. What is more, by~\eqref{cond:paired cyl} and our construction, up to flipping the role of $C$ and $D$, $C$ is defined by all of the left towers at one unbounded index, say $a(n)$ and $D$ is defined by all of the right towers $a(n)$. Moreover, this index is smaller than the indices defining $B$. Now by a similar computation to those we have done before \[ 1 < \frac{(d(n)+1)\P(n)-\nu(X)^{-1}\nu\left(\bigcup_{\ell>n}W_{a(\ell)}\right)}{(d(n)+1)(\P(n)-\Q(n))}\leq \frac{\mu(C)}{\mu(D)}<1+2\frac{\Q(n)}{\P(n)} \] and~\eqref{eqn:bad pair sets comparable} follows from Lemma~\ref{lem:indep} and the fact that the index defining $C$ and $D$ is at least $2^{2^5}$ smaller than the smallest index defining $B$. With~\eqref{eqn:bad pair sets comparable} established it suffices to find a positive proportion of $(m,\tau)$ buddies in either $A^{(i)}_j$ or $A^{(i)}_k$. How this is done depends on why the pair $h,j$ is bad. \begin{itemize} \item If $\scl(n^{(i)}_j) = \lambda^i(\scl(m))$ and $s \in \{ \lft(n^{(i)}_j), \rht(n^{(i)}_j) \}$ at a bounded scale not smaller than $\tau$ apply Theorem~\ref{thm:bounded wfriends} with $s$ in place of $r$ and note that the set $B$ therein is defined by coordinates no larger than $\scl(s) + 1$ while $A^{(i)}_j$ is defined by coordinates no smaller than $\lambda^{i-1}(\scl(m))$. We may therefore apply Lemma~\ref{lem:indep} to get \[ \mu\left(B \cap A^{(i)}_j\right) \ge \left(1 - \frac{1}{2^{2^5}} \right) \mu(B) \mu\left(A^{(i)}_j\right) \] whence a positive proportion of $A^{(i)}_j$ intersects $B$. Removing $\bad(\scl(s),8s)$ and~\eqref{eqn:bad ancestors} leaves us with a positive proportion of $(m,\tau)$ buddies, as in the proof of Corollary~\ref{cor:bounded mmix}. \item If $\scl(n^{(i)}_j) = \lambda^i(\scl(m))$ and $n^{(i)}_j < \frac{3}{4} \p(\lambda^i(\scl(m)))$ apply Proposition~\ref{prop:small friend}. \item If instead $\scl(n^{(i)}_h) = \lambda^i(\scl(m))$ the arguments are exactly the same upon replacing $j$ with $h$. \end{itemize} \item When $j$ is reducible or neither bad nor reducible, we must show that $F(j)$ has a positive proportion of $(m,j)$-buddies for $j\geq \tau$. \begin{itemize} \item We claim that it suffices to show this when $j$ is reducible, because the measure of the paired cylinders differs by less than a factor of 2 and when $h$ is neither bad nor reducible it is paired with $j$ which is reducible. Indeed, by~\eqref{cond:paired cyl} they differ by at most one unbounded symbol. We consider the other (necessarily unbounded) entries defining these sets~\eqref{cond:partition} and apply Lemma~\ref{lem:indep} to these cylinders and the cylinders defined by the (necessarily unbounded) position where they are defined to differ. \item By Lemma~\ref{lem:friends}, if $j$ is reducible the set of $(n_j^{(i)},\lambda^{i}(\scl(m)))$-friends is at least $\frac{1}{10^6}\mu(F(j))$. \item By Lemmas~\ref{lem:friend to weak} and~\ref{lem:buddies through reduc} pairs that are $(n_j^{(i)},\lambda^{i}(\scl(m)))$-friends and for which~\eqref{eq:same coords} hold for both elements of the pair are automatically $(m,\tau)$-buddies. \end{itemize} \end{itemize} Now we define $k(i+1)$ and the sets $A^{(i+1)}_1,\dots,A^{(i+1)}_{k(i+1)}$ and the times $n^{(i+1)}_1,\dots,\linebreak n^{(i+1)}_{k(i+1)}$. Write \[ G = \{ 1 \le j \le k(i) : j \text{ reducible} \} \cup \{ 1 \le h \le k(i) : h \text{ neither bad nor reducible} \} \] which consists of those $1 \le j \le k(i)$ belonging to a pair that is not bad. Temporarily write $c = c(\lambda^i(\scl(m)))$ and $d = d(\lambda^i(\scl(m)))$. There are two cases to consider. \begin{itemize} \item When $j \in G$ is reducible and its pair is too, write $e = \lst(n^{(i)}_j)$ and define the sets \begin{align*} L^{(i)}_j &= \left\{ x \in A^{(i)}_j \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [9,d-e] \right\} \\ R^{(i)}_j &= \left\{ x \in A^{(i)}_j \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [d+2,c-e] \right\} \end{align*} if $n^{(i)}_j > 0$ and \begin{align*} L^{(i)}_j &= \left\{ x \in A^{(i)}_j \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [10-e,d-1] \right\} \\ R^{(i)}_j &= \left\{ x \in A^{(i)}_j \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [d+2-e,c-2] \right\} \end{align*} if $n^{(i)}_j < 0$. In this case form the pairs $(L^{(i)}_j, \lft(n^{(i)}_j))$ and $(R^{(i)}_j, \rht(n^{(i)}_j))$. \item For neither bad nor reducible $h \in G$ its paired index $j \in G$ is necessarily reducible by~\eqref{cond:differ}. Again write $e = \lst(n^{(i)}_j)$ and in this case define \begin{align*} L^{(i)}_h &= \left\{ x \in A^{(i)}_h \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [9,d-e] \right\} \\ R^{(i)}_h &= \left\{ x \in A^{(i)}_h \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [d+2,c-e] \right\} \end{align*} if $n^{(i)}_j > 0$ and \begin{align*} L^{(i)}_h &= \left\{ x \in A^{(i)}_h \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [10-e,d-1] \right\} \\ R^{(i)}_h &= \left\{ x \in A^{(i)}_h \,:\, x\left(\scl\left(n^{(i)}_j\right)\right) \in [d+2-e,c-2] \right\} \end{align*} if $n^{(i)}_j < 0$. In this case form the pairs $(L^{(i)}_h, n^{(i)}_h)$ and $(R^{(i)}_h, n^{(i)}_h)$. \end{itemize} Take $k(i+1) = 2|G|$. Let $(A^{(i+1)}_1, n^{(i+1)}_1),\dots,(A^{(i+1)}_{k(i+1)}, n^{(i+1)}_{k(i+1)})$ be an enumeration of the $k(i+1)$ pairs that result from the above two cases. For each $1 \le j \le k(i+1)$ let $1 \le \alpha(j) \le k(i)$ be the index for which $n^{(i)}_{\alpha(j)}$ determines $n^{(i+1)}_j$ as the above pairs are formed. For each $g \in G$ the sets $F(g)$, $L^{(i)}_g$, $R^{(i)}_g$ partition $A^{(i)}_g$. It is immediate from the recursion and the definitions that~\eqref{cond:setup} and~\eqref{cond:partition} are satisfied. We have already verified~\eqref{cond:Azero friends}. We next verify~\eqref{cond:agree}. Fix $1 \le j \le k(i+1)$. If its ancestor $1 \le \alpha(j) \le k(i)$ is such that $n^{(i+1)}_j = n^{(i)}_j$ then there is nothing to check. Suppose instead that its ancestor is a value $1 \le \alpha(j) \le k(i)$ for which $n^{(i+1)}_j = \lft(n^{(i)}_{\alpha(j)})$ or $n^{(i+1)}_j = \rht(n^{(i)}_{\alpha(j)})$. Fix $x \in A^{(i+1)}_j$ with $j \ne 0$. Write $n = n_{\alpha(j)}^{(i)}$ and let $r \in \{\lft(n),\rht(n)\}$ be the reduction we are considering. By either Corollary~\ref{cor:pos reduc conseq} or Corollary~\ref{cor:neg reduc conseq} we have that $(T^n x)(\ell) = (T^r x)(\ell)$ for $\ell \leq\scl(n)-1$, so long as $x \notin \bad(\scl(n),n)$. Invoking~\eqref{cond:agree} gives $(T^n x)(\ell) = (T^m x)(\ell)$ for $\ell\leq \lambda^{i-1}\scl(m)-1$ so long as~\eqref{eqn:bad ancestors} holds. Combining the two verifies~\eqref{cond:agree}. It remains to define how $1,\dots,k(i+1)$ are paired and verify~\eqref{cond:differ}, \eqref{cond:paired numbers}, \eqref{cond:paired cyl}. There are two cases to consider: \begin{itemize} \item When $j,h \in G$ are paired and both are reducible, pair the descendants of $j$ and pair the descendants of $h$. \item When $j,h \in G$ are paired and neither is bad and only $j$ is reducible then necessarily $h$ is neither bad nor reducible. Thus $\scl(n^{(i)}_h) < \lambda^i(\scl(m))$ and since $h$ is not bad, $J(n^{(i)}_h)\leq \lambda^{i+1}(\scl(m))$. There are two sub-cases to consider: \begin{enumerate} \item\label{subcase1} If $\scl(n_h^{(i)}) = \lambda^{i+1}(\scl(m))$ then pair the descendants of $j$ and pair the descendants of $h$. \item\label{subcase2} If $\scl(n_h^{(i)}) < \lambda^{i+1}(\scl(m))$ we form the pairing so that $L^{(i)}_j$ is with $L^{(i)}_h$ and $R^{(i)}_j$ is with $R^{(i)}_h$. \end{enumerate} \end{itemize} First we verify~\eqref{cond:paired cyl}. When the descendants $h'$, $h''$ of some $1 \le h \le k(i)$ are paired we have \[ \left\{ A^{(i+1)}_{h'}, A^{(i+1)}_{h''} \right\} = \left\{ L^{(i)}_h, R^{(i)}_h \right\} \] and therefore the sets $A^{(i+1)}_{h'}$, $A^{(i+1)}_{h''}$ disagree only in the $\scl(n^{(i)}_{j'})$ coordinate, where $j' = h$ when $h$ is reducible and $j'$ is the pair of $h$ when $h$ is not reducible. In either case $\scl(n^{(i)}_{j'})$ is unbounded. It remains to verify that the pairing satisfies~\eqref{cond:paired cyl} when we are in the second subcase. In that case the additional restriction defining $L^{(i)}_j$ and $L^{(i)}_h$ are the same because both are determined by $n^{(i)}_j$ so $L^{(i)}_j$ and $L^{(i)}_h$ will only disagree in the coordinate where $A^{(i)}_j$ and $A^{(i)}_h$ disagreed. The same is true for $R^{(i)}_j$ and $R^{(i)}_h$. To conclude we check~\eqref{cond:differ} and~\eqref{cond:paired numbers} simultaneously. Fix again $1 \le j,h \le k(i)$ paired. The construction is such that each descendent is paired with exactly one other descendant. Now we check at least one descendant in each pair has its associated time at scale $\lambda^{i+1}(m)$. If both $n_j^{(i)}$ and $n_h^{(i)}$ are reducible we can apply Lemma~\ref{lem: reduction not both small} to both separately. Note that in this case~\eqref{eq:differnce in reduction} gives~\eqref{cond:paired numbers}. This is also the justification of Subcase~\eqref{subcase1}. We are left with Subcase~\eqref{subcase2} where $n_j^{(i)}$ is reducible and $n_h^{(i)}$ is neither bad nor reducible. We are done if $\scl(n_h^{(i)}) = \lambda^{i+1}(\scl(m))$, so we assume $\scl(n_h^{(i)}) \le \lambda^{i+2}(\scl(m))$. By~\eqref{cond:paired numbers} \[ n_j^{(i)}=k \q\left(\lambda^{i-1}\scl(m)\right)+n_h^{(i)} \quad\text{with}\quad 0<|k|\leq 2d(\lambda^{i})+3 \] and (applying Lemma~\ref{lem:for pairing} with $b = n^{(i)}_h$ and using that $n^{(i)}_j$ is not bad) we have that both $\lft(n_j^{(i)})$ and $\rht(n_j^{(i)})$ have the form \[ k'_L \q\left(\lambda^i \scl(m)\right) + n_h^{(i)} \quad\text{and}\quad k'_R \q\left(\lambda^i \scl(m)\right) + n_h^{(i)} \] with $0<|k'_L|,|k'_R|\leq c(\lambda^i(J(m))) + 1$. So both pairs created from $n_h^{(i)}$ and $n_j^{(i)}$ satisfy~\eqref{cond:differ}, with the terms $\lft(n_j^{(i)})$ and $\rht(n_j^{(i)})$ satisfying \[ \scl\left(\lft\left(n_j^{(i)}\right)\right) = \lambda^{i+1}\left(\scl(m)\right) = \scl\left(\rht\left(n_j^{(i)}\right)\right) \] as desired. Similarly, we have both pairs satisfy~\eqref{cond:paired numbers}. \end{proof} \let\qed\relax \end{proof} \begin{prop}\label{prop:reduction multiples} Fix $\tau=a(\beta) \ge a(10)$ a target unbounded scale and $m \in \N$ with $\scl(m)=a(b)$ unbounded and $b>\beta$. Let $I \in \N$ and the tuples $(A^{(i)}_0, A^{(i)}_1, n^{(i)}_1,\dots, A^{(i)}_{k(i)}, n^{(i)}_{k(i)})$ for $1 \le i \le I=b-\beta$ result from an application of Proposition~\ref{prop:reduction} to $m$. In particular, $\scl(n_j^{(i)}) \leq \tau$ and for at least half of the $j$ we have $\scl(n_j^{(I)})=\tau$. If $\mu(A^{(I)}_0)<\frac 1 {10^{50}}$ and tuples $(\overline{A}^{(i)}_0, \overline{A}^{(i)}_1, \overline{n}^{(i)}_1,\dots, \overline{A}^{(i)}_{\overline{k}(i)}, \overline{n}^{(i)}_{\overline{k}(i)})$ for $1 \le i \le K = b-\beta$ result from applying Proposition~\ref{prop:reduction} to $6d(\beta) m$ then $\mu(\overline{A}^{(K)}_0)>\frac 1 {10^{50}}$. \end{prop} This follows from repeated applications of the following result. \begin{lemm}\label{lem:reduc multiply} Fix $m \in \Z$ with $\scl(m) = a(b)$ an unbounded scale. If $\rht(m)$ is at an unbounded scale and $k \in \Z$ satisfies $|k|\leq 6d(b)$ and $km$ is at an unbounded scale then one of the following is true: \begin{itemize} \item $\rht(km)=k\rht(m)$. \item $\rht(km)$ is at a bounded scale. \item At least one of $\rht(m)$, $\rht(km)$ is both at scale $\scl(m)$ and less than $\frac{3}{4} \p(\scl(m))$. \end{itemize} Moreover, if $\rht(km)$ is at scale $\lambda(\scl(m))$ then $|k| < 6d(a^{-1}(\scl(\rht(m))))$. \pagebreak If $\lft(m)$ is at an unbounded scale and $k \in \Z$ satisfies $|k|\leq 6d(b)$ and $km$ is at an unbounded scale then one of the following is true: \begin{itemize} \item $\lft(km) = k\lft(m)$, \item $\lft(km)$ is at a bounded scale. \end{itemize} Moreover, if $\lft(km)$ is at scale $\lambda(\scl(m))$ then $|k| < 6d(a^{-1}(\scl(\lft(m)))$. \end{lemm} \begin{proof}[Proof of Proposition~\ref{prop:reduction multiples} assuming Lemma~\ref{lem:reduc multiply}] Consider any sequence $m, n^{(1)},\linebreak n^{(2)},\dots,n^{(I)}$ of reductions of $m$. The corresponding sequence for $km$ is $km,kn^{(1)},\linebreak kn^{(2)},\dots$ until one of $kn^{(i)}$ is in a bad pair. This follows by Lemma~\ref{lem:reduc multiply} and the definition of bad pairs in the proof of Proposition~\ref{prop:reduction}. Now if $kn^{(i)}$ is bad the corresponding points are put in $A_0$ and we have proportional friends. On the other hand $6d(\beta) n^{(I)} > \p(\tau)$ so $A^{(I-1)}$ is put in $A_0^{(I)}$ if it is not put in earlier. \end{proof} \begin{proof}[Proof of Lemma~\ref{lem:reduc multiply}] We present the proof in the case of $\rht$. The case of $\lft$ is similar. First, if $km$ is at an unbounded scale, then $\scl(km) = \scl(m)$ because the bound on $k$ gives $\scl(km) \le \scl(m) + 1$. If $\rht(km) \neq k\rht(m)$ then \begin{multline*} \Big\vert\rht(km)-k\rht(m)\Big\vert \\ \begin{aligned} &= \left\vert km- \lst(km) \Big(\p (\scl (m))-\q(\scl(m))\Big)-k\Big(m - \lst(m) \big(\p(\scl(m))-q(\scl(m))\big)\Big) \right\vert \\ &\ge \p(\scl(m))-\q(\scl(m)) \end{aligned} \end{multline*} and by Lemma~\ref{lem:reduction LR} they are both at most $\frac 3 4 \p(J(m))$. Lastly, if $\scl(m)=a(\gamma))$ and $6d(\gamma+1)\geq|k|\geq 6d(\gamma)$ then $\scl(km)$ is strictly between $a(\gamma)$ and $a(\gamma+1)$. Thus it is at a bounded scale. \end{proof} \subsection{Proof of Theorem~\ref{thm:mild mixing}}\label{subsec:mild mixing} We are ready to conclude that $(X,\mu,T)$ is mild mixing. \begin{proof}[Proof of Theorem~\ref{thm:mild mixing}] Fix a non-constant function $f$ in $\ltwo(X,\mu)$ and a sequence $ \bir$ in $\Z$ with $|r(n)| \to \infty$. Set $\rho=10^{-50}\cdot 10^{-9}$. Let $\delta$ and $\tau'$ be as in Corollary~\ref{cor:end} for our choices of $f$ and $\rho$. Fix $\tau = a(\beta) > \tau'$ an unbounded scale. It suffices to show that at least one of $ \bir$ or $6 d(\beta) \bir$ is not a rigidity sequence for $f$. Indeed if $ \bir$ is a rigidity sequence for $f$ then so is $k \bir$ for any fixed $k \in \Z\setminus \{0\}$. By Theorem~\ref{thm:bounded wfriends} and by passing to a subsequence if necessary we may first assume for all $n \in \N$ that $\scl(r(n))$ and $\scl(6d(\beta) r(n))$ are unbounded scales and that $\scl(r(n)) \ge \tau$. Fix $I(n)$ and $K(n)$ maximal with $\lambda^{I(n)}(r(n)) \ge \tau$ and $\lambda^{K(n)}(6d(\beta) r(n)) \ge \tau$ respectively. Apply Proposition~\ref{prop:reduction} to $r(n)$ and $6 d(\beta) r(n)$. Let $\Gamma_n \subset A^{(I(n))}_0$ and $\overline{\Gamma}_n \subset \overline{A}^{(K(n))}_0$ be the respective sets furnished by~\eqref{cond:Azero friends}. There are measure preserving maps $\phi_n : \Gamma_n \to X$ and $\overline{\phi}_n : \overline{\Gamma}_n \to X$ with the following properties. \begin{itemize} \item For each $z \in \Gamma_n$ we have that $(z,\phi_n(z))$ are $(r(n),\tau)$ buddies. \item For each $z \in \overline{\Gamma}_n$ we have that $(z,\overline{\phi}_n(z))$ are $(6d(\beta)r(n),\tau)$ buddies. \item $\mu(\Gamma_n) \ge 10^{-9}\mu(A^{(I(n))}_0)$ \item $\mu(\overline{\Gamma}_n) \ge 10^{-9}\mu(\overline{A}^{(K(n))}_0)$ \end{itemize} By Proposition~\ref{prop:reduction multiples} \[ \max\left\{\mu\left(A^{(I(n))}_0\right), \mu\left(\overline{A}^{(K(n))}_0\right) \right\} \geq 10^{-50} \] and thus Corollary~\ref{cor:end} gives \[ \max \left\{ \int \left|f-f\circ T^{r(n)}\right| \intd \mu,\int \left|f-f \circ T^{6d(\beta)r(n)}\right| \intd\mu \right\} > \delta \] proving Theorem~\ref{thm:mild mixing}. \end{proof} \appendix \section{Changes for the proof of Proposition~\ref{prop:ce}}\label{app:ce_proof} In this section we prove the version of~\cite[Corollary~3.3]{MR4269425} that is needed to prove Proposition~\ref{prop:ce}. The next lemma and its proof are unchanged from~\cite[Lemma~3.2]{MR4269425}. \begin{lemm}\label{lemma:to:bary} Given $c>0$ and $d \in \N$ there exists $\rho<1$, $C$ so that if $0 < \delta_i <1/2$ and $a_i, b_i$ are such that $a_i,b_i>c$ and $1\geq a_i+b_i>1-\delta_i$ and also $0 \le \zeta_i^{(\ell)}\le 1$ are sequences of real numbers for each $\ell \in \{1,\dots,d\}$ satisfying \begin{equation}\label{eq:gammas:assumption} \left|\zeta_i^{(\ell)}-\left(a_i\zeta_{i-1}^{(\ell-1)}+b_i\zeta_{i-1}^{(\ell)}\right)\right|<\delta_{i-1} \end{equation} then \[ \left|\zeta_i^{(s)}-\frac 1 d \sum_{\ell=1}^d\zeta_k^{(\ell)}\right|\leq C\sum_{j=k}^{i-1}\left(\delta_i+\frac{\delta_i}{1-\delta_i}\right)+C\rho^{i-k} \] for all $k \ge 0$, $i>k$ and $s \in \{1,\dots,d\}$. \end{lemm} \begin{coro}\label{cor:to:bary} Under the assumptions of Proposition~\ref{prop:ce} there exist $ \rho<1$, $C'>0$ so that $d_{KR}(\nu_k^{(\ell)},\frac 1 d\sum_{\ell=1}^d \nu_{b}^{(\ell)})\leq C'\sum_{j=b}^k\epsilon_j +C'\rho^{k-b}$ whenever $k\geq b$ and $\ell\in \{1,\dots,d\}$. \end{coro} \begin{proof}[Proof of Corollary~\ref{cor:to:bary}] First notice that by~\eqref{ce:nearness} we have that \begin{equation}\label{eq:bary1} \begin{aligned} d_{KR}\left(\nu_{\alpha(j)}|_{A(j)},\;\nu_{\gamma(j-1)}|_{A(j)}\right) &<\epsilon(j)\\ \text{and}\quad d_{KR}\left(\nu_{(\alpha(j)}|_{B(j)},\;\nu_{\alpha(j-1)}|_{B(j)}\right) &<\epsilon(j). \end{aligned} \end{equation} We now claim that there exists a constant $D$ so that for all $j$, \begin{equation}\label{eq:KR est2} d_{KR}\left(\left.\frac 1 {\mu(A(j))} \nu_{\gamma(j-1)}\right|_{A(j)},\;\nu_{\gamma(j-1)}\right)0:S^ix\in J_j \right\}=\ell\right\}, \] we get that there exists $E>0$ so that \begin{equation}\label{eq:KR:est:2prime} d_{KR}\left(\left.\frac 1 {\mu(B(j)} \nu_{(\alpha(j-1))}\right|_{B(j)},\;\nu_{(\alpha(j-1))}\right) b} \{x \in X : (S|Y_b)^i(x) \in W_{a(m)} \text{ for some } -|r| \le i \le |r|\}$. \end{itemize} \subsection*{Specific times} \begin{itemize} \item $\p(n)$ is the minimal natural number for which $(S|Y_n)^{\p(n)}[0^{n-1} 0] = [0^{n-1} 1]$, \item $\P(n) = \p(a(n))$ is the height of the left tower at scale $a(n)$, \item $\Q(n) = 2\P(n-1) - \Q(n-1)$ is the number of levels removed from $\Omega$ at scale $a(n)$, \item $\q(n) = \Q(a^{-1}(n))$ when $n$ unbounded and zero otherwise. \end{itemize} \subsection*{Expansion notation} \begin{itemize} \item $\lst(r)$ is the leading coefficient in the tower expansion of $r \in \Z$, \item $\scl(r)$ is the leading scale in the tower expansion of $r \in \Z$. \end{itemize} \subsection*{Reduction notation} \begin{itemize} \item $\lft(r) = r - \lst(r) \p(\scl(r))$ whenever $\scl(r)$ is unbounded, \item $\rht(r) = r - \lst(r) \p(\scl(r)) + \lst(r) \q(\scl(r))$ whenever $\scl(r)$ is unbounded, \item $\lambda(n)$ is the largest unbounded scale strictly smaller than $n$. \end{itemize} %\let\OLDthebibliography\thebibliography %\renewcommand\thebibliography[1]{ %\OLDthebibliography{#1} %\setlength{\parskip}{0pt} %\setlength{\itemsep}{5pt plus 0.3ex}} \bibliography{chaika} \end{document}