# Finite odometer factors of rank one $\mathbb{Z}^{d}$-actions

Aimee S. A. Johnson1 and  and David M. McClendon2

(Date: June 15, 2023)

###### Abstract.

In this paper, we give explicit conditions characterizing the Følner rank one
$\mathbb{Z}^{d}$-actions that factor onto a finite odometer; those that factor
onto an arbitrary, but specified $\mathbb{Z}^{d}$-odometer, and those that
factor onto an unspecified $\mathbb{Z}^{d}$-odometer. We also give explicit
conditions describing the Følner rank one $\mathbb{Z}^{d}$-actions that are
conjugate to a specific $\mathbb{Z}^{d}$-odometer, and those that are
conjugate to some $\mathbb{Z}^{d}$-odometer. These conditions are based on
cutting and stacking procedures used to generate the action, and generalize
results given in [FGHSW] for rank one $\mathbb{Z}$-actions.

###### 2010 Mathematics Subject Classification:

Primary 37A15, Secondary 37A35, 37A05

1Corresponding author: Department of Mathematics and Statistics, Swarthmore
College, 500 College Ave. Swarthmore, PA 19081; email<EMAIL_ADDRESS>
2Department of Mathematics, Ferris State University, ASC 2021, Big Rapids, MI
49307; email<EMAIL_ADDRESS>
##  1\. Introduction

In recent work, Foreman, Gao, Hill, Silva and Weiss [FGHSW] gave a
characterization of rank one transformations which factor onto an odometer
(equivalently, they characterize which rank one transformations are not
totally ergodic) in terms of parameters coming from a cutting and stacking
construction used to describe the rank one transformation. They used this
characterization to describe the rank one transformations which are conjugate
to a given odometer, and the rank one transformations that are conjugate to
some odometer.

The work in [FGHSW] can be thought of as one step towards a more ambitious
goal of classifying rank one actions up to conjugacy. In particular, the
conjugacy relation on rank one systems is known to be a Borel equivalence
relation [FRW], and so there should be an explicit way of describing all rank
one transformations that are conjugate to a specific rank one transformation.
Foreman et al. address exactly this question in the setting where the specific
rank one transformation is an odometer.

In this paper, we generalize the work in [FGHSW] to Følner rank one actions of
$\mathbb{Z}^{d}$ (though we believe our results hold for actions of groups
beyond $\mathbb{Z}^{d}$). We characterize, in Theorem 3.1, those actions which
factor onto a specified finite $\mathbb{Z}^{d}$-odometer. Using this result,
we then characterize the Følner rank one $\mathbb{Z}^{d}$-actions that factor
onto a specific odometer (Corollary 4.2) and those that factor onto an
arbitrary infinite odometer (Theorem 4.7). Finally, we describe the Følner
rank one $\mathbb{Z}^{d}$-actions that are conjugate to a specific
$\mathbb{Z}^{d}$-odometer (Theorem 5.1) and those that are conjugate to some
$\mathbb{Z}^{d}$-odometer (Theorem 5.2). We also show, via an example in
Theorem 3.2, that the results of [FGHSW] do not necessarily generalize to rank
one $\mathbb{Z}^{d}$-actions when the tower shapes do not form a Følner
sequence in $\mathbb{Z}^{d}$. In Theorem 3.3, we show that, unlike the
situation for rank one $\mathbb{Z}$-actions, for Følner rank one
$\mathbb{Z}^{d}$-actions total ergodicity does not coincide with the existence
of a finite odometer factor. Finally, in Theorem 6.1 , we give an example of a
Følner rank one $\mathbb{Z}^{2}$-action generated by transformations that each
factor onto a free odometer but yield a $\mathbb{Z}^{2}$-action that does not
itself factor onto a free odometer.

When $d=1$, our results can also be seen as a generalization of the results of
[FGHSW] to funny rank one $\mathbb{Z}$-actions whose tower shapes form a
Følner sequence; this would include, for example, funny rank one
$\mathbb{Z}$-actions constructed by Ferenczi [F1] that are known not to be
loosely Bernoulli and thus not of rank one.

_Acknowledgment:_ The authors wish to acknowledge the Little School Dynamics
research group of the American Institute of Mathematics, through which they
were initially exposed to the work in [FGHSW].

##  2\. Rank one actions and odometers

###  2.1. Dynamical systems

We begin with some standard definitions from measurable dynamics. Let $G$ be a
group and let $(X,\mathcal{X},\mu)$ be a Lebesgue probability space. We say
$(X,\mu,\mathbf{T})$ is a measure-preserving $G$-action, and write
$\mathbf{T}:G\curvearrowright(X,\mu)$, if for every $g\in G$ there is an
invertible, measurable, measure-preserving transformation $\mathbf{T}^{g}:X\to
X$, and these transformations satisfy
$\mathbf{T}^{gh}=\mathbf{T}^{g}\circ\mathbf{T}^{h}$ for every $g,h\in G$ and
also that $\mathbf{T}^{0}(x)=x$ for all $x\in X$, where $0$ denotes the
identity element of $G$.

While the measure of a Lebesgue probability space is by definition non-atomic,
in this paper we will also want to discuss systems with finite phase spaces.
With this in mind, if $X$ is a finite set, $\mathcal{X}$ is the power set of
$X$, and $\mu$ is normalized counting measure on $(X,\mathcal{X})$, we will
also decree any action of group $G$ on the finite set $X$ that preserves $\mu$
to be a measure-preserving $G$-action.

A measure-preserving $G$-action $(X,\mathbf{T})$ is called free if, for almost
every $x\in X$, $\mathbf{T}^{g}(x)=x$ implies $g=0$. If the only measurable
sets invariant under every $\mathbf{T}^{g}$ have $\mu$-measure $0$ or $1$,
then we say $(X,\mu,\mathbf{T})$ is ergodic.

We say that one measure-preserving $G$-action $(X,\mu,\mathbf{T})$ factors
onto another $(Y,\nu,\mathbf{S})$ if there is a measurable function $\phi:X\to
Y$ so that $\phi^{*}(\mu)=\nu$ and
$\phi(\mathbf{T}^{g}(x))=\mathbf{S}^{g}(\phi(x))$ for almost every $x\in X$
and every $g\in G$. Any such $\phi$ is called a factor map. Two actions
$\mathbf{T}:G\curvearrowright(X,\mu)$ and
$\mathbf{S}:G\curvearrowright(Y,\nu)$ are conjugate if there is a factor map
from $(X,\mathbf{T})$ to $(Y,\mathbf{S})$ which is almost surely one-to-one.

In this paper, we concern ourselves with actions where $G=\mathbb{Z}^{d}$ for
some $d$, but our results likely translate to actions of more general discrete
amenable groups.

###  2.2. Rank one systems

Rank one actions form a rich class of measure-preserving actions that provide
examples, and especially counterexamples, addressing many questions of
measurable dynamics. They have their roots in Chacon’s “geometric
construction” of examples of systems with particular mixing properties [C1,
C2] and in an example of a measure-preserving transformation with no square
root due to Ornstein [O]. The name “rank one” was coined in [ORW], and many
properties of rank one actions are well-known: for instance, they are ergodic
[Fr, RS], have simple spectrum [B, RS] and are of zero entropy [B, RS].

The study of rank one systems, as with the rest of ergodic theory, spread from
actions generated by a single transformation to actions of more general
groups. Rank one $\mathbb{Z}^{d}$-actions were first considered by Johnson and
Şahin in [JS], where the authors showed that any such action was loosely
Bernoulli. Their work built on the one-dimensional work of [ORW] and the
generalization of Feldman’s $\overline{f}$-metric to $\mathbb{Z}^{d}$-actions
found in [H]. More recently, Danilenko has described rank one actions of more
general groups using $(C,F)$-models [D1, D2], although these actions might
better be described as generalizing Thouvenot’s notion of a _funny rank one_
$\mathbb{Z}$-action (see [F1]).

Today there are many different approaches to defining rank one systems which
are “mostly” equivalent (see [F2, AFP] for details), including a constructive
geometric definition coming from Chacon’s work, a constructive symbolic
definition [K, dJR] and a non-constructive symbolic definition [dJ]. Our
description of rank one systems is what is formally called the non-
constructive geometric approach and informally referred to as _cutting and
stacking_. This approach has its origin in Ornstein’s example and is rooted in
the notion of periodic approximation as studied in [KS]. The following
definitions can be found in [RS] but are repeated here for completion.

To begin, we define a shape to be a nonempty finite subset of
$\mathbb{Z}^{d}$. Then a sequence $\\{F_{n}\\}$ of shapes is called a Følner
sequence for $\mathbb{Z}^{d}$ if for any $\mathbf{v}\in\mathbb{Z}^{d}$,

$\lim_{n\to\infty}\frac{\\#\left(F_{n}\,\triangle\,[F_{n}+\mathbf{v}]\right)}{\\#(F_{n})}=0.$

A shape can be associated to a measure-preserving $\mathbb{Z}^{d}$-action by
partitioning the space into subsets, arranged via the shape in the following
way:

######  Definition 2.1.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ and let
$R\subset\mathbb{Z}^{d}$ be a shape. If there exists a measurable set $B$ such
that $\mu(B)>0$ and
$\mathbf{T}^{\mathbf{v}_{1}}(B)\cap\mathbf{T}^{\mathbf{v}_{2}}(B)=\emptyset$
for every $\mathbf{v}_{1},\mathbf{v}_{2}\in R$, then the $\mathbf{T}$-tower of
shape $R$ is defined to be
$\mathcal{T}=\\{\mathbf{T}^{\mathbf{v}}(B):\mathbf{v}\in R\\}\cup E$, where
$E=\left(\cup_{\mathbf{v}\in R}\mathbf{T}^{\mathbf{v}}(B)\right)^{C}$. The
sets $\mathbf{T}^{\mathbf{v}}(B)$ are then the levels of the $\mathbf{T}$\-
tower; the set $B$ is the base of the $\mathbf{T}$-tower; and $E$ is called
the _error set_ of $\mathcal{T}$.

Note that we can assume $\mathbf{0}\in R$ by simply shifting the indices of
$R$.

If $\mathcal{T}$ is a $\mathbf{T}$-tower and $x\in X$, then we write
$x\in\mathcal{T}$ to mean that $x$ is not in the error set of $\mathcal{T}$.

We are now ready to define what it means for a $\mathbb{Z}^{d}$-action to be
rank one:

######  Definition 2.2.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ and let $\\{F_{n}\\}$
be a sequence of shapes. We say $\mathbf{T}$ is _rank one for $\\{F_{n}\\}$_
if there is a sequence $\\{\mathcal{T}_{n}\\}$ of $\mathbf{T}$-towers of shape
$F_{n}$, so that given any $A\in\mathcal{X}$, there is a sequence of
$\mathcal{T}_{n}$-measurable sets $L_{n}$ (i.e. a union of levels of the tower
$\mathcal{T}_{n}$) so that

$\lim_{n\to\infty}\mu\left(L_{n}\,\triangle\,A\right)=0.$

We say $\mathbf{T}$ is _rank one_ if it is rank one for some sequence
$\\{F_{n}\\}$ of shapes.

In Definition 2.2, we make no assumptions about the sequence of shapes. The
following definitions are more restrictive.

######  Definition 2.3.

A rank one $\mathbb{Z}^{d}$-action
$\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ is called
$\\{F_{n}\\}$-_Følner rank one_ if $\mathbf{T}$ is rank one for $\\{F_{n}\\}$
where $\\{F_{n}\\}$ is a Følner sequence for $\mathbb{Z}^{d}$. If
$\\{F_{n}\\}$ is understood, we simply say $\mathbf{T}$ is _Følner rank one_.

######  Definition 2.4.

A rank one $\mathbb{Z}^{d}$-action
$\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ is called _stacking rank
one_ if $\mathbf{T}$ is rank one for $\\{F_{n}\\}$ and the associated
$\mathbf{T}$-towers $\mathcal{T}_{n}$ are such that every $\mathcal{T}_{n+1}$
refines $\mathcal{T}_{n}$ and the error sets decrease: $E_{n+1}\subseteq
E_{n}$ for all $n$.

In other words, every level in $\mathcal{T}_{n}$ is a union of levels from
$\mathcal{T}_{n+1}$, and thus one can think of a stacking rank one system as
being one associated with a “cutting and stacking” construction. The two types
of rank one systems listed above are related, as described in the following
theorem.

######  Theorem 2.5 ([RS], Theorem 3.8).

Let $\\{F_{n}\\}$ be a Følner sequence for $\mathbb{Z}^{d}$ with
$\mathbf{0}\in F_{n}$ for all $n$. Then any $\\{F_{n}\\}$-Følner rank one
$\mathbb{Z}^{d}$-action is stacking rank one, with towers whose shapes are
given by $\\{F_{n}\\}$.

Now consider a $\mathbb{Z}^{d}$-action which is Følner rank one, and thus is
also stacking. Fix $m$ and consider the $\mathbf{T}$-tower $\mathcal{T}_{m}$
of shape $F_{m}$. Because $\mathbf{T}$ is stacking, we know the base $B_{m}$
of $\mathcal{T}_{m}$ is refined by the $\mathbf{T}$-tower $\mathcal{T}_{n}$ of
shape $F_{n}$ for any $n\geq m$. That is, for each $n\geq m$, there is a union
of levels in $\mathcal{T}_{n}$ which equal $B_{m}$, leading us to the
following definition:

######  Definition 2.6.

Given a Følner rank one $\mathbb{Z}^{d}$-action
$\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ and any $n,m\in\mathbb{N}$
with $n\geq m$, define

$I_{m,n}=\\{\mathbf{i}\in F_{n}:\mathbf{T}^{\mathbf{i}}(B_{n})\subseteq
B_{m}\\}.$

In [FGHSW], these sets are called the descendants of the base $B_{m}$.

One of the reasons to restrict to $\mathbb{Z}^{d}$-actions that are Følner
rank one is given by the following lemma, which identifies a relationship
between any Følner sequence and any finite index subgroup of $\mathbb{Z}^{d}$.
More specifically, it says that eventually, every set in the Følner sequence
intersects each coset of a finite index subgroup in roughly the same
proportion.

######  Lemma 2.7.

Let $\\{F_{n}\\}$ be a Følner sequence in $\mathbb{Z}^{d}$, and let $G$ be a
finite-index subgroup of $\mathbb{Z}^{d}$. Then for every
$\mathbf{v}\in\mathbb{Z}^{d}$,

$\stackbin[n\to\infty]{}{\lim}\frac{\\#(F_{n}\cap(\mathbf{v}+G))}{\\#(F_{n})}=\frac{1}{[\mathbb{Z}^{d}:G]}.$

###### Proof.

Suppose not. Then there is a coset $\mathbf{w}+G$ in $\mathbb{Z}^{d}/G$ and a
subsequence $\\{F_{n_{k}}\\}$ of $\\{F_{n}\\}$ so that one of two things
happens: either there is a constant $\alpha\in(0,1)$ so that for all
sufficiently large $k$,

(2.1)
$\frac{\\#(F_{n_{k}}\cap(\mathbf{w}+G))}{\\#(F_{n_{k}})}\leq\frac{\alpha}{[\mathbb{Z}^{d}:G]},$

or there is a constant $\beta>1$ so that for all large enough $k$,

(2.2)
$\frac{\\#(F_{n_{k}}\cap(\mathbf{w}+G))}{\\#(F_{n_{k}})}\geq\frac{\beta}{[\mathbb{Z}^{d}:G]}.$

In the first situation, notice that it is also the case that for every $k$,
there exists a coset $\mathbf{v}+G$ such that

$\\#\left(F_{n_{k}}\cap(\mathbf{v}+G)\right)\geq\frac{1}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k}}).$

Since there are only finitely many cosets in $\mathbb{Z}^{d}/G$, it follows
that there is a single coset $\mathbf{v}+G\in\mathbb{Z}^{d}/G$ and a
subsequence $\\{F_{n_{k_{l}}}\\}$ of $\\{F_{n_{k}}\\}$ so that for all $l$,

(2.3)
$\frac{\\#\left(F_{n_{k_{l}}}\cap(\mathbf{v}+G)\right)}{\\#(F_{n_{k_{l}}})}\geq\frac{1}{[\mathbb{Z}^{d}:G]}.$

Now consider $\mathbf{x}\in F_{n_{k_{l}}}\cap(\mathbf{v}+G)$. In particular,
this says $\mathbf{x}\in(\mathbf{v}+G)$, and so we have
$\mathbf{x}-\mathbf{v}\in G$ and
$\mathbf{x}-\mathbf{v}+\mathbf{w}\in\mathbf{w}+G$. Either
$\mathbf{x}-\mathbf{v}+\mathbf{w}$ is in $F_{n_{k_{l}}}$ or it is not: if it
is, then $\mathbf{x}-\mathbf{v}+\mathbf{w}\in
F_{n_{k_{l}}}\cap(\mathbf{w}+G)$; if it is not, then
$\mathbf{x}-\mathbf{v}+\mathbf{w}\in
F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]$.
We thus have

(2.4)
$\\#\left(\,F_{n_{k_{l}}}\cap(\mathbf{v}+G)\,\right)\leq\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)\,+\,\\#\left(F_{n_{k_{l}}}\cap(\mathbf{w}+G)\right).$

Using (2.1) on the right and (2.3) on the left, we get

$\frac{1}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k_{l}}})\leq\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)\,+\frac{\alpha}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k_{l}}}),$

yielding that for all $l$ we have

$\frac{1-\alpha}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k_{l}}})\leq\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)$

or

$\frac{1-\alpha}{[\mathbb{Z}^{d}:G]}\leq\frac{\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)}{\\#(F_{n_{k_{l}}})}.$

We thus have that

$\lim_{l\rightarrow\infty}\frac{\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)}{\\#(F_{n_{k_{l}}})}\neq
0$

which contradicts $\\{F_{n}\\}$ being a Følner sequence.

A similar argument works in the second situation: this time, we notice that
for every $k$, there is $\mathbf{v}+G$ so that

$\\#(F_{n_{k}}\cap(\mathbf{v}+G))\leq\frac{1}{[\mathbb{Z}^{d}:G]}\\#(F_{n_{k}}),$

so there is a single coset $\mathbf{v}+G$ and a subsequence
$\\{F_{n_{k_{l}}}\\}$ of $F_{n_{k}}$ so that for all $l$,

(2.5)
$\frac{\\#\left(F_{n_{k_{l}}}\cap(\mathbf{v}+G)\right)}{\\#(F_{n_{k_{l}}})}\leq\frac{1}{[\mathbb{Z}^{d}:G]}.$

Repeating the argument used to obtain (2.4) but with the roles of $\mathbf{v}$
and $\mathbf{w}$ interchanged, we obtain

$\\#\left(\,F_{n_{k_{l}}}\cap(\mathbf{w}+G)\,\right)\leq\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{v}-\mathbf{w})\right]\right)\,+\,\\#\left(F_{n_{k_{l}}}\cap(\mathbf{v}+G)\right).$

Now using (2.2) on the left and (2.5) on the right we get

$\frac{\beta}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k_{l}}})\leq\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)\,+\frac{1}{[\mathbb{Z}^{d}:G]}\,\\#(F_{n_{k_{l}}}),$

which can be rewritten as

$\frac{\beta-1}{[\mathbb{Z}^{d}:G]}\leq\frac{\\#\left(F_{n_{k_{l}}}\,\triangle\,\left[F_{n_{k_{l}}}+(\mathbf{w}-\mathbf{v})\right]\right)}{\\#(F_{n_{k_{l}}})}.$

This too contradicts $\\{F_{n}\\}$ being Følner. ∎

###  2.3. Odometers

We will especially consider a well-studied class of rank one systems called
odometers; background on odometers where the acting group is $\mathbb{Z}$ can
be found in [Do]. $\mathbb{Z}^{d}$-odometers can be defined in a variety of
ways; the approach we review here is a construction due to Cortez [Co], but
there is an equivalent characterization given by Giordano, Putnam and Skau
[GPS].

For Cortez’ construction, we begin by considering any decreasing sequence
$\mathfrak{G}=\\{G_{j}\\}_{j=1}^{\infty}$ of subgroups of $\mathbb{Z}^{d}$,
where each $G_{j}$ has finite index in $\mathbb{Z}^{d}$. For each $j\geq 1$,
let $q_{j}:\mathbb{Z}^{d}/G_{j+1}\to\mathbb{Z}^{d}/G_{j}$ be the quotient map.
Then define

$\displaystyle X_{\mathfrak{G}}$
$\displaystyle=\stackrel{{\scriptstyle[}}{{\longleftarrow}}]{}{\lim}(\mathbb{Z}^{d}/G_{j})$
$\displaystyle=\\{(\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3},...):\mathbf{x}_{j}\in\mathbb{Z}^{d}/G_{j}\textrm{
and }q_{j}(\mathbf{x}_{j+1})=\mathbf{x}_{j}\textrm{ for all }j\\}.$

$X_{\mathfrak{G}}$ is a topological group, where the topology is the product
of the discrete topologies on each $\mathbb{Z}^{d}/G_{j}$. Also, for each
$j\geq 1$ there is a natural coordinate map
$\pi_{j}:X_{\mathfrak{G}}\to\mathbb{Z}^{d}/G_{j}$, and we define
$\mathcal{X}_{\mathfrak{G}}$ on $X_{\mathfrak{G}}$ to be the $\sigma$-algebra
generated by the sets
$\\{\pi_{j}^{-1}(\mathbf{x}+G_{j}):j\in\\{1,2,3,...\\},\mathbf{x}\in\mathbb{Z}^{d}\\}$.
More importantly, there is a minimal action
$\sigma_{\mathfrak{G}}:\mathbb{Z}^{d}\curvearrowright X_{\mathfrak{G}}$ given
by

$\sigma_{\mathfrak{G}}^{\mathbf{v}}(\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3},...)=(\mathbf{x}_{1}+\mathbf{v}+G_{1},\mathbf{x}_{2}+\mathbf{v}+G_{2},\mathbf{x}_{3}+\mathbf{v}+G_{3},...).$

######  Definition 2.8.

A _$\mathbb{Z}^{d}$ -odometer_ is any $\mathbb{Z}^{d}$-action conjugate to one
of the form $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ described above, where
$\mathfrak{G}$ is some decreasing sequence of finite-index subgroups of
$\mathbb{Z}^{d}$.

We remark that $G$-odometers can be defined for any residually finite group
$G$ (not just $\mathbb{Z}^{d}$); see [CP].

The following results can be found in Theorem 2.2 of [GPS]:

######  Theorem 2.9 (Basic properties of odometers).

Let $\mathfrak{G}=\\{G_{1},G_{2},...\\}$ be a decreasing sequence of finite-
index subgroups of $\mathbb{Z}^{d}$.

  1. (1)

So long as $G_{j}\neq G_{j+1}$ for infinitely many $j$, $X_{\mathfrak{G}}$ is
a Cantor set, $(X_{\mathfrak{G}},\mathcal{X}_{\mathfrak{G}})$ is a standard
Lebesgue space and $\sigma_{\mathfrak{G}}$ is a minimal action of
$\mathbb{Z}^{d}$ by homeomorphisms on $X_{\mathfrak{G}}$;

  2. (2)

$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is free if and only if
$\stackrel{{\scriptstyle[}}{{j}}=1]{\infty}\bigcap G_{j}=\\{\mathbf{0}\\}$;

  3. (3)

$(X_{\mathfrak{G}},\mathcal{X}_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is
uniquely ergodic and its invariant Borel probability measure
$\mu_{\mathfrak{G}}$ is Haar measure on $X_{\mathfrak{G}}$. This measure
satisfies
$\mu_{\mathfrak{G}}\left(\pi_{j}^{-1}(\mathbf{x}+G_{j})\right)=[\mathbb{Z}^{d}:G_{j}]^{-1}$
for all $j\geq 1$ and all $\mathbf{x}\in\mathbb{Z}^{d}$.

From Theorem 2.9, we can classify $\mathbb{Z}^{d}$-odometers into two types.
First, if $X_{\mathfrak{G}}$ is infinite (assured in the setting of (1) in
Theorem 2.9), we will decree $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ to be
an infinite $\mathbb{Z}^{d}$-odometer. On the other hand, if the sequence
$\mathfrak{G}$ is eventually constant, we obtain an odometer with only
finitely many points in its phase space. This is equivalent to the following:

######  Definition 2.10.

Suppose $G$ is a finite-index subgroup of $\mathbb{Z}^{d}$. Let $\delta$ be
uniform counting measure on the finite set $\mathbb{Z}^{d}/G$, and for each
$\mathbf{v}\in\mathbb{Z}^{d}$, define
$\tau^{\mathbf{v}}:\mathbb{Z}^{d}/G\to\mathbb{Z}^{d}/G$ to be
$\tau^{\mathbf{v}}(\mathbf{g}+G)=\mathbf{g}+\mathbf{v}+G$. We call any
measure-preserving system conjugate to
$\tau:\mathbb{Z}^{d}\curvearrowright(\mathbb{Z}^{d}/G,\delta)$ a _finite
$\mathbb{Z}^{d}$-odometer_.

It is well known that odometers are rank one actions. That said, we will make
use of a specific rank one structure for a given $\mathbb{Z}^{d}$-odometer,
which we describe in the following result.

######  Theorem 2.11.

Let $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ be the $Z^{d}$-odometer given
by the decreasing sequence $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$ of
finite-index subgroups of $\mathbb{Z}^{d}$. Then:

  1. (1)

There exist upper-triangular matrices $A_{1},A_{2},A_{3},...\in
GL(d,\mathbb{Z})$ so that each $G_{j}$ is generated by the columns of $A_{j}$;
and

  2. (2)

$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is stacking rank one for a Følner
sequence of rectangular shapes, where the error sets of the associated towers
are empty.

###### Proof.

For each $j$, first let

$a_{j,1,1}=\min\\{n>0:n\mathbf{e}_{1}\in G_{j}\\}.$

and set $a_{j,k,1}=0$ for $k\in\\{2,...,d\\}$. (Here and throughout,
$\mathbf{e}_{j}$ is the $j^{th}$ standard basis vector of $\mathbb{R}^{d}$.)
Since $G_{j}$ has finite index, the integer $a_{j,1,1}$ exists.

Next, let

$a_{j,2,2}={\mbox{min}}\\{n>0:\exists\,i\textrm{ such that
}n\mathbf{e}_{2}+G_{j}=(i,0,...,0)+G_{j}\\}.$

and set $a_{j,1,2}$ to be any value of $i$ that works in the previous line,
meaning $a_{j,2,2}\mathbf{e}_{2}+G_{j}=(a_{j,1,2},0,...,0)+G_{j}$. Then, for
any $k\in\\{3,..,d\\}$, set $a_{j,k,2}=0$.

Continuing in this fashion, for each $l\in\\{1,...,d\\}$, define

$\displaystyle a_{j,l,l}$
$\displaystyle={\mbox{min}}\\{n>0:\exists\,i_{1},i_{2},...,i_{l-1}\textrm{
such that }$ $\displaystyle\qquad
n\mathbf{e}_{l}+G_{j}=(i_{1},i_{2},...,i_{l-1},0,...,0)+G_{j}\\};$

then choose $a_{j,1,l},a_{j,2,l},...a_{j,l-1,l}$ to be any values of
$i_{1},i_{2},...,i_{l-1}$ that work in the previous line, and finally, for any
$k\in\\{l+1,...,d\\}$, set $a_{j,k,l}=0$.

Having done this, we obtain upper-triangular $d\times d$ matrices
$A_{j}=\left(a_{j,k,l}\right)_{k,l}$ so that $G_{j}=A_{j}\mathbb{Z}^{d}$. The
columns of each of these matrices form a set of $d$ linearly independent
vectors that generate $G_{j}$.

Furthermore, by setting
$F_{j}=\\{0,1,...,a_{j,1,1}-1\\}\times\\{0,1,...,a_{j,2,2}-1\\}\times\cdots\times\\{0,1,...,a_{j,d,d}-1\\}$,
we see that the shape $F_{j}$ contains exactly one representative element from
each coset in $\mathbb{Z}^{d}/G_{j}$. So for each $j$, by setting
$B_{j}=\pi_{j}^{-1}(\mathbf{0}+G_{j})$, the sets
$\\{\pi_{j}^{-1}(\mathbf{v}+G_{j}):\mathbf{v}\in F_{j}\\}$ form a
$\sigma_{\mathfrak{G}}$-tower $\mathcal{T}_{j}$ of shape $F_{j}$ with base
$B_{j}$, where the error set is empty. The corresponding
$\sigma_{\mathfrak{G}}$-towers so obtained generate the $\sigma$-algebra
$\mathcal{X}_{\mathfrak{G}}$ by definition, meaning
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is rank one.

Finally, to show the towers are stacking, let $\mathbf{v}\in F_{j+1}$, and
then set $\widetilde{\mathbf{v}}\in F_{j}$ to be the unique representative in
$F_{j}$ of the coset $q_{j}(\mathbf{v}+G_{j+1})\in\mathbb{Z}^{d}/G_{j}$. The
level $\pi_{j+1}^{-1}(\mathbf{v}+G_{j+1})$ of $\mathcal{T}_{j+1}$ is a subset
of the level $\pi_{j}^{-1}(\widetilde{\mathbf{v}}+G_{j})$ of
$\mathcal{T}_{j}$, so the towers stack as wanted. ∎

###  2.4. Finite odometer factors

An eventual goal of this work is to determine criteria that ensure when a
Følner rank one $\mathbb{Z}^{d}$-system factors onto, or is conjugate to, a
$\mathbb{Z}^{d}$-odometer. Toward that end, it will be useful to consider when
such a system factors onto a finite odometer. For example, when the original
system is itself an odometer, given by sequence
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$, it is easy to obtain a factor map
from $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ onto each finite odometer
$(\mathbb{Z}^{d}/G_{j},\tau)$ simply by projecting $x\in X_{\mathfrak{G}}$
onto its $j^{th}$ component. In fact, we will see later that the set of $G$
for which an odometer factors onto $(\mathbb{Z}^{d}/G,\tau)$ determines the
odometer up to conjugacy. With this idea as motivation, we will examine the
set of all finite odometer factors of a general action as follows, where
$\mathcal{FF}$ refers to “finite factors”, a shortened version of “finite
odometer factors”:

######  Definition 2.12.

Given the measure-preserving system
$\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$, set

$\mathcal{FF}(\mathbf{T})=\\{G\leq\mathbb{Z}^{d}:[\mathbb{Z}^{d}:G]<\infty\textrm{
and }(X,\mu,\mathbf{T})\textrm{ factors onto
}(\mathbb{Z}^{d}/G,\delta,\tau)\\}.$

We begin with two general properties of the set $\mathcal{FF}(\mathbf{T})$:

######  Definition 2.13.

Let $\mathcal{F}$ be a set of subgroups of $\mathbb{Z}^{d}$. We say
$\mathcal{F}$ is _closed under supergroups_ if whenever $G\in\mathcal{F}$ and
$G\leq H\leq\mathbb{Z}^{d}$, then $H\in\mathcal{F}$. We say $\mathcal{F}$ is
_closed under intersections_ if $G\in\mathcal{F}$ and $H\in\mathcal{F}$ imply
$G\cap H$ must also be in $\mathcal{F}$.

######  Lemma 2.14.

For any $\mathbb{Z}^{d}$-action $(X,\mu,\mathbf{T})$,
$\mathcal{FF}(\mathbf{T})$ is closed under supergroups and closed under
intersections.

###### Proof.

We begin by showing $\mathcal{FF}(\mathbf{T})$ is closed under supergroups.
Let $G\in\mathcal{FF}(\mathbf{T})$ and suppose $G\leq H$. Thus there is a
natural quotient map
$q:(\mathbb{Z}^{d}/G,\delta_{G},\tau_{G})\to(\mathbb{Z}^{d}/H,\delta_{H},\tau_{H})$,
and therefore any factor map
$\pi:(X,\mu,\mathbf{T})\to(\mathbb{Z}^{d}/G,\delta_{G},\tau_{G})$ yields a
factor map
$q\circ\pi:(X,\mu,\mathbf{T})\to(\mathbb{Z}^{d}/H,\delta_{H},\tau_{H})$,
meaning $H\in\mathcal{FF}(\mathbf{T})$ as wanted.

We now show $\mathcal{FF}(\mathbf{T})$ is closed under intersections. Let
$G,H\in\mathcal{FF}(\mathbf{T})$, so that there are factor maps
$\pi_{G}:(X,\mu,\mathbf{T})\to(\mathbb{Z}^{d}/G,\delta_{G},\tau_{G})$ and
$\pi_{H}:(X,\mu,\mathbf{T})\to(\mathbb{Z}^{d}/H,\delta_{H},\tau_{H})$. Now
consider the partition $\mathcal{P}$ of $X$ coming from the values of
$\pi_{G}$ and $\pi_{H}$. Note that for each $\mathbf{v}\in\mathbb{Z}^{d}$,
$\mathbf{T}^{\mathbf{v}}$ permutes the atoms of this partition.

Now, given two atoms $E$ and $\widetilde{E}$ of $\mathcal{P}$, we say
$E\sim\widetilde{E}$ if there is $\mathbf{v}\in\mathbb{Z}^{d}$ so that
$\mathbf{T}^{\mathbf{v}}(E)=\widetilde{E}$; this gives an equivalence relation
on $\mathcal{P}$ where we denote the equivalence class of atom $E$ by $[E]$.
For each equivalence class under the equivalence relation $\sim$, take an atom
$E$ in that class; notice that
$\bigcup_{\mathbf{v}\in\mathbb{Z}^{d}}\mathbf{T}^{\mathbf{v}}(E)=[E]$. At the
same time, $\mathbf{T}^{\mathbf{v}}(E)=E$ if and only if $\mathbf{v}\in G\cap
H$, so each $x\in[E]$ can be written as
$x=\mathbf{T}^{\mathbf{v}}(x^{\prime})$ for $x^{\prime}\in E$ where
$\mathbf{v}$ is unique mod $(G\cap H)$. This gives us a map $\pi_{G\cap
H}:[E]\to\mathbb{Z}^{d}/(G\cap H)$ defined by

$\pi_{G\cap H}(x)=\mathbf{v}+(G\cap H)\Leftrightarrow
x=\mathbf{T}^{\mathbf{v}}(x^{\prime})\textrm{ for }x^{\prime}\in E.$

As each $x\in X$ belongs to exactly one class $[E]$, we can extend $\pi_{G\cap
H}$ to be a map from $X$ to $\mathbb{Z}^{d}/(G\cap H)$. The resulting map
$\pi_{G\cap H}$ is then a factor map from $(X,\mu,\mathbf{T})$ to
$(\mathbb{Z}^{d}/(G\cap H),\delta_{G\cap H},\tau_{G\cap H})$, meaning $G\cap
H\in\mathcal{FF}(\mathbf{T})$ as desired. ∎

##  3\. Finite odometer factors of Følner rank one actions

As was mentioned before, the Cortez definition of an odometer makes it is easy
to see that there is a factor map $\pi_{j}$ from the odometer
$X_{\mathfrak{G}}$ onto the finite odometer $\mathbb{Z}^{d}/G_{j}$. Thus in
order to explore the kinds of (general) odometers onto which a given
$\mathbb{Z}^{d}$-action can factor, we will begin by studying the types of
finite odometers onto which a $\mathbb{Z}^{d}$-action can factor.

######  Theorem 3.1.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ be Følner rank one and
let $G$ be a finite index subgroup of $\mathbb{Z}^{d}$. Then the following are
equivalent:

  1. (1)

$G\in\mathcal{FF}(\mathbf{T})$.

  2. (2)

“$\textrm{Eventually, most descendants are congruent mod }G$.” By this, we
mean that for every $\epsilon>0$, there exists $N\in\mathbb{N}$ so that for
all $n\geq m\geq N$, there is $\mathbf{g}\in\mathbb{Z}^{d}$ such that

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G\\})}{\\#(I_{m,n})}<\epsilon.$

###### Proof.

(1) $\Rightarrow$ (2): We begin with supposing that $\pi:X\to\mathbb{Z}^{d}/G$
is a factor map and recalling that since $X$ is Følner rank one, there exists
a sequence of towers $\\{\mathcal{T}_{n}\\}$ of shape $\\{F_{n}\\}$ whose
levels generate the $\sigma$-algebra. Fix $\epsilon$ such that $0<\epsilon<1$;
let $\tilde{\epsilon}=\frac{\epsilon}{2}$ and
$\eta=\frac{\tilde{\epsilon}}{2}[\mathbb{Z}^{d}:G]^{-1}$.

We will show that (2) holds with the following three steps:

  1. _(i)_

We begin by showing that there is an $N\in\mathbb{N}$ such that for every
$m\geq N$, each level $L$ of $\mathcal{T}_{m}$ has associated to it some
$\gamma+G$ such that

$\frac{\mu\left(\,L-\pi^{-1}(\gamma+G)\,\right)}{\mu(L)}<\tilde{\epsilon}.$

  2. _(ii)_

We next show that having the coset $\gamma+G$ associated to level $B_{m}$ as
described in step (i) means the same coset is associated to most
$T^{\mathbf{v}}(B_{n})$, where $\mathbf{v}\in I_{m,n}$.

  3. _(iii)_

Finally, we show that when two levels $T^{\mathbf{v}_{1}}(B_{n})$ and
$T^{\mathbf{v}_{2}}(B_{n})$ of $\mathcal{T}_{n}$ are associated to the same
coset, then $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are congruent mod $G$.

Step (i): Consider the sets
$\\{\pi^{-1}(\gamma+G):\gamma+G\in\mathbb{Z}^{d}/G\\}$. These partition $X$
into sets of equal measure $[\mathbb{Z}^{d}:G]^{-1}$. Since the levels of the
towers generate the $\sigma$-algebra, there exists $N$ such that whenever
$m\geq N$, each $\pi^{-1}(\gamma+G)$ is well-approximated by some union of
levels in $\mathcal{T}_{m}$. In other words, there is a union of levels, call
it $\Lambda_{m,\gamma}$, such that

(3.1) $\mu(\Lambda_{m,\gamma}\,\,\triangle\,\,\pi^{-1}(\gamma+G)\,)<\eta.$

If we fix $\gamma$ for a moment and let $L$ be one of the levels in
$\Lambda_{m,\gamma}$, we can use the fact that every other level
$\widetilde{L}\notin\Lambda_{m,\gamma}$ equals $\mathbf{T}^{\mathbf{v}}(L)$
for some $\mathbf{v}$ to argue that
$\widetilde{L}\in\Lambda_{m,\gamma+\mathbf{v}}$. In other words, we can assume
that every level in $\mathcal{T}_{m}$ lies in $\Lambda_{m,\tilde{\gamma}}$ for
some $\tilde{\gamma}$. We will use this to justify the following claim:

For every level $L\in\mathcal{T}_{m}$, there exists
$\gamma+G\in\mathbb{Z}^{d}/G$ such that

$\frac{\mu\left(L\,\,-\,\,\pi^{-1}(\gamma+G)\right)}{\mu(L)}<\tilde{\epsilon}.$

We prove this claim by contradiction: suppose there exists a level $L$ such
that for all $\gamma+G$,

$\frac{\mu\left(L\,\,-\,\,\pi^{-1}(\gamma+G)\right)}{\mu(L)}\geq\tilde{\epsilon}.$

But then for all $\mathbf{v}\in\mathbb{Z}^{d}$, we also have

$\frac{\mu\left(\mathbf{T}^{\mathbf{v}}(L)\,\,-\,\,\pi^{-1}(\mathbf{v}+\gamma+G)\right)}{\mu(L)}\geq\tilde{\epsilon}.$

which we can rewrite as saying that for every every
$\mathbf{v}\in\mathbb{Z}^{d}$ and $\tilde{\gamma}+G\in\mathbb{Z}^{d}/G$, we
have

$\frac{\mu\left(\mathbf{T}^{\mathbf{v}}(L)-\pi^{-1}(\tilde{\gamma}+G)\right)}{\mu(\mathbf{T}^{\mathbf{v}}(L))}\geq\tilde{\epsilon}.$

Let $\gamma$ be such that $L\subseteq\Lambda_{m,\gamma}$. Using the above, we
can say that for this particular $\gamma$ and for every
$\widetilde{L}\in\Lambda_{m,\gamma}$,

(3.2)
$\mu(\widetilde{L}-\pi^{-1}({\gamma}+G))\geq\tilde{\epsilon}\,\mu(\widetilde{L}).$

We thus have the following chain of inequalities, which ultimately yield the
contradiction $\eta>\eta$:

$\displaystyle\eta$
$\displaystyle>\mu(\Lambda_{m,\gamma}\,\triangle\,\pi^{-1}(\gamma+G)\,)\qquad\textrm{(from
(\ref{wellapprox}) above)}$
$\displaystyle\geq\mu(\Lambda_{m,\gamma}-\pi^{-1}(\gamma+G)\,)$
$\displaystyle=\sum_{\widetilde{L}\in\Lambda_{m,\gamma}}\mu(\widetilde{L}-\pi^{-1}({\gamma}+G))$
$\displaystyle\geq\sum_{\widetilde{L}\in\Lambda_{m,\gamma}}\tilde{\epsilon}\,\mu(\widetilde{L})\qquad\textrm{(from
(\ref{wellapprox2}) above)}$
$\displaystyle=\tilde{\epsilon}\,\mu(\Lambda_{m,\gamma})$
$\displaystyle\geq\tilde{\epsilon}\,\mu\left(\Lambda_{m,\gamma}\cap\pi^{-1}(\gamma+G)\right)$
$\displaystyle=\tilde{\epsilon}\,\left[\mu(\pi^{-1}(\gamma+G))-\mu(\pi^{-1}\left(\gamma+G)-\Lambda_{m,\gamma}\right)\right]$
$\displaystyle=\tilde{\epsilon}\,\left[[\mathbb{Z}^{d}:G]^{-1}-\mu(\pi^{-1}(\gamma+G)-\Lambda_{m,\gamma})\right]$
$\displaystyle\geq\tilde{\epsilon}\,\left[[\mathbb{Z}^{d}:G]^{-1}-\mu(\pi^{-1}(\gamma+G)\,\triangle\,\Lambda_{m,\gamma})\right]$
$\displaystyle\geq\tilde{\epsilon}\,\left[[\mathbb{Z}^{d}:G]^{-1}-\eta\right]\qquad\textrm{(again
using (\ref{wellapprox}))}$
$\displaystyle=\tilde{\epsilon}\,\left[[\mathbb{Z}^{d}:G]^{-1}-\frac{1}{2}\tilde{\epsilon}\,[\mathbb{Z}^{d}:G]^{-1}\right]\qquad\textrm{(by
the definition of $\eta$)}$
$\displaystyle>\tilde{\epsilon}\,\left[[\mathbb{Z}^{d}:G]^{-1}-\frac{1}{2}[\mathbb{Z}^{d}:G]^{-1}\right]\qquad\textrm{(since
$\tilde{\epsilon}<\epsilon<1$)}$
$\displaystyle=\frac{1}{2}\tilde{\epsilon}\,[\mathbb{Z}^{d}:G]^{-1}$
$\displaystyle=\eta.$

This proves the claim, and finishes the first step of the proof.

Step (ii): Note that the first step says that for the base level $B_{m}$ of
$\mathcal{T}_{m}$, where $m\geq N$, there is some coset, which we will denote
by $\widetilde{\mathbf{g}}+G$, such that

(3.3)
$\frac{\mu\left(B_{m}\,\,-\,\,\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)}{\mu(B_{m})}<\tilde{\epsilon}.$

We know that, for any $n\geq m$, $B_{m}$ is divided into subsets that become
various levels in $\mathcal{T}_{n}$. Those levels are the sets
$\mathbf{T}^{\mathbf{v}}(B_{n})$, where $\mathbf{v}\in I_{m,n}$. We want to
say that since (3.3) holds for $B_{m}$, then the same inequality with $B_{m}$
replaced by $\mathbf{T}^{\mathbf{v}}(B_{n})$ holds for most $\mathbf{v}\in
I_{m,n}$. To this end, define

$I_{\textrm{bad}}=\\{\mathbf{v}\in
I_{m,n}:\frac{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)}{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)}\geq\tilde{\epsilon}\\}.$

We will complete this step in the proof by showing that

$\frac{\\#(I_{\textrm{bad}})}{\\#(I_{m,n})}<2\tilde{\epsilon}.$

To do this, first note that $\stackbin[\mathbf{v}\in
I_{\textrm{bad}}]{}{\bigcup}\mathbf{T}^{\mathbf{v}}(B_{n})\subseteq B_{m}$ and
thus

$\bigcup_{\mathbf{v}\in
I_{\textrm{bad}}}\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)\,\subseteq\,B_{m}-\pi^{-1}(\widetilde{\mathbf{g}}+G)$

which implies

$\mu\left(\,\bigcup_{\mathbf{v}\in
I_{\textrm{bad}}}\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)\,\right)\,\leq\,\mu\left(B_{m}-\pi^{-1}(\widetilde{\mathbf{g}}+G)\,\right).$

But although $\mathbf{v}\in I_{\textrm{bad}}$ means that
$\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)/\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)\geq\tilde{\epsilon}$,
Step (i) assured us there is some $\gamma+G\in\mathbb{Z}^{d}/G$ with

$\frac{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\gamma+G)\right)}{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)}<\tilde{\epsilon}.$

For this $\gamma+G$, divide $\mathbf{T}^{\mathbf{v}}(B_{n})$ into two disjoint
subsets: the part in $\pi^{-1}(\gamma+G)$ and the part not in
$\pi^{-1}(\gamma+G)$. Since

$\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\gamma+G)\right)<\tilde{\epsilon}\,\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right),$

we have
$\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\cap\pi^{-1}(\gamma+G)\right)>(1-\tilde{\epsilon})\,\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)$
and we can then say

$\displaystyle\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)$
$\displaystyle\geq\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\cap\pi^{-1}(\gamma+G)\right)$
$\displaystyle>(1-\tilde{\epsilon})\,\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)$
$\displaystyle=(1-\tilde{\epsilon})\,\mu(B_{n}).$

Therefore

$\displaystyle\mu\left(\bigcup_{\mathbf{v}\in
I_{\textrm{bad}}}\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)\right)$
$\displaystyle=\sum_{\mathbf{v}\in
I_{\textrm{bad}}}\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)$
$\displaystyle>\sum_{\mathbf{v}\in
I_{\textrm{bad}}}(1-\tilde{\epsilon})\,\mu(B_{n})$
$\displaystyle=\\#(I_{\textrm{bad}})\cdot(1-\tilde{\epsilon})\,\mu(B_{n})$
$\displaystyle=(1-\tilde{\epsilon})\cdot\frac{\\#(I_{\textrm{bad}})}{\\#(I_{m,n})}\,\cdot\,\\#(I_{m,n})\,\mu(B_{n})$
$\displaystyle>\frac{\\#(I_{\textrm{bad}})}{\\#(I_{m,n})}(1-\tilde{\epsilon})\mu(B_{m}).$

Now suppose the claim is false, meaning
$\frac{\\#(I_{\textrm{bad}})}{\\#(I_{m,n})}\geq 2\tilde{\epsilon}$. This would
mean

$\mu\left(B_{m}-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)\geq\mu\left(\bigcup_{\mathbf{v}\in
I_{\textrm{bad}}}\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)\right)>\,2\tilde{\epsilon}\,(1-\tilde{\epsilon})\,\mu(B_{m}).$

But $\tilde{\epsilon}<\frac{1}{2}$, so $1-\tilde{\epsilon}>\frac{1}{2}$, and
we thus have

$\mu\left(B_{m}-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)>\tilde{\epsilon}\,\mu(B_{m}),$

contradicting (3.3).

Step (iii): Let $\mathbf{v}\in I_{m,n}-I_{\textrm{bad}}$ and let
$\mathbf{g}\in\mathbb{Z}^{d}$ be such that $\mathbf{v}\equiv\mathbf{g}\mod G$.
Take any other $\mathbf{i}\in I_{m,n}-I_{\textrm{bad}}$. Since we have both

$\frac{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)}{\mu\left(\mathbf{T}^{\mathbf{v}}(B_{n})\right)}<\tilde{\epsilon}\,\,{\textrm{
and
}}\frac{\mu\left(\mathbf{T}^{\mathbf{i}}(B_{n})-\pi^{-1}(\widetilde{\mathbf{g}}+G)\right)}{\mu\left(\mathbf{T}^{\mathbf{i}}(B_{n})\right)}<\tilde{\epsilon},$

we can find
$x\in\mathbf{T}^{\mathbf{v}}(B_{n})\cap\pi^{-1}(\tilde{\mathbf{g}}+G)$ such
that
$\mathbf{T}^{\mathbf{i}-\mathbf{v}}(x)\in\mathbf{T}^{\mathbf{i}}(B_{n})\cap\pi^{-1}(\tilde{\mathbf{g}}+G)$.
Thus $\pi(x)\in\tilde{\mathbf{g}}+G$ and
$\pi(\mathbf{T}^{\mathbf{i}-\mathbf{v}}x)\in\tilde{\mathbf{g}}+G$. But

$\pi(\mathbf{T}^{\mathbf{i}-\mathbf{v}}x)=\mathbf{i}-\mathbf{v}+\pi(x)=\mathbf{i}-\mathbf{v}+\tilde{\mathbf{g}}+G.$

This implies that $\mathbf{i}-\mathbf{v}\in G$, or
$\mathbf{i}\equiv\mathbf{v}$ mod $G$. So $\mathbf{i}\equiv\mathbf{g}$ mod $G$
as well. In other words, all $\mathbf{i}\in I_{m,n}-I_{\textrm{bad}}$ are
congruent to $\mathbf{g}$ mod $G$ and so

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G\\})}{\\#(I_{m,n})}\leq\frac{\\#(I_{\textrm{bad}})}{\\#(I_{m,n})}<2\tilde{\epsilon}<\epsilon,$

as wanted.

(2) $\Rightarrow$ (1): We are now assuming that for every $\epsilon$, there
exists $N\in\mathbb{N}$ such that for all $n\geq m\geq N$, there is
$\mathbf{g}\in\mathbb{Z}^{d}$ such that

(3.4) $\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G\\})}{\\#(I_{m,n})}<\epsilon.$

We want to show that $G\in\mathcal{FF}(\mathbf{T})$, i.e. that
$(X,\mu,\mathbf{T})$ factors onto $(\mathbb{Z}^{d}/G,\delta,\tau)$.

For each $i\in\\{1,2,3,...\\}$, let $\epsilon_{i}=\frac{1}{2^{i+1}}$ and apply
the assumption to obtain the corresponding $N_{i}$ and $\mathbf{g}_{i}$.
Without loss of generality the sequence $\\{N_{i}\\}$ can be taken to be
increasing, and can be chosen large enough so that the measure of stage
$N_{i}$-tower $\mathcal{T}_{N_{i}}$ is at least $1-\frac{1}{2^{i+1}}$. Also,
since $\\{F_{n}\\}$ is a Følner sequence, from Lemma 2.7, we can assume that
for all $i$, $F_{N_{i}}$ intersects every coset in $\mathbb{Z}^{d}/G$.

In order to find the wanted factor map from $(X,\mu,\mathbf{T})$ onto
$(\mathbb{Z}^{d}/G,\delta,\tau)$, we begin by first defining what might be the
most obvious map from $\mathcal{T}_{N_{i}}$ onto $\mathbb{Z}^{d}/G$. As we are
assuming the descendents of the base $B_{N_{i}}$ are mostly equal mod
$\mathbf{g}_{i}+G$, we then consider the set of points where the map on
$\mathcal{T}_{N_{i}+1}$ is not simply the map on $\mathcal{T}_{N_{i}}$
adjusted by this $\mathbf{g}_{i}+G$, and show this set has small measure. This
will then let us define a factor map on all of $X$ by using the sum of these
adjustments. Now for the details.

We begin by defining, for each $i$, a function
$\phi_{i}:\mathcal{T}_{N_{i}}\to\mathbb{Z}^{d}/G$ by setting
$\phi_{i}(x)=\mathbf{v}+G$ whenever $x\in\mathbf{T}^{\mathbf{v}}(B_{N_{i}})$.
Observe that since $F_{N_{i}}$ intersects every coset in $\mathbb{Z}^{d}/G$,
each $\phi_{i}$ is onto. Now set

$D_{i}=\\{x\in\mathcal{T}_{N_{i}}:\phi_{i+1}(x)\neq\phi_{i}(x)+(\mathbf{g}_{i}+G)\\}.$

We want to show that this has small measure.

_Claim:_ $\mu(D_{i})<\epsilon_{i}$.

_Proof of claim:_ Let $F_{N_{i}}$ be the shape corresponding to
$\mathcal{T}_{N_{i}}$. We know most $\mathbf{w}\in I_{N_{i},N_{i+1}}$ are such
that $\mathbf{w}\equiv\mathbf{g}_{i}$ mod $G$. For such $\mathbf{w}$, consider
the points at these levels and call their union $B_{\textrm{good}}$. That is,
set

$B_{\textrm{good}}=\\{x\in\mathbf{T}^{\mathbf{w}}(B_{N_{i+1}}):\mathbf{w}\in
I_{N_{i},N_{i+1}}\textrm{ and }\mathbf{w}\equiv\mathbf{g}_{i}\mod G\\}.$

Note $B_{N_{i}}=B_{\textrm{good}}\,\sqcup\,(B_{N_{i}}-B_{\textrm{good}})$, and
observe that for any $x\in\stackrel{{\scriptstyle[}}{{\mathbf{v}}}\in
F_{N_{i}}]{}{\bigsqcup}\mathbf{T}^{\mathbf{v}}(B_{\textrm{good}})$,
$x\in\mathbf{T}^{\mathbf{v}}(\mathbf{T}^{\mathbf{w}}(B_{N_{i+1}}))=\mathbf{T}^{\mathbf{v}+\mathbf{w}}(B_{N_{i+1}})$.
Thus
$\phi_{i+1}(x)=\mathbf{v}+\mathbf{w}+G=\mathbf{v}+\mathbf{g}_{i}+G=\phi_{i}(x)+\mathbf{g}_{i}+G$,
and so $x\notin D_{i}$. In other words, the set $D_{i}$ does not include any
points from $\stackrel{{\scriptstyle[}}{{\mathbf{v}}}\in
F_{N_{i}}]{}{\bigsqcup}\mathbf{T}^{\mathbf{v}}(B_{\textrm{good}})$. We can
thus say,

$\displaystyle\mu(D_{i})$ $\displaystyle\leq\mu\left(\bigsqcup_{\mathbf{v}\in
F_{N_{i}}}\mathbf{T}^{\mathbf{v}}(B_{N_{i}}-B_{\textrm{good}})\right)$
$\displaystyle=\sum_{\mathbf{v}\in
F_{N_{i}}}\mu\left(\mathbf{T}^{\mathbf{v}}(B_{N_{i}}-B_{\textrm{good}})\right)$
$\displaystyle=\sum_{\mathbf{v}\in
F_{N_{i}}}\mu\left(B_{N_{i}}-B_{\textrm{good}}\right).$

Recall that $B_{N_{i}}$ is divided into subsets and placed at levels
$I_{N_{i},N_{i+1}}$ in $\mathcal{T}_{N_{i+1}}$. Some of these levels are at
heights congruent to $\mathbf{g}_{i}$ mod $G$, and those would not be included
in the set $B_{N_{i}}-B_{\textrm{good}}$. The proportion of the remaining
levels in $I_{N_{i},N_{i+1}}$ is bounded by (3.4), and thus we can say

$\mu\left(B_{N_{i}}-B_{\textrm{good}}\right)\leq\mu(B_{N_{i+1}})\,\epsilon_{i}\,\\#(I_{N_{i},N_{i+1}})=\mu(B_{N_{i}})\,\epsilon_{i}.$

Putting this together with the above we see that

$\mu(D_{i})\leq\\#(F_{N_{i}})\mu(B_{N_{i}})\,\epsilon_{i}\leq\mu(X)\,\epsilon_{i}=\epsilon_{i},$

completing the proof of the claim.

We next adjust the functions $\phi_{i}$ by the amounts
$\mathbf{g}_{1},\mathbf{g}_{2},...\mathbf{g}_{i-1}$. More specifically, set
$\gamma_{i}\in\mathbb{Z}^{d}/G$ to be
$\gamma_{i}=\stackrel{{\scriptstyle[}}{{j}}=1]{i}{\sum}(\mathbf{g}_{j}+G)$.
Then define, for each $i$, a function
$\pi_{i}:\mathcal{T}_{N_{i}}\to\mathbb{Z}^{2}/G$ by setting

$\pi_{i}(x)=\phi_{i}(x)-\gamma_{i-1}.$

Since each $\phi_{i}$ is surjective, so is each $\pi_{i}$. Now observe

$\displaystyle x\in D_{i}$
$\displaystyle\Leftrightarrow\phi_{i+1}(x)\neq\phi_{i}(x)+(\mathbf{g}_{i}+G)$
$\displaystyle\Leftrightarrow\phi_{i+1}(x)-(\mathbf{g}_{i}+G)\neq\phi_{i}(x)$
$\displaystyle\Leftrightarrow\phi_{i+1}(x)+\gamma_{i-1}-\gamma_{i}\neq\phi_{i}(x)$
$\displaystyle\Leftrightarrow\phi_{i+1}(x)-\gamma_{i}\neq\phi_{i}(x)-\gamma_{i-1}$
$\displaystyle\Leftrightarrow\pi_{i+1}(x)\neq\pi_{i}(x).$

So

(3.5)
$\mu\left(\\{x\in\mathcal{T}_{N_{i}}:\pi_{i+1}(x)\neq\pi_{i}(x)\\}\right)=\mu(D_{i})<\epsilon_{i}=\frac{1}{2^{i+1}}.$

Finally, we can define the factor map from $X$ to $\mathbb{Z}^{d}/G$. Note
that the Borel-Cantelli Lemma says that almost every $x\in X$ belongs to only
finitely many of the sets
$\\{x\in\mathcal{T}_{N_{i}}:\pi_{i+1}(x)\neq\pi_{i}(x)\\}$. In other words,
for almost every $x\in X$ the sequence $\\{\pi_{i}(x)\\}$ eventually
stabilizes. We can thus define $\pi:X\to\mathbb{Z}^{d}/G$ by
$\pi(x)=\stackrel{{\scriptstyle[}}{{i}}\to\infty]{}{\lim}\pi_{i}(x)$.

It only remains to show that this $\pi$ is indeed a factor map. To do this,
fix $x\in X$ and $\mathbf{v}\in\mathbb{Z}^{d}$. Choose $i$ large enough so
that $x$ and $\mathbf{T}^{\mathbf{v}}(x)$ both belong to
$\mathcal{T}_{N_{i}}$, with $x$ at level $\mathbf{w}$ and
$\mathbf{T}^{\mathbf{v}}(x)$ at level $\mathbf{w}+\mathbf{v}$, and that
$\pi_{i}(x)=\pi(x)$ and
$\pi_{i}(\mathbf{T}^{\mathbf{v}}(x))=\pi(\mathbf{T}^{\mathbf{v}}(x))$. Then

$\displaystyle\pi(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=\pi_{i}(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=\phi_{i}(\mathbf{T}^{\mathbf{v}}(x))-\gamma_{i-1}$
$\displaystyle=(\mathbf{v}+\mathbf{w}+G)-\gamma_{i-1}$
$\displaystyle=[\,(\mathbf{w}+G)-\gamma_{i-1}\,]+(\mathbf{v}+G)$
$\displaystyle=[\,\phi_{i}(x)-\gamma_{i-1}\,]+(\mathbf{v}+G)$
$\displaystyle=\pi_{i}(x)+(\mathbf{v}+G)$
$\displaystyle=\pi(x)+(\mathbf{v}+G)$
$\displaystyle=\tau^{\mathbf{v}}(\pi(x))$

so $\pi$ intertwines $\mathbf{T}$ and $\tau$ as desired. ∎

We next turn to an example which shows that if one replaces “Følner rank one”
with “stacking rank one” (or just “rank one”), the $(2)\Rightarrow(1)$
direction of Theorem 3.1 fails.

######  Theorem 3.2.

There exists a $\mathbf{T}:\mathbb{Z}^{2}\curvearrowright(X,\mu)$ which is
stacking rank one and a finite-index subgroup $G\leq\mathbb{Z}^{2}$ so that
for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n\geq
m\geq N$, there is $\mathbf{g}\in\mathbb{Z}^{2}$ such that

(3.6) $\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G\\})}{\\#(I_{m,n})}<\epsilon,$

but $(X,\mathcal{X},\mu,\mathbf{T})$ does not factor onto
$(\mathbb{Z}^{2}/G,\delta,\tau)$.

###### Proof.

Our system $(X,\mathbf{T})$ will be the $\mathbb{Z}^{2}$-odometer
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ associated to the sequence of
subgroups

$\mathfrak{G}=\\{2\mathbb{Z}\times\mathbb{Z},4\mathbb{Z}\times\mathbb{Z},...,2^{n}\mathbb{Z}\times\mathbb{Z},...\\}.$

In particular, $\mathbf{T}^{(0,1)}=\sigma_{\mathfrak{G}}^{(0,1)}$ is the
identity map for this odometer. Set $G=2\mathbb{Z}\times 2\mathbb{Z}$.

Given any $\epsilon>0$, choose $N=0$, and for any $n\geq m\geq 0$, let
$\mathbf{g}=\mathbf{0}$. Note that for any $n\geq m\geq 0$,
$I_{m,n}=\\{(2^{m}c,0):0\leq c<2^{n-m}\\}$ so

$\\#\left(\mathbf{i}\in I_{m,n}:i\not\equiv\mathbf{0}\mod
G\right)=\\#(\emptyset)=0,$

ensuring (3.6).

Finally, suppose $\pi:X_{\mathfrak{G}}\to\mathbb{Z}^{2}/G$. Then, for any
$x\in X_{\mathfrak{G}}$,

$\displaystyle\pi(\mathbf{T}^{(0,1)}(x))=\pi(x)\neq\pi(x)+(0,1)=\tau^{(0,1)}(\pi(x))$

so $\pi$ cannot intertwine $\mathbf{T}$ and $\tau$. Thus no factor map from
$(X,\mathbf{T})$ to $(\mathbb{Z}^{2}/G,\tau)$ exists.

∎

In [FGHSW] the authors note that for rank one $\mathbb{Z}$-actions, the action
has a nontrivial finite factor if and only if the action is not totally
ergodic. To put this in the terminology of Theorem 3.1, we would say that a
$\mathbb{Z}$-action is not totally ergodic if and only if there is a
nontrivial finite index $G<\mathbb{Z}$ for which
$G\in\mathcal{FF}(\mathbf{T})$. Since this is essentially a consequence of the
fact that $\mathbb{Z}$ has no nontrivial subgroups of infinite index, it is
not a surprise to find that a similar characterization does not hold for
$\mathbb{Z}^{d}$-actions. We make this clear with the next example.

######  Theorem 3.3.

There is a Følner rank one action
$\mathbf{T}:\mathbb{Z}^{2}\curvearrowright(X,\mu)$ which is not totally
ergodic, but does not factor onto any nontrivial finite odometer.

###### Proof.

To construct $\mathbf{T}$, start with Chacon’s [C2] map
$T:\mathbb{Z}\curvearrowright(Y,\nu)$ and let $\mathbf{T}$ be the product of
$T$ with itself, i.e. $\mathbf{T}:\mathbb{Z}^{2}\curvearrowright(Y\times
Y,\nu\times\nu)$ is
$\mathbf{T}^{\mathbf{v}}(y,y)=\mathbf{T}^{(v_{1},v_{2})}(y,y)=(T^{v_{1}}(y),T^{v_{2}}(y))$.

$(X,\mathbf{T})$ is Følner rank one, with square tower shape
$\\{0,...,h_{m}-1\\}^{2}$ where $h_{1}=1$ and $h_{m+1}=3h_{m}+1$ for all
$m\geq 1$. It is clearly not totally ergodic; indeed, for any measurable set
$A\subseteq Y$, $A\times Y$ is $\mathbf{T}^{\mathbf{e}_{2}}$-invariant.

Now suppose that $G\in\mathcal{FF}(\mathbf{T})$. We will derive a
contradiction by applying Theorem 3.1. To do this, let $\epsilon<\frac{1}{3}$
and observe that for each $m$ and $n$ we can write $I_{m,n}=S_{m,n}\times
S_{m,n}$ where $S_{m,m+1}=\\{0,h_{m},2h_{m}+1\\}$ and for all $n\geq m+1$,

$S_{m,n+1}=S_{m,n}\sqcup\left(S_{m,n}+h_{n}\right)\sqcup\left(S_{m,n}+2h_{n}+1\right).$

Note that $\frac{2}{3}$ of the $3^{n}$ elements of $S_{m,m+n}$ come in pairs
separated by $h_{m}$, so $\frac{2}{3}$ of the $9^{n}$ elements of $I_{m,n}$
come in pairs separated by $(h_{m},0)$. Therefore
$(h_{m},0)\equiv\mathbf{0}\mod G$, as otherwise, condition (2) of Theorem 3.1
would be violated.

Consequently, $(h_{m},0)\in G$ for all $m$, and therefore
$(h_{m+1},0)=(3h_{m}+1,0)\in G$ as well. Therefore
$(1,0)=(h_{m+1},0)-3(h_{m},0)\in G$ as well. A similar argument shows
$(0,1)\in G$, so $G=\mathbb{Z}^{2}$. This means $\mathbf{T}$ has no
(nontrivial) finite factors. ∎

##  4\. Rank one $\mathbb{Z}^{d}$ systems that factor onto an odometer

The last section gives us a characterization of when a Folner rank one
$\mathbb{Z}^{d}$-action factors onto a finite odometer. We now make use of
this to describe when such a system factors onto an arbitrary odometer.

######  Theorem 4.1.

Let $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$ be a decreasing sequence of
finite-index subgroups of $\mathbb{Z}^{d}$. Let
$\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$. $(X,\mu,\mathbf{T})$
factors onto the odometer
$(X_{\mathfrak{G}},\mu_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ if and only if
for every $k$, $(X,\mu,\mathbf{T})$ factors onto the finite action
$(\mathbb{Z}^{d}/G_{k},\delta_{k},\tau_{k})$.

###### Proof.

$(\Rightarrow)$ Let
$\pi:(X,\mathbf{T})\to(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ be a factor
map. Then by composing $\pi$ with each coordinate map
$\pi_{k}:X_{\mathfrak{G}}\to\mathbb{Z}^{d}/G_{k}$, we obtain factor maps
$\pi_{k}\circ\pi:(X,\mathbf{T})\to(\mathbb{Z}^{d}/G_{k},\tau_{k})$ as wanted.

$(\Leftarrow)$ Suppose that for each $k$, there is a factor map
$\pi_{k}:(X,\mathbf{T})\to(\mathbb{Z}^{d}/G_{k},\tau_{k})$. Because
$X_{\mathfrak{G}}$ is the inverse limit of the $\mathbb{Z}^{d}/G_{k}$, if the
$\pi_{k}$ were commensurate with the quotient maps
$q_{k}:\mathbb{Z}^{d}/G_{k+1}\to\mathbb{Z}^{d}/G_{k}$ described in Section 2.3
(so $q_{k}\circ\pi_{k+1}=\pi_{k}$ for all $k\geq 1$), then we could easily use
these to define a factor map from $X$ to $X_{\mathfrak{G}}$. But our given
factor maps have no particular relationship to one another and thus we will
need to construct other factor maps $\psi_{k}:X\to\mathbb{Z}^{d}/G_{k}$ which
are commensurate with the quotient maps. We will do this by first defining a
function which, in a sense, keeps track of how far off the given factor maps
are from being commensurate with the quotient maps, and then shifting the
given factor maps by this amount. The details are as follows.

Begin by setting $\psi_{1}=\pi_{1}$ and assume that factor maps
$\psi_{1},...,\psi_{n}$ have been so constructed, so that
$q_{k}\circ\psi_{k+1}=\psi_{k}$ for all $k<n$. Define
$\mathbf{r}_{n+1}:X\to\mathbb{Z}^{d}/G_{n}$ by

$\mathbf{r}_{n+1}(x)=q_{n}\circ\pi_{n+1}(x)-\psi_{n}(x).$

Note that $\mathbf{r}_{n+1}$ is $\mathbf{T}$-invariant, because for any
$\mathbf{v}\in\mathbb{Z}^{d}$, we have

$\displaystyle\mathbf{r}_{n+1}(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=q_{n}\circ\pi_{n+1}(\mathbf{T}^{\mathbf{v}}(x))-\psi_{n}(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=q_{n}\circ\tau^{\mathbf{v}}(\pi_{n+1}(x))-\tau^{\mathbf{v}}(\psi_{n}(x))$
$\displaystyle=q_{n}\left(\pi_{n+1}(x)+(\mathbf{v}+G_{n+1})\right)-(\psi_{n}(x)+(\mathbf{v}+G_{n}))$
$\displaystyle=q_{n}(\pi_{n+1}(x))+q_{n}(\mathbf{v}+G_{n+1})-\psi_{n}(x)-(\mathbf{v}+G_{n})$
$\displaystyle=q_{n}(\pi_{n+1}(x))+(\mathbf{v}+G_{n})-\psi_{n}(x)-(\mathbf{v}+G_{n})$
$\displaystyle=q_{n}\circ\pi_{n+1}(x)-\psi_{n}(x)$
$\displaystyle=\mathbf{r}_{n+1}(x).$

We remark that if $\mathbf{T}$ is ergodic, a situation including all cases
where $\mathbf{T}$ has rank one, then $\mathbf{r}_{n+1}$ is constant.

Next define a map $c_{n+1}:\mathbb{Z}^{d}/G_{n}\to\mathbb{Z}^{d}/G_{n+1}$
which maps cosets to cosets in a way that is consistent with the quotient map.
More specifically, for each $\mathbf{s}\in\mathbb{Z}^{d}/G_{n}$, set
$c_{n+1}(\mathbf{s})\in\mathbb{Z}^{d}/G_{n+1}$ to be such that
$q_{n}(c_{n+1}(\mathbf{s}))=\mathbf{s}$. Use this to define
$\widehat{\mathbf{r}}_{n+1}:X\to\mathbb{Z}^{d}/G_{n+1}$ as

$\widehat{\mathbf{r}}_{n+1}(x)=c_{n+1}(\mathbf{r}_{n+1}(x));$

the function $\widehat{\mathbf{r}}_{n+1}$ so defined is measurable and
$\mathbf{T}$-invariant (and is therefore constant whenever $\mathbf{T}$ is
ergodic) and satisfies
$q_{n}\circ\widehat{\mathbf{r}}_{n+1}=\mathbf{r}_{n+1}$.

Finally, define $\psi_{n+1}:X\to\mathbb{Z}^{d}/G_{n+1}$ by

$\psi_{n+1}(x)=\pi_{n+1}(x)-\widehat{\mathbf{r}}_{n+1}(x).$

For any $\mathbf{v}\in\mathbb{Z}^{d}$, we see

$\displaystyle\psi_{n+1}(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=\pi_{n+1}(\mathbf{T}^{\mathbf{v}}(x))-\widehat{\mathbf{r}}_{n+1}(\mathbf{T}^{\mathbf{v}}(x))$
$\displaystyle=\tau_{n+1}^{\mathbf{v}}(\pi_{n+1}(x))-\widehat{\mathbf{r}}_{n+1}(x)$
$\displaystyle=\pi_{n+1}(x)+(\mathbf{v}+G_{n+1})-\widehat{\mathbf{r}}_{n+1}(x)$
$\displaystyle=\tau_{n+1}^{\mathbf{v}}(\,\pi_{n+1}(x)-\widehat{\mathbf{r}}_{n+1}(x)\,)$
$\displaystyle=\tau_{n+1}^{\mathbf{v}}(\psi_{n+1}(x)),$

so $\psi_{n+1}$ intertwines $\mathbf{T}$ and $\tau_{n+1}$. Next,

$\displaystyle q_{n}\circ\psi_{n+1}(x)$
$\displaystyle=q_{n}(\pi_{n+1}(x)-\widehat{\mathbf{r}}_{n+1}(x))$
$\displaystyle=q_{n}(\pi_{n+1}(x))-q_{n}(\widehat{\mathbf{r}}_{n+1}(x))$
$\displaystyle=q_{n}(\pi_{n+1}(x))-\mathbf{r}_{n+1}(x)$
$\displaystyle=q_{n}\circ\pi_{n+1}(x)-\left[q_{n}\circ\pi_{n+1}(x)-\psi_{n}(x)\right]$
$\displaystyle=\psi_{n}(x),$

so the maps $\psi_{n}$ are commensurate with one another. We can then define a
factor map $\psi:X\to X_{\mathfrak{G}}$ by
$\psi(x)=(\psi_{1}(x),\psi_{2}(x),\psi_{3}(x),...)$, as wanted. ∎

######  Corollary 4.2.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ be Følner rank one and
let
$\sigma_{\mathfrak{G}}:\mathbb{Z}^{d}\curvearrowright(X_{\mathfrak{G}},\mu_{\mathfrak{G}})$
be the $\mathbb{Z}^{d}$-odometer corresponding to the sequence of subgroups
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. Then the following are equivalent:

  1. (1)

$(X,\mu,\mathbf{T})$ factors onto
$(X_{\mathfrak{G}},\mu_{\mathfrak{G}},\sigma_{\mathfrak{G}})$.

  2. (2)

“For each $j$, eventually most descendants are congruent mod $G_{j}$.” This
means that for each $j$, and for every $\epsilon>0$, there exists
$N\in\mathbb{N}$ such that for all $n\geq m\geq N$, there is
$\mathbf{g}\in\mathbb{Z}^{d}$ such that

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G_{j}\\})}{\\#(I_{m,n})}<\epsilon.$

###### Proof.

This is immediate from Theorem 3.1 and Theorem 4.1. ∎

The above lets us say, given a specific $\mathbb{Z}^{d}$-odometer, whether or
not $(X,\mu,\mathbf{T})$ factors onto it. We next want a condition that says
if there is some $\mathbb{Z}^{d}$-odometer onto which $(X,\mu,\mathbf{T})$
will factor. Since factoring onto a finite odometer is basically taken care of
by Theorem 3.1, we focus our attention here to infinite odometers.

We first give a complete description of $\mathcal{FF}(X_{\mathfrak{G}})$.

######  Theorem 4.3.

Let $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ be the
$\mathbb{Z}^{d}$-odometer given by $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$.
Then

$\mathcal{FF}(\sigma_{\mathfrak{G}})=\\{H\leq\mathbb{Z}^{d}:[\mathbb{Z}^{d}:H]<\infty\textrm{
and }H\geq G_{N}\textrm{ for some }N\\}.$

###### Proof.

$(\subseteq)$ First, recall from Theorem 2.11 that $\sigma_{\mathfrak{G}}$ is
rank one for a sequence of rectangular towers $\\{\mathcal{T}_{j}\\}$, where
each level of tower $\mathcal{T}_{j}$ corresponds to a coset $\mod G_{j}$.

Now, let $\tilde{H}\in\mathcal{FF}(\sigma_{\mathfrak{G}})$, meaning there is a
factor map
$\pi:(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})\to(\mathbb{Z}^{d}/\tilde{H},\tau)$.
This produces a measurable partition
$\mathcal{P}=\\{\pi^{-1}(\mathbf{v}+\tilde{H}):\mathbf{v}+\tilde{H}\in\mathbb{Z}^{d}/\tilde{H}\\}$
of $X_{\mathfrak{G}}$, and since $\sigma_{\mathfrak{G}}$ is rank one, we can
therefore find an $N$ so that for each
$\mathbf{v}+\tilde{H}\in\mathbb{Z}^{d}/\tilde{H}$, there are sets
$E_{\mathbf{v}}$, each a union of levels of tower $\mathcal{T}_{N}$, so that

$\mu_{\mathfrak{G}}(E_{\mathbf{v}}\,\triangle\,\pi^{-1}(\mathbf{v}+\tilde{H}))<\frac{1}{4}[\mathbb{Z}^{d}:\tilde{H}]^{-1}=\frac{1}{4}\mu_{\mathfrak{G}}(\pi^{-1}(\mathbf{v}+\tilde{H})).$

Now let $\mathbf{g}\in G_{N}$. This means that for each
$\mathbf{v}+\tilde{H}\in\mathbb{Z}^{d}/\tilde{H}$,
$\sigma_{\mathfrak{G}}^{\mathbf{g}}(E_{\mathbf{v}})=E_{\mathbf{v}}$. Also,
$\pi^{-1}(\mathbf{g}+\tilde{H})=\pi^{-1}(\tau^{\mathbf{g}}(\mathbf{0}+\tilde{H}))=\sigma_{\mathfrak{G}}^{\mathbf{g}}(\pi^{-1}(\mathbf{0}+\tilde{H}))$,
so

$\displaystyle\mu_{\mathfrak{G}}\left(\pi^{-1}(\mathbf{g}+\tilde{H})\,\triangle\,\pi^{-1}(\mathbf{0}+\tilde{H})\right)$
$\displaystyle=\mu_{\mathfrak{G}}\left(\sigma_{\mathfrak{G}}^{\mathbf{g}}(\pi^{-1}(\mathbf{0}+\tilde{H}))\,\triangle\,\pi^{-1}(\mathbf{0}+\tilde{H})\right)$
$\displaystyle\leq\mu_{\mathfrak{G}}\left(\sigma_{\mathfrak{G}}^{\mathbf{g}}(\pi^{-1}(\mathbf{0}+\tilde{H}))\,\triangle\,\sigma_{\mathfrak{G}}^{\mathbf{g}}(E_{0})\right)$
$\displaystyle\qquad+\mu_{\mathfrak{G}}\left(\sigma_{\mathfrak{G}}^{\mathbf{g}}(E_{0})\,\triangle\,E_{0}\right)$
$\displaystyle\qquad+\mu_{\mathfrak{G}}\left(E_{0}\,\triangle\,\pi^{-1}(\mathbf{0}+\tilde{H})\right)$
$\displaystyle<\frac{1}{4}[\mathbb{Z}^{d}:\tilde{H}]^{-1}+0+\frac{1}{4}[\mathbb{Z}^{d}:\tilde{H}]^{-1}$
$\displaystyle=\frac{1}{2}[\mathbb{Z}^{d}:\tilde{H}]^{-1}$
$\displaystyle=\frac{1}{2}\mu_{\mathfrak{G}}\left(\pi^{-1}(\mathbf{0}+\tilde{H})\right).$

That means that $\pi^{-1}(\mathbf{g}+\tilde{H})$ and
$\pi^{-1}(\mathbf{0}+\tilde{H})$ intersect in a set of positive measure, which
implies that $\mathbf{g}+\tilde{H}=\mathbf{0}+\tilde{H}$, i.e.
$\mathbf{g}\in\tilde{H}$. We have proven $G_{N}\leq\tilde{H}$, as wanted.

$(\supseteq)$ Let
$\tilde{H}\in\\{H\leq\mathbb{Z}^{d}:[\mathbb{Z}^{d}:H]<\infty\textrm{ and
}H\geq G_{N}\textrm{ for some }N\\}$. As mentioned above we know
$G_{N}\in\mathcal{FF}(\sigma_{\mathfrak{G}})$. By Lemma 2.14,
$\mathcal{FF}(\sigma_{\mathfrak{G}})$ is closed under supergroups, and thus
$\tilde{H}\in\mathcal{FF}(\sigma_{\mathfrak{G}})$ as wanted. ∎

Not only does the above describe the set $\mathcal{FF}(\sigma_{\mathfrak{G}})$
associated to an odometer, it can be used to characterize the conjugacy class
of the odometer, as described in the next result.

######  Corollary 4.4.

Let $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ and
$(X_{\mathfrak{H}},\sigma_{\mathfrak{H}})$ be $\mathbb{Z}^{d}$-odometers
corresponding to sequences $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$ and
$\mathfrak{H}=\\{H_{1},H_{2},H_{3},...\\}$, respectively.
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ and
$(X_{\mathfrak{H}},\sigma_{\mathfrak{H}})$ are conjugate if and only if
$\mathcal{FF}(\sigma_{\mathfrak{G}})=\mathcal{FF}(\sigma_{\mathfrak{H}})$.

###### Proof.

The $(\Rightarrow)$ direction is immediate, so we focus on the converse here.
Assume $\mathcal{FF}(\sigma_{G})=\mathcal{FF}(\sigma_{H})$ and set $i_{1}=1$.
Clearly $G_{1}\in\mathcal{FF}(\sigma_{\mathfrak{G}})$ and so
$G_{1}\in\mathcal{FF}(\sigma_{\mathfrak{H}})$ as well. By Theorem 4.3 we then
know there is $i_{2}$ such that $G_{1}\geq H_{i_{2}}$. As
$H_{i_{2}}\in\mathcal{FF}(\sigma_{\mathfrak{H}})=\mathcal{FF}(\sigma_{\mathfrak{G}})$,
we can apply Theorem 4.3 again to obtain $i_{3}>i_{1}$ with $H_{i_{2}}\geq
G_{i_{3}}$. Continuing in this fashion, we obtain a sequence
$\mathfrak{G}^{\prime}=\\{G_{i_{1}},H_{i_{2}},G_{i_{3}},H_{i_{4}},...\\}$ of
finite-index subgroups of $\mathbb{Z}^{d}$ where

$G_{i_{1}}\geq H_{i_{2}}\geq G_{i_{3}}\geq H_{i_{4}}\geq\cdots$

Since inserting or deleting subgroups from a decreasing sequence of subgroups
defining an odometer does not change its conjugacy class, we see
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ is conjugate to
both $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ and
$(X_{\mathfrak{H}},\sigma_{\mathfrak{H}})$, and therefore
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ and
$(X_{\mathfrak{H}},\sigma_{\mathfrak{H}})$ are conjugate to one another, as
desired. ∎

We now have a way to associate to an odometer
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ the set
$\mathcal{FF}(\sigma_{\mathfrak{G}})$ of finite index subgroups of
$\mathbb{Z}^{d}$, closed under supergroups and intersections. We next want to
show the converse: given such a set of subgroups, one can construct an
associated odometer. This will be shown by the next lemma which also gives a
characterization of when the associated odometer is infinite.

######  Lemma 4.5.

Let $\mathcal{F}$ be a collection of finite-index subgroups of
$\mathbb{Z}^{d}$ that is closed under supergroups and intersections. Then:

  1. (1)

There is an odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$, unique up to
conjugacy, such that $\mathcal{F}=\mathcal{FF}(\sigma_{\mathfrak{G}})$.

  2. (2)

If $\mathcal{F}$ is infinite, then $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$
is infinite.

  3. (3)

If $\stackbin[G\in\mathcal{F}]{}{\bigcap}G=\\{\mathbf{0}\\}$, then
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is free.

###### Proof.

As there are only countably many subgroups of $\mathbb{Z}^{d}$ of finite
index, $\mathcal{F}$ must be countable, and so we can write
$\mathcal{F}=\\{H_{1},H_{2},H_{3},...\\}$. For each $k\geq 1$, let
$G_{k}=\cap_{j=1}^{k}H_{j}$. If $\mathcal{F}$ is finite and has, say, $t$
members, then for every $k>t$ set $G_{k}=G_{t}$. This gives a decreasing
sequence $\mathfrak{G}=\\{G_{1},G_{2},G_{3},..\\}$ of finite index subgroups
of $\mathbb{Z}^{d}$ yielding a corresponding odometer
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$.

To show $\mathcal{F}=\mathcal{FF}(\sigma_{\mathfrak{G}})$, let
$H\in\mathcal{F}$. Then $H=H_{i}$ for some $i$ and thus $G_{k}\leq H$ for any
$k\geq i$. From Theorem 4.3 it follows that
$H\in\mathcal{FF}(\sigma_{\mathfrak{G}})$. Next, take
$G\in\mathcal{FF}(\sigma_{\mathfrak{G}})$. By Theorem 4.3, $G$ is a supergroup
of some $G_{k}$. Because $\mathcal{F}$ is closed under intersections and
$G_{k}=\cap_{j=1}^{k}H_{j}$, we know $G_{k}\in\mathcal{F}$. Using that
$\mathcal{F}$ is also closed under supergroups, we have $G\in\mathcal{F}$ as
well, completing the proof of the first statement.

If $\mathcal{F}$ is infinite, the sequence $\\{G_{k}\\}$ is not eventually
constant, so the odometer $X_{\mathfrak{G}}$ is infinite. If
$\stackbin[G\in\mathcal{F}]{}{\bigcap}G=\stackbin[k]{}{\bigcap}H_{k}=\\{\mathbf{0}\\}$,
then $\stackbin[k]{}{\bigcap}\,G_{k}=\\{\mathbf{0}\\}$ as well, so
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is free (see Theorem 2.9). ∎

The odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ constructed in the
above result will be called the odometer generated by  $\mathcal{F}$.

We now turn to the connection between the set $\mathcal{FF}(\mathbf{T})$ and
the kinds of odometer actions onto which $(X,\mathbf{T})$ can factor:

######  Lemma 4.6.

Let $(X,\mathbf{T})$ be a measure-preserving $\mathbb{Z}^{d}$-system. Then:

  1. (1)

$(X,\mathbf{T})$ factors onto an infinite $\mathbb{Z}^{d}$-odometer if and
only if $\mathcal{FF}(\mathbf{T})$ is infinite.

  2. (2)

$(X,\mathbf{T})$ factors onto a free $\mathbb{Z}^{d}$-odometer if and only if
$\stackbin[G\in\mathcal{FF}(\mathbf{T})]{}{\bigcap}G=\\{\mathbf{0}\\}$.

###### Proof.

We begin with statement (1). Suppose first that $(X,\mathbf{T})$ factors onto
the infinite $\mathbb{Z}^{d}$-odometer
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ associated to the sequence
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. It follows that for every $k$,
$(X,\mathbf{T})$ factors onto $\mathbb{Z}^{d}/G_{k}$, i.e.
$G_{k}\in\mathcal{FF}(\mathbf{T})$. As $\sigma_{\mathfrak{G}}$ is infinite,
there must be infinitely many different $G_{k}$, so $\mathcal{FF}(\mathbf{T})$
is infinite, as wanted.

Conversely, if $\mathcal{FF}(\mathbf{T})$ is infinite, then by Lemma 4.5 we
can let $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ with
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$ be the infinite odometer generated
by $\mathcal{FF}(\mathbf{T})$. Then
$\mathcal{FF}(\mathbf{T})=\mathcal{FF}(X_{\mathfrak{G}})$ and thus
$G_{k}\in\mathcal{FF}(\mathbf{T})$ for every $k$. By definition, that means
$(X,\mathbf{T})$ factors onto every $(\mathbb{Z}^{d}/G_{k},\tau)$ and so by
Theorem 4.1$,(X,\mathbf{T})$ must also factor onto
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ as wanted.

Now for statement (2). Suppose first that $(X,\mathbf{T})$ factors onto the
free $\mathbb{Z}^{d}$-odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$
associated to the sequence $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. As
before, we have $G_{k}\in\mathcal{FF}(\mathbf{T})$ for every $k$, which
implies
$\stackbin[G\in\mathcal{FF}(\mathbf{T})]{}{\bigcap}G\subseteq\bigcap_{k}G_{k}=\\{\mathbf{0}\\}$.

Conversely, assume
$\stackbin[G\in\mathcal{FF}(\mathbf{T})]{}{\bigcap}G=\\{\mathbf{0}\\}$. Use
Lemma 4.5 to obtain a free odometer $(X_{\mathfrak{G}},\sigma_{G})$ where
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$ and
$\mathcal{FF}(\sigma_{G})=\mathcal{FF}(\mathbf{T})$. Since each
$G_{k}\in\mathcal{FF}(\sigma_{G})$, we then have that $(X,\mathbf{T})$ factors
onto every $(\mathbb{Z}^{d}/G_{k},\tau)$. Thus by Theorem 4.1 $(X,\mathbf{T})$
must also factor onto $(X_{\mathfrak{G}},\sigma_{G})$, as wanted. ∎

We can now characterize when, given a Følner rank one $\mathbb{Z}^{d}$ system,
there is some infinite odometer on which it will factor.

######  Theorem 4.7.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ be Følner rank one. The
following are equivalent:

  1. (1)

$(X,\mu,\mathbf{T})$ factors onto some infinite odometer.

  2. (2)

For all $j\in\mathbb{N}$, there is a finite-index subgroup $G_{j}$ with
$[\mathbb{Z}^{d}:G_{j}]\geq j$ so that for all $\epsilon>0$, there exists
$N\in\mathbb{N}$ for all $n\geq m\geq N$, there is
$\mathbf{g}\in\mathbb{Z}^{d}$ such that

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G_{j}\\})}{\\#(I_{m,n})}<\epsilon.$

###### Proof.

(1) $\Rightarrow$ (2): Assume $(X,\mu,\mathbf{T})$ factors onto some infinite
odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ where
$\mathfrak{G}=\\{G_{1},G_{2},...\\}$. Composing the factor map with the
projection onto the $j^{th}$ coordinate, Theorem 3.1 yields the result in (2).

(2) $\Rightarrow$ (1): By Theorem 3.1, $G_{j}\in\mathcal{FF}(\mathbf{T})$ for
every $j$. Since $[\mathbb{Z}^{d}:G_{j}]\geq j$, $\mathcal{FF}(\mathbf{T})$
must be infinite, and (1) follows from Lemma 4.6. ∎

##  5\. Rank one $\mathbb{Z}^{d}$ systems conjugate to an odometer

In this section we explore when a Følner rank one $\mathbb{Z}^{d}$-system not
only factors onto some infinite odometer but in fact is conjugate to that
odometer. We first consider the situation of when the Følner rank one system
and the odometer are given, and one wants to check if they are conjugate. This
is followed by a theorem that considers how one can check, given the cutting
and stacking procedure that generates a Følner rank one system, if there is
any odometer to which it is conjugate.

Both of these theorems will involve two properties which, when taken together,
give necessary and sufficient conditions on a Følner rank one
$\mathbb{Z}^{d}$-action to guarantee the wanted conjugacy. For instance,
consider the two properties listed in Theorem 5.1 below. Note that condition
(2b) is familiar to us from the proceeding sections: one can think of (2b) as
saying that eventually, the Følner rank one $\mathbb{Z}^{d}$-action mimics the
group structure of the odometer. More specifically, it requires that for each
subgroup $G_{j}$ in the odometer structure, eventually, most descendants of
every tower in the rank one system are congruent$\mod G_{j}$. Importantly,
condition (2b) does not depend at all on how the descendants of the first few
towers in the cutting and stacking construction behave. In order to achieve a
conjugacy between a Følner rank one action and an odometer, it is necessary
that _every_ tower is stacked in a way that does not deviate too much from the
structure of the odometer. This is captured with condition (2a), which says
that eventually, the descendants of the base of each stage $l$ tower mostly
coincide with a single subset of some $\mathbb{Z}^{d}/G_{k}$.

######  Theorem 5.1.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ be Følner rank one and
let
$\sigma_{\mathfrak{G}}:\mathbb{Z}^{d}\curvearrowright(X_{\mathfrak{G}},\mu_{\mathfrak{G}})$
be the $\mathbb{Z}^{d}$-odometer corresponding to the sequence of subgroups
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. The following are equivalent:

  1. (1)

$(X,\mathbf{T})$ is conjugate to $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$.

  2. (2)

Both of the following statements hold:

    1. (a)

For all $l\in\mathbb{N}$ and all $\epsilon>0$, there exists $k\in\mathbb{N}$
and $N\in\mathbb{N}$ such that for all $m\geq N$, there is
$D\subseteq\mathbb{Z}^{d}/G_{k}$ such that

$\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in F_{m}:\mathbf{i}+G_{k}\in
D\\}\right)}{\\#(I_{l,m})}<\epsilon.$

    2. (b)

For each $j$, and for every $\epsilon>0$, there exists $N\in\mathbb{N}$ such
that for all $n\geq m\geq N$, there is $\mathbf{g}\in\mathbb{Z}^{d}$ such that

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G_{j}\\})}{\\#(I_{m,n})}<\epsilon.$

###### Proof.

We begin with (1) $\Rightarrow$ (2). Let $\phi:X\to X_{\mathfrak{G}}$ give a
conjugacy between the systems. Since we can also think of $\phi$ as a factor
map from $X$ to $X_{\mathfrak{G}}$, Corollary 4.2 yields (2b). Thus it
suffices to establish statement (2a).

Because each level in a tower has the same measure, the inequality in (2a) is
equivalent to

$\frac{\mu\left(\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
I_{l,m}]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\right)}{\mu\left(\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
I_{l,m}]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)}<\epsilon$

which we can rewrite as

(5.1)
$\frac{\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\right)}{\mu\left(B_{l}\right)}<\epsilon.$

So fix $l\in\mathbb{N}$ and $\epsilon>0$ and then set
$\delta=\frac{1}{4}\epsilon\,\mu(B_{l})$. We first find the specific $k$ for
statement (2a): letting
$\pi_{\tilde{k}}:X_{\mathfrak{G}}\rightarrow\mathbb{Z}^{d}/G_{\tilde{k}}$ and
recalling that the sets
$\\{\pi_{\tilde{k}}^{-1}(\mathbf{g}+G_{\tilde{k}}):\mathbf{g}+G_{\tilde{k}}\in\mathbb{Z}^{d}/G_{\tilde{k}},{\tilde{k}}\in\mathbb{N}\\}$
generate the $\sigma$-algebra on $X_{\mathfrak{G}}$, let $k$ be such that
$\phi(B_{l})$ can be approximated by the $\pi_{k}$-pullback of some union of
cosets from $\mathbb{Z}^{d}/G_{k}$. More specifically, choose $k$ so that
there is a union $E$ of cosets from $\mathbb{Z}^{d}/G_{k}$ so that

$\mu_{\mathfrak{G}}\left(\,\phi(B_{l})\,\triangle\,\pi_{k}^{-1}(E)\,\right)<\delta.$

Since $\phi$ is a conjugacy, we thus have, for this $k$ and this collection of
cosets $E$,

(5.2) $\mu\left(\,B_{l}\,\triangle\,\phi_{k}^{-1}(E)\,\right)<\delta,$

where we set $\phi_{k}=\pi_{k}\circ\phi$, so that
$\phi_{k}:X\rightarrow\mathbb{Z}^{d}/G_{k}$.

Choose $N_{1}\geq l$ so that $\mu\left(\cup_{\mathbf{i}\in
F_{N_{1}}}\mathbf{T}^{\mathbf{i}}(B_{N_{1}})\right)>1-\delta$. We then find
the specific $N$ for statement (2a) by choosing $N\geq N_{1}$, similar to what
we did in part (i) in the proof of Theorem 3.1, such that for all $m\geq N$,
there exists a $\mathbf{g}\in\mathbb{Z}^{d}$ such that

$\frac{\mu\left(\,B_{m}\,-\phi_{k}^{-1}(\mathbf{g}+G_{k})\,\right)}{\mu(B_{m})}<\delta.$

Fix $m\geq N$ and let $\mathbf{g}$ be as described above. This in turn means,
for each $\mathbf{i}\in F_{m}$,

(5.3)
$\frac{\mu\left(\,\mathbf{T}^{\mathbf{i}}(B_{m})\,-\phi_{k}^{-1}(\mathbf{g}+\mathbf{i}+G_{k})\,\right)}{\mu\left(\mathbf{T}^{\mathbf{i}}(B_{m})\right)}<\delta.$

Finally, we specify the set $D\subseteq\mathbb{Z}^{d}/G_{k}$ referred to in
statement (2a) by setting $D=E-(\mathbf{g}+G_{k})$.

With $k$ and $N$ thus chosen and our set $D$ determined, we will now turn to
verifying inequality (5.1). First, observe

(5.4)
$\displaystyle\mu\left(\,B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\right)\leq$
$\displaystyle\qquad\quad\mu\left(\,B_{l}\,\triangle\,\phi_{k}^{-1}(E)\,\right)+\mu\left(\,\phi_{k}^{-1}(E)\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\right).$

As
$\phi_{k}^{-1}(E)\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]$
is the disjoint union of

$S_{1}=\phi_{k}^{-1}(E)\,-\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]$

and

$S_{2}=\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,-\,\phi_{k}^{-1}(E),$

we will next bound the measures of $S_{1}$ and $S_{2}$ separately.

First, consider points in
$S_{1}=\phi_{k}^{-1}(E)\,-\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\,\mathbf{T}^{\mathbf{i}}(B_{m})\right]$ and further divide the
points in $S_{1}$ by whether or not they also lie in $B_{l}$. For $x\in
S_{1}\cap B_{l}$, we see that since the towers stack and $m\geq l$, $x$ lies
in the $l^{th}$-tower $\mathcal{T}_{l}$, so we can find $\hat{\mathbf{i}}\in
F_{m}$ with $x\in\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})$. As $x\in
S_{1}=\phi_{k}^{-1}(E)-\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})$, it
must be that $\hat{\mathbf{i}}+G_{k}\notin D$ or equivalently,
$\hat{\mathbf{i}}+\mathbf{g}+G_{k}\notin E$. Despite the fact that (5.3) says
that most of $\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})$ is covered by
$\phi_{k}^{-1}(\hat{\mathbf{i}}+\mathbf{g}+G_{k})$, it must be that
$x\notin\phi_{k}^{-1}(\hat{\mathbf{i}}+\mathbf{g}+G_{k})$. For if it were,
then $\phi_{k}(x)=\hat{\mathbf{i}}+\mathbf{g}+G_{k}$, which we said earlier
was not in $E$; however, we also have $x\in\phi_{k}^{-1}(E)$, a contradiction.
So we have $x\in\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})$ and
$x\notin\phi_{k}^{-1}(\hat{\mathbf{i}}+\mathbf{g}+G_{k})$; by (5.3) we know
the measure of such points is less than
$\delta\,\mu(\mathbf{T}^{\hat{\mathbf{i}}}(B_{m}))$. Therefore

(5.5) $\mu\left(S_{1}\cap
B_{l}\right)<\delta\cdot\mu\left(\stackrel{{\scriptstyle[}}{{\hat{\mathbf{i}}}}\in
F_{m}]{}{\bigcup}\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})\right)<\delta.$

Turning to points in $S_{1}-B_{l}$, we see from the definition of $S_{1}$ that
such points lie in $\phi_{k}^{-1}(E)$. So

$\displaystyle\mu\left(S_{1}-B_{l}\right)$
$\displaystyle\leq\mu\left(\phi_{k}^{-1}(E)-B_{l}\right)$
$\displaystyle\leq\mu\left(\phi_{k}^{-1}(E)\,\triangle\,B_{l}\right)$ (5.6)
$\displaystyle<\delta$

by (5.2). Finally, from (5.5) and (5), we have

(5.7) $\mu(S_{1})=\mu(S_{1}\cap
B_{l})+\mu(S_{1}-B_{l})<\delta+\delta=2\delta.$

Second, we consider points in
$S_{2}=\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\,\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,-\,\phi_{k}^{-1}(E)$.
For such a point $x$, we have $\hat{\mathbf{i}}\in F_{m}$ so that
$x\in\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})$ with $\hat{\mathbf{i}}+G_{k}\in D$
or equivalently, $\hat{\mathbf{i}}+\mathbf{g}+G_{k}\in E$. Note that we must
have $\phi_{k}(x)\neq\hat{\mathbf{i}}+\mathbf{g}+G_{k}$, because otherwise we
would have $\phi_{k}(x)\in E$, contradicting the definition of $S_{2}$. So we
have $x\in\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})$ and
$x\notin\phi_{k}^{-1}(\hat{\mathbf{i}}+\mathbf{g}+G_{k})$; by (5.3) we know
the measure of such points is less than
$\delta\,\mu(\mathbf{T}^{\hat{\mathbf{i}}}(B_{m}))$. All together we then get

(5.8)
$\mu(S_{2})<\delta\cdot\mu\left(\stackrel{{\scriptstyle[}}{{\hat{\mathbf{i}}}}\in
F_{m}]{}{\bigcup}\mathbf{T}^{\hat{\mathbf{i}}}(B_{m})\right)<\delta.$

Combining (5.7) and (5.8), we obtain

$\mu\left(\phi_{k}^{-1}(E)\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)=\mu(S_{1}\sqcup
S_{2})<3\delta,$

and returning to (5.4), we then have

$\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)<\mu\left(B_{l}\,\triangle\,\phi_{k}^{-1}(E)\right)+3\delta$

which, using (5.2), gives us
$\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)<\delta+3\delta=4\delta$.
Therefore

$\frac{\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)}{\mu\left(B_{l}\right)}<\frac{4\delta}{\mu\left(B_{l}\right)}=\epsilon,$

establishing (5.1) as wanted.

We now show (2) $\Rightarrow$ (1). From statement (2b) and Theorem 3.1, we
know for that $G_{k}\in\mathcal{FF}(\mathbf{T})$ for every $k$, so there
exists a factor map $\pi_{k}:X\rightarrow\mathbb{Z}^{d}/G_{k}$. It then
follows by Theorem 4.1 that $X$ factors onto $X_{\mathfrak{G}}$ with the map
$\psi(x)=(\psi_{1}(x),\psi_{2}(x),...)$ where
$\psi_{k}(x)=\pi_{k}(x)-\hat{\mathbf{r}}_{k}(x)$. As commented on in the proof
of that theorem, $(X,\mathbf{T})$ being rank one, and hence ergodic, means
$\hat{\mathbf{r}}_{k}(x)$ is a constant function and we can write
$\psi_{k}(x)=\pi_{k}(x)-\hat{\mathbf{r}}_{k}$.

Our goal is to show that the factor map $\psi$ is in fact a conjugacy. This
will be assured if $\psi^{-1}(\mathcal{X}_{\mathfrak{G}})=\mathcal{X}$; since
these $\sigma$-algebras are generated respectively by
$\\{\psi_{k}^{-1}(\mathbf{g}):k\in\mathbb{N},\mathbf{g}\in\mathbb{Z}^{d}/G_{k}\\}$
and $\\{\mathbf{T}^{\mathbf{i}}(B_{l}):l\in\mathbb{N},\mathbf{i}\in F_{l}\\}$,
it suffices to show that for every $l\in\mathbb{N}$ and every $\epsilon>0$,
there is $k\in\mathbb{N}$ and $E\subseteq\mathbb{Z}^{d}/G_{k}$ with
$\mu\left(B_{l}\,\triangle\,\psi_{k}^{-1}(E)\right)<\epsilon$.

So let $l\in\mathbb{N}$ and $\epsilon>0$ be fixed. By statement (2a) we obtain
$k\in\mathbb{N}$ and $N_{1}\in\mathbb{N}$ (without loss of generality we can
assume $N_{1}>l$) so that whenever $m\geq N_{1}$, there is
$D\subseteq\mathbb{Z}^{d}/G_{k}$ so that

(5.9) $\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in
F_{m}:\mathbf{i}+G_{k}\in D\\}\right)}{\\#(I_{l,m})}<\frac{\epsilon}{4}.$

Since every level in the tower $\mathcal{T}_{m}$ has the same measure, (5.9)
is equivalent to

$\frac{\mu\left(\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
I_{l,m}]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\,\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)}{\mu\left(\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
I_{l,m}]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)}<\frac{\epsilon}{4},$

and by the definition of $I_{l,m}$ this inequality is the same as

(5.10)
$\frac{\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\,\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)}{\mu\left(B_{l}\right)}<\frac{\epsilon}{4}.$

Similar to what we did in part (i) of Theorem 3.1, we can find $N_{2}\geq
N_{1}$ such that for every $m\geq N_{2}$, there is some
$\gamma\in\mathbb{Z}^{d}/G_{k}$ such that
$\mu(B_{m}-\psi_{k}^{-1}(\gamma))<\frac{\epsilon}{4}\mu(B_{m})$. This in turn
means

(5.11)
$\mu(\mathbf{T}^{\mathbf{i}}(B_{m})-\psi_{k}^{-1}(\gamma+\mathbf{i}+G_{k}))<\frac{\epsilon}{4}\mu(B_{m})\hskip
28.90755pt\forall\,\mathbf{i}\in F_{m}.$

We now find the required set $E\subseteq\mathbb{Z}^{d}/G_{k}$ by fixing $m\geq
N_{2}$ so large that $\mu(\mathcal{T}_{m})>1-\frac{\epsilon}{4}$ and letting
$D$ and $\gamma$ be as specified above. We then take $E=\gamma+D$; it remains
for us to prove that
$\mu\left(B_{l}\,\triangle\,\psi_{k}^{-1}(E)\right)<\epsilon$.

Observe that

$\displaystyle\mu\left(B_{l}\,\triangle\,\psi^{-1}_{k}(E)\right)$
$\displaystyle\leq\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)$
$\displaystyle\hskip
36.135pt+\mu\left(\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\,\triangle\,\psi^{-1}_{k}(E)\right)$
(5.12)
$\displaystyle=\mu\left(B_{l}\,\triangle\,\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right)$
$\displaystyle\hskip
36.135pt+\mu\left(\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]-\psi^{-1}_{k}(E)\right)$
$\displaystyle\hskip
36.135pt+\mu\left(\left[\psi^{-1}_{k}(E)-\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]\right).$

The first addend in (5) is, by (5.10), less than
$\frac{\epsilon}{4}\,\mu(B_{l})\leq\frac{\epsilon}{4}$.

For the second addend of (5), note that

$\mu\left(\left[\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right]-\psi^{-1}_{k}(E)\right)=\sum_{\mathbf{i}\in
F_{m},\mathbf{i}+G_{k}\in
D}\mu\left(\mathbf{T}^{\mathbf{i}}(B_{m})-\psi^{-1}_{k}(E)\right).$

Note that since $\mathbf{i}+G_{k}\in D$, $\gamma+\mathbf{i}+G_{k}\in E$. So by
applying (5.11),
$\mu(\mathbf{T}^{\mathbf{i}}(B_{m})-\psi_{k}^{-1}(E))\leq\mu(\mathbf{T}^{\mathbf{i}}(B_{m})-\psi_{k}^{-1}(\gamma+\mathbf{i}+G_{k}))<\frac{\epsilon}{4}\mu(B_{m})$.
This makes the second addend of (5) less than

$\sum_{\mathbf{i}\in F_{m},\mathbf{i}+G_{k}\in
D}\frac{\epsilon}{4}\mu(B_{m})\,\leq\,\frac{\epsilon}{4}.$

For the third addend of (5), we divide $\psi^{-1}_{k}(E)$ into the part that
is not in $\mathcal{T}_{m}$ and the part that is in
$\mathcal{T}_{m}=\cup_{\mathbf{i}\in F_{m}}\mathbf{T}^{\mathbf{i}}(B_{m})$. By
our choice of $m$ we know
$\mu(\psi^{-1}_{k}(E)-\mathcal{T}_{m}\,)<\frac{\epsilon}{4}$. For the rest,
note that

$\mu\left(\left[\psi^{-1}_{k}(E)\cap\mathcal{T}_{m}\right]-\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)\leq\sum_{\mathbf{i}\in
F_{m},\mathbf{i}+G_{k}\notin
D}\mu\left(\,\psi^{-1}_{k}(D+\gamma)\cap\mathbf{T}^{\mathbf{i}}(B_{m})\,\right).$

Recall from (5.11) that we picked $\gamma$ such that
$\mu(\mathbf{T}^{\mathbf{i}}(B_{m})-\psi_{k}^{-1}(\gamma+\mathbf{i}+G_{k}))<\frac{\epsilon}{4}\mu(B_{m})$.
Here we are only concerned with those $\mathbf{i}$ such that
$\mathbf{i}+G_{k}\notin D$ or equivalently,
$\mathbf{i}+\gamma+G_{k}\notin\gamma+D=E$. This yields, for each such
$\mathbf{i}$,

$\mu\left(\,\mathbf{T}^{\mathbf{i}}(B_{m})\cap\psi^{-1}_{k}(E)\,\right)\leq\mu\left(\,\mathbf{T}^{\mathbf{i}}(B_{m})-\psi^{-1}(\gamma+\mathbf{i}+G_{k})\,\right)<\frac{\epsilon}{4}\mu(B_{m})$

and thus $\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G_{k}\notin
D]{}{\sum}\mu\left(\psi^{-1}_{k}(E)\cap\mathbf{T}^{\mathbf{i}}(B_{m})\right)<\frac{\epsilon}{4}$.
Thus the third addend of (5) is less than $\frac{\epsilon}{2}$, so (5) becomes

$\mu\left(B_{l}\,\triangle\,\psi_{k}^{-1}(E)\right)<\frac{\epsilon}{4}+\frac{\epsilon}{4}+\frac{\epsilon}{2}=\epsilon$

as wanted. ∎

######  Theorem 5.2.

Let $\mathbf{T}:\mathbb{Z}^{d}\curvearrowright(X,\mu)$ be Følner rank one. The
following are equivalent:

  1. (1)

$(X,\mu,\mathbf{T})$ is conjugate to an infinite $\mathbb{Z}^{d}$-odometer.

  2. (2)

For all $l\in\mathbb{N}$ and all $\epsilon>0$, there is a finite index
subgroup $G$ of $\mathbb{Z}^{d}$ such that

    1. (a)

there exists $N_{a}\in\mathbb{N}$ so that for all $m\geq N_{a}$, there exists
$D\subseteq\mathbb{Z}^{d}/G$ for which

$\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in F_{m}:\mathbf{i}+G\in
D\\}\right)}{\\#(I_{l,m})}<\epsilon;{\textrm{ and}}$

    2. (b)

for every $\eta>0$, there is $N_{b}\in\mathbb{N}$ so that for all $n>m\geq
N_{b}$, there exists $\mathbf{g}+G\in\mathbb{Z}^{d}/G$ where

$\frac{\\#(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G\\})}{\\#(I_{m,n})}<\eta.$

###### Proof.

We begin with (1) $\Rightarrow$ (2): Suppose $(X,\mathbf{T})$ is conjugate to
an infinite $\mathbb{Z}^{d}$-odometer
($X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ given by
$\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. Since the odometer is infinite,
there must be infinitely many distinct $G_{j}$, all of which are elements in
$\mathcal{FF}(\sigma_{\mathfrak{G}})$. Thus
$\mathcal{FF}(\sigma_{\mathfrak{G}})$ is an infinite set. Fix $l\in\mathbb{N}$
and $\epsilon>0$. By condition (2a) of Theorem 5.1, we obtain $k_{l,\epsilon}$
and $N_{l,\epsilon}\in\mathbb{N}$ so that whenever $m\geq N_{l,\epsilon}$,
there is $D\subseteq\mathbb{Z}^{d}/G_{k_{l,\epsilon}}$ so that

$\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in
F_{m}:\mathbf{i}+G_{k_{l,\epsilon}}\in D\\}\right)}{\\#(I_{l,m})}<\epsilon.$

This gives condition (2a) above with $N_{a}=N_{l,\epsilon}$ and
$G=G_{k_{l,\epsilon}}$. Now fix $\eta>0$ and use $j=k_{l,\epsilon}$ and
$\epsilon=\eta$ in condition (2b) of Theorem 5.1 to find $N_{b}\in\mathbb{N}$
so that for all $n\geq m\geq N_{b}$, there is $\mathbf{g}\in\mathbb{Z}^{d}$ so
that

$\frac{\\#\left(\\{\mathbf{i}\in I_{m,n}:\mathbf{i}\not\equiv\mathbf{g}\mod
G_{k_{l,\epsilon}}\\}\right)}{\\#(I_{m,n})}<\eta,$

giving condition (2b) above. Thus we obtain (2).

We now turn to proving (2) $\Rightarrow$ (1). This is done in three steps: we
first use the finite-index subgroups given to us by (2) to obtain an odometer
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$. We then modify
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ to obtain a
conjugate odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ whose generating
groups are more controlled. Finally, we show this new odometer is conjugate to
the original Følner rank one system $(X,\mu,\mathbf{T})$ .

For the first step, for each $l\in\mathbb{N}$ and each $\epsilon>0$, let
$\mathcal{G}(l,\epsilon)$ be the set of all finite-index subgroups of
$\mathbb{Z}^{d}$ which satisfy conditions (2a) and (2b) of this theorem. By
assumption, this is nonempty. Then define $\mathcal{F}$ to be the smallest set
of finite-index subgroups of $\mathbb{Z}^{d}$ that is closed under
supergroups, closed under intersections, and contains every member of every
$\mathcal{G}(l,\epsilon)$. Note that this is equivalent to saying

(5.13)
$\mathcal{F}=\left\\{G\leq\mathbb{Z}^{d}:\begin{array}[]{l}\textrm{there
exists }\epsilon_{1},\dots,\epsilon_{t}\in(0,1)\textrm{ and
}l_{1},\dots,l_{t}\in\mathbb{N},\\\ \textrm{and for each
}j\in\\{1,\dots,t\\}\textrm{ there exists
}G_{j}\in\mathcal{G}(l_{j},\epsilon_{j})\\\ \textrm{ so that
}G\geq\bigcap_{j=1}^{t}G_{j}.\end{array}\right\\}.$

By Lemma 4.5, there is an odometer
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ such that
$\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})=\mathcal{F}$. We will next show
that $\mathcal{F}$ is an infinite set; Lemma 4.5 will then also tell us that
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ is an infinite
odometer.

We will show $\mathcal{F}$ is an infinite set by in fact showing that
$\cup_{l,\epsilon}\mathcal{G}(l,\epsilon)$ is infinite, which we do by
contradiction. So assume that $\cup_{l,\epsilon}\mathcal{G}(l,\epsilon)$ is
finite. Then there is $t\in\mathbb{N}$ so that $[\mathbb{Z}^{d}:G]\leq t$ for
all $G\in\cup_{l,\epsilon}\mathcal{G}(l,\epsilon)$; without loss of generality
we choose such a $t$ so that $t\geq 8$. Now, choose $l$ so that
$\mu(B_{l})<\frac{1}{t^{2}}$ and let $\epsilon<1$. The inequality in (2a)
provides us with $G$, $N_{a}$ and $D$ so that for all $m\geq N_{a}$,

$\displaystyle\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in
F_{m}:\mathbf{i}+G\in D\\}\right)}{\\#(I_{l,m})}$ $\displaystyle=$ (5.14)
$\displaystyle\frac{\mu\left(B_{l}\,\triangle\,\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)}{\mu(B_{l})}$
$\displaystyle<\,\epsilon.$

Note that the collection of cosets $D$ above must be nonempty, because
otherwise (5.14) would become $1<\epsilon$, which is impossible. So $D$ must
contain at least one coset in $\mathbb{Z}^{d}/G$, say $\mathbf{w}+G$.

Applying Lemma 2.7, there exists $\tilde{N}$ such that for all
$n\geq\tilde{N}$ and every $\mathbf{v}\in\mathbb{Z}^{d}$,

(5.15)
$\frac{\\#(F_{n}\cap(\mathbf{v}+G))}{\\#(F_{n})}\geq\frac{1/2}{[\mathbb{Z}^{d}:G]}\geq\frac{1}{2t}.$

Choose $n\geq\textrm{ max}\\{\tilde{N},N_{a}\\}$ so that the measure of the
error set of the tower $\mathcal{T}_{n}$ is less than $\frac{1}{2}$. We thus
have $\mu(\mathcal{T}_{n})>\frac{1}{2}$ which means
$\mu(B_{n})>\frac{1}{2}\frac{1}{\\#(F_{n})}$. This gives us

$\displaystyle\mu\left(\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{n},\mathbf{i}+G\in D]{}{\bigcup}\mathbf{T}^{i}(B_{n})\right)$
$\displaystyle=\\#(\mathbf{i}\in F_{n},\mathbf{i}+G\in D)\cdot\mu(B_{n})$
$\displaystyle>\\#(\mathbf{i}\in F_{n},\mathbf{i}+G\in
D)\cdot\frac{1}{2}\,\frac{1}{\\#(F_{n_{k}})}$
$\displaystyle\geq\\#(\mathbf{i}\in
F_{n},\mathbf{i}\in\mathbf{w}+G)\cdot\frac{1}{2}\,\frac{1}{\\#(F_{n_{k}})}$
(since $\mathbf{w}+G\in D$)
$\displaystyle\geq\left[\frac{1}{2t}\,\\#(F_{n})\right]\cdot\frac{1}{2}\,\frac{1}{\\#(F_{n_{k}})}$
(from (5.15)) $\displaystyle=\frac{1}{4t}.$

Thus when $m=n$, the set in the numerator of (5.14) is the symmetric
difference of a set of measure less than $\frac{1}{t^{2}}$ and a set of
measure at least $\frac{1}{4t}$. The smallest possible measure of the
symmetric difference of these two sets is when the small set is a subset of
the larger one, meaning

$\mu\left(B_{l}\,\triangle\,\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)>\frac{1}{4t}-\frac{1}{t^{2}}.$

Therefore

$\frac{\mu\left(B_{l}\,\triangle\,\stackrel{{\scriptstyle[}}{{\mathbf{i}}}\in
F_{m},\mathbf{i}+G\in
D]{}{\bigcup}\mathbf{T}^{\mathbf{i}}(B_{m})\right)}{\mu(B_{l})}>\frac{\frac{1}{4t}-\frac{1}{t^{2}}}{\frac{1}{t^{2}}}=\frac{t}{4}-1\geq\frac{8}{4}-1=1.$

and this cannot be less than $\epsilon$ when $\epsilon<1$, violating (2a).
Thus $\cup_{l,\epsilon}\mathcal{G}(l,\epsilon)$, and the set $\mathcal{F}$, is
infinite as wanted, and we have obtained an infinite odometer
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$. Write
$\mathfrak{G}^{\prime}=\\{G_{1}^{\prime},G_{2}^{\prime},G_{3}^{\prime},...\\}$.

In the final two steps of the proof, we show $(X,\mathbf{T})$ is conjugate to
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ by checking that
(2a) and (2b) of Theorem 5.1 hold, but not for the sequence
$\mathfrak{G}^{\prime}$; rather, we will check these properties for a
conjugate version of
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$ coming from a
different sequence of subgroups we now describe.

For each $l$ and $j\in\mathbb{N}$, choose a particular
$G_{l,\frac{1}{j}}\in\mathcal{G}(l,\frac{1}{j})$ and then set, for each
$k\in\mathbb{N}$,

$G_{k}=G_{k}^{\prime}\cap\left(\bigcap_{i=1}^{k}\bigcap_{j=1}^{k}G_{i,\frac{1}{j}}\right).$

Consider the odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ determined by
the sequence $\mathfrak{G}=\\{G_{1},G_{2},G_{3},...\\}$. We claim that
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ is conjugate to
$(X_{\mathfrak{G}^{\prime}},\sigma_{\mathfrak{G}^{\prime}})$, and to verify
this we will show $\mathcal{F}=\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})$
coincides with $\mathcal{FF}(\sigma_{\mathfrak{G}})$: the claim then follows
from Corollary 4.4.

Toward that end, let
$\tilde{G}\in\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})$. By Theorem 4.3
this means that $[\mathbb{Z}^{d}:\tilde{G}]<\infty$ and $\tilde{G}\geq
G_{N}^{\prime}$ for some $N$. But $G_{N}^{\prime}\geq G_{N}$ and thus
$\tilde{G}\geq G_{N}$ and, again by Theorem 4.3, this means
$\tilde{G}\in\mathcal{FF}(\sigma_{\mathfrak{G}})$ and so
$\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})\subseteq\mathcal{FF}(\sigma_{\mathfrak{G}})$.
Now let $G\in\mathcal{FF}(\sigma_{\mathfrak{G}})$. Again, this means
$[\mathbb{Z}^{d}:{G}]<\infty$ and ${G}\geq G_{N}$ for some $N$. We will show
$G\in\mathcal{F}=\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})$, completing
this step in the proof. Note that every
$G_{l,\epsilon}\in\mathcal{G}(l,\epsilon)$ lies in $\mathcal{F}$ by
definition. Since every $G_{N}^{\prime}$ lies in
$\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})$ and
$\mathcal{FF}(\sigma_{\mathfrak{G}^{\prime}})=\mathcal{F}$, we have every
$G_{N}^{\prime}\in\mathcal{F}$, too. Together these tells us that, since
$\mathcal{F}$ is closed under intersections, $G_{N}\in\mathcal{F}$. But
$\mathcal{F}$ is also closed under supersets, so we have $G\in\mathcal{F}$, as
needed.

For the final step in the proof, we will show that $(X,\mathbf{T})$ is
conjugate to $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ by checking that (2a)
and (2b) of Theorem 5.1 hold.

Fix $l\in\mathbb{N}$ and $\epsilon>0$. Recall that we choose
$G_{l,\frac{1}{k}}\in\mathcal{G}(l,\frac{1}{k})$ to satisfy (2a) and (2b) of
this theorem. Thus there exists $N_{a}$ (which we can assume is $>l$) where
for all $m\geq N_{a}$, there exists
$\widetilde{D}\subseteq\mathbb{Z}^{d}/G_{l,\frac{1}{k}}$ so that

(5.16) $\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in
F_{m}:\mathbf{i}+G_{l,\frac{1}{k}}\in\widetilde{D}\\}\right)}{\\#(I_{l,m})}<\frac{1}{k}.$

Fix $m\geq N_{a}$. As $G_{k}\leq G_{l,\frac{1}{k}}$, there is a natural
quotient map $q:\mathbb{Z}^{d}/G_{k}\to\mathbb{Z}^{d}/G_{l,\frac{1}{k}}$. Set
$D=q^{-1}(\widetilde{D})$. Then $D\subseteq\mathbb{Z}^{d}/G_{k}$ and if
$\mathbf{i}\in F_{m}$ and $\mathbf{i}+G_{k}\in D$, then
$\mathbf{i}+G_{l,\frac{1}{k}}\in\widetilde{D}$. Thus

$\frac{\\#\left(I_{l,m}\,\triangle\,\\{\mathbf{i}\in F_{m}:\mathbf{i}+G_{k}\in
D\\}\right)}{\\#(I_{l,m})}<\frac{1}{k}<\epsilon,$

giving us (2a) of Theorem 5.1.

To get (2b) of Theorem 5.1, recall that
$\mathcal{F}=\mathcal{FF}(\sigma_{\mathcal{G}^{\prime}})$ where
$\mathcal{G}^{\prime}=\\{G_{1}^{\prime},G_{2}^{\prime},...\\}$. Thus for each
$j$, $G_{j}^{\prime}\in\mathcal{F}$. By the description of $\mathcal{F}$ given
in (5.13), this means
$G_{j}^{\prime}\geq\stackrel{{\scriptstyle[}}{{i}}=1]{t}{\cap}G_{l_{i},\epsilon_{i}}$,
where $G_{l_{i},\epsilon_{i}}\in\mathcal{G}(l_{i},\epsilon_{i})$ for some
$l_{1},...,l_{t}$ and $\epsilon_{1},...,\epsilon_{t}$. By definition of
$\mathcal{G}(l_{i},\epsilon_{i})$, we know $G_{l_{i},\epsilon_{i}}$ satisfies
(2b) of this theorem for $l_{i}$ and $\epsilon_{i}$. By Theorem 3.1, this says
$G_{l_{i},\epsilon_{i}}\in\mathcal{FF}(T)$. Since $\mathcal{FF}(T)$ is closed
under intersections, we have
$\stackrel{{\scriptstyle[}}{{i}}=1]{t}{\cap}G_{l_{i},\epsilon_{i}}\in\mathcal{FF}(T)$,
and since it is also closed under supergroups, we get
$G_{j}^{\prime}\in\mathcal{FF}(T)$. By similar reasoning and its definition,
we then get $G_{j}\in\mathcal{FF}(T)$ for each $j$. By the definition of
$\mathcal{FF}(T)$, this says $(X,\mu,T)$ factors onto
$(\mathbb{Z}^{d}/G_{j},\delta,\tau)$ for each $j$ and thus by Theorem 4.1,
$(X,T)$ factors onto $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$. Corollary 4.2
then yields (2b) of Theorem 5.1 as wanted.

∎

##  6\. Example

In the example of Theorem 3.3 we indicated how the two-dimensional case is
different and more complicated than the one-dimensional case. We continue with
this type of investigation here and show that a $\mathbb{Z}^{2}$-action does
not automatically inherit the factoring properties of its subactions:

######  Theorem 6.1.

There is a Folner rank one action
$\mathbf{T}:\mathbb{Z}^{2}\curvearrowright[0,1]$ so that
$\mathbf{T}^{\mathbf{e}_{1}}$ and $\mathbf{T}^{\mathbf{e}_{2}}$ each factor
onto the dyadic $\mathbb{Z}$-odometer, but $\mathbf{T}$ does not factor onto
any free $\mathbb{Z}^{2}$-odometer.

###### Proof.

This action $\mathbf{T}$ will be constructed inductively by cutting and
stacking, as described in [PR] and [JS]. We thus will define a sequence of
towers $\mathcal{T}_{n}$ and an action $\mathbf{T}_{n}$ on each tower such
that in the limit we will have a $\mathbb{Z}^{2}$-action $\mathbf{T}$ on a
space of full measure.

Begin by cutting four disjoint subintervals from $[0,1]$ of equal length
$\delta_{1}<\frac{1}{4}$, a number that will be determined later. Associate
these subintervals to a $2\times 2$ grid and label the subintervals
accordingly by
$\\{L^{1}_{\mathbf{v}}:\mathbf{v}\in\mathbb{Z}^{2}\cap\left([0,2)\times[0,2)\right)\\}$;
we think of $\mathbf{v}$ as recording the position of the subinterval in a
tower $\mathcal{T}_{1}$. Let $E_{1}$ be the remaining portion of $[0,1]$: we
consider this the spacer portion for this step of the construction.

We then define, for $j\in\\{1,2\\}$, an action
$\mathbf{T}^{\mathbf{e}_{j}}_{1}:L^{1}_{\mathbf{v}}\to
L^{1}_{\mathbf{v}+\mathbf{e}_{j}}$ to be a linear map for any $\mathbf{v}$
where both $\mathbf{v}$ and $\mathbf{v}+\mathbf{e}_{j}$ lie in
$\mathbb{Z}^{2}\cap\left([0,2)\times[0,2)\right)$. In other words, the
functions $\mathbf{T}^{\mathbf{e}_{j}}_{1}$ map the interval at position
$\mathbf{v}$ to the interval at position $\mathbf{v}+\mathbf{e}_{j}$ by simple
translation. Observe that
$\mathbf{T}^{\mathbf{e}_{1}}\circ\mathbf{T}^{\mathbf{e}_{2}}=\mathbf{T}^{\mathbf{e}_{2}}\circ\mathbf{T}^{\mathbf{e}_{1}}$
where these maps are both defined, so we obtain a tower $\mathcal{T}_{1}$ of
shape $\mathbb{Z}^{2}\cap\left([0,2)\times[0,2)\right)$ and a
$\mathbb{Z}^{2}$-action $\mathbf{T}_{1}$ defined on part of $\mathcal{T}_{1}$.
Our final $\mathbb{Z}^{2}$-action $\mathbf{T}$ will coincide with
$\mathbf{T}_{1}$ on that portion of $[0,1]$ where $\mathbf{T}_{1}$ is defined.

For the inductive step, suppose we have a tower $\mathcal{T}_{n}$, which
consists of $2^{4^{n-1}}\times 2^{n}$ subintervals of equal length
$\delta_{n}$, and labeled by
$\\{L^{n}_{\mathbf{v}}:\mathbf{v}\in\mathbb{Z}^{2}\cap\left([0,2^{4^{n-1}})\times[0,2^{n})\right)\\}$.
Also suppose that we have maps
$\mathbf{T}^{\mathbf{e}_{j}}_{n}:L^{n}_{\mathbf{v}}\to
L^{n}_{\mathbf{v}+\mathbf{e}_{j}}$ that translate subinterval
$L^{n}_{\mathbf{v}}$ to subinterval $L^{n}_{\mathbf{v}+\mathbf{e}_{j}}$ for
any $\mathbf{v}$ where both $\mathbf{v}$ and $\mathbf{v}+\mathbf{e}_{j}$ lie
in $\mathbb{Z}^{2}\cap\left([0,2^{4^{n-1}})\times[0,2^{n})\right)$. Let
$E_{n}$ be the portion of $[0,1]$ not in $\mathcal{T}_{n}$.

To construct the tower $\mathcal{T}_{n+1}$ from $\mathcal{T}_{n}$, we first
cut each interval $L^{n}_{\mathbf{v}}$ into
$\left(2^{4^{n}-4^{n-1}}-1\right)\times 2$ equal size subintervals so that we
create $\left(2^{4^{n}-4^{n-1}}-1\right)\times 2$ copies of $\mathcal{T}_{n}$
which we then place into two slightly staggered rows. This is shown in Figure
1, where $t_{n}=\left(2^{4^{n}-4^{n-1}}-1\right)$ and $\mathcal{T}_{n}^{i}$
indicates the various copies of $\mathcal{T}_{n}$. The length of each
subinterval is thus
$\delta_{n+1}=\frac{\delta_{n}}{\left(2^{4^{n}-4^{n-1}}-1\right)\times 2}$: we
cut from $E_{n}$ enough subintervals of this same length to result in a tower
$\mathcal{T}_{n+1}$ of size $2^{4^{n}}\times 2^{n+1}$, filling these new
subintervals into the locations indicated by the starred areas in Figure 1.
Each of these subintervals is then labeled by
$\\{L^{n+1}_{\mathbf{v}}:\mathbf{v}\in\mathbb{Z}^{2}\cap\left([0,2^{4^{n}})\times[0,2^{n+1})\right)\\}$
to indicate its location in $\mathcal{T}_{n+1}$ and we let $E_{n+1}\subset
E_{n}$ be the remaining portion of $[0,1]$.

$\mathcal{T}_{n}^{1}$$\mathcal{T}_{n}^{2}$$\cdots$$\mathcal{T}_{n}^{t_{n}}$$\mathcal{T}_{n}^{t_{n}+1}$$\mathcal{T}_{n}^{t_{n}+2}$$\cdots$$\mathcal{T}_{n}^{2t_{n}}$$1$$*$$*$$*$$2^{4^{n-1}}$$2^{n}$
Figure 1. Construction of tower $\mathcal{T}_{n+1}$ from tower
$\mathcal{T}_{n}$.

We then define, for $j\in\\{1,2\\}$, a map
$\mathbf{T}^{\mathbf{e}_{j}}_{n+1}:L^{n+1}_{\mathbf{v}}\to
L^{n+1}_{\mathbf{v}+\mathbf{e}_{j}}$ by simple translation, for any
$\mathbf{v}$ where both $\mathbf{v}$ and $\mathbf{v}+\mathbf{e}_{j}$ lie in
$\mathbb{Z}^{2}\cap\left(\,[0,2^{4^{n}})\times[0,2^{n+1})\,\right)$. Observe
that $\mathbf{T}^{\mathbf{e}_{j}}_{n+1}$ coincides with
$\mathbf{T}^{\mathbf{e}_{j}}_{n}$ where these both are defined, and that
$\mathbf{T}^{\mathbf{e}_{1}}\circ\mathbf{T}^{\mathbf{e}_{2}}=\mathbf{T}^{\mathbf{e}_{2}}\circ\mathbf{T}^{\mathbf{e}_{1}}$
where both are defined. We thus obtain a tower $\mathcal{T}_{n+1}$ of shape
$\mathbb{Z}^{2}\cap\left([0,2^{4^{n}})\times[0,2^{n+1})\right)$ and a
$\mathbb{Z}^{2}$-action $\mathbf{T}_{n+1}$ defined on an increasingly larger
portion of $[0,1]$.

The total of the lengths of all the new subintervals introduced into
$\mathcal{T}_{n+1}$ is bounded above by $\frac{1}{2^{n}}\delta_{1}$, and thus
the sum of the lengths of all the intervals used in this construction is
bounded by $4\delta_{1}+\sum\frac{1}{2^{n}}\delta_{1}$: we can thus choose
$\delta_{1}$ so that the sum of the lengths of all the intervals used in the
construction is exactly 1. We then define
$\mathbf{T}:\mathbb{Z}^{2}\curvearrowright[0,1]$ by setting, for a.e.
$x\in[0,1]$,
$\mathbf{T}^{\mathbf{e}_{j}}(x)=\mathbf{T}^{\mathbf{e}_{j}}_{n}(x)$ where $n$
is large enough so that $x\in L^{n}_{\mathbf{v}}$ and both ${\mathbf{v}}$ and
${\mathbf{v}+\mathbf{e}_{j}}$ lie in
$\mathbb{Z}^{2}\cap\left([0,2^{4^{n-1}})\times[0,2^{n})\right)$. We have thus
defined a Følner rank one action
$\mathbf{T}:\mathbb{Z}^{2}\curvearrowright[0,1]$.

We now want to show that this $\mathbb{Z}^{2}$-action has the desired
properties, and we will begin by showing it does not factor onto any free
odometer. Suppose that $(X,\mathbf{T})$ factors onto some
$\mathbb{Z}^{d}$-odometer $(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ given by
the sequence of subgroups $\mathfrak{G}=\\{G_{j}\\}$. We will apply Corollary
4.2 to show that eventually, most descendants _cannot_ be congruent mod
$G_{j}$ unless $(1,0)\in G_{j}$. To do this, first fix $j$ and
$\epsilon<\frac{1}{4}$. Next, for any $m\in\mathbb{N},$ look at $I_{m,m+1}$
and note that, by the construction (as visualized by Figure 1),

$\displaystyle I_{m,m+1}$ $\displaystyle=\left\\{(k\cdot
2^{4^{m-1}},0):k\in\\{0,...,t_{m}-1\\}\right\\}$
$\displaystyle\qquad\bigsqcup\left\\{(1+k\cdot
2^{4^{m-1}},2^{m}):k\in\\{0,...,t_{m}-1\\}\right\\}.$

This means we can organize the elements in $I_{m,m+1}$ into pairs as each
$\mathbf{w}$ in the bottom row can be paired with $\mathbf{w}+(1,2^{m})$ in
the top row. But the two elements of each pair are not congruent$\mod G_{j}$
unless $(1,2^{m})\in G_{j}$. Thus it is impossible for there to exist one
single $\mathbf{g}\in\mathbb{Z}^{d}$ so that $\\#(\mathbf{i}\in
I_{m,m+1}:\mathbf{i}\not\equiv\mathbf{g}\mod G_{j})$ is less than a quarter of
$\\#(I_{m,m+1})$ unless $(1,2^{m})\in G_{j}$.

At the same time, we can also think of the elements in $I_{m,m+2}$ as divided
into four rows, at heights $0$, $2^{m}$, $2\cdot 2^{m}$ and $3\cdot 2^{m}$.
The elements in the middle two rows can be organized into pairs, as each
$\mathbf{w}$ in the row at height $2^{m}$ can be paired with
$\mathbf{w}+(0,2^{m})$ in the row at height $2\cdot 2^{m}$. The two elements
of each of these pairs are not congruent$\mod G_{j}$ unless $(0,2^{m})\in
G_{j}$, and so as before there cannot exist a particular
$\mathbf{g}\in\mathbb{Z}^{d}$ so that $\\#(\mathbf{i}\in
I_{m,m+1}:\mathbf{i}\not\equiv\mathbf{g}\mod G_{j})$ is less than a quarter of
$\\#(I_{m,m+1})$ unless $(0,2^{m})\in G_{j}$.

We have $(1,2^{m})\in G_{j}$ and $(0,2^{m})\in G_{j}$. As $G_{j}$ is a group,
we can conclude $(1,0)\in G_{j}$. This holds for every $j$, meaning
$(1,0)\in\bigcap_{j}G_{j}$, so by Theorem 2.9 this odometer
$(X_{\mathfrak{G}},\sigma_{\mathfrak{G}})$ onto which $(X,\mathbf{T})$ factors
cannot be free.

We will finish this example by showing that for $j\in\\{1,2\\}$,
$\mathbf{T}^{\mathbf{e}_{j}}:\mathbb{Z}\curvearrowright X$ does factor onto
the dyadic $\mathbb{Z}$-odometer, i.e. the action
$\sigma_{\mathfrak{G}}:\mathbb{Z}\curvearrowright X_{\mathfrak{G}}$ coming
from the sequence $\mathfrak{G}=\\{G_{1},G_{2},...\\}$ of subgroups of
$\mathbb{Z}$, where $G_{k}=2^{k}\mathbb{Z}$. Note that
$\mathbf{T}^{\mathbf{e}_{j}}$ is not rank one ($\mathbf{T}^{\mathbf{e}_{1}}$
is not ergodic, for example), so we cannot apply Corollary 4.2. Rather, we
will establish the existence of appropriate factor maps by applying Theorem
4.1. In other words, we will show that for every $k$,
$(X,\mu,\mathbf{T}^{\mathbf{e}_{j}})$ factors onto the finite action
$(\mathbb{Z}/2^{k}\mathbb{Z},\delta_{k},\tau_{k})$.

We first consider the horizontal direction $\mathbf{T}^{\mathbf{e}_{1}}$. Fix
$k$; choose $N_{0}$ so that $4^{N_{0}-1}\geq k$; this ensures
$2^{4^{N_{0}-1}}\geq 2^{k}$ (i.e. the horizontal length of tower
$\mathcal{T}_{N_{0}}$ is larger than $2^{k}$) and furthermore, that $2^{k}$
divides $2^{4^{N_{0}-1}}$. For notational convenience, we will set
$h_{0}=2^{4^{N_{0}-1}}$. For $x\in\mathcal{T}_{N_{0}}$, we have $x\in
L^{N_{0}}_{\mathbf{v}}$ for some $\mathbf{v}=(v_{1},v_{2})$. We can then
define $\pi(x)=v_{1}+2^{k}\mathbb{Z}$. For all other $x\in X$, we will make
use of its eventual appearance in some $\mathcal{T}_{n}$ and its location in
this $\mathcal{T}_{n}$ in relation to the tower $\mathcal{T}_{N_{0}}$ in order
to define $\pi(x)$. This will be done by induction.

Suppose $x\in\mathcal{T}_{N_{0}+1}-\mathcal{T}_{N_{0}}$. Then $x$ lies in one
of three regions of new spacers in $\mathcal{T}_{N_{0}+1}$ (see Figure 1).

  * •

The first possibility is that $x$ is in the bottom-right corner of the tower
$\mathcal{T}_{N_{0}+1}$; in this case $\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)$
lies in $\mathcal{T}_{N_{0}}$ and we can thus define
$\pi(x)=\pi(\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x))$.

  * •

The second possibility is that $x$ is in the top-left part of
$\mathcal{T}_{N_{0}+1}$; in this case, $\mathbf{T}^{h_{0}\mathbf{e}_{1}}(x)$
is in the tower $\mathcal{T}_{N_{0}}$, and we can set
$\pi(x)=\pi(\mathbf{T}^{h_{0}\mathbf{e}_{1}}(x))$.

  * •

The last possibility is when $x$ is in the top-right portion of
$\mathcal{T}_{N_{0}+1}$, in which $\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)$ lies
in $\mathcal{T}_{N_{0}}$ and we can define
$\pi(x)=\pi(\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x))$.

This defines $\pi$ on all $x\in\mathcal{T}_{N_{0}+1}$ and we can similarly
define $\pi$ on $\mathcal{T}_{N_{0}+2}-\mathcal{T}_{N_{0}+1}$ by where the
points on the new spacers lie in relationship to the tower
$\mathcal{T}_{N_{0}+1}$, which in turn relies on its location in relation to
the tower $\mathcal{T}_{N_{0}}$. For instance, if $x$ is in the bottom-right
corner of the tower $\mathcal{T}_{N_{0}+2}$, then
$\mathbf{T}^{-h_{1}\mathbf{e}_{1}}(x)$ lies in $\mathcal{T}_{N_{0}+1}$, where
$h_{1}$ is the horizontal length of tower $\mathcal{T}_{N_{1}}$. We can thus
define $\pi(x)=\pi(\mathbf{T}^{-h_{1}\mathbf{e}_{1}}(x))$, and similarly for
the other new points in $\mathcal{T}_{N_{0}+2}$. Continuing in this vein, we
will eventually have $\pi$ defined on almost all of $X$.

This yields a measurable map $\pi:X\to\mathbb{Z}/2^{k}\mathbb{Z}$. We need to
show $\pi$ is a factor map, i.e. that
$\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\tau_{k}(\pi(x))$ where $\tau_{k}(x)=x+1$
mod $2^{k}\mathbb{Z}$. This will also be done by induction.

We begin with the situation where
$x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}}$, and let
$\mathbf{v}=(v_{1},v_{2})$ be such that $x\in L^{N_{0}}_{\mathbf{v}}$. There
are two cases:

Case 1:

The point $x$ does not sit on the right edge of the tower
$\mathcal{T}_{N_{0}}$, so $v_{1}\neq h_{0}-1$. In this case,
$\mathbf{T}^{\mathbf{e}_{1}}x\in L^{N_{0}}_{\mathbf{v}+\mathbf{e}_{1}}$ and we
have

$\displaystyle\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=(v_{1}+1)+2^{k}\mathbb{Z}$
$\displaystyle=(v_{1}+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(v_{1}+2^{k}\mathbb{Z})=\tau_{k}(\pi(x)).$

Case 2:

The point $x$ does sit on the right edge of the tower $\mathcal{T}_{N_{0}}$,
so $v_{1}=h_{0}-1$. In this case, $\mathbf{T}^{\mathbf{e}_{1}}x\in
L^{N_{0}}_{\mathbf{w}}$ where $w_{1}=0$, and, using that $2^{k}$ divides
$h_{0}=2^{4^{N_{0}-1}}$, we thus have

$\displaystyle\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=0+2^{k}\mathbb{Z}=h_{0}+2^{k}\mathbb{Z}$
$\displaystyle=(h_{0}-1+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(h_{0}-1+2^{k}\mathbb{Z})=\tau_{k}(\pi(x)).$

Next, consider the situation where
$x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}+1}$. Where these two
points lie in $\mathcal{T}_{N_{0}+1}$ determine how $\pi$ is defined on them,
so we consider the following list of possibilities:

Case 1:

If $x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}}$, then by prior work
we know $\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\tau_{k}(\pi(x))$.

Case 2:

Suppose $x\in\mathcal{T}_{N_{0}}$ but
$\mathbf{T}^{\mathbf{e}_{1}}x\notin\mathcal{T}_{N_{0}}$. Then by construction,
$\mathbf{T}^{\mathbf{e}_{1}}x$ must lie in the bottom or top right section of
spacers indicated in Figure 1 and in fact must be in the first column of one
of those sections. Then, similarly to the previous Case 2,

$\displaystyle\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\pi(\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(\mathbf{T}^{\mathbf{e}_{1}}x))$
$\displaystyle=0+2^{k}\mathbb{Z}=(h_{0}-1+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(h_{0}-1+2^{k}\mathbb{Z})=\tau_{k}(\pi(x)).$

Case 3:

Suppose $\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}}$ but
$x\notin\mathcal{T}_{N_{0}}$. Then by construction, this means $x$ lies in the
upper left portion of spacers in Figure 1 and $\mathbf{T}^{\mathbf{e}_{1}}x$
must lie in the first column of $\mathcal{T}_{N_{0}}$. We thus have

$\displaystyle\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=0+2^{k}\mathbb{Z}$
$\displaystyle=(h_{0}-1+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(h_{0}-1+2^{k}\mathbb{Z})=\tau_{k}(\pi(\mathbf{T}^{h_{0}\mathbf{e}_{1}}(x)))=\,\,\,\tau_{k}(\pi(x)).$

Case 4:

Suppose both
$x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}+1}-\mathcal{T}_{N_{0}}$.
If they both lie in the bottom or top right sections of spacers from Figure 1,
then $\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)$ and
$\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(\mathbf{T}^{\mathbf{e}_{1}}x)$ will
maintain their relative positions and thus the definition of $\pi$ will yield
the result in a straightforward manner. Less obvious is the case where $x$
lies in the upper right region while $\mathbf{T}^{\mathbf{e}_{1}}x$ lies in
the upper left region. But because the upper right region has horizontal
length $h_{0}-1$, we then have that $\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)\in
L^{N_{0}}_{\mathbf{v}}$ with $v_{1}=h_{0}-2$. Meanwhile, having
$\mathbf{T}^{\mathbf{e}_{1}}x$ lie in the upper left region means
$\mathbf{T}^{h_{0}\mathbf{e}_{1}}(\mathbf{T}^{\mathbf{e}_{1}}x)\in
L^{N_{0}}_{\mathbf{v}}$ with $v_{1}=h_{0}-1$. We thus get

$\displaystyle\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\pi(\mathbf{T}^{h_{0}\mathbf{e}_{1}}(\mathbf{T}^{\mathbf{e}_{1}}x))$
$\displaystyle=h_{0}-1+2^{k}\mathbb{Z}$
$\displaystyle=(h_{0}-2+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(h_{0}-2+2^{k}\mathbb{Z})$
$\displaystyle=\tau_{k}(\pi(\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)))=\tau_{k}(\pi(x)).$

We now have that for all points with
$x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{N_{0}+1}$,
$\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\tau_{k}(\pi(x))$. Note that we did this by
considering the points associated to $x$ and $\mathbf{T}^{\mathbf{e}_{1}}x$ by
the definition of $\pi$ (e.g. $\mathbf{T}^{-h_{0}\mathbf{e}_{1}}(x)$, etc.).
These associated points are in $\mathcal{T}_{N_{0}}$, allowing us to apply
that we had already showed that
$\pi\circ\mathbf{T}^{\mathbf{e}_{1}}=\tau_{k}\circ\pi$ holds on
$\mathcal{T}_{N_{0}}$. The exact same reasoning can be used for induction; if
we assume that for all $x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{m}$,
$\pi(\mathbf{T}^{\mathbf{e}_{1}}x)=\tau_{k}(\pi(x))$, we can then show that it
also holds for all $x,\mathbf{T}^{\mathbf{e}_{1}}x\in\mathcal{T}_{m+1}$.
Having verified that $\pi$ intertwines $\mathbf{T}^{\mathbf{e}_{1}}$ with
$\tau$, Theorem 4.1 yields that $\mathbf{T}^{\mathbf{e}_{1}}$ factors onto the
dyadic odometer.

We finish by considering the vertical direction $\mathbf{T}^{\mathbf{e}_{2}}$.
Fix $k$; similar to before we seek a factor map $\pi$ from
$(X,\mathbf{T}^{\mathbf{e}_{2}})$ to $(\mathbb{Z}/2^{k}\mathbb{Z},\tau_{k})$.
First recall from the construction that when a tower $\mathcal{T}_{n}$ is cut
into $t_{n}$ copies and positioned as in Figure 1 to create tower
$\mathcal{T}_{n+1}$, any point $x\in L^{n}_{(v_{1},v_{2})}$ is moved to a
level $L^{n+1}_{(\widetilde{v}_{1},\widetilde{v}_{2})}$ where
$\widetilde{v}_{2}$ is either $v_{2}$ or $v_{2}+2^{n}$. We can then
unambigously define $\pi$ on all $\mathcal{T}_{n}$, $n\geq k$, by setting
$\pi(x)=v_{2}+2^{k}\mathbb{Z}$ where $x\in L^{n}_{(v_{1},v_{2})}$.

To show that $\pi$ intertwines the maps $\mathbf{T}^{\mathbf{e}_{2}}$ and
$\tau_{k}$, note that for almost every $x$, there exists some $n$ such that
both $x,\mathbf{T}^{\mathbf{e}_{2}}x\in\mathcal{T}_{n}$. We then have that
$x\in L^{n}_{(v_{1},v_{2})}$ and thus $\mathbf{T}^{\mathbf{e}_{2}}x\in
L^{n}_{(v_{1},v_{2}+1)}$ . Then
$\pi(\mathbf{T}^{\mathbf{e}_{2}}x)=v_{2}+1+2^{k}\mathbb{Z}=(v_{2}+2^{k}\mathbb{Z})+(1+2^{k}\mathbb{Z})=\tau_{k}(v_{2}+2^{k}\mathbb{Z})=\tau_{k}(\pi(x))$,
as wanted. By Theorem 4.1, $(X,\mathbf{T}^{\mathbf{e}_{2}})$ factors onto the
dyadic odometer. ∎

This example gives a specific $\mathbb{Z}^{2}$-action that does not factor
onto a free odometer, despite the fact that each
$(X,\mathbf{T}^{\mathbf{e}_{j}})$ does factor onto a free odometer, and
indicates how the $\mathbb{Z}^{2}$-action carries complexity beyond what can
be seen by the subactions.

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