A Rauzy fractal unbounded in all directions of the plane
aa r X i v : . [ m a t h . D S ] F e b A Rauzy fractal unbounded in all directions of the plane
Mélodie Andrieu
Abstract
We construct an Arnoux-Rauzy word for which the set of all differences of two abelianizedfactors is equal to Z . In particular, the imbalance of this word is infinite - and its Rauzy fractalis unbounded in all directions of the plane. Résumé
Nous construisons explicitement un mot d’Arnoux-Rauzy pour lequel l’ensemble des diffé-rences possibles des facteurs abélianisés est égal à Z . En particulier, le déséquilibre de ce motest infini, et son fractal de Rauzy n’est borné dans aucune direction du plan. À l’algorithme de fraction continue soustractif décrit par l’itération de l’application (dite de Farey) ( R + ) → ( R + ) ( x, y ) ( x − y, y ) si x ≥ y , ( x, y − x ) sinon,est associée une classe particulière de mots infinis binaires appelés mots sturmiens. Rappelons qu’un mot est une suite finie ou infinie d’éléments ( lettres ) pris dans un ensemble fini ( alphabet ). Les motssturmiens jouissent de nombreuses caractérisations combinatoires, arithmétiques et géométriques(consulter [7] pour une introduction générale). En particulier, ce sont exactement les mots apé-riodiques binaires dont le déséquilibre vaut 1, c’est-à-dire dans lesquels tous les facteurs de mêmelongueur (un facteur de longueur n est un sous-mot constitué de n lettres consécutives) contiennent,à 1 près, le même nombre de 0 (et donc, à 1 près également, le même nombre de 1). Par exemple,un mot commençant par w = 001000100100010001001 . . . pourrait être sturmien, tandis qu’un motcommençant par w = 011011100 ... ne l’est pas, car il contient les facteurs et . Cette propriétégarantit en particulier que les lettres et sont uniformément distribuées par rapport à une mesurede probabilité ν sur { , } , et que l’écart entre la somme de Birkhoff /N P N − n =0 { } ( w [ n ]) , quimesure la fréquence de observée parmi les N premières lettres du mot w , et sa valeur attendue ν (0) (appelée fréquence de ) est majoré par /N . D’un point de vue géométrique, cela signifie queles points P N := P Nn =0 e w [ n ] , où ( e , e ) désigne la base canonique de R , restent à une distancebornée de la droite portée par le vecteur fréquence ( ν (0) , ν (1)) . On appelle ligne brisée associée à w la suite ( P N ) N ∈ N . En informatique, les lignes brisées associées aux mots sturmiens sont utiliséespour discrétiser les droites de pente irrationnelles.Depuis Jacobi, plusieurs algorithmes ont été proposés pour généraliser les fractions continues àdes triplets de réels positifs (on peut consulter à ce sujet le livre [9]). De tels algorithmes devraientpermettre d’approcher simultanément et efficacement deux réels par une suite de couples de nombresrationnels. 1ans ce document, nous nous intéressons aux mots d’Arnoux-Rauzy, introduits par Arnoux etRauzy dans [2], qui sont les mots ternaires associés à l’algorithme (défini sur un ensemble de mesurenulle) : F AR : ( R + ) → ( R + ) ( x, y, z ) ( x − y − z, y, z ) si x ≥ y + z , ( x, y − x − z, z ) si y ≥ x + z , ( x, y, z − x − y ) si z ≥ x + y .Parce qu’ils conservent de nombreuses propriétés combinatoires des mots sturmiens, les motsd’Arnoux-Rauzy sont souvent présentés comme leur généralisation. En particulier, on peut montrerqu’ils admettent un vecteur fréquence des lettres. Aussi, une façon d’étudier la ligne brisée (tridi-mensionnelle) associée à un mot d’Arnoux-Rauzy consiste à la projeter, parallèlement au vecteurfréquence, sur le plan diagonal ∆ : x + y + z = 0 . On appelle fractal de Rauzy de w l’adhérence decet ensemble de points.Jusqu’en 2000, on a pensé que, comme pour les mots sturmiens, le déséquilibre des mots d’Arnoux-Rauzy était borné, ou au moins fini. Cassaigne, Ferenczi et Zamboni [4] ont contredit cette conjectureen construisant un mot d’Arnoux-Rauzy de déséquilibre infini - un mot donc, dont la ligne brisées’écarte régulièrement et de plus en plus loin de sa direction moyenne, ou, dit encore autrement, unmot dont le fractal de Rauzy n’est pas borné.Aujourd’hui, on ne sait presque rien sur les propriétés géométriques et topologiques de ces fractalsde Rauzy déséquilibrés. Le théorème d’Oseledets [8] suggère toutefois que ces fractals sont contenusdans une bande du plan ; en effet, si les exposants de Lyapounov associés au produit de matricesdonné par l’algorithme existent, l’un de ces exposants au moins doit être négatif puisque leur sommeest nulle.Dans cette note, nous prouvons que cette intuition est fausse. Théorème 1.
Il existe un mot d’Arnoux-Rauzy dont le fractal de Rauzy n’est borné dans aucunedirection du plan.
La construction que nous présentons s’adapte à la classe des mots associée à l’algorithme defraction continue multidimensionnelle de Cassaigne-Selmer, introduite dans [5], ainsi qu’aux motsépisturmiens stricts, qui sont la généralisation des mots d’Arnoux-Rauzy. Rappelons qu’un mot surun alphabet contenant d lettres est épisturmien strict si son langage est clos par miroir et s’il admet,pour chaque longueur n , un unique facteur multi-prolongeable à droite, et si ce facteur peut de plusêtre prolongé par chacune des d lettres de l’alphabet. Théorème 1’.
Il existe un mot C-adique w ∞ dont le fractal de Rauzy n’est borné dans aucunedirection du plan. Théorème 1” .
Soit d ≥ . Il existe un mot episturmien strict w ∞ sur l’alphabet { , ..., d } tel quepour tout hyperplan H de R d , la distance des points de la ligne brisée (ab( p n ( w ∞ ))) n ∈ N à l’hyperplan H n’est pas bornée. Les démonstrations des Théorèmes 1’ and 1” reposent sur des techniques similaires à celles duThéorème 1, et sont intégralement rédigées dans [1].Par ailleurs, nous proposons une preuve élémentaire du :
Théorème 2.
Le vecteur fréquence des lettres d’un mot d’Arnoux-Rauzy a des coordonnées ration-nellement indépendantes.
Ce résultat, conjecturé par Arnoux et Starosta en 2013 [3], a été démontré très récemment pardes moyens plus sophistiqués par Dynnikov, Hubert et Skripchenko [6].Le Théorème 2 est en fait vrai en toute dimension (voir [1] pour une preuve complète) :2 héorème 2’.
Soit d ≥ . Le vecteur fréquence des lettres d’un mot épisturmien strict sur l’alphabet { , ..., d } a des coordonnées rationnellement indépendantes. Until 2000, it was believed that, as for Sturmian words, the imbalance of Arnoux-Rauzy wordswas bounded - or at least finite. Cassaigne, Ferenczi and Zamboni disproved this conjecture byconstructing an Arnoux-Rauzy word with infinite imbalance, i.e. a word whose broken line deviatesregularly and further and further from its average direction [4]. Today, we know virtually nothingabout the geometrical and topological properties of these unbalanced Rauzy fractals. The Oseledetstheorem suggests that these fractals are contained in a strip of the plane: indeed, if the Lyapunovexponents of the matricial product associated with the word exist, one of these exponents at leastis nonpositive since their sum equals zero. This article aims at disproving this belief.
Theorem 1.
There exists an Arnoux-Rauzy word whose Rauzy fractal is unbounded in all directionsof the plane.
Theorem 1 also holds, on one hand, for C-adic words, which are the infinite words over { , , } associated with the Cassaigne-Selmer multidimensional continued fraction algorithm introduced in[5] and, on the other hand, for strict episturmian words, which are the generalization of Arnoux-Rauzy words. We recall that a strict episturmian word is a word whose language is close by mirrorand which admits, for each length, a unique right-special factor -which is moreover prolonged byeach letter in the alphabet. Theorem 1’.
There exists w ∞ a C-adic word whose Rauzy fractal is unbounded is all directions ofthe plane. Theorem 1” .
Let d ≥ . There exists w ∞ a strict episturmian word over the alphabet { , ..., d } such that for any hyperplane H in R d , the distance between H and the broken line (ab( p n ( w ∞ ))) n ∈ N is unbounded. The proofs of Theorems 1’ and 1” are based on techniques similar to those of Theorem 1; theycan be found in [1].Besides, we propose an elementary proof of:
Theorem 2.
The vector of letter frequencies of an Arnoux-Rauzy word has rationally independententries.
This theorem completes the works of Arnoux and Starosta, who conjectured it in 2013, to provethat the Arnoux-Rauzy continued fraction algorithm detects all kind of rational dependencies [3].Note that it has been recently proved by Dynnikov, Hubert and Skripchenko using quadratic forms[6]. Again, with a similar proof (see [1]), this result holds in arbitrary dimension:
Theorem 2’.
Let d ≥ . The vector of letter frequencies of a strict episturmian word over { , ..., d } has rationally independent entries. Preliminaries
We denote by A ∗ the set of all finite words over an alphabet A . A finite word u = u [0] u [1] ...u [ n − ,where u [ k ] denotes the ( k + 1) -th letter of u , is a factor of length n of a (finite or infinite) word w if there exists a nonnegative integer i such that for all k ∈ { , ..., n − } , w [ i + k ] = u [ k ] ; in theparticular case i = 0 , we say that u is the prefix of length n of w , and denote it by u = p n ( w ) . Wedenote by F n ( w ) the set of factors of w of length n and by F ( w ) its set of factors of all lengths.A substitution is an application mapping letters to finite words: A A ∗ , that we extend intoa morphism on the free monoid for the concatenation operation A ∗ on one hand, and on the set ofinfinite words A N on the other hand. Three substitutions will be of high interest in this paper: σ , σ and σ defined over A = { , , } by: σ i : A → A ∗ i ij ij for j ∈ A \{ i } . They are called
Arnoux-Rauzy substitutions ; we denote AR = { σ , σ , σ } . The set AR can be seenas a three letter alphabet -it should not be confused with A = { , , } over which the substitutionsare defined. As much as we can, we refer to the elements of AR ∗ or AR N as "sequences" instead of"words"; nonetheless, some tools like the notions of factor and prefix will turn out to be useful forthis second alphabet as well, especially in Section 4.The set A N of infinite words over A is endowed with the distance δ : for all w, w ′ ∈ A N , δ ( w, w ′ ) =2 − n , where n = min { n ∈ N | w [ n ] = w ′ [ n ] } if w = w ′ , and δ ( w, w ′ ) = 0 otherwise. We say that asequence of finite words ( u n ) n ∈ N ∈ ( A ∗ ) N converges to an infinite word w ∈ A N if for any sequenceof infinite words ( v n ) n ∈ N ∈ ( A N ) N , the sequence of infinite words ( u n · v n ) n ∈ N ∈ ( A N ) N converges to w . If ( s n ) n ∈ N ∈ AR N is a sequence containing infinitely many occurrences of each Arnoux-Rauzysubstitution σ , σ and σ , then the sequence of finite words ( s ◦ ... ◦ s n − ( α )) , with α ∈ A , convergesto an infinite word w which does not depend on α . The infinite words w obtained this way arecalled standard Arnoux-Rauzy words . An infinite word w is an Arnoux-Rauzy word if it has the sameset of factors than a standard Arnoux-Rauzy word w . One can show that the standard Arnoux-Rauzy word w and the directive sequence ( s n ) n ∈ N associated with w are unique. This definition ofArnoux-Rauzy words is equivalent to the more usual one: an infinite word is an Arnoux-Rauzy word if it has complexity n + 1 and admits exactly one right and one left special factor of each length.Given a finite word u ∈ A ∗ and a letter α ∈ A , we denote by | u | α the number of occurrences of α in u . The abelianized vector of u , sometimes called Parikh vector of u , is the vector ab( u ) = ( | u | α ) α ∈ A ,which counts the number of times that each letter occurs in the finite word u . At this point, itis useful to order the alphabet. For the convenience of typing, we choose to represent abelianizedwords as line vectors. Observe that the sum of the entries of ab( u ) is equal to the length of the word u , that we denote by | u | . Now, given a substitution s : A → A ∗ , the incidence matrix of s is thematrix M s whose i − th row is the abelianized of the image by s of the i − th letter in the alphabet.For instance, the incidence matrices of the Arnoux-Rauzy substitutions are: M σ = , M σ = and M σ = ∈ GL ( Z ) . Abelianized words and incidence matrices are made to satisfy: ab( s ( u )) = ab( u ) M s for any substi-tution s : A → A ∗ and any finite word u ∈ A ∗ . 4f w ∈ A N is an infinite word and α ∈ A is a letter, the frequency of α in w is the limit, if itexists, of the proportion of α in the sequence of growing prefixes of w : f w ( α ) = lim n →∞ | p n ( w ) | α n . Wedenote by f w = ( f w ( α )) α ∈ A the vector of letter frequencies of w , if it exists. When the vector of letterfrequencies exists, as it is the case for any Arnoux-Rauzy word, it is natural to study the differencebetween the predicted frequencies of letters and their observed occurrences. Given an infinite word w ∈ A N for which the vector of letter frequencies is defined, we consider the discrepancy function : N → R n max α ∈ A | | p n ( w ) | α − nf w ( α ) | . The discrepancy is linked to a combinatorial property: the imbalance. The imbalance of an infiniteword w is the quantity (possibly infinite) : imb( w ) = sup n ∈ N sup u,v ∈F n ( w ) || ab( u ) − ab( v ) || ∞ . The imbalance of an infinite word w is finite if and only if its discrepancy function is bounded.Geometrically, the discrepancy is linked to the diameter of the Rauzy fractal. Let ∆ denotesthe plane of R with equation x + y + z = 0 . For w an Arnoux-Rauzy word, denote by f w itsletter frequencies vector and by π w the (oblique) projection onto ∆ associated with the direct sum: R f w ⊕ ∆ = R . The Rauzy fractal of w , denoted by R w , is the closure of the image of the set ofabelianized prefixes of w (the broken line of w ) by the projection π w : R w = ∪ k ∈ N { π w (ab( p k ( w ))) } ⊂ ∆ . Note that the statement of our main result (Theorem 1) does not depend on the choice of theplane we project onto. Lemma 1.
For any ( a, b, c ) ∈ Z , there exists s ∈ AR ∗ and there exist u, v ∈ F ( s (1)) that satisfy ab( u ) − ab( v ) = ( a, b, c ) . Remark 1 (Abuse of notation) . If s = s · ... · s n − ∈ AR ∗ , and if w ∈ A ∗ ∪ A N , then s ( w ) denotesthe image of the word w by the substitution s ◦ ... ◦ s n − .Proof. Section 5 is devoted to the proof of Lemma 1.Therefore, all standard Arnoux-Rauzy words -and thereby all Arnoux-Rauzy words- whose di-rective sequence starts with the prefix s will admit ( a, b, c ) as difference of abelianized factors. Lemma 2.
For any p ∈ AR ∗ and any ( a, b, c ) ∈ Z , there exists s ∈ AR ∗ and there exist u, v ∈F ( p · s (1)) that satisfy ab( u ) − ab( v ) = ( a, b, c ) .Proof. Let p ∈ AR ∗ and ( a, b, c ) ∈ Z . Denote by M p the incidence matrix of the substitutionassociated with p (following Remark 1), which is a product of the Arnoux-Rauzy matrices M σ , M σ and M σ , and thus belongs to GL ( Z ) . By Lemma 1, there exists s ∈ AR ∗ and there exist u and v ∈ F ( s (1)) such that ab( u ) − ab( v ) = ( a, b, c ) M − p . But then, p ( u ) and p ( v ) are factors of F ( p · s (1)) and satisfy ab( p ( u )) − ab( p ( v )) = (ab( u ) − ab( v )) M p = ( a, b, c ) .We now construct a standard Arnoux-Rauzy word for which all triplets of integers can be obtainedas a difference of two of its abelianized factors. Proposition 1.
There exists an Arnoux-Rauzy word w ∞ such that for all ( a, b, c ) ∈ Z , there exist u and v ∈ F ( w ∞ ) satisfying ab( u ) − ab( v ) = ( a, b, c ) . roof. Let ϕ : N → Z a bijection (that can be chosen explicitly). We construct an infinite word d ∈ AR N as the limit of the sequence of finite words ( p k ) k ∈ N ∈ ( AR ∗ ) N that we define by recurrenceas follows. We first set p as the prefix given by Lemma 1 for ( a, b, c ) = ϕ (0) . Now, for k ∈ N , we set p k +1 = p k .σ .σ .σ .s , where s ∈ AR ∗ is given by applying Lemma 2 to the word p k .σ .σ .σ ∈ AR ∗ and the vector ϕ ( k + 1) ∈ Z . By construction, the sequence of finite words ( p k ) k ∈ N converges toan infinite sequence d which contains infinitely many occurrences of σ , σ and σ . This guaranteesthat the sequence of finite words ( d ◦ ... ◦ d n − (1)) n ∈ N converges to an Arnoux-Rauzy word, thatwe denote by w ∞ . Finally, for any k ∈ N , since the directive sequence of w ∞ starts with the prefix p k , there exist u k , v k ∈ F ( w ∞ ) such that ab( u k ) − ab( v k ) = ϕ ( k ) . Corollary 1.
The imbalance of the word w ∞ is infinite.Proof. For any n ∈ N , there exist u n and v n ∈ F ( w ∞ ) such that ab( u n ) − ab( v n ) = ( n, , − n ) ; thisimplies both | u n | = | v n | and | u n | − | v n | = n . The imbalance of w ∞ is thus infinite.The imbalance of a word, which is a combinatorial quantity, is linked to the geometrical shape ofits associated broken line. More precisely: a word w admitting frequencies has an infinite imbalanceif and only if its Rauzy fractal is unbounded. We now propose to show that the word w ∞ actuallysatisfies a stronger property: its Rauzy fractal is unbounded in all directions of the plane. Thisrelies on the following proposition. Proposition 2.
Let w ∈ A N . If for all d ∈ Z ∩ ∆ , where ∆ denotes the plane of R with equation x + y + z = 0 , there exist u and v ∈ F ( w ) such that ab( u ) − ab( v ) = d , then, for any plane Π andfor any D ∈ R + , there exists k ∈ N such that the euclidean distance between the point ab( p k ( w )) and the plane Π is larger than D .Proof. Without loss of generality, we can assume that Π contains (0 , , .If Π = ∆ , then for any D ∈ R + , dist(ab( p k ( w )) , Π) > D , with k = ⌊ D √ / ⌋ + 1 .Let Π = ∆ . By contradiction, assume that there exists D ∈ R + such that for all nonnegativeinteger k , dist(ab( p k ( w )) , Π) ≤ D . Let d ∈ Z ∩ ∆ with dist( d , Π) > D , and factors u, v ∈ F ( w ) such that ab( u ) − ab( v ) = d . Then, without loss of generality, we have dist( ab ( u ) , Π) > D . Let t ∈ A ∗ be such that tu is a prefix of w . Then we have dist(ab( t ) , Π) > D or dist(ab( tu ) , Π) > D , acontradiction. Remark 2.
Proposition 2 and its proof remain valid by replacing ∆ by any other plane whoseintersection with Z is not trapped between two parallel lines. Theorem 1.
There exists an Arnoux-Rauzy word whose Rauzy fractal is unbounded in all directionsof the plane.Proof.
We obtain, by applying Proposition 2 to the word w ∞ described in Proposition 1 and toplanes spanned by f w and a vector of ∆ , that the Rauzy fractal associated with w ∞ cannot betrapped between two parallels lines. We consider the infinite oriented graph whose vertices are the elements of Z and whose edgesmap triplets to their images by one the 15 following applications. For δ ∈ {− , − , , , } and i ∈ { , , } , consider: 6 i,δ : Z → Z ( x j ) j ∈{ , , } ( y j ) j ∈{ , , } where y i = x + x + x + δ and y j = x j for j = i. Our aim is to show that all vertices can be reached from the triplet O = (0 , , ∈ Z , movingthrough a finite number of edges (see Definition 1 and Proposition 3 below.) The motivation lies inthe following lemma. Lemma 3.
Let d ∈ Z . If there exist n ∈ N and a finite sequence ( i l , δ l ) ≤ l ≤ n − ∈ ( { , , } ×{− , − , , , } ) n such that d = τ i n − ,δ n − ◦ ... ◦ τ i ,δ ( O ) , then there exist s ∈ AR ∗ and u, v ∈F ( σ i n − ◦ ... ◦ σ i ( s (1))) satisfying ab( u ) − ab( v ) = d .Proof. Let d ∈ Z . Assume that there exist n ∈ N and ( i l , δ l ) ≤ l ≤ n − ∈ ( { , , }× {− , − , , , } ) n such that d = τ i n − ,δ n − ◦ ... ◦ τ i ,δ ( O ) . We are going to build iteratively two finite sequences of finitewords ( u l ) and ( v l ) , where l ∈ { , ..., n } , and s ∈ AR ∗ , such that for all l , the words u l and v l arefactors of σ i l − ◦ ... ◦ σ i ( s (1)) , and such that ab( u n ) − ab( v n ) = d .First, we choose s ∈ AR ∗ that satisfies | s (1) | ≥ n , and we set u = v = s (1) . Then, assumingthat u l and v l ∈ F ( σ i l − ◦ ... ◦ σ i ( s (1))) are built, we set ˜ u l +1 = σ i l ( u l ) and ˜ v l +1 = σ i l ( v l ) . From ˜ u l +1 and ˜ v l +1 , we now define u l +1 and v l +1 according to the following table. δ choice for u l +1 choice for v l +1 u l +1 ˜ v l +1 u l +1 .i l ˜ v l +1 u l +1 .i l v l +1 such that i l .v l +1 = ˜ v l +1 ( ∗ ) − u l +1 ˜ v l +1 .i l − u l +1 such that i l .u l +1 = ˜ u l +1 ( ∗ ) ˜ v l +1 .i l The steps marked with ( ∗ ) (removal of the initial i l ) are always possible since, for all l ∈ { , ..., n − } ,the words u l and v l (and thus ˜ u l +1 and ˜ v l +1 ) are nonempty.In all of these cases, the words u l +1 and v l +1 are factors of σ l ◦ ... ◦ σ ( s (1)) and satisfy ab( u l +1 ) − ab( v l +1 ) = τ i l ,δ l (ab( u l ) − ab( v l )) . In particular, at step l = n − , the finite words u n and v n arefactors of σ i n − ◦ ... ◦ σ i ( s (1)) and satisfy ab( u n ) − ab( v n ) = τ i n − ,δ n − ◦ ... ◦ τ i ,δ ( O ) = d .In the sequel, it is convenient to introduce some vocabulary from graph theory. Definition 1.
A triplet ( a, b, c ) ∈ Z is accessible from a triplet ( d, e, f ) if there exist a nonnegativeinteger n and a finite sequence ( i l , δ l ) ≤ l ≤ n − ∈ ( { , , } × {− , − , , , } ) n such that ( a, b, c ) = τ i n − ,δ n − ◦ ... ◦ τ i ,δ (( d, e, f )) . Proposition 3.
All triplets in Z are accessible from O . The proof of Proposition 3 lies on the two following lemmas.
Lemma 4.
The triplet ( a, b, c ) ∈ Z is accessible from O if and only if ( − a, − b, − c ) is also accessiblefrom O . Similarly, ( x j ) j ∈{ , , } is accessible from O if and only if for all s ∈ S , where S denotesthe symmetric group acting on three elements, the triplet ( x s ( j ) ) j ∈{ , , } is accessible from O .Proof. For the first assertion, change δ l into − δ l in the finite sequence of edges going from O to ( a, b, c ) . For the second assertion, change i l into s ( i l ) in the finite sequence of edges going from O to ( x j ) j ∈{ , , } . 7 emma 5. Let a ∈ N . The triplet ( a, − a, − a ) ∈ Z is accessible from O .Proof. The lemma is trivially true for a = 0 . By recurrence, consider an arbitrary nonnegative integer a such that the triplet ( a, − a, − a ) is accessible from O . One can check that ( a + 1 , − a − , a + 1) = τ , ◦ ( τ , ) a +1 ◦ τ , − (( a, − a, − a )) . So the triplet ( a + 1 , − a − , a + 1) is accessible from O . Butthen, Lemma 4 indicates that ( a + 1 , − a − , − a − is accessible from O . Proof of Proposition 3.
The proof relies on the four following observations.• The vertices ( a, b, − a ) and ( a, − a, c ) are accessible from O for all a, b, c ∈ Z . Indeed, itsuffices to write ( a, b, − a ) = ( τ , ) a + b (( a, − a, − a )) and ( a, − a, c ) = ( τ , ) a + c (( a, − a, − a )) andremember that ( a, − a, − a ) is accessible from O by Lemma 5.• The vertex ( a, c, c ) = τ , (( a, − a, c )) is also accessible from O ,• If a ≥ b > c > − a , then ( a, − a + b − a, c ) = ( τ , ) − ( a, b, c ) is closer to ( a, − a, − a ) in supnorm, and we have | − a + b − c | , | c | < a ,• If a ≥ c > b > − a , then ( a, b, − a − b + c ) = ( τ , ) − (( a, b, c )) is closer to ( a, − a, − a ) than ( a, b, c ) , and we have | b | , | − a − b + c | < a. Let a, b, c ∈ Z . By Lemma 4, it suffices to deal with the case | b | , | c | ≤ | a | , and a > . Followingthe observations above, we recursively construct a finite sequence ( i l ) ≤ l ≤ n − ∈ { , } n such that ( a, b, c ) = τ i n − , ◦ ... ◦ τ i , (( a, − a, − a )) . Since ( a, − a, − a ) is accessible from O (Lemma 5), thevertex ( a, b, c ) is also accessible from O . Proof of Lemma 1.
Lemma 1 follows from Proposition 3, Definition 1 and Lemma 3.
Remark 3.
The graph G is a simplification, exploiting the remarkable properties of the substitutions σ , σ and σ , of the imbalance automaton, introduced in [1] for a much wider range of S-adic systems(ie class of words obtained from a set of substitutions through directive sequences ). w ∞ has rationally independententries We sketch an elementary proof of the much wider result:
Theorem 2.
The vector of letter frequencies of an Arnoux-Rauzy word has rationally independententries.
The proof is inspired from a similar result that holds for C-adic words [5].
Proof.
Let w an Arnoux-Rauzy word; denote by ( s n ) n ∈ N its directive sequence and by f its letterfrequencies vector. We recall that for all nonnegative integer n , s n = σ i if and only if the i − th entryof F nAR ( f ) ( F AR is defined in Section 1) is greater than the sum of the two others. By contradiction,assume that the entries of f are not rationally independent.First, observe that if for some r ∈ N , the i − th entry of F rAR ( f ) is zero, then it will remain zero;and from this point on the directive sequence will not contain the substitution σ i , which is conflictingwith the definition of Arnoux-Rauzy words and the uniqueness of the directive sequence. Thus, forall n ∈ N , all entries of F nAR ( f ) are positive. Let l a nonzero integer column vector such that f l = 0 (recall that f is a line vector). Let l m = M s m − ...M s l . The Arnoux-Rauzy matrices being invertible, l m is also a nonzero integer column vector; it satisfies F mAR ( f ) l m = f.M − s ...M − s m − l m = f l = 0 .8enote l m = ( a, b, c ) t and consider D m = max( | b − a | , | c − b | , | c − a | ) ∈ N the difference between themaximum and the minimum entry of l m , that we call spread of l m . We claim that the sequence ofnonnegative integers ( D m ) m ∈ N is non-increasing and that it furthermore decreases infinitely often -and here will be the contradiction.Indeed, the vector l m +1 is of the form M l m , where M is one the the three Arnoux-Rauzy matrices M σ , M σ or M σ , which give respectively: l m +1 = ( a, a + b, a + c ) t , l m +1 = ( a + b, b, c + b ) t and l m +1 = ( a + c, b + c, c ) t . One can easily show, observing that the extreme entries of l m have oppositesigns, that in all cases D m ≥ D m +1 . Similarly, we write l m +2 = M s m +1 M s m l m . A quick argumentshow that as soon as s m +1 = s m , which happens infinitely many times by definition of Arnoux-Rauzywords, we have D m > D m +2 . References [1] M. Andrieu. Thesis. in preparation , 2020.[2] P. Arnoux and G. Rauzy. Représentation géométrique de suites de complexité 2n+1.
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