n+2k 2k−r+1

of multi-dimensional shifts

Marie-Pierre Beal

InstitutGaspardMonge,UniversitedeMarnelaVallee,5Bd Desartes,

Champs-sur-Marne,F-77454MarnelaVallee,Frane

e{mail bealuniv-mlv.fr

Franesa Fiorenzi

LIX,

EolePolytehnique

91128Palaiseau Cedex,Frane

e{mail orenzimat.uniroma1.it

FilippoMignosi

DipartimentodiMatematiaedAppliazioni,UniversityofPalermo

ViaArhira34,90123 Palermo,Italy

e{mail mignosialtair.math.unipa.it

Abstrat

Westudywhethertheentropy(orgrowthrate)ofminimalforbidden

patternsofsymbolidynamialshiftsofdimension2ormore,isaonju-

gayinvariant. Weprovethattheentropyofminimalforbiddenpatterns

isaonjugayinvariantforuniformlysemi-stronglyirreduibleshifts. We

proveaweakerinvariantinthegeneralase.

1 Introdution

Symbolidynamialsystemsareoftendenedbyasetofforbiddenpatterns. In

dimensiontwoforinstane, ashiftis thesetof labellingsofthesquarelattie,

alledongurations,whihavoidanyforbiddenpattern. Shiftsofnitetypeare

thosewhih anbedened byanite setofforbiddenpatterns. Thisproperty

is aonjugayinvariant,see for instane [15℄ or[13℄. Many naturalexamples

oftwo-dimensional shiftsof nitetype arisefrom lattie systemsin statistial

mehanis[1℄.

The dynami of multi-dimensional shifts is muh more omplex than the

one of one-dimensional shifts. For instane the entropy of a shift, whih is

a onjugay invariant that gives the omplexity of the allowed patterns (i.e.

patternsontainedinsomeongurationoftheshift),iseasily omputablefor

evenforthesimplest examplesindimensiontwo.

In[3℄hasbeenintroduedthenotionofminimal forbiddenwordforaone-

dimensional shift: a word is minimal forbidden if it is forbidden and if both

itsproperprex and itspropersuÆxare allowed bloksof theshift. The set

of minimalforbidden words isnite for ashift ofnite type. The entropy, or

omplexity,of theset of minimal forbiddenwordsis aonjugayinvariantfor

one-dimensional shifts. Moreoverthis invariant is independent of some other

knowninvariantsliketheentropyoftheshiftorthezetafuntion forinstane.

Notethatthisinvariantisnotmeaningfulforshiftsofnitetype,oritjustsays

thatshiftsofnitetypeareonlyonjugatetoshiftsofnitetype,whihiswell

known. Minimalforbiddenwordshaveappliationsinseveralareaslikelossless

ompression(data ompression usingantiditionaries[6℄)and also reonstru-

tionofDNAsequenesfromitsfragments(the fragmentassemblyproblem[4℄,

[14℄and[16℄).

Inthispaper,westudythenotionofminimalforbiddenpatternsformulti-

dimensionalshifts,withthegoalto providesomenewonjugayinvariantsfor

multi-dimensionalshifts. We givetwo notionsof minimal forbidden patterns.

Therst oneis the diret extensionof thenotionof minimal forbiddenwords

for one-dimensional shifts. It an be onsidered for patterns with a square

or retangular shape. For instane, a square is minimal forbidden if it is a

forbiddenpatternsuhthat eah stritsubsquareisallowed. It turnsoutthat

this denition is too weak. Indeed, a shift of nite type an have aninnite

numberofminimalforbiddenpatternsofthistype. Thusweonsiderastronger

notionwhihleadstomuh lessminimal forbiddenpatterns. Indimensiontwo

for instane, a forbidden square of size n is minimal for this stronger notion

if it is ontained in a onguration suh all squares of of size n 1 are

allowed patterns of the shift. These two notions oinide in dimension one.

Anothermain dierene between the one-dimensionaland higher dimensional

ase appears with the omputational point of view. The omputation of the

setofminimalforbiddenwordsforone-dimensionalshiftsofnitetype,orone-

dimensionalsoshifts,andtheomputationofitsgrowthrateiseasilydonein

polynomialtime(see[5℄and[2℄),whiletheproblemseemstohaveatleast the

samediÆultyastheproblemofomputingtheentropyoftheshiftindimension

two. Itis also undeidableto hekwhether agivenpattern isontainedin a

ongurationofagivenshiftofnite type(see[11℄forinstane).

Foramulti-dimensionalshiftX,wedenotebyh 1

(X)theomplexityof the

strong minimal forbidden patterns and by h 2

(X)the omplexity of the weak

minimal forbiddenpatterns. We provetwopartial resultsof invariane under

onjugay. First,ifX andY aretwoonjugateshifts,h 1

(X)h 2

(Y). Seond,

thestrongentropyh 1

(X)ofminimalforbiddenpatternsisaonjugayinvariant

forshiftswhih areuniformlysemi-stronglyirreduible. Thislatterpropertyis

apropertyof irreduibilityofshiftsofnite typeapproximatingtheshiftfrom

the outside. It is always satised by one-dimensional irreduible so shifts.

Weproveour main resultsfor squareshapesand theresults are validfor any

dimension. The proofs of these results annot be generalized in the ase of

The paperis organizedas follows. InSetion 2, we reallsome baside-

nitionsfromsymbolidynamis. Thereaderis referredto [12℄or[10℄ formore

details,seealso [11℄,[15℄,and[13℄ formulti-dimensionalshifts,[7℄forshiftson

Cayleygraphs. Wedene herethenotionsof minimalforbidden patterns, the

weakerandthestronger. Weproveourmainresultsin Setion3.

2 Denitions and examples

2.1 Bakground on shifts and onjugaies

We reall here somebasi denitions and properties about multi-dimensional

shiftsandonjugaies. Wealsox somenotations.

Let A bea nite alphabet and d bea positiveinteger. The d-dimensional

fullA-shiftisthesetA Z

d

ofallfuntions:Z d

!A. Inthelanguageofellular

automatawehaveaspaeinwhihthe\universe"istheintegerlattieZ d

and

aongurationisanelementofA Z

d

,thatisafuntionassigningtoeahellof

thegridaletterofA. Onthis setwehaveanaturalmetri: if

1

;

2 2A

Z 2

are

twoongurations,wedenethedistane

dist(

1

;

2 ):=

1

n+1

;

wherenistheleastnaturalnumbersuhthat

1 6=

2 inD

n

:=[ n;n℄

d

. Ifsuh

anndoesnotexist,that isif

1

=

2

,weset theirdistaneequalto zero. This

metriinduesatopologyequivalenttotheusual produttopology,where the

topologyin Aisthedisreteone. Inthesequelwefousontheased=2.

ThegroupZ 2

atsonA Z

2

asfollows:

(

)

j :=

j+

foreah 2A Z

2

and eah ;2 Z 2

, where

j

is thevalue of at and the

additionistheusualoperationin thediret sumZ 2

.

Nowwegiveatopologialdenitionofashiftspae(brieyshift). Asstated

in Proposition 2.2, this denition is equivalent to the lassial ombinatorial

one.

Denition2.1 A subsetX of A Z

2

isalled ashiftifitis topologiallylosed

andZ 2

-invariant.

Here Z 2

-invariane means that X is invariant under the ation of Z 2

on

A Z

2

(that is

2 X for eah 2 X and eah 2 Z 2

). Notie that this is

equivalenttohave (1;0)

2X and (0;1)

2X foreah 2X.

A pattern is a funtion p : E ! A, where E is a non-empty nite subset

ofZ 2

. Theset E isalledthesupportof thepattern. Inthesequel,wedonot

distinguishbetweena pattern pwith support E and the pattern obtainedby

support. Apattern (resp.blok) of Xistherestritionof aongurationof X

to a nite (resp. a nite onneted) subset of Z 2

. Notie that, being ashift

Z 2

-invariant,thesenotionsareindependentof thepositionoftheirsupports.

Wedenote byB(X)the setof bloksof ashiftX and byB

n

(X)the setof

squarebloksof size nof X. If X isa subshiftof A Z

, aongurationis abi-

innitewordandablokofX isanitewordappearinginsomeonguration

ofX.

LetF beaset ofpatterns,wedenotebyX

F

theset ofallongurationsof

A Z

2

avoidingeahpatternofF. Itiseasytoprovethatthetopologialdenition

of ashift spae is equivalent to thefollowing ombinatorial one involving the

avoidaneofertainforbidden patterns.

Proposition 2.2 A subset X A Z

2

is ashift if andonly if there existsaset

ofpatterns F suhthat X =X

F

. In thisase,F isasetofforbiddenpatterns

ofX.

Denition2.3 Let X beasubshift ofA Z

2

. A map :X !A Z

2

isk-loal if

thereexistsÆ:B

2k +1

(X)!A suhthat forevery2X and2Z 2

(())

j

=Æ((

)

jD

k )=Æ(

j+

1

;

j+

2

;:::;

j+

m );

where

1

;:::;

m

denotetheelementsofD

k

=[ k;k℄

2

.

Inthis denition,wehaveassumedthat thealphabet ofthe shiftX is the

sameasthealphabetof itsimage(X). Inthisassumptionthere isno lossof

generalitybeauseif :X A Z

2

!B Z

2

, onean alwaysonsiderX asashift

overthealphabetA[B.

It is known from the Curtis-Lyndon-Hedlundtheorem that loal maps are

exatly the funtions whih are ontinuous and ommute with the Z 2

-ation

(i.e. for eah 2X and eah 2Z 2

, onehas(

)=()

). Hene ifaloal

mapisone-to-oneandonto, itsinverseisalsoaloalmap.

Thisresultleadsus togivethefollowingdenition.

Denition2.4 Twosubshifts X;Y A Z

2

are onjugate if there exists a lo-

albijetivemap betweenthem (namelyaonjugay). Theinvariants are the

propertiesofashiftinvariantunderonjugay.

Thebasidenitionofashiftofnitetypeisintermsofforbiddenpatterns.

Inasense wemaysaythatashiftisof nitetypeifweandeidewhether or

nota ongurationbelongs to the shift onlyby hekingits bloks of axed

(andonlydependingontheshift)size.

Denition2.5 A shift is of nite type if it admits a nite set of forbidden

bloks.

Wewillseelatersomeexamplesofshiftsofnitetype.

Denition2.6 IfX A Z

is ashift,theentropy of Xisdenedas

h(X):= lim

n!1 logjB

n (X)j

n 2

; (1)

whereB

n

(X)istheset ofsquarebloksofX ofsizen. Wewill alwaysusethe

base2forlogarithms.

The existeneof the limit in (1) is proved for instane in [12, Proposition

4.1.8℄fortheone-dimensionalaseand in[9℄forthemulti-dimensionalone.

For eah 2Z 2

, the set D

n

provides, bytranslation, aneighborhood of ,

thatisthesetD(;n):=+D

n

=[ n;+n℄

2

. GivenasubsetEZ 2

and

foreahk2N wedenote by

E +k

:=

[

2E

D(;k)andE k

:=f2EjD(;k)Eg

thek-losureof Eandthek-interiorof E,respetively.

Let :X !Y beak-loalmap. IfpisapatternofX withsupportE,the

map isdenedon pand givesapattern ofY withsupport E k

. Indeedone

andene

(p):=()

jE k;

whereisanyongurationofX extendingp.

The following well-known result guarantees that the entropy is invariant

underonjugay.

Proposition 2.7 LetX be a shift andlet :X !A Z

2

bea loal map. Then

h((X))h(X).

Proof Let bek-loalandletY :=(X). Themap :B

n+2k

(X)!B

n (Y)

issurjetiveandhene

jB

n

(Y)jjB

n+2k

(X)jjB

n (X)jjA

[1;n+2k ℄ 2

n[1;n℄

2

j:

Fromthepreviousinequalitieswehave

logjB

n (Y)j

n 2

logjB

n (X)j

n 2

+

((n+2k) 2

n 2

)logjAj

n 2

andhene,takingthemaximumlimit,h(Y)h(X). 2

IntheaseofCayleygraphstheentropyisdenedasamaximumlimitand,

asprovedin[7,Theorem2.12℄,itisaninvariantifthegroupisamenable.

We dene below several notions of minimal forbidden patterns. In the one-

dimensionalase,awordisminimalforbiddenifitisforbiddenandifeahstrit

fatoris allowed. Thenatural extension of thisproperty leadsto aforbidden

blok whose proper subbloks are allowed. This orresponds to our seond

denition below. But, as we will see later, this denition is too weak. For

instane,ashiftof nitetypedoesnotneessarilyhaveaniteset ofminimal

forbidden patterns with respet to this denition. For this reason, our rst

denition below orrespondsto astrongerpropertywhih is equivalent to the

otheroneintheone-dimensionalase. Wealsomakeadistintionbetweenthe

asesin whih these bloks are squares orretangles. Wewill see below that

thisdistintionisrelevant.

Let m;n be two nonnegative integers. We denote by F

n

(X) the set of

forbiddensquaresofX ofsizen,andbyF

m;n

(X)thesetofforbiddenretangles

ofX of sizemn. Ifit isnotspeiedapartiularset of forbiddenpatterns

forX,withforbiddenpatternwemeanapatternwhih isnotallowed.

Nowwegivefourdierentpossibledenitionsofminimalforbiddenpatterns

ofashiftX.

M

1

n

(X):=F

n

(X)\B(X

F

n 1 (X)

). That isasquare ofsize n isminimal

forbiddenifitisforbiddenandifitisontainedinaongurationinwhih

eah squareofsizen 1isallowed;

M

2

n

(X)is the set ofsquares of F

n

(X) suh that eah subsquareof size

n 1isanelementofB(X);

M

1

m;n

(X):=F

m;n

(X)\B(X

Fm 1;n(X) )\B(X

Fm;n 1(X) );

M

2

m;n

(X)istheset ofretanglesofF

m;n

(X)suhthateahpropersub-

retangleisanelementofB(X).

Itisstraightforwardthat,foranyintegersm;n,wehavetheinlusionsM 1

n;n (X)

M

1

n

(X)M 2

n

(X)and M 1

m;n

(X)M 2

m;n

(X). Wewill see that there are

examplesin whihthese inlusionsarestrit.

Wedenote withM i

(X)theset S

n M

i

n

(X)(fori=1;2). Weprovebelow

thatthesets S

m;n M

1

m;n

(X)andM 1

(X)aresetsofforbiddenpatternsforX.

Proposition 2.8 Theset S

m;n M

1

m;n

(X)isasetofforbidden patterns forX,

thatisX =X S

m;n M

1

m;n (X)

.

Proof If 2 X and p is a retangle of , p 2= F

m;n

(X) and then p 2=

S

m;n M

1

m;n

(X). Hene XX S

m;n M

1

m;n (X)

.

Supposethat 2= X. Sine X = T

m;n X

F

m;n (X)

wehave 2= X

F

m;n (X)

for

somem;n. NotiethattheshiftsX

F

m;n (X)

havethepropertythat2=X

F

m;n (X)

implies2=X

Fm+1;n(X)

and2=X

Fm;n+1(X)

. Thismeansthatin thegridof the

naturalnumbersinwhihapair(m;n)is\marked"ifandonlyif2=X

Fm;n(X) ,

there are some extremal pairs, that is, pairs whih are marked but suh that

orners of the dashed line show the extremal pairs for . Notie that sine

F

m;0

=; =F

0;n

, the pairs (m;0) and (0;n)are alwaysunmarked. Hene if

Figure1: Theextremalpairsfor.

(m;n) is an extremalpair for wehave 2X

Fm 1;n(X)

and 2 X

Fm;n 1(X) .

Sine2=X

Fm;n(X)

there existsaforbiddenretangle pofsize mnin and

thisis also apattern ofX

Fm 1;n(X)

and apattern of X

Fm;n 1(X)

. This means

thatp2M 1

m;n

(X)andhene2=X S

m;n M

1

m;n (X)

. 2

Remark. OneanseethatProposition2.8holdsindimensiond. Indeedthe

shiftsX

Fn

1

;:::n

d (X)

aresuhthat2=X

Fn

1

;:::n

d (X)

implies2=X

Fn

1

;::: ;n

k +1;:::;n

d (X)

foreahk=1;:::;d.

Proposition 2.9 The setM 1

(X) isasetof forbidden patterns for X,that is

X=X

M 1

(X) .

Proof If 2X and pis asquare of,p2=F

n

(X)andhenep2= M 1

n (X).

ThusX X

M 1

(X) .

Suppose that 2= X. If 2= X

F

1 (X)

we have 2= X

M 1

1 (X)

and hene 2=

X

M 1

(X)

(notiethatM 1

1

(X)=F

1

(X)sineF

0

(X)=;). Otherwise,sine

X

F

1 (X)

X

F

2 (X)

X

F

n (X)

X;

there exists anintegeri suh that 2X

F

i (X)

and 2= X

F

i+1 (X)

. Hene there

isapatternpof ofsize i+1whihis forbiddenin X (that isp2F

i+1 (X)).

Moreoverwehave p2 B(X

Fi(X)

) and henep 2M 1

i+1

(X)whih implies 2=

X

M 1

(X) . 2

S

m;n M

2

m;n

(X)andM 2

(X)alsoarepossiblesetsof forbiddenpatterns forX.

Proposition 2.10 AshiftX isof nitetype ifandonly ifM 1

(X)isnite.

Proof IfM 1

(X) is nite, theshift X is of nite type by Proposition 2.9.

Conversely,supposethat X isofnite typeand henethatX =X

F

,where F

isanitesetofforbiddensquares. LetnbesuhthattherearenosquaresinF

ofsizegreaterthanorequalton. IfhnandpisasquareofM 1

h

,thereexists

aonguration 2 X

Fh 1(X)

whih ontains pand thus do not belong to X.

Hene there is asquare ofF ontainedin , and the size of this square must

begreaterthan orequalto hn, whih isexluded. Hene M 1

h

=;for eah

hn. 2

ByProposition2.8, if S

m;n M

1

m;n

(X)isnitethen X is ofnitetype. We

will see with an example that the onverseis not true. Nevertheless we have

thefollowingresult.

Proposition 2.11 Ashift X isof nite typeif and only if there isa positive

integern

0

suhthat M 1

m;n

(X)=;for m;nn

0 .

Proof Suppose that X is of nite type and henethat X = X

F

where F

isaniteset offorbiddenretangles. Letn

0

beanintegersuhthat thereare

noretangles in F of size mn, when m n

0

orn n

0

. Let m and n be

twointegerssuh that m n

0

and n n

0

. If pis a retangle of M 1

m;n (X),

there exists aonguration ontainingpwhih belongs to X

Fm;n 1(X) . This

ongurationis notin X. Then ontainsaretangleof F ofsize m nwith

m > m n

0

or n n n

0

. This ontradits the fat that there are no

retanglesin F of sizemn,when mn

0

ornn

0

. Hene M 1

m;n

=;for

eahm;nn

0 . 2

Inthefollowingproposition,weprovethatthepossiblenotionsofminimal

forbiddenpatterns oinideintheone-dimensionalase.

Proposition 2.12 LetX beaone-dimensionalshift. ThenM 1

n

(X)=M 2

n (X).

Proof Let wbeaminimal forbiddenwordin M 2

n

(X),where w

1

is itsleft

prex of length n 1, and w

2

its right suÆx of length n 1. Then w

1

;w

2

areallowedwordsof X. Let lbea left-innitewordand r bearight-innite

wordsuhthatlw

1

r2X. Similarlylet

lbealeft-innitewordandrbearight-

innitewordwith

l w

2

r2X. ThenlwrbelongstoX

Fn 1(X)

. Thusw2M

1 (X).

2

Nowwegivetwoexamplesofshiftsofnitetypeforwhihthesetsofminimal

forbiddensquares M 1

and M 2

arebothnite. Other examplesanbefound

in[13℄and[15℄.

forwhih thevalue ofits entropyof allowed bloks h(X)is known. Let A be

thealphabetf0;1;2g. WedenetheshiftofnitetypeX =X

F

whereF isthe

followingsetofpatterns:

x x

x

x

withx2A. Theongurationsofthisbidimensionalshiftarethethreeolorings

ofasquarelattie. Twoadjaentellshaveadierentolor. It turnsoutthat

theexatvalueoftheentropyofthisshiftisknown(see[1℄)andequalto

h(X)= 3

2 log

4

3 : 2:

Example2.14 We now give an example of a two-dimensional shift of nite

typeX forwhihtheexatvalueofitsentropyofallowedbloksisnotknown.

LetAbethealphabetf0;1g. Wedenetheshiftofnite typeX =X

F where

F isthefollowingsetofpatterns:

1 1

1

1

Theseonstraintsare knownasthe hardsquareonstraints. Theyorrespond

tosomelattiegasmodels[1℄. 2

ForthetwoshiftsofExamples2.13and2.14,bothsetsM 1

andM 2

arenite.

Indeed, M 2

only ontains the22squares ontaininga forbidden retangle

ofF. Wenowgiveanexampleofabidimensionalshiftof nitetypeforwhih

thesets M 2

and S

m;n M

2

m;n

arenotnite.

Example2.15 LetA bean alphabet and

A:= A[fa;bg, where a andb do

not belong to A. We dene the shift of nite type X = X

F

where F is the

followingsetofpatterns:

x

a

y

b

where x6=a and y 6=b. Fornbig enoughwe haveM 1

n

(X)=;, but M 2

n (X)

ontains,ifnisodd,thefollowingsquaresofsize n:

a ::: b

:::

:::

.

.

. .

.

. ::: .

.

. .

.

.

:::

where eah * an be replaed by any letter in A. Thus jM 2

n

(X)j jAj n

2

2

.

Moreover,form;nbigenough,wehaveM 1

m;n

(X)=;. Inthisexamplewealso

havethat

m;n M

1

m;n

(X) isnotnite. IndeedM 1

1;n

(X)ontains,if nis odd,

thefollowingretangle:

a ::: b

where eah * an be replaed by any letter in A. This an be easily seen

observingthat theongurationlledofa's outsidetheretangleis ontained

inX

F

0;n (X)

\X

F

1;n 1 (X)

. HenejM 1

1;n

(X)jjAj n 2

. 2

Example2.16 Let A be the alphabet fa;b;g and F be the set of squares

borderedbyb'sasfollows,

b b ::: b b

b ::: b

.

.

. .

.

. ::: .

.

. .

.

.

b ::: b

b b ::: b b

(2)

whereeah*anbereplaedwithanaora. LetX =X

F

. WehaveM 1

n (X)=

M 2

n

(X)=M 1

n;n

(X),andaminimalforbiddensquareofsizeninX isasquare

nn bordered by b's and ontained in F. Hene jM 1

n

(X)j = jM 2

n (X)j =

jM 1

n;n

(X)j=2 (n 2)

2

. MoreoverwehaveM 1

m;n

(X)=;ifm6=n. 2

Example2.17 With aslightmodiationof Example 2.16,one ansee that

in general M 1

n 6=M

1

n;n

. Let A be thealphabet fa;b;gand F be the set of

retangles n(n+1)borderedbyb'sasin(2). Thesquare(n+1)(n+1)

b b ::: b b

b a ::: a b

.

.

. .

.

. ::: .

.

. .

.

.

b a ::: a b

b b ::: b b

a a ::: a a

| {z }

(n+1)(n+1)

isontainedin M 1

n+1

(X),butM 1

n+1;n+1

(X)=;. Inthisexampleoneanalso

seethatin generalX 6=X S

n M

1

n;n (X)

. 2

3 Entropy of minimal forbidden patterns

Inthissetion,westateandproveourmaininvarianeresultsontheentropyof

minimalforbidden patternsin theaseofsquare bloks. Wewill explainlater

whythese resultsannotbeextended totheaseofretangularshapes.

Denition3.1 Fori =1;2, we denote byh i

(X) theentropy of the sequene

(M i

n

(X))ofminimal forbidden patterns of X,that is:

h i

(X):=limsup

n!1 1

n 2

logjM i

n (X)j:

Notiethat h is always 1forshifts ofnite type. IntheExample 2.15we

haveh 1

(X)= 1andh 2

(X)log(jAj).

Let beak-loalmapdenedonX. Themap

n

iswelldenedonX

Fn(X)

ifn2k+1and

n ()

j :=(

jD(;k )

) (3)

(indeed

jD(;k )

isapattern ofX).

Lemma3.2 Let bea k-loal map denedon X. If 2X then 2 X

F

n (X)

and

n

()=().

Proof Wehave

n ()

j

=(

jD(;k )

)=()

jD(;k ) k

=()

j . 2

Lemma3.3 Let bea k-loal map denedon X. If pis apatternof X then

itis alsoapattern ofX

Fn(X) and

n

(p)=(p).

Proof Let E be the support of p and let 2 X be a onguration ex-

tending p. One has

n (p) =

n ()

jE

k. By Lemma 3.2, we have

n ()

jE k =

()

jE

k=(p). 2

Proposition 3.4 Let : X ! Y be a k-loal map. If n 2k+1, we have

n :X

Fn(X)

!X

Fn 2k(Y) .

Proof Let2X

F

n (X)

andletpasquareofsizen 2kof

n

(). Thereexists

asquare pof withsize nsuhthat

n

(p)=p. Wehavep2B(X)and hene

thereexistsa2X suhthat pisasquareof. Thismeansthat pisasquare

of( )2(X)Y. 2

Proposition 3.5 Let :X !Y beaninjetivek-loal map. Thenthereexists

an n suh that

n

is injetive on X

Fn

(and hene itis injetive on eah X

Fm

withmn).

Proof Supposethatforeahnthemap

n

isnotinjetiveonX

Fn(X) . Then

there exist twosequenes (

n )

n

and (

n )

n with

n

;

n 2 X

Fn(X)

suh that for

eah n we have

n 6=

n and

n (

n ) =

n (

n

). We an always suppose that

n 6=

n

at theenter (0;0) ofA Z

2

. Being A Z

2

ompat,there exist ;2A Z

2

andtwosubsequenes(

n

k )

k and(

n

k )

k

suhthatlim

k

n

k

=andlim

k

n

k

=.

Foreahh,thesequene(

n

k )

k h

is ontainedin X

F

n

h (X)

and itbeinglosed,

2X

F

n

h (X)

. This implies that 2X and analogously 2X. Moreoverthe

ontinuity of

n

h

impliesthat () =( ). But being dist (

n

;

n

) =1, 6= ,

whihontraditstheinjetivityof. 2

Lemma3.6 Let :X !Y be a bijetive k-loal map, let 1

be

k-loal and

let2X

Fn(X)

with n2k+2

k+1. Hene( 1

)

n 2k (

n

())=.

Proof LetbeanelementofZ . Wehave

( 1

)

n 2k (

n ())

j

= 1

(

n ()

jD(;

k) )=

1

(

n (

jD(;

k +k ) )):

ByLemma3.3, wehave

1

(

n (

jD(;

k+k ) ))=

1

((

jD(;

k+k ) ))=

1

(( )

jD(;

k) );

where 2 X extends

jD(;

k +k )

. Now 1

(( )

jD(;

k) ) =

1

(( ))

j

=

j

=

j . 2

Notie that in Proposition 3.5 wehaveproved that, if isone-to-one, the

map

n :X

F

n (X)

!

n (X

F

n (X)

)is invertible. InLemma 3.6, wehave proved

thattheinverseofthismapis((

1

)

n 2k )

j

n (X

F

n (X)

)

. Inpartiulartheonstant

ofloalityof(

n )

1

isthesameasthat of 1

(thatis

k).

Lemma3.7 Let :X !Y be a bijetive k-loal map, let 1

be

k-loal and

letpbea pattern of X

Fn(X)

with n2k+2

k+1. Hene, if E isthe support

ofp, wehave( 1

)

n 2k (

n

(p))=p

jE k

k.

Proof Let beaongurationofX

F

n (X)

extendingp. Wehave

( 1

)

n 2k (

n

(p))=( 1

)

n 2k (

n ()

jE

k)=( 1

)

n 2k (

n ())

jE k

k :

ByLemma3.6, wehave( 1

)

n 2k (

n ())

jE k

k

=

jE k

k

=p

jE k

k . 2

Observe that a shift of nite type analso be dened using the notionof

allowed patterns. Morepreisely ashift X isof nitetypeif andonly ifthere

existsanite setC ofpatternssuhthat X=X(C),where

X (C):=f2A Z

2

jeahpatternof belongstoCg:

Indeed if F is anite set of forbidden patterns, wean always suppose that

eah ofthem hasthesamesupportF andweandeneC :=A F

nF. Observe

thatinthisaseitisnotneessarythateahpatternofCisapatternofX,but

weanalwayssupposethatC=C\B(X).

With this equivalentharaterization itis possible to provethe invariane

ofthenotionofbeingofnite type. Forthisweandene

C:=f()

jF +

k j2

X and extendsp2Cgand prove,using Lemma 3.7, that Y =X (

C). In the

followingtheoremweprovethisinvarianeasaonsequeneofProposition3.5.

Proposition 3.8 Thenotionofbeingofnitetypeformulti-dimensionalshifts

isinvariantunder onjugay.

Proof Suppose that :X !Y is ak-loal onjugayandsuppose thatY

is of nite type. Hene there exists n suh that Y = X

Fn(Y)

(obviously this

ondition is also suÆient). Suppose that 2 X

Fn+2k(X)

. The onguration

n+2k

() belongs to X

Fn(Y)

= Y and then

n+2k

() = ( ) with 2 X. By

n+2k n+2k n+2k

=. ThisimpliesX =X

Fn+2k(X) . 2

The following theorem extends the one-dimensional onstrution of [2℄ to

multi-dimensionalshifts.

Theorem3.9 Let : X ! Y be a onjugay, let be k-loal and let 1

be

k-loal. Henefor n2k+2

k+1

jM 1

n

(X)jC(n) 2

k

X

r= 2k jM

2

n+r (Y)j;

whereC(n)=(2

k+2k+1) 2

jAj 4(

k+k )(n+

k k )

.

Proof If p 2 M 1

n

(X) we have p 2 B(X

Fn 1(X)

), and hene there exists

2 X

Fn 1(X)

suh that ontains p in E := [1;n℄

2

. Consider the pattern

p:=

n 1 ()

jE +

k of

n 1

(). IfpisapatternofY thereexists 2Y suhthat

p is a pattern of . Now 1

( ) 2 X and 1

()

jE

= 1

(

jE +

k) = 1

(p).

BeingpapatternofX

F

n 1 2k (Y)

,weanapplyLemma 3.3andthen 1

(p)=

( 1

)

n 1 2k

(p) = ( 1

)

n 1 2k (

n 1 ()

jE +

k ) = (

1

)

n 1 2k (

n 1 (

jE +

k+k )).

Nownotiethat

jE +

k+k

isapatterninX

F

n 1 (X)

andweanapplyLemma3.7.

Hene wehave 1

()

jE

=( 1

)

n 1 2k (

n 1 (

jE +

k+k))=

jE

=p,whih on-

traditsthefat thatp2=B(X).

Hene to eah p 2 M 1

n

(X) one an assoiate a pattern p2 F

n+2

k (Y)\

B(X

Fn

1 2k(Y)

) and being ( 1

)

n 1 2k

(p) = p this assoiation is one-to-one.

Notiethat in pthere is apattern ofM 2

n+r

(Y), where 2kr 2

k. Hene

themaximalnumberofsuh patternsontainedin pis

(2

k r+1) 2

jAj (n+2

k) 2

(n+r) 2

jM 2

n+r (Y)j;

where(2

k r+1) 2

isthe numberofpositions in whih we aninserttheleft

bottomvertexofasquareofsizen+rinasquareofsizen+2

k(seeFigure2),

and(n+2

k) 2

(n+r) 2

isthenumberoffreepositions whih weanllwith

lettersin thealphabet A. Hene

jM 1

n (X)j

2

k

X

r= 2k (2

k r+1) 2

jAj (n+2

k) 2

(n+r) 2

jM 2

n+r (Y)j

(2

k+2k+1) 2

jAj 4(

k+k )(n+

k k ) 2

k

X

r= 2k jM

2

n+r

(Y)j: 2

Remark. Inthed-dimensionalasewehave

jM 1

n

(X)jC(n) 2

k

X

r= 2k jM

2

n+r (Y)j;

whereC(n)=(2

k+2k+1) d

jAj (n+2

k) d

(n 2k ) d

.

n+r

n+2k 2k−r+1

p

Figure 2: Howapatternin M 2

n+r

(Y)anappearin p.

Corollary3.10 LetX andY betwoonjugatemulti-dimensionalshifts. Then

h 1

(X)h 2

(Y):

Proof ByTheorem3.9wehavethat

jM 1

n

(X)j(2

k+2k+1) 3

jAj 4(

k+k )(n+

k k ) 2

k

max

r= 2k jM

2

n+r (Y)j:

Hene

log(jM 1

n (X)j)

n 2

3log(2

k+2k+1)

n 2

+

+ (4(

k+k)(n+

k k))logjAj

n 2

+ logjM

2

n+r (Y)j

n 2

;

where 2kr=r(n)2

k. Bytakingthemaximumlimitsrightandleftfrom

theinequality,wehavetheonlusion. 2

From the previous result and by Proposition 2.12, we reover the known

resultforone-dimensionalshifts(see[2℄).

Corollary3.11 LetX and Y be twoonjugate one-dimensional shifts. Then

h 1

(X)=h 1

(Y).

3.1 Semi-strongly irreduible shifts

Inthissetionweprovethath 1

isaninvariantforasuitablelassofshifts.

Aone-dimensionalshiftX isirreduibleifforeveryu;v2B(X)thereexists

awordwsuhthattheonatenationuwv belongsto B(X).

Thisoneptanbegeneralizedinthemulti-dimensionalase: ashiftX

A Z

d

is alledirreduible ifforeah pairofbloksp;q2B(X)with supportsE

andF, thereexistsaonguration2X suhthat=pin Eand=qin

F,

where

F is atranslationofF ontainedin{E.

We now give the denition of semi-strong irreduibility for a shift. This

denitionisstritlyweakerthantheonegivenin[8℄neededtoproveaGarden

generatedgroup(ofnon-exponentialgrowth).

Denition3.12 AshiftX isalled(M;h)-irreduible(whereM;harenatural

numberssuhthatMh)ifforeahpairofbloksp;q2B(X)whosesupports

EandF havedistanegreaterthanM,thereexistsaonguration2X suh

that=pinE and=qin

F,where

F isatranslationofF ontainedinF +h

.

TheshiftX isalledsemi-stronglyirreduibleifitis(M;h)-irreduibleforsome

M;h2N.

A shift X is uniformly (M;h)-irreduible if the sequene (X;X

Fn(X) ) is

uniformly(M;h)-irreduible,i.e.X and X

Fn(X)

are(M;h)-irreduibleforany

nonnegativeintegern. TheshiftX isuniformlysemi-stronglyirreduibleifitis

uniformly(M;h)-irreduibleforsomeM;h2N.

Reall(seeforinstane[12℄),that aone-dimensionalshiftissoifitisthe

setoflabelsofallbi-innite pathsonanitelabeledgraph. Itisirreduible if

thisgraphanbehosenwithastronglyonnetedgraph.

Proposition 3.13 Everyone-dimensionalirreduiblesoshiftissemi-strongly

irreduible.

Proof Let X be theset of labelsof ann-state labeled stronglyonneted

graph. Let u;v 2 B(X). Letus assume that uand v appear in points uand

v ofof X with supports, E =[i;j℄ and F =[k;l℄ respetively, withj <k, at

apositive distane M. Thus uu

j+1 u

j+M

2 B(X). Sine the graphhas n

statesandisstronglyonneted, thereiswordwof lengthat mostn 1suh

that uu

j+1 u

j+M

wv 2 B(X). Sine M ju

j+1 u

j+M

wj M+(n 1),

X is(M;n 1)irreduibleforanyM>0. Thusitissemi-stronglyirreduible.

2

Theorem3.14 Let X be a uniformly semi-strongly irreduible shift. Let :

X ! Y be a onjugay, let be k-loal and let 1

be

k-loal. Hene if n

2k+2

k+1

jM 1

n

(X)jC(n) 2

k

X

r= 2k jM

1

n+r (Y)j;

whereC(n)=an 2

jAj bn+

,a;b; areonstantsdependingonlyonk;

k;M;hand

the shiftX isuniformly(M;h)-irreduible.

Proof Consider p 2 M 1

n

(X). Being X

F

n 1 (X)

an (M;h)-irreduibleshift

and being p a pattern of it, there exists a onguration 2 X

F

n 1 (X)

suh

that ontains p in E := [1;n℄

2

and a opy of p in eah translation

E of E

ontainedin squares of size n+2h at mutual distane M+1and positioned

asin Figure 3. Hene :=

n 1

() is a onguration in X

F

n 1 2k (Y)

and, as

provedin Theorem3.9,wehave2=X

F

n+2

k (Y)

. Thenthereexists anintegerr,

with 2k r 2

k suh that 2X

Fn+r 1(Y)

and 2= X

Fn+r(Y)

. This means

that ontainsa pattern p2 M 1

n+r

(Y) with asupport F whose left bottom

p

p

p

p p

p p p

n n+2h {

p E M+2

E ^+h

Figure3: Theonguration2X

Fn 1(X) .

orner belongs to some square of size n+2h+M obtained by overing the

planewithdisjointopiesof[1 h;n+h+M℄

2

(wereallthat[1;n℄

2

=E and

[1 h;n+h℄

2

=E +h

). The numberof possiblepositions of this left bottom

orner of F inside this square is then (n+2h+M) 2

. Let q be the pattern

of of size n 2k dened by q :=

n 1

(p). As one an see in Figure 4, the

patternpdetermines(atmost)fourretanglesintheopiesofqinterseting p,

andheneitdetermines (atmost) four retanglesofq. Wearegoingto ount

themaximalnumberofpointsinqwhiharenotontainedinoneofthesefour

retangles. Firstnotiethatthemaximaldistanebetweentwoopiesofqin

isM+4h+2kand henetheminimalnumberofpointsinthefourretangles

is(n+r) 2

(2(M+4h+2k)(n+r) (M+4h+2k) 2

). Itturnsoutthat the

maximalnumberof points in q whih are notin theretangles is (n 2k) 2

(n+r) 2

+(2(M+4h+2k)(n+r) (M+4h+2k) 2

). Bytherestritionsonrthis

numberislessthanorequaltoe(n):=2(M+4h+2k)(n+2

k) (M+4h+2k) 2

.

Now pdetermines q exepting for at most e(n) points and hene we an

ompleteqinatmostjAj e(n)

ways. Ontheotherside,qdeterminespexepting

forat mostf(n):=n 2

(n 2k 2

k) 2

pointsandheneweanompletepin

atmostjAj f(n)

ways. Indeedqdetermines,by( 1

)

n 1 2k

,asquareontained

inpofsizen 2k 2

k. Thus wehave:

jM 1

n

(X)j(n+2h+M) 2

jAj

e(n)+f(n) 2

k

X

r= 2k jM

1

n+r

(Y)j: 2

000 000 000 111 111 111

000 000 000 000 000 111 111 111 111 111

000000 000000 111111 111111

000 000 000 000 000 111 111 111 111 111

000000 000000 111111 111111

000 000 000 111 111 111 000 000 000 111 111 111

q

n−2k p

1 2

2 1

3 4 3

Figure4: Theonguration:=() ontainingthepatternp2M 1

n+r (Y).

Remark. Inthed-dimensionalasewehave

jM 1

n

(X)jC(n) 2

k

X

r= 2k jM

2

n+r (Y)j;

where C(n) = (n+2h+M) d

jAj

e(n)+f(n)

with e(n) := d(M +4h+2k)(n+

2

k) d 1

(M +4h+2k) d

, f(n) := n d

(n 2k 2

k) d

and the shift X is

uniformly(M;h)-irreduible.

Corollary3.15 If X and Y are two onjugate uniformly semi-strongly irre-

duible shiftsthenh 1

(X)=h 1

(Y).

Example3.16 HereisanexampleoftwoshiftsX;Y A Z

2

suhthath 1

(X)6

h 2

(Y) and hene whih are notonjugate. Consider the shift X of Example

2.16 and let Y be the shift in whih is forbidden to replae eah * with an

a in the onguration (2). As we have seen, jM 1

n

(X)j = 2 (n 1)

2

and then

h 1

(X)=log(2). Onthe otherside wehave M 1

n

(Y)=M 2

n

(Y)and aminimal

forbiddensquareofsizeninY isasquarennborderedbyb'sandonlywith

a'sinside:

b a ::: a b

.

.

. .

.

. ::: .

.

. .

.

.

b a ::: a b

b b ::: b b

| {z }

nn

HenejM 1

n

(Y)j=1=jM 2

n

(Y)jandh 1

(Y)=0=h 2

(Y). This impliesthat X

andY arenotonjugate.

Observe that in this example h(X) < h(Y) beause Y is a strongly irre-

duiblesubshift ofA Z

2

(see[7,Lemma 4.4℄). With aslightmodiationofthis

example(forinstane if

X is theshift in whih isforbidden to replaeeah *

with an aor aneven number of 's), it is still easy to proveh 1

(

X) 6h 2

(Y).

Butwegettheinequalityh(

X)<h(Y)onlybysuessiveapproximationsand

notby previous argumentbeause X 6Y. In any ase, for these shifts, the

omputationoftheentropiesh i

isquitesimplerthanthatofh. 2

Proposition 3.17 If there exists an integer n suh that for eah n n the

sequene(X

F

n (X)

)isuniformly(M;h)-irreduible thenX is(M;h)-irreduible.

Proof Letp;qbetwopatternsof X whose supportsare at distane>M.

Wehavep;q2B(X

F

n (X)

) foreah nnand henethere exists

n 2X

F

n (X)

in whih pand q simultaneouslyappear in asuitable position; weanalways

supposethat these positions arethesamefor eah

n

. Bytheompatnessof

A Z

2

, a subsequeneof (

n

) onvergesto 2 X in whih pand qappear asin

the

n 's. 2

Counterexample3.18 Nowweshowthat thereis anexampleof areduible

shift X suh that eah X

Fn(X)

is semi-strongly irreduible but the sequene

(X

Fn(X)

) is not uniformly semi-strongly irreduible. Let X be the so shift

aeptedbythelabeledgraphinthegurebelow.

q a

~

b

} b

HeneasetofforbiddenwordsforX isgivenby

fab n

ajn0g:

IntheongurationsofX

Fn(X)

thereare nowordsab asuhthat0n

n 2. Supposethat X

Fn(X)

is(M;h)-irreduible. Beinga2B(X

Fn(X) ),there

existsawordwoflengthM hiM+hsuhthatawa2B(X

Fn(X) ). We

musthavein 1andheneM+kn 1. ThisshowsthatX

Fn(X) annot

be(M;h)-irreduibleforeahn(howeverX

Fn(X)

is(n 1)-irreduibleforeah

n). 2

Proposition 3.19 Aone-dimensionalsemi-stronglyirreduibleshiftisuniformly

semi-stronglyirreduible.

Proof LetX bean(M;h)-irreduiblesubshiftofA Z

. Letp;q2B(X

F

n (X)

)

andlet mM. We an alwayssuppose that thelengthsof pand qare both

greater than n. Hene p = pu with juj = n and u 2 B(X) and q = vqwith

jvj=nandv2B(X). SineXis(M;h)-irreduible,thereexistsw2B(X)suh

thatuwv2B(X)andm hjwjm+h. Moreover,beingp;q2B(X

Fn(X) ),

thereexist ;2X

Fn(X)

suhthat ontainspandontainsq. Considerthe

followingonguration.

... p u

| {z }

p

w v q

| {z }

q

...

As one an see, in this onguration does not appear any forbidden word of

lengthnandheneitmustbein X

Fn(X)

. Thereforepwq2B(X

Fn(X) ). 2

Corollary3.20 An irreduible so subshift of A Z

is uniformly semi-strongly

irreduible.

3.2 The ase of retangles

Wenowdisusstheaseofminimalforbiddenpatternswitharetangularshape.

Theentropyofthesequene(M i

m;n

(X))ofminimalforbiddenretanglesof

X isdened as

h i

(X):=limsup

m;n!1 1

mn logjM

i

m;n (X)j:

Let be a k-loal map dened on X. In this ase also the map

m;n is

welldened onX

F

m;n (X)

ifm;n2k+1and itsdenition oinideswith the

denition(3) givenintheaseofsquares(indeed if2X

F

m;n (X)

onehasthat

jD(;k )

isapatternofX). With thisnotationLemma3.2andLemma3.3still

hold. Moreover we have

m;n : X

F

m;n (X)

! X

F

m 2k ;n 2k (Y)

and for m;n big

enough,

m;n

and areone-to-one.

Let :X !Y beabijetivek-loalmap,let 1

be

k-loal,let2X

F

m;n (X)

andletpbeapatternofX

F

m;n (X)

. AsageneralizationofLemmas3.6and3.7

wehave( 1

)

m 2k ;n 2k (

n

())= and( 1

)

m 2k ;n 2k (

n

(p))=p.

Inspite of these results,weannot generalizethe proof ofTheorem 3.9 to

getCorollary3.10. Indeedto eah pattern pofM 1

m;n

(X)oneanassoiatea

patternp2F

m+2

k ;n+2

k

(Y)\B(X

Fm

2k ;n 1 2k(Y)

)\B(X

Fm

1 2k ;n 2k(Y)

). This

meansthatinpthereisaretangleofM

m;n

(Y),wherem m+2

k,nn+2

k

andm >m 2k orn>n 2k orbothm =m 2k and n=n 2k. Hene

itould also happenthat thesize of this retangle is 1nand this does not

allowus to haveaonstantrange into thesumappearing in thestatementof

Theorem 3.9. Forthe same reasons asabove we annot generalize either the

proofofTheorem3.14.

Referenes

[1℄ R. J. Baxter, Exatly solved models in statistial mehanis, Aademi

PressIn.[HarourtBraeJovanovihPublishers℄,London,1989. Reprint

ofthe1982original.

[2℄ M.-P. B

eal, M. Crohemore, F. Mignosi, A. Restivo, and

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