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Tiling Semigroups of n-Dimensional Hypercubic TilingsDonald B. McAlister a & Filipa Soares ba Centro de Álgebra da Universidade de Lisboa (CAUL), Complexo Interdisciplinar , Lisbon,Portugaland Department of Mathematical Sciences, Northern Illinois University, DeKalb,Illinois, USAb CAUL, Área Departamenta de Matemática , Instituto Superior da Engenharia de Lisboa(ISEL) , Lisbon, PortugalPublished online: 21 Jun 2011.

To cite this article: Donald B. McAlister & Filipa Soares (2011) Tiling Semigroups of n-Dimensional Hypercubic Tilings,Communications in Algebra, 39:6, 2002-2023, DOI: 10.1080/00927872.2010.482547

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Communications in Algebra®, 39: 2002–2023, 2011Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2010.482547

TILING SEMIGROUPS OF n-DIMENSIONALHYPERCUBIC TILINGS

Donald B. McAlister1 and Filipa Soares21Centro de Álgebra da Universidade de Lisboa (CAUL),Complexo Interdisciplinar, Lisbon, Portugal and Department of MathematicalSciences, Northern Illinois University, DeKalb, Illinois, USA2CAUL, Área Departamenta de Matemática, Instituto Superior da Engenhariade Lisboa (ISEL), Lisbon, Portugal

In this work, we generalize to hypercubic tilings of dimension n the description oftiling semigroups as inverse semigroups associated to factorial languages and therepresentation of this semigroup as a P∗-semigroup. In addition, we show that, incontrast with the one-dimensional case, the tiling semigroup of any n-dimensionalhypercubic tiling is always infinitely presented (even as a strongly E∗-unitary inversesemigroup) and give a necessary and sufficient condition for two hypercubic tilingsemigroups to be isomorphic.

Key Words: n-Dimensional hypercubic tiling; n-Dimensional tiling language; Isomorphism of tilingsemigroups; Presentation of an inverse semigroup; P∗-Semigroup.

2000 Mathematics Subject Classification: 20M18; 20M05; 68Q45.

1. INTRODUCTION AND BACKGROUND

The study of tiling semigroups was initially motivated by a construction ofKellendonk in the late 1990s with a series of three papers [4–6] in connection with aproblem in solid state physics, and then established by Kellendonk and Lawson [7].The definitions we will now present can be found in these works, as well as in [8].

An n-dimensional tiling is a covering of the Euclidean space �n by closed,bounded, and connected subsets of �n with connected interiors which are theclosures of their interiors and which may carry a decoration, called tiles, withoutgaps or overlaps, except, possibly, at the boundary of the tiles. A pattern is a finiteconnected union of tiles with connected interior. A pattern class is an equivalenceclass of a pattern under translation, and a doubly pointed pattern class is a patternclass with a choice of an ordered pair of its tiles, the first of which is usually calledin-tile, and the second out-tile. An n-dimensional tiling is said to be of finite type ifthere are only finitely many doubly pointed pattern classes whose underlying pattern

Received December 24, 2008; Revised February 12, 2010. Communicated by V. Gould.In memory of W. Douglas Munn (1929–2008).Address correspondence to Filipa Soares de Almeida, Av. Prof. Gama Pinto, 2, Lisbon 1649-

003, Portugal; E-mail: falmeida@deea.isel.pt

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has two tiles or, equivalently, if for every positive real number r, the maximalnumber of tiles the tiling can fit into a ball of radius r is finite. As usual, we dealonly with finite type tilings, without further mention.

Let � be an n-dimensional tiling. If A is a pattern in � and a and b are twotiles of A, denote the doubly pointed pattern class of the pattern A, in-tile a andout-tile b by �a� A� b�. We say that the doubly pointed pattern classes �a� A� b� and�c� C� d� are composable if there exist translations x and y of �n such that �A+ x� ∪�C + y� is a pattern in � and b + x = c + y. Let S�� � be the set of all doubly pointedpattern classes together with a new element 0, and consider the following operationbetween its elements:

�a� A� b��c� C� d� =

�a+ x� �A+ x� ∪ �C + y�� d + y�� if �a� A� b� and �c� C� d�

are composable

0� otherwise

and all products involving 0 yield 0. Then we have the following theorem.

Theorem 1.1 (Theorem 9.5.1, [8]). The above binary operation is well defined andendows S�� � with the structure of an inverse semigroup with zero.

We call S�� � the tiling semigroup of the tiling � . It is easy to check that anelement �a� A� b� is a (nonzero) idempotent if and only if a = b and that �a� A� b�−1 =�b� A� a�. Also, the natural partial order has the following characterization in S�� �:

�a� A� b� ≤ �c� C� d� ⇔ ∃x ∈ �n C + x ⊆ A� c + x = a� and d + x = b�

From a semigroup theoretic point of view, a notable feature of tiling semigroups isthat they are inverse and strongly E∗-unitary.

The best studied case is that of one-dimensional tiling semigroups, that is,tiling semigroups associated with tilings of the real line. Since such tilings can beidentified with bi-infinite words over a finite alphabet, there is a natural languageassociated with each one-dimensional tiling � , which consists of all the (finite) wordsthat occur in the tiling. This is called the tiling language of � and is denoted byL�� �. It is obviously a factorial language. Most of the findings regarding one-dimensional tiling semigroups rely on the tiling language. Next, we present threeexamples of this fact that will be important for, and indeed motivated, the workthat follows. Let � be a one-dimensional tiling and L an arbitrary factorial languageover a finite alphabet �.

1) Let S�L� be the set with elements 0 and all sequences of the form

xay� with a ∈ �� x� y ∈ �∗� and xay ∈ L�

xaybz� with a� b ∈ �� x� y� z ∈ �∗� and xaybz ∈ L� and

xaybz� with a� b ∈ �� x� y� z ∈ �∗� and xaybz ∈ L�

and consider the following operation in S�L�, that clearly imitates the operationof a tiling semigroup: given two nonzero elements s and t, their product is

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nonzero when the underlying words of s and t match as the letter of s carryingthe acute accent and letter of t carrying the grave accent are superimposed (inthe case of sequences of the form xay, the letter with the check carries bothaccents) and the resulting word belongs to L; the resulting sequence inheritesthe grave accent from s and the acute accent from t. In a nonzero element,the letter carrying the grave accent, respectively, the acute accent, is called thecalled in-letter, respectively, out-letter of the element. It has been shown thatS�L� is a strongly E∗-unitary inverse semigroup and that, whenever L = L�� �,then S�� � and S�L� are isomorphic [9]. In particular, this is a construction thatgeneralizes the tiling semigroup. We call S�L� the inverse semigroup associatedwith the factorial language L.

2) A word w ∈ �∗ is a minimal forbidden word of L if w �∈ L but all proper factorsof w belong to L. Let M�L� denote the set of minimal forbidden words of L; dueto the fact that L is factorial, we have

M�L� = �w ∈ �∗\L w = axb� with a� b ∈ �� x ∈ �∗ and ax� xb ∈ L �

In [12], the authors constructed a presentation for the inverse semigroupassociated to a factorial language.

Theorem 1.2 (Theorem 3.3). Let L be a factorial language over �. Then

S�L� = Inv0 A u = v ∈ RS� z = 0 ∈ RL� �

where

A = �a a ∈ � ∪ {aiaj ai� aj ∈ �

}�

RS = �aa = a a ∈ � ∪ {aiaj = 0 ai� aj ∈ �� ai �= aj

}∪ {

aiaiaj = aiaj� aiaj aj = aiaj aiaj ∈ L}

∪ {aiaj akaj

−1 = 0 aiaj� akaj ∈ L with ai �= ak

}∪ {

aiaj−1aiak = 0 aiaj� aiak ∈ L with aj �= ak

}�

and

RL = �a1a2a2a3 � � � am−1am = 0 a1 � � � am ∈ M�L� �

In particular, this result gives a presentation for one-dimensional tilingsemigroups. Moreover, the presentation is finite if and only if the set M�L� is finite,as is the case for periodic tilings, that is, tilings whose bi-infinite word consists of therepetition of some finite word.

3) Again using the representation of a tiling semigroup as the inverse semigroupassociated with a factorial language, the authors found a representation for thissemigroup as a P∗-semigroup, which we formulate in the following theorem.

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Theorem 1.3 (Theorem 4.2, [12]). Let � = �× L and � = ��m�w� ∈ � 1 ≤ m ≤w. Then, defining

�m�w� ≤ �n� v� ⇔ ∃x� y ∈ �∗ such that w = xvy and x = m− n�

for all �m�w�, �n� v� ∈ � , and the action of � on � by k · �m�w� = �m− k�w�, for allk ∈ � and �m�w� ∈ � , we have that ����0��0� is a McAlister 0-triple and S�L� �P∗����0��0�.

In this work, we will consider a special kind of n-dimensional tiling. Ann-dimensional hypercubic tiling is a tiling of �n by hypercubes of side 1 (that is,translations of �0� 1�n), possibly coloured, and aligned in such a way that the centreof each tile belongs to �n. This means that adjacent tiles share a common face. Sincewe only consider finite type tilings, all one-dimensional tilings can be thought ofas one-dimensional hypercubic tilings: instead of regarding the tiling as a collectionof intervals of (possibly) different lengths, consider the tiles as being colouredintervals with length 1. For convenience, we will refer to the tiling semigroup of ann-dimensional hypercubic tiling as an n-dimensional hypercubic tiling semigroup.

The article is organized in the following way. In Section 2, we generalize ton-dimensional hypercubic tiling semigroups the description, given in [12] for one-dimensional tiling semigroups, as the inverse semigroup associated to a factoriallanguage (Theorem 2.2) and as a P∗-semigroup (Theorem 2.5). In Section 3, weestablish the infinite presentability of n-dimensional hypercubic tiling semigroups(Theorem 3.4 and Remark 3.5), making use of the Cayley graph expansion of agroup presentation introduced by Margolis and Meakin in [10]. Whereas the resultsin Section 2 are a close imitation of those found in the one-dimensional case, theresults in Sections 3 contrast with them. Finally, in Section 4 we give a necessaryand sufficient condition for two hypercubic tiling semigroups to be isomorphic(Theorem 4.1), using both the classical and the language representation of tilingsemigroups.

2. THE GENERALIZATION OF THE LANGUAGE REPRESENTATION

An n-dimensional hypercubic tiling � whose tiles are coloured with coloursfrom a set � can be identified with a map f �n → � in the obvious way: f�x� is thecolour of the hypercube in the tiling whose centre is x, for each x ∈ �n. Conversely,each map f �n → � gives rise to a unique n-dimensional hypercubic tiling. To takeadvantage of this observation, we define the following setup.

Let � be a finite set. By a �-coloured subset of �n, we shall mean a pair �A� g�where A is a subset of �n and g is a map from A to �. We say that �A� g� isfinite, nonempty if A is finite, nonempty; we say that it is connected if A is the vertexset of a connected subgraph of the Cayley graph of �n with respect to the usual(additive) group presentation �n = Gp X C� with X = n and C = �x + y − x −y = 0 x� y ∈ X; that is, the graph with vertex set �n and edges �g� g + x� with g ∈�n and x ∈ X. There is an obvious partial order on the set of �-coloured subsets of�n: set �A� g� ≤ �B� h� if and only if A ⊆ B and h�a� = g�a� for each a ∈ A. Underthis partial order, �A� g� and �B� h� have a common upper bound if and only if g andh agree on A ∩ B; in this case, their least upper bound is �A� g� ∨ �B� h� = �A ∪ B� k�

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where k�c� = g�c� if c ∈ A, while k�c� = h�c� if c ∈ B. The group �n acts in a naturalway—by translation—on the set of all �-coloured subsets of �n: given a �-colouredsubset �A� g� of �n and u ∈ �n, then �A� g�+ u = �A+ u� h� where h A+ u → � isdefined by h�a+ u� = g�a� for each a ∈ A. Evidently, this action preserves colours,finiteness, and connectedness.

An n-dimensional hypercubic tiling � , with colours from a set �, is thus a�-colouring ��n� f� of �n, and a pattern of � is a finite, nonempty, connected�-coloured subset �A� g� with �A� g� ≤ ��n� f�. Since the colouring on A is therestriction f A of f to A and so is determined by A, whenever there is no danger ofconfusion, we shall omit explicit mention of the colouring map of A and refer to Aas a �-coloured subset of ��n� f�.

Let � = ��n� f� be an n-dimensional tiling. Because each finite, nonempty,connected �-coloured subset of ��n� f� corresponds to a unique pattern in the tiling,we can identify the elements of the tiling semigroup S�� � with the equivalenceclasses defined by the equivalence relation

�a� A� b� ∼ �c� C� d� ⇔ ∃x ∈ �n such that A+ x = C� a+ x = c and b + x = d�

where A and C are finite, nonempty, connected �-coloured subsets of ��n� f�, a� b ∈A, and c� d ∈ C. Recall that when we speak of a �-coloured subset A of ��n� f�we implicitly consider the restriction of the map f to the set A and note that therelation above is colour-preserving; that is, when we write “A+ x = C”, this impliesthat f A+x = f C . In order to obtain the tiling semigroup S�� �, we equip this setwith the following multiplication: given �a� A� b�� �c� C� d� ∈ S�� �, let

�a� A� b��c� C� d� = �a+ x� �A+ x� ∪ �C + y�� d + y��

in case there exist translations x� y ∈ �n such that �A+ x� ∪ �C + y� is afinite, nonempty, connected �-coloured subset of ��n� f� and b + x = c + y, and�a� A� b��c� C� d� = 0 otherwise; all products involving 0 yield 0. Note that �A+x� ∪ �C + y� is obviously finite and nonempty, since both sets A and C are finiteand nonempty, and connected, provided b + x = c + y; thus, �A+ x� ∪ �C + y� is afinite, nonempty, connected �-coloured subset of ��n� f� if and only if the colouringsof A+ x and C + y agree on their overlap �A+ x� ∩ �C + y�. In other words, theproduct �a� A� b��c� C� d� is nonzero if and only if �A+ x� f� ∨ �C + y� f� exists in��n� f� for some x� y ∈ �n such that b + x = c + y.

Since comparing patterns in the tiling � now amounts to comparing finite,nonempty, connected �-coloured subsets of ��n� f�, we consider the followingwell known order in �n.

Lemma 2.1. The binary relation defined by

x≤ y⇔ x= y or there exists 1 ≤ i ≤ n such that xi < yi and xj = yj for all i < j ≤ n�

for all x = �x1� � � � � xn�� y = �y1� � � � � yn� ∈ �n, is a total order compatible with theoperation in �n.

Note that, from the fact that ≤ is a total order, every finite subset A of �n

has a minimum element, which we will denote by A0 or a0, and that the least upper

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bound x ∨ y of any two elements x and y of �n always exists. Also note that, fromthe fact that ≤ is compatible, �A+ x�0 = a0 + x, for all x ∈ �n.

We now introduce the key notion for the generalization we aim for. Given ann-dimensional hypercubic tiling � = ��n� f�, we define the language of the tiling �or tiling language of � to be the set

L�� � = �A− a0 A is a finite, nonempty, connected �-coloured subset of ��n� f� �

Note that the language of an n-dimensional hypercubic tiling semigroup is factorial,in the sense that if A is a finite, nonempty, connected �-coloured subset of ��n� f�,so that A− a0 ∈ L�� �, and B is a nonempty and connected �-coloured subset of A,then also B − b0 ∈ L�� �. In the one-dimensional case, this is equivalent to sayingthat w ∈ L�� � implies that each subword of w belongs to L�� �, and agrees withthe fact that the language of a one-dimensional tiling is always factorial. Note thattwo-dimensional tiling languages are not a particular case of the two-dimensionallanguages considered in [1], since these are sets of rectangular arrays of symbols.

Let � be an n-dimensional hypercubic tiling. Consider the set

S�L� = S�L�� �� = ��p� P� q� P ∈ L�� � and p� q ∈ P ∪ �0 �

Theorem 2.2. Let � be an n-dimensional hypercubic tiling. Equipped with themultiplication

�p� P� q� ∗ �r� R� s� = �p− q + q ∨ r� ��P − q� ∪ �R− r��+ q ∨ r� s − r + q ∨ r�

if �P − q� ∪ �R− r�+ q ∨ r ∈ L�� � and 0 otherwise, with all other products equal to0, the semigroup S�L� is isomorphic to the tiling semigroup S�� �. In particular, S�L� isan E∗-unitary inverse semigroup with zero.

Proof. Straightforward calculations show the result. �

Remark 2.3.

1. As recalled in Section 1, in [8] Lawson showed that the tiling semigroup S�� �of a one-dimensional tiling � , which is identified with a bi-infinite word overa finite alphabet, is isomorphic to the inverse semigroup S�L� associated witha factorial language, consisting of all the words that occur in the bi-infiniteword, that is, the patterns of the tiling identified up to translation in � or,equivalenty, in �, if we think of the tiles as being coloured intervals with length 1.Theorem 2.2 shows that exactly the same is true for an n-dimensional hypercubictiling semigroup. In fact, for n = 1 the construction amounts to the same asLawson’s representation: both the elements of S�L� and the operation defined arethe same, for if �p� P� q�� �r� R� s� ∈ S�L�, then ��P − q� ∪ �R− r��+ q ∨ r is theoutcome of the matching of the out-tile of �p� P� q� with the in-tile of �r� R� s�,either by shifting P by r − q in case q ∨ r = r or by shifting R by q − r in caseq ∨ r = q (and the product is nonzero if and only if the matching is a pattern inthe tiling), q ∨ r + p− q is the position of p, the in-tile of �p� P� q�, in ��P − q� ∪�R− r��+ q ∨ r, and q ∨ r + p− q the position of s, the out-tile of �r� R� s�.

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2. The outer coordinates multiply in S�L� exactly as in the bicyclic monoid B =�+

0 × �+0 —recall that, for all �p� q�� �r� s� ∈ B, we have

�p� q��r� s� = �p− q +max�q� r�� s − r +max�q� r��

—and the connection is the following: there is a pre-homomorphism fromS�L� onto the inverse semigroup with zero T = ��n × �n�0, with multiplicationdefined as in the bicyclic monoid, mapping a nonzero element �p� P� q� to�p� q� and yielding ET = ES�L��, and whereas the semilattice of idempotents ofB is isomorphic to C� = �e0� e1� e2� � � � with e0 > e1 > e2 > � � � , or, equivalently,anti-isomorphic to �+

0 , the semilattice of nonzero idempotents of T is anti-isomorphic to �n (with respect to the order of �n fixed in Lemma 2.1).

We have just observed how S�L� generalizes to higher dimensions therepresentation of a one-dimensional tiling semigroup as an inverse semigroupassociated with a factorial language. Furthermore, just as in the one-dimensionalcase, we have the following proposition.

Proposition 2.4. Let �1 and �2 be two n-dimensional hypercubic tilings. If L��1� ⊆L��2�, then

I = ��p� P� q� P ∈ L��2�\L��1� and p� q ∈ P ∪ �0

is an ideal of S�L��2�� and S�L��1�� � S�L��2��/I .

Proof. Let �1 = ��n� f1� and �2 = ��n� f2� be two n-dimensional hypercubictilings, and denote L��i� by Li and S�L��i�� by Si, with i ∈ �1� 2.

Clearly, I ⊆ S2. Let �p� P� q� ∈ I and �u� U� v� ∈ S2. If �p� P� q� ∗ �u� U� v� = 0,then trivially �p� P� q� ∗ �u� U� v� ∈ I ; if �p� P� q� ∗ �u� U� v� �= 0, then

�p� P� q� ∗ �u� U� v� = �p− q + q ∨ u� ��P − q� ∪ �U − u��+ q ∨ u� v− u+ q ∨ u��

In order to obtain a contradiction, suppose that ��P − q� ∪ �U − u��+ q ∨ u ∈ L1.Then ��P − q� ∪ �U − u��+ q ∨ u = A− a0 for some finite, nonempty, connected �-coloured subset A of ��n� f1�; that is, ��P − q� ∪ �U − u��+ q ∨ u+ a0 is a finite,nonempty, connected �-coloured subset of ��n� f1�. In particular, so is P − q + q ∨u+ a0. Then, by definition of L1, we have that P − q + q ∨ u+ a0 − �P − q + q ∨u+ a0�0 ∈ L1. Now, since P ∈ L2 implies that P = B − b0 for some finite, nonempty,connected �-coloured subset B of ��n� f2�, we have P0 = �B − b0�0 = b0 − b0 = 0;but then �P − q + q ∨ u+ a0�0 = P0 + q − q ∨ u− a0 = q − q ∨ u− a0. Therefore,P = P − q + q ∨ u+ a0 − �P − q + q ∨ u+ a0�0 ∈ L1, which is a contradiction sinceP ∈ L2\L1. A similar argument shows that I is a left ideal as well.

Now, since the nonzero elements of S1 are precisely the elements in S2\Iand since, as we have just seen, �p� P� q� ∗ �r� R� s� �= 0 in S1 if and only �p� P� q� ∗�r� R� s� �= 0 in S2, we conclude that S1 is isomorphic to the Rees quotient S2/I . �

We now generalize the description given in [12] of a one-dimensional tilingsemigroup as a P∗-semigroup to tiling semigroups of n-dimensional hypercubictilings. Note that it is as similar as it could possibly be (cf. Theorem 1.3).

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HYPERCUBIC TILING SEMIGROUPS 2009

Theorem 2.5. Let � be an n-dimensional hypercubic tiling semigroup. Let � = �n ×L�� � and � = ��p� P� ∈ � p ∈ P. Then, defining

�p� P� ≤ �q�Q� ⇔ Q− q ⊆ P − p�

for all �p� P�� �q�Q� ∈ � , and the action of �n on � by u · �p� P� = �p− u� P�, for allu ∈ �n and �p� P� ∈ � , we have that ��n��0��0� is a McAlister 0-triple and S�� � �P∗��n��0��0�.

Sketch of the Proof. Most of the proof is routine checking; we comment onlyon a few steps. It can be shown that �p� P� ∧ �q�Q� = �p ∨ q� ��P − p� ∪ �Q− q��+p ∨ q� if ��P − p� ∪ �Q− q��+ p ∨ q ∈ L�� � and 0 otherwise, for all �p� P� ∈ � and�q�Q� ∈ � (or �p� P� ∈ � and �q�Q� ∈ �), and so � is an inf-subsemilattice and anorder ideal of � . By definition of P∗-semigroup, we have

P∗��n��0��0� = ���p� P�� u� ∈ � × �n �−u� · �p� P� ∈ � ∪ �0

= ��p� P� u� ∈ �n × L�� �× �n p� p+ u ∈ P ∪ �0�

where the obvious identification between the Cartesian products � × �n and�n × L�� �× �n is made. Finally, to prove that S�� � � P∗��n��0��0� show thatS�L�� �� � P∗��n��0��0�, via the map S�L�� �� → P∗��n��0��0� defined by0 = 0 and �p� P� q� = �p� P� q − p�, for all �p� P� q� ∈ S�L�� ��\�0. �

3. INFINITE PRESENTABILITY

It is well known that the tiling semigroup S�� � associated with an arbitraryn-dimensional tiling � is a strongly E∗-unitary inverse semigroup with zero (see, forexample, [13]), by means of the 0-restricted idempotent-pure pre-homomorphism �of S�� � into the group with zero ��n�+�0

′that maps 0 to 0′ and each nonzero

element �p� P� q� to the vector vpq with initial point in the center of the tile p andterminal point in the center of q. Since we only deal with finite type tilings, whichimplies that S�� � is finitely generated, and since the group

(�d�+)

is Abelian andtorsion-free, we have that �S�� �� � is a finitely generated Abelian torsion-free group,and so isomorphic to some �m. In the case of an n-dimensional hypercubic tiling, itis easy to see that this group is �n.

This observation leads to an interesting feature of any tiling semigroup. Givennonzero elements x = �p� P� q� and y = �r� R� s� in S�� � such that xyx−1y−1 �= 0, then

(xyx−1y−1

)� = x�+ y�+ x�−1 + y�−1 = vpq + vrs − vpq − vrs = 0�

so that xyx−1y−1 is an idempotent in S�� �. In this section, we prove that anyn-dimensional hypercubic tiling semigroup S�� � is infinitely presented, even asa strongly E∗-unitary inverse semigroup that admits an idempotent-pure pre-homomorphism into an Abelian group with zero. This means that S�� � requires aninfinite number of relations beyond those that ensure that all elements of the formxyx−1y−1 are idempotents. We prove this for the two-dimensional single-colouredhypercubic tiling semigroup, but the proof presented adapts to any dimension andto any (finite) set of colours.

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Note that a two-dimensional hypercubic tiling with colours from a set � isgenerated, as an inverse semigroup, by the � single-tile idempotents and the two-tile elements

with i� j ∈ �, such that the underlying pattern occurs in the tiling. Thus, in thecase of the two-dimensional single-coloured hypercubic tiling (note that in this case,we can regard S�� � as an E-unitary inverse monoid, rather than as an E∗-unitaryinverse semigroup), the set �a� b, where we are denoting by a and b the generatorsa�1�1� and b�1�1�, respectively, generates S�� � as an inverse monoid.

When the set of colours has only one element, the tiling semigroup can berepresented as a quotient of a well known construction: the graph expansion of agroup presentation (see [10]). We recall this construction in the particular case ofthe free Abelian group FAG�X� generated by X with the usual group presentationGp X C� and its associated Cayley graph ��X�C� (cf. Section 2). Margolis andMeakin defined the Cayley graph expansion of ��X�C� to be the monoid M�X�C�with elements ��� g�, where � is a finite, nonempty, connected subgraph of ��X�C�and 0� g ∈ V���, and multiplication given by

��� g���� h� = �� ∪ ��+ g�� g + h��

The following summarizes some of the results obtained by Margolis and Meakin in[10], where � denotes, as usual, the minimum group congruence. When necessary,a subscript will be added to clarify on which monoid is the minimum groupcongruence defined. Also, �� denotes the canonical epimorphism associated with thecongruence �.

Theorem 3.1. In the notation above, M�X�C� is an E-unitary monoid generated byX as an inverse monoid with maximum group image FAG�X�. Furthermore:

i) ��� g� is idempotent if and only if g = 0;ii) ��� g��� = ��� h��� if and only if g = h;iii) M�X�C� possesses the following universal property: given any E-unitary monoid M

generated by X as an inverse monoid with maximum group image FAG�X�, thereexists a unique idempotent-pure epimorphism M�X�C� � M such that ��

M =��M�X�C�;

iv) M�X�C� is defined by the inverse monoid presentation

Inv⟨X (uvu−1v−1

)2 = uvu−1v−1(u� v ∈ (

X ∪ X−1)+)⟩

�

So let S�� � = ��2� f�, with f �2 → � and � = 1, be the two-dimensionalsingle-coloured hypercubic tiling semigroup and take X = �a� b. We have alreadynoted that S�� � is an E-unitary inverse monoid generated by X; it is not hard to seethat its maximum group image is FAG�X�, that is, �2. In fact, since every element

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of S�� � has a representative of the form �0� A� a� and since S�� � is E-unitary, given�0� A� a�� �0� B� b� ∈ S�� �, we have

�0� A� a�� = �0� B� b�� ⇔ �0� A� a�−1�0� B� b� ∈ ES�� � ⇔ �a� A ∪ B� b� ∈ ES�� � ⇔ a = b�

The next proposition describes the unique idempotent-pure epimorphism M�X�C� � S�� � with the property that ��

S�� � = ��M�X�C�.

Proposition 3.2. The mapping M�X�C� � S�� � is defined by ��� u� =�0� V���� g�, for each ��� g� ∈ M�X�C�. Moreover, its kernel is

ker = {(��� g�� ��� h�

) V��� = V��� and g = h

}�

Proof. Let ��� g� ∈ M�X�C�. Then � is a finite, nonempty, connected subgraph of��X�C�, and so V��� ⊂ �2 is a finite, nonempty, connected �-coloured subset of��2� f�. Since, by definition, g ∈ V���, then �0� V���� g� is an element of S�� �. It isstraightforward to check that is a homomorphism—simply note that, for all ��� g�,��� h� ∈ M�X�C�, we have

V�� ∪ ��+ g�� = V��� ∪ �V���+ g��

Since is clearly onto, we conclude that it is an epimorphism. It is also easy tosee that is idempotent-pure, since an element ��� g� ∈ M�X�C� is idempotent ifand only if g = 0; that is, if and only if the in- and out-tiles of ��� g� coincide,or, equivalently, ��� g� is an idempotent of S�� �. That ��

S�� � = ��M�X�C�, is an

immediate consequence of the characterization of the minimum group congruenceon M�X�C� and on S�� � and of the definition of .

Finally, the characterization of ker follows easily using the definition of and the definition of the equivalence relation that gives rise to the elements of S�� �,for given ��� g�� ��� h� ∈ M�X�C�, we have �0� V���� g� = �0� V���� h� if and only ifV��� = V��� and g = h. �

Next, we establish what will be for us the key property of the map .

Proposition 3.3. The congruence ker is infinitely generated.

Proof. In order to obtain a contradiction, assume that ker is generated by thefinite set

R = {(��i� gi�� ��i� gi�

) ∈ M�X�C�×M�X�C� i ∈ I} ⊂ ker �

Then, since each �i (and �i) in a pair belonging to R is finite, we conclude that therecannot exist ���i� gi�� ��i� gi�� ∈ R such that V��i� = V��i� has arbitrarily manyvertices; take v to be the maximum number of vertices of the graphs of elements inR. For simplicity, assume without loss of generality that R = R−1 and that all pairsin R are nontrivial.

The cases v = 1 and v = 2 and v = 3 are trivial, as there are no distinct, ker -related elements ��i� gi� and ��i� gi� such that V��i� = V��i� has that (or fewer)number of vertices.

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Thus, assume v ≥ 4. Let m = ⌊v2

⌋; that is, let m be the largest integer smaller

or equal to v2 . Since

⌊v2

⌋ = v2 if v is even and

⌊v2

⌋ = v−12 if it is odd, we have m ≥ 2.

Let � be the finite, nonempty, connected subgraph of ��X�C�

with 4m vertices. Note that, since m ≥ 2, then a+ b does not belong to V���.Consider the element ��� 0� of M�X�C�. Then ��� 0� = ��c� 0� , with �c the graph

thus, V��� = V��c�, but whereas �0� a� is an edge in �c, it is not so in � . Then��� 0� and ��c� 0� are distinct, ker -related elements of M�X�C�, and so there exista positive integer n and a nontrivial sequence

��� 0� = ��1� s1� → ��2� s2� → · · · → ��n� sn� = ��c� 0�

of elementary R-transitions; that is, there exist elements ��1� s1�� ��2� s2�� � � � ,��n� sn� in M�X�C� such that, for each i ∈ �1� � � � � n− 1,{

��i� si� = ��i� ai���i� gi���i� bi�

��i+1� si+1� = ��i� ai���i� gi���i� bi��

for some ��i� ai�� ��i� bi� ∈ M�X�C�1 and(��i� gi�� ��i� gi�

) ∈ R.

Claim 1. All graphs �i are connected subgraphs of �c with V��i� = V��c�. First,note that {

��i� si� = ��i ∪ ��i + ai� ∪ ��i + ai + gi�� ai + gi + bi�

��i+1� si+1� = ��i ∪ ��i + ai� ∪ ��i + ai + gi�� ai + gi + bi��

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Thus,

� = �1 = �1 ∪ ��1 + a1� ∪ ��1 + a1 + g1�

�2 = �1 ∪ ��1 + a1� ∪ ��1 + a1 + g1�

= �2 ∪ ��2 + a2� ∪ ��2 + a2 + g2�

�3 = �2 ∪ ��2 + a2� ∪ ��2 + a2 + g2�

= �3 ∪ ��3 + a3� ∪ ��3 + a3 + g3�

���

�n−1 = �n−2 ∪ ��n−2 + an−2� ∪ ��n−2 + an−2 + gn−2�

= �n−1 ∪ ��n−1 + an−1� ∪ ��n−1 + an−1 + gn−1�

�c = �n = �n−1 ∪ ��n−1 + an−1� ∪ ��n−1 + an−1 + gn−1��

By definition, all �i are connected graphs, since ��i� si� ∈ M�X�C� for all i. From

� = �1 = �1 ∪ ��1 + a1� ∪ ��1 + a1 + g1��

we conclude that �1, �1 + a1, and �1 + a1 + g1 are connected subgraphs of� , and so connected subgraphs of �c, and that V��1� = V��� = V��c�. Since(��1� g1�� ��1� g1�

) ∈ R implies that V��1� = V��1�, then also �1 + a1 is a connectedsubgraph of �c (although not necessarily of �), as �c contains all possible edgesbetween its vertices that a subgraph of ���a� b� C� with vertex set V��c� can contain.Therefore, �2 = �1 ∪ ��1 + a1� ∪ ��1 + a1 + g1� is a connected subgraph of �c withV��2� = V��c�

V��2� = V��1� ∪ V��1 + a1� ∪ V��1 + a1 + g1�

= V��1� ∪ V��1 + a1� ∪ V��1 + a1 + g1� = V��1� = V��c��

Now since �2 = �2 ∪ ��2 + a2� ∪ ��2 + a2 + g2�, we have that �2, �2 + a2, and �2 +a2 + g2 are connected subgraphs of �c, and we can conclude, as above, that �3 =�2 ∪ ��2 + a2� ∪ ��2 + a2 + g2� is a connected subgraph of �c with V��3� = V��c�.Proceeding in a similar manner for the remaining graphs �i, we have our claim.

Claim 2. si = 0, for all i. On the one hand, we have that si = ai + gi + bi =si+1, for all i ∈ �1� � � � � n− 1; on the other hand, the fact that ��� 0� = ��1� s1�implies that s1 = 0.

Hence, since by assumption all R-transitions are nontrivial, what happens ineach step is the addition, deletion or substitution of at least one edge in the graph

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�i + ai by replacing it by �i + ai:

To complete � to �c, we must add the edge �0� a�.

Claim 3. The edge �0� a� cannot be added using only pairs from R. Since �0� a�is an edge in �c but not in � , at some point of the sequence ��� 0� = ��1� s1� →��2� s2� → · · · → ��n� sn� = ��c� 0�, it will be added by an R-transition. Then, thereis a graph �i + ai which contains the vertices 0 and a but not the edge �0� a�. Since�i + ai is a connected graph, there exists a path in �i + ai joining 0 and a; and since�i + ai is a connected subgraph of �c, this path is a path in �c. Therefore, there is apath in �c joining the vertices 0 and a and avoiding the edge �0� a�. Now, there is aunique such path, and it contains all 4m vertices of �c. But this cannot be a path in�i + ai, since �i has at most v vertices. Therefore, it is impossible to add the edge�0� a� to � using only pairs

(��i� gi�� ��i� gi�

)from R.

Hence, we conclude that no finite set R generates ker . �

Theorem 3.4. The two-dimensional single-coloured hypercubic tiling semigroup isinfinitely presented, even as an E-unitary inverse monoid with maximum groupimage �2.

Proof. The result follows from the previous proposition and the fact that M�X�C�is defined by the inverse monoid presentation

Inv⟨X (uvu−1v−1

)2 = uvu−1v−1(u� v ∈ (

X ∪ X−1)+)⟩

�

Details can be found in [12]. �

Remark 3.5.

1. Proposition 3.2 can be generalized to any n-dimensional hypercubic tilingsemigroup S�� �, with n ≥ 3: simply take X to have n elements. Further, theanalogue to Proposition 3.3 still holds, as the graphs used in the proof are stillsubgraphs of ��X�C� in the general case (with a and b any distinct elementsof X). Thus, we can conclude that, as in Theorem 3.4 for the two-dimensionalsingle-coloured hypercubic tiling semigroup, any n-dimensional single-colouredhypercubic tiling semigroup S�� � is infinitely presented, even as an E-unitaryinverse monoid with maximum group image �n.

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2. Also, the argument used in the proof of Proposition 3.3 can be adapted to thegeneral case of an arbitrary n-dimensional hypercubic tiling with colours in afinite set �. Since the reasoning is precisely the same (only the details in theconstruction of an appropriate analogue to the semigroup M�X�C� being moreelaborate) we omit this proof; the details can be found in [14]. As a consequence,Theorem 3.4 or, more precisely, its generalization carried on in part 1 of thisremark, can be used to show that an arbitrary n-dimensional hypercubic tilingsemigroup is infinitely presented—now, even as a strongly E∗-unitary inversesemigroup that admits an idempotent-pure pre-homomorphism into an Abeliangroup with zero.

Note how these results contrast with the one-dimensional case. A one-dimensional tiling semigroup is finitely or infinitely presented according to whetheror not the set of minimal forbidden words is finite. Recall that the minimalforbidden words are the words that do not belong to the language but whose properfactors all do. Thus, for one-dimensional tilings, the part of the operation of thetiling semigroup that does not have to do with the matching of colours (or lengths)is always finitely presented. We have shown that it is not so in higher dimensions.

4. ISOMORPHISM OF HYPERCUBIC TILING SEMIGROUPS

In [6], Kellendonk remarks that two n-dimensional tilings have the same tilingsemigroup if and only if each pattern in the first appears as a pattern in the second,and vice-versa. In terms of hypercubic semigroups, this is equivalent to saying thatthe semigroups are equal if and only if they have the same tiling languages. Inthis section, we turn to the question of when two n-dimensional hypercubic tilingshave isomorphic tiling semigroups. The answer is that, under mild conditions, thesemigroups of two n-dimensional hypercubic tilings are isomorphic if and only iftheir languages are the same, subject to a bijective change of colours and a symmetryof the n-dimensional hypercube.

Let Gn denote the full symmetry group of the n-dimensional hypercube. Recallthat this group consists of all automorphisms of the n-dimensional hypercube, thatis, permutations of its vertices that preserve neighbours. For instance in �2, such anautomorphism is either the identity, a reflection about the x- or y-axis or the straightlines y = x or y = −x, or a rotation about the origin of �

2 , or � or 3�2 . See [2] for a

survey of group Gn. Note that each automorphism of the n-dimensional hypercubeextends naturally to an automorphism of �n (in fact, of the Euclidean space �n).Also note that each � ∈ Gn is represented, with respect the standard basis of �n, bya n× n matrix that has exactly one nonzero entry in each column and in each row,which is either 1 or −1, that is, a signed permutation matrix. This is a consequenceof the fact that � sends a standard basis vector uj , with j ∈ �1� � � � � n, either to astandard basis vector uk or to −uk, for some k ∈ �1� � � � � n, and that in this case nodistinct standard basis vector uj′ is also mapped to either uk or −uk. In particular,� defines a linear map from �n to �n.

Since any symmetry from Gn takes �n onto itself, there is a natural action ofGn on the set of �-coloured subsets of �n: if � is a symmetry from Gn and �A� g� isa �-coloured subset of �n, then �A� g�� = �B� h� is the �-coloured subset of �n withB = �x� x ∈ A and h B → � the map defined by h�b� = g�b�−1�, for all b ∈ B.

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Thus, the action of � may change the position of a vertex, but it does not changethe colour of the corresponding vertex. Also note that if �A� g� is finite, nonempty,or connected, then so is �A� g��.

Given two finite alphabets �1 and �2 and a bijection �1 → �2, denote by ∗ the map that naturally extends to the �1-coloured subsets of �n: for each �1-coloured subset �A� g� of �n, set �A� g� ∗ to be the �2-coloured subset �A� h� of�n with h�a� = �g�a��, for all a ∈ A. As usual, whenever the colour maps are wellunderstood by the context, we omit any mention of them, and simply write A ∗.

The main theorem of this section is the following one.

Theorem 4.1. Let �1 = ��n� f1� and �2 = ��n� f2� be two n-dimensional hypercubictilings with f1 �

n → �1 and f2 �n → �2. Let �1 → �2 be a bijection and � a

symmetry of Gn. Suppose that if A is a pattern of �1 then A ∗� + v is a pattern of �2

for some v ∈ �n, and if B is a pattern of �2 then B� −1�∗�−1 + v is a pattern of �1 forsome v ∈ �n. Then the mapping � S��1� → S��2� defined by{

0� = 0�

�a� A� b�� = �a � + v� A ∗� + v� b � + v�� �a� A� b� ∈ S��1�\�0�where v ∈ �n is such that A ∗� + v is a pattern of �2, is an isomorphism. Conversely,if both �1 and �2 contain all patterns up to four tiles, every isomorphism of S��1� toS��2� is of this form for some bijection �1 → �2 and some symmetry � ∈ Gn.

In order to prove this result, we begin by establishing a few helpful factsabout the relation between the images of the generators when the semigroups oftwo n-dimensional hypercubic tilings are isomorphic. First, recall from Section 3that a two-dimensional hypercubic tiling with colours from a set � is generated, asan inverse semigroup with zero, by the � single-tile idempotents and the two-tileelements a�i�j� and b�i�j�, with i� j ∈ �, such that the underlying pattern occurs in thetiling. In general, the tiling semigroup S�� � of an n-dimensional hypercubic tiling� with colours from a set � is generated, as an inverse semigroup with zero, by the� single-tile idempotents, which we will denote by ei (with i ∈ �), and the two-tile elements a1�i�j�� a2�i�j�� � � � � an�i�j� (with i� j ∈ �) whose underlying pattern occursin the tiling: each generator ak�i�j� admits as underlying pattern the union of twohypercubes, the in-tile with colour i and the out-tile with colour j, and such that thevector from the center of the in-tile to the center of the out-tile is the standard basisvector uk. Note that there are at most n �2 of these generators.

In Lemmas 4.2–4.4, let �1 = ��n� f1� and �2 = ��n� f2� be two n-dimensionalhypercubic tilings, suppose that �1 and �2 contain all patterns with four tiles, andthat � S��1� → S��2� is an isomorphism. For clarity, we will denote the generatorsof S��1� by ei (with i ∈ �1) and ak�i�j� (with k ∈ �1� � � � � n and i� j ∈ �1) and thegenerators of S��2� by fi (with i ∈ �2) and bk�i�j� (with k ∈ �1� � � � � n and i� j ∈ �2).Since � maps maximal idempotents to maximal idempotents, then for all i ∈ �1

we have ei� = fj , for some j ∈ �2.1 Consider the mapping �1 → �2 defined by

�i� = j if and only if ei� = fj , for all i ∈ �1. It is easy to check, using the fact that� and �−1 are bijections, that is a bijection as well.

1This observation was first made by Nick Gilbert.

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Lemma 4.2. For all ak�i�j� ∈ S��1�, there exists m ∈ �1� � � � � n for which ak�i�j�� =bm� �i�� �j�� or ak�i�j�� = b−1

m� �j�� �i��.

Proof. Since isomorphisms map �-classes onto �-classes, from the description ofGreen’s relations in tiling semigroups (see [13]) we have that for each ak�i�j� ∈ S��1�there exists bm�r�s� ∈ S��2� such that Dak�i�j�

� = Dbm�r�s�; that is,

{ak�i�j�� a

−1k�i�j�� ak�i�j�a

−1k�i�j�� a

−1k�i�j�ak�i�j�

}� =

{bm�r�s�� b

−1m�r�s�� bm�r�s�b

−1m�r�s�� b

−1m�r�s�bm�r�s�

}�

Since � maps idempotents to idempotents, we have ak�i�j�� = bm�r�s� or ak�i�j�� =b−1m�r�s�. More precisely, we must have ak�i�j�� = bm� �i�� �j�� or ak�i�j�� = b−1

m� �j�� �i��: ifak�i�j�� = bm�r�s�, then ak�i�j� = eiak�i�j� implies that

ak�i�j�� = (eiak�i�j�

)� = ei�ak�i�j�� ⇔ bm�r�s� = f �i�bm�r�s� �= 0�

which in turn implies that r = �i�. From the fact that ak�i�j� = ak�i�j�ej we concludesimilarly that s = �j�. Thus, in this case, ak�i�j�� = bm� �i�� �j��. If ak�i�j�� = b−1

m�r�s�, thenak�i�j� = eiak�i�j� now implies that

ak�i�j�� = (eiak�i�j�

)� = ei�ak�i�j�� ⇔ b−1

m�r�s� = f �i�b−1m�r�s� �= 0�

and so s = �i�. Similarly, r = �j� and therefore ak�i�j�� = b−1m� �j�� �i��. �

The following result concerns only one-dimensional (hypercubic) tilings. Forsimplicity of notation, here we denote the two-tile generators of S��1� (respectively,S��2�) by a�i�j�, with i� j ∈ �1 (respectively, b�i�j�, with i� j ∈ �2).

Lemma 4.3. If a�i�j�� = b� �i�� �j�� for some a�i�j� ∈ S��1�, then a�r�s�� = b� �r�� �s��, forall a�r�s� ∈ S��1�; if a�i�j�� = b−1

� �j�� �i�� for some a�i�j� ∈ S��1�, then a�r�s�� = b−1� �s�� �r��,

for all a�r�s� ∈ S��1�.

Proof. Without loss of generality, we will assume that = id. This way, ek� = fkfor all k ∈ �1 = �2 and Lemma 4.2 states that, for all a�i�j� ∈ S��1�, we have a�i�j�� =b�i�j� or a�i�j�� = b−1

�j�i�.Let a�i�j�� a�r�s� ∈ S��1�. Then both words ij and rs occur in the tiling language

of �1, and so there exists a word u ∈ L��1� which has both ij and rs as factors.Without loss of generality, suppose that the factor ij occurs to the left of rs.Then either j = r and u = ijs or u = ijt1t2 � � � tmrs, with t1� t2� � � � � tm ∈ �1 ∪ �1 andt1t2 � � � tm ∈ �∗

1. Let x ∈ S��1� be the element with underlying word u, in-tile inthe first letter and out-tile in the last. Suppose u = ijs. Then x� = (

a�i�j�a�j�s�

)� =

a�i�j��a�j�s�� is a nonzero element of S��2�. If a�i�j�� = b�i�j� and a�j�s�� = b−1�s�j�, then the

fact that the product x� = b�i�j�b−1�s�j� is nonzero would imply that s = i and that x�

is a two-tile idempotent (with underlying word ij and both in- and out-tiles in thefirst tile), whereas x is a non-idempotent three-tile element of S��1�, a contradiction.Thus, a�j�s�� = b�j�s�, that is, a�r�s�� = b�r�s�. Similarly, if a�i�s�� = b−1

�j�i� and a�j�s�� =b�j�s�, then again s = i and the image of the non-idempotent three-tile element x

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would be the two-tile idempotent x� (which has underlying word ji and both in- andout-tiles in the second tile). So a�r�s�� = b−1

�s�r�. Now suppose u = ijt1t2 � � � tmrs. Then

x� = (a�i�j�a�j�t1�

� � � a�tm�r�a�r�s�

)� = a�i�j��a�j�t1�

� � � � a�tm�r��a�r�s���

is a nonzero element of S��2�, and in particular, every product a�i�j��a�j�t1��,

a�j�t1��a�t1�t2�

�� � � � , and a�tm�r�� a�r�s�� is nonzero. Using the same arguments as before,

we conclude that if a�i�j�� = b�i�j�, then a�j�t1�� = b�j�t1�, and that if a�i�j�� = b−1

�j�i�, thena�j�t1�

� = b−1�t1�j�

. Proceeding in a similar way with the remaining pairs we concludethat if a�i�j�� = b�i�j�, then a�r�s�� = b�r�s�, and that if a�i�j�� = b−1

�j�i�, then a�r�s�� = b−1�s�r�,

as claimed. �

The following lemma generalizes to n-dimensional hypercubic tilings, withn≥ 2, the previous result; that is, it establishes that the image of a generator ak�i�j�

determines the image of all generators ak�r�s�. In addition, it also shows how theimage of ak�i�j� restricts the image of the generators ak′�r�s� with k′ �= k.

Lemma 4.4.

a) If ak�i�j�� = bm� �i�� �j�� or ak�i�j�� = b−1m� �j�� �i��, then, for all ak′�r�s� ∈ S��1� with k′ �= k,

we have ak′�r�s�� �= bm� �r�� �s�� and ak′�r�s�� �= b−1m� �s�� �r��.

b) If ak�i�j�� = bm� �i�� �j��, then, for all ak�r�s� ∈ S��1�, we have ak�r�s�� = bm� �r�� �s��; ifak�i�j�� = b−1

m� �j�� �i��, then, for all ak�r�s� ∈ S��1�, we have ak�r�s�� = b−1m� �s�� �r��.

Proof. Again, assume that = id. The outline of the proof is as follows:

1. If ak�i�j�� = bm�i�j� or ak�i�j�� = b−1m�j�i�, then ak′�j�r�� �= bm�j�r� and ak′�j�r�� �= b−1

m�r�j�;2. If ak�i�j�� = bm�i�j�, then ak�r�s�� = bm�r�s�, and, if ak�i�j�� = b−1

m�j�i�, then ak�r�s�� = b−1m�s�r�;

3. If ak�i�j�� = bm�i�j� or ak�i�j�� = b−1m�j�i�, then ak′�r�s�� �= bm�r�s� and ak′�r�s�� �= b−1

m�s�r�.

Claim 1. Suppose that ak�i�j�� = bm�i�j�. Take ak′�j�r� ∈ S��1� with k′ �= k (whichexists since, by assumption, n ≥ 2 and �1 contains all patterns with two tiles). It iseasy to see that ak′�j�r�� �= b−1

m�r�j�: the fact that ak�i�j�ak′�j�r� is, by assumption, a nonzeroelement of S��1� implies that

(ak�i�j�ak′�j�r�

)� is a nonzero element of S��2�, and, if

ak′�j�r�� = b−1m�r�j�, then

(ak�i�j�ak′�j�r�

)� = bm�i�j�b

−1m�r�j� would be a nonzero idempotent, a

contradiction since ak�i�j�ak′�j�r� is not an idempotent. Now suppose ak′�j�r�� = bm�j�r�.Consider the element x = ak�i�j�ak′�j�r�a

−1k�j�r�a

−1k′�i�j� of S��1�. Then x is a nonzero (four-

tile) idempotent, whose projection on the xkxk′ -plane of �n is shown in the figurebelow:

(We denoted by i© the coinciding in- and out-tiles of the element.) Since � isinjective, ak�j�r�� �= bm�j�r� = ak′�j�r��; since ak�i�j�� = bm�i�j� and ak�i�j�ak�j�r� is nonzero,then ak�j�r�� �= b−1

m�r�j�. Thus, by Lemma 4.2, ak�j�r�� = bl�j�r� or ak�j�r�� = b−1l�r�j�, with

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l �=m. However, because l �= m, the image of x cannot be an idempotent,independently of ak′�i�j��: the projection ak�i�j��ak′�j�r��ak�j�r��

−1 on the xmxl-planeof �n is

(In the figure above, ak�j�r�� = b−1l�r�j�, but the same conclusion is drawn for ak�j�r�� =

bl�j�r�.) Therefore ak′�j�r�� �= bm�j�r�. Analogously, if ak�i�j�� = b−1m�j�i�, then ak′�j�r�� �=

bm�j�r� and ak′�j�r�� �= b−1m�r�j�.

Claim 2. Suppose ak�i�j�� = bm�i�j� (again, the case ak�i�j�� = b−1m�j�i� is

analogous). Let ak�j�r� ∈ S��1�. We claim that ak�j�r�� = bm�j�r�. As in the proofof claim 1, consider the nonzero idempotent x = ak�i�j�ak′�j�r�a

−1k�j�r�a

−1k′�i�j� of S��1�.

By claim 1, ak′�j�r�� = bl�j�r� or ak′�j�r�� = b−1l�r�j� with l �= m. If ak′�j�r�� = bl�j�r�, then

ak�j�r�� �= bl�j�r� since � is injective, and so x� is an idempotent if and only ifak�j�r�� = bm�j�r�, because ak′�i�j�� only has two tiles. Similarly, if ak′�j�r�� = b−1

l�r�j�, thenthe injectivity of � implies that ak�j�r�� �= b−1

l�r�j�, so that x� is an idempotent if andonly if ak�j�r�� = bm�j�r�. Applying this argument now to ak�j�r� and ak�r�s�, we concludethat ak�r�s�� = bm�r�s�.

Claim 3. Suppose ak�i�j�� = bm�i�j� and let k′ �= k. Then the fact that ak′�r�s�� �=bm�r�s� and ak′�r�s�� �= b−1

m�s�r� is a consequence of claim 1 since, by claim 2, ak�t�r�� =bm�r�t� for any t.

This proves statements a) and b).�

The following lemma establishes a useful technical property of patterns in anhypercubic tiling.

Lemma 4.5. Let A be a connected subset of �n with at least three tiles and a andb tiles in A. Then there exist B1, B2, and B3 connected proper subsets of A such that:A = B1 ∪ B2 ∪ B3, a ∈ B1, b ∈ B3, B1 ∩ B2 �= ∅, and B2 ∩ B3 �= ∅.

Proof. If A\�a� b is connected and a = b, then we can take B1 = B3 = �a� a′,where a′ is a neighbour of a, and B2 = A\�a; if A\�a� b is connected and a �= b,then either b is the only neighbour of a (or a is the only neighbour of b) and wetake B1 = B3 = �a� b and B2 = A\�a� b, or there exist a′ neighbour of a and b′

neighbour of b and we can take B1 = �a� a′, B2 = A\�a� b, and B3 = �b� b′. IfA\�a� b is not connected and a = b or a and b are neighbours, we take B1 = C ∪�a� b, B2 = �a� b, and B3 = A\C, where C is a connected component of A\�a� b(note that there exist at least two); if A\�a� b is not connected and a and b arenot neighbours, then there exists a connected component C of A\�a� b containinga neighbour of a and a neighbour of b, and we take B1 = Ca ∪ �a, B2 = C ∪�a� b, and B3 = Cb ∪ �b, where Ca (respectively, Cb) is the union of the connectedcomponents of A\�a� b which contain a neighbour of a (respectively, of b). �

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Proof of Theorem 4.1. Suppose that �1 = ��n� f1� and �2 = ��n� f2� containall patterns with four tiles and that � S��1� → S��2� is an isomorphism. Theexistence of a bijection �1 → �2 has already been established. Consider then× n matrix M with 1 (respectively, −1) in the �k�m�-entry whenever ak�i�j�� =bm� �i�� �j��

(respectively, ak�i�j�� = b−1m� �j�� �i��

), for some i� j ∈ �1, and all other entries 0.By Lemma 4.3 for n = 1 and by Lemma 4.4 for n ≥ 2, M has exactly one nonzeroentry in each row and column, which is 1 or −1. For instance, there cannot betwo entries equal to 1 in the same row for it cannot be that ak�i�j�� = bm� �i�� �j�� andak�r�s�� = bp� �r�� �s�� with p �= m. So M is a signed permutation matrix and defines asymmetry � of �n (in fact, of �n). Clearly, 0� = 0; we show that, for all �a� A� b� ∈S��1�, we have �a� A� b�� = �a � + v� A ∗� + v� b � + v�, for some translation v ∈�n, by induction on the number t of vertices of A. This is trivially true for t = 1from the way was defined; for t = 2, it is true from the way was defined andfrom the choice of �. In fact, if A has two tiles, then �a� A� b� is either (i) ak�i�j�, or(ii) a−1

k�i�j�, or (iii) ak�i�j�a−1k�i�j�, or (iv) a

−1k�i�j�ak�i�j�, for some generator ak�i�j� of S��1�. By

Lemma 4.2, we know that:

(i) �a� A� b�� = bm� �i�� �j�� or �a� A� b�� = b−1m� �j�� �i��;

(ii) �a� A� b�� = b−1m� �i�� �j�� or �a� A� b�� = bm� �j�� �i��;

(iii) �a� A� b�� = bm� �i�� �j��b−1m� �i�� �j�� or �a� A� b�� = b−1

m� �j�� �i��bm� �j�� �i��; and(iv) �a� A� b�� = b−1

m� �i�� �j��bm� �i�� �j�� or �a� A� b�� = bm� �j�� �i��b−1m� �j�� �i��,

for some m ∈ �1� � � � � n. In any case, the underlying pattern of �a� A� b�� is the�2-coloured subset A of �n acted upon by the chosen symmetry � with coloursgiven by and in-tile (respectively, out-tile) defined in the same way from thein-tile (respectively, out-tile) of �a� A� b�. Therefore, �a� A� b�� = �a � + v� A ∗� +v� b � + v� for some translation v ∈ �n. Now assume it is true for all patterns witht′ < t tiles and let A have t > 2 tiles. Then, by Lemma 4.5, there exist B1, B2, and B3

patterns in �1 (not necessarily distinct) with at most t − 1 tiles such that: A = B1 ∪B2 ∪ B3, a ∈ B1, b ∈ B3, B1 ∩ B2 �= ∅, and B2 ∩ B3 �= ∅. Let c be a tile in B1 ∩ B2 andd a tile in B2 ∩ B3. Then �a� B1� c��c� B2� d��d� B3� b� = �a� A� b�. By assumption, thereexist translations u, v, and w such that �a� B1� c�� = �a � + u� B1

∗� + u� c � +u�, �c� B2� d�� = �c � + v� B2

∗� + v� d � + v�, and �d� B3� b�� = �d � + w�B3 ∗� +

w� b � + w�. Thus, �a� A� b�� = �a� B1� c�� �c� B2� d�� �d� B3� b�� is a nonzero element,and so

�a� A� b�� = �a � + u� B1 ∗� + u� c � + u��c � + v� B2

∗� + v� d � + v�

�d � + w�B3 ∗� + w� b � + w�

= �a � + u+ u′� �B1 ∗� + u+ u′� ∪ �B2

∗� + v+ v′�� d � + v+ v′�

�d � + w�B3 ∗� + w� b � + w�

= �a � + u+ u′ + u′′� ��B1 ∗� + u+ u′� ∪ �B2

∗� + v+ v′��

+ u′′ ∪ �B3 ∗� + w + w′′�� b � + w + w′′��

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for some translations u′, v′, u′′, and w′′ with{c � + u+ u′ = c � + v+ v′

d � + v+ v′ + u′′ = d � + w + w′′ �

But then u+ u′ = v+ v′ and v+ v′ + u′′ = w + w′′. Therefore, taking

z = u+ u′ + u′′ = v+ v′ + u′′ = w + w′′�

we get

�a� A� b�� = �a + z� �B1 ∪ B2 ∪ B3� ∗� + z� b + z� = �a + z� A ∗� + z� b + z��

with z a translation in �n, as claimed.Straightforward calculations show the reverse implication. �

The next result rephrases Theorem 4.1 in terms of the tiling languages ofhypercubic tilings, which will then lead us to a neat characterization for the one-dimensional case.

Corollary 4.6. Let �1 = ��n� f1� and �2 = ��n� f2� be two n-dimensional hypercubictilings with languages L1 and L2, respectively. Let �1 → �2 be a bijection and �a symmetry of Gn. Suppose that the mapping � L1 → L2 defined by P� = P ∗� −�P ∗��0 is a bijection. Then � S�L1� → S�L2�, defined on the nonzero elements by

�p� P� q�� = �p � − �P ∗��0� P�� q � − �P ∗��0��

is an isomorphism. Conversely, if �1 and �2 contain all patterns of up to four tiles, everyisomorphism of S�L1� onto S�L2� is of this form for some bijection �1 → �2 andsome symmetry � of Gn.

Proof. Suppose that � S�L1� → S�L2� is an isomorphism. By Theorems 2.2and 4.1, � is defined as stated in the latter for some bijection �1 → �2 andsymmetry � ∈ Gn, since by assumption, �1 and �2 both contain all patterns up tofour tiles. Since � is a linear mapping and the order ≤ of �n is compatible withaddition, then, for all P ∈ L1 and u ∈ �n such that P + u is a pattern of �1, we havethat

�P + u� ∗� − ��P + u� ∗��0 = P ∗� + u� − �P ∗� + u��0

= P ∗� + u� − �P ∗��0 − u�

= P ∗� − �P��0�

This observation can be used to show � is well defined and onto. That � is injectiveis a consequence of both ∗ and � being bijections and, therefore, invertible maps.

Conversely, assume that such a mapping � is bijective. Let A be a pattern in�1. Then �A− a0�� = A ∗� − �A��0 ∈ L2 and so A ∗� − �A��0 = B − b0 for somepattern B of �2. Therefore, A

∗� + v, with v = −�A��0 + b0 ∈ �n, is a pattern of �2.

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It is easy to check that �−1 L2 → L1 is defined by Q�−1 = Q� −1�∗�−1 − �Q�−1�0for all Q ∈ L2; now using �−1, we can show, just as before, that for all patterns Bin �2 there exists v ∈ �n such that B� −1�∗�−1 + v is a pattern in �1. �

Remark 4.7. As the group G1 has only two elements, the identity element and thereflection about the origin, in the one-dimensional case, Corollory 4.6 states that theinverse semigroups S�L1� and S�L2� associated with the tilings languages L1 = L��1�and L2 = L��2� are isomorphic if and only if there is a bijection �1 → �2 suchthat either L2 = L1

∗ or L2 = Lop1 ∗ (where L

op1 is the language consisting of the

reversals wop of the words w ∈ L1). Moreover, we can give the explicit form for theisomorphism � S�L1� → S�L2�: 0� = 0 and

�i� w� j�� ={�i� w ∗� j�� if � is the identity

�w − i+ 1� �w ∗�op� w − j + 1�� if � is the reflection�

for all nonzero �i� w� j� ∈ S�L1�. In fact, the leftmost element of w ∗� is 0 if � is theidentity and −�w − 1� if � is the reflection about the origin.

ACKNOWLEDGMENTS

The authors wish to thank J. B. Stephen for useful conversations aboutthe theory of Schützenberger graphs and the problem of presentability of tilingsemigroups in particular. The second author wishes to thank the Department ofMathematical Sciences of the Northern Illinois University for its hospitality duringa visit that took place form February 3 to 22, 2007, and to thank M. V. Lawson andN. Gilbert of the Department of Mathematics of Herriot–Watt University for theirhospitality and fruitful conversations during a visit in July, 2007. Both visits weresupported by FCT, and the second was also supported by Herriot–Watt University.The authors would also like to thank a referee for many comments and suggestionsthat substantially improved this work.

The authors gratefully acknowledge support of FCT and PIDDAC, withinthe projects ISFL-1-143 and PTDC/MAT/69514/2006 of CAUL. The first authoris also supported by Fundação Luso-Americana para o Desenvolvimento (FLAD)under grant 071/2008.

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[3] Grunbaum, B., Shephard, C. S. (1989). Tilings and Patterns—An Introduction.New York: W. H. Freeman and Company.

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[10] Margolis, S., Meakin, J. C. (1989). E-unitary inverse monoids and the Cayley graphof a group presentation. Journal of Pure and Applied Algebra 58:45–76.

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