\mathcal{S}-adic characterization of minimal ternary dendric shifts
aa r X i v : . [ m a t h . D S ] F e b S -adic characterization of minimal ternary dendric subshifts France Gheeraert , Marie Lejeune , and Julien Leroy Department of Mathematics, University of Liège, Allée de la Découverte 12 (B37), B-4000 Liège,Belgium. { france. gheeraert,m. lejeune,j. leroy}@ uliege. be Supported by a FNRS fellowship
Abstract
Dendric subshifts are defined by combinatorial restrictions of the extensions of thewords in its language. This family generalizes well-known families of subshifts such asSturmian subshifts, Arnoux-Rauzy subshifts and codings of interval exchange transforma-tions. It is known that any minimal dendric subshifts has a primitive S -adic representationwhere the morphisms in S are positive tame automorphisms of the free group generatedby the alphabet. In this paper we investigate those S -adic representations, heading to-wards an S -adic characterization ot this family. We obtain such a characterization in theternary case, involving a directed graph with 9 vertices. Dendric subshifts are defined in terms of extension graphs that describe the left and rightextensions of their factors. Extension graphs are bipartite graphs that can be roughly bedescribed as follows: if u is a word in the language L ( X ) of the subshift X , one puts an edgebetween the left and right copies of letters a and b such that aub is in L ( X ) . A subshift is thensaid to be dendric if the extension graph of every word of its language is a tree. These sub-shifts were initially defined through their languages under the name of tree sets [BDFD + + + + + + ar]. They for instance have fac-tor complexity L ( X ) ∩A n ) = ( A− n +1 [BDFD + A [BBD + ar],where A is the alphabet of the subshift. They also fall into the class of subshifts satisfying theregular bispecial condition [DF20], which implies that the number of their ergodic measures isat most A / . An important property for our work is that the derivated subshift of a minimaldendric subshift is again a minimal dendric subshift on the same alphabet, where derivationis here understood as derivation by return words (see Section 3 for definitions). This allows togive S -adic representations of such subshifts [Fer96], i.e., to define a set S of endomorphisms Key words.
Symbolic dynamics, substitutions, S-adic, dendric, interval exchange, return words..2010
Mathematics subject classification.
1f the free monoid A ∗ and a sequence σ = ( σ n ) n ≥ ∈ S N such that X = { x ∈ A Z | u ∈ L ( x ) ⇒ ∃ n ∈ N , a ∈ A : u ∈ L ( σ σ · · · σ n ( a )) } . The sequence σ is called an S -adic representation of the subshift. S -adic representations are a classical tool that allows to study several properties of subshiftssuch as the factor complexity [DLR13, DDMP20], the number of ergodic measures [BD14,BHL20b, BHL20a], the dimension group and topological rank [BBD + ar] or yet the automor-phism group [EM20]. In the case of minimal dendric subshifts, the involved endomorphismsare particular tame automorphisms of the free group generated by the alphabet [BDFD + + + ar, BHL20a].An important open problem concerning S -adic representations is the S -adic conjecturewhose goal is to give an S -adic representation of subshifts with at most linear complex-ity [Ler12], i.e., to find a stronger notion of S -adicity such that a subshift has an at mostlinear factor complexity if and only if it is “strongly S -adic”. In this article we attack thisconjecture by studying S -adic representations of minimal dendric subshifts. Our main resultis the following that gives an S -adic characterization of minimal dendric subshifts over aternary alphabet, where S is defined in Section 5.1. It involves a labeled directed graph G with 9 vertices and which is non-deterministic, i.e., a given morphism may label several edgesleaving a given vertex. Theorem 1.
A subshift ( X, S ) is a minimal dendric subshift over A = { , , } if and only ifit has a primitive S -adic representation σ ∈ S N that labels a path in the graph G representedin Figure 9. We then characterize, within this graph or its deterministic version (with 16 vertices),the well-known families of Arnoux-Rauzy subshifts (Proposition 33) and of coding of regularIET (Theorem 51). We also show that subshifts arising from the Cassaigne MCF are neverArnoux-Rauzy subshifts, nor codings of regular IET (Proposition 53).Observe that we do not focus only on the ternary case. We investigate the S -adic represen-tations of minimal dendric subshifts over any alphabet obtained when considering derivationby return words to letters. We for instance show that when taking the image Y of subshift X under a morphism in S , the extension graphs of long enough factors of Y are the imageof the extension graph of factors of X under some graph homomorphism (Proposition 13).This allows us to introduce the notion of dendric preserving morphism for X which is thefundamental notion for the construction of the graph G . We also characterize the morphisms σ of S that are dendric preserving for all X using Arnoux-Rauzy morphisms (Proposition 17).The paper is organized as follows. We start by giving, in Section 2, the basic definitionsfor the study of subshifts. We introduce the notion of extension graph of a word, of dendricsubshift and of S -adic representation of a subshift.In Section 3, we recall the existence of an S -adic representation using return words forminimal subshifts (Theorem 4) and the link between return words and Rauzy graph.In Section 4, we then study the relation between words in a subshift and in its imageby a strongly left proper morphism (Proposition 7). We deduce from it a link between theextension graphs (Proposition 13) using graph morphisms and we prove that the injective andstrongly left proper morphisms that preserve dendricity can be characterized using Arnoux-Rauzy morphisms (Proposition 17). 2n Section 5, we study the notions and results of Section 4 in the case of a ternary alphabet.We then prove the main result of this paper (Theorem 1) which gives an S -adic characterizationof ternary dendric subshifts using infinite paths in a graph.Finally, in Section 6, we focus on three sub-families of dendric subshifts: Arnoux-Rauzysubshifts, interval exchanges and Cassaigne subshifts. For interval exchanges, we first recallthe associated definitions and basic properties, then provide an S -adic characterization (The-orem 51) in the ternary case using a subgraph of the graph obtained in the dendric case. Wealso prove that the families of Cassaigne subshifts, of Arnoux-Rauzy subshifts and of regularinterval exchanges are disjoint. Let A be a finite alphabet of cardinality d ≥ . Let us denote by ε the empty word of the freemonoid A ∗ (endowed with concatenation), and by A Z the set of bi-infinite words over A . Fora bi-infinite word x ∈ A Z , and for i, j ∈ Z with i ≤ j , the notation x [ i,j ) (resp., x [ i,j ] ) standsfor x i · · · x j − (resp., x i · · · x j ) with the convention x [ i,i ) = ε . For a word w = w · · · w ℓ ∈ A ℓ ,its length is denoted | w | and equals ℓ . We say that a word u is a factor of a word w if thereexist words p, s such that w = pus . If p = ε (resp., s = ε ) we say that u is a prefix (resp., suffix ) of w . For a word u ∈ A ∗ , an index ≤ j ≤ ℓ such that w j · · · w j + | u |− = u is calledan occurrence of u in w and we use the same term for bi-infinite word in A Z . The number ofoccurrences of a word u ∈ A ∗ in a finite word w is denoted as | w | u .The set A Z endowed with the product topology of the discrete topology on each copy of A is topologically a Cantor set. The shift map S defined by S (( x n ) n ∈ Z ) = ( x n +1 ) n ∈ Z is ahomeomorphism of A Z . A subshift is a pair ( X, S ) where X is a closed shift-invariant subsetof some A Z . It is thus a topological dynamical system . It is minimal if the only closed shift-invariant subset Y ⊂ X are ∅ and X . Equivalently, ( X, S ) is minimal if and only if the orbitof every x ∈ X is dense in X . Usually we say that the set X is itself a subshift.The language of a sequence x ∈ A Z is its set of factors and is denoted L ( x ) . For a subshift X , its language L ( X ) is ∪ x ∈ X L ( x ) and we set L n ( X ) = L ( X ) ∩ A n , n ∈ N . Its factorcomplexity is the function p X : N → N defined by p X ( n ) = L n ( X ) . We say that a subshift X is over A if L ( X ) = A . Dendric subshifts are defined with respect to combinatorial properties of their language ex-pressed in terms of extension graphs. Let F be a factorial set of finite words on the alphabet A . For w ∈ F , we define the sets of left, right and bi-extensions of w by E − F ( w ) = { a ∈ A | aw ∈ F } ; E + F ( w ) = { b ∈ A | wb ∈ F } ; E F ( w ) = { ( a, b ) ∈ A × A | awb ∈ F } . The elements of E − F ( w ) , E + F ( w ) and E − F ( w ) are respectively called the left extensions , the rightextensions and the bi-extensions of w in F . If X is a subshift on A , we will use the terminologyextensions in X instead of extensions in L ( X ) and the index L ( X ) will be replaced by X or3ven omitted if the context is clear. Observe that as X ⊂ A Z , the set E X ( w ) completelydetermines E − X ( w ) and E + X ( w ) . A word w is said to be right special (resp., left special ) if E + ( w )) ≥ (resp., E − ( w )) ≥ ). It is bispecial if it is both left and right special. Thefactor complexity of a subshift is completely governed by the extensions of its special factors.In particular, we have the following result. Proposition 2 (Cassaigne and Nicolas [CN10]) . Let X be a subshift. For all n , we have p X ( n + 1) − p X ( n ) = X w ∈L n ( X ) ( E + ( w ) − X w ∈L n ( X ) ( E − ( w ) − In addition, if for every bispecial factor w ∈ L ( X ) , one has E ( w ) − E − ( w ) − E + ( w ) + 1 = 0 , then p X ( n ) = ( p X (1) − n + 1 for every n . For a word w ∈ F , we consider the undirected bipartite graph E F ( w ) called its extensiongraph with respect to F and defined as follows: its set of vertices is the disjoint union of E − F ( w ) and E + F ( w ) and its edges are the pairs ( a, b ) ∈ E − F ( w ) × E + F ( w ) such that awb ∈ F .For an illustration, see Example 3 below. We say that w is dendric if E ( w ) is a tree. We thensay that a subshift X is a dendric subshift if all its factors are dendric in L ( X ) . Note thatevery non-bispecial word is trivially dendric. By Proposition 2, we immediately deduce thatany dendric subshift has factor complexity p X ( n ) = ( p X (1) − n + 1 for every n . Example 3.
Let σ be the Fibonacci substitution defined over the alphabet { a, b } by σ : a ab, b a and consider the subshift generated by σ (i.e., the set of bi-infinite words over A whose factors belong to some σ n ( a ) ). The extension graphs of the empty word and of the twoletters a and b are represented in Figure 1. E ( ε ) ab ab E ( a ) ab ab E ( b ) a a Figure 1: The extension graphs of ε (on the left), a (in the center) and b (on the right) aretrees. S -adicity Let A , B be finite alphabets with cardinality at least 2. By a morphism σ : A ∗ → B ∗ , wemean a non-erasing monoid homomorphism (also called a substitution when A = B ). Bynon-erasing, we mean that the image of any letter is a non-empty word. We stress the factthat all morphisms are assumed to be non-erasing in the following. Using concatenation, we4xtend σ to A N and A Z . In particular, if X is a subshift over A , the image of X under σ isthe subshift Y = { S k σ ( x ) | x ∈ X, ≤ k < | σ ( x ) |} . The morphism σ is said to be left proper (resp., right proper ) when there exists a letter ℓ ∈ B such that for all a ∈ A , σ ( a ) starts with ℓ (resp., ends with ℓ ). It is strongly left proper (resp., strongly right proper ) if it is left proper (resp., right proper) and the starting letter(resp., ending letter) ℓ only occurs once in each image σ ( a ) , a ∈ A . It is said to be proper ifit is both left and right proper. With a left proper morphism σ : A ∗ → B ∗ with first letter ℓ ,we associate a right proper morphism ¯ σ : A ∗ → B ∗ by σ ( a ) ℓ = ℓ ¯ σ ( a ) , ∀ a ∈ A . Let σ = ( σ n : A ∗ n +1 → A ∗ n ) n ≥ be a sequence of morphisms such that max a ∈A n | σ ◦ · · · ◦ σ n − ( a ) | goes to infinity when n increases. We assume that all the alphabets A n are minimal,in the sense that for all n ∈ N and b ∈ A n , there exists a ∈ A n +1 such that b is a factorof σ n ( a ) . For ≤ n < N , we define the morphisms σ [ n,N ) = σ n ◦ σ n +1 ◦ · · · ◦ σ N − and σ [ n,N ] = σ n ◦ σ n +1 ◦ · · · ◦ σ N . For n ≥ , the language L ( n ) ( σ ) of level n associated with σ isdefined by L ( n ) ( σ ) = (cid:8) w ∈ A ∗ n | w occurs in σ [ n,N ) ( a ) for some a ∈ A N and N > n (cid:9) . As max a ∈A n | σ [1 ,n ) ( a ) | goes to infinity when n increases, L ( n ) ( σ ) defines a non-emptysubshift X ( n ) σ that we call the subshift generated by L ( n ) ( σ ) . More precisely, X ( n ) σ is the setof points x ∈ A Z n such that L ( x ) ⊆ L ( n ) ( σ ) . Note that it may happen that L ( X ( n ) σ ) is strictlycontained in L ( n ) ( σ ) . Also observe that for all n , X ( n ) σ is the image of X ( n +1) σ under σ n .We set L ( σ ) = L (1) ( σ ) , X σ = X (1) σ and call X σ the S -adic subshift generated by the directive sequence σ . We also say that the directive sequence σ is an S -adic representation of X σ .We say that σ is primitive if, for any n ≥ , there exists N > n such that for all ( a, b ) ∈A n × A N , a occurs in σ [ n,N ) ( b ) . Observe that if σ is primitive, then min a ∈A n | σ [1 ,n ) ( a ) | goesto infinity when n increases, L ( X ( n ) σ ) = L ( n ) ( σ ) , and X ( n ) σ is a minimal subshift (see forinstance [Dur00, Lemma 7]).We say that σ is ( (strongly) left , (strongly) right ) proper whenever each morphism σ n is((strongly) left, (strongly) right) proper. We also say that σ is injective if each morphism σ n is injective (seen as an application from A ∗ n +1 to A ∗ n ). By abuse of language, we saythat a subshift is a (strongly left or right proper, primitive, injective) S -adic subshift if thereexists a (strongly left or right proper, primitive, injective) sequence of morphisms σ such that X = X σ . S -adicity using return words and shapes of Rauzy graphs S -adicity using return words and derived subshifts Let X be a minimal subshift over the alphabet A and let w ∈ L ( X ) be a non-empty word. A return word to w in X is a non-empty word r such that w is a prefix of rw and, rw ∈ L ( X ) and rw contains exactly two occurrences of w (one as a prefix and one as a suffix). We let5 X ( w ) denote the set of return words to w in X and we omit the subscript X whenever it isclear from the context. The subshift X being minimal, R ( w ) is always finite.Let w ∈ L ( X ) be a non-empty word and write R X ( w ) = { , . . . , R X ( w )) } . A morphism σ : R ( w ) ∗ → R ( w ) ∗ is a coding morphism associated with w if σ ( R ( w )) = R ( w ) . It is triviallyinjective. Let us consider the set D w ( X ) = { x ∈ R ( w ) Z | σ ( x ) ∈ X } . It is a minimal subshift,called the derived subshift of X (with respect to w ) . We know show that derivation of minimalsubshifts allows to build left proper and primitive S -adic representations of a minimal subshift.We inductively define the sequences ( a n ) n ≥ , ( R n ) n ≥ , ( X n ) n ≥ and ( σ n ) n ≥ by• X = X , R = A and a ∈ A ;• for all n , R n +1 = R X n ( a n ) , σ n : R ∗ n +1 → R ∗ n is a coding morphism associated with a n , X n +1 = D a n ( X n ) and a n +1 ∈ R n +1 .Observe that the sequence ( a n ) n ≥ is not uniquely defined as well as the morphism σ n (even if a n is fixed). However, to avoid heavy considerations when we deal with sequencesof morphisms obtained in this way, we will speak about “the” sequence ( σ n ) n ≥ and it isunderstood that we may consider any such sequence. Also observe that as we consider derivedsubshifts with respect to letters, each coding morphism σ n is strongly left proper. Theorem 4 (Durand [Dur98]) . Let X be a minimal subshift. Using the notation defined above,the sequence of morphisms σ = ( σ n : R ∗ n +1 → R ∗ n ) n ≥ is a strongly left proper, primitive andinjective S -adic representation of X . In particular, for all n , we have X n = X ( n ) σ . In the case of minimal dendric subshifts, the S -adic representation σ can be made stronger.This is summarized by the following result. Recall that if F A is the free group generated by A ,an automorphism α of F A is tame if it belongs to the monoid generated by the permutationsof A and by the elementary automorphisms λ a,b : ( a ab,c c, for c = a, and ˜ λ a,b : ( a ba,c c, for c = a. Theorem 5 (Berthé et al. [BDFD + . Let X be a minimal dendric subshift over thealphabet A = { , . . . , d } . For any w ∈ L ( X ) , D w ( X ) is a minimal dendric subshift over A and the coding morphism associated with w is a tame automorphism of F A . As a consequence,if σ = ( σ n ) n ≥ is the primitive directive sequence of Theorem 4, then all morphisms σ n arestrongly left proper tame automorphisms of F A . Let X be a subshift over an alphabet A . The Rauzy graph of order n of X , is the directedgraph G n ( X ) whose set of vertices is L n ( X ) and there is an edge from u to v if there areletters a, b such that ub = av ∈ L ( X ) ; this edge is labeled by a . If X is minimal, then anyRauzy graph G n ( X ) is strongly connected.If w ∈ L ( X ) is non-empty, then any return word r to w in X labels a path from w to w in G n ( X ) . Indeed, the word u = rw = u u · · · u N being in L ( X ) , u i ∈ A for all i , we canassociate with r = u u · · · u N − n the path w = u [1 ,n ] u −→ u [2 ,n +1] u −→ · · · u N − n −−−→ u [ N − n +1 ,N ] = w. r being a return word, all vertices u [ i,i + n − , i / ∈ { , N − n + 1 } , are differentfrom w . Observe also that the converse does not hold, i.e., there might exist paths from w to w whose label is not a return word (see Example 6 below). However, the shape of theRauzy graph G ( X ) provides restrictions on the possible return words to a letter a in X . Forinstance, in the minimal case, for all letters a, b, c ∈ L ( X ) , b = c , there is an edge from a to b in G ( X ) if and only if there is a return word r to c in which ab occurs. Furthermore, theextension graph of the empty word having for edges the pairs ( a, b ) ∈ E X ( ε ) , it completelydetermines L ( X ) , hence the Rauzy graph G ( X ) . Therefore, the extension graph E X ( ε ) provides restrictions on the possible return words to letters in X . Example 6.
The Rauzy graph of order 2 of the subshift X generated by the Fibonacci sub-stitution is given in Figure 2. For any n ≥ , the word a ( ab ) n labels a path from aa to aa .However, as aaa is not an element of L ( X ) , neither is ( ab ) thus a ( ab ) n is not a return wordto aa if n ≥ . aaab baa ab b Figure 2: Rauzy graph of order 2 for the Fibonacci subshift S -adic subshifts S -adicsubshifts Our aim is to describe bispecial factors and their bi-extensions in an S -adic subshift X σ . Aclassical way to do this is to “desubstitute” a bispecial factor u , i.e., to find the set of “minimal”factors v i in L ( X ( k ) σ ) such that u is a factor of the words σ [1 ,k ) ( v i ) and then to deduce theextensions of u from those of the v i ’s. The easiest case is when the set of v i ’s is singleton. Thenext result states that this is the case for injective and strongly left proper directive sequences.If X is a subshift over A and σ : A ∗ → B ∗ is a morphism, then for any words v ∈ L ( X ) and x, y ∈ B ∗ , we define the sets E − X,x ( v ) = { a ∈ E − X ( v ) | σ ( a ) ∈ B ∗ x } ; E + X,y ( v ) = { b ∈ E + X ( v ) | σ ( b ) ∈ y B ∗ } ; E X,x,y ( v ) = E X ( v ) ∩ ( E − X,x ( v ) × E + X,y ( v )) . Proposition 7.
Let X be a subshift over A , σ : A ∗ → B ∗ be an injective and strongly leftproper morphism (with first letter ℓ ), Y the image of X under σ and u a non-empty word in L ( Y ) . If ℓ does not occur in u , then there is b ∈ A such that u is a non-prefix factor of σ ( b ) .Otherwise, there is a unique triplet ( s, v, p ) ∈ B ∗ × L ( X ) × B ∗ for which there exists a pair ( a, b ) ∈ E X ( v ) such that u = sσ ( v ) p with1. s a strict suffix of σ ( a ) ; . p a strict prefix of σ ( b ) and if p = ε , then v = v ′ c for some letter c such that E X,s,σ ( c ) ( v ′ ) = E − X,s ( v ) × { c } .In the latter case, the left, right and bi-extensions of u are governed by those of v throughthe relation E Y ( u ) = { ( a ′ , b ′ ) ∈ B × B | ∃ ( a, b ) ∈ E X,s,p ( v ) : σ ( a ) ∈ B ∗ a ′ s ∧ σ ( b ) ℓ ∈ pb ′ B ∗ } . (1) Proof.
Since u is a word in L ( Y ) , it is a factor of σ ( w ) for some w ∈ L ( X ) . The word u beingnon-empty, any such w is non-empty as well. We say that w is covering u if u is a factor of σ ( w ) and for any proper factor w ′ of w , u is not a factor of σ ( w ′ ) . Recall that ℓ is the firstletter of σ ( a ) for all a ∈ A . Thus, if | u | ℓ = 0 , then any word w covering u is a letter and u isa non-prefix factor of σ ( w ) .Now assume that | u | ℓ ≥ and let u = s ′ u ′ p ′ with | s ′ | ℓ = 0 , p ′ ∈ ℓ ( A \ { ℓ } ) ∗ and u ′ ∈{ ε } ∪ ℓ A ∗ . As the letter ℓ occurs only as a prefix in any image σ ( a ) , a ∈ A , any word w covering u is of the form xv ′ y with x ∈ A ∪ { ε } and y ∈ A , where one has σ ( v ′ ) = u ′ , s ′ is asuffix of σ ( x ) (which is proper if x = ε ) and p ′ is a prefix of σ ( y ) . As σ is injective, the word v ′ is uniquely defined, i.e., v ′ does not depend on w .If there is such a covering word w for which p ′ is a proper prefix of σ ( y ) , then the triplet ( s, v, p ) = ( s ′ , v ′ , p ′ ) satisfies the requirements. Indeed, either the pair ( x, y ) is in E X ( v ′ ) or,if x = ε , then we may take any pair ( a, y ) ∈ E X ( v ′ ) . It will thus remain to show that it is theunique triplet.If there is no such covering word w , then for any covering word w ′ = x ′ v ′ y ′ ( v ′ is commonto all covering words), one has p ′ = σ ( y ′ ) . As σ is injective, there is a letter c such that allcovering words w are of the form w = x ′ v ′ c with u = s ′ σ ( v ′ c ) and s ′ suffix of σ ( x ′ ) . Thisshows that with v = v ′ c , the triplet ( s, v, p ) = ( s ′ , v, ε ) satisfies the requirement. Indeed, onehas E X,s ′ ,σ ( c ) ( v ′ ) = E − X,s ′ ( v ) × { c } and either we consider any pair ( x, b ) ∈ E X ( v ) (if x = ε ),or we may consider any pair ( a, b ) ∈ E X ( v ) (if x = ε ).Let us now show that ( s, v, p ) is unique. Assuming that (˜ s, ˜ v, ˜ p ) satisfies the conditions,then | ˜ s | ℓ = 0 and σ (˜ v )˜ p starts with ℓ . Thus, ˜ s = s ′ and σ (˜ v )˜ p = σ ( v ′ ) p ′ . Since ˜ p and p ′ areboth prefixes of words in σ ( A ) (and σ is strongly left proper), there are only two possibilities:either ˜ p = p ′ (hence ˜ v = v ′ ) or ˜ p = ε . In the latter situation, since p ′ = ε , there exists a unique c ∈ A such that ˜ v = v ′ c and p ′ = σ ( c ) . To prove the uniqueness of ( s, v, p ) , it remains to showthat ( s ′ , v ′ , p ′ ) and ( s ′ , v ′ c, ε ) cannot both verify the hypotheses. If the second decompositiondoes, then E X,s ′ ,p ′ ( v ′ ) = E − X,s ′ ( v ′ c ) × { c } . Thus there is no ( a, b ) ∈ E X ( v ′ ) for which s ′ is aproper suffix of σ ( a ) and p ′ is a proper prefix of σ ( b ) . Otherwise stated, the triplet ( s ′ , v ′ , p ′ ) does not satisfy the conditions, hence the conclusion.Let us now prove Equation (1). The inclusion E Y ( u ) ⊃ { ( a ′ , b ′ ) ∈ B × B | ∃ ( a, b ) ∈ E X,s,p ( v ) : σ ( a ) ∈ B ∗ a ′ s ∧ σ ( b ) ℓ ∈ pb ′ B ∗ } is trivial. For the other one, assume that ( a ′ , b ′ ) is in E Y ( u ) . Thus we have a ′ ub ′ ∈ L ( Y ) . Let w ∈ L ( X ) be a covering word for a ′ ub ′ . By definition of v , v is a factor of w , thus one has w = xvy for some words x, y . We then have a ′ ub ′ = a ′ sσ ( v ) pb ′ , with a ′ s a suffix of σ ( x ) and pb ′ a prefix of σ ( y ) . In particular, x and y are non-empty. Let a be the last letter of x and b be the first letter of y . We have ( a, b ) ∈ E X ( v ) and, as | s | ℓ = 0 , s is a proper suffix of σ ( a ) ,from which we have σ ( a ) ∈ B ∗ a ′ s . We also have that p is a prefix of σ ( b ) . If it is strict, then pb ′ is a prefix of σ ( b ) so that σ ( b ) ℓ ∈ pb ′ B ∗ . Otherwise, p = σ ( b ) , y has length at least 2 and8 ′ is the first letter of σ ( c ) , where c is such that bc is prefix of y . Otherwise stated, b ′ = ℓ andwe indeed have σ ( b ) ℓ ∈ pb ′ B ∗ . Remark 8.
Observe that, as we have seen in the previous proof (and using the same notation),as s is a proper suffix of σ ( a ) , the letter ℓ does not occur in it. As a consequence, for any a ′ ∈ E − X,s ( v ) , s is a proper suffix of σ ( a ′ ) . Whenever u and v are as in the previous lemma with | u | ℓ ≥ , the word v is called the antecedent of u under σ and u is said to be an extended image of v . Thus, the antecedent isdefined only for words containing an occurrence of the letter ℓ and an extended image alwayscontains an occurrence of ℓ . Whenever u is a bispecial factor, the next result gives additionalinformation about s and p . We first need to define the following notation. If σ : A ∗ → B ∗ isa morphism and a , a ∈ A , let us denote by s ( a , a ) (resp., p ( a , a ) ) the longest commonsuffix (resp., prefix) between σ ( a ) and σ ( a ) . Corollary 9.
Let X be a subshift over A , σ : A ∗ → B ∗ be an injective and strongly left propermorphism (with first letter ℓ ), Y the image of X under σ and v a word in L ( X ) . A word u isa bispecial extended image of v if and only if there exist ( a , b ) , ( a , b ) ∈ E X ( v ) with a = a , b = b and u = s ( a , a ) σ ( v ) p ( b , b ) . In particular, the antecedent v of a bispecial word u isbispecial.Proof. First assume that there exist ( a , b ) , ( a , b ) ∈ E X ( v ) with a = a , b = b and u = s ( a , a ) σ ( v ) p ( b , b ) . Let us fix s = s ( a , a ) and p = p ( b , b ) . Since σ is injective, σ ( a ) = σ ( a ) , hence s is a proper suffix of one of them. Thus, as σ is strongly left proper, s does not contain any occurrence of the letter ℓ . As a consequence, s is a proper suffix of both σ ( a ) and σ ( a ) . In particular, u is left special. The same reasoning shows that p is a properprefix of σ ( b i ) for some i ∈ { , } and that u is right special, hence bispecial. Furthermore, p is non-empty since it admits ℓ as a prefix. The pair ( a i , b i ) thus satisfies Proposition 7.Now assume that u is a bispecial extended image of v with u = sσ ( v ) p . Since u is bispecial,there exist ( a ′ , b ′ ) , ( a ′ , b ′ ) ∈ E Y ( u ) with a ′ = a ′ and b ′ = b ′ . From Equation 1, there exist ( a , b ) , ( a , b ) ∈ E X,s,p ( v ) such that σ ( a i ) ∈ B ∗ a ′ i s and σ ( b i ) ℓ ∈ pb ′ i B ∗ for all i ∈ { , } . We deduce that s = s ( a , a ) and a = a . As b ′ and b ′ cannot besimultaneously equal to ℓ , we deduce that b = b and that p = p ( b , b ) . Example 10.
Consider the morphism σ : a ℓabb ℓac ℓcd ℓcde ℓce and assume that v is a bispecial factor of X ⊂ { a, b, c, d, e } Z whose extension graph is bcd abcde By Corollary 9, the word v admits σ ( v ) ℓ , σ ( v ) ℓa and σ ( v ) ℓc as bispecial extended images.Using Proposition 7, their extension graphs are given in Figure 3. E ( σ ( v ) ℓ ) bacd ac E ( σ ( v ) ℓa ) bac bℓ E ( σ ( v ) ℓc ) cd ℓde Figure 3: Extension graphs of the bispecial extended images of v Assume that X σ is an S -adic subshift where the directive sequence σ = ( σ n : A ∗ n +1 →A ∗ n ) n ≥ is primitive and contains only strongly left proper injective morphisms. The directivesequence σ being primitive, the sequence (min a ∈A n | σ [1 ,n ) ( a ) | ) n ≥ goes to infinity. Hence,iterating Proposition 7, with any word u ∈ L ( X σ ) one can associate a unique finite sequence ( u , u , . . . , u k ) such that u = u , u k ∈ L ( X ( k ) σ ) does not have any antecedent under σ k and,for i < k , u i +1 ∈ L ( X ( i +1) σ ) is the antecedent of u i under σ i . We say that u is a descendant ofeach u i , ≤ i ≤ k , and, reciprocally, that each u i , ≤ i ≤ k , is an ancestor of u . The word u k ∈ L ( X ( k ) σ ) is its oldest ancestor and it is either empty or a non-prefix factor of σ k ( b ) forsome letter b ∈ A k +1 . Observe that with our definition, u is an ancestor and a descendant ofitself.Let X be an S -adic subshift with a strongly left proper and injective S -adic representation σ . Let u be a bispecial factor of X . From Corollary 9, all ancestors of u are bispecial factorsof some X ( k ) σ . From Proposition 7, the extensions of u are completely governed by those of itsoldest ancestor. More precisely, we have the following direct corollary. Corollary 11.
For all k ≥ , there is a finite number of bispecial factors of X ( k ) σ that do nothave an antecedent under σ k . They are called initial bispecial factors of order k . Furthermore,for any bispecial factor u of X , there is a unique k ≥ and a unique initial bispecial factor v ∈ L ( X ( k ) σ ) such that u is a descendant of v . Finally, E X ( u ) depends only on E X ( k ) σ ( v ) , i.e.,if Y is a subshift such that E Y ( v ) = E X ( k ) σ ( v ) and if Z is the image of Y under σ [1 ,k ) , then E Z ( u ) = E X ( u ) . Proposition 7 shows that whenever v is the antecedent of u under σ , the extension graph E Y ( u ) is the image under a graph morphism of a subgraph of E X ( v ) (where by subgraph we mean10he subgraph generated by a subset of edges). In particular, if E − Y ( u )) = E − X ( v )) and E + Y ( u )) = E + X ( v )) , then E Y ( u ) and E X ( v ) are isomorphic. In this section, we formalizethis observation and study the behavior of a tree structure when we consider the extensiongraphs of bispecial extended images.In this section, X is a subshift over A , σ : A ∗ → B ∗ an injective and strongly left propermorphism (with first letter ℓ ), Y the image of X under σ and v a bispecial word in L ( X ) . ByCorollary 9, the bispecial extended images of v under σ are the words u of the form sσ ( v ) p where s = s ( a , a ) and p = p ( b , b ) for some ( a , b ) , ( a , b ) ∈ E X ( v ) such that a = a and b = b . For any such σ and v , we introduce the following notations: S ( σ ) = { s ( a , a ) | a , a ∈ A , a = a } ; P ( σ ) = { p ( b , b ) | b , b ∈ A , b = b } ; S v ( σ ) = { s ( a , a ) | a , a ∈ E − X ( v ) , a = a } ; P v ( σ ) = { p ( b , b ) | b , b ∈ E + X ( v ) , b = b } ; S v ( σ ) = S v ( σ ) ∪ σ ( E − X ( v )); P v ( σ ) = P v ( σ ) ∪ σ ( E + X ( v )) ℓ. The prefix strict order (resp., suffix strict order) defines a tree structure on P v ( σ ) (resp., on S v ( σ ) ), where the root p (resp., s ) is the smallest word of the set. In particular, E − X,s ( v ) = E − X ( v ) and E + X,p ( v ) = E + X ( v ) . Furthermore, the leafs of P v ( σ ) (resp., S v ( σ ) ) are exactly theelements of σ ( E + X ( v )) ℓ (resp., σ ( E − X ( v )) ) and every internal node (i.e., every element of P v ( σ ) or S v ( σ ) ) has at least two children.For ( s, p ) ∈ S v ( σ ) × P v ( σ ) , we define the subgraph E X,s,p ( v ) of E X ( v ) whose vertices arethose involved by the edges in E X,s,p ( v ) . Observe that the sets of left and right vertices of E X,s,p ( v ) are respectively included in E − X,s ( v ) and E + X,p ( v ) . Furthermore, if s ′ ∈ S σ ( v ) (resp., p ′ ∈ P σ ( v ) ) is a child of s (resp., of p ), then E X,s ′ ,p ( v ) (resp., E X,s,p ′ ( v ) ) is a subgraph of E X,s,p ( v ) . Example 12.
Using the notations from Example 10, we have the following tree structures on S v ( σ ) and P v ( σ ) : S v ( σ ) εℓaℓab ℓc ℓcd P v ( σ ) ℓℓaℓabℓ ℓaℓ ℓcℓcℓ ℓcdℓ ℓceℓ These structures help us understand the construction of the extension graphs of Figure 3. Let s ∈ S v ( σ ) and p ∈ P v ( σ ) be such that u = sσ ( v ) p . The extension graph of u can be obtainedfrom the extension graph of v as follows:1. Start by selecting the elements of E − X,s ( v ) and E + X,p ( v ) (see Figure 4). These elementsare the letters such that the corresponding leaf in S v ( σ ) (resp., P v ( σ ) ) is in the subtreewith root s (resp., p ). . Take the subgraph of E X ( v ) with only the vertices which are in these two sets and removethe isolated vertices that were created. This gives the graph E X,s,p ( v ) (see Figure 5).3. For any letter a ∈ A , merge the vertices b on the left side such that σ ( b ) ∈ A ∗ as intoa new left vertex labeled by a . In other words, for any left vertex b , map it to the leftvertex labeled by the letter a such that the leaf corresponding to b is in the subtree whoseroot is the only child of s ending by as . Do the same on the right side with the vertices b such that σ ( b ) ℓ ∈ pa A ∗ (see Figure 6). σ ( v ) ℓ σ ( v ) ℓa σ ( v ) ℓcE − X,ε ( v ) = { a, b, c, d } E − X,ε ( v ) = { a, b, c, d } E − X,ε ( v ) = { a, b, c, d } E + X,ℓ ( v ) = { a, b, c, d, e } E + X,ℓa ( v ) = { a, b } E + X,ℓc ( v ) = { c, d, e } Figure 4: Step 1 E X,ε,ℓ ( v ) E X,ε,ℓa ( v ) E X,ε,ℓc ( v ) abcd abcde abc ab cd cde Figure 5: Step 2Left side Right side σ ( v ) ℓ a bb ac cd d a ab ac cd ce cσ ( v ) ℓa a bb ac c a bb lσ ( v ) ℓc c cd d c ℓd de e Figure 6: Step 3The next result gives a more formal description of this construction and directly followsfrom Equation (1). Observe that for right extensions, we have to split the cases because if p ′ ∈ σ ( E + X ( v )) ℓ , then the set E + X,p ′ ( v ) is empty.12 roposition 13. If ( s, p ) ∈ S v ( σ ) × P v ( σ ) is such that u = sσ ( v ) p is an extended imageof v , the extension graph of u is the image of E X,s,p ( v ) under the graph morphism ϕ v,s,p : E X,s,p ( v ) → E Y ( u ) that • for every child s ′ of s in S v ( σ ) , maps all vertices of E − X,s ′ ( v ) to the left vertex with label a ∈ B such that s ′ ∈ B ∗ as ; • for every child p ′ ∈ P v ( σ ) of p , maps all vertices of E + X,p ′ ( v ) to the right vertex with label b ∈ B such that p ′ ∈ pb B ∗ ; • for every child p ′ ∈ σ ( E + X ( v )) ℓ of p , if p ′ = σ ( a ) ℓ , maps a to the right vertex with label b ∈ B such that p ′ ∈ pb B ∗ .In particular, if S v ( σ ) = { s } and P v ( σ ) = { p } , then v has a unique bispecial extended image u and the associated morphism ϕ v,s,p is an isomorphism. Observe that the morphism ϕ v,s,p of the previous result acts independently on the leftand right vertices of E X,s,p ( v ) , i.e. it can be seen as the composition of two commuting graphmorphisms ϕ ( L ) v,s and ϕ ( R ) v,p , where ϕ ( L ) v,s acts only on the left vertices of E X,s,p ( v ) and ϕ ( R ) v,p actsonly on the right vertices of E X,s,p ( v ) . Furthermore, if a letter a belongs to E − X,s ( v ) ∩ E − X,s ( v ′ ) (resp., to E + X,p ( v ) ∩ E + X,p ( v ′ ) ) then ϕ ( L ) v,s ( a ) = ϕ ( L ) v ′ ,s ( a ) (resp., ϕ ( R ) v,p ( a ) = ϕ ( R ) v ′ ,p ( a ) ). Thus we candefine partial maps ϕ ( L ) s , ϕ ( R ) p : A → B by ϕ ( L ) s ( a ) = ϕ ( L ) ε,s ( a ) whenever a ∈ E − X,s ( ε ) and by ϕ ( R ) p ( a ) = ϕ ( R ) ε,p ( a ) whenever a ∈ E + X,p ( ε ) . In this section, we use the results and notations of the previous section to understand underwhich conditions a dendric bispecial factor only has dendric bispecial extended images undersome morphism. We then characterize the morphisms for which every dendric bispecial factoronly has dendric bispecial extended images.
Proposition 14.
Let X be a subshift over A and σ : A ∗ → B ∗ be an injective and stronglyleft proper morphism. If v ∈ L ( X ) is a dendric bispecial factor, then all the bispecial extendedimages of v under σ are dendric if and only if the following conditions are satisfied1. for every s ∈ S v ( σ ) \ { s } , E X,s,p ( v ) is a tree;2. for every p ∈ P v ( σ ) \ { p } , E X,s ,p ( v ) is a tree.Proof. Let us first assume that every bispecial extended image of v is dendric. We show item 2,the other one being symmetric. Consider p ∈ P v ( σ ) . The graph E X,s ,p ( v ) is a subgraph of E X ( v ) , which is a tree. Thus, E X,s ,p ( v ) is acyclic. Let us show that it is connected. As thereis no isolated vertex, it suffices to show that for all distinct right vertices b , b of E X,s ,p ( v ) ,there is a path in E X,s ,p ( v ) from b to b . Assume by contrary that there exist two rightvertices b , b of E X,s ,p ( v ) that are not connected.For i ∈ { , } , let A i denote the set of left vertices of E X,s ,p ( v ) that are connected to b i .Consider a maximal word s ′ (for the suffix order) in { s ( a , a ) | a ∈ A , a ∈ A } . Let also B i , i ∈ { , } , denote the set of right vertices E X,s ′ ,p that are connected to vertices of A i andconsider a maximal word p ′ (for the prefix order) in { p ( b ′ , b ′ ) | b ′ ∈ B , b ′ ∈ B } . We claimthat the extension graph of the bispecial extended image u = s ′ σ ( v ) p ′ is not connected.13et Y be the image of X under σ and let ϕ be the morphism from E X,s ′ ,p ′ ( v ) to E Y ( u ) givenby Proposition 13. By maximality of s ′ and p ′ , the morphism ϕ identifies two left vertices a, a ′ (resp., right vertices b, b ′ ) only if they belong to the same A i (resp., B i ). This impliesthat E Y ( u ) is not connected, which is a contradiction.Let us now show that, under the hypothesis 1 and 2, any bispecial extended image of v isdendric. By Corollary 9, the bispecial extended images of v are of the form sσ ( v ) p ∈ L ( Y ) where s is in S v ( σ ) and p is in P v ( σ ) .For any such pair ( s, p ) , we first show that the graph E X,s,p ( v ) is a tree. It is triviallyacyclic as it is a subgraph of E X,s,p ( v ) , which is assumed to be a tree. Let us show that it isconnected. Let b , b be right vertices of E X,s,p ( v ) . As E X,s,p ( v ) is a subgraph of both E X,s ,p ( v ) and E X,s,p ( v ) which are trees, there exist a path q in E X,s ,p ( v ) and a path q ′ in E X,s,p ( v ) connecting b and b . In particular, the right vertices occurring in q belong to E + X,p ( v ) and theleft vertices occurring in q ′ belong to E − X,s ( v ) . As E X,s ,p ( v ) and E X,s,p ( v ) are both subgraphsof E X,s ,p ( v ) , which is a tree, the paths q and q ′ coincide. It means that this path only goesthrough left vertices belonging to E − X,s ( v ) and through right vertices belonging to E + X,p ( v ) .This implies that it is a path of E X,s,p ( v ) , hence that b and b are connected in E X,s,p ( v ) .We now show that if u = sσ ( v ) p is an extended image of v , then it is dendric. ByProposition 13, E Y ( u ) is the image of E X,s,p ( v ) under some graph morphism ϕ , hence itis connected. We proceed by contradiction to show that it is acyclic. Assume that c =( a , b , a , b , . . . , a n , b n , a ) , n ≥ , is a non-trivial cycle in E Y ( u ) , where a , . . . , a n are leftvertices and b , . . . , b n are right vertices. Again by Proposition 13, for every i ≤ n , there exist• a child s i of s in S v ( σ ) ;• a child p i of p in P v ( σ ) ;• left vertices a ′ i , a ′′ i of E X,s,p ( v ) , belonging to E − X,s i ( v ) ;• right vertices b ′ i , b ′′ i of E X,s,p ( v ) , belonging to E + X,p i ( v ) ;such that ϕ ( a ′ i ) = ϕ ( a ′′ i ) = a i , ϕ ( b ′ i ) = ϕ ( b ′′ i ) = b i and such that the pairs ( a ′ j , b ′ j ) , ( a ′′ j +1 , b ′′ j ) , j < n, ( a ′ n , b ′ n ) , ( a ′′ , b ′′ n ) , are edges of E X,s,p ( v ) .Observe that as E X,s,p ( v ) is a tree, for each i ≤ n there is a unique simple path q i from a ′′ i to a ′ i and a unique simple path q ′ i from b ′ i to b ′′ i . As E X,s i ,p ( v ) and E X,s,p i ( v ) are sub-trees of E X,s,p ( v ) , the path q i is a path of E X,s i ,p ( v ) and the path q ′ i is a path of E X,s,p i ( v ) (otherwisethis would contradict the fact that E X,s,p ( v ) is acyclic). In particular, each path ϕ ( q i ) (resp., ϕ ( q ′ i ) ) reduces to the length- path a i (resp., b i ). In E X,s,p ( v ) , we thus have the cycle c ′ = ( a ′′ , . . . , a ′ | {z } q , b ′ , . . . , b ′′ | {z } q ′ , a ′′ , . . . , a ′ | {z } q , b ′ , . . . , b ′′ | {z } q ′ , . . . , a ′′ n , . . . , a ′ n | {z } q n , b ′ n , . . . , b ′′ n | {z } q ′ n , a ′′ ) . It is not trivial as its image by ϕ reduces to the non-trivial cycle c of E Y ( u ) . This contradictsthe fact that E X,s,p ( v ) is a tree, which ends the proof.14e say that a strongly left proper morphism σ is dendric preserving for v ∈ L ( X ) if allbispecial extended images of v under σ are dendric. We let DP X ( v ) denote the set of suchmorphisms, or simply DP( v ) when the context is clear. Remark 15.
Due to the previous result, we have the following properties.1. If S v ( σ ) = { s } , then σ is dendric preserving for v if and only if for every p ∈ P v ( σ ) , E X,s ,p ( v ) is a tree. This is in particular the case if E − X ( v )) = 2 .2. If P v ( σ ) = { p } , then σ is dendric preserving for v if and only if for every s ∈ S v ( σ ) , E X,s,p ( v ) is a tree. This is in particular the case if E + X ( v )) = 2 . The previous result is illustrated in the preceding examples. Indeed, for σ and v as inExample 10, we have S v ( σ ) = { s } = { ε } and P v ( σ ) = { ℓ, ℓa, ℓc } . The extension graphs E X,ε,ℓa ( v ) and E X,ε,ℓc ( v ) in Figure 5 being trees, v has only dendric bispecial extended images,as already observed in Figure 3.Another consequence of Proposition 14 is given by the following corollary. Corollary 16.
Let σ be an injective and strongly left proper morphism on the alphabet A .The following properties are equivalent.1. The sets S ( σ ) and P ( σ ) both only contain one element.2. For any subshift X on the alphabet A and any dendric bispecial factor v ∈ L ( X ) , σ isdendric preserving for v .Proof. If S ( σ ) = { s } and P ( σ ) = { p } , the fact that σ is dendric preserving for any dendricbispecial word v is a direct consequence of Proposition 14 since S v ( σ ) ⊂ S ( σ ) and P v ( σ ) ⊂P ( σ ) .To prove the other implication, let us assume that S ( σ ) contains at least two elements s and s and show that there exist a subshift X and a dendric bispecial factor v ∈ L ( X ) such that σ is not dendric preserving for v . We can assume that s is not a suffix of s . Bydefinition of S ( σ ) , there exist three different letters a, b, c ∈ A such that s = s ( a, b ) and s = s ( a, c ) . Let a , . . . , a n be the elements of A \ { a, b, c } . Let us denote by X the subshift coding theregular interval exchange transformation represented below (for precise definitions and moredetails about interval exchanges, see Subsection 6.2). c b a a . . . aa c b a . . . a The extension graph of the word ε is given by acba . . .a cba a . . .a
15t is a tree thus ε is bispecial dendric. However, the graph E X,s ,p ( ε ) is not connected as itdoes not contain the vertex c on the left but both a and b are left vertices. By Proposition 14, σ is not planar preserving for ε . This proves that S ( σ ) must contain exactly one element.Similarly, P ( σ ) also contains one element.The previous result characterizes injective and strongly left proper morphisms for whichevery dendric bispecial factor has only dendric bispecial extended images. However, the con-dition does not imply that the image of a dendric subshift by such a morphism σ is againdendric. Indeed, the result gives information only on the bispecial factors that are extendedimages under σ , i.e., that have an antecedent. The next result characterizes those morphisms σ for which even the new initial bispecial factors are dendric. For any letter a ∈ A , let α a and ¯ α a denote the so-called Arnoux-Rauzy morphisms α a ( b ) = ( a if b = a,ab otherwise, ¯ α a ( b ) = ( a if b = a,ba otherwise. Proposition 17.
The injective and strongly left proper morphisms preserving dendricity, i.e.such that the image of any dendric subshift is a dendric subshift, are exactly the morphisms ¯ α a ◦ · · · ◦ ¯ α a n ◦ α ℓ ◦ π for any n ≥ , any a , . . . , a n ∈ A \ { ℓ } and any permutation π of A .Proof. Using Corollary 16, an injective and strongly left proper morphism σ for the letter ℓ preserves dendricity if and only if the following conditions are satisfied:1. the sets S ( σ ) and P ( σ ) both only contain one element;2. if F σ is the set F σ = Fac ( { σ ( a ) ℓ : a ∈ A} ) , then any word w ∈ F σ such that | w | ℓ = 0 is dendric in F σ .Let σ = ¯ α a ◦ · · · ◦ ¯ α a n ◦ α ℓ ◦ π for a , . . . , a n ∈ A \ { ℓ } . It is easily verified that σ is injectiveand strongly left proper for the letter ℓ . As conditions 1 and 2 only depend on the set σ ( A ) ,one can assume that π = id . For any letter c , ℓc is a prefix of α ℓ ( c ) ℓ thus ( ¯ α a ◦ · · · ◦ ¯ α a n ( ℓ )) c is a prefix of σ ( c ) ℓ . This shows that P ( σ ) = { ¯ α a ◦ · · · ◦ ¯ α a n ( ℓ ) } . Similarly, cℓ is a suffix of ℓ ¯ α ℓ ( c ) thus c ( α a ◦ · · · ◦ α a n ( ℓ )) is a suffix of ℓ ( α a ◦ · · · ◦ α a n ◦ ¯ α ℓ ( c )) = ( ¯ α a ◦ · · · ◦ ¯ α a n ◦ α ℓ ( c )) ℓ = σ ( c ) ℓ S ( σ ) only contains one element and σ satisfies the condition 1. To prove condition 2, letus proceed by induction on n . If n = 0 , then the only bispecial word is ε and it is easy toverify that it is dendric. If σ ′ = ¯ α a ◦ · · · ◦ ¯ α a n ◦ α ℓ satisfies condition 2, then simple adaptionsof Proposition 7 and of Proposition 14 tell us that, as ¯ α a is strongly right proper for the letter a and the images of letters by ¯ α a have a unique longest common suffix and a unique longestprefix, it suffices to prove that the words w ∈ F σ such that | w | a = 0 are dendric. These wordsare the elements of { ε } ∪ A \ { a } , of which only ε is bispecial. The conclusion follows.Let us now assume that σ ′ is a strongly left proper morphism for the letter ℓ which satisfiesconditions 1 and 2. As S ( σ ′ ) = { s } , for any letter a ∈ A , as is suffix of some σ ′ ( b ) . Inparticular, for a = ℓ , this implies that there exists b ∈ A such that ℓs = σ ′ ( b ) and that, forany letter c = b , σ ′ ( c ) is strictly longer than σ ′ ( b ) . Similarly, p ℓ is prefix of some σ ( b ′ ) ℓ thus σ ′ ( b ′ ) = p and, as σ ′ ( b ′ ) must be strictly shorter than any other σ ′ ( a ) , we obtain b = b ′ and p = ℓs . We can thus assume that σ ′ = σ ◦ π where π is a permutation of A such that ℓs a is a prefix of σ ( a ) ℓ for all a ∈ A . In particular, σ ( ℓ ) = ℓs . We have F σ = F σ ′ . By construction, for any prefix u of s ℓ and any suffix v of ℓs , E − F σ ( u ) = A and E + F σ = A . In particular, if s ℓ ∈ a A ∗ and ℓs ∈ A ∗ b , then E F σ ( ε ) = ( A × { a } ) ∪ ( { b } × A ) because ε is dendric in E F σ . Thus, any occurrence of c = b in F σ can only be followed by anoccurrence of a . Let us use this observation to prove that σ = ¯ α a ◦ · · · ◦ ¯ α a n ◦ α ℓ for someletters a , . . . , a n ∈ A \ { ℓ } . If s = ε , then a = b = ℓ thus σ ( c ) = ℓc for all c ∈ A \ { ℓ } and σ = α ℓ . If s is not empty, then a is the first letter of s and b the last one. In particular, a and b cannot be the letter ℓ and, as s ℓ is an element of F σ , a cannot only be followed byoccurrences of a thus a must be equal to b . For any letter c ∈ A , we know that σ ( c ) beginswith ℓ and that any letter d = a in σ ( c ) is followed by an a thus σ ( c ) ∈ ℓa ( { a } ∪ ( A \ { ℓ } ) a ) ∗ . Let us define the morphism τ such that σ ( c ) = ¯ α a ◦ τ ( c ) . This morphism is unique and τ ( c ) is obtained by removing an occurrence of a after each letter d = a in σ ( c ) . By construction, τ is injective and strongly left proper for the letter ℓ . Inaddition, s begins and ends with the letter a thus there exists s ′ ∈ A ∗ such that a ¯ α a ( s ′ ) = s . It is easy to check that, as ℓs c is a prefix of σ ( c ) ℓ , ℓs ′ c is a prefix of τ ( c ) ℓ and that, as cs isa suffix of σ ( c ) , cs ′ is a suffix of τ ( c ) for all c ∈ A . Thus, S ( τ ) and P ( τ ) both contain onlyone element and τ satisfies the condition 1. In addition, for all w ∈ F τ and all c, d ∈ A cwd ∈ F τ ⇔ ca ¯ α a ( w ) d ∈ F σ . c = a and, for c = a , it derives from the fact that a = ℓ thus, if awd ∈ F τ , then there exists c ′ such that c ′ awd ∈ F τ . As a consequence, the extensiongraph of w in F τ is the same as the extension graph of a ¯ α a ( w ) in F σ and τ satisfies thecondition 2. By construction, we have | s ′ | < | s | thus we can conclude by iterating the proofon τ . In Section 3, we showed that any minimal dendric subshift X over the alphabet A is S -adic with S a set of tame automorphisms of F A , a directive sequence of X being given byTheorem 4. In this section, we give an S -adic characterization of minimal dendric subshiftsover the alphabet A = { , , } . More precisely, we strengthen Theorem 4 by exhibiting a set S (several choices are possible) and by characterizing which are the sequences ( σ n ) n ≥ ∈ S N that are a directive sequence of a minimal dendric subshift. Assume that a, b, c are such that A = { a, b, c } . Let us start with an example. Assume that X is a minimal dendric subshift over A and that the extension graph of ε in X is abc abc The associated Rauzy graph G ( X ) is ab c From it, we deduce that• the return words to a are a , ab and acb ;• the return words to b are of the form ba k or ba k c with k ≥ ;• the return words to c belong to c ( ba + ) + .An additional restriction concerning the powers of a occurring in the return words to b can bededuced from the fact that X is dendric. We claim that if ba k or ba k c is a return word to b ,then the other return words can not be of the form ba ℓ or ba ℓ c for some ℓ such that | k − ℓ | ≥ .Indeed, if for instance both ba k , k ≥ , and ba ℓ , ℓ ≥ k + 2 , are return words, then by definitionof return words, the words ba k b, ba ℓ b belong to L ( X ) . This implies that there is a cycle in theextension graph of a k , contradicting the fact that X is dendric. Therefore, the set of returnwords to b is one of the following for some k ≥ : { ba k , ba k +1 , ba k c } , { ba k , ba k +1 , ba k +1 c } , { ba k c, ba k +1 c, ba k } , { ba k c, ba k +1 c, ba k +1 } . c are less easily described. Since R ( c ) ⊂ c ( ba + ) + and R ( c )) = 3 , the set R ( c ) is determined by three sequences ( k ( j ) i ) ≤ i ≤ n j , j ∈ { , , } such that R ( c ) = { cba k ( j )1 ba k ( j )2 · · · ba k ( j ) nj | j ∈ { , , }} . Similar arguments show that there must exist k ≥ such that { k ( j ) i | j ∈ { , , } , ≤ i ≤ n j } = { k, k + 1 } , but precisely describing the three sequences ( k ( j ) i ) ≤ i ≤ n j , j ∈ { , , } , ismuch more tricky. The main reason for this difference is that the letter c is not left special in X . If u is the smallest left special factor having c as a suffix, then writing u = vc , we have v R ( c ) = R ( u ) v . To better understand the possible sequences ( k ( j ) i ) ≤ i ≤ n j , we thus need theRauzy graph of order | u | of X and not just G ( X ) .With the notation of Theorem 4, any choice of sequence of letters ( a n ) n ≥ leads to adirective sequence of X . Consequently, in the sequel we will only consider return words to leftspecial letters with the “simplest” return words. In other words, if the extension graph of theempty word in X is as in the previous example, we will only consider the coding morphismsassociated with the left special letter a .The possible extension graphs of the empty word for minimal dendric subshifts on A aregiven in Figures 7 and 8. They must satisfy two conditions: E ( ε ) must be a tree and theassociated Rauzy graph G ( X ) must be strongly connected (by minimality of X ). We alwaysassume that a is a left special letter (recall that a represent any letter in A = { , , } ) andwe present these associated Rauzy graph of order 1 as well as coding morphisms associatedwith R ( a ) . Whenever some power appear in an image, we always have k ≥ . The reasonwhy we only have k and k + 1 as exponent is the same as in the previous example: a biggerdifference would contradict dendricity by inducing a cycle in some extension graph. From thislist, we consider the set S = { α a , β abc , γ abc , δ ( k ) abc , ζ ( k ) abc , η abc | { a, b, c } = A , k ≥ } and obtain the following result. abc abc abc abc abc abcab c ab c ab cα a : a ab abc ac β abc : a ab abc acb γ abc : a ab abc abc Figure 7: The case with a unique left special letter and/or a unique right special letter19 bc abc abc abc abc abcab c ab c ab cδ ( k ) abc : a ab abc k c abc k +1 ζ ( k ) abc : a abb ac k c ac k +1 η abc : a abb abcc ac Figure 8: The cases with two left special letters and two right special letters
Proposition 18.
Any ternary minimal dendric subshift X has a primitive S -adic represen-tation σ and, for each such representation, X ( n ) σ is a ternary minimal dendric subshift foreach n .Proof. For the existence, we consider the construction of a directive sequence following The-orem 4 where at each step, we choose the left special letter a n for which σ n is in S . UsingTheorem 4 and Theorem 5, we obtain that for each S -adic representation σ and each n , X ( n ) σ is a ternary minimal dendric subshift. Assume that X is a minimal subshift over A and that v ∈ L ( X ) is a bispecial factor and let Y be the image of X under some injective and strongly left proper morphism σ : A ∗ → B ∗ .Recall from Section 4.2 (and, in particular, Proposition 13) that the extension graph of anybispecial extended image of v under σ is the image under two consecutive graph morphisms ϕ ( L ) s and ϕ ( R ) p of a subgraph of E X ( v ) . In this section, we determine those graph morphismswhen σ is a morphism in S and we give necessary and sufficient conditions on the extensiongraph of v ∈ L ( X ) so that v only has dendric bispecial extended images. In this particularcase, since the alphabet has cardinality 3, P ( σ ) and S ( σ ) have cardinality at most 2. Tables 1and 2 define the (possibly partial) maps ϕ ( L ) s , ϕ ( R ) p : A → A associated with each morphism σ ∈ S .A direct application of Proposition 14 shows that whenever v is dendric, then v has onlydendric bispecial extended images if and only if the following conditions are satisfied:1. either S v ( σ ) = { s } , or both S v ( σ ) = { s , s } and E X,s,p ( v ) is a tree;2. either P v ( σ ) = { p } , or both P v ( σ ) = { p , p } and E X,s ,p ( v ) is a tree.We first give a handier interpretation of these conditions. Observe that for convenience, weactually characterize the dendric bispecial factors v ∈ L ( X ) that have a non-dendric bispecialextended images. When considering a letter a ∈ A as a vertex of E ( v ) , we respectively write a L or a R to emphasize that a is considered as a left or right vertex. For a ∈ A and v ∈ L ( X ) ,we define the following conditions on v : 20 S ( σ ) ϕ ( L ) s P ( σ ) ϕ ( R ) p α a { ε } id { a } id β abc { ε, b } ϕ ( L ) ε : ( a ab, c b { a } id ϕ ( L ) b : ( b ac cγ abc { ε } id { a, ab } ϕ ( R ) a : ( a ab, c bϕ ( R ) ab : ( b ac c Table 1: Definition of the graph morphisms ϕ ( L ) s and ϕ ( R ) p associated with the morphisms α a , β abc and γ abc C L ( a ) : the subgraph of E X ( v ) obtained by deleting a L and the induced isolated vertices(if any) is not connected;2. C R ( a ) : the subgraph of E X ( v ) obtained by deleting a R and the induced isolated vertices(if any) is not connected. Remark 19. If v ∈ L ( X ) is a dendric factor, then in the condition C i ( a ) , the consideredsubgraph is not connected if and only if a i has at least two neighbors that are not leaves, i.e.,that have degree at least 2. In particular, v is bispecial. Observe also that as E ( v ) is a tree,this implies that the i -side of E ( v ) contains three vertices. Hence, another equivalent conditionof C i ( a ) when E X ( v ) is a tree is that, writing A = { a, b, c } , the path from b i to c i has length4. Proposition 20.
Let X be a subshift over A , σ be a morphism in S and Y be the image of X under σ .If v ∈ L ( X ) is a dendric bispecial factor, then v has a non-dendric bispecial extended imageunder σ if and only if there is a letter a ∈ A for which one of the following conditions issatisfied:1. v satisfies C L ( a ) and σ ∈ { β abc , δ ( k ) abc , ζ ( k ) abc , η abc | k ≥ } ;2. v satisfies C R ( a ) and σ ∈ { γ abc , δ ( k ) abc , ζ ( k ) abc , η bca | k ≥ } .Proof. The negation of item 2 of Proposition 14 is equivalent to “there exists a ∈ E + X ( v ) such that E X,s ,p ( b,c ) ( v ) is not a tree”. As E X,s ,p ( b,c ) ( v ) is a subgraph of E ( v ) , it is acyclicand, if p ( b, c ) is a prefix of σ ( a ) it is also connected (indeed, we then have E X,s ,p ( b,c ) ( v ) = E X,s ,p ( v ) = E ( v ) ). Thus, the second condition of Proposition 14 is not satisfied if and only ifthere exists a permutation { a, b, c } of A such that p ( b, c ) is not a prefix of σ ( a ) and v satisfies C R ( a ) . Using Figures 7 and 8, we see that it is equivalent to condition 2. We proceed in asimilar way to show that the first condition of Proposition 14 is not satisfied if and only if thecondition 1 above is. Example 21.
Assume that v is a dendric bispecial factor in some minimal ternary subshift X with extension graph S ( σ ) ϕ s P ( σ ) ϕ p δ ( k ) abc { ε, c k } ϕ ( L ) ε : ( a ab, c c { a, abc k } ϕ ( R ) a : ( a ab, c bϕ ( L ) c k : ( b bc c ϕ ( R ) abc k : ( b ac cζ ( k ) abc { ε, c k } ϕ ( L ) ε : ( a bb, c c { a, ac k } ϕ ( R ) a : ( a bb, c cϕ ( L ) c k : ( b ac c ϕ ( R ) ac k : ( b ac cη abc { ε, c } ϕ ( L ) ε : ( a bb, c c { a, ab } ϕ ( R ) a : ( a, b bc cϕ ( L ) c : ( b bc a ϕ ( R ) ab : ( a ab c Table 2: Definition of the graph morphisms ϕ ( L ) s and ϕ ( R ) p associated with the morphisms δ ( k ) abc , ζ ( k ) abc and η abc abc abc Thus v satisfies C L ( c ) and C R ( a ) . If Y is the image of X under β abc , the bispecial extendedimages of v in Y are u = β abc ( v ) a and u = bβ abc ( v ) a and they have the following extensiongraphs: E Y ( u ) E Y ( u ) ab abc ac ac Similarly, if Y is the image of X under β cab , the bispecial extended images of v in Y are w = β cab ( v ) c and w = aβ cab ( v ) c and they have the following extension graphs: E Y ( w ) E Y ( w ) ac abc bc abc Assuming that v ∈ L ( X ) is a dendric bispecial factor, Proposition 20 characterizes underwhich conditions v has a non-dendric bispecial extended image under σ ∈ S or, in otherwords, under which conditions σ ∈ S is not dendric preserving for v . As we consider theternary case, the set DP( v ) of dendric preserving morphisms for v will be implicitly restricted22 α a β abc γ abc δ ( k ) abc ζ ( k ) abc η abc A L ( ε ) ∅ { a } ∅ { c } { c } { b }A R ( ε ) ∅ ∅ { a } { a } { c } { c } Table 3: Conditions for the empty wordto morphisms from S . We set A L ( v ) = { a ∈ A | v satisfies C L ( a ) } , A R ( v ) = { a ∈ A | v satisfies C R ( a ) } and, if X is a ternary dendric subshift, A L ( X ) = [ v ∈L ( X ) A L ( v ); A R ( X ) = [ v ∈L ( X ) A R ( v );DP( X ) = \ v ∈L ( X ) DP( v ) . Using Proposition 20, the sets A L ( X ) and A R ( X ) completely determine the set DP( X ) of all morphisms in S that are dendric-preserving for X , i.e. that are dendric-preserving forall v ∈ L ( X ) . In this section, we in particular show that A L ( X ) and A R ( X ) contain at mostone letter and we show that, when Y is the image of X under σ ∈ DP( X ) , A L ( Y ) (resp., A R ( Y ) ) is completely determined by A L ( X ) (resp., A R ( X ) ) and σ . The next lemma is atrivial consequence of Remark 19. Lemma 22.
Let X be a subshift over A . For every dendric bispecial factor v ∈ L ( X ) , A L ( v ) (resp., A R ( v ) ) contains at most one letter. Lemma 23.
Let X be a subshift over A which is the image under σ ∈ S of another subshift Z over A . The sets A L ( ε ) and A R ( ε ) are given in Table 3, where ε is considered as a factorof X .Proof. Indeed, the morphism σ completely determines the extension graph E X ( ε ) . The resultthus directly follows from the definition of the conditions C L ( a ) and C R ( a ) . Lemma 24. If X is a ternary dendric subshift and if Y is the image of X under somemorphism σ ∈ S , then Y is dendric if and only if σ ∈ DP( X ) .Proof. Indeed, if σ ∈ DP( X ) , every bispecial extended image of a bispecial factor of X isdendric by definition of DP( X ) . Any other bispecial factor of Y is the empty word or a non-prefix factor of an image σ ( a ) , a ∈ A (by Proposition 7). It suffices to check that any suchbispecial factor is dendric when σ belongs to S to prove that Y is dendric. Assume now that Y is dendric. If σ is not in DP( X ) then there exists v ∈ L ( X ) such that σ / ∈ DP( v ) thus v has an extended image in Y which is not dendric.We say that a morphism σ ∈ S is left-invariant (resp., right-invariant ) if S ( σ ) (resp., P ( σ ) ) is a singleton, i.e. S ( σ ) = { s } (resp., P ( σ ) = { p } ). The next lemma directly followsfrom the definition of the morphisms α a , β abc , γ abc , δ ( k ) abc , ζ ( k ) abc and η abc .23 emma 25. α a is both left-invariant and right-invariant;2. β abc is right-invariant, but not left-invariant;3. γ abc is left-invariant, but not right-invariant;4. δ ( k ) abc , ζ ( k ) abc and η abc neither are left-invariant, nor right-invariant. We let S LI and S RI respectively denote the left-invariant and right-invariant morphisms,i.e., S LI = { α a , γ abc | A = { a, b, c }} and S RI = { α a , β abc | A = { a, b, c }} . Observe that if σ ∈ S LI (resp., σ ∈ S RI ), then the associated graph morphism ϕ ( L ) s (resp., ϕ ( R ) p ) is the identity. Moreover, from Table 3, σ ∈ S LI (resp., σ ∈ S RI ) if and only if A L ( ε ) = ∅ (resp., A R ( ε ) = ∅ ). Lemma 26.
Let X be a subshift over A and v ∈ L ( X ) be a dendric bispecial factor. If σ ∈ S is not left-invariant (resp., not right-invariant), then any dendric extended image u of v is such that A L ( u ) = ∅ (resp., A R ( u ) = ∅ ).In particular, if X is dendric and σ ∈ DP( X ) is not left-invariant (resp., not right-invariant) and if Y is the image of X under σ , then A L ( Y ) = A L ( ε ) = ∅ (resp., A R ( Y ) = A R ( ε ) = ∅ ), where ε is considered as a factor of Y .Proof. Let us show the result for a non-left-invariant morphism, the other case being sym-metric. By definition of left-invariance, S ( σ ) contains two elements s and s . It suffices tocheck in Tables 1 and 2 that for each of them, the range of the associated graph morphism ϕ ( L ) s i has cardinality 2. As the left vertices of E Y ( u ) are images of the left vertices of E X ( v ) , E Y ( u ) contains at most two left vertices. From Remark 19, u thus cannot satisfies C L ( a ) forany letter a ∈ A .Now assume that σ ∈ DP( X ) . By Lemma 24, Y is a dendric subshift. Let u be a non-empty factor of Y . By Proposition 7, either u is a non-prefix factor of σ ( a ) for some letter a ∈ A , or u is an extended image of a factor v ∈ L ( X ) . In the first case, it suffices to checkthat E − Y ( u )) ≤ , which implies that A L ( u ) = ∅ . In the second case, as X is dendric, v is also dendric so by the first part of the lemma, A L ( u ) = ∅ . Thus A L ( Y ) = A L ( ε ) and, byLemma 23, it is non-empty. Lemma 27.
Let X be a subshift over A and v ∈ L ( X ) be a dendric bispecial factor. If σ ∈ DP( v ) is left-invariant (resp., right-invariant), then A L ( v ) = [ u bispecial extended image of v A L ( u ); (2) (resp., A R ( v ) = [ u bispecial extended image of v A R ( u ) ) . In particular, if X is dendric and σ ∈ DP( X ) is left-invariant (resp., right-invariant) and Y is the image of X under σ , then A L ( Y ) = A L ( X ) (resp., A R ( Y ) = A R ( X ) ).Proof. Let us assume that σ is left-invariant, the other case is symmetric. We first show thatif u is a bispecial extended image of v , then A L ( u ) ⊂ A L ( v ) . As S ( σ ) = { s } , there exists p ∈ P ( σ ) such that u = s σ ( v ) p . Assume that a ∈ A L ( u ) . Writing A = { a, b, c } , Remark 1924tates that the path q from b L to c L in E Y ( u ) has length 4. This path q is the image under ϕ s ,p of a path q ′ of length at least 4 in E X ( v ) . As E X ( v ) is a tree with at most 6 vertices andthe extremities of q ′ are left vertices, the path has length exactly 4. As ϕ ( L ) s is the identity,we conclude that q ′ is a path of length 4 from b L to c L in E X ( v ) , hence that a ∈ A L ( v ) .If A L ( v ) = ∅ , then Equality (2) is direct. If A L ( v ) = ∅ , we show that there exists exactlyone bispecial extended image u of v such that A L ( u ) = ∅ and for this one, A L ( u ) = A L ( v ) .As σ is left-invariant, we have by Lemma 25 that σ = α i or σ = γ ijk for some i, j, k such that A = { i, j, k } .If σ = α i , then as α i is also right-invariant (see Lemma 25), Proposition 13 implies that u is the unique bispecial extended image of v and that E Y ( u ) = E X ( v ) (the graph morphism isthe identity), hence that A L ( u ) = A L ( v ) .If σ = γ ijk , then as P ( σ ) = { i, ij } , (see Table 1), Corollary 9 implies that v has at mosttwo bispecial extended images u = σ ( v ) i and u = σ ( v ) ij . Let ϕ = ϕ ε,i and ϕ = ϕ ε,ij . Themorphism ϕ ( L ) ε is the identity thus we have ϕ = ϕ ( R ) i and ϕ = ϕ ( R ) ij .As A L ( v ) = ∅ , by Lemma 22 there is a letter a ∈ A such that A L ( v ) = { a } . UsingRemark 19, the path q from b L to c L has length 4 in E X ( v ) . Let us write q = ( b L , x R , a L , y R , c L ) ,with x, y ∈ A .If i ∈ { x, y } , we assume without loss of generality that i = x . Then we have ϕ ( q ) =( b L , i, a L , j, c L ) which is a path of length 4 from b L to c L in E Y ( u ) . By Remark 19 andLemma 22, one has A L ( u ) = { a } . On the other hand, i is not a right vertex of E X,ε,ij ( v ) .Thus q is not a path in E X,ε,ij ( v ) . As ϕ is a graph isomorphism that is invariant on the leftvertices, there is no path from b L to c L in E Y ( u ) . Using again Remark 19, a / ∈ A L ( u ) . Butas A L ( u ) ⊂ A L ( v ) by the first part of the proof, we get A L ( u ) = ∅ .If { x, y } = { j, k } , we assume without loss of generality that ( x, y ) = ( j, k ) . Then we have ϕ ( q ) = ( b L , i, a L , k, c L ) which is a path of length 4 from b L to c L in E Y ( u ) . By Remark 19and Lemma 22, one has A L ( u ) = { a } . On the other hand, we have ϕ ( j ) = ϕ ( k ) = j , whichshows that ( b L , j, c L ) is a path of length 2 from b L to c L in E Y ( u ) . Using again Remark 19, a / ∈ A L ( u ) . But as A L ( u ) ⊂ A L ( v ) by the first part of the proof, we get A L ( u ) = ∅ .Let us finally show that A L ( Y ) = A L ( X ) . With σ ∈ { α i , γ ijk } , the non-prefix factor of σ ( a ) , a ∈ A , are not bispecial. Hence, using Proposition 7, a bispecial factor u ∈ L ( Y ) is either empty, or a bispecial extended image of some bispecial factor v ∈ L ( X ) . As σ isleft-invariant, we have A L ( ε ) = ∅ by Lemma 23. Using Equation (2), we get A L ( Y ) = [ v ∈L ( X ) , bispecial [ u bispecial extended image of v A L ( u )= [ v ∈L ( X ) , bispecial A L ( v )= A L ( X ) , which ends the proof.The following corollary is a direct consequence of Lemmas 26 and 27. Corollary 28.
Let X be a dendric subshift over A , σ ∈ DP( X ) and Y the image of X under σ . If A L ( Y ) = ∅ , then A L ( X ) = ∅ and σ is left-invariant. Respectively, if A R ( Y ) = ∅ , then A R ( X ) = ∅ and σ is right-invariant. roposition 29. Let X be a ternary minimal dendric subshift with S -adic representation σ = ( σ n ) n ≥ . Then A L ( X ) and A R ( X ) contain at most one letter. Moreover,1. A L ( X ) = ∅ if and only if σ belongs to S N LI , if and only if X has a unique left specialfactor of each length;2. A R ( X ) = ∅ if and only if σ belongs to S N RI , if and only if X has a unique right specialfactor of each length.Proof. Using the notation of Section 2.3, we have X = X σ and for each n ≥ , X ( n +1) σ isdendric by Proposition 18 thus σ n ∈ DP( X ( n +1) σ ) by Lemma 24. We first show item 1.Assuming that A L ( X ) = ∅ , we have σ ∈ S N LI by Corollary 28.Now assuming that σ belongs to S N LI , we deduce that any bispecial factor of X is a descen-dant of the empty word in some X ( n ) σ . Using Figure 7, the bispecial factor v = ε ∈ L ( X ( n ) σ ) has a unique right extension va , a ∈ A , which is left special and it satisfies E − X ( n ) σ ( va ) = A .It then suffices to observe, using Proposition 7, that this property is preserved by taking bis-pecial extended images under some morphism σ ∈ S LI . This shows that any bispecial factor u , and hence any left special factor, of X satisfies E − X ( u ) = A . Proposition 2 then impliesthat X has a unique left special factor of each length.Finally assume that X has a unique left special factor u n of each length n . By Proposi-tion 2, we have E − X ( u n ) = A . Using Remark 19, the set A L ( u n ) is non-empty if and only ifthere are two letters x, y ∈ A such that both u n x and u n y are left special factors of X . Thisimplies that A L ( u n ) = ∅ . Any non-left-special factor u is such that A L ( u ) = ∅ by Remark 19,hence A L ( X ) = ∅ .The proof of item 2 is symmetric. We finish the proof by showing that A L ( X ) and A R ( X ) contain at most one letter. Assume by contrary that a, b ∈ A L ( X ) for some different letters a and b . By Lemmas 22, 26 and 27, all morphisms σ n are left-invariant. But then item 1 impliesthat A L ( X ) = ∅ . S -adic characterization of minimal ternary dendric subshifts By Proposition 29, any ternary minimal dendric subshift X satisfies A L ( X ) , A R ( X ) ≤ .To alleviate notations in what follows, we consider the alphabet A = A ∪ { } and we write A L ( X ) = a instead of A L ( X ) = { a } and A L ( X ) = 0 instead of A L ( X ) = ∅ (and similarlyfor A R ( X ) ). We then define the equivalence relation ∼ on the set of minimal ternary dendricsubshifts by X ∼ Y ⇔ ( A L ( X ) , A R ( X )) = ( A L ( Y ) , A R ( Y )) . For all l, r ∈ A , we let [ l, r ] denote the equivalence class of all minimal ternary dendricsubshifts satisfying ( A L ( X ) , A R ( X )) = ( l, r ) . Lemma 30.
Let X and Y be minimal ternary dendric subshifts. We have X ∼ Y if andonly if DP( X ) = DP( Y ) . Furthermore, if X ∼ Y and σ ∈ DP( X ) and if X ′ and Y ′ are therespective images of X and Y under σ , then X ′ ∼ Y ′ .Proof. The equivalence between X ∼ Y and DP( X ) = DP( Y ) follows from Proposition 20.The second part of the statement follows from Lemma 26 and Lemma 27.26sing the previous lemma, we can define, for each equivalence class C = [ l, r ] , the set DP( C ) = DP( X ) , where X ∈ C . Furthermore, for any σ ∈ DP( C ) , there is a uniqueequivalence class C ′ such that when X ∈ C and Y is the image of X under σ , then Y ∈ C ′ .We call C ′ the image of C under σ . Lemma 31.
Let C = [ l, r ] for some l, r ∈ A and let σ ∈ DP( C ) .1. If σ is left-invariant, then for all l ′ ∈ A , we have σ ∈ DP([ l ′ , r ]) .2. If σ is not left-invariant, then there is a unique l ′ ∈ A such that σ / ∈ DP([ l ′ , r ]) .3. If σ is right-invariant, then for all r ′ ∈ A , we have σ ∈ DP([ l, r ′ ]) .4. If σ is not right-invariant, then there is a unique r ′ ∈ A such that σ / ∈ DP([ l, r ′ ]) .Proof. It follows from Proposition 20 and Lemma 25.We are now ready to prove Theorem 1.
Proof of Theorem 1.
Let us first define the following directed graph G ′ . The set of verticesis the set of equivalent classes of the relation ∼ . For every vertices C, C ′ ∈ V , and everymorphism σ ∈ S , there is an edge from C to C ′ with label σ if σ ∈ DP( C ′ ) and C is theimage of C ′ under σ .Assume that X is a minimal ternary dendric subshift. By Proposition 18, X has a primitive S -adic representation σ where for each n , X ( n ) σ is a ternary minimal dendric subshift. Forevery n , X ( n ) σ is the image of X ( n +1) σ under σ n and we thus have σ n ∈ DP( X ( n +1) σ ) . Bydefinition of the graph, P = ([ A L ( X ( n ) σ ) , A R ( X ( n ) σ )]) n ≥ is a path in G ′ with label σ .Now consider a primitive sequence σ labeling a path in G ′ and let us show that the subshift X σ is minimal and dendric. It is minimal by primitiveness of σ . If it is not dendric, thereexists a bispecial factor u ∈ L ( X σ ) which is not dendric. Using Corollary 11, there is a unique k ≥ and a unique initial bispecial factor v in L ( X ( k ) σ ) such that u is a descendant of v and E X ( u ) depends only on E X ( k ) σ ( v ) . By definition of initial bispecial factors, v is either the emptyword or a non-prefix factor of σ k ( a ) for some letter a . In both cases, E X ( k ) σ ( v ) is completelydetermined by σ k and is a tree. By definition of the edges of G ′ , the morphism σ · · · σ k − isdendric-preserving for v . Thus E X ( u ) is a tree, which is a contradiction.We finally show that we can reduce G ′ to the subgraph G obtained by deleting the verticesof the form [0 , r ] and of the form [ l, . This subgraph contains 9 vertices and is representedin Figure 9. We show that for every path P in G ′ , there is a path P ′ in G with the same label.We thus consider a path P = ([ l n , r n ]) n ≥ in G ′ with label σ = ( σ n ) n ≥ (hence σ n labelsthe edge from [ l n , r n ] to [ l n +1 , r n +1 ] ). If P is not a path in G , then there exists N such that l N or r N is . Assume that N is the smallest integer such that l N = 0 . By Corollary 28, wededuce that for all n ≥ N , σ n is left-invariant and l n = 0 . Using Lemmas 26, 27 and 31, wehave that for all n ≥ N and every l ′ ∈ A , σ n is in DP([ l ′ , r n +1 ]) and labels the edge from [ l ′ , r n ] to [ l ′ , r n +1 ] . Thus for every l ′ ∈ A , the sequence P l ′ = ([ l ′ , r n ]) n ≥ N is a path in G ′ withlabel ( σ n ) n ≥ N .If N = 1 , we have found a path P ′ with label σ and that does not go through vertices ofthe form [0 , r ] . If N > , then by Lemma 27, σ N − is not left-invariant. By Lemma 31, wecan choose l ′ ∈ A such that σ N − ∈ DP([ l ′ , r N ]) . Then by Lemmas 26 and 27, σ N − labelsthe edge from [ l N − , r N − ] to [ l ′ , r N ] . We finally consider the path P ′ consisting in the prefix27f P from [ l , r ] to [ l N − , r N − ] followed by the edge from [ l N − , r N − ] to [ l ′ , r N ] , followedby the path P l ′ . This path has label σ and does not go through vertices of the form [0 , r ] .Observe that if P ′ = ([ l ′ n , r ′ n ]) n ≥ , then for all n , we have r ′ n = r n .Starting from P ′ we similarly find another path P ′′ = ([ l ′′ n , r ′′ n ]) n ≥ with label σ and suchthat for all n , l ′′ n = l ′ n = 0 and r ′′ n = 0 . This concludes the proof.The graph G represented in Figure 9 is a co-deterministic automaton, i.e., for every vertex C and every morphism σ ∈ S , there is at most one edge with label σ reaching C . This followsfrom Lemmas 23, 26 and 27. The edges are computed using Table 3, Lemmas 26 and 27 (todetermine the starting vertex of a given morphism) and Proposition 20 (to determine theallowed target vertices).This is summarized in Table 4, where for each x ∈ A , ¯ x denotes the set A \ { x } .Morphism Starting vertex Target vertex Conditions α a [ l, r ] [ l, r ] l, r ∈ ¯0 β abc [ a, r ] [ l, r ] l ∈ ¯ a , r ∈ ¯0 γ abc [ l, a ] [ l, r ] l ∈ ¯0 r ∈ ¯ aδ ( k ) abc [ c, a ] [ l, r ] l, r ∈ ¯ aζ ( k ) abc [ c, c ] [ l, r ] l, r ∈ ¯ aη abc [ b, c ] [ l, r ] l ∈ ¯ a, r ∈ ¯ c Table 4: List of edges of the graph G represented in Figure 9We now build a deterministic automaton G d which is equivalent to G (where all vertices of G are initial and final). The states of the automaton are the sets ¯ l × ¯ r = { [ x, y ] | x ∈ ¯ l, y ∈ ¯ r } , l, r ∈ A . The initial state is ¯0 × ¯0 and all states are final. For each morphism σ ∈ S , there is a transitionfrom ¯ l × ¯ r to ¯ l ′ × ¯ r ′ whenever ¯ l ′ × ¯ r ′ is the set of vertices reachable from ¯ l × ¯ r in G by reading σ . Thus the edges are as given in Table 5.Morphism Starting vertex Target vertex Conditions α a ¯ l × ¯ r ¯ l × ¯ r l, r ∈ ¯0 β abc ¯ l × ¯ r ¯ a × ¯ r l ∈ ¯ a ∪ { } , r ∈ A γ abc ¯ l × ¯ r ¯ l × ¯ a l ∈ A , r ∈ ¯ a ∪ { } δ ( k ) abc ¯ l × ¯ r ¯ a × ¯ a l ∈ ¯ c ∪ { } , r ∈ ¯ a ∪ { } ζ ( k ) abc ¯ l × ¯ r ¯ a × ¯ a l, r ∈ ¯ c ∪ { } η abc ¯ l × ¯ r ¯ a × ¯ c l ∈ ¯ b ∪ { } , r ∈ ¯ c ∪ { } Table 5: List of transitions in the deterministic automaton G d The automaton G d has 16 vertices and except for the fact that it is not complete (i.e., thesink state has been removed), it is the minimal automaton. It has four strongly connected28 ,
1] [1 ,
2] [1 , ,
1] [2 ,
2] [2 , ,
1] [3 ,
2] [3 , ββ βγ γγ α (a) Every α a labels a loop on every vertex; β abc and β acb label the vertical edges leaving anyvertex of the form [ a, x ] ; γ abc and γ acb labelthe horizontal edges leaving any vertex of theform [ x, a ] . [1 ,
1] [1 ,
2] [1 , ,
1] [2 ,
2] [2 , ,
1] [3 ,
2] [3 , (b) The morphism δ ( k ) abc labels the edges from [ c, a ] to [ x, y ] for all x, y ∈ { b, c } . Only theedges with label δ ( k )123 (solid), δ ( k )132 (dashed) and δ ( k )213 (dotted) are represented. [1 ,
1] [1 ,
2] [1 , ,
1] [2 ,
2] [2 , ,
1] [3 ,
2] [3 , (c) The morphism ζ ( k ) abc labels the edges from [ c, c ] to [ x, y ] for all x, y ∈ { b, c } . Only theedges with label ζ ( k )132 (solid) and ζ ( k )312 are rep-resented. [1 ,
1] [1 ,
2] [1 , ,
1] [2 ,
2] [2 , ,
1] [3 ,
2] [3 , (d) The morphism η abc labels the edges from [ b, c ] to [ x, y ] for all ( x, y ) ∈ { b, c } × { a, b } .Only the edges with label η (solid), η (dashed) and η (dotted) are represented. Figure 9: A subshift on A is minimal and dendric if and only if it has a primitive S -adicrepresentation labeling an infinite path in this graph denoted G . For the sake of clarity, werepresent 4 copies of the graph and show only a part of the edges in each copy, assuming that A = { a, b, c } . 29omponents that are C = { ¯0 × ¯0 }C = { ¯0 × ¯ r | r ∈ A }C = { ¯ l × ¯0 | l ∈ A }C = { ¯ l × ¯ r | l, r ∈ A } The component C thus consists in one vertex with three loops, one for each morphism α a .By Proposition 29, any minimal dendric subshift whose S -adic representation labels a paththat stays in C has exactly one left and one right special factor of each length. This class ofminimal dendric subshifts exactly corresponds to the class of Arnoux-Rauzy subshifts [AR91].The component C consists in three vertices and any edge in this component is labeled bya left-invariant morphism. This component is accessible only from C and any edge from C to C is labeled by a morphism γ abc . By Proposition 29, any minimal dendric subshift whose S -adic representation labels a path that ends in C has exactly one left and two right specialfactors of each length.The component C is symmetric to C where the involved morphisms are the right-invariantmorphisms. By Proposition 29, any minimal dendric subshift whose S -adic representationlabels a path that ends in C has exactly two left and one right special factors of each length.The component C consists in 9 vertices. It is reachable from C only by morphisms thatare neither right- nor left-invariant, from C only by morphisms that are not left-invariantand from C only by morphisms that are not right-invariant. By Proposition 29, any minimaldendric subshift whose S -adic representation labels a path that ends in C has exactly two leftand two right special factors of each length. Observe that if we restrict G d to the component C in which we consider that all states are initial, then this automaton is again equivalent tothe initial one. Thus, except for the fact that all states are initial, we have a deterministicautomaton with 9 states. Remark 32.
A very similar S -adic characterization of minimal subshifts with factor complex-ity ≤ p ( n + 1) − p ( n ) ≤ was given by the third author [Ler14]. This S -adic characterizationalso involves a graph with four strongly connected components, each one corresponding to hav-ing exactly one or two left or right special factors of each length. G of well-known families of minimal ternarydendric subshifts The class of minimal dendric ternary subshifts contains several classes of well-known familiesof subshifts, namely Arnoux-Rauzy subshifts, codings of regular 3-interval exchange transfor-mations and Cassaigne subshifts. In this section, we study the S -adic representations of theseparticular families in the light of G and G d . A subshift X ⊂ A Z is an Arnoux-Rauzy subshift if it is a minimal subshift with factorcomplexity p ( n ) = 2 n + 1 that has exactly one left and one right special factor of eachlength. Another equivalent definition is that X admits a primitive S -adic representation σ ∈ { α , α , α } N [AR91]. As already mentioned, we get the following result.30 roposition 33. A subshift X ⊂ A Z is an Arnoux-Rauzy subshift if and only if it has aprimitive S -adic characterization that labels a path in the component C of G d . Let us recall the definition of an interval exchange transformation. A semi-interval is a non-empty subset of the real line of the form [ α, β [= { z ∈ R | α ≤ z < β } . Thus it is a left-closedand right-open interval. For two semi-intervals ∆ , Γ , we denote ∆ < Γ if x < y for any x ∈ ∆ and y ∈ Γ .Let ( A , ≤ ) be an ordered set. A partition ( I a ) a ∈A of [0 , in semi-intervals is ordered if a < b implies I a < I b .Let A be a finite set ordered by two total orders ≤ and ≤ . Let ( I a ) a ∈A be a partitionof [0 , in semi-intervals ordered for ≤ . Let λ a be the length of I a . Let µ a = P b ≤ a λ b and ν a = P b ≤ a λ b . Set ξ a = ν a − µ a . The interval exchange transformation relative to ( I a ) a ∈A isthe map T : [0 , → [0 , defined by T ( z ) = z + ξ a if z ∈ I a . Observe that the restriction of T to I a is a translation onto J a = T ( I a ) , that µ a is the rightboundary of I a and that ν a is the right boundary of J a . We additionally denote by ι a the leftboundary of I a and by κ a the left boundary of J a . Thus I a = [ ι a , µ a [ , J a = [ κ a , ν a [ . Note that a < b implies ν a < ν b and thus J a < J b . This shows that the family ( J a ) a ∈A is apartition of [0 , ordered for < . In particular, the transformation T defines a bijection from [0 , onto itself.An interval exchange transformation relative to ( I a ) a ∈A is also said to be on the alphabet A . The values ( ξ a ) a ∈A are called the translation values of the transformation T . Example 34.
Let R be the interval exchange transformation corresponding to A = { a, b } , a < b , b < a , I a = [0 , − α [ , I b = [1 − α, . The transformation R is the rotation of angle α on the semi-interval [0 , defined by R ( z ) = z + α mod 1 . a − α b b α a Since ≤ and ≤ are total orders, there exists a unique permutation π of A such that a < b if and only if π ( a ) < π ( b ) . Conversely, ≤ is determined by ≤ and π and ≤ isdetermined by ≤ and π . The permutation π is said to be associated with T .If we set A = { a , a , . . . , a s } with a < a < · · · < a s , the pair ( λ, π ) formed by thefamily λ = ( λ a ) a ∈A and the permutation π determines the map T . We will also denote T as T λ,π . The transformation T is also said to be an s -interval exchange transformation. Example 35. A -interval exchange transformation where the associated permutation is thecycle π = ( abc ) . a b c b c a .2.1 Regular interval exchange transformations The orbit of a point z ∈ [0 , is the set { T n ( z ) | n ∈ Z } . The transformation T is said to be minimal if, for any z ∈ [0 , , the orbit of z is dense in [0 , .Set A = { a , a , . . . , a s } with a < a < . . . < a s , µ i = µ a i and κ i = κ a i . The points , µ , . . . , µ s − form the set of separation points of T , denoted Sep( T ) . Note that the singularpoints of the transformation T (that is the points z ∈ [0 , at which T is not continuous) areamong the separation points but that the converse is not true in general (see Example 35).An interval exchange transformation T λ,π is called regular if the orbits of the nonzeroseparation points µ , . . . , µ s − are infinite and disjoint. Note that the orbit of cannot bedisjoint of the others since one has T ( µ i ) = 0 for some i with ≤ i ≤ s − . A regular intervalexchange transformation is also said to be without connections or to satisfy the idoc condition(where idoc stands for infinite disjoint orbit condition). Note that since κ = T ( µ ) , . . . , κ s = T ( µ s − ) , T is regular if and only if the orbits of κ , . . . , κ s are infinite and disjoint. As anexample, the -interval exchange transformation of Example 34 which is the rotation of angle α is regular if and only if α is irrational. The following result is due to Keane. Theorem 36 (Keane [Kea75]) . A regular interval exchange transformation is minimal.
The converse is not true. Indeed, consider the rotation of angle α with α irrational, as a -interval exchange transformation with λ = (1 − α, α, α ) and π = (132) . The transformationis minimal as any rotation of irrational angle but it is not regular since µ = 1 − α , µ = 1 − α and thus µ = T ( µ ) .The following necessary condition for minimality of an interval exchange transformationis useful. A permutation π of an ordered set A is called decomposable if there exists anelement b ∈ A such that the set B of elements strictly less than b is non-empty and suchthat π ( B ) = B . Otherwise it is called indecomposable . If an interval exchange transformation T = T λ,π is minimal, the permutation π is indecomposable. Indeed, if B is a set as above, theset S = ∪ a ∈B I a is closed under T and strictly included in [0 , . The following example showsthat the indecomposability of π is not sufficient for T to be minimal. Example 37.
Let A = { a, b, c } and λ be such that λ a = λ c . Let π be the transposition ( ac ) .Then π is indecomposable but T λ,π is not minimal since it is the identity on I b . Let T be an interval exchange transformation relative to ( I a ) a ∈A . We say that a word w = b b · · · b m − ∈ A ∗ is admissible for T if the set I w = I b ∩ T − ( I b ) ∩ . . . ∩ T − m +1 ( I b m − ) is non-empty. The language of T is the set L T of admissible words for T . It uniquely definesthe subshift X T = { x ∈ A Z | L ( x ) ⊂ L T } that we call as the natural coding of T . If T is regular, then X T is minimal, hence X T = { x ∈A Z | L ( x ) = L T } . Ferenczi and Zamboni gave the following combinatorial characterization ofnatural codings of regular interval exchange transformations.32 heorem 38 (Ferenczi and Zamboni [FZ08]) . A subshift X over A is the natural coding of aregular interval exchange transformation with orders ≤ and ≤ if and only if it is minimal, A ⊂ L ( X ) and it satisfies the following conditions:1. for every w ∈ L ( X ) , E − X ( w ) is an interval for ≤ and E + X ( w ) is an interval for ≤ ;2. for every w ∈ L ( X ) and all ( a , b ) , ( a , b ) ∈ E X ( w ) , if a < a , then b ≤ b ;3. for every w ∈ L ( X ) and all a , a ∈ E − X ( w ) , if a and a are consecutive for ≤ , then E + X ( a w ) ∩ E + X ( a w ) is a singleton. This result can be reformulated as follows. Let ≤ L and ≤ R be two total orders on A . Afactor w of a subshift X ⊂ A Z is said to be planar for ( ≤ L , ≤ R ) if for all ( a , b ) , ( a , b ) ∈ E X ( w ) , if a < L a , then b ≤ R b . This implies that, placing the left and right vertices of E X ( w ) on parallel lines and ordering them respectively by ≤ L and ≤ R , the edges of E X ( w ) maybe drawn as straight noncrossing segments, resulting in a planar graph. A minimal dendricsubshift over A is said to be planar if there exists two total orders ≤ L and ≤ R on A suchthat every w ∈ L ( X ) is planar for ( ≤ L , ≤ R ) . Theorem 38 states that a subshift X over A isthe natural coding of a regular interval exchange transformation if and only if X is a minimalplanar dendric subshift such that A ⊂ L ( X ) [BDFD + Example 39.
Let X be the Tribonacci subshift obtained with the substitution σ defined by σ (1) = 12 , σ (2) = 13 , σ (3) = 1 . The extension graphs of ε , and are given below and itis easily seen that it is not possible to find two orders on A making the three graphs planar. E X ( ε ) E X (1) E X (121)123 321 231 132 312 213 Observe that a regular interval exchange provides orders for which the natural coding isplanar dendric: ( ≤ L , ≤ R ) = ( ≤ , ≤ ) . It is also clear that if a minimal dendric subshift X isplanar for the orders ( ≤ L , ≤ R ) , then it is also planar for the dual orders ( ≤ ∗ L , ≤ ∗ R ) , where thedual order ≤ ∗ of ≤ is defined by x ≤ ∗ y ⇔ y ≤ x . We now show that the pairs ( ≤ L , ≤ R ) and ( ≤ ∗ L , ≤ ∗ R ) are the only possibilities. Proposition 40.
Let X be a minimal dendric subshift which is planar with respect to theorders ( ≤ L , ≤ R ) .1. Any long enough left special factor w is such that E − X ( w ) is equal to { a, b } , where a, b are consecutive for ≤ L . Furthermore, for every two ≤ L -consecutive letters a, b and forall n ∈ N , there is a (unique) left special factor w of length n such that { a, b } ⊂ E − X ( w ) .2. The same holds on the right, i.e., any long enough right special factor w is such that E + X ( w ) is equal to { a, b } , where a, b are consecutive for ≤ R . Furthermore, for every two ≤ R -consecutive letters a, b and for all n ∈ N , there is a (unique) right special factor w of length n such that { a, b } ⊂ E + X ( w ) . roof. By Theorem 38, X is the natural coding of a regular interval exchange transformation T relative to ( I a ) a ∈A . For every n ∈ N , the transformation T n is a regular interval exchangetransformation relative to ( I w ) w ∈L n ( X ) [BDFD + w , we set J w = T | w | ( I w ) and κ w the left boundary of J w . By definition, for every w ∈ L ( X ) , the word w is admissible andfor all a, b ∈ A , we have a ∈ E − X ( w ) ⇔ J a ∩ I w = ∅ ; b ∈ E + X ( w ) ⇔ J w ∩ I b = ∅ . Therefore, w is left special if and only if I w contains a point κ a , a ∈ A , as an interior point. Inthis case, a and its ≤ L -predecessor belong to E − X ( w ) . Similarly, it is right special if and onlyif J w contains a point ι b , b ∈ A , as an interior point. In this case, b and its ≤ R -predecessorbelong to E + X ( X ) . By minimality, we have min w ∈L n ( X ) | I w | = min w ∈L n ( X ) | J w | → whenever n goes to infinity. Therefore, for all large enough n , every interval I w or J w , w ∈ L n ( X ) ,contains at most one point κ a and at most one point ι b , a, b ∈ A , as interior points, showingthat E − X ( w ) , E + X ( w ) ≤ and that the (left or right) extensions of w are consecutive. Sinceboth ( I w ) w ∈L n ( X ) and ( J w ) w ∈L n ( X ) are partitions of [0 , , every κ a and every ι b belong tosome intervals I w and J w ′ and they are interior points because T is regular. This ends theproof. Corollary 41.
Let X be a minimal dendric subshift. If X is planar with respect to the orders ( ≤ L , ≤ R ) and ( (cid:22) L , (cid:22) R ) , then either ( ≤ L , ≤ R ) = ( (cid:22) L , (cid:22) R ) or ( ≤ L , ≤ R ) = ( (cid:22) ∗ L , (cid:22) ∗ R ) .Proof. This follows from Proposition 40. The left extensions of the long enough left specialfactors determine the letters that are consecutive for ≤ L . Then, the dual order ≤ ∗ L is theunique order that has exactly the same set of consecutive letters. The same holds for ≤ R -consecutive letters. It then suffices to observe that if X is planar with respect to ( ≤ L , ≤ R ) ,then it is not planar with respect to ( ≤ ∗ L , ≤ R ) , nor to ( ≤ L , ≤ ∗ R ) . Indeed, let x m , y m (resp., x M , y M ) denote the minimal (resp., maximal) elements for ≤ L and ≤ R . As E X ( ε ) is planar for ( ≤ L , ≤ R ) , both ( x m , y m ) and ( x M , y M ) are edges of E X ( ε ) thus it is not planar for ( ≤ ∗ L , ≤ R ) ,nor for ( ≤ L , ≤ ∗ R ) as x m < L x M and y m < R y M .Notice that, on the ternary alphabet A , if the pair of orders ( ≤ L , ≤ R ) can correspond toa regular interval exchange, then the set { ( ≤ L , ≤ R ) , ( ≤ ∗ L , ≤ ∗ R ) } is completely determined bythe middle letters of ≤ L and ≤ R , i.e. the letters l, r ∈ A for which there exist a, b, c, d ∈ A such that a < L l < L b and c < R r < R d. Indeed, if l, r are the middle letters of two orders ( ≤ L , ≤ R ) that can correspond to a regularinterval exchange T λ,π , then, for any x ∈ A \ { l, r } , one cannot have ( x < L l and x < R r ) or ( l < L x and r < R x ) , as otherwise the permutation π would be decomposable. Thus, if { y L } = A \ { x, l } and { y R } = A \ { x, r } , the two possible pairs of orders are ( x < L l < L y L , y R < R r < R x ) and ( y L < L l < L x, x < R r < R y R ) and the set { ( ≤ L , ≤ R ) , ( ≤ ∗ L , ≤ ∗ R ) } is unique. Let us denote it o ( l, r ) . Remark that, for everypair ( l, r ) , the set o ( l, r ) is non empty as we can always find a regular interval exchange withthe middle intervals labeled by these letters. 34 .2.3 S -adic representations of regular 3-interval exchange transformations From Theorem 38, we know that the natural coding of a regular 3-interval exchange trans-formation is a minimal ternary dendric subshift. By Theorem 1, it thus admits a primitive S -adic representation which labels a path in G . In this section, we characterize those labeledpaths in G . Lemma 42. If X is a minimal planar dendric subshift over A , then its S -adic representationdoes not belong to S ∗ ( S N LI ∪ S N RI ) . In particular, X belongs to [ l, r ] for some l, r ∈ A .Proof. Let σ be an S -adic representation of X . By Propositions 29 and 40, σ does not belongto S N LI ∪ S N RI . To prove that σ does not belong to S ∗ ( S N LI ∪ S N RI ) , we observe that for all N ≥ , X ( N +1) is a minimal planar dendric subshift over A . Indeed, X ( N +1) is a derived subshiftof X ( N ) and, by [BDFD + A is a minimal planar dendric subshift over A . Thesecond part of the statement directly follows from Proposition 29.The next lemma relates the orders of a minimal planar dendric subshift X to its equivalenceclass [ A L ( X ) , A R ( X )] . Lemma 43.
Let X be a subshift over A and v ∈ L ( X ) a dendric bispecial factor which isplanar for the orders ( ≤ L , ≤ R ) .1. If A L ( v ) = l for l ∈ A , then ( ≤ L , ≤ R ) ∈ o ( l, r ) for some r ∈ A ;2. if A R ( v ) = r for r ∈ A , then ( ≤ L , ≤ R ) ∈ o ( l, r ) for some l ∈ A .In particular, if X is a planar dendric subshift of [ l, r ] , l, r ∈ A , then X is only planar forthe orders of o ( l, r ) .Proof. Let us first treat the case where A L ( v ) = l , the other case is symmetric. Consider a, b such that A = { a, b, l } . Since A L ( v ) = l , by Remark 19, there is a simple path of length 4 from a L to b L in E X ( v ) ; we denote it by ( a L , x R , l L , y R , b L ) . Thus we have ( a, x ) , ( l, x ) , ( l, y ) , ( b, y ) ∈ E X ( v ) . Since v is planar for ( ≤ L , ≤ R ) , we have ( a < L l and x < R y ) or ( l < L a and y < R x ) and ( l < L b and x < R y ) or ( b < L l and y < R x ) . Thus, we have the two following possibilities ( a < L l < L b and x < R y ) or ( b < L l < L a and y < R x ) . In both cases, l is the middle letter for ≤ L , which implies that ( ≤ L , ≤ R ) ∈ o ( l, r ) for some r ∈ A .If X is in the class [ l, r ] then there exist v, v ′ ∈ L ( X ) such that A L ( v ) = l and A R ( v ′ ) = r .Using the first part of the result, any pair of orders for which X is planar is in o ( l, r ) . Corollary 44.
Let Y be any subshift over A such that L ( X ) = A and let X be the imageof Y under some morphism σ ∈ S . . If σ ∈ { α a | a ∈ A } , then up to taking the dual orders, there are four pairs of ordersfor which ε is planar (in X ); they are represented in Table 6.2. If σ / ∈ { α a | a ∈ A } , then up to taking the dual orders, there is a unique pair of ordersfor which ε is planar (in X ); they are represented in Table 7. In addition, the otherinitial bispecial factors of X are also planar for these orders. abc cba acb cba abc bca acb bca Table 6: Planar extension graphs of ε in images under α a β abc γ abc δ ( k ) abc ζ ( k ) abc η abc bac acb acb bac acb bac bca acb abc bca Table 7: Planar extension graphs of ε in images under σ for σ / ∈ { α a | a ∈ A } Proof.
It is clear that the pairs of orders given in Tables 6 and 7 make E X ( ε ) planar.Whenever σ = α a , it is easily seen that for ε to be planar, the letter a must be minimalfor ≤ L and maximal for ≤ R or vice-versa. It then suffices to check that Table 6 gives all suchpairs of orders (when also considering the dual orders).Whenever σ = β abc , { a, b, c } = A , Table 3 shows that A L ( ε ) = a thus, by Lemma 43, a must be the middle letter of the ≤ L . One can easily check that ≤ R is then completelydetermined. The case σ = γ abc is symmetric.Whenever σ / ∈ { α a , β abc , γ abc | { a, b, c } = A } , we have A L ( ε ) , A R ( ε ) = 0 . This case thusdirectly follows from Lemma 43.Recall that the non-empty initial bispecial factors of X are non prefix factors of a word in σ ( A ) . The only such factors that are bispecial are• c l , l ≤ k when σ = δ ( k ) abc ;• c l , l ≤ k when σ = ζ ( k ) abc .Their extension graphs are represented in Table 8 and one can see that they are indeed planarfor the orders that make E X ( ε ) planar. δ ( k ) abc ζ ( k ) abc cb ac ca ac Table 8: Planar extension graphs of c l , l < k , in images under δ ( k ) abc and ζ ( k ) abc σ ∈ S , we let o ( σ ) denote the set of pairs of orders for which ε is planar inimages under σ . The following is a direct consequence of Lemmas 42 and 43 and Corollary 44. Proposition 45. If X is a minimal dendric planar subshift over A , then it has an S -adicrepresentation ( σ n ) n ≥ labeling a path ([ l n , r n ]) n ≥ in G such that, for each n ≥ , X ( n ) isplanar for the orders of o ( l n , r n ) and o ( l n , r n ) ⊂ o ( σ n ) . It is clear that not every path in G has a label ( σ n ) n ≥ which is the S -adic representationof a minimal ternary planar dendric subshift X . One reason is given by Proposition 45: themorphism σ ∈ S may label edges whose starting vertex [ l, r ] does not satisfy o ( l, r ) ⊂ o ( σ ) .For instance, the morphism β labels two edges leaving the vertex [1 , , two edges leavingthe vertex [1 , and two edges leaving the vertex [1 , but only [1 , satisfies o ( l, r ) ⊂ o ( β ) .We will prove that another reason is that, even though the morphism σ may label severaledges leaving the vertex [ A L ( X ) , A R ( X )] , the class of the subshift X (2) is uniquely defined by σ and by [ A L ( X ) , A R ( X )] .This motivates the following definitions. Observe that they are valid on an alphabet ofany size.If (cid:22) and ≤ are two total orders on A , a partial map ϕ : A → A is order preserving from (cid:22) to ≤ if, for all distinct x, y ∈ A and for all x ′ ∈ ϕ − ( { x } ) , y ′ ∈ ϕ − ( { y } ) , x ′ ≺ y ′ ⇔ x < y. An injective and strongly left proper morphism σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) if1. for all s ∈ S ( σ ) , ϕ ( L ) s is order preserving from (cid:22) L to ≤ L ;2. for all p ∈ P ( σ ) , ϕ ( R ) p is order preserving from (cid:22) R to ≤ R .In this definition, the graph morphisms ϕ ( L ) s and ϕ ( R ) p are seen as applications on the letters.From the graph point of view, saying that ϕ ( L ) s is order preserving from (cid:22) L to ≤ L meansthat, starting from a bipartite graph ordered by (cid:22) L on the left and applying ϕ ( L ) s , the verticesin the image graph will be ordered by ≤ L on the left. Example 46.
Let σ = η abc . By Table 2, the associated morphisms ϕ ( L ) s and ϕ ( R ) p are given by ϕ ( L ) ε : ( a bb, c c , ϕ ( L ) c : ( b bc a , ϕ ( R ) a : ( a, b bc c , ϕ ( R ) ab : ( a ab c . Let us consider the following orders on A : b ≺ L c ≺ L a, c ≺ R a ≺ R b, c < L b < L a and b < R a < R c. By definition, both ϕ ( L ) ε and ϕ ( L ) c are order preserving from (cid:22) L to ≤ L and ϕ ( R ) ab is orderpreserving from (cid:22) R to ≤ R but ϕ ( R ) a is not. Thus σ is not planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) .If v is a word whose extension graph is given by bca cab hen the extension graphs of its extended images σ ( v ) a , σ ( v ) ab , cσ ( v ) a and cσ ( v ) ab are givenrespectively by cb bc cb ac ba bc ba ac It is easy to see that v is planar for ( (cid:22) L , (cid:22) R ) and that only two of its extended images areplanar for ( ≤ L , ≤ R ) . The terminology planar preserving comes from the fact that the situation of the previousexample does not occur if σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) . This is formalizedin the following result. Proposition 47.
Let X be a dendric subshift, v ∈ L ( X ) and σ ∈ DP( v ) which is planarpreserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) . Then v is planar for ( (cid:22) L , (cid:22) R ) if and only if everybispecial extended image of v under σ is planar for ( ≤ L , ≤ R ) .Proof. Let Y denote the image of X by σ . Let us first assume that E X ( v ) is a tree but notplanar for ( (cid:22) L , (cid:22) R ) . There exist x ′ , x ′ , y ′ , y ′ ∈ A such that ( x ′ , y ′ ) , ( x ′ , y ′ ) belong to E X ( v ) and are crossing edges in E X ( v ) , i.e., they satisfy x ′ ≺ L x ′ and y ′ ≺ R y ′ . If s = s ( x ′ , x ′ ) and p = p ( y ′ , y ′ ) , then the extension graph of the bispecial extended image u = sσ ( v ) p of v is the image of a subgraph of E X ( v ) by the morphisms ϕ ( L ) s and ϕ ( R ) p . If wedenote x = ϕ ( L ) s ( x ′ ) , x = ϕ ( L ) s ( x ′ ) , y = ϕ ( R ) p ( y ′ ) and y = ϕ ( R ) p ( y ′ ) , then ( x , y ) and ( x , y ) belong to E Y ( u ) and, since σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) , they are crossing edges of E Y ( u ) , i.e., x < L x and y < R y . Thus v has a bispecial extended image whose extension graph is not planar for ( ≤ L , ≤ R ) .Assume now that there exists a bispecial extended image u of v whose extension graph isnot planar for ( ≤ L , ≤ R ) . As σ is in DP( v ) , the graph E Y ( u ) is a tree. If E Y ( u ) is not planarfor ( ≤ L , ≤ R ) , there exist x , x , y , y ∈ A such that ( x , y ) , ( x , y ) ∈ E Y ( u ) are crossingedges in E ( u ) , i.e., x < L x and y < R y . By Proposition 13, the graph E Y ( u ) is the image of a subgraph of E X ( v ) by some morphisms ϕ ( L ) s and ϕ ( R ) p . Thus there exist x ′ , x ′ , y ′ , y ′ ∈ A such that ( x ′ , y ′ ) , ( x ′ , y ′ ) belong to E X ( v ) and satisfy ϕ ( L ) s ( x ′ ) = x , ϕ ( L ) s ( x ′ ) = x , ϕ ( R ) p ( y ′ ) = y and ϕ ( R ) p ( y ′ ) = y . Since σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) , we must have, x ′ ≺ L x ′ and y ′ ≺ R y ′ ,which contradicts the fact that E X ( v ) is planar for ( (cid:22) L , (cid:22) R ) .38bserve that σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) if and only if it is planarpreserving from ( (cid:22) ∗ L , (cid:22) ∗ R ) to ( ≤ ∗ L , ≤ ∗ R ) .Thus, in the ternary case we say that σ is planar preserving from o ( l, r ) to o ( l ′ , r ′ ) ifit is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) for some ( (cid:22) L , (cid:22) R ) ∈ o ( l, r ) and some ( ≤ L , ≤ R ) ∈ o ( l ′ , r ′ ) The next result will imply that the subgraph of G representing the S -adicrepresentations of minimal ternary planar dendric subshifts is deterministic. Lemma 48.
For every σ ∈ S and every l, r ∈ A , there exists a unique pair ( l ′ , r ′ ) ∈ A such that σ is planar preserving from o ( l ′ , r ′ ) to o ( l, r ) .Proof. We actually prove the following stronger result: for every σ ∈ S and every total order ≤ L on A , there is a unique total order (cid:22) L on A such that for all s ∈ S ( σ ) , ϕ ( L ) s is orderpreserving from (cid:22) L to ≤ L . Similarly, for every total order ≤ R on A , there is a unique totalorder (cid:22) R on A such that for all p ∈ P ( σ ) , ϕ ( R ) p is order preserving from (cid:22) R to ≤ R .This follows from the fact that either S ( σ ) = { s } , in which case ϕ ( L ) s is the identity, or S ( σ ) = { s , s } and ϕ ( L ) s ( A ) and ϕ ( L ) s ( A ) both contain two letters (and not the same). Thesame holds on the right.The next example shows how we can build ( (cid:22) L , (cid:22) R ) from ( ≤ L , ≤ R ) and σ ∈ S such that σ is planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) . Example 49.
The morphism δ ( k ) abc labels edges from [ c, a ] to [ x, y ] for all x, y ∈ { b, c } (seeFigure 9). Up to considering the dual orders, the pair of orders in o ( c, a ) is a < L c < L b and b < R a < R c. From Table 2, we have S ( δ ( k ) abc ) = { ε, c k } , P ( δ ( k ) abc ) = { a, abc k } and ϕ ( L ) ε : ( a ab, c c , ϕ ( L ) c k : ( b bc c , ϕ ( R ) a : ( a ab, c b , ϕ ( R ) abc k : ( b ac c . We have ϕ ( L ) ε ( b ) = ϕ ( L ) ε ( c ) = c , ϕ ( L ) ε ( a ) = a and a < L c . Thus for ϕ ( L ) ε to be order preservingfrom (cid:22) L to ≤ L , we must have a ≺ L b, c . Then, from ϕ ( L ) c k ( c ) = c < L b = ϕ ( L ) c k ( b ) , we deducethat we must have c ≺ L b , so that a ≺ L c ≺ L b. We similarly deduce that for δ ( k ) abc to be planar preserving from ( (cid:22) L , (cid:22) R ) to ( ≤ L , ≤ R ) , we musthave b ≺ R c ≺ R a. Therefore, δ ( k ) abc is planar preserving from o ( c, a ) to o ( x, y ) if and only if ( x, y ) = ( c, c ) . The list of all edges ([ l, r ] , σ, [ l ′ , r ′ ]) of G for which o ( l, r ) ⊂ o ( σ ) and σ is planar preservingfrom o ( l ′ , r ′ ) to o ( l, r ) is given in Table 9. Lemma 50.
Let X ∈ [ l, r ] and Y ∈ [ l ′ , r ′ ] be two ternary dendric planar subshifts such that X is the image of Y under σ for σ ∈ S . Then σ is planar preserving from o ( l ′ , r ′ ) to o ( l, r ) . α a β abc γ abc δ ( k ) abc ζ ( k ) abc η abc [ l, r ] [ b, b ] [ b, c ] [ c, b ] [ c, c ] [ a, c ] [ c, a ] [ c, a ] [ c, c ] [ b, c ][ l ′ , r ′ ] [ b, b ] [ b, c ] [ c, b ] [ c, c ] [ c, c ] [ c, c ] [ c, c ] [ c, c ] [ c, a ] Table 9: List of edges ([ l, r ] , σ, [ l ′ , r ′ ]) of G for which o ( l, r ) ⊂ o ( σ ) and σ is planar preservingfrom o ( l ′ , r ′ ) to o ( l, r ) Proof. As X is planar, it is for the orders of o ( l, r ) thus o ( l, r ) ⊂ o ( σ ) . By the previous lemma,there exists a unique pair of letters ( l ′′ , r ′′ ) such that σ is planar preserving from o ( l ′′ , r ′′ ) to o ( l, r ) . For all v ∈ L ( Y ) , its bispecial extended images are in L ( X ) thus are planar for theorders of o ( l, r ) . Using Proposition 47, v is planar for the orders of o ( l ′′ , r ′′ ) . As it is true forany v ∈ L ( Y ) , Y is planar for the orders of o ( l ′′ , r ′′ ) and, by Lemma 43, l ′′ = l ′ and r ′′ = r ′ ,thus σ is planar preserving from o ( l ′ , r ′ ) to o ( l, r ) .We can now prove a result similar to Theorem 1 but in the case of interval exchanges.The graph that we obtain is the subgraph of the graph G in Figure 9 where we only kept theedges ([ l, r ] , σ, [ l ′ , r ′ ]) such that o ( l, r ) ⊂ o ( σ ) and σ is planar preserving from o ( l ′ , r ′ ) to o ( l, r ) .These edges are given by Table 9. Theorem 51.
A subshift X over A is a coding of a regular interval exchange if and only ifit has a primitive S -adic representation σ ∈ S N that labels a path in the graph represented inFigure 10.Proof. Let us assume that X is the coding of a regular interval exchange. From Proposition 45, X has an S -adic representation σ = ( σ n ) n ≥ labeling a path ([ l n , r n ]) n ≥ in the graph G suchthat, for each n ≥ , X ( n ) σ is planar exactly for the orders of o ( l n , r n ) and o ( l n , r n ) ⊂ o ( σ n ) .By Lemma 50, σ n is planar preserving from o ( l n +1 , r n +1 ) to o ( l n , r n ) . Thus, all the edges ofthe path are preserved in the subgraph represented in Figure 10.Assume now that X has a primitive S -adic representation σ ∈ S N labeling a path ([ l n , r n ]) n ≥ in the graph of Figure 10. As this graph is a subgraph of the graph G , X isminimal dendric. Let us show that X is planar for the orders of o ( l , r ) . Let ( ≤ L , ≤ R ) be anelement of o ( l , r ) . The initial bispecial factor of X are planar for ( ≤ L , ≤ R ) by Corollary 44as ( ≤ L , ≤ R ) ∈ o ( l , r ) ⊂ o ( σ ) . Let u ∈ L ( X ) be a non-initial bispecial factor of X . UsingCorollary 11, there is a unique n > and a unique initial bispecial factor v in L ( X ( n ) σ ) suchthat u is a descendant of v . From Corollary 44, v is planar for any order of o ( l n , r n ) ⊂ o ( σ n ) .By construction of the subgraph and by Proposition 47, the extension graph of u is planar for ( ≤ L , ≤ R ) . A subshift X over A is a Cassaigne subshift if it has a primitive C -adic representation, where C = { c , c } and c : and c : . ,
1] [1 ,
2] [3 , ,
1] [2 ,
2] [2 , ,
3] [3 ,
2] [3 , β β β β β β γ , δ γ , δ γ , δ γ , δ γ , δ γ , δ ζ , ζ α , α ζ , ζ α , α ζ , ζ α , α η η η η η η α α α α α α Figure 10: A subshift over A is the coding of an interval exchange if and only if it has aprimitive S -adic representation labeling an infinite path in this graph. We stress the factthat, compared with Figure 9, the vertices [1 , and [3 , are exchanged.41assaigne subshifts are minimal ternary dendric subshifts and a directive sequence ( σ n ) n ≥ ∈C N is primitive if and only if it cannot be eventually factorized over { c , c } , i.e., there is no N ∈ N such that for all n ≥ N , c N +2 n = c N +2 n +1 [CLL17].By considering products of morphisms, we obtain that a subshift is a Cassaigne subshift ifand only if it has a primitive C ′ -adic representation where C ′ = { c , c , c , c , c , c } and c = c :
13 ; c = c c : c = c c c :
12 ; c = c : c = c c :
23 ; c = c c c : . Thus a subshift is a Cassaigne subshift if and only if it has a primitive S -adic representationusing morphisms from the set { α , α , γ π , β π , η π , η π } or, equivalently, if and only if it has a primitive S C -adic representation where S C = { α , α , γ π , β π , η π , η π } . Lemma 52.
Let X and Y be two dendric subshifts over A such that Y = π ( X ) where π isa permutation on A . For any l, r ∈ A , if X ∈ [ l, r ] , then Y ∈ [ π ( l ) , π ( r )] where we take theconvention π (0) = 0 .Proof. It directly follows from Remark 19.
Proposition 53.
There is no Cassaigne subshift which is an Arnoux-Rauzy or the coding ofa regular interval exchange.Proof.
Let X be a Cassaigne subshift and σ = ( σ n ) n ≥ be a primitive S C -adic representationof X . Firstly, it is clear that X is not an Arnoux-Rauzy subshift. Using Proposition 29, itwould indeed require σ to contain only left-invariant and right-invariant morphisms (extendingthese notions to morphisms in S C ), hence to be in { α , α } N . It then suffice to observe thatthere is no primitive sequence in { α , α } N .For each n , let us denote σ n = µ n π n where µ n ∈ S and π n is a permutation on A . The sequence σ ′ = ( σ ′ n ) n ≥ where σ ′ n − = µ n and σ ′ n = π n is also a primitive S -adic representation of X . Let us assume that X is the coding of a regularinterval exchange and prove that we reach a contradiction. By [BDFD + X ( n +1) σ ′ is either a derived subshift of X ( n ) σ ′ if n is odd or a permutation of X ( n ) σ ′ if n is even,it is the coding of a regular interval exchange for all n ≥ . Using Lemma 42, let us denote X ( n ) σ ′ ∈ [ l n , r n ] , l n , r n ∈ A .
42y Lemma 50, µ n is planar preserving from o ( l n − , r n − ) to o ( l n , r n ) and, by Lemma 52, l n +1 = π n ( l n ) and r n +1 = π n ( r n ) . Thus, ( l n +1 , r n +1 ) can be fully determined by ( l n − , r n − ) and σ n using Table 9. In con-clusion, the sequence σ labels an infinite path in the following graph: [1 ,
1] [1 ,
2] [1 , ,
1] [2 ,
2] [2 , ,
1] [3 ,
2] [3 , β π γ π α α , α α η π η π α α α α This is absurd as the only infinite paths in this graph belong to the set S ∗C { α , α } N anddo not correspond to primitive sequences. The problem of finding an S -adic characterization of minimal dendric subshift over largeralphabets is still open. For any fixed alphabet A k = { , , . . . , k } , one could define a set S k ofinjective and strongly left proper morphisms over A k such that any minimal dendric subshifthas a primitive S k -adic representation, thus extending Proposition 18. It is likely that thereexists a graph G k that allows to extend Theorem 1, but its definition is much more tricky.Indeed, a key point in the ternary case is that we were able to interpret Proposition 14 withthe conditions C L ( a ) and C R ( a ) (see Section 5.2), which then allows to define the vertices of G . What could be an equivalent interpretation on A k is not so clear.Another interesting question would be to study the S -adic representations obtained byfactorizing the morphisms of S k into elementary automorphisms. Such factorizations alwaysexist (see Theorem 5) and therefore yield to another graph G ′ k where the edges are labeledby elementary automorphisms of F A k . Numerical computations show that the minimal graphobtained from the graph G of Theorem 1 has more than 200 vertices. Of course, this graphdepends on the choices that we made for the set S of morphisms. Finding a handier graphin the ternary case is an open question. References [AR91] Pierre Arnoux and Gérard Rauzy. Représentation géométrique de suites decomplexité n + 1 . Bull. Soc. Math. France , 119(2):199–215, 1991.43Arn63] V. I. Arnold. Small denominators and problems of stability of motion in classicaland celestial mechanics.
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