Follower, Predecessor, and Extender Entropies
FFOLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES
THOMAS FRENCH AND RONNIE PAVLOV
Abstract.
Using the follower/predecessor/extender set sequences defined in[3], we define quantities which we call the follower/predecessor/extender en-tropies, which can be associated to any shift space. We analyze the behaviorof these quantities under conjugacies and factor maps, most notably showingthat extender entropy is a conjugacy invariant and that having follower entropyzero is a conjugacy invariant. We give some applications, including examplesof shift spaces with equal entropy which can be distinguished by extender en-tropy, and examples of shift spaces which can be shown to not be isomorphicto their inverse by using follower/predecessor entropy. Introduction
Topological entropy is one of the most well-studied and useful invariants of topo-logical dynamical systems. In the specific case of (one-dimensional) shift spaces,topological entropy is precisely the exponential growth of the number of words inthe language of the shift space, as the length n grows.In [3], the first author used the classical follower/predecessor sets and the exten-der sets introduced in [5] to create analogues of the word complexity function, calledthe follower set sequence, predecessor set sequence, and extender set sequence. Inthis work, we use these sequences to create analogues of entropy, which we callfollower/predecessor/extender entropy.These turn out to have some interesting properties in terms of behavior un-der conjugacy/factor maps. In particular, extender entropy is a conjugacy invari-ant (Theorem 3.3), but it need not decrease under factors (Theorem 3.7). Fol-lower/predecessor entropy is not a conjugacy invariant (Theorem 3.6), but havingzero follower/predecessor entropy is a conjugacy invariant (Corollary 3.5).The properties we prove for extender entropy are reminiscent of the (left/right)constraint entropy defined in [1], which is also a conjugacy invariant which mayincrease under factor maps. We define constraint entropy in Section 2, and therecompare and contrast it with extender entropy.Finally, we give two applications of our results. The first is a way to show thata shift space is not conjugate to its inverse. Specifically, our results show that anyshift space with zero follower entropy and positive predecessor entropy cannot beconjugate to its inverse. We then use results from [4] and [8] to show that for all β outside a meager set of Lebesgue measure zero, the β -shift X β is not conjugate toits inverse (Theorem 4.1).Our second application is that extender entropy can theoretically help to dis-tinguish between shift spaces with the same topological entropy; in particular, we Mathematics Subject Classification. a r X i v : . [ m a t h . D S ] N ov THOMAS FRENCH AND RONNIE PAVLOV prove that the only restrictions on extender and topological entropy is that the firstmust be less than or equal to the second (Theorem 4.3).2.
Definitions and preliminaries
Definition 2.1. A topological dynamical system is a pair ( X, T ) where X is acompact metric space and T : X → X is continuous. If T is also a homeomorphism,then ( X, T ) is invertible .In this paper, the topological dynamical systems considered will be symbolicallydefined. Let A denote a finite set, which we will refer to as our alphabet. Definition 2.2. A word over A is a member of A n for some n ∈ N . We denotethe length of a word w by | w | and the set of all words on A by A ∗ . Definition 2.3.
For any words v ∈ A n and w ∈ A m , we define the concatenation vw to be the pattern in A n + m whose first n letters are the letters forming v andwhose next m letters are the letters forming w . Definition 2.4.
A word w is a prefix of a right-infinite sequence z if the first | w | -many letters of z are the letters forming w . We denote the n-letter prefix of asequence z by ( z ) n . Definition 2.5. A shift space is a set X ⊂ A G ( G = Z or N ) which is closed andinvariant under the left shift map σ defined by ( σx )( n ) = x ( n + 1) for n ∈ G . Ashift space is called two-sided if G = Z and one-sided if G = N . All shift spaceswe consider are two-sided unless explicitly specified otherwise.For any shift space X , ( X, σ ) is a topological dynamical system, and it is in-vertible if X is a two-sided shift space. In that case, ( X, σ − ) is also a topologicaldynamical system. Definition 2.6.
The language of a shift space X , denoted by L ( X ), is the set ofall words which appear in points of X . For any finite n ∈ N , L n ( X ) := L ( X ) ∩ A n ,the set of words in the language of X with length n . The complexity sequence ofa shift space X is {| L n ( X ) |} n ∈ N . That is, the complexity sequence is the sequencewhich records the number of words of length n appearing in some point of X forevery length n . Definition 2.7.
For any shift space X over the alphabet A , and any word w inthe language of X , we define the follower set of w in X , F X ( w ), to be the set ofall finite words u ∈ L ( X ) such that the word wu occurs in some point of X . The predecessor set of w in X , P X ( w ), is defined to be the set of all finite words s ∈ L ( X ) such that the word sw occurs in some point of X . In some works, thefollower and predecessor sets have been defined to be the set of all one-sided infinitesequences (in A N or A − N for followers and predecessors, respectively) which mayfollow/precede w . This definition is equivalent for followers, and in the case of atwo-sided shift, for predecessors as well. For a one-sided shift, of course, no infinitesequence may precede a word w . The results of this paper will apply for eitherdefinition in any case where that definition makes sense. Definition 2.8.
For any shift space X over the alphabet A , and any word w inthe language of X , we define the extender set of w in X , E X ( w ), to be the setof all pairs ( s, u ) where s, u ∈ L ( X ) and the word swu occurs in some point of X . OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 3
Again, a definition replacing finite words with infinite sequences is equivalent in thetwo-sided case.
Remark 2.9.
For any word w ∈ L ( X ), define the surjective projection function f w : E X ( w ) → F X ( w ) by f ( s, u ) = u . Any two words w, v with the same extenderset would have the property then that f w ( E X ( w )) = f v ( E X ( v )), that is, that w and v have the same follower set. Similarly, words which have the same extenderset also have the same predecessor set. Definition 2.10.
For any positive integer n , define the sets F X ( n ) = { F X ( w ) | w ∈ L n ( X ) } , P X ( n ) = { P X ( w ) | w ∈ L n ( X ) } , and E X ( n ) = { E X ( w ) | w ∈ L n ( X ) } . Definition 2.11.
Given a shift space X , the follower set sequence of X isthe sequence {| F X ( n ) |} n ∈ N , the predecessor set sequence of X is the sequence {| P X ( n ) |} n ∈ N , and the extender set sequence of X is the sequence {| E X ( n ) |} n ∈ N .These sequences measure the number of distinct follower/predecessor/extender setsof words of length n in X .By Remark 2.9, for any X and n ,(2.1) | E X ( n ) | ≥ | P X ( n ) | and | E X ( n ) | ≥ | F X ( n ) | . Example 2.12.
The full shift on the alphabet A is just the shift space X = A Z .Then any word w ∈ L ( X ) may be followed legally by any word v ∈ L ( X ), and thusthe follower sets of all words are the same. Hence there is only one follower set ina full shift. Similarly, there is only one predecessor and one extender set in a fullshift. Then {| F X ( n ) |} n ∈ N = {| P X ( n ) |} n ∈ N = {| E X ( n ) |} n ∈ N = { , , , ... } . Example 2.13.
The even shift is the sofic shift space X with alphabet { , } defined by forbidding runs of 0 symbols of odd length between two nearest 1 sym-bols. It is a simple exercise to show that the even shift has exactly three followersets: F X (0), F X (1), and F X (10). It also has six extender sets: E X (0), E X (1), E X (00), E X (01), E X (10), and E X (010). The follower set sequence of the evenshift is {| F X ( n ) |} n ∈ N = { , , , , ... } (the predecessor set sequence is the same bysymmetry of X ) and the extender set sequence is {| E X ( n ) |} n ∈ N = { , , , , ... } . Example 2.14.
The context-free shift C is the shift space with alphabet { a, b, c } consisting of all x where any two nearest c symbols must have a word of the form a n b n between them. The follower set of a word in the context-free shift dependsonly on its final letter, the location of the final c appearing in the word (if any), andthe location after that c where the first b appears (if any). Thus we have the upperbound | F C ( n ) | < n . Similarly, the extender set of a word in the context-free shiftdepends only on its first and last letters, the locations of the first and last c in theword (if any), as well of the locations of the last a before the first c , and the first b after the final c (if any), yielding the upper bound | E C ( n ) | < n . Definition 2.15. A factor map from a topological dynamical system ( X, T ) to atopological dynamical system (
Y, S ) is a surjective map ϕ : X → Y where for any x ∈ X , Sϕ ( x ) = ϕ ( T x ). If ϕ is a bijection, then it is called a conjugacy and wesay that ( X, T ) and (
Y, S ) are conjugate .The well-known Curtis-Lyndon-Hedlund theorem states that a factor map φ from a two-sided shift space ( X, σ ) to another two-sided shift space (
Y, σ ) must bea so-called sliding block code, i.e. there exists r (called the radius of φ ) so that THOMAS FRENCH AND RONNIE PAVLOV ( φ ( x ))( i ) depends only on x ( i − r ) . . . x ( i + r ) for all x ∈ X and i ∈ Z . A factormap with radius 0 is called 1 -block , because it is induced by a map between thealphabets of the shifts. Definition 2.16.
Given a shift space X , the topological entropy of X is givenby h ( X ) = lim n →∞ n log | L n ( X ) | . It is well-known that topological entropy is a conjugacy invariant.
Definition 2.17.
Given a shift space X , the extender entropy of X is given by h E ( X ) = lim n →∞ n log | E X ( n ) | . The existence of this limit is given by Theorem 3.1.
Definition 2.18.
Given a shift space X , the follower entropy of X is given by h F ( X ) = lim sup n →∞ n log | F X ( n ) | . The predecessor entropy h P ( X ) is defined in an analogous fashion using prede-cessor sets.By (2.9), for any X , h E ( X ) ≥ h P ( X ) and h E ( X ) ≥ h F ( X ). Remark 2.19.
It is clear from the upper bounds observed in Example 2.14 thatfor the context-free shift C , we have h F ( C ) = h P ( C ) = h E ( C ) = 0. Definition 2.20.
For any two-sided shift space X , define (cid:98) X to be its reversedshift , i.e. . . . x ( − x (0) x (1) . . . ∈ (cid:98) X iff . . . x (1) x (0) x ( − . . . ∈ X .The following lemmas are immediate from definitions and are presented withoutproof. Lemma 2.21.
For any two-sided shift space X , ( X, σ − ) and ( (cid:98) X, σ ) are conjugate. Lemma 2.22.
For any two-sided shift space X , h F ( X ) = h P ( (cid:98) X ) and h P ( X ) = h F ( (cid:98) X ) . Example 2.23.
Given β >
1, let d β : [0 , → { , . . . , (cid:100) β (cid:101) − } N be the map whichsends each point x ∈ [0 ,
1) to its expansion in base β . That is, if x = ∞ (cid:88) n =1 x n β n , then d β ( x ) = .x x x ... . (In the case where x has more than one β -expansion, we takethe lexicographically largest expansion.) The closure of the image, d β ([0 , β -shift , denoted X β and originally defined in [7].An equivalent characterization of the β -shift is given by the right-infinite sequence d ∗ β (1) := lim x (cid:37) d β ( x ). For any sequence x on the alphabet { , ..., (cid:98) β (cid:99)} , x ∈ X β if andonly if every σ n x is lexicographically less than or equal to d ∗ β (1) (see [6]).Though β -shifts are one-sided shift spaces as defined, each X β has a two-sidedversion defined as the set of all sequences on the same alphabet where every subwordis a subword of some point in the one-sided version. (This is just the so-callednatural extension.) Unless otherwise stated, β -shifts in this paper refer to thosetwo-sided versions. OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 5
Remark 2.24.
There is yet another equivalent definition of β -shifts. If T β : [0 , → [0 ,
1) is given by T β ( x ) = βx (mod 1), then the one-sided β -shift X β is a symboliccoding of T β : for any i ∈ N , x ∈ [0 , σ i ( d β ( x )) = k if and only if T iβ ( x ) ∈ [ kβ , k +1 β ).So, X β is the smallest shift space containing such codings of all x ∈ [0 , β -shifts, we will notuse it further in this work.The following results about follower/predecessor set sequences for β -shifts wereproved in [4]. Theorem 2.25.
For any β -shift X β and any n , | F X β ( n ) | ≤ n + 1 . Theorem 2.26.
For any β -shift X β and any n , | P X β ( n ) | is equal to the numberof n -letter subwords of d ∗ β (1) . The latter is particularly useful when combined with the following result ofSchmeling.
Theorem 2.27 ([8], Theorems B and E) . For all β outside a meager set of Lebesguemeasure , all words in L ( X β ) appear as subwords of d ∗ β (1) . Since h ( X β ) = log β for all β , the following corollary is immediate and will beuseful in several later arguments. Corollary 2.28.
For all β , h F ( X β ) = 0 . For all β outside a meager set of Lebesguemeasure , h E ( X β ) = h P ( X β ) = h ( X β ) = log β . Finally, we briefly recall some definitions from [1] which are somewhat similar toours.
Definition 2.29.
Given a shift space X , a word w ∈ L ( X ) is a left constraint ifwe can write w = av for a ∈ A and F X ( av ) (cid:54) = F X ( v ). Definition 2.30.
Given a shift space X , the left constraint entropy is h C ( X ) = lim sup n →∞ log |{ w ∈ L n ( X ) : w is a left constraint }| n . Buzzi also defined right constraints and right constraint entropy, and provedmany useful structural properties on X under the hypothesis that either constraintentropy is strictly smaller than the topological entropy; such shift spaces were called subshifts of quasi-finite type in [1]. Any β -shift, for instance, is a subshift ofquasi-finite type.More directly relevant to our definitions is the fact that left constraint entropyis a conjugacy invariant which may increase under factor maps, just like extenderentropy. One interesting observation is that the definition of left constraint entropyinvolves only follower sets, and our follower entropy is not a conjugacy invariant.This can be explained by another difference; in the definition of left constraintentropy, a single follower set corresponding to multiple words is counted multipletimes, whereas in our definitions it would be counted only once.3. The behavior of extender and follower entropy under products,conjugacies, and factors
Theorem 3.1.
The limit in the definition of extender entropy exists for every shiftspace X . THOMAS FRENCH AND RONNIE PAVLOV
Proof.
We claim that for every m, n ∈ N , | E X ( n + m ) | ≤ | E X ( n ) | · | E X ( m ) | . Oncethis is verified, the existence of the limit follows from usual submultiplicativityarguments (i.e. the application of Fekete’s Lemma.)To this end, we will define an injection g : E X ( n + m ) → E X ( n ) × E X ( m ). Thedefinition of g is as follows: for each extender set E in E X ( n + m ), arbitrarily choosea word w ∈ L n + m ( X ) for which E = E X ( w ), then write w = uv for u ∈ L n ( X )and v ∈ L m ( X ), and define g ( E ) = ( E X ( u ) , E X ( v )).It’s clear that g has the claimed domain and co-domain. We must only checkthat it is injective. To see this, we assume that g ( E ) = g ( E (cid:48) ) = ( F , F ) for E, E (cid:48) ∈ E X ( n + m ). Then E = E X ( u u ) and E (cid:48) = E X ( u (cid:48) u (cid:48) ) where E X ( u ) = E X ( u (cid:48) ) = F and E X ( u ) = E X ( u (cid:48) ) = F . Then, for any ( (cid:96), r ) ∈ E , (cid:96)u u r ∈ X .Since E X ( u ) = E X ( u (cid:48) ), (cid:96)u (cid:48) u r ∈ X . Since E X ( u ) = E X ( u (cid:48) ), (cid:96)u (cid:48) u (cid:48) r ∈ X , andso ( (cid:96), r ) ∈ E X ( u (cid:48) u (cid:48) ). Since ( (cid:96), r ) was arbitrary, E = E X ( u u ) ⊂ E X ( u (cid:48) u (cid:48) ) = E (cid:48) .A trivially similar proof shows that E (cid:48) ⊂ E , so E = E (cid:48) , and we’ve verified injectivityof g . This implies the claimed inequality | E X ( n + m ) | ≤ | E X ( n ) | · | E X ( m ) | , and sothe existence of extender entropy as outlined above. (cid:3) Theorem 3.2.
Extender entropy is additive under products, i.e. if X , X are shiftspaces, then h E ( X × X ) = h E ( X ) + h E ( X ) .Proof. Any word in L n ( X × X ) can be written as a pair ( w , w ), where w i isgiven by the i th coordinate of letters in w . It is immediate from the definitionof Cartesian product and extender set that E X × X ( w , w ) = { (( s, s (cid:48) ) , ( u, u (cid:48) )) :( s, u ) ∈ E X ( w ) , ( s (cid:48) , u (cid:48) ) ∈ E X ( w ) } . Then, ( w , w ) and ( w (cid:48) , w (cid:48) ) have the sameextender set in X × X iff w and w (cid:48) have the same extender set in X and w and w (cid:48) have the same extender set in X . Therefore | E X × X ( n ) | = | E X ( n ) || E X ( n ) | , and taking logarithms, dividing by n , and letting n → ∞ yields h E ( X × X ) = h E ( X ) + h E ( X ). (cid:3) Theorem 3.3.
Extender entropy is a conjugacy invariant, i.e. if X and Y areconjugate shift spaces, then h E ( X ) = h E ( Y ) .Proof. Suppose that X and Y are conjugate shift spaces, via a conjugacy φ : X → Y . Since φ is a sliding block code, it has a radius r , meaning that x ( − r ) . . . x ( r )uniquely determines ( φ ( x ))(0). We can then decompose φ = ψ ◦ f [ − r,r ] , where f [ − r,r ] : X → X [ − r,r ] is the canonical conjugacy from X to its higher-block presenta-tion X [ − r,r ] and ψ : X [ − r,r ] → Y is a 1-block conjugacy. We will prove Theorem 3.3by showing first that h E ( X ) = h E ( X [ − r,r ] ) and then that h E ( X [ − r,r ] ) ≥ h E ( Y ),implying that h E ( X ) ≥ h E ( Y ); reversing the roles of X and Y then shows that h E ( X ) = h E ( Y ). For notational convenience, we from now on use Z to denote X [ − r,r ] and f to denote f [ − r,r ] . We need the following auxiliary fact: Fact 1:
For n ≥ r and w, w (cid:48) ∈ L n ( X ), E Z ( f ( w )) = E Z ( f ( w (cid:48) )) if and only if E X ( w ) = E X ( w (cid:48) ) and w ( i ) = w (cid:48) ( i ) for all 1 ≤ i ≤ r and n − r < i ≤ n .= ⇒ : Assume that E Z ( f ( w )) = E Z ( f ( w (cid:48) )). Note that by the rules defining Z = X [ − r,r ] , any letter a ∈ A [ − r,r ] (the alphabet of Z ) for which f ( w ) a ∈ L ( Z ) OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 7 must have first 2 r letters (in A ) equal to the final 2 r letters (in A ) of w . Therefore,it must be the case that w and w (cid:48) agree on their final 2 r letters, and a triviallysimilar argument shows the same about the first 2 r letters.Now, choose any ( u, v ) ∈ E X ( w ). Then uwv ∈ X , and so clearly f ( uwv ) ∈ Z .We can then write f ( uwv ) = f ( u ) sf ( w ) tf ( v ), where s is determined by the final2 r letters of u and initial 2 r letters of w , and t is determined by the final 2 r lettersof w and initial 2 r letters of v . Since w, w (cid:48) agree on their first and last 2 r letters, f ( uw (cid:48) v ) = f ( u ) sf ( w (cid:48) ) tf ( v ). But then since E Z ( f ( w )) = E Z ( f ( w (cid:48) )), and f ( uwv ) = f ( u ) sf ( w ) tf ( v ) ∈ Z , it must be the case that f ( uw (cid:48) v ) = f ( u ) sf ( w (cid:48) ) tf ( v ) ∈ Z . Butthen since f is invertible, uw (cid:48) v ∈ X and so ( u, v ) ∈ E X ( w (cid:48) ). Since ( u, v ) was anarbitrary element of E X ( w ), we’ve shown that E X ( w ) ⊆ E X ( w (cid:48) ), and a triviallysimilar argument shows the reverse, so E X ( w ) = E X ( w (cid:48) ). ⇐ =: Assume that E X ( w ) = E X ( w (cid:48) ) and that w ( i ) = w (cid:48) ( i ) for all 1 ≤ i ≤ r and n − r < i ≤ n . Choose any ( u, v ) ∈ E Z ( f ( w )), meaning that uf ( w ) v ∈ Z .Then clearly f − ( uf ( w ) v ) ∈ X , and can be written as uwv . Similar arguments tothose used in the reverse direction show that since w, w (cid:48) share the same first andlast 2 r letters, f − ( uf ( w (cid:48) ) v ) = uw (cid:48) v . Since E X ( w ) = E X ( w (cid:48) ) and f − ( uf ( w ) v ) = uwv ∈ X , it must be the case that f − ( uf ( w (cid:48) ) v ) = uw (cid:48) v ∈ X as well. But then uf ( w (cid:48) ) v ∈ Z . Since u, v were arbitrary, we’ve shown that E Z ( f ( w )) ⊆ E Z ( f ( w (cid:48) )),and a trivially similar argument shows the reverse, so E Z ( f ( w )) = E Z ( f ( w (cid:48) )), com-pleting the proof of Fact 1.Fact 1 implies that for every n > r , | E X ( n ) | ≤ | E Z ( n ) | ≤ | E X ( n ) | · | A | r . Upontaking logarithms, dividing by n , and letting n → ∞ , we see that h E ( X ) = h E ( Z ).It remains to prove that h E ( Z ) ≥ h E ( Y ), for which we will use only the fact that ψ : Z → Y is a 1-block conjugacy. Since ψ − is a sliding block code, it has a radius s , meaning that ( ψ ( x ))( − s ) . . . ( ψ ( x ))( s ) uniquely determines x (0). The final stepin our proof is the following auxiliary fact: Fact 2: If E Z ( w ) = E Z ( w (cid:48) ) and u, v ∈ A s are such that uwv ∈ L ( Z ) (meaningthat uw (cid:48) v ∈ L ( Z ) also), then E Y ( ψ ( w )) = E Y ( ψ ( w (cid:48) )).To see this, suppose that E Z ( w ) = E Z ( w (cid:48) ) and u, v ∈ A s are such that uwv, uw (cid:48) v ∈ L ( Z ), and recall that ψ is a 1-block map. Our key observation is that(3.1) E Y ( ψ ( uwv )) = (cid:91) x ∈ ψ − ( ψ ( uwv )) ψ ( E Z ( x )) = (cid:91) y ∈ ψ − ( u ) ,z ∈ ψ − ( v ) s.t. ywz ∈ L ( Z ) ψ ( E Z ( ywz )) . Here, the second equality uses the fact that any x ∈ ψ − ( uwv ) must have a w atits center since s is the radius of ψ − .We now note that since E Z ( w ) = E Z ( w (cid:48) ), the sets { y ∈ ψ − ( u ) , z ∈ ψ − ( v ) s.t. ywz ∈ L ( Z ) } and { y ∈ ψ − ( u ) , z ∈ ψ − ( v ) s.t. yw (cid:48) z ∈ L ( Z ) } are the same. Forthe same reason, given any pair y, z in this set, E Z ( ywz ) = E Z ( yw (cid:48) z ). Combiningwith (3.1), we see that indeed E Y ( ψ ( uwv )) = E Y ( ψ ( uw (cid:48) v )), completing the proof.By Fact 2, for every n > s , | E Y ( n ) | ≤ | E Z ( n − s ) | · | A | s . Upon taking loga-rithms, dividing by n , and letting n → ∞ , we see that h E ( Z ) ≥ h E ( Y ). Combining THOMAS FRENCH AND RONNIE PAVLOV with the fact that h E ( X ) = h E ( Z ), we see that h E ( X ) ≥ h E ( Y ), and since wecould repeat this proof with the roles of X and Y reversed, we see that indeed h E ( X ) = h E ( Y ), completing the proof of Theorem 3.3. (cid:3) Unfortunately, we will show that follower entropy is not conjugacy-invariant.However, the following result bounds the amount by which it can increase under aconjugacy.
Theorem 3.4. If φ : X → Y is a conjugacy, r is the radius of φ , and s is theradius of φ − , then h F ( Y ) ≤ r + s ) h F ( X ) .Proof. Suppose that φ : X → Y is a conjugacy with radius r and whose inversehas radius s . As before, we decompose φ = ψ ◦ f [ − r,r ] , where f [ − r,r ] : X → X [ − r,r ] .We will prove Theorem 3.4 by showing first that h F ( X ) = h F ( X [ − r,r ] ) and thenthat h F ( Y ) ≤ r + s ) h F ( X [ − r,r ] ). We again use Z to denote X [ − r,r ] and f todenote f [ − r,r ] . We need the following auxiliary fact: for any n ≥ r and w, w (cid:48) ∈ L n ( Z ), F Z ( f ( w )) = F Z ( f ( w (cid:48) )) if and only if F X ( w ) = F X ( w (cid:48) ) and w ( i ) = w (cid:48) ( i ) for n − r < i ≤ n . We omit the proof, as it is trivially similar to that of Fact 1 in theproof of Theorem 3.3.Then, for every n > r , | F X ( n ) | ≤ | F Z ( n ) | ≤ | F X ( n ) | · | A | r . Upon takinglogarithms, dividing by n , and taking the limsup as n → ∞ , we see that h F ( X ) = h F ( Z ). It now suffices to show that h F ( Y ) ≤ r + s ) h F ( Z ). We note that ψ : Z → Y is a 1-block map, and that its inverse ψ − has radius r + s . Our key observationis that for any w ∈ L ( Y ) and u, v ∈ A r + s for which uwv ∈ L ( Y ),(3.2) F Y ( ψ ( uwv )) = (cid:91) x ∈ ψ − ( ψ ( uwv )) ψ ( F Z ( x )) = (cid:91) y ∈ ψ − ( u ) ,z ∈ ψ − ( v ) s.t. ywz ∈ L ( Z ) ψ ( F Z ( ywz )) . This means that for n > r + s ), each follower set of an n -letter word in Y isdetermined by a collection of at most | A | r + s ) follower sets of n -letter words in Z .Therefore, | F Y ( n ) | ≤ r + s ) (cid:88) i =1 (cid:18) | F X ( n ) | i (cid:19) ≤ r + s ) (cid:88) i =1 | F X ( n ) | i ≤ r + s ) | F X ( n ) | r + s ) . Taking logarithms, dividing by n , and taking the limsup as n → ∞ yields h F ( Y ) ≤ r + s ) h F ( Z ) = 2( r + s ) h F ( X ), completing the proof of Theorem 3.4. (cid:3) As an immediate corollary, we see that having zero follower entropy is a conjugacy-invariant condition.
Corollary 3.5.
Having zero follower entropy is a conjugacy invariant, i.e. if X and Y are conjugate shift spaces and h F ( X ) = 0 , then h F ( Y ) = 0 . Follower entropy itself is, however, not conjugacy invariant.
Theorem 3.6.
There exist a pair of conjugate shift spaces
X, Y with h F ( X ) (cid:54) = h F ( Y ) . OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 9
Proof.
We begin by defining a shift space X with alphabet { , } and h F ( X ) > X β with 1 < β < h P ( X β ) = log β and define X = (cid:99) X β . Then by Lemma 2.22, h F ( X ) = h P ( X β ) =log β >
0. Then, define X with alphabet { , , (cid:48) , (cid:48) } to be the disjoint union X ∪ X (cid:48) , where X (cid:48) consists of points of X with primes placed above every letter.Clearly | F X ( n ) | = 2 | F X ( n ) | , and so h F ( X ) = h F ( X ).Now, we will define X with alphabet { , , (cid:48) , (cid:48) , a, b, c } . Firstly, letters from { a, b, c } cannot appear consecutively or at distance 1 in points of X , i.e. we declareall words xy and xzy for x, y ∈ { a, b, c } and any z to be forbidden. Secondly, forany word of the form w = w d w d w . . . d n w n , with d i ∈ { a, b, c } and w i ∈ { , , (cid:48) , (cid:48) } ∗ ( w may be the empty word), definean auxiliary word η ( w ) in { , , (cid:48) , (cid:48) } ∗ as follows: whenever d i = a , use the odd-indexed letters from w i , whenever d i = b , use the even-indexed letters from w i , andwhenever d i = c , ignore w i entirely. Then, η ( w ) is formed by concatenating thecorresponding words for each i . As an example, η (1 (cid:48) a (cid:48) (cid:48) b (cid:48) c (cid:48) a (cid:48)
1) = 01 (cid:48) (cid:48) .We say that w ∈ L ( X ) iff η ( w ) ∈ L ( X ). Note that this includes some degeneratecases, for instance { , , (cid:48) , (cid:48) } ∗ ⊂ L ( X ) because for any w ∈ { , , (cid:48) , (cid:48) } ∗ , η ( w ) isthe empty word, which is vacuously in L ( X ).We note that for any w ∈ L n ( X ), F X ( w ) is determined entirely by its right-most letter from { a, b, c } and by η ( w ) ∈ L ( X ) of length at most n/
2. Therefore, | F X ( n ) | ≤ (cid:80) (cid:98) n/ (cid:99) i =0 | F X ( i ) | , and taking logarithms, dividing by n , and taking thelimsup as n → ∞ shows that h F ( X ) ≤ h F ( X ) = h F ( X ).We now define X to be the higher-block recoding X [0 , . By the proof of The-orem 3.4, h F ( X ) = h F ( X ) ≤ h F ( X ). The alphabet of X is now L ( X ) ⊆{ , , (cid:48) , (cid:48) , a, b, c } . Finally, we define a 1-block map ψ on X as follows: all lettersof the alphabet of X of the form az or bz are sent under ψ to a new symbol ∗ , andon all other letters of the alphabet of X , ψ acts as the identity. Define X = ψ ( X ).We claim that ψ is injective, and therefore that X and X are conjugate. To seethis, simply note that for any x ∈ X , ( ψ ( x ))( − ψ ( x ))(0)( ψ ( x ))(1) uniquely deter-mines x (0); if ( ψ ( x ))(0) (cid:54) = ∗ , then ( ψ ( x ))(0) = x (0), and if ( ψ ( x ))(0) = ∗ , then since a, b cannot appear consecutively or separated by distance 1 in X , both ( ψ ( x ))( − ψ ( x ))(1) are not ∗ . Therefore, ( ψ ( x ))( −
1) = x ( −
1) and ( ψ ( x ))(1) = x (1),and by the definition of X as a higher-block recoding, x ( −
1) and x (1) determine x (0). We have then shown that X and X are conjugate, and it remains only toshow that h F ( X ) > h F ( X ).To see this, we begin by defining an injection from F X ( n ) × F X (cid:48) ( n ) to F X (2 n ).For any pair ( F, F (cid:48) ) ∈ F X ( n ) × F X (cid:48) ( n ), arbitrarily choose v ∈ L n ( X ) and v (cid:48) ∈ L n ( X (cid:48) ) for which F = F X ( v ) and F (cid:48) = F X (cid:48) ( v (cid:48) ). Then, define a word π ( F, F (cid:48) ) by π ( F, F (cid:48) ) = ∗ ( v (1) v (cid:48) (1))( v (cid:48) (1) v (2))( v (2) v (cid:48) (2)) . . . ( v ( n ) v (cid:48) ( n )) . Firstly, we claim that π ( F, F (cid:48) ) ∈ L n ( X ); this is because v ∈ L ( X ), and so av (1) v (cid:48) (1) . . . v ( n ) v (cid:48) ( n ) ∈ L ( X ). Next, we claim that for any pairs ( F , F (cid:48) ) (cid:54) =( F , F (cid:48) ) ∈ F X ( n ) × F X (cid:48) ( n ), F X ( π ( F , F (cid:48) )) (cid:54) = F X ( π ( F , F (cid:48) )). Either F (cid:54) = F or F (cid:48) (cid:54) = F (cid:48) ; we begin with the former case. Let’s say that v i ∈ L n ( X ) and v (cid:48) i ∈ L n ( X (cid:48) ) ( i = 1 ,
2) were used in the definition of π ( F i , F (cid:48) i ), meaning that F i = F X ( v i ) and F (cid:48) i = F X (cid:48) ( v (cid:48) i ). Without loss of generality, we assume that thereis a word w = w (1) . . . w ( m ) ∈ F \ F . Now, we claim that u = ( v ( n )0)(0 w (1))( w (1)0)(0 w (2))( w (2)0) . . . (0 w ( m ))( w ( m )0)is in F X ( π ( F , F (cid:48) )), but is not in F X ( π ( F , F (cid:48) )). To see that π ( F , F (cid:48) ) u ∈ L ( X ),we just note that v w ∈ L ( X ), and so av (1) v (cid:48) (1) v (2) v (cid:48) (2) . . . v ( n ) v (cid:48) ( n ) w (1)0 w (2)0 . . . w ( m )0 ∈ L ( X ) . Assume for a contradiction that π ( F , F (cid:48) ) u ∈ L ( X ). Then, by definition of ψ and X as a higher-block presentation, there must be some word of the form tv (1) v (cid:48) (1) . . . v ( n ) v (cid:48) ( n ) w (1)0 w (2)0 . . . w ( m )0in L ( X ), where t ∈ { a, b } . However, clearly t (cid:54) = b , since if it were, the word v (cid:48) m would have to be in L ( X ), which is impossible since the letters of v (cid:48) are in { (cid:48) , (cid:48) } . Therefore, t = a , and so by definition of X , v w ∈ L ( X ), a contradictionto w / ∈ F = F X ( v ). This implies that F X ( π ( F , F (cid:48) )) (cid:54) = F X ( π ( F , F (cid:48) )) when F (cid:54) = F . The case for F (cid:48) (cid:54) = F (cid:48) is trivially similar, and so the map sending ( F, F (cid:48) )to F X ( π ( F, F (cid:48) )) is injective.This implies that | F X (2 n ) | ≥ | F X ( n ) | · | F X (cid:48) ( n ) | = | F X ( n ) | . Taking loga-rithms, dividing by 2 n , and taking the limsup as n → ∞ shows that h F ( X ) ≥ h F ( X ). Since we showed earlier that h F ( X ) ≤ h F ( X ), h F ( X ) > h F ( X ), andthe theorem is proved. (cid:3) We will show that under factor maps, nothing similar can be said, and alsothat even though extender entropy is a conjugacy invariant, it does not necessarilydecrease under factor maps.
Theorem 3.7.
There exist a shift space X and a factor map φ : X → Y such that h E ( X ) = h F ( X ) = 0 , but h E ( Y ) , h F ( Y ) > .Proof. The alphabet of X will be { , , , a, b, c } . The points of X are exactly thosesequences . . . a x (1)0 x (2)0 x (3)0 a x (1)1 x (2)1 x (3)1 a . . . with a n ∈ { , , } for every n ∈ Z , x ( j ) n ∈ { a, b, c } for all n ∈ Z and j ∈ { , , } ,and for which the point ( x ( a n ) n ) n ∈ Z is in the context-free shift C .First, we claim that h E ( X ) = 0, which will trivially imply that h F ( X ) = 0. Tosee this, we first claim that for v ∈ L ( X ) of the form a x (1)1 x (2)1 x (3)1 . . . a n x (1) n x (2) n x (3) n , E X ( v ) is completely determined by E C ( κ ( v )), where κ ( v ) denotes x ( a )1 . . . x ( a n ) n ∈ L ( C ); the proof is left to the reader. Then, for an arbitrary word w ∈ L ( X ), wecan decompose as w = pvs , where v has the form above, p has length at most3 and consists of letters in { a, b, c } , and s has length at most 3, has first letterin { , , } , and all remaining letters in { a, b, c } . Then, E X ( w ) is a projection of E X ( v ) determined by p and s , meaning that E X ( w ) is completely determined by E C ( κ ( v )), p , and s . Therefore, | E X ( n ) | ≤ (27 + 9 + 3 + 1) E C ( (cid:100) n/ (cid:101) ) . Since h E ( C ) = 0, taking logarithms, dividing by n , and letting n → ∞ shows that h E ( X ) = 0, trivially implying that h F ( X ) = 0 as well.We now define a simple 1-block factor map on X : φ maps 1, 2, and 3 to a newsymbol ∗ and each of a, b, c to themselves. We denote φ ( X ) by Y , a shift space OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 11 with alphabet {∗ , a, b, c } . We claim that h F ( Y ) >
0, which will obviously implythat h E ( Y ) > n , all words of theform ∗ w ∗ w . . . ∗ w n , where each w i is either aba or abc , are in L ( Y ), and havedistinct follower sets in Y .The first claim is fairly simple: clearly, since a ∞ ∈ C , for any choice of the words w n (between aba and abc ), the point . . . w w w . . . is in X , and so every pointof the form . . . ∗ w ∗ w ∗ w . . . is in Y . To see the second claim, consider any n and unequal words v = ∗ w ∗ w . . . ∗ w n and v (cid:48) = ∗ w (cid:48) ∗ w (cid:48) . . . ∗ w (cid:48) n . Since v (cid:54) = v (cid:48) ,there exists i so that w i (cid:54) = w (cid:48) i ; without loss of generality we assume that w i = abc and w (cid:48) i = aba .We claim that the word u = ( ∗ aaa ) i ( ∗ bbb ) n ∗ ccc is in F Y ( v ) \ F Y ( v (cid:48) ). To seethat vu ∈ L ( Y ) (i.e. that u ∈ F Y ( v )), we simply note that2 w w . . . w i − w i w i +1 . . . w n (1 aaa ) i (1 bbb ) n ccc is in L ( X ), since the word spelled out by the indicated letters is b i − ca n b n c , whichis in L ( C ). The φ -image of this word is vu , which is therefore in L ( Y ).Assume for a contradiction that v (cid:48) u ∈ L ( Y ). Then, some word of the form a w (cid:48) a w (cid:48) . . . a n w (cid:48) n a n +1 aaa . . . a n + i aaaa n + i +1 bbb . . . a n + i bbba n + i +1 ccc would have to be in L ( X ), where each a i ∈ { , , } . In particular, this implies thata word of the form b b . . . b n a i b n c is in L ( C ), where each b i ∈ { a, b, c } . Note that since w (cid:48) i = aba , b i (cid:54) = c . However,by definition of C , a i b n c for n > i must be immediately preceded by ca n − i in anypoint of C . This is a contradiction since b i (cid:54) = c . Therefore, v (cid:48) u / ∈ L ( Y ). Wehave then shown that F Y ( v ) (cid:54) = F Y ( v (cid:48) ), and since v and v (cid:48) were arbitrary, that | F Y (4 n ) | ≥ n . We take logarithms, divide by n , and take the limsup as n → ∞ tosee that h F ( Y ) ≥ log 24 >
0; clearly h E ( Y ) ≥ h F ( Y ) >
0, and our proof is complete. (cid:3) Applications
The key to both applications mentioned in the introduction is Corollary 2.28.
Theorem 4.1.
For all β outside a meager set of zero Lebesgue measure, the β -shift ( X β , σ ) is not conjugate to its inverse.Proof. For any β outside the meager set of Lebesgue measure 0 from Corollary 2.28, h F ( β ) = 0 and h P ( X β ) >
0. Then by Lemma 2.22, h F ( (cid:99) X β ) = h P ( X β ) >
0, andso by Corollary 3.5, ( X β , σ ) and ( (cid:99) X β , σ ) are not conjugate. Then by Lemma 2.21,( X β , σ ) and ( X β , σ − ) are also not conjugate. (cid:3) Remark 4.2.
This fact could be proved in a similar fashion by using the left andright constraint entropies from [1].
Theorem 4.3.
For any x ≤ y , there exists a shift space X with h E ( X ) = x and h ( X ) = y .Proof. It is clearly sufficient to treat the cases x = 0 and x = y only, since h ( X )and h E ( X ) are additive under products (Theorem 3.2), and one can write ( x, y ) as(0 , y − x ) + ( x, x ). Consider any x >
0. By Corollary 2.28, there exists α > x where x/α / ∈ Q andwhere h E ( X β ) = h ( X β ) = α . Take S to be a Sturmian shift space with rotationnumber x/α ∈ (0 , \ Q . (We will not define Sturmian shift spaces here, but theonly properties we will need is that for S Sturmian with rotation number η , thealphabet is { , } and every word in L n ( S ) contains either (cid:98) nη (cid:99) or (cid:100) nη (cid:101) Z as follows. Choose ∗ / ∈ A and define thealphabet of Z to be ( ∗ , (cid:116) ( A × { } ). If we write a sequence z on this alphabetas z = (( z ( n )) , ( z ( n )) ) n ∈ Z , then the shift space Z is the set of all such z where(( z ( n )) ) n ∈ Z ∈ S and (( z ( n )) ) { n : z ( n ) =1 } ∈ X β , i.e. the first coordinates of z atlocations where the second coordinate has a 1, taken in order, comprise a point of X β .We claim that h ( Z ) = h E ( Z ) = x . Let’s first show that h ( Z ) = x . For any v ∈ L n ( S ), denote by | v | the number of 1 symbols in v . Then, for each u ∈ L | v | ( X β ),we can create w ( u, v ) ∈ L n ( Z ) by placing v in the second coordinate, ∗ in the firstcoordinate at each location where v has a 0, and placing u in the first coordinatealong the set of indices where v has 1s. By definition, L n ( Z ) is in fact the set of allsuch w ( u, v ), and the map ( u, v ) (cid:55)→ w ( u, v ) is a bijection. To estimate | L n ( Z ) | , wetherefore need only to estimate the number of such pairs ( u, v ).Recall that since S is Sturmian, | v | is always either (cid:98) nx/α (cid:99) or (cid:100) nx/α (cid:101) , and that | L n ( S ) | = n + 1. Therefore,( n + 1) | L (cid:98) nx/α (cid:99) ( X β ) | ≤ | L n ( Z ) | ≤ ( n + 1) | L (cid:100) nx/α (cid:101) ( X β ) | . But then, taking logarithms, dividing by n , and letting n → ∞ yields h ( Z ) =( x/α ) h ( X β ) = x .Then h E ( Z ) ≤ h ( Z ) = x , so we need only show that h E ( Z ) ≥ x . For this, wefix v ∈ L n ( S ), and claim that if u, u (cid:48) ∈ L | v | ( X β ) and E X β ( u ) (cid:54) = E X β ( u (cid:48) ), then E Z ( w ( u, v )) (cid:54) = E Z ( w ( u (cid:48) , v )). To see this, assume that E X β ( u (cid:48) ) (cid:54)⊂ E X β ( u ), choose apair ( s, t ) ∈ E X β ( u (cid:48) ) \ E X β ( u ), and choose ( a, b ) ∈ E S ( v ) where a has | s |
1s and b has | t | p with a in the second coordinate, ∗ in the first coordinate atall locations of 0s in a , and s spelled out along the first coordinate at indices where a has 1 symbols. Similarly create q from b and t . It’s immediate from the definitionof z that ( p, q ) ∈ E Z ( w ( u (cid:48) , v )) \ E Z ( w ( u, v )), and so E Z ( w ( u, v )) (cid:54) = E Z ( w ( u (cid:48) , v )).Then, all distinct extender sets among u ∈ L | v | ( X β ) yield distinct extender sets inwords of L n ( Z ), i.e. | E Z ( n ) | ≥ | E X β ( | v | ) | . Since | v | = (cid:98) nx/α (cid:99) or | v | = (cid:100) nx/α (cid:101) , taking logarithms, dividing by n , andletting n → ∞ yields h E ( Z ) ≥ ( x/α ) h E ( X β ) = x , completing the proof that h ( Z ) = h E ( Z ) = x .It remains only to construct Y with h E ( Y ) = 0 and h ( Y ) = x . First, choose n with log n > x and x/ log n / ∈ Q , and define F to be the full shift on n symbols.Then, h ( F ) = log n and h E ( F ) = 0. Perform exactly the same procedure as above,where F replaces X β and S is chosen to have rotation number x/ log n . Then,exactly as above, h E ( Y ) = ( x/ log n ) h E ( F ) = 0, and h ( Y ) = ( x/ log n ) h ( F ) = x ,completing the proof. (cid:3) OLLOWER, PREDECESSOR, AND EXTENDER ENTROPIES 13
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Thomas French and Ronnie Pavlov, Department of Mathematics, University of Den-ver, 2390 S. York St., Denver, CO 80208
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