# Large normalizers of $\mathbb{Z}^{d}$-odometers systems and realization on
substitutive subshifts

Christopher Cabezas  Departament of Mathematics, Université de Liège, Allée de
la découverte 12 (B37), 4000 Liège, Belgium<EMAIL_ADDRESS>and  Samuel
Petite  LAMFA, CNRS, UMR 7352, Université de Picardie Jules Verne, 80 000
Amiens, France<EMAIL_ADDRESS>
###### Abstract.

For a $\mathbb{Z}^{d}$-topological dynamical system $(X,T,\mathbb{Z}^{d})$, an
_isomomorphism_ is a self-homeomorphism $\phi:X\to X$ such that for some
matrix $M\in GL(d,\mathbb{Z})$ and any ${\bm{n}}\in\mathbb{Z}^{d}$, $\phi\circ
T^{{\bm{n}}}=T^{M{\bm{n}}}\circ\phi$, where $T^{\bm{n}}$ denote the self-
homeomorphism of $X$ given by the action of ${\bm{n}}\in{\mathbb{Z}}^{d}$. The
collection of all the isomorphisms forms a group that is the normalizer of the
set of transformations $T^{\bm{n}}$. In the one-dimensional case, isomorphisms
correspond to the notion of _flip conjugacy_ of dynamical systems and by this
fact are also called _reversing symmetries_.

These isomorphisms are not well understood even for classical systems. We
present a description of them for odometers and more precisely for constant-
base $\mathbb{Z}^{2}$-odometers, which is surprisingly not simple. We deduce a
complete description of the isomorphisms of some minimal
$\mathbb{Z}^{d}$-substitutive subshifts. This enables us to provide the first
example known of a minimal zero-entropy subshift with the largest possible
normalizer group.

###### Key words and phrases:

multidimensional substitutive subshift, odometer, normalizer, automorphism

###### 2020 Mathematics Subject Classification:

Primary: 37B10; Secondary: 52C23, 37B52, 20H15, 20E18

The authors acknowledge the financial support of ANR project IZES
ANR-22-CE40-0011 and ECOS projet ACEDic C21E04.

##  1\. Introduction

This article concerns $\mathbb{Z}^{d}$-odometers and their symbolic
extensions. More specifically, for such dynamical system
$(X,T,\mathbb{Z}^{d})$, we study their _isomorphisms_ , which are self-
homeomorphisms $\phi:X\to X$ such that for some linear transformation $M\in
GL(d,\mathbb{Z})$ and any ${\bm{n}}\in\mathbb{Z}^{d}$, $\phi\circ
T^{{\bm{n}}}=T^{M{\bm{n}}}\circ\phi$, where $T^{\bm{n}}$ denotes the self-
homeomorphism of $X$ given by the $\mathbb{Z}^{d}$ action $T$.

Isomorphisms associated with the linear transformation $M=\textrm{Id}$ are, in
essence, nothing more than self-conjugacies of the system. They are commonly
referred to as _automorphisms_ of dynamical systems. Therefore, an isomorphism
can be thought of as a self-conjugacy up to a
$\textrm{GL}(d,\mathbb{Z})$-transformation. In the one-dimensional case
($d=1$), isomorphisms correspond to the notion of _flip conjugacy_ in
dynamical systems [6]. Because of this, they are also known as _reversing
symmetries_ (see [19, 4]). The automorphisms can be algebraically defined as
elements of the centralizer of the group $\left\langle T\right\rangle$,
considered as a subgroup of all self-homeomorphisms $\operatorname{Homeo}(X)$
of $X$. From this algebraic perspective, isomorphisms can be seen as elements
of the normalizer group of $\left\langle T\right\rangle$ within the group
$\operatorname{Homeo}(X)$. Consequently, the automorphism group is a normal
subgroup of the normalizer. The quotient is isomorphic to a linear subgroup of
$\textrm{GL}(d,\mathbb{Z})$, called the _linear representation group_ ,
achieved through the identification of an isomorphism $\phi$ with its matrix
$M$. The automorphism group is never trivial (it always contains the
transformations $T^{\bm{n}}$, $\bm{n}\in\mathbb{Z}^{d}$), but generally, the
existence of an isomorphism for a given $M\in GL(d,\mathbb{Z})$ remains an
open problem, which is significant in the context of higher-rank actions.

There are few general results on these groups so natural questions on the
algebraic properties of the group of isomorphisms (is it finite? is it
amenable? what are the subgroups? etc.), on their relations with the dynamics,
or on a description of their actions, are widely open. In this article, we
explore these issues using classic and widely studied examples of odometers
and substitutive Toeplitz subshifts.

Even though the dynamics of an odometer is one of the most well understood, a
partial description of their isomorphisms has only been initiated in [18] for
$d=2$ and pursued for higher rank in [22, 25, 26]. These works were
essentially motivated by the fact that isomorphisms define specific
transformations of orbit equivalence. We complete these studies, but mainly
for dimension $d=2$, by describing explicitly the group structure of their
normalizers. This leads to a complete classification of them (Theorem 3.3) for
constant-base odometer systems, i.e., whose base is given by the iteration of
the same expansive matrix. The classification depends on arithmetical
properties of the expansive matrix. In addition, we provide computable
arithmetic conditions for deciding whether a given matrix arises as an
isomorphism of a constant-base odometer. Our techniques use elementary
arithmetic, but extending our results to higher ranks would require
sophisticated notions of number theory.

Interest in the study of the normalizer group for multidimensional subshifts
has increased in recent years. These objects represent geometrical and
combinatorial symmetries, such as rotations and reflections, that a particular
subshift possesses. Similar studies were initiated in the framework of tilings
and Delone sets (see e.g., [24]), where the existence of some point set
symmetries is at the heart of the discovery of aperiodicity of quasicrystals
[27]. For instance, the Penrose tiling serves as a model with ten-fold
symmetry [23]. However, still few results and no systematic description of the
analogous normalizer group for $\mathbb{R}^{d}$-flows exist. For a self-
similar tiling, its linear representation group is known to be countable and
isomorphic to its mapping class group [21]. Recently, classical examples of
multidimensional subshifts have been considered to study their normalizers [2,
5, 7, 8]. These works have yielded sufficient conditions (of a combinatorial
nature in [7] and a geometrical nature in [8]) on a substitutive subshift so
that the quotient of the normalizer by the automorphism group is finite, and
each group is virtually $\mathbb{Z}^{d}$. The article [2] concerns different
classes of non-minimal subshifts with positive entropy having a normalizer not
virtually isomorphic to its automorphism group.

However, the question remains open whether an infinite linear representation
group is also possible for minimal, deterministic (zero-entropy) subshifts. In
this article, we provide a positive answer to this problem by completely
describing the normalizer group of a family of minimal subshifts (Theorem
4.1). The examples are substitutive Toeplitz subshifts that can be presented
in an effective way, in the algorithmic sense. These subshifts are highly
deterministic meaning they have zero entropy. More precisely, the rate of
growth of their complexity is polynomial of degree the dimension $d$. In
particular, we exhibit examples where the linear representation group is the
largest one, i.e., equal to $\textrm{GL}(d,\mathbb{Z})$. Together with the
results of [2], this shows that the normalizer is not restricted by the
complexity. More surprisingly, low complexity is not enough to ensure the
amenability of the normalizer group in dimension $d>1$. This phenomenon
differs from what happens in dimension one [11, 12, 13, 14, 15, 16], where the
amenability of the automorphism group (of finite-index in the normalizer
group) is shown for a large class of zero-entropy subshifts and any Toeplitz
subshift.

This article is organized as follows: Section 2 is devoted to the background
on topological and symbolic dynamics. In particular, relations between the
normalizer of a system with the one of its maximal equicontinuous factor are
exhibited. The next section concerns the study of the normalizers of odometers
in order to describe them (Theorem 3.3). Finally, we construct examples of
multidimensional subshifts with an odometer as their maximal equicontinuous
factors in Section 4. We characterize their infinite linear representation
groups (Theorem 4.1) by using our study of normalizers of odometers and the
relations with their extensions.

##  2\. Definitions and basic properties

###  2.1. Topological dynamical systems

We recall that a _topological dynamical system_ $(X,T,\mathbb{Z}^{d})$ is
given by a continuous left-action $T\colon\mathbb{Z}^{d}\times X\to X$ on a
compact metric space $X$ equipped with a distance. This provides a family of
self-homeomorphisms of $X$: $\\{T^{\bm{n}}\colon{\bm{n}}\in\mathbb{Z}^{d}\\}$,
also denoted by $\langle T\rangle$, such that $T^{\bm{m}}\circ
T^{\bm{n}}=T^{\bm{m}+\bm{n}}$ for any ${\bm{m},\bm{n}}\in\mathbb{Z}^{d}$. In
particular, the homeomorphisms $T^{\bm{n}}$ commute with each other. The
_orbit_ of a point $x\in X$ is the set
$\mathcal{O}(x,T)=\\{T^{\bm{n}}(x)\colon{\bm{n}}\in\mathbb{Z}^{d}\\}$. We will
be mainly concerned by topological dynamical systems that are minimal, i.e.,
where any orbit is dense. In this case, there is no topological way to
classify the orbits.

An important type of topological dynamical systems are the equicontinuous
ones. A topological dynamical system $(X,T,\mathbb{Z}^{d})$ is said to be
_equicontinuous_ if the set of maps
$\\{T^{{\bm{n}}}\colon{\bm{n}}\in\mathbb{Z}^{d}\\}$ forms an equicontinuous
family of homeomorphisms. The equicontinuous systems are, in some sense, the
simplest dynamical systems.

In the following, we define the endomorphisms and isomorphisms of a
topological dynamical system, which are the central objects of study of this
article. Isomorphisms represent internal symmetries of a given system that do
not commute with the action, such as rotations and reflections.

######  Definition 2.1.

Let $(X,T,\mathbb{Z}^{d})$, $(Y,S,\mathbb{Z}^{d})$ be two topological
dynamical systems and ${M\in\textrm{GL}(d,\mathbb{Z})}$. An $M$-_epimorphism_
is a continuous surjective map ${\phi:X\to Y}$ that for any
${\bm{n}}\in\mathbb{Z}^{d}$ satisfies $\phi\circ
T^{{\bm{n}}}=T^{M{\bm{n}}}\circ\phi$. When
$(X,T,\mathbb{Z}^{d})=(Y,S,\mathbb{Z}^{d})$, it is called an
$M$-_endomorphism_. Moreover, if $\phi$ is invertible, it is called an
$M$-_isomorphism_.

We simply call a $\textrm{GL}(d,\mathbb{Z})$-_endomorphism_ (or
$\textrm{GL}(d,\mathbb{Z})$-_isomorphism_), any $M$-endomorphism (resp.
isomorphism) for some $M\in\textrm{GL}(d,\mathbb{Z})$.

In other terms, the $\textrm{GL}(d,\mathbb{Z})$-isomorphisms are conjugacies
of $\mathbb{Z}^{d}$-actions, up to a
$\textrm{GL}(d,\mathbb{Z})$-transformation, i.e., an orbit equivalence with a
constant orbit cocycle. In the special case where $M$ is the identity matrix
${\rm Id}$, the ${\rm Id}$-endomorphisms and ${\rm Id}$-isomorphisms are
called _endomorphisms_ (or _self-factor maps_) and _automorphisms_ (or _self-
conjugacies_), respectively.

When the action is _aperiodic_ , i.e., the stabilizer of any point is trivial
or $T^{\bm{n}}x=x$ only for ${\bm{n}}={\bm{0}}$, the matrix $M$ associated
with a $\textrm{GL}(d,\mathbb{Z})$-endomorphism $\Phi$ is unique: $\Phi$ is an
$M_{1}$\- and $M_{2}$-endomorphism if and only if $M_{1}=M_{2}$. In the
following, we fix different notations that we will use throughout this
article:

  * •

$N_{M}(X,T)$, with $M\in\textrm{GL}(d,\mathbb{Z})$, denotes the set of all
$M$-endomorphisms of $(X,T)$;

  * •

$N(X,T)$ denotes the set of all the $\textrm{GL}(d,\mathbb{Z})$-endomorphisms
of the dynamical system $(X,T,\mathbb{Z}^{d})$, i.e.

$N(X,T)=\bigcup\limits_{M\in\textrm{GL}(d,\mathbb{Z})}N_{M}(X,T).$

The set of $\textrm{GL}(d,\mathbb{Z})$-isomorphisms, is denoted by
$N^{*}(X,T)$. In algebraic terms, this set is a group and, when the action is
aperiodic, corresponds to the normalizer of the group action $\left\langle
T\right\rangle$ in the group of self-homeomorphisms $\operatorname{Homeo}(X)$
of $X$, that is, the set $\phi\colon X\to X$ such that $\phi\circ
T^{\bm{n}}\circ\phi^{-1}\in\\{T^{\bm{m}}:{\bm{m}}\in\mathbb{Z}^{d}\\}$ for any
${\bm{n}}\in\mathbb{Z}^{d}$;

  * •

$\operatorname{End}(X,T)$ and $\operatorname{Aut}(X,T)$ denotes respectively
the set of all endomorphisms and automorphisms of $(X,T,\mathbb{Z}^{d})$.

  * •

We define the _linear representation semigroup_ $\vec{N}(X,T)$ as the
semigroup of all matrices $M\in\textrm{GL}(d,\mathbb{Z})$ with
${N_{M}(X,T)\neq\emptyset}$.

Notice that for $\phi\in N_{M_{1}}(X,T)$ and $\psi\in N_{M_{2}}(X,T)$, the
composition $\phi\circ\psi$ belongs to $N_{M_{1}M_{2}}(X,T)$. Moreover, if
$\phi$ is an $M$-isomorphism, then its inverse is an $M^{-1}$-isomorphism, so
the sets $N_{M}(X,T)$ are not semigroups (except when $M$ is the identity
matrix).

Concerning the linear representation semigroup $\vec{N}(X,T)$, it is direct to
check that the isomorphism class of $\vec{N}(X,T)$ is invariant under
conjugacy. However, it is not necessarily a group, since the existence of an
$M$-endomorphism associated with a matrix $M\in\textrm{GL}(d,\mathbb{Z})$ does
not necessarily imply the existence of an $M^{-1}$-endomorphism. Nevertheless,
we have the following result when a dynamical system is coalescent. Recall
that a topological dynamical system $(X,T,\mathbb{Z}^{d})$ is _coalescent_
when any endomorphism of $(X,T,\mathbb{Z}^{d})$ is invertible.

######  Proposition 2.2.

Let $(X,T,\mathbb{Z}^{d})$ be a coalescent system. If the linear
representation semigroup $\vec{N}(X,T)$ is a group, then any
$\textrm{GL}(d,\mathbb{Z})$-endomorphism in $N(X,T)$ is invertible, i.e.,
$N(X,T)=N^{*}(X,T)$.

Equicontinuous systems are examples of coalescent systems [1].

###### Proof.

Let $\phi,\psi$ be respectively an $M$\- and $M^{-1}$-endomorphism onto
$(X,T,\mathbb{Z}^{d})$. Then $\phi\circ\psi$ is an endomorphism onto
$(X,T,\mathbb{Z}^{d})$. Since $(X,T,\mathbb{Z}^{d})$ is coalescent, then
$\phi\circ\psi$ is invertible. It follows that $\phi$ and $\psi$ are
invertible maps. ∎

A particular case is when the linear representation semigroup of a coalescent
system is finite. In this case, it is always a group, so any
$\textrm{GL}(d,\mathbb{Z})$-endomorphism is invertible.

The groups $\left\langle T\right\rangle$ and $\operatorname{Aut}(X,T)$ are
normal subgroups of $N^{*}(X,T)$ (the group of isomorphisms). More precisely,
for aperiodic systems the following exact sequence holds:

(1) $\displaystyle 1$ $\displaystyle\to$
$\displaystyle\operatorname{Aut}(X,T)\quad$ $\displaystyle\to\quad N^{*}(X,T)$
$\displaystyle\xrightarrow{j}\quad\vec{N^{*}}(X,T)\quad\to$
$\displaystyle\quad 1,$

where all the maps are the canonical injections and $j(\phi)=M$ whenever
$\phi\in N_{M}(X,T)$.

A _factor map_ $\pi:(X,T,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ between two
topological dynamical systems $(X,T,\mathbb{Z}^{d})$ and
$(Y,S,\mathbb{Z}^{d})$ is a continuous onto map commuting with the action,
i.e., $\pi\circ T^{\bm{n}}=S^{\bm{n}}\circ\pi$ for any
${\bm{n}}\in\mathbb{Z}^{d}$. The system $(Y,S,\mathbb{Z}^{d})$ is said to be a
_factor_ of $(X,T,\mathbb{Z}^{d})$, and $(X,T,\mathbb{Z}^{d})$ is an
_extension_ of $(Y,S,\mathbb{Z}^{d})$. If $|\pi^{-1}(\\{y\\})|\leq K<\infty$
for all $y\in Y$, we say that $\pi$ is _finite-to-1_. Sometimes there exists a
$G_{\delta}$-dense subset $Y_{0}\subseteq Y$ where any $y\in Y_{0}$ satisfies
$|\pi^{-1}(\\{y\\})|=1$. In this case, the factor map $\pi$ is said to be
_almost_ _1-to-1_. We recall that when the system $(Y,S,\mathbb{Z}^{d})$ is
minimal, the existence of one point with only one preimage is enough to ensure
that the map is almost 1-to-1.

For every topological dynamical system, there exists at least one
equicontinuous factor, which is the system given by one point. Furthermore,
any topological dynamical system admits a _maximal equicontinuous factor_ ,
i.e., a factor
$\pi_{eq}:(X,T,\mathbb{Z}^{d})\to(X_{eq},T_{eq},\mathbb{Z}^{d})$ where
$(X_{eq},T_{eq},\mathbb{Z}^{d})$ is an equicontinuous system, satisfying the
following universal property: for any other equicontinuous factor
$\pi:(X,T,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ there exists a factor map
$\phi:(X_{eq},T_{eq},\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ such that
$\pi=\phi\circ\pi_{eq}$ [1]. Also, in the particular case where
$\pi:(X,T,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ is an almost 1-to-1 factor
on an equicontinuous system $(Y,S,\mathbb{Z}^{d})$, this factor is the maximal
equicontinuous factor of $(X,T,\mathbb{Z}^{d})$. As typical examples of this
case, are the odometer systems (see next section) that are almost 1-to-1
factors of particular symbolic systems [9, 10, 17].

A factor map $\pi:(X,T,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ is _compatible_
if any endomorphism $\phi\in\operatorname{End}(X,T)$ preserves the
$\pi$-fibers, i.e., $\pi(\phi(x))=\pi(\phi(y))$ for any $x,y\in X$, such that
$\pi(x)=\pi(y)$. With the same spirit, we say that a factor $\pi$ is
_compatible with $\textrm{GL}(d,\mathbb{Z})$-endomorphisms_ if for any
$\textrm{GL}(d,\mathbb{Z})$-endomorphism $\phi\in N(X,T)$,
$\pi(\phi(x))=\pi(\phi(y))$ for any $x,y\in X$, such that $\pi(x)=\pi(y)$.

The compatibility property allow us to relate
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms between factor systems. The next
lemma follows the ideas from [11, Theorem 3.3] but in a larger context.

######  Lemma 2.3.

Let $(X,T,\mathbb{Z}^{d})$, $(Y,S,\mathbb{Z}^{d})$ be two minimal systems,
such that $\pi:(X,T,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ is a compatible
factor. Then, there is a semigroup homomorphism
$\hat{\pi}:\operatorname{End}(X,T)\to\operatorname{End}(Y,S)$ such that

  1. 1.

$\hat{\pi}(\phi)(\pi(x))=\pi(\phi(x))$ for all
$\phi\in\operatorname{End}(X,T)$ and $x\in X$.

  2. 2.

$\hat{\pi}(\operatorname{Aut}(X,T))\leqslant\operatorname{Aut}(Y,S)$.

  3. 3.

For all $\psi\in\operatorname{End}(Y,S)$,
$|\hat{\pi}^{-1}(\\{\psi\\})|\leq\min\limits_{y\in Y}|\pi^{-1}(y)|$.

Moreover, if $\pi$ is compatible with
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms, there is an extension of
$\hat{\pi}:N(X,T)\to N(Y,S)$ defined as in Item 1. for all $\phi\in N(X,T)$,
satisfying the following properties

  1. i)

$\hat{\pi}(N_{M}(X,T))\subseteq N_{M}(Y,S)$, for any $M\in GL(d,\mathbb{Z})$.

  2. ii)

For any $M\in\textrm{GL}(d,\mathbb{Z})$, the map $\hat{\pi}:N_{M}(X,T)\to
N_{M}(Y,S)$ is at most $\min\limits_{y\in Y}|\pi^{-1}(\\{y\\})|$-to-1.

  3. iii)

For any $\phi\in\hat{\pi}^{-1}(N^{*}(Y,S))$, the cardinality of the
$\pi$-fiber is non decreasing under the
$\textrm{GL}(d,\mathbb{Z})$-isomorphism $\hat{\pi}(\phi)^{-1}$. In other
terms, for any integer $n\geq 1$, the map $\hat{\pi}(\phi)$ satisfies

$\\{y\in Y\colon|\pi^{-1}(\\{y\\})|\geq
n\\}\subset\hat{\pi}(\phi)\left(\\{y\in Y\colon|\pi^{-1}(\\{y\\})|\geq
n\\}\right).$

###### Proof.

Set $\phi\in\operatorname{End}(X,T)$. By definition, the map
$\hat{\pi}(\phi):Y\to Y$ given by $\hat{\pi}(\phi)(\pi(x))=\pi(\phi(x))$ is
well defined and is surjective by minimality of $(Y,S,\mathbb{Z}^{d})$. So
$\hat{\pi}(\phi)$ is an endomorphism of $(Y,S,\mathbb{Z}^{d})$. Moreover, if
$\phi$ is an automorphism of $(X,T,\mathbb{Z}^{d})$, then $\hat{\pi}(\phi)$ is
invertible. Indeed,
$\hat{\pi}(\phi)\circ\hat{\pi}(\phi^{-1})\circ\pi=\pi\circ\phi\circ\phi^{-1}=\pi$,
so we conclude that
$\hat{\pi}(\phi)\circ\hat{\pi}(\phi^{-1})=\textrm{id}_{Y}$.

To prove Item 3., fix any $\psi\in\operatorname{End}(Y,S)$ and suppose that
$\min\limits_{y\in Y}|\pi^{-1}(\\{y\\})|=c<\infty$ (if not, then there is
nothing to prove). Let $x_{0}\in X$ and $y_{0}\in Y$ be such that
$|\pi^{-1}(\\{y_{0}\\})|=c$, and $y_{0}=\psi(\pi(x_{0}))$. Assume there exists
$c+1$ endomorphisms $\phi_{0},\ldots,\phi_{c}$ of $(X,T,\mathbb{Z}^{d})$, in
$\hat{\pi}(\\{\psi\\})^{-1}$. The compatibility then implies that
$y_{0}=\psi(\pi(x_{0}))=\pi(\phi_{0}(x_{0}))=\cdots=\pi(\phi_{c}(x_{0}))$. So,
by the pigeonhole principle, there must exist two different indices $0\leq
i,j\leq c$ such that $\phi_{i}(x_{0})=\phi_{j}(x_{0})$. The minimality of
$(X,T,\mathbb{Z}^{d})$ then gives that $\phi_{i}=\phi_{j}$.

The proofs concerning the items i and ii on
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms use similar arguments and are left
to the reader.

To prove Item iii, we consider any $y\in Y$ with $n$ preimages
$x_{1},\ldots,x_{n}\in X$. Since $\phi$ is onto, there are
$x^{\prime}_{1},\ldots,x^{\prime}_{n}\in X$ such that
$\phi(x^{\prime}_{i})=x_{i}$. Notice that,
$\hat{\pi}(\phi)(\pi(x^{\prime}_{i}))=\pi(\phi(x^{\prime}_{i}))=\pi(x_{i})=y$.
It follows that
$\hat{\pi}(\phi)(\pi(x^{\prime}_{i}))=\hat{\pi}(\phi)(\pi(x^{\prime}_{j}))$
for any indices $i,j=1,\ldots,n$. Since $\hat{\pi}(\phi)$ is invertible, all
the element $x^{\prime}_{i}$ belongs to the same $\pi$-fiber of $z\in Y$. Thus
$z$ admits at least $n$ preimages by $\pi$ and satisfies
$\hat{\pi}(\phi)(z)=y$. The claim follows. ∎

It is known that factor maps between equicontinuous systems are compatible
[1], but as we will see in the next section, they are not necessarily
compatible with $\textrm{GL}(d,\mathbb{Z})$-endomorphisms (see Remark 3.5).
Nevertheless, the maximal equicontinuous factor is an example of a factor
compatible with $\textrm{GL}(d,\mathbb{Z})$-endomorphisms as proved in [5,
Theorem 5 and Corollary 3]. This can also be seen by using the universal
property of the maximal equicontinuous factor.

######  Lemma 2.4.

[5, Corollary 3] Let $(X,T,\mathbb{Z}^{d})$ be a minimal topological dynamical
system and let
${\pi_{eq}:(X,T,\mathbb{Z}^{d})\to(X_{eq},T_{eq},\mathbb{Z}^{d})}$ denote its
maximal equicontinuous factor. Then $\pi_{eq}$ is compatible with
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms.

Lemma 2.3 and Lemma 2.4 illustrate that to study
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms, a first step is to understand the
equicontinuous systems. This will be done in Section 2.2 for the class of
minimal equicontinous Cantor systems: the odometer systems.

###  2.2. $\mathbb{Z}^{d}$-Odometer systems

Odometer systems are the equicontinuous minimal Cantor systems. Hence, they
are the maximal equicontinuous factor for a large family of symbolic systems,
such as Toeplitz flows and some substitutive subshifts. We refer to [9] for
the study of odometer systems and $\mathbb{Z}^{d}$-Toeplitz sequences. We use
the same notations.

In this section, we briefly recall the basic definitions of odometers.
Subsequently, we describe the linear representation semigroup of odometer
systems (Lemma 2.6), which we then use to completely characterize it for
constant-base $\mathbb{Z}^{2}$\- odometer systems (Theorem 3.3). It is worth
noting that the initial exploration of
$\textrm{GL}(2,\mathbb{Z})$-endomorphisms between $\mathbb{Z}^{2}$-odometer
systems was initiated in [18] and later extended to higher dimensions in [22].

####  2.2.1. Basic definitions

Let $Z_{0}\geqslant Z_{1}\geqslant\ldots\geqslant Z_{n}\geqslant
Z_{n+1}\geqslant\ldots$ be a nested sequence of finite-index subgroups of
$\mathbb{Z}^{d}$ such that $\bigcap\limits_{n\geq 0}Z_{n}=\\{{\bm{0}}\\}$ and
let $\alpha_{n}:\mathbb{Z}^{d}/Z_{n+1}\to\mathbb{Z}^{d}/Z_{n}$ be the function
induced by the inclusion map. We consider the inverse limit of these groups

$\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}=\lim\limits_{\leftarrow
n}(\mathbb{Z}^{d}/Z_{n},\alpha_{n}),$

i.e., $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$ is the subset of the product
$\prod\limits_{n\geq 0}\mathbb{Z}^{d}/Z_{n}$ consisting of the elements
${\overleftarrow{g}=({g}_{n})_{n\geq 0}}$ such that
$\alpha_{n}({g}_{n+1})={g}_{n}\ (\text{mod}\ Z_{n})$ for all $n\geq 0$. The
odometer $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$ is a compact 0-dimensional
topological group, whose topology is spanned by the cylinder sets

$[{\bm{a}}]_{n}=\left\\{\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}:{{\bm{g}}}_{n}={{\bm{a}}}\right\\},$

with ${{\bm{a}}}\in\mathbb{Z}^{d}/Z_{n}$, and $n\geq 0$. Now, consider the
group homomorphism $\kappa_{(Z_{n})}:\mathbb{Z}^{d}\to\prod\limits_{n\geq
0}\mathbb{Z}^{d}/Z_{n}$ defined for ${\bm{n}}\in\mathbb{Z}^{d}$ as

$\kappa_{(Z_{n})}({\bm{n}})=[{\bm{n}}\ \text{mod}\ Z_{n}]_{n\geq 0}.$

The image of $\mathbb{Z}^{d}$ under $\kappa_{(Z_{n})}$ is dense in
$\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$, allowing us to define the
$\mathbb{Z}^{d}$-action
${\bm{n}}{\bm{+}}\overleftarrow{g}=\kappa_{(Z_{n})}({\bm{n}})+\overleftarrow{g}$,
where ${\bm{n}}\in\mathbb{Z}^{d}$ and
$\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$. This action is
well-defined and continuous. The resulting topological dynamical system
$(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})},{\bm{+}},\mathbb{Z}^{d})$ is called
a $\mathbb{Z}^{d}$-_odometer system_ , and for the rest of this article, we
will simply use the notation $\overleftarrow{\mathbb{Z}^{d}}_{(Z{n})}$, i.e.,
denoted just by its phase space. Similarly, its set of automorphisms,
endomorphisms, and linear representation semigroup will be denoted as
$\operatorname{Aut}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$,
$N(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$, and
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$, respectively.

Notice that the “return times” of the action to a cylinder set
$[{\bm{a}}]_{n}$ is a finite-index subgroup of $\mathbb{Z}^{d}$, or more
precisely:

(2)
$\displaystyle\forall\overleftarrow{g}\in[{\bm{a}}]_{n},\quad\\{{\bm{n}}\in\mathbb{Z}^{d}:{\bm{n}}{\bm{+}}\overleftarrow{g}\in[{\bm{a}}]_{n}\\}=Z_{n}.$

This observation is the base to show that an odometer system is a minimal
equicontinuous system. This also shows that the action is aperiodic since the
intersection of all the $Z_{n}$ is trivial.

We will be particularly concerned by a special case of odometers: namely the
_constant-base_ ones. In these systems, $Z_{n}=L^{n}(\mathbb{Z}^{d})$ for each
$n\geq 0$, where $L\in\mathcal{M}(d,\mathbb{Z})$ is an expansion matrix.
Recall that an integer matrix $L\in\mathcal{M}(d,\mathbb{Z})$ is an
_expansion_ if the norm of each eigenvalue is greater than 1. To simplify
notation, we denote the constant-base odometer
$\overleftarrow{\mathbb{Z}^{d}}_{(L^{n}(\mathbb{Z}^{d}))}$ as
$\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$.

The next result characterizes the factor odometer systems of a fixed odometer
system.

######  Lemma 2.5.

[9, Lemma 1] Let $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{j})}$ be two
odometer systems $(j=1,2)$. There exists a factor map
$\pi:\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})}\to\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{2})}$
if and only if for every $Z_{n}^{2}$ there exists some $Z_{m}^{1}$ such that
$Z_{m}^{1}\leqslant Z_{n}^{2}$.

####  2.2.2. Normalizer condition

The proof of Lemma 2.5 can be modified to provide a characterization for the
matrices $M\in GL(d,\mathbb{Z})$ defining a
$\textrm{GL}(d,\mathbb{Z})$-epimorphism between two odometer systems
$\phi:\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})}\to\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{2})}$.

######  Lemma 2.6.

Set $M\in GL(d,\mathbb{Z})$. There exists a continuous map
$\phi:\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})}\to\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{2})}$,
such that

$\forall{\bm{n}}\in\mathbb{Z}^{d},\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})},\quad\phi({\bm{n}}{\bm{+}}(\overleftarrow{g}))=M{\bm{n}}{\bm{+}}\phi(\overleftarrow{g}),$

if and only if

(Epimorphism Condition) $\forall n\in\mathbb{N},\exists
m(n)\in\mathbb{N}\textrm{ s.t. }MZ_{m(n)}^{1}\leqslant Z_{n}^{2}.$

It follows from Lemma 2.6 that a matrix $M\in GL(d,\mathbb{Z})$ belongs to the
linear representation semigroup
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ of an odometer system if
and only if it satisfies the following condition, that we called _Normalizer
condition_

(NC 1) $\forall n\in\mathbb{N},\exists m(n)\in\mathbb{N}\textrm{ s.t.
}MZ_{m(n)}\leqslant Z_{n}.$

###### Proof.

For the sufficiency, assume that $M\in GL(d,\mathbb{Z})$ satisfies
(Epimorphism Condition). Since the sequences $\\{Z_{n}^{i}\\}_{n>0}$, $i=1,2$
are decreasing, we may assume that $m(n)\leq m(n+1)$ for all $n\in\mathbb{N}$.
Thus, we have homomorphisms
$\phi_{m(n)}^{M}:\mathbb{Z}^{d}/Z_{m(n)}^{1}\to\mathbb{Z}^{d}/Z_{n}^{2}$,
given by $\phi_{m(n)}({\bm{m}}\ (\text{mod}\ Z_{m(n)}^{1}))=M{\bm{m}}\
(\text{mod}\ Z^{2}_{n})$. To finish the proof, we only have to remark that
${\phi^{M}:\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})}\to\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{2})}}$
defined as
${\phi(({\bm{g}}_{n})_{n\in\mathbb{N}})}=(\phi_{m(n)}({\bm{g}}_{m(n)}))_{n\in\mathbb{N}}$
is an $M$-epimorphism.

We now prove the necessity. Let
$\phi:\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{1})}\to\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n}^{2})}$
be an $M$-epimorphism. By continuity, for any $n\in\mathbb{N}$ and
${\bm{g}}\in\mathbb{Z}^{d}/Z_{n}^{2}$, there exists $m\in\mathbb{N}$ and
${\bm{f}}\in\mathbb{Z}^{d}/Z_{m}^{1}$ such that
${[{\bm{f}}]_{m}\subseteq\phi^{-1}([{\bm{g}}]_{n})}$. Set ${\bm{h}}\in
Z_{m}^{1}$. For all $\overleftarrow{f}\in[{\bm{f}}]_{m}$, we have by (2) that
${\bm{h}}{\bm{+}}\overleftarrow{f}\in[{\bm{f}}]_{m}$, which implies that
$\phi({\bm{h}}{\bm{+}}\overleftarrow{f})=M{\bm{h}}{\bm{+}}\phi(\overleftarrow{f})\in[{\bm{g}}]_{n}$.
Since $\phi(\overleftarrow{f})$ is in $[{\bm{g}}]_{n}$, the set
$\left\\{{\bm{m}}\in\mathbb{Z}^{d}\colon{\bm{m}}{\bm{+}}\phi(\overleftarrow{f})\in[{\bm{g}}]_{n}\right\\}$
is equal to $Z_{n}^{2}$ (by (2)), which implies that $M{\bm{h}}\in Z_{n}^{2}$.
∎

Notice that since the sequences $\\{Z_{n}^{i}\\}$, $i=1,2$ are decreasing, the
(Epimorphism Condition) is satisfied for any large enough $m\in\mathbb{N}$
provided it is true for one constant. This remark implies that the set of
matrices $M$ (not necessarily invertible) satisfying the condition (NC 1) is
stable under product and sum, so it is a ring. By applying this remark we get
the following result

######  Corollary 2.7.

The semigroup ${N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ of
$\textrm{GL}(d,\mathbb{Z})$-endomorphims of an odometer
$\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$ is a group.

In particular any $\textrm{GL}(d,\mathbb{Z})$-endomorphim is invertible and
the linear representation semigroup
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ is a group.

###### Proof.

Recall that odometer systems are equicontinuous and, hence, are coalescent
[1]. Thus, from Corollary 3.2, to show that any
$\textrm{GL}(d,\mathbb{Z})$-endomorphism of an odometer system is invertible,
it is enough to show that $\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$
is a group.

Since $\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ is a semigroup, we
only have to prove that any element
$M\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$ admits an inverse
inside. Since the sets of matrices satisfying (Epimorphism Condition) is a
ring, any integer polynomial of $M$ also satisfies (Epimorphism Condition).
Now, the Cayley-Hamilton theorem implies that
$M^{d}=\sum\limits_{k=0}^{d-1}b_{k}M^{k}$, where $b_{k}\in\mathbb{Z}$ are the
coefficients of the characteristic polynomial of $M$. Notice that
$b_{0}=(-1)^{d}\det(M)$, so that $b_{0}\in\\{-1,1\\}$. Multiplying by
$M^{-1}$, we conclude that $M^{-1}$ can be written as an integer polynomial on
$M$. Hence, $M^{-1}$ satisfies (Epimorphism Condition), and by Lemma 2.6, we
conclude that $M^{-1}\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$. ∎

Recall that the automorphisms of the odometer system are well known: they are
the translations on it [1].

######  Lemma 2.8.

For any odometer system we have that

$\operatorname{Aut}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})=\\{\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}\mapsto\overleftarrow{g}+\overleftarrow{h}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}:{\overleftarrow{h}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}}\\}.$

In particular $\operatorname{Aut}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$
is an abelian group isomorphic to $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$.

As a direct consequence of Corollary 2.7 we get the following algebraic
structure of the normalizer group of odometer systems.

######  Corollary 2.9.

The normalizer group $N(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ of an
odometer system is isomorphic to a semidirect product between the odometer
system $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$ and the linear
representation group $\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$.

###### Proof.

Recall from Lemma 2.6 that for each
$M\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$, one can associate a
$M$-isomorphism $\phi^{M}$ of ${\mathbb{Z}^{d}}_{(Z_{n})}$ defined for any
$({\bm{g}}_{m(n)}\ (\text{mod}\
Z_{m(n)}))_{n\in\mathbb{N}}\in\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$ by

${\phi^{M}(({\bm{g}}_{m(n)}\ (\text{mod}\
Z_{m(n)}))_{n\in\mathbb{N}})}=(M{\bm{g}}_{n}\ (\text{mod}\
Z_{n}))_{n\in\mathbb{N}}.$

Notice that the set $\\{\phi^{M}\colon
M\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})\\}$ is a group and
defines a group homomorphism
$h:\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})\to
N(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ such that $j\circ h$ is the
identity in $\\{\phi^{M}\colon
M\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})\\}$, so the exact
sequence (1) is split exact. ∎

So, to study the normalizer group of an odometer system, we just have to
determine its linear representation group. Actually, all these results lead to
wonder the following question:

######  Question 2.10.

Give a characterization of the groups of the form
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ for any odometer
$\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$.

We do not answer to this question, but we provide necessary conditions for
specific odometers: the universal and the constant-base ones in Sections 3.1
and 3.2 respectively.

###  2.3. Symbolic dynamics

We recall here classical definitions and we fix the notations on
multidimensional subshifts.

####  2.3.1. Basic definitions

Let $\mathcal{A}$ be a finite alphabet and $d\geq 1$ be an integer. We define
a topology on $\mathcal{A}^{\mathbb{Z}^{d}}$ by endowing $\mathcal{A}$ with
the discrete topology and considering in $\mathcal{A}^{\mathbb{Z}^{d}}$ the
product topology, which is generated by cylinders. Since $\mathcal{A}$ is
finite, $\mathcal{A}^{\mathbb{Z}^{d}}$ is a metrizable compact space. In this
space, the group $\mathbb{Z}^{d}$ acts by translations (or shifts), defined
for every ${\bm{n}}\in\mathbb{Z}^{d}$ as

$S^{{\bm{n}}}(x)_{{\bm{k}}}=x_{{\bm{n}}+{\bm{k}}},\
x=(x_{\bm{k}})_{\bm{k}}\in\mathcal{A}^{\mathbb{Z}^{d}},\
{\bm{n},\bm{k}}\in\mathbb{Z}^{d}.$

Let $P\subseteq\mathbb{Z}^{d}$ be a finite subset. A _pattern_ is an element
$\texttt{p}\in\mathcal{A}^{P}$. We say that $P$ is the _support_ of p, and we
denote $P=\operatorname{supp}(\texttt{p})$. A pattern _occurs in_
$x\in\mathcal{A}^{\mathbb{Z}^{d}}$, if there exists
${\bm{n}}\in\mathbb{Z}^{d}$ such that $\texttt{p}=x|_{{\bm{n}}+P}$
(identifying ${\bm{n}}+P$ with $P$ by translation). In this case, we denote it
$\texttt{p}\sqsubseteq x$ and we call this ${\bm{n}}$ an _occurrence in_ $x$
of p.

A _subshift_ $(X,S,\mathbb{Z}^{d})$ is given by a closed subset
$X\subseteq\mathcal{A}^{\mathbb{Z}^{d}}$ which is invariant by the
$\mathbb{Z}^{d}$-action. A subshift defines a _language_. For a finite subset
$P\Subset\mathbb{Z}^{d}$ we define

$\mathcal{L}_{P}(X)=\\{\texttt{p}\in\mathcal{A}^{P}:\exists x\in X,\
\texttt{p}\sqsubseteq x\\}.$

The _language_ of a subshift $X$ is defined as

$\mathcal{L}(X)=\bigcup\limits_{P\Subset\mathbb{Z}^{d}}\mathcal{L}_{P}(X).$

Let $(X,S,\mathbb{Z}^{d})$ be a subshift and $x\in X$. We say that
${\bm{p}}\in\mathbb{Z}^{d}$ is a _period_ of $x$ if for all
${\bm{n}}\in\mathbb{Z}^{d}$, $x_{{\bm{n}}+{\bm{p}}}=x_{{\bm{n}}}$. The
subshift $(X,S,\mathbb{Z}^{d})$ is said to be _aperiodic_ if there are no
nontrivial periods.

Let $\mathcal{B}$ be another finite alphabet and
$Y\subseteq\mathcal{B}^{\mathbb{Z}^{d}}$ be a subshift. For
$P\Subset\mathbb{Z}^{d}$, we define a $P$-_block map_ as a map of the form
$\Phi:\mathcal{L}_{P}(X)\to\mathcal{B}$. This induces a factor map $\phi:X\to
Y$ given by

$\phi(x)_{{\bm{n}}}=\Phi(x|_{{\bm{n}}+P}).$

The map $\phi$ is called the _sliding block code_ induced by $\Phi$ and $P$ is
the support of the map $\phi$. In most of the cases we may assume that the
support of the sliding block codes is a ball of the form $B({\bm{0}},r)$,
where $r$ is a positive integer. We define the _radius_ , denoted as
$r(\phi)$, as the smallest positive integer $r$ for which a
$B({\bm{0}},r)$-block map can be defined to induce $\phi$.. The next theorem
characterizes the factor maps between two subshifts.

######  Theorem 2.11 (Curtis-Hedlund-Lyndon).

Let $(X,S,\mathbb{Z}^{d})$ and $(Y,S,\mathbb{Z}^{d})$ be two subshifts. A map
$\phi:(X,S,\mathbb{Z}^{d})\to(Y,S,\mathbb{Z}^{d})$ is a factor map if and only
if there exists a $B({\bm{0}},r)$-block map
$\Phi:\mathcal{L}_{B({\bm{0}},r)}(X)\to\mathcal{L}_{1}(Y)$, such that
$\phi(x)_{{\bm{n}}}=\Phi(x|_{{\bm{n}}+B({\bm{0}},r)})$, for all
${\bm{n}}\in\mathbb{Z}^{d}$ and $x\in X$.

For $\textrm{GL}(d,\mathbb{Z})$-epimorphisms there is a similar
characterization. See [8, Theorem 2.7] for a proof.

######  Theorem 2.12 (Curtis-Hedlund-Lyndon theorem for
$\textrm{GL}(d,\mathbb{Z})$-epimorphisms).

Let $(X,S,\mathbb{Z}^{d})$ and $(Y,S,\mathbb{Z}^{d})$ be two subshifts and
${M\in\textrm{GL}(d,\mathbb{Z})}$. A map
$\phi:(X,S,\mathbb{Z}^{d})\to(X,S,\mathbb{Z}^{d})$ is an $M$-endomorphism if
and only if there exists a $B({\bm{0}},r)$-block map
$\Phi:\mathcal{L}_{B({\bm{0}},r)}(X)\to\mathcal{L}_{1}(Y)$, such that
$\phi(x)_{{\bm{n}}}=\Phi(x|_{M^{-1}{\bm{n}}+B({\bm{0}},r)})$, for all
${\bm{n}}\in\mathbb{Z}^{d}$ and $x\in X$.

This means, for any $\textrm{GL}(d,\mathbb{Z})$-epimorphism $\phi$ we can
define a _radius_ (also denoted by $r(\phi)$), as the infimum of
$r\in\mathbb{N}$ such that we can define a $B({\bm{0}},r)$-block map which
induces it. In the case $r(\phi)=0$, we say that $\phi$ is induced by a
_letter-to-letter map_.

####  2.3.2. Substitutive subshifts

We provide a brief overview of multidimensional substitutive subshifts of
constant-shape that will be used throughout this article. We refer to [8] for
basic properties on this topic, where we follow the same notations. Let
$L\in\mathcal{M}_{d}(\mathbb{Z})$ be an integer expansion matrix,
$F\subseteq\mathbb{Z}^{d}$ be a fundamental domain of $L(\mathbb{Z}^{d})$ in
$\mathbb{Z}^{d}$, i.e., a set of representative classes of
$\mathbb{Z}^{d}/L(\mathbb{Z}^{d})$ (with ${\bm{0}}\in F$) and $\mathcal{A}$ be
a finite alphabet. A _constant-shape substitution_ is a map
$\zeta:\mathcal{A}\to\mathcal{A}^{F}$. We say that $F$ is the _support_ of the
substitution. Since any element $\bm{n}\in\mathbb{Z}^{d}$ can be expressed
uniquely as $\bm{n}=L(\bm{j})+\bm{f}$, with $\bm{j}\in\mathbb{Z}^{d}$ and
$\bm{f}\in F$, the substitution extends to $\mathcal{A}^{\mathbb{Z}^{d}}$ as

$\zeta(x)_{L(\bm{j})+\bm{f}}=\zeta(x_{\bm{j}})_{\bm{f}}.$

For any $n>0$, we define the $n$-th iteration of the substitution
$\zeta^{n}:\mathcal{A}\to\mathcal{A}^{F_{n}}$ by induction
$\zeta^{n+1}=\zeta\circ\zeta^{n}$, where the supports of these substitutions
satisfy the recurrence $F_{n+1}=L(F_{n})+F_{1}$ for all $n\geq 1$. We will
always assume that the sequence of supports $(F_{n})_{n>0}$ is _Følner_ ,
i.e., for all ${\bm{n}}\in\mathbb{Z}^{d}$ we have that

$\lim\limits_{n\to\infty}\dfrac{|F_{n}\triangle(F_{n}+{\bm{n}})|}{|F_{n}|}=0.$

The supports do not need to cover all the space. Nevertheless, up to adding a
finite set and taking its images under power of the expansion matrix $L$, they
cover the space. This property is explained in the following proposition. It
is similar to the notion of rest in numeration theory and will be technically
useful.

######  Proposition 2.13.

[8, Proposition 2.10] Let $\zeta$ be a constant-shape substitution. Then, the
set
$K_{\zeta}=\bigcup\limits_{m>0}((\textrm{id}-L^{m})^{-1}(F_{m})\cap\mathbb{Z}^{d})$
is finite and satisfies

$\bigcup\limits_{n\geq 0}L^{n}(K_{\zeta})+F_{n}=\mathbb{Z}^{d},$

using the notation $F_{0}=\\{0\\}$.

The _language_ of a substitution is the set of all patterns that occur in
$\zeta^{n}(a)$, for some $n>0$, $a\in\mathcal{A}$, i.e.,

$\mathcal{L}_{\zeta}=\\{\texttt{p}\colon\texttt{p}\sqsubseteq\zeta^{n}(a),\
\text{for some }n>0,\ a\in\mathcal{A}\\}.$

A substitution $\zeta$ is called _primitive_ if there exists a positive
integer $n>0$, such that for every $a,b\in\mathcal{A}$, $b$ occurs in
$\zeta^{n}(a)$. If $\zeta$ is a primitive constant-shape substitution, the
existence of _periodic points_ is well-known, i.e., there exists at least one
point $x_{0}\in X_{\zeta}$ such that $\zeta^{p}(x_{0})=x_{0}$ for some $p>0$.
In the primitive case, the subshift is preserved by replacing the substitution
with a power of it; that is, $X_{\zeta^{n}}$ is equal to $X_{\zeta}$ for all
$n>0$. Consequently, we may assume the existence of at least one fixed point.
In other words, there exists a point $x\in X_{\zeta}$ such that $x=\zeta(x)$.
As in the one-dimensional case, it is important to note that the number of
periodic points (or, if we consider proper iterations, the number of fixed
points) is finite.

The substitutive subshift $(X_{\zeta},S,\mathbb{Z}^{d})$ is the topological
dynamical system, where $X_{\zeta}$ is the set of all sequences
$x\in\mathcal{A}^{\mathbb{Z}^{d}}$ such that every pattern occurring in $x$ is
in $\mathcal{L}_{\zeta}$. When the substitutive subshift
$(X_{\zeta},S,\mathbb{Z}^{d})$ is aperiodic, the substitution satisfy a
combinatorial property called _recognizability_ [8, 28].

######  Definition 2.14.

Let $\zeta$ be a primitive substitution and $x\in X_{\zeta}$ be a fixed point.
We say that $\zeta$ is _recognizable on $x$_ if there exists some constant
$R>0$ such that for all ${\bm{i}},{\bm{j}}\in\mathbb{Z}^{d}$,

$x|_{B(L_{\zeta}({\bm{i}}),R)\cap\mathbb{Z}^{d}}=x|_{B({\bm{j}},R)\cap\mathbb{Z}^{d}}\implies(\exists{\bm{k}}\in\mathbb{Z}^{d})(({\bm{j}}=L_{\zeta}({\bm{k}}))\wedge(x_{{\bm{i}}}=x_{{\bm{k}}})).$

The recognizability property implies some topological and combinatorial
properties of the substitutive subshift that we summarize in the following:

  * •

The substitutive subshift $(X_{\zeta},S,\mathbb{Z}^{d})$ is aperiodic.

  * •

For any $n>0$, the map $\zeta^{n}:X_{\zeta}\to\zeta^{n}(X_{\zeta})$ is a
homeomorphism.

  * •

For any $n>0$, every $x\in X_{\zeta}$ can be written in a unique way
$x=S^{{\bm{f}}}\zeta^{n}(x_{1})$ with ${\bm{f}}\in F_{n}$ and $x_{1}\in
X_{\zeta}$.

It follows that the map $\pi_{n}\colon X_{\zeta}\to F_{n}$ by
$\pi_{n}(x)={\bm{f}}$ when $x\in S^{\bm{f}}\zeta^{n}(X_{\zeta})$ is well-
defined, continuous and can be extended to a factor map
$\pi\colon(X_{\zeta},S,\mathbb{Z}^{d})\to(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})},{\bm{+}},\mathbb{Z}^{d})$
defined as $\pi(x)=(\pi_{n}(x))_{n}$ [8].

##  3\. Description of the linear representation group of odometer systems

In this section, we describe the linear representation group and its elements
for several odometers, specifically the universal and constant-base
$\mathbb{Z}^{2}$-odometer systems. It is worth noting that we did not find a
similar result in the existing literature. One of our main tools involves the
characterization (NC 1) for matrices in the linear representation group of an
odometer system $\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})}$.

We begin by presenting an equivalent formulation of (NC 1) in terms of
arithmetical equations. For any given $n\in\mathbb{N}$, let
$L_{n}\in\mathcal{M}(d,\mathbb{Z})$ be a matrix such that
$L_{n}(\mathbb{Z}^{d})=Z_{n}$. It is important to note that this matrix is
unique, up to a composition with a matrix in $\textrm{GL}(d,\mathbb{Z})$.
Then, the condition (NC 1) is equivalent to: for all $n\in\mathbb{N}$, there
exists $m_{M}(n)\in\mathbb{N}$ such that $L_{n}^{-1}ML_{m_{M}(n)}$ is an
endomorphism in $\mathbb{Z}^{d}$. Since $\det(L)L^{-1}=\operatorname{adj}(L)$,
where $\operatorname{adj}(L)$ is the _adjugate matrix_ of $L$, we can express
(NC 1) equivalently as:

(NC 2) $\forall n\in\mathbb{N},\exists m_{M}(n)\in\mathbb{N},\
\operatorname{adj}(L_{n})ML_{m_{M}(n)}\equiv 0\ (\text{mod}\ \det(L_{n})).$

###  3.1. The universal $\mathbb{Z}^{d}$-odometer case

Let $(\Gamma_{n})_{n\in\mathbb{N}}$ be an enumeration of all finite-index
subgroups of $\mathbb{Z}^{d}$. We define the _universal_ $d$-_dimensional
odometer system_ as follows: Start with $\Lambda_{0}=\Gamma_{0}$, and for any
$n\geq 1$ set $\Lambda_{n}=\Lambda_{n-1}\cap\Gamma_{n}$. Since the
intersection of finite-index subgroups remains a finite-index subgroup, we can
define the universal $d$-dimensional odometer as
$\overleftarrow{\mathbb{Z}^{d}}_{(\Lambda_{n})}$. This odometer is universal
in the sense that, by Lemma 2.5, any odometer system is a topological factor
of the universal odometer. For example, the universal 1-dimensional odometer
is equal to $\overleftarrow{\mathbb{Z}}_{(n!\mathbb{Z})}$. With respect to its
linear representation group, (NC 2) leads to the following result.

######  Proposition 3.1.

The linear representation group of the $d$-dimensional universal odometer is
equal to $\textrm{GL}(d,\mathbb{Z})$.

###### Proof.

Consider $L_{n}\in\mathcal{M}(d,\mathbb{Z})$ such that
$L_{n}(\mathbb{Z}^{d})=\Lambda_{n}$. A matrix $M\in\textrm{GL}(d,\mathbb{Z})$
is in $\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(\Lambda_{n})})$ if and only if
$M$ satisfies (NC 2). Now, for any $n\in\mathbb{N}$, we can choose
$m(n)\in\mathbb{N}$ large enough such that
$\Lambda_{m(n)}\leqslant\det(L_{n})\mathbb{Z}^{d}$. This implies that
$\operatorname{adj}(L_{n})ML_{m(n)}\equiv 0\ (\text{mod}\ \det(L_{n}))$ for
any matrix $M\in\textrm{GL}(d,\mathbb{Z})$. We then conclude that
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(\Lambda_{n})})=\textrm{GL}(d,\mathbb{Z})$.
∎

###  3.2. The constant-base $\mathbb{Z}^{2}$-odometer case

We will be mainly interested in $\textrm{GL}(d,\mathbb{Z})$-endomorphisms of
constant-base odometers, i.e., when $Z_{n}=L^{n}(\mathbb{Z}^{d})$ for each
$n\in\mathbb{N}$ and for some expansion matrix $L$. In this case we get the
following direct corollary of Lemma 2.6 and the condition (NC 2).

######  Corollary 3.2.

Let $L\in\mathcal{M}(d,\mathbb{Z})$ be an expansion matrix.

  1. (1)

A matrix $M\in\textrm{GL}(d,\mathbb{Z})$ is in
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$, if and only if

(NC 3) $\forall n\in\mathbb{N},\exists m(n)\in\mathbb{N},\
\operatorname{adj}(L^{n})ML^{m(n)}\equiv 0\ ({mod}\ \det(L^{n})).$

  2. (2)

If $M\in\textrm{GL}(d,\mathbb{Z})$ commutes with some power of the expansion
matrix $L$, then $M$ is in the linear representation semigroup
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$.

  3. (3)

For any $M\in\textrm{GL}(d,\mathbb{Z})$ we have that
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(ML^{n}M^{-1})})=M\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})M^{-1}$.

In the next theorem, we present the structure of the linear representation
group of constant-base $\mathbb{Z}^{2}$-odometer systems based on computable
arithmetical conditions of the expansive matrix $L$. Within this family, we
obtain a bifurcation phenomenon at the level of the linear representation
group, depending on arithmetic relations of the coefficients of the matrix
$L$. To describe the different cases, we introduce some additional notations.
For any positive integer $n>1$, the _radical_ $\operatorname{rad}(n)$ _of_ $n$
is defined as the product of the distinct prime numbers that divide $n$. If
$n<-1$, we define $\operatorname{rad}(n)$ just as $\operatorname{rad}(-n)$.
The _centralizer_ $\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L)$ _of a
matrix_ $L$ _in_ $\textrm{GL}(2,\mathbb{Z})$ is defined as the subgroup
consisting of all matrices in $\textrm{GL}(2,\mathbb{Z})$ commuting with $L$.
Recall that, as established in Corollary 3.2, the centralizer
$\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L)$ is always a subgroup of
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$.

######  Theorem 3.3.

Let $L\in\mathcal{M}(2,\mathbb{Z})$ be an integer expansion matrix.

  1. (1)

If $\operatorname{rad}(\det(L))$ divides $\operatorname{trace}(L)$, then the
linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is equal to
$\textrm{GL}(2,\mathbb{Z})$.

  2. (2)

Otherwise

    1. (a)

If the spectrum of the matrix $L$ is disjoint from the integers, then the
linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is the centralizer
$\operatorname{Cent}_{GL(2,\mathbb{Z})}(L)$.

Moreover, if the spectrum of $L$ is disjoint from the real line, then
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is a finite group.

    2. (b)

When the spectrum of $L$ contains an integer value, the linear representation
group $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is finite or
virtually $\mathbb{Z}$.

More precisely, under explicit arithmetical properties of $L$,
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n}(\mathbb{Z}^{2}))})$ isomorphic
to $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}^{2}/(2\mathbb{Z}\times
2\mathbb{Z})$, or its abelianization is finite, and its commutator subgroup is
cyclic.

Along the proof, the group structure of
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n}(\mathbb{Z}^{2}))})$ is
specified in terms of the arithmetical properties of the coefficient of $L$.

The following examples illustrate the different cases of Theorem 3.3 according
to the expansion matrix $L$.

######  Example 3.4 (Different results for Theorem 3.3).

  1. (1)

As we will see in the proof of Theorem 3.3, the case (1) can be easily
generalized for higher dimensions in the following way: If
$\operatorname{rad}(\det(L))$ divides every coefficient of the characteristic
polynomial of $L$, then the linear representation semigroup
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$ is equal to
$\textrm{GL}(d,\mathbb{Z})$. In particular, if $L=pM$, with $p\in\mathbb{Z}$
and $M\in\textrm{GL}(d,\mathbb{Z})$, then the linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{L^{n}})$ is
$\textrm{GL}(d,\mathbb{Z})$.

  2. (2)

The matrix $L_{1}=\begin{pmatrix}2&-1\\\ 1&5\end{pmatrix}$ illustrates the
case (2)(2)(a): $\operatorname{trace}(L_{1})=7$, $\det(L_{1})=11$, and $L_{1}$
has real eigenvalues (which are equal to $7/2\pm\sqrt{5}/2$). The matrices in
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L_{1}^{n})})$ are the ones commuting
with $L_{1}$, which is an infinite group containing $\begin{pmatrix}2&1\\\
-1&-1\end{pmatrix}$.

  3. (3)

The matrix $L_{2}=\begin{pmatrix}2&-1\\\ 1&3\end{pmatrix}$ also illustrates
the case (2)(2)(a) but with a spectrum disjoint from the real line:
$\operatorname{trace}(L_{2})=5$ and $\det(L_{2})=7$, and $L_{2}$ has complex
eigenvalues $5/2\pm i\sqrt{3}/2$. The linear representation semigroup
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{L_{2}^{n}})$ is equal to
$\operatorname{Cent}_{GL(2,\mathbb{Z})}(L_{2})$, which corresponds to the set

$\left\\{\begin{array}[]{ccc}\begin{pmatrix}1&1\\\
-1&0\end{pmatrix},&\begin{pmatrix}-1&-1\\\
1&0\end{pmatrix},&\begin{pmatrix}0&-1\\\ 1&1\end{pmatrix}\\\
\begin{pmatrix}0&-1\\\ 1&-1\end{pmatrix},&\begin{pmatrix}1&0\\\
0&1\end{pmatrix},&\begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix}\\\
\end{array}\right\\}.$

  4. (4)

The matrix $L_{3}=\begin{pmatrix}6&1\\\ 0&2\end{pmatrix}$ illustrates the case
(2)(2)(b). This is an upper triangular matrix which is not diagonalizable by
$\textrm{GL}(2,\mathbb{Z})$. It will be proved that
$\vec{N}(\overleftarrow{\mathbb{Z}}_{(L_{3}^{n})})$ is conjugate to
$\left\\{\begin{pmatrix}m_{11}&m_{12}\\\
0&m_{22}\end{pmatrix}\colon|m_{11}m_{22}|=1,m_{12}\in\mathbb{Z}\right\\}$ via
the matrix $\begin{pmatrix}1&0\\\ 4&1\end{pmatrix}$, so it is virtually
$\mathbb{Z}$. It can be directly checked that the linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$ associated with a matrix
$L$ diagonalizable by $\textrm{GL}(2,\mathbb{Z})$ also has the same group
structure, being isomorphic to a set of invertible upper triangular matrices.

  5. (5)

The matrix $L_{4}=\begin{pmatrix}3&1\\\ 0&5\end{pmatrix}$ also concerns the
case (2)(2)(b). This matrix has eigenvalues $3$ and $5$. Along the proof of
Theorem 3.3 it will be shown that a matrix $M$ is in
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}_{(L^{n}}})$ if and only if $M$ commutes
with $L_{4}$, so $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L_{4}^{n})})$ is
isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

  6. (6)

The last example illustrating the case (2)(2)(b) is the matrix
$L_{5}=\begin{pmatrix}2&1\\\ 0&3\end{pmatrix}$ with eigenvalues $2$ and $3$.
It will be also shown that
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L_{5}^{n})})=\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L_{5})$,
which is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{2}$.

######  Remark 3.5.

Note that Theorem 3.3 implies that the factor map between equicontinuous
systems is not necessarily compatible with
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms. Consider $X$ as the universal
$\mathbb{Z}^{2}$-odometer, and set
$Y=\overleftarrow{\mathbb{Z}^{2}}_{(L_{1}^{n})}$. Hence
$(Y,{\bm{+}},\mathbb{Z}^{d})$ is an equicontinuous factor of
$(X,{\bm{+}},\mathbb{Z}^{d})$. Now, by Proposition 3.1, we can define an
isomorphism associated with the matrix $\begin{pmatrix}2&1\\\
1&1\end{pmatrix}$ in $X$, However, in contrast, Theorem 3.3 and Lemma 2.3
establish that such isomorphism is not possible in $Y$.

###  3.3. Proof of Theorem 3.3

In this subsection, we will prove Theorem 3.3. We decompose this proof
according to its spectral properties. We will get more precise results as the
ones stated in Theorem 3.3. We start with the case where the expansion matrix
has integer eigenvalues.

From now on, an integer expansion matrix $L$ will be denoted as
$L=\begin{pmatrix}p&q\\\ r&s\end{pmatrix}$, its powers as
$L^{n}=\begin{pmatrix}p(n)&q(n)\\\ r(n)&s(n)\end{pmatrix}$ and a matrix $M$ in
$\textrm{GL}(2,\mathbb{Z})$ as $M=\begin{pmatrix}m_{11}&m_{12}\\\
m_{21}&m_{22}\end{pmatrix}$.

####  3.3.1. The triangular case

We now consider the case where $L$ is a triangular matrix. We focus only on
the upper triangular case, i.e., $q\neq 0$ and $r=0$. The lower triangular
case can be deduced from this, thanks to Corollary 3.2 via conjugation with
the matrix $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$. For all $n\in\mathbb{Z}$
we have $L^{n}=\begin{pmatrix}p^{n}&q(n)\\\ 0&s^{n}\end{pmatrix},$ where
$q(n)=q(p^{n}-s^{n})/(p-s)=q\sum\limits_{i=0}^{n-1}p^{i}s^{n-1-i}$. Since
$\det(L)=ps$ and $\operatorname{trace}(L)=p+s$, so
$\operatorname{rad}(\det(L))$ divides $\operatorname{trace}(L)$ if and only if
$\operatorname{rad}(p)=\operatorname{rad}(s)$. In this case we get a more
precise result about the linear representation group as the one mentioned in
Theorem 3.3.

######  Proposition 3.6.

Let $L\in\mathcal{M}(2,\mathbb{Z})$ be an expansion upper triangular matrix
such that $\operatorname{rad}(\det(L))$ does not divide
$\operatorname{trace}(L)$. Then, we have one of the following:

  1. (1)

If $\operatorname{rad}(p)$ does not divide $s$ and $\operatorname{rad}(s)$
divides $p$, then a matrix $M\in GL(2,\mathbb{Z})$ is in
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{L^{n}})$ if and only if
$(p-s)^{2}m_{12}=m_{21}q^{2}+(p-s)(m_{11}-m_{22})q$. Moreover,
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is virtually $\mathbb{Z}$.

  2. (2)

Assume that $\operatorname{rad}(p)$ divides $s$ and $\operatorname{rad}(s)$
does not divide $p$. Then $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$
is virtually $\mathbb{Z}$.

  3. (3)

If $\operatorname{rad}(p)$ does not divide $s$ and $\operatorname{rad}(s)$
does not divide $p$, we have two cases:

     * •

If $2q\in(p-s)\mathbb{Z}$, then
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is isomorphic to
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

     * •

Otherwise, $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is isomorphic
to $\mathbb{Z}/2\mathbb{Z}$.

###### Proof.

Let $M$ be in $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$. Define the
matrix $\overline{M}=(p-s)M-(m_{11}-m_{22})L-(p\cdot m_{22}-m_{11}\cdot
s)\textrm{id}_{\mathbb{R}^{2}}$. Then, $\overline{M}$ satisfies (NC 3).
Moreover, note that $\overline{M}$ has the form
$\overline{M}=\begin{pmatrix}0&\overline{m_{12}}\\\
\overline{m_{21}}&0\end{pmatrix}$, where
$\overline{m_{12}}=(p-s)m_{12}-(m_{11}-m_{22})q$ and
$\overline{m_{21}}=(p-s)m_{21}$, with
$\overline{m_{12}},\overline{m_{21}}\in\mathbb{Z}$. Now, (NC 3) implies that
for all $n,m>0$,

(3)
$\displaystyle\begin{pmatrix}-\overline{m_{21}}p^{m}q(n)&\overline{m_{12}}s^{n+m}-\overline{m_{21}}q(m)q(n)\\\
\overline{m_{21}}p^{n+m}&\overline{m_{21}}pq(m)\end{pmatrix}\equiv\begin{pmatrix}0&0\\\
0&0\end{pmatrix}\ (\text{mod}\ p^{n}s^{n}).$

Suppose that $\operatorname{rad}(s)$ does not divide $p$. Then, there exists a
prime number $t$ dividing $s$ such that for all $n>0$ and $m>0$, $p^{m}$ is an
invertible element in $\mathbb{Z}/t^{n}\mathbb{Z}$. Hence,
$\overline{m}_{21}\equiv 0\ (\text{mod}\ t^{n})$ for any $n>0$, which implies
that $\overline{m}_{21}=0$, so $m_{21}=0$. Now, by (3), we get that

(4) $\displaystyle\forall n\geq 0,\exists m\geq 0,\quad\overline{m}_{12}s^{m}$
$\displaystyle\equiv 0\ (\text{mod}\ p^{n}).$

There are two cases:

  * •

If $\operatorname{rad}(p)$ does not divide $s$, then (4) implies that
$\overline{m}_{12}=0$. We conclude that $\overline{M}=\begin{pmatrix}0&0\\\
0&0\end{pmatrix}$, i.e., $(p-s)M=(m_{11}-m_{22})L+(p\cdot m_{22}-m_{11}\cdot
s)\textrm{id}_{\mathbb{R}^{2}}$. Since $m_{21}=0$, then $M$ has the form

$M=\begin{pmatrix}m_{11}&m_{12}(m_{11},m_{22})\\\ 0&m_{22}\end{pmatrix},$

where $m_{12}(m_{11},m_{22})$ satisfies
$(p-s)m_{12}(m_{11},m_{22})=(m_{11}-m_{22})q$.

    * –

Note that $m_{11}=m_{22}$ if and only if $m_{12}=0$.

    * –

If $m_{11}\neq m_{22}$, then $m_{11}-m_{22}\in\\{-2,2\\}$, so $(p-s)m_{12}=\pm
2q$. Since $M$ has integer coefficients, this necessarily implies that
$2q\in(p-s)\mathbb{Z}$. If this condition is satisfied, then $M$ has the form

$M=\begin{pmatrix}m_{11}&\frac{(m_{11}-m_{22})q}{p-s}\\\
0&m_{22}\end{pmatrix}.$

It is not difficult to see that $M^{2}$ is the identity matrix. We conclude
that $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is isomorphic to
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. If
$2q\notin(p-s)\mathbb{Z}$, then
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is isomorphic to
$\mathbb{Z}/2\mathbb{Z}$.

  * •

If $\operatorname{rad}(p)$ divides $s$, then any
$\overline{m}_{12}\in\mathbb{Z}$ satisfies (4). Thus, any matrix
$M=\begin{pmatrix}m_{11}&m_{12}\\\ 0&m_{22}\end{pmatrix}$ with
$|m_{11}m_{22}|=1$ satisfies (NC 3).

Finally, if $\operatorname{rad}(s)$ divides $p$, then for any $n>0$ and any
$m$ large enough $s^{n}$ divides $p^{m}$ and $q(m)$. Let $t$ be a prime number
dividing $p$ that does not divide $s$. Then, by (3) we obtain that

(5)
$\displaystyle(p-s)^{2}\overline{m}_{12}s^{n+m}\equiv\overline{m}_{21}q^{2}s^{n+m}\
(\text{mod}\ t^{n}).$

Since $t$ does not divide $s$, for any $n,m>0$, $s^{n+m}$ is an invertible
element in $\mathbb{Z}/t^{n}\mathbb{Z}$. So (5) is reduced to

(6) $\displaystyle\forall n,\ (p-s)^{2}\overline{m}_{12}$
$\displaystyle\equiv\overline{m}_{22}q^{2}\ (\text{mod}\ t^{n}).$

This implies that $(p-s)^{2}\overline{m}_{12}=\overline{m}_{21}q^{2}$. Thus,
we get that

(7) $\displaystyle(p-s)^{2}m_{12}=m_{21}q^{2}+(p-s)(m_{11}-m_{22})q.$

This implies that if $M\in\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$,
then $M$ is in
$\operatorname{span}_{\mathbb{Q}}\left\\{L,\textrm{id},\begin{pmatrix}0&1\\\
(p-s)^{2}/q^{2}&0\end{pmatrix}\right\\}$. We separate here in two cases:

  * •

If $(p-s)$ divides $q$, we write $q=k(p-s)$ for some $k\in\mathbb{Z}$. By (7),
we have that

$m_{12}=m_{21}k^{2}+k(m_{11}-m_{22}).$

Since $|\det(M)|=1$ and $\det(M)=(m_{11}+m_{21}\cdot k)(m_{22}-m_{21}\cdot
k)$, we get that $|m_{11}+m_{21}\cdot k|=1$ and $|m_{22}-m_{21}\cdot k|=1$. We
can parameterize the matrices in
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ as follows:

$\left\\{\begin{array}[]{cl}\begin{pmatrix}1-m\cdot k&-mk^{2}\\\ m&1+m\cdot
k\end{pmatrix},\begin{pmatrix}1-m\cdot k&2k-mk^{2}\\\ m&m\cdot
k-1\end{pmatrix}\\\ \begin{pmatrix}-1-m\cdot k&-2k-mk^{2}\\\ m&1+m\cdot
k\end{pmatrix},\begin{pmatrix}-1-m\cdot k&-mk^{2}\\\ m&-1+m\cdot
k\end{pmatrix}&\colon m\in\mathbb{Z}\end{array}\right\\}.$

Note that this group is virtually $\mathbb{Z}$, since the quotient by
$\left\langle\begin{pmatrix}1-k&-k^{2}\\\ 1&1+k\end{pmatrix}\right\rangle$ is
finite.

  * •

If $(p-s)$ does not divide $q$, we will find a matrix
$P\in\textrm{GL}(2,\mathbb{Z})$ such that
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is conjugate to the group
of matrices $\left\\{\begin{pmatrix}m_{11}&m_{12}\\\
0&m_{22}\end{pmatrix}\colon
m_{11},m_{12},m_{22}\in\mathbb{Z},|m_{11}m_{22}|=1\right\\}$ and we conclude
that $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is virtually
$\mathbb{Z}$. Indeed, set $c=\gcd(p-s,q)$ and $g=(p-s)/c$, $h=q/c$. Since
$\gcd(g,h)=1$, Bézout’s lemma implies the existence of two numbers
$e,f\in\mathbb{Z}$ such that $eh-gf=1$. A standard computation shows that
$P=\begin{pmatrix}e&f\\\ g&h\end{pmatrix}$ is such a matrix.

∎

####  3.3.2. The general case

We are ready to prove Theorem 3.3.

###### Proof of Theorem 3.3.

We continue to use the notations introduced in Section 3.3. Since we already
proved the triangular case, we assume that the coefficients of the expansion
matrix $L$ satisfy $q\cdot r\neq 0$.

It will be useful to note that the Cayley-Hamilton theorem implies that

(8) $\displaystyle
L^{2}=\operatorname{trace}(L)L-\det(L)\textrm{id}_{\mathbb{R}^{2}}.$

First assume that $\operatorname{rad}(\det(L))$ divides
$\operatorname{trace}(L)$. By (8), we can conclude that ${L^{2}\equiv 0\
(\text{mod}\ \operatorname{rad}(\det(L)))}$. Hence, for all $n\in\mathbb{N}$
there exists $m(n)\in\mathbb{N}$ large enough such that $L^{m(n)}\equiv 0\
(\text{mod}\ \det(L)^{n})$. Therefore, any matrix in
$\textrm{GL}(2,\mathbb{Z})$ satisfies (NC 3) and we can deduce that
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})=GL(2,\mathbb{Z})$.

Now we deal with the case when $\operatorname{rad}(\det(L))$ does not divide
$\operatorname{trace}(L)$. In dimension $2$, this implies that $L$ is
diagonalizable. Let $M=\begin{pmatrix}m_{11}&m_{12}\\\
m_{21}&m_{22}\end{pmatrix}$ be in
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$, so that it satisfies (NC
3) . Define the matrix $\overline{M}=rM-m_{21}L-(r\cdot m_{11}-p\cdot
m_{21})\textrm{id}_{\mathbb{R}^{2}}$. The matrix $\overline{M}$ also satisfies
(NC 3) and has the form $\overline{M}=\begin{pmatrix}0&\overline{m_{12}}\\\
0&\overline{m_{22}}\end{pmatrix}$, with
$\overline{m}_{12},\overline{m}_{21}\in\mathbb{Z}$.

Suppose first that $L$ has integer eigenvalues $t_{1},t_{2}\in\mathbb{Z}$,
i.e., we can write

$(eh-fg)L=P\begin{pmatrix}t_{1}&0\\\
0&t_{2}\end{pmatrix}\operatorname{adj}(P),\quad\text{ for some integer matrix
}P=\begin{pmatrix}e&f\\\ g&h\end{pmatrix}.$

If ${|eh-fg|=1}$, then we can use Proposition 3.6 with Corollary 3.2 to
conclude that $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is conjugate
(via $P$ in $\textrm{GL}(2,\mathbb{Z})$) to the linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(t_{1}^{n}\mathbb{Z}\times
t_{2}^{n}\mathbb{Z})})$. The same conclusion holds when $L$ is conjugate (via
a $\textrm{GL}(2,\mathbb{Z})$-matrix) to a triangular matrix. We then assume
that $|eh-fg|>1$, $e,f,g,h\in\mathbb{Z}$, and $\gcd(e,g)=\gcd(f,h)=1$. For any
$n>0$, the coefficients of $L^{n}$ are given by:

$\begin{array}[]{ll}p(n)=\dfrac{eht_{1}^{n}-fgt_{2}^{n}}{eh-
fg}&q(n)=\dfrac{ef(t_{1}^{n}-t_{2}^{n})}{eh-fg}\\\ &\\\
r(n)=\dfrac{gh(t_{1}^{n}-t_{2}^{n})}{eh-
fg}&s(n)=\dfrac{eht_{2}^{n}-fgt_{1}^{n}}{eh-fg}.\end{array}$

So, (NC 3) can be rewritten as:

(9) $\displaystyle
gh(t_{1}^{m}-t_{2}^{m})[\overline{m}_{12}(eht_{2}^{n}-fgt_{1}^{n})-\overline{m}_{22}ef(t_{2}^{n}-t_{1}^{n})]\equiv
0\ (\text{mod}\ t_{1}^{n}t_{2}^{n})$ $\displaystyle
gh(t_{1}^{m}-t_{2}^{m})[\overline{m}_{22}(eht_{1}^{n}-fgt_{2}^{n})-\overline{m}_{12}gh(t_{1}^{n}-t_{2}^{n})]\equiv
0\ (\text{mod}\ t_{1}^{n}t_{2}^{n})$
$\displaystyle(eht_{2}^{m}-fgt_{1}^{m})[\overline{m}_{12}(eht_{2}^{n}-fgt_{1}^{n})-\overline{m}_{22}ef(t_{2}^{n}-t_{1}^{n})]\equiv
0\ (\text{mod}\ t_{1}^{n}t_{2}^{n})$
$\displaystyle(eht_{2}^{m}-fgt_{1}^{m})[\overline{m}_{22}(eht_{1}^{n}-fgt_{2}^{n})-\overline{m}_{12}gh(t_{1}^{n}-t_{2}^{n})]\equiv
0\ (\text{mod}\ t_{1}^{n}t_{2}^{n}).$

Since $\operatorname{rad}(\det(L))=\operatorname{rad}(t_{1}t_{2})$ does not
divide $\operatorname{trace}(L)=t_{1}+t_{2}$, the following three cases hold:

  1. Case 1.

Suppose that $\operatorname{rad}(t_{1})$ divides $t_{2}$, but there exists a
prime number $t$ dividing $t_{2}$ that does not divide $t_{1}$. Then (9) can
be reduced to

(10) $\displaystyle fgt_{1}^{m+n}[\overline{m}_{22}\cdot
eh-\overline{m}_{12}\cdot gh]\equiv 0\ (\text{mod}\ t^{n}).$

Since $t$ does not divide $t_{1}$, for any $n,m\geq 0$, $t_{1}^{n+m}$ is an
invertible element in $\mathbb{Z}/t^{n}\mathbb{Z}$. We can also choose $n$
large enough so that $t^{n}$ does not divide any of the coefficients
$e,f,g,h\in\mathbb{Z}$. We then conclude that
$\overline{m}_{12}g=\overline{m}_{22}e$. This implies that

$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})\subseteq\left\\{aL+b\textrm{id}+c\begin{pmatrix}0&e\\\
0&g\end{pmatrix},a,b,c\in\frac{1}{r}\mathbb{Z}\right\\}\cap\textrm{GL}(2,\mathbb{Z}).$

Since $P^{-1}\begin{pmatrix}0&e\\\ 0&g\end{pmatrix}P=\begin{pmatrix}g&h\\\
0&0\end{pmatrix}$, the set
$P^{-1}\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})P$ is a subgroup of
unimodular upper triangular matrices in $G$, where

$G=\left\\{\begin{pmatrix}a&b\\\ 0&c\end{pmatrix},\quad
a,b,c\in\frac{1}{r}\mathbb{Z},|ac|=1\right\\}.$

Notice that any commutators of $G$ are of the form $\begin{pmatrix}1&b\\\
0&1\end{pmatrix}$. So, the derived subgroup $G^{\prime}$ (generated by the
commutators) is isomorphic to $\mathbb{Z}$. Moreover, the abelianization
$G/G^{\prime}$ of $G$ is finite. Therefore, the abelianization of
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is finite, and its derived
subgroup is isomorphic to a subgroup (eventually trivial) of $\mathbb{Z}$.
Conversely, a direct computation shows that the matrix
$P^{-1}\begin{pmatrix}1&\det(P)\\\ 0&1\end{pmatrix}P$ satisfies (NC 3),
proving that the derived subgroup of
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is nontrivial.

  2. Case 2.

The case where $\operatorname{rad}(t_{2})$ divides $t_{1}$ but
$\operatorname{rad}(t_{1})$ does not divide $t_{2}$ is symmetric to the former
one.

  3. Case 3.

Neither $\operatorname{rad}(t_{1})$ divides $t_{2}$ nor
$\operatorname{rad}(t_{2})$ divides $t_{1}$. The former computations in the
two cases provide that

$\displaystyle\overline{m}_{12}g=\overline{m}_{22}e,\quad\wedge\quad\overline{m}_{12}h=\overline{m}_{22}f.$

Since $eh-fg\neq 0$, this implies that $\overline{m}_{12}=0$ and
$\overline{m}_{22}=0$, so $\overline{M}=0$. We conclude that $M$ commutes with
$L$, i.e., the linear representation group
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is equal to
$\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L)$, which can be isomorphic
to $\mathbb{Z}/2\mathbb{Z}$ or $(\mathbb{Z}/2\mathbb{Z})^{2}$.

Now we suppose that $L$ does not have integer eigenvalues. A direct induction
on (8) gives, for any $n>0$, that $L^{n}\equiv\operatorname{trace}(L)^{n-1}L\
(\text{mod}\ \det(L))$. Since $\operatorname{rad}(\det(L))$ does not divide
$\operatorname{trace}(L)$, there exists a prime number $t$ dividing $\det(L)$
that does not divide $p$ or $s$. Without loss of generality (up to a
conjugation with $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$) we may assume that
$t$ does not divide $s$. As $s(n)\equiv\operatorname{trace}(L)^{n-1}s\
(\text{mod}\ \det(L))$, then, for all $n>0$ and $m>0$, $s(m)$ is an invertible
element in $\mathbb{Z}/t^{n}\mathbb{Z}$. Hence, (NC 3) implies that

(11) $\displaystyle\overline{m_{12}}s(n)-\overline{m_{22}}q(n)$
$\displaystyle\equiv 0\ (\text{mod}\ t^{n})$ (12)
$\displaystyle-\overline{m_{12}}r(n)+\overline{m_{22}}p(n)$
$\displaystyle\equiv 0\ (\text{mod}\ t^{n}),$

which is equivalent to

(13)
$\displaystyle\operatorname{adj}(L^{n})\dbinom{\overline{m_{12}}}{\overline{m_{22}}}\equiv\dbinom{0}{0}\
(\text{mod}\ t^{n}).$

Consider the set
$E=\left\\{\dbinom{\overline{m_{12}}}{\overline{m_{22}}}\in\mathbb{Z}^{2}\colon\
\text{satisfying}\ \eqref{eq7generalcase}\ \text{for all}\ n>0\right\\}$. This
set is $\operatorname{adj}(L)$-invariant, and if $\overline{m_{22}}=0$, then
(11) implies that $\overline{m_{12}}=0$.

Now, set
$\dbinom{\overline{m_{12}}^{(1)}}{\overline{m_{22}}^{(1)}},\dbinom{\overline{m_{12}}^{(2)}}{\overline{m_{22}}^{(2)}}\in
E$. Note that

$\overline{m_{22}}^{(2)}\dbinom{\overline{m_{12}}^{(1)}}{\overline{m_{22}}^{(1)}}-\overline{m_{22}}^{(1)}\dbinom{\overline{m_{12}}^{(2)}}{\overline{m_{22}}^{(2)}}=\dbinom{\overline{m_{22}}^{(2)}\overline{m_{12}}^{(1)}-\overline{m_{22}}^{(1)}\overline{m_{12}}^{(1)}}{0},$

hence, by the former remark
$\overline{m_{22}}^{(2)}\dbinom{\overline{m_{12}}^{(1)}}{\overline{m_{22}}^{(1)}}=\overline{m_{22}}^{(1)}\dbinom{\overline{m_{12}}^{(2)}}{\overline{m_{22}}^{(2)}}$.
So, $E$ is an $\operatorname{adj}(L)$-invariant $\mathbb{Z}$-module of rank at
most 1.

Since $L$ does not have integer eigenvalues, $E$ must have rank 0. This
implies that $\overline{m_{12}}=\overline{m_{22}}=0$. Hence $M$ commutes with
$L$. We conclude that $\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L^{n})})$ is
equal to $\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L)$.

We claim that the centralizer
$\operatorname{Cent}_{\textrm{GL}(2,\mathbb{Z})}(L)$ is finite when $L$ has no
real eigenvalues. Indeed a matrix $M$ commuting with $L$ has to satisfy

(14) $\begin{array}[]{cc}r\cdot m_{12}-q\cdot m_{21}&=0\\\ r\cdot
m_{22}-s\cdot m_{21}-r\cdot m_{11}+p\cdot m_{21}&=0.\end{array}$

  * •

Suppose $p=s$. In this case, (14) implies that $m_{11}=m_{22}$ and
$m_{21}=m_{11}\cdot r/q$. Note that $L$ has complex eigenvalues if and only if
$qr<0$, as determined by the condition $(2p)^{2}-4(p^{2}-qr)<0$. Since
$|\det(M)|=1$, then $|m_{11}^{2}-m_{12}^{2}\cdot r/q|$ equals $1$. Therefore,
when $qr<0$, there exists a finite number of points
$(m_{11},m_{12})\in\mathbb{Z}^{2}$ satisfying (14).

  * •

If $p\neq s$, then (14) implies that $m_{12}=q(m_{11}-m_{22})/(p-s)$ and
$m_{21}=r(m_{11}-m_{22})/(p-s)$. Since $M\in GL(2,\mathbb{Z})$, we get that

(15) $m_{11}m_{22}-(m_{11}-m_{22})^{2}\dfrac{qr}{(p-s)^{2}}=\pm 1.$

In this case, there is a finite number of solutions if
$\operatorname{trace}(L)^{2}-4\det(L)<0$, which is equivalent to $L$ having no
real eigenvalues.

∎

######  Remark 3.7.

In the particular case when $\gcd(\operatorname{trace}(L),\det(L))=1$, we can
simplify the proof noting that (11), (12) imply the existence of two sequences
$(k_{n}^{1})_{n>0},(k_{n}^{2})_{n>0}\subseteq\mathbb{Z}$ such that
$\det(L)^{n}k_{n}^{1}=\overline{m_{12}}s(n)-\overline{m_{22}}q(n)$ and
$\det(L)^{n}k_{n}^{2}=-\overline{m_{12}}r(n)+\overline{m_{22}}p(n)$, i.e.,

(16)
$\begin{array}[]{cl}k_{n}^{1}&=\overline{m_{12}}\dfrac{s(n)}{\det(L)^{n}}-\overline{m_{22}}\dfrac{q(n)}{\det(L)^{n}}\\\
&\\\
k_{n}^{2}&=-\overline{m_{12}}\dfrac{r(n)}{\det(L)^{n}}+\overline{m_{22}}\dfrac{p(n)}{\det(L)^{n}}.\end{array}$

Since $L$ is an expansion matrix, then $L^{-1}$ is a contraction, so we have
that

$\lim\limits_{n\to\infty}\dfrac{p(n)}{\det(L)^{n}}=\lim\limits_{n\to\infty}\dfrac{q(n)}{\det(L)^{n}}=\lim\limits_{n\to\infty}\dfrac{r(n)}{\det(L)^{n}}=\lim\limits_{n\to\infty}\dfrac{s(n)}{\det(L)^{n}}=0,$

this implies that for all $n\in\mathbb{N}$ large enough,
$k_{n}^{1}=k_{n}^{2}=0$, and we conclude that
$\overline{m_{12}}=\overline{m_{22}}=0$.

Theorem 3.3 implies that the linear representation group of constant-base
$\mathbb{Z}^{2}$-odometer systems is computable. But the techniques developed
in this article may not be directly applicable to higher dimensions. This
raises the following question:

######  Question 3.8.

Regarding the linear representation group of higher dimensional constant-base
odometer systems, are its elements computable? Is its group structure
computable?

By “computable elements”, we mean that if there exists an algorithm to decide
whether a matrix $M$ belongs to the linear representation group or not. The
second question involves finding an algorithm to determine the linear
representation group, up to isomorphism, as a function of the base matrix $L$.

##  4\. Minimal subshifts with infinite linear representation group

In this section, we present minimal substitutive subshifts with infinite
linear representation groups, thereby providing a positive response to a
question posed in [2]. Their normalizer groups are fully explained. We prove
the following result.

######  Theorem 4.1.

For any expansion matrix $L\in{\mathcal{M}}(d,\mathbb{Z})$ with $|\det L|\geq
3$, there exists an aperiodic minimal substitutive $\mathbb{Z}^{d}$-subshift
$X$ with expansion matrix $L$ such that

  * •

It is coalescent.

  * •

It is an almost 1-to-1 extension of
$\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$.

  * •

Its automorphisms are reduced to the shift transformations.

  * •

Its linear representation semigroup $\vec{N}(X,S)$ is equal to

$\\{M\in\bigcup_{k\geq 0}\bigcap_{n\geq
k}L^{n}\textrm{GL}(d,\mathbb{Z})L^{-n}:\exists
n_{0},L^{-n}ML^{n}=L^{-p}ML^{p}\ (\text{mod }L(\mathbb{Z}^{d})),\forall
n,p\geq n_{0}\\}.$

  * •

Its normalizer group is a semidirect product of $\mathbb{Z}^{d}$ with
$\vec{N}(X,S)$.

The isomorphisms are explicit.

In particular, when $L$ is proportional to the identity, the former result
provides an example of a minimal subshift with a linear representation
semigroup equal to $\textrm{GL}(d,\mathbb{Z})$.

To describe these explicit examples, we will briefly introduce some notions
coming from (aperiodic) tiling theory. Most of the references come from [3].
From a tiling perspective, the _half-hex inflation_ is a well-known inflation
rule analogous to a symbolic substitution (for more properties about this
tiling substitution, see [3, Section 6.4]). The tiles consist of 6 regular
half-hexagons, each one being the image under a 6-fold rotation of a single
one. The inflation rule, up to rotation, is described in Fig. 1.

Figure 1. Tile-substitution of the half-hex tiling.

In tiling terminology, it is an _edge-to-edge inflation_ , which means that
each inflated tile is precisely dissected into copies of the tiles, and the
vertices of any tile only intersect with the vertices of the adjacent tiles.
This inflation defines an aperiodic tiling of the plane (see [3, Example
6.4]). Since the largest edge of any half-hex can only meet the largest edge
of the adjacents half-hexes, two half-hexes always join to form a regular
hexagon via their largest edges. By applying this procedure, the half-hex
tiling can be decomposed into three hexagons, each distinguished by a single
diagonal line, as shown in Fig. 2 (see [3]).

Figure 2. The three tiles as a new alphabet for the half-hex tiling.

Using these full hexagons, we can define a _pseudo inflation_ (using the
vocabulary on [3]), which is conjugated to the half-hex tiling as in Fig. 3.

$u$$v$ Figure 3. New tile-substitution conjugate to the half-hex tiling, with
a discrete 2-dimensional tranlsation-invariant subaction in $\mathbb{R}^{2}$.

From this pseudo inflation, we construct a tiling substitution with only the
four shaded hexagons in Fig. 3. In this tiling substitution, there is an
invariant discrete lattice $\Lambda\subseteq\mathbb{R}^{2}$ generated by the
centers of these hexagons, using the vectors ${\bm{u}}$ and ${\bm{v}}$ as
depicted in Fig. 3. The discrete translation $\Lambda$-subaction is conjugate
to the substitutive subshift associated with the following constant-shape
substitution, called _half-hex substitution_ , $\zeta_{hh}$ with an expansion
matrix $L_{hh}=2\cdot\textrm{id}_{\mathbb{Z}^{2}}$ and support
$F_{1}^{hh}=\\{(0,0),(1,0),(0,1),(1,-1)\\}$

$\begin{array}[]{llllllllllllll}&&0&&&&&0&&&&&0&\\\
0&\mapsto&0&2&&1&\mapsto&1&2&&2&\mapsto&2&2\\\ &&&1&&&&&1&&&&&1.\\\
\end{array}$

We recall that the notations and the notions we will use are summarized in
Section 2.3.2. A straightforward computation, based on [20, Theorem 4.8],
reveals that the extreme points of the convex hull of $F_{n}^{hh}$ are
$\\{(0,0),(0,2^{n}-1),(2^{n}-1,0),(2^{n}-1,1-2^{n})\\}$. Since $F_{n}^{hh}$ is
a fundamental domain of $2^{n}\mathbb{Z}^{2}$, it has cardinality $4^{n}$.
Furthermore,
$F_{n}^{hh}\subset\operatorname{conv}(F_{n}^{hh})\cap\mathbb{Z}^{2}$, where
$\operatorname{conv}(F_{n}^{hh})$ denotes the convex hull of $F_{n}^{hh}$.
Actually, the Pick formula provides that the cardinality of
$\operatorname{conv}(F_{n}^{hh})\cap\mathbb{Z}^{2}$ and $F_{n}^{hh}$ are the
same, so $F_{n}^{hh}=\operatorname{conv}(F_{n}^{hh})\cap\mathbb{Z}^{2}$. It
then follows that $(F_{n}^{hh})_{n\geq 0}$ is a Følner sequence.

$x$$y$$F_{1}$$x$$y$$F_{2}$ Figure 4. The first two supports of the half-hex
substitution.

Inspired by the half-hex substitution, we consider an integer expansion matrix
$L\in\mathcal{M}(d,\mathbb{Z})$ with $|\det(L)|\geq 3$, a fundamental domain
$F_{1}$ of $L(\mathbb{Z}^{d})$ in $\mathbb{Z}^{d}$, and set the finite
alphabet $\mathcal{A}=F_{1}\setminus\\{\bm{0}\\}$. We define the substitution
$\sigma_{L}\colon\mathcal{A}\to\mathcal{A}^{F_{1}}$ as follows:

(19) $\displaystyle\forall
a\in\mathcal{A},\quad\sigma_{L}(a)_{{\bm{f}}}=\left\\{\begin{array}[]{cl}a&\text{
when }{\bm{f}}={\bm{0}},\\\ {\bm{f}}&\text{ when
}{\bm{f}}\neq{\bm{0}}.\end{array}\right.$

Under the hypothesis that the sequence of supports $(F_{n})_{n>0}$ is a Følner
sequence, we get the substitutive subshift
$(X_{\sigma_{L}},S,\mathbb{Z}^{d})$. It is important to notice that all the
patterns $\sigma_{L}(a)$ coincide except at the origin, where the letter is
uniquely determined.

For computational purposes, we introduce the map

(20)
$\displaystyle\tau\colon\bm{n}\in\mathbb{Z}^{d}\setminus\\{{\bm{0}}\\}\mapsto{\bm{f}}\in
F_{1}\setminus\\{\bm{0}\\},$

where ${\bm{n}}=L^{p+1}({\bm{z}})+L^{p}({\bm{f}})$ with
${\bm{z}}\in\mathbb{Z}^{d}$, ${\bm{f}}\in F_{1}\setminus\\{\bm{0}\\}$ and $p$
is the smallest integer such that $\bm{n}\not\in L^{p+1}(\mathbb{Z}^{d})$. The
value $p$ serves as a multidimensional $L$-adic valuation of $\bm{n}$. A
motivation to introduce this map is due to the next formula that enables to
compute the value of a $\sigma_{L}$-fixed point $\bar{x}$ at some position
only by the knowledge of this position. More precisely, it is straightforward
to check that

(21)
$\displaystyle\forall{\bm{n}}\neq{\bm{0}}\in\mathbb{Z}^{d},\quad\bar{x}_{\bm{n}}=\tau(\bm{n}).$

This property is a typical one of automatic sequences. As a consequence,
$\sigma_{L}$ has exactly $|\mathcal{A}|=|\det L|-1$ fixed points in
$X_{\sigma_{L}}$, and they all coincide except at the origin. Moreover, we
have the following standard recognizability property.

######  Lemma 4.2.

Let $\overline{x}$ be a fixed point of $\sigma_{L}$, then for any integer
$n>0$ and any ${\bm{a}},{\bm{b}}\in\mathbb{Z}^{d}\setminus\\{0\\}$,

$\overline{x}_{{\bm{a}}+F_{n}}=\overline{x}_{{\bm{b}}+F_{n}}\implies{\bm{a}}\equiv{\bm{b}}\
(\text{mod}\ L^{n}(\mathbb{Z}^{d})).$

In particular, if the sequence of supports of the iterations $\sigma_{L}^{n}$
is Følner, then the substitution $\sigma_{L}$ is recognizable on any fixed
point $\overline{x}$ of ${\sigma_{L}}$, so $\sigma_{L}$ is aperiodic.

###### Proof.

We prove the claim by induction on $n>0$. We start with the base case $n=1$.

Suppose ${\bm{a}}\notin L(\mathbb{Z}^{d})$, i.e.,
${\bm{a}}=L({\bm{c}})+{\bm{g}}$ with ${\bm{g}}\in
F_{1}\setminus\\{{\bm{0}}\\}$. If $b\notin L(\mathbb{Z}^{d})$, then
${\bm{b}}=L({\bm{d}})+{\bm{h}}$ with ${\bm{h}}\in F_{1}\setminus{\bm{0}}$.
Since $\overline{x}$ is a fixed point of $\sigma_{L}$, we have that
$\overline{x}_{{\bm{a}}}=\sigma(x_{{\bm{c}}})_{{\bm{g}}}={\bm{g}}=\sigma(x_{{\bm{d}}})_{{\bm{h}}}$,
so ${\bm{g}}={\bm{h}}$, which implies that ${\bm{a}}\equiv{\bm{b}}\
(\text{mod}\ L(\mathbb{Z}^{d}))$. If $b\in L(\mathbb{Z}^{d})$, then for any
${\bm{f}}\in F_{1}\setminus\\{{\bm{0}}\\}$ we have that
$\overline{x}_{{\bm{b}}+{\bm{f}}}={\bm{f}}=\overline{x}_{{\bm{a}}+{\bm{f}}}$.
We consider ${\bm{f}}\in F_{1}\setminus\\{{\bm{0}}\\}$ such that
${\bm{a}}+{\bm{f}}\notin L(\mathbb{Z}^{d})$, i.e.,
${\bm{a}}+{\bm{f}}=L({\bm{e}})+{\bm{h}}$, so
$\overline{x}_{{\bm{a}}+{\bm{f}}}={\bm{h}}$, and ${\bm{h}}={\bm{f}}$, i.e.,
${\bm{a}}\in L(\mathbb{Z}^{d})$ which is a contradiction.

Now, suppose there exists some $n>0$ such that
$\overline{x}_{{\bm{a}}+F_{n}}=\overline{x}_{{\bm{b}}+F_{n}}\implies{\bm{a}}\equiv{\bm{b}}\
(\text{mod}\ L^{n}(\mathbb{Z}^{d}))$. Let ${\bm{a}},{\bm{b}}\in\mathbb{Z}^{d}$
be such that

$\overline{x}_{{\bm{a}}+F_{n+1}}=\overline{x}_{{\bm{b}}+F_{n+1}}.$

Since $F_{n}\subseteq F_{n+1}$, by the induction hypothesis we have that
${\bm{a}}\equiv{\bm{b}}\ (\text{mod}\ L^{n}(\mathbb{Z}^{d}))$. We recall that
$F_{n+1}=F_{n}+L^{n}(F_{1})$, so we write

${\bm{a}}=L^{n+1}({\bm{c}})+{\bm{f}}+L^{n}({\bm{g}}),\quad{\bm{b}}=L^{n+1}({\bm{d}})+{\bm{f}}+L^{n}({\bm{h}})$

for some ${\bm{f}}\in F_{n}$, ${\bm{g}},{\bm{h}}\in F_{1}$ and
${\bm{c}},{\bm{d}}\in\mathbb{Z}^{d}$. We prove that ${\bm{g}}={\bm{h}}$. If
${\bm{f}}={\bm{0}}$ we can use a similar argument as for the case $n=1$ to
conclude that ${\bm{g}}={\bm{h}}$. Suppose then ${\bm{f}}\neq{\bm{0}}$. We
consider ${\bm{j}}\in F_{n+1}$ such that ${\bm{f}}\equiv-{\bm{j}}\
(\text{mod}\ L^{n}(\mathbb{Z}^{d}))$, so

${\bm{a}}+{\bm{j}}=L^{n+1}({\bm{c}}_{1})+L^{n}({\bm{g}})\quad{\bm{b}}+{\bm{j}}=L^{n+1}({\bm{d}}_{1})+L^{n}({\bm{h}}),$

for some ${\bm{c}}_{1},{\bm{d}}_{1}\in\mathbb{Z}^{d}$. Since $\overline{x}$ is
a fixed point of $\sigma_{L}$, we get that

$\displaystyle\overline{x}_{{\bm{a}}+{\bm{j}}}$
$\displaystyle=\sigma_{L}^{n-2}(\sigma_{L}(\sigma_{L}(\overline{x})_{{\bm{c}}_{1}})_{{\bm{g}}})_{{\bm{0}}})_{{\bm{0}}}={\bm{g}}$
$\displaystyle\overline{x}_{{\bm{b}}+{\bm{j}}}$
$\displaystyle=\sigma_{L}^{n-2}(\sigma_{L}(\sigma_{L}(\overline{x})_{{\bm{d}}_{1}})_{{\bm{h}}})_{{\bm{0}}})_{{\bm{0}}}={\bm{h}}.$

Recall that
$\overline{x}_{{\bm{a}}+{\bm{j}}}=\overline{x}_{{\bm{b}}+{\bm{j}}}$, hence
${\bm{g}}={\bm{h}}$ which implies that ${\bm{a}}\equiv{\bm{b}}\ (\text{mod}\
L^{n+1}(\mathbb{Z}^{d}))$.

∎

Recall that the map
$\pi:(X_{\sigma_{L}},S,\mathbb{Z}^{d})\to(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})},{\bm{+}},\mathbb{Z}^{d})$
is defined in Section 2.3.2.

######  Proposition 4.3.

If the sequence of supports of the iterations $\sigma_{L}^{n}$ is Følner, then
$\sigma_{L}$ is an aperiodic, primitive constant-shape substitution and the
factor map
$\pi:(X_{\sigma_{L}},S,\mathbb{Z}^{d})\to(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})},{\bm{+}},\mathbb{Z}^{d})$
is almost 1-to-1.

More precisely, we have

$|\pi^{-1}(\\{\overleftarrow{g}\\})|=\left\\{\begin{array}[]{cl}|\mathcal{A}|&\text{
if }\overleftarrow{g}\in\mathcal{O}(\overleftarrow{0},{\bm{+}}),\\\ 1&\text{
otherwise.}\end{array}\right.$

In particular, the subshift $X_{\sigma_{L}}$ is a substitutive Toeplitz
subshift and its maximal equicontinuous factor is
$\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$. As an explicit example, the
substitutive subshift $X^{hh}$ associated with the half-hex substitution
$\zeta_{hh}$ is an almost 1-to-1 extension of the constant-base odometer
$\overleftarrow{\mathbb{Z}^{2}}_{(2^{n}\mathbb{Z}^{2})}$.

###### Proof.

Since $\tau$ is a bijection, $\sigma_{L}$ is a primitive substitution. The
aperiodicity follows from the recognizability given by Lemma 4.2.

Now we study the fibers $\pi^{-1}(\\{\overleftarrow{g}\\})$ for
$\overleftarrow{g}=(g_{n})_{n}\in\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$.
Suppose $|\pi^{-1}(\\{\overleftarrow{g}\\})|\geq 2$ and set
$x_{1},x_{2}\in\pi^{-1}(\\{\overleftarrow{g}\\})$, i.e., for any $n>0$ there
exists $y_{1}^{(n)},y_{2}^{(n)}\in X_{\sigma_{L}}$ such that
$x_{i}=S^{{g}_{n}}\sigma_{L}^{n}(y_{i}^{(n)})$, for $i\in\\{1,2\\}$. Let
${\bm{a}}\in\mathbb{Z}^{d}$ such that $x_{1{\bm{a}}}\neq x_{2{\bm{a}}}$. This
implies that
$\sigma_{L}^{n}(y_{1}^{(n)})_{{\bm{a}}+{g}_{n}}\neq\sigma_{L}^{n}(y_{2}^{(n)})_{{\bm{a}}+{g}_{n}}$.
For every $n>0$, we write ${\bm{a}}+{g}_{n}=L^{n}({\bm{b}}_{n})+{\bm{f}}_{n}$,
with ${\bm{b}}_{n}\in\mathbb{Z}^{d}$ and ${\bm{f}}_{n}\in F_{n}$. Since for
$i\in\\{1,2\\}$ we have that
$\sigma_{L}^{n}(y_{i}^{(n)})_{{\bm{a}}+{\bm{g}}_{n}}=\sigma_{L}^{n}(y_{i}^{(n)}({\bm{b}}_{n}))_{{\bm{f}}_{n}}$
and these letters are different, then for any $n>0$, ${\bm{f}}_{n}$ must be
${\bm{0}}\in F_{n}$. This implies that for every $n>0$

$\displaystyle{\bm{g}}_{n}\equiv-{\bm{a}}\ (\text{mod}\
L^{n}(\mathbb{Z}^{d})).$

Hence $\overleftarrow{g}=\kappa_{(L^{n})}(-{\bm{a}})$, i.e.,
$\overleftarrow{g}\in\mathcal{O}(\overleftarrow{0},{\bm{+}})$. It follows that
$x_{1}$ and $x_{2}$ are in the orbit of two fixed points of $\sigma_{L}$. In
this case, $|\pi^{-1}(\overleftarrow{g})|$ has cardinality $|\mathcal{A}|$ and
they only differ in the coordinate
$-\kappa_{(L^{n})}^{-1}(\overleftarrow{g})$. If $\overleftarrow{g}$ is not in
$\mathcal{O}(\overleftarrow{0},{\bm{+}})$, then $\pi^{-1}(\overleftarrow{g})$
has cardinality 1. We conclude that the factor map
$\pi:(X_{\sigma_{L}},S,\mathbb{Z}^{d})\to(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})},{\bm{+}},\mathbb{Z}^{d})$
is almost 1-to-1. ∎

As a consequence of Proposition 4.3, we get the following property on the
$\textrm{GL}(d,\mathbb{Z})$-endomorphisms of $X_{\sigma_{L}}$.

######  Corollary 4.4.

Assume the sequence of supports of the iterations $\sigma_{L}^{n}$ is Følner.
Then any $\textrm{GL}(d,\mathbb{Z})$-endomorphism $\phi\in
N(X_{\sigma_{L}},S)$ maps a $\sigma_{L}$-fixed point onto a shifted of a
$\sigma_{L}$-fixed point.

###### Proof.

Since the factor map
$\pi:X_{\sigma_{L}}\to\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$ is almost
1-to-1 (Proposition 4.3), then the odometer system
$\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}$ is the maximal equicontinuous
factor of the substitutive subshift $(X_{\sigma_{L}},S,\mathbb{Z}^{d})$. So
there exists a semigroup homomorphism $\hat{\pi}:N(X_{\sigma_{L}},S)\to
N(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$ which is injective (Lemma 2.3).
Recall that any equicontinuous system is coalescent, so any endomorphism is
invertible and by Item iii of Lemma 2.3, any endomorphism $\phi$ satisfies

$\displaystyle\left\\{\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}\colon|\pi^{-1}(\\{\overleftarrow{g}\\})|=|\mathcal{A}|\right\\}\subseteq\hat{\pi}(\phi)\left(\left\\{\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}\colon|\pi^{-1}(\\{\overleftarrow{g}\\})|=|\mathcal{A}|\right\\}\right).$

Since
$\left\\{\overleftarrow{g}\in\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})}\colon|\pi^{-1}(\\{\overleftarrow{g}\\})|=|\mathcal{A}|\right\\}$
is the orbit
$\mathcal{O}(\overleftarrow{0},{\bm{+}})=\kappa_{(L^{n})}(\mathbb{Z}^{d})$, it
implies that $\phi$ maps the $\pi$-fiber of the orbit
$\mathcal{O}(\overleftarrow{0},{\bm{+}})$ onto itself. This $\pi$-fiber
consists of the orbits of $\sigma_{L}$-fixed points. It follows that, up to
compose $\phi$ with a shift map, the image of a $\sigma_{L}$-fixed point
$\bar{x}$ by $\phi$ is also a $\sigma_{L}$-fixed point. ∎

We also characterize the $\textrm{GL}(d,\mathbb{Z})$-endomorphisms of the
substitutive subshift $(X_{\sigma_{L}},S,\mathbb{Z}^{d})$ by the following
results. The first one concerns endomorphisms and automorphisms.

######  Lemma 4.5.

Let $\sigma_{L}$ defined as (19) and assume the sequence of supports of the
iterations $\sigma_{L}^{n}$ is Følner. Then the subshift
$(X_{\sigma_{L}},S,\mathbb{Z}^{d})$ satisfies

  * •

it is coalescent,

  * •

the automorphism group $\operatorname{Aut}(X_{\sigma_{L}},S)$ is trivial,
i.e., consist only on the shifted transformations $S^{\bm{n}}$,
${\bm{n}}\in\mathbb{Z}^{d}$,

###### Proof.

First we prove that $\operatorname{End}(X_{\sigma_{L}},S)=\left\langle
S\right\rangle$. We keep the notations of Lemma 2.3. Set
$\phi\in\operatorname{End}(X_{\sigma_{L}},S)$. According to Corollary 4.4,
$\hat{\pi}(\phi)$ is an endomorphism of the odometer, which means it is a
translation, as proven in Lemma 2.8. Moreover, since it preserves the
$\overleftarrow{0}$-orbit, $\hat{\pi}(\phi)$ is a translation by some element
$\kappa_{(L^{n})}({\bm{n}})$ with ${\bm{n}}\in\mathbb{Z}^{d}$. By definition
of $\hat{\pi}$, this translation is $\hat{\pi}(S^{\bm{n}})$, thus equal to
$\hat{\pi}(\phi)$. We conclude, by the injectivity of $\hat{\pi}$, that
$\operatorname{End}(X_{\sigma_{L}},S)=\left\langle S\right\rangle$. As a
consequence, $(X_{\sigma_{L}},S,\mathbb{Z}^{d})$ is a coalescent system. ∎

Now, we characterize the elements of their linear representation semigroups.
For this, we introduce the following notations. Let $N_{L}$ be the group

$\\{M\in\bigcup_{k\geq 0}\bigcap_{n\geq
k}L^{n}\textrm{GL}(d,\mathbb{Z})L^{-n}:\exists
n_{0},L^{-n}ML^{n}=L^{-p}ML^{p}\ (\text{mod }L(\mathbb{Z}^{d})),\forall
n,p\geq n_{0}\\}.$

The crucial properties of an element $M$ of $N_{L}$ are the following: for
each integer $n>0$, there is a $\mathbb{Z}^{d}$-automorphism $M_{n}$ such that
$L^{n}M_{n}=ML^{n}$. Moreover, each automorphism $M_{n}$ permutes the
$L(\mathbb{Z}^{d})$-cosets. With an abuse of notation, we will denote these
permutations on $F_{1}\setminus\\{\bm{0}\\}$ by $M_{n}\ (\text{mod
}L(\mathbb{Z}^{d}))$. These permutations are ultimately all the same, for
large enough $n$. Linked with the computation of the digits of fixed points,
we have the following relation

(22) $\displaystyle\tau\circ M(\bm{n})=M_{p}\circ\tau(\bm{n})\ (\text{mod
}L(\mathbb{Z}^{d})),$

where $p$ is the smallest integer such that $\bm{n}\not\in
L^{p+1}(\mathbb{Z}^{d})$.

######  Lemma 4.6.

Assume that the sequence of supports of the iterations $\sigma_{L}^{n}$ is
Følner. Then the linear representation semigroup $\vec{N}(X_{\sigma_{L}},S)$
is the linear group $N_{L}$.

###### Proof.

We start to show that $\vec{N}(X_{\sigma_{L}},S)\leqslant N_{L}$. Set
$M\in\vec{N}(X_{\sigma_{L}},S)$, and let $\phi\in N(X_{\sigma_{L}},S)$ be an
$M$-endomorphism with radius $r(\phi)$. Up to compose $\phi$ with a shift, we
can assume that it preserves the set of $\sigma_{L}$-fixed points (Corollary
4.4).

Since $\pi$ is compatible with $\textrm{GL}(d,\mathbb{Z})$-endomorphisms
(Lemma 2.4), Lemma 2.3 provides that
$M\in\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$. This set is a group
(Corollary 2.7), so $M^{-1}$ also belongs to
$\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(L^{n})})$, i.e., for any $n>0$,
there exists $m>0$ such that $L^{-n}M^{-1}L^{m}$ is an endomorphism of
$\mathbb{Z}^{d}$ (see Section 3). We define
$m(n)=\min\limits\\{m>0:L^{-n}M^{-1}L^{m}\text{\ is an endomorphism of}\
\mathbb{Z}^{d}\\}$. Since the determinant of a $\mathbb{Z}^{d}$-endormorphism
is an integer, we have that $m(n)\geq n$.

We will show that actually $m(n)=n$ for any large enough $n$, so that $M^{-1}$
belongs to $\bigcup_{k>0}\bigcap_{n\geq
k}L^{n}\textrm{GL}(d,\mathbb{Z})L^{-n}$. Since this set is stable by taking
the inverse, this is also true for $M$.

We prove the claim by contradiction, i.e., we assume there is an infinite set
of integers $j$ such that $m(j)>j$. Choose $n$ large enough integer so that
the ball of radius $r(\phi)$ centered at the origin is included in
$L^{n}(K_{\sigma_{L}})+F_{n}$, see Proposition 2.13. For such $n$,
$L^{j}(\mathbb{Z}^{d})\cap(L^{n+1}(K_{\sigma_{L}})+F_{n})=\\{\bm{0}\\}$, for
any $j$ large enough. Since the sequence $(m(j))_{j\in S}$ goes to infinity,
one can moreover assume that $m(j+1)>m(j)$. With this convention, the group
$L^{-j}M^{-1}L^{m(j)}(\mathbb{Z}^{d})$ can not be a subset of
$L(\mathbb{Z}^{d})$, since otherwise this implies $m(j)\geq m(j+1)$. So there
is some $\bar{\bm{g}}\in\mathbb{Z}^{d}$ such that
$L^{-j}M^{-1}L^{m(j)}(\bar{\bm{g}})\neq 0\ (\text{mod }L(\mathbb{Z}^{d}))$.
Moreover, the set $L^{-m(j)-1}ML^{j+1}(\mathbb{Z}^{d})\setminus\mathbb{Z}^{d}$
is not empty, since otherwise the matrix $L^{-m(j)-1}ML^{j+1}$ would have
integer coefficients, which is impossible since its determinant $\det
L^{j-m(j)}$ is not an integer. This provides an element $\bm{h}_{0}\in
L^{-m(j)-1}ML^{j+1}(\mathbb{Z}^{d})\setminus\mathbb{Z}^{d}$. Set
$\bm{h}_{1}=L(\bm{h}_{0})$, we have then $\bm{h}_{1}\not\in L(\mathbb{Z}^{d})$
and $L^{m(j)}(\bm{h}_{1})=ML^{j+1}(\bm{h}_{2})$ for some element
$\bm{h}_{2}\in\mathbb{Z}^{d}$. These elements will enable us to provide a
contradiction.

Set $\bm{g}_{1}=L^{m(j)}(\bar{\bm{g}})$ and
$\bm{g}_{2}=\bm{g}_{1}+ML^{j+1}\bm{h}_{2}$. By construction, such elements
satisfy

$\displaystyle
M^{-1}\bm{g}_{1}=L^{j}(L^{-j}M^{-1}L^{m(j)}(\bar{\bm{g}}))=M^{-1}\bm{g}_{2}\
(\text{mod }L^{j}(\mathbb{Z}^{d})),$

so that

$\displaystyle\tau(M^{-1}\bm{g}_{1})=\tau(M^{-1}\bm{g}_{2})=L^{-j}M^{-1}L^{m(j)}\bar{\bm{g}}\
(\text{mod }L(\mathbb{Z}^{d})).$

Moreover the same relation and the very choice of $j$ give for any $\bm{k}\in
K_{\sigma_{L}}$, $\bm{f}\in F_{n}$

(23)
$\displaystyle\tau(M^{-1}\bm{g}_{1}+L^{n+1}(\bm{k})+\bm{f})=\tau(M^{-1}\bm{g}_{2}+L^{n+1}(\bm{k})+\bm{f}).$

On the other way, the very choice of $\bm{h}_{2}$ implies that

$\tau(\bm{g}_{2})=\tau(\bar{\bm{g}}+\bm{h}_{1})\qquad\text{ whereas
}\tau(\bm{g}_{1})=\tau(\bar{\bm{g}}).$

Since $\bm{h}_{1}\not\in L(\mathbb{Z}^{d})$, we get

(24) $\displaystyle\tau(\bm{g}_{1})\neq\tau(\bm{g}_{2}).$

Consider $\overline{x}\in X_{\sigma_{L}}$ a fixed point of $\sigma_{L}$. The
relation (23) implies that

$\bar{x}_{|M^{-1}\bm{g}_{1}+L^{n+1}(K_{\sigma_{L}})+F_{n}}=\bar{x}_{|M^{-1}\bm{g}_{2}+L^{n+1}(K)+F_{n}}.$

By the choice of $n\in\mathbb{N}$ and the Curtis-Hedlund-Lyndon theorem
(Theorem 2.12), we also have that
$\phi(\bar{x})_{\bm{g}_{1}}=\phi(\bar{x})_{\bm{g}_{2}}$. Since $\phi$
preserves the set of $\sigma_{L}$-fixed points, we get by (21),
$\phi(\bar{x})_{\bm{g}_{1}}=\tau(\bm{g}_{1})$ and
$\phi(\bar{x})_{\bm{g}_{2}}=\tau(\bm{g}_{2})$, contradicting (24). So $m(n)=n$
for any large enough $n$.

We still have to show that $L^{-p}ML^{p}\ (\text{mod }L(\mathbb{Z}^{d}))$ is
uniform in $p$ for any large enough integer $p$. Let $K_{\sigma_{L}}$ be the
finite set provided by Proposition 2.13 and let $n$ be large enough so that
$L^{n}(K_{\sigma_{L}})+F_{n}$ contains the ball $B_{r(\phi)}({\bm{0}})$.
Consider $\bm{f}\in F_{1}\setminus\\{0\\}$ and integers $p,q>n$. We claim that

(25)
$\displaystyle\bar{x}_{|L^{p}({\bm{f}})+L^{n}(K_{\sigma_{L}})+F_{n}}=\bar{x}_{|L^{q}({\bm{f}})+L^{n}(K_{\sigma_{L}})+F_{n}}.$

Indeed, by the equality (21) and since $p>n$ we get the following for any
$\bm{k}\in K_{\sigma_{L}}$, $\bm{f}_{n}\in F_{n}$

$\displaystyle\bar{x}_{L^{p}({\bm{f}})+L^{n}(k)+\bm{f}_{n}}=\begin{cases}\tau(\bm{f}_{n})&\text{
if }\bm{f}_{n}\in F_{n}\setminus\\{0\\}\\\ \tau(\bm{k})&\text{ if
}\bm{k}\neq\bm{0}\wedge{\bm{f}}_{n}={\bm{0}}\\\ \tau({\bm{f}})&\text{
otherwise}.\end{cases}$

In particular, notice that
$\bar{x}_{|L^{p}({\bm{f}})+L^{n}(K_{\sigma_{L}})+F_{n}}$ is independent of $p$
and so the equality (25) follows.

Moreover, the equality (25) implies, by Curtis-Hedlund-Lyndon theorem (Theorem
2.12), that $\phi(\bar{x})_{ML^{p}{\bm{f}}}=\phi(\bar{x})_{ML^{q}{\bm{f}}}$.
Recall that $M_{k}$ denotes $L^{-k}ML^{k}$ for any integer $k\geq 0$. Since
$\phi(\bar{x})$ is also fixed by $\sigma_{L}$, the same computation as before
provides $\phi(\bar{x})_{ML^{p}{\bm{f}}}=\tau(ML^{p}{{\bm{f}}})=M_{p}\bm{f}\
(\text{mod }L(\mathbb{Z}^{d}))$, by (22). Similarly, we also have that
$\phi(\bar{x})_{ML^{q}{\bm{f}}}=M_{q}\bm{f}\ (\text{mod }L(\mathbb{Z}^{d}))$.
Hence $M_{p}\bm{f}=M_{q}\bm{f}(\text{mod }L(\mathbb{Z}^{d}))$ for any $p,q>n$
and $\bm{f}\in F_{1}\setminus\\{0\\}$. It follows that $M$ is in $N_{L}$.

We will show now the converse inclusion, that is
$N_{L}\leqslant\vec{N}(X_{\sigma_{L}},S)$, so that the two sets are actually
equal.

Recall that for a matrix $M\in N_{L}$, all the matrices $M_{p}=L^{-p}ML^{p}$
permute $L(\mathbb{Z}^{d})$-cosets, so they define an isomorphism on
$F_{1}\setminus\\{\bm{0}\\}$. Moreover, this isomorphism is independent of
$p$, for any $p$ no greater than some $n_{0}\in\mathbb{N}$. The
recognizability property of $\sigma_{L}$ enables us to define the truncation
of a “$L$-adic” valuation as a local map. More precisely, from Lemma 4.2, we
can define the local map
$v\colon{\mathcal{L}}_{F_{n_{0}}}(X_{\sigma_{L}})\to\\{0,1,\ldots,n_{0}\\}$ by

$\displaystyle v(\bar{x}_{|\bm{n}+F_{n_{0}}})=\min\\{n_{0},q\text{ where
}\bm{n}\in L^{q}(\mathbb{Z}^{d})\setminus L^{q+1}(\mathbb{Z}^{d})\\}.$

We then set the map $\phi_{M}:X_{\sigma_{L}}\to\phi(X_{\sigma_{L}})$ induced
by the local map

$\phi_{M}(x)_{{\bm{n}}}=M_{v(x_{|\bm{M}^{-1}\bm{n}+F_{n_{0}}})}x_{M^{-1}{\bm{n}}}\
(\text{mod }L(\mathbb{Z}^{d})),\quad\text{for any }x\in
X_{\sigma_{L}},{\bm{n}}\in\mathbb{Z}^{d}.$

Notice that $\phi_{M}$ is an $M$-epimorphism onto the subshift
$\phi_{M}(X_{\sigma_{L}})$ by Curtis-Hedlund-Lyndon theorem (Theorem 2.12).
Actually, the two subshifts are the same
$\phi_{M}(X_{\sigma_{L}})=X_{\sigma_{L}}$ so that $\phi_{M}$ is an
$M$-endomorphism. To prove it, it is enough to show that $\phi_{M}$ maps a
$\sigma_{L}$-fixed point $\overline{x}\in X_{\sigma_{L}}$ to another fixed
point of $\sigma_{L}$ within $X_{\sigma_{L}}$, so that
$\phi_{M}(X_{\sigma_{L}})\cap X_{\sigma_{L}}\neq\emptyset$. The minimality of
the subshift $X_{\sigma_{L}}$ enables us to conclude that $\phi_{M}$ is an
$M$-endomorphism.

Indeed, Equation (21) provides for any ${\bm{n}}\neq{\bm{0}}\in\mathbb{Z}^{d}$
that

$\displaystyle\phi_{M}(\bar{x})_{M\bm{n}}$
$\displaystyle=M_{v(\bar{x}_{|\bm{n}+F_{n_{0}}})}\bar{x}_{{\bm{n}}}\
(\text{mod }L(\mathbb{Z}^{d}))$
$\displaystyle=M_{v(\bar{x}_{|\bm{n}+F_{n_{0}}})}\tau(\bm{n})\ (\text{mod
}L(\mathbb{Z}^{d}))$ $\displaystyle=\tau(M\bm{n})\hfill\text{ by relation
}\eqref{eq:CommutationTauM}.$

So $\phi_{M}(\bar{x})$ is fixed by $\sigma_{L}$, and the claim follows, i.e.,
$\phi_{M}$ is an $M$-endomorphism of $X_{\sigma_{L}}$. Hence
$\vec{N}(X_{\sigma_{L}},S)=N_{L}$, and it is a group. Proposition 2.2 ensures
then that $N(X_{\sigma_{L}},S)$ is a group. ∎

######  Lemma 4.7.

Assume the sequence of supports of the iterations $\sigma_{L}^{n}$ is Følner,
then the normalizer semigroup $N(X_{\sigma_{L}},S)$ is isomorphic to a
semidirect product between $\mathbb{Z}^{d}$ and the linear group $N_{L}$.

###### Proof.

From preceding Lemma 4.6 and Lemma 4.5, Proposition 2.2 ensures then that
$N(X_{\sigma_{L}},S,\mathbb{Z}^{d})$ is a group. We have to prove that the map

$M\in N_{L}\mapsto\phi_{M}\in\vec{N}(X_{\sigma_{L}},S)$

is a group embedding. This will show that the exact sequence (1) splits and
${N}(X_{\sigma_{L}},S,\mathbb{Z}^{d})$ is a semidirect product between
$\mathbb{Z}^{d}$ and the linear group $N_{L}$. To prove it is a group
morphism, the only nontrivial point to check is the composition relation
$\phi_{MM^{\prime}}=\phi_{M}\circ\phi_{M^{\prime}}$ for any $M,M^{\prime}\in
N_{L}$. Since the maps $\phi_{MM^{\prime}}$ and
$\phi_{M}\circ\phi_{M^{\prime}}$ have the same linear part, the closed set
$\\{x\in
X_{\sigma_{L}}:\phi_{MM^{\prime}}(x)=\phi_{M}\circ\phi_{M^{\prime}}(x)\\}$ is
$S$-invariant. By minimality, we only have to prove it is nonempty. We will
show it contains any $\sigma_{L}$-fixed point $\bar{x}$. Since in the previous
part, we have shown that the fixed points are preserved under the maps
$\phi_{M}$, we only have to check that the images under the two maps have the
same $\bm{0}$ coordinate.

Let $n_{0}$ be the integer associated with $M$ such that the transformation
$M_{p}$ coincides mod $L^{p}(\mathbb{Z}^{d})$ for $p\geq n_{0}$. Define
$n^{\prime}_{0}$ similarly for $M^{\prime}$. It is direct to check that the
integer $\max(n_{0},n^{\prime}_{0})$ plays a similar role for $MM^{\prime}$.
By definition we have
$\phi_{M^{\prime}}(\bar{x})_{\bm{0}}=M^{\prime}_{n^{\prime}_{0}}\bar{x}_{\bm{0}}\
(\text{mod }L(\mathbb{Z}^{d}))$ and
$\phi_{M}\circ\phi_{M^{\prime}}(\bar{x})_{\bm{0}}=M_{n_{0}}M^{\prime}_{n^{\prime}_{0}}\bar{x}_{\bm{0}}\
(\text{mod }L(\mathbb{Z}^{d}))$. Now, a direct computation gives that
$(MM^{\prime})_{\max(n_{0},n^{\prime}_{0})}=M_{n_{0}}M^{\prime}_{n^{\prime}_{0}}\
(\text{mod }L(\mathbb{Z}^{d}))$ and this shows the claim, i.e., the two images
have the same ${\bm{0}}$ coordinate.

To prove that the morphism is injective, consider a matrix $M$ in its kernel,
i.e., such that $\phi_{M}=\textrm{Id}$. Composing this relation with the shift
map $S^{\bm{z}}$, with $\bm{z}\in\mathbb{Z}^{d}$, and since $\phi_{M}$ is an
$M$-endomorphism, we get that $S^{M\bm{z}}=S^{\bm{z}}$ for any
$\bm{z}\in\mathbb{Z}^{d}$. By aperiodicity of the subshift, $M$ has to be the
identity matrix. ∎

Theorem 4.1 resumes all the former results. In particular, if the expansion
matrix $L$ is a multiple of the identity, then
$\vec{N}(X_{\sigma_{L}},S)=\operatorname{Cent}_{\textrm{GL}(d,\mathbb{Z})}(L)=\textrm{GL}(d,\mathbb{Z})$.
As a consequence, we get the following direct corollary:

######  Corollary 4.8.

The normalizer semigroup of the half-hex substitution $N(X_{hh},S)$ is a group
and it is isomorphic to a semidirect product between $\mathbb{Z}^{2}$ and
$\textrm{GL}(2,\mathbb{Z})$. Moreover, its automorphism group
$\operatorname{Aut}(X_{hh},S)$ is trivial.

This implies that the half-hex substitutive subshift is a minimal subshift
with an infinite linear representation group. In fact, since
$\vec{N}(X_{hh},S)=\textrm{GL}(2,\mathbb{Z})$, its linear representation group
is the largest possible. As an other example, consider the matrix
$L_{6}=\begin{pmatrix}2&0\\\ 0&4\end{pmatrix}$. By Theorem 3.3, we have that
$\vec{N}(\overleftarrow{\mathbb{Z}^{2}}_{(L_{6}^{n})})=\textrm{GL}(2,\mathbb{Z})$,
but Theorem 4.1 and a standard analysis provide that
$\vec{N}(X_{\sigma_{L_{6}}},S)$ is the set of matrices
$\left\\{\begin{pmatrix}a&2b\\\ 0&d\end{pmatrix}\colon
a,d\in\\{-1,1\\},b\in\mathbb{Z}\right\\}$. In particular,
$\vec{N}(X_{\sigma_{L_{6}}},S)$ is virtually $\mathbb{Z}$: its quotient by the
group generated by the matrix $\begin{pmatrix}1&2\\\ 0&1\end{pmatrix}$ is
finite.

It is then natural to wonder what is the collection of all the groups
$\vec{N}(X_{\sigma_{L}},S)$ that appear for all the matrices $L$ and in
particular if any subgroup of $\textrm{GL}(2,\mathbb{Z})$ can be realized like
this. This question can be very difficult because it requires precise control
of the combinatorics which can be difficult to manage. A more manageable way
could be by realizing linear representation groups of specific odometers (see
Question 2.10). With this, we may expect to get an answer of the following:

######  Question 4.9.

Does there exist for any odometer system
$(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$ an almost 1-to-1 Toeplitz
extension $(X,S,\mathbb{Z}^{d})$ such that
$\vec{N}(X,S)=\vec{N}(\overleftarrow{\mathbb{Z}^{d}}_{(Z_{n})})$?

Together with Question 2.10 this will enable us to enrich that family of
examples with a large linear representation group.

In the more restrictive class of subshifts given by the finite data of a
constant-shape substitution, it is natural to ask whether the elements of the
normalizer are computable. There is a body of evidence indicating that their
automorphisms can be described by an algorithm. But, as illustrated by the
characterization in Theorem 4.1, nothing is clear concerning the elements of
the linear representation group. Related to Question 3.8, we ask the
following:

######  Question 4.10.

Regarding the linear representation group for substitutive constant-shape
subshifts, are its elements computable? Is its group structure computable?

Here we mean “computable elements” in the sense that there is an algorithm
deciding whether or not a matrix $M$ belongs to the linear representation
group. The second question is to find an algorithm for determining the linear
representation group, up to isomorphism, as a function of the substitution.

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