Factors of Pisot tiling spaces and the coincidence rank conjecture

0.23 MB·English

by  Marcy Barge

FACTORS OF PISOT TILLNG SPACES AND THE COINCIDENCE RANK
CONJECTURE
MARCYBARGE
3
1
ABSTRACT. WeconsiderthestructureofPisotsubstitutiontilingspaces,inparticular,thestructure
0
ofthosespacesforwhichthetranslationactiondoesnothavepurediscretespectrum. Suchaspace
2
is always a measurable m-to-onecoverof an action by translation on a groupcalled the maximal
n
equicontinuous factor. The integer m is the coincidence rank of the substitution and equals one
a
J if and only if translation on the tiling space has pure discrete spectrum. By considering factors
9 intermediate between a tiling space and its maximal equicontinuous factor, we establish a lower
2 bound on the cohomology of a one-dimensional Pisot substitution tillng space with coincidence
ranktwoanddilationofoddnorm. TheCoincidenceRankConjecture,forcoincidenceranktwo,is

]
acorollary.
S
D
.
h
1. INTRODUCTION
t
a
m A subset D of Rn is a Delone set if it is relatively dense and uniformly discrete. Such a set has
[ finite local complexity (FLC) if, for each r > 0, there are, up to translation, only finitely many

subsets of D of diameter less than r and is repetitive if, for each r there is an R so that every
1
v Euclidean ball of radius R intersected with D contains a translated copy of every subset of D of
4
diameter less than r. Repetitive FLC Delone sets (RFLC) are commonly employed as models of
9
0 the atomic structure of materials and there is a well-developed diffraction theory to go with these
7 models ([BMRS],[L], [Le]). Of course if a Delone set is a periodic lattice, then its diffraction
.
1 spectrum is pure point. But there is also a vast menagerie of non-periodic repetitive FLC Delone
0
setswithpurepointdiffractionspectra(theso-calledquasicrystals)and itisa verysubtleproblem
3
1 topredict whichsets willhavethisproperty.
v: An effective method of encoding the structure of a single Delone set D is to form the space,
i calledthehullofDanddenotedΩ ,consistingofallDelonesetsthatare,uptotranslation,locally
X D
indistinguishablefrom D. The hull has a metric topology in which two Delone sets are close if a
r
a small translate of one is identical to the otherin a large neighborhood of the origin. If D is RFLC
thenΩ iscompactandconnectedandRn actsminimallyonΩ bytranslation. Ahighlightofthis
D D
approachisthat(inthesubstitutivecontextconsideredhere)D haspurepointdiffractionspectrum
ifand onlyiftheRn-actionon Ω has purediscretedynamicalspectrum ([D],[LMS]).
D
In this article we consider Delone sets that arise from a substitution rule. In this case, the
eigenfunctions of the Rn-action on Ω can all be taken continuous ([S1]) and then collectively
D
determine a continuous map g factoring the Rn action on Ω to a translation action on a compact
D
abeliangroup. Themapg iscalledthemaximalequicontinuousfactormapanditturnsoutthatthe
Rn-actionon Ω has purediscretespectrum ifand onlyif g isa.e. one-to-one([BK]).
D
Date:October14,2012.
2000MathematicsSubjectClassification. Primary68R15&05D10.
Keywordsandphrases. Pisotsubstitution,tilingspace.
1
2 M.BARGE
Two ingredients make up a substitution: a linear expansion Λ : Rn → Rn; and a subdivision
rule. FortheRn-actiononΩ tohaveanydiscretecomponentinitsspectrum,theeigenvaluesofΛ
D
mustsatisfyaaPisotfamilycondition([LS]). Inparticular,ifΛissimplyadilationbythenumber
λ > 1 (the “self-similar” case), the Rn-action will have a discrete component if and only if λ is a
Pisotnumber. (RecallthataPisotnumberisanalgebraicinteger,allofwhosealgebraicconjugates
lie strictly inside the unit circle.) For the purpose of determining which substitution Delone sets
D have pure point diffraction spectrum, we are thus led to consider factors of the spaces Ω for
D
whichtheexpansionis Pisotfamily.
Forconvenience,wewillconsidersubstitutiontilingsratherthansubstitutionDelonesets. (These
notionsareessentiallyequivalent-see[LMS]and[LW].) Atileisacompact,topologicallyregular
subset of Rn (perhaps “marked”); fora given substitution,there are only finitely many translation
equivalence classes of tiles and each such class is a type. A substitution expands tiles by a linear
mapΛ,thenreplacestheexpandedtilebyacollectionoftiles. Thesubstitutionmatrixisthed×d
matrix M (d being the number of distinct tile types) whose ij−th entry is the number of tiles of
type i replacing an inflated tile of type j and the tiling space associated with a substitution Φ is
the collection Ω of all tilings T of Rn with the property that each finite patch of T occurs as a
Φ
sub patch of some repeatedly inflated and substitutedtile. To pass from a tiling T to a Delone set
D, simply pick a point from each tile in T: if the points picked are “control points”(see [LS]), D
will be a substitution Delone set with linear expansion Λ. For a thorough treatment of the basics
ofsubstitutiontilingspaces, see[AP].
A 1-dimensional Pisot substitution is irreducible if the characteristic polynomial of its substi-
tution matrix is irreducible over Q, and is unimodular if that matrix has determinant ±1. The
followinghas becomeknownas thePisotSubstitution Conjecture:
Conjecture 1. (PSC) If Φ is a 1-dimensional irreducible, unimodular, Pisot substitution then the
R-actionon Ω haspurediscretespectrum.
Φ
The dimension of the first (Cˇech , with rational coefficients) cohomology of a 1-dimensional
substitutiontilingspace is at least as large as thedegree oftheassociated dilation. Replacing irre-
ducibilityinthePSCbyminimalityofcohomologyresultsintheHomologicalPisotConjecture:
Conjecture2. (HPC)IfΦisa1-dimensionalsubstitutionwithdilationaPisotunitofdegreedand
the first Cˇech cohomology H1(Ω ) has dimension d, then the R-action on Ω has pure discrete
Φ Φ
spectrum.
As noted above for substitution Delone sets, the Rn-action on Ω has pure discrete spectrum
Φ
if and only if the maximal equicontinuous factor map g is a.e. one-to-one. It is proved in [BK]
that, for Pisot family Φ, there is cr(Φ) ∈ N (called the coincidence rank of Φ) so that g is a.e.
cr(Φ)-to-one. The following Coincidence Rank Conjecture ( see [BBJS]) extends the HPC to
thenon-unitcase.
Conjecture 3. (CRC) If Φisa 1-dimensionalsubstitutionwithPisotdilationλandthedimension
ofH1(Ω ) equalsthedegree ofλ, thencr(Φ) dividesthenormof λ.
Φ
FordiscussionsofthevariousPisotconjectures,and extensionstohigherdimensions,see[BS],
[BBJS], [BG], and,particularly,[ABBLSS].
Conjecture 3 is established in [BBJS] for λ of degree 1. The main result of this article verifies
Conjecture 3 for λ of any degree in case the coincidence rank is 2. We prove (Theorem 22 in the
text):
FACTORSOFPISOTTILINGSPACES 3
Theorem: Suppose that Φ is a 1-dimensional substitution with Pisot dilation λ of degree d. If
thenormof λ isoddand cr(Φ) = 2, thendim(H1(Ω )) ≥ 2d−1.
Φ
Toprovetheabove,wewillshowthatthemaximalequicontinuousfactormap
g
Ω → T
Φ max
for a Pisot family tiling space Ω with coincidence rank 2 (and with no almost automorphic sub
Φ
actions)factors as
Ω →πs Ω →πp Ω →g′ T
Φ s p max
withΩ andΩ alsosubstitutiontilingspaces,π and g′ measureisomorphisms,and π atopolog-
s p s p
ical2-to-1coveringprojection. Thespace Ω isthemaximalpurediscretefactorofΩ -examples
p s
havebeenconsideredin[BGG]whereitsutilityinanalyzingthenon-discretepartofthespectrum
ofΩ isexplored. Restrictingtodimension1,themap π inducesan injectionincohomologyand
Φ s
the lower bound in the above theorem follows from arithmetic associated with the double cover.
The number two seems to be quite special in this argument; it would be extremely interesting to
seean extensionto highercoincidencerank.
2. PRELIMINARIES
Inthisarticle,allsubstitutionswillbeprimitive(somepowerofthesubstitutionmatrixisstrictly
positive), non-periodic (no element of the tiling space has a non-zero translational period), and
FLC. If Φ is an n-dimensionalsubstitution,the tilingspace Ω with the tilingmetricis then com-
Φ
pact, connected, and locally the product of a Cantor set with an n-dimensional disk. The action
of Rn on Ω is minimal and uniquely ergodic, the substitution induced map Φ : Ω → Ω is a
Φ Φ Φ
homeomorphism, and Φ(T − v) = Φ(T) − Λv for all T ∈ Ω and v ∈ Rn, Λ being the linear
Φ
inflationassociatedwithΦ.
If φ : A = {1,...,m} → A∗ is a symbolic substitution with matrix M (the ij-th entry is the
number of i’s occurring in φ(j)), there is an associated 1-dimensional tile substitution Φ on a set
ofmprototiles: thei-thprototileistheinterval[0,ω ],markedwiththesymboli(formally,thei-th
i
prototile is the pair ([0,ω ],i)), where ω = (ω ,...,ω ) is a positiveleft eigenvectorfor M. The
i 1 m
linearexpansionfor Φ isthePerron-Frobenius eigenvalue λ ofM: ωM = λω.
We will say that the substitution tiling spaces Ω and Ω are isomorphic, and write Ω ≃
Φ Ψ Φ
Ω , if there is a homeomorphism of Ω with Ω that conjugates both the Rn-actions and the
Ψ Φ Ψ
substitutioninduced homeomorphisms. The stableset of T ∈ Ω is the set Ws(T) = {T′ ∈ Ω :
Φ Φ
d(Φn(T′)Φn(T)) → 0as n → ∞}. We write T ∼ T′ if T′ ∈ Ws(T) and we say that T and T′
s
are stablyequivalent.
The eigenvalues of the linear expansion Λ of a substitution are necessarily algebraic integers.
Given k ∈ N and α ∈ C, let n(k,α) denote the number of k × k Jordan blocks in the Jordan
form of Λ that have eigenvalue α. Following [LS] and [K], we will say that Φ is a Pisot family
substitution if the eigenvalues, spec(Λ), of Λ satisfy the condition: if λ ∈ spec(Λ) and η is an
algebraic conjugate of λ with |η| ≥ 1, then n(k,η) = n(k,λ) for all k ∈ N. A substitution is
self-similarifits expansionis a scalar, Λ = λI: such a substitutionis Pisot family if and onlyif λ
isaPisot numberand wewillcall suchsubstitutionssimplyPisotsubstitutions.
4 M.BARGE
Rather generally, group actions on compact spaces have maximal equicontinuous factors. For
theRn-actionon asubstitutiontilingspaceΩ , themaximalequicontinuousfactorisa Kronecker
Φ
actionon asolenoidalgroup(an inverselimitoflinearhyperbolicendomorphismsoftori):
g : Ω → T .
Φ max
The“maximality”propertyof g isthatiff : Ω → X isanymapthatfactorstheRn-actiononΩ
Φ Φ
ontoanequicontinuousRn-actiononX,thenf factorsthroughg: thereisacontinuousfactormap
h : T → X withf = h◦g. IfΦisan n-dimensionalPisotsubstitutionwithexpansionλofal-
max
gebraicdegreed,T hasdimensionnd. Ifλisaunit,thenT isjustthend-torus. Forgeneral
max max
abelian group actions on compact spaces, the equicontinuous structure relation (the quotient by
whichgivesthemaximalequicontinuousfactor)istheregionalproximalrelation. ForPisotfamily
substitutiontilingspaces, thisrelationtakes thestrongerformgivenby(6)ofTheorem 4 below.
The following theorem 1 collects results that we will require later. We denote by B [T] the
r
collection of all tiles of T whose supports meet the closed ball of radius r centered at 0 and we
definethecoincidencerankofΦtobe
cr(Φ) = min{♯g−1(z) : z ∈ T }.
max
(In[BK],thisiscalledtheminimalrankofg : Ω → T . In[BKw],cr(Φ)isthemaximumcar-
Φ max
dinalityofacollectionoftilingsinthesamefiberofg,notwoofwhichshareatile(i.e.,coincide).
Theminimalrank and coincidence rank are shownto bethesame forPisot familysubstitutionsin
[BK]-see(2)ofTheorem4 below.) The“a.e.” inthetheorem refers to Haarmeasureon T .
max
Theorem 4. Let Φ be a Pisot family substitution and let g : Ω → T be the map onto the
Φ max
maximalequicontinuousfactor. Then:
(1) g isfinite-to-oneanda.e. cr(Φ)-to-one.
(2) For each z ∈ T there are T ,...,T ∈ g−1(z) so that Φk(T ) ∩ Φk(T ) = ∅ for
max 1 cr(Φ) i j
i 6= j,and allk ∈ Z, andif T ∈ g−1(z) then T ∩T 6= ∅for somei.
i
(3) T ∼ T′ if andonlyif thereisk ∈ N withB [Φk(T)] = B [Φk(T′)].
s 0 0
(4) Up to translation, there are only finitely many pairs of the form (B [T]),B [T′]) with
0 0
g(T) = g(T′).
(5) TheRn-action onΩ haspurediscretespectrum⇐⇒ g is a.eone-to-one⇐⇒cr(Φ) = 1.
Φ
(6) g(T) = g(T′) if and only if for every r > 0 there are S ,S′ ∈ Ω and v ∈ Rn so that
r r Φ r
B [T] = B [S ], B [T′] = B [S′], andB [S −v ] = B [S′ −v ].
r r r r r r r r r r r r
(7) If T −v ∼ T′ −v forallv ∈ Rn, thenT = T′.
s
Proof. (1)gisfinite-to-onebyTheorem5.3of[BK]anda.e. cr(Φ)-to-onebytheproofofTheorem
2.25of[BK].
(2)FromTheorem5.4andLemma2.14of[BK],thereisδ > 0sothat,foreachz ∈ T ,there
max
areT ,...,T ∈ g−1(z)withinf d(T −v,T −v) ≥ δ fori 6= j. Moreover,ifT ∈ g−1(z),
1 cr(Φ) v∈Rn i j
theninf d(T −v,T −v) = 0forsomei ∈ {1,...,cr(φ)}. If,foreverysuch{T ,...,T },
v∈Rn i 1 cr(Φ)
therearei 6= j sothatΦk(T )∩Φk(T ) 6= ∅forsomek,then,forsufficientlylargektherewouldnot
i j
beT′,...,T′ ∈ g−1(z′) = Φk(g−1(z))withinf d(T′−v,T′−v) ≥ δ,fori 6= j,asrequired.
1 cr(Φ) v∈Rn i j
1The definition of Pisot family used in [BK], where the results on which Theorem 4 is based are established, is
narrowerthanthatusedinthepresentpaper. By[K]thoseresultsextendtothecontexthere.
FACTORSOFPISOTTILINGSPACES 5
Furthermore,if T ∈ g−1(z)weredisjointfromalltheT ,theninf d(T −v,T −v) > 0,by(4)
i v∈Rn i
of this theorem, and then, by minimality of the Rn-action, g would be at least (cr(Φ)+1)-to-one
everywhere,in contradictionto(1).
(3)SeeLemma3.6of[BO]
(4)SeeCorollary 5.8 of[BK].
(5)See[BK].
(6)Theconditionthatforeveryr > 0thereareS ,S′ ∈ Ω andv ∈ RnsothatB [T] = B [S ],
r r Φ r r r r
B [T′] = B [S′],andB [S −v ] = B [S′ −v ]iscalledstrongregionalproximalityin[BK]: that
r r r r r r r r r
thisisequivalentto g(T) = g(T′) followsfrom Theorems5.6and 3.4 ofthat paper.
(7)SeeLemma3.10of[BO]. (cid:3)
3. PISOT FACTORS
We will write T ≈ T′, provided {v : T − v ∼ T′ − v} is dense in Rn and g(T) = g(T′).
s s
It follows from (3) of Theorem 4 that {v : T − v ∼ T′ − v} is open and, from this, that ≈ is
s s
an equivalence relation. Since whether or not T ∼ T′ is entirely dependent on (B [T],B [T′]),
s 0 0
it follows from (4) of Theorem 4 that the relation ≈ is also closed. The cardinalities of the ≈ -
s s
equivalenceclassesareuniformlyboundedbycr(Φ). ItisclearthatT ≈ T′ ⇐⇒ Φ(T) ≈ Φ(T′)
s s
and T −v ≈ T′ −v forall v ∈ Rn. ThustheZ-and Rn-actionson Ω induceZ-and Rn-actions
s Φ
on thequotientΩ / ≈ .
Φ s
Theorem5. Ω / ≈ isthemaximalequicontinuousfactoroftheRn-actiononΩ ifftheRn-action
Φ s Φ
on Ω haspurediscretespectrum.
Φ
Proof. Suppose that the Rn-action on Ω has pure discrete spectrum and suppose that g(T) =
Φ
g(T′). If there are v and ǫ > 0 so that T −v −v ≁ T′ − v −v for each v ∈ B (v ), choose
0 0 s 0 ǫ 0
n → ∞ such that Φni(T − v ) → S ∈ Ω and Φni(T′ − v ) → S′ ∈ Ω . Then g(S) = g(S′)
i 0 Φ 0 Φ
and S ∩S′ = ∅. But thismeans thatthecoincidence rank of Φ is at least 2, so thespectrumofthe
Rn-actionisnotpurediscrete((5)ofTheorem4). Thustherearenosuchv ,ǫandT −v ∼ T′−v
0 s
foradensesetofv. Thusg(T) = g(T′) =⇒ T ≈ T′. Thatis,≈ istheequicontinuousstructure
s s
relationand Ω / ≈ is themaximalequicontinuousfactor.
Φ s
Conversely,supposethatthespectrumoftheRn-actiononΩ doesnothavepurediscretespec-
Φ
trum. Thencr(Φ) ≥ 2and,inparticular,therearetilingsT,T′ ∈ Ω thatareperiodicunderΦwith
Φ
g(T) = g(T′)andT ∩T′ = ∅. ButthenΦk(T)∩Φk(T′) = ∅forallk ∈ NsothatT −v ≁ T′−v
s
foranyv. Thusitisnotthecasethatg(T) = g(T′) =⇒ T ≈ T′,so≈ isnottheequicontinuous
s s
structurerelation and Ω / ≈ isnotthemaximalequicontinuousfactor.
Φ s
(cid:3)
WewillseethatifΦisa1-dimensionalPisotsubstitution,andtheR-actiononΩ doesnothave
Φ
pure discrete spectrum, then Ω / ≈ is also a 1-dimensional Pisot substitution tiling space. The
Φ s
moststraightforwardgeneralizationofthisisnottrueinhigherdimensionsasthefollowingsimple
exampleshows.
Example6. Let ψ betheThue-Morsesubstitution
ψ : a 7→ ab, b 7→ ba
andlet φ betheFibonaccisubstitution
φ : a 7→ ab, b 7→ a.
6 M.BARGE
The corresponding tile substitutions, Ψ and Φ, are Pisot and the product Ψ × Φ defines a 2-
dimensional Pisot family substitution on rectangular tiles. The relation ≈ is trivial on Ω while
s Ψ
Ω / ≈ isthe2-torus. Itis nothardto seethatΩ / ≈ ishomeomorphicwith Ω ×T2,which
Φ s Ψ×Φ s Ψ
isnot a2-dimensionalsubstitutiontilingspace.
Intheaboveexample,eventhoughtheR2-actionon thetillngspacedoesnothavepurediscrete
spectrum,thereare minimalsets under1-dimensionalsub-actionsthatdo havepurediscretespec-
trum. A difficultyin tryingto capturethissort ofphenomenonin ageneral settingisthat arbitrary
subactionsmightnotcarryuniqueinvariantmeasures,makingitdifficulttomeaningfullyspeakof
theirdynamicalspectra. Rather,wecanmeasurehowcloseasubactionistobeingequicontinuous.
TheminimalrankofaminimalRk-actiononaspaceX isthetheminimumcardinalityofafiber
g−1(g(x)) of the maximal equicontinuous factormap g : X → X for the action. Forthe full
X X max
Rn-action on an n-dimensionalPisotfamily substitutiontilingspacetheminimalrank isthesame
as the coincidence rank and hence equals 1 if and only if the spectrum is pure discrete (Theorem
4). Letus say thatthen-dimensionaltilingspace Ω
has an almost automorphic sub action if there is a nontrivial subspace V ⊂ Rn and a minimal
set X ⊂ Ω under the V-action Ω × V ∋ (T,v) 7→ T − v on which this V-action has minimal
rank 1. In Example 6, the V-action determined by V = {(0,t) : t ∈ R} on X := {T}×Ω has
Φ
minimal rank 1 for every T ∈ Ω . We will prove (Theorem 12), for general Pisot family Φ, that
Ψ
Ω / ≈ is asubstitutiontilingspaceifand onlyifΩ has noalmostautomorphicsub actions.
Φ s Φ
RecallthatasubstitutionΦforcesitsborderprovidedthereism ∈ Nsothatif0 ∈ τ ∈ T∩T′for
any T,T′ ∈ Ω , then B [Φm(T)] = B [Φm(T′)]. By, for example, introducing collared tiles, we
Φ 0 0
may always replace a substitutionby onethat forces its borderand has an isomorphictiling space
(see [AP]). The Anderson-Putnam complex, X , for Φ is an n-dimensional CW-complex whose
Φ
n-cells are theprototileswith prototilesglued along an n−1 face ifthey haverepresentativetiles
occurring in a tiling T ∈ Ω and intersecting along that face. More precisely, X = Ω / ∼ with
Φ Φ Φ
∼ the smallest equivalence relation for which 0 ∈ τ ∈ T ∩T′ =⇒ T ∼ T′. Let π : Ω → X
Φ Φ
bethe quotientmap. Thereis then a map f : X → X so that f ◦π = π ◦Φ and hence amap
Φ Φ Φ Φ
πˆ : Ω → limf that semi-conjugatesΦ with the shift homeomorphism fˆ : limf → limf . If
Φ ←− Φ Φ ←− Φ ←− Φ
Φforces itsborder, πˆ is ahomeomorphism([AP]).
Wewishtotransfertherelation ≈ toX . Let R betherelationon X defined by
s Φ Φ
′ ′ ′ ′ ′
xRx ⇐⇒ ∃T,T ∈ Ω withT ≈ T , π(T) = x, andπ(T ) = x.
Φ s
Therelation Risclearlyreflexiveandsymmetricandisalsotopologically closed(seetheproofof
Lemma 11), but it may not be transitive. And the smallest equivalencerelation containing R may
notbeclosed (fortheFibonaccisubstitution,all equivalenceclasses are dense).
Givenpatches(collectionsoftiles)P andP′ andx ∈ Rn,letuswriteP −x ∼ P′−xprovided
s
x is in the interior of the intersection of the supports of P and P′ and T − x ∼ T′ − x for all
s
(equivalently,some)tilingsT,T′ withP ⊂ T andP′ ⊂ T′. IfP −x ∼ P′−sforasetofxdense
s
in the interior of the intersection of the supports of P and P′, we’ll say that P and P′ are densely
stablyequivalentonoverlap.
Theorem 7. Suppose that Φ is an n-dimensional substitution of Pisot family type and that there
are 0 6= v ∈ Rn with v → 0 and a tile τ so that τ and τ −v are densely stably equivalent on
k k k
overlapforeach k. Then Ω has analmostautomorphicsubaction.
Φ
The proof of Theorem 7 will require some buildup. Given a nontrivial subspace V of Rn, let’s
say that∅ =6 X ⊂ Ω is V-minimalifX is aminimalsetforthe V-action.
Φ
FACTORSOFPISOTTILINGSPACES 7
Lemma8. IfΦisann-dimensionalsubstitutionofPisotfamilytypeandV isanontrivialsubspace
of Rn, then the union of all V-minimal sets, Y := ∪ X, has full measure with respect to
XV-minimal
theuniqueergodic(under theRn-action)measureon Ω .
Φ
Proof. Let(z,v) 7→ z−vdenotetheRn-actiononT ,sothatg(T−v) = g(T)−vforallT ∈ Ω ,
max Φ
v ∈ Rn. Foreach z ∈ T , the set ω(z) := cl{z −v : v ∈ V} is V-minimal. The set g−1(ω(z))
max
is closed and V-invariant, hence contains a V-minimal set, call it X . It must be the case that
z
g(X ) = ω(z). Thus g(Y) ⊂ g(∪ X ) = T . Let G = {z ∈ T : ♯g−1(z) = cr(Φ)}.
z z∈Tmax z max max
G has full measure in T by (1) of Theorem 4. It is proved in [BK] (see the proof of Theorem
max
2.25 there) that G′ := g−1(G) has full measure in ΩΦ and g|G′ : G′ → G is a topological cr(Φ)-
to-1 covering map. Furthermore, if U is a (nonempty) relatively open set of G evenly covered by
the relatively open sets U ,...,U in G′, then g| is a measure isomorphism of normalized
1 cr(Φ) Ui
measures foreach i. Since g(Y ∩g−1(U)) = U and U has positivemeasure, it followsthat Y has
positivemeasure. Note that if X is a V-minimal set and w ∈ Rn, then X −w is also V-minimal.
ThusY is Rn-invariantand by ergodicitymusthavefullmeasure. (cid:3)
GivenaV-minimalsetX inΩ ,letg : X → X denotethemaximalequicontinuousfactor
Φ X max
(with respect to the V-action). Since g| : X → g(X) ⊂ T is also an equicontinuous factor,
X max
g| factors through g : there is p : X → T with g| = p ◦g . That is, if T,T′ ∈ X
X X X max max X X X
are regionally proximal with respect to the V-action, then T and T′ are regionally proximal with
respect tothe Rn-action.
Lemma 9. SupposethatΦ is an n-dimensionalsubstitutionof Pisotfamilytype,V is a nontrivial
subspaceofRn,andX ⊂ Ω isaV-minimalset. IfT,T′ ∈ X areregionallyproximalwithrespect
Φ
totheV-action,thenthereareS,S′ ∈ X andv ∈ V sothatB [T] = B [S],B [T′] = B [S′], and
0 0 0 0
B [S −v] = B [S′ −v].
0 0
Proof. Let ρ ,...,ρ be the prototiles for Φ and choose c ∈ spt(ρ ) for each i. Let Ξ := {v =
1 m i i
(y −c )−(y −c ) : thereis T ∈ Ω withρ +y ,ρ +y ∈ T}. Then Ξ is a Meyerset ([LS]).
i i j j Φ i i j j
That is, Ξ − Ξ is uniformly discrete, as is Ξ ±Ξ ±···±Ξ for any choice of signs ([M]). Since
T and T′ are also regionallyproximalwithrespect to thefull Rn-action, thereare P,P′ ∈ Ω and
Φ
w ∈ Rn so that B [T] = B [P], B [T′] = B [P′], and B [P −w] = B [P′ −w] ((6) of Theorem
0 0 0 0 0 0
4). Supposethatρ +y ,ρ +y ,andρ +y aretilesinT,T′,resp. S−w,withsupportcontaining
i i j j k k
0. Then ((y + c ) − (y + c )) − ((y + c ) − (y + c )) = (y + c ) − (y + c ) ∈ Ξ − Ξ.
j j k k i i k k j j i i
Also,sinceT,T′ areregionallyproximalwithrespecttothe V-action,bydefinitionoftheregional
proximal relation, there are, for each k ∈ N, tilings Sk,Sk′ ∈ X, vk ∈ V and wk ∈ B1(0) with
B [T] = B [S ], B [T′] = B [S′], and B [S − v ] = B [S′ − v ] − w . From this wke deduce
0 0 k 0 0 k 0 k k 0 k k k
that (y +c )−(y +c )+w ∈ Ξ−Ξ. Thus the vectors w all lie in the uniformly discrete set
i i j j k k
Ξ−Ξ+Ξ−Ξ. As w → 0 it must be that w = 0 for large k: let S = S , S′ = S′, and v = v
k k k k k
forsuch k. (cid:3)
Proof. (Of Theorem 7) Let v ∈ Rn be as hypothesized and let E := ∩ span {v ,v ,...}.
k k R k k+1
There is then k so that E = span {v ,v ,...} and without loss of generality we may assume
R k k+1
k = 1. Nowchoosen → ∞sothatΛniE → V. Bythiswemeanthattherearebases{ei,...,ei}
i 1 b
forΛniE sothatei → e and{e ,...,e }isabasisforV. WewillshowthatthereisaV-minimal
j j 1 b
set X ⊂ Ω onwhich theV-actionis almostautomorphic.
Φ
For each δ > 0 there is a subset W = {w (δ),...,w (δ)} of {v ,v ,...} that spans E with
δ 1 b 1 2
|w (δ)| < δ foreach i. By translatingτ we may assume that there is r > 0 with B (0) contained
i 3r
8 M.BARGE
inthesupportofτ. Givenw ∈ Rn,letU := {x ∈ Rn : x,x+w ∈ int(spt(τ)) and τ −w−x ∼
w s
τ −x}. If τ −w −x ∼ τ −x, then τ −w −y ∼ τ −y forall y sufficiently near x, so the sets
s s
U areopen.
w
Forafixedδ,0 < δ < r,weseethatU isdenseinB (0)forallw ∈ W . Also,U = U +w
w 2r δ −w w
isdenseinB2r(0)forsuchw. Supposethatw′ issuchthat|w′| < r andUw′ isdenseinB2r(0)and
letw ∈ W . thenτ −w′−x ∼ τ −xforadenseopensetofxinB (0)andτ −w−x ∼ τ −x
δ s 2r s
for a dense open set of x in B (0). Thus τ − (w′ − w) − x ∼ τ − x for a dense set of x in
2r s
B2r(0) so that Uw′−w is dense in B2r(0). Also, U−w′ = Uw′ +w′ is dense in B2r(0), so U−w′−w is
dense in B2r(0) (as above) and hence Uw′+w = U−w′−w + (w′ + w) is dense in B2r(0), provided
|w′ + w| < r. In this way, we see that if w ,...,w is any sequence in W with the property
i1 ik δ
that ±w ± w ± ··· ± w ∈ B (0) for l = 1,...,k, then U is dense in B (0), where w =
i1 i2 il r w 2r
±w ±···±w . This set of w contains all points of the lattice span W that lie in B (0). Thus
i1 ik Z δ r
wehaveestablishedthat :
(3.1) {w ∈ E : U isdenseinB (0)} isdenseinE ∩B (0).
w r r
ByLemma8andthefactthat{T : ♯g−1(g(T)) = cr(Φ)}hasfullmeasure,thereisaV-minimal
set X and a T ∈ X so that ♯g−1(g(T)) = cr(Φ). We now argue that if g : X → X is the
X max
maximalequicontinuousfactormapfortheV-actiononX,theng−1(g (T)) = {T}. Supposethat
X X
g (T′) = g (T). ThenT′ isregionallyproximalwith T underthe V-actionsothat, byLemma9,
X X
thereareS,S′ ∈ X andv ∈ V withB [T] = B [S],B [T′] = B [S′],andB [S−v] = B [S′−v].
0 0 0 0 0 0
Fix R > |v|. By repetitivity of patches, there is R′ large enough so that if P is any tiling in Ω ,
Φ
thenBR′[P]containstranslatesofBR[S]andBR[S′]. Forsufficientlylargei,ΛniBr(0) ⊃ B2R′(0).
Sinceτ −x ∼ τ =⇒ Φni(τ)−Λnix ∼ Φni(τ), wehavefrom 3.1abovethat:
s s
(3.2) {w ∈ ΛniE : Uwi isdenseinB2R′(0)}isdensein ΛniE ∩B2R′(0)
where Ui := {x ∈ Rn : x,x+w ∈ int(spt(Λniτ)) and Φni(τ)−w −x ∼ Φni(τ)−x}.
w s
Tosimplifynotation,letB := B [S],B′ := B [S′],andQ := B [S−v]+v = B [S′−v]+v.
R R 0 0
Forlarge ithereare then y = y and y′ = y′ so thatthepatchesB +y and B′ +y′ are subpatches
i i
of Φni(τ) and have supports contained in BR′(0). Note that y + v ∈ int(spt(Q + y)). Since
ΛniE → V, we may take i large enough so there is w ∈ ΛniE ∩BR′(0) close enough to v so that
y + w ∈ int(spt(Q + y)); that is, w ∈ int(spt(Q)). Moreover, by 3.2, we can choose this w so
thatUwi isdenseinB2R′(0). SinceUwi isalsoopenand|y|,|y′| ≤ R′,(Uwi −y)∩(Uwi −y′)isdense
inBR′(0)and thereis x sothat:
(1) x ∈ Ui −y,
w
(2) x ∈ Ui −y′,
w
(3) x ∈ int(spt(B [T])),
0
(4) x ∈ int(spt(B [T′])),and
0
(5) x ∈ int(spt(Q−w)).
Let P := Φni(τ). Since Q + y and Q + y′ are both patches in P and 0 ∈ int(spt(Q − w)), we
havefrom(5)that:
′
P −w −y −x ∼ P −w−y −x.
s
From (1)weget:
P −y −x ∼ P −w−y −x
s
and from(2):
′ ′
P −y −x ∼ P −w −y −x.
s
FACTORSOFPISOTTILINGSPACES 9
Hence
′
P −y −x ∼ P −y −x
s
andfrom(3)and(4)wehaveB [T]−x ∼ B [T′]−x;thatis,T−x ∼ T′−x. ButT′ ∈ g−1(g(T))
0 s 0 s
and ♯g−1(g(T)) = cr(Φ). From (2) ofTheorem 4 it mustbethe casethat T′ = T. In otherwords,
♯g−1(g (T)) = 1 and theV-action onX hasminimalrank1.
X X
(cid:3)
Wereturn nowtoconsiderationoftherelation R.
Lemma 10. Suppose that Φ is an n-dimensional substitution of Pisot family type that forces its
border. If Ω has no almost automorphic sub actions, there is B < ∞ so that x Rx for i =
Φ i i+1
1,...,n−1 =⇒ ♯{x ,...,x } ≤ B.
1 n
Proof. LetussaythatxR¯yiftherearen ∈ Nandz ,...,z sothatz = x,z = yandz Rz for
1 n 1 n j j+1
j = 1,...,n−1. Forsuchz ,letT ,T′ ∈ Ω besuchthatT ≈ T′,π(T ) = z andπ(T′) = z .
j j j Φ j s j j j j j+1
Since Φ forces its border, there is ǫ > 0 so that if π(T) = π(T′), then T −v ∼ T′ −v for all v
s
with|v| < ǫ. Thus,T −v ∼ T′ −v foreach j > 1and |v| < ǫ, andT −v ∼ T′−v foreach
j s j−1 j s j
j ≥ 1andanopendensesetofv ∈ B (0). WeseethatxR¯y,π(T) = xandπ(T′) = y impliesthat
ǫ
T −v ∼ T′ −v foran open densesetof v inB (0).
s ǫ
Now, if there is no such B as in the statement of the lemma, we may find two sequences
{x } ,{y } with the properties: x and y are in the interior of the same n-cell of X for
i i∈N i i∈N i i Φ
all i; x 6= y for all i; x R¯y for all i; and x ,y → x ∈ X as i → ∞. Since the x ,y are
i i i i i i Φ i i
in the interior of the same n-cell, there is T ∈ Ω with π(T) = x and r ,s ∈ spt(B [T]) with
Φ i i 0
π(T−r ) = x andπ(T−s ) = y . Withoutlossofgenerality,assumethat|s | < ǫ/2. Wehavethat
i i i i i
T−r −v ∼ T−s −v foradenseset(andopen)setofv ∈ B (0). ThenT−(r −s )−v ∼ T−v
i s i ǫ i i s
for a set S of v that is open and dense in B (0). Then T −(r −s )−v ∼ T −v for a dense
i ǫ/2 i i s
set of v in B (0) (namely, for v in B (0) ∩ ∩ S ). Let N be large enough so that the support
ǫ/2 ǫ/2 i i
of Q := B [ΦN(T)] is contained in ΛNB (0). Then Q and Q−v are densely stably related on
0 ǫ/2 i
overlapfor v := ΛN(r −s ). By Theorem 7,Ω has an almostautomorphicsubaction. (cid:3)
i i i Φ
SupposethattheRn-actiononΩ doesnothaveanalmostautomorphicsubaction. Define≍ on
Φ s
X byx ≍ x′ iffthereare x ,...,x ∈ X withx = x , x′ = x ,andx Rx ,i = 1,...,n−1.
Φ s 1 n φ 1 n i i+1
(That is,≍ isthetransitiveclosureof R.)
s
Lemma 11. If the Pisot family tiling space Ω does not have an almost automorphic sub action,
Φ
then≍ isa closedequivalencerelationonX .
s Φ
Proof. By definition, ≍ is an equivalence relation. To see that it’s closed, we first show that the
s
relation R is closed. Supposethatxi,yi ∈ X are such that xi → x, yi → y,and xiRyi for i ∈ N.
Φ
There are then T ,T′ ∈ Ω with T ≈ T′ and π(T ) = xi,π(T′) = yi. Passing to a subsequence,
i i Φ i s i i i
wemaysupposethatT → T,T′ → T′. Thenπ(T) = x,π(T′) = y,andsince,asobservedearlier,
i i
≈ is closed,T ≈ T′. That is,xRy,and R isclosed.
s s
Nowsupposethatxi,yi,x,y are asabove,except xi ≍ yi. ByLemma10 thereare B ∈ Nand
s
zi, j ∈ {1,...,B},i ∈ N, with xi = zi, yi = zi and ziRzi for 1 ≤ j ≤ B − 1. Passing to
j 1 B j j+1
a subsequence, we may suppose that zi → z for 1 ≤ j ≤ B as i → ∞. Then, from the above,
j j
z Rz for1 ≤ j ≤ B −1, sox ≍ y and ≍ is closed. (cid:3)
j j+1 s s
Ifτ is atileforΦ and v ∈ spt(τ), by π(τ −v) we mean π(T −v) forany T ∈ Ω with τ ∈ T.
Φ
We call a point x ∈ X an inner point if each element of [x] is an interiorpoint of its n-face, and
Φ
10 M.BARGE
if v ∈ spt(τ), τ a tile for Φ , and π(τ − v) is an inner point, we’ll say that v is an inner point
of τ. A stack is a collection τ¯ = {τ ,...,τ } of tiles with the property that spt(τ¯) := ∩k τ has
1 k i=1 i
nonemptyinteriorandforeach v ∈ int(spt(τ¯)),[π(τ −v)] = {π(τ −v),...,π(τ −v)}forsome
j 1 k
(equivalently, all) j ∈ {1,...,k}. Note that each inner point v of a tile τ uniquely determines a
stacks(τ,v) := { tilesσ : v ∈ spt(σ) and π(σ−v) ≍ π(τ −v)}. Itfollowsfrom(4)ofTheorem
s
4 and Lemma 10 that there are only finitely many translation equivalence classes of stacks. Each
tileτ forΦisthustiledby thefinitelymanystackss(τ,v), v an innerpointof τ.
We now define a substitution Φ on stacks. Let τ¯ = {τ ,...,τ } be a stack with support e =
s 1 k
∩k spt(τ ) and let Φ(τ ) = {σ ,...,σ }. Then
i=1 i 1 1 l
Φ (τ¯) := {s(σ ,v) : v ∈ Λe,v an innerpointof σ ,j = 1,...,l}.
s j j
Note that since f preserves ≍ the definition of Φ is not altered by replacing τ by τ for any
Φ s s 1 i
i ∈ {1,...k}. Let Σ : Ω → Ω by
Φ Φs
Σ(T) := {s(τ,v) : τ ∈ T, v an innerpointof τ}.
Theorem 12. Supposethat Φ is an n-dimensional Pisot family substitutionthat forces the border
and has convex tiles. Suppose also that Ω does not have an almost automorphic sub action. Let
Φ
g : Ω → T bethemapontothemaximalequicontinuousfactor. Then:
Φ max
(1) Φ is Pisotfamily,forcestheborder,and isprimitiveandnon-periodic;
s
(2) The map Σ : Ω → Ω is a.e. one-to-one, and semi-conjugates Φ with Φ and the Rn-
Φ Φs s
actionon Ω with thatonΩ ;
Φ Φs
(3) Ω ishomeomorphicwithΩ / ≈ byahomeomorphismthatconjugatesbothtranslation
Φs Φ s
andsubstitutiondynamics andtherelation≈ onΩ istrivial;
s Φs
(4) The map g factors as g = g ◦Σ, where g : Ω → T is the maximal equicontinuous
s s Φs max
factormapof Ω .
Φs
Proof. Suppose that T,S ∈ Ω are such that π(T) ≍ π(S). There are then T ,S ∈ Ω with
Φ s i i Φ
π(T) = π(T ), T ≈ S , π(T ) = π(S ), i = 1,...,k − 1, and π(S ) = π(S). Since Φ
1 i s i i+1 i k
forces the border, there is ǫ > 0 and m ∈ N so that if T′,S′ ∈ Ω and π(T′) = π(S′), then
Φ
π(Φm(T′ −v)) = π(Φm(S′ −v)) for all v ∈ B (0). Together with the fact that Φ preserves ≈ ,
ǫ s
thismeansthat
π(T) ≍ π(S) =⇒ π(Φm(T −v)) ≍ π(Φm(S −v))
s s
forall v ∈ B (0).
ǫ
Towardsprovingthat Φ forcestheborder,let’sexaminewhatitmeansforstackstobeadjacent
s
in an element of Ω . First supposethat stacks s(σ ,v ) and s(σ ,v ) in Φ (τ¯) (see the definition
Φs 1 1 2 2 s
of Φ ) are distinct and meet along an (n−1)-cell. Say v ∈ spt(s(σ ,v ))∩spt(s(σ ,v )). Then
s 1 1 1 1
thereis atiling T ∈ Ω and tilesη ∈ T, η ∈ s(σ ,v ), i = 1,2, withv ∈ spt(η )∩spt(η ). Note
Φ i i i i 1 2
thats(σ ,v ) = s(η ,v ),i = 1,2. Thisstructurepersistsunderapplicationof Φ and wesee,since
i i i i s
every allowed adjacency occurs in some Φk(τ¯), that if σ¯ and σ¯′ are adjacent stacks in an element
of Ω with v in the intersection oftheirsupports,then there is T ∈ Ω and tiles η,η′ ∈ T so that
Φs Φ
σ¯ = s(η,w),σ¯′ = s(η′,w′),withw,w′ as closetov as wewish.
SupposenowthatS¯ ,S¯ ∈ Ω aresuchthatη¯∈ S¯ ∩S¯ ,andtherearestacks σ¯ ∈ S¯ ,σ¯ ∈ S¯
1 2 Φs 1 2 1 1 2 2
thatmeetinan(n−1)-cellcontainedintheboundaryofthesupportof η¯;sayv isinthiscommon
boundary. There are then T,S ∈ Ω and η ,σ ∈ T, η ,σ ∈ S, with s(η ,w) = s(η ,w) = η¯,
Φ 1 1 2 2 1 2
s(σ ,w′) = σ¯ , s(σ ,w′) = σ¯ , and w,w′ as close to v as desired. Since, w is as close to v
1 1 2 2
as we wish, π(T − w) = π(η − w) ≍ π(η − w) = π(S − w), and ≍ is closed, we have
1 s 2 s