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ADMISSIBLE REVERSING AND EXTENDED SYMMETRIES FOR

BIJECTIVE SUBSTITUTIONS

´

ALVARO BUSTOS, DANIEL LUZ, AND NEIL MA˜

NIBO

Abstract. In this paper, we deal with reversing and extended symmetries of shifts generated

by bijective substitutions. We provide equivalent conditions for a permutation on the alphabet

to generate a reversing/extended symmetry, and algorithms how to check them. Moreover, we

show that, for any ﬁnite group Gand any subgroup Pof the d-dimensional hyperoctahedral

group, there is a bijective substitution which generates an aperiodic hull with symmetry group

Zd×Gand extended symmetry group (ZdoP)×G.

1. Introduction

The study of symmetry groups, often also known as automorphism groups, is an important

part of the analysis of a dynamical system, as it can oﬀer insight on the behaviour of the system,

as well as allowing classiﬁcations of distinct families of dynamical systems (acting as a conjugacy

invariant). In particular, symmetry groups of shift spaces have been thoroughly studied (see e.g.

the analysis of the symmetry group of the full shift [BLR88], the series of works on symmetries

in low-complexity subshifts [CK16,CQY16,DDMP16], and recent works on shifts of algebraic

and number-theoretic origin [BBH+19,FY18]).

Symmetries of subshifts can be algebraically deﬁned as elements of the topological centraliser

of the group hσigenerated by the shift, seen as a subgroup of the space Aut(X) of all self-

homeomorphisms of Xonto itself. Thus, a natural question at this point is whether the cor-

responding normaliser has an interesting dynamical interpretation as well. This leads to the

concept of reversing symmetries (for d= 1); see [BR06,Goo99,BRY18], the monograph [OS15]

for a group-theoretic exposition, and [LR98] for a more physical background. These are special

types of ﬂip conjugacies; see [BM08]. In higher dimensions, one talks of extended symmetries;

see [Baa18,BRY18], which are examples of GL(d, Z)-conjugacies; compare [Lab19,BBH+19].

These kinds of maps are related to phenomena such as palindromicity and several proper-

ties of geometric and topological nature, which is more evident in the higher-dimensional set-

ting [BRY18,Bus20].

High complexity is often (but not always, see for instance the square-free subshift [BBH+19])

linked to a complicated symmetry group. For instance, determining whether the symmetry

groups of the full shifts in two and three symbols are isomorphic has consistently proven to

be a diﬃcult question [BLR88]. The low-complexity situation, thus, often allows for a more

in-depth analysis and more complete descriptions, up to and including explicit computation of

these groups in many cases.

The particular case of substitutive subshifts has gathered signiﬁcant attention and here a

lot of progress has been made; see [MY21,KY19]. Unsurprisingly, the presence of non-trivial

symmetries is also tied to the spectral structure of the underlying dynamical system; see [Que10,

2010 Mathematics Subject Classiﬁcation. 52C23, 37B10, 37B52, 20B27.

Key words and phrases. Extended symmetries, automorphism groups, substitution subshifts, aperiodic tilings.

1

Fra05]. In this work, we restrict to systems generated by bijective substitutions, both in one

and in higher dimensions. These substitutions are typically n-to-1 extensions of odometers and

generate coloured tilings of Zdby unit cubes, where one usually identiﬁes a letter with a unique

colour; see [Fra05]. We compile and extend known properties about this family of substitutive

subshifts regarding symmetries. Some natural questions in this direction are:

(1) What kinds of groups can appear as symmetry groups/extended symmetry groups of

speciﬁc substitutive subshifts?

(2) Given a speciﬁc group G, can we construct a substitution whose associated subshift has

Gas its symmetry group/extended symmetry group?

Both questions are accessible for bijective substitutions. For symmetry groups, the second

question is answered in full in [DDMP16], which extends to higher dimensions with no addi-

tional assumptions because the result does not depend on the geometry of the substitution;

see [CP20] for realisation results for more general group actions. We add to such known results

in Theorem 14. Aperiodicity also plays a key role here, which can easily be conﬁrmed in the

bijective setting; see Propositions 5and 33.

On the other hand, the existence of non-trivial reversing or extended symmetries depends

heavily on the geometry and requires more in terms of the relative positions of the permu-

tations in the corresponding supertiles, the expansive maps, and the shape of the supertiles

themselves. In Theorem 22, we provide equivalent conditions for the existence of non-trivial re-

versing symmetries, which we generalise to higher dimensions in Theorem 30 to cover extended

symmetries.

As a corollary, in any dimension d, given a ﬁnite group Gand a subgroup Pof the hy-

peroctahedral group P, we provide a construction in Theorem 35 of a bijective substitution

whose underlying shift space has symmetry group and extended symmetry group Zd×Gand

(ZdoP)×G, respectively. A similar construction with a diﬀerent structure of the extended

symmetry group is done in Theorem 41. We also provide algorithms on how one can check

whether there exist non-trivial symmetries and extended symmetries for a given substitution ;

see Sections 2.2 and 3.1.

2. Bijective constant-length substitutions

2.1. Setting and basic properties. Let Abe a ﬁnite alphabet and A+=SL>1ALbe the set

of ﬁnite non-empty words over A; we shall write A∗=A+∪ {ε}, where the latter is the empty

word. A substitution is a map :A→A+. If there exists an L∈Nsuch that (a)∈ ALfor

all a∈ A,is called a constant-length substitution. If there exists a power ksuch that k(a)

contains all letters in A, for every a∈ A, we call primitive.

The full shift is the set AZof all functions (conﬁgurations) x:Z→ A. More generally, we

deﬁne the d-dimensional full shift as the set AZd. To this space, we assign the product topology,

giving Athe discrete topology. This is a particular version of the local topology used in tiling

spaces and discrete point sets, in which two tilings (or point sets) xand yare said to be ε-close

if a small translation of x(of magnitude less than ε) matches yon a large ball (of radius at least

1/ε) around the origin; this can be used to deﬁne a metric d(x, y). In the particular case of shift

spaces seen as tiling spaces, since tiles are aligned with Zd, we can disregard the translation and

get, e.g., the following as an equivalent metric:

d(x, y)=2−infnn:x|[−n,n]d6=y|[−n,n]do.

2

This space is endowed with the shift action of Zdon AZd, which is the action of Zdover

conﬁgurations by translation, and can be deﬁned via the equality (σn(x))m=xn+mfor all

x∈ AZd,m,n∈Zd(in particular, in one dimension we have the shift map σ=σ1, which

completely determines the group action).

Asubshift is a topologically closed subset X⊆ AZdwhich is also invariant under the shift

action. Thus, a subshift combined with the restriction of this group action to Xdeﬁnes a

topological dynamical system, which can be endowed with one or more measures to obtain a

measurable dynamical system. In the one-dimensional case, the language (or dictionary ) of a

subshift Xis the set of all words that may appear contained in some x∈X, that is:

L(X) = {x|[0,n]:x∈X, n >0}∪{ε}.

We may verify that any nonempty set of words Lwhich is extensible (that is, any w∈ L

is a subword of a longer word w0∈ L) and closed under taking subwords is the language of a

subshift, and two subshifts are equal if and only if they share the same language.

Higher-dimensional subshifts have a similar combinatorial characterisation, where the role

of words is taken by patterns, ﬁnite conﬁgurations of the form P:U⊂Zd→ A,|U|<∞; we

identify a pattern with any of its translations. In most cases1(and, in particular, in the rest of

this work), it makes no diﬀerence to allow arbitrary “shapes” Uor to restrict ourselves to only

rectangular patterns, i.e. products of intervals of the form U=Qd

i=1[0, ni−1]. Regardless of

our chosen convention, we collect all valid patterns x|Uthat appear in some x∈Xinto a set

L(X) as above, which we once again call the language of X. As in the one-dimensional case, a

language closed under taking subpatterns and where every pattern of shape Uis contained in

a pattern of shape V⊃Ufor any larger (ﬁnite) Vdeﬁnes a unique subshift, and vice versa.

Thus, given that iterating a primitive substitution :A → ALof constant length L > 1

over a symbol a∈ A produces words of increasing length, the set L%of all words that are

subwords of some k(a) for some k>1 and a∈ A is the language of a unique subshift that

depends only on , which we shall call the substitutive subshift deﬁned by and denote by X%.

This deﬁnition extends to d-dimensional rectangular substitutions :A→AR(where Ris a

product of intervals), which are higher-dimensional analogues of constant-length substitutions;

see [Fra05,Que10,Bar18]. It is well known that the primitivity of implies that X%is strictly

ergodic (uniquely ergodic and minimal); see [Que10,BG13]. We refer the reader to [MR18] for

a treatment of substitutions which are non-primitive.

Deﬁnition 1. A constant-length substitution :A→ALis called bijective if the map which

is given by j:a7→ (a)jis a bijection on A, for all indices 0 6j6L−1. Equivalently,

is bijective if there exist L(not necessarily distinct) bijections 0, . . . , L−1:A → A such that

(a) = 0(a). . . L−1(a) for every a∈ A. We shall refer to the mapping jas the j-th column

of the substitution .

Consider jL−1

j=0 ⊂S|A|. Let Φ : S|A| →GL(|A|,Z) be the representation via permutation

matrices. One then has the following; compare [Fra05, Cor. 1.2].

1Pattern shapes do matter when studying certain generalisations of topological mixing in the d-dimensional

setting, where either restricting ourselves to speciﬁc shapes (rectangles, L-shapes, hollow rectangles, etc.) or

allowing arbitrary ones may be preferrable depending on context. However, we are not concerned with these

kinds of properties here.

3

Fact 2. Let be a primitive, bijective substitution, whose columns are given by 0, . . . , L−1.

Then the substitution matrix Mis given by M=PL−1

j=0 Φ(−1

j). Moreover, (1,1,...,1)Tis

a right Perron–Frobenius eigenvector of M, so each letter has the same frequency for every

element in the hull X%, i.e., νa=1

|A| for all a∈ A and all x∈X%.

Deﬁne the n-th column group G(n)to be the following subgroup of the symmetric group of

bijections A → A:

G(n):= h{j1◦ · · · ◦ jn|06j1, . . . , jn6L−1}i.

As it turns out, the groups G(n)generated by the columns give a good description of the

substitution in the bijective case; see [KY19] for its relation to the corresponding Ellis semi-

group of X%. The primitivity of may be characterised entirely by this family of groups, as

seen below. Recall that a subgroup G6Snof the symmetric group on {1, . . . , n}is transitive

if for all 1 6j, k 6nthere exists τ∈Gsuch that τ(j) = k. Here, we let N∈Nbe the minimal

power such that N

j= id for some 0 6j6LN−1; compare [Que10, Lem. 8.1]. In [KY19], G(N)

is called the structure group of .

Proposition 3. Let :A→ALbe a bijective substitution. Then, the following are equivalent:

(1) The substitution is primitive.

(2) All groups G(n), n ∈N, are transitive.

(3) The group G(N)is transitive.

Proof. Evidently, (2) =⇒(3), so we only need to prove (3) =⇒(1) =⇒(2). To see the ﬁrst

implication, note ﬁrst that the columns of the iterated substitution Nare compositions of the

form j1,...,jN:= j1◦ · · · ◦ jN,06j1, . . . , jN6L−1, that is, for any a∈ A the following holds:

N(a) = 0,...,0,0(a)0,...,0,1(a). . . 0,...,0,L−1(a)0,...,1,0(a). . . L−1,...,L−1,L−1(a).

Since, by (3), the group G(N)is transitive, the substitution matrix M%Nis irreducible, i.e. it is

the adjacency matrix of a strongly connected digraph. In other words, for all a, b ∈ A, there

exists a composition of columns q, q 0, .. . , q00 of Nsuch that q◦q0◦ · · · ◦ q00(a) = b, which may be

identiﬁed with a path in the graph whose vertices are the letters of Aand with one edge from

cto r(c) for any c∈ A and column r. The choice of Nalso shows that M%Nhas a non-zero

diagonal, since one of the columns of Nis the identity. These two conditions immediately

imply that M%Nis a primitive matrix (see [LM95, Ch. 2]) which in turn implies primitivity of

, as desired.

To prove (1) =⇒(2), note that primitivity of implies that, for some k > 0 and for

all a∈ A, the word k(a) contains all symbols of the alphabet A, including aitself. Since

the columns of kgenerate G(k), this implies that for all a, b ∈ A there is some generator of

this group that maps ato b, i.e. G(k)is transitive. Since k(a) contains aas a subword, this

implies that 2k(a) contains k(a) as a subword, and, by induction, that mk(a) contains k(a)

as a subword for all m>1; thus, all groups G(mk)are transitive. Now, it is easy to see that

G(n)6G(d)if d|n. Then, for all n∈N,G(n)has G(nk)as a transitive subgroup and hence it

is transitive.

The bijective structure of can also be exploited to conclude the aperiodicity of X%by just

looking at simple features of . Below, we provide several criteria for aperiodicity in terms of

|A|,L, and the existence of certain legal words.

4

Proposition 4. Let X%be the hull of a primitive, bijective substitution of length Lon a ﬁnite

alphabet A. If X%is periodic with least period p, then it has to satisfy the following conditions:

(1) |A| divides p

(2) Ldoes not divide |A|.

Proof. From Fact 2we know that every letter has the same frequency. An element of a periodic

shift is just a concatenation of its periods and thus every letter has the same frequency in every

period. This is only possible if every letter appears equally often in the period and thus the

period length has to be a multiple of the alphabet size, which settles the ﬁrst claim.

Let us assume that w∞is a periodic word with pas least period. Without loss of generality,

choose a power ksuch that its ﬁrst column is the identity, and so w∞is ﬁxed by k. We choose

cand dminimal such that:

cL =dp ⇐⇒ cb|A| =da|A|.(1)

We apply −1to w∞|[0,cL−1] of length cL, which has a unique pre-image in Lsince is injective

on letters, i.e., (a) = (b) if and only if a=b. Then this segment must be of the form x1,· · · xc.

Applied to w∞, it yields · · · xcx1,· · · xcx1· · · , which means that w∞is c-periodic. Since pis the

least period c=ep but then c=ea|A|, making the factor |A| in Eq. (1) redundant, and thus

contradicting the minimality of c.

Another way to get aperiodicity is through the existence of proximal pairs; see [DDMP16,

Sec. 3.2.1] and [BG13, Cor. 4.2 and Thm. 5.1]. Two elements x6=y∈(X, σ) are said to

be proximal if there exists a subsequence {nk}of Nor −Nsuch that d(σnkx, σnky)→0 as

k→ ∞. A stronger notion is that of asymptoticity, which requires d(σnx, σny)→0 as n→ ∞

or −∞. For bijective substitutions, these two notions are equivalent, and asymptotic pairs are

completely characterised by ﬁxed points of ; see [KY19].

Consider a one-dimensional substitution and a ﬁxed point warising from a legal seed a|b,

i.e., w=∞(a|b). Here, the vertical bar represents the location of the origin, and the letter

agenerates all the letters at the negative positions, while bdoes the same for all non-negative

ones. Two ﬁxed points w1, w2∈X%generated by a1|b1and a2|b2are left-asymptotic if they agree

at all negative positions and disagree for all non-negative positions. Right-asymptotic pairs are

deﬁned in a similar manner. We have the following equivalent condition for aperiodicity in

terms of existence of certain legal words; compare [KY19, Prop. 4.1]

Proposition 5. Let be a primitive, bijective substitution on a ﬁnite alphabet Ain one di-

mension. Then the hull X%is aperiodic if and only if there exist distinct legal words of length 2

which either share the same starting or ending letter.

Proof. Let := 0· · · L−1, with i∈G. Choosing k= lcm(|0|,|L−1|), we get that the ﬁrst

and the last columns of kare both the identity, i.e., k

0(a) = k

Lk−1(a) = afor all a∈ A. If

there exist ab, ac ∈ L%with b6=c, the bi-inﬁnite ﬁxed points ∞(a|b) and ∞(a|c) they generate

under kcoincide in all negative positions and diﬀer in at least one non-negative position, and

hence are left-asymptotic and proximal. Since X%is minimal and admits a proximal pair, all of

its elements must then be aperiodic. Now suppose that every letter has a unique predecessor

and successor in A. This means that every element x∈X%is uniquely determined by the letter

at the origin. From the ﬁniteness of A, one gets x=w∞and hence is periodic, from which the

periodicity of the hull follows.

5

Example 6. The substitution :a7→ aba, b 7→ bab is primitive, bijective and admits a periodic

hull. Here, the only legal words of length 2 are ab and ba. Note that is of height 2 and

generates the same hull as the substitution 0:a, b 7→ ab.♦

2.2. Symmetries. In the following sections, we deal with the symmetry groups of our subshifts

of interest, which are certain homeomorphisms of the shift space which preserve the dynamics

of the shift action in a speciﬁc sense.

Deﬁnition 7. Let Xbe a Zd-subshift. The symmetry group (often called automorphism group2)

is the set S(X) of all homeomorphisms X→Xwhich commute with the shift action, i.e.,

(2) (∀n∈Zd): σn◦f=f◦σn.

That is, S(X) is the centraliser of the set of shift maps in the group of all self-homeomorphisms

of the space X. In this context, every symmetry f∈ S(X) is entirely determined by its local

function, which is a mapping F:AU→ A, with U⊂Zdﬁnite, such that for every n∈Zd,

f(x)n=F(x|n+U). This fact is known as the Curtis–Hedlund–Lyndon (or CHL) theorem;

see [LM95]. We say that fhas radius r>0 if this is the least non-negative integer such that

U⊆[−r, r]d.

Symmetry groups of one-dimensional bijective substitutions are a thoroughly studied sub-

ject, both in the topological and ergodic-theoretical contexts. Complete characterisations of

these groups are known, as seen in e.g. [Cov71] for a two-symbol alphabet, or [LM88] for a

characterisation in the measurable case; see also [Fra05,CQY16] for further elaboration in the

description of the symmetries in this category of subshifts. The following theorem summarizes

this classiﬁcation:

Theorem 8. Let X%be the hull generated by an aperiodic, primitive, bijective substitution

on Zd. Then, the symmetry group S(X%)is isomorphic to the direct product of Zd, generated

by the shift action, with a ﬁnite group of radius-0sliding block codes τ∞:X%→X%given by

τ∞((xj)j∈Zd)=(τ(xj))j∈Zdfor some bijection τ:A→A.

Furthermore, let Nbe any integer such that N

jis the identity for some j(note that such an

Nalways exists). Then, τ:A → A induces a symmetry if and only if τ∈centS|A|G(N).

As a consequence, every symmetry on X%is a composition of a shift map and a radius-zero

sliding block code as above. These conditions arise as a consequence of such a symmetry having

to preserve the supertile structure of any x∈X%at every scale, which in particular implies

that a level-ksupertile k(a), a ∈ A has to be mapped to some k(b) for some other b∈ A by

the “letter exchange map” τ. The choice of Nabove ensures that, when kis a multiple of N,

the equality a=bholds, which implies that τcommutes with the columns of N, and thus

N◦τ∞=τ∞◦N. This in turn implies Eq. (2). For further elaboration on the proof of the

above result, the reader may consult [Fra05,CQY16], among others.

Example 9. Consider the following substitution on the three-letter alphabet A={a, b, c}:

:a7→ abc,

b7→ bca,

c7→ cab.

2In this work, we follow the notational conventions of [BRY18], and thus we avoid the term “automorphism

group” as it may be understood as the set of all homeomorphisms f:X→X.

6

The columns correspond to the three elements of the cyclic group generated by τ= (a b c). It

is not hard to verify that the only elements of S3=D3that commute with τare the powers

of τthemselves, and thus S(X%)'Z×C3, with the ﬁnite subgroup C3being generated by the

symmetries induced by the powers of τ.♦

As it turns out, Theorem 8provides an algorithm to compute S(X%) explicitly. To introduce

this algorithm, let us recall some easily veriﬁable facts from group theory [Hall76, Ch. 1 and 5]:

Fact 10. Let Gbe any group and H=hSi6Ga subgroup generated by S⊂G. Then,

centG(H) = {c∈G|(∀h∈H): ch =hc}=\

s∈S

centG(s).

Fact 11. Any permutation decomposes uniquely (up to reordering) as a product of disjoint

cycles. Conjugation by some τ∈Sncan be computed from this decomposition using the identity:

τ(a1a2. . . an)τ−1= (τ(a1)τ(a2). . . τ (an)).

A permutation τ∈Snbelongs to centSn(π)if and only if τ πτ −1=π, and thus:

π= (a1a2. . . ak1)(b1b2. . . bk2)· · · (c1c2. . . ckr)

= (τ(a1)τ(a2). . . τ (ak1))(τ(b1)τ(b2). . . τ (bk2)) · · · (τ(c1)τ(c2). . . τ (ckr)).

Hence, the uniqueness of this decomposition implies that every cycle in the second decomposition

is equal to a cycle of the same length in the ﬁrst one.

Thus, to compute the letter exchange maps that determine S(X%), we need to ﬁnd all per-

mutations τthat preserve certain cycle decompositions. We obtain the following procedure:

Algorithm. Assuming that is a primitive, bijective, aperiodic substitution, the following algorithm

computes S(X%) explicitly.

•Input: is a length-Lbijective substitution, which may be represented as a function (dictio-

nary) :A → ALor a set of Lpermutations 0, 1, . . . , L−1:A → A, corresponding to each

column.

•Output: A (ﬁnite) set of permutations Cforming a group, so that S(X%) = Zd×C.

(1) Compute the least positive integer Nsuch that N

jis the identity on Afor some column of the

substitution .Nequals the least common multiple of all cycle lengths in the decomposition of

the columns jinto disjoint cycles (and is thus ﬁnite).

(2) Determine all columns j1◦ · · · ◦ jNof the iterated substitution N. This is a generating set

for the group G(N).

(3) For every column computed in (2), compute Gj1,...,jN= centSn(j1◦ ·· · ◦ jN) by taking the

cycle decomposition of this permutation (in where we identify Awith the set {1,2,...,|A|})

and employing the characterisation above.

(4) Let C=Tj1,...,jnGj1,...,jN. As Ccan be biunivocally identiﬁed with the set of valid letter

exchange maps modulo a shift, return S(X%) = Zd×Cas output.

Example 9above corresponds to a simple case in which G(N)=G(1) is a cyclic group, and

we derive an abelian subgroup of S3corresponding to the valid letter exchange maps. We can

use the above procedure to construct examples with more complicated symmetry groups, see

Example 12.

7

Example 12. We take as alphabet the quaternion group Q={e, i, j, k, ¯e,¯ı, ¯, ¯

k}(see [Hall76]

for the multiplication table and basic properties of this group, which is generated by the two

elements iand j). With this, we construct a length-3 bijective substitution deﬁned by right

multiplication, x7→ (x·i)(x·j)(x·k), given in full by:

e7→ ijk, ¯e7→ ¯ı¯¯

k,

i7→ ¯ek¯, ¯ı7→ e¯

kj,

j7→ ¯

k¯ei, ¯7→ ke¯ı,

k7→ j¯ı¯e, ¯

k7→ ¯ie.

By direct computation, G(n)=G(1) 'Qfor all n, making the substitution primitive (as Qacts

transitively on itself in an obvious way). The three columns which generate G(1) are:

Ri:= (e i ¯e¯ı)(j¯

k¯ k),

Rj:= (e j ¯e¯)(i k ¯ı¯

k),

Rk:= (e k ¯e¯

k)(j i ¯¯ı).

Thus, the substitution 3has as columns Rxyz (g) = g·xyz with x, y, z ∈ {i, j, k}; in particular,

since jik =e,3must have an identity column. Also, since G(3) =G(1) , this group is the right

Cayley embedding of Qinto S8. By applying the above algorithm, we obtain that the group of

letter exchange maps is generated by the following two permutations:

π0:= (e i ¯e¯ı)(j k ¯

j¯

k),

π1:= (e j ¯e¯)(i¯

k¯

i k).

We can verify that these permutations generate the left Cayley embedding of Qinto S8. Al-

ternatively, if we consider the transposition ν= (k¯

k), we can use Fact 11 above to see that

π0=νRiν−1and π1=ν Rjν−1, which in turn implies that the group generated by π0and π1is

conjugate to the group generated by Riand Rj, the latter being isomorphic to Q. This shows

that S(X%)'Z×Q.♦

It is well known that symmetry groups of aperiodic minimal one-dimensional subshifts are

virtually Z. The following result gives a full converse for shifts generated by bijective substitu-

tions.

Theorem 13 ( [DDMP16, Thm. 3.6]).For any ﬁnite group G, there exists an explicit primitive,

bijective substitution , on an alphabet on |G|letters, such that S(X%)'Z×G.

The proof, which may be consulted in [DDMP16], follows a similar schema to the analysis

done in Example 12 above. In [Fra05, Sec. 4.1], it was shown that the number of letters needed

in Theorem 13 is actually a tight lower bound. Below, we actually prove something stronger.

Theorem 14. Let be an aperiodic, primitive, bijective substitution on the alphabet A. If

S(X%)'Z×G, then Gmust act freely on A, and the order of Ghas to divide |A|.

Proof. As seen in [Fra05, Sec. 4.1], if we replace with a suitable power, we may ensure that

the word q(a) starts with aand contains every other symbol, for all a∈ A. Thus, for any

π∈Sn, the equality π(a) = bimplies π(q(a)) = q(b), which in turn determines the images of

every symbol in the alphabet; the bound |G|6|A| follows from here.

8

Note as well that, since is bijective, if π(a)6=a, then π(c)6=cfor every c∈ A as the words

q(a) and q(b) are either equal or diﬀer at every position. This implies that if πhas any ﬁxed

point then it must be the identity, i.e. that, if we identify Gwith the corresponding group

of permutations over A, the action of Gon the alphabet is free. Equivalently, the stabiliser

Stab(c) of any c∈ A is the trivial subgroup.

The elements of Gcommute with every column of q. Due to primitivity, there always exists

a column ∗=j1◦ · · · ◦ jqwhich maps this ato any desired c∈ A. Since ∗commutes

with every π∈G(i.e., it is an equivariant bijection for the action of Gon A), we have that

Orb(c) = ∗[Orb(a)], i.e. the orbit of cunder Gis necessarily the image of the orbit of aunder

∗.

Thus, every orbit is a set of the same cardinality. This means that Ginduces a partition of

Ainto disjoint orbits of the same cardinality , which then must divide |A|. By the freeness of

the group action and the orbit-stabiliser theorem, we have |G|=|Orb(a)|·|Stab(a)|=, and

thus |G|divides |A|.

Remark 15. It follows from Theorem 14 that the substitution in Example 12 is a minimal one

in the sense that for one to get a Q-extension in S(X%), one needs at least eight letters. ♦

Remark 16. At no point in the proof of Theorem 13 found in [DDMP16] nor in Theorem 14

above the fact that the substitution was one-dimensional is actually used. Thus, since Theorem

8is known to be valid for general rectangular substitutions, the two theorems above must be

valid in this more general setting as well, provided that the substitution is aperiodic in Zd,

which one can always guarantee; see Propositions 5and 33.♦

Corollary 17. For any ﬁnite group G, there exists an explicit primitive, bijective d-dimensional

rectangular substitution , on an alphabet of |G|letters, such that S(X%)'Zd×G. Furthermore,

this is the least possible alphabet size: for any bijective, primitive and aperiodic d-dimensional

rectangular substitution on the alphabet A, if S(X%)'Zd×G, then Gacts freely on A, and

|G|divides |A|.

3. Extended and reversing symmetries of substitution shifts

3.1. One-dimensional shifts. Since the term symmetry group does not cover everything that

can be thought of as a symmetry (in the geometric sense of the word) we introduce the notion of

the reversing symmetry group; see [BRY18] for a detailed exposition. We will exclusively look

at shift spaces X%which are given by a bijective, primitive substitution and we will exploit

this additional structure in determining the reversing symmetry group for this class.

Deﬁnition 18. The extended symmetry group of a shift space Xis given as

R(X) := normAut(X)(G) = {H∈Aut(X)|HG=GH}

where Gis the group generated by the shift. In the case where the shift space is one-dimensional,

we call R(X) the reversing symmetry group given by

R(X) = {H∈Aut(X)|H◦σ◦H−1=σ±1}.

A homeomorphism H∈Aut(X) which satisﬁes H◦σ◦H−1=σ−1is called a reversor or a

reversing symmetry. A Curtis–Hedlund–Lyndon-type characterisation of reversing symmetries,

which incorporates the mirroring component (GL(d, Z)-component in higher dimensions) can

be found in [BRY18].

9

In what follows, we investigate the eﬀect of a reversor fon inﬂated words. Given a bijective

substitution :A → AL, := 01· · · L−1, the mirroring operation macts on the columns of

via m((a)) = L−1(a)· · · 2(a)0(a). We may extend this to inﬁnite conﬁgurations over Zin

two non-equivalent ways, given by m(x)k=x−kand m0(x)k=x1−k, respectively; we shall refer

to both as basic mirroring maps.

Proposition 19. Let be an aperiodic, primitive, bijective substitution. Then, any reversor is

a composition of a letter exchange map π∈Sn, where n=|A|, a shift map σkand one of the

two basic mirroring maps mor m0(depending only on whether the substitution has odd or even

length, respectively).

See [BRY18, Prop. 1] and Theorem 8. This result, while desirable, is not immediately obvious

(and can indeed be false for non-bijective substitutions, which may have reversors whose local

functions have positive radius), and thus we show this result as a consequence of bijectivity.

Proof. Suppose f:X%→X%is a reversor of positive radius r>1, i.e., x|[−r,r]=y|[−r,r]implies

that one has f(x)0=f(y)0. There is some power k>1 such that the words k(a) of length Lk

are longer than the local window of f, which has length 2r+ 1 (say, k=dlog(2r+ 1)/log(L)e).

Any point of X%is a concatenation of words of the form k(a), a ∈ A, which is unique up to

a shift because of aperiodicity; see [Sol98]. In particular, if we choose a ﬁxed x∈X%and let

y=f(x), both points have such a decomposition.

Now, suppose that the value Lk= 2+ 1 is odd (the case where Lis even is dealt with

similarly). By composing fwith an appropriate shift map (say ˜

f=f◦σh), we can ensure that

the central word k(a) in the aforementioned decomposition has support [−, ] for both xand

y(note that we employ the uniqueness of the decomposition here, to avoid ambiguity in the

chosen h). Since Lk= 2+ 1 >2r+ 1, we must have >r, and thus y0is entirely determined

by x|[−`,`], which is a substitutive word k(a). But, since is bijective, this word is in turn

completely determined by its central symbol x0.

Figure 1. A reversor festablishes a 1-1 correspondence between words k(a)

in a point xand its image f(x).

A similar argument shows that, for any n∈Z, if n∈mLk+ [−, ], then yndepends only

on the word x|−mLk+[−`,`], which contains (and is thus entirely determined by) x−n. Since any

point in X%is transitive, ˜

fis entirely determined by the points xand y, and thus, ˜

fis a map

of radius 0. Equivalently, for some bijection π:A → A, we have ˜

f(x)−n=π(xn), that is,

˜

f=f◦σh=π◦m(identifying πwith the letter exchange map AZ→ AZ). We conclude that

fis a composition of a letter exchange map, a mirroring map and a shift map.

Remark 20. With some care, it can be shown that the same argument applies in the higher-

dimensional case, where an element of the normaliser is a composition of a letter exchange map,

a map of the form f(x)n=xAn, with Aa linear map from the hyperoctahedral group (see

Theorem 26, below), and a shift map; see [BRY18, Prop. 3] for a more general formulation. ♦

10

This result leads to the following criterion for the existence of a reversor in terms of the

columns i.

Proposition 21. Let be an aperiodic, primitive, bijective substitution of length Lon a ﬁnite

alphabet Aof nletters. Suppose that there exists a letter-exchange map π∈Sn, π :A→A

which gives rise to a reversing symmetry. Then one has

(3) π−1◦i◦−1

j◦π=L−(i+1) ◦−1

L−(j+1)

for all 06i, j 6L−1, where iis the i-th column of seen as an element of Sn.

Proof. Let a∈ A. Let mbe the mirroring operation and suppose that there exists π∈Snsuch

that m◦πextends to a reversor f∈ R(X%). One then has

(a) = 0(a)· · · L−1(a)m

7−→ L−1(a)· · · 0(a)π

7−→ π◦L−1(a)· · · π◦0(a).

Since Proposition 19 guarantees that this must result to mapping substituted words to substi-

tuted words, one gets

(4) π◦L−1(a)· · · π◦0(a) = 0(b)· · · L−1(b) = 0◦τ(a)· · · L−1◦τ(a),

where the permutation τdescribes precisely this induced shuﬄing of inﬂation words. This yields

τ=−1

j◦π◦L−(j+1)

for all 0 6j6L−1. Equating the corresponding right hand-sides for some pair i, j yields

Eq. (3). The claim follows since this must hold for all 0 6i, j 6L−1.

Theorem 22. Let be as in Proposition 21. Suppose further that i=L−(i+1) = id for some

06i6L−1. Then, given a permutation (letter exchange map) π∈Sn, π :A → A, the

following are equivalent:

(i) The letter exchange map πgives rise to a reversing symmetry f∈ R(X%)\ S(X%)given

by either f(x)n=π(x−n)or f(x)n=π(x1−n).

(ii) The permutation πsatisﬁes the system of equations

(5) π−1◦i◦π=L−(i+1)

for all 06i6L−1.

(iii) There exist κ0, κ1, . . . , κL−1∈Sn, where each κisatisﬁes κ−1

i◦i◦κi=L−(i+1), such

that the following intersection of cosets is non-empty:

(6) K=

L−1

\

i=0

centSn(i)κi,

and π∈K.

Proof. It is clear that Eq. (5) implies Eq. (3). Note that it is suﬃcient to satisfy Eq. (3) for

j=i+ 1 mod Las any term can be obtained by multiplying suﬃcient numbers of succeeding

terms. Under the extra assumption that there exist a column pair which is the identity, Eq. (3)

simpliﬁes to Eq. (5). This shows that (i) =⇒(ii).

For the other direction, we show that if Eq. (5) is satisﬁed at by the level-1 inﬂation words,

then these sets of equations must also be fulﬁlled by any power kof . Remember that,

from any arbitrary bijective substitution , we may derive another bijective substitution 0

that satisﬁes the additional condition of having two identity columns in opposing positions by

choosing k= lcm(|0|,|L−1|) and replacing by its k-th power, 0:= k. This makes no

11

diﬀerence when studying R(X%), because and kdeﬁne the same subshift and the group of

reversing symmetries is a property of the hull.

First, we prove an important property of the columns of powers. Fix a power k∈Nand pick

a column iof k, where 0 6i6Lk−1. One then has i=i0· · · ik−1where i0i1· · · ik−1is

the L-adic expansion of iand i`are columns of the level-1 substitution .

The corresponding L-adic expansion of Lk−(i+ 1) is then given by

Lk−(i+ 1) = (L−(i0+ 1)) · · · (L−(ik−1+ 1)).

This can easily be shown via the following direct computation

k−1

X

j=0

(L−(ij+ 1))Lj=

k−1

X

j=0

(Lj+1 −Lj)−

k−1

X

j=0

ijLj=Lk−(i+ 1).

This implies that if one considers the corresponding column Lk−(i+1) one gets that

(7) Lk−(i+1) =L−(i0+1) · · · L−(ik−1+1).

This has two consequences. First, if has an identity column pair, then all powers of admit at

least one identity column pair. For each power kone just needs to choose jwith j=iii · · · i,

which implies j=k

i= id. By Eq. (7), we also get that Lk−(j+1) = (L−(i+1))k= id. In fact,

kcontains at least 2k−1pairs of identity columns.

Second, this property allows one to prove that if satisﬁes the system of equations in Eq. (5),

then it is satisﬁed at all levels, i.e., by all powers of . To this end, choose 0 6i6Lk−1 with

L-adic expansion i0i1· · · ik−1. From Eq. (5) one then obtains

π−1◦i◦π=π−1◦i0· · · ik−1◦π=π−1◦i0ππ−1· · · ππ−1ik−1π

=L−(i0+1) · · · L−(ik−1+1) =Lk−(i+1).

Since iis chosen arbitrarily and πinduces a permutation of the substituted words at all levels,

this means it extends to a map f=σn◦m◦π:X%→X%, which by Proposition 19 is a reversor.

This shows (ii) =⇒(i).

To prove the remaining equivalences, note that if π1, π2∈Snare two permutations satisfying

the equality π−1◦i◦π=L−(i+1), then we have:

π1◦L−(i+1) ◦π−1

1=i=⇒(π2◦π−1

1)−1◦i◦(π2◦π−1

1) = i,

that is, (π2◦π−1

1)∈centSn(i). As a consequence, π1belongs to the right coset centSn(i)π2

for any choice of π1, π2, and, since right cosets are either equal or disjoint, this means that all

solutions of Eq. (5), for a ﬁxed i, lie in the same right coset of centSn(i). Reciprocally, if π

satisﬁes Eq. (5) and γ∈centSn(i), it is easy to verify that γ◦πsatisﬁes Eq. (5) as well. Thus,

the set of solutions of this equation is either empty or the aforementioned uniquely deﬁned right

coset.

Thus, suppose that πsatisﬁes Eq. (5) for all 0 6i6L−1. The set of solutions for each i

equals the unique coset centSn(i)π, and thus the set of all permutations that satisfy Eq. (5)

for all iis exactly the intersection of all these cosets, i.e. TL−1

i=0 centSn(i)π. Taking κi=πfor

all i, we see that this is exactly the set Kfrom (6). Evidently, πbelongs to this intersection,

and so we conclude that (ii) =⇒(iii).

As stated before, our choice of κiensures that the set centSn(i)κiis exactly the set of

solutions of Eq. (5) for a given i; thus, any permutation πthat satisﬁes all of these equalities

12

must be in all of these cosets and thus in the intersection (6), which is therefore non-empty.

This shows that (iii) =⇒(ii), concluding the proof.

The following general criterion on when a letter-exchange map generates a reversor is given

in [BRY18].

Lemma 23 ( [BRY18, Lem. 2]).Let be a primitive constant-length substitution of height 1

and column number c%. Suppose that is strongly injective. Then, a permutation π:A → A

generates a reversor f∈ R(X%)if and only if

(1) ab ∈ L2

%=⇒π(ba)∈ L2

%

(2) (π◦c%!)(ab)=(c%!◦π)(ba)

for each ab ∈ L2

%.

For a primitive, aperiodic and bijective , one has c%=|A|. Moreover, is always strongly

injective. Note that Theorem 22 implies conditions (1) and (2) in Lemma 23. The ﬁrst one

immediately follows from primitivity, and the fact that any legal word ab appears in some level-n

superword, which is sent to another level-nsuperword by π◦m, which guarantees the legality

of π(ba). The second follows from the fact that πis compatible with superwords of all levels.

In fact, one has π◦n(ab) = n◦π(ba) for all n∈N.

Remark 24. It is a known fact from group theory that, if g1, . . . , grare elements of a group

Gand H1, . . . , Hrare subgroups of this group, the intersection of cosets Tr

i=1 giHiis either

empty or a coset of Tr

i=1 Hi. In this case, the latter intersection is exactly the group of non-

trivial standard symmetries modulo a shift (letter exchanges), and thus, if there exist non-

trivial reversing symmetries, these must all belong to a single coset of the group of valid letter

exchanges. This is consistent with the fact that R(X%) is at most an index 2 group extension

of S(X%). ♦

Item (3) in Theorem 22 provides an explicit algorithm to compute the group of permutations

πwhich deﬁne extended symmetries, which is a counterpart to that in Section 2.2 for standard

symmetries. As stated previously, the centralisers centSn(j) can be computed for each column

using Fact 11, and thus the problem reduces to obtaining a suitable candidate for each κi, which

once again can be done by an application of Fact 11. The algorithm is as follows:

Algorithm. Assuming that is a primitive, bijective, aperiodic substitution, the following algorithm

computes the set Kof permutations that induce reversors, which determines R(X%).

•Input: is a length-Lbijective substitution, represented either as a function or a set of columns.

•Output: A (ﬁnite) set of permutations K, either empty or a coset of the group Ccomputed

by the previous algorithm, so that R(X%)/hσi ' C∪K(i.e. R(X%)'Z oϕ(C∪K), with

ϕ(g, n) = nif g∈C, and −nif g∈K).

(1) Let Nbe the least positive integer which ensures that two opposite columns of Nare the

identity map. This can be computed as:

N= min nlcm(ord(i),ord(L−(i+1))) : 0 6i6N/2o.

(2) For each 0 6i6N/2, compute κivia the following subroutine:

(2.i) If iand L−(i+1) are non-conjugate (i.e., their cycle decomposition has a diﬀerent number

of cycles of some length), stop the algorithm, as reversors do not exist (see Theorem 22).

13

(2.ii) Sort the cycles from the disjoint cycle decomposition of iby increasing order of length.

Using this as a basis, by appropriately sorting the elements of each cycle in this decom-

position, deﬁne a total order relation <on A, given by, say, a1<· · · < an, such that all

of the elements of a given cycle come before the elements of the following cycle, in the

sorting by left. Do the same for L−(i+1), deﬁning a corresponding total order <0given

by b1<0· · · <0bn. This ensures that there are cycle decompositions of both permutations

such that the corresponding cycles, ordered from left to right, have the same length, as

follows:

i= (a1. . . aj)(aj+1 . . . aj0)· · · (aj00 +1 . . . an),

L−(j+1) = (b1. . . bj)(bj+1 . . . bj0)· · · (bj00 +1 . . . bn),

with 1 6j6j06. . . 6j00 6n.

(2.iii) Deﬁne:

κi= a1a2· · · an

b1b2· · · bn!, κL−(i+1) =κ−1

i.

(3) Compute each centraliser C(i)= centSn(i), using the same procedure as in the computation

of S(X%).

(4) Return K=TN

i=1 C(i)κi. Any element of Kinduces a reversor; if Kis empty, reversors do not

exist.

Any programming environment with suitable data structures (e.g. computer algebra systems

such as Sagemath R

or Mathematica R

) is amenable to the implementation of this algorithm,

providing eﬀective procedures to entirely characterise the groups S(X%) and R(X%) from a

suitable description of the substitution , e.g. using a dictionary.

3.2. Higher-dimensional subshifts. Now, we turn our attention to the situation in higher

dimensions. The extended symmetry group of a Zd-shift is deﬁned as R(X) = normAut(X)(G),

where now G=hσe1, . . . , σedi ' Zd; see [BRY18,Bus20,BBH+19]. In this more general context,

an extended symmetry is a element f∈ R(X)\ S(X).

Similar to standard symmetries, there is a direct generalisation of the characterisation of

extended symmetries from Proposition 21 and the subsequent theorem to the higher-dimensional

setting, which is given by the following.

Proposition 25. Let be an aperiodic, primitive, bijective, block substitution in Zd. Then

any extended symmetry f∈ R(X%)\ S(X%)must be (up to a shift) a composition of a letter

exchange map and a rearrangement function fAgiven by fA(x)n=xAn, where A∈GL(d, Z),

with A6=I.

For shifts generated by bijective rectangular substitutions one has the following restriction

on the linear component Aof an extended symmetry f.

Theorem 26 ( [Bus20, Thm. 18]).Let an aperiodic, primitive, bijective rectangular substi-

tution in Zd. One then has

R(X%)/S(X%)'P6Wd,

where Wd'Cd

2oSdis the d-dimensional hyperoctahedral group, which represents the symmetries

of the d-dimensional cube.

With this, one can show that all extended symmetries of such subshifts are of ﬁnite order.

The proof of the following result is patterned from [BR06, Prop. 2], which deals with the

14

order of reversors of an automorphism hof a general dynamical system with ord(h) = ∞;

compare [Goo99].

Proposition 27. Let X%be the same as above with symmetry group S(X%) = Zd×G. Let

f∈ R(X%)\ S(X%)be an extended symmetry, whose associated matrix is A∈Wd. Then ord(f)

divides ord(A)· |G|. Moreover, ord(f)62|G| · max {ord(τ)|τ∈Sd}.

Proof. Under the given assumptions, f◦σm◦f−1=σAmholds for all m∈Zd, which yields

f`◦σm◦f−`=σA`m

(8)

f◦σnm◦f−1=σnAm

(9)

for all , n ∈N. Choosing = ord(A), Eq. (8) gives ford(A)∈ S(X%). From Theorem 8,

ford(A)=σp◦π, for some p∈Zdand letter-exchange map π. From the direct product structure

of the symmetry group, one has σp◦π=π◦σp, which implies ford(A)·|G|=σ|G|p◦π|G|=σ|G|p.

Using the two equations above, one gets ford(A)·|G|=σ|G|A`(p)for all ∈N. Since fis an

extended symmetry, A6=I. Next we show that pcannot be an eigenvector of A.

Suppose Ap=pwith p6=0. Note that f−ord(A)|G|=σ−|G|p. From Eqs. (8) and (9), one also

has f−1◦σ|G|A−1p◦f=σ−|G|p, which implies A−1p=−p, contradicting the assumption on p.

Since ord(σp) = ∞, this forces p=0and hence ford(A)·|G|= id from which the ﬁrst claim is

immediate. The upper bound for the order follows from the upper bound for the order of the

elements of the hyperoctahedral group Wd; see [Baa84].

Due to the fact that R(X%) is (possibly) a larger extension of S(X%) (that is, the corresponding

quotient can have up to 2dd!−1 non-trivial elements instead of just one), we would end up

with a much larger number of equations of the form of Eq. (3), one for each element of the

hyperoctahedral group Wdexcept the identity. This leads us to another problem of diﬀerent

nature: if the rectangle R, which is the support of the level-1 supertiles of , is not a cube in Zd,

some symmetries from Wdmay not be compatible with R, i.e., they may map Rto a diﬀerent

rectangle that is not a translation of R, so the corresponding equation does not have a proper

meaning (as it may compare an existing column with a non-existent one).

7→ 7→

Figure 2. A non-square substitution that generates the two-dimensional Thue-

Morse hull.

This could be taken as a suggestion that such symmetries cannot actually happen, imposing

further limitations on the quotient R(X%)/S(X%). Interestingly, this is not actually the case. For

instance, consider the two-dimensional rectangular substitution from Figure 2. As the support

for this substitution is a 4×2 rectangle, we could guess that this substitution is incompatible with

rotational symmetries or reﬂections along a diagonal axis, which would produce a 2×4 rectangle

instead. However, further examination shows that the hull generated by this substitution is

actually the same as the hull of the two-dimensional Thue–Morse substitution as seen in e.g.

[Bus20], which is compatible with every symmetry from W2=D4. Thus, only geometrical

considerations are not enough to exclude candidates for extended symmetries.

Fortunately, there is a subcase of particular interest in which this geometrical intuition is

actually correct, which involves an arithmetic restriction on the side lengths of the support

15

rectangle R. It turns out that coprimality of the side lengths is a suﬃcient condition (although

it can be weakened even further) to rule out such symmetries, e.g. there are no extended

symmetries compatible with rotations when Ris a, say, 2 ×5 rectangle. To be precise:

Theorem 28. Let :A → ARbe a bijective rectangular substitution with faithful associated

shift action. Suppose that R= [0,L−1]with L= (L1, . . . , Ld)(that is, Ris a d-dimensional

rectangle with side lengths L1, L2, . . . , Ld) and that for some indices i, j there is a prime psuch

that p|Ljbut p-Li, i.e. Liand Ljhave diﬀerent sets of prime factors. Let A∈Wd6GL(d, Z)

and suppose that Ais the underlying matrix associated to an extended symmetry f∈ R(X%).

Then Aij =Aji = 0.

The underlying idea is that, if A∈Wdinduces a valid extended symmetry for some sub-

stitution with support U, we can ﬁnd another substitution ηwith support A·U(up to an

appropriate translation) such that X%=Xη, and then we use the known factor map from an

aperiodic substitutive subshift onto an associated odometer to rule out certain matrices A.

Similar exclusion results have been studied by Cortez and Durand [CD08].

Proof. Let ϕ:X%ZL1× · · · × ZLd=ZLbe the standard factor map from the substitutive

subshift to the corresponding product of odometers. It is known [BRY18, Thm. 5] that, for any

extended symmetry f:X%→X%with associated matrix A, there exists kf= (k1, . . . , kd)∈ZL

and a group automorphism αf:ZL→ZLsatisfying the following equation:

(10) ϕ(f(x)) = kf+αf(ϕ(x)),

where αfis the unique extension of the map n7→ An, deﬁned in the dense subset Zd, to ZL.

In particular, for any n∈Zd, if f=σnis a shift map, then kσn=nand ασn= idZL.

Now, consider the sequence hm=Lm

iei, and suppose Aji =±1. Equivalently, Aei=±ej,

since Ais a signed permutation matrix. Without loss of generality, we may assume the sign to be

+. One has Lm

i

m→∞

−−−−→ 0 in the Li-adic topology, and thus ϕ(σhm(x)) = hm+ϕ(x)m→∞

−−−−→ ϕ(x),

as it does so componentwise. By compactness, we may take a subsequence hβ(m)such that

σhβ(m)(x) converges to some x∗; then, as the factor map ϕis continuous, we have ϕ(x∗) = ϕ(x).

Eq. (10) and this last equality imply that ϕ(f(x)) = ϕ(f(x∗)) as well. Writing x∗as a limit,

we obtain from continuity that

ϕ(x∗) = lim

m→∞ ϕ(f(σhβ(m)(x))) = lim

m→∞ ϕ(σAhβ(m)(f(x)))

=ϕ(x) + lim

m→∞ Ahβ(m)=ϕ(x) + lim

m→∞ Lβ(m)

iAei

=⇒lim

m→∞ Lβ(m)

iej=ϕ(x∗)−ϕ(x) = 0.

The last equality implies that, in the topology of ZLj, the sequence Lβ(m)

iconverges to 0.

However, since there is a prime pthat divides Ljbut not Li, due to transitivity we must have

Lj-Ln

ifor all n, as otherwise p|Ln

iand thus p|Li. Thus, in base Lj, the last digit of

Lβ(m)

iis never zero, and thus Lβ(m)

iremains at ﬁxed distance 1 from 0(in the Lj-adic metric),

contradicting this convergence. Thus, Aji cannot be 1 and must necessarily equal 0. For Aij ,

the same reasoning applies to f−1. Since Ais a signed permutation matrix, Aij =±1 would

imply (A−1)ji =±1, again a contradiction.

We now proceed to the generalisation of Theorem 22 in higher dimensions. As before, for

a block substitution , we have R=Qd

i=1[0, Li−1], with Li>2 and the expansive map

16

Q= diag(L1, L2, . . . , Ld). Let A∈Wd6GL(d, Z) be a signed permutation matrix. First, we

assume that the location of a tile in any supertile is given by the location of its centre. Deﬁne

the aﬃne map A(1) :R→Rvia A(1)(i) = A(i−x1) + |A|x1where i∈Rand x1=Qv−vwith

v=1

2(1,1,...,1)T. Here, (|A|)ij =|Aij |. The vector |A|x1is the translation needed to shift the

centre of the supertile to the origin, which we will need before applying the map Aand shifting

it back again. We extend A(1) to any level-ksupertile by deﬁning the map A(k):R(k)→R(k)

given by

(11) A(k)(i) = A(i−xk) + |A|xk,

with i∈R(k)and xk=Qkv−v. Here R(k):= Qd

i=1[0, Lk

i−1] is the set of locations of tiles in

a level-ksupertile.

Example 29. Let be a two-dimensional block substitution with Q= ( 2 0

0 2 ) and Abe the

counterclockwise rotation by 90 degrees, with corresponding matrix A=0−1

1 0 . Consider the

level-3 supertile and let i= (7,3)T∈R(3), with Q-adic expansion ib=i2i1i0. Here one has

i0=i1=e1+e2and i2=e1. One then gets A(3)(i) = (4,7)T; see Figure 3. One can check

that P2

j=0 Qj(A(1)(ij)) = A(3) (i). ♦

Figure 3. The transformation of a marked level-3 location set R(3) under the

map A(3).

The following result is the analogue of Theorem 22 in Zd.

Theorem 30. Let be an aperiodic, primitive, bijective block substitution :A→AR. Let Wd

be the d-dimensional hyperoctahedral group and let A∈Wd. Suppose there exists `∈Rsuch

that `0= id for all `0∈OrbA(`). Assume further that [A, Q]=0and |A|x1=x1. Then π,

together with A, gives rise to an extended symmetry f∈ R(X%)if and only if

(12) π−1◦i◦π=A(1)(i)

for all i∈R.

17

Proof. Most parts of the proof mimics those of the proof of Theorem 22, where one replaces the

mirroring operation mwith a more general map A∈Wd. One then gets an analogous system

of equations, as in those coming from Eq. (4). Using this, one can show the necessity direction.

To prove suﬃciency, we show that if Eq. (12) is satisﬁed for all i∈R, then it also holds for

all positions in any level-ksupertile. Let i∈R(k), which admits the unique Q-adic expansion

given by ib=ik−1ik−2· · · i1i0, i.e., i=Pk−1

j=0 Qj(ij). We now show that the Q-adic expansion

of A(k)(i) is given by A(k)(i)b=A(1)(ik−1)A(1)(ik−2)· · · A(1) (i0). Plugging in the expansion of i

into Eq. (11), one gets A(k)(i) = Pk−1

j=0 AQj(ij)−Axk+xk. On the other hand, one also has

k−1

X

j=0

Qj(A(1)(ij)) =

k−1

X

j=0

QjA(i−Qv+v) + Qv−v

=

k−1

X

j=0

QjA(ij) +

k−1

X

j=0 −AQj+1v+AQjv+

k−1

X

j=0 Qjv−Qjv

=

k−1

X

j=0

AQj(ij)−AQkv+Av

| {z }

−Axk

+Qkv−v

| {z }

xk

=A(k)(i),

where the penultimate equality follows from [A, Q] = 0 and the evaluation of the two telescoping

sums. As in Theorem 22, one then obtains

π−1◦i◦π=π−1◦ik−1◦ik−2◦ · · · i0◦π=A(k)(i),

whenever ib=ik−1ik−2· · · i0and π−1◦is◦π=A(1)(is)for all is∈R, which ﬁnishes the

proof.

Remark 31. The conditions [A, Q] = 0 and |A|x1=x1in Theorem 30 are automatically

satisﬁed if is a cubic substitution, i.e., Li=Lfor all 1 6i6d, which means one can use

Eq. (12) to check whether a given letter-exchange map works for any A∈Wd. For general

, these relations are only satisﬁed for certain A∈Wd, e.g. reﬂections along coordinate axes,

which means one needs a diﬀerent tool to ascertain whether it is possible for other rigid motions

to generate extended symmetries. For example, one can use Theorem 28 to exclude some

symmetries. ♦

Before we proceed, we need a higher-dimensional generalisation of Proposition 5regarding

aperiodicity. For this, we use the following result, which is formulated in terms of Delone sets.

Here, Sd−1is the unit sphere in Rd.

Theorem 32 ( [BG13, Thm. 5.1]).Let X(Λ)be the continuous hull of a repetitive Delone set

Λ⊂Rd. Let bi∈Sd−1|16i6dbe a basis of Rdsuch that for each i, there are two distinct

elements of X(Λ)which agree on the half-space {x| hbi|xi> αi}for some αi∈Rd. Then one

has that X(Λ)is aperiodic.

The proof of the previous theorem relies on the generalisation of the notion of proximality

for tilings and Delone sets in Rd, which is proximality along s∈Sd−1; see [BG13, Sec. 5.5] for

further details. Note that from a Zd-tiling generated by a rectangular substitution, one can

derive a (coloured) Delone set Λby choosing a consistent control point for each cube (usually

one of the corners or the centre). Primitivity guarantees that Λis repetitive and the notion

18

of proximality extends trivially to coloured Delone sets using the same metric. The two hulls

X(Λ) and X%are then mutually locally derivable, and the aperiodicity of one implies that of the

other. We then have a suﬃcient criterion for the aperiodicity of X%in higher dimensions.

Proposition 33. Let :A → ARbe a d-dimensional rectangular substitution which is bijective

and primitive. If there exist two legal blocks u, v ∈ L of side-length 2in each direction such that

uand vdisagrees at exactly one position and coincides at all other positions, then the hull X%

is aperiodic.

Proof. The proof proceeds in analogy to Proposition 5. Here we choose the appropriate power

to be k= lcm n|r|:r=Pd

i=1 riei, ri∈ {0, Li−1}o. If we then place uand vat the origin, the

resulting ﬁxed points x=∞(u) and x0=∞(v) which cover Zdwill coincide at every sector

except at the one where uj6=vj. One can then choose bi=eiand αi= 0 in Theorem 32, and

for each i,xand x0to be the two elements which agree on a half-space, which guarantees the

aperiodicity of X%. More concretely, xand x0are asymptotic, and hence proximal, along eifor

all 1 6i6d.

Remark 34. Obviously, one can have a lattice of periods of rank less than din higher dimen-

sions. An example would be when =1×2, where 1is the trivial substitution a7→ aa, b 7→ bb

and 2is Thue–Morse. Although 1is itself not primitive, the product is and admits the legal

blocks given in Figure 4, which generate ﬁxed points that are Ze1-periodic. If one requires that

the shift component in S(X%) is Zd, one needs all elements of X%to be aperiodic in all cardinal

directions, hence the stronger criterion in Proposition 33.♦

Figure 4. The image of two distinct blocks under coincide in the upper

half-plane and are distinct in the lower half-plane. In the limit, these legal seeds

generate two ﬁxed points which are neither left nor right asymptotic with respect

to σe1.

The next result is the analogue of Theorem 13 for extended symmetries, which holds in any

dimension.

Theorem 35. Given a ﬁnite group Gand a subgroup Pof the d-dimensional hyperoctahedral

group Wd, there is an aperiodic, primitive, bijective d-dimensional substitution whose shift

space satisﬁes

S(X%)'Zd×G

R(X%)'(ZdoP)×G.

19

Proof. We start by taking a cursory look at the proof of Theorem 3.6 in [DDMP16]. For a

given ﬁnite group G, we choose a generating set S={s1, . . . , sr}that does not contain the

identity, and build a substitution whose columns correspond to the left multiplication maps

Lsj(g) = sj·g, seen as permutations of the alphabet A=G. These permutations generate

the left Cayley embedding of Gin the symmetric group on |G|elements, whose corresponding

centraliser, which induces all of the letter exchanges in S(X%), is the right Cayley embedding of

Ggenerated by the maps Rsj(g) = g·sj.

In what follows, we shall assume ﬁrst that the group Gis non-trivial, as the case in which

Gis trivial requires a slightly diﬀerent construction. We also assume that the rectangular

substitution we will construct engenders an aperiodic subshift, so that the group generated by

the shifts is isomorphic to Zd. We delay the proof of this until later on, to avoid cluttering our

construction with extraneous details.

Since S(X%) depends only on the columns of the underlying substitution and not their relative

position, we shall construct a d-dimensional rectangular substitution with cubic support whose

columns correspond to copies of the aforementioned Lsj, placed in adequate positions along the

cube. We start with a cube R= [0,2|S|+ 2d+ 1]dof side length 2|S|+ 2d+ 2, where the

additional layer corresponding to the term 2 will be used below to ensure aperiodicity. This

cube is comprised of N=|S|+d+ 1 “shells” or “layers”, which are the boundaries of the inner

cubes [j, 2|S|+ 2d+ 2 −j]d; we shall denote each of them by Λj, where jcan vary from 0 to

N−1.

Fill the i-th inner shell ΛN−iwith copies of the column Lsi, for all 1 6i6r. This ensures

that, as long as every other column is a copy of Lsjfor some jor an identity column, the

symmetry group S(X%) of the corresponding subshift will be isomorphic to G, because in our

construction the 2dcorners of the point will always be identity columns.

Now, note that Nis chosen large enough so that the point p= (0,1, . . . , d −1) lies in the

outer N−r>dshells and, moreover, the cube [0, d −1]dis contained in these outer shells

as well. Thus, any permutation of the coordinates maps the cube [0, d −1]dto itself and, in

particular, two diﬀerent permutations map this point to two diﬀerent points in this cube, that

is, the orbit of phas d! diﬀerent points. Combining this with the fact that the mirroring maps

send this cube to one of 2ddisjoint cubes (translations of [0, d −1]d) in the corners of the larger

cube [0, N −1]d, it can be seen that Wdacts freely on the orbit of the point p, that is, there is

a bijection between the hyperoctahedral group Wdand the set Orb(p).

Next, choose a ﬁxed sj∈Sthat is not the identity element of G, so that Lsjis not an identity

column. As Pis a subgroup of Wd, it is bijectively mapped to the set P·p={g·p:g∈P}.

Place a copy of Lsjin each position from P·p, and an identity column in every other position

from Orb(p). Fill every remaining position in the cube with identity columns. This ensures

that the group of letter exchanges will remain isomorphic to G, and, for each matrix A∈Wd

associated with some element g∈P, the map fAgiven by the relation fA(x)n=xAnwill be a

valid extended symmetry, as a consequence of Theorem 30.

Since every other extended symmetry is a product of such an fAwith some letter-exchange

map that has to satisfy the conditions given by Eq. (12) due to our construction, and Lsjcannot

be conjugate to the identity column, the only other extended symmetries are compositions of

the already extant fAwith elements from S(X%), i.e. R(X%)/S(X%) has the equivalence classes

of each fAas its only elements. As the set of all fAis an isomorphic copy of Pcontained in

R(X%), we conclude that R(X%) is isomorphic to the semi-direct product S(X%)oP. However,

20

Figure 5. Examples of substitutions obtained by the above construction, for the

Klein 4-group C2×C2, the cyclic group C4and the whole W2=D4, respectively.

The thicker lines mark the layer of identity columns separating the inner cube

from the outer shell.

since every letter exchanges from Gcommutes with every fAtrivially, this semi-direct product

may be written as R(X%)'(ZdoP)×G, as desired.

In the case where Gis trivial, we may choose an alphabet with at least three symbols (to

ensure that S|A| is non-Abelian) and repeat the construction above with a collection of columns

0, . . . , r−1that generates some subgroup of S|A| with trivial centraliser (e.g. the two generators

of S|A| itself). The rest of the proof proceeds in the same way.

To properly conclude the proof, we need to verify that the constructed substitution generates

an aperiodic shift space. We focus on the case d > 1, as the one-dimensional case is a straight-

forward modiﬁcation of the construction from Theorem 13. Since our d-dimensional cube has

at least d+ 1 >3 outer layers, we see that there is a 2 × · · · × 2 cube R0contained in the

outer layers that does not overlap any of the 2dcubes of size d× · · · × don the corners nor

the inner cube of size 2|S| × · · · × 2|S|. As a consequence, this cube R0contains only identity

columns. Since we have a layer Λdconsists only of identity columns directly enveloping the inner

cube Λd+1 ∪ · · · ∪ Λd+|S|, the layer immediately following Λdis comprised only of non-identity

columns, which are copies of the same bijection π:A→A.. Thus, the 2dcorners of the hollow

cube Λd∪Λd+1 are 2 × · · · × 2 cubes R1,· · · , R2dhaving exactly one non-identity column each,

with this non-identity column τbeing placed in every one of the 2dpossible positions on these

cubes.

Since τis not the identity, there must exist some a∈ A such that τ(a)6=a. The previous

discussion thus implies that there is an admissible pattern Paof size 2 × · · · × 2 comprised only

of copies of the symbol a, and 2d+ 1 other admissible patterns P(n)

athat diﬀer from Paonly in

the position n∈[0,1]d. Using the proximality criterion from Proposition 33, we conclude that

the subshift obtained is indeed aperiodic, as desired.

Remark 36. An alternative Cantor-type construction, which produces the prescribed symme-

try and extended symmetry groups, involves putting the non-trivial columns on the faces of R

and labelling all columns in the interior to be the identity. Let Gand Pbe given. From Theo-

rem 13, there exists a substitution on Awith S(X%) = Z×G. Let 0, . . . , r−1be the non-trivial

columns of . Pick Lto be large enough such that Wdacts freely on the faces of R= [0, L −1]d.

Choose j0∈Rand consider the orbit of j0under P, i.e., O0:= P·j0={A·j0|A∈P}

where A·j=A(1)(j) as in Eq. (11) . Label all the columns in O0with 0. We then expand

Rvia Q= diag(L, . . . , L) to get the d-dimensional cube Q(R) of side length L2. Consider

B1:= Q(O0) + R, pick j1∈ B1and let O1=P·j1. Relabel all columns in B1\ O1with 0and

21

all columns in O2with 1. One can continue this process until all needed column labels appear;

see Figure 6for a two-dimensional example.

(a) O0in blue (b) B1\ O1in

blue, O1in red

(c) B2\ O2in green

Figure 6. An example in Z2with three non-trivial columns 0(blue), 1(red)

and 2(green). Here, one has G= centS|A| h0, 1, 2iand P'V4, where

V46D4=W2is the Klein-4 group.

Note that one has i=A(1)(i)for all A∈Pand i∈R= [0, L −1]dby construction, which

means π= id gives rise to an element of R(X%) for all A∈Pby Theorem 30. No other extended

symmetries can occur because all the location sets Bionly contain non-trivial labels and are

P-invariant, whereas if A /∈Pinduces an extended symmetry, one must have `= id for some

`∈ Br.

The resulting block substitution is primitive, since reordering the columns does not aﬀect

primitivity. It is also aperiodic because one has enough identity columns, and hence one can

ﬁnd the legal words required in Proposition 33. For example, in the constructed substitution in

Figure 6, the legal seeds can be derived from the 2 ×2 block consisting of all identity columns

(i.e. all white squares), and another one with all columns being the identity except at exactly

one corner, where it is blue. This completes the picture and one has S(X%)'Zd×Gand

R(X%)'(ZdoP)×G.♦

We now turn our attention to examples where the letter-exchange map πthat generates

f∈ R(X%) is not given by the identity. In particular, in these examples, πdoes not commute

with the letter-exchanges which correspond to the standard symmetries in S(X%). To avoid

confusion, we will use letters for our substitution and the action of the hyperoctahedral group

will be given by numbers, seen as permutations of the coordinates. Mirroring along a hyperplane

will be denoted by mi, where iis the respective coordinate.

Example 37. We explicitly give a substitution whose symmetry group is S(Xε) = Zd×C3

and build another C3component in R(Xε), which produces reversors of order 9. With the

requirement on R(Xε)/S(Xε), the space has to be at least of dimension 3.

ε0= (a d g)(b e h)(c f i)ε2= (a b c)(d e f )(g h i)ε5= id

ε1= (a g d)(b h e)(c i f )ε3= (b c d)(e f g)(h i a)

ε4= (c d e)(f g h)(i a b)

22

Here one has S(Xε) = Z3×C3, which is generated by (a d g)(b e h)(c f i). Depending on the

positioning of the columns, R(Xε) can either be Z3oC9,Z3oC3×C3or Z3×C3. The group

Z3oC3×C3can be realised using the construction from Theorem 35. On the other hand,

Z3×C3is obtained if one orbit of maximal size is labelled with just one non-identity εionce,

and the rest with ε0.

Note that π= (a b c d e f g h i) sends ε2→ε3→ε4→ε2and ε0→ε0, ε1→ε1. Taking

the cube of (a b c d e f g h i) gives (a d g)(b e h)(c f i)∈centS9(G(1)), where G(1) is the group

generated by the columns. This is consistent with the bounds calculated in Proposition 27.

We will illustrate the positioning of a few elements following the construction in Theorem 35.

We look at a position that has the maximum orbit size under W3, for example (0,1,2) ∈R.

The orbit under C3is (0,1,2),(1,2,0),(2,0,1), which is obtained by cyclically permuting the

coordinates. We place ε2at position (0,1,2), ε3at position (1,2,0) and ε4at position (2,0,1).

Since ε0, ε1∈centS9(G), we will position them each along a diﬀerent orbit. All remaining

positions will be ﬁlled with the identity to ensure that we cannot have additional symmetries.

We use Proposition 33 to ensure aperiodicity. It is easy to see that one gets the required patches

by choosing the 2 ×2×2 cube in the upper right corner from the ﬁrst and second slices and

the other one from the second and third. For this conﬁguration, one has R(Xε) = Z3oC9.♦

Figure 7. The gray cubes are ﬁlled with ε5(the identity). Yellow and brown

can be ﬁlled by either ε0, ε1, respectively. Lastly, ε2is blue, ε3is green and ε4

is red, where one has the obvious freedom in choosing the colours due to the

C3-symmetry.

Remark 38. As a generalisation of Example 37, for any given cyclic groups Cnand Ck, we can

construct a substitution in Zn, such that X%has the symmetry group Zn×Ckand its extended

symmetry group is given by (Zn×Ck)oCn. More precisely, since the extended symmetry group

contains an element of order nk,R(X%) = ZnoCnk. The substitution can be realised by the

following columns

ε0= (a1ak+1 · · · a(n−1)k+1 )· · · (ak· · · ank)

εi= (aiai+1 · · · ak−1+i)(ak+iak+i+1 · · · a2k+i−1)· · · (a(n−1)k+ia(n−1)k+i+1 · · · ank−1+i)

εn+1 = id

where iruns from 1 to n, where the values are seen modulo nk.

From the columns εiwith i6= 0 we can see that the centraliser can only be the permutation

of the cycles limiting the centraliser to Sk, while ε0limits it further to be Ck, since this copy

of Skoperates on the cycles independently and the centraliser of a cycle is just the cycle itself.

23

The extended symmetry is realised by the permutation (a1· · · akn) which maps εito εi+1. Its

orbit is determined by the action of Cn6Wnon the positioning of the columns. ♦

In the next example we illustrate how important it is to choose compatible structures for the

letter-exchange map and the corresponding action in Wd.

Example 39. We look at a four-letter alphabet with the following columns in Eq. (13) which

generate S4as a subgroup of S4, thus implying that the shift space to have a trivial centraliser.

We plan to have S3' R(Xε)/S(Xε), so we place the columns in a three-dimensional cube.

ε0= id ε1= (a b c d)ε4= (a c d b)

ε2= (a b d c)ε5= (a d b c)(13)

ε3= (a c b d)ε6= (a d c b)

The symmetry group is trivial since the columns generate S4. Conjugation with τ= (c d) maps

ε1to ε2, just as any τκ, with κ∈centS4(ε2).

Figure 8. The columns assigned to the colors are as follows: ε1(blue),

ε2(yellow), ε3(green) ε4(purple), ε5(black) and ε6(red).

Here C3oC2'S3is realised by (b c d)(0 1 2) and (c d)(0 1). The transposition (c d) cannot

be realised in Wdby mirroring along an axis in the cube since that is not consistent with the

interaction between (b c d) and (c d). This can be easily be seen by looking at mirroring along

all hyperplanes.

(a b c d)(2,1,0) (a c d b)(0,2,1)

(a b d c)(1,2,3) (a d c b)/(a c b d)(3,1,2)

(bcd)

(0 1 2)

m012

(c d)m012

(c d)

(bcd)

(0 1 2)

We see that the diagram does not commute, thus there is no way to assign a single column to

the vertex (3, 1, 2). One can do this for all axes, which rules out the C3

2component in W3, thus

yielding R(X%) = ZdoS3.♦

24

Remark 40. One can also ask whether, starting with a group G, one can build the centraliser

S(X%) and normaliser R(X%) organically from G, under a suitable embedding of G. Consider

the Cayley embedding G →S|G|as in Example 12. We know that centS|G|(G)'Gand

normS|G|(G)'GoAut(G); see [Seh89]. Since the automorphisms of Gare given by conjugation

in S|G|, they deﬁne letter-exchange maps which are compatible with reversors in R(X%). By

choosing the dimension appropriately, one can construct a substitution on A=Gsuch that

the extended symmetry group is given by

R(X%) = Zd(G)×GoAut(G),

where we choose d(G) such that Aut(G)6Wd(G). This can always be done for d(G) = |G|, but

depending on Aut(G), a smaller dimension is possible. Let π∈Aut(G) and let Aπ∈Wd. The

construction from the proof of Theorem 35 can be applied. Here, the orbits of Aπwill not be

ﬁlled with the same element, but with columns that are determined by π, i.e., Aπ(i)=π◦i◦π−1,

where πis seen as an element of S|G|.♦

These series of examples with more complicated structure can be generalised for arbitrary

groups Gand P. Here, we have the following version of Theorem 35 where the letter exchange

map is no longer π= id, which we build from a speciﬁc set of columns.

Theorem 41. Let H, P be arbitrary ﬁnite groups. Then for all >c(P), where c(P)is a

constant which depends only on the group P, there is a shift space X%originating from an

aperiodic, primitive and bijective substitution such that

S(%) = `×H

R(%) = ( `×H)oP.

Proof. The proof will be divided into two parts, beginning with a manual for the construction

of the substitution and a second part where we verify the claims made in the construction and

check if the subshift has the desired properties.

•We ﬁrst turn our attention to the construction of Pwhich later is supposed to be

isomorphic to R(%)/S(%). For that purpose we embed P →S`which is certainly

possible for some . It is clear that there is a minimal c(P)∈Nfor which this embedding

is possible, and that every >c(P) gives a valid embedding as well. This means the

choice of has a lower bound, but can be increased arbitrarily. This chosen determines

the dimension of the space Z`where the subshift is constructed. Let us now ﬁx a suitable

, excluding = 2,3,6 since we use want to use AutS`(S`) = InnS`(S`)'S`which does

not hold for these values of ; see [Seg40].

•Next, we look for suitable columns for our substitution. Choose the set T={1,· · · k}

of all transpositions in S`, together with the identity column as the set of columns. T

generates S`and the action of S`(viewed as the automorphism group) acts faithfully

on T. From this, we get that P⊂S`'InnS`(S`)⊂normS`({1,· · · , k}). This is

enough for now, since P⊂normS`({1,· · · , k}) and we can exclude the surplus later.

•Now, we compute the centraliser of the column group. In our current construction

the centraliser is trivial, which is why we need to modify our columns. We do this by

extending our alphabet {a, · · · , }to {a1,· · · , a|H|, b1,· · · , b|H|,· · · , 1,· · · , |H|}. We

simply duplicate the cycles in each column: The permutations of the columns are

mapped by ρ→ρ0sending i= (x y)7→ εi= (x1y1)· · · (x|H|y|H|).

25

•We embed G →S|H|with the usual Cayley embedding. This group is only acting on

the indices of the letters in the new alphabet. The action on the indices is applied to

every {a, . . . , }, giving the ﬁnal set of columns {η1, . . . , ηm}added to the substitution

ρ0giving a new substitution .

•The Cayley embedding guarantees that centS|H|(G%)'H, where G%is the column

group of . We can decrease the size of R(%)/S(%) with the same arguments as in

Theorem 35. This way we achieve a group R(%)/S(%)'Pwhere the letter exchange

component πof the extended symmetries are not in centS|H|(G%).

Aperiodicity of X%can be easily obtained via proximal pairs. Regarding primitivity, it is

suﬃcient to check the transitivity of G%and use Proposition 3. For any pair (xj, yk) of letters

with indices chosen from the alphabet we need to ﬁnd a g∈G%such that gxj=yk. Note that

the permutation (x1y1)· · · (xjyj)· · · (x|H|y|H|)∈G%and maps xjto yj. Now we need to map

yjto yk, which is an action solely on the indices. The mapping on the indices can be realized

by the right embedding copy of Hin S|H|and thus by an element composed of the columns

{η1,· · · , ηm}.

Let us prove that the centraliser is indeed isomorphic to G. The centraliser of G{ε1,···εk}can

only contain elements that are pure index permutations, since those columns generate S`. Since

the structure of the cycles in each column are independent of the index, any index permutation

is an element of centS`(G{ε1,···εk}) = S`.

We continue by determining centS`|H|(G{η1,··· ,ηm})TS`. The group S`are the pure index

switches and since η1,· · · , ηmare the columns generated by the Cayley embedding of Hinto S`

their centraliser is isomorphic to H.

The following rule lifts an automorphism h0on Gρto hon G%.

h(i) = h0()i

Thus S|H|6AutS`|H|(G{ε1,··· ,εk}). It is suﬃcient to prove that the automorphism group did not

decrease in size by the addition of the columns (η1,· · · , ηm). Then we can use the geometric

placement of the columns in Theorem 35 in the substitution to exclude any unwanted Wd-

component. Any lifted automorphism hstill only maps the letters and ﬁxes the indices. Since

the cycles in any η1,· · · , ηmcontain only the same letter with diﬀerent indices and the index

structure is independent of the letter, every his in centS|H|`(G{η1,··· ,ηm}) and surely legal. Thus

it is an automorphism on the whole of G%.

Remark 42. Theorems 35 and 41 fall under realisation theorems for shift spaces. The most

general current result along this vein known to the authors is that of Cortez and Petite, which

states that every countable group Gcan be realised as a subgroup G6R(X, Γ ), where R(X, Γ )

is the normaliser of the action of a free abelian group Γon an aperiodic minimal Cantor space

X; see [CP20]. ♦

4. Concluding remarks

While the higher-dimensional criteria in Theorems 30 and 28, which conﬁrm or rule out the

existence of extended symmetries, are rather general, it remains unclear how to ﬁnd a way to

extend this to a larger (possibly all) class of systems, with no constraints on the geometry of the

supertiles. This is related to a question of determining whether, given a substitution in Zd(or

Rd), one can come up with an algorithm which decides whether there is a simpler substitution

which generates the same or a topologically conjugate hull, which is easier to investigate. This

26

is exactly the case for the two-dimensional Thue–Morse substitution in Figure 2. Such an issue

is non-trivial both in the tiling and the subshift context; see [CD08,DL18,HRS05].

Note that the letter-exchange map π∈S|A| in Theorem 30 always induces a conjugacy

between columns whenever it generates a valid reversor. It would be interesting to know whether

outer automorphisms in this case can yield valid reversors for a bijective substitution subshift

in Zd, for example for those whose geometries are not covered by Theorems 30 and 28. For

instance, Aut(S6) contains elements which are not realised by conjugation.

Another natural question would be to determine other possibilities for S(X%) and R(X%)

outside the class of bijective, constant-length substitutions. Here, the higher-dimensional gen-

eralisations of the Rudin–Shapiro substitution would be good candidates; see [Fra03]. There

are also substitutive planar tilings with |R(X)/S(X)|=D6, which arises from the hexagonal

symmetry satisﬁed by the underlying tiling. For these classes, and in the examples treated

above, the simple geometry of the tiles introduces a form of rigidity which leads to R(X) being

a ﬁnite extension of S(X); see [BRY18, Sec. 5] for the notion of hypercubic shifts. There are

substitution tilings whose expansive maps Qare no longer diagonal matrices, and whose super-

tiles have fractal boundaries; compare [Fra20, Ex. 12], which allows more freedom in terms of

admissible elements of GL(d, Z) which generate reversors. This raises the following question:

Question 43. What is the weakest condition on the shift space/tiling dynamical system X

which guarantees [R(X) : S(X)] <∞?

This is always true in one dimension regardless of complexity, since either the subshift is

reversible or not, but is non-trivial in higher dimensions because |GL(d, Z)|=∞for d > 1, so

inﬁnite extensions are possible; see [BBH+19]. We suspect that this is connected to the notions

of linear repetitivity, ﬁnite local complexity, and rotational complexity; compare [BRY18, Cor. 4]

and [HRS05]. For inﬂation systems, the compatibility condition [A, Q] = 0 in Theorem 30 might

also be necessary in general when the maximal equicontinuous factor (MEF) has an explicit form.

5. Acknowledgements

The authors would like to thank Michael Baake for fruitful discussions and for valuable

comments on the manuscript. AB is grateful to ANID (formerly CONICYT) for the ﬁnan-

cial support received under the Doctoral Fellowship ANID-PFCHA/Doctorado Nacional/2017-

21171061. NM would like to acknowledge the support of the German Research Foundation

(DFG) through the CRC 1283.

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Departamento de Ingenier

´

ıa Matem´

atica, Universidad de Chile,

Beauchef 851, Santiago, Chile

Email address:abustos@dim.uchile.cl

Fakult¨

at f¨

ur Mathematik, Universit¨

at Bielefeld,

Postfach 100131, 33501 Bielefeld, Germany

Email address:{dluz,cmanibo}@math.uni-bielefeld.de

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