J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
L
OUIS
-S
ÉBASTIEN
G
UIMOND
Z
UZANA
M
ASÁKOVÁ
E
DITA
P
ELANTOVÁ
Combinatorial properties of infinite words associated
with cut-and-project sequences
Journal de Théorie des Nombres de Bordeaux, tome 15, n
o3 (2003),
p. 697-725.
<http://www.numdam.org/item?id=JTNB_2003__15_3_697_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Combinatorial
properties
of
infinite words
associated
with
cut-and-project
sequences
par
LOUIS-SÉBASTIEN
GUIMOND,
ZUZANAMASÁKOVÁ
et EDITAPELANTOVÁ
RÉSUMÉ. Le but de cet article est d’étudier certaines
propriétés
combinatoires des suites binaires et ternaires obtenues par"coupe
et
projection".
Nous considérons ici le processus de coupe etpro-jection
en dimension deux où les sous-espaces deprojection
sont enposition
générale.
Nous prouvons que les distances entre deuxter-mes
adjacents
dans une suite ainsi obtenue prennenttoujours
soitdeux soit trois valeurs. Une suite obtenue par coupe et
projection
détermine ainsi de manière naturelle une suite
symbolique
(mot
infini)
sur deux ou trois lettres. En fait ces suites peuvent aussi être obtenues commecodages d’échanges
de deux ou trois inter-valles. Dupoint
de vue de lacomplexité,
la construction par coupe etprojection
donne des mots decomplexité n
+1, n
+ constante et 2n + 1. Les mots binaires ont unecomplexité égale
à n + 1 et sont donc sturmiens. Les mots ternaires ont unecomplexité
égale
à n + constante ou 2n + 1. Une coupe etprojection
atrois
paramètres
dont deuxspécifient
les sous-espaces deprojec-tion,
le troisième déterminant la bande de coupe. Nous classifions lestriplets qui correspondent à
des mots infinis combinatoirementéquivalents.
ABSTRACT. The aim of this article
is
tostudy
certain combinato-rialproperties
of infinitebinary
and ternary words associated tocut-and-project
sequences. We consider here thecut-and-project
scheme in two dimensions with
general
orientation of theproject-ing subspaces.
We prove that acut-and-project
sequencearising
in such a
setting
hasalways
either two or three types of distances betweenadjacent
points.
Acut-and-project
sequence thusdeter-mines in a natural way a
symbolic
sequence(infinite word)
in twoor three letters. In
fact,
these sequences can be constructed alsoby
a
coding
of a 2- or 3-intervalexchange
transformation.According
to the
complexity
thecut-and-project
construction includes words withcomplexity n
+1, n
+ const. and 2n + 1. The words on twoletter
alphabet
havecomplexity
n+1 and thus are Sturmian. Theternary words associated to the
cut-and-project
sequences havecomplexity n
+ const. or 2n + 1. Acut-and-project
scheme has three parameters, two of themspecifying
theprojection subspaces,
the third onedetermining
thecutting strip.
Weclassify
thetriples
thatcorrespond
tocombinatorially equivalent
infinite words.1. Introduction
In this paper we
study
certainternary
generalizations
ofbinary
Stur-mian sequences
[14,
15].
Sturmian sequences have severalequivalent
char-acterizations. One of them isusing
the so-calledcut-and-project
scheme.Consider a lattice
Z2 ,
astraight
line p : y = ax with an irrationalslope
a, and a
strip parallel
to the line p of such a width that the intersectionof the
strip
with the x-axis is aline-segment
oflength
1. Theorthogonal
projection
of latticepoints
in thestrip
on the line p determines apoint
set in R withonly
twopossible
distances betweenneighbours.
If werepresent
these distances
by
letters A and Brespectively
and arrange these lettersaccording
to the order of distances in thepoint
set,
we obtain a Sturmianinfinite word.
Replacing
theorthogonal
projection
in the above schemeby projection
in any direction may cause thechange
inordering
ofprojected
points
onthe
straight
line p and thus the distances betweenneighbours
maypossibly
take more than two values. Our
generalization
of Sturmian words stems inallowing
any width of thestrip
and anyprojection
in thecut-and-project
scheme.Sequences
ofpoints
obtained in the described way are calledcut-and-project
sequences.In our paper we use the notion of
geometrical
similarity.
Twopoint
setsare said to be
geometrically similar,
if one is an affineimage
of the other.Geometrical
similarity
is anequivalence
relation. One of the main resultsof the
present
paper is the characterization of classes of thisequivalence,
cf. Theorem 6.3.’
Our construction of
cut-and-project
sequences is aspecial
case of thedefinition of the so-called model sets.
Properties
ofgeneral
model sets have beenextensively
studied becausethey
are suitable models for materialswith
long-range
aperiodic
order. Forexample,
it is well known[18]
that every model set has a finite number of localconfigurations.
Inparticular,
for our one-dimensional case it means that there are
only
finitely
manydifferent distances between
adjacent points.
In our article we show that the distances between
neighbours
in acut-and-project
sequence take at least two and at most three values. Similarproblem
was studiedby Langevin
in[13]
as ageneralization
of the wellknown three-distance theorem
[21].
The theorem is asimple
consequenceTo a
cut-and-project
sequence we can associate asymbolic
sequence in a two- or three-letteralphabet. Obviously,
geometrically
similarcut-and-project
sequences determine identicalsymbolic
sequences. Theopposite
statement is however not true. Even twocut-and-project
sequences thatare not
geometrically
similar mayprovide
coinciding
words. It is thereforepractical
to introduce the notion of combinatorialequivalence.
InTheo-rem 7.5 we describe the
equivalence
classes.Another result of this paper is the
description
ofcomplexity
of words associated tocut-and-project
sequences(Theorem 5.3).
We show that thebinary
cut-and-project
words havecomplexity
n + 1 and therefore areStur-mian ;
theternary
wordscorrespond
tocodings
of three-intervalexchange
transformations withpermutation
(3, 2,1).
2. Words
A finite
alphabet
is a set ofsymbols
.A = A concatenationw of letters is called a word. The
length
of a word z,v is the number of lettersfrom which w is formed. We denote
by
,A.* the set of words in thealphabet
,~4. A one-way infinite word u is a sequence u =
(Un)nEN =
ulu2u3 ... ·with values in .A. One may consider also bidirectional infinite words as
sequences u =
(un)nEZ-
In our article we work with bidirectional infinitewords. Two words in an
alphabet
,A and(vn)nEz
in analphabet
x3 are
combinatorially equivalent
if there exists abijection
h : an no E Z such thath(un)
=vn+no for all n E Z.
Let i E
Z,
n E N. A concatenation uiuz+1... uZ+n_1 is called a factor of u oflength
n. We define thedensity
of a factor w of u oflength n
asif the limit exists.
The function that
assigns
to apositive integer n
the number of differentfactors of
length
n in an infinite word u is called thecomplexity
of u,usually
denoted
by C,
For the
complexity
of an infinite word in a finitealphabet
.,4 one hasThe
complexity
function is thus a measure of disorder in the infinite word.For more about the
complexity
function see[3,
7,
10].
Let us recall some facts about one-way infinite words. We say that an
infinite word U = ulu2u3 ... in a finite
alphabet
A iseventually periodic,
ifthere exist finite words wo, wl in ,~1.* such that w =
WOWIWIWI .... A word
which is not
eventually periodic
is calledaperiodic.
It is well known that asuch that
G(n)
n,(see [17, 9]).
It follows that the mostsimple aperiodic
one-way infinite words are those ofcomplexity
C(n)
= n + 1. Such wordsare called Sturmian. A nice survey of Sturmian and related sequences may
be found in
[17,
15,
14].
Algebraically,
every Sturmian word can be writtenas a sequence of 0’s and l’s
given by
one of thefollowing
formulasfor
of,/3
E~0,1),
a irrational. Note that the aboveexpressions
can beextended for n E Z so that
they
defineaperiodic
bidirectional infinite words.J.
Cassaigne
defines in[8]
thequasisturmian
sequences as such infinitesequences for which there exist
integers
k and no such that thecomplexity
function is
C(n)
= n + k for n > no. Anexample
of aquasisturmian
sequence is a sequence that arises from a Sturmian word if one substitutes
every 0
by
a finite word wo and every 1by
a finite word wl.Cassaigne
hasshown that every
quasisturmian
sequenceis,
up to a finiteprefix,
of thiskind.
Using
thepowerful
notion of theRauzy
graphs,
the words withcomplex-ity
2n+ 1 can be divided into four classesaccording
to the maximalindegree
andoutdegree
in the associated sequence of directedgraphs.
Arnoux andRauzy
gavegeometrical
characterization of infinite words oftype
3-3. We show thatcut-and-project
sequences aregeometrical
reprezentations
ofin-finite words of
type
2-2.In order to
explain
the classification of sequences withcomplexity
2n + 1into four groups, let us recall the definition of a
Rauzy
graph.
ARauzy
graph
for agiven
infinite word u andgiven integer n
is an orientedgraph
rn
=(Vn, En),
where the vertices arerepresented by
factors oflength
nand the oriented
edges
are determinedby
factors oflength
n+1. Anedge
estarts in a vertex vi and terminates in a vertex v2 if the factor vl of
length
~ is a
prefix
of the factor e and v2 is its suffix. Thus#Vn
=C(n)
and#En
=C(n
+1).
The number ofedges
that start(end)
in a vertex v is called theoutdegree
(indegree)
of v. For a bidirectional infiniteword,
everyvertex of every
graph
has both in- andoutdegree
at least one. For wordsof
complexity
2n + 1 in everygraph
rn
the number ofedges
isequal
to the number of verticesplus
2. Therefore the maximaloutdegree
for thegraph
is either 2 or 3. The same statement is true for maximalindegree.
Thepair
maximalindegree -
maximaloutdegree
for words ofcomplexity
2n + 1can therefore have
only
four values. This classification holds for sequences for which theindegrees
arealways
non-zero. This is the case if the word isbidirectional or one-way recurrent. Arnoux and
Rauzy
in[4]
show that if for some no the maximaloutdegree
in thegraph
is2,
then it is 2 for everygraph
> no. The same statementagain
holds forindegree.
Inthe article the authors find
geometrical
representation
of words withRauzy
graphs
of thetype
3-3 for every n E N. The infinite words considered inthis article are of the
type
2-2,
which means thatstarting
from a certainno both
maximal indegree
and maximaloutdegree
areequal
to 2.3.
Cut-and-project
sequencesGenerally, cut-and-project
sets are defined asprojections
of lattices ofarbitrary
dimensions[18].
In this paper westudy
thesimpliest
case. Thecut-and-project
scheme in1I~2
isgiven by
two one-dimensionalsubspaces
Vi
and
I§
andby projections
7r,: and 7r2:}R2 ~ V2
whichsatisfy:
1)
xi restricted toZ~
is aninjection.
2)
1I"2(Z2)
is dense inV2.
The scheme is illustrated
by
thefollowing
picture.
In this scheme
1rl(Z2)
and1r2(Z2)
are additive abelian groups. Thebi-jection
betweenthem,
~ri 10 ~r2
isusually
denotedby
* and called the starmap. Its inverse is denoted
by
-*.Let
Ví
be the linear span of a vector~1
=(1, é)
andV2
the linearspan of a
vector x2
=(-1, ~7).
In order tosatisfy
conditions1)
and2)
we choose,-7 17 irrational -7/.
Any
vector(p, q)
E7G2
can be written asFor
simplicity
we omit the common factor1/(e + 7~) (which
corresponds
to different normalization of
vectors -
and~2)
and consider the abelian groups’
The star map *:
Z[77]
-7G(e~,
given by
is an
isomorphism
of the two groups.Using
this formalism we caneasily
define a
cut-and-project
sequence aswhere f2 is a bounded
interval,
called theacceptance
window. We denote thelength
of an interval f2by
101.
Anexample
of the construction of aFIGURE 1. Construction of a
cut-and-project
sequence.It is well known
[18]
that anycut-and-project
set is Delone with finite number ofpolygons
in Voronoitiling
of space. For C R itim-plies
that there exists anincreasing
sequence(Xn)nEZ
such thatE~~~(SZ) _
E
Z}
and the set of tiles T =Itn
= EZ}
is finite. If weassign
to each tile t E T a letterh(t)
from analphabet
A,
thesequence can thus be viewed as a bidirectional infinite word in
the finite
alphabet
A.
For the
description
of we can use a differentincreasing
sequencexn
:= xn+p forarbitrary
fixed p E Z. Thecorresponding
sequence has then theproperty
=tn+p
for all n E Z. The word(h(in))nEZ
is thus ap-shift
of the infinite word(h(tn))nEZ.
In order to avoidspecification
of thepoint
indexedby
0,
we introduce thesymbol
for the entire class ofall bidirectional infinite
words,
that are associated toE~~~ (SZ).
In[17]
the infinite bidirectional words withoutspecific origin
are calledtrajectories.
The
terminology
pointed/non-pointed
words is also used.We are interested in combinatorial
properties
of words Inpar-ticular,
westudy
- the
cardinality
of thealphabet
ET~
C,A.,
in case whendiffer-ent tiles are
assigned
to differentletters, i.e.,
themap h
isinjective,
- the subwordcomplexity
of-
the
triples
ofparameters
ë,11, !1
thatgive
the same bidirectionalinfinite
words,
up to the choice of the letterassignment h
into thealphabet
A.4. Distances in
In this section we first determine the number of different tiles in
and then we derive the
lengths
of these tiles.Let us first focus on
cut-and-project
sequences withsymmetric
accep-tance intervaj f2 =
(-d, d),
with d > 0. Let 0 x 1 X2 ...de-note the
positive
elements of the setE,,,7(-d,
d).
Let xk be the minimalpositive
element ofE,,,7(-d,
d)
withsign
xk =1=
sign
xi.
Since the im-agesx*
coverdensely
the interval(-d, d),
such an Xk must exist. Then0 xZ - Xl xi xk for all i =
2, 3, ... ,
k - 1 and(zj - xl)*
E(-d, d).
Thus Xi -
xl arepositive
elements ofE(-d,
d)
which ispossible only
asxi - x1 = Xj for
some j
=1, 2, ... ,
i - 1. Thisimplies
thatMoreover,
Indeed,
suppose thatxk - x1 - (zk -
(-d, d),
Le.,
xk - zi EEë,1J( -d, d).
Since Xk > ~ 2013 xi >0,
according
to(2)
there must exist aj
E{0,1,..., A? - 1},
such that Xk - xi =jxi,
whichgives
xk = ( j
+ The latter is a contradiction with theassumption
sign
x¡ =1=
sign
xi .
Theorem 4.1. For any irrational numbers
~, r~, ~ ~
2013~~ and any bounded interval 11 =[c, c-I-d),
there existpositive
numbersAi
andA2
such that the distances between consecutivepoints
in take at most three values amongA2,
Ai
+Proof.
Thepoints
of +d)
form anincreasing
sequencei.e.,
+d) -
I
n EZ}.
Apositive
number A is a distancebetween consecutive
points
of c +d)
if there exists such that0 = For
any tile A > 0 we have l:1 E
E~,~(-d, d).
Moreover,
letboth 6 > 0 and
16,
i EN,
i>_
2,
belong
toE-d, d).
Then 16 is not a tilein
Eë,1J[C,
c +d),
since if x and x + 16 are elements of c +d)
then alsoz + 6
belongs
to and hencez + 16
is not the closestneighbour
of x.’
Denote
by Ai
the smallestpositive
element of~E~~(-d, d)
withði
>0,
~2
0.Formally,
The values
ill
andA2
are the two smallest candidates to belengths
of tiles in c +d).
According
to(3)
we haveLet xl E
E~,,~ ~c,
c +d)
such thatxi
E[c,
c + d -AT). 1
Sincexi
+Di
E[c,
c +d) and xl
+A*
c + d -A*
+A*
c(compare (4)),
thepoint
xj +
O1
belongs
to c +d)
and xl +Ai
is the closestright neighbour
of xl.
’
Let X2 E such that
xi
ESimilarly,
Ec, c
+d) and x*
+[c, c
+d),
therefore x2 +A2
is the closestright
neighbour
Of
X2-Now let X3 E
Ec,
c +d)
such thatx3
E[c
+ d - c - Thenand the
point
X3 +Ai
+A2
belongs
to c +d).
We want to show thatx3 +
O1
+A2
is the closestright neighbour
of thepoint
x3.Suppose
that there exists a tileA,
such thatCombining
theinequalities
we obtain
Consider the
point 6
:=Ai
+A2 - A
> 0. Since A >Ai, A2
(as
~1,
A2
are the two smallest candidates for
lengths
oftiles),
it holds that 0 6.nIA1,
A2}
butusing
(5)
weget
It means that J E
F,,,,7(-d,
d),
which contradicts the definition of~1
andA2
asbeing
the smallest elements ofEE~~ (-d,
d).
As aresult,
the closestright neighbour
of X3 Eê,[C,
c +d), with x*
E[c
+ d - c -2)’
is thepoint
x3 +A1
+02.
By
that we have determined theright
neighbours
of all the elements of~ê,r¡[C,
c +d)
and theproof
iscompleted.
0 Let us make several comments on the above theorem:~ From the
proof
of the above theorem it follows that~1
andA2
de-pend only
on thelength
d of the interval 0 =[c,
c +d)
and not onthe
position
of thepoint
c. Thelengths
of tiles~1
andA2
satisfy
A]A]
0 and we shallalways
denoteby
~1
the tile withpositive
star
image
A]
>0, similarly,
~2
0.~ If the
acceptance
window were an interval f2 =(c,
c +d~
or Q =(c,
c +d)
theproof
of Theorem 4.1 could berepeated identicaly;
for the case S2 =[c, c+d]
a minor amendment is needed. Hence for everyacceptance
interval 12 withnon-empty
interior the setE,,,7(f2)
has atmost three
types
of tiles.e Theorem 4.1
implies
that everysegment
of the set has threetypes
of distances~1, A2,
~1
+A2
betweenneighbouring
points.
Moreprecisely,
the sethas at most three distances between
neighbours
forarbitrary
irra-tional c,t7, - 0
-77, andarbitrary
bounded intervalsJ,
f2. For agiven
positive
irrational a and apositive integer n
wedefine 17
= -a,ê
= -~,
J =(0,1~,
and Q =~1, K),
where K =[a(n
+1)]
+a(n+1)+1.
ThenThe condition 0 a - ba 1
implies
a = 1 +(ba),
hence a - ba =1 -
{ba}.
We also haveWe thus obtain M =
{1 -
1~
_ ~0,1, ... , n}.
The set M hasat most three distances between
neighbours.
The sameproperty
issatisfied
by
1- M =1~
=0,1, ... ,
n},
which is the well known three-distance theorem[21].
From the
proof
of Theorem 4.1 we may derive aprescription
to find theright neighbour
of agiven
point x
in thecut-and-project
sequence,according
to theposition
of x* in theacceptance
interval.Corollary
4.2. Let {1 =[c,c
-I-d)
be a boundedinterval,
and letA2,
and~1
+A2
be the tiles in such thatLli
> 0 >02.
The closestIn the
remaining
part
of this section we derive thelengths
of tiles infor a
given
interval {1 =[c,c
+d)
andpositive
e, q. As it will beseen from the
study
ofgeometrical
similarities ofcut-and-project
sets,
theassumption
of c, 77positive
does not cause a loss ofgenerality.
Changing
continuously
thelength
d of theacceptance
intervalS2 =
[c,c
+d)
causes discretechanges
of thetriplet
of distancesA2,
01
+A2).
Let c +d)
be ageneric cut-and-project
sequence, and letA*
>0,
OZ
0 andA*
+02
be the star mapimages
of its tiles.Denote
by
d+
the smallest number such that dd+
and at least one ofthe distances
Ai ,
A2,
O1
+A2
does not occur in c +d+).
Similarly,
denoteby
d- thelargest
number such that d- d and at least one ofthe distances
Ai , A2,
O1
+A2
does not occur in c +d-).
Due toCorollary
4.2,
we haved+
=A* - 02
and thusE~~~(c,
c +d+)
hasonly
twodistances,
namely
Ai
andA2.
Hencegrowing
thelength
of theacceptance
interval,
thelargest
distanceO1
+A2
disappears.
Similar situation occurs in c +
d-).
One of the distancesill,
A2,
O1
+A2
disappears,
but ithappens
in such a way that theremaining
distances have their star map
images
ofopposite sign.
Thus[c,
c +d_ )
has distancesFrom
Corollary
4.2 we obtainrespectively.
Starting
from agiven
initial valuedo,
for which the set c +do)
has 2tiles,
we can determineby
recurrence alllengths
dn, n
EZ,
of theacceptance
windows for which c +dn)
hasonly
two tiles. The initial valuedo
is determined below inExample
1.Let > 0 and 0 be the star
images
of distancesoccurring
in the sequencedn),
i.e., dn
=~~2.
Similarly,
thealgorithm
which from thetriple
dn,
An2
finds thetriple
7 A(n-1)2
has the inverse formWe have
yet
to determine some initial values for the recurrences(6)
and
(7).
It turns out that it is reasonable to start with acut-and-project
sequence whose
acceptance
interval is of unitlength.
Since the distances do notdepend
on theposition
of theinterval,
we can focus on theacceptance
interval
[0, 1).
Example
1. Let -,17 be fixedpositive
irrational numbers. Let usstudy
the sequence
E~[0,1) =
{p
p,
q EZ, 0 p -
qé1}.
Elements of this set have tosatisfy
qé :5 p
1 + qe. Therefore we can writeFrom the above formula one may
easily
observe that the tiles in a~ê,f1[O, 1)
form a Sturmian word with
slope
ê.Indeed,
since is anincreasing
sequence, the
lengths
of tiles inE,,,7[0, 1)
can becomputed
asThe sequence
E~~~[0,1)
has thus two distances ordered as a Sturmian wordwith
slope
a = e and shiftintercept /3
=0,
cf. definition of Sturmian words(1).
We denote thelength
of theacceptance
intervalby
do
= 1. In the notation of the recurrence relations(6)
there isIt is now obvious that for the determination of the
lengths
of tiles ina
generic
cut-and-project
sequence c +d)
with ê, 1/, d >0,
it suf-fices to find n E ~ such that ddn.
Then numbersAni , An2,
A(n+1)2
take three different values that are thelengths
of tiles in c +d).
Proposition
4.3. Let ê, 1/ > 0 be irrational numbers and let E2 =[c,
c +d)
satisf y
- - - - --. - - - -.
The sequence
Ee,l1(O)
has threetypes
of
tiles and thelengths of
these tilesare
Let us realize that in this case
A*
andOZ
do notdepend
on 17 > 0. Therefore also thealgorithm
(6),
corresponding
toshortening
the accep-tanceinterval,
does notdepend
on q(unlike
thealgorithm
(7)).
The valueof q
influencesonly
thelength
of tiles and not theirordering.
Thereforewe can choose the
parameter 17
> 0according
to ourneeds,
forexample
we canset 17
=c-1
whichcorresponds
to the case when theprojection
in thecut-and-project
scheme isorthogonal.
Proposition
4.4. Let c, 77 > 0 and let fl be an mtervalof
length d,
In case that the
acceptance
window haslength
d >1,
theparameter 77
plays
animportant role,
as it is shownby
thefollowing
assertion,
whichsummarizes the results of this section. Its
proof
is a consequence ofalgo-rithms
(6)
and(7)
and uses mathematical induction.Proposition
4.5. Let ê, ’11 > 0 be irrational numbers with continuedfrac-tion
[ao,
ai,a2, ... ~
and~bo,
bi , b2 , ...
]
respectively.
Denoteby
andthe sequences
of
theconvergents
associated to - and ’11, that iso Let 0 d 1. c +
d)
is acut- and-project
sequences with twotiles there exist
k E No,
and s E 1 s ak+l, such that d = =1(8 -1)(Pk -
eqk)
+Pk-1 -
eqk-ll.
In this case thelengths
of
tiles areo Let 1 d. c +
d)
is acut-and-project
sequence with two tiles there existk E No,
and s E 1 sbk+l’
such that d =Pk,s =
(S
+1)(rk
-I-etk)
+ rk-1 -i-etk-l.
In this case thelengths
of
tiles areNote that we have described all
acceptance
windowsf2,
for whichhas
only
twotypes
of distances betweenneighbouring
points.
Thishappens
if thelength
of nbelongs
to one of two sequences ofvalues, Jk,,
>1,
6k,s
=0,
and thelengths
ofcorresponding
tiles increase to oo. Forthe other sequence we have
limk-+oo
pk,s =oo, and the
lengths
ofcorresponding
tiles tend to 0.5.
Complexity
ofcut-and-pro ject
sequencesIt is useful to introduce a function that allows to determine the
neighbour
of apoint x
Eaccording
to theposition
of ~* in f2. Its definition is based onCorollary
4.2.Definition 5.1. Let Q =
[c,
c +d)
be a boundedinterval,
and letAi , A2,
andO1
+A2
be the tiles inE,,,7(Q),
such thatA*
> 0 >il2.
LetThe function is called the
stepping
function ofEE~~(S2).
If there is nomisunderstanding
possible,
we shall omit thesubscripts.
The map
f
isactually
anexchange
of three intervals.Codings
ofthree-interval
exchange
transformations are studied for instance in[11].
Thethree-interval
exchange corresponding
to the mapf
from the above defini-tion is associated with thepermutation
(3, 2,1).
Every
exchange
of three intervals withpermutation
(3, 2,1)
can be realized as a mapNote that the
stepping
function isobviously
invertible. Theright
neigh-bour of x EEe,1J(O)
is~fE,~(x*)~ *.
Similarly,
then-tuple
of itsright
neighbours
isgiven by
( f (x*)) *,
( f ~2>(x*)) *,
... ,( f ~"~(x*)) *.
Inpar-ticular,
for anygiven Yo EOn
7G(e) we
have
Let us denote the tile
O1
+A2
by
letterA,
the tileO1
by
letter Band the tile
A2
by
letter C.According
to the first nright neighbours,
we can associate to everypoint
E a word oflength
n in thealphabet
A =IA, B, Cl.
This word will be denotedby
word (x, n).
Forxi, x2
E-1
the wordsword (xl, n)
andword(x2, n)
are different if andonly
if there exists an i =1, 2, ... ,
n, such that at least onediscontinuity
point
off
lies between andf(i-l)(x2).
Therefore the set of allpoints
x* Esatisfying
word(x,
n)
=w, has the form where
by Qw
and without loss ofgenerality
we shall consider 9 semi-closed. Inparticular,
we haveIt is known that the
density
ofpoints
in acut-and-project
set isproportional
to the volume of the
acceptance
window[19].
It follows that thedensity
of the word w isproportional
to thelength
of intervalf2,,.
Inparticular,
wehave the
following
proposition.
Proposition
5.2.Let e, q
be irrationad -"1, and let S2 bea bounded interval. Let w be a
factor
in the word Thenfor
thedensity
owof
thefactor
w we haveIn
particular,
for
the densitiesof
lettersA, B,
C we haveRecall that the
complexity
C(n)
of an infinite word u is the number ofdifferent factors in u of
length
n. In our case it isgiven by
the number of allnon-empty
intervalsQw
that cover Q. The number of such intervals isdetermined from the number of left
end-points
of these intervals. Apoint
z is a left
end-point
of anon-empty
intervalQw
if andonly
if it is either a leftend-point
of 0itself, i.e., z
=c, or there exists
i E f 0,1, 2 ... n - 1 ~
such that
f i (z)
is adiscontinuity point
of the functionf ,
i.e.,
fi (z)
E{c
+ d -0 i ,
c -~2}.
Therefore we have thefollowing prescription
for thecomplexity,
where we denote for
simplicity
of notation a = c + d -ili
andQ
= c -~2.
We can now determine thecomplexity
of c +d).
Theorem 5.3. Let C denote the
complexity ,function
+d).
. then
9
If
d EZ[e],
then there exists aunique
non-negative k E No
such thatf ~k~ (a) _
(3
or(p)
= cx, wherea, Q
are thediscontinuity
points
Proof.
and arestepping
functionscorresponding
toacceptance
intervals 92 and f2 + t
respectively,
then =t)
+ t forany translation t of the interval 11. This means that blocs
occurring
in the wordc+ d)
occur also inc+d+t)
and vice versa. Two words withsuch
property
are said tobelong
to the same localisomorphism
class. Thewords and must have the same
complexity.
Without loss of
generality
we can thus consider f2 =[0, d).
Recall that~i,
~2
E Z[e],
and that thediscontinuity points
off
for such an intervalS~ are a = d -
A*
and{3
=-A*.
Our aim is to determine the number ofelements in the set
The function
f ~~~
has no fixedpoint
for anyk,
whichimplies
thefollowing
facts:
(i)
Thepoints
a, f ~-1~ (a), ... , f ~-~n-1» (a)
aremutually
distinct. The same holds if wereplace
aby Q.
(ii)
One can never havesimultaneously
,8
andf’(,8)
= a for somek, i
E N.(iii)
Sincef((3)
= 0 = c, thepoint
0 is the closestright
neighbour
of(3-*
and we havef -~ (,~)
~
c for all i ENo.
~
Suppose
that d 0
Z[c].
Then a = d -Z[e]
and (3 =
-A*
EZ [,-].
Sincef(y)
E if andonly
if y E we havef ~~‘~~ (a) ~
{O, {3,
f ~-1~
(a)7....
¡( -en-I»~ ((3)}.
Together
withproperties
(i)
and(iii)
this means that the set M has 2n + 1 elements.~ Let now d E
?~ ~~J,
i.e.,
the star mapimages
a-*, (3-*
of bothdiscon-tinuity points
off
and thepoint
0belong
toE~~,~(S~).
Since thepoint
0 is the closestright neighbour
of(3-*
we have either a-*~i-*
or0 a-*. In the first case it means that there exists a
non-negative
k such that a-* is the k-th left
neighbour
of(3-*,
i.e.,
# a
for i k and
f ~-~} (,~)
= a. Inthe
second case that there existsa
non-negative
k such that 0 is the k-th leftneighbour
ofa-*, i.e.,
f t-i~ (a) ~
{O, {3}
for i k and/(-~)(~)
= 0 =f((3).
Starting
from th value n -1= k thepairs
of elements in the setcoincide,
whichinflu-ences the
cardinality
of this set andconsequently
also thecomplexity
of the infinite word.0
Since infinite words with
complexity
n+ const.(quasisturmian
sequences,see
[8])
are welldescribed,
we shall now focus on words ofcomplexity
2n+1.According
to the result of Alessandri and Berth6[2, 7J,
for agiven n
E N the densities of factors oflength n
can take at most3 (C(n
+ 1) -
C(n))
= 6 values. Infact,
there areonly
fivefrequencies
for factors ofgiven
length.
This can be
proved
following
the ideas in[2]
andusing
the fact that thewords associated to
cut-and-project
sequences arecodings
of three intervalexchange.
Let us determine the maximal in- and
outdegree
in theRauzy graphs
corresponding
to the infinite words in consideration. Let e be a factor oflength n
+ 1 in the infinite wordu,,,7[c,
c +d)
and let v be theprefix
of e oflength
n.Obviously
S2e C 011.
Therefore in theRauzy graph
rn
the vertex v is the
starting point
of theedge
e. If52e
=011,
then theoutdegree
of the vertex v isequal
to one. Thishappens
only
if thepoints
f (-n) (a),
f ~-n~(~3)
do notbelong
to the interval011.
If the interval011
containsexactly
one of thepoints
f -"
(a),
f -" (,Q),
thenf2v=
52e1
where el, e2 are the
only
two factors oflength
n + 1 withprefix v
and theoutdegree
of the vertex v isequal
to 2. In case that bothpoints
f (-n) (a),
f ~-"~
(p)
happen
tobelong
to the interval011,
theoutdegree
of the vertex vis
equal
to 3 and all other vertices in theRauzy
graph
rn
haveoutdegree
1. Since the set{ f ~-"~ (a), f ~-"~ (~i) ~ n
EN}
covers theacceptance
intervalSZ
densely
anduniformly
[22],
thelengths
of intervals011
tend to 0 as n tends toinfinity.
If for every n E N both of thepoints
f (-n) (a)
andf ~-n~ (/~)
fall into the same interval011
for some factor v oflength
n, then= 0. Take c
large
enough
so thatI f (n) (a)
-f -’ (,)
I
j. From theproperties
of thestepping function,
aspiecewise
linear function with
slope
1,
it canhappen
thatI =
I f ~-n~ (a) - f ~-"~ (,Q) I
.However,
this cannot be true for every n. In theopposite
case, we havef ~-n-1~ (a)
I
»I f~-n> (a) -
f~-n~ (,(3)
I
and thuspoints
f ~-’~-1~ (a),
f ~-n-1~ (/3)
cannot be in the same interval011.
Thus we have
proved
thefollowing
statement.Proposition
5.4. Let c, 77 be irrationalnumbers,
20137?~ and let S2 =[c,
c +d)
be a nonempty
interval. Then there exists an no such thatfor
all n > no the maximale
indegree
and the maximaleoutdegree
in theRauzy
graph
rn
of
’UE,1J(O)
areequal
to 2.6.
Geometrically
similarcut-and-project
sequencesTwo sets A and
A
C R aregeometrically
similar,
if there existp, 1b
ER,
cp =1=
0,
such thatA
=OA
+ ~.
We shall denote thisproperty
by
A 19
A.
Remark 6.1. Note that if twocut-and-project
sequences andare
geometrically
similar with apositive
similarity
factor cp, thenthe
corresponding
infinite bidirectional words and coincide.A converse statement is not
true,
cf.Proposition
4.4.In order to describe classes of
mutually
geometrically
similarWe have the translations
7~2
+ ( g )
=Z~,
fora, b
EZ,
and thegroup of rotation
symmetries
of?~2.
They
aregiven by
allinteger
valued matrices Awith determinant + I . Then
AZ
=72.
Thesesymmetries
of Z2
correspond
to the
following
transformations ofcut-and-project
sequences.Proposition
6.2. Let e, r~ be irrational 2013y?~ and let S~ be abounded interval.
~ For x we have
o Let a,
b,
c, d beintegers,
such that ad - bc = ThenProof.
First,
let x = a EZ[77].
ThenNow
let A = ( g [ )
be aninteger
valued matrix with determinant ±1. ThenThe group of
integer
valued matrices with determinant :J:l isgenerated
by
matricesAi =(~),A2
= ~ o -°i ~ ~
A3 =
(~ å ).
Elementary
transforma-tions ofcut-and-project
sequences for these matrices areUsing
the aboveelementary
transformation we show that we may limit our considerations to c, 77 and 11 with certainproperties,
withoutloosing
any infinite bidirectional word. The aim of this section is to prove thefollowing
theorem.Theorem 6.3. Let e, q are irrationat
numbers, - :A
2013~. Let Q be a boundedinterval. Then there exist irrational, numbers
6,17
and an intervalfl,
suchthat
The
proof
of the theorem will be divided into two lemmas. Beforestating
thelemmas,
let us recall certainproperties
of continued fractions[12].
If ~
> 0 is an irrational number with continued fraction[ao, ~i,~2! -"
]
the sequence of the
convergents
associatedto ~,
then for everyLemma 6.4. For any irrational El
~
-77, and bounded intervalí2,
there exist irrational numbers~, ~
and a bounded intervalS2
such thatProo f .
Using
transformation(12)
we may assume without loss ofgenerality
that 1] ’ > 0. We first find
convergents
q2n
and associated to 77, so thatq2n 1 "
2013c [S2s.
From theproperty
(i)
of continued fractions we have’
Let us define the matrix A =
( a b )
= fortransforma-c d -92n-E-1 9’2n
tion
( 10) .
Then from(14)
we haveand from
(15)
we haveFrom the
property
(iii)
of continued fractions we obtainIn the transformation
(10)
with the matrix A we obtain a newacceptance
window Ô
:= Q. Let us find an estimate of the denominator in thefraction
Note that in the estimate we have used
property
(ii)
of continued fractions.Since +oo and
117
+-1
~
0,
it ispossible
to choose asufficiently
large
n, so that thelength
of the intervalf2
is smaller orequal
to 1. Cl
Lemma 6.5. Let ê, 17 >
0,
ê,17 irrational and let S2 be an intervalof
length
1. Then there exist irrational numbers
~, ~
and an intervalf2
satis-lying
Proof.
Without loss ofgenerality
we may.assume that - 1. Otherwise wewould use the transformation
(11)
toget
where - -
[e]
1. We first show that there exist0 ~ 1, ~ > 0
andC2
such that
- -
-The case of
JOI I
> e istrivial,
because it suffices toput t = e, ij =
1/and
C2
= Q. Assume that forgiven
f2 we haveInl
e. Let(~-) ~
be theconvergents
associated to e.According
toproperty
(ii)
of continuedfractions,
the sequence(IPn -
isdecreasing
and thus it ispossible
to find n E N so that
Let us transform the
cut-and-project
setusing
thetransforma-tion
(10)
with matrix A =( qn_ 1 qn ) ~
We obtainE~,~ (SZ),
whereAccording
toproperties
(i)
and(ii)
of the continued fractions we have0 t 1. The
number 7y
isclearly
positive.
Since forInl
we have(16),
forthe
length of C2
we may write- I IV ,..
what was to be shown.
If moreover
101
I
> 1-ë,
we may set % _ë, f¡
:=~,
and f2 :
=C2,
and thelemma is
proved.
It thus remains to solve the caseThis, however,
ispossible only
1,
i.e.,
its continued fraction has the form t =[0,
cl , c2, ...with
2. Sinceit is
possible
to find a minimal s6 {1,2,...,
cl -1}
so thatNow we use the transformation
(10)
with the toget
Let us
verify
that6,7y
andS2
satisfy
theinequalities required by
the lemma.The
paralneter fi
ispositive
as it is a ratio ofpositive
numbers. For e we use theinequality
(18)
to obtain 0 e 1. For the estimateof Ö
we derivefrom
(17)
and(19)
thatIn order to
complete
theproof
of thelemma,
it remains to show thaif2i
> 1 - e. If the minimal s satisfies 5 cl -2,
we have 1 -(s
+l)e
1 - st which
implies
- --
-If the minimal s is s =
ci - I,
thenautomatically e
> 1 - êwhich,
together
with(20)
implies
in both cases 1 >if2i
>max(ê,1 -
e)
and proves thelemma. 0
Remark 6.6. Note that in Theorem 6.3 we may choose the
interval f2
insuch a way that the
similarity
factor between and ispos-itive,
thus infinite bidirectional words associated to thesecut-and-project
sets are identical. If ithappens
that = for some p0,
we can use the transformation matrix A =
(Õl ~1)
to obtainA I1
Therefore
using
the interval-s1
insteadof Ö
gives
us apositive
similarity
factor -Sp.
7. Combinatorial classification of
cut-and-pro ject
sequences Let us recall that two words u =(un)nez
and u =(fill),,EZ
inalphabets
,r4 and
4,
respectively,
arecombinatorially equivalent,
if there exists abi-jection
h:A - /l
and an no E Z such thatfin =
for all n E Z. We denote this û. Note that if abijection h
shouldexist, certainly
the
density
of a letter a in the word u must be the same as thedensity
ofh (a)
in Û.Remark 7.1. Recall that two words associated to
cut-and-project
setswhich are
geometrically
similar withpositive similarity
factor arecombi-natorially equivalent,
see Remark 6.1.Using
Theorem 6.3 and Remark 6.6 we may thus limit ourconsideration,
without loss ofgenerality,
to wordswith
parameters
satisfying
and thus we have
For the densities of tiles
A, B,
and C we have in this caserespectively,
cf.Proposition
5.2.Proposition
4.4 shows that twogeometrically
non-similarcut-and-project
sequences maycorrespond
tocombinatorially equivalent infinite
words. Letus show
yet
anotherexample
ofcombinatorially
equivalent
words.Example
2. Consider Sturmian words u, u in thealphabet
10, 11
with theslope
a, and a = 1 - arespectively,
where 0 a 1.Obviously, u
and t arecombinatorially equivalent
since fi arises from uby
replacing
O’s withl’s and vice versa. Such Sturmian words
correspond
tocut-and-project
sequences with Q a semi-closed interval of
length
1,
seeExample
1. For77 > 0 we have
Both of the sets have two tiles of
lengths 77 and 77
+ 1. Moreprecisely,
the distances areWe can see that
The above
example
illustrates that foracceptance
window oflength
1 theinterchange -
++1 -,-, fl
++ -0gives
combinatorially equivalent
words. Thefollowing
lemma states that the sameproperty
holds forgeneral
acceptance
window as well.
Lemma 7.2. Let
0 e 1, ~ > 0 be
irrationalnumbers,
and let S2 be aninterval
of length
max(e,1 - e)
1. Then, , , . , , ..
Proof.
Although
=7G (1-
eJ,
the star maps from7G r)
toZ[e]
=are different for the
cut-and-project
sequencesEe,l1 ( -0)
andE1-e,17(0).
We shall thereforedistinguish
themby indices,
According
toProposition
4.3 in bothcut-and-project
sets the three dis-tances are q, 1+ 1/
and1 + 2r¡.
Theirimages
under the star map are howeverdifferent for * 1 and *2,
namely,
For the
proof
we use two facts:Let Then the fact
(F2)
says that the sequence of distances tothe left from the
point
in the setE~~~ (SZ)
is the same as the sequence of distances to theright
from thepoint - xo * 1
in the setE ( - SZ ) .
The fact
(Fl)
says that the sequence ofsteps
to theright
from thepoint
in the set is the same as the sequence ofsteps
to the left from thepoint
xo *1 in
the setE~~~(SZ),
only
thelength
ofsteps 77
and 1+ ~
are
interchanged.
0Note that the
density
of the shortest tile 77 in the setEE~,~ (-S~)
isequal
to1- a,
whereas thedensity
of the shortesttile 77
in the set isequal
to 1 - It is therefore obvious that the
cut-and-project
setsand are not
geometrically
similar.Nevertheless,
the associated infinite bidirectional words arecombinatorially equivalent.
From every Sturmian sequence in the
alphabet
~A, B}
we can form aspecial
word in thealphabet
B,
C}
in the way that in between everytwo letters of the Sturmian sequence we insert k-times the letter C. Such a sequence we call a
k-padded
Sturmian sequence.Definition ?’.3. Let u be a Sturmian word in the
alphabet
(A, B).
Letk E N and h be a
B } -~ IA, B, C } * ,
given by
h (A)
=ACk ,
h(B)
=BCk.
The wordh(u)
is called ak-padded
Sturmian word.The
following
proposition
shows thatk-padded
Sturmian sequences canalso be
cut-and-project
sequences.Proposition
7.4.Every k-padded
Sturmian word is a word associated toa
cut-and-project
sequence.Proof.
Let u be a Sturmian word withslope
cx, 0 a 1.Similarly
as inExamples
1 and 2 we can consider u to be the word c +1),
for some c E Rand p
> 0. For agiven
k e N we form ak-padded
Sturmian sequenceand want to find
parameters
~, TJ,SZ,
so that thek-padded
Sturmian wordis a word We
put
Note that f2 can be written as 0 =
[c,
c +j)
for c = ccand i
=(k
+1)e.
Since
1
c1
we havei
1 and therefore stark+2
images
of tiles in are,1-
-, and 1 - 2e.For such
parameters
thestepping
function isgiven by
We can
verify
easily
thatTherefore every letter A or B is followed
by
thestring
of J~ lettersC,
whereas the
(1~
+1)-th
letter is different from C.Moreover,
the restriction ofto OA U
Q B
is a scaledstepping
functionf a ~ ~ 1.
0Theorem 7.5.
. For any irrational numbers ê,
~
-77, and a bounded interval 0there exist irrational t and an interval
Ö satisfying
such that
u,,,7(ft)
for
anyirrational
> 0..
0 Let e,
t, q, fi
bepositive
irrational numbers such that 0e, e
2
Let
f2,
f2
be intervalssatisfying
1 - E1,
1-IÖI
1,
u S c u
,.
Then
-Then
,
-
either 6 = e and there exists x E
Z[e]
such =Z[e],
i. e.,
=-
or there exists
k E N,
and an irrational a, 0 a1,
such thatis
k-padded
Sturmian word withslope
a and is ak-padded
Sturmian word withslope
1- a.The
proof
of the first statement of the theorem is a direct consequenceof Lemma 7.2 and Remark 7.l. In the
remaining
part
of the section we aregoing
to prove the second statement.Let us determine