Algèbre/Algebra
(Combinatoire/Combinatorics)
On invertible substitutions with two fixed points *
Zhi-Xiong Wen
a, Zhi-Ying Wen
b, Jun Wu
aaNonlinear Science Center, Department of Mathematics, Wuhan University, Wuhan 430072, China bDepartment of Mathematics, Tsinghua University, Beijing 10084, China
Received 24 October 2001; accepted 3 December 2001 Note presented by Jean-Pierre KAHANE.
Abstract Letϕbe a primitive substitution on a two-letter alphabet{a, b}having two fixed pointsξa andξb. We show that the substitutionϕis invertible if and only if one hasξa=abξ and ξb=baξ. To cite this article: Z.-X. Wen et al., C. R. Acad. Sci. Paris, Ser. I 334 (2002) 727–731.2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS
Sur les substitutions inversibles ayant deux points fixes
Résumé On considère une substitution primitiveϕsur l’alphabet{a, b}ayant deux points fixesξa etξb (commençant respectivement para etb). Nous montrons que la substitutionϕ est inversible si et seulement si l’on aξa=abξ etξb=baξ. Pour citer cet article : Z.-X.
Wen et al., C. R. Acad. Sci. Paris, Ser. I 334 (2002) 727–731. 2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS
Version française abrégée
SoitS= {a, b}un alphabet de deux lettres. Nous désignons parS∗ etF2 le monoïde libre et le groupe libre engendrés parS. PosonsS+=S∗\ {ε}oùεest le mot vide et notonsSωl’ensemble des mots infinis surS.
Siwest un mot, nous désignons par|w|sa longueur et par|w|a (resp.|w|b) le nombre de fois que la lettrea(resp.b) figure dansw.
Un motv∈S∗est dit facteur du motw(ce que l’on notev≺w) s’il existe deux motsuetutels qu’on aitw=uvu. Dans le cas oùu=ε(resp.u=ε), nous disons quevest un préfixe (resp. suffixe) dew. Les notions de facteur et de préfixe gardent un sens même siwest un mot infini.
Une substitutionϕsurSest un endomorphisme deS∗. Dans la suite, nous identifierons la substitutionϕ au couple de mots(ϕ(a), ϕ(b))et nous supposerons toujoursϕ(a)etϕ(b)différents deε. Une substitution surSagit également surSω(par concatenation des images des lettres composant un mot infini).
S’il existec∈S tel que c soit un préfixe deϕ(c) et tel que|ϕ(c)|2, alors la suite des mots finis (ϕn(c))n1converge vers un point fixeϕω(c)∈Sωdeϕ.
E-mail addresses: zhxwen@whu.edu.cn (Z.-X. Wen); wenzy@tsinghua.edu.cn (Z.-Y. Wen);
wujunyu@public.wh.hb.cn (J. Wu).
Z.-X. Wen et al. / C. R. Acad. Sci. Paris, Ser. I 334 (2002) 727–731
À chaque substitution ϕ sur S, on associe la matrice Mϕ indexée par S×S dont le coefficient correspondant à(x, y)est le nombre de fois que la lettrexfigure dansϕ(y). Si la matriceMϕest primitive, la substitutionϕest également dite primitive.
Une substitutionϕest dite inversible si elle définit un automorphisme deF2.
Notre résultat principal donne la structure des points fixes d’une substitution primitive inversible : THÉORÈME 1. – Siξa et ξb sont deux points fixes d’une substitution primitive inversible, alors nous avonsξa=abξetξb=baξ. De plus, les trois suitesbξ,aξetξsont points fixes de substitutions primitives inversibles. Réciproquement si les points fixesξa etξbd’une substitutionϕ sont de la formeξa=abξet ξb=baξ, alorsϕest une substitution inversible.
1. Introduction
LetS= {a, b}be a two-letter alphabet. LetS∗(resp.F2) be the free monoid (resp. the free group) with empty wordεas neutral element generated byS. For anyu, v∈S∗, we callua prefix ofv (and then we writeu≺v) ifv=uwfor somew∈S∗.
Ifw∈S∗is a word,|w|denotes its length and|w|a(resp.|w|b) the number of times the lettera(resp.b) appears in the wordw.
An endomorphismϕ ofS∗is called a substitution overS. We identify a substitutionϕ with the couple (ϕ(a), ϕ(b)).
One associates with a substitutionϕ overS aS×S-matrixMϕ: its entry corresponding tox, y is the number of times thatx appears in ϕ(y). WhenMϕ is primitive (i.e., one of its power has only positive coefficients), the substitutionϕis also said to be primitive.
A substitutionϕ is said to be invertible if it defines an automorphism ofF2. It is proved in [7] that the setGof invertible substitutions is generated as a monoid by the three following substitutionsσ=(ab, a), τ =(ba, a)and π =(b, a). We define two subclasses ofG:Gσ = σ, π, generated byσ and π, and Gτ= τ, π.
It is known [5] that one hasϕ(bab−1a−1)=u(bab−1a−1)±1u−1 (whereu∈S∗ oru−1∈S∗) for any invertible substitutionϕ. Furthermore [2], ϕ is an invertible substitution if and only if ϕ(bab−1a−1)= u(bab−1a−1)±1u−1for someu∈S∗oru−1∈S∗.
Invertible substitutions and their properties has been studied by many authors (for instance, see Berstel [1], Mignosi [3], Mignosi and Séébold [4], Séébold [6], Wen and Wen [7–9]). They studied both combinatorial and arithmetic properties of fixed points of primitive invertible substitutions. In this note, we give a thorough study of the structure of fixed points of primitive invertible substitutions.
2. Results and proofs
Letϕ∈Gσ, thenϕ(a)andϕ(b)begin by the same letter, soϕhas only one fixed point.
Now considerϕ ∈Gτ and we write ϕ=πα◦τn1 ◦π◦τn2 ◦ · · · ◦π◦τnk ◦πβ. We call the number L(ϕ)=ik=1ni+(k−1)+α+βthe length ofϕ.
It is easy to see that, ifL(ϕ)is even, thena andbare prefixes ofϕ(a)andϕ(b)respectively. Therefore, in this case, the substitutionϕhas two fixed pointsξaandξb(beginning byaandbrespectively).
WhenL(ϕ)is odd, thenaandbare prefixes ofϕ(b)andϕ(a)respectively, soϕhas no fixed point.
Finally we considerϕ /∈Gτ. Remark that for anyϕ∈G, we can writeϕ=ϕ1· · ·ϕn withϕi ∈ {π, σ, τ}. Sinceϕ /∈Gτ, there exists at leastϕi =σ. Then the first letters ofϕ(a)andϕ(b)coincide, hence in this case,ϕhas only one fixed point.
Denote byGτ,eandGτ,othe subclass of elements ofGτof length even and odd respectively. The following lemma sums up the preceding discussion.
LEMMA 1. – Letϕbe a primitive invertible substitution. Then (1) ϕhas two fixed points if and only ifϕ∈Gτ,e,
(2) ϕhas only one fixed point if and only ifϕ /∈Gτ, (3) ϕhas no fixed point if and only ifϕ∈Gτ,o.
COROLLARY 1. – Letξa,ξbbe fixed points ofϕ∈Gτ,e. Thenξa=abη1andξb=baη2.
Proof. – We haveab≺τ2(ab),ab≺τ ◦π(ab),ab≺π ◦τ (ab),ba≺τ2(ba),ba≺τ ◦π(ba), and ba≺π◦τ (ba). This allows a proof by induction on the length ofϕ.
Now we characterize the elements inGτ
PROPOSITION 1. – Letϕbe a substitution. Then (1) ifϕ(aba−1b−1)=aba−1b−1, thenϕ∈Gτ,e, (2) ifϕ(aba−1b−1)=(aba−1b−1)−1, thenϕ∈Gτ,o, (3) ifϕ∈Gτ, thenϕ(aba−1b−1)=(aba−1b−1)±1.
Proof. – Assertions (1) and (2) follow fromπ(aba−1b−1)=τ (aba−1b−1)=(aba−1b−1)−1.
Assumeϕ(aba−1b−1)=(aba−1b−1)±1. By Ei–Ito’s theorem [2], we haveϕ∈G. Buta≺ϕ(a)and b≺ϕ(b)(ora≺ϕ(b)andb≺ϕ(a)). Thus by Lemma 1,ϕ∈Gτ.
THEOREM 1. – Letξaandξbbe the fixed points ofϕ∈Gτ,e, then there existsηsuch thatξa=abηand ξb=baη.
Proof. – We know (Corollary 1) thatξa=abη. We have ϕ(abη)=abη. Then by Proposition 1, we obtain:
ϕ(baη)=ϕ
bab−1a−1abη
=ϕ
bab−1a−1
abη=bab−1a−1abη=baη.
This means thatbaηis also a fixed point ofϕ. It follows thatξb=baη.
The next theorem shows that the converse holds.
THEOREM 2. – Supposeξa andξb are two fixed points of the primitive substitutionϕ withξa=abξ, ξb=baξ, thenϕ∈Gτ,e.
Proof. – By hypothesis, we have ϕ(abξ )=abξ and ϕ(baξ )=baξ. However |ϕ(ab)| = |ϕ(ba)|. So there existsw∈S∗such thatϕ(ab)=abwandϕ(ba)=baw. Henceϕ(aba−1b−1)=ϕ(ab)ϕ(a−1b−1)= aba−1b−1,thus by Proposition 1,ϕ∈Gτ,e.
Letϕ∈Gτ,e be a primitive invertible substitution, by Theorem 1,ϕ has two fixed pointsabηandbaη.
This raises a natural question: are the infinite words aη, bη and η fixed points of primitive invertible substitutions? The answer is given by Theorem 3 below.
LEMMA 2. – Letϕ∈Gτ,ebe a primitive substitution and letϕ0, ϕaandϕbbe defined as:
ϕ0=
b−1a−1ϕ(a)ba, a−1b−1ϕ(b)ab , ϕa=
b−1ϕ(b)ϕ(a)ϕ(b−1)b, b−1ϕ(b)b , ϕb=
a−1ϕ(a)a, a−1ϕ(a)ϕ(b)ϕ(a−1)a , thenϕb,ϕa,ϕ0are invertible primitive substitutions.
Proof. – We only considerϕ0, since the proof of the assertions forϕaandϕbfollows the same lines.
Sinceϕis primitive,ϕ=π◦τ. One can easily see thatabandbaare prefix ofϕ(a)andϕ(b), henceϕ0 is a substitution overS. The primitivity ofϕ0comes from the primitivity ofϕ.
An easy calculation, using the relationϕ(aba−1b−1)=aba−1b−1, gives ϕ0
aba−1b−1
=
a−1b−1ϕ(b)ϕ(a)
aba−1b−1
a−1b−1ϕ(b)ϕ(a)−1 ,
Z.-X. Wen et al. / C. R. Acad. Sci. Paris, Ser. I 334 (2002) 727–731
which yieldsϕ0∈Gdue to Ei–Ito’s theorem.
THEOREM 3. – Letϕ be a substitution inGτ,e, andabηandbaηits two fixed points. Then we have ϕb(bη)=bη,ϕa(aη)=aη, andϕ0(η)=η, whereϕa,ϕb,ϕ0are defined as in Lemma 2.
Proof. – We only proveϕ0(η)=η. The other two cases can be proved in a similar way.
First by using the definition ofϕ0and taking into account the relationϕ(aba−1b−1)=aba−1b−1, we obtain:
ϕ0(a)=
a−1b−1ϕ(b)ϕ(a) ϕ(a)
a−1b−1ϕ(b)ϕ(a)−1
, ϕ0(b)=
a−1b−1ϕ(b)ϕ(a) ϕ(b)
a−1b−1ϕ(b)ϕ(a)−1
, from which it follows that, for anyw∈S∗, we have:
ϕ0(w)=
a−1b−1ϕ(b)ϕ(a) ϕ(w)
a−1b−1ϕ(b)ϕ(a)−1
. (∗)
By taking forwlonger and longer prefixes ofη, we get:
ϕ0(η)=a−1b−1ϕ(b)ϕ(a)ϕ(η)=a−1b−1ϕ(baη)=η.
Now, we are going to show some further relations betweenϕ0andϕ.
LetT be the endomorphism ofGsuch that T (σ )=τ, T (τ )=σ, andT (π )=π. This means that, if η=η1◦η2◦ · · · ◦ηnwithηi∈ {σ, τ, π}(1in), then
T η=T η1◦T η2◦ · · · ◦T ηn. (1) THEOREM 4. – For anyϕ∈Gτ,e, we haveϕ0=T ϕ.
In other words,ϕ0is obtained by exchangingτ andσ inϕ.
Proof. – We have to prove
T ϕ=
b−1a−1ϕ(a)ba, a−1b−1ϕ(b)ab
. (2)
This is done by induction on the length ofϕ.
First suppose that the length ofϕ is 2. Since neitherτ π norπ τ are primitive,ϕ =τ2. Hence σ2= (b−1a−1τ2(a)ba, a−1b−1τ2(b)ab), i.e., the conclusion holds fork=1.
Now assume the conclusion holds whenL(ϕ)=2kand consider the caseL(ϕ)=2(k+1).
Writeϕ=ψ2◦ψ1, whereψ1∈ {τ2, τ◦π, π◦τ},soL(ψ1)=2, L(ψ2)=2k. Now we distinguish three cases depending onψ1.
(i)ψ1=τ2=(aba, ba). Sinceψ2∈Gτ,e, we haveψ2(aba−1b−1)=aba−1b−1. From this equality and the hypothesis of induction, we get:
T ϕ(a)=T (ψ2)◦T τ2
(a)=T (ψ2)◦σ2(a)=T ψ2(aba)
=
b−1a−1ψ2(a)ba
a−1b−1ψ2(b)ab
b−1a−1ψ2(a)ba
=b−1a−1ψ2(aba)ba
=b−1a−1ψ2◦ψ1(a)ba=b−1a−1ϕ(a)ba, T ϕ(b)=T ψ2◦σ2(b)=T ψ2(ab)
=
b−1a−1ψ2(a)ba
a−1b−1ψ2(b)ab
=b−1a−1ψ2(a)ψ2(b)ab
=a−1b−1ψ2(b)ψ2(a)ab=a−1b−1ψ2(ba)ab
=a−1b−1ψ2◦ψ1(b)ab=a−1b−1ϕ(b)ab.
(ii)ψ1=π◦τ =(ab, b). Proceeding as in (i), we have:
T ϕ(a)=T ψ2◦π◦σ (a)=T ψ2(ba)
=
a−1b−1ψ2(b)ab
b−1a−1ψ2(a)ba
=a−1b−1ψ2(b)ψ2(a)ba
=a−1b−1ψ2◦ψ1(a)ba=b−1a−1ϕ(a)ba, T ϕ(b)=T ψ2◦π◦σ (b)=T ψ2(b)
=a−1b−1ψ2(b)ab=a−1b−1ϕ(b)ab.
(iii)ψ1=τ◦π=(a, ba). As in (i) we have:
T ϕ(a)=T ψ2◦σ◦π(a)=T ψ2(a)
=b−1a−1ψ2(a)ba=b−1a−1ϕ(a)ba, T ϕ(b)=T ψ2◦σ◦π(b)=T ψ2(ab)
=b−1a−1ψ2(a)baa−1b−1ψ2(b)ab=b−1a−1ψ2(a)ψ2(b)ab
=a−1b−1ψ2(b)ψ2(a)ab=a−1b−1ϕ(b)ab.
This completes the proof of Theorem 4.
Remark 1. – Letϕ=τ2, thenτ has two fixed pointsξa=abξandξb=baξ. Due to Theorems 3 and 4, we know thatξ is the Fibonacci sequence. In [10] we show that for anyu∈S∗,u=ε,
(1) neitheruabξnorubaξcan be a fixed points of an invertible substitution, (2) u−1ξ cannot be a fixed point of an invertible substitution.
*Supported by the Special Funds for Major State Basic Research Projects of China and Morningside Center of Mathematics (CAS).
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