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Conjugacy of morphisms and Lyndon decomposition of standard Sturmian words
Gwenaël Richomme
To cite this version:
Gwenaël Richomme. Conjugacy of morphisms and Lyndon decomposition of standard Sturmian words.
2005, pp.341-351. �hal-00013457�
Université dePiardie Jules Verne CNRSFRE2733
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Conjugay of morphisms
and Lyndon deomposition of
standard Sturmian words
G. Rihomme
a
LaRIARESEARCHREPORT:LRR 2005-09
(Otober2005)
Sturmian words
∗
G. Rihomme
LaRIA, Université de PiardieJules Verne
33, Rue SaintLeu,
F-80039 Amiens edex 1
(gwenael.rihommeu-piardie.fr)
Otober28,2005
Abstrat
Usingthenotionsofonjugayofmorphisms,weansweraquestionofG.Melançononern-
ingthedeompositioninLyndonwordsofstandardSturmianwords. Weshowsomeonnetions
withmorphismspreservingLyndonwords
1 Introdution
Finite (or innite) Lyndon words an be enountered inmany studies (see for instane [8 , 9 ,10 ℄).
Theyarethenonemptywordswhiharesmallerinlexiographiorderthanalltheirpropersuxes.
The Lyndon fatorization theorem [4℄ states that any nite word an be deomposed uniquely in
a produt of noninreasing (in lexiographi order) Lyndon words. This result was extended to
innitewords[19℄ (Insuhaase, thedeomposition anend withan inniteLyndonword). Thus
some works onern the deomposition in Lyndon words of some innite words (see for instane
[3, 5,11,12,18 ℄for suh results).
In [12 ℄, G. Melançon gives a deomposition inLyndon words of standard Sturmian words. He
asksthefollowing question: inwhih ases,thesequeneof noninreasing Lyndonwordsappearing
in the deomposition of a standard Sturmian word an be written
(g n (ℓ 0 )) n≥0
withℓ 0
a Lyndonword and
g
amorphism. InSetion 5,we answerthisquestion.Forthis,weuseresultsabout morphismspreservingLyndonwords[14 ℄ andaboutonjugay of
morphisms[13℄. Inpartiular,weshowthatwhenapositiveanswerexiststo thepreviousquestion,
g
preserves Lyndon words and is the onjugate of a morphismf
that generates the deomposedstandard Sturmianword.
In Setion 2, we reall notions on Sturmian words and morphisms. Setion 3 realls both
the deomposition in Lyndon words of standard Sturmian words obtained by G. Melançon, and
his question. This setion also reall notions on morphisms preserving Lyndon words. Setion 4
presentsnotionsononjugayofmorphismsandintroduesanewpartiularase,namelythestrong
∗
Thispaperwas presentedat the 5thInternational ConfereneonWords whihheld inMontréalonseptember
2005(PubliationsduLaCIMnuméro36, page341-351(S.Brlek,C.Reutenauereds.).)
that for any standard Sturmian words
w
over{a < b}
,aw
isan innite Lyndon word [3℄. Finally,inSetion 5, we answerG. Melançon. Note that at arst step, we expressthe deomposition of a
standard Sturmianword using only morphisms.
2 Sturmian words and morphisms
We reallhere notions onwords (see for instane[8,9 ℄ formore details).
An alphabet
A
isa setofsymbolsalledletters. Hereweonsideronly nitealphabets. A wordover
A
is a sequene of letters fromA
. The empty wordε
isthe emptysequene. Equipped withtheonatenation operation,theset
A ∗
ofnitewordsoverA
isafreemonoid withneutralelementε
andsetofgeneratorsA
. WedenotebyA ω
thesetofinnitewordsoverA
. Asusually,for aniteword
u
andanintegern
,then th
powerofu
,denotedu n
,isthewordε
ifn = 0
andthewordu n−1 u
otherwise. If
u
isnottheemptyword,u ω
denotes theinniteword obtainedbyinnitelyrepeatingu
. A nite wordw
is said primitive if for any wordu
, the equalityw = u n
(withn
an integer)implies
n = 1
. Any word isthepowerofa unique primitive word alledtheprimitiveroot ofw
.Given anonemptyword
u = u 1 . . . u n
withu i ∈ A
,thelength|u|
ofu
istheintegern
. Onehas|ε| = 0
. Iffor some wordsu, v, p, s
(possiblyempty),u = pvs
,thenv
is a fator ofu
,p
is a prexof
u
ands
is a sux ofu
. Whenp 6= u
(resp.s 6= u
),we saythatp
isa proper prex (resp.s
is aproper sux) of
u
. By|u| a
we denotethenumberof ourrenes of thelettera
inthewordu
.Sturmianwordsmaybedenedinmanyequivalentways(see[1℄forinstane). Theyareinnite
binary words. Here we will onsider them as the innitebalaned non ultimately periodi words.
We reallthata (niteor innite)word
w
over{a, b}
isbalaned ifforanyfatorsu
andv
of samelength
||u| a − |v| a | ≤ 1
,andthataninnitewordw
isultimatelyperiodi ifw = uv ω
for somenitewords
u
andv
.Many studies of Sturmian words use Sturmian morphisms. Let
A, B
be two alphabets. Amorphism (endomorphism if
A = B
)f
fromA ∗
toB ∗
isamappingfromA ∗
toB ∗
suh thatfor allwords
u, v
overA
,f (uv) = f (u)f (v)
. We also saythatf
is a morphism onA
or thatf
isdenedon
A
(withoutanyotherpreisionwhenB
hasnoimportane). AmorphismonA
isentirelyknownby theimages oftheletters of
A
. A morphism extendsnaturally oninnitewords. We denotejustbyjuxtapositiontheompositionofmorphisms. Givenanendomorphism
f
,iflim n→∞ f n (a)
exists,thenthis limit isdenoted
f ω (a)
and isa xedpoint off
: thewordf ω (a)
is saidgenerated byf
.Sturmian morphisms arethemorphisms in
{E, L a , L b , R a , R b } ∗
whereE, L a , L b , R a , R b
aretheendomorphisms dened on
{a, b}
byE(a) = b
,E(b) = a
,L a (a) = a
,L a (b) = ab
,L b (a) = ba
,L b (b) = b
,R a (a) = a
,R a (b) = ba
,R b (a) = ab
,R b (b) = b
. Many relations existsbetween Sturmianwords and Sturmian morphisms. For instane, it is known [2 , 6℄ that any Sturmian word an be
dened asan inniteprodutofSturmian morphisms.
A partiular ase of Sturmian words is thestandard (or harateristi) one. For any standard
Sturmian words, there existsa sequene
(d n ) n≥0
ofintegers, alled the diretive sequene verifyingd 1 ≥ 0
andd k ≥ 1
for allk ≥ 2
,suhthatw = lim
n→∞ s n
where thesequene
(s n ) n≥−1
ofwordsisdened by:s −1 = b
,s 0 = a
ands n = s n−1 d n s n−2
forn ≥ 1
.Letus observe thatfor every
n ≥ 0
,s 2n
endswitha
. Moreover[1 ℄,s 2n = L d a 1 L d b 2 . . . L d a 2 n− 1 L d b 2n (a)
= L d a 1 L d b 2 . . . L a d 2n− 1 L d b 2n L d a 2n+1 (a) s 2n+1 = L d a 1 L d b 2 . . . L a d 2n− 1 L d b 2n L d a 2n+1 (b)
= L d a 1 L d b 2 . . . L a d 2n− 1 L d b 2n L d a 2n+1 L d b 2n+2 (b)
3 Lyndon words and morphisms
From now on we onsider ordered alphabets. We denote
{α 1 < . . . < α n }
then
-letter alphabet{α 1 , . . . , α n }
with orderα 1 < . . . < α n
. Given an ordered alphabetA
, we denote by thelexiographi order whenever used on
A ∗
or onA ω
. Let reall that for two dierent (nite orinnite) words
u
andv
,u ≺ v
if and only ifu = xay
,v = xbz
witha, b ∈ A
,a < b
,x ∈ A ∗
,y, z ∈ A ∗ ∪ A ω
,or if(whenu
isnite)u
isaprex ofv
.A nonempty niteword
w
isaLyndon word iffor allnonemptywordsu
andv
,w = uv
impliesw ≺ vu
. Equivalently[4,8℄,anonemptywordw
isaLyndonwordifallitsnonemptypropersuxesaregreaterthan itfor thelexiographi order. For instane,on theone-letteralphabet
{a}
,onlya
is a Lyndon word. On
{a < b}
the Lyndonwords oflength at most 5 area
,b
,ab
,aab
,abb
,aaab
,aabb
,abbb
,aaaab
,aaabb
,aabab
,aabbb
,abbbb
. Lyndon wordsareprimitive.The seonddenitionofLyndonwordsextendsto innitewords: An innitewordisan innite
Lyndonword ifallits propersuxesaregreaterthanitfor thelexiographi order. Ausefulresult
ofG. Melançon[12℄statesthatan inniteword isa Lyndonwordifandonly ifithasaninnityof
prexesthat areLyndonwords. See for instane[7 ℄ for areent exampleof inniteLyndon word.
Any nonempty nite or innite Lyndon words an be deomposed asa noninreasing produt
of Lyndonwords. First,R.C.Lyndon proved (see[8 ℄ for instane):
Anyword
w ∈ A +
maybe writtenuniquely asanoninreasingprodutof Lyndonwords:w = ℓ 1 ℓ 2 . . . ℓ n
where for eahi
,ℓ i
isa Lyndon word andℓ n ℓ n−1 . . . ℓ 1
.This resultwasgeneralized to innitewords [19 ℄:
Anyright inniteword
w
maybe uniquely expressed asa noninreasingprodutof Lyn- don words, nite or innite, in one of the two following forms: either there exists aninnite noninreasingsequene of nite Lyndon words
(ℓ k ) k≥0
suhthatw = Y
n≥0
ℓ n = ℓ 0 ℓ 1 . . .
or there exist nite Lyndon words
ℓ 0 , . . . , ℓ m−1
(m ≥ 0
) and an innite wordℓ m
suhthat
ℓ m ≺ ℓ m−1 ℓ m−2 . . . ℓ 0
andw = ℓ 0 . . . ℓ m−1 ℓ m .
some innite words. In[12℄, G. Melançon obtains the deomposition of standard Sturmian words.
We onsider these words here on thealphabet
{a < b}
. For any wordw
ending withthe lettera
,let usdenote
w
theword suhthatw = wa
.Theorem 3.1 [12℄ Let
s
be a standard Sturmian word with diretive sequene(d n ) n≥1
. Letℓ n = as d 2n 2n+1 −1 s 2n−1 s 2n
. (ifd 1 = 0
thenℓ 0 = b
).The words
(ℓ n ) n≥0
form a stritly dereasing sequene of Lyndon words and the unique fator-izationof
s
as a noninreasingprodut of Lyndonwords iss = Y
n≥0
ℓ d n 2n+1 .
G. Melançon wrote[12, Remark3.7℄ :
Whenisthesequene
(ℓ n ) n≥0
morphi? Morepreisely,isitpossibletogiveamorphismϕ : {a, b} ∗ → {a, b} ∗
and a Lyndon wordℓ 0
suh thatℓ n+1 = ϕ(ℓ n )
? This questionhas a positive answer inthe ase where the diretivesequene isonstant. For instane,
if
d n = 2
for alln ≥ 0
, then we may setℓ 0 = aab
and use the morphism mappinga 7→ aaabaab
andb 7→ aab
.AharateristiSturmianwordmaybeitselfmorphi. Thatis,maybethelimit
lim n ϕ n (a)
of a (nonerasing) morphism(satisfying
ϕ(a) ∈ aA ∗
). Itis knownthat this isessentially equivalent to the fat that its diretive sequene is periodi. Unfortunately, even whena harateristi Sturmianword
s
has a periodidiretive sequene, it seems that the se-quene
(ℓ n ) n≥ 0
is notalways morphi, althoughit ispossibleto desribe patterns in thefatorization.
Theaimofthispaperisto answerthisquestion. Themainideasofourproofaregeneralizations
of the following remarks: the morphism
a 7→ aaabaab
andb 7→ aab
is the Sturmian morphismL 2 a R 2 b
and preserves Lyndonwords. MoreoverL 2 a L 2 b
is a onjugate ofL 2 a R 2 b
,L a 2 L 2 b (a) = (l 0 ) 2 a
andL 2 a R 2 b (a) = a(l 0 ) 2
. Let us notethat, in[14℄, similarremarks aremade about thedeomposition of the Fibonai word (the standard Sturmian word of diretive sequene(1) n≥0
). In Setion 4, wereallnotions on onjugay ofmorphisms.
Let us now reall some results on morphisms preserving (nite) Lyndon words. These mor-
phismsare studied in[14℄. Bydenition, a morphism
f
preserves Lyndon word iffor eah Lyndonword
w
,f (w)
is a Lyndon word. Eetive haraterizations of suh morphisms are given in [14 ℄.Consequently Sturmianwords preservingLyndon wordsareknown:
Proposition 3.2 [14 ℄ A Sturmian morphism on
{a < b}
is a Lyndon morphism if and only if itbelongs to
{L a , R b } ∗
.Toend this setion, letus observe thata study ofmorphisms preservinginniteLyndon words
isgiven in[15 ℄.
Inthissetion, wereallthenotionofonjugay(see,e.g.,[9,13℄). Wealsointroduethepartiular
aseof strong onjugay whihwill beuseful to answer G.Melançon.
Let
A
andB
betwoalphabetsandletf
andg
betwomorphismsfromA ∗
toB ∗
. Themorphismg
isa(right)onjugate off
ifthereexistsawordu
suhthatforanywordx
overA
,f(x)u = ug(x)
.We will also saythat
f
andg
areu
-onjugated, and we will denotef ⊳ u g
. Moreover iff (a) = ua
and
g(a) = au
for alettera
,f
andg
willbe alledstrongly (ona
)u
-onjugated.Let us reall that any morphism
f
has at least one onjugate: itself (f ⊳ ε f
). The Fibonaimorphism
ϕ = L a E
dened byϕ(a) = ab
andϕ(b) = a
has exatlytwo onjugates, itselfand themorphism
ϕ ˜ = R a E
(ϕ(a) = ˜ ba
,ϕ(b) = ˜ a
). A lotofrelations between onjugayof morphismsandSturmianmorphismsweregivenbyP.Séébold[17℄andgeneralized to alargerfamily ofmorphisms
in[13 ℄.
Sine
ϕ(a)
does not end with the lettera
, no morphism is strongly onjugate (ona
) to theFibonai morphism. Nevertheless we an observe that
ϕ 2
(a 7→ aba
,b 7→ ab
) is stronglyab
-onjugated to
ϕ ϕ ˜
(a 7→ aab
,b 7→ ab
). More generally, for all integersx
andy
(y 6= 0
), the mor-phism
L x a L y b
isstrongly onjugated to themorphismL x a R y b
. This follows immediatlythe formulas:L x a L y b (a) = (a x b) y a
,L x a L y b (b) = a x b
,L x a R y b (a) = a(a x b) y
,L x a R y b (b) = a x b
(L a x L y b ⊳ (a x b) y L x a R y b
).Abasipropertyofonjugayis[9 ,13 ℄: formorphisms
f
,f ′
,g
,g ′
,andwordsu
,u ′
,iff ⊳ u g
andf ′ ⊳ u ′ g ′
thenf f ′ ⊳ f (u ′ )u gg ′
(ofoursef (u ′ )u = ug(u ′ )
). Thispropertyextendsto strongonjugay:Lemma 4.1 Let
f, f ′ , g, g ′
, (a
a letter) andu, u ′
wordssuh thatf
is strongly(ona
)u
-onjugatedto
g
andf ′
isstrongly (ona
)u ′
- onjugated tog ′
. Thenf f ′
is strongly (ona
)[f (u ′ )u]
-onjugatedto
gg ′
.Proof. We already know
f f ′ ⊳ f(u ′ )u gg ′
. By hypothesis,f (a) = ua
,g(a) = au
,f ′ (a) = u ′ a
etg ′ (a) = au ′
. Thusf f ′ (a) = f (u ′ a) = f (u ′ )ua
andgg ′ (a) = g(au ′ ) = aug(u ′ ) = af (u ′ )u
. Sof f ′
isstrongly
[f(u ′ )u]
-onjugated togg ′
.We end this setion with a rst use of strong onjugay onerning Sturmian words. One
partiular property of any standard Sturmian word
w
over{a < b}
is that bothaw
andbw
areSturmian words [16℄. Words
aw
(withw
standard Sturmian) arealso known as Christoelwords.In[3℄, itisshown, thatChristoelwordsareinniteLyndon words:
Proposition 4.2 [3 ℄ For any standard Sturmian word
w
over{a < b}
,aw
is an innite Lyndonword.
Proof. Let
w
be a standard word with diretive sequene(d n ) n≥1
. We have already said that astandard word an be viewed as
w = lim n→∞ s n
for some wordss n
dened in Setion 2. In fat,wean verifythatthen
w = lim n→∞ s 2n
. Letn ≥ 1
. Weknowthats 2n = L d a 1 L d b 2 . . . L a d 2n− 1 L d b 2n (a)
.Asa onsequene of Lemma 4.1and of thefat thatfor all integers
x
andy
,the morphismL x a L y b
is strongly onjugated to themorphism
L x a R y b
,we an verifythatL a d 1 L d b 2 . . . L d a 2n− 1 L d b 2n
is stronglyonjugated to
L d a 1 R d b 2 . . . L d a 2n− 1 R d b 2n
.In partiular,
aL d a 1 L d b 2 . . . L a d 2n− 1 L d b 2n (a) = L a d 1 R d b 2 . . . L d a 2n− 1 R d b 2n (a)a
. By Proposition 3.2, the morphismL d a 1 R d b 2 . . . L a d 2n− 1 R b d 2n
preservesLyndonwords. HeneL a d 1 R d b 2 . . . L a d 2n− 1 R b d 2n (a)
isaLyn-don word. Consequently the word
w
has an innity of Lyndon words as prexes. It is a Lyndonword.
we let thereader prove:
Proposition 4.3 Let
A
be an alphabet anda
a letter inA
. Letf, g
be two nonerasing endomor-phisms on
A
and letu
be a word overA
suh thatf
isu
-strongly onjugatetog
. Thenf ω (a)
andg ω (a)
existandaf ω (a) = g ω (a)
.Thusif
g
generatesona
aninniteLyndonword(whihistheaseifitpreservesLyndonwordsor ifitpreservesinniteLyndonwords(see [15℄)),
af ω (a)
isan inniteLyndon word.ThesituationofProposition4.3anbemetfor morphismsthatarenot Sturmian. For instane,
this isthease withthemorphisms:
f :
( a 7→ aba
b 7→ abb g :
( a 7→ aab b 7→ bab
Moreoveroneanseethat
g
preservesinniteLyndonwordsandgeneratesaninniteLyndonword.5 An answer to G. Melançon
In this setion, we onsider a standard Sturmian word
w
over theordered alphabet{a < b}
withdiretive sequene
(d n ) n≥1
(Letreallthatd 1 ≥ 0
andd n ≥ 1
for alln ≥ 2
). Thesequeneofwords(s n ) n≥0
and(ℓ n ) n≥0
are those dened respetively at theend of Setion 2 and inTheorem 3.1. Inpartiular,
w = lim n→∞ Q
n≥0 ℓ d n 2 n
isthe deomposition inLyndon words ofw
(for eahn ≥ 0
,ℓ n
isa Lyndonword and
ℓ n+1 ℓ n
). Ourresult is:Theorem 5.1 Withthehypothesesofthissetion,thereexistsamorphism
g
suhthatforalln ≥ 0
,ℓ n+1 = g(ℓ n )
ifand onlyif one of the two following ases hold:• 1 ≤ d 1 ≤ d 3
, and for alln ≥ 1
,d 2n = d 2
andd 2n+1 = d 3
. In this ase,ℓ 0 = a d 1 b
andg = L d a 1 R b d 2 L d a 3 −d 1
.• d 1 = 0
,1 ≤ d 2 ≤ d 4
, and for alln ≥ 1
,d 2n+2 = d 4
andd 2n+1 = d 3
. In this ase,ℓ 0 = b
andg = R d b 2 L a d 3 R d b 4 −d 2
.We observe that in eah ase, the morphism
g
is a Sturmian morphism that preserves Lyn-don words (see Proposition 3.2). Moreover the word
w
is generated by a Sturmian morphism(
L d a 1 L d b 2 L d a 3 −d 1
orL d b 2 L d a 3 L d b 4 −d 2
).Inordertoprovetheprevioustheorem,usingthestrongonjugay,werstexpresseahLyndon
word
ℓ n
withmorphisms. Forn ≥ 0
,we denote:f n = (L d a 1 L d b 2 ) . . . (L d a 2 n− 1 L d b 2n )
g n = (L d a 1 R d b 2 ) . . . (L d a 2n− 1 R d b 2 n )
The interestof the morphisms
f n
is immediate sine we have already seenrelations between themandthewords
s n
(s 2n = f n (a)
,s 2n+1 = f n+1 (b)
). Wealsoobservethateahg n
isamorphism thatpreservesLyndon words. Asaonsequene ofLemma4.1andof thefatthatforall integers
x
andy
,themorphismL x a L y b
isstrongly onjugated to themorphismL x a R y b
,we have:Lemma 5.2 For all
n ≥ 1
,f n
isstrongly (ona
) onjugated tog n
.Nowwegive anew formula forthewords
(ℓ n ) n≥0
:Lemma 5.3 For all
n ≥ 0
,ℓ n = g n L d a 2n+1 (b)
Proof.
ℓ n a = as d 2n 2n+1 −1 s 2n−1 s 2n a
= as d 2n 2n+1 −1 s 2n−1 s 2n
= af n (a d 2n+1 −1 ba).
If
n = 0
,ℓ n a = a d 1 ba = L d a 1 (b)a = g 0 L d a 1 (b)a
.When
n ≥ 1
,letu n
be theword suh thatf ⊳ u n g n
. By Lemma 5.2,f n (a) = u n a
,g n (a) = au n
.Thus
ℓ n a = af n (a d 2 n +1 −1 b)u n a
= au n g n (a d 2n+1 −1 b)a
= g n (a d 2n+1 b)a
= g n L d a 2n+1 (b)a
Consequently for all
n ≥ 0
,ℓ n = g n L d a 2n+1 (b)
.Let us observe that Lemma 5.2 allows to give a new proof of the fat that the words
(ℓ n ) n≥0
form a stritly dereasing sequene of Lyndon words. Indeed, by Proposition 3.2, eah morphism
g n L d a 2n+1
isaLyndonmorphism,heneg n L d a 2n+1
isaLyndonword. MoreoverR b d 2 n L d a 2n+1 (b)
foreahn ≥ 1
, thenR d b 2 n L d a 2n+1 (b) ≺ b
whih impliesℓ n = g n L d a 2n+1 (b) ≺ g n−1 L d a 2n− 1 (b) = ℓ n−1
(sine anymorphism preserving Lyndon words also stritly preserves the lexiographi order on nite words
[14 ℄).
Proof of Theorem 5.1. Note that the if part of the theorem is immediate. Assume the
sequene
(ℓ n ) n≥0
ismorphi. Letg
bethemorphismsuhthat,foralln ≥ 0
,g(ℓ n ) = ℓ n+1
. Observethatthemorphism
g
annotbeerasingsineotherwisethisontraditsthefatthatℓ 2
isaprimitiveword (as aLyndon word).
Werst onsiderthease
d 1 ≥ 2
. Observeℓ 0 = a d 1 b
andg(a d 1 b) = ℓ 1 = [a(a d 1 b) d 2 ] d 3 a d 1 b.
Assume
g(a) = a
, and sog(b) = ab(a d 1 b) d 2 −1 [a(a d 1 b) d 2 ] d 3 −1 a d 1 b
. The wordℓ 2 = g(ℓ 1 )
hasg(a d 1 +1 b)
asprex. Thusthewordsa d 1 +2
andba d 1 b
arefatorsofℓ 2
. Thisontraditsthefatthatℓ 2
,asafator of aSturmian word,is balaned. Heneg(a) 6= a
.Sine
d 1 ≥ 2
andg(a d 1 b)
starts witha d 1 +1 b
,theworda d 1 +1 b
isa prexofg(a)
. Morepreisely,a(a d 1 b) d 2
must be a prex ofg(a)
. Finally, we an verify thatg(a) = (a(a d 1 b) d 2 ) k
for an integerk ≥ 1
. Itfollowsg(b) = (a(a d 1 b) d 2 ) d 3 −kd 1 a d 1 b
whih impliesd 3 ≥ kd 1
.Assume
k ≥ 2
. The wordℓ 2 = g(ℓ 1 )
ontainsg(ba d 1 b)
andg(a d 1 +1 b)
as fators. The wordg(ba d 1 b)
ends withbub
whereu = (a d 1 b) d 2 [a(a d 1 b) d 2 ] d 3 a d 1
. Furthermore the wordg(a d 1 +1 b) =
[a(a d 1 b) d 2 ] d 3 +k a d 1 b
startswithaua
. Thisontradits thefatthatℓ 2
is balaned.Hene
k = 1
,d 3 ≥ d 1
,g(a) = a(a d 1 b) d 2
,g(b) = [a(a d 1 b) d 2 ] d 3 −d 1 a d 1 b
. We observe thatg = L a d 1 R d b 2 L a d 3 −d 1
andthatit isan injetive morphism.Nowweanprovethat,forall
n ≥ 1
,d 2n = d 2
andd 2n+1 = d 3
. Weatbyindutiononn
. Thereisnothingtodofor
n = 1
. Letn ≥ 1
. Assumethatwehavealreadyprovedd 2p = d 2
andd 2p+1 = d 3
for allintegers
p
with1 ≤ p ≤ n
. Wehaveℓ n+1 = g n+1 L d a 2n+3 (b) = L d a 1 (R d b 2 L a d 3 ) n R d b 2n+2 L d a 2n+3 (b) = (L d a 1 R b d 2 L d a 3 −d 1 ) n L d a 1 R d b 2n+2 L d b 2n+3 (b) = g n (L d a 1 R d b 2n+2 L d b 2n+3 (b))
. Moreoverℓ n+1 = g n (ℓ 1 )
. Sineg
isinjetive,
ℓ 1 = L d a 1 R b d 2 n +2 L b d 2 n +3 (b)
. This impliesd 2n+2 = d 2
andd 2n+3 = d 3
.Now we onsider the ase
d 1 = 1
. We haveℓ 0 = ab
andℓ 1 = [a(ab) d 2 ] d 3 ab
. As in ased 1 ≥ 2
, we annot haveg(a) = a
. Heneg(a)
starts withaa
. We observe thatg(a)
annot endswith
a
, sine otherwise the balaned wordℓ 2 = g(ℓ 1 )
ontainsaaa
andbab
. We observe also thatg(a) 6= [a(ab) d 2 ] i a(ab) k
for any integerk, i
suh that1 ≤ k < d 2
andi ≥ 0
. Indeed otherwise theword
ℓ 2
ontaining bothg(aa)
andg(ab)
should ontains thefatorsa(ab) k aa
andb(ab) k ab
(sine(ab) d 2 +1
endsg(ab)
): thisontradits thefatthatℓ 2
isbalaned. Itfollows thatg(a) = [a(ab) d 2 ] k
with
1 ≤ k ≤ d 3
andg(b) = [a(a d 1 b) d 2 ] d 3 −k a d 1 b
. Exatly asinased 1 ≥ 2
,we an thenprove thatk = 1
,g = L a R b d 2 L d a 3 −1
and for allintegersn ≥ 1
,d 2n = d 2
andd 2n+1 = d 3
.From now on, we onsider the ase
d 1 = 0
. we haveℓ 0 = b
and sog(b) = ℓ 1 = (ab d 2 ) d 3 b
.Moreover
ℓ 2 = R d b 2 L d a 3 R d b 4 L d a 5 (b)
,that isℓ 2 = [ab d 2 [(ab d 2 ) d 3 b] d 4 ] d 5 (ab d 2 ) d 3 b.
Furthermore
ℓ 2 = g 2 (b) = g((ab d 2 ) d 3 b)
. It followsthatg((ab d 2 ) d 3 ) = [ab d 2 g(b) d 4 ] d 5
Sine the word
ab d 2 g(b) d 4 = ab d 2 [(ab d 2 ) d 3 b] d 4
is a primitive word,g(ab d 2 ) = [ab d 2 g(b) d 4 ] x
andxd 3 = d 5
for anintegerx ≥ 1
. Sineab d 2
is not asuxofg(b)
,d 2 ≤ d 4
.Let us prove that
x = 1
, that is,d 3 = d 5
. Assume by ontradition thatx ≥ 2
. The wordℓ 2
has(ab d 2 ) d 3 +1
as a prex and[(ab d 2 ) d 3 b] 2
as a sux. Letu = ab d 2 g(b) d 4
:g(ab d 2 ) = u x
.The word
ℓ 3 = g(ℓ 2 )
ontains the fatorg((ab d 2 ) d 3 +1 ) = u (d 3 +1)x = uu d 5 uu x−2
whih ontainsthe fator
ab d 2 g(b) d 4 u d 5 ab d 2 (ab d 2 ) d 3 b
whih starts withab d 2 g(b) d 4 u d 5 (ab d 2 ) d 3 a
. Observe now thatg((ab d 2 ) d 3 b) = [ab d 2 g(b) d 4 ] d 5 g(b)
endswithb d 2 +1 g(b) d 4
. Consequently thewordℓ 3
also ontainsthefator
b d 2 +1 g(b) d 4 g(((ab) d 2 ) d 3 b) = bb d 2 g(b) d 4 u d 5 (ab d 2 ) d 3 b
. We have a ontradition with the fat thatℓ 3
isa balanedword.From what preedes,
g(ab d 2 ) = ab d 2 g(b) d 4
and sog(a) = ab d 2 g(b) d 4 −d 2 = ab d 2 ((ab d 2 ) d 3 b) d 4 −d 2
.Moreover
g(b) = (ab d 2 ) d 3 b
. We observeg = R d b 2 L a d 3 R d b 4 −d 2
. Asin ased 1 ≥ 2
, we an state that,for all integers
n ≥ 2
,d 2n = d 4
andd 2n−1 = d 3
.6 Conlusion
This paper shows the interest of onjugay of morphisms and of morphisms preserving Lyndon
words as tools to takle problems onerning Sturmian words and/or Lyndon words. We are now
working to nd other situations where these tools an be useful. Inpartiular, we are looking for
thedeompositioninLyndon wordsof any Sturmianwords.
ManythankstoP.Sééboldfor useful disussionson strong onjugay.
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