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THE PISOT CONJECTURE FOR β-SUBSTITUTIONS

MARCY BARGE

ABSTRACT. We prove the Pisot Conjecture forβ-substitutions: Ifβ is a Pisot number then thetiling dynamical system(Ωψβ ,R) associated with theβ-substitution has pure discrete spectrum. Ascorollaries: (1) Aritmetical coding of the hyperbolic solenoidal automorphism associated with thecompanion matrix of the minimal polynomial of any Pisot number is a.e. one-to-one; and (2) AllPisot numbers are weakly finitary.

1. INTRODUCTION.

A substitutionis a map that takes letters of some finite alphabet to finite words in that alphabet.The space of all bi-infinite words that can arise from infinitely repeating a particular substitution,equiped with the product topology and the shift operation, is called asubstitutive system. ThegroupR acts on the suspension (there is a natural choice of roof function) of a substitutive system,resulting in thetiling dynamical systemassociated with the substitution. One would like to knowhow close the tiling dynamical system is to being simply an action ofR by translation on a compactabelian group. ThePisot Substitution Conjecturesassert that, under various hypotheses, the tilingdynamical system should be an almost everywhere one-to-oneextension of an action ofR bytranslation on a finite dimensional torus or solenoid.

As long as the substitution is primitive (under some power ofthe substitution, every letter ap-pears in the image of a single letter), long words are, on average, stretched under substitution bya constant multiplier called theinflation of the substitution. The one key hypothesis for the PisotSubstitution Conjectures is that the inflation be a Pisot-Vijayaraghavan (or, simply, Pisot) number- a real algebraic integer greater than 1, all of whose (other) algebraic conjugates lie strictly insidethe unit circle. For the tiling dynamical system to have a nontrivial group rotation as a factor itis necessary that the inflation be Pisot ([S2]). The tiling dynamical system of any primitive sub-stitution with Pisot inflation (let’s call such a substitution aPisot substitution) is always a finiteextension of a translation action on a torus or solenoid ([BKw, BBK]) but maybe not an almosteverywhere one-to-one extension of such an action. For example, the tiling dynamical system as-sociated with the Thue-Morse substitution,a → ab, b → ba, with inflationβ = 2, is, at best, ana.e. two-to-one extension of translation on the 2-adic solenoid.

For tiling dynamical systems, equivalent to being an almosteverywhere one-to-one extension ofan action by rotation on a torus or solenoid is the property ofhavingpure discrete spectrum(moreon this below). There are two main versions of the Pisot Substitution Conjecture:

(1) If β is Pisot, the tiling dynamical system associated with theβ-substitution has pure discretespectrum.

Date: May15, 2015.2000Mathematics Subject Classification.Primary 37B50, 37B10, 11K16, 37P99, 37D40.Key words and phrases.Pisot substitution, tiling space,β-numeration, hyperbolic automorphism.

1

http://arxiv.org/abs/1505.04408v2

2 M. BARGE

(2) The tiling dynamical system of anirreducible1 Pisot substitution has pure discrete spectrum.

Given a real numberβ > 1, the β-transformation, Tβ : [0, 1] → [0, 1], is given byTβ(x) =βx−⌊βx⌋, where⌊ ⌋ denotes the greatest integer function. Ifβ is Pisot, the orbit of 1 underTβ isfinite and so breaks[0, 1] into finitely many subintervals. These subintervals map over each otherunderTβ in a pattern described by theβ-substitution, ψβ (see Section 2); it is to this substitutionthat version (1) above refers.

The two Pisot Substitution Conjectures are independent: most irreducible Pisot substitutionsdon’t come from theβ-transformation, and mostβ-substitutions are not irreducible. We proveversion (1) of the Pisot Substitution Conjecture in the present article.

In [Ra] G. Rauzy constructed 2-dimensional fractal tiles, calledRauzy tiles, that tile the planeboth periodically and non-periodically. His constructionis based on the ‘tribonacci substitution’ -theβ-substitution associated with the largest root ofx3−x2−x−1. Using arithmetical propertiesof β-numeration, Thurston ([T]) described a general method forobtaining a self-similar ‘shingling’of the plane based on any degree three Pisot unitβ. His shingling is a multi-tiling, ‘Galois dual’to the tilings ofR provided directly by theβ-substitution, in which almost every point in theplane lies in exactlyr tiles for somer ≥ 1. The degree,r, of the shingling is 1 if and only ifthe shingling is a tiling, and, it turns out, if and only if thetiling dynamical system associatedwith the β-substitution has pure discrete spectrum ([A2]). Thurstonmentions that when theseshinglings are, in fact, tilings, they are associated with Markov partitions for hyperbolic toralautomorphisms. This procedure for constructing Markov partitions was subsequently developedby Bertrand-Mathis ([B-M]) and Pragastis ([Pr]) and underlies the ‘arithmetical coding’ of Kenyon,Vershik, Sidorov, Schmidt, and others ([KV, Si1, Si2, Sch2,Ver1, Ver2, Bor] and see Section 3below).

Solomyak showed in [S1] that a purely arithmetical condition - the so-called Property (F), thatevery positve element of the ringZ[1/β] have finiteβ-expansion - is enough to guarantee pure dis-crete spectrum of the tiling dynamical system associated with theβ-substitution and, with Frougny([FS]), described a collection of polynomials with dominant root a Pisot number satifying Property(F). Akiyama and Sadahiro ([AS]) and Akiyama ([A1]) showed that Thurston’s shingling producesa tiling when the Pisot unitβ satisfies Property (F). In [H] Hollander introduced the weaker Prop-erty (W) (see Section 3 below), which Akiyama ([A2]) subsequently proved to be equivalent topure discrete spectrum for the 1-d tiling system (and to the Galois dual shingling being a tiling). In[Si1], Sidorov proves that, for Pisot units, Property (W) isequivalent to arithmetical coding (withfundamental homoclinic point) being a.e. 1-1.

Interest in nonperiodic tilings and spectral properties ofsubstitutions was stimulated by Schect-man’s discovery, in the early 80’s, of quasicrystalline materials. Self-similar point sets that havegood diffractive properties, and so provide mathematical models of quasicrystals, can be createdwith substitutions of Pisot type: if one places an ‘atom’ at acharacteristic position in each tileof a Pisot substitution tiling (of whatever dimension) the diffraction of the resulting arrangementwill have Bragg peaks. In fact, the diffraction spectrum of the arrangement will be pure point (asfor a perfect quasicrystal) if and only if the tiling dynamical system has pure discrete spectrum,meaning that the eigenfunctions of the dynamical system span the spaceL2 of square-integrable

1 A substitution is irreducible if the characteristic polynomial of its abelianization (or substitution matrix - seeSection 5) is irreducible over the rationals.

THE PISOT CONJECTURE 3

functions on the tiling space ([D],[LMS], and see [Le] for a survey). For substitution tiling dynam-ical systems, it is a consequence of the Halmos- von Neumann theory (and Solomyak’s result thateigenfunctions are continuous ([S4])) that pure discrete spectrum of the tiling dynamical system isequivalent to the system being an almost everywhere one-to-one extension of a translation actionon a torus or solenoid (see [BK] for more detail and related characterizations).

For more on Pisot substitutions and the Pisot Substitution Conjectures, see the surveys [BS] and[ABBLS]. A general introduction to the dynamical and topological properties of substitution tilingspaces can be found in [S2], [Ro], and [AP]. The direct antecedent to the current article is [B1] inwhich the author dealt withβ-substitutions forβ a Pisot simple Parry number. As outlined below,we extend the approach taken there to all Pisot numbers. In the summary that follows,θ is anarbitrary (primitive, nonperiodic) substitution with Pisot inflationβ, ψβ is theβ-substitution, andΘ denotes the substitution-induced homeomorphism of the substitution tiling spaceΩθ. Here arethe key steps in our approach:

(1) There is an almost everywherecr-to-1 mapπmax : Ωθ → T̂d that factors theR-action onΩθ onto a Kronecker action on ad-dimensional solenoid̂Td. Hered is the algebraic degreeof β andcr = cr(θ) 1, then(Ωθ/ ≈s) ≃ Ωθ̃ is a substitution tiling space for a substitutionθ̃, cr(θ̃) = cr(θ), and≈s is trivial onΩθ̃.

(7) There is some leeway in choosing a substitution to generate an isomorphism class of tilingspaces. Ifcr(ψβ) > 1, the substitutionψ̃β may be chosen to have the special property:there is ann ∈ N and a letterb in the alphabetAψ̃β for ψ̃β so thatψ̃nβ (a) = b · · · a for alla ∈ Aψ̃β .

(8) If θ is any substitution for which there aren ∈ N andb ∈ Aθ with θn(a) = b · · · a for alla ∈ Aθ, then≈s is nontrivial onΩθ.

Pure discrete spectrum for(Ωψβ ,R) then follows: If the spectrum were not pure discrete, thesubstitutionψ̃β would be such that≈s is trivial onΩψ̃β (by (6)). But this is contradicted by (7) and(8). Versions of items (1)-(6) hold quite generally for substitutions of Pisot type in any dimension.For (1)-(3) see [BKw, BBK, BK, B2], (4) comes from [BO], a version of (5) first appears in [BBK]and in the form stated here in [B1] with a proof derived from [BSW], and (6) is from [B1]. Item (8)is from [B1]. Section 4 is devoted to proving item (7) for the particular case ofβ-substitutions. The

2The coincidence rank is equal to the degree of the ‘Galois dual’ shingling, or multi-tiling, of Rd−1 mentionedpreviously.

4 M. BARGE

argument is somewhat technical and is based on the topological structure ofΩψβ , as we describenow.

There are tilings inΩψβ , corresponding to the fixed point 0 ofTβ and to the point of the periodicorbit, of periodp, ofTβ on which theTβ-orbit of 1 eventually lands. These tilings are periodic underΨβ of periodp and pairwise strong regionally proximal. Using the basic technique of [BD2], andthe special ‘monotonic’ nature ofψβ, we show that these tilings are entangled: forT, T ′ any two ofthemT − t ∼s T ′− t for a set oft dense inR+. But also, for suchT 6= T ′, we show thatT andT ′are not asymptotic onR+ (there are arbitrarily larget so that the tiles ofT andT ′ at t are distinct).From this it follows (again using the monotonicity ofψβ) that there isǫ > 0 so that ifτ andτ ′ areany two tiles having the same initial vertexx, thenτ − t ∼s τ ′− t for a set oft dense in[x, x+ ǫ),and there are tilesτ 1−, . . . , τ

p− so that ifτ is any tile, there is exactly one of theτ

i− so that, witht0

chosen such that the terminal vertices ofτ − t0 andτ i− are both equal toy, τ − t0− t ∼s τ i−− t fora set oft dense in(y − ǫ, y]. Item (7) then follows.

As mentioned earlier, Akiyama (in [A2]) established the equivalence of the arithmetical Prop-erty (W) with pure discrete spectrum of(Ωψβ ,R), and Sidorov (in [Si1]) proved the equivalenceof Property (W) with a.e.1-1-ness of arithmetical coding (at least for Pisot units). In the final Sec-tion 5 we reprove these results (and extend the Sidorov result to non-units) from a unified point ofview. We think it is of interest to note that arithmetical coding occurs in the context of hyperbolicdynamics, while the map onto the maximal equicontinuous factor is determined by properties ofthe translation action. We will tie these two viewpoints together by showing the structure rela-tion for the maximal equicontinuous factor map (strong regional proximality) is the same as the‘global shadowing’ relation defined in terms of the the hyperbolic homeomorphismΨβ (such anequivalence holds also in higher dimensions - see [BG]).

Finally, Property (W) will be interpreted as a statement about homoclinic return times. Therelation between the structure of these return times and pure discreteness of the translation actionis a general phenomenon (see, for example, Corollary 4.5 of [B2]).

2. BACKGROUND ONβ-NUMERATION AND SUBSTITUTION TILINGS.

Forβ > 1, theβ-transformation,Tβ : [0, 1]→ [0, 1] is given byTβ(x) := βx− ⌊βx⌋,

where⌊ ⌋ denotes the greatest integer function. Ifβ is a Pisot number, then theTβ-orbit of 1,OTβ(1) := {T nβ (1) : n ≥ 0}, is finite ([B-M],[Sch1]). Such aβ is a simple Parry numberif0 ∈ OTβ(1) and otherwise is anon-simple Parry number. Everyx ∈ [0, 1) has anitenerary

κ(x) = x1x2 · · · ,with xn := i ∈ {0, . . . , ⌊β⌋} providedT n−1β (x) ∈ [i/β, (i+ 1)/β). For suchx

x =

∞∑

n=1

xnβ−n

is the (greedy) β-expansion ofx. Thekneading invariantor characteristic sequenceof β is givenby

κ(1) = c1c2 · · · := limxր1

κ(x),

the limit being taken in the product topology on the sequences. If β is a simple Parry number, thenκ(1) is periodic of some (least) periodp: κ(1) = c1 · · · cp. If β is a non-simple Parry number, then

THE PISOT CONJECTURE 5

κ(1) is strictly preperiodic: there arem ≥ 1 andp ≥ 1 with κ(1) = c1 · · · cmcm+1 · · · cm+p, wherewe takem, and thenp, as small as possible.

Theone-sidedβ-shift is the Cantor dynamical system

(X+β , σ),

with Xβ+ := cl({(x1, x2, . . .) : x1x2 · · · = κ(x) for somex ∈ [0, 1]}) andσ((x1, x2, x3, . . .)) :=(x2, x3, . . .). One has thatXβ consists precisely of the sequences of digits from{0, . . . , ⌊β⌋} all ofwhose shifts are lexicographically no larger than the sequence(c1, c2, . . .) corresponding toκ(1)([Pa]). The (two-sided)β-shift is obtained by taking inverse limits:Xβ := {(. . . , x−1, x0, x1, . . .) :(xk, xk+1, . . .) ∈ X+β for all k ∈ Z}. For eachx ∈ R+ there is a corresponding point

x := (. . . , 0, 0, x−k, . . . , x−1x0, x1, . . .) ∈ Xβwherex =

∑∞i=−k xiβ

−i, and∑∞

i=1 x−k+iβ−i is the greedyβ-expansion ofβ−kx ∈ [0, 1). For

more detail on theβ-shift, see [R, Pa, IT, Bl, LM].For a given Pisotβ, let {0, 1} ∪ {T iβ(1) : i = 1, . . . , m + p} = {0 = s0 < s1 < · · · <

sm+p = 1}. The prototilesare the marked intervalsτi := [si−1, si] × {i}, i = 1, . . . , m + p;their verticesaremin(τi) := si−1 andmax(τi) = si. Thesupportof τi is spt(τi) := [si−1, si].We call τ = τi + t := (spt(τi) + t) × {i} a tile of typei with supportspt(τi) + t and verticesmin(τ) := si−1+ t andmax(τ) := si+ t. Theβ-substitutionon the alphabetA = {1, . . . , m+p}is the mapψβ : A → A∗, A∗ being the collection of finite nonempty words onA, given byψβ(i) = i1i2 · · · in providedTβ(x) passes through the support ofτi1 , then the support ofτi2 ,...,then the support ofτin asx increases fromsi−1 to si.

3 The correspondingtile substitutionΨβis defined on prototiles byΨβ(τi) := {τi1 −min(τi1) + βmin(τi), τi2 −min(τi2) +max(τi1) −min(τi1) + βmin(τi), . . . , τin −min(τin) + (

∑n−1j=1 ((max(τj)−min(τj))) + βmin(τi)}. (That

just says thatΨβ multiplies the support ofτi by a factor ofβ and tiles the resulting interval bytheτij , following the pattern ofψβ .) The tile substitutionΨβ is then defined on arbitrary tiles byΨβ(τi+t) := Ψβ(τi)+βt. By apatchP we will mean a finite collection of tiles with the properties:if τ 6= τ ′ ∈ P , thenint(spt(τ)) ∩ int(spt(τ ′)) = ∅; andspt(P ) := ∪τ∈P spt(τ) is connected.Ψβextends to patches byΨβ(P ) := ∪τ∈PΨβ(τ).

A tiling T of R is a collection of tiles with the properties: ifτ 6= τ ′ ∈ T thenint(spt(τ)) ∩int(spt(τ ′)) = ∅ ; and∪τ∈T spt(τ) = R. A patchP is allowedfor Ψβ if there is a tileτ = τi + tand ann ∈ N so thatP ⊂ Ψnβ(τ). Forβ /∈ N, thetiling spaceassociated withΨβ is the collectionof tilings ofR, all of whose patches are allowed forΨβ:

Ωψβ := {T : every patchP ⊂ T is allowed forΨβ}.Forr ≥ 0 andT ∈ Ωψβ , let

Br[T ] := {τ ∈ T : spt(τ) ∩ [−r, r] 6= ∅}.So, for example,B0[T ] is the collection of tiles ofT whose supports contain the origin.

The groupR acts onΩψβ by translation,T − t := {τ − t : τ ∈ T}, and there is a metricd (calledthetiling metric) onΩψβ with the property thatd(T, T

′) is small if there arer, t, t′ with r large and|t|, |t′| small, so thatBr[T−t] = Br[T ′−t′] (see, for example, [AP]). With the topology induced byd, Ωψβ is compact and the translation action onΩψβ is continuous. It is well-known that ifβ /∈ N,

3Typically, the prototiles are taken to be[0, si] × {i}, resulting in a different ‘β-substitution’. It is straightforwardto pass from one to the other by a rewriting procedure (see [D]or [BD1]) and the resulting tiling dynamical systemsare isomorphic.

6 M. BARGE

thenψβ is non-periodic(meaningT − t = T for someT ∈ Ωψβ impliest = 0) andprimitive(therearen andi so that all letters occur inψnβ (i)) and it follows that the translation dynamical system(Ωψβ ,R) is minimal and uniquely ergodic ([S2]). We will denote the unique Borel translationinvariant measure onΩψβ by µ. Moreover,Ψβ : Ωψβ → Ωψβ by Ψβ(T ) := ∪τ∈TΨβ(τ) is ahomeomorphism withµ invariant and ergodic ([S3]). Note that the translation- and Ψβ- dynamicsare intertwined by:

Ψβ(T − t) = Ψβ(T )− βt.If β = n is an integer, the substitutionψβ is the periodic substitution1 7→ 1n and, as usually

defined, the tiling space is the circleT1 and theΨn is then-fold covering mapx+Z 7→ nx+Z. Itwill be convenient (looking ahead to the discussion of arithmetical coding) to instead defineΩψnto be the solenoidlim←−Ψn, interpretΨn to be the shift homeomorphism, and take theR-action tobe the natural Kronecker action(x1 + Z, x2 + Z, . . .)− t := (x1 − t + Z, x2 − t/n+ Z, . . .).

A continuous mapf : Ωψβ → T1 := R/Z is a continuous eigenfunctionof (Ωψβ ,R) if thereis γ ∈ R (the associated eigenvalue) so thatf(T − t) = f(T ) − (γt + Z) for all t ∈ R. Thesystem(Ωψβ ,R) is said to havepure discrete spectrumif the continuous eigenfunctions spanL2(µ). An alternative formulation of this property, in terms of equicontinuous factors, is closerto the spirit of this article. A continuous system(Y,R) (that is, a continuous action ofR on themetric spaceY ) is equicontinuousprovided the collecton of homeomorphisms{y 7→ y · t}t∈Ris equicontinuous. Amaximal equicontinuous factorof a system(X,R) is an equicontinuousfactor (Xmax,R) of (X,R) with the property that every equicontinuous factor of(X,R) is alsoa factor of (Xmax,R). Compact minimal systems always have a (unique up to isomorphism)maximal equicontinuous factor ([Aus]). In the case at hand,the maximal equicontinuous fac-tor of (Ωψβ ,R) is a Kronecker action on ad-dimensional torus or solenoid̂T

dβ , with d the al-

gebraic degree ofβ ([BBK],[B2]). Here T̂dβ is the inverse limitT̂dβ := lim←−FM with M the

companion matrix of the minimal polynomial ofβ andFM : Td := Rd/Zd → Td the hy-perbolic toral endomorphismFM(v + Zd) := Mv + Zd and the Kronecker action is given by(v1 +Zd, v2 +Zd, v3 +Zd, . . .)− t := (v1− tω+Zd, v2− β−1tω, v3− β−2tω+Zd, . . .), whereωis a right positive eigenvector for the eigenvalueβ of M .

The structure relation for the maximal equicontinuous factor mapπmax : Ωψβ → T̂dβ is givenby strong regional proximality4 ([BK]): Tilings T, T ′ are strong regionally proximal, denotedT ∼srp T ′, provided for allR > 0 there areSR, S ′R ∈ Ωψβ abdtR ∈ R so that

BR[T ] = Br[SR],

BR[T′] = Br[S

′R], and

BR[SR − tR] = Br[S ′R − tR].There isr ∈ N, called thecoincidence rankof ψβ so thatπmax : Ωψβ → T̂ dβ is a.e.r-to-1 and

for eachz ∈ T̂ dβ there areS1, . . . , Sr ∈ π−1max(z) with Si ∩ Sj = ∅ for i 6= j. The system(Ωψβ ,R)has pure discrete spectrum if and only if the coincidence rank of ψβ equals 1, that is, if and only ifπmax is a.e. 1-1 (see [BKw, BBK, BK]).

Conjecture 1. (Pisot conjecture forβ-substitutions) Ifβ is a Pisot number, then(Ωψβ ,R) has purediscrete spectrum.

4In the general context of an abelian group acting minimally on a compact metric space, the structure relation forthe maximal equicontinuous factor map is regional proximality ([V]). For Pisot substitution tiling spaces, the Meyerproperty is responsible for the strong version we use here.

THE PISOT CONJECTURE 7

3. ARITHMETICAL CODING AND PROPERTY (W)

Given a Pisot numberβ of algebraic degreed andM the companion matrix of its minimalpolynomial, letEu andEs denote the 1-dimensional and(d − 1)-dimensional unstable and stablespaces ofM . (SoEu is spanned by the positive right eigenvectorω andEs = {v ∈ Rd : Mnv →0 asn → ∞}.) ThenRd = Eu ⊕ Es and fory ∈ Rd we writey = yu + ys with yu ∈ Eu andys ∈ Es. A homoclinic pointȳ = y + Zd ∈ Td, by which we meany = zu for somez ∈ Zd,is fundamentalif {z,Mz, . . . ,Md−1z} is a basis forZd. (Such a point always exists.) If̄y is afundamental homoclinic point, let̄̂y := (y + Zd,M−1y + Zd, . . .) ∈ T̂dβ be the correspondingpoint, homoclinic tô̄0 under the shift automorphism̂FM . The map

hȳ : Xβ → T̂dβgiven by

hȳ((xi)) :=∑

i∈Z

xiF̂−iM (ˆ̄y)

is called anaritmetical codingof the hyperbolic automorphism̂FM : T̂dβ → T̂dβ . Such a coding hasthe felicitous properties:

hȳ(βx) = F̂M(hȳ(x))

for all x ∈ R+, more generallyhȳ ◦ σ = F̂M ◦ hȳ,

andhȳ(x+ x

′) = hȳ(x) + hȳ(x′)

for all x, x′ ∈ R+.The pointx ∈ R+ is said to havefinite β-expansionif x = (. . . , 0, x−k, . . . , xn, 0, 0, . . .) for

somek ∈ N. LetFin(β) := {x ∈ R+ : x has finiteβ-expansion}. The Pisot numberβ is said tobefinitary if Z[1/β] ∩ R+ ⊂ Fin(β) and is said to beweakly finitaryif for all z ∈ Z[1/β] ∩ R+there arex, y ∈ Fin(β), with y as small as desired, such thatz = x − y. The so-calledProperty(W) is thatβ is weakly finitary.

Akiyama proves in [A1] thatβ satisfies Property (W) if and ony if(Ωψβ ,R) has pure discretespectrum and Sidorov ([Si1]) shows that, at least for Pisotβ that are algebraic units, Property(W) is equivalent to arithmetical coding (with fundamentalhomoclinic point) being a.e. 1-1. It isshown in [ARS] that all cubic Pisot units, and higher degree Pisot numbers satisfying a ‘dominantcondition’ are weakly finitary. We prove Conjecture 1 in the next section and provide an alternativeargument for the equivalence of Conjecture 1 with a.e. one-to-oneness of arithmetical coding forPisotβ and satisfaction of Property (W) for all Pisotβ in the final section. (In fact we will prove thestronger form of Property (W): Ifβ is Pisot andz ∈ Z[1/β]∩R+, then{y : y ∈ Fin(β) andy+z ∈Fin(β)} is dense inR+.)

4. PROOF OF THEPISOT CONJECTURE FORβ-SUBSTITUTIONS.

The languageof ψβ is defined byL(ψβ) := {w ∈ A∗ : w is a factor ofψnβ (i) for somen ∈N, i ∈ A}.

The following ‘monotonicity’ properties ofL(ψβ) will be used repeatedly.

Property 1: If ab is a two letter word inL(ψβ) andb ∈ {2, . . . , m+ p}, thena = b− 1.

8 M. BARGE

Property 2: If ab andac are two letter words inL(ψβ) andb 6= c, then eitherb = 1 or c = 1.

Property 3: If ac andbc are two letter words inL(ψβ) anda 6= b, thenc = 1.

The first property is a simple consequence of the fact thatTβ increases monotonically from 0 oneach of its maximal intervals of continuity. Properties 2 and 3 follow immediately from Property1. Note that Property 1 implies that ifτk + t ∈ T ∈ Ωψβ , then{τ1, τ2, . . . , τk}+ t ⊂ T .

We have labeled the prototiles according to the order of their occurance on the interval[0, 1]. Itwill be convenient to also have labelings of the prototiles according to where, along the orbit of1underTβ, their maximum and minimum vertices lie: Letzi := T

i−1β (1) for i = 1, . . . , m+ p; then

if max(τi) = zj , defineτj− := τi, and ifmin(τi) = z

j setτ j+ := τi. If m > 0, then correspondingto each of theTβ-periodic pointszm+i, i = 1, . . . , p, there are twoΨβ-periodic tilings:

Ti := ∪∞n=0Ψnpβ ({τ i+m− − zm+i, τm+i+ − zm+i})and

T 0i := ∪∞n=0Ψnpβ ({τm+i− − zm+i, τ1}).Note thatT 0j ∼srp T 0i andT 0k ∼srp Tk for all i, j, k ∈ {1, . . . , p}, so alsoTi ∼srp Tj for all

i, j ∈ {1, . . . , p}.GivenT, T ′ ∈ Ωψβ we write T ∼s T ′ providedd(Ψkβ(T ),Ψkβ(T ′)) → 0 ask → ∞. Since

d(Ψkβ(T ),Ψkβ(T

′)) → 0 ask → ∞ if and only if there isk ∈ N so thatB0[Ψkβ(T )] = B0[Ψkβ(T ′)]([BO]), we see thatT ∼s T ′ =⇒ T − t ∼s T ′ − t for all |t| < ǫ, for someǫ > 0. We’ll saythatT ∼s T ′ densely on the intervalJ if T − t ∼s T ′ − t for a dense set oft ∈ J . Note that thenT − t ∼s T ′ − t for an open dense subset oft in J . If P, P ′ are allowed patches, we’ll say thatP ∼s P ′ densely onJ if T ∼s T ′ densely onJ for all T, T ′ ∈ Ωψβ with P ⊂ T , P ′ ⊂ T ′.Lemma 2. Suppose thatm > 0 andp > 1. There are theni 6= j so thatTi ∼s Tj densely onR+.If m > 0 andp = 1, thenT1 ∼s T 01 densely onR+.Proof. We give a variant of the basic argument of [BD2]. Letr be the coincidence rank ofψβ . Wemay assume thatr > 1, otherwiseTi ∼s Tj ∼s T 0k densely onR for all i, j, k. (Ti ∼srp Tj ∼srp T 0kfor all i, j, k, as noted above, and, in the context of any Pisot substitution tiling, T ∼srp T ′ =⇒T ∼s T ′ densely onR - see [BK].) There are tilingsS1, . . . , Sr with the properties (see [BK]):

(1) Si ∼srp Sj for 1 ≤ i, j ≤ r,(2) Si is Ψβ-periodic for eachi ∈ {1, . . . , , r}, and(3) Si ∩ Sj = ∅ for all i 6= j ∈ {1, . . . , r}.

ReplacingΨβ by an appropriate power, we may assume that theSi are all fixed byΨβ. For eachi, let Vi be the collection of all the vertices of tiles inSi and letV = {· · · v−1 < v1 < v2 · · · } :=∪ri=1Vi. For eachj ∈ Z andi ∈ {1, . . . , r}, let ηij be the tile ofSi with (vj + vj+1)/2 ∈ spt(ηij).We call the collectionCj := {ηij : i = 1, . . . , r} a configuration. For eachj ∈ Z, let mj :=max{min(ηij) : i = 1, . . . , r} andMj := min{max(ηij) : i = 1, . . . , r}.

Up to translation, there are only finitely many configurations (this is the Meyer property, see[BK]). Thus, since theSi are not translation-periodic, there arek, l ∈ Z andw ∈ R with Ck =Cl + w butCk+1 6= Cl+1 + w. For eachi ∈ {1, . . . , r}, let i′ ∈ {1, . . . , r} be (uniquely) defined byηi

′

k = ηil +w. Note that ifmax(η

ij) > Mj , thenη

ij+1 = η

ij . Hence, lettingC′l := {ηil : ηi

′

k+1 6= ηil+1}we see thatmax(ηil ) = M

il for all η

il ∈ C′l. Furthermore, ifηil ∈ C′l , then eitherηil+1 is of type

THE PISOT CONJECTURE 9

1 (that is, is a translate ofτ1) or ηi′

k+1 is of type 1 (this is Property 2). Since theSi are pairwisedisjoint, it follows that♯C′l ∈ {1, 2}.

We consider two cases:Case 1: ♯C′l = 1.Let {ηil} = C′l. We claim thatηil+1 + w ∼s ηi

′

k+1 densely on[0, ǫ), whereǫ := min{Ml+1 −ml+1,Mk+1−mk+1}. If not, there ist0 ∈ (0, ǫ) andδ > 0 so thatB0[Ψnβ(Si−t)]∩B0[Ψnβ(Si′−w−t)] = ∅ for all n ∈ N and|t−t0| < δ. That is,Bβnδ/2[Ψnβ(Si−t0)]∩Bβnδ/2[Ψnβ(Si′−w−t0] = ∅ forall n ∈ N. Choosenk →∞ so thatΨnkβ (Sj− t0)→Wj ∈ Ωψβ andΨnkβ (Sj+w− t0)→ Uj ∈ Ωψβask → ∞, for all j = 1, . . . , r (such{nk} exists by compactness ofΩψβ ). Then theWj , j =1, . . . , r, are pairwise disjoint (theSj are pairwise disjoint and fixed byΨβ) and stongly regionallyproximal (∼srp is closed and preserved by both translation andΨβ), as are theUj . Furthermore,Wi ∩ Ui′ = ∅. Since∅ 6= {Wj : j 6= i} = {Uj : j 6= i′},Wi ∼srp Ui′ . We now haver + 1 pairwisedisjoint and strongly regionally proximal tilings (W1, . . . ,Wr, Ui′), contradicting coincidence rankr.

Soηil+1+w ∼s ηi′

k+1 densely on[0, ǫ). Again, sinceηil +w = η

i′

k , it follows from Property 2 thatone ofηil+1 andη

i′

k+1 must be of type 1: sayηil+1 = τ1+Ml andη

i′

k+1 = τj−min(τj)+Ml+w. Wehave thatτ1 ∼s τj −min(τj) densely on[0, ǫ) (andj ∈ {2, . . . , m + p} sinceCk+1 6= Cl+1 + w).Let q ∈ {1, . . . , p} be such thatq andj are in phase, that is,p|(q − j). Thenτ1 ∼s τq −min(τq)densely on[0, ǫ). SinceΨpβ(τ1) = {τ1, . . .} andΨpβ(τq −min(τq)) = {τq −min(τq), . . .} we seethatT 0n ∼s Tq densely onR+ (for all n). ThenTj = Ψj−qβ (Tq) ∼s Ψj−qβ (T 0n) = T 0n+j−q ∼s Tqdensely onR+ for all j = 1, . . . , p.

Case 2: ♯C′l = 2.SayC′l = {ηi1l , ηi2l } and, without loss of generality:

ηi1l+1 = τ1 +Ml,

ηi2l+1 = τi+ − zi +Ml,

ηi′1k+1 = τ

j+ − zj +Ml + w, and

ηi′2k+1 = τ1 +Ml + w,

for some1 6= i 6= j 6= 1. Suppose thati andj are in phase and, say,0 < i < j. There is thenqso thati + q = m andj + q = m + sp, which we may take to bem + p. Now, from Property 1,τ i−−zi+Ml ∈ Cl andτ j−−zj+Ml+w ∈ Ck = Cl+w. Thusτ j−−zj+Ml ∈ Cl. Sayτ i−−zi+Ml ∈ Saandτ j− − zj +Ml ∈ Sb. Thenτm+1− − zm+1 + βq+1Ml ∈ Ψq+1β (Sa) ∩ Ψq+1β (Sb) = Sa ∩ Sb, incontradiction to the disjointness ofSa andSb.

Thus,i andj are not in phase (sop > 1). It follows as in Case 1 thatτ i+− zi ∼s τ j+− zj denselyon [0, ǫ) for someǫ > 0 and then thatTi′ ∼s Tj′ densely onR+, with i′ 6= j′ ∈ {1, . . . , p} suchthati′ +m is in phase withi andj′ +m is in phase withj. �

Tilings T, T ′ areasymptotic onR+ providedlimt→∞ d(T − t, T ′ − t) = 0. It follows from theMeyer property, which all Pisot substitution tilings enjoy, thatT, T ′ ∈ Ωψβ are asymptotic onR+if and only if there ist0 so thatB0[T − t] = B0[T ′− t] for all t > t0 (see [BO]). For a discussion ofthe fundamental role asymptotic tilings play in the classification of one-dimensional substitutiontilings, and an algorithm for finding them, see [BD1].

The tilingsT 0j are all pairwise asymptotic onR+. At least for someβ, Ωψβ has additional

asymptotic pairs.

10 M. BARGE

Example 3. Considerβ = (3 +√5)/2. Thenκ(1) = 21 and the substitution is

1→ 121; 2→ 21.The tilingsT 01 andT1 are asymptotic onR

+.

Example 4. Considerβ with κ(1) = 2201 and substitution

1→ 12; 2→ 34; 3→ 2341; 4→ 23.The tilings corresponding to theTβ-periodic points with iteneraries21 and20 are asymptotic onR+.

For an arbitrary substitutionφ with prototilesρ1, . . . , ρd, the roseRφ is the wedge of circlesobtained by identifying the vertices in the disjoint union of the prototiles:Rφ = ∪di=1ρi/ ∼. Letgφ : Ωφ → Rφ be given by

(4.1) gφ(T ) = [(x, i)]

providedx ∈ spt(ρi) andρi − x ∈ T , and let

(4.2) fφ : Rφ →Rφbe the continuous map induced byΦ; that is,fφ ◦ gφ = gφ ◦ Φ. For β substitutions, it will beconvenient to also have the (discontinuous) map

(4.3) gI : Ωψβ → I = [0, 1]given bygI(T ) = x providedmin(τj) ≤ x < max(τj) andτj−x ∈ T . Note thatTβ ◦gI = gI ◦Ψβandgψβ = q ◦ gI , whereq : I → Rψβ ≃ I/{0 = s0 < s1 < · · · < sm+p = 1} is the naturalquotient map and{si : i = 0, . . . , sm+p} = {0, 1} ∪ OTβ(1).

The following somewhat technical result is crucial to our analysis.

Proposition 5. If T 6= T ′ ∈ Ωψβ are asymptotic onR+ and are bothΨβ-periodic of (least) periodp, then either{T, T ′} ⊂ {T 0j : j = 1, . . . , p} or T andT ′ are in distinctΨβ-orbits.Proof. Let’s suppose thatT 6= T ′ ∈ Ωψβ are asymptotic onR+ andΨβ-periodic of periodp. Lett0 = t0(T, T

′) := inf{t′ : B0[T − t] = B0[T ′ − t] for all t > t′}. Then there arek 6= k′ so thatB0[T−t0] = {τk−−zk+t0, τ1+t0} andB0[T ′−t0] = {τk

′

− −zk′

+t0, τ1+t0}. (The occurance of theτ1 in both patches follows from Property 3.) Ift0(T, T ′) = 0, then{T, T ′} ⊂ {T 0j : j = 1, . . . , p}.We will assume from now on thatt0(T, T ′) > 0. Thenk, k′ must bein phase; that is,p|(k − k′),for otherwise,t0(T, T ′) = t0(Ψ

pβ(T ),Ψ

pβ(T

′)) = βpt0(T, T′) > t0(T, T

′). Sincetψβ(r) = tψβ(s)for r 6= s only whenmax(τr),max(τs) ∈ {zm, zm+p}, t0(n) := t0(Ψnβ(T ),Ψnβ(T ′)) will increasewith n until, for somen0, τm− − zm + t0(n0), τm+p− − zm+p + t0(n0) ∈ B0[Ψn0β (T ) − t0(n0)] ∪B0[Ψ

n0β (T

′) − t0(n0)]. Let us examine the effect of applyingΨβ to the patches inΨn0β (T ) andΨn0β (T

′) that lie over[0, t0(n0)].First let’s suppose thatzm+p < zm and, without loss, thatτm+p− − zm+p + t(n0) ∈ Ψn0β (T ) and

τm− − zm + t0(n0) ∈ Ψn0(T ′). Let k, k′ be such thatτk = τm+p− andτk′ = τm− . From Property 1we have that the patchesP := {τ1, . . . τk}− zm+p + t0(n0) andP ′ := {τ1, . . . , τk′} − zm + t0(n0)are contained inΨn0(T ) andΨn0(T ′), respectively. SinceTβ(zm+p) = Tβ(zm), we have that

THE PISOT CONJECTURE 11

Tβ(t + zm − zm+p) = Tβ(t) for 0 ≤ t ≤ zm+p and it follows thatΨβ(P ) ⊂ Ψβ(P ′). Let

τ l−−zl−zm+p+t0(n0) be the tile inΨn0β (T ) that immediately precedesP . We have thatt0(n0+1) =βt0(n0)− βzm+p andτ l+1− , τ 1− ∈ B0[Ψn0+1β (T )− t0(n0 + 1)] ∪B0[Ψn0+1β (T ′)− t0(n0 + 1)]. Nowt0(n) must increase forn = n0 + 1, . . . , n0 +m− 1 and this means thatm ≤ p. Thus, sincel + 1and1 must be in phase,l = p.

An identical analysis with the same conclusion applies to the casezm < zm+p. To ease notation,we may then (replacingT andT ′ by Ψn0+1β (T ) andΨ

n0+1β (T

′)) assume thatB0[T − t0(0)] ={τ p+1− − zp+1 + t0(0), τ1 + t0(0)} andB0[T ′ − t0(0)] = {τ 1− − 1 + t0(0), τ1 + t0(0)}. We seenow thatB0[Ψnβ(T ) − t0(n)] = {τ p+1− − zp+1 + t0(n), τ1 + t0(n)} andB0[Ψnβ(T ′) − t0(n)] ={τ 1− − 1 + t0(n), τ1 + t0(n)} if and only if n = km for somek. Hencem|p.

Let us next rule outzm < zm+p. Suppose that this is the case. The patchP ′ := {τ1, τ2, . . . , τm− }−zm + t0(m − 1)} ⊂ Ψm−1β (T ′) is immediately preceded (inΨm−1β (T ′)) by a translate of sometile τ l−. ThenB0[Ψ

mβ (T

′) − t0(m)] = {τ l+1− − zl+1 + t0(m), τ1 + t0(m)} while B0[Ψmβ (T ) −t0(m)] = {τ 1− − 1 + t0(m), τ1 + t0(m)}. Sincezm < zm+p, m must be greater than 1 and thenzm < βm−1z1. It follows that the tileτ1 − zm + t0(m − 1) in P ′ can’t be the first tile in a patchΨm−1β (τ1 + β

1−m(t0(m − 1) − zm) with τ1 + β1−m(t0(m − 1) − zm) ∈ T ′. Thus it must be thecase that there isk, 1 ≤ k < m− 1, andτ ∈ Ψkβ(T ′) with Ψβ(τ) = {. . . , τ 1−− z1 + t′, τ1 + t′, . . .}andΨm−2−kβ (τ1+ t

′) = {τ1− zm+ t0(m− 1), . . .}. But thenl = m− 1− k so that1 < l+1 < mand then1 andl + 1 are definitely out of phase.

Thuszm+p < zm.For eachn ∈ N we have:

(1) B0[Ψnmβ (T )− t0(nm)] = {τ p+1− − zp+1 + t0(mn), τ1 + t0(nm)},(2) B0[Ψnmβ (T

′)− t0(nm)] = {τ 1− − z1 + t0(mn), τ1 + t0(nm)},(3) {τ p− − zp, τ1, τ2, . . . , τm+p− } − zm+p + t0(mn− 1) ⊂ Ψmn−1β (T ), and(4) {τ1, τ2, . . . , τm− } − zm + t0(mn− 1) ⊂ Ψmn−1β (T ′).

Claim 1: t0(nm) < 1 for all n.To see that this must be true, suppose thatb = t0(nm) − 1 > 0 for somen. Then the patch

Pn = {τ1, τ2, . . . , τ 1−} − z1 + t0(nm) in Ψnmβ (T ′) has the property thatΨmβ (P ) contains the patchPn+1 in Ψ

(n+1)mβ (T

′), so thatt0((n + 1)m) ≥ βmb + 1. That is,t0((n + 1)m) − 1 ≥ βmb. In-ductively, t0((n + k)n) ≥ βkb for k ∈ N. But t0 is periodic. If t0(nm) = 1 for somen thenτ1 ∈ Ψnmβ (T ′) andτ1 ∈ Ψkmβ (T ′) = T ′, soT ′ 6= Tj for anyj. We have established Claim 1.

Let y := gI(T ′) (recall definition 4.3) and letr := ⌊βzm+p⌋. In other words,y = 1 − t0(0) =z1 − t0(0) andr is the number of discontinuities ofTβ on [0, zm+p].

Claim 2: κ(y) = a1a2 · · · am−1(am − r − 1).To prove the claim, we first suppose that there isj < m with (κ(y))j < aj . Then the

patch{τ1, τ2, . . . , τ j+1− } − zj+1 + t0(j) in Ψjβ(T ′) has support contained in[0,∞). The patch{τ1, τ2, . . . , τm− }− zm+ t0(m− 1) of Ψm−1β (T ′) then also has support in[0,∞) and hence so doesthe patch{τ1, τ2, . . . , τ 1−} − z1 + t0(m). This means thatt0(m) ≥ 1, which has been ruled outby Claim 1. Thus(κ(y))j = aj for j = 1, . . . , m − 1. It follows that t0(j) = zj+1 − T jβ(y) forj = 0, . . . , m− 2

12 M. BARGE

If |t0(m−1)−t| < zm+p, thenB0[Ψmβ (T )−βt] = B0[Ψmβ (T ′)−βt]. Consequently,t0(m−1) >zm+p, which means thatTm−1β (y) < z

m − zm+p. For t ∈ [0, zm+p], Tβ(t) = Tβ(t + zm − zm+p).ThusTβ hasr discontinuities on the interval(zm−zm+p, zm]; it’s also discontinuous atzm−zm+p.ThusTβ has at leastr + 1 discontinuities betweenT

m−1β (y) andz

m and we have that(κ(y))m ≤am − r − 1. Suppose that(κ(y))m < am − r − 2. Thent0(m − 1) > zm − zm+p − 1/β andt0(m) ≥ 1, contrary to Claim 1. Thus,(κ(y))j is as claimed forj = 1, . . . , m. Repeating theargument periodically givesκ(y) = a1a2 · · · am−1(am − r − 1), as claimed.

From Claim 2,y isTβ-periodic of periodm. ThusT ′ isΨβ-periodic of periodm, som = p and wehave thatκ(y) = a1a2 · · · ap−1(ap − r − 1). Let’s now determine iteneraryκ(x) of x := gI(T ).

Consider the patch{τ p−−zp, τ1}−z2p+ t0(p−1) ⊂ Ψp−1β (T ) (see item (3), withm = p, n = 1)and note thatτ1 − z2p + t0(p) has support contained inR+. This can only come from a patch{τ p−1− −zp−1, τ1}−β−1(z2p−t0(p−1)) inΨp−2β (T ), which can only have come from a patch{τ p−2− −zp−2, τ1}−β−2(z2p− t0(p− 1)) in Ψp−3β (T ),... . We see that the patch{τ 1−− 1, τ1, τ2, . . . .τ p+1− }−zp+1+ t0(0) is contained inT and the support ofτ1−zp+1+ t0(0) is contained inR+. The origin isthen located in the interior of the support of the patch{τ1, τ2, . . . , τ 1−}− 1− zp+1+ t0(0) ⊂ T (seeClaim 1) at relative positionx. That is,x = 1+zp+1−t0(0). Since, under aplication ofΨp−1β , the tileτ 1−−1−zp+1+t0(0) produces the patch{. . . , τ p−−zp−z2p+t0(p−1)} in Ψp−1β (T ) and the supportof the latter patch is contained in(−∞, t0(p − 1)], we haveβj(max(τ 1− − 1 − zp+1 + t0(0))) =βj(t0(0)− zp+1) ≤ t0(j) for j = 0, . . . , p− 1. If there were a discontinuity ofTβ betweenT jβ (x)andzj+1 for somej ∈ {0, . . . , p− 2}, thent0(j + 1) would be at least 1, which is not allowed byClaim 1. Thus(κ(x))j = aj for j = 1, . . . , p− 1.

One can show thatκ(x) = a1a2 · · · ap−1(ap − 1) (as in Example 4) but at this point we haveall we need to know that the tilingsT 6= T ′ are not in the sameΨβ-orbit. Indeed, sinceΨβ ◦gI = gI ◦ Tβ, Ψkβ(T ) = T ′ would imply σk(κ(gI(T ))) = κ(gI(T ′)), which would mean that∑p

j=1(κ(gI(T )))j =∑p

j=1(κ(gI(T′)))j . The latter equality can only hold ifgI(T ) = gI(T ′) (since

we have shown that(κ(gI(T )))j = (κ(gI(T ′)))j for j = 1, . . . , p− 1), but thenT = T ′.�

Corollary 6. If i 6= j, thenTi andTj are not asymptotic onR+.Lemma 7. If m > 0 thenTi ∼s Tj ∼s T 0k densely onR+ for all i, j, k ∈ {1, . . . , p}.Proof. If p = 1 this is the second statement of Lemma 2. Ifp > 1, let i 6= j be so thatTi ∼s Tjdensely onR+ (from the first statement of Lemma 2). From Corollary 6 we knowthat Ti andTj are not asymptotic onR+. Thus there are patches{τ, τ ′} ⊂ Ti and{τ, τ ′′} ⊂ Tj with supportcontained inR+, τ ′ 6= τ ′′, andmin(τ ′) = min(τ ′′) = max(τ). From Property 2, one ofτ ′, τ ′′ is oftype 1: sayτ ′ = τ1+t andτ ′′ = τr−min(τr)+t, with r ∈ {2, . . . , m+p}. Let l ∈ {m+1, . . . , m+p} be in phase withr (that is,p|(l−r)). ThenΨqpβ (τ ′′) = {τl−min(τl)+βqpt, . . .} for someq. Letǫ = min{length(τ1), length(τr), β−qp · length(τl)}. Thenτ1 ∼s τr −min(τr) ∼s τl −min(τl)densely on[0, ǫ). It follows thatTl ∼s T 0k densely onR+ for all k ∈ {1, . . . , p}. Thus, withsubscripts taken mod(p), Tl+n = Ψnβ(Tl) ∼s Ψnβ(T 0k ) = T 0k+n densely onR+ for all n ∈ N. �

Thestable equivalence relation, ≈s, is defined for a substitution tiling spaceΩφ by T ≈s T ′ ifand only ifT ∼s T ′ densely onR andT ∼srp T ′.Corollary 8. If m > 0 thenTj ≈s T 0j for eachj ∈ {1, . . . , p}.

THE PISOT CONJECTURE 13

The following is Theorem 5 of [B3].

Theorem 9. If φ is a primitive, non-periodic Pisot substitution, then(Ωψβ ,R) has pure discretespectrum if and only if stable equivalence equals strong regional proximality onΩψβ .

Given a morphismγ from letters of an alphabetA to nonempty words on an alphabetA′, letiγ , tγ : A → A′ be the initial and terminal letter maps: Ifγ(i) = a · · · z then iγ(i) = a andtγ(i) = z.

The stable equivalence relation is pushed down to a relationR on the roseRφ by xRy if andonly if there areT1, . . . , Tn, T ′1, . . . , T

′n ∈ Ωφ with Ti ≈s T ′i for i = 1, . . . , n, gφ(Ti+1) = gφ(T ′i )

for i = 1, . . . , n− 1, gφ(T1) = x, andgφ(T ′n) = y. The following lemma is proved in [B1] underthe slightly stronger assumption that eithertφ or iφ is eventually constant (see Lemmas3.5 and 3.6there), but the proof goes through virtually without changeunder the condition we assume here ofdense terminal or dense initial stable relation.

Lemma 10. Suppose thatφ is a nonperiodic Pisot substitution with the properties:(Ωφ,R) doesnot have pure discrete spectrum and there isǫ > 0 so that eitherτ − min(τ) ∼s τ ′ − min(τ ′)densely on[0, ǫ) for all tiles τ, τ ′, or τ −max(τ) ∼s τ ′ −max(τ ′) densely on(−ǫ, 0] for all tilesτ, τ ′. ThenR is a closed equivalence relation onRφ with boundedly finite equivalence classes.

Recall (definition 4.2) thatfφ : Rφ → Rφ is the map induced byΦ : Ωφ → Ωφ (so thatfφ◦gφ = gφ◦Φ). ThenxRy =⇒ fφ(x)Rfφ(y) and there is an induced map̃fφ : Rφ/R→ Rφ/R.Under the assumptions of Lemma 10,Rφ/R is a wedge of circles,̃fφ defines a substitutioñφ, andwe may identifyRφ/R with Rφ̃. If A andà are the alphabets forφ andφ̃ then the quotient mapfromRφ toRφ̃ induces a morphismα : A → Ã∗ so thatα ◦ φ = φ̃ ◦ α.

The following is distilled from [B1] (see, in particular, Theorem 3.9 there).

Theorem 11. Suppose thatφ is a nonperiodic Pisot substitution with the properties:(Ωφ,R) doesnot have pure discrete spectrum and there isǫ > 0 so that eitherτ − min(τ) ∼s τ ′ − min(τ ′)densely on[0, ǫ) for all tiles τ, τ ′; or τ −max(τ) ∼s τ ′ −max(τ ′) densely on(−ǫ, 0] for all tilesτ, τ ′. Thenφ̃ is a nonperiodic Pisot substitution and(Ωφ̃,R) does not have pure discrete spectrum.

To apply Theorem 11 toβ-substitutions we will require the following lemma.

Lemma 12. If m > 0, there isǫ > 0 so thatτ −min(τ) ∼s τ ′ −min(τ ′) densely on[0, ǫ) for alltiles τ, τ ′ for ψβ.

Proof. For eachi > 1 there isj ∈ {1, . . . , p} so thatzi is in phase withzm+j . This means thatΨm−i+1β (τ

i+) = {τm+1+ , . . .} andΨm−i+1β (τm+j+ ) = {τm+1+ , . . .}. Henceτ i ∼s τm+j densely on

[0, ǫi), with ǫi := βi−m−1(length(τm+1)). SinceTi ∼s Tj ∼s T 0k densely onR+ (Lemma 7) wehave thatτ1 ∼s τ l+ densely on[0, ǫ), with ǫ := min{β−m−1(length(τm+1+ )), length(τ1)}, for alll ∈ {2, . . . , m+ p}. �Lemma 13. Suppose that(Ωψβ ,R) does not have pure discrete spectrum andm > 0. Theniψ̃β iseventually constant andtψ̃β is injective.

Proof. Pick i ∈ {1, . . . , p} and letP ⊂ Ti andP ′ ⊂ T 0i be patches, both supported on[0, t0], withP∩P ′ = ∅ and so thatτ1+t0 ∈ B0[Ti−t0]∩B0[T 0i −t0] (that is,P, P ′ are the initial patches ofTi, T 0ipreceding their first agreement). ThenΨpβ(P ) = P∪{τ1+t0}∪Q andΨpβ(P ′) = P ′∪{τ1+t0}∪Q′.Let k > 0 be minimal with the property thatψkβ(P ) ∩ Ψkβ(P ′) 6= ∅. We may as well assume

14 M. BARGE

(after replacingTi andT 0i by Ψk−1β (Ti) andΨ

k−1β (T

0i ), andP, P

′ by the initial disjoint patches ofΨk−1β (Ti),Ψ

k−1β (T

0i )) thatk = 1. ThenΨβ(P ) = P1∪{τ1+t}∪P2 andΨβ(P ′) = P ′1∪{τ1+t}∪P ′2

for somet > 0, with P1 ∩ P ′1 = ∅ andspt(P1) = [0, t] = spt(P ′1). Consider the (disjoint) patchesB0[Ti − t/β] ⊂ P andB0[T 0i − t/β] ⊂ P ′. Let us first observe that these patches can’t bothbe doubletons. Indeed, ifB0[Ti − t/β] = {ρ1 < ρ2} andB0[T 0i − t/β] = {ρ′1 < ρ′2}, thenρ2 = τ1 + t/β = ρ

′2 (sinceτ1 is the only prototile whose image underΨβ begins with a tile of type

1), contradicting the disjointness ofP andP ′. If B0[Ti−t/β] andB0[T 0i −t/β] are both singletons,then these must be of the form{τ}, {τ ′} with τ a translate of someτl andτ ′ a translate of someτl′with l, l′ such thatk/β ∈ int(spt(τl)) andk′/β ∈ int(spt(τl′)) for somek, k′ ∈ {1, . . . ⌊β⌋}. Butthenτm+p + t − 1 ∈ Ψβ(τ) ∩ Ψβ(τ ′) is in bothP1 andP ′1, contradicting the disjointness of thesepatches. Thus, exactly one ofB0[Ti− t/β] andB0[T 0i − t/β] is a doubleton and one of the patchesP1, P

′1 ends inτm+p+t−1 and the other ends inτk+t−max(τk) for somek ∈ {1, . . . , m+p−1}.

SinceΨβ(Ti) ≈s Ψβ(T 0i ), we see from the definition of the relationR thatq(1−s)Rq(max(τk)−s) in the roseRβ for all positives less than the smaller of the lengths ofτk andτm+p. This meansthat tα(k) = tα(m + p). Let max(τk) = zn+1. Then tψn

β(m + p) = k and it follows from

α ◦ ψβ = ψ̃β ◦ α thattψ̃nβ(tα(m+ p)) = tα(tψn

β(m+ p)) = tα(k) = tα(m + p). Now leta be any

letter in the alphabet for̃ψβ and letl ∈ {1, . . . , m+ p} be so thattα(l) = a. Let r ∈ N be so thattψr

β(m+ p) = l. Then

tψ̃nβ(a) = tψ̃n

β(tα(l))

= tψ̃nβ(tα(tψr

β(m+ p)))

= tψ̃rβ(tα(tψn

β(m+ p)))

= tψ̃rβ(tα(m+ p))

= tα(tψrβ(m+ p))

= tα(l)

= a.

Hencetψ̃β is injective.Pick j ∈ {1, . . . , p} (so Tj ≈s T 0j by Corollary 8) and letk ∈ {2, . . . , m + p} be so that

min(τk) = zm+j (thus τk is the initial tile of Tj). Then iα(k) = iα(1). Now, for any l ∈

{2, . . . , m+p} there isn ∈ N so thatiψnβ(l) = k, and theniψ̃n

β(iα(l)) = iα(iψn

β(l)) = iα(k) = iα(1).

Sinceiψβ(1) = 1, iψ̃β(iα(1)) = iα(1). Thatiψ̃β is eventually constant (with valueiα(1)) followsfrom surjectivity ofiα.

�

The following is Theorem 3.12 of [B1].

Theorem 14. Suppose thatφ is a nonperiodic Pisot substitution with the properties:iφ is eventu-ally constant andtφ is injective. Then(Ωφ,R) has pure discrete spectrum.

Theorem 15. If β is Pisot, then(Ωψβ ,R) has pure discrete spectrum.

THE PISOT CONJECTURE 15

Proof. This is clearly true ifβ = n ∈ N. Suppose thatβ is Pisot, and not an integer. Ifm = 0 it iseasy to check thatiψβ is eventually constant andtψβ is injective. Hence(Ωψβ ,R) has pure discretespectrum by Theorem 14. Ifm > 0 and(Ωψβ ,R) does not have pure discrete spectrum, then thesubstitutionψ̃β is non-periodic Pisot and(Ωψ̃β ,R) does not have pure discrete spectrum (Lemma

12 to getψ̃β and Theorem 11). But(Ωψ̃β ,R) must have pure discrete spectrum by Lemma 13 andTheorem 14. �

Remark 16. If the degreed of the Pisot numberβ equalsm + p (so that theβ-substitution isirreducible), then it follows from Theorem 15 and[CS] (see, also,[Sa]) that the substitutive system(not to be confused with theβ-shift) also has pure discrete spectrum. There are, however, examplesof Pisotβ for which the substitutive system does not have pure discrete spectrum ([EI] . It followsfrom Theorem 15 and Proposition 8.1 of[BBK] that, for Pisot units, the substitutive system as-sociated withψβ has, as an a.e. one-to-one factor, an isometric exchange of domains (the Rauzypieces) induced as a first return of a minimal translation on the(d− 1)-torus.Corollary 17. All Pisot numbers are weakly finitary.

Corollary 18. If β is Pisot,M is the companion matrix for the minimal polynomial ofβ, andȳ is afundamental homoclinic point for the solenoidal automorphismF̂M , then the arithmetical codinghȳ : Xβ → T̂dβ is a.e. one-to-one.

We prove Corollaries 17 and 18 in the next section.

5. THE CONNECTION BETWEEN PURE DISCRETE SPECTRUM, ARITHMETICAL CODING , ANDPROPERTY (W).

For this section, fix a Pisot numberβ of algebraic degreed. Thesubstitution matrixA associatedwith ψβ is the(m+ p)× (m+ p) matrix with ij-th entryaij given by the number of occurances ofthe letterj in the wordψβ(i). Thenβ is a simple eigenvalue ofA and the characteristic polynomialof A factors overZ as

pA(x) = pβ(x)q(x)

with pβ(x) the minimal polynomial ofβ andq(x) relatively prime withpβ(x). There are thens1(x), s2(x) ∈ Q[x] with s1(x)pβ(x) + s2(x)q(x) = 1 and anA-invariant splitting

Rm+p = V ⊕W,whereV := ker(s1(A)pβ(A)) andW := ker(s2(A)q(A)) are rational (that is, spanned by rationalvectors). ThePisot subspaceV is d-dimensional and the linear transformationL := A|V hascharacteristic polynomialpβ(x).

Let V = Es ⊕ Eu be the splitting ofV into (d − 1)-dimensional stable and 1-dimensionalunstable spaces, letπV : Rm+p → V be projection alongW , and, forv ∈ V , write v = vs + vuwith vs,u ∈ Es,u. By πs (resp.,πu) we will denote both projection ofRm+p andV alongW ⊕ Eu(resp.,W ⊕Es) and ofV alongEu (resp.,Es) ontoEs (resp.,Eu). The subgroup

Γ := πV (Zm+p)

of V is ad-dimensional lattice.It is convenient to view tilingsT ∈ Ωψβ as maps ofR into the roseRψβ , rather than as collections

of tiles: GivenT ∈ Ωψβ , letT (t) := gψβ(T − t) (recall the definition 4.1). TheR-action on tilings-as-maps is then given by(T − t)(s) := T (s+ t); Ψβ translates asΨβ(T )(t) := fψβ(T (t/β)) (with

16 M. BARGE

fψβ as in 4.2); and the tiling metric induces the topology of unifom convergence on compact sets.We view the prototiles as parameterizations of the corresponding petals of the rose viaτi(t) :=T (t−min(τi)) whereτi ∈ T ∈ Ωψβ .

For eachi ∈ {1, . . . , m+ p}, let ei ∈ Zm+p denote the standard unit vector and letσi := {tei :t ∈ [0, 1]} ⊂ Rm+p. Let R̃ψβ ⊂ Rm+p be the grid

R̃ψβ := ∪a∈Zm+p ∪i=1,...,m+p (a+ σi).The map̃π : R̃ψβ →Rψβ given by

π̃(a+ tei) := τi(t)

is theuniversal abelian coverofRψβ .We denote byf̃ψβ : R̃ψβ → R̃ψβ the unique lift offψβ with f̃ψβ(0) = 0 and, if T ∈ Ωψβ ,

T̃ : R→ R̃ψβ will denote any lift ofT : R→Rψβ . We then definẽΨβ on such lifts byΨ̃β(T̃ )(t) := f̃ψβ ◦ T̃ (t/β).

Note that sinceA is the ‘abelianization’ offψβ , f̃ψβ(a) = Aa for a ∈ Zm+p.By astrandγ corresponding toT ∈ Ωψβ we will mean any

γ = x+ πV ◦ T̃ ,with x ∈ V andT̃ a lift of T . ForT ∈ Ωψβ , let

(5.1) t∗(T ) := sup{t ≤ 0 : τ1 + t ∈ T}.SoT (t∗(T )) = ∗ is the branch point ofRψβ andt∗(T ) is the largest non-positive occurance of theinitial vertex of a tile of type 1 inT . For a strandγ corresponding toT we define itsinitial vertexto be

(5.2) a(γ) := γ(t∗(T ))

and the (abelian)prefixof T , p(T ), to be

(5.3) p(T ) := γ(t∗(T ))− γ(βt∗(Ψ−1β (T ))).Let us extendΨβ to strands by settingΨβ(x + πV ◦ T̃ ) := Lx + πV ◦ Ψ̃β(T̃ ). Then, if γ is

a strand corresponding toT , Ψβ(γ) is the unique strand corresponding toΨβ(T ) having initialvertexa(γ′) = p(Ψβ(T )) + Lγ(t∗(T )). It is clear thatΨβ is invertible on strands (even though, ifdet(L) 6= ±1, Ψ̃β is not invertible), and

p(T ) = a(γ)− La(Ψ−1β (γ))for any strandγ corresponding toT . We extend theR-action to lifts and strands by(T̃ − t)(s) :=T̃ (s + t) and (γ − t)(s) := γ(s + t). One easily checks that̃Ψβ(T̃ − t) = Ψ̃β(T̃ ) − βt andΨβ(γ − t) = Ψβ(γ)− βt.

Given a strandγ corresponding toT ∈ Ωψβ , consider the sequence(ai(γ))i∈Z of initial verticesof the strands in theΨβ-orbit of γ:

ai(γ) := Ψiβ(γ)(t∗(Ψ

iβ(T ))).

THE PISOT CONJECTURE 17

We see that|ai(γ) − Lai−1(γ)| = |p(Ψiβ(T ))| is bounded. From the hyperbolicity ofL it followsthat(ai(γ)) is globally shadowedby a unique point in V:

Lemma 19. Given a strandγ, there is a unique pointG(γ) ∈ V with the property that|LiG(γ)−ai(γ)|, i ∈ Z, is bounded. Furthermore,a(γ)−G(γ) is uniformly bounded as a function ofγ.Proof. We haveL−iaui (γ) = a

u0(γ) + L

−1pu(Ψ0β(T )) + · · · + L−ipu(Ψi−1β (T )) andLias−i(γ) =as0(γ)− ps(Ψ−1β (T ))− · · · − Li−1ps(Ψ−iβ (T )). Since thep(Ψiβ(T )) are bounded, the limits in thefollowing formula forG(γ) exist:G(γ) = limi→∞ L−iaui (γ) + limi→∞ L

ias−i(γ). �

Note that ifγ′ = x + γ is another strand corresponding toT , thenai(γ′) = Lix + ai(γ), soG(γ′) = G(γ) + x. Thus

γT := γ −G(γ)is the unique strand corresponding toT with the property that(ai(γT )) is bounded. We defineπ : Ωψβ → V/Γ by

(5.4) π(T ) := a0(γT ) + Γ.

Let l = (l1, . . . , lm+p) with li := length(τi) the lenngth of thei-th prototile. Thenl is a lefteigenvector ofA (for the eigenvalueβ) and is orthogonal toEs. Thus, settingω to be the rightpositive eigenvector ofA, normalized with〈lt, ω〉 = 1, we havevu = πu(v) = 〈lt, v〉ω for allv ∈ V . It follows that if γ is any strand, then

(5.5) γu(t)− γu(t′) = (t− t′)ωfor all t, t′ ∈ R.Lemma 20. π(T − t) = π(T )− (tω + Γ) for all T ∈ Ωψβ andt ∈ R.Proof. SinceΨiβ(γ − t) = Ψiβ(γ) − βit for any strandγ, i ∈ Z andt ∈ R, |ai(γ − t) − ai(γ)|is bounded (for fixedγ, t) for i ≤ 0. Now |ai(γ) − Ψiβ(γ)(0)| and |ai(γ − t) − Ψiβ(γ − t)(0)|are bounded and(Ψiβ(γ) − βit)u(0) − (Ψiβ(γ))u(0) = βitω. Thus|aui (γ − t) − aui (γ) − βitω| isbounded fori ≥ 0. It follows thatG(γ − t) = G(γ)− tω. Note thata0(γ − t) ≡ a0(γ) mod (Γ).We haveπ(T − t) = a0(γ(T−t))+Γ = a0((γT − t)−G(γT − t))+Γ = a0(γT )−G(γT )− tω+Γ =π(T )− (tω + Γ) (usingG(γT ) = 0). �Lemma 21. π : Ωψβ → V/Γ is a continuous surjection.Proof. Continuity is clear from the explicit formula forG given in the proof of Lemma 19. Sur-jectivity is a consequence of Lemma 20 and the irrationalityof ω: If cl{tω : t ∈ R} were a propersub-torusT of dimensionk < d, then there would be a rationalk-dimensional subspaceU of V ,invariant underL. But then the degree ofβ would be at mostk, and notd. �

The previous two lemmas combine to say thatπ is factor map of(Ωψβ ,R) onto (Td,R), with

theR-action onTd ≃ V/Γ given by(x + Γ) − t := x − tω + Γ. It is easy to check thatπ alsosemiconjugatesΨβ with FL : V/Γ→ V/Γ by FL(x+ Γ) = Lx+ Γ. Thusπ induces

(5.6) π̂ : Ωψβ → T̂d := lim←−FL,which factors(Ωψβ ,R) onto(T̂

d,R).

18 M. BARGE

ForT, T ′ ∈ Ωψβ , we say thatT globally shadowsT ′, and writeT ∼gs T ′, provided there are liftsT̃ , T̃ ′ of T, T ′ so that the strandsγ := πV (T̃ ), γ′ := πV (T̃ ′) satisfy:|ai(γ)− ai(γ′)| is bounded.

If β is a unit, thendet(L) = ±1, Ψ̃β is invertible on{T̃ : T ∈ Ωψβ}, and the above definition canbe rephrased as:T ∼gs T ′ provided there are lifts̃T , T̃ ′ so that|πV ◦Ψ̃iβ(T̃ )(t∗,i)−πV ◦Ψ̃iβ(T̃ ′)(t′∗,i)|is bounded, wheret∗,i := t∗(Ψiβ(T )) andt

′∗,i := t∗(Ψ

iβ(T

′)). To cover the ‘nonunimodular’ case aswell, we will need the following awkward (but straightforward) reformulation.

Lemma 22. T ∼gs T ′ if and only if there areB < ∞ and liftsT̃ , T̃ ′, so that for eachi ∈ N thereare vi ∈ Γ and lifts T̃i, T̃ ′i of Ψ−iβ (T ),Ψ−iβ (T ′), such thatΨ̃iβ(T̃i) = vi + T̃ , Ψ̃iβ(T̃ ′i ) = vi + T̃ ′,and |πV ◦ Ψ̃kβ(T̃i)(t∗,k−i)− πV ◦ Ψ̃kβ(T̃ ′i )(t′∗,k−i)| ≤ B for all k ∈ N, wheret∗,j := t∗(Ψjβ(T )) andt′∗,j := t∗(Ψ

jβ(T

′)).

Lemma 23. π̂(T ) = π̂(T ′) if and only ifT ∼gs T ′.Proof. Suppose that̂π(T ) = π̂(T ′) and fix i ≥ 0. Let γΨ−i

β(T ) = x + πV ◦ T̃i andγΨ−i

β(T ′) =

x′+πV ◦S̃i with T̃i andS̃i lifts of Ψ−iβ (T ) andΨ−iβ (T ′), resp. Thenπ(Ψ−iβ (T )) = π(Ψ−iβ (T ′))meansthatx− x′ ∈ Γ. There isB so that|a(γT )| < B/2 for all T (Lemma 19), so|(ak(γΨ−i

β(T ))− x)−

(ak(γΨ−iβ

(T ′))−x)| = |(ak(γΨ−iβ

(T )−x))−(ak(γΨ−iβ

(T ′)−x))| < B for all i, k ≥ 0. Lety ∈ Zm+p besuch thatπV (y) = x−x′ and set̃T ′i := y+S̃i. Then, witht∗,j := t∗(Ψjβ(T )) andt′∗,j := t∗(Ψjβ(T ′)),we have|πV ◦Ψ̃kβ(T̃i)(t∗,k−i)−πV ◦Ψ̃kβ(T̃ ′i )(t′∗,k−i)| = |(ak(γΨ−i

β(T )−x))−(ak(γΨ−i

β(T ′)−x))| < B

for k ∈ N. Letting T̃ = T̃0 andT̃ ′ = T̃ ′0 we haveΨ̃iβ(T̃i) = T̃ + vi andΨ̃iβ(T̃ ′i ) = T̃ ′ + vi for somevi ∈ Γ; henceT ∼gs T ′.

Now suppose thatT ∼gs T ′ and letB > 0, vi, T̃ , T̃ ′, T̃i and T̃ ′i be as in the reformulationof ∼gs (Lemma 22). Letγ := πV ◦ T̃ andγ′ := πV ◦ T̃ ′. ThenΨ−iβ (γ) = xi + πV ◦ T̃i andΨ−iβ (γ

′) = xi + πV ◦ T̃ ′i with xi = −1/βiπV (vi). Thena(γ) ≡ a(γ′) mod (Γ) and from|πV ◦Ψ̃kβ(T̃i)(t∗,k−i)−πV ◦Ψ̃kβ(T̃ ′i )(t′∗,k−i)| ≤ B for all k ∈ N, we see thatG(γ) = G(γ′). Thus,a(γT ) ≡a(γ)−G(γ) ≡ a(γ′)−G(γ′) ≡ a(γT ′) mod (Γ), soπ(T ) = π(T ′). It is clear from the definitionof ∼gs thatT ∼gs T ′ =⇒ Ψkβ(T ) ∼gs Ψkβ(T ′) for all k ∈ Z. Thusπ(Ψ−kβ (T )) = π(Ψ−kβ (T ′)) forall k ≥ 0 andπ̂(T ) = π̂(T ′). �

Translated into the context of tilings-as-maps, the strongregional proximal relation reads asfollows: T ∼srp T ′ provided for eachR > 0 there areSR, S ′R ∈ Ωψβ and tR ∈ R so thatT (t) = SR(t), T ′(t) = S ′R(t), andSR(t− tR) = S ′R(t− tR) for all |t| ≤ R.Theorem 24. If β is Pisot then the strong regional proximal and global shadowing relations onΩψβ are the same.

This can be deduced (at least, in the unimodular case) as a corollary to Propositions 24 and 25of [BG] which combine for a similar result in the higher dimensional setting. We give a simplifiedargument for the one-dimensional case at hand.

Let’s say that strandsγ, γ′ corresponding toT, T ′ ∈ Ωψβ are strong regionally proximal, andwrite γ ∼srp γ′, provided there are: lifts̃T , T̃ ′ of T, T ′ andx ∈ V with γ = x + πV (T̃ ), γ′ =x + πV (T̃

′); and, for eachR > 0, there aretR ∈ R andSR, S ′R ∈ Ωψβ with lifts S̃R, S̃ ′R so thatT̃ (t) = S̃R(t), T̃

′(t) = S̃ ′R(t), S̃R(t− tR) = S̃ ′R(t− tR), for all |t| ≤ R.Lemma 25. There isB ∈ R so that ifγ ∼srp γ′, then|γ(0)− γ′(0)| ≤ B.

THE PISOT CONJECTURE 19

Proof. Letn ∈ N andρ < 1 be so that|(Lx)s| ≤ ρ for all x ∈ X with |xs| ≤ 1. LetB1 be such thatif j : [a, b]→Rψβ is any parameterization of a petal ofRψβ with lift j̃, then|πs(πV (f̃nψβ(j̃(t1))))−πs(πV (f̃

nψβ(j̃(t2))))| ≤ B1 for all t1, t2 ∈ [a, b]. Now takeB large enough so thatB/2 − ρB/2 ≥

B1. Then if γ is any strand corresponding to an element ofΩψβ , |γs(t1) − γs(t2)| ≤ B/2 forall t1, t2 ∈ R (since any compact piece of the image ofγ is contained in a translation of somef̃kn(j̃([a, b])). Now if γ andγ′ are strong regionally proximal, withS1, S ′1 as in the above definitionandR = 1, then |(γ(0))s − (πV (S̃(t1)))s| ≤ B2 and |(γ′(0))s − (πV (S̃ ′(t1)))s| ≤ B2 . SinceS̃(t1) = S̃

′(t1), we have(γ(0))u = (γ(0))u and|(γ(0))s − (γ′(0))s| ≤ B.�

Lemma 26. If γ ∼srp γ′ thenΨβ(γ) ∼srp Ψβ(γ′) andΨ−1β (γ) ∼srp Ψ−1β (γ′).

Proof. Suppose thatγ andγ′ correspond toT, T ′ ∈ Ωψβ andγ ∼srp γ′. LetR > 0 be given and letSβR, S

′βR ∈ Ωψβ , tβR ∈ R, be as in the above definition of∼srp, withβR replacingR. We may then

lift Ψ−1β (T ),Ψ−1β (SβR),Ψ

−1β (S

′βR), andΨ

−1β (T

′), successively, so that̃Ψ−1β (T )(t) =˜Ψ−1β (SβR)(t),

˜Ψ−1β (SβR)(t − (1/β)tβR) = ˜Ψ−1β (SβR)(t − (1/β)tβR), andΨ̃−1β (T ′)(t) = ˜Ψ−1β (S ′βR)(t) for all|t| ≤ R. Projecting to strands, and appropriately translating, shows thatΨ−1β (γ) ∼srp Ψ−1β (γ′).The argument thatΨβ(γ) ∼srp Ψβ(γ′) is similar.

�

Recall thatσi := {tei : 0 ≤ t ≤ 1} ⊂ Rm+p, whereei is thei-th standard unit vector. Bythesegment of typei atx ∈ Γ we will mean(x+πV (σi), i). LetΣ := {(πV (y+σi), i) : y ∈ Zm+p, i =1, . . . , m + p} be the collection of allsegments. If γ = πV ◦ T̃ is a strand with vertices inΓ, wemay view the image ofγ as a collection of segments. Let’s say thatwe can get from segmentσ tosegmentσ′ in one stepif there is a strandγ with σ andσ′ in the image ofγ, and let’s saywe canget fromσ to σ′ in n stepsif there are segmentsσ0 = σ, . . . , σn = σ′ so that we can get fromσi−1

to σi in one step fori = 1, . . . , n.Ψβ induces a map that takes segments to collections of segments: if σ is a segment defined by

the image ofπV ◦ T̃ on the interval[t1, t2], thenΨβ(σ) is the collection of segments defined by theimage ofπV ◦ Ψ̃β(T̃ ) on the interval[βt1, βt2]. This map has the property that ifσ′ ∈ Ψβ(σ) thenΨβ(Σσ) ⊂ Σσ′ .Lemma 27. If β > 1 is Pisot then there isK ∈ N so that for eachσ, σ′ ∈ ∪k≥KΨkβ(Σ) there isn ∈ N so that we can get fromσ to σ′ in n steps.Proof. Given a segmentσ, letΣσ be the collection of allσ′ ∈ Σ for which there isn ∈ N such thatwe can get fromσ to σ′ in n steps. It is clear that:

(1) theΣσ partitionΣ,(2) for eachσ there areσ′, σ′′ ∈ Σσ so that the terminal vertex ofσ′ is the initial vertex ofσ

and the initial vertex ofσ′′ is the terminal vertex ofσ, and(3) if σ′ ∈ Ψβ(σ), thenΨβ(Σσ) ⊂ Σσ′ .

Let’s takeσ = (πV (σ1, 1)) so thatΨβ(Σσ) ⊂ Σσ. Note that ifβ is a simple Parry number (m=0),there isK ∈ N so that ifi ∈ {1, . . . , p} thenσ ∈ Ψkβ(πV ((σi, i))) for all k ≥ K; and ifβ is non-simple, it follows from Lemma 12 that there isK ∈ N so thatΨkβ(πV ((σi, i)))∩Ψkβ(πV ((σj , j))) 6=∅ for all i, j ∈ {1, . . . , m + p} andk ≥ K. From this (and items (1)-(3) above) it follows that if

20 M. BARGE

σ′ andσ′′ have a common vertex, then there isσ′′′ so thatΨKβ (Σσ′) ∪ΨKβ (Σσ′′) ⊂ Σσ′′′ with K asabove. From connectedness ofspt(Σ) := πV (R̃ψβ) we haveΨkβ(Σ) ⊂ Σσ for all k ≥ K. �

Given the segmentσ = (x+πV (σi), i), let’s callx+πV (σi) thesupportof σ (denotedspt(σ)) andi its type. We’ll say that the strandγ = πV ◦ T̃ parameterizesσ on [s, s′] if γ([s, s′]) = spt(σ) andT ([s, s′]) is thei-th loop ofRψβ . Note that ifγ parameterizesσ on [s, s′], thenγ− t parameterizesσ on [s− t, s′ − t]. Suppose now that we can get fromσ to σ′ in n steps. There are then segmentsσ = σ0, . . . , σn = σ′ and strandsγ1, . . . , γn so that (on some intervals)γi parameterizes bothσi−1

andσi for i = 1, . . . , n. By replacingγi by γi−ti, with the appropiately chosenti for i = 2, . . . , n,we may arrange that:γ1 parameterizesσ0 on [s0, s′0] andσ

1 on [s1, s′1]; γ2 parameterizesσ1 on

[s1, s′1] and σ

2 on [s2, s′2]; ... ; andγn parameterizesσn−1 on [sn−1, s′n−1] and σn on [sn, s

′n].

Then if γi = xi + πV ◦ S̃i, the tilingsSi have the property:Si(t) = Si+1(t) for t ∈ [si, s′i] andi = 1, . . . , n− 1.Lemma 28. Suppose thatT, T ′ ∈ Ωψβ and there isn ∈ N so that, for eachR > 0, there areSi = Si(R) ∈ Ωψβ , i = 1, . . . , n, andti = ti(R) ∈ R, i = 1, . . . , n + 1, so that for all|t| < R:(T − t1)(t) = (S1− t1)(t); (Si− ti+1)(t) = Si+1− ti+1)(t), i = 1, . . . , n−1; and(Sn− tn+1)(t) =(T ′ − tn+1)(t). ThenT ∼srp T ′.Proof. Let ǫ > 0 be given. There is thenR > 0 so that ifS, S ′ ∈ Ωψβ are such thatS(t) = S ′(t) for|t| < R, thend(πmax(S), πmax(S ′)) < ǫ/(n+ 1). SinceR acts by isometries on̂Tdβ andπmax(S −t) = πmax(S)− t, we have, for suchS, S ′, thatd(πmax(S − t), πmax(S ′ − t)) < ǫ/(n + 1) for allt ∈ R. For thisR, let theSi(R) andti(R) be as hypothesized. Then:d(πmax(T ), πmax(S1)) <ǫ/(n + 1); d(πmax(Si)πmax(Si+1)) < ǫ/(n + 1), i = 1, . . . , n− 1; andd(πmax(Sn), πmax(T ′)) <ǫ/(n+ 1). Henced(πmax(T ), πmax(T ′)) < ǫ, soπmax(T ) = πmax(T ′) andT ∼srp T ′.

�

Proof. (Of Theorem 24) Suppose thatT, T ′ ∈ Ωψβ with T ∼rp T ′. There are then correspondingstrandsγ, γ′ that are strong regionally proximal. By Lemma 26,Ψkβ(γ) ∼srp Ψkβ(γ′) for all k ∈ Zand by so Lemma 25,T ∼gs T ′.

Now suppose thatT ∼gs T ′. LetB t0(S) :

S(t) = ∗}. As in Lemma 22 we lett∗,j = t∗(Ψjβ(T )) andt′∗,j = t∗(Ψjβ(T ′)); correspondingly, wesett1,j = t1(Ψ

jβ(T )) andt

′1,j = t1(Ψ

jβ(T

′)). For each(k, i) ∈ N2 let γ(k,i) andγ′(k,i) be the strands:

γ(k,i) := πV ◦ Ψ̃kβ(T̃i)and

γ′(k,i) := πV ◦ Ψ̃kβ(T̃ ′i ).Thenγ(k,i) parameterizes a segment, call itσ(k,i), on [t∗,k−i, t1,k−i] andγ′(k,i) parameterizes a seg-mentσ′(k,i) on [t

′∗,k−i, t

′1,k−i]. We observe:

(1) The pair of segments(σ(k,i), σ′(k,i)) depends, up to translation by an element ofΓ, only onk − i.

(2) σ(k,i), σ′(k,i) ∈ Ψkβ(Σ), so ifK is as in Lemma 27 andk ≥ K, then there isn so that we canget fromσ(k,i) to σ′(k,i) in n steps.

THE PISOT CONJECTURE 21

From (1) it follows that there areK ≤ i1 < i2 < i3 · · · so that the pairs(σ(K,ij), σ′(K,ij)) are allthe same, up to translation by elements ofΓ, and then, by (2), there isn, independent ofj, so thatwe can get fromσ(K,ij) to σ

′(K,ij)

in n steps.Fix j ≥ 1 and letσ0 = σ(K,ij), σ1, . . . , σn = σ′(K,ij) be such that we can get fromσi−1 to

σi in one step fori = 1, . . . , n. There are then tilingsSi, i = 1, . . . , n, and intervals[si, s′i],i = 0, . . . , n, with [s0, s′0] = [t∗,K−ij , t1,K−ij ], and liftsS̃i so that the strandπV ◦ S̃i parameterizesσi−1 on [si−1, s′i−1] andσ

i on [si, s′i], i = 1, . . . , n.

Claim: [sn, s′n] = [t′∗,K−ij

, t′1,K−ij ].To see this, note that for any strandγ, we have:πu(γ(t)) = πu(γ(0)) + tω for all t ∈ R. Thus, ifγ andγ′ are strands for whichπu(γ(t)) = πu(γ′(t)) for somet thenπu(γ(t)) = πu(γ′(t)) for allt. Hence,πu(πV ◦ Ψ̃Kβ (T̃ij )(t)) = πu(πV ◦ S̃i(t)) for all t ∈ R andi = 1, . . . , n. We also knowthatπV ◦ Ψ̃Kβ (T̃ij )(0)− πV ◦ Ψ̃Kβ (T̃ ′ij )(0) ∈ Es (if γ andγ′ are such thatγ(0)− γ′(0) /∈ Es, then|Ψnβ(γ)(0)−Ψnβ(γ′)(0)| grows without bound asn→∞, and then so also does|an(γ)− an(γ′)|).Thusπu(πV ◦ Ψ̃Kβ (T̃ij )(t)) = πu(πV ◦ Ψ̃Kβ (T̃ ′ij )(t)) for all t. NowπV ◦ Ψ̃Kβ (T̃ ′ij ) andπV ◦ S̃n param-eterize the same segmentσ′(K,ij) = σn on the intervals[t

′∗,K−ij

, t′1,K−ij ] and [sn, s′n], resp., while

πu(πV ◦ Ψ̃Kβ (T̃ ′ij )(t)) = πu(πV ◦ S̃n)(t) for all t, so these intervals must be the same, as claimed.

If r = min{length(τi) : i = 1, . . . , m + p}, then each of the intervals[si, s′i] occurring abovehas length at leastr. GivenR > 0, let j be large enough so thatβij−Kr > 2R. The hypothe-ses of Lemma 28 are then satisfied, withΨij−Kβ (Si) playing the role ofSi in the lemma, and

ti(R) := (si−1+s′i−1

2)βij−K . �

Corollary 29. π̂ = πmax : Ωψβ → T̂d ≃ T̂dβ.

To connect the foregoing with arithetical coding, letΠ : X+β → I by Π((xi)) :=∑∞

i=1 xiβ−i.

ThenΠ ◦ σ = Tβ ◦ Π, Π is a.e. one-to-one, and so iŝΠ : xβ → lim←−Tβ . (The unique measureη of maximal entropy forTβ is equivalent to Lebesgue measure ([Hof1, Hof2]), forlim←−Tβ weuse the measure induced byη.) The map fromΩψβ to [0, 1) given byT 7→ −t∗(T ) is surjectiveand from−t∗(Ψβ(T )) = Tβ(−t∗(T )), there is an induced map̂−t∗ : Ωψβ → lim←−Tβ. The imageof −̂t∗ is lim←−Tβ|[0,1), which is a full measure subset oflim←−Tβ . For T ∈ Ωψβ and i ∈ Z, letxi(T ) := ⌊−βt∗(Ψiβ(T ))⌋. Then(xi(T )) ∈ Xβ and−̂t∗(T ) = Π̂((xi(T )). It follows that{xi(T )) :T ∈ Ωψβ} has full measure inXβ.

Recall thatω is the positive right eigenvector ofL, normalized with〈lt, ω〉 = 1, l = (l1, . . . , lm+p)being the left eigenvector of the substitution matrixA with li = length(τi) (note that

∑m+pi=1 li =

1). Thenπu(πV (ei)) = liω for eachi ∈ {1, . . . , m + p} and sinceω is ‘totally irrational’ in V ,πu : Γ → Eu is one-to-one and we haveΓu := πu(Γ) = 〈l1ω, . . . , lm+pω〉Z = 〈l1, . . . , lm+p〉Zω.We claim that〈l1, . . . , lm+p〉Z = Z[β]. To see this, recall the notationzj = T j−1β (1). Clearly,zj ∈ Z[β]. Since the endpoints of theτi are contained in{0} ∪ {zj : j = 1, . . . , m + p},we have〈l1, . . . , lm+p〉Z ⊂ Z[β]. On the other hand, for eachj = 1, . . . , d ≤ m + p, there isi ∈ {1, . . . , m + p} with max(τi) = zj . We havez1 = 1 and , forj ≥ 2, zj = βj−1 + qj−2(β),whereqj−2(x) ∈ Z[x] has degreej−2. Thenβj = zj+1−qj−1(β) = l1+l2+· · ·+lj+1−qj−1(β) ∈〈l1, . . . , lm+p〉Z for all j, inductively. We thus haveΓu = Z[β]ω.

22 M. BARGE

Now let e := −πV (∑m+p

i=1 ei) ∈ Γ. From πu ◦ L = βπu and the above claim, we see thate, Le, . . . , Ld−1e generatesΓ, that is,ȳ := y + Γ, y := eu, is a fundamental homoclinic point forFL : V/Γ→ V/Γ.

For eachi ∈ Z andT ∈ Ωψβ let pi(T ) := p(Ψiβ(T )) = γ(t∗(Ψiβ(T )))− γ(βt∗(Ψi−1β (T )), γ anystrand corresponding toΨiβ(T ), be the abelian prefix ofΨ

iβ(T ) and letai := ai(γT ). Then(ai) is

bounded andai = pi+Lai−1 for all i ∈ Z. We havea0 = p0+La−1 = p0+Lp−1+L2a−2 = · · · .Since theai are bounded,

as0 =∞∑

i=0

Lips−i,

Similarly,

au0 = −∞∑

i=1

L−ipui .

Sincepi(T ) ∈ Γ, Lips−i ≡ −Lipu−i mod (Γ) for i ≥ 0, and we have

a0 = a0(T ) ≡∞∑

i=−∞

Li(−pu−i(T )), mod (Γ).

Thus, sincepui (T ) = xi(T )ω,

π(T ) = a0(γT ) + Γ = −∞∑

i=−∞

(xi(T )β−iω + Γ).

On the other hand, with fundamental homoclinic pointȳ = eu = −ω + Γ (and ˆ̄y = (−ω +Γ,−β−1ω + Γ, . . .)), we havehȳ((xi(T )) =

∑∞i=−∞ xi(T )F̂

−iL (−ω + Γ,−1/βω + Γ, . . .) =

−(∑∞−∞(xi(T )β−iω + Γ),∑∞

i=−∞(β−i−1ω + Γ), . . .). Thus we see that

π̂(T ) = hȳ((xi(T )))

for all T ∈ Ωψβ .From Theorem 15 and Corollary 29,hȳ, restricted to{(xi(T )) : T ∈ Ωψβ} is a.e one-to-one.

Since{(xi(T )) : T ∈ Ωψβ} has full measure inXβ, we have proved Corollary 18.We finish with an argument that a stronger formulation than Property (W) is equivalent to pure

discrete spectrum of(Ωψβ ,R). ForT ∈ Ωψβ we writeW s(T ) for {T ′ ∈ Ωψβ : d(Ψnβ(T ′),Ψβ(T ))→0 asn → ∞} and for ˆ̄z ∈ T̂dβ, W s(ˆ̄z) := {ˆ̄z′ : d(F̂ nM(ˆ̄z′), F̂ nM(ˆ̄z)) → 0 asn → ∞}. (HereF̂M : T̂dβ → T̂dβ is the hyperbolic automorphism induced on the maximal equicontinuous factor ofΩψβ .)

Lemma 30. If (Ωψβ ,R) has pure discrete spectrum andT, T′ ∈ Ωψβ are such thatT ∼s T ′, then

{t : T − t ∼s T ′ − t} is open and dense inRProof. We have previously observed thatT − t ∼s T ′ − t is an open property. IfT ∼s T ′ thend(πmax(Ψ

kβ(T − t)), πmax(Ψkβ(T ′ − t))) = d(F̂ kM((πmax(T ))− βkt, F̂ kM (πmax(T ′))− βkt)→ 0 as

k → ∞ uniformly in t. Suppose there is an intervalJ = (t0 − ǫ, t0 + ǫ) with T − t ≁s T ′ − tfor all t ∈ J . Takeki → ∞ so thatΨkiβ (T − t0) → S ∈ Ωψβ andΨkiβ (T ′ − t0) → S ′ ∈ Ωψβ .ThenS ∼srp S ′ (sinced(πmax(S), πmax(S ′)) = 0) andS ∩ S ′ = ∅ (if B0[S − t] = B0[S ′ − t] forsomet thenB0[Ψ

kiβ (T − (t0 − t/βki))] = B0[Ψkiβ (T ′ − (t0 − t/βki))] for all largei, contradicting

THE PISOT CONJECTURE 23

T − t′ ≁s T ′− t′ with t′ = t0 − t/βki ∈ J). But then the coincidence rank ofψβ is at least two, so(Ωψβ ,R) does not have pure discrete spectrum. �

A few observations:

(1) For anyT, T ′ ∈ Ωψβ , {t : T ′ − t ∈ W s(T )} is dense inR. (This is a consequence of theprimitivity of ψβ.)

(2) d(tβkω,Γ) → 0 ask → ∞ if and only if t ∈ Z[1/β]. Hence, ifˆ̄0 = (0̄, 0̄, . . .), ˆ̄0 − t ∈W s(ˆ̄0) if and only if t ∈ Z[1/β].

(3) −t∗(Ψβ(T )) = Tβ(−t∗(T )) for all T ∈ Ωψβ .(4) Sinceπ−1max(

ˆ̄0) is finite ([BBK], or [BK]) andΨβ-invariant there isN > 0 so thatΨNβ (T ) =

T for all T ∈ π−1max(ˆ̄0).(5) GivenT, T ′ ∈ Ωψβ with T ′ ∈ W s(T ), there isǫ > 0 so thatT ′ − t ∈ W s(T − t) for

all |t| < ǫ. (This is a consequence of ‘local product structure’ - see [AP]. Or, invoke theopenness of the relation∼s, which is proved in [BO].)

Proposition 31. If β is Pisot, then(Ωψβ ,R) has pure discrete spectrum if and only if for allt ∈Z[1/β] ∩ R+ the set{t′ ∈ Fin(β) : t+ t′ ∈ Fin(β)} is dense inR+.

Proof. Suppose that(Ωψβ ,R) has pure discrete spectrum and pickt ∈ Z[1/β] ∩ R+. For giveni ∈ {1, . . . , p}, there isT ∈ π−1max(ˆ̄0) so thatT i0 − t ∈ W s(T ) (this follows from item (2)). Now,{t′ > 0 : T i0 − t′ ∈ W s(T i0)} is dense inR+ (from (1)) and{t′ ∈ R : T i0 − t′ ∼s T − t′} is openand dense inR (Theorem 9). ThenC := {t′ > 0 : T − t′, T i0− t′ ∈ W s(T i0)} is dense itR+ and alltheset′ are also inFin(β). Indeed, ift ≥ 0 andT i0 − t′ ∈ W s(T i0), let k be large enough so thatt′/βkp < 1. ThenT kp+nβ (t

′/βkp) = T kp+nβ (−t∗(T i0 − t′/βkp)) = −t∗(Ψnβ(T i0 − t′)) = 0 for n largeenough thatτ1 ∈ Ψnβ(T i0 − t′), sot′/βkp, and hencet′, is inFin(β). We haveT i0 − t ∼s T , so byLemma 30, the setD := {t′ : T i0 − t− t′ ∼s T − t′} is open and dense inR. NowC ∩D is densein R+ and fort′ ∈ C ∩ D we haveT i0 − t − t′ ∼s T − t′ ∼s T i0, sot + t′ ∈ Fin(β) (as above).Thus if t′ ∈ C ∩D thent′ andt+ t′ are both inFin(β).

Now suppose thatF (t) := {t′ ∈ Fin(β) : t + t′ ∈ Fin(β)} is dense inR+ for all t ∈Z[1/β]∩R+ and letT ∈ π−1max(ˆ̄0). Fix i ∈ {1, . . . , p} and lett ∈ R+ be so thatT i0 − t ∼s T . Thent ∈ Z[1/β]. Let ǫ > 0 be small enough so that if|t′| < ǫ, thenT i0 − t − t′ ∼s T − t′. Note that ift′ ∈ Fin(β), thent′/βkp ∈ Fin(β) and we have0 = T n+kpβ (t′/βkp) = T n+kpβ (−t∗(T i0−t′/βkp)) =−t∗(Ψnβ(T i0 − t′)) for largen andk big enough so thatt′/βkp ∈ [0, 1). Thus,T i0 − t′ ∈ W s(T j0 ) forsomej ∈ {1, . . . , p}. Then fort′ ∈ F (t) ∩ [0, ǫ) we haveT i0 − t′ ∼s T j0 , T i0 − (t + t′) ∼s T − t′andT i0 − (t + t′) ∼s T k0 for somej, k ∈ {1, . . . , p}. SinceT j0 − t′′ ∼s T k0 − t′′ for all t′′ > 0,we see thatT i0 ∼s T densely on[0, ǫ). Using theΨβ-periodicity of T i0 andT , this means thatW (T ) := {t′ > 0 : T i0 − t′ ∼ T − t′} is open and dense inR+. Thus, for eachT, T ′ ∈ π−1max(ˆ̄0),W (T ) ∩W (T ′) 6= 0, and thenT ∩ T ′ 6= ∅. Thus the coincidence rank ofψβ is 1 and(Ωψβ ,R) haspure discrete spectrum.

�

It follows from Theorem 15 and Proposition 31 that all Pisotβ are weakly finitary, provingCorollary 17.

24 M. BARGE

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DEPARTMENT OFMATHEMATICS, MONTANA STATE UNIVERSITY, BOZEMAN, MT 59717-0240, USA.E-mail address: barge@math.montana.edu

http://arxiv.org/abs/0802.3242http://arxiv.org/abs/1501.05328

1. Introduction.2. Background on -numeration and substitution tilings.3. Arithmetical coding and Property (W)4. Proof of the Pisot Conjecture for -substitutions.5. The connection between pure discrete spectrum, arithmetical coding, and Property (W).References