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Admissible reversing and extended symmetries for bijective substitutions

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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 $G$ and any subgroup $P$ of the $d$-dimensional hyperoctahedral group, there is a bijective substitution which generates an aperiodic hull with symmetry group $\mathbb{Z}^{d}\times G$ and extended symmetry group $(\mathbb{Z}^{d} \rtimes P)\times G$.
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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.
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
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 LNsuch 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) 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)=2infnn:x|[n,n]d6=y|[n,n]do.
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,nZd(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 xX, that is:
L(X) = {x|[0,n]:xX, 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:UZd→ 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, ni1]. Regardless of
our chosen convention, we collect all valid patterns x|Uthat appear in some xXinto 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 VUfor 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 6j6L1. Equivalently,
is bijective if there exist L(not necessarily distinct) bijections 0, . . . , L1:A → A such that
(a) = 0(a). . . L1(a) for every a∈ A. We shall refer to the mapping jas the j-th column
of the substitution .
Consider jL1
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.
Fact 2. Let be a primitive, bijective substitution, whose columns are given by 0, . . . , L1.
Then the substitution matrix Mis given by M=PL1
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 xX%.
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, . . . , jn6L1}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 τGsuch that τ(j) = k. Here, we let NNbe the minimal
power such that N
j= id for some 0 6j6LN1; 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, . . . , jN6L1, that is, for any a∈ A the following holds:
N(a) = 0,...,0,0(a)0,...,0,1(a). . . 0,...,0,L1(a)0,...,1,0(a). . . L1,...,L1,L1(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 qq0◦ · · · ◦ 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 nN,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.
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 wis a periodic word with pas least period. Without loss of generality,
choose a power ksuch that its first column is the identity, and so wis fixed by k. We choose
cand dminimal such that:
cL =dp cb|A| =da|A|.(1)
We apply 1to w|[0,cL1] 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 wis 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, w2X%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· · · L1, with iG. Choosing k= lcm(|0|,|L1|), we get that the first
and the last columns of kare both the identity, i.e., k
0(a) = k
Lk1(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 xX%is uniquely determined by the letter
at the origin. From the finiteness of A, one gets x=wand hence is periodic, from which the
periodicity of the hull follows.
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 XXwhich commute with the shift action, i.e.,
(2) (nZd): σnf=fσn.
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 UZdfinite, such that for every nZd,
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)jZd)=(τ(xj))jZdfor 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 τcentS|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 xX%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
Nτ=τN. 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:XX.
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 SG. Then,
centG(H) = {cG|(hH): ch =hc}=\
Fact 11. Any permutation decomposes uniquely (up to reordering) as a product of disjoint
cycles. Conjugation by some τSncan be computed from this decomposition using the identity:
τ(a1a2. . . an)τ1= (τ(a1)τ(a2). . . τ (an)).
A permutation τSnbelongs 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, . . . , L1:A → A, corresponding to each
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.
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¯¯
i7→ ¯ek¯, ¯ı7→ e¯
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¯
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 ¯
π1:= (e j ¯e¯)(i¯
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=νRiν1and π1=ν Rjν1, 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-
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
πSn, 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.
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 πG(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) = {HAut(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) = {HAut(X)|HσH1=σ±1}.
A homeomorphism HAut(X) which satisfies HσH1=σ1is 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].
In what follows, we investigate the effect of a reversor fon inflated words. Given a bijective
substitution :A → AL,  := 01· · · L1, the mirroring operation macts on the columns of
via m((a)) = L1(a)· · · 2(a)0(a). We may extend this to infinite configurations over Zin
two non-equivalent ways, given by m(x)k=xkand m0(x)k=x1k, 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 πSn, 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 xX%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=fσh), 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 nZ, if nmLk+ [, ], then yndepends only
on the word x|mLk+[`,`], which contains (and is thus entirely determined by) xn. 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=fσh=πm(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.
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 πSn, π :A→A
which gives rise to a reversing symmetry. Then one has
(3) π1i1
jπ=L(i+1) 1
for all 06i, j 6L1, 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 πSnsuch
that mπextends to a reversor f∈ R(X%). One then has
(a) = 0(a)· · · L1(a)m
7−L1(a)· · · 0(a)π
7−πL1(a)· · · π0(a).
Since Proposition 19 guarantees that this must result to mapping substituted words to substi-
tuted words, one gets
(4) πL1(a)· · · π0(a) = 0(b)· · · L1(b) = 0τ(a)· · · L1τ(a),
where the permutation τdescribes precisely this induced shuffling of inflation words. This yields
for all 0 6j6L1. 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 6L1.
Theorem 22. Let be as in Proposition 21. Suppose further that i=L(i+1) = id for some
06i6L1. Then, given a permutation (letter exchange map) πSn, 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(xn)or f(x)n=π(x1n).
(ii) The permutation πsatisfies the system of equations
(5) π1iπ=L(i+1)
for all 06i6L1.
(iii) There exist κ0, κ1, . . . , κL1Sn, where each κisatisfies κ1
iiκi=L(i+1), such
that the following intersection of cosets is non-empty:
(6) K=
and πK.
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|,|L1|) and replacing by its k-th power, 0:= k. This makes no
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 kNand pick
a column iof k, where 0 6i6Lk1. One then has i=i0· · · ik1where i0i1· · · ik1is
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(ik1+ 1)).
This can easily be shown via the following direct computation
(L(ij+ 1))Lj=
(Lj+1 Lj)
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(ik1+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 2k1pairs 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 6i6Lk1 with
L-adic expansion i0i1· · · ik1. From Eq. (5) one then obtains
π1iπ=π1i0· · · ik1π=π1i0ππ1· · · ππ1ik1π
=L(i0+1) · · · L(ik1+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=σnmπ:X%X%, which by Proposition 19 is a reversor.
This shows (ii) =(i).
To prove the remaining equivalences, note that if π1, π2Snare two permutations satisfying
the equality π1iπ=L(i+1), then we have:
π1L(i+1) π1
1) = i,
that is, (π2π1
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 γcentSn(i), it is easy to verify that γπsatisfies Eq. (5) as well. Thus,
the set of solutions of this equation is either empty or the aforementioned uniquely defined right
Thus, suppose that πsatisfies Eq. (5) for all 0 6i6L1. 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. TL1
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
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) (πc%!)(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 πm, 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 πn(ab) = nπ(ba) for all nN.
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 ' CK(i.e. R(X%)'Z oϕ(CK), with
ϕ(g, n) = nif gC, and nif gK).
(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).
(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
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) =κ1
(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
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 AGL(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
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
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 AWd. Then ord(f)
divides ord(A)· |G|. Moreover, ord(f)62|G| · max {ord(τ)|τSd}.
Proof. Under the given assumptions, fσmf1=σAmholds for all mZd, which yields
for all , n N. Choosing = ord(A), Eq. (8) gives ford(A)∈ S(X%). From Theorem 8,
ford(A)=σpπ, for some pZdand letter-exchange map π. From the direct product structure
of the symmetry group, one has σpπ=πσp, which implies ford(A)·|G|=σ|G|pπ|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 ford(A)|G|=σ−|G|p. From Eqs. (8) and (9), one also
has f1σ|G|A1pf=σ−|G|p, which implies A1p=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
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,L1]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 AWd6GL(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 AWdinduces 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:ZLZLsatisfying 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 nZd, 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
0 in the Li-adic topology, and thus ϕ(σhm(x)) = hm+ϕ(x)m→∞
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 xas 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)
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
ifor all n, as otherwise p|Ln
iand thus p|Li. Thus, in base Lj, the last digit of
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 f1. Since Ais a signed permutation matrix, Aij =±1 would
imply (A1)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, Li1], with Li>2 and the expansive map
Q= diag(L1, L2, . . . , Ld). Let AWd6GL(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) :RRvia A(1)(i) = A(ix1) + |A|x1where iRand x1=Qvvwith
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(ixk) + |A|xk,
with iR(k)and xk=Qkvv. Here R(k):= Qd
i=1[0, Lk
i1] 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=01
1 0 . Consider the
level-3 supertile and let i= (7,3)TR(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 AWd. Suppose there exists `Rsuch
that `0= id for all `0OrbA(`). 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) π1iπ=A(1)(i)
for all iR.
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 AWd. 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 iR, then it also holds for
all positions in any level-ksupertile. Let iR(k), which admits the unique Q-adic expansion
given by ib=ik1ik2· · · i1i0, i.e., i=Pk1
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)(ik1)A(1)(ik2)· · · A(1) (i0). Plugging in the expansion of i
into Eq. (11), one gets A(k)(i) = Pk1
j=0 AQj(ij)Axk+xk. On the other hand, one also has
Qj(A(1)(ij)) =
QjA(iQv+v) + Qvv
QjA(ij) +
j=0 AQj+1v+AQjv+
j=0 QjvQjv
| {z }
| {z }
where the penultimate equality follows from [A, Q] = 0 and the evaluation of the two telescoping
sums. As in Theorem 22, one then obtains
π1iπ=π1ik1ik2◦ · · · i0π=A(k)(i),
whenever ib=ik1ik2· · · i0and π1isπ=A(1)(is)for all isR, which finishes the
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 AWd. For general
, these relations are only satisfied for certain AWd, 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
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, Sd1is the unit sphere in Rd.
Theorem 32 ( [BG13, Thm. 5.1]).Let X(Λ)be the continuous hull of a repetitive Delone set
ΛRd. Let biSd1|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 αiRd. 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 sSd1; 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
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, Li1}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
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
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
Fill the i-th inner shell ΛNiwith 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 Nr>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 sjSthat 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:gP}.
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 AWd
associated with some element gP, 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,
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, . . . , r1that 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 ΛdΛd+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
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, . . . , r1be the non-trivial
columns of . Pick Lto be large enough such that Wdacts freely on the faces of R= [0, L 1]d.
Choose j0Rand consider the orbit of j0under P, i.e., O0:= P·j0={A·j0|AP}
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
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 APand iR= [0, L 1]dby construction, which
means π= id gives rise to an element of R(X%) for all APby 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
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)
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 ε2ε3ε4ε2and ε0ε0, ε1ε1. 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, ε1centS9(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
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(n1)k+1 )· · · (ak· · · ank)
εi= (aiai+1 · · · ak1+i)(ak+iak+i+1 · · · a2k+i1)· · · (a(n1)k+ia(n1)k+i+1 · · · ank1+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.
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 κcentS4(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)
(0 1 2)
(c d)m012
(c d)
(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.
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 πAut(G) and let AπWd. 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)=πiπ1,
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 PS`'InnS`(S`)normS`({1,· · · , k}). This is
enough for now, since PnormS`({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 ρρ0sending i= (x y)7→ εi= (x1y1)· · · (x|H|y|H|).
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 gG%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,··· m})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,··· k}). 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,··· m}) 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
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 πS|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
at f¨
ur Mathematik, Universit¨
at Bielefeld,
Postfach 100131, 33501 Bielefeld, Germany
Email address:{dluz,cmanibo}
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