OON THE EXISTENCE OF OPEN AND BI-CONTINUING CODES
UIJIN JUNG
Abstract.
Given an irreducible sofic shift X , we show that an irreducible shiftof finite type Y of lower entropy is a factor of X if and only if it is a factor of X by an open bi-continuing code. If these equivalent conditions hold and Y ismixing, then any code from a proper subshift of X to Y can be extended to anopen bi-continuing code on X . These results are still valid when X is assumed tobe only an almost specified shift, i.e., a subshift satisfying an irreducible versionof the specification property. Introduction and Preliminaries
Let X and Y be shift spaces. Suppose that there exists a factor code from X to Y . If we denote by h ( X ) the topological entropy of X , then h ( X ) ≥ h ( Y ). Also,whenever x is a periodic point of X , there exists a periodic point of Y whose perioddivides the period of x . We denote this condition by P ( X ) (cid:38) P ( Y ) and call itthe periodic condition . These two conditions (on entropies and periodic points) arenecessary for X to factor onto Y . To find satisfactory sufficient conditions for X tofactor onto Y (by a code with certain properties) is one of the important problemsin symbolic dynamics.In [5], Boyle proved the following theorem: Given irreducible shifts of finite type X and Y with h ( X ) > h ( Y ), Y is a factor of X if and only if the periodic condition P ( X ) (cid:38) P ( Y ) holds. One can consider two directions of generalizing this result:One is to impose certain properties on the factor code, and the other is to weakenthe finite-type constraints on X and Y .The first type of generalization is due to Boyle and Tuncel. In [7], they showedthat the entropy and the periodic conditions also imply that Y is a factor of X by aright continuing code. A right continuing code is a code which is surjective on eachunstable set (see Definition 2.1). It is a natural dual version of a right closing codeand plays a fundamental role in the class of infinite-to-one codes [7, 6].The other direction of generalization is considered by Thomsen [16], who investi-gated the existence of factor codes between irreducible sofic shifts. In particular, hegave a necessary and sufficient condition for an irreducible shift of finite type to bea factor of a given synchronized system of greater synchronized entropy (see § open if it sends an open set to an open set. Finite-to-one open Mathematics Subject Classification.
Primary 37B10; Secondary 37B40, 54H20.
Key words and phrases. open, continuing, sofic shift, shift of finite type, almost specified, spec-ification property. a r X i v : . [ m a t h . D S ] F e b UIJIN JUNG codes between shifts of finite type have been studied in several contexts [9, 14, 10].A finite-to-one factor code between irreducible shifts of finite type is open if and onlyif it is bi-closing [14]. A natural generalization exists: A factor code from a shift offinite type to a shift space is open if and only if it is bi-continuing (see Theorem 2.6).In § shifts with the almost specification property (or almost specifiedshifts ) which are subshifts satisfying an irreducible version of the specification prop-erty (see Definition 3.1). Every irreducible sofic shift is almost specified, and almostspecified shifts behave much like shifts with the specification property except beingmixing. Our main theorem is stated as follows. Theorem 5.2.
Let X be an almost specified shift and Y an irreducible shift of finitetype such that h ( X ) > h ( Y ) . Then the following are equivalent. (1) Y is a factor of X by a bi-continuing code with a bi-retract. (2) Y is a factor of X by an open code with a uniform lifting length. (3) Y is a factor of X .If X is of finite type, then the above conditions are also equivalent to: (4) P ( X ) (cid:38) P ( Y ) . As a corollary, it follows that given an irreducible shift of finite type X , a shiftspace Y is a lower entropy open factor of X exactly when h ( X ) > h ( Y ), P ( X ) (cid:38) P ( Y ) and Y is an irreducible shift of finite type (see Corollary 5.4).It is not surprising that factor problems usually involve extension problems [1,5, 7]. Even in the case where X and Y are irreducible shifts of finite type and P ( X ) (cid:38) P ( Y ), not every code can be extended (see Example 5.6). In §
5, we give anecessary condition for a code on a proper subshift of X to be extended to a codeon X and show that it is the only obstruction to extend it to a code on X . We givean extension theorem as follows. Theorem 5.5.
Let X be an almost specified shift and Y an irreducible shift of finitetype such that h ( X ) > h ( Y ) and Y is a factor of X . Then any code from a propersubshift of X to Y , which can be extended to a code on X , can be extended to anopen bi-continuing code of X onto Y . In particular, if Y is mixing, then any code from a proper subshift of X to Y canbe extended to an open bi-continuing code of X onto Y (see Theorem 4.5). Thetheorem says that the openness and the bi-continuing property of a code are bothglobal in the sense that these properties cannot be ruled out by any property a codemight exhibit on a proper subshift.We introduce some notations and definitions used. If X is a shift space with theshift map σ , denote by B n ( X ) the set of all words of length n appearing in the pointsof X and B ( X ) = (cid:83) n ≥ B n ( X ). Also let A ( X ) = B ( X ) and B Xn ( a, b ) = { u . . . u n ∈B n ( X ) : u = a, u n = b } for each a, b ∈ A ( X ). A shift space X is called irreducible if for all u, v ∈ B ( X ), there is a word w with uwv ∈ B ( X ). It is called mixing if forall u, v ∈ B ( X ), there is an integer N ∈ N such that whenever n ≥ N , we can find PEN AND BI-CONTINUING CODES 3 w ∈ B n ( X ) with uwv ∈ B ( X ). If there is such an N which works for all u, v ∈ B ( X ),then we call N + 1 a transition length for X .Given u ∈ B ( X ), let l [ u ] denote the open and closed set { x ∈ X : x [ l,l + | u |− = u } ,which we call a cylinder . If | u | = 2 l + 1, then − l [ u ] is called a central l + 1 cylinder .When u ∈ B ( X ) and l = 0, we usually discard the subscript 0.A code φ : X → Y is a continuous σ -commuting map between shift spaces. It iscalled a factor code if it is onto. In this case we say that Y is a factor of X . Anycode can be recoded to be a 1-block code, i.e. a code for which x determines φ ( x ) .A subshift X is called a shift of finite type if there is a finite set F of words suchthat X consists of all points on A ( X ) in which there is no occurrence of words from F . Any shift of finite type is conjugate to an edge shift, i.e., a shift space whichconsists of all bi-infinite trips in a directed graph G . If A is the adjacency matrixfor G , denote by X A the edge shift on G . A shift space is called sofic if it is a factorof a shift of finite type. A mixing sofic shift always has a transition length.A word v ∈ B ( X ) is synchronizing if whenever uv and vw are in B ( X ), we have uvw ∈ B ( X ). Following [4], an irreducible shift space X is a synchronized system if it has a synchronizing word. For a given set of words W , let X W be the smallestshift space containing all the sequences obtained by concatenating words in W . Wecall X W a coded system . A synchronized system is coded [4].For a shift space X with periodic points, let per( X ) be the greatest commondivisor of periods of all periodic points of X . Let X be an irreducible edge shift and p = per( X ). Then there exists a unique partition { A , A , · · · , A p − } of A ( X ) suchthat whenever ab ∈ B ( X ) and a ∈ A i , then b ∈ A i +1 (mod p ) . Also there is n ∈ N with the property that B Xn ( a, b ) (cid:54) = ∅ for any a, b ∈ A i and i = 0 , , · · · , p −
1. This n is called a weak transition length for X .The entropy of a shift space X is defined by h ( X ) = lim n →∞ (1 /n ) log |B n ( X ) | ,which equals the topological entropy of ( X, σ ) as a dynamical system. If X is anirreducible shift of finite type and p = per( X ), then for any a, b ∈ A ( X ) we havethe following equalities: h ( X ) = lim n →∞ np log | p np ( X ) | = lim n →∞ np log | q np ( X ) | = lim sup n →∞ n log |B Xn ( a, b ) | , where p k ( X ) (resp. q k ( X )) denotes the number of periodic points of X with period k (resp. least period k ). If X is mixing, then limsup becomes limit.For more details on symbolic dynamics, see [12]. For a perspective for open mapsbetween shift spaces, see [10].2. Properties of continuing codes
A code φ : X → Y is called right closing if whenever x ∈ X , y ∈ Y and φ ( x )is left asymptotic to y , then there exists at most one ¯ x ∈ X such that ¯ x is leftasymptotic to x and φ (¯ x ) = y . Right closing codes form an important class of UIJIN JUNG finite-to-one codes, especially between irreducible shifts of finite type. The followingnotion, which can be thought as a dual of the above property, first appeared in [7].
Definition 2.1.
A code φ : X → Y between shift spaces is called right continuing if whenever x ∈ X , y ∈ Y and φ ( x ) is left asymptotic to y , then there exists at leastone ¯ x ∈ X such that ¯ x is left asymptotic to x and φ (¯ x ) = y . A left continuing codeis defined similarly. If φ is both left and right continuing, it is called bi-continuing .A right continuing code into an irreducible sofic shift must be onto. In general itneed not be onto. The Perron-Frobenius theory implies that a factor code betweenirreducible shifts of finite type with the same entropy is right continuing exactlywhen it is right closing. Definition 2.2.
An integer n ∈ Z + is called a ( right continuing ) retract of a rightcontinuing code φ : X → Y if, whenever x ∈ X and y ∈ Y with φ ( x ) ( −∞ , = y ( −∞ , ,we can find ¯ x ∈ X such that φ (¯ x ) = y and x ( −∞ , − n ] = ¯ x ( −∞ , − n ] .Note that having a retract is a conjugacy invariant. If φ has both left and rightcontinuing retracts n , we simply say that φ has bi-retract n .A code φ : X → Y between shift spaces is open if and only if for each k ∈ N ,there is m ∈ N such that whenever x ∈ X, y ∈ Y and φ ( x ) [ − m,m ] = y [ − m,m ] , wecan find ¯ x ∈ X with ¯ x [ − k,k ] = x [ − k,k ] and φ ( x ) = y . We say that φ has a uniformlifting length if for each k ∈ N , there exists m satisfying the above property suchthat sup k | m − k | < ∞ . In this case, we can assume that m − k is constant andnonnegative. Note that having a uniform lifting length is also a conjugacy invariant. Lemma 2.1.
Let φ : X → Y be a code between shift spaces. If φ is open with auniform lifting length, then it is bi-continuing with a bi-retract. When Y is of finitetype, then the converse holds.Proof. First, suppose that φ is open with a uniform lifting length. We can assume φ is a 1-block code. Choose l ≥ k + 1 cylinder in X ,its image consists of central 2 k + 2 l + 1 cylinders. Let x ∈ X and y ∈ Y satisfy φ ( x ) ( −∞ , = y ( −∞ , . By considering φ ( − k − l [ x − k − l · · · x − l ]) for k ∈ N , we obtainpoints z ( k ) such that z ( k )[ − k − l, − l ] = x [ − k − l, − l ] and φ ( z ( k ) ) = y . Let z be a limit pointof { z ( k ) } k ∈ N . Then z ( −∞ , − l ] = x ( −∞ , − l ] and φ ( z ) = y , as desired. Thus φ is rightcontinuing with retract l . Similarly φ is left continuing with retract l .Next, assume that φ is bi-continuing with bi-retract n ∈ N and Y is of finite type.We can assume that Y is an edge shift and φ is a 1-block code. Suppose that l ≥ u ∈ B l +1 ( X ). Choose x ∈ − l [ u ] and let y ∈ Y with y [ − l − n,l + n ] = φ ( x ) [ − l − n,l + n ] .The point ¯ y , given by ¯ y i = φ ( x ) i for i ≤ y i = y i for i ≥
0, is in Y . Since n is a right continuing retract and φ ( x ) ( −∞ ,l + n ] = ¯ y ( −∞ ,l + n ] , there is z ∈ X suchthat z ( −∞ ,l ] = x ( −∞ ,l ] and φ ( z ) = ¯ y . Then φ ( z ) [ − l − n, ∞ ) = ¯ y [ − l − n, ∞ ) = y [ − l − n, ∞ ) .Since n is also a left continuing retract, there is an ¯ x ∈ X such that ¯ x [ − l, ∞ ) = z [ − l, ∞ ) and φ (¯ x ) = y . Note that ¯ x [ − l,l ] = z [ − l,l ] = x [ − l,l ] , hence ¯ x is in − l [ u ]. So − l − n [ φ ( x ) [ − l − n,l + n ] ] ⊂ φ ( − l [ u ]). Thus φ is open with uniform lifting length n . (cid:3) Remark . Note that the proof of the second statement in Lemma 2.1 is still validunder a weaker assumption on Y , that is, when there is a shift of finite type Z suchthat φ ( X ) ⊂ Z ⊂ Y and Z is open and closed in Y . PEN AND BI-CONTINUING CODES 5
Lemma 2.3. [10]
Let X be a shift of finite type and φ an open factor code from X to a shift space Y . Then Y is of finite type. In [7], the first statement of the following proposition is proved when both shiftspaces are of finite type. We show that the codomain need not be of finite type.
Proposition 2.4.
Let φ : X → Y be a code between shift spaces and X of finite type. (1) If φ is right continuing, then it has a retract. (2) If φ is open, then it has a uniform lifting length.Proof. By recoding, we can assume that X is an edge shift and φ is 1-block.(1) It suffices to show the case where φ is onto. We claim that there exists n ∈ N such that whenever x ∈ X and y ∈ Y with φ ( x ) ( −∞ , = y ( −∞ , , we can find ¯ x ∈ X such that φ (¯ x ) ( −∞ , = y ( −∞ , and x ( −∞ , − n ] = ¯ x ( −∞ , − n ] . Suppose not. Then foreach k ∈ N , there exist x ( k ) ∈ X and y ( k ) ∈ Y such that φ ( x ( k ) ) ( −∞ , = y ( k )( −∞ , andthere is no ¯ x ∈ X with φ (¯ x ) ( −∞ , = y ( k )( −∞ , and ¯ x ( −∞ , − k ] = x ( −∞ , − k ] . By choosinga subsequence, we can assume that there are x ∈ X and y ∈ Y with x ( k ) → x , y ( k ) → y . Since φ is right continuing and φ ( x ) ( −∞ , = y ( −∞ , , there exist z ∈ X and m ≥ φ ( z ) = y and z ( −∞ , − m ] = x ( −∞ , − m ] . Take k > m satisfying x ( k )[ − m, = x [ − m, and y ( k )[ − m, = y [ − m, . Define ¯ x ∈ X by letting¯ x i = (cid:40) ( x ( k ) ) i if i ≤ − mz i if i ≥ − m Then φ (¯ x ) ( −∞ , = y ( k )( −∞ , and ¯ x ( −∞ , − k ] = x ( k )( −∞ , − k ] , which is a contradiction. Thusthe claim holds.This n in the claim is indeed a retract. Suppose x ∈ X and y ∈ Y satisfy φ ( x ) ( −∞ , = y ( −∞ , . Let z (0) = x . By inductive process, for each k ∈ N thereexists z ( k ) ∈ X such that φ ( z ( k ) ) ( −∞ ,k ] = y ( −∞ ,k ] and z ( k )( −∞ ,k − n − = z ( k − −∞ ,k − n − .Since z ( k )( −∞ , − n ] = x ( −∞ , − n ] for all k ∈ N , any limit point z of { z ( k ) } k ∈ N satisfies z ( −∞ , − n ] = x ( −∞ , − n ] and φ ( z ) = y .(2) Choose l ≥ φ ([ a ]) consists of central 2 l +1 cylinders for all a ∈ B ( X ).Let x ∈ X and y ∈ Y satisfy φ ( x ) ( −∞ , = y ( −∞ , . Consider φ ( − l [ x − l ]). Since itcontains φ ( x ) and φ ( x ) [ − l, = y [ − l, , there exists z ∈ − l [ x − l ] with φ ( z ) = y . Nowdefine ¯ x by ¯ x i = x i for i ≤ − l and ¯ x i = z i for i ≥ − l . Then ¯ x is left asymptotic to x and φ (¯ x ) = y , so φ is right continuing with retract l . Similarly φ is left continuingwith retract l .Since φ is open, Z = φ ( X ) is open and closed in Y . Also Z is a shift of finite typeby Lemma 2.3. Thus by Remark 2.2, φ is open with a uniform lifting length. (cid:3) Remark . A code φ : X → Y between shift spaces is called right continuing a.e. (almost everywhere) if whenever x ∈ X is left transitive in X and φ ( x ) is leftasymptotic to a point y ∈ Y , then there exists ¯ x ∈ X such that ¯ x is left asymptoticto x and φ (¯ x ) = y . Similarly we have the notions called left continuing a.e. andbi-continuing a.e. A slight modification of the proof of Lemma 2.4 (2) shows thatan open code from a synchronized system is bi-continuing a.e. UIJIN JUNG
Theorem 2.6.
Let φ be a factor code from a shift of finite type X to a sofic shift Y . Then φ is open if and only if it is bi-continuing.Proof. Suppose first that φ is open. By Lemma 2.3, Y is of finite type. It followsfrom Proposition 2.4 (2) and Lemma 2.1 that φ is bi-continuing.Suppose φ is bi-continuing. Since Y is sofic, there are a shift of finite type Z anda factor code π : Z → Y . Consider the fiber product (Σ , ψ , ψ ) of ( φ, π ) [12]. ThenΣ is of finite type. By a usual fiber product argument, it is easy to show that ψ isalso bi-continuing. Since Σ and Z are of finite type, it follows from Proposition 2.4(1) and Lemma 2.1 that ψ is open. Since fiber product pulls down the openness ofa code, it follows that φ is open [10, Lemma 2.4]. (cid:3) Remark . Recently, J. Yoo [17] showed us that a right continuing factor of a shiftof finite type is also of finite type. This result, combined with Proposition 2.4 (1)and Lemma 2.1, implies the ‘if’ part of Theorem 2.6.
Remark . It is well known that a finite-to-one factor code between irreducibleshifts of finite type is open if and only if it is bi-closing [14]. Thus Theorem 2.6 canbe thought as an infinite-to-one version of this fact. In the finite-to-one case, we alsohave the following relation: A finite-to-one factor code between irreducible shifts offinite type is open exactly when it is constant-to-one. By virtue of this relation, itis natural to ask whether an infinite-to-one code between irreducible shifts of finitetype is open if and only if it is infinite-to-one everywhere, i.e., | φ − ( y ) | = ∞ for all y . However, it turns out that each implication is not true.We summarize the implications between factor codes in the following diagram.When X is of finite type, then all conditions but bi-continuing a.e. are equivalentby Proposition 2.4 and Theorem 2.6. We present examples to show that in generalthese conditions are different. open with auniform lifting length (a) (cid:11) (cid:19) (cid:43) (cid:51) bi-continuing witha bi-retract (b) (cid:11) (cid:19) Y =SFT (cid:110) (cid:118) open X =synchronized(c) (cid:34) (cid:42) (cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76) (cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76)(cid:76) bi-continuing (d) (cid:117) (cid:125) (cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115)(cid:115) bi-continuing a.e. PEN AND BI-CONTINUING CODES 7
Example 2.9. (1) Let X = X W and Y = X W , where W = { ab k c k : k ≥ } and W = { a, abc } . Define Φ : B ( X ) → A ( Y ) as follows: Φ( abc ) = b, Φ( bca ) = c andΦ( u ) = a otherwise. Let φ : X → Y be a code defined by φ ( x ) i = Φ( x [ i − ,i +1] ). Notethat φ replaces each b k c k , k > a k and also b ∞ , c ∞ with a ∞ .First we show that φ is not right continuing. Take x = b ∞ ∈ X and y = a ∞ . ( abc ) ∞ ∈ Y . If z (cid:54) = b ∞ is left asymptotic to x , then z is in the orbit of b ∞ .c ∞ ,and we have φ ( z ) = a ∞ (cid:54) = y . Thus φ is not right continuing. Similarly it is not leftcontinuing.Next we show that φ is open. Let U be an open set of X and x ∈ U . Since U isopen, there is l ∈ N with V = − l [ x − l · · · x l ] ⊂ U . We claim that there is ¯ x ∈ V suchthat φ (¯ x ) = φ ( x ) and there are i ≤ − l and j ≥ l with ¯ x i = ¯ x j = a . If x already hasthis property, we are done. Otherwise, there are two cases (up to symmetry): Case
1. There is i ≤ − l with x i = a and x j (cid:54) = a for all j ≥ l . Choose ¯ j ∈ Z suchthat x ¯ j = a is the rightmost occurrence of a in x . Let ¯ x = x ( −∞ ,l ] b c l − ¯ j +2 a ∞ . Since x (¯ j, ∞ ) must be of the form b ∞ , it follows that ¯ x ∈ V and φ (¯ x ) = φ ( x ). Case a does not occur in x ( −∞ , − l ] and x [ l, ∞ ) . Suppose first that a occurs in x ( − l,l ) . Let ¯ i and ¯ j be the leftmost and rightmost occurrence of a in x , respectively.Then ¯ x = a ∞ b ¯ i − l +2 c x [ − l,l ] b c l − ¯ j +2 a ∞ ∈ X is the desired point. Otherwise, x mustbe one of the following form: b ∞ , c ∞ , or b ∞ c ∞ . If x = b ∞ , then ¯ x = a ∞ b l .b l +1 c l +1 a ∞ satisfies the property. Other cases can be proved similarly. This proves the claim.Take i ≤ − l and j ≥ l such that ¯ x i = ¯ x j = a . For each y ∈ Y with y [ i,j ] = φ ( x ) [ i,j ] ,let z = y ( −∞ ,i ) ¯ x [ i,j ] y ( j, ∞ ) . Then we have z ∈ V and φ ( z ) = y . So i [ φ ( x ) [ i,j ] ] ⊂ φ ( U ).Thus φ is an example of an open code which is neither right nor left continuing. ByLemma 2.1, φ has no uniform lifting length. Thus the converse of (a) does not holdin general.(2) This example is due to J. Yoo [17]. Let X be a shift space on the alphabet { , ¯1 , , } defined by forbidding { ¯12 n n ≥ } , and Y the full 3-shift { , , } Z .Define φ : X → Y by letting φ (¯1) = 1 and φ ( a ) = a for all a (cid:54) = ¯1.We first show that φ is right continuing: Suppose x ∈ X and y ∈ Y satisfy φ ( x ) ( −∞ , = y ( −∞ , . If x (1) = x ( −∞ , − .x y [1 , ∞ ) is in X , then we are done. Otherwise,the word ¯12 n x (1) . Let x (2) be the point obtained from x (1) by replacing ¯12 n n
3. Then φ ( x (2) ) = y and x (2) is left asymptotic to x . Butconsidering x = ¯1 ∞ n . ∞ and y = 1 ∞ n . ∞ for each n ∈ N , one can see that φ has no (right continuing) retract.Now we show that φ is left continuing with a (left continuing) retract. If x ∈ X and y ∈ Y satisfy φ ( x ) [0 , ∞ ) = y [0 , ∞ ) , then by letting ¯ x = y ( −∞ , − .x [0 , ∞ ) , we have¯ x ∈ X , φ (¯ x ) = y and ¯ x [0 , ∞ ) = x [0 , ∞ ) . Thus φ is left continuing with retract 0, henceit is bi-continuing but does not have a right continuing retract. So a continuing codemay not have a retract and the converse of (b) does not hold.Moreover, φ is an example of a bi-continuing code which is not open. For, if φ is open, then φ ([¯1]) = (cid:83) ki =1 C i for central 2 l + 1 cylinders C i in Y . Since 1 ∞ . ∞ ∈ φ ([¯1]), there exists C i of the form − l [1 l +1 l ]. Let y = 1 ∞ . l ∞ ∈ C i . Then y / ∈ φ ([¯1]),a contradiction comes. Indeed X is an irreducible strictly sofic shift. UIJIN JUNG (3) As is easily seen, if Y is the even shift and φ : X A → Y is its canonical cover(i.e., a code given by its minimal right resolving presentation [12]), then φ is bi-continuing a.e. but neither open nor bi-continuing. Thus the converses of (c) and(d) are false even if the domain is of finite type.3. Almost specified shifts
To investigate the existence of continuing codes, we consider the almost specifiedshifts. In this section we present several properties of almost specified shifts definedas follows. These will be used in subsequent sections.
Definition 3.1.
Let X be a shift space.(1) X has the specification property if there exists N ∈ N such that for all u, v ∈ B ( X ), there exists w ∈ B N ( X ) with uwv ∈ B ( X ).(2) X has the almost specification property (or X is almost specified ) if thereexists N ∈ N such that for all u, v ∈ B ( X ), there exists w ∈ B ( X ) with uwv ∈ B ( X ) and | w | ≤ N .Note that an irreducible sofic shift is almost specified. Lemma 3.1. [3] If X is an almost specified shift, then it is a synchronized system.Proof. Let N satisfy the condition in Definition 3.1 (2). For given u, v ∈ B ( X ),let B ( u, v ) be the set of words w such that uwv ∈ B ( X ) and | w | ≤ N . Note that B ( u, v ) is a nonempty finite set. Fix two words u (0) and v (0) in B ( X ). We define u ( n ) and v ( n ) inductively as follows. Suppose that u (0) , · · · , u ( n − and v (0) , · · · , v ( n − aregiven. If there are u, v ∈ B ( X ) such that ∅ (cid:54) = B ( uu ( n − · · · u (0) , v (0) · · · v ( n − v ) (cid:32) B ( u ( n − · · · u (0) , v (0) · · · v ( n − ) , then let u ( n ) = u and v ( n ) = v . Since B ( u, v )’s are nonempty finite sets, thisprocess eventually terminates in the sense that there is m ∈ N such that by letting¯ u = u ( m ) · · · u (0) and ¯ v = v (0) · · · v ( m ) we have B ( u ¯ u, ¯ vv ) = B (¯ u, ¯ v ) for all u, v ∈ B ( X )with u ¯ u, ¯ vv ∈ B ( X ). Then ¯ uw ¯ v is a synchronizing word for each w ∈ B (¯ u, ¯ v ). (cid:3) The following lemma shows that the almost specification property is a naturalirreducible version of the specification property.
Lemma 3.2.
Let X be a mixing shift space. Then X has the almost specificationproperty if and only if it has the specification property.Proof. Suppose X is almost specified. Let N be given as in Definition 3.1 (2) andby using Lemma 3.1 take a synchronizing word w ∈ B ( X ). Since X is mixing, thereis L > l ≥ L , we can find u ∈ B l ( X ) with wuw ∈ B ( X ). Let M = 2 N + 2 | w | + L . For any u, v ∈ B ( X ), there exist ¯ u, ¯ v ∈ B ( X ) with | ¯ u | , | ¯ v | ≤ N and u ¯ uw, w ¯ vv ∈ B ( X ). Also we can find ¯ w ∈ B L +2 N −| ¯ u |−| ¯ v | ( X ) with w ¯ ww ∈ B ( X ).Let ˜ w = ¯ uw ¯ ww ¯ v . Then u ˜ wv ∈ B ( X ) and | ˜ w | = M . Thus M satisfies the conditionin Definition 3.1 (1) and X has the specification property. (cid:3) PEN AND BI-CONTINUING CODES 9
For a shift space X , denote by X − ( X + , resp.) the set of left (right, resp.) infinitesequences of X . For x ∈ X , we denote by x − (resp. x + ) the sequence x ( −∞ , − ∈ X − (resp. x [0 , ∞ ) ∈ X + ). If x − ∈ X − and u ∈ B ( X ), their concatenation x − u isnaturally defined. We write x − u ∈ X − when there exists a sequence y ∈ X + suchthat x − uy ∈ X . A similar operation is defined on B ( X ) and X + . Then we have thefollowing lemma whose proof is obvious by compactness. Lemma 3.3.
Let X be a shift space. (1) X has the specification property if and only if there is N ∈ N such that when-ever x − ∈ X − and u ∈ B ( X ) , there is w ∈ B N ( X ) such that x − wu ∈ X − . (2) X is almost specified if and only if there is N ∈ N such that whenever x − ∈ X − and u ∈ B ( X ) , there is w ∈ B ( X ) such that | w | ≤ N and x − wu ∈ X − . Example 3.4.
Let S = { n , n , . . . } ⊂ Z + with n i < n i +1 (possibly, finite) and X = X ( S ) the coded system generated by { n : n ∈ S } . This X ( S ) is called the S -gap shift [12]. Note that X is a synchronized system. Then we have the followingcharacterizations.(1) X is almost specified if and only if sup i | n i +1 − n i | < ∞ .(2) X is mixing if and only if gcd { n + 1 : n ∈ S } = 1.(3) X has the specification property if and only if sup i | n i +1 − n i | < ∞ andgcd { n + 1 : n ∈ S } = 1. Proof. (1) If S is finite, then X is of finite type (hence almost specified) andsup i | n i +1 − n i | < ∞ , thus the equivalence is clear. So we may assume S is infi-nite. Suppose that X is almost specified. Let N be given as in Definition 3.1 (2).Then for each n i ∈ S , there exists w such that 10 n i +1 w ∈ B ( X ) and | w | ≤ N .Thus | n i +1 − n i | ≤ N + 1 for all i . Conversely, if L = sup i | n i +1 − n i | < ∞ , then let N = max( n , L ). For any v, w ∈ B ( X ), there exist i, j ≤ N with v i , j w ∈ B ( X ).Then v i j w ∈ B ( X ) and | i j | ≤ N + 1. Thus X is almost specified.(2) If X is mixing, then there exists N > n ≥ N , there exists w ∈ B n ( X ) with 1 w ∈ B ( X ). Thus there are words of length N + 1 and N + 2 ofthe form 10 i i . . . i m with i k ∈ S , implying that gcd { n + 1 : n ∈ S } = 1.On the other hand, if gcd { n + 1 : n ∈ S } = 1 then for all sufficiently large n ,there is a word of length n of the form 10 i i . . . i m
1. Since 1 is synchronizingand X is irreducible, the result follows.(3) This follows from (1), (2) and Lemma 3.2. (cid:3) Example 3.5.
We list some examples reflecting strict inclusions between classes ofshift spaces.(1) Let X = X ( S ) be the S -gap shift, where S = (2 N + 1) \ (cid:20) ∞ (cid:91) k =1 (cid:0) k N ∩ (10 k − , k ) (cid:1)(cid:21) . Then it is a nonsofic almost specified shift without the specification property.(2) For any X with the specification property, define its root (cid:101) X by (cid:101) X = { (¯ x i ) i ∈ Z : ∃ x ∈ X with ¯ x i +1 = a, ¯ x i = x i for all i ∈ Z } , where a is a symbol which is not in B ( X ). Then Y = (cid:101) X ∪ σ ( (cid:101) X ) is almostspecified, but does not have the specification property.(3) Let W = { ab k c k : k ≥ } and X = X W . It is easy to see that X is mixingand synchronized. But if x − = b ∞ , then there is no w ∈ B ( X ) such that x − wa ∈ X − . By Lemma 3.3, X cannot be almost specified.A synchronized system has a global periodic structure similar to an irreducibleshift of finite type. The collection of the subsets D i in the following theorem is called the cyclic cover of X . It is unique up to cyclic permutation. Theorem 3.6. [16]
Let X be a synchronized system. Then there exist a unique p ∈ N and closed sets D i ⊂ X , i = 0 , , · · · , p − , such that i) X = (cid:83) p − i =0 D i , ii) σ ( D i ) = D i +1 (mod p ) , iii) σ p | D i is mixing for all i = 0 , , · · · , p − , and vi) D i ∩ D j has empty interior when i (cid:54) = j . If X is an irreducible shift of finite type, then this decomposition is well known(cf, [12]). In this case p = per( X ) and D i ∩ D j = ∅ for i (cid:54) = j .Let X p be the p th higher power shift of X and γ : X → X p the p th higherpower code [12] given by γ ( x ) i = x [ ip,ip + p − . If u ∈ B pM ( X ), then γ ( u ) ∈ B M ( X p )is naturally defined. This γ is a topological conjugacy between ( X, σ p ) and ( X p , σ ).When X is a synchronized system, we define X i = γ ( D i ). Then X i is an irreduciblesubshift of X p and γ is also a topological conjugacy between ( D i , σ p ) and ( X i , σ )for each 0 ≤ i < p . Note that for a block u ∈ B pM ( X ), γ ( u ) ∈ B ( X i ) if and only ifthere exists a point x ∈ D i with x [0 ,pM − = u . From the construction of the cycliccover of X , it follows that the sets { γ ( u ) ∈ B ( X i ) : u is a synchronizing word for X } are disjoint. This is because X i is the closure of an ‘ irreducible subshift ’ of X p in thesense of Thomsen (see [16, § u ∈ B pM ( X ), thereexists a unique i with γ ( u ) ∈ B ( X i ). We use this fact to prove the following result. Lemma 3.7.
Let X be an almost specified shift. If { D , · · · , D p } is the cyclic coverof X , then σ p | D i has the specification property.Proof. It suffices to show the case i = 0. Note that σ p | D has the specificationproperty if and only if X has the specification property. Since X is almost specified,there exists N satisfying the condition in Definition 3.1 (2). We claim that X isalso almost specified.Suppose that ˜ u = γ ( u ) , ˜ v = γ ( v ) ∈ B ( X ). By extending ˜ u to the left and ˜ v tothe right, we may assume that u, v are synchronizing words for X . Note that | u | , | v | are multiples of p . Since X is almost specified, there exists a word w ∈ B ( X ) with l = | w | ≤ N such that uwv ∈ B ( X ). Let x ∈ X be a point with x [0 , | uwv |− = uwv .Since u is synchronizing and γ ( u ) = ˜ u ∈ B ( X ), we have γ ( x ) ∈ X (by the remarkfollowing Theorem 3.6). Thus γ ( σ | u | + l ( x )) ∈ X l (mod p ) , and we get ˜ v ∈ B ( X l (mod p ) ).Since v is synchronizing and ˜ v ∈ B ( X ), it follows that l = 0 (mod p ) and we have˜ uγ ( w )˜ v = γ ( uwv ) ∈ B ( X ) . PEN AND BI-CONTINUING CODES 11
Thus X is almost specified, as desired. Since X is mixing, the result follows fromLemma 3.2. (cid:3) Proposition 3.8.
Let X be an almost specified shift with h ( X ) > . For given (cid:15) > , there exist an irreducible shift of finite type Z ⊂ X and k ∈ N such that h ( Z ) > h ( X ) − (cid:15) and any k -block of Z is a synchronizing word of X . If X ismixing, then Z can be chosen to be mixing.Proof. The proof essentially follows the lines in [5, 13]. First, we prove the casewhere X has the specification property. Using Lemma 3.1 find a synchronizingword w ∈ B ( X ). For each k > | w | , let X k be the ( k − k -blocks are a . . . a k ∈ B k ( X ) such that w occurs in a . . . a k . Then X k ⊂ X .Let N be given in Definition 3.1 (1). Fix k > N + 2 | w | . For each l = mk with m ∈ N , we have |B l ( X k ) | ≥ |B k − N − | w | ( X ) | m , since for each m pair of words v (1) , · · · , v ( m ) ∈ B k − N − | w | , there exists a word in B l ( X k ) whose initial sequence is of the form wu (1) v (1) ¯ u (1) wu (2) v (2) ¯ u (2) w · · · v ( m ) ¯ u ( m ) w ,with u ( i ) , ¯ u ( i ) ∈ B N ( X ). It follows that h ( X k ) ≥ k log |B k − N − | w | ( X ) | . Thuslim k →∞ h ( X k ) = h ( X ). It is easy to see that X k is mixing for all large k . Take k large so that X k is mixing, h ( X k ) > h ( X ) − (cid:15) and let Z = X k .Next, we prove the case where X is almost specified. Let { D , · · · , D p } be thecyclic cover of X and take X = γ ( D ) as in the remark following Theorem 3.6.Since X has the specification property by Lemma 3.7, by applying the previousresult to X , we get Z ⊂ X satisfying the properties. Define Z = p − (cid:91) i =0 σ i ( γ − ( Z )) . Then Z satisfies all the desired properties. (cid:3) Remark . There is a synchronized system X such that sup Y ⊂ X h ( Y ) < h ( X ),where the supremum is taken over all sofic subshifts Y of X [15]. Thus Proposition3.8 does not hold when X is merely synchronized.4. Extension Theorem: A mixing case
In this section, we prove that when X is almost specified and Y is mixing andof finite type, the entropy and the periodic conditions guarantee the existence of abi-continuing factor code from X to Y with a bi-retract. With little extra work,in fact we prove further that every code from a proper subshift of X to Y can beextended to an open bi-continuing code. First we need some lemmas. Lemma 4.1.
Let X and Y be irreducible shifts of finite type with h ( X ) > h ( Y ) .Then there exist an irreducible shift of finite type Z ⊂ X and a bi-closing factorcode π : Z → Y . Proof.
We can assume Y = X B with B irreducible. Let m be the size of B . For each n ∈ N , define Y n = X B n , where B n is the nm × nm irreducible matrix given by B n = B · · ·
00 0 B · · · · · · BB · · · . Note that per( Y n ) = n · per( Y ) and p kn ( Y n ) = n · p kn ( Y ) for each k ∈ N . For each n ∈ N , it is easy to see that the natural projection code from Y n to Y is bi-resolving(e.g., see [14]). So it suffices to show that Y n embeds into X for large n .Let (cid:15) = ( h ( X ) − h ( Y )) /
3. Take n large such that per( X ) | n , n − log n < (cid:15), kn log p kn ( Y ) < h ( Y ) + (cid:15) for all k ∈ N , and1 kn log q kn ( X ) > h ( X ) − (cid:15) for all k ∈ N . Then for each k ∈ N ,1 kn log n + 1 kn log p kn ( Y ) < h ( Y ) + 2 (cid:15) = h ( X ) − (cid:15) < kn log q kn ( X )so we have n · p kn ( Y ) < q kn ( X ) , and therefore q kn ( Y n ) ≤ p kn ( Y n ) = n · p kn ( Y ) < q kn ( X ) . Since p j ( Y n ) = 0 if j / ∈ n N , we have q j ( Y n ) ≤ q j ( X ) for all j ∈ N . Now applying theKrieger’s Embedding Theorem [11] gives the result. (cid:3) Lemma 4.2. [8, Theorem 26.17]
Let X be a mixing shift of finite type and (cid:101) X aproper subshift of X . For given h < h ( X ) , there is a mixing shift of finite type Z ⊂ X such that h ( Z ) > h and Z ∩ (cid:101) X = ∅ . Lemma 4.3 (Extension Lemma) . [5] Let X be a shift space and Y a mixing shiftof finite type such that P ( X ) (cid:38) P ( Y ) . Then any code from a proper subshift of X to Y can be extended to a code from X to Y . It is the following theorem which we extend in this section. By usual reductionto the mixing case, Theorem 4.4 shows that the entropy and the periodic conditionsguarantee the existence of a right continuing factor code between two irreducibleshifts of finite type. However, to extend an arbitrary code on a proper subshift to acode on the whole domain, the mixing condition on Y is crucial (see Example 5.6). Theorem 4.4. [7]
Let X be an irreducible shift of finite type and Y a mixing shiftof finite type such that h ( X ) > h ( Y ) and P ( X ) (cid:38) P ( Y ) . Then any code from aproper subshift of X to Y can be extended to a right continuing code on X . Now we are ready to prove the main theorem in this section. As indicated in § § PEN AND BI-CONTINUING CODES 13
Theorem 4.5.
Let X be an almost specified shift and Y a mixing shift of finite typesuch that h ( X ) > h ( Y ) and P ( X ) (cid:38) P ( Y ) . Then any code from a proper subshiftof X to Y can be extended to an open bi-continuing code on X with a bi-retract.Proof. Suppose that (cid:101) X is a proper subshift of X and ˜ φ : (cid:101) X → Y is a code. Wewill construct a bi-continuing code φ : X → Y with a bi-retract such that φ | e X = ˜ φ .We divide the proof into three parts. In Part I, we construct an extension code inthe case where X has the specification property. In Part II, we prove that the codeconstructed in Part I is bi-continuing with a bi-retract. In Part III, by using theresults in Part I and II we prove the case where X is almost specified. Part I. In this part, we prove the case where X has the specification property.By Proposition 3.8, there exists a mixing shift of finite type Z with h ( Z ) > h ( Y )in which all sufficiently long blocks are synchronizing for X . By Lemma 4.2, we canfind a mixing shift of finite type Z ⊂ Z disjoint from (cid:101) X with h ( Z ) > h ( Y ). Alsoby Lemma 4.1, there exist an irreducible shift of finite type Z ⊂ Z and a biclosingfactor code π : Z → Y . By Lemma 4.3, we can find a factor code ψ : X → Y suchthat ψ | Z = π and ψ | e X = ˜ φ . Finally find a mixing shift of finite type V ⊂ Z disjointfrom Z with h ( V ) > h ( Y ) by using Lemma 4.2.By passing to higher block shifts, we can assume that(a) Z , Z, V and Y are edge shifts,(b) A ( Z ) ∩ A ( (cid:101) X ) = ∅ and A ( Z ) ∩ A ( V ) = ∅ ,(c) ψ is a 1-block code,(d) if a, b ∈ A ( Z ) and ab ∈ B ( X ), then ab ∈ B ( Z ),(e) each a ∈ A ( Z ) is a synchronizing word for X .Let D be the right and left closing delay of π [12]. Take N large such that N is atransition length for X , Y and a weak transition length for Z . Fix a synchronizingword α ∈ A ( V ). Since X has the specification property, by Lemma 3.3 for each x ∈ X there exist w, ¯ w ∈ B N ( X ) such that x − wα ∈ X − and α ¯ wx + ∈ X + .Fix i (cid:29) N . For each a ∈ A ( X ) , b ∈ A ( Z ) and w ∈ B N ( X ), define HL i ( a ; w ; b ) = { u ∈ B Xi ( a, b ) : u [1 ,N +2] = awα, u [ N +2 ,i − N ) ∈ B ( V ) ,u i − N / ∈ A ( Z ) , and u ( i − N,i ] ∈ B ( Z ) } ; LH i ( b ; w ; a ) = { u ∈ B Xi ( b, a ) : u [1 ,N ] ∈ B ( Z ) , u N +1 / ∈ A ( Z ) ,u (2 N +1 ,i − N − ∈ B ( V ) , and u [ i − N − ,i ] = αwa } . (Note that there is no explicit restriction on subinterval [ i − N, i − N ) in thedefinition of HL i ( a ; w ; b ). It guarantees that for a fixed a ∈ A ( X ) and w ∈ B N ( X )with awα ∈ B ( X ), there exists b ∈ A ( Z ) with HL i ( a ; w ; b ) (cid:54) = ∅ . Similarly for LH i ( b ; w ; a ).) Since h ( V ) > h ( Y ), there is I ∈ N such that |HL I + N ( a ; w ; b ) | ≥ |B YI + N ( ψa, ψb ) | and |LH I + N ( b ; w ; a ) | ≥ |B YI + N ( ψb, ψa ) | for all a , b and w ∈ B ( X ) whenever these sets are nonempty. (This is possible sinceif HL I + N ( a ; w ; b ) (cid:54) = ∅ , then the asymptotic cardinality of this set is greater than ce ( I + N ) h ( V ) for some c > LH I + N ( b ; w ; a ).) For each a ∈ B ( X ) and b ∈ A ( Z ), define surjections Ψ a,bHL from B XI + N ( a, b ) onto B YI + N ( ψa, ψb ) such that the restriction Ψ a,bHL | HL I + N ( a ; w ; b ) is surjectivefor each w ∈ B N ( X ) with HL I + N ( a ; w ; b ) (cid:54) = ∅ . Similarly define surjections Ψ b,aLH .Finally for each 2 N ≤ j ≤ N + 2 I , define a map Φ j : A ( X ) → B j ( Y ) such thatΦ j ( c, d ) ∈ B Yj ( ψc, ψd ). This is possible since N is a transition length for Y . Thesemaps will be used as marker fillers.For given x ∈ X , we divide x ∈ X into low and high-stretches as in [7]. Call asegment of x a low-stretch if it is a Z -word of length > N + D , and not precededor followed by a symbol from A ( Z ), i.e., a maximal Z -word of length > N + D .Remaining stretches of maximal length are called high-stretches (of x ). By thecondition (d), low-stretches of x cannot overlap and hence x is uniquely decomposedas low and high-stretches. Also, if a high-stretch of x is of length greater than 2 I ,then it is called a long high-stretch. Otherwise, call it a short high-stretch. Thefigure below shows a typical decomposition of x .Now we define a code φ : X → Y . Let x ∈ X .i) low-stretches. If x [ i − N,i + N ] is in a low-stretch, then define φ ( x ) i = ψ ( x i ).ii) long high-stretches. If x [ i − I,i + I ] is in a long high-stretch, let φ ( x ) i = ψ ( x i ).iii) short high-stretches. If x [ i,j ] is a short high-stretch, then j − i + 1 ≤ I .Define φ ( x ) [ i − N,j + N ] = Φ N + j − i +1 ( x i − N , x i + N ).iv) high-low transition. If x [ i,i + I ) is the end of some long high-stretch and x [ i + I,i + N + I ) is the beginning of some low-stretch, then define φ ( x ) [ i,i + N + I ) = Ψ x i ,x i + N + I − HL ( x [ i,i + N + I ) ) . v) low-high transition. Similarly as in (iv), using Ψ LH .The figure above describes the action of φ on parts of a point corresponding tosome of these cases. Note that these cases cover all parts of x and φ is a well-definedcode from X to Y . Indeed φ has memory and anticipation 2 N + 2 I + D . Since x ∈ Z consists of a single low-stretch and x ∈ (cid:101) X consists of a single high-stretch, wehave φ | Z = π and φ | e X = ˜ φ and hence φ is a factor code which is an extension of ˜ φ . PEN AND BI-CONTINUING CODES 15
Part
II.
We show that φ is bi-continuing with bi-retract n = 2 I + 6 N + 3 D .Suppose x ∈ X , y ∈ Y satisfy φ ( x ) ( −∞ , = y ( −∞ , . Case
1. Suppose there exists an i ∈ [ − n, − N − D ] such that x [ i,i +2 N + D ] is part ofa low-stretch. Then, since π is a right closing factor code, there exists a one-sidedsequence z [ i + N, ∞ ) in Z + such that z i + N = x i + N and π ( z [ i + N, ∞ ) ) = y [ i + N, ∞ ) . Definea point ¯ x by ¯ x k = x k if k ≤ i + N , and ¯ x k = z k if k ≥ i + N . Note that ¯ x ∈ X since x i + N is synchronizing. Also note that ¯ x [ i, ∞ ) is a part of low-stretch of ¯ x andtherefore rule i) applies for k ≥ i + N and we have φ (¯ x ) = y . Case
2. If Case 1 does not holds, then x [ − n +2 N + D, − N − D ] is part of a high-stretch(note that there may be a part of a low-stretch at the left end of the interval). Thisinterval is of length 2 I + 2 N + D and thus it is a part of a long high-stretch of x .Since there is no Z -word of length greater than 2 N + D in this part, there exists a ∈ A ( X ) \ A ( Z ) and − N − D − I ≤ i ≤ − N − D − I with x i = a (by (d) in therecoding step). Since X has the specification property, by Lemma 3.3 there exists w ∈ B N ( X ) such that x ( −∞ ,i ] wα ∈ X − . Take b ∈ A ( Z ) with ψ ( b ) = y i + I + N − and HL I + N ( a ; w ; b ) (cid:54) = ∅ . Claim.
Such b exists. Proof.
Let p = per( Z ). Then there exists a partition { A , A , · · · , A p − } whoseunion is A ( Z ) with the property that whenever ab ∈ B ( Z ) and a ∈ A k , we have b ∈ A k +1(mod p ) . Since Y is mixing and π : Z → Y is a factor code, for each kπ | A k : A k → A ( Y ) is onto. Take any block v of length I − N − v [1 ,N +2] = awα and v [ N +2 ,I − N ) ∈ B ( V ). There exist d ∈ A ( Z ) and c ∈ A ( Z ) \ A ( Z ) such that cd ∈B ( X ). If we take ¯ k with d ∈ A ¯ k , then there exists b ∈ A ¯ k such that π ( b ) = y i + I + N − .Since N is a weak transition length for Z , there exists a block u ∈ B N ( Z ) with u = d and u N = b . As Z is an edge shift containing Z , we have cu ∈ B ( X ). Since N is a transition length for X , we can find a block w ∈ B N ( X ) with vwcu ∈ B ( X ).Then vwcu ∈ HL I + N ( a ; w ; b ), which completes the proof. (cid:3) Since Ψ a,bHL | HL I + N ( a ; w ; b ) is onto, there exists u [ i,i + I + N ) ∈ HL I + N ( a ; w ; b ) with Ψ a,bHL ( u )= y [ i,i + I + N ) . Since u i + I + N − ∈ A ( Z ) and π is right closing, there exists z [ i + I + N − , ∞ ) in Z + such that z i + I + N − = u i + I + N − and π ( z [ i + I + N − , ∞ ) ) = y [ i + I + N − , ∞ ) . Let¯ x k = x k if k ≤ iu k if i ≤ k ≤ i + I + N − z k if k ≥ i + I + N − x ∈ X since α and u i + I + N − are synchronizing. Note that ¯ x [ − n +2 N + D,i + I ) is apart of long high-stretch and ¯ x [ i + I, ∞ ) is a low-stretch (our chosen block ¯ x i = x i = a / ∈ A ( Z ) guarantees no occurrence of a Z -block of length greater than 2 N + D in¯ x [ − n +2 N + D,i + I ) ). Therefore rules ii), iv) and i) apply to ¯ x [ − n +2 N + D + I, ∞ ) and we have φ (¯ x ) = y . So φ is right continuing with retract n . Similarly φ is left continuing withretract n , which completes the proof when X has the specification property. Part
III.
We prove the general case where X is merely almost specified. Let { D , D , · · · , D p − } be the cyclic cover of X . Define Z = (cid:91) i (cid:54) = j D i ∩ D j . Then Z is a proper subshift of X . By using Lemma 4.3, we may assume that Z is contained in (cid:101) X . Note that σ p | D ∩ e X is (conjugate to) a shift space. Also σ p | D has the specification property by Lemma 3.7. The code ˜ φ naturally induces a code˜ ψ : ( D ∩ (cid:101) X, σ p ) → ( Y, σ p ) by restriction. By applying Parts I and II, we have a σ p -commuting bi-continuing code ψ : ( D , σ p ) → ( Y, σ p ) with a bi-retract such that ψ | D ∩ e X = ˜ ψ . Define a code φ : X → Y as follows: Given x ∈ X , there exists0 ≤ i < p with x ∈ D i . Then we let φ ( x ) = σ i ψ ( σ − i ( x )). If x ∈ D j for some j (cid:54) = i ,then x ∈ Z ⊂ (cid:101) X and therefore σ j ψ ( σ − j ( x )) = σ j ˜ ψ ( σ − j ( x )) = σ j ˜ φ ( σ − j ( x )) = ˜ φ ( x )= σ i ˜ φ ( σ − i ( x )) = σ i ˜ ψ ( σ − i ( x )) = σ i ψ ( σ − i ( x )) . Thus φ is well defined. These equalities also show that φ | e X = ˜ φ . Note that φ iscontinuous. It is easy to see that φ is indeed σ -commuting (hence a code) and bi-continuing with a bi-retract. By Lemma 2.1, φ is also an open code with a uniformlifting length. (cid:3) Remark . Suppose X is assumed to be of finite type in Theorem 4.5. Then wecan take Z = X in Part I. Also Case 2 in Part II can be proved as easily as in Case1 (since any word is synchronizing) and by using a usual reduction to the mixingcase, Part III can be removed. Thus in this case the proof gives an alternative proofof bi-continuing version of Theorem 4.4 in which the definition of φ is much simpler.This is because by using the extension lemma, we need not consider the marker setsas in [7] and thus φ on low-high and high-low transitions can be defined more easily.5. Lower entropy factor theorems
For a synchronized system X , fix a synchronized word w ∈ B ( X ). Let C n ( X ) bethe set of words v ∈ B n ( X ) such that wvw ∈ B ( X ). Then a synchronized entropy h syn ( X ) is defined by h syn ( X ) = lim sup n →∞ n log | C n ( X ) | . This value is independent of w and h ( X ) ≥ h syn ( X ). In general h ( X ) (cid:54) = h syn ( X ).If X is almost specified, then by Proposition 3.8 and Theorem 3.2 of [16] we have h syn ( X ) = h ( X ). By using the cyclic cover of a synchronized system, Thomsen hasgiven a generalization of the lower entropy factor theorem in [5] for the case where Y is merely an irreducible shift of finite type. Theorem 5.1. [16]
Let X be a synchronized system and Y an irreducible shift offinite type such that h syn ( X ) > h ( Y ) . Then Y is a factor of X if and only if thefollowing hold: PEN AND BI-CONTINUING CODES 17 i) P ( X ) (cid:38) P ( Y ) . ii) If q = per( Y ) and { D , D , · · · , D p − } is the cyclic cover of X , then q | p and (cid:18) pq − (cid:91) j =0 D i + jq (cid:19) ∩ (cid:18) pq − (cid:91) j =0 D k + jq (cid:19) = ∅ when i, k ∈ { , , . . . , q − } , i (cid:54) = k . Note that when X is of finite type, then the condition (ii) is trivial. Using thisresult, we have our main theorem. Theorem 5.2.
Let X be an almost specified shift and Y an irreducible shift of finitetype such that h ( X ) > h ( Y ) . Then the following are equivalent. (1) Y is a factor of X by a bi-continuing code with a bi-retract. (2) Y is a factor of X by an open code with a uniform lifting length. (3) Y is a factor of X .If X is of finite type, then the above conditions are also equivalent to: (4) P ( X ) (cid:38) P ( Y ) .Proof. (1) ⇒ (2) follows from Lemma 2.1. (2) ⇒ (3) is clear. Now suppose (3).Then we have two conditions in Theorem 5.1. Let { E , E , · · · , E q − } be the cycliccover of Y . Let C = (cid:83) pq − j =0 D jq . Note that { C, σ ( C ) , · · · , σ q − ( C ) } is a partitionof X and ( C, σ q ) is (conjugate to) a shift space. As in the proof of Lemma 3.7,it is easy to see that σ q | C is an almost specified shift. Since q = per( Y ), σ q | E is a mixing shift of finite type. By Theorem 4.5, there exists a ( σ q -commuting)code ψ : ( C, σ q ) → ( E , σ q ) which is bi-continuing with a bi-retract. As usual, wecan define a factor code φ : X → Y as follows: For x ∈ X , there exists a unique0 ≤ i < q with x ∈ σ i ( C ). Define φ ( x ) = σ i ψ ( σ − i ( x )). It is easy to check that φ isa ( σ -commuting) code and bi-continuing with a bi-retract. Thus (1) holds.Finally, when X is of finite type, then by using a usual reduction to mixing case(cf. [12]), the problem can be reduced to Proposition 4.5. Thus (4) implies (1). (cid:3) Remark . The equivalence of (1), (2) and (3) in Theorem 5.2 fails under thehypothesis of Theorem 5.1, that is, when X is a synchronized system and Y is anirreducible shift of finite type with h syn ( X ) > h ( Y ).For example, let X = X W and Y = X W , where W = { ab k c k : k ≥ } and W = { a, abc } . Note that X is a synchronized system and Y is an irreducible shiftof finite type. Since Y is a proper subsystem of a mixing sofic shift X W , where W = { ab k c k : k ≤ } , it follows that h ( Y ) < h ( X W ) ≤ h syn ( X ). Since Y has theunique fixed point a ∞ , we have P ( X ) (cid:38) P ( Y ) and Y is a factor of X .We claim that there exists no right continuing code from X to Y . Indeed, if φ : X → Y is a code, then let x = b ∞ and y = a ∞ . · · · where y [0 , ∞ ) is right transitive.Then φ ( x ) and y are left asymptotic. But when z (cid:54) = b ∞ is left asymptotic to x , then z is in the orbit of b ∞ .c ∞ , so φ ( z ) must be of the form a ∞ · · · a ∞ , since a ∞ is theonly fixed point in Y . Thus φ ( z ) (cid:54) = y and φ cannot be right continuing. By Lemma2.1, φ cannot be open with a uniform lifting length. As we have shown in Example 2.9 (1), there is an open code from X to Y . Atthis time, it is an open question whether there exists an open factor code from asynchronized system X to an irreducible shift of finite type Y if there is a factorcode from X to Y .From Lemma 2.3 and Theorem 5.2, the following result is immediate. Corollary 5.4.
Let X be an irreducible shift of finite type and Y a shift space. Then Y is a lower entropy open factor of X if and only if h ( X ) > h ( Y ) , P ( X ) (cid:38) P ( Y ) and Y is an irreducible shift of finite type. Let X and Y be synchronized systems with their cyclic covers { D , D , · · · , D p − } and { E , E , · · · , E q − } , respectively. If φ : X → Y is a code, then there exists aninteger 0 ≤ j < q with φ ( D ) ⊂ E j since σ pq acts transitively on D i . So for acode ˜ φ : (cid:101) X → Y from a proper subshift (cid:101) X of X to be extended to a code on X , anecessary condition is that there exists an integer 0 ≤ j < q with˜ φ ( (cid:101) X ∩ D ) ⊂ E j . When this holds, we say that ˜ φ satisfies the cyclic condition for X . Note that if Y ismixing, then any code from a proper subshift of X to Y satisfies the cyclic conditionfor X .The following extension theorem says that if the conditions in Theorem 5.2 hold,then the above cyclic condition is the only obstruction to extend a code definedon a proper subshift of X to a code from X to Y . It is this obstruction which isresponsible for the failure of Theorem 4.4 and 4.5 when the target is not mixing. Theorem 5.5.
Let X be an almost specified shift and Y an irreducible shift of finitetype such that h ( X ) > h ( Y ) and Y is a factor of X . For a code ˜ φ : (cid:101) X → Y definedon a proper subshift (cid:101) X of X , the following are equivalent. (1) ˜ φ satisfies the cyclic condition for X . (2) There is a code ψ : X → Y with ψ | e X = ˜ φ . (3) There is a factor code φ : X → Y with φ | e X = ˜ φ which is open with a uniformlifting length and which is bi-continuing with a bi-retract.Proof. It is clear that (3) ⇒ (2) ⇒ (1). So assume (1) and let { E , E , · · · , E q − } be the cyclic cover of Y . Since Y is a factor of X , we have the two conditionsin Theorem 5.1. Let C = (cid:83) pq − j =0 D jq . Then σ q | C is an almost specified shift. Let˜ ψ be the restriction of ˜ φ to C . Since ˜ φ satisfies the cyclic condition for X , thereexists an element E j in the cyclic cover of Y with ˜ φ ( D ∩ (cid:101) X ) ⊂ E j . Then we have˜ φ ( C ∩ (cid:101) X ) ⊂ E j . Note that σ q | E j is a mixing shift of finite type.By Theorem 4.5, we have a ( σ q -commuting) code ψ : ( C, σ q ) → ( E j , σ q ) which isan extension of ˜ ψ and bi-continuing with a bi-retract. Remainder of the proof goesas in that in Theorem 5.2. (cid:3) We remark that in [2], Barth and Dykstra considered related conditions in thecase of reducible shifts of finite type using phase matrices . PEN AND BI-CONTINUING CODES 19
Our last example shows that even when both shift spaces are of finite type, thecyclic condition is very crucial to extend a code defined on a proper subshift of thedomain. In contrast to the existence theorems, extension theorems do not directlyfollow from the mixing case.
Example 5.6.
Let X = X A with A = ( ) and Y = { ( ab ) ∞ , ( ba ) ∞ } . The edges for X are given as in the following figure. Then P ( X ) (cid:38) P ( Y ) and Y is a factor of X .Let (cid:101) X = { (13) ∞ , (31) ∞ , (24) ∞ , (42) ∞ } . We define ˜ φ : (cid:101) X → Y by˜ φ ((13) ∞ ) = ( ab ) ∞ , ˜ φ ((31) ∞ ) = ( ba ) ∞ , ˜ φ ((24) ∞ ) = ( ba ) ∞ , ˜ φ ((42) ∞ ) = ( ab ) ∞ . Suppose φ : X → Y is a code with φ | e X = ˜ φ . Write X = D ˙ ∪ D where D = { x ∈ X : x = 1 or 2 } and D = σ ( D ). Then (13) ∞ , (24) ∞ ∈ D . Since σ | D isirreducible, σ | φ ( D ) is also irreducible, hence φ ( D ) must be a single point. This isa contradiction. Thus ˜ φ cannot be extended to a code on X . Acknowledgment.
This paper was written as part of the author’s doctorial thesis,under the guidance of Prof. Sujin Shin. I would like to thank for all her suggestionsand good advice. Thanks also to Jisang Yoo for valuable comments. Also I amgrateful to Kim Bojeong Basic Science Foundation for generous help. Partial supportwas provided by the second stage of the Brain Korea 21 Project, The DevelopmentProject of Human Resources in Mathematics, KAIST in 2008.
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Department of Mathematical Sciences, Korea Advanced Institute of Science andTechnology, Daejeon 305-701, South Korea
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