Sturmian Words
2.0. Introdution
Sturmian words are innite words over a binary alphabet that have exatly
n+1fators of length n for eah n 0. It appears that these wordsadmit
severalequivalentdenitions,andanevenbedesribedexpliitlyinarithmeti
form. Thisarithmetidesriptionisabridgebetweenombinatorisandnumber
theory. Moreover,the denition byfators makesthat Sturmian wordsdene
symbolidynamial systems. The rst detailed investigations of these words
were done from this point of view. Their numerousproperties andequivalent
denitions, andalso thefat that theFibonaiwordisSturmian,hasleadto
agreat development,undervariousterminologies,oftheresearh.
TheaimofthishapteristopresentbasipropertiesofSturmianwordsand
of theirtransformation by morphisms. Thestyleof exposition relies basially
onombinatorialarguments.
The rst setion is devoted to the proof of the Morse-Hedlund theorem
stating the equivalene of Sturmian wordswith the set of balaned aperiodi
wordandthesetofmehanialwordsofirrationalslope. Wealsomentionseveral
otherformulationsofmehanialwords,suhasrotationsanduttingsequenes.
We next give properties of the set of fators of one Sturmian word, suh as
losureunder reversal,theminimalityoftheassoiateddynamialsystem,the
fatthatthesetdependsonlyontheslope,andwegivethedesriptionofspeial
words.
In the seond setion, we give a systemati exposition of standard pairs
and standardwords. Weprovetheharaterizationbythe doublepalindrome
property,desribetheonnetionwithFineandWilf'stheorem. Then,standard
sequenesareintroduedtoonnetstandardwordstoharateristiSturmian
words. Therelation to Beatty sequenesis in the exerises. This setion also
ontains the enumeration formula for nite Sturmian words. It ends with a
shortdesriptionoffrequenies.
ThethirdsetionstartsbyprovingthatthemonoidofSturmianmorphisms
is generated by three well-known morphisms. Then, standard morphisms are
investigated. A desription of all Sturmian morphisms in terms of standard
morphisms is given next. The setion ends with theharaterization of those
algebrainumbersthatyieldxedpointsbystandardmorphisms.
Someproblems arejust exerises,but mostontainadditionalpropertiesof
Sturmianwords,withappropriatereferenes. ItisdiÆulttotraebakmanyof
thepropertiesofSturmianwords,beauseofthesatteredorigins,terminology
and notation. When we quote a referene in the Notes setion, we are only
relativelyertainthat itisthesoureoftheresult.
Inthis hapter,wordswillbeoverabinaryalphabetA=f0;1g.
2.1. Equivalent denitions
This setionisdevoted totheproofofatheorem(Theorem2.1.13)statingthe
equivalene ofthree properties, alldening what weall Sturmian words. We
startbydeningSturmianwordsto haveminimalomplexityamongaperiodi
innite words. We rst provethat Sturmian words are exatly the aperiodi
balaned words. Wethenintrodueso alledmehanialwordsandprovethat
these yieldanother haraterizationof Sturmianwords. Other formulationsof
the mehanial denition,by rotationand utting sequenes, are given in the
seond paragraph. Thethird paragraphontainsseveral propertiesonerning
thesetoffatorsofaSturmianword.
2.1.1. Complexityand balane
TheomplexityfuntionofaninnitewordxoversomealphabetAwasdened
inChapter1. Itisthefuntion thatounts,foreahintegern0,thenumber
P(x;n)offatorsoflengthnin x:
P(x;n)=Card(F
n (x)):
ASturmianwordisaninnitewordssuhthatP(s;n)=n+1foranyinteger
n 0. Aording to Theorem 1.3.13, Sturmian words are aperiodi innite
wordsofminimalomplexity. SineP(s;1)=2,anySturmianwordisovertwo
letters. A right speialfatorof awordx isawordusuh that u0and u1are
fatorsof x. Thus,s isaSturmianwordifandonlyifithasexatlyoneright
speialfatorofeahlength.
AsuÆxofaSturmianwordis aSturmianword.
Example 2.1.1. WeshowthattheFibonai word
f =0100101001001010010100100101001001
denedinChapter1isSturmian. Itwillbeonvenient,inthishapter,tostart
thenumerationofniteFibonaiwordsdierently,andtosetf
1
=1,f
0
=0.
Sinef ='(f),itisaprodutofwords01and0. Thus,theword11isnot
a fator of f and onsequently P(f;2) = 3. The word000 is not a fator of
'(f),sineotherwiseit isaprex ofsome'(x)for afatorx off, andx has
to startwith11.
Toshowthat f isSturmian, weprovethat f hasexatlyonerightspeial
We start by showingthat, for no wordx, both0x0and 1x1are fatorsof
f. This islear ifx isthe emptywordand ifx is asingle letter. Arguing by
indution onthe length,assumethat 0x0and1x1are in F(f). Then x starts
and ends with 0,and x =0y0for somey. Sine 00y00 and 10y01haveto be
fators of '(f), there exists a fator z of f suh that '(z) = 0y. Moreover,
00y0='(1z1) and010y01='(0z0), showingthat 1z1and 0z0arefatorsof
f. Thisisaontraditionbeausejzjj'(z)j<jxj.
We show now that f has at most one right speial fator of eah length.
Assume indeed that uand v are right speial fators of thesame length,and
letx be thelongestommonsuÆxofuand v. Then thefour words0x0,0x1,
1x0,1x1arefatorsoff,whih ontraditsourpreviousobservation.
Toshow that f has atleast onerightspeial fator ofeahlength, weuse
therelation
f
n+2
=g
n
~
f
n
~
f
n t
n
(n2) (2.1.1)
whereg
2
="andforn3
g
n
=f
n 3 f
1 f
0
; t
n
=
01 ifnisodd,
10 otherwise.
Observethat therstletterof
~
f
n
istheoppositeof therstletterof t
n . This
provesthat
~
f
n
isarightspeialfatorfor eahn2. SineasuÆxof aright
speialfatorisitselfarightspeialfator,thisprovesthatrightspeialfators
ofanylengthexist.
Equation (2.1.1) is provedby indution. Indeed, f
4
= "(010)(010)10and
f
5
=0(10010)(10010)01. Next,isiteasilyhekedbyindution that
'(u )0~ =0('(u))
(2.1.2)
foranywordu. Itfollowsthat '(
~
f
n t
n )=0
~
f
n+1 t
n+1
andsine'(g
n )0=g
n+1 ,
onegets(2.1.1).
We now start to give another desription of Sturmian words, namely as
balaned words. Theheight of awordx isthenumberh(x)oflettersequalto
1in x. Giventwowordsxandyofthesamelength,theirbalaneÆ(x;y)isthe
number
Æ(x;y)=
h(x) h(y)j
Asetof wordsX isbalanedif
x;y2X; jxj=jyj ) Æ(x;y)1
Aniteorinnitewordisitself balanediftheset ofitsfatorsisbalaned.
Proposition2.1.2. LetX beafatorialsetofwords. IfX isbalaned,then
foralln0,
n
Proof. The onlusion is lear for n = 0;1, and it holds for n = 2 beause
X annot ontain both 00 and 11. Arguing by ontradition, let n 3 be
the smallestinteger for whih the statement is false. Set Y =X\A n 1
and
Z = X \A n
. Then Card(Y) n and Card(Z) n+2. For eah z 2 Z,
its suÆxof lengthn 1isin Y. Bythepigeon-holepriniple,there exist two
distint wordsy;y 0
2Y suhthatallfour words0y;1y;0y 0
;1y 0
areinZ. Sine
y 6=y 0
there existsawordx suh thatx0 andx1areprexesofy and y 0
. But
then, both0x0and1x1arewordsinX,showingthatX isunbalaned.
Theargumentusedin theproofanberenedasfollows.
Proposition2.1.3. LetX be afatorial set of words. The set X is unbal-
aned ifand onlyif there existsapalindrome wordwsuh that 0w0and 1w1
arein X.
Proof. Theonditionis learlysuÆient. Conversely,assumethat X is unbal-
aned. Considertwowordsu;v2X ofthesamelengthnsuhthatÆ(u;v)2,
and takethemof minimallength. Therstlettersof uandv aredistint, and
soare thelast letters. Assuming that ustartswith 0and v with1, there are
fatorizations u = 0wau 0
and v = 1wbv 0
for some words w;u 0
;v 0
and letters
a6=b. Infat a=0andb=1sineotherwiseÆ(u 0
;v 0
)=Æ(u;v),ontraditing
theminimalityofn. Thus,againbyminimality, u=0w0andv=1w1.
Assumenextthatwis notapalindrome. Thenthereisaprexzof wand
alettera suh thatzais aprex ofw, z~isasuÆxofw buta~z is notasuÆx
of w. Then ofourseb~z isasuÆxofw, whereb istheotherletter. Thisgives
a proper prex 0za of u and a proper suÆx b~z1 of v. If a = 0 and b = 1,
then Æ(0z0;1~z 1)=2, ontraditing theminimality ofn. Butthen u=0z1u 00
and v =v 00
1~z0fortwowordswith Æ(u 00
;v 00
)=Æ(u;v),ontraditing againthe
minimality. Thuswisapalindrome.
Remark 2.1.4. Inthe proofthat theFibonaiwordf is Sturmiangivenin
Example2.1.1,weatuallystartedbyshowingthatf isbalaned.
Theorem 2.1.5. Letxbeaninniteword. Thefollowingonditionsareequiv-
alent.
(i) xisSturmian,
(ii) xisbalaned andaperiodi.
Proof. If x is aperiodi, then P(x;n) n+1for all n byTheorem 1.3.13. If
x is balaned,then by Proposition 2.1.2,P(x;n)n+1for alln. Thus x is
Sturmian.
Toprovetheonverse,weassumex isSturmianandunbalaned,andshow
thatxiseventuallyperiodi. Sinexisunbalaned,thereisapalindromeword
w suh that0w0,1w1arefatorsof x. Thisshowsthat wisrightspeial. Set
n=jwj+1. SinexisSturmian,thereisauniquerightspeialfatoroflength
n, whihis either0w or1w. Wesupposethat0w isrightspeial,so1w isnot,
Any ourrene of1w in x isfollowed by theletter 1. Letv bea wordof
lengthn 1suhthatu=1w1visinF(x). Theworduhaslength2n. Weprove
thatallfatorsoflengthnofuareonservative. InviewofProposition1.3.14,
x iseventuallyperiodi.
Toshow thelaim, it suÆesto provethat theonly right speial fatorof
lengthn,thatis0w,isnotafatorofu. Assumetheontrary. Thenthereexist
fatorizationsw=s0t;v=yz;w=t1y.
u
1 w 1 v
0 w
1 s 0 t 1 y z
Sinew isapalindrome, therstfatorization impliesw=
~
t0~s, and theletter
followingtheprextin wisbotha0anda1.
Theslopeofanonemptywordx isthenumber(x)= h(x)
jxj .
Example 2.1.6. The height of x = 0100101 is 3, and its slope is 3=7. The
wordx anbedrawn onagridbyrepresentinga0(resp. a1) asahorizontal
(resp. adiagonal)unit segment. Thisgivesapolygonallinefrom theoriginto
thepoint(jxj;h(x)),and theline from theoriginto thispoint hasslope(x).
SeeFigure 2.1.
r r
r r r
r r r
0 1 0 0 1 0 1
(7,3)
Figure2.1. Heightandslopeoftheword0100101.
It iseasilyhekedthat
(xy)= jxj
jxyj (x)+
jyj
jxyj (y)
Proposition2.1.7. A fatorialset ofwordsX isbalaned ifand onlyif,for
allx;y2X,x;y6=",
(x) (y)
<
1
+ 1
: (2.1.3)
Proof. Assume rst that (2.1.3) holds. For x;y 2 X of the same length, the
equationgives
h(x) h(y)
<2
showingthat X isbalaned.
Conversely,assume that X is balaned,and letx;y bein X. If jxj= jyj,
then(2.1.3)holds. Assumejxj>jyj,andsetx=zt,withjzj=jyj. Arguingby
indution onjxj+jyj,wehave
(t) (y)
<
1
jtj +
1
jyj
andsineX is fatorial,jh(z) h(y)j1,whene
(z) (y)
1
jyj . Next,
(x) (y)= jzj
jxj (z)+
jtj
jxj
(t) (y)
= jzj
jxj
(z) (y)
+ jtj
jxj
(t) (y)
thus
(x) (y)
<
1
jxj +
jtj
jxj
1
jyj +
1
jtj
= 1
jxj +
1
jyj :
Corollary 2.1.8. Let x be an innite balaned word, and for eah n 1,
let x
n
be the prex of lengthn of x. The sequene ((x
n ))
n1
onvergesfor
n!1.
Proof.Indeed,(2.1.3)showsthat ((x
n ))
n1
isaCauhysequene.
Thelimit
= lim
n!1 (x
n )
istheslopeoftheinnitewordx.
Example 2.1.9. Toomputetheslopeofaninnitebalanedword,itsuÆes
toomputethelimitoftheslopesofaninreasingsequeneofprexes(oreven
fators,asshownbythenextproposition). FortheFibonaiinniteword,the
slopesof theniteFibonaiwordsf
n
are easilyomputed. Indeed,jf
n j=F
n
andh(f
n )=F
n 2
,whene
(f)= lim
n!1 F
n 2
F
n
= 1
2
;
where =(1+ p
5)=2.
Proposition2.1.10. Let x be an innitebalaned word with slope. For
everynonemptyfatoruofx,onehas
(u)
1
: (2.1.4)
Morepreisely,oneofthefollowingholds: either
juj 1<h(u)juj+1 forallu2F(x) (2.1.5)
or
juj 1h(u)<juj+1 forallu2F(x) (2.1.6)
Ofourse,theinequalitiesin (2.1.5)and(2.1.6)arestrit ifisirrational.
Proof. Let x
n
bethe prex of length n of x. Given some ", onsider n
0 suh
thatforallnn
0 ,
(x
n
)
":
Then,using(2.1.3),
(u)
(u) (x
n )
+
(x
n
)
<
1
juj +
1
n +"
Forn!1andthen"!0,theinequalityfollows. Equation(2.1.4)meansthat
juj 1h(u)juj+1
Iftheseondlaimwerewrong,therewouldexistu;vinF(x)suhthat juj
1 = h(u) and jvj+1 = h(v). But then j(u) (v)j = 1=juj+1=jvj, in
ontraditionwith(2.1.3).
Proposition2.1.11. Letx beaninnitebalanedword. Theslopeofxis
arationalnumberifandonlyifxiseventuallyperiodi.
Proof.Ifx=uy
!
, then
(uy n
)=
h(u)+nh(y)
juj+njyj
!(y)
forn!1,showingthattheslopeisrational.
Fortheonverse,wesupposethat(2.1.5)holds. Theotheraseissymmetri.
Theslopeofx isarationalnumber=q=pwithqand prelativelyprime. By
(2.1.5),anyfatoruofxoflengthphasheightqorq+1. Thereareonlynitely
manyourrenesoffatorsoflengthpandheightq+1,sineotherwisethere
isafatorw=uzvofx withjuj=jvj=pandh(u)=h(v)=q+1. Inviewof
(2.1.5)
2+2q+h(z)=h(uzv)1+p+jzj+p=1+2q+jzj
wheneh(z)jzj 1,in ontraditionwith(2.1.5).
Bythepreedingobservation,thereisafatorizationx=tysuhthatevery
wordinF
p
(y)hasthesameheight. Considernowanourreneazbofafator
in y of lengthp+1,with aandb letters. Sine h(az)=h(zb), onehasa=b.
This means that y is periodi with period p. Consequently, x is eventually
2.1.2. Mehanial words,rotations
Giventworealnumbersandwith01,wedenetwoinnitewords
s
;
:N !A; s 0
;
:N !A
by
s
;
(n)=b(n+1)+ bn+
s 0
;
(n)=d(n+1)+e dn+e
(n0)
Itis easyto hekthats
;
(n)ands 0
;
(n)indeedarein f0;1g. Thewords
;
is thelower mehanial wordands 0
;
istheupper mehanial wordwith slope
andinterept. (Thisslopewillbeshowninamomenttobethesameasthe
slopeofabalanedword.) Itislearthatif 0
isaninteger,thens
;
=s
; 0
and s 0
;
=s 0
; 0
. Thus we may assume0<1or0<1(both will be
useful).
r r
r r r
r r r
r r
r r r
r r r
0 1 0 0 1 0 0
y=x+
P
n P
0
n
Figure2.2. Mehanialwordsassoiatedwiththeliney=x+.
Theterminologystemsfromthefollowinggraphialinterpretation(seeFig-
ure2.2). Considerthestraightlinewith equationy=x+. Thepointswith
integeroordinatesjustbelowthislineareP
n
=(n;bn+). Twoonseutive
pointsP
n and P
n+1
are joined bya straight line segmentthat is horizontal if
s
;
(n)=0anddiagonalifs
;
(n)=1.
Thesameobservationholds forthe pointsP 0
n
=(n;dn+e) loatedjust
abovetheline.
r r
r r r
r
r r r
r r r
s;
s 0
;
n
0 1 0 0
0 0 1 0
y=x+
!
!
!
!
!
!
!
!
!
!
!
!
Clearly,
s
0;
=s 0
0;
=0
!
; s
1;
=s 0
1;
=1
!
Let 0 < < 1. Sine 1+bn+ = dn+e whenever n+ is not an
integer,onehass
;
=s 0
;
exeptedwhenn+isanintegerforsomen0.
Inthisase(seeFigure 2.3),
s
;
(n)=0; s 0
; (n)=1
and,ifn>0,
s
;
(n 1)=1; s 0
;
(n 1)=0
Thus, if isirrational, s
; and s
0
;
dierby at mostonefator of length2.
Amehanialwordisirrational orrational aordingtoitsslopeisrationalor
irrational.
A speial asedeservesonsideration, namelywhen 0<<1and =0.
Inthisase,s
;0
(0)=b=0,s 0
;0
(0)=de=1,and ifisirrational
s
;0
=0
; s 0
;0
=1
wheretheinniteword
isalledtheharateristiwordof.
Remark 2.1.12. The ondition 0 1 in the denition of mehanial
wordsisnotarestrition,but asimpliation. Oneouldindeedusethesame
denition of s
;
without anyondition on. Sine bs
;
(n)1+b,
thenumberss
;
(n)thenanhavethetwovalueskandk+1where k=b.
Thusthewordss
; ands
0
;
areoverthetwoletteralphabet fk;k+1g. This
alphabet anbetransformedbakintof0;1gbyusingtheformula
s
;
(n)=b(n+1)+ bn+ b
Mehanialwordsanbeinterpretedin severalotherways. Consideragain
a straight line y = x+, for some > 0 not restrited to be less than 1,
andnotrestritedtobepositive. Considertheintersetions ofthisline with
thelines of the gridwith nonnegativeintegeroordinates. We geta sequene
Q
0
;Q
1
;:::of intersetion points. Weall Q
n
=(x
n
;y
n
)horizontal if y
n is an
integer,andvertialifx
n
isaninteger. Ifbothareintegers,weinsertbeforeQ
n
asiblingQ
n 1 ofQ
n
with thesameoordinates,andweagreethatthe rstis
horizontalandtheseond isvertial(orvie-versa,but wedoalwaysthesame
hoie). InFigure2.4below,Q
0
isvertial,beauseispositive.
Writing a 0 for eah vertial point and a 1 for eah horizontal point, we
obtain an inniteword K
;
that is alled the (lower) utting sequene (with
theotherhoieforlabelingsiblings,onegetsanupperuttingsequeneK 0
; ).
To eah Q
n
= (x
n
;y
n
), we assoiate a point I
n
= (u
n
;v
n
) with integer
oordinates. ThepointI
n
isthepointbelow(belowand totherightof)Q
n if
Q
n
isvertial(horizontal). Formally,
(u
n
;v
n )=
(dx
n e;y
n
1) ifQ
n
ishorizontal,
(x ;by ) ifQ isvertial
y=x+
r r
r
r r
r r r
r r
r
r r
r r
r r
I
0 I
1 I
2 J
0 J
1 J
2
Q0 Q1 Q
2
011 0111 01
y=
1+
x+
1+
r r
r
r r
r r
r r
r r
r
r r
r r
r r
I 0
0 I
0
1 I
0
2 J
0
0 J
0
1 J
0
2
0 1 1 0 1 1 1 0
Figure 2.4. Cuttingsequeneandorrespondingmehanialsequene.
SimilarpointsJ
n
aredenedabovetheline(seeFigure2.4). Itiseasytohek
that u
n +v
n
=nforn0,andthat
K
; (n)=v
n+1 v
n
=1+u
n u
n+1
Inthespeialase=0and irrational, weagaingetthesameinniteword
uptotherstletter. There isawordC
suhthat
K
;0
=0C
; K
0
;0
=1C
ObservethatQ
n
ishorizontalifandonlyif
1+v
n u
n
+<1++v
n
(2.1.7)
andQ
n
isvertialifandonlyif
v
n u
n
+<1+v
n
(2.1.8)
Wenowhekthat
K
;
=s
=(1+);=(1+)
Indeed, thetransformation (x;y)7!(x+y;x) of theplane maps theline y =
x+toy==(1+)x+=(1+),andapointI
n
=(u
n
;v
n )toI
0
n
=(n;v
n ).
It remainsto showthat
v
n
=
n+
(2.1.9)
Usingu
n +v
n
=n,wegetfrom(2:1:7)that
v
n
+1=(1+)=(1+)n+=(1+)<1+v
n
andfrom(2:1:8) that
v
n
=(1+)n+=(1+)<v
n
+1=(1+)
Thus,(2:1:9)holdsforhorizontalandforvertialsteps. Thus,uttingsequenes
arejust anotherformulationofmehanialwords.
Mehanialwordsanalsobegenerated byrotations. Let0<<1. The
rotationofangleisthemappingR=R
from [0;1[intoitselfdened by
R (z)=fz+g
wherefzg=z bzisthefrationalpartofz. IteratingR ,onegets
R n
()=fn+g
Moreover,astraightforwardomputationshowsthat
b(n+1)+=1+bn+ () fn+g1
Thus,deningapartitionof[0;1[by
I
0
=[0;1 [; I
1
=[1 ;1[;
onegets
s
; (n)=
0 ifR n
()2I
0
1 ifR n
()2I
1
(2.1.10)
Itwillbeonvenienttoidentify[0;1[withthetorus(ortheunitirle). For0
b<a<1,theset [a;1℄[[0;b[isonsideredasanintervaldenoted[a;b[. Then,
for any subinterval I of [0;1[, the sets R (I) and R 1
(I) are always intervals
(evenwhenoverlappingthepoint0).
As anexample of theuse of rotations, onsider awordw = b
0 b
1 b
m 1 ,
with b
0
;b
1
;::: letters. Wewantto knowwhether wis afatorof somes
;
=
a
0 a
1
,witha
0
;a
1
;:::letters. By(2.1.10),a
n+k
=b
i
ifandonlyifR n+i
()2
I
b
i
,orequivalently,ifandonlyifR n
()2R i
(I
b
i
). Thus, forn0,
w=a
n a
n+1 a
n+m 1
() R n
()2I
w
(2.1.11)
whereI
w
istheinterval
I
w
=I
b0
\R 1
(I
b1
)\\R m+1
(I
bm
1 )
TheintervalI
w
isnon emptyifandonlyifw isafatorofs
;
. Observethat
thispropertyisindependentof,andthuswordss
; ands
;
0 havethesame
setoffators. Aombinatorialproofwill begivenlater(Proposition2.1.18).
Mehanial words are quite naturally dened as two-sided innite words.
However,itappearsthatseveralproperties,suhasTheorem2.1.13below,only
Theorem 2.1.13. Letsbeaninniteword.Thefollowingareequivalent:
(i) sisSturmian;
(ii) sisbalanedandaperiodi;
(iii) sisirrationalmehanial.
The proof will bea simpleonsequene of twolemmas. In theproofs, we
willuseseveraltimestheformula
x 0
x 1<bx 0
bx<x 0
x+1:
Lemma 2.1.14. Letsbeamehanialwordwithslope. Thensisbalaned
ofslope. Ifisrational,thensispurelyperiodi. Ifisirrational,thensis
aperiodi.
Proof. Lets=s
;
bealowermehanialword. Theproofissimilarforupper
mehanialwords. Theheightofafatoru=s(n)s(n+p 1)isthenumber
h(u)=b(n+p)+ bn+, thus
juj 1<h(u)<juj+1 (2.1.12)
This implies bjuj h(u)1+bjuj, and shows that h(u) takesonly two
onseutivevalues,whenurangesoverthefatorsofaxed lengthofs. Thus,
sisbalaned. Moreover,by(2.1.12)
(u)
<
1
juj
Thus(u)!forjuj!1andistheslopeofsasitwasdenedforbalaned
words. Thisprovestherststatement.
Ifisirrational,thewordsisaperiodibyProposition2.1.11. If=q=pis
rational,thenb(n+p)+=q+bn+,foralln0. Thuss(n+p)=s(n)
foralln,showingthatsispurelyperiodi.
Lemma 2.1.15. Let sbeabalaned inniteword. Ifsis aperiodi,then sis
irrationalmehanial. Ifs ispurely periodi,thensisrationalmehanial.
Proof.InviewofCorollary2.1.8,shasaslope,say. Denotebyh
n
theheight
oftheprexoflengthnofs.
Foreveryrealnumber,oneat leastofthefollowingholds:
{h
n
bn+foralln;
{h
n
bn+foralln.
Indeed,ontheontrarythereexistarealnumber andtwointegersn;n+ksuh
that h
n
< bn+ and h
n+k
> b(n+k)+ (or the symmetri relation).
This implies that h
n+k h
n
2+b(n+k)+ bn+ > 1+k, in
ontraditionwith(2.1.4).
Set
=inff jh bn+forallng
ByProposition 2.1.10,onehas1,and<1ifisirrational. Observethat
foralln0
h
n
n+h
n
+1 (2.1.13)
sine otherwisethere is an integern suh that h
n
+1 < n+, and setting
=h
n
+1 n, onehas <and n+ =h
n
+1>h
n
, in ontradition
withthedenitionof.
Ifsisaperiodi,thenisirrationalbyProposition2.1.11,andn+isan
integerforatmostonen. By(2.1.13),eitherh
n
=bn+ foralln,andthen
s=s
; ,orh
n
=bn+ for allbut onen
0 , andh
n0
+1=n
0
+. Inthis
ase,onehash
n
=dn+ 1eforallnands=s 0
; 1 .
If s = u
!
is purely periodi with period juj =p, then = q=p with q =
h(u)=h
p
. Againh
n
=bn+ifn+isneveraninteger(thisdepends on
).
If h
n
=n+for somen, welaimthat h
n
=bn+for alln. Assume
the ontrary. Then by (2.1.13), 1+h
m
= m+, for some m and we may
assume n < m < n+p. Consider the words y = s(n+1)s(m) and z =
s(m+1)s(n+p). Then(y)=(h
m h
n
)=(m n)= 1=jyjand(z)=
(h
n+p h
m
)=(n+p m)=+1=jzj,whene
(y) (z)
=1=jyj+1=jzj,in
ontraditionwithProposition2.1.7. Similarly, if1+h
n
=n+forsomen,
thenh
n
=dn+e foralln.
Proof of theorem2.1.13. Weknowalreadyby Theorem2.1.5that (i)and (ii)
are equivalent. Assume that s is irrational mehanial. Then s is balaned
aperiodibyLemma2.1.14. Conversely,ifsisbalanedandaperiodi,thenby
theLemma2.1.15sisirrationalmehanial.
Example 2.1.16. To show that a balaned innite word is not always me-
hanialwhen theslope isrational (sotheonverse isfalsein Lemma 2.1.14),
onsider theinnitebalaned word01
!
. It isnot amehanialword. Indeed,
ithasslope1,andallmehanialwordss
1;
areequalto 1
!
.
Let us onsider mehanial wordswith rationalslopein somemoredetail.
For arational number =p=q with 0 1and p;q relativelyprime, the
innitewordss
;0 ands
0
;0
arepurelyperiodi. Denenitewords
t
p;q
=a
0 a
q 1
; t 0
p;q
=a 0
0 a
0
q 1
by
a
i
=
(i+1) p
q
i p
q
; a 0
i
=
(i+1) p
q
i p
q
Clearly,t
p;q andt
0
p;q
haveheightp. Theyareprimitivewordsbeause(p;q)=1.
Inpartiular, t
0;1
=0 andt
1;1
=1. These wordsare alled Christoel words.
In any ase, s
p=q;0
= t
!
p;q and s
0
p=q;0
= t 0
p;q
!
. Moreover, if 0 < p=q < 1, the
word t
p;q
starts with0 andends with 1(and t 0
p;q
starts with 1and ends with
0). Thereisawordz
p;q
suhthat
t
p;q
=0z
p;q 1; t
0
=1z
p;q
0 (2.1.14)
The wordz
p;q
is easily seento beapalindrome. Later, wewill seethat these
words,alled entralwords,haveremarkableombinatorialproperties.
Thefollowingresultdealswithnitewords.
Proposition2.1.17. A nite word w is a fator of some Sturmian word if
andonlyifitisbalaned.
Proof. Clearly a fator of a Sturmian word is balaned. For the onverse,
onsider abalanedwordw,and dene
0
=max((u) 1=juj);
00
=min ((u)+1=juj)
wherethemaximumandtheminimumistakenoverallnonemptyfatorsuof
w. Sinewisbalaned,onegetsfromProposition2.1.10that
(u) 1=juj<(v)+1=jvj
forallnonemptyfatorsuandv ofw. Thus 0
<
00
.
Takeanyirrationalnumberwith 0
<<
00
. Thenbyonstrution,for
everynonemptyfatoruofw,
(u)
<1 (2.1.15)
Let w
n
be the prex of length n of w. By(2.1.15), there exists a real
n
suhthat
h(w
n
)=n+
n
; j
n j<1
Moreover,for n > m, setting w
n
= w
m
u, onegets h(w
n
) h(w
m
) = h(u)=
(n m)+(
n
m
),showingthatj
n
m
j<1. Set
= max
1njwj
n :
Then
n+h(w
n
)=n++(
n
)>n+ 1
wheneh(w
n
)=bn+. ThisprovesthatwisaprexoftheSturmianword
s
; .
2.1.3. The fators ofone Sturmian word
Theaim ofthis paragraphisto givepropertiesof theset offatorsofasingle
Sturmianword.
Proposition2.1.18. LetsandtbeSturmianwords.
1. Ifsandthavesameslope,thenF(s)=F(t).
Proof. Let be the ommon slope of s and t. By Proposition 2.1.10, every
fatoruofsveries
j(u) j <
1
juj
(indeed,equalityisimpossiblebeauseisirrational). Next,foreveryfatorv
oft,
j(v) j <
1
jvj
LetX=F(s)[F(t). Theset X isfatorial. Itisalsobalanedsine
j (u) (v)j j(u) j+j(v) j <
1
juj +
1
jvj
InviewofProposition2.1.2
Card(X\A n
)n+1
foreveryn. ThusF(s)=X =F(t).
Let nowbetheslopeofsand betheslopeoft. Wemaysupposethat
>. Foranyfatoruofssuhthat( )2=juj,onehas(u) > 1=juj
byProposition2.1.10whene(u) =((u) )+( )1=jujshowing
thatuisnotafatoroft.
Proposition2.1.19. Theset F(s) offatorsof aSturmianwords islosed
underreversal.
Proof. Set
~
F(s) = f~x j x 2 F(s)g. The set X = F(s)[
~
F(s) is balaned.
In view of Proposition 2.1.2, Card(X \A n
) n+1, for eah n, and sine
Card(F(s)\A n
)=n+1,onehasX =F(s). Thus
~
F(s)=F(s).
WenowompareSturmianwords,withrespettotheirslopeandinterept.
ThelexiographiorderdenedinChapter1extendstoinnitewordsasfollows,
withtheassumptionthat 0<1. Giventwoinnitewordsx=a
0 a
n and
y = b
0 b
n
, we say that x is lexiographially less than y, and we write
x<y ifthereisanintegern suhthat a
i
=b
i
fori=0;:::;n 1and a
n
=0,
b
n
=1.
Proposition2.1.20. Let0<<1beanirrationalnumberand let; 0
be
realnumberswith 0; 0
<1. Then
s
;
<s
; 0
() <
0
:
Proof.Sineisirrational,thesetoffrationalpartsfngforn0isdensein
theinterval[0;1[. Thus<
0
ifand onlyifthereexistsanintegern1suh
that1 0
fng<1 ,andthisisequivalenttobn+ 0
=1+bn+.
If nis thesmallestinteger forwhih this equality holds, thens
;
(n 1)=0
ands
; 0
(n 1)=1ands
; 0
(k)=s
;
(k)fork<n 1.
Observethatthispropositiondoesnotholdforrationalslopes,sineindeed
s =0
!
forall.
Lemma 2.1.21. Let 0<; 0
<1be irrationalnumbersand let; 0
be real
numbers. Anyoftheequalitiess
;
=s
0
; 0,
s
;
=s 0
0
; 0
ors 0
;
=s 0
0
; 0
implies
=
0
and 0
mod1.
Proof. Anyof theequalitiesimpliesthat = 0
beauseequalwordshavethe
sameslope. Next,s
;
=s
; 0
implies 0
mod1bythepreviousproposition.
Finally, onsider the equality s
;
=s 0
; 0
. If n+ 0
is notan integerfor all
n 1, then s 0
; 0
= s
; 0
and the onlusion holds. Otherwise, let n be the
uniqueintegersuhthatn+
0
isaninteger. Thens
;+(1+n)
=s 0
; 0
+(1+n) ,
showingagainthat 0
mod1.
Sturmianwordswithinterept0havemanyinterestingproperties. Weob-
served already that, for an irrational number0 < < 1, the words s
;0 and
s 0
;0
dieronlybytheirrstletter,andthat
s
;0
=0
; s 0
;0
=1
where
istheharateristiwordofslope. Equivalently,
=s
;
=s 0
;
Thefollowingproposition statesaombinatorial haraterizationof harater-
istiwordsamongSturmianwords.
Proposition2.1.22. ForeverySturmianwords,either0sor1sisSturmian.
ASturmianwordsisharateristiifandonlyif0sand1sarebothSturmian.
Proof. The rst laim follows from the fat that s
;
= as
;
, for some
a2f0;1g.
Ifs=s
;
=s 0
;
istheharateristiwordofslope, then0s=s
;0 and
1s=s 0
;0
areSturmian.
Conversely,theSturmianwords0sand 1shavesameslope,say. Denote
by and 0
theirinterept. Thentheir ommonshifts hasinterept +=
0
+, andbyLemma 2.1.21, 0
mod1and we maytake0= 0
<1.
Thus 0s =s
;
and 1s= s 0
;
. Assume > 0. The rst letter of 0sis gives
0=b+ b=b+andtherstletterof1sis1=d+e de. Then
2=d+e,aontradition. Thus=0.
Wearenowabletodesriberightspeialfators.
Proposition2.1.23. The set of right speial fators of a Sturmian word is
thesetofreversalsoftheprexesof theharateristiwordofsameslope.
CallafatorwofaSturmianwordsleftspeialifboth0wand1warefatorsof
s. Clearly,wisleftspeialifandonlyifw~isrightspeial. Thustheproposition
statesthatthesetofleftspeialfatorsofaSturmianwordisthesetofprexes
Proof. LetsbeaSturmianwordofslope. ByProposition2.1.22,theinnite
words0
and1
areSturmianandlearlyhaveslope. Thus
F(s)=F(
)=F(0
)=F(1
)
by Proposition 2.1.18. Consequently, for eah prex p of
, 0p and 1p are
fators of s. Sine F(s) is losed under reversal, this shows that p~is right
speial. Thusp~istheuniquerightspeialfatoroflengthjpj.
Example 2.1.24. Consider again the Fibonai word f. We have seen in
Example 2.1.1 that its right speial fators are the reversals of its prexes.
Thuseah prexof f is leftspeial. This showsthat F(f)=F(0f)=F(1f).
Consequently,f is harateristiofslope1=
2
.
Proposition2.1.25. Thedynamial system generated by a Sturmianword
isminimal.
Proof.LetsbeaSturmianword,andletxbeaninnitewordsuhthatF(x)
F(s). Clearly,xisbalaned. Also,xhasthesameirrationalslopeass. Thusx
isaperiodiandthereforeisSturmian. ByProposition 2.1.18(1),F(x)=F(s).
Thisshowsthatsandx generatethesamedynamialsystem.
Observe that Proposition 2.1.18(2)is aonsequene of Proposition 2.1.25.
Indeed,theintersetionoftwodistintminimaldynamialsystemsisthetrivial
system.
2.2. Standard words
Thissetionisonernedwithafamilyofnitewordsthat arebasibriksfor
onstrutingharateristiSturmianwords,in the sense that everyharater-
istiSturmianwordisthelimit ofasequeneof standardwords. Thiswill be
shownin Setion2.2.2.
2.2.1. Standard words and palindromewords
After basi denitions, wegivetwoharaterizations of standardwords. The
rstisbyaspeialdeomposition intopalindromewords(Theorem2.2.4),the
seond (Theorem 2.2.11) by anextremal propertyon the periods of the word
thatisloselyrelatedtoFineandWilf'stheorem. Wegivethena\mehanial"
haraterization of entral and standard words(Proposition 2.2.15). We end
withanenumerationformulaforstandardwords.
Considertwofuntions and fromf0;1g
f0;1g
intoitselfdened by
(u;v)=(u;uv); (u;v)=(vu;v)
The set of standard pairs is the smallestset of pairs of words ontaining the
pair(0;1) andlosed under and . Astandard wordisanyomponentofa
(0;1)
(0;01) (10;1)
(0;001) (010;01) (10;101) (110;1)
(0;0001) (0010;001)(010;01001)(01010;01) (10;10101)
(01001010;01001)
Figure2.5. Thetreeofstandardpairs.
Example 2.2.1. Figure2.5showsthebeginningofthetreeofstandardpairs.
Consideringtheleftmostandrightmostpaths,onegetsthepairs
(0;0 n
1); (1 n
0;1) (n1)
Next tothemarethepairs
(0(10) n
;01); (10;(10) n
1) (n1)
These arethepairswithoneomponentoflength1or2.
FiniteFibonaiwordsarestandard,sine(f
0
;f
1
)=(0;1), andforn1,
(f
2n+2
;f
2n+1
)= (f
2n
;f
2n 1 ).
Everystandardwordwhih isnotaletter isaprodut oftwostandardwords
whih aretheomponentsofsomestandardpair. Thenextpropositionstates
someelementaryfats.
Proposition2.2.2. Letr=(x;y)beastandardpair.
1. Ifr6=(0;1)thenoneofx ory isaproperprexoftheother.
2. Ifx(resp.y)isnotaletter,thenx endswith10(resp.y endswith01).
3. Onlythelast twolettersofxyandyxaredierent.
Proof.Weprovethelast laimbyindutiononjxyj. Assumeindeedthat xy=
p01andyx=p10.Then (r)=(x;xy)andxxy=xp01,(xy)x=x(yx)=xp10,
sothelaimistruefor (r). Thesameholdsfor(r).
Everystandardpairisobtainedinauniquewayfrom(0;1) byiterateduse
of and . Indeed, if (x;y) is a standardpair, then it is an image through
(resp. ) ifand only ifjxj <jyj (resp. jxj>jyj). Thus, there is aunique
produtW =
1
Æ:::Æ
n
,with
i
2f ;gsuh that
Considertwomatries
L=
1 0
1 1
; R=
1 1
0 1
and deneamorphism from the monoidgenerated by and into theset
of22matriesby
( )=L; ()=R ;
and(
1 Æ :::Æ
n
)=(
1
)(
n
). If(x;y)=W(0;1),thenastraightforward
indutionshowsthat
(W)=
jxj
0 jxj
1
jyj
0 jyj
1
(2.2.1)
Observethateverymatrix(W)hasdeterminant1. Thusif(x;y)isastandard
pair,
jxj
0 jyj
1 jxj
1 jyj
0
=1 (2.2.2)
showingthattheentriesinthesamerow(olumn)of(W)arerelativelyprime.
From(2.2.2),onegets
h(y)jxj h(x)jyj=1: (2.2.3)
(reallthath(w)=jwj
1
istheheightofw). Thisshowsalsothatjxjandjyjare
relativelyprime. A simpleonsequeneis thefollowingproperty.
Proposition2.2.3. A standardwordisprimitive.
Proof.Let wbeastandardwordwhih is notaletter. Then w=x orw=y
for some standard pair (x;y). From (2.2.3), one gets that h(w) and jwj are
relativelyprime. This impliesthat wisprimitive.
The operations and anbeexplained throughthree morphismsE, G,
Donf0;1g
whihweintroduenow. Thesewillbeusedalsointhesequel. Let
E: 07!1
17!0
; G: 07!0
17!01
; D:
07!10
17!1
It is easily heked that E ÆD = GÆE = '. We observe that, for every
morphismf,
(f(0);f(1))=(fG(0);fG(1)); (f(0);f(1))=(fD(0);fD(1))
ForW =
1
Æ:::Æ
n
,with
i
2f ;g,dene
^
W =
^
n Æ:::Æ
^
1 ,with
^
=G,
^
=D. Then
W(0;1)=(
^
W(0);
^
W(1)): (2.2.4)
Standardwordshavethefollowingdesription.
Theorem 2.2.4. Awordwisstandardifandonlyifitisaletterorthereexist
palindromewordsp,qandrsuhthat
w=pab=qr (2.2.5)
Example 2.2.5. Theword01001010isstandard(seeFigure2.5)and
01001010=(010010)10=(010)(01010):
Westarttheproofwithalemmaofindependentinterest.
Lemma 2.2.6. If aprimitiveword isa produtof twononempty palindrome
words,thenthis fatorizationisunique.
Proof. Let w be a primitive word and assume w = pq = p 0
q 0
for palindrome
wordsp;q;p 0
;q 0
. Wesuppose jpj> jp 0
j, so that p=p 0
s(=sp~ 0
), sq =q 0
(= q~s)
forsomenonemptywords. Thussp~ 0
q=pq=p 0
q 0
=p 0
q~s,showingthatp 0
qand
~
s are powersof somewordz. But thenw =pq = ~sp 0
q =z n
forsome n2,
ontraditingprimitivity.
Observethat (2.2.5)impliesthefollowingrelations.
Lemma 2.2.7. Ifw=pab=qrforpalindromewordsp,q,r,andlettersa6=b,
thenoneofthefollowingholds
(i) r=",p=(ba) n
b,q=(ba) n+1
b=wforsomen0;
(ii) r=b,p=a n
,q=a n+1
,w=a n+1
b forsomen0;
(iii) r=bab,p=b n+1
,q=b n
,w=b n+1
ab forsomen0;
(iv) r=basab,p=qbas,w=qbasabforsomepalindromewords.
Weneedanotherlemma.
Lemma 2.2.8. Letx;ybewordswithjxj;jyj2. Thepair(x;y)isastandard
pairifandonlyifthere existpalindromewordsp,q,rsuhthat
x=p10=qr and y=q01 (2.2.6)
or
x=q10 and y=p01=qr: (2.2.7)
Proof.Assumethat(2.2.6)holds(theotheraseissymmetri). Ifristheempty
word,thenbythepreviouslemma
(x;y)=((01) n+1
0;(01) n+1
001)= ((01) n+1
0;01)
showingthat thepair(x;y)isstandard.
Ifr=0,then(x;y)=(1 n
0;1 n
01)= (1 n
0;1), andifr=010,then(x;y)=
(0 n
10;0 n
1)=(0;0 n
1).
Thus, we may assume that r = 01s10 for some palindrome word s. By
(2.2.6),if followsthat y is aprexof x,sox=yz forsomewordz. Weshow
that (z;y)is standard. From p=q01s= s10qit followsthat q 6=s. Assume
jqj<jsj (the otherase issymmetri). Then s=qt for somewordt, andthe
equationp=qt10q showsthat thewordr 0
=t10isapalindrome. Thus
0
and(z;y)satises(2.2.6).
Conversely, let(x;y)beastandardpair,and assume(x;y)= (x;z),that
isy=xz. Ifzisaletter,then(x;z)=(1 n
0;1)forsomen1and
x=q10; y=p01=qr
forq=1 n 1
,p=1 n
,r=101.
Thuswemayassumethat forsomepalindromewordsp,q,r,either
x=p10=qr; z=q01
or
x=q10; z=p01=qr:
Intherstase,
x=p10; y=xz=(qrq)01=p(10q01)
Intheseond ase,
x=q10; y=xz=q(10p01)=(qrq)01
beause10p=rq. Thus(2.2.7)holds.
Proof ofTheorem2.2.4. Letwbeastandardword,jwj2. Thenthereexists
astandardpair(x;y)suh that w=xy (or symmetriallyw=yx). If x=0,
then y = 0 n
1 for somen 0, and xy = 0 n+1
1has the desired fatorization.
A similar argument holds for y = 1. Otherwise, either (2.2.6) or (2.2.7) of
Lemma2.2.8holds. Intherstase,xy=p(10q01)=qrq01andin theseond
ase,xy=q(10p01)=qrq01beause10p=rq. Thefatorizationisuniqueby
Lemma2.2.6beauseastandardwordisprimitive.
Conversely,ifw=p10=qr(orw=p01=qr)forpalindromewordsp,q,r,
then byLemma 2.2.8, theword wis aomponentof somestandardpair,and
thusisastandardword.
A wordwisentralifw01(orequivalentlyw10)isastandardword. As we
shallsee,entral wordsplayindeedaentralrole.
Corollary 2.2.9. Awordisentral ifandonlyifitisintheset
0
[1
[(P\P10P)
whereP isthe setofpalindrome words. Thefatorizationofaentralwordw
asw=p10q withp;qpalindromewordsisunique.
Observethat P\P10P =P\P01P.
Proof. Let w 20
[1
[(P \P10P). By theprevious haraterization,w01
is a standard word, so w is entral. Conversely, if w01 is standard, then w
is a palindrome and w01 = qr for some palindrome words q and r. Either
w 2 0
[1
, or by Lemma 2.2.7, r =" and w = (10) n
1 for some n 1, or
w=q10sforsomepalindrome s,asrequired.
Corollary 2.2.10. A palindromeprex(suÆx)ofaentralwordisentral.
Proof.Weonsidertheaseofaprex. Letpbeaentralword.Ifp20
[1
,the
resultislear. Letxbeastandardwordsuhthatx=pab,withfa;bg=f0;1g.
Thenx=yzforastandardpair(y;z)or(z;y). Sety=qbaandz=rab,where
q;rareentralwords. Thenp=qbar=rabq andbysymmetrywemayassume
that jrj<jqj.
Letwbeapalindromeprexofp. Ifjwjjqj,theresultholdsbyindution.
If w= qb then wis apowerof b. Thus set w=qbat where t is aprex of r.
Siner isaprexofq, thewordtisaprexofq, andsinew=
~
tabq, onehas
t=
~
t. Thus,byCorollary2.2.9,w=qbatisentral.
Thenextharaterizationrelatesentral wordstoperiods inwords. Reall
from Chapter 1that givena word w= a
1 a
n
, where a
1
;:::;a
n
are letters,
anintegerkis aperiodofwif k1and a
i
=a
i+k
forall1in k. Any
integerknisaperiodwiththisdenition.
An integerk with 1 k jwj is aperiod of w if and only if there exist
wordsx,y,andz suhthat
w=xy=zx; jyj=jzj=k:
Fine andWilf's theoremstatesthat ifawordwhastwoperiods kand `, and
jwjk+` gd(k;`),then gd(k;`) isalso aperiod ofw. Inpartiular,if k
and `arerelativelyprime,andjwjk+` 1,thenwisthepowerofasingle
letter. Theboundis sharp,andthe questionarisesto desribethewordsw of
length jwj= k+` 2having periods k and `. This isthe objetof thenext
theorem.
Theorem 2.2.11. Awordwisentralifandonlyifithastwoperiodskand
` suh that gd(k;`) =1and jwj =k+` 2. Moreover,if w 2= 0
[1
, and
w =p10q with p,q palindrome words, thenfk;`g=fjpj+2;jqj+2gand the
pairfk;`gisunique.
The proof will show that any word w having two periods k and ` suh that
gd(k;`)=1andjwj=k+` 2isoveranalphabetwithat mosttwoletters.
Proof. Let w be a entral word. Then w01 is a standard word, and there is
a standard pair (x;y) suh that w01 = xy. If x = 0 ory = 1, then w is a
powerof 0 resp. of 1, and w has periods k = 1and ` = jwj+1. Otherwise,
x =p10and y =q01 for somepalindrome wordsp, q, and w =p10q =q01p
has two periods k = jxj and ` = jyj whih are relatively prime by Equation
(2.2.3). Assume that w has also periods fk 0
;` 0
g, with k 0
+` 0
2 =jwj. We
maysupposek<k 0
<` 0
<`. Sinek+` 0
1jwj,Fine andWilf's theorem
applies. Sowhasalsotheperiodd=gd(k;` 0
). Similarly,whasalsotheperiod
d 0
=gd(k;k 0
). Soithas theperiod gd(d;d 0
)=1. This provesthat thepair
fk;`gisunique.
Conversely,ifwisapowerofaletter,theresultistrivial. Thusweassume
Sine whasperiod k,there isawordx oflengthjxj=` 2thatis botha
prexandasuÆxofw. Similarly,thereisawordyoflengthjyj=k 2thatis
bothaprexand asuÆxofw. Consequently,thereexist wordsuandv,both
oflength2,suhthat
w=yux=xvy
We provebyindution onjwj that x, y, w are palindrome words, that uand
v are omposed of distint letters, and that no other letters than those of u
appear inw(thatiswisoveranalphabet oftwoletters).
Ifk=2,theny istheemptyword. Thusux=xv,and `isodd. Therefore
u=ab, v =ba, x =(ab) n
a, w =(ab) n+1
a for lettersa 6=b and somen 0.
Theresultholdsin thisase.
Ifk=` 1,thenx=ya=by forlettersaandb. Butthena=b andwis
apowerofaletter,aasethatwehaveexluded.
Thus weassumek` 2. Then yuis aprex ofx. Dene z byyuz=x.
Then
x=yuz=zvy
showingthat x hasperiodsjyuj=k and juzj=` k. Sinegd(k;` k)=1
andjxj=k+(` k) 2,wegetbyindutionthatx isapalindrome,andthat
itsprex oflengthk 2,that isy, anditssuÆxof length` k 2,thatis z
also are palindromes. Moreover,u =ab for letters a6=b, and u~= v beause
yuz=z~uy=zvy. Also, thewordx (andy,andthereforealso w)isomposed
onlyofa'sandb's. Thuswisentral.
Theorem 2.2.11assoiates,to everyentral wordof lengthm,apair fk;`g
of relativelyprime integerssuh that k+` 2 =m. We now show that, for
eahpairfk;`gofrelativelyprime integers,thereexists indeed aentralword
oflengthk+` 2andperiods kand`.
Leth;mberelativelyprimeintegerswith1h<m. Deneaword
z
h;m
=a
1 a
2 a
m 2 (a
n
2f0;1g)
by
a
n
=
(n+1) h
m
n h
m
:
Thesewordshavealreadybeenmentionedinourdisussionofrationalmehan-
ial words (Equation 2.1.14). Eah word z
h;m
has length m 2 and height
h 1.
Proposition2.2.12. Foreveryouple1h<mofrelativelyprimeintegers,
the word z
h;m
is entral. It has the periods k and ` where k+` = m and
kh1modm.
Proof.Denekby1km 1,andsetkh=1+m. Observethatkexists
beausehandm arerelativelyprime. Let`=m k. Then `h 1modm,
and`istheuniqueintegerintheinterval[0:::;m 1℄withthisproperty. Next
(n+k) h
=+
nh+1
Sinenh6 1modmfor1n` 1,itfollowsthat
nh+1
m
=
nh
m
(1n` 1)
Consequently,a
n+k
=a
n
for 1n` 2. Asimilar argumentholdswhen k
isreplaedby`and 1ishangedinto1.
Assume that some integerddivides k and `. Then d divides also m. But
k and` arerelativelyprimeto m, sod=1andgd(k;`)=1. Thisproves,by
Theorem2.2.11,thatz
h;m
isentral.
Example 2.2.13. The wordsz
1;m
= 0 m 2
and z
m 1;m
= 1 m 2
are entral.
Inpartiular,z
1;2
=".
Example 2.2.14. Forh=5,m =18, onegetsz
5;18
=0010001001000100,a
wordoflength16. Byinspetion,onendstheperiods7and11. Theprevious
propositionallowsto omputethem, sine1151mod18.
Proposition2.2.15. Leth;mberelativelyprime integerswith 1h<m.
Thereexistexatlytwostandardwordsofheighthandlengthm,namelyz
h;m 10
andz
h;m
01. Thesewordsarebalaned.
Proof.ByProposition2.2.12,thewordsz
h;m
10andz
h;m
01arestandardwords
of height handlengthm. Theyarefatorsof theSturmianwordss
h=m;0 and
s 0
h=m;0
andthereforearebalaned. Weprovethatthereexistsonlyonestandard
wordof heighthandlengthm endingin10. Assume therearetwo,saywand
w 0
. Then
w=xy; w 0
=x 0
y 0
forsomestandardpairs(x;y),(x 0
;y 0
). Byformula(2.2.3),
h(x)jyj h(y)jxj=1; h(x 0
)jy 0
j h(y 0
)jx 0
j=1
Sinem=jxj+jyjandh=h(x)+h(y),thisgives
h(x)m jxjh=1; h(x 0
)m jx 0
jh=1
whene
(h(x) h(x 0
))m=(jx 0
j jxj)h
Sine gd(m;h) =1, m divides jx 0
j jxj. Thus jxj =jx 0
j, that is x =x 0
and
y=y 0
.
Reall that Euler's totient funtion is dened for m 1 as thenumber
(m) ofpositiveintegerslessthanmandrelativelyprime tom
Corollary 2.2.16. Thenumberofstandardwordsoflengthmis2(m),the
number of entral words of length m is (m+2), where is Euler's totient
2.2.2. Standard sequenesand harateristi words
Inthissetion,weusepartiularmorphismsthatwillalsobeonsideredin the
next setion. Three of them, namely E, G, and D, were already introdued
earlier. Here, these morphisms are used to relate standard words to hara-
teristi words, and bothto theontinued fration expansion of theslopeof a
harateristiword. Considerthemorphisms
E: 07!1
17!0
; ':
07!01
17!0
; '~:
07!10
17!0
Fromthese, wegetothermorphisms,denotedG,
~
G,D,
~
Dand denedby
G='ÆE: 07!0
17!01
;
~
G='~ÆE: 07!0
17!10
D=EÆ':
07!10
17!1
;
~
D=EÆ'~:
07!01
17!1
Ofourse,'=GÆE=EÆDand'~=
~
GÆE=EÆ
~
D.
Lemma 2.2.17. Foranyrealnumber,thefollowingrelationshold: E(s
; )=
s 0
1 ;1
andE(s 0
; )=s
1 ;1 .
Proof.Forn0,
s 0
1 ;1
(n)=d(1 )(n+1)+1 e d(1 )n+1 e
=1 (d n e d (n+1) e)=1 s
; (n)
beause d re=brfor everyreal numberr. This provestherst equality,
andtheseondissymmetri.
Lemma 2.2.18. Let0<<1. For0<1,
G(s
; )=s
1+
;
1+
;
~
G(s
; )=s
1+
; +
1+
; '(s
; )=s
0
1
2
; 1
2
andfor0<1,
G(s 0
; )=s
0
1+
;
1+
;
~
G(s 0
; )=s
0
1+
; +
1+
; '(s 0
;
)=s1
2
; 1
2 :
Proof.Lets=a
0 a
1 a
n
beaninniteword,thea
i
beingletters. Aninteger
nis theindex ofthe k-thourreneof theletter1in sif a
0 a
n
ontainsk
letters1anda
0 a
n 1
ontainsk 1letters1. Ifs=s
;
and0<1,this
meansthat
b(n+1)+=k; bn+=k 1
whihimpliesn+<k(n+1)+,that is
n=
k
1
:
Similarly,ifs=s 0
;
and0<1,then
d(n+1)+e=k+1; dn+e=k
andn= j
k
k
.
SetG(s
; )=b
0 b
1 b
i
, with b
i
2f0;1g. Sine everyletter1in s
; is
mapped to 01in G(s
;
), the prex a
0 a
n of s
;
(where n is theindex of
thek-thletter 1)is mappedontotheprex b
0 b
1 b
n+k of G(s
;
). Thusthe
index ofthek-thletter1inG(s
; )is
n+k=
&
k
1+
1+
1 '
Thisprovestherstformula.
Next,weobservethat,foranyinnitewordx,onehas
G(x)=0
~
G(x)
Indeed,the formulaG(w)0=0
~
G(w)iseasily shown tohold fornite wordsw
by indution. Furthermore,if aSturmianwords
;
starts with 0and setting
s
;
= 0t, onegets t = s
;+
. Altogether
~
G(s
; )= s
=(1+);(+)=(1+) for
0<1. Theproofoftheotherformulais similar. Finally, sine'=GÆE,
'(s
;
)=G(s 0
1 ;1 )=s
0
(1 )=(2 );(1 )=(2 ) .
Corollary 2.2.19. For any Sturmianwords, theinnite wordsE(s),G(s)
~
G(s), '(s),'(s),~ D(s)
~
D(s)areSturmian.
Formulas similar to those of Lemma 2.2.18hold for ' ;~ D;
~
D (Problem 2.2.6).
Reallthat theharateristiwordofirrationalslopeisdenedby
=s
;
=s 0
; :
Thepreviouslemmasimply
Corollary 2.2.20. Foranyirrationalwith0<<1,onehas
E(
)=
1
; G(
)=
=(1+)
Form1,deneamorphism
m by
m :
07!0 m 1
1
17!0 m 1
10
It iseasilyhekedthat
m
=G m 1
ÆEÆG:
Corollary 2.2.21. Form1,onehas
m (
)= .
Proof. Sine E ÆG(
) =
1=(1+)
, the formula holds for m = 1. Next,
G(
1=(k +) )=
1=(1+k +)
,sothelaimistruebyindution.
We use this orollary for onneting ontinued frations to harateristi
words. Reall that everyirrational number admits aunique expansionasa
ontinuedfration
=m
0 +
1
m
1 +
1
m
2 +
1
(2.2.8)
where m
0
;m
1
;::: are integers,m
0
0, m
i
>0for i1. If(2.2.8) holds, we
write
=[m
0
;m
1
;m
2
;:::℄:
The integersm
i
are alled the partial quotientsof . If thesequene (m
i )is
eventuallyperiodi,andm
i
=m
k +i
forih,thisisreportedbyoverliningthe
purelyperiodi part,asin
=[m
0
;m
1
;m
2
;:::;m
h 1
;m
h
;:::;m
h+k 1
℄:
Let =[0;m
1
;m
2
;:::℄ betheontinued fration expansionof an irrational
with0<<1. If,forsome with0< <1,
=[0;m
i+1
;m
i+2
;:::℄
weagreetowrite
=[0;m
1
;m
2
;:::;m
i +℄:
Corollary 2.2.22. If =[0;m
1
;m
2
;:::;m
i
+℄for someirrationaland
0<;<1,then
=
m1 Æ
m2
ÆÆ
mi (
)
Let (d
1
;d
2
;:::;d
n
;:::) beasequene of integers, with d
1
0 and d
n
>0 for
n>1. Tosuhasequene,weassoiateasequene(s
n )
n 1
ofwordsby
s
1
=1; s
0
=0; s
n
=s dn
n 1 s
n 2
(n1) (2.2.9)
Thesequene(s
n )
n 1
isastandardsequene, andthesequene(d
1
;d
2
;:::)is
itsdiretive sequene. Observethat ifd
1
>0,thenanys
n
(n0)starts with
0;ontheontrary,ifd
1
=0,thens
1
=s
1
=1,ands
n
startswith1forn6=0.
Everys
2n
endswith0,everys
2n+1
endswith1.
Example 2.2.23. Thediretivesequene(1;1;:::)givesthestandardsequen-
e dened by s
n
= s
n 1 s
n 2
, that is the sequene of nite Fibonai words.
Observethatthediretivesequene(0;1;1;:::)resultsinthesequeneofwords
Every standard word ours in some standard sequene, and every word
ourringin astandardsequeneisastandardword. Thisresultsbyindution
from thefat that,fors
n
=s d
n
n 1 s
n 2
,onehas
(s
n
;s
n 1 )=
dn
(s
n 2
;s
n 1
); (s
n 1
;s
n )=
dn
(s
n 1
;s
n 2 )
Thus
(s
2n
;s
2n 1 )=
d2n
Æ d2n
1
ÆÆ d1
(0;1)
(s
2n
;s
2n+1 )=
d2n+1
Æ d2n
Æ d2n
1
ÆÆ d1
(0;1)
ByEquation2.2.4,thisgivestheexpressions
s
2n
=G d
1
ÆD d
2
ÆÆD d
2n
(0)=G d
1
ÆÆD d
2n
ÆG d
2n+1
(0)
s
2n+1
=G d
1
ÆD d
2
ÆÆD d
2n+2
(1)=G d
1
ÆÆD d
2n
ÆG d
2n+1
(1)
Proposition2.2.24. Let=[0;1+d
1
;d
2
;:::℄ betheontinuedfrationex-
pansion of some irrational with 0 < < 1, and let (s
n
) be the standard
sequeneassoiatedto (d
1
;d
2
;:::). Theneverys
n
isaprexof
and
= lim
n!1 s
n :
Proof.Bydenition,s
n
=s dn
n 1 s
n 2
forn1. Dene morphismsh
n by
h
n
=
1+d
1 Æ
d
2
ÆÆ
d
n :
Welaimthat
s
n
=h
n (0); s
n s
n 1
=h
n
(1); n1
This holdsforn=1sineh
1 (0)=0
d1
1=s
1 andh
1 (1)=0
d1
10=s
1 s
0 . Next,
forn2,
h
n (0)=h
n 1 (
d
n
(0))=h
n 1 (0
dn 1
1)=s d
n 1
n 1 s
n 1 s
n 2
=s
n
and
h
n (1)=h
n 1 (0
dn 1
10)=s
n s
n 1
Forany innite word x, the innite word h
n
(x) starts with s
n
beauseboth
h
n
(0) andh
n
(1) startwiths
n
. Thus,setting
n
=[0;d
n+1
;d
n+2
;:::℄, one has
=h
n (
n
) byCorollary2.2.22andthus
startswith s
n
. This provesthe
rstlaim. Theseond isanimmediateonsequene.
Itiseasilyhekedthat
1+d
1 Æ
d
2
ÆÆ
d
r
=G d1
ÆEÆG d2
ÆEÆÆG dr
ÆEÆG
=
G d
1
ÆD d
2
ÆÆD d
r
ÆG ifriseven,
G d1
ÆD d2
ÆÆD dr
ÆDÆE otherwise.
Example 2.2.25. ThediretivesequenefortheFibonaiwordis (1;1;:::).
The orresponding irrational is 1=
2
= [0;2;1;1;:::℄, and indeed the innite
2
Example 2.2.26. Sine 1= = [0;1;1;1;:::℄, theorresponding standardse-
quene iss
1
=1,s
2
=10,s
3
=101,.... Thesequene isobtainedfrom theFi-
bonaisequenebyexhanging0'sand1's,inonordanewithLemma2.2.17,
sineindeed1=+1=
2
=1.
Example 2.2.27. Consider=( p
3 1)=2=[0;2;1;2;1;:::℄. Thediretive
sequene is(1;1;2;1;2;1;:::), and thestandardsequene startswith s
1
=01,
s
2
=010,s
3
=01001001,...,whene
( p
3 1)=2
=010010010100100100101001001001
Duetotheperiodiityofthedevelopment,wegetforn2thats
n+2
=s 2
n+1 s
n
ifnisodd,ands
n+2
=s
n+1 s
n
ifniseven.
Corollary 2.2.28. Every standard word is a prex of some harateristi
word.
Thus,everystandardwordisleftspeial.
Corollary 2.2.29. A wordisentralifand onlyifitisapalindrome prex
ofsomeharateristiword.
Proof.Aentralwordisaprexofsomestandardword,soalsoofsomehara-
teristiword. Conversely,apalindromeprexofaharateristiwordisaprex
ofanysuÆientlylongwordinitsstandardsequene,soalsoofsomesuÆiently
longentralword. Thus theresultfollowsfromProposition2.2.10.
Proposition2.2.24hasseveralinterestingonsequenes. Therelationtox-
pointsislefttosetion2.3.6. Wefousontwoproperties,rstthepowersthat
mayappear in aSturmian word, and then the omputation of thenumberof
fatorsofSturmianwords.
Letx beaninniteword. Forw2F(x),theindexofwin xisthegreatest
integerdsuhthatw d
2F(x),ifsuhanintegerexists. Otherwise,wissaidto
haveinniteindex.
Proposition2.2.30. EverynonemptyfatorofaSturmianwordshasnite
indexin s.
Proof. Assume the ontrary. There exist a Sturmianwords and a nonempty
fator uof s suh that u n
is afatorof s for everyn1. Consequently, the
periodiwordu
!
is inthedynamialsystemgeneratedbys. Sinethissystem
isminimal,F(s)=F(u
!
),aontradition.
An innite wordx hasbounded index ifthere existsan integerdsuhthat
everynonemptyfatorofx hasanindexlessthanorequalto d.
Theorem 2.2.31. ASturmianwordhasboundedindexifandonlyiftheon-
Westartwithalemma.
Lemma 2.2.32. Let (s
n )
n 1
be the standardsequene of the harateristi
word
, with=[0;1+d
1
;d
2
;:::℄. Forn3,thewords 1+d
n+1
n
isaprexof
, ands
2+dn+1
n
isnotaprex. Ifd
1
1,this holdsalsoforn=2.
Example 2.2.33. For the Fibonai word f = 0100101001001, we have
s
n
=f
n andd
n
=1foralln. Thelemmalaimsthatforn2,thewordf 2
n is
aprexoftheinnitewordf,andthatf 3
n
isnot. Asanexample,f 2
2
=010010
is a prex and f 3
2
=010010010is not. Observealso that f 2
1
= 0101 is nota
prexoff.
Proof.Weshowthatforn3(andforn2ifd
1
1),onehas
s
n 1 s
n
=s
n t
n 1
; with t
n
=s d
n 1
n 1 s
n 2 s
n 1
Indeed
s
n 1 s
n
=s
n 1 s
dn
n 1 s
n 2
=s dn
n 1 s
d
n 1
n 2 s
n 3 s
n 2
=s d
n
n 1 s
n 2 s
dn
1 1
n 2 s
n 3 s
n 2
=s
n t
n 1
provided d
n 1
1. Observe that t
n 1
is not a prex of s
n
, sine otherwise
s
n
=t
n 1
uforsomewordu,ands
n 1 s
n u=s
2
n ands
n
isnotprimitive.
Clearly,s
n+1 s
n
isaprexoftheharateristiword
. Sine
s
n+1 s
n
=s dn+1
n s
n 1 s
n
=s 1+dn+1
n t
n 1
thewords 1+d
n+1
n
isaprexof
,andsinet
n 1
isnotaprexofs
n
,theword
s 2+d
n+1
n
isnotaprexof
.
Proof of Theorem 2.2.31.Sine aSturmian word has the samefators asthe
harateristiwordofsameslope,itsuÆestoprovetheresultforharateristi
words. Let be the harateristi word of slope = [0;1+d
1
;d
2
;:::℄. Let
(s
n )
n 1
betheassoiatedstandardsequene.
To provethat the ondition is neessary, observe that s dn+1
n
is a prex of
for eah n 1. Consequently, if the sequene (d
n
) of partial quotients is
unbounded, theinnitewordhasfatorsofarbitrarilygreatexponent.
Conversely,assumethat thepartialquotients(d
n
)areboundedbysomeD
and arguing by ontradition, suppose that has unbounded index. Letr be
some integersuh that F() ontainsaprimitivewordof length r with index
greaterthanD+4. Amongthosewords,letwbeawordoflengthrofmaximal
index. Letd+1betheindexofw. ThendD+3. Theproofisinthreesteps.
(1)Theharateristiword hasprexesofthe formw d
, with dD+3.
Indeed,ifw d+1
isaprexof,wearedone. Otherwise,onsideranourrene
of w d+1
. Set w = za with a a letter, and let b be the letter preeding the
ourrene ofw d+1
. If b =a, replae w byaz and proeed. The proess will
stopafteratmostjwj 1stepsbeauseeitheraprexofisobtained,orbeause
Thusb(za) d+1
isafatorof. Thisimpliesthata(za) d
andb(za) d
arefators,
sow d
isarightspeialfator,andthereforeitisaprexof.
(2)Ifw d
isaprexoftheharateristiword,thenwisoneofthestandard
wordss
n
. Indeed,sete=d 2,sothateD+1. Letnbethegreatestinteger
suh that s
n
is aprex of w e+1
. Then w e+1
is aprex of s
n+1
=s dn+1
n s
n 1 ,
thusalsoofs 1+dn+1
n
. Thisshowsthat
(1+D)jwj(1+e)jwj(1+d
n+1 )js
n
j(1+D)js
n j
whene jwj js
n
j. Now, sine both w e+2
and s 1+d
n+1
n
are prexesof , one
is a prex of the other. If w e+2
is the shorter one, then jw e+2
j = jw e+1
j+
jwj js
n
j+jwj. Thus, w e+2
and s 1+dn+1
n
share a ommon prex of length
js
n
j+jwj. Consequently, wand s
n
arepowersof thesameword,and sine
theyareprimitive,theyareequal.
Ifs 1+d
n+1
n
is theshorteronethen, sine(1+e)jwj(1+d
n+1 )js
n j,
s 1+d
n+1
n
=js
n j+d
n+1 js
n jjs
n j+
d
n+1
1+d
n+1
(1+e)jwjjs
n j+jwj
andthesameonlusionholds.
(3)Iffollowsthats 1+e
n
isaprexofand,sineeD+1d
n+1
+1,also
s 2+d
n+1
n
isaprexof,ontraditingLemma 2.2.32.
We onludethis setionwith theomputation ofthenumberoffatorsof
Sturmian words. Another haraterization of entral words will help. Reall
that a nite word is balaned if and only if it is a fator of some Sturmian
word. Moreover,everybalanedwordw,asafatorofsomeuniformlyreurrent
inniteword,an beextended to theright andto the left, that is waand bw
arebalanedforsomelettersa;b.
Proposition2.2.34. Foranywordw,thefollowingareequivalent:
(i) thewordwisentral;
(ii) thewords0w0,0w1,1w0,1w1arebalaned;
(iii) thewords0w1and1w0arebalaned.
Proof. (i) ) (ii). The words w01 and w10 are standard, and therefore are
prexesof someharateristiwords and 0
. By Proposition 2.1.22thefour
innitewords0,1, 0 0
and1 0
areSturmian,andonsequentlytheirprexes
0w0,0w1,1w0,1w1arebalaned. (ii))(iii)istrivial.
(iii))(i). Weproverstthatwisapalindromeword. Assumetheontrary.
Thenthere arewordsu,v,v 0
andletters a6=b suh thatw=uav=v 0
bu.~ But
then awb = auavb = av 0
b~ub has fators aua and b~ub with height satisfying
jh(aua) h(b~ub)j=2,ontradition.
Let be aharateristi wordsuh that 0w12F(). Sine F() islosed
underreversal(Proposition2.1.19),andwisapalindrome,1w02F(),showing
thatwisarightspeialfatorof. Thusitsreversal(thatiswitself)isaprex
