Geometric realization for substitution tilings
aa r X i v : . [ m a t h . D S ] N ov GEOMETRIC REALIZATION FOR SUBSTITUTION TILINGS
M. BARGE AND J.-M. GAMBAUDO
Abstract.
Given an n -dimensional substitution Φ whose associated linear expan-sion Λ is unimodular and hyperbolic, we use elements of the one-dimensional integerˇCech cohomology of the tiling space Ω Φ to construct a finite-to-one semi-conjugacy G : Ω Φ → T D , called geometric realization, between the substitution induced dynam-ics and an invariant set of a hyperbolic toral automorphism. If Λ satisfies a Pisotfamily condition and the rank of the module of generalized return vectors equals thegeneralized degree of Λ, G is surjective and coincides with the map onto the maximalequicontinuous factor of the R n -action on Ω Φ . We are led to formulate a higher-dimensional generalization of the Pisot substitution conjecture: If Λ satisfies thePisot family condition and the rank of the one-dimensional cohomology of Ω Φ equalsthe generalized degree of Λ, then the R n -action on Ω Φ has pure discrete spectrum. Contents
1. Introduction 22. Main results 43. Abelian covers of Ω Φ and global shadowing 74. Generalized return vectors 95. From substitutions to toral automorphisms 116. Geometric realization 137. Non-triviality of the global shadowing relation 158. Global shadowing and regional proximality for Pisot family substitutions 169. Pisot family substitutions 2010. Connections with the traditional Pisot Substitution Conjecture 24References 27 Date : January 14, 2018. Introduction
Let ∆ be a subset of the n -dimensional Euclidean space R n , n ≥ . A tiling of∆ is a countable collection T = { t j } j ∈ J of topological closed n -balls in ∆, called tiles , that cover ∆ and have pairwise disjoint interiors. Consider a finite collection ofpolyhedra P = { ρ , . . . , ρ m } in R n , called prototiles . We say that P generates a tiling T = { t j } j ∈ J of ∆ if each t i is a translated copy of one of the ρ j and the t i ’s meet fullface to full face in all dimensions (so each t i has finitely many faces in each dimension,and if t i and t j meet in a point x that is in the relative interior of a face of t i or t j ,then they meet in that entire face). We denote by Ω P (∆) the set of all tilings of ∆generated by P , and by Ω P the set Ω P ( R n ). When n >
1, it is not always the casethat, for a given family P , the set Ω P is not empty. In fact it is well known that theproblem of whether or not Ω P is empty is not decidable [B]. When Ω P = ∅ , the group R n acts on Ω P by translation: for any tiling T = { t j } j ∈ J in Ω P , and for any u ∈ R n wedefine: T − u = { t i − u } j ∈ J . The set Ω P has a natural metric defined as follows. For each r ≥
0, let ¯ B r (0) denotethe closed ball of radius r centered at 0 ∈ R n and, for T ∈ Ω P , let B r [ T ] := { t j ∈ T : t j ∩ ¯ B r (0) = ∅} be the collection of tiles in T that meet ¯ B r (0). Given T, T ′ ∈ Ω P , let A denote the set of ǫ ∈ (0 ,
1) such that there are u, u ′ ∈ R n , with | u | , | u ′ | < ǫ/
2, so that B /ǫ [ T − u ] = B /ǫ [ T ′ − u ′ ]. Then: d ( T, T ′ ) = (cid:26) inf A if A = ∅ T and T ′ are close if, up to small translation, they agree exactly in a largeneighborhood of the origin. When Ω P is equipped with this metric, the translationaction is continuous. If Ω P has finite local complexity (that is, there are only finitelymany patterns of tiles in elements of Ω P of any fixed finite radius - see below), Ω P iscompact and has the structure of a lamination whose n -dimensional leaves are theorbits of the translation action and with totally disconnected transverse direction.Substitutions provide an important method for constructing tilings. Consider a poly-hedral family P = { ρ , . . . , ρ m } of prototiles in R n and suppose there is an expandinglinear map Λ : R n → R n (called inflation ) so that for each j ∈ { , . . . , m } , there existsa tiling S j of Λ( ρ j ) generated by P . The collection of tilings S = {S , . . . , S m } is calleda substitution rule . The incidence matrix associated with Λ and S is the m -by- m matrix M S = ( m i,j ) i,j , where for each i, j ∈ { , . . . , m } , the entry m i,j is the number of EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 3 translated copies of ρ i in S j . Recall that a matrix M is primitive if there exists k > M k are positive.By iterating inflation followed by substitution infinitely many times, one sees thatΩ P = ∅ . On the one hand, the inflation Λ induces a natural continuous map K fromΩ P to Ω Λ P , where Λ P = { Λ p | p ∈ P} . On the other hand, the substitution rule S induces a continuous map J from Ω Λ P to Ω P , which is defined by subdividing the tilesof a tiling in Ω λ P according to the substitution rule. The composition of these twomaps yields a self-map Φ := J ◦ K : Ω P → Ω P which we call the substitution map .We will say that a finite collection P = { t , . . . , t k } of tiles is an allowed patch for Φif there is a prototile ρ j and a k ∈ N so that P is contained in some translate of thepatch Φ k ( ρ j ) obtained by k iterations of inflation and substitution applied to ρ j . The tiling space associated with Φ is the collectionΩ Φ := { T ∈ Ω P : B r [ T ] is an allowed patch for Φ for all r ≥ } . It is clear that Ω Φ is invariant under both translation and the substitution map.. Throughout this paper we will make the following three assumptions: • the map Φ is primitive , which means that the associated substitution matrix M S is primitive; • the tiling space Ω Φ has finite local complexity (FLC), which means that foreach r > B r [ T ] := { t ∈ T : t ∩ ¯ B r (0) = ∅} , T ∈ Ω Φ ; and • Ω Φ is translationally non-periodic which means that if there exist a tiling T in Ω Φ and u ∈ R n such that T − u = T , then u = 0.Despite the fact that the substitution map is basically a linear inflation, it turnsout the laminated structure of the phase space Ω φ induces a very rich dynamics. Theaction of R n by translation on Ω Φ is minimal and uniquely ergodic ([AP]); the mapΦ is a homeomorphism ([S1]) and is ergodic with respect to the unique R n -invariantmeasure µ ; and the dynamics on Ω Φ interact by Φ( T − v ) = Φ( T ) − Λ v .Our first goal in this paper is to provide a way to understand the dynamical sys-tem (Ω Φ , Φ) in terms of the standard geometric theory of dynamical systems whereone studies iteration of maps on compact manifolds. This is what we call geometricrealization . More precisely, we will show that, under some assumptions on the hyper-bolicity of the eigenvalues of the inflation Λ, there exists a finite-to-one continuous map It is often desirable to mark, or color, prototiles and tiles so that tiles may occupy the sameunderlying set but nonetheless be distinct. Formally, then, a tile is a pair τ = ( t, m ) where t is aclosed topological n -ball and m is one of finitely many possible marks. We will stick to the simplerunmarked language in this paper, though all of the results are to be understood in the more generalsetting. M. BARGE AND J.-M. GAMBAUDO from Ω Φ to some D -dimensional torus T D that factors the dynamics of Φ into those ofa linear hyperbolic map. The basic idea is as follows. The one-dimensional cohomologyof Ω Φ supplies a map into the D -torus that, on the level of cohomology, conjugatesΦ with a hyperbolic toral automorphism. The technique of global shadowing is thenapplied to improve the map into an actual dynamical semi-conjugacy of hyperbolicsystems. (This technique originated with [F] and [Fr], and is used also in [BKw] to a.e.embed pseudo-Anosovs into hyperbolic toral automorphisms.)The second goal is to establish a link between this geometric realization and thetraditional Pisot Substitution Conjecture ([BS]) regarding pure discreteness of the R n -action. On Ω Φ we have Φ( T − v ) = Φ( T ) − Λ v and a like relation holds between thehyperbolic action on T D and a Kronecker action on the u -dimensional leaves of itsunstable foliation. Under an assumption on the “Pisotness” of the inflation Λ, u = n and the semi-conjugacy of hyperbolic systems becomes also a semi-conjugacy of R n -actions. The coordinate functions of the semi-conjugacy are thus eigenfunctions of R n -action and their associated eigenvalues constitute a generating set for the discrete partof the spectrum of the R n -action on Ω Φ . In essence, elements of the first cohomology ofΩ Φ are converted into eigenfunctions of the R n -action by means of global shadowing.2. Main results
The eigenvalues of the inflation Λ are all algebraic integers ([KS], or see Lemma 16below): let us partition the spectrum, spec (Λ), into families spec (Λ) = F ∪ · · · ∪ F k of algebraic conjugacy classes. For each i ∈ { , . . . , k } , let d i be the algebraic degreeof the elements of F i , let m i := max λ ∈F i m ( λ ), where m ( λ ) is the multiplicity of λ asan eigenvalue of Λ, and let the degree of Λ be defined by D (Λ) := P ki =1 m i d i . Wewill say that Φ is unimodular if each element of spec (Λ) is an algebraic unit, and hyperbolic if no element of spec (Λ) has an algebraic conjugate on the unit circle.If Φ is unimodular, there is a finitely generated subgroup GR (Φ) of R n , the groupof generalized return vectors (the definition is given below), that has rank D = D ( GR ) ≥ D (Λ) (Lemma 16) and is invariant under Λ. Let A denote the D × D matrixrepresenting Λ : GR (Φ) → GR (Φ) with respect to some basis, and let F A : T D → T D be the toral automorphism x + Z D Ax + Z D induced by A . Whenever Φ is hyperbolic, F A is also hyperbolic. Theorem 1.
Suppose that Φ is unimodular and hyperbolic. There is then a finite-to-onecontinuous map G : Ω Φ → T D so that G ◦ Φ = F A ◦ G . Furthermore, G is µ -a.e. r -to-onefor some r ∈ N and G is homologically essential in that G ∗ : H ( T D ; Z ) → H (Ω Φ ; Z ) is injective. EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 5
Let Φ ∗ : H (Ω Φ ; Z ) → H (Ω Φ ; Z ) denote the isomorphism induced by Φ on theinteger ˇCech cohomology of Ω Φ . We will see that there is a Φ ∗ -invariant subgroup H of H (Ω Φ ; Z ) restricted to which Φ ∗ is conjugate with the dual isomorphism Λ ∗ : Hom ( GR (Φ); Z ) → Hom ( GR (Φ); Z ). It often happens that H is proper (see example12 below), in which case it may be possible to lower the r of Theorem 1 by increasingthe size of the torus. Let H hyp denote the largest subgroup of H (Ω Φ ; Z ) that contains H , is invariant under Φ ∗ , and restricted to which Φ ∗ is unimodular and hyperbolic.Let D ′ be the rank of H hyp and let A ′ be the transpose of a matrix representingΦ ∗ : H hyp → H hyp in some basis. Theorem 2.
Suppose that Φ is unimodular and hyperbolic. There is then a finite-to-one, and µ -a.e. r ′ -to-one for some r ′ ≤ r , map G ′ : Ω Φ → T D ′ so that G ′ ◦ Φ = F A ′ ◦ G ′ .Furthermore, ( G ′ ) ∗ : H ( T D ′ ; Z ) → H (Ω Φ ; Z ) is injective with range H hyp . Theorems 1 and 2 are proved in Section 6.
Conjecture 3. If Φ is hyperbolic and H hyp = H (Ω Φ ; Z ) (i.e., Φ ∗ is unimodular andhyperbolic on all of H (Ω Φ ; Z )), then r ′ = 1 (that is, G ′ is a.e. one-to-one). A family F i of eigenvalues of Λ is a Pisot family provided all the elements of F i have the same multiplicity as eigenvalues of Λ and, if λ is an algebraic conjugate ofthe elements of F i with | λ | ≥
1, then λ ∈ F i . We will say that Φ is a Pisot familysubstitution if each F i is a Pisot family.For any group action on a space there is a maximal factor (unique up to conju-gacy) on which the group acts equicontinuously. In the setting here, of R n -actions ontiling spaces Ω, the maximal equicontinuous factor is a Kronecker action on a torus orsolenoid. We will denote the maximal equicontinuous factor map by g . It is a conse-quence of the Halmos - von Neumann theory that the R n -action on Ω has pure discretespectrum if, and only if, g is a.e. one-to-one ([BK]). Theorem 4.
Suppose that Φ is a unimodular Pisot family substitution with linearexpansion Λ . If D (Λ) = D ( GR ) = D , then G : Ω Φ → T D is surjective, semi-conjugatesthe R n -action on Ω Φ with a Kronecker action of R n on T D , and the latter action isthe maximal equicontinuous factor of the R n -action on Ω Φ . That is, we may take G = g : Ω Φ → T D . Conjecture 5. If Φ is a unimodular Pisot family substitution and rank ( H (Ω Φ ; Z )) = D (Λ) then the R n -action on Ω Φ has pure discrete spectrum. There are counterexamples to Conjecture 5 if the assumption of unimodularity isdropped (see [BBJS], where a one-dimensional version of this conjecture is discussed).Our proofs of Theorems 1, 2, and 4 route through a certain non-compact cover of Ω Φ M. BARGE AND J.-M. GAMBAUDO for which there doesn’t appear to be an adequate analog in the non-unimodular case.J. Kwapisz has pointed out that results for non-unimodular hyperbolic substitutionscan be obtained using sufficiently large, but compact, covers.Theorem 4, and the Corollaries below, are proved in Section 8.Given a tiling T = { τ i } , a puncture map is a function p : T → R n so that p ( τ i ) ∈ τ i and if τ i = τ j + v , then p ( τ i ) = p ( τ j ) + v . A set Γ ⊂ R n is a Meyer set if Γ is relativelydense and Γ − Γ is uniformly discrete. A tiling T is said to have the Meyer property iffor any (and hence all) puncture map(s) p , p ( T ) is a Meyer set. Corollary 6.
Under the assumptions of Theorem 4: • g is finite-to-one and a.e. cr -to-1; • the eigenvalues of the R n -action on Ω Φ are relatively dense in R n ; • the tilings T ∈ Ω Φ have the Meyer property. Corollary 7.
If Conjecture 3 is true, then so is Conjecture 5.
A Pisot family substitution is said to be of ( m, d ) -Pisot family type if Λ is di-agonalizable over C and spec (Λ) = F consists of a single family with m = m and d = d . Note in this case that D (Λ) = md . It is a result of [BK] that, if Φ is of( m, d )-Pisot family type, then the maximal equicontinuous factor of the R n -action onΩ Φ is a Kronecker action on an md -dimensional torus (or solenoid, in case Φ is not uni-modular). Moreover, the factor map g : Ω Φ → T md is finite-to-one and a.e. cr -to-onefor cr = min { ♯g − ( x ) : x ∈ T md } . Theorem 8. If Φ is unimodular of ( m, d ) -Pisot family type, then G : Ω Φ → T D is themaximal equicontinuous factor of the R n -action. That is, D (Λ) = D ( GR ) = md and G = g . See Section 9 for the proof.
Remark 9.
The traditional Pisot Substitution Conjecture is: If Φ is a 1-dimensionalsubstitution with irreducible and unimodular incidence matrix and Pisot inflation fac-tor, then the R -action on Ω Φ has pure discrete spectrum. There are substitutions thatsatisfy these hypotheses and not those of Conjecture 5, and vice-versa. In the finalsection of this paper the conditions in Conjectures 3 and 5 are relaxed a bit to pro-duce conjectures that do generalize the traditional Pisot Substitution Conjecture. Therelaxation involves replacing H (Ω Φ ; Z ) by the potentially smaller H ess (Ω Φ ; Z ) . Remark 10.
Lee and Solomyak ( [LS1] ) show that if Φ is of ( m, d ) -Pisot family type,then the eigenvalues of the R n -action are relatively dense. The difficult part of theirproof lies in demonstrating that the return vectors are contained in a Z -module ofrank md = D (Λ) . This is an assumption of Theorem 4 above. But the condition EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 7 D (Λ) = D ( GR ) is easily verified in practice: It follows from Lemma 16 that if themultiplicity of λ as an eigenvalue of f ∗ : H ( X ; Z ) → H ( X ; Z ) is the same as itsmultiplicity as an eigenvalue of Λ , for each λ ∈ spec (Λ) , then D (Λ) = D ( GR ) . (Here f : X → X is the map on the collared A-P complex, see below.) The cohomologies (and essential cohomologies - see Section 10) in the followingexamples are easily computed by the methods of [BD1].
Example 11.
Let φ be the substitution on letters: a ab ′ , b a, a ′ a ′ b, b ′ a ′ . The corresponding one-dimensional substitution Φ is of (1 , -Pisot family type. H hyp = H ≃ Z , so G ′ = G = g . These maps are a.e. 2-to-1 and H hyp is proper in H ess (Ω Φ ; Z ) ≃ Z . (Here H (Ω Φ ; Z ) ≃ Z .) Example 12.
Let φ be the substitution on letters: a aba ′ , b ab, a ′ a ′ b ′ a, b ′ a ′ b ′ .The corresponding one-dimensional substitution Φ is of (1 , -Pisot family type so G = g : Ω Φ → T and these maps are a.e. 2-to-1. H (Ω Φ ; Z ) ≃ Z , but H hyp = H ess (Ω Φ ; Z ) ≃ Z and G ′ : Ω Φ → T is a.e. 1-to-1. Abelian covers of Ω Φ and global shadowing To simplify notation, let us fix a (primitive, FLC, non-periodic) n -dimensional sub-stitution Φ and let Ω := Ω Φ . Let X be the Anderson-Putnam complex (A-P complex)for Φ (see [AP]): X is a cell complex with one n-cell ρ i × { i } for each prototile ρ i , andthese n-cells are glued along faces according to the following scheme. Suppose that ρ i and ρ j are prototiles and u, v ∈ R n , T ∈ Ω, are such that ρ i + u, ρ j + v ∈ T . Set ρ i × { i } ∋ ( x, i ) ∼ ( y, j ) ∈ ρ j × { j } if x + u = y + v and extend ∼ to an equivalencerelation on ∪ i ρ i × { i } . The A-P complex is the quotient X := ∪ i ρ i × { i } / ∼ . There isa natural map p : Ω → X given by p ( T ) = [( x, i )] ∼ provided 0 = u + x ∈ u + ρ i ∈ T and an induced map f : X → X with p ◦ Φ = f ◦ p . The map ˆ p : Ω → lim ←− f given byˆ p ( T ) := ( p ( T ) , p (Φ − ( T )) , . . . ) and the shift homeomorphism ˆ f : lim ←− f → lim ←− f satisfy: • ˆ p ◦ Φ = ˆ f ◦ p ; and • ˆ p is an a.e. (with respect to µ on Ω) one-to-one surjection.Moreover, when Φ has the property that it “forces the border”, which can be arrangedby replacing the tiles of tilings in Ω by their collared versions, the map ˆ p is a homeo-morphism ([AP]).Let π : ˜ X → X be the universal cover of the A-P complex. The group of decktransformations of ˜ X can be identified with the fundamental group of X . We formthe abelian cover π ab : ˜ X ab → X by quotienting out the action of the commutatorsubgroup, C , of the fundamental group of X : ˜ X ab := ˜ X/ ∼ ab with ˜ x ∼ ab ˜ y if and only ifthere is γ ∈ C with γ (˜ x ) = ˜ y . Since the map f ♯ induced by f on the fundamental group M. BARGE AND J.-M. GAMBAUDO of X takes C into C , f lifts to ˜ f ab : ˜ X ab → ˜ X ab . The group of deck transformations of˜ X ab can be identified with the first homology H ( X ; Z ). If h ∈ H ( X ; Z ) and ˜ x ∈ ˜ X ab we write ˜ x + h , or sometimes h (˜ x ), for the image of ˜ x under the deck transformationcorresponding to h , and we have ˜ f ab (˜ x + h ) = ˜ f (˜ x ) + f ∗ ( h ) with f ∗ the homomorphisminduced on H ( X ; Z ) by f .Suppose that K is any subgroup of H ( X ; Z ). There is then a corresponding cover π K : ˜ X K → X where ˜ X K is the quotient of ˜ X ab by the action of K . The group ofdeck transformations of ˜ X K is H ( X ; Z ) /K and if K is invariant under f ∗ , f lifts to˜ f K : ˜ X K → ˜ X K , and ˜ f K (˜ x + [ h ]) = ˜ f K (˜ x ) + [ f ∗ ( h )], where [ h ] denotes the coset h + K . Lemma 13.
Suppose that the subgroup K of H ( X ; Z ) is invariant under f ∗ and that ¯ f ∗ defined by ¯ f ∗ ([ h ]) := [ f ∗ ( h )] is an isomorphism of H ( X ; Z ) /K . Then the naturalmap ˆ π K : lim ←− ˜ f K → lim ←− f , given by ˆ π K ((˜ x ) i ) := ( π K (˜ x i )) , is a covering map with groupof deck transformations equal to H ( X ; Z ) /K .Proof. The issue is surjectivity. Given x, y ∈ X with f ( x ) = y , and ˜ y ∈ ˜ X K with π K (˜ y ) = y , pick ˜ x ′ ∈ ˜ X K with π K (˜ x ′ ) = x . Then π K ( ˜ f K (˜ x ′ )) = y so there is h ∈ H ( X ; Z ) with ˜ f K (˜ x ′ ) + [ h ] = ˜ y . Let ˜ x := ˜ x ′ + ¯ f − ∗ ([ h ]). Then ˜ f K (˜ x ) = ˜ y andsurjectivity of ˆ π K follows. (cid:3) Let us give another description of the covering ˆ π K : lim ←− ˜ f K → lim ←− f . Let Ω = Ω Φ and let ˜Ω K := { ( T, ˜ x ) : p ( T ) = π K (˜ x ) } ⊂ Ω × ˜ X K , with the product topology. Let π : ˜Ω K → Ω and π : ˜Ω K → ˜ X K be projections onto first and second factors. Lemma 14.
Suppose, as in Lemma 13, that K is invariant under f ∗ and that ¯ f ∗ is anisomorphism. Suppose also that X is the collared A-P complex of Φ . Then π : ˜Ω K → Ω is isomorphic with ˆ π K : lim ←− ˜ f K → lim ←− f .Proof. Since X is collared, the map ˆ p : Ω → lim ←− f , ˆ p ( T ) := ( p ( T ) , p (Φ − ( T ) , . . . ), is ahomeomorphism. The continuous map ˜ˆ p defined by ( T, ˜ x ) (˜ x i ), where ˜ x i satisfies:˜ x = ˜ x and π K (˜ x i ) = p (Φ − i ( T )) has continuous inverse (˜ x i ) (ˆ p − (( π K (˜ x i ))) , ˜ x ).Moreover, ˆ π K ◦ ˜ˆ p = ˆ p ◦ π . (cid:3) Under the hypotheses of Lemma 14: • We can lift the homeomorphism Φ on Ω to a homeomorphism ˜Φ K : ˜Ω K → ˜Ω K defined by ˜Φ K (( T, ˜ x )) := (Φ( T ) , ˜ f K (˜ x )). This homeomorphism is conjugatedwith ˆ˜ f K , the shift homeomorphism on lim ←− ˜ f K , by ˜ˆ p . • We may also lift the R n -action from Ω to ˜Ω K as follows. Given T ∈ Ω let p T : R n → X be defined by p T ( v ) := p ( T − v ). Given ˜ x ∈ ˜ X K with π K (˜ x ) = EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 9 p ( T ), let ˜ p T ˜ x : R n → ˜ X K be the unique lift of p T satisfying ˜ p T ˜ x (0) = ˜ x . For˜ T = ( T, ˜ x ) ∈ ˜Ω K and v ∈ R n define ˜ T − v := ( T − v, ˜ p T ˜ x ( v )). Note that˜Φ K ( ˜ T − v ) = ˜Φ K ( ˜ T ) − Λ v . • Finally we equip ˜Ω K with a metric ¯ d as follows. First we select a metric ˜ d on ˜ X K with two important properties. Suppose that { [ h i ] } Ni =1 is a basis for H ( X ; Z ) /K . For [ h ] = P Ni =1 b i [ h i ], let | [ h ] | := P Ni =1 | b i | . We take ˜ d so that:(1) ˜ d (˜ x + [ h ] , ˜ y + [ h ]) = ˜ d (˜ x, ˜ y ) for all ˜ x, ˜ y ∈ ˜ X K , [ h ] ∈ H ( X ; Z ) /K ; and(2) ˜ d (˜ x, ˜ y + [ h ]) → ∞ as | [ h ] | → ∞ for all ˜ x, ˜ y ∈ ˜ X K .For the metric on ˜Ω K we set ¯ d (( T, ˜ x ) , ( S, ˜ y )) := d ( T, S ) + ˜ d (˜ x, ˜ y ).For T, S ∈ Ω, we say that T globally shadows S (with respect to K), and write T ∼ gsK S , if there are ˜ T = ( T, ˜ x ) and ˜ S = ( S, ˜ y ) in ˜Ω K so that { ¯ d ( ˜Φ kK ( ˜ T ) , ˜Φ kK ( ˜ S ) } k ∈ Z ,is bounded. 4. Generalized return vectors
A vector v ∈ R n is called a return vector for Φ if there is T ∈ Ω Φ so that p ( T − v ) = p ( T ). Let us call a vector v ∈ R n a generalized return vector for Φ if there are v i ∈ R n and T i ∈ Ω, i = 1 , . . . , k , with v = v + · · · + v k , so that p ( T i − v i ) = p ( T i +1 ) for i = 1 , . . . , k −
1, and p ( T k − v k ) = p ( T ). The collection of generalized return vectors isa subgroup of R n which we denote by GR (Φ). If Φ is unimodular, one can show that GR (Φ) equals the subgroup of R n generated by the return vectors.Any path α : [0 , → X in the Anderson-Putnam complex X for Φ can be “lifted”to a curve ˜ α : [0 , → R n (think of unfolding the tiles that α runs through). Let l ( α ) := ˜ α (1) − ˜ α (0). Lemma 15. ( [BSW] ) If α is a path in the A-P complex X for Φ , the vector l ( α ) is awell-defined (independent of lift) element of GR (Φ) and depends only on the homotopyclass (rel. endpoints) represented by α . Furthermore, if α is a loop, l ( α ) dependsonly on the homology class of α . The resulting function l : H ( X ; Z ) → GR (Φ) is asurjective group homomorphism and l ◦ f ∗ = Λ l . Let K Λ be the kernel of l : H ( X ; Z ) → GR (Φ). By Lemma 15, K Λ is invariant under f ∗ ; let ¯ f ∗ : H ( X ; Z ) /K Λ → H ( X ; Z ) /K Λ be the homomorphism induced by f ∗ . Thegroup H ( X ; Z ) /K Λ ≃ GR (Φ) is a finitely generated ( H ( X ; Z ) is finitely generated)free abelian group ( GR (Φ) is a subgroup of R n ) so, with the choice of some basis, ¯ f ∗ isrepresented by an integral matrix A . The following lemma is an adaptation of a resultin [KS]. Lemma 16.
Let spec (Λ) = ∪ ki =1 F i be the partition of spec (Λ) into families of alge-braically conjugate eigenvalues and let m i be the maximum multiplicity of the elements of F i . Then λ is an eigenvalue of A of multiplicity m if and only if λ has a conjugatein F i for some i . Moreover, m ≥ m i .Proof. Let { [ h ] , . . . , [ h k ] } be a basis for the Z -module H ( X ; Z ) /K Λ with respect towhich ¯ f ∗ is represented by A and let v i := l ( h i ) ∈ GR (Φ), i = 1 , . . . , k . Let L : R k → R n be the linear map that takes the standard basis vector e i to v i for each i . Then LA = Λ L and, since the return vectors span R n (a consequence of minimality of the R n -action onΩ), L is surjective. Suppose that λ has a conjugate µ in F i for some i and that µ is areal eigenvalue of Λ with multiplicity m i . There is then a Λ t -invariant m i -dimensionalsubspace V ⊂ R n , restricted to which (Λ t − µI ) m i is zero. Then ( A t − µI ) m i is zeroon the m i -dimensional subspace L t V of R k . Since L t is injective, this means that µ isan eigenvalue of A of multiplicity at least m i , and thus that λ is an eigenvalue withmultiplicity at least m i of A . We leave it to the reader to reach the same conclusionwhen λ is complex.Now suppose that λ is an eigenvalue of A . Let p ( t ) = r ( t ) q ( t ) be the characteristicpolynomial of A factored with r ( t ) , q ( t ) ∈ Z [ t ] so that all roots of q ( t ), and no rootsof r ( t ), are algebraic conjugates of λ . Then W := ker ( q ( A )) is a non-trivial subspaceof R k , invariant under A , restricted to which all eigenvalues of A are conjugates of λ .Moreover, since q and A are integer, W has a basis { w , . . . , w d } with w i ∈ Z k foreach i . Let w i = P dj =1 w ij e j with w ij ∈ Z . Then the elements [ h ′ i ] ∈ H ( X ; Z ) /K Λ , h ′ i := P dj =1 w ij h i , span a ¯ f ∗ -invariant submodule of H ( X ; Z ) /K Λ of rank d >
0. Since h ′ i / ∈ K Λ , l ( h ′ i ) = 0 and it follows that Lw i = 0. Thus L ( W ) is a nontrivial subspace of R n , invariant under Λ. Applying the argument of the previous paragraph to A | W , L | W and Λ | L ( W ) , we conclude that Λ has an eigenvalue that is also an eigenvalue of A | W ,and is hence a conjugate of λ . (cid:3) From Lemma 16 we have:
Corollary 17. D ( GR ) ≥ D (Λ) . Question 18.
It is a consequence of Theorem 3.1 of [LS1] that, if Λ is diagonalizableover C and all of its eigenvalues are algebraic conjugates of the same multiplicity, then D ( GR ) = D (Λ) . Is it ever the case that D ( GR ) > D (Λ) ? We see from Lemma 16 that if Φ is unimodular and hyperbolic, then ¯ f ∗ is unimodularand hyperbolic on H ( X ; Z ) /K Λ . It is sometimes possible to increase the size of the quo-tient module while retaining the unimodularity and hyperbolicity of the correspondingquotient isomorphism (example 12). Let T denote the torsion subgroup of H ( K ; Z ), let f ′∗ denote the induced homomorphism f ′∗ : H ( X ; Z ) /T → H ( X ; Z ) /T of the finitelygenerated free abelian group H ( X ; Z ) /T , and let A be the matrix representing f ′∗ insome basis. We may then factor the characteristic polynomial p ( t ) of A , over Z , as EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 11 p ( t ) = q ( t ) r ( t ) so that all roots of q ( t ) are algebraic units and none have modulus 1; andall roots of r ( t ) are not units, or have a conjugate of modulus 1. Let F := Ker ( r ( f ′∗ ))and let K hyp := π − ( F ) ⊂ H ( X ; Z ), where π : H ( X ; Z ) → H ( X ; Z ) /T is the quo-tient homomorphism. Then K hyp ⊂ K Λ is invariant under f ∗ , is independent of thechoice of basis, and is minimal with respect to the property that the induced isomor-phism ¯ f ∗ : H ( X ; Z ) /K hyp → H ( X ; Z ) /K hyp is unimodular and hyperbolic.The group H ( X ; Z ) is (naturally) isomorphic with Hom ( H ( X ; Z ) , Z ) so the ˇCechcohomology H (lim ←− f ; Z ) = lim −→ f ∗ : H ( X ; Z ) → H ( X ; Z ) is isomorphic with the directlimit of the dual of f ∗ . Let H hyp := { c ∈ H ( X ; Z ) : c ( h ) = 0 for all h ∈ K hyp } . Then f ∗ | H hyp : H hyp → H hyp is an isomorphism so that H hyp ≃ lim −→ f ∗ | H hyp . By means ofˆ p ∗ : lim −→ f ∗ → H (Ω Φ ; Z ), H hyp can be viewed as a subgroup of H (Ω Φ ; Z ); if X iscollared (so that ˆ p ∗ is an isomorphism), H hyp ⊂ H (Ω Φ ; Z ) is the largest subgroup of H (Ω Φ ; Z ) that is invariant under Φ ∗ , contains H , and on which Φ ∗ is unimodular andhyperbolic. 5. From substitutions to toral automorphisms
The first cohomology group H ( X ; Z ) is naturally isomorphic with the Bruschlinskigroup consisting of homotopy classes of maps from X to the additive circle group T := R / Z (that is, T is a K ( Z , γ ]) = γ ∗ ( ),where is the fundamental class of H ( T ; Z ) and γ ∗ ( ) is its pullback to H ( X ; Z ).Let us fix a subgroup K of H ( X ; Z ), invariant under f ∗ , and define H K := { c ∈ Hom ( H ( X ; Z ) , Z ) : c ( h ) = 0 for all h ∈ K } ⊂ Hom ( H ( X ; Z ) , Z ) ≃ H ( X ; Z ).Let { c , . . . , c N } be a basis for H K and, for each i = 1 , . . . , N , choose a map γ i with[ γ i ] := Θ − ( c i ). Then, for each i , Θ([ γ i ]) = ( γ ∗ i ) − ( ) = c i annihilates K .For the following proposition, let A = ( a i,j ) denote the transpose of the matrix repre-senting the homomorphism induced on H K by f with respect to the basis { c , . . . , c N } .Thus γ j ◦ f = P Ni =1 a j,i γ i , up to homotopy. We also assume that X is collared so thatlim ←− f ≃ Ω Φ (although the proposition would still be true in general, replacing Ω Φ bylim ←− f and making the appropriate adjustments). Let F A : T N → T N be the linear torusmap F A ( x + Z N ) := Ax + Z N , let Γ : X → T N be given by Γ( x ) := ( γ ( x ) , . . . , γ N ( x )) t ,and let G := Γ ◦ p : Ω Φ → T N . Proposition 19.
Suppose that Φ is unimodular and hyperbolic, and that K is a sub-group of H ( X ; Z ) that is invariant under f ∗ and lies between K hyp and K Λ . Thenthe map F A is a hyperbolic toral automorphism and there exists a continuous map G : Ω Φ → T N with the properties:(i) G ∗ = G ∗ : H ( T N ; Z ) → H (Ω Φ ; Z ) ;(ii) G ◦ Φ = F A ◦ G ; (iii) G ( T ) = G ( S ) if and only if T ∼ gsK S . For the following lemma, let ˜Ω → Ω be a covering map with group of deck transfor-mations H . Lemma 20.
Suppose that G i : Ω → T N , i = 1 , , are continuous maps with lifts ˜ G i : ˜Ω → R N . If | ˜ G − ˜ G | is bounded, then G ∗ = G ∗ : H ( T N ; Z ) → H (Ω; Z ) .Proof. There are homomorphisms α i : H → Z N so that ˜ G i (˜ x + h )) = ˜ G i (˜ x ) + α i ( h )for all h ∈ H and i = 1 ,
2. Clearly, | ˜ G − ˜ G | bounded implies that α = α =: α .Suppose that γ : T N → T is continuous. Let ˜ γ : R N → R be a lift. Define ˜ H (˜ x, t ) := t (˜ γ ◦ ˜ G (˜ x )) + (1 − t )(˜ γ ◦ ˜ G (˜ x )). Then ˜ H (˜ x + h, t ) = · · · = ˜ H (˜ x, t ) + α ( h ), so that ˜ H descends to a homotopy from γ ◦ G to γ ◦ G . Thus G ∗ ( γ ∗ ( )) = G ∗ ( γ ∗ ( )). That is, G ∗ ◦ Θ = G ∗ ◦ Θ, so G ∗ = G ∗ , as Θ : [ T N , T ] → H ( T N ; Z ) is an isomorphism. (cid:3) Proof. (of Proposition 19) Suppose that α is a loop in ˜ X K . Then π K ◦ α is a loop thatrepresents a homology class [ π K ◦ α ] in K and c i ([ π K ◦ α ]) = 0. That is, ( γ ∗ i ) − ( )(([ π K ◦ α ])) = (( γ i ) ∗ ([ π K ◦ α ]) = ([ γ i ◦ π K ◦ α ]) = 0. This means that the loop γ i ◦ π K ◦ α in T is null homotopic and it follows that the map γ i : X → T lifts to ˜ γ i : ˜ X K → R . Recallthat Γ : X → T N is given by Γ( x ) := ( γ ( x ) , . . . , γ N ( x )) t ; then ˜Γ := (˜ γ , . . . , ˜ γ N ) t :˜ X K → R N is a lift of Γ and ˜Γ := ˜Γ ◦ π : lim ←− ˜ f K → R N is a lift of G = Γ ◦ p (here π is projection onto the zeroth coordinate and we have identified ˜Ω K with lim ←− ˜ f K viaLemma 14).We know from Lemma 16 that the integer matrix A is unimodular and hyperbolic andthus the linear torus map F A is a hyperbolic toral automorphism. The homomorphismΓ ∗ takes K to 0 and the induced homomorphism ¯Γ ∗ : H ( X ; Z ) /K → H ( T N ; Z ) isan isomorphism. Moreover, the choice of A guarantees that ¯Γ ∗ ◦ ¯ f ∗ = ( F A ) ∗ ◦ ¯Γ ∗ .Since ˜Γ is a lift of Γ, we have ˜Γ(˜ x + [ h ]) = ˜Γ(˜ x ) + ¯Γ ∗ ([ h ]) for ˜ x ∈ ˜ X K and [ h ] ∈ H ( X ; Z ) /K . Thus A ¯Γ ∗ = ¯Γ ∗ ◦ ¯ f ∗ (we have identified Z N ⊂ R N with H ( T N ; Z )). Wesee then that A ˜Γ ((˜ x i ) + [ h ]) = A (˜Γ ((˜ x i )) + ¯Γ ∗ ([ h ])) = A ˜Γ ((˜ x i )) + ¯Γ ∗ ◦ ¯ f ∗ ([ h ]), while˜Γ ◦ ˆ˜ f K ((˜ x i ) + [ h ]) = ˜Γ ( ˆ˜ f K ((˜ x i ) + ¯ f ∗ ([ h ])) = ˜Γ ◦ ˆ˜ f K ((˜ x i )) + ¯Γ ∗ ◦ ¯ f ∗ ([ h ]). It follows that | A ˜Γ − ˜Γ ◦ ˆ˜ f K | is uniformly bounded on lim ←− ˜ f K (by its bound on a single fundamentaldomain).Let E s and E u denote the stable and unstable linear subspaces of R N under applicationof A . Then R N = E s ⊕ E u , and E s and E u are invariant under A . For each z ∈ R N ,let z s ∈ E s and z u ∈ E u be so that z = z s + z u . There are C and η , 0 < η <
1, with || A k z s || ≤ Cη k || z s || and || A − k z u || ≤ Cη − k || z u || for all z ∈ R N and k ∈ N . Now, given(˜ x i ) ∈ lim ←− ˜ f K and k ∈ Z , let y k := ˜Γ ( ˜ f kK ((˜ x i ))). By the above, b k := y k +1 − Ay k is EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 13 bounded. For k ∈ N we have A − k y k = y + P ki =1 A − i b i − and A k y − k = y − P k − i =0 A i b i .Define z = z ((˜ x i )) by z = z u + z s where z u := lim k →∞ ( A − k y k ) u = y u + ∞ X i =1 A − k b uk − and z s := lim k →∞ ( A k y − k ) s = y s − ∞ X i =0 A k b s − k . It is clear that z depends continuously on (˜ x i ), that z ((˜ x i ) + [ h ]) = z ((˜ x i )) + ¯Γ ∗ ([ h ]),and that | ˜Γ ((˜ x i )) − z ((˜ x i )) | is bounded. Thus the map ˜Γ ′ : lim ←− ˜ f K → R N givenby ˜Γ ′ ((˜ x i )) := z ((˜ x i )) is the lift of a continuous map Γ ′ : lim ←− f → T N . Let ˜ˆ p :˜Ω K → lim ←− ˜ f K be the isomorphism of Lemma 14, let ˜ G := ˜Γ ′ ◦ ˜ˆ p : ˜Ω K → R N andlet G := Γ ′ ◦ ˆ p : Ω → T N . By Lemma 20, G ∗ = G ∗ . Furthermore, (˜Γ ′ ( ˆ˜ f K ((˜ x ) i ))) u =(˜Γ ′ ((˜ x i +1 ))) u = lim k →∞ ( A − k y k +1 ) u = A lim k →∞ ( A − ( k +1) y k +1 ) u = A (˜Γ ′ ((˜ x i ))) u . Simi-larly, (˜Γ ′ ( ˆ˜ f K ((˜ x i )))) s = A (˜Γ ′ ((˜ x ) i ))) s , so that ˜Γ ′ ◦ ˆ˜ f K = A ˜Γ ′ , whence Γ ′ ◦ ˆ f = F A ◦ Γ ′ .Since ˆ f ◦ ˆ p = ˆ p ◦ Φ, we have G ◦ Φ = F A ◦ G .It remains to prove (iii). First suppose that T ∼ gsK S . Let ˜ T and ˜ S , lying over T and S , have the property that ¯ d ( ˜Φ kK ( ˜ T ) , ˜Φ kK ( ˜ S )) is bounded for k ∈ Z . As ˜ G is alift of a continuous function on a compact space, and the metric ¯ d is equivariant, ˜ G is uniformly continuous and hence | ˜ G ( ˜Φ kK ( ˜ T )) − ˜ G ( ˜Φ kK ( ˜ S )) | = | A k ( ˜ G ( ˜ T ) − ˜ G ( ˜ S )) | isalso bounded. Since A is hyperbolic, this can only happen if ˜ G ( ˜ T ) = ˜ G ( ˜ S ). Thus G ( T ) = G ( S ).Conversely, if G ( T ) = G ( S ), let ˜ T and ˜ S lie over T and S . There is then h ∈ Z N = H ( T N ; Z ) so that ˜ G ( ˜ T ) + h = ˜ G ( ˜ S ) and there is h ′ ∈ H ( X ; Z ) /K so that ¯Γ ∗ ( h ′ ) = h .Let ˜ T ′ := ˜ T + h ′ . Then ˜ T ′ also lies over T and ˜ G ( ˜ T ′ ) = ˜ G ( ˜ S ). From ˜ G ◦ ˜Φ K = A ˜ G it follows that ˜ G ( ˜Φ kK ( ˜ T ′ )) = ˜ G ( ˜Φ kK ( ˜ S )) for all k ∈ Z . Were ¯ d ( ˜Φ kK ( ˜ T ′ ) , ˜Φ kK ( ˜ S )) notbounded, there would be k j ∈ Z and h j ∈ H ( X ; Z ) /K with ¯ d ( ˜Φ k j K ( ˜ T ′ ) + h j , ˜Φ k j K ( ˜ S ))bounded and | h j | → ∞ (see the second of the assumptions on the nature of ˜ d ), hence | ¯Γ ∗ ( h j ) | → ∞ . But (by uniform continuity of ˜ G and equivariance of ¯ d ), | ˜ G ( ˜Φ k j K ( ˜ T ′ ) + h j ) − ˜ G ( ˜Φ k j K ( ˜ S ) | = | ¯Γ ∗ ( h j ) | is bounded. Thus it must be the case that T ∼ gsK S . (cid:3) Geometric realization
Proof. (of Theorems 1 and 2) Let K = K Λ for Theorem 1 or K = K hyp for Theorem2 and let A , G , N be as in Proposition 19, with N = D or N = D ′ , depending onwhether K = K Λ or K = K hyp . Both ( T N , F A ) and (Ω , Φ) are Smale spaces - the actions are hyperbolic with local product structure (see [AP]). A lemma of Putnam([P]) asserts that a factor map (such as G ) between Smale spaces that is injective onunstable manifolds is globally finite-to-one.Thus we are reduced to proving that G is injective on unstable manifolds. Given T ∈ Ω, the unstable manifold of T under Φ is the set W u ( T ) := { T ′ ∈ Ω : d (Φ k ( T ) , Φ k ( T ′ )) → k → −∞} . It is easy to see that W u ( T ) = { T − v : v ∈ R n } .Suppose that G ( T − v ) = G ( T ) for some T ∈ Ω and 0 = v ∈ R n . Accordingto Proposition 19, T ∼ gsK T − v . Let us first show that the lifts ˜ T = ( T, ˜ x ) and˜ T − v = ( T − v, ˜ p T ˜ x ( v )) of T and T − v to ˜Ω K are such that ¯ d ( ˜Φ kK ( ˜ T ) , ˜Φ kK ( ˜ T − v )) → ∞ as k → ∞ . Note that π ( ˜Φ kK ( ˜ T )) = ˜ p Φ k ( T )˜ f kK (˜ x ) (0) and π ( ˜Φ kK ( ˜ T − v )) = ˜ p Φ k ( T )˜ f kK (˜ x ) (Λ k v ).We may choose return vectors v k so that | v k − Λ k v | is bounded and B r [Φ k ( T )] = B r [Φ k ( T ) − v k ] with r twice the maximum diameter of all tiles (recall that B r [ T ] denotesthe collection of all tiles in T that meet the closed ball centered at 0 with radius r ).Then p (Φ k ( T )) = p (Φ k ( T ) + v k ) in the collared Anderson-Putnam complex X . Thereis [ h k ] ∈ H ( X ; Z ) /K with l ( h k ) = v k . Note that there are infinitely many distinctsuch v k , and hence infinitely many distinct [ h k ] for k ∈ N . It follows that˜ d ( p Φ k ( T )˜ f kK (˜ x ) (0) , p Φ k ( T )˜ f kK (˜ x ) (0) + [ h k ]) = ˜ d ( p Φ k ( T )˜ f kK (˜ x ) (0) , p Φ k ( T )˜ f kK (˜ x ) ( v k ))is unbounded for k ∈ N . Hence˜ d ( p Φ k ( T )˜ f kK (˜ x ) (0) , p Φ k ( T )˜ f kK (˜ x ) (Λ k v )) ≥ ˜ d ( p Φ k ( T )˜ f kK (˜ x ) (0) , p Φ k ( T )˜ f kK (˜ x ) ( v k )) − ˜ d ( p Φ k ( T )˜ f kK (˜ x ) ( v k ) , p Φ k ( T )˜ f kK (˜ x ) (Λ k v )) ≥ ˜ d ( p Φ k ( T )˜ f kK (˜ x ) (0) , p Φ k ( T )˜ f kK (˜ x ) ( v k )) − B, for some finite B independent of k , is also unbounded for k ∈ N .Suppose now ˜ T and ˜ T ′ are lifts of T and T − v so that ¯ d ( ˜Φ kK ( ˜ T ) , ˜Φ kK ( ˜ T ′ )) is bounded for k ∈ Z and ˜ T ′ = ˜ T − v . Since ¯ d ( ˜Φ kK ( ˜ T ) , ˜Φ kK ( ˜ T − v )) → k → −∞ , ¯ d ( ˜Φ kK ( ˜ T ′ ) , ˜Φ kK ( ˜ T − v )) must be bounded as k → −∞ . But ˜ T ′ and ˜ T − v are not equal and both lie over T − v so there is a nonzero [ h ] ∈ H ( X ; Z ) /K with ˜ T ′ + [ h ] = ˜ T − v . Then the distancebetween ˜Φ kK ( ˜ T − v ) = ˜Φ kK ( ˜ T ′ ) + ¯ f k ∗ ([ h ]) and ˜Φ kK ( ˜ T ′ ) is certainly unbounded as k → −∞ since ¯ f k ∗ ([ h ]), by hyperbolicity, takes on infinitly many different values in H ( X ; Z ) /K .This establishes that G is one-to-one on unstable manifolds and hence G is globallyfinite-to-one.That G is µ -a.e. r -to-1 (or r ′ -to-1) is a consequence of the ergodicity of Φ with respectto the unique invariant probability measure µ of the R n -action and the measurabilityof the function f : Ω → R given by f ( T ) := ♯G − ( G ( T )). (cid:3) EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 15 Non-triviality of the global shadowing relation
To get some idea of what tilings are ∼ gsK -related, consider K = K Λ : Proposition 21.
Suppose that Φ is unimodular and hyperbolic and that T, T ′ ∈ Ω are Φ -periodic and share a tile. Then T ∼ gsK Λ T ′ .Proof. For simplicity, assume that T and T ′ are fixed by Φ. Let K = K Λ and supposethat τ ∈ T ∩ T ′ . Let v ∈ int ( τ ), so that B [ T − v ] = B [ T ′ − v ]. Choose ˜ T = ( T, ˜ x ) ∈ ˜Ω K to be fixed by ˜Φ K . There is then ˜ x ′ ∈ ˜ X K so that π ( ˜ T ′ − v ) = π ( ˜ T − v ), where˜ T ′ = ( T ′ , ˜ x ′ ). Consider the loop α in the A-P complex X: α ( t ) = (cid:26) p T (2 t Λ v + (1 − t ) v ) if 0 ≤ t ≤ / ,p T ′ (2 t Λ v + (1 − t ) v ) if 1 / ≤ t ≤ . Then l ( α ) = 0 so the homology class of α is in K and α lifts to a loop ˜ α in ˜ X K with˜ α (0) = π ( ˜ T − v ) = ˜ α (1). The lift ˜ α is of the form:˜ α ( t ) = (cid:26) ˜ p T ˜ x (2 t Λ v + (1 − t ) v ) if 0 ≤ t ≤ / , ˜ p T ′ ˜ y (2 t Λ v + (1 − t ) v ) if 1 / ≤ t ≤ y is determined by continuity. Since ˜ α (1) = ˜ p T ′ ˜ y ( v ) = ˜ α (0) = ˜ p T ˜ x ( v ) = ˜ p T ′ ˜ x ′ ( v ),it must be the case that ˜ y = ˜ x ′ . Now π ( ˜ T − v ) = π ( ˜ T ′ − v ) ⇒ π ( ˜ T − Λ v ) = π ( ˜Φ K ( ˜ T − v )) = π ( ˜Φ K ( ˜ T ′ − v )) = π ( ˜Φ K ( ˜ T ′ ) − Λ v ). We have π ( ˜ T ′ − Λ v ) = ˜ p T ′ ˜ x ′ (Λ v ) =˜ α (1 /
2) = π ( ˜ T − Λ v ) = π ( ˜Φ K ( ˜ T ′ ) − Λ v ) and hence ˜Φ K ( ˜ T ′ ) = ˜ T ′ . Since ˜ T and ˜ T ′ areboth fixed by ˜Φ K , T ∼ gsK T ′ . (cid:3) It is proved in [BO] that, if Φ is an n -dimensional self-similar substitution (thatis, Λ = λI ), there are Φ-periodic T = T ′ ∈ Ω that agree in a half-space: there is0 = u ∈ R n and R so that B [ T − v ] = B [ T ′ − v ] for all v ∈ R n with h u, v i ≥ R . Forsuch T, T ′ , { T, T ′ } is called an asymptotic pair . Corollary 22.
Suppose that Φ is self-similar. There are then Φ -periodic T = T ′ ∈ Ω and = u ∈ R n so that T ∼ gsK Λ T ′ and T − v ∼ gsK Λ T ′ − v for all v ∈ R n with h u, v i > .Proof. Let
T, T ′ be an asymptotic pair with direction vector u , as above. Then T ∼ gsK Λ T ′ by Proposition 21. There are then lifts ˜ T , ˜ T ′ that are both fixed by ˜Φ mK for some m >
0. Then ˜ T − v and ˜ T ′ − v are lifts of T − v and T ′ − v with the properties:˜Φ kmK ( ˜ T − v ) → ˜ T and ˜Φ kmK ( ˜ T ′ − v ) → ˜ T ′ as k → −∞ . Also, if h u, v i >
0, there is k ′ so that h u, λ k ′ v i ≥ R , and it follows that d (Φ mk ( T − v ) , Φ mk ( T ′ − v )) → k → ∞ .We saw in the proof of Proposition 21 that if ˜ T is fixed by ˜Φ mK , T ′ is fixed by Φ m , and π ( ˜ T − w ) = π ( ˜ T ′ − w ), then ˜ T ′ is also fixed by ˜Φ mK . It follows from hyperbolicity of ˜Φ K on deck transformations that, conversely, if ˜ T and ˜ T ′ are both fixed by ˜Φ K and B [ T − w ] = B [ T ′ − w ] (so that p T ( w ) = p T ′ ( w )) then π ( ˜ T − w ) = π ( ˜ T ′ − w ). If h u, v i >
0, then B [ T − w ] = B [ T ′ − w ] for w = λ k v and k > k ′ . Consequently, for such v , ¯ d ( ˜Φ mkK ( ˜ T − v ) , ˜Φ mkK ( ˜ T ′ − v )) = d (Φ mk ( T − v ) , Φ mk ( T ′ − v )) + ˜ d ( π ( ˜ T − λ mk v ) , π ( ˜ T ′ − λ mk v )) → k → ∞ , so T − v ∼ gsK Λ T ′ − v . (cid:3) Global shadowing and regional proximality for Pisot familysubstitutions
A fundamental result of Auslander ([A]) asserts that the structure relation of themaximal equicontinuous factor for an abelian group action on a compact metrizablespace is given by regional proximality. In the context of the R n -action on a tilingspace Ω, two tilings T, T ′ ∈ Ω are regionally proximal provided, for every ǫ >
S, S ′ ∈ Ω and v ∈ R n so that: (i) d ( T, S ) < ǫ ; (ii) d ( T ′ , S ′ ) < ǫ ; and (iii) d ( S − v, S ′ − v ) < ǫ . If T and T ′ are regionally proximal, we will write T ∼ rp T ′ . Thus T ∼ rp T ′ if and only if g ( T ) = g ( T ′ ), where g is the factor map onto the maximalequicontinuous factor of the translation action on Ω.We thus have two closed equivalence relations on a substitution tiling space Ω Φ :regional proximality and global shadowing. The aim of this section is to compare thesetwo relations. The following Proposition 24 shows that for unimodular hyperbolicsubstitutions, global shadowing is stronger than regional proximality. Proposition 25establishes that if the substitution is Pisot family, and D = D (Λ) = D ( GR ), then theregional proximal and global shadowing relations coincide. From this we will easilydeduce Theorem 4, and Corollaries 6 and 7.First, a lemma about regional proximality. Lemma 23.
Suppose that
T, S ∈ Ω Φ , k i → ∞ , and v i ∈ R n are such that Φ k i ( T ) → ¯ T ∈ Ω Φ , Φ k i ( S ) → ¯ S ∈ Ω Φ , and d (Φ k i ( T − v i ) , Φ k i ( S − v i )) → . Then ¯ T ∼ rp ¯ S .Proof. Let g be the map of Ω Φ onto the maximal equicontinuous factor Ω Φ / ∼ rp . If d (Φ k i ( T − v i ) , Φ k i ( S − v i )) = d (Φ k i ( T ) − Λ k i v i , Φ k i ( S ) − Λ k i v i ) →
0, then d ( g (Φ k i ( T )) − Λ k i v i , g (Φ k i ( S )) − Λ k i v i ) → g . Then, by equicontinuity ofthe R n -action on Ω Φ / ∼ rp , d ( g (Φ k i ( T )) , g (Φ k i ( S ))) →
0. Thus d ( g ( ¯ T ) , g ( ¯ S )) = 0 and¯ T ∼ rp ¯ S . (cid:3) Given ˜ x, ˜ x ′ ∈ ˜ X = ˜ X K Λ and a path ˜ γ in ˜ X from ˜ x to ˜ x ′ , let l (˜ γ ) := l ( γ ), where γ is the path γ = π ◦ ˜ γ in X (see Lemma 15). If ˜ γ ′ is any other such path, then theconcatenation of ˜ γ ′ with the reverse of ˜ γ is a loop in ˜ X and hence must project toan element of K Λ , that is, to a loop of displacement 0. Thus the displacement l (˜ γ )depends only on ˜ x and ˜ x ′ , and not on ˜ γ . EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 17
Proposition 24.
Suppose that Φ is unimodular and hyperbolic. Then the global shad-owing relation is contained in the regional proximality relation: T ∼ gsK Λ T ′ = ⇒ T ∼ rp T ′ .Proof. We suppose that the A-P complex X is collared. In this situation, if S, S ′ ∈ Ω Φ are such that p ( S ) = p ( S ′ ), then d (Φ k ( S ) , Φ k ( S ′ )) → k → ∞ .Suppose that ˜ T = ( T, ˜ x ) , ˜ T ′ = ( T ′ , ˜ x ′ ) ∈ ˜Ω are such that ¯ d ( ˜Φ k ( ˜ T ) , ˜Φ k ( ˜ T ′ )), k ∈ Z ,is bounded. There are then: a sequence k i → −∞ ; [ h i ] in the group H ( X ; Z ) /K Λ ofdeck transformations of ˜ X ; and n -cells ˜ τ , ˜ τ ′ of ˜ X so that ˜ x i ∈ [ h i ](˜ τ ) and ˜ x ′ i ∈ [ h i ](˜ τ ′ )for all i , where ( ˜ T i , ˜ x i ) := ˜Φ k i ( ˜ T ) and ( ˜ T ′ i , ˜ x ′ i ) := ˜Φ k i ( ˜ T ′ ). Let ˜ γ i be a path in ˜ X from˜ x i to ˜ x ′ i . We may take ˜ γ i = ˜ γ i ∗ · · · ∗ ˜ γ il with each ˜ γ ij a lift of a path γ ij in X of theform γ ij ( t ) = p ( S ij − tv ij ), t ∈ [0 , S ij ∈ Ω Φ and v ij ∈ R n with S i = T i and S il = T ′ i . (It is key to this argument that l is constant, independent of i .) We have p ( S ij − v ij ) = p ( S ij +1 ) for j = 1 , . . . , l − i . Passing to a subsequence, we mayassume that Φ | k i | ( S ij + v i + · · · + v ij − ) → ¯ S j ∈ Ω Φ as i → ∞ for j = 2 , . . . , l . Let¯ S = T . We conclude from Lemma 23 that ¯ S j ∼ rp ¯ S j +1 for j = 1 . . . , l − | k i | ◦ ˜ γ i is a path from ˜ x to ˜ x ′ , the vector Λ | k i | ( v i + · · · + v il − ) =: v isconstant (this is the displacement vector l ( ˜Φ | k i | ◦ ˜ γ i ) of the path ˜Φ | k i | ◦ ˜ γ i in ˜ X ). Thus¯ S l = T ′ + v . We have a path in ˜ X , call it ˜ γ , from ˜ x to ˜ x ′ with displacement v . Thenany path in ˜ X from ˜ f k (˜ x ) to ˜ f k (˜ x ′ ) has displacement Λ k v : since ¯ d ( ˜Φ k ( ˜ T ) , ˜Φ k ( ˜ T ′ )) isbounded, and ˜Φ k ( ˜ T ) = (Φ k ( T ) , ˜ f k (˜ x )) and ˜Φ k ( ˜ T ′ ) = (Φ k ( T ′ ) , ˜ f k (˜ x ′ )), v must be zero -that is, ¯ S l = T ′ . Since ∼ rp is a translation invariant equivalence relation, T ∼ rp T ′ . (cid:3) Proposition 25.
Suppose that Φ is a unimodular Pisot family n -dimensional substi-tution with linear expansion Λ . If D (Λ) = D ( GR ) = D , then the global shadowingrelation contains the regional proximal relation: T ∼ rp T ′ = ⇒ T ∼ gsK Λ T ′ .Proof. Let f : X → X be the substitution induced map on the collared A-P com-plex for Φ. Let A represent ¯ f ∗ : H ( X ; Z ) /K Λ → H ( X ; Z ) /K Λ in some basis, say { [ h ] , . . . , [ h D ] } , which we fix for the remainder of this proof. A is then hyperbolic,and, as an isomorphism of R D , has invariant stable and unstable spaces E s and E u with E s ⊕ E u = R D . There are η ∈ (0 ,
1) and C so that | A k x | ≤ Cη k | x | for x ∈ E s , k ∈ N and | A − k x | ≤ Cη k | x | for x ∈ E u , k ∈ N . Let x = x s + x u be the decompositionof x ∈ R D into stable and unstable parts. Thus, for each [ h ] ∈ H ( X ; Z ) /K Λ , there arecorresponding [ h ] s ∈ E s , [ h ] u ∈ E u .Given T ∈ Ω Φ and ˜ x ∈ ˜ X := ˜ X K Λ with π (˜ x ) = p ( T ), we will call the subset S ( T, ˜ x ) := { ˜ p T ˜ x ( v ) : v ∈ R n } of ˜ X a sheet . Claim 26.
There is B ∈ R so that if S is any sheet in ˜ X and [ h ] ∈ H ( X ; Z ) /K Λ issuch that [ h ]( S ) ∩ S contains an n -cell of ˜ X , then | [ h ] s | ≤ B . To prove the claim, let m ∈ N and α < | A mk x | ≤ α k | x | for all k ∈ N and x ∈ E s . We may assume that m is large enough so that, for each tile τ , Φ m ( τ )contains, in its interior, a tile of every type. Let B be large enough so that if ˜ τ isany n -cell in ˜ X , ˜ τ , ˜ τ are two n -cells in ˜ f m (˜ τ ) of the same type, and [ h ](˜ τ ) = ˜ τ ,then | [ h ] s | ≤ B . Now let T ∈ Ω Φ be fixed by Φ m and let ˜ x ∈ ˜ X be fixed by ˜ f m , with π (˜ x ) = p ( T ). The sheet S := { ˜ p T ˜ x ( v ) : v ∈ R n } is invariant under ˜ f m . We may assumethat ˜ x is in the interior of an n -cell ˜ τ of ˜ X so that S = ∪ k ∈ N ˜Φ km (˜ τ ). Let B be largeenough so that αB + B ≤ B . Subclaim:
If [ h ](˜ x ) ∈ S then | [ h ] s | ≤ B .To see that this is the case, let S k := ∪ j =0 ,...,k ˜ f mj (˜ τ ) for k = 0 , , . . . . We provethe claim, with S replaced by S k , by induction on k . If [ h ](˜ x ) ∈ S , then [ h ] s =0 and if [ h ](˜ x ) ∈ S , then | [ h ] s | ≤ B < B . Suppose the claim is true with S replaced with S k , k ≥
1, and suppose that [ h ](˜ x ) ∈ S k +1 . Let ˜ τ ′ be the n -cell of S k with [ h ](˜ x ) ∈ ˜ f m (˜ τ ′ ). There is then an n -cell of the same type as ˜ τ in ˜ f m (˜ τ ′ ) ⊂S k . Thus there is [ h ] with [ h ](˜ x ) ∈ ˜ f m (˜ τ ′ ): by hypothesis, | [ h ] s | ≤ B . Then˜ f m ([ h ](˜ x )) = [ f m ∗ ( h )](˜ x ) ∈ ˜ f m (˜ τ ′ ), so that | [ h ] s − [ f m ∗ ( h )] s | = | [ h − f m ∗ ( h )] s | ≤ B .Also, | [ f m ∗ ( h )] s | ≤ α | [ h ] s | ≤ αB , so that | [ h ] s | ≤ αB + B ≤ B , completing theinductive argument and establishing the subclaim.For each of the finitely many different types of n -cell in ˜ X , represented, say, by˜ τ , . . . , ˜ τ r , there is thus a sheet S i = S ( T i , ˜ x i ) so that if the n -cell ˜ τ ⊂ S i has the sametype as ˜ τ i , and [ h ](˜ τ ) ⊂ S i , then | [ h ] s | ≤ B := B . Now if S = S ( T, ˜ x ) is any sheetwith ˜ τ ⊂ S ∩ [ h ]( S ), let ˜ τ i be of the same type as ˜ τ . Let Q be a connected finite patch of T containing the tiles τ and τ − l ([ h ]), with ˜ p T ˜ x ( τ ) = ˜ τ , and hence also ˜ p T ˜ x ( τ − l ([ h ])) =[ − h ](˜ τ ). There is then v with Q − v ⊂ T i . Then [ h ](˜ p T ˜ x i ( τ − l [ h ] − v ) = ˜ p T ˜ x i ( τ − v ), so | [ h ] s | ≤ B . This establishes Claim 26.Suppose that T, T ′ ∈ Ω Φ are such that T ∼ rp T ′ . For each r = 1 , , . . . there are S r , S ′ r ∈ Ω Φ and v r , w r ∈ R n so that: B r [ T ] = B r [ S r ]; B r [ S r − v r ] = B r [ S ′ r − v r ]; B r [ T ′ − w r ] = B r [ S ′ r ]; and w r → r → ∞ . Pick a lift ˜ T = ( T, ˜ x ) ∈ ˜Ω, then chooselifts ˜ S r = ( S r , ˜ x ) , ˜ S ′ r = ( S ′ r , ˜ x ′ r ) ∈ ˜Ω so that ˜ p S r ˜ x ( v r ) = ˜ p S ′ r ˜ x ′ r ( v r ). Claim 27.
For sufficiently large r , ˜ x ′ r =: ˜ x ′ is constant and w r = 0 . To see this, let R be large enough so that every R -patch contains a tile of everytype. Let ˜ τ be an n -cell of ˜ X with ˜ x ∈ ˜ τ ⊂ S ( T, ˜ x ) and for each r ≥ R , let ˜ τ r bean n -cell of ˜ X with ˜ τ r ⊂ S ( S r , ˜ x ) ∩ S ( S ′ r , ˜ x ′ r ) of the same type as ˜ τ . Pick also n -cells˜ τ ′ r ⊂ S ( S ′ r , ˜ x ′ ), of the same type as ˜ τ , in ˜ p S ′ r ˜ x ′ r ( B R (0)). As B r [ S ′ r ] = B r [ T ′ − w r ], we maytake ˜ τ ′ r = ˜ p S ′ r ˜ x ′ r ( τ ′ − w r ) with τ ′ ∈ T ′ . Let L : R D → R n be the linear map given by L (( a , . . . , a D )) := P Di =1 a i l ([ h i ]). Then LA = Λ L and, since D = D (Λ), L | E u : E u → EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 19 R n is an isomorphism. Also, L is one-to-one on Z D (recall that l : H ( X ; Z ) → GR has kernel K Λ ). Thus Γ := L ( { a ∈ R D : | a s | ≤ B } ), B as in Claim 26, is a regularmodel set. In particular, Γ is uniformly discrete, relatively dense, and has the Meyerproperty (see [M]). Let [ h r ](˜ τ ) = ˜ τ r and [ h r ](˜ τ ′ r ) = ˜ τ r . Then | [ h r ] s | ≤ B and | [ h r ] s | ≤ B by Claim 26, so l ([ h r ]) , l ([ h r ]) ∈ Γ and { l ([ h r ]) − l ([ h r ]) : r ≥ R } is uniformly discrete.But ( l ([ h r ′ ]) − l ([ h r ′ ])) − ( l ([ h r ]) − l ( h r ])) = w r − w r ′ , and since w r →
0, it must bethat w r = 0 for all sufficiently large r . Thus, l ([ h r − h r ]) is constant for large r , andsince l is injective on H ( X ; Z ), [ h r − h r ] := [ h ] is constant for large r . Thus the n -cells˜ τ ′ r = [ h ](˜ τ ) and the patches B R [ S ′ r ] are constant for all large r and it follows that ˜ x ′ r isalso constant for large r , establishing Claim 27.We can deduce more from the above. For large r let ˜ γ ( t ) := ˜ p S r ˜ x ( tv r ), ˜ γ ( t ) :=˜ p S ′ r ˜ x ′ ((1 − t ) v r ), and ˜ γ ( t ) := ˜ p T ′ ˜ x ′ ( tw ), 0 ≤ t ≤
1, where w is such that ˜ p T ′ ˜ x ′ ( w ) = [ h ](˜ x ).Then | l ([ h ]) | = | l (˜ γ ∗ ˜ γ ∗ ˜ γ ) | = | v r − v r + w | ≤ R . Let ¯ R := sup { ˜ d (˜ y, [ h ′ ](˜ y )) : ˜ y ∈ ˜ X, [ h ′ ] ∈ Γ , | [ h ′ ] | ≤ R } . Then ¯ R < ∞ since Γ is a discrete subset of R n and the metric˜ d is equivariant with respect to the deck transformations. Let ¯ R := sup { ˜ d (˜ p S ˜ y ( w ) , ˜ y ) : | w | ≤ R, ( S, ˜ y ) ∈ ˜Ω } . ¯ R is also finite by equivariance. We have: ˜ d (˜ x, ˜ x ′ ) ≤ ¯ R + ¯ R .Now let ˜ T ′ = ( T ′ , ˜ x ′ ). For k ∈ N , it is clear that the sheets ˜ f k ( S ( T, ˜ x )) = S (Φ k ( T ) , ˜ f k (˜ x )), ˜ f k ( S ( S r , ˜ x )) = S (Φ k ( S r ) , ˜ f k (˜ x )), ˜ f k ( S ( S ′ r , ˜ x ′ )) = S (Φ k ( S ′ r ) , ˜ f k (˜ x ′ )),and ˜ f k ( S ( T ′ , ˜ x ′ )) = S (Φ k ( T ′ ) , ˜ f k (˜ x ′ )) overlap in the same manner as those sheets with k = 0, and the above argument yields ˜ d ( ˜ f k (˜ x ) , ˜ f k (˜ x ′ )) ≤ ¯ R + ¯ R . Now fix k < r , there is r ′ = r ′ ( r ) big enough so that if S, S ′ ∈ Ω Φ satisfy B r ′ [ S ] = B r ′ [ S ′ ],then B r [Φ k ( S )] = B r [Φ k ( S ′ )]. (This consequence of the invertibility of Φ is referredto as “recognizability”, see, for example, [S1].) For r > r ′ = r ′ ( r ), con-sider ˜Φ k (( T, ˜ x )) = (Φ k ( T ) , ˜ y ) and ˜Φ k (( S r ′ , ˜ x )) = (Φ k ( S r ′ ) , ˜ y ′ ). Since B [Φ k ( S r ′ )] = B [Φ k ( T )], π (˜ y ) = π (˜ y ′ ) and there is [ h ] ∈ H ( X ; Z ) /K Λ so that [ h ](˜ y ′ ) = ˜ y . Then˜ x = ˜ f k (˜ y ′ ) = ˜ f k ([ h ](˜ y )) = ¯ f k ∗ ([ h ])( ˜ f k (˜ y )) = ¯ f k ∗ ([ h ])(˜ x ), so [ h ] = 0 ( ¯ f ∗ is invert-ible on H ( X ; Z ) /K Λ ). Thus ˜ y ′ = ˜ y . In this way, we see that the sheets deter-mined by ˜Φ k ( ˜ T ), ˜Φ k ( ˜ S r ′ ), ˜Φ k ( ˜ S ′ r ′ ), and ˜Φ k ( ˜ T ′ ), for r ′ large, overlap as above so that¯ d ( ˜Φ k ( ˜ T ) ′ ˜Φ k ( ˜ T ′ )) ≤ d (Φ k ( T ) , Φ k ( T ′ )) + ¯ R + ¯ R . We conclude that ¯ d ( ˜Φ k ( ˜ T ) , ˜Φ k ( ˜ T ′ )), k ∈ Z , is bounded; that is, T ∼ gsK Λ T ′ . (cid:3) Proof. (Of Theorem 4.) By Proposition 19, G ( T ) = G ( T ′ ) if and only if T ∼ gsK Λ T ′ ,and, by the fundamental result of Auslander ([A]), g ( T ) = g ( T ′ ) if and only if T ∼ rp T ′ .By Propositions 24 and 25, T ∼ gsK Λ T ′ if and only if T ∼ rp T ′ . Thus we may identify G with g . (cid:3) Proof. (Of Corollary 6.) The first statement follows immediately from Theorems 1 and4 and the characterization of cr in [BK]. For the second statement: the eigenvaluesform subgroup of R n so are relatively dense if and only if their linear span is all of R n .Suppose there is a vector v = 0 perpendicular to the linear span of the eigenvalues E . If f : Ω Φ → T is a continuous eigenfunction with eigenvalue β and T ∈ Ω Φ , then f ( T − tv ) = exp (2 πi h β, tv i ) f ( T ) = f ( T ) for all t ∈ R . This means that g ( T − tv ) = g ( T )for all t ∈ R (the Halmos - von Neumann theory asserts that the map g is determinedby the continuous eigenfunctions - see [R] or [BK]). But then g is not finite-to-one.It is shown in [LS2] that, in this context, the third statement is equivalent to thesecond. (cid:3) Proof. (Of Corollary 7.) The spectrum of the R n -action on Ω Φ is pure discrete if andonly if g is a.e. one-to-one (see, for example, [BK]). (cid:3) Pisot family substitutions
Proof. (of Theorem 8) Suppose that Φ is unimodular of ( m, d )-Pisot family type. Let X be the collared A-P complex for Φ so that lim ←− f is identified with Ω via ˆ p . We provethat, for T, T ′ ∈ Ω, g ( T ) = g ( T ′ ) if and only if G ( T ) = G ( T ′ ).Suppose first that G ( T ) = G ( T ′ ). Let ˜ X = ˜ X K Λ and ˜ f = ˜ f K Λ : ˜ X → ˜ X . We willargue that g : Ω = lim ←− f → T md lifts to ˜ g : ˜Ω K Λ = lim ←− ˜ f → R md . We may supposethat the origin is in the interior of each prototile. For each k ∈ N , let X k denote theA-P complex made of k -th order supertiles. That is, the n -faces of X k are the patchesΦ k ( ρ i ), with face Φ k ( ρ i ) glued to face Φ k ( ρ j ) along ( n − k e in X k if and only ifthe prototile ρ i is glued to the prototile ρ j along the ( n − e in X . As each tilingin Ω is uniquely tiled by k -th order super tiles, there is a natural map p k : Ω → X k ( p k ( T ) = [ v ] in the face Φ k ( ρ j ) of X k if the k -th order super tile of T containing theorigin is Φ k ( ρ j ) − v ). The decomposition of k -th order supertiles into ( k − f k : X k → X k − so that Ω ≃ lim ←− f k . Furthermore, there aresubstitution induced maps f k : X k → X k with f k ◦ f k = f k − ◦ f k so that Φ on Ω isconjugated with the homeomorphism induced by ( f k ) on lim ←− f k by ( p k ). In order tolift g we approximate g by maps g k ◦ p k , where the g k : X k → T md are constructed asfollows. For each k -th order super tile ρ k := Λ k ( ρ ), ρ a prototile for Φ, and ǫ k >
0, let N ( ρ k , ǫ k ) be the ǫ k -neighborhood of the boundary of ρ k . Choose (arbitrarily ) T ρ k ∈ Ωso that the supertile ρ k occurs in the decomposition of T ρ k into k -th order supertiles(we mean this supertile occurs exactly, with 0 translation). Define g k on ρ k \ N ( ρ k , ǫ k )by g k ( p k ( T ρ k − v )) := g ( T ρ k − v ) for v ∈ ρ k \ N ( ρ k , ǫ k ). Since we have collared Φ,there is δ = δ ( k, ǫ k ) so that if ρ ki and ρ kj are adjacent faces in X k , glued along an EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 21 l -face e , and e ∋ x = p k ( T ρ ki − v ) = p k ( T ρ kj − w ), with v ∈ ρ ki and w ∈ ρ kj , then d ( T ρ ki − v, T ρ kj − w ) < δ , and δ → k → ∞ . Choosing ǫ k sufficiently small, and usingthe local convexity of T md , we extend g k continuously to all of X k so that, for each i ,if v, w ∈ ρ ki are such that | v − w | < ǫ k , then d ( g k ( p k ( T ρ ki − v )) , g k ( p k ( T ρ ki − w ))) < δ .Now for T ∈ Ω, let T ρ ki and v ∈ ρ ki be such that p k ( T ) = p k ( T ρ ki − v ). Since wehave collared, and 0 ∈ int ( ρ i ), T and T ρ ki − v agree in an r k -ball about the origin,with r k → ∞ as k → ∞ . Thus d ( g ( T ) , g k ◦ p k ( T )) = d ( g ( T ) , g k ◦ p k ( T ρ ki − v )) ≤ d ( g ( T ) , g ( T ρ ki − v )) + d ( g ( T ρ ki − v, g k ◦ p k ( T ρ ki − v )) → k → ∞ . That is, g k ◦ p k → g uniformly as k → ∞ .With the goal still of lifting g , we show now that g k : X k → T md lifts to ˜ g k : ˜ X k → R md . Let E k be a dual 1-skeleton in X k . That is, we choose a vertex in the interior ofeach n -face, a vertex in the (relative) interior of each ( n − X k , and we makea straight line edge from the vertex interior to every n -face to each of the vertices onthe ( n −
1) -subfaces. Let ˜ γ : [0 , → ˜ X k be a loop in ˜ X k . Then the loop π k ◦ ˜ γ in X k ishomotopic to a piecewise affine loop γ lying on E k . We can write γ as a concatenation γ = c ∗ c ∗ · · · ∗ c l with each c i of the form c i ( t ) = p k ( T ρ kj − ( v i + tw i )), t i ≤ t ≤ t i +1 ,for some j = j ( i ) , v i , w i . Let η := g k ◦ γ : [0 , → T md and let ˜ η : [0 , → R md be a liftof η . Let ι : R n → R md be the linear embedding so that g ( T − v ) = g ( T ) − ι ( v ). Wehave˜ η (1) − ˜ η (0) = l − X i =0 (˜ η ( t i +1 ) − ˜ η ( t i )) ≈ l − X i =0 ( ^ ι ( v j ( i ) ) + t i +1 ι ( w i )) − ( ^ ι ( v j ( i ) ) + t i ι ( w i ))= l − X i =0 ( t i +1 − t i ) ι ( w i ) , with the approximation improving as k gets larger. The covering space ˜ X k is thequotient of the abelian cover ( ˜ X k ) ab by the action of those deck transformations lyingin the kernel K k of the homomorphism l k : H ( X k ; Z ) → R n . That ˜ γ is a loop in ˜ X k means that the homology class [ γ ] lies in K k . That is, l k ([ γ ]) = P l − i =0 ( t i +1 − t i ) w i = 0.Thus P l − i =0 ( t i +1 − t i ) ι ( w i ) = 0 also. As ˜ η (1) − ˜ η (0) ∈ Z md , it must be that, for sufficientlylarge k , ˜ η (1) − ˜ η (0) = 0. The lifting criterion is satisfied: g k ◦ ˜ π ◦ ˜ γ is homotopic to aloop pushed down from R md to T md .Let us record a property of ˜ g k for later use. From the displayed approximation of˜ η (1) − ˜ η (0) above, we see that, for sufficiently large k and any loop γ in X k , and η := g k ◦ γ , ˜ η (1) − ˜ η (0) = ι ◦ l k ([ γ ]). It follows that for ˜ x ∈ ˜ X k , [ h ] ∈ H ( X k ; Z ) /K k , and k sufficiently large:(1) ˜ g k (˜ x + [ h ]) = ˜ g k (˜ x ) + ι ◦ l k ( h ) . In the constructions above we have identified Ω with lim ←− f (by means of ˆ p : T ( p (Φ − i ( T )))) and lim ←− f with lim ←− ( f k ) by means of a rescaling: since ρ k +1 i = Λ ρ ki and X = X there are natural maps Λ k : X → X k with Λ k ◦ f = f k ◦ Λ k +1 . Thenlim ←− f is identified with lim ←− ( f k ) by ( x i ) (Λ i ( x i )). The spaces lim ←− ˜ f and lim ←− ( ˜ f k ) arethen identified by first choosing a lift of f to ˜ f , then choosing lifts ( ˜Λ k ) , ( ˜ f k ) so that˜Λ k ◦ ˜ f = ˜ f k ◦ ] Λ k +1 . Now choose lifts ˜ g k : ˜ X k → R md so that | ˜ g k ◦ ˜ f k +1 (˜ x k ) − ˜ g k +1 (˜ x k ) | → x k ) ∈ lim ←− ( ˜ f k ) and let ˜ g : lim ←− ˜ f → R md be defined by˜ g ((˜ x i )) := lim k →∞ ˜ g k ◦ ˜Λ k (˜ x k ). Convergence of this limit to a lift of g is assured by theconvergence of g k ◦ p k to g .It is proved in [BK] that there is a hyperbolic and unimodular md × md integer matrix B so that g ◦ Φ = F B ◦ g , F B ( x + Z md ) := Bx + Z md being the corresponding hyperbolicautomorphism of T md . It follows that ˜ g ◦ ˆ˜ f = B ˜ g . We have assumed that G ( T ) = G ( T ′ )with the objective of showing that then g ( T ) = g ( T ′ ). Let ( x i ) , ( x ′ i ) ∈ lim ←− f be suchthat ˆ p ( T ) = ( x i ) , ˆ p ( T ′ ) = ( x ′ i ). By Proposition 19, ( x i ) ∼ gsK Λ ( x ′ i ), so there are(˜ x i ) , (˜ x ′ i ) ∈ lim ←− ˜ f , living over ( x i ) , ( x ′ i ), so that ¯ d ( ˆ˜ f k ((˜ x i )) , ˆ˜ f k ((˜ x ′ i ))), k ∈ Z , is bounded.Then | ˜ g (( ˆ˜ f k ((˜ x i ))) − ˜ g ( ˆ˜ f k ((˜ x ′ i ))) | = | B k ˜ g ((˜ x i )) − B k ˜ g ((˜ x ′ i )) | , k ∈ Z , is also bounded,and hyperbolicity of B implies that ˜ g ((˜ x i )) = ˜ g ((˜ x ′ i )). Thus g ( T ) = g ( T ′ ).Now suppose that g ( T ) = g ( T ′ ). We wish to show that T ∼ gsK Λ T ′ . From Theorem3.1 of [LS1] we have (for Φ of ( m, d )-Pisot family type) that there are v , . . . , v m ∈ R n with GR (Φ) ⊂ Z [Λ] v + · · · + Z [Λ] v m . Thus D ( GR ) ≤ D (Λ), and hence D ( GR ) = D (Λ)since the opposite inequality is always satisfied (Lemma 16). Let K be the kernel of l : H ( X ; Z ) → GR (Φ), let π K : ˜ X K → X be the corresponding cover of the collaredA-P complex X , and let ˜ f K : ˜ X K → ˜ X K be a lift of f . Let π : ˜Ω → Ω, with˜Ω := { ( T, ˜ x ) : p ( T ) = π K (˜ x ) } the cover isomorphic with ˆ π K : lim ←− ˆ˜ f K → lim ←− f as inLemma 14. Let ¯ d be the metric in ˜Ω K given by ¯ d (( T, ˜ x ) , ( T ′ , ˜ y )) = d ( T, T ′ ) + ˜ d (˜ x, ˜ y ).We will show that there are ˜ T , ˜ T ′ ∈ ˜Ω K lying over T, T ′ so that ¯ d ( ˜Φ k ( ˜ T ) , ˜Φ k ( ˜ T ′ )), k ∈ Z , is bounded.It is proved in [BK] that g ( T ) = g ( T ′ ) if and only if T and T ′ are strongly regionallyproximal , i.e., for each k ∈ N there are S k , S ′ k ∈ Ω and v k ∈ R n so that B k [ T ] = B k [ S k ], B k [ T ′ ] = B k [ S ′ k ], and B k [ S k − v k ] = B k [ S ′ k − v k ]. Pick ˜ T = ( T, ˜ x ) ∈ ˜Ω K . Let ˜ S k :=( S k , ˜ x ) ∈ ˜Ω K . There is then ˜ S ′ k = ( S ′ k , ˜ y k ) ∈ ˜Ω K such that π ( ˜ S k − v k ) = π ( ˜ S ′ k − v k ).Let ˜ T ′ k := ( T ′ , ˜ y k ) ∈ ˜Ω K . EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 23
Claim 28.
There are only finitely many distinct ˜ y k , k ∈ N . The proof of Claim 28 will follow from the
Claim 29.
There is R so that if ˜ T , ˜ S ∈ ˜Ω K and ¯ d ( ˜ T , ˜ S ) < diam (Ω) , then ¯ d ( ˜ T − v, ˜ S − v ) < R for every v ∈ R n . To prove Claim 29, let ˜ g : ˜Ω K → R md be a lift of g as constructed above. Under theidentification of ˜Ω K with lim ←− ( ˜ f k ) used above in the construction of ˜ g , the point ˜ T =( T, ˜ x ) ∈ ˜Ω K corresponds to (˜ x , ˜ x , . . . ) ∈ lim ←− ( ˜ f k ) with: ˜ x = ˜ x ; ˜ x such that ˜ f (˜ x ) =˜ x and π (˜ x ) = p (Φ − ( T )); . . . (here π : ˜ X → X is the covering projection). Then,for [ h ] in the group H ( X ; Z ) /K of deck translations of ˜Ω K , ˜ T + [ h ] (by this we meanthe deck translation [ h ] applied to ˜ T ) corresponds to (˜ x + [ h ] , . . . , ˜ x k + [(( f ) ∗ ◦ · · · ◦ ( f k ) ∗ ) − ( h )] , . . . ) (we have used unimodularity to invert ( f i ) ∗ on H ( X ; Z ) /K ) . Thus,using equation 1 for k large, we have˜ g ( T + [ h ]) ≈ ˜ g k (˜ x k + [(( f ) ∗ ◦ · · · ◦ ( f k ) ∗ ) − ( h )])= ˜ g k (˜ x k ) + ι ◦ l k ((( f ) ∗ ◦ · · · ◦ ( f k ) ∗ ) − ( h )) . Using the easily checked fact that l k = l k − ◦ ( f k ) ∗ , we have l k ((( f ) ∗ ◦· · ·◦ ( f k ) ∗ ) − ( h )) = l ( h ). Thus ˜ g ( ˜ T + [ h ]) ≈ ˜ g ( ˜ T ) + ι ◦ l ( h ), with an improving approximation as k → ∞ .That is, ˜ g ( ˜ T + [ h ]) = ˜ g ( ˜ T ) + ι ◦ l ( h ) . Now [ h ] → l ( h ) is an isomorphism of H ( X ; Z ) /K with GR (Φ) ⊂ R n and ι : R n → R md is a linear embedding, so ι ◦ l : H ( X ; Z ) /K → Z md is injective. It follows fromthis and equivariance of ¯ d that, given C , there is an R so that if ¯ d ( ˜ T , ˜ S ) > R then | ˜ g ( ˜ T ) − ˜ g ( ˜ S ) | > C .From g ( T − v ) = g ( T ) − ι ( v ) we deduce ˜ g ( ˜ T − v ) = ˜ g ( ˜ T ) − ι ( v ). Since ˜ g is a lift and ¯ d isequivariant, there is C so that ¯ d ( ˜ T , ˜ S ) < diam (Ω) implies | ˜ g ( ˜ T ) − ˜ g ( ˜ S ) | < C . We thenhave ¯ d ( ˜ T , ˜ S ) < diam (Ω) = ⇒ | ˜ g ( ˜ T ) − ˜ g ( ˜ S ) | < C = ⇒ | (˜ g ( ˜ T ) − ι ( v )) − (˜ g ( ˜ S ) − ι ( v )) | 0. We have π ( ˜Φ j ( ˜ T )) = π ((Φ j ( T ) , ˜ f jK (˜ x ))) = ˜ f jK (˜ x ) = π ( ˜Φ j ( ˜ S ))and similarly, π ( ˜Φ j ( ˜ T ′ )) = π ( ˜Φ j ( ˜ S ′ )). Also, π ( ˜ S − v ) = π ( ˜ S ′ − v ) = ⇒ π ( ˜Φ j ( ˜ S − v )) = π ( ˜Φ j ( ˜ S ′ − v )). That is, π ( ˜Φ j ( ˜ S ) − Λ j v ) = π ( ˜Φ j ( ˜ S ′ ) − Λ j v ), and by Claim29 we have ¯ d ( ˜Φ j ( ˜ S ) , ˜Φ j ( ˜ S ′ )) < R . Thus ˜ d ( ˜ f jK (˜ x ) , ˜ f jK (˜ x ′ )) < R , so ¯ d ( ˜Φ j ( ˜ T ) , ˜Φ j ( ˜ T ′ )) Given R there is R so that if ˜ T , ˜ S ∈ ˜Ω K satisfy π ( ˜Φ( ˜ T ) − v ) = π ( ˜Φ( ˜ S ) − v ) for all v ∈ B R (0) , then π ( ˜ T − v ) = π ( ˜ S − v ) for all v ∈ B R (0) . Proof of Claim 30: Let R be given. It is a consequence of invertibility of Φ that thereis an R so that if T, S ∈ Ω satisfy B R [Φ( T )] = B R [Φ( S )], then B R [ T ] = B R [ S ].(This is “recognizability”.) Now let ˜ T = ( T, ˜ x ) and ˜ S = ( S, ˜ y ). Then ˜ T − v =( T − v, ˜ p T ˜ x ( v )) and ˜ S − v = ( S − v, ˜ p S ˜ y ( v )). Now suppose that π ( ˜Φ( ˜ T ) − v ) = π ( ˜Φ( ˜ S ) − v ) for all | v | < R . This means that ˜ f K (˜ p T ˜ x (Λ − v )) = ˜ f K (˜ p S ˜ y (Λ − v )), which is tosay ˜ p Φ( T )˜ f K (˜ x ) ( v ) = ˜ p Φ( S )˜ f K (˜ y ) ( v ), for all such v . From this we have ˜ f K (˜ x ) = ˜ f K (˜ y ) and p (Φ( T ) − v ) = p (Φ( S ) − v ) for | v | < R . So B R [Φ( T )] = B R [Φ( S )] so that B R [ T ] = B R [ S ]. In particular, p ( T ) = p ( S ) and ˜ y = ˜ x + [ h ] for some [ h ] ∈ H ( X ; Z ) /K . Then˜ f K (˜ x ) = ˜ f K (˜ y ) = ˜ f K (˜ x + [ h ]) = ˜ f K (˜ x ) + [ f ∗ ( h )] = ⇒ [ f ∗ ( h )] = 0. Since f ∗ is invertibleon H ( X ; Z ) /K (unimodularity), [ h ] = 0 and ˜ x = ˜ y . Together with B R [ T ] = B R [ S ],this last implies that ˜ p T ˜ x ( v ) = ˜ p S ˜ y ( v ), that is, π ( ˜ T − v ) = π ( ˜ S − v ), for all v ∈ B R (0).To finish the proof of the theorem, let R = R ( R ) be as in Claim 30 and for m ∈ N let R m := R ( R ( · · · ( R (1)) · · · )), iterated m times. Pick j < k > R | j | .Then π ( ˜ T − v ) = π ( ˜ S k ) − v for | v | < k implies that π ( ˜Φ − ( ˜ T ) − v ) = π ( ˜Φ − ( ˜ S k ) − v )for | v | < R | j |− , which implies that ..., which gives π ( ˜Φ j ( ˜ T ) − v ) = π ( ˜Φ j ( ˜ S k ) − v )for | v | < 1. Similarly, for this k , we have π (Φ j ( ˜ S k − v k ) − v ) = π (Φ j ( ˜ S ′ k − v k ) − v )and π (Φ j ( ˜ S ′ k − v ) = π (Φ j ( ˜ T ′ ) − v ) for | v | < 1. Using Claim 29 (as in the j > d ( ˜Φ j ( ˜ T ) , ˜Φ j ( ˜ T ′ )) < diam (Ω) + R . Thus G ( T ) = G ( T ′ ) . (cid:3) Connections with the traditional Pisot Substitution Conjecture The traditional Pisot Substitution Conjecture is as follows: Conjecture 31. If Φ is a one-dimensional substitution with unimodular and irreducibleincidence matrix M (that is, the characteristic polynomial of M is irreducible over Q ) and with inflation λ a Pisot number, then the R -action on Ω Φ has pure discretespectrum. EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 25 We will see below that this is a special case of a slight strengthening of Conjecture5. Given a substitution Ψ with A-P complex X (not necessarily collared), substitutioninduced map f : X → X , and natural semiconjugacy p : Ω Ψ → lim ←− f , as definedpreviously, let H ∗ f (Ω Ψ ) := p ∗ ( H ∗ (lim ←− f ; Z )) . Let us say that a substitution Ψ is compatible with a substitution Φ if there is ahomeomorphism h Ψ : Ω Φ → Ω Ψ that conjugates some positive powers of the substi-tution homeomorphisms Φ and Ψ. We define the essential cohomology of Ω Φ tobe H ∗ ess (Ω Φ ; Z ) := ∩ Ψ h ∗ Ψ ( H ∗ f (Ω Ψ )) , the intersection being over all Ψ compatible with Φ. Conjecture 32. If Φ is unimodular and hyperbolic, and H hyp (Ω Φ ; Z ) = H ess (Ω Φ ; Z ) ,then G ′ : Ω Φ → T D ′ is a.e. 1-to-1. Conjecture 33. If Φ is a unimodular Pisot family substitution and rank ( H ess (Ω Φ ; Z )) = D (Λ) then the R n -action on Ω Φ has pure discrete spectrum. Example 34. Consider the one-dimensional substitution Φ generated by the substitu-tion on letters: ; . The incidence matrix is unimodular andirreducible and the inflation λ is the fourth power of the golden mean, a Pisot number.Thus Φ satisfies the hypotheses of the traditional Pisot Substitution Conjecture (and,in fact, the spectrum is pure discrete as the traditional Pisot conjecture is correct if theincidence matrix is × - [HS] ). But D (Λ) = deg ( λ ) = 2 while rank ( H (Ω Φ ; Z )) = 3 ,so the hypotheses of Conjecture 5 are not met. (The cohomology is easily computedby the techniques of [BD1] .) On the other hand, the A-P complex for Φ is a wedge oftwo circles; it follows that H ess (Ω Φ ; Z ) = H f (Ω Φ ) has rank two, and the hypotheses ofConjecture 33 are satisfied. Proposition 35. Suppose Φ is a one-dimensional substitution with unimodular andirreducible d × d incidence matrix M and whose inflation λ is a Pisot number. Then rank ( H ess (Ω Φ ; Z )) = d . Hence any substitution satisfying the hypotheses of the traditional Pisot SubstitutionConjecture also satisfies the hypotheses of Conjecture 33. Lemma 36. Suppose that Φ is an n -dimensional substitution. Let f : X → X bethe substitution induced map on the A-P complex for Φ and let p : Ω Φ → X be theusual map, inducing ˆ p : Ω Φ → lim ←− f . Then the map g from Ω Φ onto the maximalequicontinuous factor factors through lim ←− f via ˆ p . Proof. Let f c : X c → X c be the substitution induced map of the collared A-P complexfor Φ, let π : X c → X be the map that forgets collars, and let p c : Ω Φ → X c be as usual.It suffices to show that if ( x i ) , ( x ′ i ) ∈ lim ←− f c are such that ˆ π (( x i )) = ˆ π (( x ′ i )) then thetilings T = ˆ p c − (( x i )) and T ′ = ˆ p c − (( x ′ i )) are regionally proximal. Suppose then thatˆ π (( x i )) = ˆ π (( x ′ i )). If x i , and hence x ′ i are in the interior of an n -cell for infinitely many(hence all) i , let i j → ∞ , ρ , and v j be such that B [Φ − i j ( T )] = ρ − v j = B [Φ − i j ( T ′ )].Then the patches Φ i j ( ρ ) − Λ i j are subsets of both T and T ′ . That is, T and T ′ sharepatches of arbitrarily large internal diameter, and hence T and T ′ are proximal.If, on the other hand, x i and x ′ i are in the n − X c for all i , π ( x i ) = π ( x ′ i ) =: y i is in the n − X for all i . From the definition of the equivalencerelation that specifies the way in which the prototiles are glued along their boundariesto form X , there are, for each i ∈ N , T i , . . . , T ik i ∈ Ω Φ and tiles τ ij = τ ij +1 ∈ T ij so that0 ∈ τ ij ∩ τ ij +1 for all i, j , T i = Φ − i ( T ), and T ik i = Φ − i ( T ′ ). The k i are bounded and wemay select i l → ∞ so that k = k i l is constant and the collection { τ i l j : j = 1 , . . . , k } is constant up to translation. Using compactness of Ω Φ we may pass to a subsequence I l s → ∞ so that Φ i ls ( T i ls j ) converges, say to T j , for j = 1 , . . . , k . As above, T j and T j +1 share patches of arbitrarily large internal diameter, and are hence proximal, foreach j . As regional proximality is an equivalence relation and contains the proximalityrelation, T = T and T ′ = T k are regionally proximal. (cid:3) Suppose that Φ is a one-dimensional (primitive) substitution whose inflation is adegree d Pisot unit. Let X c be the collared A-P complex for Φ, with vertex set S c and substitution induced map f : ( X c , S c ) → ( X c , S c ). In an appropriate basis, f ∗ : H ( X c , S c ; Z ) → H ( X c , S c ; Z ) is represented by the transpose, M t , of the incidencematrix for Φ. There is then a unique f ∗ -invariant subgroup P of H ( X c , S c ; Z ) of rank d with the properties that f ∗ | P is represented by an integer matrix with eigenvalue λ , and P is maximal in the sense that H ( X c , S c ; Z ) / P is torsion free. By taking thedirect limit by f ∗ of the short exact sequence for the pair0 → ˜ H ( S c , Z ) → H ( X c , S c ; Z ) → H ( X c ; Z ) → f ∗ : ˜ H ( S c ; Z ) → ˜ H ( S c ; Z )are 0 or roots of unity, we see that H (Ω Φ ; Z ) = lim −→ f ∗ : H ( X c ; Z ) → H ( X c ; Z ) alsocontains a unique subgroup, invariant under Φ ∗ , with the properties of P . We will callthis subgroup the Pisot subgroup of H (Ω Φ ; Z ). Proof. (of Proposition 35) Let Φ be as in the proposition, let f : X → X be thesubstitution induced map on its A-P complex, and let S be the vertex set of X . Thehomomorphism f ∗ : H ( X, S ; Z ) → H ( X, S ; Z ) is represented by the transpose M t of the incidence matrix (in the basis dual to the prototile edges of X ). As M is EOMETRIC REALIZATION FOR SUBSTITUTION TILINGS 27 invertible over Z , the direct limit of this homomrphism is Z d . Moreover, the subgroup δ ∗ ( H ( S ; Z )) ⊂ H ( X, S ; Z ) is f ∗ -invariant, so, by irreducibility of M , this subgroupmust be trivial. From the exact sequence for the pair we see that f ∗ : H ( X, S ; Z ) → H ( X ; Z ) is conjugate with f ∗ : H ( X ; Z ) → H ( X ; Z ), and hence H (lim ←− f ; Z ) is also Z d .Let X c denote the collared A-P complex for Φ with vertex set S c and map f c :( X c , S c ) → ( X c , S c ) and let p c : Ω Φ → X c and p : Ω Φ → X be the standard maps ontothe A-P complexes. Let π : ( X c , S c ) → ( X, S ) be the map that forgets collars. Then π ∗ : H ( X, S ; Z ) → H ( X c , S c ; Z ) is injective and it follows that the image of H ( X ; Z )in H ( X c ; Z ) under π ∗ is an f ∗ c -invariant copy of Z d on which f ∗ c acts like M t . Thus H f (Ω Φ ) has rank d and rank ( H ess (Ω Φ ; Z )) is at most d .Suppose that Ψ is a 1-d substitution with inflation η such that Ψ n is conjugate withΦ m , by means of h Ψ , for some n, m ∈ N . It is easily checked that the topologicalentropies of the homeomorphisms Ψ and Φ are log( η ) and log( λ ), hence η m = λ n ,as conjugacies preserve entropy. Thus η m is a Pisot unit, also of degree d (as M n is irreducible). Now, by Lemma 36, if f : X → X is the substitution inducedmap on the A-P complex for Ψ, and p : Ω Ψ → X is the usual map, the map g : Ω Ψ → T d onto the maximal equicontinuous factor factors through lim ←− f via ˆ p .Thus H f (Ω Ψ ) = ˆ p ∗ ( H (lim ←− f ; Z )) ⊃ g ∗ ( H ( T d ; Z )). 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