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Lecture Notes by Fields Medalist Bill Thurston on Groups, Tilings and Finite State Automata
Ciroups,Tings and Fttlite State Automata
Sg/989AS C?Lccr"Wliam P.Thttston
Research Report GCG l
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Groups,tilings,and anite stato automataSunIner 1989AMS Colloquium lectures
(VerSiOn l.5,Jdy 20,1989)
WILLIAM P.THuRSTON
Sl. INTRODUCTION
These four lectures will deve10p some ideas involving the geOmetry of grOups, tilings(primarily of the plaFle),nttte state automattt arld dynattcal systems.They tte groupedinto three related subjects which are tied together by coHllnon themes,but are scientlyindependent that it should be possible to understand them independently.The subject of the flrst lecture is a cOnnection between tilings of the pl[me and the
geometry of grOups discovered by Conway a number of ycars agobut Only recently becndiscussed in phnt(tTh01,tCOnWay Lagariasl).It develops a necessary condition for aregion in the plane to be tiled by a given collectiOn of tiles,in tes Of COmbinatorial grouptheory.Thc second lecture also concerns tilings,but frOm a difFerent point of viewi the subject is
the theory of selfsilnilar tilings Of the plane and Of other Euclidean spaces. Many examplesand constructions will be discussed, The lnain result is a characteAzation of the complxexpansion constttlts for selfsirnilar tilings,This subject is c10sely related to the theOry ofMttkov partitions for dynamcal systems atrld flnite state automata. In a certain sense itmay be thought of as a cOmplexincation ofthe PerrOnFrobenius theorerll and its tconverse'of D.Lind.
Word processing on groups,or the theOry of automatic groups,is the subject of the lasttwo lectures.This theory has beedeVe10ped over the last few yetts pttmarily in jointwork of Jim Cannon,David Epstin,Derek HOlt,Mike Paterson,and me(lCEHPTl).An automatic grOups adlmts an algorithm of a rather sirnple type which will tell whentwo words in generators for the group represent the sme element Of the group(C,anangodthm for the wOrd problem ofthe grOup.)MoreOVer,the ttgottthm is sO special and sOsimplc that questions about the angorithrll can be algOrithncally handledi in particular,there is an algorithm which,given a presentation for an automatic grOup,ill constructangothna as above fOr the wOrd prOblem.Automatic groups are ciosely tied tO the thcOry of anite state automata, and the in
vestigation of them is partly mOtivated by the successful applicatiOns lvhich anite stateautomata have found tO practical and theoretical problems in computer science,combinedwith the need tO be able to handle algOdthnically actual flnitelypresented inanite grOups(in pa Cular,fundamenttt grOups of 3-marxFolds,)Many WOrd processors fOr exttnP
the unix utilities grepegrepsed, vi, ctc. cOnstrtlct a flnitestate automatolt wixPyou ask it tO search for a certtttn pattern, and mmy compllers directly use the thcr)r}f nnite state automata at early stages Of thdr tasks(lexical and syntactical antty6)Besides theoretically analyzing the issues invOlved in autOmatic groups,wc have beerl developing computer progrs tO carry Out`wordprocessing on groups'. Automatic groups
are more general than hyperb01ic grOups in the sensc of Gromov. At least most of thesma11-cancellation groups arc automatic.Anautomatic structllre for a group in general produces a kind Of seifsinilar tiling of a
certaan tsphere at ininity'for the group;in particular examples,this space is actually a2-sphere.
These notes are preliminary.Although some PortiOns have becn lwritten carefuny and infttr detadl,there are other portiOns are sketchy and hastily written,and some toPiCS havebeen left out altogether.
The next portion of this text,concerning COnway's tihng grOups(S2-S7)is substantianly a reprint from an tttrticle to appear in the Janua1990 iSSuc of the Attcr,canatcttcary,tThOlThiS Will be a special issuc on gcometry.
S2.CoNWAY'S TILING GROUPS
The problem of deciding whether a given flnite set Of tiles A/11l tile the plane is anundecidable question that is, there is nO general welldenned procedure Rwhich willanswer the question.The same question for a nnite rgiOn in the plane,when appropriatelyformulated,is decidable,but it is nOt easy: it ishat computer scientists call an NPcOmplete question. In practice,it is Often hard to do.John Conway discovered a techniquc using ininite,initely presented groups that in a
number of interesting cases resolves the qutstion of whether a region in the plane can bctessellated by given tiles.The idca is that the tiles can be interpreted as describing relatorsin a group,in such a way that the plane region can be tiled,9nly if the group elementwhich describes the boundattr of the regiOn is the trivial element l.Of course,the word problem for a initelypresented group(the problem Of deciding
whether or not two given words represent identical elements in the group)is anSO anundccidable question. The ability tO answer the tiling qucstions depends in part on theability to understand particular grOup presentations..
s3. GROUP CRAPHS
A convenient way to describe the cOnstruction is by lneans of the CayrcyTPOr gTaPOf a group.If C is a group,then its graph(C)With respect to generators1,92,gn isa directed graph whose vertices tte the elements of the group.FOr each vertex F(C),there will beoutgoing edges,labeled by the generators,and n incormng edgesi the edgelabeled gf connects to gf,As a flrst exaaple,the graph of Z2 with respect to standard generatOrs(,yly
~ly~1)
is the standard gd in the plarle(as in graph paper)The graph of a group is arl answer to the question,`what dOes a group look like?'which
generally is carefutty avoided in introductory courses, Note however that the graph Gf tigroup depends on the choice of generators,and the appearance can change cOnsideli)1,4ywith a change of generatorsi the group graph tes what a grOup lwith a little cxtra structtrelooks like.It is convenient to make a slight lnodincation of this Picture when a generator 9f llas
order 2. In that case,instead of drawing an arrow froHl to gt and another arow from
Version l.5 Jdy 20,1989
gi back to ,we draa single undirected edge labeled gf. Thus,in a drav/ing ofthe graphOf a grotlpif there are undirected edgesit is understood that the cOrresponding generatorhas order 2.The graph of a group is automatically hOmOgeneousi for every element g c the
transformation g is an automorphisim Of the graph. Every automorphism Of thelabeled graph has this form. This prOperty charactedzes gTaphs of groupsi a graph whoseedges are labeled by a inite set ir such that there is exactly one incOIIling and Onc outgoingedge with each label at each vertex is the graph of a group if and Only if it adHts anautomorPhism taking any vertex to any other.
Whenever tt is a relator for the group,that is,a word in the generators which representsl, then if you start from c P and trace Out , you get back to again. If C haspresentati?n
C=(gl,g2,,g.11=1,2=1,,=1) (fthe graph r(c)extends to a 2-complex2(c):seWdisks at each vertex c F(,onefor each relator jRf,so that its boundary traces out the word rif. An exception is madehere for relations of the form=1,since this relation is already incorporated by drawing
gf as tt undirected edge.The 2-complex2(c)is Simplyconnectedi that is,every loOpin2(c)can be contracted to a point. In fact,if the loop is an edge path,the sequence
of edges it f01los descttbes a word in the generators, The fact that the path retus toits starting point lneans that the word represents the identity. A proof that this wordrepresents the identity by lnaking substitutions using the relationsR.can be transiatedgeometdcally into a homotopy of the path in2(c).As a very silnple example,the sy_etric grOup S3 iS generated by the transpositions
=(12)and b=(23).They Satisfy the relation(ab)3=1.The graph is a hexagon,withundirected edges,alternately labeled a and b.
Figure 3.1 Soccerball. A30CCerbarr 13 C,3truCtc2 Pcn3, baby Ttatg an3rng c ttacC3 a rcgttrarccacT n, gctcT 2cttapcentereatc criccJ o/c CCacttr.
A shghtly more complicated example is S4It iS generated by threc elements a=(12),b=(23),nd C=(34)A presentation is
S4=(a,b,Cla2=b2=c2_1,(b)3=(bc)3(aC)2_1).
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To construct its graph, flrst mae sOme cOpies of the b hexagOn fOr the S3 Subgroupgenerated by a and b, and silnilarly make sOme cOpics of bc hexagons. The subgroupgenerated by a and c is Z2Z2,and its graph is a squewith edges iabelediternately and c. Make copies also of csquares, Take one cOpy of each pOlygOn, and nt themtogether around a vertex,glung an a edge tO an a edge, ctc. Around the perilneter Ofthis igure,keep gllng on a copy of the polygon that fltso lf you do this systematically,layer by layer,you will have constructed a polyhedronit is a truncated octalledron. Allthe edges froIII the underlying octahedron are labeled b,lwhile the squares produced bytruncating the vertices are labeled acc.The readcr may enjoy working out the graph of the alternating groupt5, uSing genh
PratOrs a=(12)(34),and b=(12345)Note that they satisfy the relatiOns b5=l andtab=(135)3=1.Try kicking arOund the construction,with white ababab hexagOns andblack bbbbb pentagons.Of course,graphs Of groups don't always ttwork out sO niccly or so casily,but often,for
sirnple presentations,they can be worked out, and they tend to have a nice geometricnavor.
S4.LozENGES'
We will begin with a relatively easy tiling probicm. Suppose we have a plane ruledintO cquilateran triangles,and a certadn region jR bounded by a polygon 7WhOSe edges areedges of the equilateran triangle netwOrk.When can jtt be tiled by ngllres,let us call themlozenges,formed from twO attaCent equilateral triangles?
Figure 4.1. region30?c Pranc,tf!c
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tiled by lozenges. PTlaby rzcnge3,
c?jraerar tTiag!T 3bj
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To analyze this problern, we flrst establish a labeling convention. We arrange the triangulation of the plane so that one set of edges is paranlel to theis,or at O. Labelthese directed edges a,label b the directed edges pointing at 120,and c the edges pointingat 240. This labeling is homogeneous,so it is the graph of a group A. e can read ofFrelators forA by tracing out the boundary curves of triangies: A satislles abc= l andcb:=1. If desired,the nrst relation could be used to elimnate c;the second relation thensays that ba:=b.The grouP A is Z tt Z,as we could have seen anyway by its action onthe plane.The shape of the polygon T is determned by the sequence of edges it traces Outi this is
a word in the generators a,b,c of A. Rather than thinking of it as a word,we prefer tOthink ofit as an element (T)in the free group F with generators a,b,c.The fact that Tcloses up is cquivalent to the condition that the homomorphism FA send (T)tO theidentity. If a lozenge is pled in the triangular network, its boundary can be traced by one
of threc elements,depending on its orientationi that elemerlt is cither1 = aba~lb~1,2==bCb~lc~1,r3=CaC~l
~1 The precise word depends on the starting Point On thcboundary of the lozenge,but starting froHl a difFerent vertex only changes the word by acircdar permuationi the two choices give conjugate elements Of F,The rc` CTttP L iS
deaned by these relators,that is
=(a,b,C1=2=3=1)
Actually, the three relatiOns say that the three generators commutevith(3aCh Othersothat=Z3.We clttm that if the region tt can be tiled by lozenges,then the image r(T)f(T)in
mlst be trivial. In fact,suppose that we have such a tiling.If Ftt consists of a singletilethe clm iSilediate. C)therwise,IInd a silnple arc inR which cuts rtinto two tiledsubregions Rl ind2By inductiO,we may assume that r(Tl)and r(T2)are bOth trivial,where Tt is a polygonal curve tracing arOund:.But r(T)=r(1)*r(T2),SO r(T)isalso trivial.
There is a very direct geometc interpretationi thinkfthe graph()aS the lskeletonof a cubical tesselation of space,0ented so that cubes are on their corners:Inore precisely,
l so that the two endpoints Of atrly path labeled abc are On the same vertical line. The 2-complex r2()is the union of the faces Of the cubes.A lozenge in the pltte is the
l may or may not come back to the starting point in().The invaant r(T)is tileending vertex. This invanant of necessity lies in the kerne1 0f the mapA, which isisomOrphic to Z: it can be described silnply as the nct rise in height.Iftt can be ttled by bzenges,the tthng itsdf can be hfted,L by tile,intO r2(),that is,
into the 2-skeleton of the cubical tesselation. This gives ttother prOOf that the invariantr(T)must be l if tt ctt be tiled.In fact,if you look at a tiling by lozenges,you canimagine it so that it springs out at yOu in a thrcedimensional picture.
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Figure 4.2. Threedilnensional interpretation of 10zenge tiling. jtt Tcgbctrctt by rccs,tec rc,c PatCrr"3c23CrCa cttb'Car tfrrtcntcagarry tc Pra,c/c rcc3. !
ca
R3,
Figure 4.3. Nontileable region. c regfc Pranc c,cr3Ctt by tc P!arcurc c,obCfrea by rcgc3,31CCc 131Ctc cubc,cT t!3 t
Cr3C.
Algebraically,given the word representing 7r,the net rise in height is silnply the suttl ofthe cxponents. The conditiOn is that 7T heads at a bcaring Of O,120or 240the samelength of time it heads at a bearing of 60,180or 300.This condition can be seen in an alternativc way using a colong argument.The triangic]
in the planc have an ttiternating co10ring,with abc triangles co10red white and cb triallgicscO10red black.Each iozenge covers one triangle of each cOlortherefOre,if tt can l)c tild,
VersiOn l.5 Jdy 20,1989
Figure 4.4. Potentially tileable regiono Pc bary c ?|13 Trv r3 ac!sccure,3 ft 7CCt3CTP_tcTec tfr ctt. 4actuarfrrr bc 3: ., rcttgCfrfttg.
the number of white triangles must equal the number of black trianglei).rrhe difFerence infact can be shOwn tO be the net risein height Of ,8Ineured in rnn diagOnals Of cubes.The co10ng cOnsideratiOn reauy gives a more elementary dettvatiOn thtr(T).nuSt Vanishfor a tiling tO be possible. However,this md related co10ring[rgument,s in general cannotgive as much infoation as r(T).one way tO think ofit is that c010dng arguments arethe abelian part ofthe grOup theory.Ifthe group is abelian as in the present case,or rnoregenerally if the subgroup consisting ofinwtttants r()for Cr3tt paths is abelian,then thatinfottation is scient.The angebrttc cOndition that r(T)=l is not suttcient to guarantee a tiling,by 10zenges.
Therc are curves T which go around nearly a full circle,with the lift in(j)riSing considerably,and then instead of c10sing,they circle arOund another 100p which brings themdown tO the starting height. If rcOuld be tiled by lozenges,it could be divided intO twOregions by a fairly shOrt path a10ng edges Of 10zenges; but the rise in height for One sidewould be fbrced to be still positive,which wOuld be a cOntradictiOn. IVe Rwill return laterto give a necessary and scient cOIlditiOn fOr a tiling by lozenges,along with a fOrmulafor a tiling if such exists.
s5. TRIBONE TILINGS
Here is another exttple,fOr which Other methOds seem inadequatc. I nrst hcard thisproblem in an electronic mttl inquiry from Cari W.Lee(ms.ukyedu!lec)in Kentucky.
Last selmester,a number Of us here becttme interested in a cOmbinatOrialproblem that was mtting the rounds. I'm sure youiready have heard Ofit, and we heard a rttOr thatOhn COnway had sOlved it. It cOncerned atriangular aay of dots. The prOblem was tO pack in as many segments as
Version l.5 uly 20, 1989
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pOssiblc,wherc each segment covered threc adjacent dots in Onc Of the threcdirections, and no wO segments were aniowed to tOuch.Is there any size
conflguratiOn that admits a packing such that each dOt is covered? Do youknOw anything abOut the status of this prOblem? Thanks in advance.
I hadn't hettd Ofit,but l asked COnway abOutit. /e sat dOn tOgether,and heorkedout.
Figure 5.. IIiangle Of hex4gons.,3 bc tfrcbyrfb,cs P
Ttgurar arTay?ctta3, C 3tdC,
This question can be alternately formulated in teS Of a triangular array Of hexagOns.The problem is to show that one cannot tesselate the regiOn using tiles made Of threehexagons hoOked linearly together. MOre generally,One can ask fOr the minilnum numberof holcs left in an attempt to tile the region by these tiles.If the region has side length,then the number of hexagOns is(+1)/2.A flrst,
necessary condition is that tt Oris divisible by 3,that is,is cOngruent to O or 2mod 3. Note that ifit is ever possible tO sOlve the problem whenis cOngrucnt to 2 111od3,one can extend the solutiOn by adding a row of tiles a10ng one side,tO derive a sOlutionfOr+1.Label each side in the hexagonal gd with an a,b,Or c,according to the directiOn Of thc
edge:a r it is parttlel to the a axis,b if the angle ttom the is tO the edge(meaSurtdcounterclockwise)is 60,and c if this angle is 120.Thus,the sides Of every hexagon alelabeled abcabc.This iabeling gives the lskeletOn Of the grid the structure of a grOup graph,where the
group lsA=(a,b,Cl2=b2=c2=(abC)2=1).
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The grOup is a grOup OfisOmetries of the planegenerated by 180rev01utiOns about thecenters of the edgesiit alsO cOntalns the 180rev01utiOns abOut the centers of the hexagons.
The grOupis sOmetimes called the(2,2,292)groupA path 7r in the l,skeletOn Of the hexagOnan gdd now is determned by a wOrd in thegenerators Of Ao We prefer tO think of this in a slightly difFerent way: T deterrmnes anelement a(T)in the free product F=z2Z2Z2We are particularly interested in c10sedpaths, that is,elements Of the kerne1 0f j A. lunfOrtunately,this kernel is inanitelygenerated: it is a free grOup whOse generatOrs are given by arbitrary paths P
,fOl10wed bya circuit arOund One Of the threc hexagons at the endPOint Of Pl,f0110wed btt the p1.The standard tile, let us call it a rfbc, can be laid in the plane in lhree difFerent
orientatiOns.Circuits around the tribOnes in these three OdentatiOns trace Otit the elements
r=(ab)3c(ab)3c
ri= (bc)3(bc)3 =(ca)3b(ca)3b.
;litgli::,[Ftfgi:l)S::lri88ng18[lded by7r Can be
Figure 5.2. inribones irl three OrieFltatiOns./OT arfbc,lin aarray orcaag3,71i ttTTccTCay3,
r=(a,
must be trivial.The relatiOn ttL says that c
COEIjugate (ab)3 to its inversegenerates a nOIIal SubgrOup,
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PcTc arc Tcc P33brr OT,caa, n3rabcrg Cc,, cy Tc rabcrc
b,cl2=b2=c2== = 1)
cOIIjugates(b)3 to its inverse. Observe that a and b alsOin fact,this is already truc in F.In Other wOrds,(ab)3and it commutcs with every word Of even length. similarly,
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Figure 53. Second hexagonal group. rc grOP att ar3a3a plaP3TPc
tc egc3 9ettagar ttrgc Pra,c. ,
(bc)3 and(Ca)3 generate nomal subgrOups.TOgether,the three elements gencrate a noalabelian subgroup J of T.
To form a picture of T,let us arst ik at the quotient grOup = T/J =(a,b,Cl2=b2=c2=(ab)3=(bc)3=(c)3=1).The graph of tt ctt readily be constructedi take an ininite collection of three types Of hexagOns,with their edges iabeledby the relations(91,C2and93 TheSe glue tOgether to forlrrl a hexagonal pattern in theplane,where cach vertex has one c edge,one b edge,and one c edge incident to it. Thegroup tt acts fttithfully as a group of isometries of the plme,generated by renections inthe edges of this hexagonan tiling: it is a triangle group. It is curious that even though thegroups A and tt and the labeled graphs(A)and F()e dfFerent,when the labels arestripped they become isottorphic.If the region tt can be tiled by tribOnes,then ()must map to the trivial element of
T,so it maps to the trivial element of b.In our case,the region is a triangular aray ofhexagons,and its boundary can be ttten as (T)=(ab)(Ca)(bC) Obviously,ifis a multiple Of 3,the image r()/J in tt is trivial.In the Other case,
that n is 2 more than a multiple of 3,it is also trivial. This is casily seen by tracing outthe curve in our array of hexagons,or by noticing that One can add additiOnal tribonesalong onc edge to form a triangular region with side length+1,Which is a lnultiple of 3.Since wc have pushed T only across tribones,r(T)is the Same for the twO cases.
Since tt was nOt Sttcient to detect the nontriviality of r(T),we need to nnish ourjob,and build a picture of tr. First,loOk at the path in the graph of T deterrmned bythe eleIIlent.Start at a vertexwhere the circuit Cl=ababab goes cOunterclocttwiscaround a hexagon. Thengoes counterclockwise around this hexagon,then a10ng the cedge,clockwise around the91 hexagOn thrOugh that vertex,and back a10ng the c edge to
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Figure 5.4. Alternate image of tribone.3at'3'Ctt linc gTttP3 T anc,ce=P/. T3c graPo/tc gTttP.cattcc cCrcrc,3C.
Jy c3tTuctF, c T,bcrrat'o.3 aTCr3c tttagcCc tTjbc TcracclTcre3 cttag3,0CC CrC,3C
Figure 5.5.Alternate image Of a trianglec rc tTttrcT(ab)(Ca)(bC)n,31ZC =3m rn=3 +2 P3 c tTfar ercc f" . c ttTatt ab i =3 ,trace ttc ttT3T ga tc cccr. =3 +2,5taTb cc cT,
close.In particular,the signed tota1 0f l_1lexagons enclosed(cOunted accOrding to degree
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of winding With counterclockwise circuits cOunted positively),iS OIt is not hard to dcscribe now the full grOup r, which is an extension of the form
_z3 .IVe can interpret tt clement Of r to be a vertex in the graphof tt, tOgether with a path P from tt to , subject tO the cquivalence rclation that ifis another path from to%then Pq if the signed totans of91,(92, and C3 heXagonsare a11 0(Of COurse,if we pick One path such as P frOmto ,then other paths from*to are deterIIllned by three arbitrary integers,which specif37 theSe signed totals,)WViththis dcinition,the relations 71re ObViously satisded,hence the grOup so constructed isat least a quotient group of tr. ]But we have adready seen that the kernel J of the mapT is abelian,and generated byf. In the cOnstruction,this kernelis the free abeliangroup on the t,so it lnust in fact give r.Once wc know T,we can read r(7r)by inSpection.As we saw,it sces to consider the
case =3 ;the in iS CFCtt,Which is obviously not l,so the tiling is impossible`//
One can ask whether this lnethod gives a lower bound On the number Of holes one isforced to leave,in a partial tiling of jR by tribOnes, To study this question, we shouldexamine the subgroup r Of T generated by elements ofthe fOrm r(7),Where T is a path inthe graph of A going fromto some point ,circmnavigating a hexagOn,and retuming. Inother words,I is the kernel ofthe map rA.Note that ()has the form gbcabcg~1,where g is arbitra=y.In the group 7,abcabc acts as a transiation.The cottugates Ofabcabc in tt are translatiOns in three difFerent directions spaced at 120 arlgles, and thesubgroup they generate is isomorphic to Z2. Inf,there are actually an innnite numberof difFerent cottugates of abcb if g acts as a transiation in,then the commutatorgabcabcg~lcbacba is trivial in tl),but it lnight not be trivial in tri this path may enclosean arbitrary number rn of hexagons Of type(1,and an equal number Oftype Calld(3The subgroup F is therefore a nilpotent group,generated by s i:abcabc,t==bcabca,
and u=l 23,With presentation
r=(S,,ul,l=,ul=1,,=u3).
It is casy to chcck that every element Of r is rezed as r(T),fOr some simple ciosed curve7r in the plane.Even though the invanants associated with triangular regions take larger arld larger
valucs in r,this does not give any infoation limiting the number Of holesi for instance,threc holes gfbcabcgl Can yield u, for arbitranly high In fact, it is possible totesselate the triangdar region of sizeWith tribones except for l h01c,if=1(3),byplacing the hole exactly in the Hliddle,and then aranging concentric triangular layers oftribones around this holc. From these cxamples, tribone tilings with 3 holes are casilyconstructed when=0(3)or 2(3).It dOes give some infoation,hOwevert in the cttethat2(3)or=0(3),the COttugacy Class changes(inCreases")with,which impliesthat the length of the minilnuHl ciosed ioop enclosing all the hOles has to go to ininitywith . In the case = 1(3), the COnJugacy class of r(7r)iS COnstant since the region
carl always be tiled with a single hexagon missing,r(T)iS COnjugate to abcabc.HOwevcr,thc actual word changes lwith,WhiCh implics that the IIlissing hOlcannOt be tOO ciose
to the bound . Perhaps a careful ttalysis would shOw that if there is a single hole,itmust be exactly in the center of the triangle.
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s6, DoMINOES AND LOZENGES REVISITED
Conway's tiling groups are quite versatile, provided you can work out the group deteined by the tileso Even when(or perhaps especially when)the invariant r(7r)giveSnO infoatiOn which could not have been easily obtained by other rneans,the geometricPicture of the graph of the grOup can sometilnes be exPloited to give not just an angebraiccriterion,but a precise geometric criterion for the existence of a tiling.When C is ahng group(with presentatnven by a set ofles),We dettc a measureof area in2(c)to be the ttea detted by prOjection to the planet the area of a 2-cell is thearea of a correspondngWhen the ttgebrttc invariant r(T)1,the curve T boundnglitts to a ciosed tt in(C).We Can ask,whatis the minimum area of a surface S in F2(c)with bounda?This area is necessarily at least as great as the area of.If it is equal,then the images ofthe 2-cells of S must be didoint,SO that they form a tiling of.Thereare severan approaches which are sometilnes successful for cttctllating this lninimal area,but there is one particular situation when there is a really deJttitive solution:when2(c)
can be enlarged,by adding 3-cells,tO make a contractible 3-Inanifold. In this situation,there is a``max now min cut''principle which guarantees an efncient algorithrn for flndinga minlman surface.Rather than going on with the general theory,we will illustrate this with two examples.First we revisit the lozenge question.If tt is a union of triangles in the plane,and if andare vertices in iR,Possibly o
the boundary,defhe(,)tO be the minirrlurltl length of a positively directed edgepathin(pOSSibly going on the bound)jOining to,Thisdistance"function d is notsyetric,since we cannot silnply reverse an edge path. Any closed Positively directededge path has length a multiple of 3,so the(,)iS defhed moddo 3 independent ofpath, The three vertices of a triangle take the three distinct values lnodulo 3. IfR isconnccted,it is always possible to flnd at least one positively directed path froHl to,SO(,)iS Well_deined. tConsider the lifting of any tiling of tt by 10zenges to the cubicai network,2().ThiS iS
determined by a height function()fOr the vertices . We can choose the vertital scaleso thatis integer-11lued,and cach edge Of a lifted 10zenge increases in height by li the
!:13:2illtetti)
!'::::irifi:lllitf1:fiFyCOnditionthatjRcanbetiledi foranytwoverticesandIf 7r Satisflcs thisccessaryconditiOn, then there is a unique maxilnally high 10zenge
(3)=p(,)}tiling: deane
To produce the actual tiling, place a 10zenge so as to cover an edge where the hci1lLchanges by 2. Since the three vertices of a triangle take distinct values mOdulo 3, :ttisinceincreases by at lnost l along any edge,each triangle has exactly one edge whercchanges by 2t therefOre,the collectiOn of iozenges is a tiling
13VersiOn l.5 uly 20,1989
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r/,
Figure 6.1. igh lozenge tiling.bay cuttC
rc lgc3'' rzcttgC frg cttPat'brc c
There is a simple angodthm for quickly computing,and the tilingi rather than spell itout,we will describe the analogous angorithm fOr dominoes.
A closed path T in a square gd can be descbed by an element a(T)Of the free groupF(,v),WhiCh maps to the trivial element of the A=Z2.If the region tt bounded by Tcan be nlled with dOminoes,then the image r(T)of a(T)in the dominO grOup
G=(",ylV2=72,y2=32y
must be trivial. (What does the graph of C look like? We can construct a picture in R3, aS f01lows.
Fill theyplane with a black arld white checkerboard pattern. Above the black squarep,11p,11,construct a righthanded hehx,jdning(0,0,0)by a hne segmentto(0,1,1),tO(1,1,2),(0,1,3),(0,0,4),and SO Oni the r and y coordinates here marching forever aroundthe bouttdary of the square,while thecOOrdinate increases by l each move. Silnilarly,(0,0,0)iS COnnected to(0,1,-1),CtCo COnstruct a sinilar helix above cach black square.Label each edge Or y,accorng to to its image in the planeo Note that this creates lefthandcd helices above the white squares.The boundary of any dOmnO in the plane lifts toa closed path in this graph wc have constructed. Since the graph has a silnplytransitivegroup of isometrics,it is the graph of a grOup. Since it satisfles the domnO relations,it isat least a quotient group of the dOmino group G.It is not hard(and strictly speing,itis not logictty necessary)to Verify that tllis grapll is indeed the graph Of G.The curve 7r lifts to a curve tt in thc graph of G. A cOnvenicnt way to denote this,ill
the plane,is to record the height of the lift next tO cach vertex Of T in the plane. Tllc ruleis silnple: one can start lwith O at some ttbitrvertex. Along any edge of 7r WhiCh has ablack square to its left,the heightincreases by l. Along any edge with avhite square to its
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Figure 62. The doHinO grOupe rttccrtccs cr tc3?arC3 a CcccTbar,c Prac r5 3 gTP, 3artng atroTctgbT,gcrfcc3.
graP0/c grttP J3acTacc33, y Pol. rj3 rrutttrat'
a un/3?arC ttycTc3 ttj c!3
Figure 6.3. Dornlno tiling.gTttP,
ttrling by p ttttlincs,r"cc graPc ttt
left,the height decreases by l. A necessary conditiOn thatR can be fllled lwith dOnOesis that thc height after traversing once around the cutte is O.There is a criterion and cOnstruction fOr a dOrmno tiling,ana10gOus tO the construction
for iozenges. Here is how the formula can be workld out,on a sheet of grid paper. Begin,as above,by labeling thc height Of each vertex of 7r. The heights cOnsist of the integers insome intema,,lWe will cOnstruct a hdght funcOn On di verhces Of,benningWith+1,and working upo Suppose,inductively,that we have nnished with all verticesOf height icss than or equan to For each vertex Of height,and fOr each edge c icading
15Version l.5 July 20,1989
Figure 6.4. Donuno root a 16 x 16 3aTC rin grP or a,y tfritt by
rf315cfrtls terc7:,OcJ
,c ca T "c 3, C PPr,c
a3e,gcst rc graPo/c
Figure 6.5 ]DoHo bubble. rfrrttJrati33 bce3tc rttestling byf"eJa3acccrbar.rcy are f30TPfC,,crry by a90Ttattorccccrbarrt,tercattling crj.rc uPPcTfr33lin tcPPCT Pranc as ttcrr a3c ttPPer 3rraCc o/tc bubbre,c rttcTfr,. c rttcT Prac
a tc rttcr sTFaFC O/C bttbbre,c bubbrc tcy rOTtt ccr3C3er q/yfrling
byo7c3,P03brC tfring3 are FriC'P3CliJzctt31 c3?TC fPscitz
C3ta=,a3m ea3TCcaan ttctric.rc rlf30rinfriing3,riincc graP0/tC grP, jC grftt jc gc3 tCro, arc cttactry sucIPscz c,.
from which has a black square on its left,consider the second endPoint lof e. If the
16(VersiOn l.5 dy 20,1989
height Of tt has been previously denned)and ifit is not greater than+l leave it as is. Ifthe height is denned and greatcr than+1,then a dorrllnO tiling is impOssiblei give up.Otherwisettdenne the height Of tt to be+1.If this procedure reaches a successful conclusiOn, cach edge of has a difFerencc of
heights Ofits two endpoints of either 1 0r 3.(Note that the height mOdui0 4 is determnedby the pdntin the plane.)Erase ttl the edges whOse endpdnts have a dfFerencc of hdghtof 3. /hat is left is a picture Of a tiling by dOmnoes.
s7.TRIANCLES~ Here is a related sequcnce of tiling prObleHls which are resistant tO direct attempts at
generttt solutiOn,but translate nicely intO the reattn Of group theory.Consider, again, a triangular array of dots, withdOts On each side. Is it possible
to subdivide this aray into dittoint triangular arays of dots with nf on each side? /csuggest the reader indulge in experilnentation with a few cases,beforc reading further, Forexmple,the cases y=2 1withr ranging from 2 to 12 are interesting.
{
t
As in the case of the tribones,this transiates intO a tiling prObicrn: given a triangular ay of hexagons withNhexagons per side,can one tile it by tiles rM which are triangularaHays of hexagons Aper side?We can cxpress this with notation as in the casc OftribOnes:label the edges of the underlying hexagonal tiling by a's,b's,and c's. Given a path 7r in theplalle,it is dcscbed by an element (T)of F=(?,b,Cla2=b2=c2=1).If the regionbOunded by 7T Can be tiled by the copies Of ttu,then the image r(7r)Of (7r)iS triVial inthe group
GM=(ab,Cla2=b2=c2=1,M=1),where tM represents the boundtty curve of the tile TM,
tM=(b)M(C)M(bC)M
A parallel of hexagons with Ahexagons on one side and+1 0n the otheF Canbe tiled by two copies ofThiS implies that(b)M COmmutes with(bC)+l and with
(Ca)M+1,d sorth. These relations imply that(ab)M COmmutes with(bc)+1),and they also imply that
(ab)M+l COmmutes with(bC)M(M+1)combining these twO facts,it follows that(ab)commutes with(bc)y(+1)Geometttcay,One canle an+1)paranelogram
Prq:&parttdograEIt will simplify the picture at this pointifwe pass to the subgroups FC and Ctt generated
by words of even length. Since all relatiOns have even length,the wOrdlength modul(2describes a homomorphisIIl of jand Cttf to Z2,and these subgrOups have index 2. TgrouPFC is the frec grOup on 2 generators,but a more syretric description is
Fe=(,y,|y=1),
17Version l.5 July 20,1989
where r=b)y=bc,and=ca.A presentation for the group Ctt iS Obtained byattOining relations coming fromto FC: it requires two relatiOns,one obtained bytranscribingdirectly,and the other transcribing the cottugate of tby an element ofodd length. Using==l and byb==1,wc obn
MyMZM= Mtty zM =
Gtt haS an interesting anternate generating seti X=,X="~(M+1),together withy,y,z and Zdeined silnilarly,clearly generate, We have already seen that,y,andZ commute with,y'and z.The elements s=M+1,=yM+1,and u=ZM+commute with everything in G,
sO they generate a central subgroup J which is Z3r a quotient.Let C=G/JWewill andyze the structllre of G,d ttOm that constnlct C.In C,X,y,and Z sasfy relatns
i yZ=1,M+1=yM+1=z+1=1.
These relatns descbe the odentatnpreserving+1,M+1,7+1)tde group,which acts as a discrete group of isometries on the Euclidean plane if A:=2 and on thehyperbOlic plane if A>2. We have not checked that these generate arr the relatiOns onX,y,and Z,but weiediately deduce that the subgroup r of Ctt generated by,yand Z is a quotient of this triangle group.But there is a homomorphismof the originalgroup CM to the full triangle grOup(including renections),defhed by sending a,b,and ctorenections in the sides Of a T/ +1),T/ +1),T/ +1)triangle.The relationM=1is satisned,since ih this group(b)M=ba SO that(b)M(Ca)M(bC)y=(ba)(ac)(Cb)=1
Note that r Sends x to ba,y tO ac and Z to cb,that is,to the standard generators ofthe+1,M+1,M+1)triangle group,and it sends s,arld u to O.Threfore,Ir isisomorphic to the odentable+1,y+1,+1)tdangle group.A siHlilar analysis shows that the subgroup ttr generated by, y and Z is theoHentable(f,,Jttr)triangle grOup.ThgrOup acts on the sphere,the Euclidean plane,or the hyperbolic Plane whenF=2,f=3,or ttf4.The ana10gous hom01morphismr maps cM to the full(M,M,M)triangle grOup,mapping ,b,and c tO the standardgenerators.The two subgrOupsand rintersect trivially(aS SCen from the efFects ofand),they generate C,and they commute with each other.Therefore,Gtt iS the productrrf the two triangle groups.Now we nced to determine the kemel y ofthe qllohent a G,and the structure
of the central extension. As in the tribone case,we can do this geometcally,in teAS Ofareas enciosed by curves.The graph r ofthe full+1,M+1,M+1)thangle group isformed from copies of threc kinds of 2+1)gonS,With perimeters labeled(ab)M,(Ca)Mand(bC),witht one of eacl kind meeting at each vertex. Arrange the orientation so that lis an``even'vertex that is,the a,,and c edges emanating from l are in countercl(ckiseorder.Then the relation tbtted at encloses positively one cOpy of each type of polygon,while the cottugate bb encloses negatively onc copy of each type Of polygon.
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Si1ly,the graphr of the fun(,7,M)tttantte group is made ttOm threc hnds Of2A/gons. Starting at the l,which He suppose is an even vertex,the relatiOn tr enciOSespositivelysone copy of each tyPc Of polygOn,while bb enc10ses negatively One cOpy ofeach. However,in the case n:=2,there the entire graph is flnite: it is the l_skeletOn Of acube,and the number of polygons enclosed by a curve is well_deined Only mOdu10 2.
First letdett with the case>2.We can dettc an cxtenSon rf of ctt as anequivalence relation On elements Of Fe,f1lowso An clement g of FC de,errlllnes pathsP(g)in r and P(g)in.We dettc g to be cquivalent toif P(0)ends at the same pointas P(),P(g)PndS at the same point as P(),and if the c10sed iOOp P(g)P~1()enciOSesthe sttne numbers of abpolygOns,bcPOlygons,and cpolygons as P(g)P
~1().
In particular,an element of the kernel of the lnap off tormaps to c10sed loopsin both pictures,and is determned by the triple Of difFerences of the number Of POlygonsenclosed.The cicments s,t md u mapto(1,0,0),(0,1,0),and(0,0,1)It f0110ws that =C,and J=Z3(prOVided>2.)The boundary of the size Ar triangic rN can be deschbed by the lerrlent =
(ab)N(Ca)N(bC)NThe path P()inC10Ses Only when F is 0 0r-l mod+1,whilethe path(t)C10Ses only when NO Or-l mod.nce M and+l are relahvepme,there are four soluOns mOddOM+1):0,,M2_1,_1.For vttucs of Nsasfying one of these cOngrucnce condtn,the invariant in Ctt iS O,SO the invahantin J;it is a positive multiPle Of(1,1,1)in ani but the trivial case N=M.
THEOREM(CONWAY).When N>>2,the tHttguFar array tt oF hexavgons cannotbe tiFed by rM.
This analysis has an interesting vanatiOn case n:=2. Civen two elements g andofFe,we cm defhe them to be equivalentifP(g)and P()haVe the same endpoints,P(g)andP'()have the salnc endpoints,and if the numbers Of polygons Of the three types enciosedby the path P(g)P()Tl iS a multipleof(1,1,1)WhiCh has the same paty as the numberof polygons enclosed by P'(g)P()~1This dettes a central extensiOn Of rby Z310ddo the subgroup generated by s22u2=1.To justify that this grOup is in fact C,Wemust prove that s22u2=(ab)12(ca)12(bc)12=l in this grOup,or even better,that it isPOssible tO tile12 SuCh a tiling can be found fairly casilysee flgllre 5.1,the 12-stackby 2stacks. The computation of the mod 2 invanant for tilings by's can be rather annoying
when done dircctly. However,there is a ncat trick,which enables one to sce this invariantgcometrically: most regions which have a multiple of 3 hexagons can be tiled casily by 's a10ng with tribones. The boundary abbbcbbabc of a tribOne lnaps to ciosed Pathsin both r and In F,it encioses a net of O of each type Of hexagon,as we saw before.In,this curve winds counterclockwise l.5 revolutiOns about arl abfacc Of the cube,goedown a cacdge to thc opposite face,winds l,5 revolutions counterciockwise(with respcctO the odentatn of the square),and gOes up agttn to close.Ittherere eqdventintes of which kinds of stuares it enciOscs,to abcabc,which is an Odd multiple Of(1,1,1)Therefore,if a region can be tiled with a collectiOn of2L's together with an odd number of
tribonesi it cannot be tiled with t12'S For ON
Figure .1.The 12stack b2stcks.Ttrfagre2 Cabcrctt by ttL.
:i3iCkly nnds htt h he cases,,and's,while,,and rll can be tiled.Gen my tthng Or ttihng,!t|:l:ti:2x6d3x6P_ttdOgam
),,
JexagOns c byr
( S8. SELFSIMILAR TILINCS
&Fbr any Tuhng machine,it is possible tO cOnstruct a Flnite set Of tiles such that these tilesf[l tO tile if and only if the Tung machine eventually cOmes tO a halt. The Output ofthe Tung lnachine is recOrded,in teS Of the tiling,by the tiles at a certaan sequence Ofit:a ttndwhtt happentt One::1iieiFthe ttSdttng ttngOmorphtt by a simiLttyemeEventually we will prOve a rather generan thereom,in which we characterize the set Of
similarities fOr selfsimilar tilings of the plttlc(Or of higher_dimensiOnal spaces.)ThiS iS
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closely akin to the constructions for AttarkOv partitiOns in dynan ical systems. !Hovever,what is anso interesting abOut this subject is the particular constni3tiOnSat issuc is hOwsilnple andhow nice can sclfsilnilar tilings be. The gencrd construtions be very 100se,andyield estimates forimmense numbers Oftiles(or elementsa Markov partitiOn.)Therefore,we will take the tilne tilne tO discuss sevcral particular e,mples and cOnstructions.Picces of this mateal are to be fOund in a number Of sOurces, and l have not been
orgatrlized enough tO work out the appropriate attributionthishting shOuld be thOughtof as semexpositOry9 and many things l say here are not ottginal with me. The theOryof Markov partitions underlies mOst of this,and there is a very extensive mathematicalliterature. Some of the selfsiHlilar tilings can be derived frGim AnosOv maps of the torus tOitself;Vo Arnold deve10ped some connections anong this line,particularly fOr the PenrOsetilings. The theory of exotic number bases has a literature with which l am not very filiar;in particular,though,it is discussed in]Knuth's Arof computer programmngRick Kenyon,cunrently a graduate student at Pttnceton,has wOrked Out some beautiful
additional constructions lor selfsilnilar tilings which l will n,Dt discuss here.
Figure 8.. Tlling of the plane by trionilnoese r13 13 a PTtt?0PCTttttcSClr31rar tfrlingc Prane by3aPctt tT es,rc ctt3: CtT 13 2.rc
trtjcs cttc l4 oTicatatt3.
There are rather trivial examples Of this phenomenoni for instance, the tiling of theplane by squares is selfsimilar,with the subdivision rde that each tile subdivides into 4subsquares.A slightly lnore complicated exalnple is anshaped trionilno,Inade frOrn three squares
glued together. It can be subdivided into 4 silnilar flgures of half the size. If you expa=dthis shape,arld then subdivide agttn,and cOntinuc indeflnitely,the limit of this prOcessyields a scifsilnilar tiling of the plarle bytrionilno's
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Figure 82. Second stage in hexagonal fractal tiles. rc array cag3,TCby rcPcacry scc._crrg,cregrttPgtfrc3 arc brttc ttresf 3CC,rccr
This sme process does not quite wOrk with hexagons.A hexagon,together with its 6neighbors,looks roughly like a hexagonbut not exactly.Let's modify the shape a bit untilit works.As a second approximation to a selfsimilttr
tiling,instead Of a hexagon,let's use a hexagon together with its six iHInediate ncighborsas a tiling of R2. TheSe tiles can be used to tile the plane in a hexagonal patterni to doit,use a periodic sevencoloring of the hexagonal tiling. If onc of the colors is blue,theneach tile is either blue,or touches exactly one bluc tile. Use the clusters centered at bluetiles to tile the plalle.To continue the process,ihelpS tO renoHnalize the new tiling,so that the latticc of
center points of the blue tiles is lnapped to the lattice of center points Of ali tiles. Thuswe get a seven coloring of the new tiling of the plane. Group the new tiles by sevens,andrenormalize. This process,iterated, converges to a tile of a certaln fractal shape. rhelimiting shape is homeomorphic tO a disk,and it tiles the plane in the same combinatorialpatteas thc original hexagonal tilingbut nOw the tiling is selfsilnilare When wetransfothe plane,considered as C,by the transformation z where =(5+ )/2,then the image of anyh thc hmit pattem is the utton of seven dles.This cxmple has the feature that all tiles are congment,so there is only one rule for
subdivision.There is a rich collection of examples which can be cOnstructed similarly,butthere are even lnore tilings which have several tile types.We will construtt a nrst exampleon the real line. The sirnplest possibility is that there are only twO types of tiles,whichare intels:let us call them A andWe specify that when am A tile is enlargeditsubdivides into an A arid aB, and when a tile is enlarged,it becomes an A. Tllclthe secOnd subdivision of A consists of ttBA,which goesto ABAAAABAAABAABABAAAAB ctc.The nurnbers of Atiles and tiles,
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A is subdivided,are(1,0)(1,1)(2,1)
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Figure 8.3. Fourth stageby TcPcatcttry scc_crrt,
(3,2)(5,3)(8,5)..constant rnust be the golden
satisfy = ttb and b=cigenvaluc ,say = andworks.To construct tt actual selsimilar tiling with tilis pattem,we can start tt A tile p,l,
The exp ion Of this tile implies that there is ti J tile attaCent to it,l,+1 T},jeexpansion of the B tile deines another A tile... and so On. Eventually we get a tiling ofthe positive real linc. The pattern does nOt actually cxtend to a strictly selfsimilar tilingof the negative real line.If wc Put a tile t-1,01,it eXpttnds and subdivides into a singleA,which subdivides back into an Ao We get a pattern which rcPcats With PcriOd 2.Ifwe used instead the rde for the second subdivisioll,AABA and ,the tilingwould be strictly selfsimilar.
There is a famous 2-dimensional generalization Of the preceding example,due tO]RogerPenrosc,and known as the Penrose tiling. Thc Penrose tiles come in several vattants Wewill describe a version using two shapes Of isosceles triangles,the two triangles havintt sidelengths in the golden ratioo To properly deanc a rde for subdivision,however,we considerthe twO triangies to come in two types,a lefthanded form and a righthanded fOrm. Thisis graphically represented by putting dark edges on two of the sides Of each trianglQ. Inthe tiling,dark edges will lnatch with dark edges.Across any undarkened edge is a mirror image Of the given triangle. The union of twc
such mirror image thin triangles is a,te;the uniOn of twO mirror image fat trian31es is aar. 'The rules for subdivision can be applied recursively tO get flner and flner subdivisiOns
in hexagonal frilctal tiles. rc arTayteTcgrOPtt tfrc3 TC brttc
q/c3brttCfre3f/0rtt rccr???????????????????
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Figure 8.4. Pnrose tiles . e rOur b31C trfgre3,1,1,2,at2, a"ca TanccTsf3. rhCC,C33 13 33ctt by tc Tcttpes Or 3eS Cact tTfattrec Dar c3frr arttayS ttatc aTcttcs,tc tfre a arng a 3c fta arcttc ttfrr arttay3 bCc ttfrTT ttc.
Figure 8.5Penrose tiles 2.ntre rOr jttbJjc/0T b5tc tT,attrc3.Trc Jubttfc3riiagre3 aa rar trfattFcaTfare 3b,c31o,c rat tnia,ca,Cf,niac.
of a given triangle, orthe plane.To get a selfsimdlar
of a righthanded thintriangle by a cOmplexof the subdivision to
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they can be rescaled tO give tilings of larger arld larger regions in
tiling of the entire plane,we can exploit the fact that the subdivisiOntriangle contaans a rightahanded thin triangle. Map the subdividedattle transformatn(a ttmnarity)WhiCh sends the sman tanethe Odginal. If this prOcess,subdivisiOn f01lowed by expansion, is
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Figure 8.6. rescaling triangles. rtT,arcraTe3 matCttc slnarrer.
Figure 8.7.T,gre.
iterated,we
3ac3blj ,30/tcrescare
eXPanding subdivisions. rc subf31 Tgccraf TcscarC
obtadn a tiling of thc entire plane.
One construction for selfsilnilarsolitaire. First we will go over the
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S9. SOLITAIRE
tilings can be descdbed in teHis oftheory for real numberst it is a nice
a kind Of galnc Ofspecial case,and
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Figure 8.8.Penrose tiles 3.A jer/_jffrar Pcrsc tfr f 3 ctryc cTar
gTP or Ter 10 caa bc c,3RLCtcby bc, 10 TFaprcs artcTa a_
C"C33 arTaJcab c rf|. rc scc 3bfj3 Or3 a cTaf c,3
c or,r cttguraf c ccttter, 73 PICturc J c ttT353b313,iit
tc cpe3 0r trfa,re3 at c ani3 3tagcs ag cc33 3Care 3' e recttTs'cstttcture.
enableS us tO indirectly construct some tilings of the plane.Consider any real number greater th1.Thcre is a clonical way to cOnstnict a
baSe SyStem for the real numbers,which cOincides with the usuan dettnition if is aninteger. The deanitions are vry simple: A not necessarily proper representation in base meanS a sees
_l~1++0+_l~1+.,with>0,altt written
_1.0._1..
Such a series need not even converge,but in practice the dittttsf will be bounded,sothat it cOnverges to a positive real number which it represents. Improper representationshave a lexicographical orderingi they are Ordered according to the flrst digit in which theydisagrec.A representation is stActtt PrPcr if the dits are bounded,and if it is lexicographically
the greatest representation of the positive real number it represents. It is ttcary prPcr ifeach anite truncation is 3niCtry PTPcr.Thusin base =10,.999.is wely but notstrictly prOper,since l.000. is greater in lexicographica1 0rder. ImprOper representationiare frowned upon in school,but they have their place. Once,a niece Of IIline in kindergaI`tcntold lne she knew what 3 tilnes ll was:33.I asked`Well then,what's7times ll?"
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``It's seventyseven.''``Okay,I bet you don't know what 12 times ll is."TwelvetWelve",she gleefully replied.IVe agreed that twelvetytwelve is a perfectly gOOd number,we knOw whatit rneals,but
if we were talking to other pcople we wOuld teu them onc hundred and thirty twOfOrus,tweivetytwelvc is just nne.
The key to understanding any base iS itS carry,c?cnCC,The carry sequence Ilay bedescribed as the sequence of digits Of the representation of l which is weakly prop,r,butnot strictly proper.The carry sequence carry()may be constructeby a dynamictt process,as follows:start with =1.Repeatedly multiply,by ,and Fubtract the largest integerf strictlyless than the result. The sequenceso Obttttned is the carry sequence,
PROPOSITION 9.1. CARRY CHARACTERIZES. A rcPr(SCllatFon fn base iS StrictFy f)rOperff aflonFy ff the sequencl oF dts Startfng at any Pcint Fs rexicographicarryiress tl tt thecarry scquence carry()
PROOF: This is pretty obvious t 191,Cttry characiPROPOSITION 9.2.cARRY SHIFTS LESS.A sequcllce f Posffve rntegers(cf}iS r car7scqucnce fF anorlFy fF ft has tt fnttfe number oF cf0,and no sequence obttilebyoppfng a SILite number oFFnflfaF eFements is FexicOgrt phicaFFy greater tharl it,
The operation Of dropping the flrst element Of a sequerlce is knon as the BcrnOuli shift,or onesided Bernoulli shift.
PROOF: It is clear that the carly sequence for any basehas these prOperties.NOtc thatif ct were eventually O, then o would have arived at O in the dynarmcal process above,which is impossibleTo prove the converse,consider the lnap from real numbers to carry sequences. It takes
natural order of the real line to lexicographica1 0rderThe set of all sequences of positive integers with a given bound has a natural topologof a Cantor set,with the compactopen topology.If}an inCreasing convergentsequence of real numbers,then carry()alSO COnverges to carry(lim({.)))H9Wevcr,carry is discontinous at those so that the dynamcal process above eventually arrives ata discOntinuity of the greatest integer function(l integer)On the next step, is therl l,so that carry()iS periodico At such a point,one sces from the dynamcal process tht ifthe cartt sequcnce is.(CoCC-lC),the limit frOm above is.cO.c(C+1)000Consider now the closure of the set of anl sequences satisfying the condition Of the
proposition,that is,all such sequences together with a11 11ts frorrl above at periodic carrysequcnces. Foli the quotient topology,identifying each periodic carry sequenceth itshHt froHl above.It is not hard to sce that this toPologican space is homeOmorphic to tllcreal line,using the linear ordering.The ttap Cttry()iS monotone and continuOus inthis tOPology,so by the inteediate value theorem it is sllrjective.
9.2,carry shifts icss
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Tlhesc observations are ciosely related tO the theory of kneading sequences, tMilnOrThurstonl,WhiCh arise in the theory Of iterated(nonhOmcOmorphic)mapS Of intervals.
A Pi3t number is tm algebraic integer such that all its Galois cOttugates are strictlyinside the unit circle. In lnore downtO_ettrth teAS,a Pisot numberis real number a whichis a root of a polynoman71 t _1,71-1_+O With integer coettcients and leadingcoefncient l such that all the roots except a are inside the unit circle in the complex plane.
'PROPOSITION 9.3. PIsoT cARRY PERIODIC. The cttry sequence For arly PFsot basc >1
t/:
ttC Fs an erement of the ScF Q(), hen the represenation F n base Fs
eventuaFFy peoFc.
Examples. The most famous example is the golden ratiO =1.618.t lts carrysequence is1010101.. This lneans that a base repreSentation is weakly prOper if andonly if the digits e O's and l's, and each l is followed by at least one O. It can beseen by cottputation(or inspection)that the carry sequence for the cubic number_22t_1,L1465571231876768 is.100100100100.:cach l must be followed by at icasttwo zerOS, Computation shOws that the carry sequence for the cubic r3 = +1, 1,324717957244746,is.100001000010000... . From this example one sces that the length
)2q7 Of the period can be longer than the degree. The cubic number3 = 32 _22+1,-72.546818276884,is.201111111..This example shows that the carry scquenceiodic but not periodic.All these numbers are Pisot numbers.PROOF:The base representation of a positive real nurrlber O
number Of valdes. TherefOre, its Orbit eventually arries back at a lireViOus point, andfrOEl then On it repeats.If r is inthe fleld but nOt an angebralc integer,then there is sOme integer 7n SuCh that
is an algebrtttc integer.MIultiplicatiOn by preservet this prOperty.TherefOre the Orbitof tt remns in the lattice of(1/m)times algebrttc inteerS,and again it must eventuanlyrepeat.
93,Pisot carly periodic
The converse ofthis propOsition is alsO truc,butitis not true that every angebradc numberwith an eventually periodic carry sequence is Pisoti numb,rs with periOdic carry sequencesare densc,but Pisot numbers form a countable ciOsed subet Of R. For a random exaHlI)le,the sequence (32123012310)satiSnes the hypothescs Of IropOsitiOn 92, carry shifts icss,so it is the carry sequence of some number .Simple algebrttc marlipdation shOws tllat muSt be a root of the irreducible polynomial
11-310-29-E8_273-36_r4._2t3_32_E_1,i
whose largest root(compare 10.1,Largest integers expand tilings)is,
3.67558944232794403943248035257568326'74643931....
Computation shows that its carry sequence is indeed
.32123012310321230123103212301231032123012311032123012310....
This polynomal has two other r00ts,340861998669(1Oddus l.05524)outside theunit circle,so it is not a Pisot number
One iediately obttttns selfsilnilar tilings of R froHl allyTcal number with a periodiccarry seqdencei tile the positive reals accOrding to the twhol,'pOrtion of its base representation,that is,the portion to the left of the decimal ptint, MultiplicatiOn by isa shift Of the decimal point,so that each tile is taken tO a ni!ite uniOn of tiles. The factthat thereIc only a flnite number of difFerent tiles up to coni:ruence is equivanent to thefact that carry()iS eventually periOdic:in fact,the lengths(f intervals which occur areexactly the orbit of l under the dynarmcal process for the carrr sequencc.
It is curious that one ansbtaans a seifsilnilar tiling of the planc or a higherdilnensionalspace from this constrtlction,if iS a cubic Pisot number which is also an ttgebraic unit(thiS means that the constant te of the minimal polynolmal is1).The niCe c iswhen the degree is 3,when we will obt:n a selfsilnilar tiling of the plane,a kind of`(3aloisCOttugate'of the original tiling.Supposethen that >l is a PiSOt unit.For any algebrttc integer">lin Q(),letbethe greatest integer such that ~o1.The Sequcncc of digits fOr(shiftedPositions)is obtned by the dynamical process starting with ~.Note that since is arl algebraicunit,its inverse is also an algebrttc unit,sO ~ is an dgebraic integer.Therefore, iseventually periodic,tc.inating in onc Of a nnite set of repeating patterns.
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For any ttgebrttc integerQ(),let tt COnsist of anl other algebraic integers whichagree with r after the decimal pOint.The difFerence of tty two elements Ofttt is a seriesin Positive powers of . These act as cOntracting linear maps in the hyperplme S,sO theprojection of tt to S has bounded diameter,independent Of r. Each tt has at least Onerepresentative in the slab S x p,11.It f01lows that the closures IE Of the projections Ofthe trr to s overlap with bounded multiplicity.Note that ~1iS a nnite dittOint union ofTherefore,one can express ~lFasa anite union,probably not dittOint,Ofy's(a SubdivisiOn rdc)(ThiS iS a crucipointwhere it is important that be a unit.)If the dimensiOn Of S is 2,that is, iS a cubicnumber,then multiplication by is a genuine silnilarity of S. Otherwise,~l Stretchesthe shape ofy difFerentianly in difFerent directions.We clm that there are Only a flnite number Of the tXiy's,up to translatiOn in S. In
ordetO see this cladm clearly,it is a good time to introduce the language of anite stateautomata.AJC Stac au tta or"fC3C ttac,cPf over an alphabet A is annite set sM,(the SCt Of states ofi),a map A X S SM(the state transition map for ),tOgether with a distinguished element r c SM(the initial state),ld a distinguishedsubset OKSM(the accepting states).It iS Often convenient to visualize ttr as a labeleddirected graph,with a bit of extra structllret r and OK. Aoperates by starting in itsinitial state;it iS fed elements of its alphabet one_byone,it goes to the state indicatedby its state transition map. //'0Given a word7 in the alphabet A,7 is acccPCtt byif when yOu start at/,nd go
along the directions given by W,you end up in OK.The set of words(y)&epted by is called the ragagC or()is Pr"_Cr3Cttif every prenx of a wra in(M)is
: 1such an A,to omt the fail state and all roads leading to it, Whenever a word givesyou directions where there is no cOrrespondingHroyou iediately fl with no chancefor reinstatement.The dettitiOn for acceptance of an inttite wordnot so clear in general,but F(M)
is prenxciosed,there is an obviOus deanitioni an inanite word is accepted if and only ifeach inite prenx is accepted.
PROPOSITION 9.5. PERIODIC CARRY FSA. Trle scF ycaHy proper base rcPreSentafns fs thsct accepted by a fnite state machfnc,ff annFy fFcarry()fs eVentuttFype=fodic
PROOFi(See agure 9.4,proper FSA.)If c=Carry()iS eventually periOdic,so thatC+P=CfOr all>q,then let SM be integers O,,P?-l tOgether with a fl stateF.The initial state is Oo Frolm state
Figure 9.4.proper FSA. rJiaTtt frrTttJ C3C ttac c r TcCjzii ctt a bsc TcPTc3Ctati PTPcT, CTc carry()= abc(cg ), aC arTtt bac
t 0 3,3T a Crrecti/arT3,0e roT cacteper res3accatctt at. TT3t linlcctt rca ctr 3tC r,5.
digit accepted by A, Sincehas Only a flnite number Of states,the choices lnust eventually repeat.
9.5,periodic carry FSA
Given the right pttt r of(aFter the decimal point),then the questiOn Of which lehalves r satisfy that rr is a weaklyprOper base expanSion depends only On the statewhich is in aFter reading r.Let()be the set Of states after which ttf accepts T:then F()deternes the shape Of Fr,sO there are only initely many possibilitiOs.It dOes not quite follow that ther determne a tiling of the S, fOr they could in
pnciplc have substantial overiap.In fact,we have not given a dennition of a tiling,let usdo it: Asgrlg Of a 10cally compact space X is a covering by a cOuntable collectiOn ofcompact sets(shingles)Ff,caCh equal tO the closure of its intettor,such that any compactsubset Only intersccts initely manyff. Arg ofis a shingling such that theintersection of the interiors of any two shingles is empty.Note that the dennition does nOt imposc other tOpological restrictions On the shinlcs
or tilesi they nced not be connected,or locally cOnnected,Or silnplycOnnccted.However,in many cases of this cOnstruction,the shinglings are tilings,and the tiles are
disks.We shall see later that every selisilnilar shingling is closely related to a selfsilnilar tiling.
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_
__> , /=
|)
t
Figure 966. Pisot tiling of planee rjfrfg OF a PT' c Pra a30bja3c Fcar13 ar'Or tc basc frling Or R,cre a3=+1, 324717957244T46,3C Catty jcttece f3(10000).r13P'Cre 3ttJc PT"c6 CO/a CbT'C lin
gers 3 1Q( )3uC a a3at tt54 crOf cT qc ccjar P ,
r.,.,.,.,.at3abt30er"fy arc sactt accTcPT,t te nigc ttecar P.c serr3frar,fra3cttTacc"T C?uart c orc Carls cgac3,l=.662358978656227951206.,r recliProcar,ca,30r_o877438833+07448617666..ttParC t"gttTctO pf"C,PTctic ttc tfres as 3aC30rc PSA.
Now we will generalize to cOmplex numbers.If is a COmplex number of ttOdulus>1,how can we defhe a base eXpansion for C?First choose a nnite set D=1,2,,.}of`digits',with O c D.These could be0,1,,},Or any Other set of cOmplex numbers.Consider(,D)SOLtaire,defhed tt f01lows.At the beginning you areiVen a cOmplex
number(ThiS iS like a shutted deck Of cards.)You can subtract any element of Dfrom zi then it is multiplied by to get the next z. If z ever grows large enough,fromthen On,no lnatter what yOu dO,it will grow iarger,since lnultiplicatiOn by dOmnatessubtractiOn of elements of D for large. In this event,you lose.Let /be the set of initial z for which there exists a sequence of lnoves itthich dO nol,
lose.It is easy to sec thatcOnSsts Of all sms of sehes 0~, CD.7
compact,contttns O,and satisnes the equatiOn=D+~1this is a charactedzatiollofF.If D is too small,thenwill be small.However,for any there exist sets D such
that/cOntalns a ncighborhood of the ottgin. We can,fOr instance,choose an arbitrary
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Figure 9.7. Pisot FSA. rc PSAcrecgze3rPCr b3C TCPTCscatt3,CTC li3c3C a3 1in"TC p..
y Of the Origin, and then make sure that D is large enOugh that ly.jDThis guartttees that y c,since we ctt moVe back into E/after each
neighbOrhOodcontains y.move.
If7 does contn a neighborhood Of the origin,theIR every complex number admits a(prObably nonunique)base(,D)representatn,.c,n eXpression E:fHow can we select a preferred representation?First ci oose an Ordering of D. /ith thischoice,we can add arlother element of skill to(,D)SOlitttrei nOw the object is to avoidshooting to ininity,while s91ecting thc greatest possible(ligit at each stage(giVen previouschoices.)In Other words,the preferred or proper represelltatiOn of a complex numberisthe One which is greatest in icxicographica1 0rder. A repr(sentation is weakly proper if,forevery inite initial segment,there is an cxtension which is preferred.Note that this dellition agrecs with the previous deition in the case >l is a realnumber,and D=0,1,Where n is the greatest integer less than ,with the naturalhncar ordering.PROPOSITION 9.8.soLITAIRE FSA. SupPosc that c C iS an argebraFc frltegersuch thataFF fts CaFofs cottugates excep are FnSe eucfrcFe Fn C.rD Fsan oreresct Of agebrttc integers in Q(),then there ettsts ate stae machine(,D)WhiCh H recttnizQ whl,her a sequcnce f =( f}Ves a VettFy preFered representtttFon forsome demett z c.~~
f. .ReEnarkS: rrhe elementin question is oi
WVe have nOt assumed that F contaans a neighborhood of O,cven though that is thecasc of interest,It is ilnportant that we are dealing here with series which begin with the O te1.
If we aniowed pOSitive powers, there ttlight not bc aFly lexicographically latest elemntrepresenting a complex number.This proposition is closely related to ll.9,hyperbolic automatic(Cannon).
PROOFi COnSider any other sequence of moves=(3,and COmpare the result ofwith
that of r,starting at the sEEle point.
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i
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difFerencesThe difFerence of the twO trajectodes is the sttmeif we usedDD fclr digits,d started at O,applying mOvesff.AssuHling that moves remain bOunded, then the nInovls remain bOunded if and
only if o stays bOunded using mOvesf-7,. Let's now use the fact that we are in analgebraic number neld Q(),and consider the multi_dimensiOnal picture y=RQ().The difFerence moves begin at O and the mOves are always algebraic integers,so it awaysremttns an algebraic integero MdtiplicatiOn by has a 2-dimensiOnal invanant expandingsubspace y,which we can identify ttith c,d a complementary invadant cOntractingsubspace S(by the hypothesis On .)NO matter what sequence of moves are applied fr6mDD,the point anwayS remaans in a bounded neighborhoOd Of y.On the Other hand,ifis a cOmpetitOr withl fOr representing z,then the point lnust
also remaan in a bounded neighborhOOd Of s, that is, bOunded in the cOmplex plane`TherefOr,in making cOmparisons with,we can restdct tO a compact subset Of y,inwhich there is Only a nnite set tt Of algebraic integers.~~T bf 6hathitt ill ttttsiSt~bf Subtf ttr atteeq s,the state will be the subset Af of positiOns which are attainable by sequcnces of lnoveslexicOgraphicanly greater tharl tt such that the difFerence sequence is always(until thismOment)in A.The subsetl is clearly determned by the subset,tOgether with themove1.The initian state is o.Every set Af which contains O is a fail state.There maybe Other fail states in additi6ni in generan,defhe A tO be a fail state if7 ()+(in C).
9.8,solitaire FSA
Let us now agttn suppOse that cOntaans a neighborhOOd Of the Ottgin. There is asequencc of tilings 76f,where the tiles are labeled by the initialtes of weaklyprefered sequences of II10ves,and a tile cOnsists Of the complex numbers represented byall weakly preferred sequenceP Of mOVes beginning with thOseteHs.Each Of these setshas nOnempty interior,arld its shape up to similarity depends Only on the state of themachne(,D)after readng its iabel.The shape up to transiadOn depends ontt On thestate,together with.If we cxpand,multiplying by ,the shapes Oftiles Only depend On a state of7(,D).Each tile has a rde fOr subdivisiOn,given by the state trttasition rdes forf(,D)To Obttn a self_similar tiling,choose aand a tile which Occurs in the intedOr Of theth subdivisiOn of itseli Expand this th subdivisiOn by , and transiate it sO thatthe chOsen tile coincides with the original.Repeat this subdivisiOn/expansion/trSlationprocess indeinitely,tO Obtaan a self_similar tiling Of the plane with expansiOn constant It Elay nOt be possible tO Obtain a selfsiEttlar tiling with expsion cOnstttnt with thiS
particular set of tiles.HOwever,if we chOOse the linear ordering Of the digits tO mtte 0greate3t,then the Oarrow from the initial state(the empty Set Of competitors)leadS backto the initial state. In this case,the sequence of tilings 7agree where they overlap,sOtheir union is a selfsilnilar tiling of C,
There is another picture assOciated13, 3. dyncal systems. Let us
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with these tilings, which helps put it in cOntcxtconsider the case nrst that iS an algebraic unit,
34 uly 20,1989
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so that lnultiplication by aCts as 8EI automorphisln Of the lattice Of algebraicntegersAy.Therefore induces a difFeOmOrphism Of the quotient space,y/A,hich isa torus. Since no eigenvalues Of the linear transformatiOn are on the unit circle this iswhat is called an Anosov difFeOmorphisE1 0f the tOrus, The invahant subspaces and Stogether with the planes parallel tO them map into the tOrus tO dcane two f01iati,ns,juof dimension 2 and iFS of dilnension2,of the_tOrus.Tnese foliations are invaliant by
. The theory of hyperbolic dynamcal systerns tetts us that in this situatiOn, thcre is aMarkov partition for ,that is,a inite cOver by ciOsed setseach of which is a rroductin local coordinates of a set ofleaves Of Fand a set Ofleaves Of Fu,such that the rthavedittOint interiors,arld when the interior of SI(f)intersectsit stretchels clearacrossRin theu direction and squcezes insidein the3 directiOn. Anothervay tosay this is that the intrsections of the sets iwith a generic leaf of ju or a generic leaFof3 deines a tiling of the leaf;the Markov property says that aCting on an unstableleaF rnaps each tile to a uniOn Of tiles, and On a stable lcaf maPs it tO a subtile. rrhereare only a nnite number of tile types,since th9 tile type is deterIIuned by. If thc,eis a`generic'unstable leaf r which is lnapped to itself,its induced tiling is a selfsilniltilingof the plane. More generly,if is an angebraic integer but not necessarily a unit,there is an tttsociated mtt ofthe toruS T=7/A to itself,as befOre,butit may be n_to-1.HoweR/er,wecan form the inverse linut of the sequence of lnaps
.T T T
to obtn a compact space a(a Cantor set bundle over the torus),on WhiCh th,inverselimit map aCtS as a homcomOrphism.The actiOn Of iS Still hyperbOlic,all! has
two foliations,u and jF3 which are invanant. The leaves of Fu are homeOmorphic to thecomplex planes,but the leaves of Fa are homeomorphic toR~2cwhere c is a Cantorset. Agn,the generan theory of dynamcal systems implies that a Markov partition fOr eXiStS.It yields almost selfsimilar tilings of C,and with some added ctte,actualselfsimilar tilings.
S10, CIARACTEttIZATION oF EXPANSION CONSTANTS
Wc have been discussing and dcining selfsilnilar tilings by example and bylontext,butnow we give a lnore fornlan dennition.A tiling ttof the complex plane C is serr31:rar with ePan3fC3c C if
(a):The tiles of r can be divided intO a anite number of distinct ttypes',such that tilesin a given type difFer only by transiatiOns Of the plane.(b):When T is matpped by multiplication by,the image of each tile is a uldon Of tiles,(C):The pattern(relative positions,shapes,and types)Of Subdivision of themage of anytile under lnultiplication by tt dcPends only on the type of the tile.(d):The tiling is quasihomogencOus,that is,fOr any r>O there is an>O such thatfor every disk)ofradius r in C and every disk,of radius R,an isomOrphic copy Of9(inCludng types)Can be found within.
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SOme Of these points bear discussiOn.It wOuld be interesting to relax cOnditiOn(a),to say that tiles Of the sme type are congruent,but not necessarily by a transiatiOn Ofthe plane.I wOuld cOtteCture that nO more examples are Obtttned by this relaxatiOn.The types of tiles are not to be regarded as part Of the structure of the tilingi they area convenience fOr dealing with the rdes Of subdivisiOn. hen twO tiles have the sametype,it implies that they are cOngment,that their subdivis10ns are congruent,and thatali subsequent subdivisiOns are anso conttent.ConditiOn(d)is impOSed tO avOid prOblems One encounters in certttn caseS that are nOt
of real interest anyway. FOr instance,consider a tiling Of the plane by squares Of twO sizes,say l and T,with tiles Of size l tO the left Of the y_axis and size T tO the right. This canbe cOnstructed so that exPansiOn by a factOr of 2 takes each tile tO a uniOn Of 4 tiles. It
satisnes(a),(b),(C)but not(d).There are also many examples of tilings where a tiling is not strictly self_silnilar,but
where there is a cycle Of tihngs, ,.,_l Such that+P iS a subdividOn of theexpansiOn of,using rules depending only on tile types,as above. such a tiling will be
canled Pcrficarry scrr3fran There are alsO interesting still weaker cOnditiOns which wewill nOt adess now.In this section we will prOvei
THEOREM 101. LARCEST INTEGERS EXPAND TILINCS, 7compFex numbertt OFmOusbftter tFlan l Fs tt exPansfon cOnstant For sOme seFfsfniFar tfFfngf anOnFy_fsan aFgebrttc Fnteger,hfch Fs sttctry Farger than arF fts CaFOFs cOibs bther thitscmpFex cttuttte.
Remarks: This theOrem gives cxmples much mOre genertti thansectiOn.
in the preceding
The cOrresponding cOnditiOn fOr periOdically selfsilnilar tilings,which we will not prove,is that all Ga10is cOttugates Of tt have modulus less than or cqual to that Of
,and thatthosc Of the same IIloddus have a ratio with tt Or withwhich is a roOt Of unity.This theorem and its proof generalizes tttrly easily tO arbitrary dilnensiOns by talking
abOut linear transformatiOns d their Ca10is cOttugates.The dimensiOn l case isessentially the PerronFrobenius theOrem and its`cOnverse'Of Doug Lind(ILindl).The rate Ofowth of area is,which is a real number larger than anl its GOis
COttugatesi this fact is actually an easy cOnsequence Of thc PerrOnFrObenius thcOrem.It
is not enOugh to guarantee a self_silnialr tiling,for there are examples Of angebraic integerssuch thatis larger than its Ga10is cOnjugates,but tt has Ca10is cOnjugates bigger thanitself. In such a case,the(Galois grOup is necessarily slrlaller than the symlnetric grOup.The minimum number Of tiles fOr a selfsilnilar tiling Of expansiOn cOnstant tt is at least
the maximum Of the degref tt and the degrec Of. This is nOt sharp 10wer bOund,hOwever, |!PROOF:The casier directiOn is the tOnly if'directiOn,sO we will dO that arst.
Let 7'be a selfsilnilar tiling with expansiOn cOnstant. The prOOf can be thOugllt OfinteAS Of eStablishing a system of gOvernance and a systeE1 0f rOads fOr the cOuntryside Of
T
We wili arst chse a caPtar(r capital)fOr each tile,in such a way that the capital
VersiOn l.5 36 uly 20,1989
Of any tile maps to the capitttt of anOther tile under I41ltiplicatiOn ly,and sO that thepositiOn of the capital relative to a tile depends Only ol its type.To dO this,lwc can graphicatty represent the rde fol subdivisiOn()f types Of tiles as a
directed graph. The nodes Of r are labeled by the tyF es of tiles,if a given type a occurstilnes in the subdivisiOn of another type y,edges Oflead frOm y tO,with cach edgecorresponding to a relative positiOn of Onc Of the rtiles within the expansion Of y.For each node of r,chsc one distinguished OutgOintt edge.We ilnposc the condition that the capita1 0f any tile milps under expansion to the capital
of the tile pointed to by its distinguished OutgOing edge.This determnes the capital c()uniquely for every tile t,since in the sequence of subdi
visions of a tile,the subtiles necessarily shrink tO pOints.Now denne a sct D to consist of all difFerences of capital cities Of townships and cOunties,
D= (C(3)C()}
where s is a tile contaaned in.These difFerences are determned by edges of the graph r,so there are only a nnite number. Labeling the edges of r iy the appropriate elements ofD,we allnost have a linite state rnachine:we lnake such tt lilachineby adding a specialinitial state r,a special fl state F,and attoining to th3 alphabet D special symbolsbegin,where t ranges over the tiles which contttn O.(If O iS in only one tile,this is notnecessaryo ln the decimal system,andplay an analogot s role tO begint,fOr the tilingof R by intervals between integers,=10)As usual,we use the convention that if there is no outgoil gHrow with labelfrOHl a
nodc t,thenleads to the fail state,and that ani non_fall steS are accept states.
It is worth observing that the tiling can casily be reconstt ucted from. In fact,AdeterIIllnes a baseosysteEl for C,where
is proper if and only if the sequencecOmplex nmber
of digits is accepted bs
==E
The tiles are labeled by the twhole'part of this cxpansion(tO the left of the decimpOint)and cOnsist of ali complex numberssharing the whole portion.
So far we have developed enough structllre to connect the capitals in a hierarchicalgrouping,but there are no provisions for tOudsm or coerce. Next we need to buildenough roads to connect the capitals of neighboring tiles in a reasonably efncient lnanner.TO this end,we would like to flnd a nnite set R of`enough'difFcrences between capital,enough so that you can go lttom any capital to any other along roads of type R quasicfncientlyMOre formally,we stipdate that ttis sttciently large that there exists a constantsuchthat for any two capitals c and cin C,there is a nnite sequencec=Co,,Cn=C},
A, and it represents the
Version l.5 37 Jdy 20,1989
TherefOre the Lipschitz cOndition is unifOrm in tD.A Lipschitz map r iS difFerentiable dmost everywhere.Let z be a point where it is
difFerentiable. If we expand a neighbOrh00d Of z by a high pOwerof(,), then theilnage point has a neighborhood of a given radius r where the graph Ofis Very closc tobeing linear. By the quasihomogeneity of the tiling,it fol10ws that there is sOme pointin a neighborhood of radius(r)of the origin.(ThiS f01lows because thl portiOn of above any tile is determined by the type of the te.)Fixhg r atrld taking the httt as ,We Obtttn a point which has a neighborhOOd where r is exactly linear.If r islarge enough,then the disk of radius l abOut the Origin is c6ntained in this disk Of radiusr.Therefore,is linear.It follows that or .This completes the only if POrtion Of the proOf.
The second httr of the proof witt be,given an algebradc inteser tt such that all its GaloisCOttugates except tt and tt are smaller,to cOnstruct a selfsimilar tiling.The construction wili naturally make use of the vector space y=R tt Q()The ttm is to constnlct a subset cA y,where A denotes the lattice of algebraic
integers in Q(), Such that c iS Selfsimilar in sOme appropriate sense, and so that itprojects to a discrete,quasihomogeneous set in C.We lnay as wellstart with O c c. NoMpick a few more elements of A to be in c,cnOugh
sO that O is in the convex hull of the projection of the given points tO C. Deflne this setto be cO.(SketChy at the moment.)Iteratively expand and interpolate.. Use the Delaunay triangulation to decide when to
interpolate. Make a determnistic rde,depending only on the shapes of the]DeLaunaytriangles,together with the tilne since creationthat is,wttit a while before subdivi(ng,then subdivide thoroughly.Makc a hierarchical structure:each ncw vertex is associatedwith the vertices of the previous triangulation nearest The resulting tiles have good qualityif one wttits a long tilne before subdividing,but there may be very lnany ofthem. However,this picture is not yet quasihomogeneous..modify the construction,by choosing a cycttcordering of the vertex types that occur(with SOme bounds on shapes and eccentricities ofDelaunay triangles): and Put a tiny copy of the successor vertex type,as the nrst step inIIlaking the choice for cach vertex type.
101,Largest integers cxpand1lingS
sll.AuToMATIC GROUPS
Likc lnany things in mathematics,groups can be difncult to get a handic on. In fact,a celebrated result of Novikov and Boone says that there is no general algothm,giventwo presentations lor groups, to tett whether or not they are isomorphic: it is not evenpossible to tett whether a presentation deschbes the trivial group. Furthermore,there are
particular presentations for which,given two words in the generators,it is not possible totett whether or not they represent equal elements in the groupor equivalently,giverA asingic word,it is not possible to tewhether Or not it eqtlals l.It is worth emphasizing that the difnculty is not in flnding an algOttthm which will
answer tyes,they are cqual'if they(the grOupS,or the words)are Cqual.Such angorithms,
Version l.5 39 Jdy 20,1989
in fact,are easy to construct,although they tend tO be stupid and incredibly siOwi the idcais sirnply to try all possibilities. The difttctllty is in flnding algodthHls lwhich will answer`nO they are not equal'if they are not.
Despite the fact that intFaCtable groups exist,we dO not need to be discOuraged abOutflnding techniques which lnight apply to the lnany particulcr grOups which Re would liketo understand better.The theory of automatic groups is one attempt to delineate a reasonably large class of
groups,including lnany that anse in real mathematical contexts,where it is indeed possible(but nOt necessarily etty)to`SCe'what they look hke and to anyze themgothttcby computer.Here is the forman dcanition of arl autOmatic structure for a groupi we will illustrate it
by examples and interpret it lnore geolnetrically lateriAf attac3rthCtTc for a group C is
(a).a Set of generators g for C,(b)a Set of wOrds tt accepted by some nnite state automatonA(the WOrd acceptor)with alphabet g,containing at least one word representing each elerrlent Of C,such that(C)fOr each element g c g=Cctt us},there exists a nttte state automaton (thegcomparator)th alphabet gxg.Given a wOrd=(1,l)(u2,2)(un,.),let ube the word ult42un,and let =l 271 Then o accepts ttt if and only if u and are Sfrprenxes uand
follOwed by(pOSSibly empty)sthngS Of the pad symbol S,
where uand are cach accepted byA and =ugin C,The word acceptor automaton(b)ShOuld be thought of as picking out canonical forms
for group elements,although this canOnical for]m need not be unique. The comparatorautomata of(c)ctt be thOught of tt knitting the canottcforms tOgether,to constructthe the group.Note that, according to the dennition, the automatic structure is deined by the set
of generators together with the sct of words iR: a wOrd acceptor/A and comparatorsdescbed in(c)muSt exist,but they don't have to be produced tO dette the structure.Part(C)Of the dcttlition in particular may seem technical and Opaque now,but we
will soon deduce an equivalent condition which is lnOre intuitively comprehensible. Beforedoing that,however,let's ioOk at icast at a trivial exmple using these deiniti?ns.First consider Z,with generators a=l and A=-1. The wOrd acceptOr has thrce
nonfail statesSince only one word is accepted byA for each element of C,the coHlparator l just
reCOEttHZes whether two word are cqual(and accepted by A)The conParatOr Ca(11.2)has four nonfl states,of which only one is an accept state.
What sotts of canonictt forms(spedned by w)ctt there be? Many of you maybe familiar with hem in another guisei the set of words(the rangttgC)aCCepted by annite state automaton is what is called a reurar raagC Or a regular set. Regular setsare coHlmonly used in many wordprocessing applicatiOns On computers. Typically yOuspecify a regdtt set by a rcgurar cPre350Or pattern,and the progran construct,,flnite state automaton,sets it running on your flle,flnds lnatches(that is,strings lltti118your pattern),dPnts them out,mttes substitutiOns,or whatever you asked it to
Version l.5 40 Jdy 20,1989
Figure ll.1. Z word acceptor. 17PrCitry a33Ctac gcacratg3CtctescTsc, reS3 CT13Cte.
tt acccPT TZ, PTcJcntaj,(al) "
3 CFsccTcT3,tat cange or ca3C
Figure ll.2.Z comparatore r ,stc cttParatT a roTc grttP Z,gencratctt by
.rt 30 ry ,c acccPt 3tatC,a tTce er 3tatC3 ,c 3p033tbrC Tcac
tc acccP'3C. y arr 3 rca3 tOcr stac,3,rT ,ct,3
1P0331brC to cscaPc.
A good example is the lUnix utility egrep. The word acceptor for Z,for instance,couldbe specifled by the regular expression
aIA
where the sybol tt denotes zero or more repetitions of the preceding object, and thesymbol i means`o.The commarld
pttnts out a11 lines of itsthe beginning of a line,$
Version l.5
egreP 'aIA*$'
inputs which are accepted by WA.The symbol here denotesdenotes thc end of a line,and parantheses are used for grouping.
41
start
A,S)(S,a)
dy 20,1989
Figure ll.3. ee word acceptor.grP)
tt acccPt T T TCCc c/Tcc
(The quotes'protect all the special symbOls from being interpreted by the shell(cOmmandParSer),bere egrep gets them)It is alsO easy to construct an automatic structure by inspectiOn fOr the group on
generators. For instance,a word acceptOr which accepts only wOrds in reduced form forthe ice group(abl)iS illustrated in ll.3.The correspondng egrep commalxd is
egreP '(bIB)?((aIA)(bIB))(aIA+)?$' ,
Here some more notation has been introduced.Parentheses'o'are uSed for grOuping.The operatOr? means zero 6r one Occurrence:(cttPTcJJl)? is equivalent to ((cttPTc8
3101)) rrhe operator+means one or lnore occurrencesi(caPTc551)+is equivalent to
(ePTC331)(cPre33).The egrep commarld asks for a word whOse mttn part cOnsists
of repeated stnngs of a's Or A's followed by a strings Of b's or B's, and is matched by((aIA+)(bIB+)). However,the beginning and ending Hlight be in a difFerent phase,hence the extra stufF enciOsing it.The reader might ettOy cOnstnlcting the egrep expression for reduced words in the free
group on three generators,Of cOurse,this usc of egrep is nOt the use for which it was designed, and the regular
expressions for a group tend to bc a bit longwinded. Nonetheless,the ettciency and thesuccess of egrep and other related code is an insPiration and a guide to what we may beable tO accomplish with groups.
Ccometricatty,a word in the generators of a group C is equivalent tO a silnplicial pathin the group graph(C)(See s3,Group graphs)starting at the base point.The set ofwords accepted by a wOrd acceptor autOmaton thus deanes a family of paths in the graphOf the group beginning at the base point,at least ending at each vertex.
h/ersion l.5 42 July 20,1989
The fact that difFerent paths have difFerent dOmains is m incOnvenience here. If :Af is a path,whereA is an inter,let:R_Xi be the extensiOn tO al1 0f Rhich isconstant in the cOmponents Of the cOmplement Of.IVe can denne a cttbOf a metric spac
,Fily Ff paths inbeginning
at the base pOint,including the trivian const
(a):there is a cOnstant r>O such that fOr any E cat icast One path ends within adistmcc Of rf of r,arld(b):fOr any>O there is anf such that whenever twO paths end within a distance Of,then they are within distance tt fOr all time.A combing,in Other wOrds,is apprOxilnately a unifOry continuOus right inverse to themap which takes a path tO its endPOint,using the unifbrm metric On pathso HOwever,the
inverse need be deined Only sketdlily,and it need nOt be cOntinous: its discOntinuities arebOunded(by),hoWevero lf we didn't al10w discOntinuities,cOmbings cOuld exist Only fOrcontractible sPaccs(We could then try tO wOrk with classifying spaces fOr grOups,ratherthan graphs Of grOups,but we have no guarantee that the groupseill deal/ith havecompact classifying spaces.)
f ombf r(C).PROOF:SuppOse,arst,that the sct Of paths derttled by tt is a cOmbing of
(C).Let7A beaflnite state automaton which accepts wOrds in jR,with padding by$at the end perrmtted.
,sttthi e:Deflnc a inite state machine DWith ttphabet g'xg,whose set Of states is the set ofgroup elements within a distance Of Afrom the identity tOgether with al state. Onreading a pttr of wOrds(,)(COmbined,as in thc PreviOus discussiOn,tO make a singleword in g'xgthe state of Dat any time is Failif either Of the cOmponent wOrds isnot accepted byA Or if the wOrds at sOme tilne have been at a distance greater thanf from each otheri Otherwise,the state of Dr after readingsymbOls is the difFerence,ulWheredenotes the lengthpreax Of a word.The nOnFail transitiOns Of Don input(,b)gO frOm state g to state ~19b ComparatOr machines can be obtained from Djust by chOice Of the which states are
acceptedi the only accept state for Cro is g. This shOws that if jR deines a cOmbing ofr(c),then tt gives an automatic structure.SuppOse,cOnversely,that itt determnes an automatic structllre fOr C. TO shOw thatR
deines a cOmbing,it will sce tO prove that any twO accepted wOrds ending within adistancc Of l from each Other remadn a bOunded distance apart,since wOrds whOse endsare EIOre distant than l can be jOined by a chttn Of wOrds ending l apart.Thus,we necdto show that fOr each cOmparatOr,the pairs Of words accepted by cg remain a b01lndeddistance apart.If there are states in twhi