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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�

(2)

Université dePiardie Jules Verne CNRSFRE2733

33, rueSaint Leu,80039Amiensedex 01, Frane

Tel: (+33)[0℄322 82 8877

Fax: (+33)[0℄0322 8254 12

http://www.laria.u-piardie.fr

Conjugay of morphisms

and Lyndon deomposition of

standard Sturmian words

G. Rihomme

a

LaRIARESEARCHREPORT:LRR 2005-09

(Otober2005)

(3)

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

with

0

a Lyndon

word 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 morphism

f

that generates the deomposed

standard 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.).)

(4)

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 word

over

A

is a sequene of letters from

A

. The empty word

ε

isthe emptysequene. Equipped with

theonatenation operation,theset

A

ofnitewordsover

A

isafreemonoid withneutralelement

ε

andsetofgenerators

A

. Wedenoteby

A ω

thesetofinnitewordsover

A

. Asusually,for anite

word

u

andaninteger

n

,the

n th

powerof

u

,denoted

u n

,istheword

ε

if

n = 0

andtheword

u n−1 u

otherwise. If

u

isnottheemptyword,

u ω

denotes theinniteword obtainedbyinnitelyrepeating

u

. A nite word

w

is said primitive if for any word

u

, the equality

w = u n

(with

n

an integer)

implies

n = 1

. Any word isthepowerofa unique primitive word alledtheprimitiveroot of

w

.

Given anonemptyword

u = u 1 . . . u n

with

u i ∈ A

,thelength

|u|

of

u

istheinteger

n

. Onehas

|ε| = 0

. Iffor some words

u, v, p, s

(possiblyempty),

u = pvs

,then

v

is a fator of

u

,

p

is a prex

of

u

and

s

is a sux of

u

. When

p 6= u

(resp.

s 6= u

),we saythat

p

isa proper prex (resp.

s

is a

proper sux) of

u

. By

|u| a

we denotethenumberof ourrenes of theletter

a

intheword

u

.

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 ifforanyfators

u

and

v

of same

length

||u| a − |v| a | ≤ 1

,andthataninniteword

w

isultimatelyperiodi if

w = uv ω

for somenite

words

u

and

v

.

Many studies of Sturmian words use Sturmian morphisms. Let

A, B

be two alphabets. A

morphism (endomorphism if

A = B

)

f

from

A

to

B

isamappingfrom

A

to

B

suh thatfor all

words

u, v

over

A

,

f (uv) = f (u)f (v)

. We also saythat

f

is a morphism on

A

or that

f

isdened

on

A

(withoutanyotherpreisionwhen

B

hasnoimportane). Amorphismon

A

isentirelyknown

by theimages oftheletters of

A

. A morphism extendsnaturally oninnitewords. We denotejust

byjuxtapositiontheompositionofmorphisms. Givenanendomorphism

f

,if

lim n→∞ f n (a)

exists,

thenthis limit isdenoted

f ω (a)

and isa xedpoint of

f

: theword

f ω (a)

is saidgenerated by

f

.

Sturmian morphisms arethemorphisms in

{E, L a , L b , R a , R b }

where

E, L a , L b , R a , R b

arethe

endomorphisms dened on

{a, b}

by

E(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 Sturmian

words 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 verifying

d 1 ≥ 0

and

d k ≥ 1

for all

k ≥ 2

,suhthat

w = lim

n→∞ s n

(5)

where thesequene

(s n ) n≥−1

ofwordsisdened by:

s −1 = b

,

s 0 = a

and

s n = s n−1 d n s n−2

for

n ≥ 1

.

Letus observe thatfor every

n ≥ 0

,

s 2n

endswith

a

. 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 }

the

n

-letter alphabet

1 , . . . , α n }

with order

α 1 < . . . < α n

. Given an ordered alphabet

A

, we denote by

the

lexiographi order whenever used on

A

or on

A ω

. Let reall that for two dierent (nite or

innite) words

u

and

v

,

u ≺ v

if and only if

u = xay

,

v = xbz

with

a, b ∈ A

,

a < b

,

x ∈ A

,

y, z ∈ A ∪ A ω

,or if(when

u

isnite)

u

isaprex of

v

.

A nonempty niteword

w

isaLyndon word iffor allnonemptywords

u

and

v

,

w = uv

implies

w ≺ vu

. Equivalently[4,8℄,anonemptyword

w

isaLyndonwordifallitsnonemptypropersuxes

aregreaterthan itfor thelexiographi order. For instane,on theone-letteralphabet

{a}

,only

a

is a Lyndon word. On

{a < b}

the Lyndonwords oflength at most 5 are

a

,

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 12 . . . ℓ n

where for eah

i

,

i

isa Lyndon word and

nn−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 an

innite noninreasingsequene of nite Lyndon words

(ℓ k ) k≥0

suhthat

w = Y

n≥0

n = ℓ 01 . . .

or there exist nite Lyndon words

0 , . . . , ℓ m−1

(

m ≥ 0

) and an innite word

m

suh

that

m ≺ ℓ m−1m−2 . . . ℓ 0

and

w = ℓ 0 . . . ℓ m−1m .

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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 word

w

ending withthe letter

a

,

let usdenote

w

theword suhthat

w = wa

.

Theorem 3.1 [12℄ Let

s

be a standard Sturmian word with diretive sequene

(d n ) n≥1

. Let

n = as d 2n 2n+1 −1 s 2n−1 s 2n

. (if

d 1 = 0

then

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 is

s = Y

n≥0

d n 2n+1 .

G. Melançon wrote[12, Remark3.7℄ :

Whenisthesequene

(ℓ n ) n≥0

morphi? Morepreisely,isitpossibletogiveamorphism

ϕ : {a, b} → {a, b}

and a Lyndon word

0

suh that

n+1 = ϕ(ℓ n )

? This question

has a positive answer inthe ase where the diretivesequene isonstant. For instane,

if

d n = 2

for all

n ≥ 0

, then we may set

0 = aab

and use the morphism mapping

a 7→ aaabaab

and

b 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 when

a 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 the

fatorization.

Theaimofthispaperisto answerthisquestion. Themainideasofourproofaregeneralizations

of the following remarks: the morphism

a 7→ aaabaab

and

b 7→ aab

is the Sturmian morphism

L 2 a R 2 b

and preserves Lyndonwords. Moreover

L 2 a L 2 b

is a onjugate of

L 2 a R 2 b

,

L a 2 L 2 b (a) = (l 0 ) 2 a

and

L 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, we

reallnotions 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 Lyndon

word

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 it

belongs to

{L a , R b }

.

Toend this setion, letus observe thata study ofmorphisms preservinginniteLyndon words

isgiven in[15 ℄.

(7)

Inthissetion, wereallthenotionofonjugay(see,e.g.,[9,13℄). Wealsointroduethepartiular

aseof strong onjugay whihwill beuseful to answer G.Melançon.

Let

A

and

B

betwoalphabetsandlet

f

and

g

betwomorphismsfrom

A

to

B

. Themorphism

g

isa(right)onjugate of

f

ifthereexistsaword

u

suhthatforanyword

x

over

A

,

f(x)u = ug(x)

.

We will also saythat

f

and

g

are

u

-onjugated, and we will denote

f ⊳ u g

. Moreover if

f (a) = ua

and

g(a) = au

for aletter

a

,

f

and

g

willbe alledstrongly (on

a

)

u

-onjugated.

Let us reall that any morphism

f

has at least one onjugate: itself (

f ⊳ ε f

). The Fibonai

morphism

ϕ = L a E

dened by

ϕ(a) = ab

and

ϕ(b) = a

has exatlytwo onjugates, itselfand the

morphism

ϕ ˜ = R a E

(

ϕ(a) = ˜ ba

,

ϕ(b) = ˜ a

). A lotofrelations between onjugayof morphismsand

SturmianmorphismsweregivenbyP.Séébold[17℄andgeneralized to alargerfamily ofmorphisms

in[13 ℄.

Sine

ϕ(a)

does not end with the letter

a

, no morphism is strongly onjugate (on

a

) to the

Fibonai morphism. Nevertheless we an observe that

ϕ 2

(

a 7→ aba

,

b 7→ ab

) is strongly

ab

-

onjugated to

ϕ ϕ ˜

(

a 7→ aab

,

b 7→ ab

). More generally, for all integers

x

and

y

(

y 6= 0

), the mor-

phism

L x a L y b

isstrongly onjugated to themorphism

L 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

,andwords

u

,

u

,if

f ⊳ u g

and

f u ′ g

then

f f f (u ′ )u gg

(ofourse

f (u )u = ug(u )

). Thispropertyextendsto strongonjugay:

Lemma 4.1 Let

f, f , g, g

, (

a

a letter) and

u, u

wordssuh that

f

is strongly(on

a

)

u

-onjugated

to

g

and

f

isstrongly (on

a

)

u

- onjugated to

g

. Then

f f

is strongly (on

a

)

[f (u )u]

-onjugated

to

gg

.

Proof. We already know

f f f(u ′ )u gg

. By hypothesis,

f (a) = ua

,

g(a) = au

,

f (a) = u a

et

g (a) = au

. Thus

f f (a) = f (u a) = f (u )ua

and

gg (a) = g(au ) = aug(u ) = af (u )u

. So

f f

is

strongly

[f(u )u]

-onjugated to

gg

.

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 both

aw

and

bw

are

Sturmian words [16℄. Words

aw

(with

w

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 Lyndon

word.

Proof. Let

w

be a standard word with diretive sequene

(d n ) n≥1

. We have already said that a

standard word an be viewed as

w = lim n→∞ s n

for some words

s n

dened in Setion 2. In fat,

wean verifythatthen

w = lim n→∞ s 2n

. Let

n ≥ 1

. Weknowthat

s 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

and

y

,the morphism

L x a L y b

is strongly onjugated to themorphism

L x a R y b

,we an verifythat

L a d 1 L d b 2 . . . L d a 2n− 1 L d b 2n

is strongly

onjugated 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 morphism

L d a 1 R d b 2 . . . L a d 2n− 1 R b d 2n

preservesLyndonwords. Hene

L 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 Lyndon

word.

(8)

we let thereader prove:

Proposition 4.3 Let

A

be an alphabet and

a

a letter in

A

. Let

f, g

be two nonerasing endomor-

phisms on

A

and let

u

be a word over

A

suh that

f

is

u

-strongly onjugateto

g

. Then

f ω (a)

and

g ω (a)

existand

af ω (a) = g ω (a)

.

Thusif

g

generateson

a

aninniteLyndonword(whihistheaseifitpreservesLyndonwords

or 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}

with

diretive sequene

(d n ) n≥1

(Letreallthat

d 1 ≥ 0

and

d n ≥ 1

for all

n ≥ 2

). Thesequeneofwords

(s n ) n≥0

and

(ℓ n ) n≥0

are those dened respetively at theend of Setion 2 and inTheorem 3.1. In

partiular,

w = lim n→∞ Q

n≥0 ℓ d n 2 n

isthe deomposition inLyndon words of

w

(for eah

n ≥ 0

,

n

isa Lyndonword and

n+1n

). Ourresult is:

Theorem 5.1 Withthehypothesesofthissetion,thereexistsamorphism

g

suhthatforall

n ≥ 0

,

n+1 = g(ℓ n )

ifand onlyif one of the two following ases hold:

• 1 ≤ d 1 ≤ d 3

, and for all

n ≥ 1

,

d 2n = d 2

and

d 2n+1 = d 3

. In this ase,

0 = a d 1 b

and

g = L d a 1 R b d 2 L d a 3 −d 1

.

• d 1 = 0

,

1 ≤ d 2 ≤ d 4

, and for all

n ≥ 1

,

d 2n+2 = d 4

and

d 2n+1 = d 3

. In this ase,

0 = b

and

g = 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

or

L d b 2 L d a 3 L d b 4 −d 2

).

Inordertoprovetheprevioustheorem,usingthestrongonjugay,werstexpresseahLyndon

word

n

withmorphisms. For

n ≥ 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 them

andthewords

s n

(

s 2n = f n (a)

,

s 2n+1 = f n+1 (b)

). Wealsoobservethateah

g n

isamorphism that

preservesLyndon words. Asaonsequene ofLemma4.1andof thefatthatforall integers

x

and

y

,themorphism

L x a L y b

isstrongly onjugated to themorphism

L x a R y b

,we have:

Lemma 5.2 For all

n ≥ 1

,

f n

isstrongly (on

a

) onjugated to

g n

.

(9)

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

,

n a = a d 1 ba = L d a 1 (b)a = g 0 L d a 1 (b)a

.

When

n ≥ 1

,let

u n

be theword suh that

f ⊳ u n g n

. By Lemma 5.2,

f n (a) = u n a

,

g n (a) = au n

.

Thus

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

,

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,hene

g n L d a 2n+1

isaLyndonword. Moreover

R b d 2 n L d a 2n+1 (b)

foreah

n ≥ 1

, then

R d b 2 n L d a 2n+1 (b) ≺ b

whih implies

n = g n L d a 2n+1 (b) ≺ g n−1 L d a 2n− 1 (b) = ℓ n−1

(sine any

morphism 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. Let

g

bethemorphismsuhthat,forall

n ≥ 0

,

g(ℓ n ) = ℓ n+1

. Observe

thatthemorphism

g

annotbeerasingsineotherwisethisontraditsthefatthat

2

isaprimitive

word (as aLyndon word).

Werst onsiderthease

d 1 ≥ 2

. Observe

0 = a d 1 b

and

g(a d 1 b) = ℓ 1 = [a(a d 1 b) d 2 ] d 3 a d 1 b.

Assume

g(a) = a

, and so

g(b) = ab(a d 1 b) d 2 −1 [a(a d 1 b) d 2 ] d 3 −1 a d 1 b

. The word

2 = g(ℓ 1 )

has

g(a d 1 +1 b)

asprex. Thusthewords

a d 1 +2

and

ba d 1 b

arefatorsof

2

. Thisontraditsthefatthat

2

,asafator of aSturmian word,is balaned. Hene

g(a) 6= a

.

Sine

d 1 ≥ 2

and

g(a d 1 b)

starts with

a d 1 +1 b

,theword

a d 1 +1 b

isa prexof

g(a)

. Morepreisely,

a(a d 1 b) d 2

must be a prex of

g(a)

. Finally, we an verify that

g(a) = (a(a d 1 b) d 2 ) k

for an integer

k ≥ 1

. Itfollows

g(b) = (a(a d 1 b) d 2 ) d 3 −kd 1 a d 1 b

whih implies

d 3 ≥ kd 1

.

Assume

k ≥ 2

. The word

2 = g(ℓ 1 )

ontains

g(ba d 1 b)

and

g(a d 1 +1 b)

as fators. The word

g(ba d 1 b)

ends with

bub

where

u = (a d 1 b) d 2 [a(a d 1 b) d 2 ] d 3 a d 1

. Furthermore the word

g(a d 1 +1 b) =

[a(a d 1 b) d 2 ] d 3 +k a d 1 b

startswith

aua

. Thisontradits thefatthat

2

is balaned.

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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 that

g = 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

and

d 2n+1 = d 3

. Weatbyindutionon

n

. There

isnothingtodofor

n = 1

. Let

n ≥ 1

. Assumethatwehavealreadyproved

d 2p = d 2

and

d 2p+1 = d 3

for allintegers

p

with

1 ≤ p ≤ n

. Wehave

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))

. Moreover

n+1 = g n (ℓ 1 )

. Sine

g

is

injetive,

1 = L d a 1 R b d 2 n +2 L b d 2 n +3 (b)

. This implies

d 2n+2 = d 2

and

d 2n+3 = d 3

.

Now we onsider the ase

d 1 = 1

. We have

0 = ab

and

1 = [a(ab) d 2 ] d 3 ab

. As in ase

d 1 ≥ 2

, we annot have

g(a) = a

. Hene

g(a)

starts with

aa

. We observe that

g(a)

annot ends

with

a

, sine otherwise the balaned word

2 = g(ℓ 1 )

ontains

aaa

and

bab

. We observe also that

g(a) 6= [a(ab) d 2 ] i a(ab) k

for any integer

k, i

suh that

1 ≤ k < d 2

and

i ≥ 0

. Indeed otherwise the

word

2

ontaining both

g(aa)

and

g(ab)

should ontains thefators

a(ab) k aa

and

b(ab) k ab

(sine

(ab) d 2 +1

ends

g(ab)

): thisontradits thefatthat

2

isbalaned. Itfollows that

g(a) = [a(ab) d 2 ] k

with

1 ≤ k ≤ d 3

and

g(b) = [a(a d 1 b) d 2 ] d 3 −k a d 1 b

. Exatly asinase

d 1 ≥ 2

,we an thenprove that

k = 1

,

g = L a R b d 2 L d a 3 −1

and for allintegers

n ≥ 1

,

d 2n = d 2

and

d 2n+1 = d 3

.

From now on, we onsider the ase

d 1 = 0

. we have

0 = b

and so

g(b) = ℓ 1 = (ab d 2 ) d 3 b

.

Moreover

2 = R d b 2 L d a 3 R d b 4 L d a 5 (b)

,that is

2 = [ab d 2 [(ab d 2 ) d 3 b] d 4 ] d 5 (ab d 2 ) d 3 b.

Furthermore

2 = g 2 (b) = g((ab d 2 ) d 3 b)

. It followsthat

g((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

and

xd 3 = d 5

for aninteger

x ≥ 1

. Sine

ab d 2

is not asuxof

g(b)

,

d 2 ≤ d 4

.

Let us prove that

x = 1

, that is,

d 3 = d 5

. Assume by ontradition that

x ≥ 2

. The word

2

has

(ab d 2 ) d 3 +1

as a prex and

[(ab d 2 ) d 3 b] 2

as a sux. Let

u = ab d 2 g(b) d 4

:

g(ab d 2 ) = u x

.

The word

3 = g(ℓ 2 )

ontains the fator

g((ab d 2 ) d 3 +1 ) = u (d 3 +1)x = uu d 5 uu x−2

whih ontains

the fator

ab d 2 g(b) d 4 u d 5 ab d 2 (ab d 2 ) d 3 b

whih starts with

ab d 2 g(b) d 4 u d 5 (ab d 2 ) d 3 a

. Observe now that

g((ab d 2 ) d 3 b) = [ab d 2 g(b) d 4 ] d 5 g(b)

endswith

b d 2 +1 g(b) d 4

. Consequently theword

3

also ontainsthe

fator

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 that

3

isa balanedword.

From what preedes,

g(ab d 2 ) = ab d 2 g(b) d 4

and so

g(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 observe

g = R d b 2 L a d 3 R d b 4 −d 2

. Asin ase

d 1 ≥ 2

, we an state that,

for all integers

n ≥ 2

,

d 2n = d 4

and

d 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.

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ManythankstoP.Sééboldfor useful disussionson strong onjugay.

Referenes

[1℄ J. BerstelandP.Séébold. Sturmian words, hapter2 in[9 ℄.

[2℄ V. Berthé,C. Holton,and L.Q.Zamboni. Initial powers ofsturmian sequenes. To appear in

Ata Informatia,seewww.lirmm.fr/

berthe.

[3℄ J.P.BorelandF.Laubie. Quelquesmots surladroiteprojetiveréelle. JournaldeThéoriedes

Nombres deBordeaux, 5:2351,1993.

[4℄ K.T. Chen, R.H. Fox,and R.C.Lyndon. Freedierential alulusIV thequotient groups of

the lowerentral series. Ann. Math. 68,68:8195, 1958.

[5℄ A. Ido and G. Melançon. Lyndon fatorization of the Thue-Morse word and its relatives.

Disret. Math.andTheoret. Comput.Si., 1:4352,1997.

[6℄ J.JustinandG.Pirillo. Episturmianwordsandepisturmianmorphisms. Theoretial Computer

Siene, 276(1-2):281313, 2002.

[7℄ J. Justin and G. Pirillo. On a harateristi property of Arnoux-Rauzy sequenes. RAIRO

Theoretial Informatisand Appliations, 36:385388, 2002.

[8℄ M. Lothaire. Combinatoris on words, volume 17 of Enylopedia of Mathematis. Addison-

Wesley,1983.Reprintedin1997byCambridgeUniversityPressintheCambridgeMathematial

Library,Cambridge,UK, 1997.

[9℄ M. Lothaire. Algebrai Combinatoris on words, volume 90 of Enylopedia of Mathematis.

Cambridge UniversityPress, Cambridge,UK, 2002.

[10℄ M. Lothaire. Applied Combinatoris on Words. To appear.

(see www-igm.univ-mlv.fr/

berstel).

[11℄ G.Melançon. Lyndonfatorizationofinnitewords. InSTACS'96,volume1046ofLet.Notes

in Comp. Si.,pages 147154,1996.

[12℄ G. Melançon. Lyndon fatorization of Sturmian words. Disrete Mathematis, 210:137149,

2000.

[13℄ G. Rihomme. Conjugay and episturmian morphisms. Theoretial ComputerSiene, 302:1

34, 2003.

[14℄ G. Rihomme. Lyndon morphisms. Bulletin of the Belgian Mathematial Soiety, 10:761785,

2003.

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tomata, Languages and Combinatoris. An extended abstrat appeared in theproeedings of

theonferene"JournéesMontoisesd'InformatiqueThéorique",Liège,Belgium,2004(Prépub-

liation 04.006, Institutde Mathématique, Université de Liège),p325-333.

[16℄ P.Séébold. FibonaimorphismsandSturmianwords. Theoretial ComputerSiene, 88:365

384, 1991.

[17℄ P.Séébold. Ontheonjugation ofstandardmorphisms. Theoretial ComputerSiene,195:91

109, 1998.

[18℄ P.Séébold.LyndonfatorizationoftheProuhetwords. TheoretialComputerSiene,307:179

197, 2003.

[19℄ R.Siromoney,L.Mathew,V.R.Dare,andK. G.Subramanian. InniteLyndonwords. Infor-

mation Proessing Letters,50:101104, 1994.

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