# Rational self-affine tiles associated to standard and nonstandard digit
systems

Lucía Rossi  ,  Wolfgang Steiner  and  Jörg M. Thuswaldner  Montanuniversität
Leoben, Franz Josef Straße 18, A-8700 Leoben, Austria  Université de Paris,
CNRS, IRIF, F-75006 Paris, France<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

We consider digit systems $(A,\mathcal{D})$, where $A\in\mathbb{Q}^{n\times
n}$ is an expanding matrix and the _digit set_ $\mathcal{D}$ is a suitable
subset of $\mathbb{Q}^{n}$. To such a system, we associate a self-affine set
$\mathcal{F}=\mathcal{F}(A,\mathcal{D})$ that lives in a certain
representation space $\mathbb{K}_{A}$. If $A$ is an integer matrix, then
$\mathbb{K}_{A}=\mathbb{R}^{n}$, while in the general rational case
$\mathbb{K}_{A}$ contains an additional solenoidal factor. We give a criterion
for $\mathcal{F}$ to have positive Haar measure, i.e., for being a _rational
self-affine tile_. We study topological properties of $\mathcal{F}$ and prove
some tiling theorems. Our setting is very general in the sense that we allow
$(A,\mathcal{D})$ to be a nonstandard digit system. A standard digit system
$(A,\mathcal{D})$ is one in which we require $\mathcal{D}$ to be a complete
system of residue class representatives w.r.t. a certain naturally chosen
residue class ring. Our tools comprise the Frobenius normal form and character
theory of locally compact abelian groups.

###### Key words and phrases:

Self-affine set, tiling, digit system

###### 1991 Mathematics Subject Classification:

11A63, 28A80

The doctoral position of the first author is supported by the Austrian Science
Fund (FWF) as part of the Discrete Mathematics Doctoral Program, project
W1230. This work was supported by the PHC Amadeus / ÖAD Amadée project
“Topology, dynamics and number theory of fractal structures”. The second
author was supported by the Agence Nationale de la Recherche through the
project CODYS (ANR-18-CE40-0007).

##  1\. Introduction

This paper is a contribution to the theory of self-affine tiles whose
foundations were established in the early 1990s by Bandt [3], Kenyon [12],
Gröchenig and Haas [8], as well as Lagarias and Wang [16, 17, 18] and which
has gained a lot of attention in the past decades.

We recall the definition of a self-affine tile. Let $A\in\mathbb{R}^{n\times
n}$ be an expanding matrix (i.e., all its eigenvalues lie outside the unit
circle) with integer determinant, and let $\mathcal{D}\subset\mathbb{R}^{n}$
be a digit set with $|\mathcal{D}|=|\det A|$. Then we call the pair
$(A,\mathcal{D})$ a digit system. By Hutchinson [9], there exists a unique
nonempty compact subset $\mathcal{F}=\mathcal{F}(A,\mathcal{D})$ of
$\mathbb{R}^{n}$ that satisfies the set equation

(1.1) $A\mathcal{F}=\bigcup_{d\in\mathcal{D}}(\mathcal{F}+d).$

If $\mathcal{F}$ has positive Lebesgue measure, it is called a self-affine
tile. Of special interest are the so-called integral self-affine tiles (see
[16]), which are obtained when the matrix and the digits have integer
coefficients. By Bandt [3], the Lebesgue measure of an integral self-affine
tile $\mathcal{F}$ is certainly positive if $(A,\mathcal{D})$ is a standard
digit system, meaning that $\mathcal{D}$ is a complete set of residue class
representatives of $\mathbb{Z}^{n}/A\mathbb{Z}^{n}$. One very famous example
of such an integral self-affine tile is Knuth’s twin dragon (see [14, p.
206]), whose boundary is a fractal set. Lagarias and Wang [16, 17] also
regarded the matter of $(A,\mathcal{D})$ being nonstandard. In this case, for
a given matrix $A\in\mathbb{Z}^{n\times n}$ it is a highly nontrivial problem
to characterize all digit sets $\mathcal{D}$ for which
$\mathcal{F}(A,\mathcal{D})$ has positive Lebesgue measure (cf. [2, 19]).

Rational self-affine tiles are introduced by the second and thirds authors in
[26]. They constitute a natural generalization of integral self-affine tiles
to rational matrices which no longer need to have an integer determinant. In
particular, a rational self-affine tile is defined in terms of an expanding
matrix in $\mathbb{Q}^{n\times n}$ with irreducible characteristic polynomial
and a digit set taken from a $\mathbb{Z}$-module defined in terms of this
matrix. In the present paper we extend the theory of rational self-affine
tiles by taking arbitrary expanding rational matrices (this includes the
“reducible case”, as is referred to in [26]), and allowing nonstandard digit
systems, as well as defining a representation space in a somewhat more general
way. We provide results in the spirit of Lagarias and Wang [16, 17] for this
setting.

Let $A\in\mathbb{Q}^{n\times n}$ be an expanding matrix and let

$\mathbb{Z}^{n}[A]:=\bigcup_{k=1}^{\infty}\left(\mathbb{Z}^{n}+A\mathbb{Z}^{n}+\dots+A^{k-1}\mathbb{Z}^{n}\right)$

be the smallest nontrivial $A$-invariant $\mathbb{Z}$-module containing
$\mathbb{Z}^{n}$. We first define a digit system $(A,\mathcal{D})$ where $A$
acts as a base and $\mathcal{D}\subset\mathbb{Z}^{n}[A]$ is some finite digit
set, and explore its properties. In this digit system, the set
$\mathbb{Z}^{n}[A]$ plays the role of $\mathbb{Z}^{n}$ in the sense that
$(A,\mathcal{D})$ can be used to represent elements of $\mathbb{Z}^{n}[A]$. We
will always assume that the digit set $\mathcal{D}$ satisfies
$|\mathcal{D}|=|\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]|$, which turns out to be
a natural size for a digit set (such digit systems are studied for instance in
[10]). After that, in order to set up the definition of our tiles, we
introduce a representation space of the form
$\mathbb{K}_{A}:=\mathbb{R}^{n}\times\mathbb{Z}^{n}((A^{-1}))$, where
$\mathbb{Z}^{n}((A^{-1}))$ is a valuation ring of certain Laurent series of
powers of $A^{-1}$ with coefficients in $\mathbb{Z}^{n}$. The ring
$\mathbb{Z}^{n}((A^{-1}))$ is a solenoid, and in the one-dimensional case it
is isomorphic to the ring of $b$-adic numbers for some $b\in\mathbb{N}$ (which
is even a field when $b$ is a prime number). We establish a suitable
“diagonal” embedding $\varphi$ that maps the elements of $\mathbb{Z}^{n}[A]$
into $\mathbb{K}_{A}$ in a natural way. This allows us to define the main
object of study of this paper: the rational self-affine tile
$\mathcal{F}=\mathcal{F}(A,\mathcal{D})\subset\mathbb{K}_{A}$, which arises as
the unique nonempty compact solution of the set equation

$A\mathcal{F}=\bigcup_{d\in\mathcal{D}}(\mathcal{F}+\varphi(d)),$

and can be interpreted as the set of “fractional parts” of expansions in base
$A$ with digits in $\mathcal{D}$ (embedded in $\mathbb{K}_{A}$). Since we want
to study tilings of $\mathbb{K}_{A}$ induced by $\mathcal{F}$, we require
rational self-affine tiles to have positive measure, which always holds when
$\mathcal{D}$ is a complete set of residue class representatives of
$\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]$, and in this case we call
$(A,\mathcal{D})$ a standard digit system. However, we also allow
$(A,\mathcal{D})$ to be nonstandard, and give a criterion in terms of the
digits to guarantee positive measure of $\mathcal{F}$ in this general setting.
We prove some topological properties of rational self-affine tiles, as well as
the existence of two tilings given by translations of $\mathcal{F}$: the first
one has a translation set defined in terms of the digits $\mathcal{D}$, and
the second one is a multiple tiling where the translation set is a lattice
obtained from the ring $\mathbb{Z}^{n}[A]$ by embedding it into the space
$\mathbb{K}_{A}$. Before arriving to the proof of the existence of the
multiple tiling, we present a careful analysis of the character group of the
locally compact abelian group $\mathbb{K}_{A}$. The article [24] deals with a
special one-dimensional case of what we present in this paper. It is
accessible to a general audience and serves as an introduction to the topic.

Our theory presents three main features that make its study difficult and
rich. First of all, we treat the $n$-dimensional case: when dealing with
matrices, computing quotients is not always so simple, and we make use of some
machinery of linear algebra (like the Frobenius normal form) to solve some of
these issues. Secondly, we deal with a space that has a $A^{-1}$-adic factor,
and on it we define an ultrametric. More challenges arrive when we study the
character group of the representation space. Finally, we consider nonstandard
digit sets. This implies that sometimes we lose the group structure when
considering certain digit expansions, which makes it harder to define tilings.

##  2\. Setting and definitions

In this section we introduce digit sets and present a way to compute their
cardinality. After that we define the representation space, and finally arrive
at the definition of a rational self-affine tile.

What we are doing generalizes well-known facts on “ordinary” $\alpha$-ary
expansions. In particular, let $\alpha$ be an integer with $|\alpha|>1$. Every
real number has an expansion

$\pm(d_{m}d_{m-1}\ldots
d_{0}.d_{-1}d_{-2}\ldots)_{\alpha}:=\pm\sum_{j=-\infty}^{m}d_{j}\alpha^{j}$

in base $\alpha$, where every $d_{j}$ is a digit taken from the set
$\\{0,1,\ldots,|\alpha|-1\\}$ (the sign “$\pm$” is required only for positive
$\alpha$). We can associate to this digit system the set of integer
expansions, that is, the set of numbers that can be expressed using only
nonnegative powers of the base $\alpha$; it equals the lattice $\mathbb{Z}$.
Besides that there is the set of fractional parts, namely, the numbers that
can be expressed using only negative powers of the base $\alpha$ (excluding
the added $``-"$ sign at the beginning). This set is a compact interval given
by $J=[0,1]$ for $\alpha>0$ and
$J=\big{[}\tfrac{\alpha}{1-\alpha},\tfrac{1}{1-\alpha}\big{]}$ for $\alpha<0$.
The collection $\\{J+z\,|\,z\in\mathbb{Z}\\}$ forms a tiling of $\mathbb{R}$,
and this property geometrically reflects the fact that almost every real has a
unique $\alpha$-ary expansion. In our general setting, $\mathbb{Z}^{n}[A]$
will play the role of $\mathbb{Z}$, the rational self-affine tile
$\mathcal{F}$ will play the role of $[0,1]$, and $\mathbb{K}_{A}$ will play
the role of $\mathbb{R}$.

###  2.1. Digit systems with rational matrices

Many generalizations of $\alpha$-ary expansions (also known as radix
expansions or radix representations) have been studied. Kempner [11] and later
Rényi [22] proposed expansions with respect to nonintegral real bases. Knuth
[13] introduced complex bases and related them to fractal sets. Tilings of
$\mathbb{R}^{n}$ arising form radix expansions were studied by Vince [27], and
Kovács [15] considered digit systems in finite dimensional Euclidean spaces.
Akiyama et al. [1] introduced a type of expansion where the base is a rational
number, and generalizations have been studied in [4] and [25]. In [26], the
authors considered digit expansions where the base is an algebraic number,
which is equivalent to taking expansions w.r.t. rational matrices with
irreducible characteristic polynomial. The setting from the present paper is a
generalization of the one in [26] for arbitrary expanding rational matrices.

Following Kovács [15], given an integer matrix $A\in\mathbb{Z}^{n\times n}$,
we can consider expansions in base $A$ with digits in a certain set
$\mathcal{D}\subset\mathbb{Z}^{n}$, meaning we look at ways of expanding a
vector $x\in\mathbb{R}^{n}$ in the form

(2.1) $x=\sum_{j=-\infty}^{k}A^{j}d_{j},\quad d_{j}\in\mathcal{D},$

and this is linked to the study of integral self-affine tiles that we have
mentioned before (see [16]). We need the assumption that $A$ is expanding in
order for this series to converge.

If we consider a rational matrix $A\in\mathbb{Q}^{n\times n}$, then
$\mathbb{Z}^{n}[A]$ is the natural generalization of $\mathbb{Z}^{n}$ (we
refer to [10] for more on rational matrix digit systems). This leads to the
following definition.

######  Definition 2.1 (Digit system).

Let $A\in\mathbb{Q}^{n\times n}$ be an expanding matrix and let
$\mathcal{D}\subset\mathbb{Z}^{n}[A]$ be such that
$|\mathcal{D}|=|\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]|.$ Then we say that
$(A,\mathcal{D})$ constitutes a digit system, where $A$ is the base and
$\mathcal{D}$ is the digit set. When $\mathcal{D}$ is a complete set of
residue class representatives of $\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]$, we
say that $(A,\mathcal{D})$ is a standard digit system (following [16, p.
163]). Otherwise, we say that $(A,\mathcal{D})$ is a nonstandard digit system.

In connection with digit systems, finite expansions are desirable. For digit
systems $(A,\mathcal{D})$, the _finiteness property_ , stating that every
vector $x\in\mathbb{Z}^{n}[A]$ has a finite expansion of the form
$x=A^{k}d_{k}+\dots+Ad_{1}+d_{0}$ has been studied extensively (see [10] and
the references given there). It is easily seen that the requirement that
$\mathcal{D}$ is a complete system of residue class representatives of
$\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]$ is a necessary, but in general not
sufficient, condition for $(A,\mathcal{D})$ to have the finiteness property.
Eventually periodic expansions have also been investigated.

Computing the size of a digit set for a given expanding matrix
$A\in\mathbb{Q}^{n\times n}$ amounts to computing the order of the quotient
group $\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]$, and this is not always
straightforward. From here onwards, set

(2.2) $a:=|\mathbb{Z}^{n}[A]/A\mathbb{Z}^{n}[A]|,\quad
b:=|\mathbb{Z}^{n}[A^{-1}]/A^{-1}\mathbb{Z}^{n}[A^{-1}]|.$

We now show how to make use of the Frobenius normal form of $A$ to compute $a$
and $b$, and we prove that $|\det A|=\frac{a}{b}$, which will be crucial
later. Let $A\in\mathbb{Q}^{n\times n}$ with characteristic polynomial
$\chi_{A}$ be given. Consider the space $\mathbb{Q}^{n}$ regarded as a
finitely generated $\mathbb{Q}[t]$-module with the action of $t$ given by
multiplication by $A$, that is, if $v\in\mathbb{Q}^{n}$ then $t\cdot v:=Av$,
and the action can be linearly extended to all elements in $\mathbb{Q}[t]$.
According to the structure theorem for finitely generated modules over
principal ideal domains (see [5, Chapter 12, Theorem 6]), there exists an
isomorphism of the form

$\mathbb{Q}^{n}\simeq\bigoplus_{i=1}^{k}\mathbb{Q}[t]/(p_{i}),$

where $p_{i}\in\mathbb{Q}[t]$ are the so-called invariant factors of
$\mathbb{Q}^{n}$, with the divisibility properties $p_{1}\mid
p_{2}\mid\ldots\mid p_{k}\mid\chi_{A}$. The polynomials $p_{i}$ are assumed to
be monic, and with this assumption they are unique. This implies that $A$ is
similar to a block diagonal matrix $F=\mbox{diag}(C_{1},\ldots.C_{k})$, where
$C_{i}$ is the companion matrix of $p_{i}$ ($1\leqslant i\leqslant k$). $F$ is
the well-known Frobenius normal from of $A$, also called rational canonical
form, see [5, Section 12.2]. Using this notation, we get the following result
which shows how to compute the value of $a$.

######  Proposition 2.2.

Let $A\in\mathbb{Q}^{n\times n}$ be given, let $p_{i}$ $(1\leqslant i\leqslant
k)$ be the corresponding invariant factors, and consider the integer
polynomials $q_{i}=c_{i}p_{i}\in\mathbb{Z}[t]$, where each
$c_{i}\in\mathbb{Z}$ is chosen so that $q_{i}$ has coprime coefficients. Let
$q_{i}^{*}\in\mathbb{Z}[t]$ be the reciprocal polynomial of $q_{i}$, namely
$q_{i}^{*}(t):=t^{\deg{(q_{i})}}q_{i}(t^{-1})$. Then

(2.3) $a=\prod_{i=1}^{k}|q_{i}(0)|,\quad b=\prod_{i=1}^{k}|q_{i}^{*}(0)|,$

and $|\det A|=\frac{a}{b}$.

###### Proof.

Let $F=\mbox{diag}(C_{1},\ldots,C_{k})$ be the Frobenius normal form of $A$.
Then it is clear that

(2.4)
$a=\prod_{i=1}^{k}|\mathbb{Z}^{\deg(q_{i})}[C_{i}]/C_{i}\mathbb{Z}^{\deg(q_{i})}[C_{i}]|.$

Let $C\in\mathbb{Q}^{m\times m}$ be the companion matrix of some polynomial
$p\in\mathbb{Q}[t]$ and let $q=cp\in\mathbb{Z}[t]$, where $c\in\mathbb{Z}$ is
chosen so that $q$ has coprime coefficients. We claim that

(2.5) $\mathbb{Z}^{m}[C]\simeq\mathbb{Z}[t]/(q).$

To prove this, let $v\in\mathbb{Z}^{m}[C]$ be given; then it can be expressed
as $v=\sum_{j=0}^{L}C^{j}v_{j}$ with $v_{j}\in\mathbb{Z}^{m}$, $L\geqslant 0$.
For $1\leqslant j\leqslant m$, denote by $\mathbf{e}_{j}\in\mathbb{Q}^{m}$ the
$j$-th canonical basis vector. Clearly, each $v_{j}$ can be expressed in the
canonical basis with integer coefficients. Because $C$ is a companion matrix,
it follows that $C\mathbf{e}_{j}=\mathbf{e}_{j+1}$ for $1\leqslant j<m$, so
all this yields that $v$ is of the form

(2.6) $v=\sum_{j=0}^{\ell}b_{j}C^{j}\mathbf{e}_{1},$

for some $\ell\in\mathbb{N}$ minimal and $b_{0},\dots,b_{\ell}\in\mathbb{Z}$.
If $\ell\geqslant m$, we will show that we can take
$b_{m},\ldots,b_{\ell}\in\\{0,\ldots,|q^{*}(0)|-1\\}$ (here,
$q^{*}\in\mathbb{Z}[t]$ is the reciprocal polynomial of $q$, so $q^{*}(0)$
corresponds to the leading coefficient of $q$). Indeed, note that $q(C)=0$
because $C$ is the companion matrix of $q$. Suppose that $b_{\ell}$ is not in
$\\{0,\ldots,|q^{*}(0)|-1\\}$; then one can add or subtract $C^{\ell-m}q(C)=0$
in order to obtain another expression on the right side of (2.6), without
altering the value of $v$. This can be done the appropriate number of times,
so we can assume w.l.o.g. that $b_{\ell}\in\\{0,\ldots,|q^{*}(0)|-1\\}$.
Repeating this for $\ell-1,\ell-2,\dots,m$, one arrives at
$b_{m},\ldots,b_{\ell}\in\\{0,\ldots,|q^{*}(0)|-1\\}$, and the representation
(2.6) with this property and $\ell$ minimal is unique.

By [25, Lemma 4.1] each polynomial $r\in\mathbb{Z}[t]/(q)$ can be expressed
uniquely as $r=r^{\prime}+\sum_{j=m}^{\ell}r_{j}t^{j}\mod q$, with
$r^{\prime}\in\mathbb{Z}[x]$, $\deg(r^{\prime})<m$, $\ell\in\mathbb{N}$ and
$r_{m},\dots,r_{\ell}\in\\{0,\dots,|q^{*}(0)|-1\\}.$ Using this, one easily
checks that

$h:\mathbb{Z}^{m}[C]\to\mathbb{Z}[t]/(q);\quad\sum_{i=0}^{\ell}b_{i}C^{i}\mathbf{e}_{1}\mapsto\sum_{i=0}^{\ell}b_{i}t^{i}$

is an isomorphism and the claim in (2.5) is proved.

Because $t\,h(G)=h(CG)$ for any $G\in\mathbb{Z}^{m}[C]$, this isomorphism
implies that

$\mathbb{Z}^{m}[C]/C\mathbb{Z}^{m}[C]\simeq\mathbb{Z}[t]/(q,t).$

It is easy to check that $|\mathbb{Z}[t]/(q,t)|=|q(0)|$ (see [25, p. 1460])
and, hence, we have that

(2.7) $|\mathbb{Z}^{m}[C]/C\mathbb{Z}^{m}[C]|=|q(0)|.$

Applying (2.7) for $q=q_{i}$ $(1\leqslant i\leqslant k)$ in (2.4), the left
equation of (2.3) follows. The right equation of (2.3) is proved in the same
way by replacing $A$ by $A^{-1}$. The assertion $\det A=\frac{a}{b}$ follows
from (2.3) (recall the definition of $q_{i}$ and the fact that each $p_{i}$ is
monic), because

$|\det A|=\prod_{i=1}^{k}|\det
C_{i}|=\prod_{i=1}^{k}|p_{i}(0)|=\prod_{i=1}^{k}\frac{|q_{i}(0)|}{|q_{i}^{*}(0)|}=\frac{a}{b}.\qed$

###  2.2. The representation space

Properties of digit systems and digit expansions can be reflected
geometrically via self-affine sets and tilings. The space $\mathbb{K}_{A}$,
defined in what follows for a given $A\in\mathbb{Q}^{n\times n}$, will turn
out to be a natural space where these sets and tilings can be defined. Suppose
that we wanted to define a set
$\mathcal{F}(A,\mathcal{D})\subset\mathbb{R}^{n}$ satisfying (1.1). In order
to define a tiling, we would require the union on the right side of (1.1) to
be essentially disjoint. We know that the action of $A$ scales the Lebesgue
measure of a set by $|\det A|$, so if $A$ has a nonintegral determinant and
$\mathcal{F}$ has positive measure, we would need $\mathcal{D}$ to have a
nonintegral amount of digits, which is of course not doable. We will show that
the action of $A$ in $\mathbb{K}_{A}$ multiplies the Haar measure of a set by
$a$, where $a\in\mathbb{N}$ is as in (2.2).

Standard digit systems $(A,\mathcal{D})$ where the characteristic polynomial
$\chi_{A}$ of $A\in\mathbb{Q}^{n\times n}$ is irreducible are considered in
[26]. Rational self-affine tiles are introduced as subsets of a representation
space of the form $\mathbb{R}^{n}\times\prod_{\mathfrak{p}}K_{\mathfrak{p}}$,
where each $K_{\mathfrak{p}}$ is a completion of a number field $K$ (defined
in terms of $\chi_{A}$) with respect to a certain absolute value
$|\cdot|_{\mathfrak{p}}$. The representation space $\mathbb{K}_{A}$ from the
present paper is a generalization of this, and can be defined in a simpler
way. In the irreducible case, both settings are isomorphic.

Let $A\in\mathbb{Q}^{n\times n}$ be expanding. For convenience of notation,
from here onwards we set

$B:=A^{-1}.$

Consider the ring

$\mathbb{Z}^{n}(B)=\bigcup_{k\geqslant
1}(B^{-k}\mathbb{Z}^{n}+B^{-k+1}\mathbb{Z}^{n}+\dots+B^{k}\mathbb{Z}^{n})$

(note that $\mathbb{Z}^{n}(A)=\mathbb{Z}^{n}(B)$). We define on
$\mathbb{Z}^{n}(B)$ the $B$-adic valuation
$\nu:\mathbb{Z}^{n}(B)\rightarrow\mathbb{Z}\cup\\{\infty\\}$ as

(2.8) $\nu(y):=\begin{cases}\min\\{k\in\mathbb{Z}\;|\;y\in
B^{k}\mathbb{Z}^{n}[B]\setminus B^{k+1}\mathbb{Z}^{n}[B]\\},&y\neq 0,\\\
\infty,&y=0.\end{cases}$

On $\mathbb{Z}^{n}(B)$ the $B$-adic metric is defined by

(2.9) ${\bf d}_{B}(y,y^{\prime}):=b^{-\nu(y-y^{\prime})},$

for $b$ as in (2.2) and $y,y^{\prime}\in\mathbb{Z}^{n}(B)$, with the
convention that $b^{-\infty}=0$.

######  Definition 2.3 ($B$-adic series).

We define the space $\mathbb{Z}^{n}((B))$ of $B$-adic series as the completion
of $\mathbb{Z}^{n}(B)$ with respect to the metric ${\bf d}_{B}$.

We extend the metric ${\bf d}_{B}$ to the completion $\mathbb{Z}^{n}((B))$,
and hence we extend the $B$-adic valuation $\nu$ to $\mathbb{Z}^{n}((B))$ so
that it satisfies (2.9). Then, every nonzero $y\in\mathbb{Z}^{n}((B))$ can be
expressed as Laurent series

(2.10) $y=\sum_{j=\nu(y)}^{\infty}B^{j}y_{j},\quad y_{j}\in\mathbb{Z}^{n},$

of powers of $B$ with coefficients in $\mathbb{Z}^{n}$, which converges w.r.t.
the metric ${\bf d}_{B}$. Then $\nu(y)$ is the smallest index such that $y$
has an expansion (2.10) with $y_{\nu(y)}\neq 0$. We denote by
$\mathbb{Z}^{n}[[B]]$ the subring of $\mathbb{Z}^{n}((B))$ consisting of
points $y\in\mathbb{Z}^{n}((B))$ with $\nu(y)\geqslant 0$, the ring of power
series in $B$ with coefficients in $\mathbb{Z}^{n}$. The $B$-adic metric
satisfies the ultrametric inequality, namely

${\bf d}_{B}(y,y^{\prime})\leqslant\max\\{{\bf d}_{B}(y,y^{\prime\prime}),{\bf
d}_{B}(y^{\prime\prime},y^{\prime})\\}$

for every $y,y^{\prime},y^{\prime\prime}\in\mathbb{Z}^{n}((B))$. This metric
turns $\mathbb{Z}^{n}((B))$ into a complete separable space, which is also a
locally compact topological group. Thus there is a Haar measure $\mu_{B}$ on
$\mathbb{Z}^{n}((B))$ which is normalized in a way that
$\mu_{B}(\mathbb{Z}^{n}[[B]])=1$, and we call it the $B$-adic measure. If
$M\subset\mathbb{Z}^{n}((B))$ is a measurable set, then

(2.11) $\mu_{B}(A^{k}M)=b^{k}\mu_{B}(M).$

######  Remark 2.4.

Let $n=1$. In this case, $A=\frac{a}{b}$ where $a$ and $b$ are coprime
integers. Suppose $b\geqslant 2$. We obtain that
$\mathbb{Z}^{n}((B))\simeq\mathbb{Q}_{b}$, where $\mathbb{Q}_{b}$ is the ring
of $b$-adic numbers, and $\mathbb{Z}^{n}[[B]]\simeq\mathbb{Z}_{b}$, where
$\mathbb{Z}_{b}$ is the ring of $b$-adic integers.

######  Definition 2.5 (The representation space).

Given an expanding matrix $A\in\mathbb{Q}^{n\times n}$, define the
representation space $\mathbb{K}_{A}$ as

$\mathbb{K}_{A}:=\mathbb{R}^{n}\times\mathbb{Z}^{n}((B)).$

We endow the space $\mathbb{K}_{A}$ with the following structures:

  1. (1)

It inherits the structure of an additive group from its cartesian factors.

  2. (2)

Consider the group of matrices given by

$\mathbb{Z}[A]:=\bigcup_{k\geqslant
1}(\mathbb{Z}A^{-k}+\mathbb{Z}A^{-k+1}+\dots+\mathbb{Z}A^{k}).$

Then $\mathbb{Z}[A]$ acts on $\mathbb{K}_{A}$ by multiplication, i.e.,
$G\,(x,y)=(Gx,Gy)$ if $G\in\mathbb{Z}[A]$ and $(x,y)\in\mathbb{K}_{A}$.

  3. (3)

We define the metric

$\textbf{d}((x,y),(x^{\prime},y^{\prime})):=\mbox{max}\\{\|x-x^{\prime}\|,{\bf
d}_{B}(y,y^{\prime})\\},$

for $(x,y),(x^{\prime},y^{\prime})\in\mathbb{K}_{A}$, where $\|\cdot\|$
denotes the Euclidean norm in $\mathbb{R}^{n}$ and ${\bf d}_{B}$ is the
$B$-adic metric in $\mathbb{Z}^{n}((B))$. This turns $\mathbb{K}_{A}$ into a
locally compact topological group. It is easy to check that, for every closed
ball ${\bf B}_{r}(x,y)$ of radius $r>0$ and center $(x,y)\in\mathbb{K}_{A}$,
there is a decomposition

${\bf B}_{r}(x,y)={\bf B}_{r}(x)\times{\bf B}_{r}(y),$

into closed balls on each respective space. This characterizes the topology of
$\mathbb{K}_{A}$.

  4. (4)

We define a measure $\mu$ in $\mathbb{K}_{A}$ as the product measure

$\mu:=\lambda\times\mu_{B},$

with $\lambda$ being the Lebesgue measure in $\mathbb{R}^{n}$ and $\mu_{B}$
the $B$-adic measure in $\mathbb{Z}^{n}((B))$. Then $\mu$ is the Haar measure
on $\mathbb{K}_{A}$ satisfying $\mu([0,1]\times\mathbb{Z}^{n}[[B]])=1$.

######  Remark 2.6.

Note that, when $A\in\mathbb{Z}^{n\times n}$ is an integer matrix, the space
$\mathbb{Z}^{n}((B))$ is trivial and plays no role, and hence
$\mathbb{K}_{A}=\mathbb{R}^{n}$. However, $\mathbb{K}_{A}=\mathbb{R}^{n}$ may
also happen in the noninteger case: for example, if
$A=\begin{pmatrix}2&\frac{1}{2}\\\ 0&3\end{pmatrix}$ we have $b=1$. Suppose
$b=1$, i.e., that $\mathbb{Z}^{n}[B]/B\mathbb{Z}^{n}[B]$ is trivial. Then by
Lemma 2.2 $\det A=a$ is an integer. The results presented in Section 3 are
proven by Lagarias and Wang in [17] for real expanding matrices with integer
determinant. For this reason, in all that follows we assume $b\geqslant 2$.

######  Lemma 2.7.

If $M\subset\mathbb{K}_{A}$ is a measurable set, then $\mu(AM)=a\,\mu(M)$.

###### Proof.

Consider a measurable subset of $\mathbb{K}_{A}$ of the form $M_{1}\times
M_{2}$, where $M_{1}\subset\mathbb{R}^{n}$ and
$M_{2}\subset\mathbb{Z}^{n}((B))$ are both measurable sets. We have seen in
Proposition 2.2 that $\det A=\frac{a}{b}$. Then

$\mu(A(M_{1}\times
M_{2}))=\lambda(AM_{1})\,\mu_{B}(AM_{2})=\frac{a}{b}\,\lambda(M_{1})\,b\,\mu_{B}(M_{2})=a\,\mu(M_{1}\times
M_{2}).$

Since $\mu=\lambda\times\mu_{B}$ is a product measure, the $\sigma$-algebra of
$\mu$-measurable sets is generated by sets of the form $M_{1}\times M_{2}$.
Therefore, if $M\subset\mathbb{K}_{A}$ is measurable we have
$\mu(AM)=a\,\mu(M).$ ∎

The previous lemma implies that, in some sense, the base $A$ has “enough
space” for $a$ digits when the digit system is embedded in $\mathbb{K}_{A}$.
In fact, when considering a rational matrix $A$ with integer determinant, it
suffices to take $|\mathcal{D}|=|\det A|$, because $A$ acts in some sense like
an integer matrix: the action of $A$ scales the measure of a set by an integer
factor. When $A$ has nonintegral determinant, it turns turns out that $A$ acts
in $\mathbb{K}_{A}$ like an integer matrix, and that is why this space is
appropriate for our purposes. This relation between the measure and the
cardinality of the digit set will be important in all that follows.

###  2.3. Rational self-affine tiles

We proceed to introduce a set $\mathcal{F}$ associated to the digit system
$(A,\mathcal{D})$ that reflects features of its structure. It can be regarded
as the set of “fractional parts” of the digit system, in the sense that it
plays the same role as the interval $[0,1]$ does for the decimal digit system.
In Section 3 we will study topological properties of $\mathcal{F}$.

We need the following lemma, which is in the spirit of Lind [20].

######  Lemma 2.8.

Let $A\in\mathbb{Q}^{n\times n}$ be expanding and assume $b\geqslant 2$, with
$b$ as in (2.2). Then there exists a metric $\boldsymbol{\ell}$ on
$\mathbb{K}_{A}$ w.r.t. which the action of $B=A^{-1}$ is a contraction. In
particular, there exists $0\leqslant\kappa<1$ such that

(2.12)
$\boldsymbol{\ell}(B\,(x,y),B\,(x^{\prime},y^{\prime}))\leqslant\kappa\,\boldsymbol{\ell}((x,y),(x^{\prime},y^{\prime}))\quad((x,y),(x^{\prime},y^{\prime})\in\mathbb{K}_{A}).$

###### Proof.

Let $\mathop{Spec}(A)$ denote the set of eigenvalues of $A$. Since $A$ is
expanding, there exists $\rho\in\mathbb{R}$ such that
$1<\rho<\min\\{|\eta|\;|\;\eta\in\mathop{Spec}(A)\\}$. For
$x\in\mathbb{R}^{n}$, define

(2.13) $\|x\|^{\prime}:=\sum_{k=0}^{\infty}\rho^{k}\,\|B^{k}x\|.$

Since all the eigenvalues of $\rho\,B=\rho A^{-1}$ are strictly smaller than
$1$ in modulus, the series on the right hand side of (2.13) converges and
$\|\cdot\|^{\prime}$ becomes a norm in $\mathbb{R}^{n}$ that satisfies

$\|B\,x\|^{\prime}=\frac{1}{\rho}\sum_{k=1}^{\infty}\rho^{k}\,\|B^{k}\,x\|\leqslant\frac{1}{\rho}\,\|x\|^{\prime}.$

Also, for all $y,y^{\prime}\in\mathbb{Z}^{n}((B))$, it follows from the
definition of the $B$-adic metric that ${\bf
d}_{B}(By,By^{\prime})=\frac{1}{b}\,{\bf d}_{B}(y,y^{\prime}).$ Let
$(x,y),(x^{\prime},y^{\prime})\in\mathbb{K}_{A}$. Define on $\mathbb{K}_{A}$
the metric $\boldsymbol{\ell}$ given by

$\boldsymbol{\ell}((x,y),(x^{\prime},y^{\prime})):=\mbox{max}\\{\|x-x^{\prime}\|^{\prime},{\bf
d}_{B}(y,y^{\prime})\\}.$

Then (2.12) follows with $\kappa:=\max\\{\frac{1}{\rho},\frac{1}{b}\\}<1$. ∎

Note that ${\bf d}$ and $\boldsymbol{\ell}$ are equivalent, because
$\|\cdot\|^{\prime}$ is equivalent to $\|\cdot\|$.

We now introduce a suitable way to embed our digit system into the
representation space $\mathbb{K}_{A}$. Define the diagonal embedding $\varphi$
as

$\varphi:\mathbb{Z}^{n}(B)\rightarrow\mathbb{K}_{A};\quad x\mapsto(x,x).$

######  Definition 2.9 (Rational self-affine tile).

Let $(A,\mathcal{D})$ be a digit system. Define
$\mathcal{F}=\mathcal{F}(A,\mathcal{D})\subset\mathbb{K}_{A}$ as the unique
nonempty compact set satisfying the set equation

(2.14) $A\mathcal{F}=\bigcup_{d\in\mathcal{D}}(\mathcal{F}+\varphi(d)).$

If $\mu(\mathcal{F})>0$, then $\mathcal{F}$ is called a rational self-affine
tile.

Because $A$ is expanding, Lemma 2.8 implies that the mapping

$\mathbb{K}_{A}\to\mathbb{K}_{A};\quad(x,y)\mapsto A^{-1}((x,y)+\varphi(d))$

is a contraction for each $d\in\mathcal{D}$. Let $\mathcal{H}(K)$ be the
family of nonempty compact subsets of $\mathbb{K}_{A}$, and consider the map

(2.15) $\varPsi:\mathcal{H}(K)\to\mathcal{H}(K);\quad
X\mapsto\bigcup_{d\in\mathcal{D}}A^{-1}(X+\varphi(d)).$

By Hutchinson [9], there is a unique nonempty compact set which is a fixed
point of $\varPsi$, hence $\mathcal{F}$ is well defined. This set is the
attractor of an iterated function system, meaning that it is the Hausdorff
limit of the sequence of compact sets $\\{\varPsi^{k}(X)\\}_{k\geq 1}$, for
any compact set $X$.

The set $\mathcal{F}$ can be interpreted as the set of “fractional parts” of
the digit system $(A,\mathcal{D})$ embedded in $\mathbb{K}_{A}$; that is,
every point of $\mathcal{F}$ can be expressed in base $A$ with digits in
$\varphi(\mathcal{D})$ using only negative powers of the base. In fact,
$\mathcal{F}$ is given explicitly by

(2.16)
$\mathcal{F}=\Big{\\{}\sum_{j=1}^{\infty}A^{-j}\varphi(d_{j})\;|\;d_{j}\in\mathcal{D}\Big{\\}}.\
$

Indeed, it is easy to see that $\mathcal{F}$ is nonempty, bounded, and
satisfies (2.14). The fact that $\mathcal{F}$ is closed follows by a Cantor
diagonal argument.

Note that $\mathcal{F}$ is self-affine in the sense that it can be written as
the union of $a=|\mathcal{D}|$ contracted affine copies of itself, because
(2.14) is equivalent to
$\mathcal{F}=A^{-1}(\mathcal{F}+\varphi(\mathcal{D}))$. When $\mathcal{F}$ has
zero measure, it is a (generalization of a) Cantor set.

Suppose $\mathcal{F}$ has positive measure. In order to define a tiling, we
want the union $\bigcup_{d\in\mathcal{D}}(\mathcal{F}+\varphi(d))$ to be
essentially disjoint (that is, disjoint up to a $\mu$-measure zero set); since
multiplication by $A$ on $\mathbb{K}_{A}$ enlarges the measure by a factor of
$a$, then it is necessary for $\mathcal{D}$ to have exactly $a$ elements. We
will show in the next section that, if $(A,\mathcal{D})$ is a standard digit
system, then $\mathcal{F}(A,\mathcal{D})$ has positive measure.

######  Remark 2.10.

W.l.o.g. we will always assume that $0\in\mathcal{D}$. This can be done
because replacing $\mathcal{D}$ by $\mathcal{D}-v$, where
$v\in\mathbb{Z}^{n}[A]$ is a constant vector, means that
$\mathcal{F}(A,\mathcal{D}-v)$ is a translation of
$\mathcal{F}(A,\mathcal{D})$, hence it is equivalent when we study the
existence of tilings.

##  3\. Results on rational self-affine tiles

Let $(A,\mathcal{D})$ be a digit system and let $\mathbb{K}_{A}$ be the
representation space with metric d and Haar measure $\mu$ as before. In this
section, we give some equivalent topological and combinatorial conditions for
the set $\mathcal{F}(A,\mathcal{D})$ to have positive measure and we prove
that, whenever $\mathcal{F}(A,\mathcal{D})$ is a rational self-affine tile, it
induces a tiling. We also study some topological properties and present an
example.

###  3.1. Properties of rational self-affine tiles and a tiling theorem

We introduce some definitions before stating the results. To denote blocks of
digits, let

(3.1)
$\mathcal{D}_{k}:=\big{\\{}d_{0}+Ad_{1}+\dots+A^{k-1}d_{k-1}\;|\;d_{0},\ldots,d_{k-1}\in\mathcal{D}\big{\\}}\quad\hbox{and}\quad\mathcal{D}_{\infty}:=\bigcup_{k\geqslant
1}\mathcal{D}_{k}.$

From the set equation (2.14), we deduce the iterated set equation

(3.2)
$A^{k}\mathcal{F}=\bigcup_{d\in\mathcal{D}_{k}}(\mathcal{F}+\varphi(d)).$

######  Definition 3.1 (Uniform discreteness).

We say that a set $M\subset\mathbb{K}_{A}$ is uniformly discrete if there
exists $r>0$ such that every open ball of radius $r$ in $\mathbb{K}_{A}$
contains at most one point of $M$.

Our first result is a criterion for $\mathcal{F}(A,\mathcal{D})$ to have
positive measure formulated in terms of $\mathcal{D}$. It is an extension of
[17, Theorem 1.1] and [12, Theorem 10] to the case of rational self-affine
tiles.

######  Theorem 3.2.

Let $(A,\mathcal{D})$ be a digit system, and let
$\mathcal{F}=\mathcal{F}(A,\mathcal{D})\subset\mathbb{K}_{A}$. Then
$\mathcal{F}$ has positive measure if and only if for every $k\geqslant 1$,
all $a^{k}$ expansions in $\mathcal{D}_{k}$ are distinct, and
$\varphi(\mathcal{D}_{\infty})$ is a uniformly discrete subset of
$\mathbb{K}_{A}$.

###### Proof.

We omit some details of the proof, since the one in [17, p. 32 – 34] is
similar, although it is provided in the setting of self-affine tiles in
$\mathbb{R}^{n}$.

Assume first that $\varphi(\mathcal{D}_{\infty})$ is a uniformly discrete set
and that all the elements in $\mathcal{D}_{k}$ are distinct for every
$k\geqslant 1$. Recall the metric $\boldsymbol{\ell}$ and the constant
$0\leqslant\kappa<1$ defined in Lemma 2.8. Then
$\boldsymbol{\ell}(B\,(x,y),0)\leqslant\kappa\,\boldsymbol{\ell}((x,y),0)$ for
every $(x,y)\in\mathbb{K}_{A}$. Consider the closed ball ${\bf
B}^{\prime}_{r}(0):=\\{(x,y)\in\mathbb{K}_{A}\;|\;\boldsymbol{\ell}((x,y),0)\leqslant
r\\},$ and let $(x,y)\in{\bf B}^{\prime}_{r}(0)$. Let $\varPsi$ be the map
defined in (2.15); then by Hutchinson [9], $\mathcal{F}$ is the Hausdorff
limit of the sequence $\\{\varPsi^{k}({\bf B}^{\prime}_{r}(0))\\}_{k\geqslant
1}$. Suppose that
$r\geqslant\frac{\kappa}{1-\kappa}\max_{d\in\mathcal{D}}\\{\boldsymbol{\ell}(\varphi(d),0)\\}$;
then $\varPsi({\bf B}^{\prime}_{r}(0))\subset{\bf B}^{\prime}_{r}(0)$.
Consequently, by Lebesgue’s dominated convergence theorem, we get

$\mu(\mathcal{F})=\lim_{k\to\infty}\mu(\varPsi^{k}({\bf B}^{\prime}_{r}(0))).$

It suffices to find a set of positive measure contained in every
$\varPsi^{k}({\bf B}^{\prime}_{r}(0))$. Since $\varphi(\mathcal{D}_{\infty})$
is uniformly discrete, there exists $\delta>0$ such that for every
$(x,y)\neq(x^{\prime},y^{\prime})$ in $\varphi(\mathcal{D}_{\infty})$ it holds
that $\boldsymbol{\ell}((x,y),(x^{\prime},y^{\prime}))>\delta$. For
$0<\varepsilon<\min\\{\frac{\delta}{2},r\\}$, consider the closed ball ${\bf
B}^{\prime}_{\varepsilon}(0)$. By hypothesis, $\mathcal{D}_{k}$ has $a^{k}$
distinct elements for all $k$, and all sets of the form ${\bf
B}^{\prime}_{\varepsilon}(0)+\varphi(d)$ for $d\in\mathcal{D}_{k}$ are
pairwise disjoint. Therefore $\mu(\varPsi^{k}({\bf
B}^{\prime}_{\varepsilon}(0)))=\mu({\bf B}^{\prime}_{\varepsilon}(0))$ for
every $k$, and hence

$\mu(\mathcal{F})\geqslant\lim_{k\to\infty}\mu(\varPsi^{k}({\bf
B}^{\prime}_{\varepsilon}(0)))=\mu({\bf B}^{\prime}_{\varepsilon}(0))>0.$

For the converse, assume $\mu(\mathcal{F})>0$. Then

$a^{k}\mu(\mathcal{F})=\mu(A^{k}\mathcal{F})=\mu\Big{(}\bigcup_{d\in\mathcal{D}_{k}}(\mathcal{F}+\varphi(d))\Big{)}\leqslant\sum_{d\in\mathcal{D}_{k}}\mu(\mathcal{F}+\varphi(d))\leqslant
a^{k}\mu(\mathcal{F}),$

so all the terms are equal. Hence, $|\mathcal{D}_{k}|=a^{k}$ and the union is
essentially disjoint, meaning that for $d\neq d^{\prime}$ in
$\mathcal{D}_{k}$, we get

(3.3) $\mu((\mathcal{F}+\varphi(d))\cap(\mathcal{F}+\varphi(d^{\prime})))=0.$

It remains to show that $\varphi(\mathcal{D}_{\infty})$ is uniformly discrete.
Suppose this was not the case, then we can find a sequence
$\\{(d_{l},d^{\prime}_{l})\\}_{l\geqslant 1}$ where, $d_{l}$ and
$d^{\prime}_{l}$ are distinct elements of some $\mathcal{D}_{k_{l}}$ for each
$l\geqslant 1$, and such that

$\lim_{l\to\infty}\textbf{d}(\varphi(d_{l}),\varphi(d^{\prime}_{l}))=0.$

If $\mu(\mathcal{F})>0$, then by Federer [7, page 156, Corollary 2.9.9], there
exists a Lebesgue point $(x^{*},y^{*})\in\mathcal{F}$. Given $\varepsilon>0$,
this implies the existence of a sufficiently small $r$ for which

(3.4) $\mu({\bf
B}_{r}(x^{*},y^{*})\cap\mathcal{F})\geqslant(1-\varepsilon)\,\mu({\bf
B}_{r}(x^{*},y^{*})).$

Here, ${\bf B}_{r}(x^{*},y^{*})$ denotes the closed ball of center
$(x^{*},y^{*})$ and radius $r$ w.r.t. the metric ${\bf d}$. Let
$0<\varepsilon^{\prime}<r$, and consider $(x,y)\in\mathbb{K}_{A}$ such that
$\textbf{d}((x,y),0)<\varepsilon^{\prime}<r$. Then

(3.5) $\begin{split}\mu({\bf
B}_{r}(x^{*},y^{*})\cap(\mathcal{F}+(x,y)))&\geqslant\mu({\bf
B}_{r-\varepsilon^{\prime}}((x^{*},y^{*})+(x,y))\cap(\mathcal{F}+(x,y)))\\\
&\geqslant(1-\varepsilon)\mu({\bf
B}_{r-\varepsilon^{\prime}}(x^{*},y^{*})).\end{split}$

Recall that $b$ satisfies (2.11). Note that ${\bf B}_{r}(y^{*})={\bf
B}_{b^{\lfloor\log_{b}r\rfloor}}(y^{*})$, and define
$K:=\lfloor\log_{b}(r)\rfloor-\lfloor\log_{b}(r-\varepsilon^{\prime})\rfloor>0$.
Then $y\in{\bf B}_{r}(y^{*})$ if and only if $B^{K}y\in{\bf
B}_{r-\varepsilon^{\prime}}(B^{K}y^{*})$. Hence,

$\mu({\bf
B}_{r-\varepsilon^{\prime}}(x^{*},y^{*}))=\lambda\Big{(}\frac{r-\varepsilon^{\prime}}{r}{\bf
B}_{r}(x^{*})\Big{)}\,\mu_{B}(B^{K}{\bf
B}_{r}(y^{*}))=\Big{(}\frac{r-\varepsilon^{\prime}}{r}\Big{)}^{n}b^{-K}\mu({\bf
B}_{r}(x^{*},y^{*})),$

and thus for the appropriate value of $\varepsilon^{\prime\prime}>0$ it
follows from (3.5) that

(3.6) $\mu({\bf
B}_{r}(x^{*},y^{*})\cap(\mathcal{F}+(x,y)))>(1-\varepsilon^{\prime\prime})\mu({\bf
B}_{r}(x^{*},y^{*})).$

By inclusion-exclusion and combining (3.4) and (3.6), we get

$\begin{split}\mu((\mathcal{F}+(x,y))\cap\mathcal{F})&>(1-\varepsilon-\varepsilon^{\prime\prime})\mu({\bf
B}_{r}(x^{*},y^{*})>0\end{split}$

for $(x,y)$ sufficiently close to 0. This implies that, for large enough $l$,

$\mu((\mathcal{F}+\varphi(d_{l}))\cap(\mathcal{F}+\varphi(d_{l}^{\prime}))=\mu(\mathcal{F})\cap(\mathcal{F}+\varphi(d_{l}^{\prime}-d_{l}))>0,$

which is a contradiction. ∎

######  Corollary 3.3.

If $(A,\mathcal{D})$ is a standard digit system, then
$\mathcal{F}(A,\mathcal{D})$ has positive measure.

The second result of this section gives some topological equivalences for
$\mathcal{F}$ being a rational self-affine tile. It is in the spirit of [17,
Theorem 1.1].

######  Theorem 3.4.

Let $(A,\mathcal{D})$ be a digit system and let
$\mathcal{F}=\mathcal{F}(A,\mathcal{D})\subset\mathbb{K}_{A}$. The following
assertions are equivalent:

  * $(i)$

$\mathcal{F}$ has positive measure.

  * $(ii)$

$\mathcal{F}$ has nonempty interior.

  * $(iii)$

$\mathcal{F}$ is the closure of its interior, and its boundary
$\partial\mathcal{F}$ has measure zero.

###### Proof.

$(iii)\Rightarrow(ii)\Rightarrow(i)$ is trivial.

$(ii)\Rightarrow(iii)$ is analogous to the proof found in [17, p. 30 – 32].

$(i)\Rightarrow(ii):$ Because $\mathcal{F}$ has positive measure, it has a
Lebesgue point $(x^{*},y^{*})$ satisfying (3.4). This implies that we can
consider a sequence $\varepsilon_{k}\searrow 0$ together with a sequence of
radii $r_{k}\searrow 0$ such that, for every $l>0$,

(3.7) $\mu(A^{l}({\bf
B}_{r_{k}}(x^{*},y^{*})\cap\mathcal{F}))\geqslant(1-\varepsilon_{k})\,\mu(A^{l}{\bf
B}_{r_{k}}(x^{*},y^{*})).$

Claim 1: For every index $k$ there exists a large enough $l_{k}>0$ and
$(u^{(k)},v^{(k)})\in\mathbb{K}_{A}$ such that ${\bf
B}_{1}(u^{(k)},v^{(k)})\subset A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*})$ with

$\mu({\bf B}_{1}(u^{(k)},v^{(k)})\cap
A^{l_{k}}\mathcal{F})\geqslant(1-C\,\varepsilon_{k})\,\mu({\bf
B}_{1}(u^{(k)},v^{(k)})),$

where $(x^{*},y^{*})$ is a Lebesgue point of $\mathcal{F}$ and $C>0$ is a
constant depending only on the space $\mathbb{K}_{A}$.

To prove the claim, we draw on ideas from [17, p. 35]. With a slight abuse of
notation, we will use ${\bf B}_{r}(\cdot)$ to denote closed balls of radius
$r$ in each respective space. Fix $k$, and note that, since $A$ is expanding,
$A^{l}{\bf B}_{r_{k}}(x^{*})\subset\mathbb{R}^{n}$ is an ellipsoid whose
shortest axis’ length goes to infinity as $l$ goes to infinity. Consider
$l_{k}>0$ large enough so that $b^{-l_{k}}\leqslant r_{k}$, and such that the
ellipsoid $A^{l_{k}}{\bf B}_{r_{k}}(x^{*})$ has a shortest axis greater than
$4$. Define $E_{k}:=\\{x\in A^{l_{k}}{\bf
B}_{r_{k}}(x^{*})\;|\;d(x,\partial(A^{l_{k}}{\bf B}_{r_{k}}(x^{*})))\geqslant
1\\}$ as the set of points of $A^{l_{k}}{\bf B}_{r_{k}}(x^{*})$ whose
(Euclidean) distance from the boundary is at least $1$. Consider the set
$2E_{k}-x^{*}$, obtained by doubling $E_{k}$ and centering around $x^{*}$.
Then $E_{k}\subsetneq A^{l_{k}}{\bf B}_{r_{k}}(x^{*})\subsetneq 2E_{k}-x^{*}$
and $\lambda(A^{l_{k}}{\bf
B}_{r_{k}}(x^{*}))\leqslant\lambda(2E_{k}-x^{*})=2^{n}\lambda(E_{k}).$

Consider the compact subset of $\mathbb{K}_{A}$ given by
$\mathcal{U}_{k}:=E_{k}\times A^{l_{k}}{\bf B}_{r_{k}}(y^{*})$, which is
trivially covered by the collection of unit balls $\mathcal{G}:=\\{{\bf
B}_{1}(u,v)\;|\;(u,v)\in\mathcal{U}_{k}\\}.$ Let $\\{{\bf
B}_{1}(u_{1},v_{1}),\ldots,{\bf B}_{1}(u_{s},v_{s})\\}$ be a maximal disjoint
subcollection of $\mathcal{G}$. Then
$\mathcal{U}_{k}\subset\bigcup_{j=1}^{s}{\bf B}_{2}(u_{j},v_{j})$ because of
the following reason: let $(x,y)\in\mathcal{U}_{k}$. If ${\bf
B}_{1}(x,y)\notin\mathcal{G}$, then by maximality there exists
$j\in\\{1,\dots,s\\}$ with ${\bf B}_{1}(x,y)\cap{\bf
B}_{1}(u_{j},v_{j})\neq\varnothing$. Take $(x^{\prime},y^{\prime})\in{\bf
B}_{1}(x,y)\cap{\bf B}_{1}(u_{j},v_{j})$. Then

$\textbf{d}((x,y),(u_{j},v_{j}))\leqslant\textbf{d}((x,y),(x^{\prime},y^{\prime}))+\textbf{d}((x^{\prime},y^{\prime}),(u_{j},v_{j}))\leqslant
2.$

This yields

(3.8) $\begin{split}\mu(A^{l_{k}}{\bf
B}_{r_{k}}(x^{*},y^{*}))&=\lambda(A^{l_{k}}{\bf
B}_{r_{k}}(x^{*}))\,\mu_{B}(A^{l_{k}}{\bf B}_{r_{k}}(y^{*}))\\\ &\leqslant
2^{n}\lambda(E_{k})\,\mu_{B}(A^{l_{k}}{\bf B}_{r_{k}}(y^{*}))\\\
&=2^{n}\mu(\mathcal{U}_{k})\leqslant 2^{n}\sum_{j=1}^{s}\mu({\bf
B}_{2}(u_{j},v_{j})).\end{split}$

Note that $\lambda({\bf B}_{2}(u_{j}))=2^{n}\lambda({\bf B}_{1}(u_{j}))$ while
$\mu_{B}({\bf B}_{2}(v_{j}))=b^{\lfloor\log_{b}2\rfloor}\mu_{B}({\bf
B}_{1}(v_{j}));$ hence

$\sum_{j=1}^{s}\mu({\bf
B}_{2}(u_{j},v_{j}))=2^{n}b^{\lfloor\log_{b}2\rfloor}\mu\Big{(}\bigsqcup_{j=1}^{s}{\bf
B}_{1}(u_{j},v_{j})\Big{)}$

because the unit balls are disjoint. Combining this with (3.8), we have

(3.9) $\mu(A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*}))\leqslant
C\,\mu\Big{(}\bigsqcup_{j=1}^{s}{\bf B}_{1}(u_{j},v_{j})\Big{)}$

for $C:=4^{n}b^{\lfloor\log_{b}2\rfloor}$. We show next that all the balls
${\bf B}_{1}(u_{j},v_{j})={\bf B}_{1}(u_{j})\times{\bf B}_{1}(v_{j})$ are
contained in $A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*})$. Fix
$j\in\\{1,\dots,s\\}$. For the real part, $u_{j}\in E_{k}$, meaning it is a
point of $A^{l_{k}}{\bf B}_{r_{k}}(x^{*})$ which is at distance at least one
from its boundary, hence ${\bf B}_{1}(u_{j})\subset A^{l_{k}}{\bf
B}_{r_{k}}(x^{*})$. For the $B$-adic part, consider $y\in{\bf B}_{1}(v_{j})$
and recall that $A^{-l_{k}}v_{j}\in{\bf B}_{r_{k}}(y^{*})$. From the
ultrametric inequality it follows

(3.10) $\begin{split}{\bf d}_{B}(A^{-l_{k}}y,y^{*})&\leqslant\max\\{{\bf
d}_{B}(A^{-l_{k}}y,A^{-l_{k}}v_{j}),{\bf d}_{B}(A^{-l_{k}}v_{j},y^{*})\\}\\\
&=\max\\{b^{-l_{k}}{\bf d}_{B}(y,v_{j}),r_{k}\\}\\\
&\leqslant\max\\{b^{-l_{k}},r_{k}\\}=r_{k},\end{split}$

since we assumed $b^{-l_{k}}\leqslant r_{k}$. Therefore, $A^{-l_{k}}y\in{\bf
B}_{r_{k}}(y^{*})$ for every $y\in{\bf B}_{1}(v_{j})$, and so ${\bf
B}_{1}(v_{j})\subset A^{l_{k}}{\bf B}_{r_{k}}(y^{*})$. From (3.7) it follows
that

(3.11) $\mu(A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*})\setminus(A^{l_{k}}{\bf
B}_{r_{k}}(x^{*},y^{*})\cap
A^{l_{k}}\mathcal{F}))\leqslant\varepsilon_{k}\mu(A^{l_{k}}{\bf
B}_{r_{k}}(x^{*},y^{*})).$

Equations (3.9) and (3.11) and the fact that $\bigsqcup_{j=1}^{s}{\bf
B}_{1}(u_{j},v_{j})\subset A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*})$ imply

$\mu\Big{(}\bigsqcup_{j=1}^{s}{\bf
B}_{1}(u_{j},v_{j})\setminus\Big{(}\bigsqcup_{j=1}^{s}{\bf
B}_{1}(u_{j},v_{j})\cap
A^{l_{k}}\mathcal{F}\Big{)}\Big{)}\leqslant\varepsilon_{k}\,C\,\mu\Big{(}\bigsqcup_{j=1}^{s}{\bf
B}_{1}(u_{j},v_{j})\Big{)}.$

Since the balls ${\bf B}_{1}(u_{j},v_{j})$ are pairwise disjoint and contained
in $A^{l_{k}}{\bf B}_{r_{k}}(x^{*},y^{*})$, then for at least one
$j_{k}\in\\{1,\dots,s\\}$ it holds that

$\mu({\bf B}_{1}(u_{j_{k}},v_{j_{k}})\cap
A^{l_{k}}\mathcal{F})\leqslant(1-\varepsilon_{k}\,C)\mu({\bf
B}_{1}(u_{j_{k}},v_{j_{k}})),$

which yields Claim 1 with $(u^{(k)},v^{(k)})=(u_{j_{k}},v_{j_{k}})$.

Back to the main proof, Claim 1 together with the iterated set equation (3.2)
implies that, for every $k$, there exists $l_{k}>0$ and
$(u^{(k)},v^{(k)})\in\mathbb{K}_{A}$ such that

(3.12) $\mu\Big{(}{\bf
B}_{1}(u^{(k)},v^{(k)})\cap\Big{(}\bigcup_{d\in\mathcal{D}_{l_{k}}}\mathcal{F}+\varphi(d)\Big{)}\Big{)}\leqslant(1-\varepsilon_{k}\,C)\mu({\bf
B}_{1}(u^{(k)},v^{(k)})).$

Define the finite sets

$\mathcal{V}_{k}:=\\{\varphi(d)-(u^{(k)},v^{(k)})\;|\;d\in\mathcal{D}_{l_{k}},\;(\mathcal{F}+\varphi(d)-(u^{(k)},v^{(k)}))\cap\mathcal{F}\neq\varnothing\\}.$

Then shifting the arguments inside the measures in (3.12) by
$-(u^{(k)},v^{(k)})$ and restricting to translates contained in
$\mathcal{V}_{k}$ yields

$\mu({\bf
B}_{1}(0)\cap(\mathcal{F}+\mathcal{V}_{k}))\leqslant(1-\varepsilon_{k}\,C)\mu({\bf
B}_{1}(0)).$

Note that $\mathcal{F}\cap{\bf B}_{1}(e)\neq\varnothing$ for every
$e\in\mathcal{V}_{k}$. Thus, because $\mathcal{F}$ is bounded, all
$\mathcal{V}_{k}\subset{\bf B}_{R}(0)$ for a sufficiently large constant $R$.
Recall that $\varphi(\mathcal{D}_{\infty})$ is a uniformly discrete set by
Theorem 3.2, and $\mathcal{D}_{l_{k}}\subset\mathcal{D}_{\infty}$, hence there
exists $\delta>0$ such that $\textbf{d}(e,e^{\prime})\geqslant\delta$ for
every $e,e^{\prime}\in\mathcal{V}_{k}$ for every $k$. This implies that the
sequence of cardinalities $\\{|\mathcal{V}_{k}|\\}_{k\geqslant 1}$ is bounded.
Therefore, $\\{\mathcal{V}_{k}\\}_{k\geqslant 1}$ has a convergent subsequence
$\\{\mathcal{V}_{k_{j}}\\}_{j\geqslant 1}$ whose limit, denoted by
$\mathcal{V}$, is a finite set. Then

(3.13) $\begin{split}\mu({\bf
B}_{1}(0)\cap(\mathcal{F}+\mathcal{V}))&\geqslant\liminf_{j\to\infty}\mu({\bf
B}_{1}(0)\cap(\mathcal{F}+\mathcal{V}_{k_{j}})\\\
&\geqslant\liminf_{j\to\infty}(1-C\,\varepsilon_{k_{j}})\mu({\bf
B}_{1}(0))=\mu({\bf B}_{1}(0)).\end{split}$

Because $T$ is closed this implies that $(\mathcal{F}+\mathcal{V})\cap{\bf
B}_{1}(0)={\bf B}_{1}(0)$. Thus $\mathcal{F}+\mathcal{V}$ is a finite union of
translates of the compact set $\mathcal{F}$ containing inner points. Baire’s
theorem implies that $\mathcal{F}$ has nonempty interior. ∎

In what follows, we will restrict ourselves to the case where $\mathcal{F}$
has positive measure. We have referred to $\mathcal{F}$ in this case as a
tile, because we will show that there exists a tiling of the space
$\mathbb{K}_{A}$ by translates of $\mathcal{F}$.

######  Definition 3.5 (Tiling, self-replicating tiling, multiple tiling).

Assume $\mu(\mathcal{F})>0$. Let $\mathcal{S}\subset\mathbb{K}_{A}$ and
consider the collection $\\{\mathcal{F}+s\;|\;s\in\mathcal{S}\\}$, which we
denote as $\mathcal{F}+\mathcal{S}$ with a slight abuse of notation.

  1. (1)

$\mathcal{F}+\mathcal{S}$ is said to be a tiling of $\mathbb{K}_{A}$ if it is
a covering of $\mathbb{K}_{A}$ such that, for any $s\neq s^{\prime}$ in
$\mathcal{S}$, it holds that
$\mu((\mathcal{F}+s)\cap(\mathcal{F}+s^{\prime}))=0$, or, equivalently, if
$\mathcal{F}+s$ and $\mathcal{F}+s^{\prime}$ have disjoint interiors.
$\mathcal{S}$ is called a tiling set for $\mathcal{F}$. We say that
$\mathcal{F}+\mathcal{S}$ tiles $\mathbb{K}_{A}$.

  2. (2)

$\mathcal{F}+\mathcal{S}$ is said to be a self-replicating tiling if there
exists an expanding linear map $T$ on $\mathbb{K}_{A}$ such that, for each
$s\in\mathcal{S}$, there exists a finite subset $J(s)\subset\mathcal{S}$ with

$T(\mathcal{F}+s)=\bigcup_{s^{\prime}\in J(s)}(\mathcal{F}+s^{\prime}).$

  3. (3)

$\mathcal{F}+\mathcal{S}$ is said to be a multiple tiling of $\mathbb{K}_{A}$,
if there exists $k\in\mathbb{N}$ such that $\mu$-almost every point of
$\mathbb{K}_{A}$ is contained in exactly $k$ distinct sets of the form
$\mathcal{F}+s$ with $s\in\mathcal{S}$.

It follows that a self-replicating tiling is completely determined by the set
of tiles that touch the origin 0. We call a self-replicating tiling atomic if
the origin touches exactly one tile.

We want to study the nature of the tilings of $\mathbb{K}_{A}$ obtained using
self-affine tiles. For any $k\geqslant 1$, consider the difference sets

$\mathcal{D}_{k}-\mathcal{D}_{k}=\\{d-d^{\prime}\;|\;d,d^{\prime}\in\mathcal{D}_{k}\\},$

and define

$\Delta:=\bigcup_{k=1}^{\infty}\varphi(\mathcal{D}_{k}-\mathcal{D}_{k}).$

######  Theorem 3.6.

Suppose that $\mathcal{F}$ contains an open set. Then:

  1. $(i)$

There exists a set of translations $\mathcal{S}\subset\Delta$ such that
$\mathcal{F}+\mathcal{S}$ tiles $\mathbb{K}_{A}$. Furthermore, there exists a
translate $\mathcal{S}^{\prime}$ of $\mathcal{S}$ such that
$\mathcal{F}+\mathcal{S}^{\prime}$ is an atomic self-replicating tiling of
$\mathbb{K}_{A}$, and the expanding linear map associated to it is of the form
$T=A^{k}$ for some sufficiently large $k$.

  2. $(ii)$

If $\Delta$ is a group, then $\mathcal{F}+\Delta$ is a tiling.

###### Proof.

$(i)$ The proof is analogous to the one in [17, Theorem 1.2].

$(ii)$ Assume that $\Delta$ is a group. Since by $(i)$ there is a subset
$\mathcal{S}\subset\Delta$ for which $\mathcal{F}+\mathcal{S}$ tiles
$\mathbb{K}_{A}$, $\mathcal{F}+\Delta$ is a covering of $\mathbb{K}_{A}$.
Given any $s\neq s^{\prime}$ in $\Delta$, then it suffices to show

$0=\mu((\mathcal{F}+s)\cap(\mathcal{F}+s^{\prime}))=\mu(\mathcal{F}\cap(\mathcal{F}+s^{\prime}-s)).$

Because $\Delta$ is a group, then $s^{\prime}-s\in\Delta$. That means that
there exists $k\geqslant 1$ and $d\neq d^{\prime}$ in $\mathcal{D}_{k}$ such
that $s^{\prime}-s=\varphi(d-d^{\prime})$, so the assertion is equivalent to

$\mu((\mathcal{F}+\varphi(d))\cap(\mathcal{F}+\varphi(d^{\prime})))=0,$

which holds by (3.3). ∎

###  3.2. Example

We give now an example of a standard and a nonstandard digit system, together
with an illustration of a rational self-affine tile.

######  Example 3.7.

Let

$A=\begin{pmatrix}2&1\\\ 0&\frac{5}{3}\end{pmatrix}\in\mathbb{Q}^{2\times 2}.$

Since its characteristic polynomial is $\chi_{A}(x)=(x-2)(x-\tfrac{5}{3})$,
$A$ is expanding. We proceed to find the representation space
$\mathbb{K}_{A}$. We have

$B=A^{-1}=\begin{pmatrix}\frac{1}{2}&-\frac{3}{10}\\\
0&\frac{3}{5}\end{pmatrix},$

and we get

$\mathbb{Z}^{2}[B]=\left\\{\begin{pmatrix}\frac{s}{2^{n}5^{m}}\\\
\frac{t}{5^{l}}\end{pmatrix}\;|\;n,m,l\in\mathbb{N},\;s,t\in\mathbb{Z}\right\\}$

and

$B\mathbb{Z}^{2}[B]=\left\\{\begin{pmatrix}\frac{s}{2^{n}5^{m}}\\\
\frac{3t}{5^{l}}\end{pmatrix}\;|\;l,m,n\in\mathbb{N},\;s,t\in\mathbb{Z}\right\\}.$

Then, a residue set $\mathcal{E}$ for the quotient
$\mathbb{Z}^{n}[B]/B\mathbb{Z}^{n}[B]$ is given by

$\mathcal{E}=\left\\{\begin{pmatrix}0\\\ 0\end{pmatrix},\begin{pmatrix}0\\\
1\end{pmatrix},\begin{pmatrix}0\\\ 2\end{pmatrix}\right\\},$

hence $b=3$. It is not hard to check that, for this example, it is possible to
establish an isomorphism $\mathbb{Z}^{n}((B))\simeq\mathbb{Q}_{3}$, and thus
we regard the elements of $\mathbb{K}_{A}$ as points in
$\mathbb{R}^{2}\times\mathbb{Q}_{3}$. We will consider a digit set
$\widetilde{\mathcal{D}}$ and in Figure 1, we will illustrate a rational self-
affine tile associated to $(A,\widetilde{\mathcal{D}})$, and represent it in
$\mathbb{R}^{3}$ by embedding $\mathbb{Q}_{3}$ in $\mathbb{R}$.

We will first find a standard digit system $(A,\mathcal{D})$, and for that we
compute the quotient $\mathbb{Z}^{2}[A]/A\mathbb{Z}^{2}[A]$. Powers of $A$ are
of the form

$A^{k}=\begin{pmatrix}2^{k}&c_{k}\\\
0&(\frac{5}{3})^{k}\end{pmatrix}\in\mathbb{Q}^{2\times 2},$

where $c_{k}=\sum_{i,j:\,i+j=k-1}{2^{i}\left(\frac{5}{3}\right)^{j}}.$ We have

$\mathbb{Z}^{2}[A]=\left\\{\begin{pmatrix}\frac{s}{3^{n}}\\\
\frac{t}{3^{m}}\end{pmatrix}\;|\;n,m\in\mathbb{N},\;s,t\in\mathbb{Z}\right\\},$

hence, an element of $A\mathbb{Z}^{2}[A]$ is of the form

$\begin{pmatrix}2&1\\\
0&\frac{5}{3}\end{pmatrix}\begin{pmatrix}\frac{s}{3^{n}}\\\
\frac{t}{3^{m}}\end{pmatrix}=\begin{pmatrix}\frac{2s}{3^{n}}+\frac{t}{3^{m}}\\\
\frac{5t}{3^{m+1}}\end{pmatrix}.$

Note that $\frac{2s}{3^{n}}+\frac{t}{3^{m}}\equiv\frac{5t}{3^{m+1}}\mod 2$ in
$\mathbb{Q}$, because if we multiply both sides by $3^{n+m+1}$ we get
$3^{m+1}2s+3^{n+1}t\equiv t\mod 2$, and $3^{n}5t\equiv t\mod 2$. This yields

$A\mathbb{Z}^{2}[A]=\left\\{\begin{pmatrix}\frac{s}{3^{n}}\\\
\frac{5t}{3^{m}}\end{pmatrix}\;|\;n,m\in\mathbb{N},\;s,t\in\mathbb{Z},\;s\equiv
t\mod 2\right\\}.$

A complete set of residue class representatives of
$\mathbb{Z}^{2}[A]/A\mathbb{Z}^{2}[A]$ is given by

$\mathcal{D}=\left\\{\begin{pmatrix}0\\\ 0\end{pmatrix},\begin{pmatrix}0\\\
1\end{pmatrix},\begin{pmatrix}0\\\ 2\end{pmatrix},\begin{pmatrix}0\\\
3\end{pmatrix},\begin{pmatrix}0\\\ 9\end{pmatrix},\begin{pmatrix}1\\\
0\end{pmatrix},\begin{pmatrix}1\\\ 1\end{pmatrix},\begin{pmatrix}1\\\
2\end{pmatrix},\begin{pmatrix}1\\\ 3\end{pmatrix},\begin{pmatrix}1\\\
9\end{pmatrix}\right\\},$

and so $(A,\mathcal{D})$ is a standard digit system. Note that $a=10$ and so
$|\det A|=\frac{10}{3}=\frac{a}{b}$, as expected.

Next, we want to find a nonstandard digit system. Note that we can write
$\mathcal{D}=\mathcal{R}_{1}+\mathcal{R}_{2}$, where

$\mathcal{R}_{1}=\left\\{\begin{pmatrix}0\\\
0\end{pmatrix},\begin{pmatrix}0\\\ 1\end{pmatrix},\begin{pmatrix}0\\\
2\end{pmatrix},\begin{pmatrix}0\\\ 3\end{pmatrix},\begin{pmatrix}0\\\
9\end{pmatrix}\right\\},\quad\mathcal{R}_{2}=\left\\{\begin{pmatrix}0\\\
0\end{pmatrix},\begin{pmatrix}1\\\ 0\end{pmatrix}\right\\},$

and the decomposition of the digits as a sum is unique. Consider

$\widetilde{\mathcal{D}}:=\mathcal{R}_{1}+A\mathcal{R}_{2}=\left\\{\begin{pmatrix}0\\\
0\end{pmatrix},\begin{pmatrix}0\\\ 1\end{pmatrix},\begin{pmatrix}0\\\
2\end{pmatrix},\begin{pmatrix}0\\\ 3\end{pmatrix},\begin{pmatrix}0\\\
9\end{pmatrix},\begin{pmatrix}2\\\ 0\end{pmatrix},\begin{pmatrix}2\\\
1\end{pmatrix},\begin{pmatrix}2\\\ 2\end{pmatrix},\begin{pmatrix}2\\\
3\end{pmatrix},\begin{pmatrix}2\\\ 9\end{pmatrix}\right\\},$

which is not a residue set for $\mathbb{Z}^{n}[A]\mod A$. We show that
$\mathcal{F}(A,\widetilde{\mathcal{D}})$ has positive measure.

Figure 1. The tile $\mathcal{F}(A,\widetilde{\mathcal{D}}).$

Let

$E_{1}:=\Big{\\{}\sum_{j=1}^{\infty}A^{-j}\varphi(e_{j})\;|\;e_{j}\in\mathcal{R}_{1}\Big{\\}}\quad\mbox{
and }\quad
E_{2}:=\Big{\\{}\sum_{j=1}^{\infty}A^{-j}\varphi(e^{\prime}_{j})\;|\;e^{\prime}_{j}\in\mathcal{R}_{2}\Big{\\}}.$

Now, the unique decomposition of the elements of $\widetilde{\mathcal{D}}$
yields that

$\begin{split}\mathcal{F}(A,\widetilde{\mathcal{D}})&=E_{1}+A\,E_{2}=E_{1}+\Big{\\{}\sum_{j=0}^{\infty}A^{-j}\varphi(e^{\prime}_{j+1})\;|\;e^{\prime}_{j}\in\mathcal{R}_{2}\Big{\\}}\\\
&=E_{1}+E_{2}+\varphi(\mathcal{R}_{2})=\mathcal{F}(A,\mathcal{D})+\varphi(\mathcal{R}_{2}).\\\
\end{split}$

Since $(A,\mathcal{D})$ is a standard digit system, the set
$\mathcal{F}(A,\mathcal{D})$ has positive measure (see Corollary 3.3), and
hence so does $\mathcal{F}(A,\widetilde{\mathcal{D}})$, meaning it is rational
self-affine tile associated to a nonstandard digit system. We illustrate it in
Figure 1. Recall that the representation space is not Euclidean, so we
embedded the points of $\mathbb{K}_{A}$ into $\mathbb{R}^{3}$ in order to draw
the picture. We did that so the figure still reflects some of the properties
of the set, but it is not a completely faithful representation, since this is
not possible due to the $B$-adic factor.

##  4\. Characters and multiple tiling

The main result of this section states that, whenever
$\mathcal{F}(A,\mathcal{D})$ is a tile, it gives a multiple tiling of
$\mathbb{K}_{A}$. Before arriving to the proof, we introduce some definitions
in order to find the Pontryagin dual of $\mathbb{K}_{A}$, and we give a
complete description of its characters. We make use of the character theory of
locally compact abelian groups to show that the multiplication by $A$ is
ergodic on a certain torus in $\mathbb{K}_{A}$, and use this to prove the
existence of the multiple tiling.

###  4.1. Some basic results and definitions

The module $\mathbb{Z}^{n}[A]$ plays a principal role in the study of tilings
by rational self-affine tiles. We will prove first that, embedded into the
representation space $\mathbb{K}_{A}$, this module becomes a lattice, and we
show later that it is a translation set for a multiple tiling given by copies
of $\mathcal{F}$. First, we formalize the notion of lattice in our setting.

######  Definition 4.1 (Lattice).

A subset $\Lambda$ of $\mathbb{K}_{A}$ is a lattice if it satisfies the three
following conditions:

  1. (1)

$\Lambda$ is a group.

  2. (2)

$\Lambda$ is uniformly discrete, meaning there exists $r>0$ such that every
open ball of radius $r$ in $\mathbb{K}_{A}$ contains at most one point of
$\Lambda$.

  3. (3)

$\Lambda$ is relatively dense, meaning there exists $R>0$ such that every
closed ball of radius $R$ in $\mathbb{K}_{A}$ contains at least one point of
$\Lambda$.

We show next that $\varphi(\mathbb{Z}^{n}[A])$ satisfies these properties. We
state a lemma first.

######  Lemma 4.2.

There exists an integer $K\geqslant 1$ such that

(4.1) $\mathbb{Z}^{n}\cap
B\mathbb{Z}^{n}[B]=\mathbb{Z}^{n}\cap\left(B\mathbb{Z}^{n}+B^{2}\mathbb{Z}^{n}+\dots+B^{K}\mathbb{Z}^{n}\right).$

###### Proof.

For $k\geqslant 1$, define the lattices

$\mathcal{L}_{k}[B]:=\sum_{j=1}^{k}B^{j}\mathbb{Z}^{n}.$

Since $\mathcal{L}_{k}[B]$ contains $B\mathbb{Z}^{n}$ and $B$ is invertible,
the lattice $\mathcal{L}_{k}[B]$ has full rank. Consider a nonzero integer
$m_{k}$ such that $m_{k}\mathcal{L}_{k}[B]\subset\mathbb{Z}^{n}$. Then
$m_{k}\mathcal{L}_{k}[B]$ has finite index in $\mathbb{Z}^{n}$. From this
fact, one deduces that the intersection $\mathbb{Z}^{n}\cap\mathcal{L}_{k}[B]$
has finite index in $\mathbb{Z}^{n}$ for every $k\geqslant 1$. Therefore, the
chain of nested lattices

$\left(\mathbb{Z}^{n}\cap\mathcal{L}_{1}[B]\right)\subset\left(\mathbb{Z}^{n}\cap\mathcal{L}_{2}[B]\right)\subset\dots\subset\left(\mathbb{Z}^{n}\cap
B\mathbb{Z}^{n}[B]\right)\subset\mathbb{Z}^{n}$

must eventually stabilize after some $K\geqslant 1$. ∎

######  Proposition 4.3.

The set $\varphi(\mathbb{Z}^{n}[A])$ is a lattice in $\mathbb{K}_{A}$.

###### Proof.

The fact that $\varphi(\mathbb{Z}^{n}[A])$ is a group follows from the
additive group structure of $\mathbb{Z}^{n}[A]$ because $\varphi$ is a group
homomorphism.

To prove the uniform discreteness of $\varphi(\mathbb{Z}^{n}[A])$, we claim
that there exists $0<r\leqslant 1$ such that
$\textbf{d}(\varphi(z),0)\geqslant r$ for every nonzero
$z\in\mathbb{Z}^{n}[A]$. If $\textbf{d}(\varphi(z),0)\geqslant 1$ then we are
done. Suppose on the contrary that $\textbf{d}(\varphi(z),0)<1$. Since
$z\in\mathbb{Z}^{n}[A]$, we can write it as

(4.2) $z=\sum_{j=0}^{k}A^{j}z_{j},\quad z_{j}\in\mathbb{Z}^{n},$

with $z_{k}\neq 0$, and there is a minimal index $k$ with this property. If
$z_{k}\notin B\mathbb{Z}^{n}[B]$, then $B^{k}z\in\mathbb{Z}^{n}[B]\setminus
B\mathbb{Z}^{n}[B]$, so one has
$\textbf{d}_{B}(z,0)=b^{k}\textbf{d}_{B}(B^{k}z,0)\geqslant 1$, in
contradiction to $\textbf{d}(\varphi(z),0)<1$. Thus,
$z_{k}\in\mathbb{Z}^{n}\cap B\mathbb{Z}^{n}[B]$. By Lemma 4.2, there are
vectors $w_{1}$, $w_{2}$, $\dots$, $w_{K}\in\mathbb{Z}^{n}$ such that

$z_{k}=Bw_{1}+B^{2}w_{2}+\dots+B^{K}w_{K}.$

We claim that, in such case, one must have $k\leq K$. Suppose that $k>K$. Then

$z=A^{k}z_{k}+\sum_{j=0}^{k-1}A^{j}z_{j}=\sum_{j=0}^{k-K-1}A^{j}z_{j}+\sum_{j=k-K}^{k-1}(z_{j}+w_{k-j})A^{j}=\sum_{j=0}^{k-1}A^{j}z_{j}^{\prime},\quad
z_{j}^{\prime}\in\mathbb{Z}^{n}.$

However, that would contradict the minimality of $k$. Therefore, $k\leq K$.
Now, let $m$ denote the least common multiple of the denominators of the
entries of $A$. Since $z$ is a sum of integer vectors multiplied by $A^{k}$,
$0\leqslant k\leqslant K$, the non-zero entries of $z$ are at least $1/m^{K}$
in absolute value. Hence, $\textbf{d}(\varphi(z),0)\geqslant\|z\|\geqslant
1/m^{K}=r$.

We now turn to the proof of relative denseness. Let $(x,y)\in\mathbb{K}_{A}$
be arbitrary. Choose $x^{\prime}\in\mathbb{Z}^{n}$ to be the closest integer
vector to $x$, so that $\|x-x^{\prime}\|<1$, and choose
$y^{\prime}\in\mathbb{Z}^{n}[A]$ such that ${\bf d}_{B}(y,y^{\prime})\leqslant
1$ (this holds by taking $y^{\prime}=\\{y\\}_{B}$, see Definition 4.6 below).
Hence, ${\bf d}((x,y),(x^{\prime},y^{\prime}))\leqslant 1$. Choose
$y^{\prime\prime}\in\mathbb{Z}^{n}$ to be the closest integer vector to
$x^{\prime}-y^{\prime}\in\mathbb{R}^{n}$. Let
$z:=y^{\prime}+y^{\prime\prime}\in\mathbb{Z}^{n}[A]$. Then
$\|x^{\prime}-z\|=\|x^{\prime}-y^{\prime}-y^{\prime\prime}\|<1$. Moreover,
${\bf d}_{B}(z,y^{\prime})={\bf d}_{B}(y^{\prime\prime},0)\leqslant 1$ because
$y^{\prime\prime}\in\mathbb{Z}^{n}$. Therefore, ${\bf
d}((x^{\prime},y^{\prime}),\varphi(z))\leqslant 1.$ This yields
$\textbf{d}((x,y),\varphi(z))\leqslant 2$, which proves the Lemma. ∎

The next step is to define a space $\mathbb{Z}^{n}((B^{*}))$ that will be
crucial later when we study the characters of $\mathbb{K}_{A}$. Prior to that,
we prove the following Lemma.

######  Lemma 4.4.

The group $\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]$ is a lattice in
$\mathbb{R}^{n}$.

###### Proof.

We show that
$\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]\subset\mathbb{Z}^{n}+A\mathbb{Z}^{n}+\dots+A^{K}\mathbb{Z}^{n}$
for some $K\geqslant 1$. Let $z\in\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]$.
Since $z\in\mathbb{Z}^{n}[A]$, write $z=\sum_{j=0}^{k}A^{j}z_{j}$, with
$z_{j}\in\mathbb{Z}^{n}$, $z_{k}\neq 0$ and $k$ minimal. If $k=0$, then
$z\in\mathbb{Z}^{n}$. Assume $k\geqslant 1$. Since $z\in\mathbb{Z}^{n}[B]$,
from solving for $z_{k}$ it follows that $z_{k}\in\mathbb{Z}^{n}\cap
B\mathbb{Z}^{n}[B]$. By Lemma 4.2, one can find vectors
$w_{1},w_{2},\dots,w_{K}\in\mathbb{Z}^{n}$ such that

(4.3) $z_{k}=Bw_{1}+B^{2}w_{2}+\dots+B^{K}w_{K},$

and by proceeding like in the proof of Proposition 4.3, one shows $k\leqslant
K$. Therefore the inclusion follows. This implies that
$\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]$ is contained in a lattice, and also
it trivially contains the lattice $\mathbb{Z}^{n}$. Since it is a group, it is
itself a lattice. ∎

For an arbitrary lattice $\Lambda\subset\mathbb{R}^{n}$, define its dual
lattice by

$\Lambda^{*}:=\\{x\in\mathbb{Z}^{n}\;|\;\langle x,z\rangle\in\mathbb{Z}\mbox{
for every }z\in\Lambda\\},$

where $\langle\cdot,\cdot\rangle$ denotes the usual scalar product in
$\mathbb{R}^{n}$. Denote by ${A^{*}}$ the transpose of $A$, and let $\Lambda$
and $\Gamma$ be full rank lattices such that

(4.4) $\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]\subset\Lambda\quad\mbox{ and
}\quad\Gamma[{A^{*}}]\cap\Gamma[{B^{*}}]\subset\Lambda^{*}\subset\mathbb{Z}^{n}.$

This is possible because $\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]$ is a lattice
by Lemma 4.4, and the proof that $\Gamma[{A^{*}}]\cap\Gamma[{B^{*}}]$ is also
a lattice is analogous.

######  Definition 4.5 ($B$-adic and $B^{*}$-adic expansions).

Let $\mathcal{E}\subset\mathbb{Z}^{n}$ be a complete set of residue classes of
$\mathbb{Z}^{n}[B]/B\mathbb{Z}^{n}[B]$ with $0\in\mathcal{E}$. Then every
$y\in\mathbb{Z}^{n}((B))$ has a unique expansion of the form

(4.5) $y=\sum_{j=\nu(y)}^{\infty}{B}^{j}y_{j},\quad y_{j}\in\mathcal{E},$

which we call the $B$-adic expansion of $y$ with coefficients in
$\mathcal{E}$. Recall that $\nu(0)=\infty$, so the $B$-adic expansion of $0$
is the empty sum.

Let $B^{*}$ denote the transpose of $B$. Consider the full rank integer
lattice $\Gamma$ satisfying (4.4) and let
$\mathcal{E}^{*}\subset\Gamma\subset\mathbb{Z}^{n}$ be a complete set of coset
representatives of $\Gamma[B^{*}]/{B^{*}}\Gamma[B^{*}]$ with
$0\in\mathcal{E}^{*}$. Consider the space $\mathbb{Z}^{n}((B^{*}))$ defined
analogously to $\mathbb{Z}^{n}((B))$. Then every $s\in\mathbb{Z}^{n}((B^{*}))$
has a unique expansion of the form

(4.6) $s=\sum_{j=\nu^{*}(s)}^{\infty}{B^{*}}^{j}s_{j},\quad
s_{j}\in\mathcal{E}^{*},$

where $\nu^{*}$ is the valuation in $\mathbb{Z}^{n}((B^{*}))$ defined in the
same way as $\nu$. We call this the $B^{*}$-adic expansion of $s$ with
coefficients in $\mathcal{E}^{*}$.

######  Definition 4.6 ($B$-adic and $B^{*}$-adic fractional and integer
part).

Given $y\in\mathbb{Z}^{n}((B))$ with $B$-adic expansion 4.5, we define the
$B$-adic fractional part and the $B$-adic integer part of $y$, respectively,
as

$\\{y\\}_{B}:=\sum_{j=\nu(y)}^{-1}{B}^{j}y_{j},\quad\lfloor
y\rfloor_{B}:=\sum_{j=0}^{\infty}{B}^{j}y_{j}.$

Given $s\in\mathbb{Z}^{n}((B^{*}))$ with $B^{*}$-adic expansion 4.6, we define
the $B^{*}$-adic fractional part and the $B^{*}$-adic integer part of $s$,
respectively, as

$\\{s\\}^{*}_{B}:=\sum_{j=\nu^{*}(s)}^{-1}{B^{*}}^{j}s_{j},\quad\lfloor
s\rfloor^{*}_{B}:=\sum_{j=0}^{\infty}{B^{*}}^{j}s_{j}.$

From here onwards, whenever we have a $B$-adic series (resp. $B^{*}$-adic
series), we assume the coefficients to lie in $\mathcal{E}$ (resp.
$\mathcal{E}^{*}$).

######  Remark 4.7.

Recall that $b=|\mathcal{E}|$. We claim that, if $b=1$, then the multiple
tiling theorem holds. Note that, in this case, $\det A=a$ is an integer.
Indeed, an analogous version of Theorem 4.16 is proven by Lagarias and Wang in
[16] for integer matrices. However, they show in [16, Lemma 2.1] that this
results also holds for self-affine tiles associated to expanding real matrices
$A\in\mathbb{R}^{n\times n}$ with integer determinant, as long as there exists
an $A$-invariant lattice in $\mathbb{R}^{n}$ containing the difference set
$\mathcal{D}-\mathcal{D}$. If $\mathcal{F}(A,\mathcal{D})$ is a rational self-
affine tile and $b=1$, then $\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B]$ is a
lattice by Lemma 4.4, and it is $A$-invariant because
$\mathbb{Z}^{n}[B]/B\mathbb{Z}^{n}[B]$ is trivial and hence
$A\mathbb{Z}^{n}[B]\subset\mathbb{Z}^{n}[B]$. Consider
$c\in\mathbb{Z}\setminus\\{0\\}$ such that
$c\mathcal{D}\subset\mathbb{Z}^{n}$; then
$\mathcal{D}-\mathcal{D}\subset\frac{1}{c}(\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[B])$,
which is an $A$-invariant lattice. Therefore, the assumption that $b\geqslant
2$ made in Remark 2.6 also applies to this section.

######  Remark 4.8.

We can assume without loss of generality that $\mathcal{E}$ has a subset
$\\{c_{1},\dots,c_{n}\\}$ such that the lattice $\varTheta:=\langle
c_{1},\dots,c_{n}\rangle_{\mathbb{Z}}$ has full rank in $\mathbb{R}^{n}$. To
show this, suppose first that $b\geqslant n$. Take a matrix
$R\in\mathbb{Z}^{n\times n}$ whose columns are distinct elements
$\\{c_{1},\dots,c_{n}\\}$ of $\mathcal{E}$. Consider the integer matrix
$N(t):=tB-R$, where $t\in\mathbb{N}$ is chosen so that
$tB\in\mathbb{Z}^{n\times n}$. Its determinant is
$\det{(N(t))}=\det{(B)}\,\det(t\cdot Id-AR)=\det{(B)}\,\chi_{AR}(t)$, where
$\chi_{AR}(t)\in\mathbb{Q}[t]$ is the characteristic polynomial of $AR$. For
all but finitely many $t\in\mathbb{N}$, it holds that $\chi_{AR}(t)\neq 0$.
Hence, we can choose $t$ in a way that the column vectors
$\\{\widetilde{c_{1}},\dots,\widetilde{c_{n}}\\}$ of $N(t)$ are linearly
independent, and hence they span a full rank integer lattice. Note that
$c_{j}-\widetilde{c_{j}}\in B\mathbb{Z}^{n}$ for $j=1,\dots,n$, so we can
replace each $c_{j}$ by $\widetilde{c_{j}}$, and this produces a new residue
set $\widetilde{\mathcal{E}}$ with the required property.

Suppose now that $1<b<n$. Choose $k\geqslant 1$ so that $b^{k}\geqslant n$.
Note that a complete set of residues for
$\mathbb{Z}^{n}[B^{k}]/B^{k}\mathbb{Z}^{n}[B^{k}]$ is given by
$\mathcal{E}+B\mathcal{E}+\dots+B^{k-1}\mathcal{E}$, so
$|\mathbb{Z}^{n}[B^{k}]/B^{k}\mathbb{Z}^{n}[B^{k}]|=b^{k}\geqslant n$. Suppose
that we consider the digit system $(A^{k},\mathcal{D}_{k})$ with
$\mathcal{D}_{k}$ as in (3.1); then $\mathcal{E}_{k}$ satisfies the assumption
of the previous paragraph. Note that from the iterated set equation (3.2) it
follows that $\mathcal{F}(A^{k},\mathcal{D}_{k})=\mathcal{F}(A,\mathcal{D})$.
Hence, whenever $|\mathcal{E}|<n$ we can work with $(A^{k},\mathcal{D}_{k})$
instead of $(A,\mathcal{D})$ and with
$\mathcal{E}+B\mathcal{E}+\dots+B^{k-1}\mathcal{E}$ instead of $\mathcal{E}$.

###  4.2. Character theory

In order to prove the multiple tiling theorem, we use some results on the
characters of $\mathbb{K}_{A}$. For more on the topic of character theory on
locally compact abelian groups, we refer the reader to [21, Chapter 4].

######  Definition 4.9 (Character).

A character $\chi$ on an abelian group $G$ is a continuous function
$\chi:G\rightarrow\mathbb{S}^{1}$ such that $\chi(x+y)=\chi(x)\chi(y)$ for all
$x,y\in G$.

######  Lemma 4.10.

The set of all characters of $G$ constitutes a group, called the Pontryagin
dual of $G$, denoted by $\widehat{G}$. It satisfies the following properties:

  1. (1)

The Pontryagin dual of $\widehat{G}$ is isomorphic to $G$.

  2. (2)

The Pontryagin dual of the product $G_{1}\times G_{2}$ is isomorphic to
$\widehat{G}_{1}\times\widehat{G}_{2}$ and the characters are of the form
$\chi=\chi_{1}\cdot\chi_{2}$ with $\chi_{1}\in\widehat{G_{1}}$ and
$\chi_{2}\in\widehat{G_{2}}.$

  3. (3)

Given a subgroup $H\subset G$, define the annihilator of $H$ on $G$ as

$\mbox{Ann}(H):=\\{\chi\in\widehat{G}\;|\;\chi(H)=1\\}.$

Then $\widehat{(G/H)}\simeq\mbox{Ann}(H)$ and
$\widehat{H}\simeq\widehat{G}/\mbox{Ann}(H)$.

###### Proof.

See [21, Chapter 4]. ∎

It is well known that the characters on $\mathbb{R}^{n}$ are given by

$\chi_{r}:\mathbb{R}^{n}\rightarrow S^{1};\quad x\mapsto\exp(2\pi i\langle
x,r\rangle),$

where $r\in\mathbb{R}^{n}$, and $\widehat{\mathbb{R}}^{n}\simeq\mathbb{R}^{n}$
via the isomorphism $r\mapsto\chi_{r}$.

For any
$s=\sum_{j=\nu^{*}(s)}^{\infty}{B^{*}}^{j}s_{j}\in\mathbb{Z}^{n}((B^{*}))$
with $s_{j}\in\mathcal{E}^{*}$, define the map

(4.7) $\chi_{s}:\mathbb{Z}^{n}((B))\rightarrow S^{1};\quad y\mapsto\exp(2\pi
iS_{s}(y)),$

where

(4.8)
$S_{s}(y):=\sum_{j=\nu^{*}(s)}^{\infty}\langle\\{B^{j}y\\}_{B},s_{j}\rangle.$

The map is well defined because $\\{B^{j}y\\}_{B}=0$ for all but finitely many
indices $j$. We show next that this map is indeed a character.

######  Proposition 4.11.

For every $s\in\mathbb{Z}^{n}((B^{*}))$, the map $\chi_{s}$ defined in (4.7)
is continuous and multiplicative, that is,
$\chi_{s}(y+y^{\prime})=\chi_{s}(y)\chi_{s}(y^{\prime})$.

###### Proof.

Fix $s=\sum_{j=\nu^{*}(s)}^{\infty}{B^{*}}^{j}s_{j}\in\mathbb{Z}^{n}((B^{*}))$
with $s_{j}\in\mathcal{E}^{*}$. For the multiplicativity, if suffices to show
that $S_{s}(y+y^{\prime})=S_{s}(y)+S_{s}(y^{\prime})\mod\mathbb{Z}$. We prove
first the following claim: given
$\omega,\omega^{\prime}\in\mathbb{Z}^{n}((B))$, it holds that

(4.9)
$\\{\omega\\}_{B}+\\{\omega^{\prime}\\}_{B}-\\{\omega+\omega^{\prime}\\}_{B}\in\Lambda,$

where $\Lambda$ is the lattice satisfying (4.4). By definition of the $B$-adic
fractional part,
$\\{\omega\\}_{B}+\\{\omega^{\prime}\\}_{B}-\\{\omega+\omega^{\prime}\\}_{B}\in\mathbb{Z}^{n}[A].$
Also,

$\begin{split}\\{\omega\\}_{B}+\\{\omega^{\prime}\\}_{B}-\\{\omega+\omega^{\prime}\\}_{B}&=(\\{\omega\\}_{B}-\omega)+(\\{\omega^{\prime}\\}_{B}-\omega^{\prime})+(\omega+\omega^{\prime}-\\{\omega+\omega^{\prime}\\}_{B})\\\
&=-\lfloor\omega\rfloor_{B}-\lfloor\omega^{\prime}\rfloor_{B}+\lfloor\omega+\omega^{\prime}\rfloor_{B}\in\mathbb{Z}^{n}[A^{-1}].\end{split}$

Since $\mathbb{Z}^{n}[A]\cap\mathbb{Z}^{n}[A^{-1}]\subset\Lambda$ by
definition of $\Lambda$, this yields the claim. Now, for any $y,y^{\prime}$ we
have

(4.10)
$S_{s}(y)+S_{s}(y^{\prime})-S_{s}(y+y^{\prime})=\sum_{j=\nu^{*}(s)}^{\infty}\langle\\{B^{j}y\\}_{B}+\\{B^{j}y^{\prime}\\}_{B}-\\{B^{j}(y+y^{\prime})\\}_{B},s_{j}\rangle$

where
$\\{B^{j}y\\}_{B}+\\{B^{j}y^{\prime}\\}_{B}-\\{B^{j}(y+y^{\prime})\\}_{B}\in\Lambda$
by (4.9). The summands in (4.10) are nonzero only for a finite number of
$j$’s. For every index $j$, we have
$s_{j}\in\mathcal{E}^{*}\subset\Gamma\subset\Lambda^{*}$, thus by definition
of dual lattice,

$\langle\\{B^{j}y\\}_{B}+\\{B^{j}y^{\prime}\\}_{B}-\\{B^{j}(y+y^{\prime})\\}_{B},s_{j}\rangle\in\mathbb{Z}$

and the multiplicativity of $\chi_{s}$ is established.

For the continuity, let $y,y^{\prime}\in\mathbb{Z}^{n}((B))$ such that ${\bf
d}_{B}(y,y^{\prime})\leqslant b^{\nu^{*}(s)}$. Then, for every
$j\geqslant\nu^{*}(s)$, it holds that ${\bf
d}_{B}(B^{j}y,B^{j}y^{\prime})\leqslant 1$, and so
$B^{j}(y-y^{\prime})\in\mathbb{Z}^{n}[[B]]$ which implies
$\\{B^{j}(y-y^{\prime})\\}_{B}=0$. Then $S_{s}(y-y^{\prime})=0$ and hence, by
multiplicativity, $\chi_{s}(y)=\chi_{s}(y^{\prime})$. Thus $\chi_{s}$ is
locally constant and, hence, continuous. ∎

We will show that the Pontryagin dual of $\mathbb{K}_{A}$ is isomorphic to
$\mathbb{R}^{n}\times\mathbb{Z}^{n}((B^{*})).$ To do so, we prove some lemmas
first.

######  Lemma 4.12.

Let $y=\sum_{k=\nu(y)}^{\infty}B^{k}y_{k}\in\mathbb{Z}^{n}((B))$ with
$y_{k}\in\mathcal{E}$ and
$s=\sum_{j=\nu^{*}(s)}^{\infty}{B^{*}}^{j}s_{j}\in\mathbb{Z}^{n}((B^{*}))$
with $s_{j}\in\mathcal{E}^{*}$ be given. Then

$S_{s}(y)=\sum_{j=\nu^{*}(s)}^{\infty}\langle\\{B^{j}y\\}_{B},s_{j}\rangle=\sum_{k=\nu(y)}^{\infty}\langle
y_{k},\\{{B^{*}}^{k}s\\}^{*}_{B}\rangle.$

###### Proof.

From direct calculation, we obtain

$\begin{split}S_{s}(y)&=\sum_{j=\nu^{*}(s)}^{\infty}\langle\\{B^{j}y\\}_{B},s_{j}\rangle=\sum_{j=\nu^{*}(s)}^{\infty}\langle\sum_{k=\nu(y)}^{-j-1}B^{j+k}y_{k},s_{j}\rangle\\\
&=\sum_{j=\nu^{*}(s)}^{\infty}\sum_{k=\nu(y)}^{-j-1}\langle
y_{k},B^{*j+k}s_{j}\rangle=\sum_{k=\nu(y)}^{\infty}\sum_{j=\nu^{*}(s)}^{-k-1}\langle
y_{k},B^{*j+k}s_{j}\rangle\\\ &=\sum_{k=\nu(y)}^{\infty}\langle
y_{k},\sum_{j=\nu^{*}(s)}^{-k-1}B^{*j+k}s_{j}\rangle=\sum_{k=\nu(y)}^{\infty}\langle
y_{k},\\{{B^{*}}^{k}s\\}^{*}_{B}\rangle.\qed\end{split}$

Our next step is to establish a Pontryagin duality between
$\mathbb{Z}^{n}((B))$ and $\mathbb{Z}^{n}((B^{*}))$. For that purpose, we
express both sets in terms of projective limits. For more on the topic we
refer the reader to [23]. For each $k\in\mathbb{N}$, consider the quotients

$\mathcal{E}_{k}:=\mathbb{Z}^{n}[B]/B^{k}\mathbb{Z}^{n}[B].$

Clearly, $\mathcal{E}_{k}\subset\mathcal{E}_{k+1}$ for every $k$, so we can
define the canonical projections

$\pi_{k}:\mathcal{E}_{k+1}\rightarrow\mathcal{E}_{k},\quad x\mapsto x\mod
B^{k}.$

Therefore, we have a projective system

$\dots\longrightarrow\mathcal{E}_{k+1}\stackrel{{\scriptstyle\pi_{k}}}{{\longrightarrow}}\mathcal{E}_{k}\stackrel{{\scriptstyle\pi_{k-1}}}{{\longrightarrow}}\mathcal{E}_{k-1}\longrightarrow\dots\stackrel{{\scriptstyle\pi_{1}}}{{\longrightarrow}}\mathcal{E}_{1},$

which entitles the existence of the projective limit

$\varprojlim_{k\in\mathbb{N}}\mathcal{E}_{k}=\\{(M_{k})_{k\in\mathbb{N}}\;|\;M_{k}\in\mathcal{E}_{k}\mbox{
and }\pi_{k}(M_{k+1})=M_{k}\mbox{ for every }k\\},$

and it holds that

(4.11)
$\mathbb{Z}^{n}((B))\simeq\varinjlim_{j\in\mathbb{N}}\varprojlim_{k\in\mathbb{N}}B^{-j}\mathcal{E}_{k}.$

Analogously, for $k\in\mathbb{N}$ consider

$\mathcal{E}^{*}_{k}:=\Gamma[B^{*}]/B^{*k}\Gamma[B^{*}].$

Then

$\mathbb{Z}^{n}((B^{*}))\simeq\varinjlim_{j\in\mathbb{N}}\varprojlim_{k\in\mathbb{N}}{B^{*}}^{-j}\mathcal{E}^{*}_{k}.$

######  Proposition 4.13.

The characters of $\mathbb{K}_{A}$ are of the form

$\chi_{r,s}:\mathbb{K}_{A}\rightarrow
S^{1},\quad\chi_{r,s}(x,y)=\chi_{r}(x)\chi_{s}(y)=\exp(2\pi i\langle
x,r\rangle)\exp(2\pi iS_{s}(y)),$

for $r\in\mathbb{R}^{n}$, $s\in\mathbb{Z}^{n}((B^{*}))$, with $S_{s}(y)$ as in
(4.8). Moreover, there is a group morphism given by $(r,s)\mapsto\chi_{r,s}$.

###### Proof.

In view of Lemma 4.10 we have the isomorphism
$\widehat{\mathbb{K}_{A}}\simeq\widehat{\mathbb{R}^{n}}\times\widehat{\mathbb{Z}^{n}((B))}$.
It is known that there is an isomorphism
$\mathbb{R}^{n}\simeq\widehat{\mathbb{R}^{n}}$ given by $r\mapsto\chi_{r}$.
Consider the map
$\mathbb{Z}^{n}((B^{*}))\rightarrow\widehat{\mathbb{Z}^{n}((B))}$,
$s\mapsto\chi_{s}$; we show that it is an isomorphism. We first prove that the
group operations are compatible on both sets, that is,
$\chi_{s+s^{\prime}}(y)=\chi_{s}(y)\chi_{s^{\prime}}(y)$ for every
$y=\sum_{j=\nu(y)}^{\infty}B^{j}y_{j}\in\mathbb{Z}^{n}((B))$ with
$y_{j}\in\mathcal{E}$. It is enough to show that
$S_{s+s^{\prime}}(y)=S_{s}(y)+S_{s^{\prime}}(y)\mod\mathbb{Z}$. Applying Lemma
4.12, we get

$S_{s}(y)+S_{s^{\prime}}(y)-S_{s+s^{\prime}}(y)=\sum_{j=\nu(y)}^{\infty}\langle
y_{j},\\{{B^{*}}^{j}s\\}^{*}_{B}+\\{{B^{*}}^{j}s^{\prime}\\}^{*}_{B}-\\{{B^{*}}^{j}(s+s^{\prime})\\}^{*}_{B}\rangle.$

Proceeding in analogy to the proof of Proposition 4.11 and using the
definition of $B^{*}$-adic fractional part, we can see that, for every index
$j\geqslant\nu(y)$,

$\\{{B^{*}}^{j}s\\}^{*}_{B}+\\{{B^{*}}^{j}s^{\prime}\\}^{*}_{B}-\\{{B^{*}}^{j}(s+s^{\prime})\\}^{*}_{B}\subset\Gamma[B^{*}]\cap\Gamma[{B^{*}}^{-1}]\subset\Lambda^{*}\subset\mathbb{Z}^{n},$

with $\Lambda^{*}$ as in (4.4). Since
$y_{j}\in\mathcal{E}\subset\mathbb{Z}^{n}$ for every $j\geqslant\nu(y)$ and is
finite only for a finite number of indices, this yields the first part of the
proof.

Next, we show the injectivity. In view of the first part of the proof, it
suffices to show that $\chi_{s}\neq 1$ for $s\neq 0$. Let
$s=\sum_{j=\nu^{*}(s)}^{\infty}B^{*j}s_{j}\in\mathbb{Z}^{n}((B^{*}))\setminus\\{0\\}$
with $s_{j}\in\mathcal{E}^{*}$, and consider a point of the form
$B^{-l}c\in\mathbb{Z}^{n}((B))$ for $0\neq c\in\mathcal{E}$ and
$l\in\mathbb{N}$. Note that $\\{B^{j-l}c\\}_{B}=0$ whenever $j\geqslant l$.
Therefore, applying again Lemma 4.12 we get

$S_{s}(B^{-l}c)=\sum_{j=\nu^{*}(s)}^{l-1}\langle
B^{j-l}c,s_{j}\rangle=\sum_{j=\nu^{*}(s)}^{l-1}\langle
c,{B^{*}}^{j-l}s_{j}\rangle=\bigg{\langle}c,\sum_{j=\nu^{*}(s)}^{l-1}{B^{*}}^{j-l}s_{j}\bigg{\rangle}.$

Suppose $\chi_{s}=1$, then $S_{s}(B^{-l}c)\in\mathbb{Z}$ for every
$l\in\mathbb{N}$ and every $c\in\mathcal{E}$. Recall that in Remark 4.8 we
assumed w.l.o.g. that $\mathcal{E}$ has a subset $\\{c_{1},\dots,c_{n}\\}$
such that $\varTheta:=\langle c_{1},\dots,c_{n}\rangle_{\mathbb{Z}}$ is a full
rank lattice in $\mathbb{R}^{n}$. Thus
$S_{s}(B^{-l}\varTheta)\subset\mathbb{Z}$ and hence, by the definition of dual
lattice, $\sum_{j=\nu^{*}(s)}^{l-1}{B^{*}}^{l-j}s_{j}\in\varTheta^{*}$ holds
for all $l\in\mathbb{N}$. Since $s\neq 0$, we have that

$\sum_{j=\nu^{*}(s)}^{l-1}{B^{*}}^{l-j}s_{j}\in(B^{*l-\nu^{*}(s)}\Gamma+\dots+B^{*}\Gamma)\setminus(B^{*l-\nu^{*}(s)-1}\Gamma+\dots+B^{*}\Gamma).$

Then $(B^{*l-\nu^{*}(s)}\Gamma+\dots+B^{*}\Gamma)_{l\geqslant 1}$ is a
strictly nested sequence of lattices in $\mathbb{Q}^{n}$. Given $l$, consider
the entries of all the vectors of $B^{*l-\nu^{*}(s)}\Gamma+\dots+B^{*}\Gamma$
expressed as irreducible fractions, and define $m_{l}$ to be the maximum of
the denominators of these fractions. It is clear that $m_{l}\leqslant
m_{l+1}$. However, note that $(m_{l})_{l\geqslant 1}$ does not stabilize: this
would imply that, for some lattice $\Lambda_{l}$, there are infinitely many
steps in which we can add a point and form a strictly larger lattice, while
not increasing the bound on the denominators, which is not possible. This
means that $\varTheta^{*}$ contains points with entries having arbitrarily
large denominators, which is a contradiction.

For the surjectivity, consider a character
$\chi\in\widehat{\mathbb{Z}^{n}((B))}$. By classical arguments following [21,
p. 139], we obtain from (4.11) that

(4.12)
$\widehat{\mathbb{Z}^{n}((B))}\simeq\varprojlim_{j\in\mathbb{N}}\varinjlim_{k\in\mathbb{N}}\widehat{B^{-j}\mathcal{E}_{k}}.$

Since $B^{-j}\mathcal{E}_{k}=B^{-j}\mathbb{Z}^{n}[B]/B^{k-j}\mathbb{Z}^{n}[B]$
is a finite group of cardinality $b^{k}$, so is its dual. Consider the group

$G_{j,k}:=B^{*j-k}\Gamma[B^{*}]/B^{*j}\Gamma[B^{*}],$

and note that $|G_{j,k}|=b^{k}$ because $\Gamma$ is a full rank lattice (and
hence, isomorphic to $\mathbb{Z}^{n}$). Given $s\in G_{j,k}$, we can regard it
as an element in $B^{*j-k}\mathcal{E}^{*}+\dots+B^{*j-1}\mathcal{E}^{*}$ and
consider the character $\chi_{s}$ as in (4.7). Note that
$\chi_{s}\in\mbox{Ann}(B^{k-j}\mathbb{Z}^{n}[B])$ (see Lemma 4.10); hence,
$\chi_{s}$ is a character of ${B^{-j}\mathcal{E}_{k}}$. Also, for $s\neq 0$ in
$G_{j,k}$, there is $y\in{B^{-j}\mathcal{E}_{k}}$ such that $S_{s}(y)\neq 0$:
in fact, there exists $l$ with $j-k\leqslant l\leqslant j-1$ with $s_{l}\neq
0$; since $\mathcal{E}$ spans a full rank lattice, find $c\in\mathcal{E}$ such
that $\langle c,s_{l}\rangle\neq 0$ and take $y=B^{-l}c$. Hence all the
characters $\chi_{s}$ for $s\in G_{j,k}$ are distinct over
$B^{-j}\mathcal{E}_{k}$, and since $|B^{-j}\mathcal{E}_{k}|=b^{k}=|G_{j,k}|$,
this implies that $\widehat{B^{-j}\mathcal{E}_{k}}\simeq G_{j,k}$, so (4.12)
yields

$\widehat{\mathbb{Z}^{n}((B))}\simeq\varprojlim_{j\in\mathbb{N}}\varinjlim_{k\in\mathbb{N}}G_{j,k}$

It is not hard to establish an isomorphism $G_{j,k}\simeq
B^{*-j}\mathcal{E}^{*}_{k}$, and so

$\chi\in\varprojlim_{j\in\mathbb{N}}\varinjlim_{k\in\mathbb{N}}\\{\chi_{s}\;|\;s\in
G_{j,k}\\}\simeq\varprojlim_{j\in\mathbb{N}}\varinjlim_{k\in\mathbb{N}}B^{*-j}\mathcal{E}^{*}_{k}\simeq\mathbb{Z}^{n}((B^{*})),$

therefore every character $\chi$ is of the form $\chi=\chi_{s}$ for some
$s\in\mathbb{Z}^{n}((B^{*}))$. ∎

###  4.3. Multiple tiling theorem

In this final section, we will prove that rational self-affine tiles give a
multiple tiling of the representation space. We will make use of the character
theory we have developed in the previous section.

Recall that the set $\varphi(\mathbb{Z}^{n}[A])$ is a lattice by Lemma 4.3,
hence the torus $\mathbb{T}:=\mathbb{K}_{A}/\varphi(\mathbb{Z}^{n}[A])$ is
well defined and compact. We endow it with the normalized quotient measure
$\bar{\mu}$, which is the Haar measure on $\mathbb{T}$. Denote the
multiplication by $A$ on $\mathbb{T}$ as

$\tau_{A}:\mathbb{T}\rightarrow\mathbb{T};\quad(x,y)\mapsto
A(x,y)\mod\varphi(\mathbb{Z}^{n}[A]),$

which is well defined because $\varphi(\mathbb{Z}^{n}[A])$ is $A$-invariant.
We will prove that this map is ergodic by using the following lemma.

######  Lemma 4.14.

If $G$ is a compact abelian group with normalized Haar measure and
$\tau:G\rightarrow G$ is a surjective continuous endomorphism of $G$, then
$\tau$ is ergodic if and only if the trivial character $\chi=1$ is the only
character of $G$ that satisfies $\chi\circ\tau^{k}=\chi$ for some $k\geqslant
1$.

###### Proof.

See [28, Theorem 1.10.1]. ∎

We are in position to prove the following result.

######  Lemma 4.15.

The map $\tau_{A}$ is ergodic.

###### Proof.

Because $A$ is an invertible matrix, $\tau_{A}$ is a continuous surjective
homomorphism. We first prove that it is measure preserving. Note that
$\varphi(\mathbb{Z}^{n}[A])$ is a sublattice of index $a$ of
$A^{-1}\varphi(\mathbb{Z}^{n}[A])$. This implies that, for any measurable set
$E\subset\mathbb{T}$,

$\begin{split}\bar{\mu}(\tau_{A}^{-1}(E))&=\bar{\mu}(\\{(x,y)\in\mathbb{T}\;|\;A(x,y)\mod\varphi(\mathbb{Z}^{n}[A])\in
E\\})\\\ &=\bar{\mu}(\\{(x,y)\in\mathbb{T}\;|\;(x,y)\mod
A^{-1}\varphi(\mathbb{Z}^{n}[A])\in A^{-1}E\\})\\\
&=a\,\bar{\mu}(\\{(x,y)\in\mathbb{T}\;|\;(x,y)\mod\varphi(\mathbb{Z}^{n}[A])\in
A^{-1}E\\})\\\ &=a\,\bar{\mu}(A^{-1}E)=\bar{\mu}(E).\\\ \end{split}$

By Proposition 4.13, the characters of $\mathbb{K}_{A}$ are of the form
$\chi_{r,s}$ for $r\in\mathbb{R}^{n}$, $s\in\mathbb{Z}^{n}((B^{*}))$. Since
$\mathbb{T}=\mathbb{K}_{A}/\varphi(\mathbb{Z}^{n}[A])$, by Lemma 4.10 there is
an isomorphism
$\widehat{\mathbb{T}}\simeq\mbox{Ann}(\varphi(\mathbb{Z}^{n}[A]))$; this means
that a character of $\mathbb{T}$ is of the form $\chi_{r,s}$ and satisfies
$\chi_{r,s}(\varphi(z))=1$ for every $z\in\mathbb{Z}^{n}[A]$. Suppose there
exists a character $\chi_{r,s}\in\widehat{\mathbb{T}}$ satisfying
$\chi_{r,s}\circ\tau^{k}_{A}=\chi_{r,s}$ for some $k\geqslant 1$. In view of
Lemma 4.14, it suffices to show that $\chi_{r,s}$ is constantly equal to $1$.
We have that, for every $(x,y)\in\mathbb{T}$,

$\chi_{r,s}\circ\tau^{k}_{A}(x,y)=\chi_{r,s}(x,y),$

meaning

$\exp(2\pi i(\langle r,A^{k}x\rangle+S_{s}(A^{k}y)))=\exp(2\pi i(\langle
r,x\rangle+S_{s}(y)))$

with $S_{s}$ defined in (4.8). Letting $y=0$ implies

$\langle r,A^{k}x\rangle=\langle r,x\rangle\mod\mathbb{Z}$

and so, for every $x\in[0,1]^{n}$,

$\langle{A^{*}}^{k}r-r,x\rangle\in\mathbb{Z},$

which can only be true if ${A^{*}}^{k}r-r=0$. Suppose $r\neq 0$; this implies
that ${A^{*}}^{k}$ has $1$ as an eigenvalue, and so therefore $A$ has $1$ as
an eigenvalue, which contradicts the fact that $A$ is expanding. Hence, $r=0$.

Next, we prove that $\chi_{(0,s)}\in\widehat{\mathbb{T}}$ implies $s=0$. For
every $z\in\mathbb{Z}^{n}[A]$ we have
$1=\chi_{(0,s)}(\varphi(z))=\chi_{0}(z)\chi_{s}(z)=\chi_{s}(z)$. Consider
points of the form $z=A^{l}c=B^{-l}c$ with $l\geqslant 1$, $c\in\mathcal{E}$.
Then $S_{s}(B^{-l}c)=\langle c,\\{B^{*-l}s\\}^{*}_{B}\rangle\in\mathbb{Z}$.
Recall that $\mathcal{E}$ contains a subset that spans a full rank lattice
$\varTheta$. By the definition of dual lattice, this implies that
$\\{B^{*-l}s\\}^{*}_{B}\in\varTheta^{*}$ for every $l\geqslant 1$. Suppose
$s\neq 0$, then the leading $B^{*}$-adic coefficient satisfies
$s_{\nu^{*}(s)}\in\Gamma\setminus B^{*}\Gamma$, because the coefficients live
in $\mathcal{E}^{*}\subset\Gamma$ which is defined to be a residue set for
$\Gamma[B^{*}]/B^{*}\Gamma[B^{*}]$. Moreover, we have that

$\\{B^{*-l}s\\}^{*}_{B}\in(B^{*-l}\Gamma+\dots+B^{*-1}\Gamma+\Gamma)\setminus(B^{*-l+1}\Gamma+\dots+B^{*-1}\Gamma+\Gamma).$

Define inductively $\Lambda_{0}:=\Gamma$ and $\Lambda_{l+1}$ the lattice
spanned by $\Lambda_{l}$ and $\\{B^{*-l}s\\}^{*}_{B}$. Then
$(\Lambda_{l})_{l\geqslant 1}$ is a strictly nested sequence of lattices in
$\mathbb{Q}^{n}$, all of which are contained in $\varTheta^{*}$. Given $l$,
consider the entries of all the vectors of $\Lambda_{l}$ expressed as
irreducible fractions, and define $m_{l}$ to be the maximum of the
denominators of these fractions. It is clear that $m_{l}\leqslant m_{l+1}$.
However, note that $(m_{l})_{l\geqslant 1}$ does not stabilize: this would
imply that, for some lattice $\Lambda_{l}$, there are infinitely many steps in
which we can add a point and form a strictly larger lattice, while not
increasing the bound on the denominators, which is not possible. Hence the
sequence of denominators $(m_{l})_{l\geqslant 1}$ tends to infinity. This
contradicts the uniform discreteness of the lattice $\varTheta^{*}$. ∎

We arrive at our final result. The proof is based on the one of [16, Theorem
1.1]. For completeness, we include it here. We recall the reader of the
definition of multiple tiling given in Definition 3.5.

######  Theorem 4.16.

Let $\mathcal{F}=\mathcal{F}(A,\mathcal{D})$ be a rational self-affine tile.
Then $\mathcal{F}+\varphi(\mathbb{Z}^{n}[A])$ is a multiple tiling of
$\mathbb{K}_{A}$.

###### Proof.

Let $\bar{\mu}$ be the normalized Haar measure on the torus
$\mathbb{T}=\mathbb{K}_{A}/\varphi(\mathbb{Z}^{n}[A])$. Consider the canonical
projection $\pi:\mathbb{K}_{A}\rightarrow\mathbb{T}$, and define the function
$\varPhi:\mathbb{T}\rightarrow\mathbb{Z}_{\geqslant 0}$ as

$\varPhi(x,y):=|\pi^{-1}(x,y)\cap\mathcal{F}|,$

meaning $\varPhi$ counts the points on $\mathcal{F}$ that are congruent to
$(x,y)$ modulo $\varphi(\mathbb{Z}^{n}[A])$. Since $\mathcal{F}$ is compact,
$\varPhi$ is finite everywhere, hence it is well defined. Also, $\varPhi$ is
positive in a set of positive measure since $\mu(\mathcal{F})>0$. If we prove
that $\varPhi(x,y)$ is equal almost everywhere to some $k\in\mathbb{N}$, this
implies the statement of the theorem: it means that almost every point of
$\mathbb{K}_{A}$ gets covered by exactly $k$ translates of $\mathcal{F}$ when
translating via the set $\varphi(\mathbb{Z}^{n}[A])$. Note that $\varPhi$ is
constant almost everywhere if and only if there exists $k$ such that every
$S\subset\mathbb{T}$ satisfies

(4.13) $\int_{S}\varPhi(x,y)\,d\bar{\mu}(x,y)=k\,\bar{\mu}(S).$

To obtain this, we show first that $\varPhi$ satisfies

(4.14)
$\varPhi(x,y)=\frac{1}{a}\sum_{(x^{\prime},y^{\prime})\in\tau_{A}^{-1}(x,y)}\varPhi(x^{\prime},y^{\prime}),$

where $\tau_{A}$ is the multiplication by $A$ on the torus.

Using the set equation (2.14) of $\mathcal{F}$ and the fact that
$\varphi(\mathcal{D})\subset\varphi(\mathbb{Z}^{n}[A])$, we have

(4.15) $\begin{split}|\pi^{-1}(x,y)\cap
A\mathcal{F}|&=\Big{|}\pi^{-1}(x,y)\bigcap\Big{(}\bigcup_{d\in\mathcal{D}}(\mathcal{F}+\varphi(d))\Big{)}\Big{|}\\\
&=\Big{|}\bigcup_{d\in\mathcal{D}}(\pi^{-1}(x,y)-\varphi(d))\cap\mathcal{F}\Big{|}\\\
&=\sum_{d\in\mathcal{D}}|\pi^{-1}(x,y)\cap\mathcal{F}|=a\,\varPhi(x,y).\end{split}$

Recall that the index of the sublattice $A\varphi(\mathbb{Z}^{n}[A])$ of
$\varphi(\mathbb{Z}^{n}[A])$ is $a$, which implies that
$|\tau_{A}^{-1}(x,y)|=a$, and so

(4.16) $|\pi^{-1}(x,y)\cap
A\mathcal{F}|=\frac{1}{a}\sum_{(x^{\prime},y^{\prime})\in\tau_{A}^{-1}(x,y)}|\pi^{-1}(\tau_{A}(x^{\prime},y^{\prime}))\cap
A\mathcal{F}|.$

Also, for any $(u,v)\in\mathbb{K}_{A}$, it holds that
$|((u,v)+\varphi(\mathbb{Z}^{n}[A]))\cap
A\mathcal{F}|=a|((u,v)+A\varphi(\mathbb{Z}^{n}[A]))\cap A\mathcal{F}|$, and
hence, for each $(x^{\prime},y^{\prime})\in\tau_{A}^{-1}(x,y)$,

(4.17) $|\pi^{-1}(\tau_{A}(x^{\prime},y^{\prime}))\cap
A\mathcal{F}|=a\,|A\pi^{-1}(x^{\prime},y^{\prime})\cap
A\mathcal{F}|=a\,|\pi^{-1}(x^{\prime},y^{\prime})\cap\mathcal{F}|=a\,\varPhi(x^{\prime},y^{\prime}).$

Thus from (4.16) and (4.17) follows that

(4.18) $\begin{split}|\pi^{-1}(x,y)\cap
A\mathcal{F}|=\sum_{(x^{\prime},y^{\prime})\in\tau_{A}^{-1}(x,y)}\varPhi(x^{\prime},y^{\prime}).\end{split}$

Combining (4.15) and (4.18) yields (4.14). Applying (4.14) and doing the
change of variables $(x,y)\mapsto\tau_{A}(x,y)$ we get

(4.19)
$\begin{split}\int_{S}\varPhi(x,y)d\bar{\mu}(x,y)&=\int_{S}\frac{1}{a}\sum_{(x^{\prime},y^{\prime})\in\tau_{A}^{-1}(x,y)}\varPhi(x^{\prime},y^{\prime})\,d\bar{\mu}(x,y)\\\
&=\int_{\tau_{A}^{-1}(S)}\frac{1}{a}\sum_{\tau_{A}(x^{\prime},y^{\prime})=\tau_{A}(x,y)}\varPhi(x^{\prime},y^{\prime})\,a\,d\bar{\mu}(x,y)\\\
&=\int_{\tau_{A}^{-1}(S)}\varPhi(x,y)\,d\bar{\mu}(x,y).\\\ \ \end{split}$

Because the map $\tau_{A}$ is ergodic by Lemma 4.15, iterating (4.19)
$j\geqslant 1$ times and afterwards applying the ergodic theorem (see [6,
Theorem 3.20]), yields

(4.20)
$\begin{split}\int_{S}\varPhi(x,y)d\bar{\mu}(x,y)&=\int_{\tau_{A}^{-j}(S)}\varPhi(x,y)d\bar{\mu}(x,y)\\\
&=\int_{\mathbb{T}}1_{s}(\tau_{A}^{j}(x,y))\varPhi(x,y)d\bar{\mu}(x,y)\\\
&=\int_{\mathbb{T}}\Big{(}\frac{1}{N}\sum_{j=1}^{N}1_{s}(\tau_{A}^{j}(x,y))\Big{)}\varPhi(x,y)d\bar{\mu}(x,y)\\\
&\xrightarrow[N\to\infty]{}\bar{\mu}(S)\int_{\mathbb{T}}\varPhi(x,y)d\bar{\mu}(x,y)=k\,\bar{\mu}(S),\end{split}$

and this concludes the proof. ∎

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