Percolation

Percolation Percolation them y was initiated some 50 years ago as a umthernatical ft aurework for the study of random physical processes such as flow through a disordered porous medium. It has proved to be a remarkably rich theory, with applications beyond natural phenomena to topics such as the theor y of networks Mathematically, it has many deep and difficult theorems, with a host of open problems remaining The aims of this hook are twofold. Fitst to present classical results. including the fundamental theorems of Harris. liesten, IVIerishikra, Aizenruan and Newman. in a way that is accessible to non-specialists. These results are presented with relatively simple proofs, making use of combinatorial techniques Second. the authors describe, fox the first time in a book recent results of Stith/70V on conformal invariance, and outline the proof that the critical probability for Voronoi percolation in the plane is 1/2 Throughout. the presentation is streantlined, with elegant and straightforward 'nook requiring minimal background in probability and graph theory. so that readers can quickly get up to speed. Numerous examples illustrate the important concepts and enrich the arguments All in all. the book will be an essential purchase lot mathematicians probabilists, physicists, electrical engineers and computer scientists alike

y

Percolation BELA BOLLOBAS University of Cambridge and UM versify of Memphis

OLIVER RIORDAN Uni vet sit y of Comb; idge

.6(= CAMBRIDGE IL .‘cuy UNIVERSITY PRESS

caauuunora UNIVERSII i PRESS Candwidge New York . Melbourne . Nlathid. Cape Town. Singapore,Silo Paulo Cambridge University Press The Edinburgh Building . Cambridge C132 2B U.. UK Published in t he United States of America by Cam bridge University Press . New York www cambridge org Information on this title: www cambridge mg/978052187232,1 Cambridge University Press 2006 this publication is in copyright Subject to statutory exception and to the pro \ owns of relevant collective licensing agreements no reproduction of any part may take place without t he written permission of Cambridge University Press. First published 2006 Printed in the United Kingdom at the University Press Cambridge A catalogue teToui for this publication is available ban: the

ilish Lanaia

.1SBN-13 078-0-521-87232-4 11 ■ 11(11MCI: ISBN-I0 0-521-87232-4 hardback

Cambridge University Press has no responsibility flu the persistence or accuracy of URts for external or third-party intemnet websites referred to in this publication, mid does not guarantee that:my content oil such websites is or will remain, accurate or appropt hate

To Gabriella and Gesine

Contents

page ix

Preface

1

Basic concepts and results

2

Probabilistic tools

3

Bond percolation on V — the Harris-Kesten Theorem 50 3.1 The Russo--Sem i an-Welsli method (65:13 3 2 Hai 1 is is Them mu 3.3 A sharp transition 67 3.4 Kesten's Theorem Dependent percolation and exponential decay 3.5 70 76 :3 6 Sulr-exponential decay

4

Exponential decay and critical probabilities - theorems of Menshikov and Aizenman Sz Barsky 78 The van den Berg-Kesten inequality and per colation 78 4 1 80 4 9 Or iented site percolation 4 3 Almost exponential decay of the adius Menshilaw's Theorem 90 104 4.4 Exponential decay of the radius Exponential decay of the volume the Aizenman-Newnmn Theorem 107

5

Uniqueness of the infinite open cluster and critical probabilities 117 5 1 Uniqueness of the infinite open cluster - the Aizenman-Nesten-Newman Theorem 117 5. 9 The Mann is-Kesten Theorem revisite d! 124 5 3 Site per colation on the triangular and square lattices 129 5.4 Bond percolation on a lattice and its dual 1:36

1 36

vii

Contents

viii 5.5 6

8

The star-delta transformation

148

Estimating critical probabilities The substitution method 6.1 Comparison with dependent percolation 6.2 6.3 Oriented percolation on Z 2 6..1 Non-rigorous bounds

156 156 162 167 175

Conformal invariance — Smirnov's Theorem 7 1 Crossing probabilities and conformal invariance Smirnov's Them em 72 7 3 Critical exponents and Sch in n-Loewner evolution

178 178 187 232

Continuum percolation The Gilbert disc model 8 1 8. 9 Finite random geometric graphs Bandon' Voronoi percolation 8 3 Bibliography Index List of notation

240 241 254 263 299 319 322

Preface

Percolation theory was founded by Broadbent. and Hammer sley [1957] almost half a century ago; by now, thousands of papers and many hooks have been devoted to the subject The original aim was to open up to mathematical analysis the study of random physical processes such as the flow of a fluid through a disordered pm MIS medium These bone fide problems in applied mathematics have attracted the attention of many physicists as well as pure mathematicians, and have led to the accumulation of much experimental and heuristic evidence for many remarkable phenomena. Mathematically, the subject has turned out to be much more difficult than might have been expected, with several deep results proved and many more conjectured The first spectacular mathernatical result in percolation theory was moved by ICesten: in 1980 he complemented Harris's 1960 lower bound on the critical probability for bond percolation on the square lattice, and so proved that this critical pr obability is 1/2 To present this result, and numerous related results. Kesten [1982] published the first monograph devoted to the rnathematical theory of percolation, concentrating on chscrete two-dimensional percolation A little later, Chayes and Chayes [1986b] came close to publishing the next book on the topic when they wrote an elegant and very long review article on percolation theory understood in a much broader sense For nearly two decades, Grimmett's 1989 book (with a second edition published in 1999) has been the standard refer ence for !larch of the basic theor y of percolation on lattices Other notable books on various aspects of percolation theory have been published by Smythe and 'Wierman [1974 Durrett [19881, Hughes [1995; 1996] and Meester and Boy [19961; valuable survey articles have been \\ ridden by Durrett [1984],

Preface

Chaves. Puha and Sweet (1990]. I:esten [200:31 tun! Gti uu uett 12001] among others. Out aims in this book me two-fold 11 hst. we aim to present the results of percolation in a way that is accessible to the nonspecialist To get straight to the point. we shut with the best known tesult in the subject, the fundamental theme/It of limits and ICesten. even though this is a special case of late' and mule genet al results given in subsequent chaplets The moot of the Han is Kesten Theorem. in pal t iculat the tippet hound clue to kesten. was a peat achievement, and his pool not simple Since them however, especiall y with the advent of new tools in probabilistic combinatmi CS aunty Sininle moors hate been B er ard.. a fact that most non-specialists ale not awn' e of Pot some of these at gutnents. all the pieces lute been published some time ago, but pet haps not in one place. in only as comments that ale easy to ndss Here. we In ing toReilly' these ions pieces and also mole tecent. ver y simple proofs of the Han Nesten Theorem In Chapters -1 and 5 we desct the t he vety genet trite:tuns of Nlenshikov, and of Aizemnan kesten and Newmn: these Jesuits ale again classical Oin aim hew is to present them in the greatest genetalit \ that does not complicate the proofs Out second aim is to present recent results that have not let appeared in book fie.i t s Nit give a complete moo( of Sinn no\ 's famous confonnal inva t bu t te result: to out knowledge no such account. tv itlt Hie Is clotted a n al t fl ossed. has previously appeared We finish by pwsenting an outline of the recent moot that the critical probabilit y for random Votonoi percolation in the plane is 1/2 As is often the case. we have flied to mite the kind of book we should like to wad I3' now percolation theory is an immensel y rich subject with enough material lot a dozen books. so it is not smptisiug that the choice of topics sttonglv reflects out tastes and intetests We have striven to give stleatulined moots and to Ming oat the elegance of the arguments. To make the book accessible to as wide a ! cadetship as possible. we have assumed NT/ 1 little little mathematical background and have illustrated the impol taut concepts and main arguments with mune/ ous extunples In niit ing this book we lime receiNed help fron t ma i n people Paul Balistet. Gesine Ciosche Henn Liu. Robert tklol r is. Anthem Sat kat and NIB/ k lkaltels were kind enough to tend parrs of t he manuscr ipt and to correct many misprints: lot the many that iemain. we apologize

1 Basic concepts and results

Percolation thecny was founded by Broadbent and Flantnansley f19571. in cruder to model the flow of fluid in a porous mediurn with randomly blocked channels. Interpreted narrowly, it is the study of the component structure of random subgraphs of graphs Usually, the underlying graph is a lattice or a lattice-like graph, which may or may not be oriented, and to obtain our landon, subgraph we select vertices or edges independently with the same probability p Iu the quintessential examples, the underlying graph is Ed The aim of this chapter is to introduce rile basic concepts of percolation theory, and some easy fundamental results concerning them. \\k_r shall use the definitions and notation of graph theory' in a standard way, as in Bollobtis [19981. for example In particular, if A is a graph, then V(A) and E.:(A) denote the sets of vertices and edges of A. respectively "re write E A for :r E V(A) We also use standard notation for the limiting behaviour of functions: for / = /(n) and g = g(o). we write = o(g) if //q as = 0(g) it fl g is bounded, = n ( g ) for g = 0(f), and f = e(g) if / = 0(g) and q= 0(.1') The standard terminology of percolation theory differs horn that of graph thew y: vertices and edges ale called sites and bonds, mid components are called clusters When our random subgraph is obtained by selecting vertices, we speak of site percolation; when we select edges, bond pen:olation In either case, the sites w bonds selected are called open and those not selected are called closed; the state of a site or bond is open if it is selected, and closed otherwise (In some of the early papers, the term 'atom' is used instead of 'site', and 'dammed' mid 'undammed' for 'closed' and 'open' ) In site percolation, the open sabgroph is the subgraph induced by the open sites: in bond percolation, the open subgraph is Mimed by the open edges and all vertices; see Figure 1

2

Basic concepts and results

o

o

•

0

o

•

•

o

•

o

o

0

0

0

o

•

•

•

I I I

0

Figure 1 Parts of the open subgraphs in site percolation (left.) and bond percolation (right)) on the square lattice 1 2 On the left, the filled circles are the Open sites; the open subgraph is the subgraph of 2: 2 induced by these For bond percolation. the open subgraph is the spanning subgraph containing all the open bonds

To streamline the discussion, we shall concentrate on mu)/ iented percolation, i e on (bond and site) percolation on an unoriented graph A \Nre assume that A is connected, infinite, and locally finite (i.e., every vertex has finite degree) In general, A is a in ulti-gurph, so multiple edges between the same pair of vertices me allowed, but not loops Most of the interesting examples will be simple graphs Often, we shall choose bonds or sites to be open with the SUMO probability p, independently of each other This gives us a probability measure on the set of subgraphs of A; in bond percolation we write Pen for this measure, mid in site percolation P sA p More often than not, we shall suppress the dependence of these measures on some or all parameters, and write simph IF or Pp Similarly, A l ; is the open subgraph in bond percolation, and in site percolation Formally, given a. graph A with edge-set E, a (bond) configuration is a function c w e ; we write Q = {0,1} E lot the set of : E {0, all (bond) configurations A bond e is open in the configuration w if and only if = 1, so configurations correspond to open subgraphs Let E be the a-field on Q generated by the cylindrical sets C(P, a) = {co E Q

f

= a f lot

17} ,

where F is a finite subset of E and a E 10, I F Let p = • with 0 < p c < 1 fin every bond c We denote by Pn p the probability measure

I Basic concepts and results on (12 E) induced by P,1,(C(F,e7))

PI fl (— p f) IEr a f =1



(1)

fEb

af=0

When pc = p for every edge e, as before, we write P,I4 for PbAp In the measure P!‘) ,p , the states of the bonds are independent, with the probability that e is open equal to p„; thus, lot two disjoint sets Eh and F 1 of bonds, Pi Ithe bonds

re open and those in Fa are closed)

=

Pf

11

fe ro

We call I1114 an independent bond percolation IIICatilln on A The special case where P. = p for every bond c is exactly the measure P IA' ,1, clefined informally above. The formal definitions for independent site percolation are similar Let us remark that site percolation is more general, in the sense that bond percolation on a graph A is equivalent to site percolation on L(A), the tine graph of A This is the graph whose vertices are the edges of A; two vet tices of L (A) are adjacent if the corresponding edges of A share a vet tex: see Figure 2 Although in this chapter we shall make some remarks about general infinite graphs, the Main applications are always to lattice-like' graphs These graphs have a finite number of 'types' of vertices and of edges Occasionally, we may select vertices or edges of different types with different probabilities For a fixed underlying graph A, there is a natural coupling of the measures il-^1 .1„ 0 < p < 1: take independent random variables „Y„ for each bond c of A, with X„ uniformly distributed on [0, We may realize Ai; as the spanning subgraph of A containing all bonds e with X„ < p. In this coupling, if p i < p), then is a subgraph of A. A similar coupling is possible for site percolation An open path is a path in the open subgraph For sites 17 and if we write 'II or {x —4 yl for the event that there is an open path horn x to y, and P(x y) for the probability of this event in the measure under consideration We also write ',/r for the event that there is an infinite open path starting at x. An open cluster is a component of the open subgraph As the graphs

3usic Concepts and rcvalts

Figure 2 Part of t he squaw lattice L 2 (solid circles and lines) and its line graph /3(2) (hollow circles and dotted lines) Note that ii,( 2:2 ) is isomorphic to the non-planar graph obtained from E 2 by adding both diagonals to every other face we consider are locall y finite.. an open cluster is infinite if and only if, for every sites in tile cluster the et- cut holds. Given a site we r\ rite Cr for the open cluster containing ;r.. if there is one; otherwise. we take Cs to be erupt) Ilms Crix = {q E A: a! ---rr y} is the set of sites q for which thew is an open Path Clearly, in bond percolation, ex always contains x, and in site percolation. = N if and 011/1' if ,rt is closed. Let Ox (n) beet he p t obahilih that C.,. is infinite. so 0,r (p) = Pvt./7 Needless to sax, O x (p) depends on the underlying graph A. and whether we take bond or site percolation More finmalty, Eat bond percolation, for example, = 0,, (A

= 0,1; (A: p) =11') 1,(1Cr i = c.c.)

where 11 ' ter is the ntunbe t of sites in C r We shall use whichever !bun of the notation is clearest in an y given context In future. we shall introduce such self-explanatory \ rat rants of our notation without blither comment: we believe that t Iris will not lead to confusion Two sites r and y of a graph A rue equivalent if thew is an automolphisni of A mapping .r to a \ hen all sites ate equivalent (i.e. the svmmet tS

.1 Basic concepts mid orsalt 5

5

group of the graph A acts transitively on the vet tires). we write 0(p) for Or (p) fa any site x The quantity 0(p), or Ox (p), is sometimes known as the percolation probability Clearly, if a, and y are sites at distance il, then (E(p) > p''U,(p), so either Orr (p) = 0 for every site x, or 0,.(p) > 0 for every x Trivially. from the coupling described above. 0,.(p) is an increasing function of p Thus there is a critical probability pn, 0 < < 1. such that if p < prr, then ()Apr = 0 For every site and if p > p H , then (E(p) > 0 for every ,r The notation p H is in honour of thimmersle) \Viten the model tinder consideration is not clear horn the context, we write pill (A) for site percolation on A and 14).1 (A) for bond percolation The component str ucture of the open subgrirph undergoes a dramatic change as p increases past pH : if p < p i] then the probability of the event E that there is an infinite open duster is 0, while for p > pp this probability is 1 To see this, note that the event E is independent of the states of /WV finite set of bonds or sites. so Kohnogorey 's 0-I law (see Theolem 1 in Chapter 2) implies that Pp (E) is either 0 or I. If p < pH, So that 0 ( p) = 0 for O y er X, then filv (E)

0,(p)-= 0

(and so for P > Th r • the" Py( E ) > O AP) > 0 lo t some site all sites), implying that. Pp (E) = I. One SUNS that percolation in a certain model if 0,(p)> 0. so P11 (E) = t \kith a slight abuse of terminoloro, we use the SlUtle word both for this particulat event and for the measures studied; this is not ideal. but. as in so man) subjects. the historical terminolog y is now entrenched To start with, the theory of percolation was concentric) mostly Willi tile Stull of critical probabilities, i e with the question of when percolation occurs NOW, howevet, it encompasses tire study of much mole detailed proper ties of the random graphs arising fron t percolation measures In fact, great efforts are made to describe the structure of these random graphs at or near the critical probability. eve" Wile/1 we cannot pin down the critical meltability itself In Chapter. 7, we shall get a glimpse of the huge amount of work done in this area, although in a setting in which the critical probability is known The theoo of percolation deals with infinite graphs.. and 110111V of the basic events stitched (such as the occurrence of percolation) involve the states of infinitel y UtiUle bonds Never tireless, it always suf[ices to consider events in finite probability spaces, since, for example. "" d

6

Basic concepts and results

b r (p)

= Pp(IC,) > a). In this book, almost all the time., even the definition of the infinite product measure will be irrelevant Fm p < p H , the open cluster Cr is finite with probability 1, but its expected size need not lie finite This leads us to another critical probability, p i , named in honour of Temper ley Again, we write K(A) or p1/4 (A) for site or bond percolation on A. For a site rr, set NAP) = Es(ICA), where E t , is the expectation associated to IF,, If all sites are equivalent, we write simply ixr(p). Trivially, x x (p) is increasing with p, and, as before, v r (p) is finite for some site rr if and only if it is finite for all sites. Hence theme is a critical probability Pr

= suP{p: iVr(P) < oc } =

(p) = cob

which does not depend on X. By definition, p i . < One of our aims will be to prove that p r = pp for many of the most interesting ground graphs, including the lattices Z d , d > 2 There rue very few cases in which p it and p t are easy to calculate The prime example is the d-regulal infinite fate, otherwise known as the .Bethe lattice (see Figure 3). For the pm poses of calculation, it is more

Figure 3 The 3-regular nee, for which A = 1. ), = A = 1ii i = 1/2 Deleting an edge (e, for example), this tree falls into two components, each of which is a 2-brunching tree convenient to consider the k-Inanching treeTk This is the rooted tree in

which each vertex has k children, so all sites but one have degree k 1 Wm Ring am for the root of I. let „ be the section of this tree up to

I Basic concepts and results

7

Figure 1 The tree J .1 w i th root vo

height (or, following the common mathematical convention of planting Taking the trees with the root at the top, depth) o, as in Figure bonds to be open independently with probability p, let 7" = (p) he the probability that Tk , „ contains an open path of length o from the root to a leaf Since such a path exists if and only if, for some child in of vo, the bond nom is open and there is an open path of length n — 1 horn v i to a leaf, we have IT„ = - (1 - int„ i) k = jk pkn-I )



(2)

On the inter val [0,4 t he function Tr; p (x) is increasing and concave, with fk,i,(0) = 0 and fk, ,,(1) < 1, so h..1,(a 0 ) = iro for some 0 < :co < 1 if and only if r (0) = kp> 1; fur thermore, the fixed point yo is tmique when it exists (see Figure 5) Thus, if p > 1/k, then, appealing to (2) we see

Figure 5 For k = 2 and p = 2/3, the increasing concave function fix) = = — ill' satisfies = 4 1(0) = 0, J(3/1) = 3/1 and 1(1) = 8/9

S

Basic concepts and results

• implies tr„ > .to Since rut = I, it follows that rr„ > .rp Thal )r„_ > up fbi ewes' so 0;1 (,)/ k : > ./ .0 > 0. implying that ph (Tr) < 1/k Also if p < 1/k. then 7„ ()olive/gem to 0 11w unique fixed point of 1 k,i,(3). and so 0/),),,(Thip) Hence. the critical probability p li 'I (Ik ) is equal to 1/k Tinning to p i note that the probability that a site p at graph distance I from the toot up belongs to is exactl y ji Thus \ I:.„(11

E

E(1C/01

„Erk

y; opt. ,=0

which is finite tot p < 1/k and infinite lot p 1/k Thus the critical mobabiliB p!; (TA ) is also equal to 1/k Fin an y infinite Bite. fillet conditioning on the toot .1 being open. the open clusters containing r in site and bond percolation have exactly the same distribution. Indeed, each child of a site in the open cluster lies in the open cluster with probabilit y p for the k-branching Bee T), we have p7 i = = 1/k It is easy to show similatly = lrlr = of indeed to deduce. that the loin critical probabilities associated to the (A) -I-I )))1 egulat nee ate also equal to 1//, The a t gu i nea' about innounts to a comparison between percolation on Tr, and a «ti lain branching process; we shall give a slighth less trivial example of such a compaiison shot tl y If A is il/IN graph with maximum son wit blanching !actress shows i degree then a one-wit\ com par that all ethical l a °liabilities associated to A a l e at least I,/(A – 1) To see this mo t e easil y. note that tot every p E C,, tin g e is at least one open path in A Irotu i` to p Thus ) (p) = is at most the expected number of open (finite) paths in A starting at .r Them are at most A(A – 1 paths in A of length I slatting at ,r. so V,'( p )

+

– 1)//5/ 1st

and 1)(-1p1+1

fin bond and site petcohttion espect ivelv Both sums converge lo t any p < l/(L\ – Pit)(A) /4 ( A ) > – As PH > t i . the c o ne sponding inequalities loi m i follow This shows that among all giaphs with maximum degree the A-1(4mila' tree has the lowest et itical probabilities Thew all' VW lolls tit ivial chan ges we can mid«) to a gi aph whose effect

1 Basic cancel& curd results

9

on the critical probability is easy to calculate. For example, if A is any graph and A ( ' is obtained from A by subdividing each edge I. - 1 Also, if where pl.' is p lA or times, then 1 4::(A (0 ) = is obtained from A by replacing each edge by k parallel edges. then = (1 -pi,1(A))Uk, where p eli is ph or Of comse, pst,(A ild ) = pas (A). Combining these operations. we may replace each bond of a graph by k independent paths of length (..• to obtain a new graph For bond percolation, the critical probabilities p o w and p„„„. satisfy 1 - (1 - mcm „.

= p„rd

In this way, by a trivial operation on the graph, a critical probability in the interval (0,1) can be moved very close to any point of (0,1)

If we know the critical probability for a graph A, then we knor, instantly the critical probabilities Mr a family of graphs A' obtained by sequences of trivial operations flour A, as in Figure 6

Pignut ti tornsfor wing one bond petcolation model into another, and their into a site percolation model If the first (the hexagonal lattice) has critical probability p, then the second has critical probability I sat isfying 1 3 (2— r ) = p which is also the critical probability For site percolation on the third graph It is easy to show that any 0 < < 1 is the critical probabilit y for

some graph, indeed, for some tree Let T be a finite rooted tree with height (depth) h, with (" leaves. Let T I = T. and let T" be the rooted tree of height ha formed from T"- 1 by identifying each leaf with the root of a copy of For example, if T is a star with k edges, then T" is the tree Tk ,, defined above Let 1' be the 'limit' of the tacos T". defined in the obvious way. Taking the bonds of 7' to be open independently with probabilit y p. the nunibm of leaves of T joined to the toot by open paths has a certain distribution X with expectation ph ( Now suppose that the bonds of T' are open independently with probability P. and let X„ be the number

Basic concepts and resu lts

LU

of sites of Tux at; distance fern horn the root joined to the root by open paths Then the sequence (X0 , Xr, .) is a branching process: we have X0 1, and each X„ is the sum of X„_ i independent copies of the distribution X As X is bounded, excluding the trivial case p = C = 1, it is easy to show (arguing as above flu Tk ) that percolation occurs if and only if E(X) > 1, i e., if and only if p fi e > 1: this is a special case of the fundamental result of the theory of branching processes In fact, one obtains 1 4. (I')

) = 1,11 .'(Tcc ) =

(3)

Suppose now that k > 1 and 1/(k 1) < 7 < 1/k Define (1 < a < 1 by (k-r-1) a k 1 - 0 = 1/7 Let a = La i be the 0-1 sequence with density a constructed as follows: whenever divides i but 2-) does not. set = 1 if and only if the )th hit in the binary expansion of a is 1. Let 'ZO be the tooted tree in which each site at distance i lion/ the root has k+arr. i children R. is easy to check that, for each n, we can find trees 77! and T" of height C = 2" such that (21's c C (T")" , where T" has (k ± 1)/k times as Mai/ \ leaves as Using (3), one can easily deduce that p„(Ta ) = 7, where p c denotes any of the four critical probabilities we have defined Alternatively, let 7 be the t andom tooted tree in which each site has k + 1 children with probability r and /ir children with probability 1 - r, with the choices made independently for each site It is easy to show that with probability 1 this random tree has p i .(T) = +1). In general, it is eas y to calculate the various critical probabilities for a graph that is 'sufficiently tree-like'. For example, for C > > 3, let Ckj be the cactus shown it/ Figure 'i This graph is formed by replacing each vortex of the k-regidar tree Tr; by a complete map(' on C vertices, and joining each pair of complete graphs col esponding to an edge of TA. by icier/allying a vertex of one with a vertex of the other, using no vertex in more than one identification, We call the vertices resulting from these identifications attachtrient vet Nees Although Gek,t contains many cycles, it still has the global structure of a tree, and percolation on CiL i may again be compiled with a blanching process Indeed, let K; be a complete graph with k distinguished (attachment) vertices v l . Taking the edges of lid to be open independently with probability p. let .Vi , be the random number of vertices among . irk that may be reached hum lir by open paths Let us explore the open cluster of a given initial site .r of Cr c by working outwards from :r Except at t he first step, how each attachment vertex that we

. Basic concepts and results

11

Figure 7 Part of the cactus C 3 r: each circle represents a complete graph on vertices Where two circles touch, the corresponding complete graphs shale an 'attachment' vertex.

step, the number of further attachment vertices that. we reach at. the next step has the same distribution as X i„ and these numbers are independentt. It follows, by considering a branching process as above, that reach at; a given

P`i'l ( C 5c,O =

( C k = lid (1) E(X11) >

This quantity may be easily calculated for given k and The critical probabilities for site percolation on Ckj may be calculated in a corresponding way Furthermore, there is nothing special about complete graphs: there me many similar constructions of 'tree-like' graphs for which the critical probabilities can he calculated using a branching process. In the rest of this chapter, we shall prove some easy bounds on the various critical probabilities for 7L", d > 2 Of course. percolation on Z is trivial: all the critical probabilities we have defined are equal to 1 In studying 7 2 , we shall make use of simple properties of plane graphs As all such graphs we consider will be piecewise-linear (in fact, they will have subdivisions that are subgraphs of 2Z 2 , if we wish), there are no topological difficulties In fact, all we shall need is that every polygon (in the graph Z-, say) separates the plane into two components, the

Basic

concepts

and

results

Thief in; and the el:lei-On, with the interior bounded (This is easily seen by considering the winding number of the polygon. which changes 1w 1 when we cross a side ) Fl our this, Euler's [bra n ch follows easily by induction (see Chapter 1 of Boflobtis (19981 for the details), which in nun implies that IS:5 and A.3 3 are non-planar Thus, for example, if C is a cycle in the plane and a, b. c, d are four vertices of C appearing in this order around C', then neither the interior nor the exterior of C can contain disjoint a- c and b-d paths; see Figure 8

Figure 8 Neither t he interim tun t he exterior of a cycle visiting a, b c d in t his older Can contain disjoint a--c and b d paths (solid lines) Both statements can be deduced either fron t the non-planarity of h, or hont t hat of ICA a by consideting t he clashed lines

Let us point out an important : sell-duality property of This will be very important later: now we shall use it in a trivial way The dual A t of a graph A drawn in the plane has a vertex for each face of A, and an edge e' fin each edge c of A: this edge joins the two ver tices of A' corresponding to the faces of A in whose boundm y e lies When A = T2 , it is custormuy to take = (3, ± 1/2, y±1/2) as the dual -Vet tex corresponding to the face with vertices (x.y). (:r+ Ty), (:r+ 1, y -k I), y+ 1): see Figure 9 dual lattice is then E 2 (1/2, 1/2), which is iti011101phic to A. The dual graph A* is important in the context of bond percolation on A In this context, the sites and bonds of A' are known as dual sites and duel bonds. and a dual bond e ." is usually taken to he open when c is closed and vice versa One trivial use of planar duality is to prolific an alternative way to visualize site percolation Let A be a plane graph: in face percolation on A. we assign a state. open 01 closed, to each face of A The faces of A form the vertex set of a graph in which two faces are adjacent if the ) share an edge: this graph is precisely A.*: see Figure 10 Thus. face

.1 Basic cconcept—, s and re5ults

MIEN II= MEM MIN

13

t

0-

0-

0-

t

°

Figure 9 Portions el t li e latti c e A lattice A • (dashed lines)

(solid law s ) awl the isomorphic dual

Figure ID lace percolation on the hexagonal lattice (left): the open faces are shaded. The corresponding open subgraph in the site percolation on the triangular lattice is shown on the right (to the saute scale!).

percolation on A is equivalent: to site percolation on A' Thinking of open faces o 1 sites as colotned black, fin example, and closed faces or sites as white, the face percolation picture is easier to visualize Returning to the study of percolation on Z2 , if H is a finite connected submaph of A = Z2 with vertex set C. then there is a unique infinite the submaph of A induced by the vertices component C of A — outside H. By the external boanda i y C of C we mean the set of bonds of A* dual to bonds of A joining C and Cc.; see Figure 11. Out first task is to show that D"C has the pope/ties we expect of a boundary

14



Basic con its and results 0

0

0

0

0

0

0

0

0

0

0

Figure 11 A finite connected subgiaph H of Z 2 with vertex set C (solid lines andcircles). Vertices in finite components of 2 2 — C are shown with crosses. (some of ) those in the infinite component with hollow circles- The dotted lines me the C--C, bonds, and the dashed lines the external boundary of C

Lemma 1. If C is the vertex set of a finite connected subgraph of V, then 0"C i5 a cycle with C" rn ils inle t iot

_ _

Proof. Let F be the set of C-C, bonds, oriented flow C to Ct. Em f = ab E F, orient the dual bond so that a is on its left, to obtain an oriented dual bond f = Equivalently, f is f ca ged counter, — clockwise through r/2 about: its midpoint Let a-0={ 7 E F}, so O'C' is an orientation of O'C. We claim that if f = T; tr° then there is a unique bond of O'C leaving v. Let the vertices of 2 2 in the face corresponding to v be a, b, c, d in cyclic order, as in Figure 12 Note that f = ab, so a E C and b E Let I?, 22 - C - Ct. be the rest of 22 , i e, the set of vertices of 22 -C in bounded components Note that, as C, is a component of 2 2 - C. there is no C,-R bond in 22. Suppose first that d E C. Then e E C or c E since c is adjacent to b E C„, and so cannot lie in R. In the first case the bond ct is the unique bond of 0"0 leaving v; in the second, de

I. Basic concepts and results ti

a E C

c

b E Coc

Figure l2 The oriented dual bond f corresponding to au oriented bond f from C to Cr,: Suppose next that c E C, Then d 0 R. so either dEemde C, and again the claim holds. We may thus suppose that c C„, so c (which is adjacent to b E lies in C, and that d C. If d E I? the claim again holds, leaving only the case a, c E C, b, d E C0 But C and C„ are disjoint connected subgraphs of Z 2 , so there are disjoint paths in Z-, one joining a to c, and one joining b to d As abed is a cycle in Z2 with no edges in its interior, both paths must lie in the exterior of this cycle, which is impossible; see Figure 8. As every edge in D = C' has a unique successor, the underlying lino/ iented graph D'C' contains a cycle S. This cycle separates the plane, and crosses only C,-Cm bonds; thus C., lies outside 5, and C is inside S. If f* E 0-"C with f = ob then, since one of a, b is in C and the other iu C„ the cycle S must cut the bond f Hence f* E S In other words, ever y bond of DE C is in S. so ac consists of a single cycle ❑ In the rest of this chapter we present some basic results concerning critical probabilities First, following Broadbent and Hammersley [1957] and Hanunersley [1957a; 1959], we show that the phenomenon of bond percolation in V is non-trivial: the critical probabilities are neither or 1. To do so, we consider the number p„ (A; r) of self-avoiding walks ilk A starting at :r, where in graph terminology, a self-avoiding walk is simply a path All vertices of Z 2 are equivalent, and, as V is =!-iegulai, pm = p.,(2?), pm(Z2;0) 1/3. For the ti pper hound we consider A = E 2 together with its dual A' = ± (1/2,1/2) defined ear hem taking a dual bond c- to be open if e is closed, and vice versa An open dual cycle is a cycle in the dual graph consisting of dual bonds that are open Suppose now- dun p > 2/3. Let Lk be the line segment joining the might to the point (k,0), and let S be a dual cycle surrounding Lk of length 21 Then .9 must contain' a dual bond c" crossing the positive x-axis a some coordinate lat: Veen k 1/2 and (2( — 3)/2; thus we have fewer than t choices fin e' As the rest of S is a path of length 2( — 1 in the dual lattice, win kIn is isomorphic to 2 2 , there are at most: f'p9m I possibilities fon S. Let Yk be the number of open dual cycles surrounding Lk Since dual bonds are open independently with probability 1 — p.

c, --It (3(1 — p))2(

( Pam — Pr _5_

Eli( Yin < >k+2

>k +2

9

As 3(1 — p) < 1, the final sum is convergent, so E v (1"",.) —" 0 as k DO, and there is some k with Ei,(4) < 1 Let il k be the event that Yk = 0; since lE p Ork ) < I. we have 1111,(A k ) > 0. Let Bc be the event that the k bonds in Lk are open Note that AR- and Bp a ye independent Also, if both hold, then there is no open dual cycle surrounding tire origin so, by Lemma 1, the open duster containing the or igin is infinite Hence, 0(p) = 00 (p) > P(A k n By) = 1P,,(AMIP 0 (80 = Pp (il k )ph;

This shows that p H < p Since p > 2/3

wi5

>

0

arbitrary. ptr < 2/3 follows

The second put of t he argument: above, bounding the critical probability from above be estimating the number of separ at ing cycles, is sometimes called a Peiells argument. after Pearls H9361. hi higher dimensions, we

I. Basil, concepts and results

17

estimate the number of separating surfaces: even the easiest bounds on the number of these surfaces suffice to show that the critical probability is strictly less than 1. Comparing Z d with Z2 and with the (2d)-regular tree, we see hat 1/(2d — 1) < p11' (DI ) < pll yZ(1 ) 1/A, the expected number of open paths of length a in Z2 starting at the or igin tends to infinity as n cx:. Thus, in analogy with numerous phenomena in probabilistic combinatorics, one might: expect that, with probability bounded away from 0, there are arbitrarily long open paths starting at the origin This would suggest that = 1/A, as is indeed the case for the k-regular trees. The trouble is that, as there are (A ± o(1)) 2 ” dual cycles of length 2n surrounding the origin (see Hantmersley (1961b1), the same intuition would indicate that for p < 1 — 1/A there are open dual cycles surrounding the origin, implying that p it = 1 —1/A. Thus, for both intuitions to be correct, A would have to be 2 As we shall see now, this is riot the case Any walk irr which every step goes up or to the right is self-avoiding. so p„ > 2" and A > 2. Let us say that a path P is a building block if P starts at 0, ends on the lbw .c +y = 2, and ever y inter mediate vertex of P lies in the region 0 < .r g < 2: see Figure 13 Then any sequence of building blocks P I , Pr, , 'nay be concatenated to make a self-avoiding

18

Bask concepts and results

0

0

'a

Figure 13 A selection of building blocks that m a y be put together in any order to create a self-avoiding walk

walk IV star ting at 0. Star ting with If, each P i is the part of If lying in the region 2k— 2 < 2+y < 2i (plus the vertex on 2:+y = 2i -- 2 where II enters this region), translated to start at 0, so distinct, sequences give distinct walks.. Let ur„ be the number of walks If of length n that may be obtained in this way As there are four building blocks with two edges, and four with four edges, we have to.1 > 4 2 +1 = 20, and > //y.r„ > = 20 1', so A > 20 111 = 2 111 In fact., we have ur n >

4w„- 2

-I- 4 Wu –.I

Solving the recurrence relation, it follows that A is at least the positive root of x 1 — 4t 2 — = 0, namely, 2.197.... Much research has been done on calculating y„ and bounding A Fisher and Sykes [1959] showed that Ir i s = 17,215,332 They also obtained the hounds 2.5767 < A < 2.712, by considering respectively a special sub-class of paths, and walks with no short cycles More sophisticated algorithms and the use of computers have enabled these results to be greatly extended For example, / 5r was calculated by Conway and Guttmann [1996] using a supercomputer (see also Guttmarm and Conway [2001]), and Jensen [200‘la] has found fin The best published bounds on A are A < 2 6792, due to Penitz and Tittmann [2000], and A > 2.6256, obtained by Jensen [2001b] using the method of irreducible bridges introduced by Mester, [1963] Connective constants of other lattices have also been studied. In particular, writing A d for the connective constant of Kesler/ [1961]

19

.1. Basic concepts and results

showed that Ad = 2d - 1 - 1/(2d) + 0(d- 2 ) More recently, Haraand Slade [1995] showed that Ad =

(2d) - 1 - (2d) -1 - 3(2d) -2 -16(2d) -3 - 102(2d) -I + 0(d-5),

and that a corresponding expansion exists to any order. Returning to general graphs A, we next show that percolation is 'more likely' in the bond model than the site model. In the arguments to conic., we shall consider step by step explorations of the states of the sites, say. Suppose that the states X„ of the sites are independent, and each site n is open with probability p„. In each step, the next site v to he explored will depend only on the states of the previously explored sites Hence, given the history of the exploration so far, the conditional probability that a is open is just N. The reticle/ may well feel that the observation above needs no justification. Note, however, that as a random variable, the histor y of the exploration up to step I depends on the states of all sites that might be explored in the first I steps But the event that this history takes a particular value (i e the event that, for 1 < s < 1, at step s we tested site as and found that X„ „ = i s ) is independent of X„ for any v { , to.} The next result is due to Hammersley [1961a]; certain special cases were proved by Fisher [196 Theorem 3. Let A be a connected, infinite locally finite multi-graph. Then

Pii(A) ^ Ail ( A ) and IP, (A) >

(A)

Both inequalities follow from the assertion that, for ever y site x of A, ever y integer n > 1, and every probability 0 < p < 1, we have Proof.

(lex I Indeed, letting T1

0

^



5_P

p

(I Cr > II)

(5)

it follows that Or( A ) < P03(4)

Thus, if 0,„(A lb,) = 0, then 0„,(Ap = 0, so Ai (A) > plA(A).

(6)

20



Basic concepts and sults

Also. if (5) holds, t /IP\ 1,(1C,) = n

< EPLP! u=1

Hence, if tx„,(A ii ) < ix, then x„,(N) < ,x, so p r (A) > (A) It remains to prove (5) As C.'„, is empty in At;, if x is closed, inequality (5) is equivalent to „(1C,4 > n

37

is open) <

p H C,1 >

(7)

In proving this, we may and shall replace A by the finite subgiaph A„ of A induced by the vertices within distance 12 of X, since the event that I C; I > n depends only on the states of sites or bonds in A„. Let us explore Ct.. the open cluster in the site percolation on A„ containing conditioning throughout on being open We shall construct a random sequence 7 DI, U1 ) (.=1 of tripartitions of the vertex set 17 (A„) of A„; this sequence will be such that the final set B1, obtained after a random number I of steps, will be the cluster The notation indicates that the sites in R t have been 'reached' by step t, those in DI are 'dead' (known to be closed).. and those in U, are 'untested' To define T, set = Ur = 1". (A„) \ {a'}. and D i = 0 Given (R t . D,,U1 ), if there is no [1,41, bond, set I = l and stop the sequence Otherwise, pick a bond e t = yi z t \vith yt E E Utz and set (l i n = \ { Now test whetherr the site z t is open If so, set R1.4.1 = Di = De. Otherw ise. set D14. 1 U }, 1? 1+1 = Bt . Note that. at each step, the conditional probability that z t is open is p The process terminates as A„ is finite. L3y construction, for ever is a connected set of open sites, and all sites in D, are closed Hence, as no site in Rc has a neighbour in Uf = 17 (A„) \ (R, U the set Bt is precisely the open cluster Cs,. To compare the distribution of rt,1 to that of let us explore in a similar manner, using a random sequence I' = 0 as above This sequence 71 is constructed as T. except that, having picked e, = Th z t , we test whether the bond e t is open As this is the first (and only) time we test e t , conditional on the sequence '7" , up to step the probability that e t is open is just p. Consequently, the sequences '7- and have the same distribution. In particular. 1C.7.1 = 1/41 and XI have I he same distil ibution This implies (7) since Iry is contained in the open

Basic: concepts mi d results

21

cluster C1.1! of ,r in A l :: it. is the vertex set of the subgraph spanned by the set of bonds we have tested in the process T' and found open. ❑ In tire argument above we could have worked directly in the infinite graph, continuing the exploration indefinitely if (1;.. is infinite One can also consider meted which sites are open with probability p and bonds with probability p', independently of each other Al r iting 0 2.(A; p, p') for the pi obabilth that there is an infinite path starting at x all of whose sites and bonds are open, Haunnersley (1980) noted that the more general inequality

pi ) < Os (A: p. rip') for 0 < p, < 1 is air immediate consequence of its special case (6) proved above The inequalities in Theorem 3 need not be strict, as shown by the k-regular tree. In most interesting examples, however, they are strict Strict inequalities have been proved by Higuchi 119821, kesten [1984 Menshilwv [1987] and others For a general result that includes the graphs r' as special cases, see Grimmett and Stacey [1998] Aizemnan and Orlin/nett [1991] proved that a certain essential enhancement' of a percolation model leads to strictly smaller critical values: under suitable conditions, adding edges to the graph strictly decreases the critical probability This result has been extended la Beznidenhout, Griumrett and Kesten [1993] and GI immett [199-1) One Irriglit, expect that, if ) < 141 (\0), their ph(A ) < (An) However, it is easy to construct examples to show that this is not tire case, by modifying a graph in a way t hat decreases ph while leaving MI unchanged; see Wie g man 12003a1. Our next result, due to Chinnuett and Stacey 119981, gives air inequality bounding A above by a function of p1 The proof will be a little less pleasant than that of Theorem 3, clue to a minor complication introduced to improve the exponent below from A to A — 1 Later, we shall give a related result, for oriented percolation, where this complication does not arise

Theorem 4. Let A he a connected, infinite graph with A < x Then 141(A) C 1 — –Ph(A)Y1-' The satire inequality holds I'm pi

U1(13:i71111111

degree (8)

9')

Basic concepts and results

Proof It suffices to show that there is a constant It K(A) such that, for every site :r of A, every integer n > 1, and every 0 < p < 1, we have P V) (I CtI > it

is open) > ED /1 i i s(0 (1,1 > Ku),

(9)

where r = 1 — (1 — We shall prove this with K = A + 1 In doing so, we may. replace A by the finite submaph A,, induced by the sites within distance (A + On of x. Let Cirb be the open cluster of :r in the bond percolation on A„ in which each edge is open independently with probability p, and let C.11:, be the open cluster in the site percolation in which x is open, and the other sites of A„ are open independently with probability ./ In order to compare the distributions of 1(1,/ 11 and 1C1'„1, we first give an algorithm that finds a significant fraction of C.1,14 and then show that a slight variant of tins algorithm finds a subset of C1:4,. This time, we star t by exploring (41,4 using a random sequence T = Di ,(1i ) f of tripartitions of the set E(A„ ) of bonds. The notation indicates that the bonds in L, are live' (known to be open), those in .DI are 'dead' (known to be closed), and those in Ur are 'untested' At each stage, the bonds of L, will for m the edge set of a tree containing r; we write Bi for the vertex set of this tree (Thus Id, is the set of sites that may be reached from x along paths consisting of bonds in L t .) The set Ur will shrink as we proceed, while the sets L, and D, will grow. We shall also work with a set: Pt of si/cs growing with L namely the of 'peripheral' sites that will not contribute to the set Pt C V (A„ ) \ growth of li t The final sets Bur and P1, obtained after a random number t of steps. will be such that R 1 C C R1 U P1 To define T. set tt, = E(A„), and D i = L 1 = 0 Thus Given (L i , Di .dri ), let P, be the set of sites 'e 0 li t such that Ur contains a single bond flour Bi to v. If U, contains no bonds from Ri to 14 (A„ )\ U Pt ), set = t and stop tire sequence Otherwise, let e, = .//1 2:1 E Bi U P, Set U1 + 1 = Ur\ fedi, be any such bond, with y, E ci and test whether the bond e t is open. If so, set L,+ = U {e,} and so i = U {e,} and +1 = D,. R + = U {z,} Other wise, set Di + i D, L 14. 1 = noting that lit+ , = A possible state of this exploration is shown in Figure 14 Since we never test the same bond twice, at each step tire conditional probability that e i is open is p. As A„ is finite, the process ternrinates By construction, the bonds in L, are open, while those in D, are closed. Thus Rt , the set of sites that may be reached from 4: along paths consisting of bonds in L 1 , is a subset of Ct..,1 Also, if c = yz is

1 Basic concepts and results

:1

0

4

23

0

0

•

O

0

O

o

0

•

0

0

ylt

6

Figure 11 A possible value of D,, U,) when expiating the open cluster of the site re in bond percolation on 2S 2 Bonds in L,. DI and U1 are represented by solid, dashed and dotted lines, respectively. Tire set B1 is shown by filled circles The site z lies in Pt , so the bond yz will never be tested; there is only one choice inzt for the next bond to lest

an open bond with y E Ri and z Ff .R1 . then c

L i and e 0Di , so Since we stopped at step I. we must have E Pi , so the site z e E is incident: with no other bonds of As z 0 Rt . the site z is incident with no bonds of L i . so all other bonds incident with z are in De, and hence closed We have shown that all open bonds leaving 14 terminate in site s that me incident with no oldier open bonds Thus, every site z E C1,1;\Rt is adjacent to a site in and, crudely. 161;l1 5, (A ± 1)IRd

(10)

We now turn to the site percolation on A„ in which 1 is open, and the other sites are open independently with probability r = (1 — To realize this probability measure, let X II y E V(A„ ) {t}, 1 < ( < A — 1, be a set of i i d random variables. with P(X,, ,i = I) p and P(Xyi = 0) = 1 — p We decline y to be open if and only if at least one of the X,, is equal to 1 Let its explote . a subset of as follows, using a tandom sequence V = (LC, DC,Un i _f of tripartitions of the set of bonds.. This sequence V is constructed as T. except that, having picked e l = yt zi , we test one of the variables Mole precisely if this is the ith time that we have chosen the satire site z, i.e if there are i — 1 values 1 < t with z s = z, then we test whether = 1 Note that i < A — 1: initially, z

Bas i c cancrids and results

z IS incident v1/4-ith at most A untested bonds. each time we choose .7.„ we test au untested bond hidden!' with z, and we cannot choose zs if there is only one such bond remaining Since each variable X,, is tested at most once, each test succeeds with conditional probability p, and the sequence has exactly the distribution of By the construction of I', if e i y i z i was the bowl chosen at step t and e, E L b then the test at step t succeeded, so one of the X, is it follows by induction on equal to 1, and z i is open. As m E that. Rr c Using (10) and the fact that R4 and 14, have the same ❑ distribution, (9) follows, completing the proof For graphs A which are not regular, the proof above gives a slightly stronger result than Theorem 4, showing that if 0, > 0 Mr the bond percolation on A in which each bond is open independently with probability p, then 0, > 0 for the site percolation in which the states of the sites are independent, and the probability p„ that a site v is open satisfies 1 - p„ = (1 - p)`r( " )-1 , where d(v) is the degree of r The bound in (8) can be improved when pl.' i (A) is close to 1; in this case, the exponent A - I may be replaced by an exponent a little larger than A/2 This is shown by the much mote difficult result of Chaves and Schomnann [2000] that. for any infinite connected graph A of maximum degree A > 3, AO) 5 1

1

92[4/21+1 (1

(A))LV2J

while there are such graphs with /4i (A) > t -

- pil'i(A))t4/2j

In prov ing (11). Chaves and Schonniann made use of the concept of mixed percolation mentioned above. where states 11(:, assigned to the sites and bonds of A. So far, we have considered percolation on ordinary graphs, in which bonds are two-way. The basic concepts of percolation make just as good sense for oriented graphs. where each edge is or iented horn one endpoint to the other 'We could also consider threefed graphs, where there may be an edge in each direction between the same pair of vertices, or indeed, undirected or chiected In ulti- g raph s where there ma y be several edges (in one or both directions) between the same pair of vertices In a I gated percolating the underl y ing (multi-)graph A is oriented We assign states, open or closed, to the sites (vertices) or bonds (oriented

I Bosh: concepts and results

25

edges) as usual, and define the open subgraph, 01 –ATI„ as before Now, however, the open subgraph is of course oriented. An oriented path is a (finite or infinite) path P =,rox i x, in A with edges zuhti+1 oriented born x i to x i±i An open path in the oriented percolation is an oriented path in the open submaph, i e an oriented path all of whose sites (for oriented site percolation) or bonds (for oriented bond percolation) are open. Given a site', we write T..7 for the open out-cluster of x namely the set: of sites y reachable from by au open path: C

ty E A : there is an open path horn ,r to :111

C the open in–cluster of is the set of sites y nom ghich x can be reached, so y if and only if :u E CT,- Note that Cl"," is a subglaph of the out-subtpstph A t of c!, i.e , the set of all sites y reachable born x by oriented paths in A Since almost always we consider wily CI, we write fig C' define 0, =01-, and x r ;\;:i as berme, using ea, = CC+. As in the tutor iented case.. Or (p) and ;:v, (p) = are increasing functions of it, so we rues- define critical probabilities pn( ; ;g)and y-r( A ; JD as before Of course, all these quantities depend on whether we are considering site or bond percolation Sometimes we shall indicate this by writing tds (p), or plli ( A ;M t and so on In tlw oriented case, these critical probabilities may well depend on the site in consider, fig example, the tooted flee T shown in Figure 15, in which the root .. .co has

1)

tiN

7I \ >No VV4 6VJ VV9

Figwe IS An oriented rooted tree

T in which = pc (1:y)

pc(Mz).

two children, y and z.. all descendants of y have two children, and all descendants of z have thiee children Orienting the edges away nom xy to obtain au oriented graph F. we have pc (T ; = 1/2 for y and all its

26

Basic concepts and iesults

descendants, and p c (i i ;:t) 1/3 for the remaining vet tices. where is any of li ' , p!I For most graphs we shall consider, we do have pH (A ; a) = pri (X; y) for all sites x and y. and the same for pr; when this holds, we write pu(X) and p i (711- ) for the C0111/11011 values Indeed, if A is strongly connected, i.e , for event panr x and y of sites there is an oriented path horn a to y, then, as in the 11110i iented case, it is immediate that 0„.(p)> 0 if and only if 0, and v„(p) < cc if and only if x y (p) < (x. so p H and p t rue well defined 11 A is am, oriented graph obtained by or ienting the edges of a simple graph A. then the paths in A are a subset of the paths in A Thus percolation in A dominates that in A, so, for example, 0 y (9) >

j4(

^ A;: r') 91(A)

for all sites X The inequalities between site and ond critical probabilities presented above carry over to oriented percolation, with a slightly different form for one of the bounds Theorem 5. Let A be an infinite. connected locally finite oriented multi-graph., and let a' be any site of A Then ( A 1./) when? ./..\ 1 „ is the equalities hold for

/4 1 (

A : , 1') S 1 — ( t—

111(0')11111171

A '''))s'n

m. - degree of a cei te3 in A

(12) The same in-

Proof Let pc denote eiabei p H or P with all occurrences of p c to be interpreted in the saute way The first inequality. /OA; x) > p {I 1( A -,x), is proved in the same way as Them ear 3, mutat/5 mulandis: it suffices to show that fia all sites .r, integers IL and 0 < p 1, we have

1Pfk. p(ICt

n) 5_ p PI* p (ICI

1),

where here Cy.,, = Cit In showing this, we work in the finite subgraph A„ induced by the vertices reachable from by oriented paths in A of length at most rn The construction of the sequences and I' proceeds exactly as in the proof of Theorem 3, except that we take the orientation into auctorial: at each step t we select an edge ?TT = if-( 7-4 with ig e a r E 1-1/. ]'he proof of the second inequality, pg A; < — —pir !( A ; ./te'"



1 Basic concepts and ves tals

27

similar to, but simpler than. the proof of Theorem It; suffices to show that lot evert' site x of A. every integer n > 1, and every 0 < p 1, we have , (iCrl

x is open)

Pi• p (tC; ( 1 ?. a),

where r = 1 — (I — p) 411.. This time, we explore the open cluster CtIt ) of the bond percolation containing in the natural way, using a tandem sequence 7- = (L t .M,Ul )i =1 of tripartitions of the set E( A „ ) of bonds, such that; the bonds of I4 form the edge set of a tree oriented away from x. We write Br for the vet lex set of this tree; thus, Rt iti the set of sites that may he reached from .1: along ot hinted paths consisting of bonds in Li To define T. set U, = E( A „). and Dr = Lr = 0. Thus R, = fal Given (L t . D,. U,), if III contains no oriented bonds front It, to rrt V(X„) \ R t . set t I and stop the sequence. Otherwise, let 77 be any such bond, so Ut, E Br. and zi B., Set U1 4-1 = Ur\ 177 1, and test whether the bond e7 is open. If so, set Len = L, U {FI} and = Rr U {z t } Other wise, set D b.. 1 D I URI and Di + 1 = p t. so ft = L i . noting that RNA = HI Since we never test the same hood twice, at each step the conditional probability that T1 is open is p As A„ is finite, the process terminates By construction. the final set; R t is exact lv Tfulning to the site percolation, we take i i d minion/ vatiables X„ t, y E i t ( A „) \ 1 < < A t „, with P(X„ = p and P(X„ = ()) = — p. and declare (r to he open if and only if at least one of the X„,i x a t e open independently, each with is equal to 1 Thus the sites ir obribility r = I — (1 — We take a! to be always open As in tire moot of Theorem we can use the K 11. 1 to construct a process Tv with the same distribution as T Indeed, the definiticm is the same as that. of 7' except that, at step 1, having chosen 77 = Fri , we test whether = 1, whew = : 1 < s < z s = zell As each 1.0 = if.,Ts is oriented towards z,, we have i < /..\;„ Tire rest of the proof is as before: the final set 14, is a subset of Cis. The bound , (12) in Theorem 5 is tight Mr oriented multi-graphs Indeed, let A be an infinite binary tree with each edge oriented away from the toot Then MI (A) = = 1/2 (In other words, 41 (A ; d) = pit'r(A;x) = 1/2 lot every site ,r ) Replacing each oriented edge by A. parallel edges with the same calculation, we obtain a graph A1' 1 with



Basic emicepts and insults

maximum in-degree k, with —40 1d ) = — ( W )

—

/)1(7kI))k,

and the same for If we do not allow malt ple edges, then (12) is still fairly tight when A ) is close to 1; in particular, the correct exponent is indeed A i „ For Ai(A) Mt(A) vent' close to 1, this can be shown using the oriented analogue of the construction of Chaves and Schonmann [2000] To get good bounds in a wider range scents to requite a atom complicated construction, which we now describe By a (d, k)-mitt. we mean the oriented graph shown in Figure 16; this has an initial site and a final site r The site v is the root of a 4-art'

Figure 16 A (3, k) wilt 'The initial site u s joined to all :3 k leaves of oriented tree with toot II

/I

tree of height k with ever y edge or ionted towards o (In the figure, the solid lines show such a tree with k = 2 and d = 3 ) The initial site a sends an edge to all (.11' leaves of this tree Suppose that the bonds of a (4, k)-unit me open independently with \\'e shall take C > 2 to be a large constant, and let probabilit p = 4 ! pc Ignoring a for the moment, the numbers X; of sites at distance [tom n that are joined to v by an open path for in a certain branching process: for I < i < A', is the sum of X i _ t independent copies of a Omaha) As d — x with C fixed, this binomial binomial BUJ, p) dish Omaha/ converges to the Poisson distribution with mean C

.1 Basic concepts and results continued to infinity, the till/ vivid probability of this branching process is a certain function p(C) of C. We have p(C) = -c- c' -0 (CC- 2c ) as C — cc: the process is highly supercritical, so given that it sin VIVOS at the first step, it is very likely to survive forever. In fact, as k oc, with probability p(C) - 0(1) we have XI; > C kl" , say Let Rd, k, p) be the probability that there is an open path horn v to v It follows that Inn lim 1(1, k,C I d)

d--cc k—x

p(C),

so choosing d and then k large enough, we have f (d. k , C d) 1- (1 -C ± 0 (ce-2c) A similar but much simpler argument ha site percolation shows that if a is open, and the other sites are open independently with probability p. then the probability g(d, p) that there is an open path hour v to v is between p and p - 0 ((1 - p) 2 ) for p close to 1, d > 3, and any k Let 0 < < 1, and let T be a tree with critical probability pp(T) p 1 (T) = 7r. Choose an arbitrary site 3: 0 as the root, and or lent each edge of T away horn to to obtain an oriented tree T with maximum in-degree 1 which also has critical probability 1) 11 ( ; x0 ) = PT (7' ; ra) = (Depending on the choice of T the critical probabilities MT; t) may or may not depend on the site ) Let us replace each bond of T by a (d. k)-unit, by identifying it with 7 and v with y and then deleting Fa. The resulting graph A ( A ; to), has bond and site critical probabilities p 0, and let Bk be the event that the k ± 1 sites in Lk are open. Note that A i,. and Bk are independent Also, if both hold, then there is rro blocking cycle surrounding the origin, so c 0 is infinite Hence, > P(A k

n

13k)

P(A)P(Bk) = P(ilk)p k+I > 0.

32

BOSic conceply and msalts

This shows that 4(172 ) < p Since p > /10/81 was at bit/ at ‘, p Ii r (2 2 ) < ❑ 80/81 follows. as desired Fin oriented bond peniolation, a cycle S of length 2( sin rounding the (nigh/ is blocking if even , bond (tossing S horn the inside to the outside is closed As there ate exactly 1 such bonds, this event has pi obability (1 — p) t , and the atgamed- above shows that ph (2 2 ) < 8/9 Just as in the unor ionic(' case, one CM/ obtain better bounds by mote caudal counting. Not every dual cycle surrounding the origin can arise as the extemal bou v d' Co of the open clustet Cy: ha example, the cycle S shown in Figure 17 cannot. as Hume is no WaV to reach the two sites att he bottom right along talented bonds inside S Balister, Bollobtis and Stacd, 110991 counted cycles that can arise as id 'Co mote carefully, showing that there ale (K o(1)) ( such cycles of length 2t, where K is a constant satisfying 5 1269 < h < 5 2623 The constant I = K(22) plays one of the «Acts of the connectiveconstant A = A(2 2 ): the argument above shows that Ai' ( 2 2 ) < 1 — 1/K < 0 81. which is analogous to the Impel bound on di'l (252 ) given in (-I) As we shall see in later chapters, much better bottncls on this critical pi @liability may be obtained in other MINS

it to the asymptotic behaviour of the vat hats et itical probabilities associated to percolation on E d and Z il As noted canter. for each fixed (I, we have di nned eight et itical probabilities, of which pi; (Zit) is the smallest and pC1 ( d ) the hugest. Since 2' 1+i includes 2 4 as a subgt mill, each of these eight functions decreases with d. \VP IlOW bn

Theorem 7. For any d > 2 toe have

2d — 1 — Ph (cid ) 5 1 4 (2:d)0(1/(1)

Proof The fi t st inequality is just as easy as the special case d = 2 moved p„(4) of in Lemma. 2, Indeed, as 2( / is 2d-regular, the number p„ paths in E d starting at 0 and having length n is at most 2d(24 — Fm bond percolation with p < 1/(2d — 1), the expected number of open so ;A(p) is finite and pp > p As paths slatting at 0 is ) p„p" <

1/(2d — 1) was arbitrary, pr > 1/(24 — 1) follows Fat the tippet bound on 14 1 (2 a ) we shall in fact prove that

p
1 Then lim„_, a„lir crisis

Lemma 2,

(I,,;

<

Proof Let e. li t n int „_, a„/(1. so 0 < e< or Given > 0 there is a 11 -= kg + 0 < < then k such that a k < e + a„ < ya k + < ga k +bk. where b k = inax i < k a k aci Hence

a„ < a k / + bk la < c+ 2E, ❑

if a is large enough

a„ > 0, then the conclusion is again that a„la exists, but we must allow the limit to take the value

If we do not assume that

— 0c Taking login Ulm's, it follows that if (a„) is a snlmuthiplicative sequence of positive /cal numbers, i e if a„+„, < a„a „, for I I, m > then ihri„_, a), /n exists Many sequences occulting in percolation theory do not satisE N the conditions of subadditivity or submultiplicativity dearth, and one needs the following extension of Fekete's Lemma proved by de Bruiju and Er dos [19521 Lemma 2'.

Let Lt.)

+

Taff+

Ii

be an increasing Lunen° with

:r(t)t -2 dt b imply that b E D Equivalently, D C P(X) is a decreasing set system or a down-set if A C B E D implies that A E Clearly, U C (2" is monotone increasing if and only if its complement, .D = Q" \II, is monotone decreasing The fundamental correlation inequality in the cube, proved by Harris [1960) in the context of percolation and rediscovered by Meitman [1964 states the intuitively obvious fact that increasing events are positively col related. Given a set A C (2", for t = 0,1 define

= {(ai)72 1 1 : (al,

,a„)) 1 ,1)E

C

(2"

II A is monotone increasing then Ao C A 1 , and if A is monotone decreasing then Ar C Ao Let Q p",–I be the weighted cube whose probability measure is induced by the sequence = (m)72 / 1 . With a slight (and usual) abuse of notation, let us write P for two different measures, namely the probability measures in Q p" and (47 1 Note that P( A ) = ( 1 –

)P(-410) +Th iP ( A r)

(3)

for every set; A C Q" We are ready to state and prove Harris's Lemma Lemma 3. Let A and B be subsets of Q p" If both are up-sets or both ale down-sets then P(A

n B) > P(A)P(B).

(4)

Probabilistic tools

-10 If

A is 01) up-set and 13 is a down.-set thco P(A n B)< INA)P(B)



(t)

Let us prove (1) by induction on II Pot n = I ((a, indeed, a = 0) the inequality is ti kris'. so suppose that I] > 2 and that (l) holds for —I Suppose that A and B ale both up-sets

P(Ai)P(A2) P(A))

(7)

me all up-sets, °I all down-sets whenever A I . rue A simple consequence of Hailis's Lemma is that if At, increasing events in (2 1 ,2 whose union A has very high probability. then one of the A; must hate high probability Indeed, the complements Ay

2 Probabilistic tools

41

are decreasing, or down-sets, so from (7) we have

H N I ") .5- NA') 1=1

It follows that fot some i we have P O) 5- (11)(Ac))11?

— PO I

IP(*) I —

U U AO)

(8)

In the case where each A; has the same probability.. inequality (8) holds Cot ever) i. For t = 2, this observation is sometimes known as the 'square-root trick; for t = this is the ''nth-toot trick' If p : Q" R is any function, then we may think of measure on Q" Thus, For E C Q n we have p(E).=

II

as a signed

Fr(x).

and the integral of a function h : Q" — R with respect to p is h

ip=

h(3)/1(3) = (HO(Q") EQ.

In pat ticultu, wining l f for the diameter istic function of a set E C we have It^ dt = p(E) In this notation. Harris's Lemma states t hat for the probabilit.v measure Pp on the cube we have (.2 p "

drinp II 1.A1.13 (Pp >

A

Li

dIPp (9)

whenever A. B C Q" me up-sets Inequality (9) yields a more high-blow formulation of Harr is's Lenarta Although the terminology is self-explanatory, we note that a function It : Q" R is (monotone) increasing if h(r) G h(g) whenever < y Lemma

4. Let f and q be increasing functions on

I

f g dPp >

Q"

Then

()Pp g drp (10)



Probalribstie tools

42

Pivot Adding a constant C to f increases both sides of (10) b y the f is positive for some C', we may same a/1101111i, C' g As C Similarly, we may assume that q > 0 Then thus assume that f > = rift and g = j= where the ef and di are positive constants and the functions h, gi are characteristic functions of up-sets Thus. cid;

I Ill

Lg.)

c i d; I L

11,

gj

❑

where the inequality is Lemma 3

The van den Berg-Kesten inequality I1985I fin monotone events is partial converse of Harris's Lemma: it also states an intuitively obvious fact Let .4 C (2; be the ineteasing event of having at least one inn of three heads, and let B C Q p" be the increasing event that there ate at least five heads Let AO /3 be the event that there is a tun of three heads and there ate at least five Ow/ heads It is Mud not to he convinced that IP(A 0 B) < INA)P(B) operation Let us give two locum definitions fin the 'square' or b B C P(.5) for set systems First, we define AO B fin S such that

C S thole ate disjoint sets

A 0

,--cnv

Dnz=cnz

implies D E A. and c B. fin any D

C

S}

Let us write / I x for the restriction of a function to a domain X If means d = ((.1 [ )i 1 . and I C [al. then the condition c hi = = tot all i E I For A, B C Q" = {0,1}", we may exactly that = define A D B by A ❑ B

e

(2' thew are disjoint sets 1 ./

C

[a] such that

d ii = e ll implies d E A, and = / implies d E B. for any d E C"2"} This definition is equivalent to that fin set-systems given above Note that A0 B is a subset of A ft /3 In fact, it may be a ratlar small subset of A n 13: fin example, if neither A not B contains a subenbe of dimension at least 42. then A ❑ B 0 di/ = q 1 implies that d E A, then we call c ti a witness on a cc/iificate 161 A with suppoll 1 Thus AUB is the set of points c fin which there ale disjoint sets I and .1 stab that c it is a witness lot A and e l [ is a witness



2 Probabilistic tools

13

tht B Note that the square operation, which is obviously commutative, is also associative: both (A l 0 AO 0.4 3 and .4 1 0 (A9 0 A 3 ) consist of the set of points c for which there ate disjoint I I , 1) and 13 such that el]; is a witness for A. = 1,2,3 For increasing events (and for decreasing events), the definition of the square operation can be simplified considerably Indeed, if A, B C P(S) are hICI easing set systems, then

A D B = {21 UB:An13 (4, A E A, B E B} Lt the context of inmeasing set-systems, we may take a witness for A to be simply a set A E A. /abet than the function that is I on .4 Then A 0 B is the set of sets C such that. C contains disjoint witnesses lot A and Mt B identifying (2" = {0,0" with the algebra 272{, so that for a = b (1)) )7_, we have (a i )7=1 and = (a + 14)7- 1 and oh = (oihi);1-1, with 1 1 defined to be (I, we have I he following simple description of A 0 B for increasing subsets A, B of Q": A 0 B =

b: ob = 0, E

b e 131

In other words, for increasing events A and B C (2", the event A0 B happens if sonic (minimal) elements of A and B occur . disjointly: A OB s the up-set generated by disjoint ly supported elements of A and B. as in the example we star tied out with. Ebb then, is the van den Berg-I:08ton inequality The simple proof below is due to Bollokis and Leader Theorem 5. Let A, B C Vr; be increasing events Then P(.4

0

B) C P(A)1P(B)

(11)

Proof We shall prove (11) by induction on n; the case it 1 (or, indeed, the case n = (1 which makes perfect: sense) is trivial As before, let Q p"7 I be the weighted (n - 1)-dimensional cube with probability measure defined k = ,p„_,) Let C = A0 B It is easily checked that, as A and B me up-sets, Co = Ao Bo and C', = (.4 0 0

U (A I 0 DO

(12)

41

Probabilistic tools

As A0 C A, and Bo C

it follows t hat

Co C (.4 0 0

) n (A 1 0 B0)



(13)

and that C A Bi Thus, by induction, P ( C0) = P1A0 0130) P(Ar)IP(BoT and P(C, ) < F(.4, ❑ B,) < P(

)

Furtheinnne, front (12) and (13). F(Co) INGO < LP ( (,4 0 0 B I ) n ( A 0 Bon +P((.4 0 ❑ /3 1 )1T (A, ❑ 130)) P(A 0 0 B 1 ) + PHA 0 B0) < IF ( :1 0) IPTB D F( A , )Fd131) Multiplying the last three inequalities b y (1 — p„) 2 pY, and p„ (1 — p„) tespectivel y and summing. we find that (1 -1,,,)P(co) Puflet { — PO P( Ain + Pit P ( T t )1

— fin ) 121 Bol +PaLNBI/}

Using (3) three times, this is just) < P(A)P(B), as required

❑

Van den Berg and kesten [1985] cot/lectured that their inequality holds lot all events, not only monotone events Van den Berg and Fiebig [1987] proved that the inequality holds lot Coll Vel events, that is, intersections of increasing and decreasing events Tate lull conjecture resisted all attempts until Rchnet [2000] moved it Theorem 6. /A A B c QV, Then P(±1 ❑ B) < P(A)P(B)

❑

The proof is much hander than that or the van den Berg—Nester inequality Han is's Lemma has many extensions, including some cal 'elation inequalities of thiffiths (1967a:1967bl concerning Ising lenomagners, and extensions by Kelly and Shetman [1968] The PEG inequality of Ratuin, kastelevn and Ginibre [1971] extends these to a cot telation inequality On pat tinily °Rioted sets. with applications to statistical mechanics;

45

9. .Probabilistie tools

the EKG inequality is the roost often quoted correlation inequality in physics: even Hair is's Lemma tends to be called the 'EKG inequality' Ahlswecle and Davkin (19781 extended the EKG inequality to a very general correlation inequality on lattices What is amazing is that such an inequality could be true: its proof, although fa.r flour trivial, is not ver y difficult. Given points a = (at)(_, and b (b 1 )7_, of the cub e Q" = {0,1}", their jour is a V b = (ei ) iL l , where ei = a i V h; = maxfa i ,bi l, and their meet is a A b = (d1 )7_,, where di = a; A b i = nrinfa i ,b 1 1 (Thus, identifying (2" with the power set 'POrd), so that a and b become subsets of H. we have aVb=aUb and a Ab=anb) For A. B C Q". define the join of A and B as E_-1. b E

B}

A b:a EA. b E

B}

A V 13 = la V b: a and their nee/ as A A B = (a

As usual, we shall identify a function * Q" = (0, II" signed measure it defines on Q". so that if E C Q" then

=

IR with the

=)((') cEE

Here, then, is the Foul Functions Diemen' of Ahlswede and Dar kin, stating that a certain trivial necessary condition for an inequality is also sufficient Theorem 7. Let o.

Q" — Pr'` = (0. yr) be such that

n(0)/1(b) < 1(a V b)6(a A h) for all a b E Q" Then el(A)/3(B) G 2(A V .8)6(A A B)

0

fat all subsets A. B C Q"

Choosing appropriate functions o. 3, 2 and 6, Theorem 7 iruplies host of inequalities. In roost of these results, a, (3, 2: and 6 ate chosen IR T We may choose any to be the same function (measure) p: Q" function p, provided it is log-supermodulm, i e satisfies p(a)p(b)

p(0 V b)p(a A b)

Probabilistic tools

16

for all a b e Q".. (Occasionally, such a function p is said to be logmonotone ) By the Ahlswede-Daykin Four Functions Theorem, if p is modular , then log-super. p(A)p(B) C p(A V B)p(A A B)

(14)

for all A, B C Q n . The very special case where ft is the normalized counting measure implies Harris's Lemma, since, when A and B are increasing, A V B is just A fl B The simple proof of Lemma 4 shows that (14) implies the FIKG inequality, which may be stated as follows. Theorem 8. Let p : Q"

sure, and let f. g: (2" --.,

F', +

P. ± be a log-supermodular probability meabe increasing functions Then

f g dit

^

1 1(11!!gdtr

Our final aim in this chapter is to present, some fundamental results concerning thresholds for events, stating that under rather general conditions, the probability Pp (A) of an increasing event A C 1111C/C1 goes a sharp transition as p passes through a critical value These sharpthreshold results will be of fundamental importance in several of the major results on percolation. Given a point w E Let A be an event in the weighted cube Q" = the it II variable w i is pivotal for A if precisely one of , w„) is in A w= (uo, . co„) and i (w) = . — wi+1 , Note that: whether the dIr coordinate is pivotal depends on the point w anal on the event A The influence of the ith variable on A is Qp "

di (A) = ii p

(A) = Pp({w:

a); is pivotal for AD,

or, in the usual notation of probability theory, simply /3i (A) = Pp (w i

is

pivotal for A)

The fundamental lemma of Margolis [1974] connects the derivatives of (A) with the influences of the variables; this lemma was rediscovered some years later by Russo (19811. Pp

Lemma 9. Let A be an increasing event in, the weighted cube Q p", where p

= ( p i•

, p„) Then p;

Pp ( A ) = /51(A)



2 Probabilistic tools

17

In particular, = dlp r ( fl )

/3r (A).

Proof It suffices to prove the first statement with i = a Given a point ,:rk) E Q k , where k < n, write

x =

f

= i:

Pi H - Pi) =0

so that if k = n then Pp ({x}) = px Also, for x E Q"- 1 , set x+ = (x ) , , 1) and x_ = (r h . -1,0), so that x i., E 2" Note that px+ + px - = For an up-set A C Q", let A„ IxE(2" -1

x+ E

A, x

fi b z=

x+ E

A,

E

AI

and E (2 "-I

A},

and note that Pp( A ) =

E (p x

Px-)

E p x -}- Pm )11 Px



(15)

xEslb

Hence

Op„

l?p (A)

Px

At the point x' = (x,x„), the nth coordinate A i„ so this last expression is exactly /3„(A)

is

xE

pivotal if and only if ❑

To prove sharp-threshold results, we wish to find lower bounds on O i (A) In the unweighted cube Q„, /31(A)2"-1 is precisely the edge-boundary of A Thus the edge-isoperimetrie inequality in Q„ tells us that if P 1t1A) f then 11

0 1 (A)

> t2"(n

logo (t2"))/2" -I = 2t;log2(1/t),

SO

max/31(A)

2t log; (11t)In

Probabilistic tools

S

Ben-Or and Linial {1985: 1990] conjectured that this last inequality can be improved substantially: up to a constant factor, logn(1/t) can be replaced by logo 'this conjecture was proved by Kahn, Dalai and Linial 11988] with the aid of the Bonarni-Becknea inequality from harmonic analysis For a combinatorial proof, see Falik and Samorodnitsky [2005] Theorem 10. Let A be a subset of the n-dimensional discrete cube (2„ = {0,1}" with probability t = 1A1/2". Then ( log 02

i (A) 2 > et 2 (1 _

fi

(16)

11

where > 0

hi

1171

❑

absolute constant

The bound above is often written with min{ t 2 , (1 — 0 2 } instead of t 2 (1— 0 2 ; apar t Rom a change in the constant, this makes no difference A similar comment: applies to the bounds below Relation (16) immediately implies that max

>

t) (log n)In

Ben-Or and Linial gave an example showing that this is best possible up to an absolute constant In order to apply Lemma 9, we need bounds on influences in weighted cubes Such an extension of Theorem 10 was first proved by Dour gain. Kahn Kalai, Katznelson and Linial [1992]; a simpler proof has since been given by Fr ieclgut [2004] Theorem 11. Let A be a subset of the rueiglrted cube (2 1; with probability P p (A) = I Then max/j; where c > 0

an absolute

(A) > d(1. — CO 11

stant

!off n

❑

Friedgut and Kaki [1996] noticed that a slight variant of the proof of Theorem 11 gives a stronger Jesuit (apartt from the constant): if the maximal influence is not much larger than the bound in Theorem 11 then there are many variables of comparably large influence, so the stun of the influences is huge

2 Probabilistic tools

49

Theorem 12. Let. .4 be a subset, al the weig hied cube Q" with Pp (A) = If 11;(.4)< S fat every i. then

— ) log(1/6),

,d i (A) > c i.1

where c > 0 is

an

❑

absolute constant

There is a simple condition uncle/ which one large influence guar antees C Q'' is signmettic if that the sum of the influences is large A set t i s • wariant under the action of sonic group acting transitively on [n] Thus, A C Q" is symmetric if, for all 1 < j < h < n, there is a pet mutation rr of [n] such that 7(j) = h and if x = CIA is in A then so is 7 (x) = (.H o)i All we shall need about a symmetric event is that ever y variable has the same influence on it Using this observation and Lemma 9. Ft iedgut and Kahl [1996] deduced from Theorem 11 that ever y svn u netnc increasing property ifi has a sharp threshold: an 0(1/ logo) increase in the probability p suffices to increase the probability of the property from close to 0 to close to 1 Theorem 13, There is an absolute constant c i such that if .1 C Q" is symmetric and increasing. 0 < z < 1/2, and IPp (.4) > then Pg (51) > 1 — e Whell CM] —p >

log(1/(25))

log o

❑

1'riedgut and Kalai [1996] showed that, for events whose threshold occurs at very small values of p. this result can be strengthened considerably Theorem 14. There is an absolute wnstant c, such that if rl C Q" is symmetric and increasing. 0 < e < 1/2, and Pp (A) > then TVA) > — e whenever —

e2P

(1//)

loo(1/(25))

log

0

It is not: hard to adapt the proof of this result to powers of probability spaces with 3, 4, 5 elements, where all elements but one have small probability

3

Bond percolation on Z 2 - the Harris--Kesten Theorem

From the publication ' of the first papers on percolation theory in the late 1950s, for over two decades one of the main challenges of the them) was the rigorous determination of p it = the critical probabilityfor bond percolation on the square lattice. Hammersley's Monte Carlo experiments suggested that the value of p H might be 1/2; [rather evidence for this was Own by Bomb (1959], Elliott., Heap, Morgan and Rushbrooke [19501, and Dumb and Sykes (1961] The fist major result on this topic was due to Harris [1960], whit) moved that p H > 1/2 In the light of this result: and the Monte Carlo evidence, Hammersley conjectured that the critical probability is indeed 1/2. Sykes and Essam [NMI] gave a non-rigorous justification of ( 4.72 ) = 1/2, fur ther supporting this conjecture. the next important step towards proving the conjecture was taken by Russo (INS] and by Seymour and \Wish [1978] who, independen(t)s proved that A (Z,12 )+ glij (Z2 ) = Kesten [1980] finally settled the conjecture: building on the Russo-Seymour--Welsh result, he gave an ingenious and intricate proof that. p H = 1/2. By now, there are many proofs of this famous Harris-Kesten Theorem; in fact, there are a number of global strategies based on various different ingredients, and frequently even the same ingredients have sever al different. proolS. :Here we shall give several variants of what is essentially one proof, using some of the basic probabilistic results presented in Chapter 2 NA,T hat follows will be heavily based on the presentation in Bollobtis and Riordan [2006c] Later, in Chapter 5, we shall indicate another proof, based on Menshikov's exponential decay theorem and the uniqueness theorem of Aizemmm, Kesten and Newman 1/2 is the fact that, for p = Intuitively, the main 'reason why' p it 1/2, the probability that there is an open path crossing an n by n —

Band percolation on Z2 - the Hann—Hester'. Theorem 51

rectangle the 'long way' is 1/2 As we shall now see, this is an immediate consequence of self-duality However, haying proved this simple fact, we ale still lather far from proving that pu = 1/2: in some sense, the real p robleru struts only then Recall that the dual of A V is the lattice A* with vertex set {(a - F 1/2, b 1/2) : (a,b) E Z2 1, in which sites at distance 1 ate adjacent Thus there is one dual bond e* for each bond e of V; this bond is the bond of A* that crosses e A rectangle R in V is a subgraph induced by a set of sites of the form [0 ,14 x where a < b and e < d are integer's We shall use the same notation for the vertex set [a, x dI and for the rectangle it induces. If b— a +1 and I'. = — 1 then we call R a k by C rectangle; note that such a rectangle has 0 sites and 2k1 — k — C bonds. A rectangle in A* is defined as in A = V; equivalently, R.' is a 'octangle in A* if P,.`+ (1/2, 1/2) is a 'octangle in A Although we have defined rectangles as suligiaphs of V and its dual, it is natural to define an 'abstract' k by C rectangle as a 'grid graph' with he vertices and 2/cf. — k — (edges. Cleat ly, a suligraph of 23 2 isomorphic to a k by C abstract rectangle with k, C > 2 is a k by C rectangle as defined above Turning to bond percolation on A = Z 2 , let us consider a configuration w E {0,1} E(A) on A. For the moment:, the measure on the space of configurations will be inelevant, lmt, we shall still call a set of configurations an event As in Chapter 1, we define a bond e" inr A" to be open if e is closed and vice versa; thus the configuration w specifies the states of the bonds in A and in A* The hot izontal dual of a rectangle R = [a, x [c, di in A or A' is the rectangle Rh = [a + 1/2, b — 1/2] x [c — 1/2, d + 1/2] in the dual lattice Analogously, the tun beat dual of R is the rectangle R" = [a x [e+ 1/2, d — 1/2]; see Figure I. Somewhat: artificially, the horizontal dual of a 1 by l rectangle is the 'empty rectangle', as is the vertical dual of a k by 1 rectangle. For k, f > 2, the horizontal dual of a k by C. rectangle is a — 1 by C + 1 rectangle, and the vertical dual is a k + 1 by C — 1 rectangle Also, (Rh )" (R'') h R. Given a configuration ro, an open horizontal erossin«A i a rectangle = [a, x [c, 4] in A or A' is an open path P C R joining a site (my) to a site (I), 2), i e., a path in the appropriate lattice, all of whose edges are open., joining the left-hand side of R to the right-hand side. We write 11(R) for the event that R has such a crossing Similarly, we write 17(R) for the event that I? has an open. VC/ heat crossing, defined analogously.

52

Bond ingeolaboa O n Z -- the flail

Theorem

M I MI I M S11W 111111 1111111111111 Figure 1 A rectangle 1? (solid litand its horizontal dual Rh (dashed lin R h is the Nettie& dual of B As a minimal open houzontal crossing of a rectangle I? does not contain anv bond joining two vet flees on the same vertical side (left or right) of I?, the event fi(R) depends only on the states of those bonds in I? whose duals appear in R h Similarly V(R) depends only on the states of bonds with duals in RE. The next lemma is the pit ()wised 'reason why' pn (2 2 ) = 1/2 The result is obvious, and it is tempting to state it without a proof However. it is not entirely t t ivial to pr ove, and it usually takes quite a while to clot the i's; for a poof of it closely /elated result see Kesten [1982. pp 386:3021 Here, we shall give a simple proof needing no topology Logically. the proof is equivalent to that given in Bollobas and Riordan [2006cd but the presentation is mote like that of a related result in Bollobas and Riordan [20061]. Lemma 1. Let I? be a rectangle in E li of its dual. Whatever the states of the bonds in B. entelly one of the events 11(R) MI (1 V(R h ) holds

Proof . Consider the partial tiling of the plane by octagons and squares shown in Figure 2.. (This is, in fact, part of the Atchinuidean lattice 8-); see Figure 18 of Chapter 5 ) We take a black octagon for each site of R. and a white one for each site of R h The bonds of R and of R h ate represented ha squares. with the sante square representing a bond e and its dual e* A square representing a bond e of R and its dual C " in H " is coloured black if c is open. so g* is closed, and white if c is closed and e* is open In the first case„ the black square joins

Bond petrolation on Z 2 Harris

lteodem

53

Figure 2 The upper figure shows the open bonds of a rectangle B (solid) and its horizontal dual Rh (dashed) In the lower figure each site of I? is drawn as a black (shaded) octagon, and each site of R h as a white octagon The central squares correspond to bonds e E I? with duals e' C R fi ; such a square is black if c is open and white if e* is open There are additional black/white squares around the edges to connect; the sides of the rectangles

the two black octagons corresponding to the sites e i0111,9 In the second, this white square joins the two white octagons corresponding to sites of

54

Bond pocalgtion on 27,2 - the Hartis-Kesten Theorem

the dual lattice that e" joins The squares corresponding to the bonds in the vertical sides of I? me coloured black, to 'join up' each side of R; the states of the corresponding bonds ate irrelevant to the event 11(R) Similarly, the squares corresponding to the (dual) bonds in the hot izontal sides of Rh ale coloured white Note that II(R) holds if and only if there is a black path of squares and octagons from the left of the figure to the right, and 11 (1?),) if and only if there is a white path of squares and octagons front top to bottom. In particular, 11(1?) and V(Rh ) cannot both hold: otherwise, these black and white paths contain disjoint (piecewise linear) curves in the plane lying in the Ulterior of a cycle C (the boundary of the partial tiling), and joining two pairs of points at, and bd, with a. b. c and d appearing in this cyclic order around C As noted in Chaplet 1 (see Figure 8), Kr, could then be drawn in the plane Let I be the interface graph. formed by taking those edges of octagons/squares that separate a black region front a (bounded) white one, with the endpoints of these edges as the vertices Then every vertex of I has degree exactly 2 except for the four vertices x, y, z and at. which have degree 1. Thus the component of I containing a" is a path Ili, ending at another vertex of degree I. Walking along I1 7 front r, there is always a black region on the right and a white one on tire left. Thus 147 cannot end at: 4, so IV ends either at y or at If II' ends at y, as in Figure 2, Own the black squares and octagons on the tight of IV give an open horizontal crossing of I?. More precisely, these squares and octagons correspond to a connected subgraph S of R joining the left. of .1? to the right, all of whose bonds ate open, except possibly lot some vertical bonds in the sides of I? Let P be a minimal connected subgtaph of 5' connecting the left of I? to the right Then P is a path. and P uses no vertical bonds in the sides of B. so P is an open horizontal crossing of R. and 11(R) holds Similarly, if IF ends at w, then the white squares and octagons on the left of II' give an open vertical crossing of R h in the dual lattice. Thus at least one of I1(R) and V(Ie) holds. ❑ The path II' in the interface graph I described above may be found step 1w step: we enter the tiling at r, and at each vertex we 'test' the two edges leaving tins vet tex and follow one of them. If II(R) holds, so t hat IV leaves the tiling at y, then the path P we find is the the top-most open hot )contal crossing of I? This has the very useful property that it can be found without examining t he states of bonds below P

Bond percolation on Z 2 - the Hands-Kesten Theorem

55

Algorithms such as this, that test for crossings by examining 'interfaces', are sometimes of practical significance, as they can be much faster than exploring, say, the set of all vet tices of R reachable from the left, and testing whether this set contains a vet tex on the tight Lemma 1 may be worded as a statement about a plane graph and its dual: contract each vertical side of B to a single vertex, and add a single edge I joining these vertices to form a graph G. Also, contract each horizontal side of Rh to a single vet tex, and add an edge f* joining these vet tices, obtaining a graph 6* Then (.7 and G* are planar duals; see Figure 3

Figure 3 A rectangle /? with its left and right sides pinched to single vertices (solid lines), and the rectangle R h with its top and bottom sides pinched (dashed lines) With additional edges f f' as shown, the solid and dashed plane graphs are chinl to each other

Figure 3 shows that Lemma 1 is a special case of the following result for a general plane graph and its dual Lemma 2. Let CI be a graph drawn in the plane, and let GI * be its dual, with edge set {e* : e E E(G)} Let f be any edge of G, and suppose that each edge e 7,4 f of G' s assigned a state, open or closed Taking an. edge e* to be open. if e is closed and vice yen sa, Mlle, there is a path in C; consisting of open. edges and 'joining the endpoints of f on there is a path

56

1301h1percolotani

Oil

Ea

the Hari is- Kasten Theorem

consisting of open (dent) CINGS and joining the endpoints of G" — f* Paths of both types cannot (I:tit:I shin/Una cously

in

The proof is the same as that of Lemma 1: replace each vertex Proof of G with degree et 1w a Hack (topological) 2d-gon, and each degree Mertes of G* by a white 2d-goo Replace each pair le, el of dual edges with a 4-gon, shining edges with the polygons corresponding to the endpoints of e and c*, respecting the cyclic order of the edges arid faces around each vertex of Gi and Cr, as in Figure -1 Colour all 4-

Figure -I A plane graph C (solid circles and lines) di awn with its dual CI (hollow circles and clashed lines) There is a black 2d-gon for each degree d vertex of C. and a white 2d-gon for each degree d vertex of a' There is a -I-gon km each edge/c hu rl-edge pair {c. this is black or white according to whether c is open or closed, apart from the hatched 4-gon, corresponding to the distinguished edge f and its dual f• goes except, that colitis/muffing to { i l"} black or white irr ai t y way Then in the inter face graph hunted by the sides of polygons separating a black region horn a white one, ever y vertex has degree 2 except the font vertices of tine -4igont coliesponding to f. f*}: the rest of the 1)1001 is as before

❑

= FPI! , p in which each bond of is open with in obabilit y p independently of t he other bonds, Lemma 1 has the follow ing immediate consequence cm fl ing to the probabilit y meastue

8 I The Russo-Segnomm Welsh method

Corollary 3.

57

0) If I? and R' are k by / - 1 UM/ k - 1 by C rectangles

in Z2 . respectively, then Pt aH(R)) -F E I-v( / 01 1) = I Ii is an It + I by II rectangle then P t12( H ( R )) =11 2

(1)

(iii) If 8 is an n by it square. then [2( 1? (S)) =P tp2( H ( S )) ^ 112 Pi vot' For part (i), recall that each bond of A = Z 2 is open indepe ndently with probability p, and the dual e' of c is open if and only if is closed By Lemma 1, every configuration w lies in exactly one of H(R) and V(R h ), so Pp (H(R))+ Illip (V(R h )) = 1 But Rh is a k - I by t] rectangle in A. where bonds are open independently with probability 1 - p, so Pp(17(/?h)) = Ruts (ii) and (iii) follow immediately hour p eat (i) ❑

It is easy to deceive oneself into thinking that (1) shows that par = 1/2 Although self-duality is of course the reason -why' Pu = 1/2, a rigorous deduction is far from easy, and took twenty years to accomplish

3.1 The Russo-Seymour--Welsh method The next ingredient we shall need in our proof of the Harris-Kesten Theorem is some toms of the Russo-Seymour Welsh Theorem relating crossings of squares to crossings of rectangles The proof we present is horn Bollobtis and Riordan [2006c] Lemma 4. Let I? = x [2,11 nn > n be an in by 2n rectangle. Let X(R) be the event that them are paths P 1 and Pt of open bonds, such that Pi CIOSSCS then by n square S = En] x (Id from top to bottom, and P, tics within I? and loins some site on P 1 to some site on the light-hand side of 2 Then IP1,(X(R))> Pp(H(R))11",,(1-(5))/2 Proof Suppose that V(S) holds, so there is a path Pr of open bonds crossing S horn top to bottom Note that any such Po separates 5 into two pieces, one to the left of Po and one to the right Let LI: (S) be the left-most open vertical crossing, when one exists.. defined analogoush to the top-most open horizontal crossing discussed in the remark after

58

Bond pemolation

OP V -

the Hartis--Kesten Theorem

Lemma 1 By that remark, for any possible value P I of L17 (.9), the event {LII (S) = } does not depend on the states of bonds of S to the right of PI We claim that, for any possible value Pi of EV (8), we have Pp (X(R)ILV(S)= PO Pp(H(R))/2 To see this, let P be the (not necessarily open) path formed by the union of 13 and its reflection PI' in ti re horizontal symmetry axis of .R, with one additional bond joining P1 to P(; see Figure 5 This path

Figure 5 A rectangle 1? a al square S inside it, drawn with paths (solid curves) whose presence as open laths would imply X(R). The path P formed by Pi its reflection P;, and th r single bond joining then, crosses 1? from top to trot tom

crosses [Id x [il] from top to bottom With (unconditional) probability Pp (11(R)) there is a path Pa of open bonds crossing 17 from right to left this path must meet P By symmetry, the (unconditional) probability that some such path first meets P at a site of PI is at least Pp(11(R))12. Hence the event Y "(PI ) that there is an open path P in 1? to the right of P joining some site on P i to the right-hand side of 1?- has probability at least Pp (H(R))/2 But Y(P1) depends only on the states of bonds to the right of P All such bonds in S are to the right of P1 in 8 As the states of these bonds ale independent of {L1 7 (8) = P1 1, we have Pp(Y(Pi) I LV(S) = P1) = Pu(Y(P1)) Pp(H(R))/2 If It (PI ) holds and LI/ (S) = Ph then X(R) holds Thus IFF,(X(/?) I Lli (S) =

IP,„(H(R))/ 2

As the event V (S) is a disjoint union of events of the form {L1 7 (S) =

3.1 The Russa-SeyMaill -IVelsh Method

59

we thus have P (X (R) I V(8)) > INH(R))/2, and the result follows

Lemma 4 allows us to bound from below the crossing probability of some non-square rectangle; in particular, of a 3n by 2,1 rectangle. Let h i,(rn,n) = (11(R)), where R is any in by n rectangle in Z 2 and let h(tn,n) = h 1/2 (711, n)

Corollary 5. Fat all n > 1 we have 143, 1 ,20> 2-7

li

Figure 6 'if - Wu 2„ by 2n (square) rectangles R, and an a by 11 square 5 in their intersection Zile solid lines indicate paths witnessing the event A(/?), and the corresponding reflected event for R.'

Proof Consider two 2n by 2n (situate) rectangles R' arranged as in Figure 6, and then by n square S in their intersection. Let X'(R!) be the

event defined analogously to X(R) but reflected horizontally Applying Leunna 4 to the rectangle 1? (which happens to be square), PiLX (Jr )) = P„ (X (T)) Pi,(11(R))Pp(1/(S))12 The events X (R), X' (R') and H(S) ate increasing, and hence positively correlated by Harris's Lemma (Lemma 3 of Chapter 2) If all three events hold, so does H (R U R') Thus, h(3n, 2n)



P 1/2 (H(R U R')) •

1111/2(X'U1'))P112(X(R))P112(H(S))

•

Prp2(H(R))2P1/2(1/(S))2P1/2(H(S))/4.

But 1? and S are squares, so. by Corollary 3. h(3n,211)

(1/2) 2 (1/2) 2 (1 /2)/4 = 2 -7

❑

Bond percolation an E2

60

the Han

hi - Kasten

Theorem

It turns out: that the difficult step is getting horn scrim/es to elongated rectangles: ft QM COI ollm V 5 it is vent' easy to deduce Hatt is's Theorem Indeed, considering in ) by 2n and mo by 2n rectangles intersecting in a 2n by 2a square as in Figure 7, by Harris's LCIUMI/1 we have

Figure 7 Two rectangles intersecting in a square II the rectangles have open horizontal crossings, and the square an open vertical one, then the union of the iect angles has an open horizontal crossing.

h(m ) - 2)1.2n) > h(a 2n)h(ne).2012

(2)

> 2n hi particular,

Ion ill

20> h(a).2n)l)(3m2t))/2

Nat

>

2-sh(m.2n),

h(lta. 2a) > 2 17-8k lot all k > :3 and n > 1 As h(m.2n h(n).2n) it Follows that thew are constants h k > 0 such that

so

1) >

li(km a): Il k (3) for all k > 2 and a > 1 Alternativeh, starting with an nt i by 2n rectangle R and tun no, b ■ the proof of Lemma 1 actually shows that 2ti rectangle Mtn ) T ne)- a. 2n) > h(m).201)(m)). 20/25 lot

in tn, >

2u

Plats

h(511.20> h(311.2n) 2 / 9r1 > 9-19 rind 1461).20 > 11(511.2014 9 m 9 0/ 95 > 9-19-1-5

=

0 -

(1)

As 2- 1 < /1(3n. 2n) < 1/2 lot ever y n. it is natural to expect that More gemnalh, one would 11 (31 1 .2a) coin el ges to a limit as n — Nun. fro) = (a lb) lot sonic function /(:c) with 0 < expect f(r) < 1 Indeed. it would be astonishing if this were not the case Surprisingly this conjecture is still open In fact. this is a very special

Hurt is 's Theorem

61

case of the general conformal invariance conjecture of Aizemnan and Langlands. Pouliot and Saint-Aubin f10941 This conjecture was proved Surnnov (2001a1 for site percolation in the triangular lattice; see Chapter 7

3.2 Harris's Theorem From crossings of long, thin rectangles to Harris's Theorem is a very short step Let us wr ite r(Co) for the radius of the open cluster containing the origin, so (C0) = sup{ d(ri. (I) r E where d(x. y) denotes the graph distance between two vertices of

Z2

Theorem 6. Fm bond ',emulation in V 0(1/2) = 0 Moot' In

fact. we shall prove slightly !now. : namely that P i/2 (1 (Ci d

fri

(5)

fin all n > 1.: where c > 0 is a constant At, p = 1/2„ the bonds of the dual lattice A" are open independently with probability 1/2.: so. hour (4), the probability that a tin by 2r i rectangle in A' has an open crossing is at least 2- =5 Consider two 6[1 by 211 and two 21i by 6n rectangles in A-, arranged to form a 'square annulus as in Figure 8 By Harris's Lemma, with probability at least

Figure S Four reel angles Fuming a square annulus

.r• = 2 -i00 > 0 each rectangle is crossed the long way by an open (dual) path Il this happens, then the union of these paths contains an open dual cycle surrounding the centre of the annulus; see Figure 8. For > 1. let A t; be the square annulus centred on (I/2,1/2) with

G2

Bond percolation on 7S 2 - the Harris-K oleo Theorem

inner and outer radii 4 k and 3 x 4 k , and let Ea he the event that Ak contains air open dual cycle surrounding the interior of A k , and hence the origin. Then P(Ek) > e for every P. As the A k are disjoint, the events Ea me independent If Ea holds, then no point inside .4 k can be joined to a point outside A k by an open path in V. so r (Co) < 1 +3 x 4 k < 4 k+ I Thus, P I/2 ( I

4c±1)

P1/2

n

\ k=

E =

H P112( Ed ^ 0

k=

and (5) follows Also, for any 0(1/2), IP 1/2 (I (Co) = x•)) 5- Pr/_ (r (Co)

n-c 0

so 0(1/2) = 0

Let S he an n by n square If H(S) holds, then at least one of the a sites v on the left of S is joined by an open path to a site at graph distance at least rr from r' It follows that Pr/.,(H(S)) < tiP i r)(1(C0 ) > Hence, It/" e (Co)) > a) > 1/(2n) for all a > 1. It is natural to expect decays as a power of n, i e., that the limit that P I 1 .)(1 (Co) > lim

log P 1/2 (I (C0) ^ (1) log n

exists Once again, this natural conjecture is still open, although a corresponding result for site per colation on the triangular lattice is known: see Chapter 7 The limit above, if it does exist, is often denoted by 11p, or 1/(5,. It is one of the critical rap(111111115 associated to percolation; see Chapter 7. Looking back, the most difficult step in the proof of Theorem (i is Lemma 4 or the equivalent results of Russo [1978] and Seymour and Welsh (19781 Although Harris's proof is very different from that presented here, a key step is similar to a step in the proof of Lemma 4: roughly speaking, given an 'outermost open semi-circle 8 around the origin' within a certain region, one reflects this to form a cycle S U . Then a path meeting this cycle from inside is as likely to meet the 'real' part .5' (all of whose bonds are open) as the reflected part S' It is pr rssible to prove Hall is's Theorem using Lemma 1, Harris's Lemma, and the independence of disjoint regions, without ever considering left-most crossings or then equivalent; such a proof has applications in other contexts, where the proof of Lemma 4 does not work Indeed, as we shall see in Chapter 8, a (long) proof of this kind was given for random



3 A sharp transition

63

\Tor onoi percolation by Bollohas and Riordan (2006a] In that setting, no result: directly equivalent to Lemma 4 or to the Russo-Seymour-Welsh Theorem is known

3.3 A sharp transition Fahly ear ly on it was recognized that there is an absolute constant e < 1 for which the following statement easily implies Kesten's Them ern: for any p > 1/2, there is an T1 = a(p) such that Pp (H(R))> c for a 211 by n rectangle R. For example, Chaves and Chaves 11986b1 gave an argument showing that c = 0.921 will do The factor 2 here is not important.: recalling that h p (nt,n) is the Pp-probability that an or by n rectangle has an open horizontal crossing, it follows from Harris's Lemma that if h 1,((1 + 5)n, 1 as n a for some E > 0. then h p (Clf n) 1 as for every C > 0. n We shall give art explicit lower bound on h p (tma), using a sharpthreshold result of Ft iedgut and Kalai, Theorem 12 of Chapter 2, and the Almgulls-Russo formula, Lemma 9 of Chapter 2.. We start by bounding the influence of a bond in a rectangle I? on the event II(R) In the context of percolation, a bond e is pivotal for an event E in a configuration w if precisely one of w and ur is in E, where ,./r± are the configurations that agree with w on all bonds other than e, with e open in to + trod closed in co- . In other words, e is pivotal if changing the state of e changes whether F holds or not The influence of e on E is Ip (e,E) = Pp ( e is pivotal for E )

If E is increasing, then Ip (e,E) =

( w 6 E. co--

Lenona 7. Let R be an rn by a rectangle in 27.? and let e be a bond in I? Then. 11,(e, MR))

2P 1/2 ( r(Co) _^

4 11 1 12

, (n



)/21)

(6)

for all 0 < p < 1 Proof Throughout the proof we work entirely within R. considering only the states of bonds e in I? Suppose that, a bond e in I? is pivotal for the increasing event 11(I?). As LO + E II(R), in tire configuration (d4 there is am open horizontal crossing of R Since ur 0 1-1(R), a n y such

Band

per rotation on

the Hal is - esten Thew um

mussing must use c Hence. in the configuration w one endpoint of is joined I A an open path to the left of R. and the (Whet to Hie tight: see Figine 9 Thus at least one endpoint of e is the shut of an open path

igwe 9 A bond e (joining the solid discs) that is pivotal lilt 11(1?): the dual bond c' (endpoints shown as ( l osses) is pivotal for Die) of length at least 10 /2 – L so 0 (e 11(11)) < 2P(1(C0 ) > 111/2 – I)

(7)

As Lt.– ii-I 11(R). IA Lemma 1 we have Lc – E V(Rh ) Siinilm iv. 1 ,), so in y t lime is au open dual path ()tossing Rh vertically and using the edge e* dual to e Hence, in it). one endpoint of c' is the :Wm t of an open dual pat h of length at least (a – 1)/2 Since dual edges ate open with pi obabilih 1 – p. it Follows that (CO>(rt– 1)f2) 11(1?)) < (8) Fin ant a the event 1(C0 ) > a is incteasing. so P p () (Co) > a) is an inmeasing function of p Thus (6) Follows immediately bou t (7). tot < 1/2, and flow (8). p > 1/2 Lemma 8. Let p > 1/2 and an allege, p > 1 be 1i.:ecd There art constants 7 = 7 (p) > 0 and 0 0 = 0 0 (p, p) such that 11

0 (pir 1 )> 1 – 11 – "'

(9)

fat all 0 > au Proof Let 1? 1) 0 a pa ht n iectangle nom (3). we have Pt/2(1-1(R))

h

(10)

3.3

.1 slung transition

65

> 0 depending only on p Rom Lemma 7 awl (5).

ha some constant for > 2 we have

1„e(e,1-I(R))<

=

for ever y bond e of I? and all p E [1/2,p], where a > 0 is an absolute constant Writing f (pi ) for Pp,(11(R)), flour the Friedgut-Kalai result Theorem 12 of Chapter 2 it follows that (e,

(100(1 — (0)

H ( R ))

log(1/6)

tell( TO all p' E [1/2, p]. where e > 0 is an absolute constant the Margolis lbws ° formula (Lemma 9 of Chapter 2). the stun above is exactly the derivative of I (p') with respect to p' 'Thus, writing for

By

g ( p9

lei l og ( V)/( 1 — 1(0)),

d i/

-



1(7)(1 — (0)

100> clog( 'WO) = en: log a..

"

horn (10), g(1/2) is bounded below by ir constant that depends on p Hence. for rt > ag(p), we have g(p) > ac(p - 1/2) log n, say awl (9) ❑ follows Using Iris weaker 'approximate 0-1 law ! instead of the mom iecient Friedgut Kalai result. [1982] proved a weaker form of Lemma 8: this weak form is more than enough to deduce Heston s Theorem Following Bollobas and Riordan (2006ci. we give an alternative proof of Lemma 8, using a version of the friedgut- Falai result for symmetric events.. Theorem l3 of Chapter 2 The other ingredients of this proof are Harris's Lemma and (3) tins. rather than (5) The idea is that. if II(1?) were a s t innwtric event. then Lemma S would follow immediately from (3) and the Fr iedgut-lialai result Of course, 11(1?) is not symmetric, but it is very eas y to convert it into a suitable symmetric went Alternative proof of tetanal 8. Fix p > 1/2 Rom (3), there is an absolute constant 0 < < 1/2 such that P i p,(11(R)) > c for any -fn by rt rectangle angle I? obtained from Z 2 1w idenFor n > 3. let T2, be tire graph C'„ x tifying all pairs of vertices for which the corresponding coordinates are congruent, modulo This graph is often known as then by a discrete has 112 vertices and 2n 2 edges torus Note that For 1 < t < n-2, a Ai by I rectangle 1? in is all induced subgraph corresponding to a k by I rectangle 1?' = [0+1, ads x [5+1. b+11 of



66

Band percolation on Z2 - the Ilan is- Kesten Theorem

in 22 Out rectangles in the torus are always too small to 'wrap ill Olin( II, so, as in the plane, any such subgraph is an abstract k by t! rectangle. We shall work in T2 = 'ffiL, taking the bonds to be open independently with probability p. We write PI, Pi 2 for the corresponding probability measure Let E„ be the event that: 1 2 ti„ contains sonic 4o by rectangle with an open horizontal crossing, or some T/ by 1n rectangle with an open vertical crossing Then E„ is symmetric as a subset of 1)(X), where X is the set of all 50n 2 edges of Considering one fixed -In by o rectangle Pr in 1 2 , we have P I: 22 ( EO ?_Pi;2(H(E)) = P o2( E (E)) > (.12 Let /5 = (p - 1/2)/(25c,), where c 1 is the constant in -Theorem 13 of Chapter 2, and set E = I/-" (16 As (5 depends only on p„ there is an no = ro ) (p) such that E < e2 < 1/2 lb/ all a > no Now 1

log(1/5-)

—9 = 25c,(5=

„ > log(n-)

log(1/(2E))

log(50n2)

Hence. by Theorem 13 of Chapter 2 . as P ,(E„ ) > Pil-/(E„) > I - 2 = 1 - /1-50

> E we have (11)

for all n > Let R I Mr, be the 3 1 I by 2/r rectangles in 1- whose bottom-left coordinates are all possible multiples of 11 Then any 4o by a rectangle R. iu 12 crosses one of the Er in the sense that the intersection of 1? and .R.; is a 3n by ri subrectartgle! of R.; Similarly, there are 211 by 3n rectangles 000, R50 so that any n by 41/. / rectangle in 12 crosses one of these hen/ top to bottom It floflovks that if E„ holds, then so does one of the events E„ ,i , i = 50, that Ri is crossed the long way by an open path Thus E;;.. tire complement of E. contains the intersection of the El; Applying Harris's Lerruna (Lemma 3 of Chapter . 2) to the product measure PI' , For each i the decreasing events E,t, and r) are rP 1=1

Thus, from (11), for n > no we have Pi; ( Eli

<

/I/5(

Kesten's Theorem

67

so Pi; (E„ ) > 1 — Now E„ is the event that there is an open horizontal crossing of a fixed 3n by 2n rectangle R in the torus. which we may identify with a

corresponding rectangle in

V

P„(H(R)) =

Thus Fp( H ( R ))

> 1—

enough. 'Using whenever R. is a 3n by 2n rectangle in 27,2 and a is large (2), it is easy to deduce (9) ❑ The proofs above involved various explicit bounds. These ate not really relevant As noted earlier, the following much weaker form of Lenrnra S is enough to deduce Kestert's Theorem. Lemma 9. Let p > 1/2 be fixed If 1?„ is a 3 1 1 by n rectangle to V, pc, their P„(II(R„)) 1 as n This may be proved by either of the methods above Indeed, it follows from Harris's 'Theorem that the influences of the bonds e on the event 11(R) tend uniformly to zero as the shorter side of R tends to infinity. Using a qualitative form of Theorem 12 of Chapter 2 (that the stun of the influences is at least 1(S) if all ale at most O. with (6) (I), or Russo's approximate 0-1 law, it follows that the stun of as t > 0 (from (3)), the influences tends to infinity Using inf„ h lt )(3n, Lemma 9 follows Alternatively, one may use (3) and a qualitative for in of the symmetric Friedgut -Kalai result, working in the torus as above Either argument shows that, unlike Lemma 8, Lemma 9 does not depend on the particular form of the Fr iedgrit-Kalai bound.

34 Kesten's Theorem As noted earlier, it is well known that Kesten's Theorem follows easily from Lemma 9; inn fact, from the quantitative form of this lemma, Lemma 8, Kesten's Theorem is more or less immediate Let E, — denote the event that there is an infinite open cluster Recall that P„(E„ ) > 0 implies that = 1 and 6(p) > 0 Theorem 10. Fat hand percolation in V. p > 1/2 then 1P 0 ( Etc ) = Fix p > 1/2, and let, 7 = -l (p) and no = no(p,2) be as in > P. 0 be an integer to be chosen below. For k = Lemma 8 Let 0,1,2, , let Rk be a rectangle whose bottom-left corner is at the origin, Proof

68

Bond pnroloboa on V the Han Pe-AR:sten Theorem

with side-lengths 2 /1 11 and 2 k -11 n. whew the longer side is vet ical if k is even and hot izantal if k is odd: see Figure 10.

Figure In 11- he mciangles En to / (00 not labelled) drama cut responding to Lhe OVentS Er;

o f en paths

Let .E / be the event, that ./? k is crossed the long way by an open path Note that it'll two such ctossings of Ilk and RAH' must meet, so if all the E h. hold then so does E x . If it is huge enough then, b y Lemma 8.

= 6>a

so P ,(Ex. ) >

F,An k ,.„

k

—9

< 1,

>0

Together Fhetnems 6 and 10 show that //n(Z 2 ) = 1/2 Staffing nom the \wake( Lemma 9, a little mo l e work is needed Again (hoe ale several possible arguments One is a 9enormalization' tugument due to Aizenman, Chaves, Chayes, Frohlich and Russo [1984 see also Chaves and Climes [19861)] Proof of Theorem 10 strum/ vet (inn, Fix p > 1/2, and recall that h p (m. n) denotes the Pr,—probability that MI m by a rectangle in :12 has an open Inn izootal classing Consider three 2n by n rectangles aye/lapping in two it by n squares as in Figure 11 Noting that the j nobabiliN that nun In a squaw has an open vertical classing is just nom Ilatris's Lemma we lnae h ijn l /.(-Lm >

031),,(n. n) 2 > h p (211.0 5

(12)

Placing No disjoint in b y n rectangles side b y side to form it -In by



liesten s Thcomn

69

Figure Four 0 by rr squares in 2 2 Each consecutive pail fin ins a 2a by rectangle It these three rectangles have open horizontal crossings, and the middle two squares have open vertical crossing,, then the union of all Emu squares has au open izoutal mussing

ELIM

11111411 1 'gum 12 ly



lis

by n

formicu e I



iv 2n rectangle

2 p rectangle as in Fig u re 12 t he elect s that eaclr has an open hotizoul al mussing ale independent Hence, h p (-1n, 2 0 ) > 1 - (1 -

›i_(1_hp(2„,„),)-

(13)

Writing h p (20, n) as 1 - 5. from (13) we lime h p (-1n, 2n) > I - 2552, which is at least I - ,F/2 if 5 < 1/50 By Lemma 9 there is an n with /i,,(2 p , > o) > 0.98: it hallows that hp(2kijii n,2/'M) > - 2 -k /50 tot all k > 0. and liPi,(E) > 0 Wilms as in the first proof of 'Theorem 10 ❑ The argument above shows that if for a single value of n we have 0) > 0 951 (a loot of :r = 1 - (1 - ii 9 ) 2 ), then 0(p) > 0. In fact, arguing as in Chaves and Chaves 11.9866], one eau do a little better Dividing a 2n by it rectangle into two ve/ t es-disjoint squates, if the rectangle has an open horizontal flossing, so do both squares As the squares ale disjoint, the events that they have open hot izmital crossings ate independent,. so h i,(2m n) < h p (n, n) 2 Therefore the first inequality > in (12) implies that h p (-bb 0)1, and the first inequality in

Bond percolation on. Z2

the Harris-Kesten Theorem

(13) gives h p (tn.2n) > 1 – (1 – /1 11 (2m0-1 ) 2 Thus the value 0.951 may be replaced by 0.920 , a root of x = 1– (1 –1, 1 )2

3.5 Dependent percolation and exponential decay In the final section of this chapter we shall prove that for any p < 1/2 we have P1,(1(0 > n) < exp(--an) for some a = a(p)> 0 This result is due to Kesten [1980; 1981], although, as we shall see later, it inflows easily front Kesten's result that p! (Z2 ) = 1/2 by an argument of Hammersley [195Th] Once again, (Mete are several possible proofs; we shall use the concept of dependent pet colation -- this will be usellil in later chapters as well Let G be a graph, and let 115 be a site percolation measure on G. i e a probability measure on the set of assignments of states (open or closed) to the VCI flees of G The 11leatill/ e P is k-independent if, whenever 8 and T are sets of vertices of separated by a graph distance of at least le, the states of the vertices in ,5 ate independent of the states of the vertices in T 'The tenn k-dependent is also used by some authors For k 1, the condition is exact l y that the states of the vertices ate independent, so the first non-trivial case is k = 2 Liggett, Schorunann and Stacey [1997] proved a general testa comparing k-independent measures with product measures We shall need two simple consequences, which ate essentially trivial to prove (Meetly. Recall that, when considering site percolation on a gtaph G, we write C„ tot the open cluster containing e, i e , ha the set of sites of CI connected to !' by paths all of whose vertices are open 1

Lemma 11. Let. k > 2 and A > 2 be positive integer s. and let G he a (finite a t graph wtih matint arn degree at most A There are constants P i = pr (k. A) and n = a (k. A) such that if P is a k-independent site percolation ineas iti r all G in which each site is open with probability at MOW

then 11;

exp(–an)

for all vet trees v of C and all n > 1. FM tit er more, fa; It t (2) we may take any constant such that the quantity eil(p i /(41+1 ) is less than 1 Proof If > n. then the submaph of 6' induced by the open vertices contains a tree T with n vett ices. one of which is ti It is easy to check IC,

3 5 Dependent percolation and exponential decay

71

that the number of such trees in G is at most (eA)"- 1 ; see Problem 45 in Bollobris [2006] Fix such a tree T If w is any vertex of U. t h en at most 1 + A + + < A k vertices of G are within graph distance k -1 of to. Hence, there is a subset S of at least n/A k vertices of T such that any E S are at graph distance at least I-; indeed, one can find such a set by a greedy algorithm, choosing vertices one by one The vertices of S are open independently, so the probability that every vertex of T is open is at most 14' 91 . Hence, NA]

< (eA)

p;'/

< (cAp

) 7/

I/Ak

= GA T), < 1, the resit!t, follows, Choosing pm small enough that with a = - log;. For the second statement, note the quantity A k above is in fact an upper bound for 1 -f- A + • + A t I Hence, when /c = 2, it may be fj.] replaced by A + 1

In the proof above, we could have assumed without loss of generality that k = 2, replacing the graph G by its (k - pawl, i e., the graph (7 (k -' ) on 17(G) in which two vet tices are adjacent if their graph distance in C; is at most k - 1 Using Lemma 11, one can easily deduce Reston's exponential decay result for bond percolation on 1 2 from Lemma 9. The basic idea is to associate a large square 5, to each v E 1 2 , and to assign a state to 0 depending on the states of the bonds in (and near) S,,, in such a way that the states of sites v separated by at least a certain constant distance k ate independent, each v is unlikely to be open. and percolation in the bond model implies an infinite connected set of open sites

Theorem 12. In bond percolation on 1 2 , fat evecy p < 1/2 there is a constant a = o(p) > 0 suck that IP0 (]C0 1 > JO< exp(-an) for all n > 0

Fix p < 1/2, let pr = (5,4) > 0 be a constant for which Proof Lemma 11 holds with k = 5, A = 4, and set c = (1 - pl)1/1. Let A = 1 2 and let A" be its dual, so the bonds of A* me open independently with probability 1-p As 1- p > 1/2, by Lemma 9 there is an in such that hm_ p (3111,1n) > c, so the probability that a 3m by or rectangle in A'' has an open horizontal crossing (of dual edges) is at least c Set s = nr +1, and let 5' be an s by s square in 72. We can arrange four 3t11 by or rectangles in A" overlapping in in by ta squares, so that

72

Bond percolation on 2: 2 - the Halt issICeslen Theorem

then onion is an annulus as in Figure 8 The inter for of A contains square in A with in + 1 = s sites 00 a side, which we shall take to be ,9 13v Harris's Lemma, t he probability that each of the Wm leo angles is crossed the long way by a path of open dual edges is at least e t = 1 —pu The union of font such ctossings contains an open dual cycle sin !omitting S. Let B(S) he the event that some site in S is connected by au open path to a site at L, distance s horn 5' Such an open path in A cannot. (loss an open dual cycle in .1 surrounding .S. so we have Lau,(/3(S)) < Informall y, we define a site pet eolatitm measure F on Z 2 k: taking each v = (di it ) E Z2 to be open if and only if B(S) holds fin the s x Is// + sit + Mot e for walk, let Al s squats = 1st' + 1. sr denote the independent bowl percolation !node! on Z- in which emit bowl is open with probability p \\e define a site petcolatam model Al on 222 as follows Let j: 2 E1(7:2) — 2" :1211 be the function [tom the state space of A/ to the state space of Al that we have just defined. so (w))(v) = 1 if and out il, in the configtuation w. the event B(8, ) holds The function , J" and the measure 1111, induce a probability measure [ —I LA)) 111 on 2 1 CL1 ' given by 1111 (A) = a . This measme on 2 1 C1:2) gives us a site percolation model A/ on 2 1 I:1-) Since the event B(S,.) depends out on the states of bonds within 1„-distance s of So .. Hirt/treasure F is 5-independent: nu thei mole each eE.= is open with P-plobabilit ■ Pi,(B(5,)) < g, Let co be the open cluster of the origin in out original howl pet colation Al. and let (17, he the open cluster of the origin in the 5-independent site percolation model Al II ■ Lemma 11 thew is i1/1 a > (I such that
Is + 1) 2 . then el. el \ site n o 1 Co is joined by au open path to some site at 1.,-distance 2s flow w If in E 8, then it follows that B(.9,.) holds Tints, if rid > (-1s+ 1) 2 . t hen B(S,,) holds tot every c such that contains sites of Co 1 he const uction of the model Il go es us a mutual coupling of Al and Al: a site c is open in A/ if and outs if FL- 0 is open in Al. S(B,.) holds ill Al. III particular. event ' with n so the set of such v forms an open cluster iu Al. and is thus a subset 01 Co Hence. as each So contains onl y s 2 sites. fi g n > (-Is + 1) 2 we have of

F,,000

g )

111 0(.7,1

completing the woof of Thement 12

g /s2 )

exit( --ust/s2).

d 5 Dependent percolation and exponential decay

cc... for p < 1/2. so Theorem 12 immediatel y implies that \ (p) > 1/2 Together with Theorem 19. this implies t i re Harris- Kesten Theorem Theorem 13. 14 (22 ) = pli I/ (Z2 ) = 1/2

❑

Later. we shall see that pr = m i holds in a very general context In fact. Merishikov [1986] proved exponential decay of the radius (and, under an additional assumption. of the volume) of Co for p < p H . again in a very genet al context 117e finish this chapter by noting that dependent percolation gives yet another way of deducing Nesten's :(irearm/1 hour Lemma 9: again, the Ices lemma will be useful in other contexts. This time we consider bond percolation A bond percolation measure on a graph Cis li-independent if the states of sets S and T of bonds are independent, WIRMICVC/ .9 and T. ate at graph distance at least A . This time, the case k = 1 is already non-trivial. Indeed. E01 A! 1 the separation condition is exactly (hat no bond in S shares a site with a bond in Lemma 14. There is a po < 1 such that if is a I-independent bond percolation ni ewiare on 23 2 in which each bond is open with probability at least po then 1111(70 1= )>0 Proof Suppose that the open cluster Co containing the origin is finite Then by Lemma 1 of Chaplet 1 there is an open dual cycle 8 1 sin I oursling the origin, of length 21, say As shown in t i re proof of Lemma 2 of Chaplet there are at most

(po i _ I < — 39 dual cycles of length 21: sou rounding fl. where I r k < -1 x 3 k - I is the number of paths of length lt ill 2 2 starting at 0 Let S denote the set of duals of the bonds in 5', so S is a set of bonds of A = Z 2 As the graph 2 2 is 1-edge colourable. there is a set of at least 1/2 Vertex-disjoint bonds hr diet-I.-probability that S' is open. i e the II-probability that all bonds in S me closed, is at most (1 - po)112. It Follows that the probability that Co is finite is at most the expected number of open dual cycles surrounding the origin. which is at most v 3210 - pot12 9 t>2

7-1



Band percolation on V the Harris-Kesten Theorem

If po is close enough to 1 (for example. po = 0 997), then this sum is less than 1 ❑ As stated, Lemma 11 is essentially trivial; it is also immediate from the general comparison result of Liggett, Schonmann and Stacey [1997]. In applications, the value of po is frequently impor taut Currently, the best known bound is given by the following result of Baster, Bollobas and Walters [2004 Lemma 15. If Pis a 1-independent bond percolation Incas 1171! on Z2 in which each bond i 9 open with probability at least 0.8639, then Pa(,'ol :Tic) > 0 ❑ Bollobas and Riordan [2006b] pointed out that Lemma 11 may be used to give yet another proof of Kesten's Theorem; lot this, the value of po is irrelevant.

Proof of Theorem 10 third version. Let p > 1/2 be fixed, let. po < 1 li)e a constant for which Lemma 11 holds, and set c p(1/ 3 Given a 3n by a rectangle H. let S' and S" be the two end squares when 1? is cut into three squares. Note that 11(1?) certainly implies II(S' ) so, by Lemma 9. Pp(1/ (S" )) =

= P 0 ( 11 ( Si ))

Pp(H(R))

if n is large enough. which we shall assume born now on Let 0(1?) be the mem H(1?) f1 fl V(.9"): see Figure 13 By

a

Figure la A au by n recta rectangle It such that

il) holds.

Harris's Lemma. Pri (G(R)) > P„(//(R))ifii,(17(51))Pp(V(S"))

c3 = po

Define 0(1e) similarly for an n by :31/ rectangle, so that by symmetry we have Pp (C(R')) = Pv(0(1)) po Writing Al for the bond percolation model in which each bond of V is open independently with probability p, let us define a new bond

3 5 Dependent percolation and exponential decay

75

percolation model Al on V as follows: the edge from (x,//) to (x +1, y) is open in Al if and only if C(R) holds in Al for the 3n by n rectangle Prix +1,21/x-1-3n] x [2ny + 1, 2ny + id Similar ly, the edge from (.e, y) to (:e,p 1) is open in Al if and only if G(D') holds in Al for the n by x [2ny 1, 2ny 311 rectangle [2nx + 1, 2rix +

Figure IA A set of open edges in A/ (left) drawn with C(1?) holding in Al

Hid



responding rectangles R.

Let F be the probability measure on 2 1(:) associated to AI •Phen is indeed 1-independent, as C(R) depends only on the states of edges in R, and vertex-disjoint edges of Z2 correspond to disjoint rectangles By Lemma 14, P l I Col = > 0. However, we have defined 0(11) in such a way that an open path in Al guarantees a corresponding (much longer) open path in the original bond percolation AT, using the fact that horizontal arid vertical crossings of a square roust meet; see Figure 14. Hence, 1111,(E,) > PaCor = ) > 0, completing the proof of Theorem 10 ❑

The ungurnent above works with 2n by n rectangles; the only reason for using 3n by rt was to make the figure clearer Also, in addition to the long crossings, it is enough to require a vertical crossing of the left-hand end square of each rectangle I?, and a horizontal crossing of the bottom square of each R'. As hp (rx n) > 11,,(2t, n) 1/2 , to prove percolation it thus suffices to find an n with h p (2n,n) 3/2 > Po, where pa is a constant for which Lemma 14 holds Using the wane go = 0 8639 from Balister, BollobAs and Walters [2005], li p (2Thn)>. 0 907 . will do.

'16

Bond percolabon an

22

thc. Harris-lticsten Theorem

3.6 Sub-exponential decay It is not hard to show that. for p > 1/2, with probability 1 there is a unique infinite open cluster; we shall present a very general result, of this type in Chapter 5. fu this range, all other open clusters are 'small' with probability close to 1; more precisely, the probability that n < < cc decays rapidly as n increases. In analogy with the situation for the random graph C7(n.p), cure might expect that this decay minors that of open clusters below the critical probability, i e., that P p < [C10 1 < (x) is approximately exp(—ai n) for some a' > 0 This turns out not to be the case We shall present only the vett' simplest: result in this direction; stronger more general results have been proved by Aizenman, Deleon and Souillatd [1980-1 among others: see Cilium/eft (19991 Theorem 16. In bond percolation on V for raw p > 1/2 there arr constants b = b(p)> 0 card = c(p) > 0 such that

exp(—bli7) for all n

<

<

S.: ext)(--- e 00

(LI)

1

Tire upper hound is essentially immediate horn Theorem 12 Indeed, suppose that [Col =Ir. and let O'C la be the external boundarr of Co, as defined in Chaplin .1 By LC1111118 .1 of Chapter 1. O'C'n is a cycle in the dual lattice + (1/2. 1/2) containing Co in its interior Thus, the area enclosed b y O'Co is at least II, so its length is at least 2 \,iti But: era s (dual) edge in O'C'ü is open As O sc Ca crosses the positive x-axis at sm il e coordinate 1/2 < :r < n — 1/2, we have shown there is some dual site v = (T. —1/2), 1/2 < H u tt whenever = < n — 1/2, such that the open dual duster containing r' contains at least 207 sites. As dual bonds are open independently with probability 1— p < 1/2. b y Theorem 12 the probability of the latter event is at most (II— 1) exp(-2aiii) for some a > 0 Thus Moat

BIRO ) ICH
(I and let 5' be the I by C FM the lower bound. suppose that n = square [0, f — x C — For each of the n sites ,r in 5', let: Es be the event that there is an open path horn :r to some site in the boundary of .5; see Figure 1.5. Note that I.L1 holds triviall y for all sites in the holds, so I"„ (E,) > boundary of Whenever Cr is infinite, then

36' Sub-6 nmenbal decal' 77

Figure 15 A 9 by 9 square .5 with bottom-left coiner tire aright. drawn together with all the open hands in its interior 'The filled circles are the sites .r for which E„ i e.. those joined to DS by an open path. If the bonds in [IS are open, and the dual bonds surrounding .9 (dashed lines) ale also open (night consists precisely of time filled circles then the open cluster ()I' the Let N be tire number of sites :It E ,5' for. which E . holds Then 0(p) > 2 t (!fir) = 5— s E s Pn( E.,) > 110(p) As A < n always holds, it fellows that PAN >

n0 (P)/2 ) 0(p)/2

Let F he time event that each of the -1(1 — I) boundary bonds of S is open, and that each of the if bonds from .9 to its complement is closed, as in Figure 15. Then

Pn1 F ) = 04 —

> (Phi — /d).'

Since each event Es depends only on the states of bonds in the interior of S. the random variable N is independent of the event F. and Pp (n0(p)12 roi nO(p)12})

Pp (11 )Pi,(1N (PG —

nO(p)/20

(0)12

this proves the towel bound in (14) for 11 a square Since As f = there is a squats between n and 3 11 for all o > tire lower bound fru all n lid lows In higher dimensions, i.e lot bond (or site) percolation in ✓d t it is not too hard to guess that Pp (n < ECM < cc) Nvill decay t oughly as exp(-10- 101 ), but this result is not so easy to prove; see Section 8 6 of Gr itnuett (1999(

4

Exponential decay and critical probabilities - theorems of Menshikoy and Aizeninan Barsky

Out ainr in this chapter is to show that, for a wide class of percolation models, when p < prr tie cluster size distribution has an exponential tail.. Such a result certainly implies that pr = vu; in fact, it will turn out that exponential decay is relatively easy to prove when p < p r (at least if -size' is taken to mean radius, rather than number- of sites) Thus the tasks of proving exponential decay and of showing that: p = are closely related. Results of the latter t y pe were proved independently by Nlenshikov [1986] (see also Menshikev, Alolchanov and Sick)/ wilco [1986]) and by Aizemnan and Barsk y [1981 under different assumptions Here we shall present Menshikov's ingenious argument in detail, and say only a re words about the Aizennian-Bar sky approach. As we shall see, 1\ fenshikoy 's proof makes essential use of the Margulis-Russo [Mrada and the van den Bel g-ICesten inequality

4.1 The van den Berg-Kesten inequality and percolation Let us briefly recall the van den Berg-Kesten inequality, Theorem 5 of Chapter 2, which has a particularly attractive inter pretation in the context of percolation In the context of site (respectively bond) percolation, an increasing event E is one which is preserved by changing the states of one or more sites (bonds) from closed to open, and a witness for an increasing event E is just a set If of open sites (bonds) such that the fact that all sites (bonds) in are open guarantees that E holds For example, considering hoed percolation on 52 , as in Chapter 3, let 1? be a. rectangle, and let 11(1?) be the increasing event that 1? is crossed horizontally by an open path P Then a witness for . 11(R) is simply an

/ I The van den Bety-Resten inequality] and percolation

79

open path P crossing I? horizontally, or a set of open bonds containing such a path The box product ED F of two increasing events is the event that there are disjoint witnesses for E and F For example, 11(R) 0 H(R) is the event that there are two edge-disjoint open paths crossing R horn left to right. (For site percolation, the paths must be vertex-disjoint ) As we saw in Chapter 2, van den Berg and Kesten [1985] proved that for increasing events in a product probability space, P(E 0 F) a, which is exactly the event {:c 41 that there is an oriented path from a given site

4 .2 Oriented site percolation.

87

Figure -I The or halted graph X on 71, 1 in which two bonds leave each site (a, b) one going to (a + 1, b), and the other to (2a. b = 1) ibis example was given by Paul Balistel

to sonic site y E ,9,11Jit) Note that, as A is locally finite, (Q) is finite if and only if C, is finite. Note also that the length of the shortest open path from :r to 8,T, (w) may he much longer titan a X

In many of the arguments, it will be convenient to wor k with a closer related event that does not depend on the state of X. Let i?„(,r) he the event that there is an or iented path P front r to some site y E S,t(r), with every site of P other than :r open. This event is illustrated in Figure 5 Equivalently, R„(:r) is the event that there is an open path P' from some (out-)neighbom of X to a site in 8,;(3). (It makes no difference whether or not we constrain .P' not to use the site x: if there is such a path P' visiting ,c, and y is the site after x along P' then the portion P" of P' star ting at y is a path horn an out-neighbour of 'a, to S,t(a) not visiting :r ) In the light of the latter definition, we will sometimes write {x + 4-} for R.„(x). We shall study the quantity Pc ( 1, , P) = 111)s)(Bc(39) As the event R„ (a) does not depend on the state ofand if and only if :r is open and R„(x) holds, we have IF;(1(C„.) > a) =

(3, 224) = piR;,(R „CO) = pp „Cr p)

241 holds

Exponential decay and ruling, probabilities

88

Figure 5 An Must radon of the event U t eri Solid circles represent open sites As a) — as it — oc.. Suppose flint CT is strongl y connected Then, IwJ Lemma 2, throe is a si tgle ethical ptobability p L1'( A) independent of t he p„ (:r,. p) = initial site a For a fixed p < p i ( A ). we thus have lot twin site 1 7. In the i ugurnents below. w(:` shall need bounds on ( p ) 811 1)

(abli)

J7E7C

obtain such a bound, we shall assume that CT. is finite. Lemma 3, ltd A be an infinite locally finite ()nettled multi-graph with C7v finite and strongly connected. and let p E (0 I) Then there is a constant a > 0 sack that

/Pa( Cji,

"Pit °Cul

for all sites .c and p and integers ti > p 0 such that the order of the pivots along P' starts with . where s 1 But then the MUGU of the initial segnrenb of P' up to lit and the final segment of P starting at bt contains an open path from at to S,t(x) avoiding 14 4. 1 , which is impossible; see Figure 6 In fact, the more detailec.1 picture is as follows: the set of sites on open paths from to 87,(v) forms a graph in which br, i br are cutvertices; see Figure 7 Taking ho x, for 1 < i < let 7-11 denote the set of sites on open paths horn 1)1 _ 1 to 14, excluding 14_, and b, Also, let Tr+ , denote the set of sites other than b, on open paths horn b, to the set T.; U .5-, ,f "(x) Then, for 1 < contains two paths



.3 Alenshileov's Theorem



91 .5;ti (r)

Figure 6 An impossible configuration: the pivots(); appear in different orders along two open paths P. P' from 3. to Stec). The path P is the sit algid line segment from to y: the section of the path P' from to 1, 1 is drawn with thick lines The path P' cannot 'jump' front I), to I), t > s+ since P UP' would then contain a path from t to p avoiding

(a)

Figure 7 The set of sites on open paths from 1 to 9;4; ( r), shown with a possible artangement of the pivots ha for the event. R„ (:r) Note that bib.. LT2 63 mid br,b6 are edges of A er

front 1)1 _ 1 to h 1 that are vertex-disjoint apart front their endpoints; in the case whole E E( A ), this holds trivially: both paths consist of this single bond Also, T,. + , contains two paths front 1.4 to St (3,) that are vertex-disjoint except at b r . For i there is no (oriented) bond front a site in T1 to a site in Ti Let ho a Whenever .11.„(x) holds, let D i , = 1,2, . denote the distance horn hi_ i to h i , where we set a = x if there are fewer than + pivots. Let I < k < v. Since D i is the distance from a to the first pivot, if D I > k then either there is no pivot, i.e., there are (at least) two disjoint open paths front neighbours of x to (4 or the first pivot h i is at distance at least k + 1 horn a In the latter case, there are two

92

Eepoio

duo q and c i it Ica! probabdlLies

and au open path how disjoint open paths fron t neighbouts of a' to b i to 5 11` (r) disjoint flow T 1 and hence how these paths: see Figure S

figure

litst to

then it there are any pivots for the event 1? ” (1)• the > . is at disun i ty mot e than k frpm r Thus thine ate open paths P it owl ) and P ' how S i±r) that are disjoint except at .r S II DI

l i t china case. there ate pat hs P and P' front .1 . to S,+, (r) and to tespecti eI N with P 111(1 P' disjoint except at .1 and all sites of P and > k} is contained in I' tit het titan x open illus. the event R „(x)fi { I lie erect .1 „( Pk(.r) Hence. by t he ran den BEng-Rester inequality. Theorem 5 of Chaplet 2. tic lime Pip (I?„ (3).

>

< Pp (R„(x))Ei,(111, Tr)),

ie.

P p ( D I > k I Rs (l i )) 5_ 14.(1', The wain tool in Nlenshiltov's proof is a genet alizatiou of tins observation When 1?„(r) holds and there are at least t pivots, we WI Re 11 lot the tilt into-1o, el u.stel meaning the set of sites z such that there is a path P born to c with I) P and all sites of .P other than ./ open: see Figure 9 Thus, if c is open, then II is the set of sites teachable fron t Jr by open paths not passing through b, We use bold font fin the random variable I/ because it will be partic u larly impottant to distinguish It we horn its possible values It ot h er random variables, such as do not bother.) When lo = y and II = It , then every site in I 1 U 0/1

4 :t itIen gbikov's Theorem

93

Figure 'the 2nd interior cluster Io consists of the thick hues including all of their endpoints except b 2 Note that z 6 1 1 , oven though z is not on an open path Irian i t neighbour of 3: to 51(x)

open, and all sites in d1 1 other than y ate closed. where 01, is the set of out-neighbours of . U 1:0 With this pteparation, we me ready to present the key lemma for denotes the probability mealenshikov's main theorem As usual. sure in which each site of A is open with of the sites are independent

ptobabilitv p. and the states

A and let 11. I nod f Lemma 4. Let x be a Ole of an unfilled !pupil di. eve hare < i < r. be positive inteye l For 1 < k < /I -

FE, (D, > lc[

E) < supl i n(y.p): C.r.

A}.

(5)

where E is the event E and as

R„(.r)

n {D i =

D2 =

(12.

Dr-1 = -1}

pk(y. p)=F;,(R),(y))

Proof Throughout this proof, we work within the finite subgr aph of A most n hour a We write IP„ For Fis, We shall pat tition the event E according to the location y of the For and according to the ulterior cluster I= (r— 1) 5 ` pivot b, let any possible value I of I consistent W ith E

consisting of sites at distance at

Ely

= R.„(e)

n

=

n =

so the events E„ r Iona = Feu 1 < i < then I = pal tit ion of E If E„ I holds them as noted above, so does the event /7„ e that all sites in ./u yl rue open. and all sites in RI\ {y} ale closed In tact, Ey e holds

ErponenHal decay and critical probabilities

91

it and onl y if Fr, , holds and there is an open path P horn a neighbour of y to .9,1-(x) disjoint from I U DI (Note that y E DI.) Let us write G y.i for the event that there is such a path P Let X = (/u0/) c . and let P IA, denote the product probability measure in which each site of X is open independently with probability p We may regard the event D ili as an event in this probability space. Note that, in the measure P„, if we condition on Ft, , ,. then every site of x is open independently with probability p (The event Flo' is defined 'without looking at' sites in X.) Hence, P„(E mi

F Fp .1 )

= WI;

al I ts 1Pyt.E„ 1 )

=

P y (p,

1 )1? (Gm!)

Let H y denote the event that X contains an open path horn a nrighbout of y to b' (p). Suppose that E l, j holds and that D, > k Then F;, r holds, and, as in the case; = 1 above, X U{y} contains open paths P. P ' . disjoint except at y, with P joining y to 8,;(a.). and P' .joining y to St (y); see Figure 10 Thus, X contains disjoint witnesses fin the f

Figure 10 The shaded region represents the interior cluster I If E y I holds, then I = I and ti is the — 1)“ pivot 14 _ 1 for the event R.,( 1! ) Ift in addition. D r > k with d(r y)+ k < then there are disjoint paths P. P' horn y tr,) (x) and to Si ;F OI L with P. P ' C = uill)'



/

illerishiltov's Theorem

events Cy r and H„ r. so Crl „ i ❑ H„ r holds As E y( subset of Fil l, we have shown that

95

n {D, > k} is a

Py(E, i )P;\,(C / 0 H y 1).

P„(E„ ,1 ri{D, >

Applying the van den Be t g-Kesten inequality Theorem 5 of Chapter 2, to the product probability measure P IA, WC have

F

(Cy

0 H„ ,1 ) < IA, (6?„,r)Plx,(H„,r)

g the three relations above, Combinin P p (E,, ,1 n{D,. > lc}) G Pp(E„ Now for any possible ti and p k (y,p) Thus

we

have

P;\ (11„ , r) =


0 ) 5 Pp 4E. „ 1)pk(p, p),

ie

Pp (D, > k E y 1 )

pk(q.p)

As the events E„ .r par tition E. the bound (5) follows

❑

Lemma -, is the key ingredient in the proof of Nlenshikov's main theorem Although this lemma gives detailed information about: the distribution of the distances to successive pivots. all we shall need is the simple consequence that, on average, there ate many pivots if 11 is large.. To state this lemma, let N(E) denote the number of sites that are pivotal for an event E

Let .r be a site if alt leafed graph A. and let a and P be positive integers Then Lemma 5.

E„

HMO - sup ppOi ,p)) LikRj

(R.„ GO) 1?„(x))

Proof Let D,, D2,

be as in Lemma -1 As the event

D, _

> • by < • • • and pa > and p 14. 1 = Pr — p i log(l/pd,

=

as long as p i n > 0; in fact, as we shall see later, p i > p for all p Let and p_ = p1+1 Thus = p+ = us apply (9) with k = pk p„,(pi)== p i , so (9) gives pd, and (p+ ) = = 1/1 /k) -= p ix:(m+ ) 5 P,,,

/ ex l) HP/ — lb+1)Ll iPd ( 1 — Pi)ri

The sequence p i is decreasing., so p i < pa < 1/100 for every i from which it follows that (I — pni 1 /P , 1 > 1/3 Thus we have p„ , (m +1 ) 5 pi exp Hp; log(1 /p i )11 /9 1 1 /3) pi exp(— log(1/ 9 0/3) = p:+113 As n • •

'u, the event II,,,„(x) is a subset of H,, , (:c) for ally (Pift) < Pa.(Pt+. 1) < Pi1/3 91-1-1 =

so

(

we have pi < pr r so, very udely, < pal for every As x log(I./T) is increasing on (0,1/e),

Fr om (10)

Pa — Pi

og(i/pj) 5



c —J pa log(e-ilp0)

pa(7 + log( 1 /p0 ) ) 5 2p0 log(1/90 ) + pa .< (pa — p)12,

where the second last inequality uses E i>0 )c < 1 and the final inequality holds by out choice of no. it follows that the construction of the pi continues indefinitely, and p i > (Th -1- p)/2 ha ever y sequences

l.3

Theorem

99

Let p' = (m p)I2 > p At this point we have constructed an increasing sequence n it and a decreasing sequence p i , such that < Pu,(Pi)= for i = 0,1,2, To understand the significance of this bound, we should compare r1a and pa Let s; = 11/Thl, so = no {L c se, and 5o = 11/pol > 100. Since s; > so > 100, the rounding in the definition of Si makes little difference Iu particular, (10) implies that sa± i > say. Hence, for 1 < j < we have < 4 /5) , which gives 5i--; 5

S—,:>,(1/9)J Yt

< st.ino Let n > no be 111bitutiv Then them is an

SO /11+1 =

Iii < 11
p i n, and solr7= Pat Theorem 7, Let A

be on infinite. locally finite oriented with C t finite and stiongly connected If there is a constant C' such that

exp(C l /r/(log 0 3 )

1.5,4(01 fm fret

i ti ligi



t hen M(

(IA)

= Pi1(7)

Prof By Lemma 2, the critical probabilities pi i (X :3) and I/ ( A: ") are independent of the site Let p < ICH ( A ). and let .r he any site of

A Thiel' (A :

=

E

= ,r

11/n;,(// E C's)

I

H- ( 3 11firi (.1.

S ;':(.1 I

The final stun converges In Theorem 6, so p < pq( A ; :r) = p (A) As P < Pi t / " as at hitral A it 1°Ikms that (//t ( A ) Pit( A ): so 11/41 ( A ❑ p i ( A ) Using the equivalence between percolation models discussed in Section 2, Theorems 6 and 7 immediately imply corresponding statements for bond percolation on A. and for 1 u/oriented site and bond percolation. Mann, of the most interesting percolation models satisfy the assumptions of Theorem 7 (alter the appropriate translation to oriented site percolation) [ indeed, in many cases (Ion example site or bond percolation on Ed ), the class-graph has only a single vertex and the sizes of the neighbourhoods .51,1"(x) in A grow polk nomially iu n It is tempting to think that for any finite-type graph, cipher the

)

s Tluore tti

101

'growth function' 2 (u) = Sup , , : ‘ 15„(.01 will be bounded by a polymania'. or 2 (0> exp(ur) fin sonic n > 0 Indeed. Milma 119681 conjectured that this assertion holds in tilt' special case of Cayle y graphs of finitel y generated groups. Surprisingly. even this is not true: Gtigorchnk 11983: 198-0 gave a counterexample Fut the/mo t e. solving in the negative a problem of Gramm' 119811. AFilson [2004] proved that a glom of exponential growth need not be of unifianth exponential growth (See also Arachnile rind Pak [2001]. Pvber 1200-11 and Eskin. Mozes and Oh [2005]) The condition that C–c be finite is essential fin Theorem 7. We illustrate this lie untalented bond percolation Let G i he it 2' •gt. id i c . a square subgt twit of r with 2 2 ' vertices. and let A be the graph obtained b y stringing together tire gtaphs G; as in Figure II

corners Figure II. A series of grid-graphs strung toget he t by I heir opposite lot the example, t he sizes of t he grids grow super-exponentially.

Note that A may be embedded into E 2 in a way that preserves graph distance, so 1.5„(A; < 1.5„(72; = 4n for n > 1, and the growth condition (IA) is satisfied If p> 1/2. then it follows easily horn the results in Chapter 3 that the probability that there is an open path horn one collier of Cr to the opposite comet is bounded below by a constant (This may be shown by using exponential decay for p < 1/2 and considering the dual lattice ) Also, the expected flambe! ' of sites of that YilaV he reached by open paths horn a given cot net is 0(16l 1) Since Pi] , grows super exponentially. it follows that pi; (A) < 1/2, so. as A C V. we have I); (A) = 1/2 On the other hand. Fir (A) = 1. since. for p C 1, the probability that each of the infinitely many cut vertices is incident with at least two open edges is (I A slight variant of this construction

102

Exponeatial decay and critical probabilities

works for oriented site percolation: take the line graph, and then replace e each bond by two oriented bonds. In the unmiented case, we do require the underlying graph A to be of finite type. However, there is little reason to consider the graph structure of CA, defined analogously to Cdr : the graphs A we study are always connected, so CA is also connected. Note that if A = ips(A) is obtained front A by replacing each bond by two oriented bonds, as in Section 2, then two sites a; and y are out-like in A if and only if they are equivalent; in A Thus, if A is a connected finite-type graph, then C1 is finite and strongly connected For locally finite graphs, the transformation mapping an oriented or unmiented graph to its line graph also preserves the finiteness of the class-graph Finally, these transformations also preserve the growth rate of the neighbourhoods, so Theorem 7 does indeed imply cm responding results for bond percolation, and for unmiented percolation In the statement and proofs of Theorems 6 and 7, we took e ye' y site to have the same pr obability p of being open These results extend easily to a sornew bat more general setting, in which different sites :r have different probabilities pd. of being open, with the states of the sites independent Recall that the definition of the class-graph C-v. or the equivalence of sites in the man iented case, involves isomolphisms between subgraphs of A or amount ' phisms of A Naturally, we now requite any such isomorphism to preser'e the 'weights', i e., the probabilities that the sites are open. Thus, out-like sites still behave in the same way for percolation As we require C-Tc. to be finite, there is one probability pi for each out-like class, so we obtain a per colation model parametrized by o a vector p = (p i , . pk ) For example, one could take A = with alternate sites having pr obabilities p i and pi, of being open, resulting in two oil-like classes Theorem 6 carries over to this 'finite-type weighted context in a natural way: using self-explanatory notation, the result is that if for some . hr; We have . .pk )= 0 then, whenever pi for each an assertion equivalent to (7) holds in the probability measure Pp, Hew Pp , is the measure in which sites of class i are open with probability flit with the states of all sites independent. There are two ways to see this One is to note that the proof given above carries over mutabs unitantlis In particular. the Margolis–Russo fornmla (Lemma 9 of Chapter 2) states that for tar increasing event E. the stun of the partial derivatives of IPp (E) is exactly the expected number 01(N(E)) of sites

/3 Alenskikov's Theorem



103

that are pivotal for E If we decrease each p i at the same rate, then the calculations in the proofs above are exactly the same as for the uniform case. Alternatively, one can realize an arbitrarily good approximation to the weighted model as an unweighted model, by choosing p close to 1 and replacing each site open with probability p r by an oriented path of length C chosen so that pd is within a factor p of pa,. Using this idea it is easy to deduce the weighted version of Menshikov's Theorem from the umveighted one. Note that we require g p i for all i: in general, it is not enough to require p < p i for all i and pie < p i for some i. This may be seen by considering any graph in which sites of one type are ' useless ' , for example, the graph A obtained from Z - by adding a directed cycle to each site, where the sites of Z 2 are open with probability p h and tine new sites with probability Po Marry of the other results we shall present also have natural weighted versions Most of the time, we shall present only the unweighted version, and shall not discuss the simple modifications or deductions needed for the weighted versions Aizemnan and Barsky [1087] gave a result closely related to Theorem 7. Their proof uses yen v differeM methods to those of Menshikov, although it also relies on fire Vail den Berg-Kesten inequality and the Margulis-Russo for mula A key idea of the proof is the introduction of a 2-variable generalization of the percolation probability 0(p) Considering bond percolation on a graph A in which all sites are equivalent, this may be written as

moi,

= 1—

E

p( 7;

II),

rt=I

where p = 1 — 6- 0 The reason for the reparamettization is that: Aizennmn and Barsky consider a more general model of percolation on Ed, where 'long-range' bonds are allowed: each edge xy of the complete graph on 27,(1 is open with probability 1 — exp(-0./(x — y)), where .1 is a non-negative symmetric function on Z d . Taking It = 0, the sum above gives exactly Ilt,OC„. < = 0(p) Writing VI for .1(X), Alain/Mill arid Barsky prove two differential inequalities for AI = Al(13,11), namely 8111

08

< 1.111

0.111

104

Exponential dung and cIilical pi ° hybrid ies

and <

,r + :11" •fli Ur i

Oh

The latter inegt tlitv genet alizes an eat lie/ tesult of Chaves and Chaves 1986a] that (() < 0(i()2

0(p)i)(e)11))

lot hood pet colation on Z il (For site pet colation. the Lena 0(p)2 replaced by li t 0(p) 2 ) Using these differential inequalities Aizentnan and Barsky show that 11/(el l > ch i/2 for some c > 0. where ;IT corresponds to p i Math the) deduce that p i /in Note that the It) has a simple combinatorial description: if we extend quantity the pet colation model la adding a single /KM vertex G (the -ghost' vertex). and join each site of E d to G independentl y probability — then 111(13.11) is the to obabilit v that thew is no open path (tour 0 to G

4.4 Exponential decay of the radius Out aim in this section is to show t hat. undo mild conditions < I he distribution of t he 'whits of the open dust et containing a given site has all exponential tail Floiment 6 does not quite give this. although it conies close Flowevet exponential decav follows from Theo/tin t 7 by tesnit of Elatumetslev )19574 a special case of this result was described at the start ()I the chaplet The statement and pool that follow appear considerabl y tome complicated than those lin the special case of bond petcolation on D I considered at the stint of the chapter Then) a l e two 'fiasc 's: first. thew is a noun ' technical complication that.vises in the case of site petcolation Second we shall separate out the heart of the result as a lemma, and we wish to state a teasonalth silo/1g form of this lemma ha hame 'detente If out aim we t° only to move flanuocuslev's result. I Iwo/ em 9 below. then a weaker km to of the lemma would suffice In tact. the much greater genetaffix consideted here int t oduces no essential complications Given a site t of a directed graph A. let

13'1 ( 3 ')

U

if ( ,I )



be the not-boll of r.adius r centred at

e 71 • (I(:r. y) :)); t Let A ,• (.r) denote the numbe t of

Exponential decay

105

of the Indira;

sites in5';r(r) that ma y lie leached In an open path P iu B7( t ) starting at all out-neighbour y of c Lemma 8. Let A be ml oriented multi-graph r > 1 an integer. mai < 1 a real nymber. If Elis.,(N,)+T.r)) < for every site .r of A. then d fat every site ;I

Of

A um/ CM, y 11> 1

Proof As usual. we write tr,, for IP); Let 1,n > 1 Recall that I?„, = tr. + -14. 1 denotes the event that tin g e is an open path P From an outneighbour of x to a site y e S,t(:r), and that a„,(r.p) = Pp(R„,(r)). By splitting the path P at the point it first reaches Sit ()), we see that Rrr.„(x) is the union of the events E„, y E .9;(,r), where E y is the event: that there is an open path P in a,t,..„(r) horn an out-neighbour of r to 87: „(: y ) that first meets .9)± (x) at q (This is not, in general. a disjoint union. since there m ay be many such paths P ) Splitting such a path P at y. the initial segment P' Rom a neighbour of 1 to y lies in 8,5(r). and is thus a witness For the event E,C that theme is a path in 13)/(t) flow an out-neighbour of x to y; see Figure 12. The remainder, .P", of P is a path star ting at an ontmeighbout of y and ending in S it i. „(x). Thus P" must contain a site of 5,1)(q). so P'' (ot an initial segment of this path) is a witness for 17„(y) As P' and P'' are disjoint, we have /7„(y)

C

Writing p„(p) for sup y p„(y, p), as before. by the van den 3eig inecpuditv it follows that

P,±n(x,



PpuLT ;,FP,,(R„ on)

Lii )„(E„)< (J)

itC-b;!-(x)

E IFIY( E/y) p a ye sr))

Err(N;3-(I))aa(a) 5 -IM(t)

)

we have p, ÷ „(p) < p„(p), As this holds fin every lot every a As Py (x 22+) = pp„(r), the result Follows

so p„(p)

< -siln/r1

Note that we work with paths star ting at a neighbour of a given site to avoid -re-using' a site when we concatenate paths This complication does not arise for bond percolation, where the corresponding result is that, if the expected number of sites of 5 + (t) that may be reached

1(16

Exponential decay and critical probabilities

Figure 12 The (Wick lines show an open path P witnessing the event R.,-E.n(:c)• The sub-paths P' from it ' to y and P" front y ' to z are disjoint witnesses For the events E,/, attd R.,,(y)= fy + 22- 1 respectively

nom x b y open Paths within B,+ (x) is at most i for P i ,(3, 11-) < -1 1" h -I holds for ever y kite's and integer a Mom Lemma 8, it: is very easy to deduce Hammeislev's result

every 7 then

Theorem 9. Let A be an oriented multi-graph with C hi finite and SilDlIgly

connected. and let p < p4-( A ). Bum there is an a

> 0

such

that 11F7,(3: fin

exp

Im)



(15)

every s ite 3, and integer a >

Proof Let t' be a site of A By Lemma 2 we have so ivj, ( p) < x Now

E ()))

PE

E

IIESt (x)

p4

a;

-

=

4.5 Exponential decay of the volume

107

where Ix + yl is the event that them is an open path flour an out(x)al Also, as neiglthour of x to y (Thus, I?, = Uaes;t-cofx-i(x) to y before, if {x + y} holds then there is an open path fiom are independent; ) ± --r is open' and {x not visiting:, so the events Let "o(c)=

E

P r+

acis;•(3,) Then E r 7, (x) is convergent for each x, so 7,.(r) a As 7, (a:) depends only on the out-class [x] of x, and there are only finitely many outclasses, there is an r such that 7, (x) < 1/2 for every site x Since ❑ 4,(N,+ (x)) < (x), the result then follows how Lem a n 8. Let us remark that, under the assumptions of Theorem 9, if the expected untidier of open paths star ting at each site x is finite, then (l5) can be deduced without using the van den Bet g-Kesten inequality. Flowever, it is pet reedy possible for this expectation to be infinite even when x =i,(p) is finite, as shown by the graph in Figure 13 Indeed, for this

7N 7N 7N N7 N/ = I in V.111(711 there are exponentially many paths Figure 1$ A graph With of length C from a given site.

graph Pit = = 1, but taking each site to he open independently with probability p, the expected number of open paths of length C starting at xo is ,ri which tends to infinity as C. for any p > lb,F5 Performing the saute construction star ting with a binary tree rather than a path gives a similiu example with pit = pis! < 1

4,5 Exponential decay of the volume - the Aizenman-Newman Theorem Om aim in this section is to strengthen Theorem 9: under suitable conditions, not only does the radius of the open cluster . C„, containing a given site a: decay exponentially, but so does its ' volume ' , IC1I Aizenman and Newman (19811 gave a very general result of this form, winch we shall come to later However, a special case of their theorem may

108

Exponential decay and critical probabilitic5

he proved in a ver dilleren t way, using 2-independence: we present this Hest shall for initiate the next result for until iented site percolation on graph A Recall that, in this context, two sites are equivalent if there an autorumphisur of A mapping one into the other (This corresponds to being out-like if we replace each edge by two oriented edges ) A graph is of finite type if there are finitely many equivalence classes of sites under this relation As usual, we write S, (c) For the set of sites at graph distance a from a \VC write B„(x) for U7--u S i(/1 ) , e for the hall o radius a centred at a' The conditions in the result below are generous enough to ensure that it applies to the 1/10tit hale'fir g tiltieS, including site and bond percolation on D I and on Atchimedean lattices The proof is based on Theorem 6 or Theorem 9, and the concept: of 2-independent site percolation Although it is ratlwr easy, and strongly ierniniscent of the pool . of Theorem 12 of Chapter '3, hem we have to soil( hauler to get a covering corresponding to the cover of 2 2 with large squares Theorem 10. Let A la a connected, infinite locolly-fruityfinite-type wan tented graph S l ippage that sup 12,011

< t (hig t Vino

(16)

■

fa all soffit amity tangy t o = u(11 p) > 0 such that

net fat meet// p < /)11(-‘)

;MCH

Lot all sites . t and all

ill( I

> a) < exp(—on)

>0

Proof The (nem!l plan of the proof is as follows: NW shall cover 11(A) In a set of balls Bo, OIL u E II' that are reasonabl y \\ ell spread out': binning a graph (F) on II" by joining a'. a' ' if they ale at distance at most tit . say we shall show that the maximum degree of D(11) is not too large We then construct a 2-independent site percolation measure on D(II ) as follows: a site la E II will be active if some E gi,(w) is joined In an open path to a : distant' site i e a site at distance at least r limn .r: these events are independent if d(W. a') > 6/ Also. an y site in a large enough open utast:el is ,joined to a distant site. so a large open cluster in A implies a l a rge active cluster in D(IJ ) From Theorem 6 (\lenshiky . 's rliemein) and inequalit y (16). each a' is yen in /likel y to



4

5 Exponential decay qf the volume

109

active if t is chosen huge enough: it will then Follow lion Lemma I Chapter 3 that open clusters in this 2-independent measure are small, of giving c, - the result To carry out this plan, we shall first show that balls of /minis Tr in A ale not too much huger titan those of radius r Let A < c be the maximum degree of A As A is connected.. (di each pair ([4. [y]} of E Ix] to some equivalence classes of sites thew is a path P hour some E [y]. Thus there is an integer L such that for am .r and (1 , there are E and ij E (y] with d(3: ( (/) < L. It follows that. crudely. )e

i p rODi = i 13,C ri )i

(I +A+

i r3 ,+1.0a =

±AL)18,,(y)l

: :.r E A} and bi = for every t > I Let left' = ntaxilB, where C' depends A}. Then we have I < < C' for every only on A. \\'e claim that <

(18)

holds Mt infinitel y many t Otherwise. there is an r n s uch that

1.11

> t 0 we hate let 11/:"

) > \/7: IC'

Tin t s. lin r large enough, adding log? to = log y increases log/; by at least (1/3)logt = (/3, which implies that log ft- > 1 2 /100 lot huge t . contradicting (lay and proving the claim F l om now on we consider onl y r for which (IS) holds Let C 1'(A) he a maximal (infinite) set (}1 sites subject to the halls (w) : w E 111 being disjoint (Such a set 11' exists in am graph Lv Zonfs Lemma In fact. since A is connected and locally finite, A is countable. so IV may be constructed step b y step ) Note that the balls {Tin, (w) : w E It } covet V(A): if sonic site y (lid riot belong to Wc it -8.), ( 11 ') . then q could have been added to IF. We define an auxiliar y graph D(11') with vertex set 11 - as follows: two sites W. a E H ate adjacent in D(11 7 ) if d(w. n) < . where t1(.t.. te) denotes the graph distance in A If r„ denotes the set of neighbours of 0 , in D(11"), then the balls (13,06 : u ate disjoint subsets of 137 ,(w): see Figure Hence lb,- 5_

Li T3, (u) < [37, (o . )1 < Er„

Therefore. using (18). we have degree A i of D(11") is at most

bt, \AT


0 such that, for any z E IV and any nr > 0, i13 0C.(D(11 7 ))]

in) 5 exp(—cm),

where Cz (D(IF)) is the open (active) cluster of DO V) containing z To complete the proof, note that if x is a site of A and I CA > then for every site y E C„, the event {y =} holds. (There is no room for all of Cr inside B, _ i (y)) Let IV' = fw E FP : (w) n 01 Then every site w E IV' is active Also, as the balls Bo, (w), w E TV, cover 1 7 (11), the balls Bo, (w), E IV', cover Ci„ We claim that 11 7 ' induces a connected subgraph of D(11) Indeed, E then there are sites y, y' E with y E .80 1 (w) and if w, As C'„, is connected, there is a path (al) y = !pry° • • yt = E yj Iv( E IV' with Eh E Bo, (wi), with every y i in C . We nilw choose and to, = w, air = i v'. For 1 < i < t — 1 we have death < witvi+, is an = 4r ± 1 < 6r, so either mi = edge of D(IV) He/we, w and iv' are joined by a path in the subgraph of D(11") induced by IV' Fixing E A, let E II be any site such that :r E B2r (2) We have shown that II' is a set of active sites that is connected in D(II), so TV' is a subset of Cz (D(W)).. Also. as C . is covered by balls of radius 2t, we have > Hence, for n > > n) <

HA ) Ill'OC',;(.0(1V))1 ^ n/q)

As r is constant, the proof is complete

exp(—cnnib ) ❑

In the p oof of Theorem 10, we used Menshikov's Theorem. Theorern 6, to obtain a.) good bound for on ner 4 p < Mi . Instead, we could have used Hanunersley's much simpler result, Theorem 9. Modifying the proof in this way would give the satire conclusion, (17), but only for p < . (Of course, under the assumptions of Theorem 10, by Menshilcov's 'Theorem ) =p Next we shall present the general result of Aizernnau and Newman [1984] mentioned at the beginning of the section.

Exponential decay and critical probabilities

I 12

Let A he infinite lacally-finilc oriented multi- graph with and strongly connected. and le! p < pq ( A ) Then th in is art o > 0 such that Theorem 11.

r71111(

E l ;(1Cli I

ir)

fat all sites re and integ e rs n >

exp(—on)

(19)

1

Pox"' As p < pl.(71.) is fixed vve suppress the dependence On pin of

notation Suppose first that Csc has only a single vertex, i e that all sites arc out-like this is the wain par t of the proof: the extension to [-had many out-classes is a minor variation plr denote the event that there is an open path As berme let fr y+ from an out-neighbour of x to y interpret {.r + — .r} as the event which always holds. Thus — It holds if amid only if .1' ± — II and :1' is open. Note that El

E P(.1. !t)/P= yr(r)/P

PLr li pc.,

1r

ye:\

\\ le shall estimate the moments of ICH in terms of using the van den start with the second moment Berg rkesten inequalit \ We mk, not assume that z. fir and yo ate Suppose that . yr. E distinct. although the Heinle is clearer if we do. As um E there is an open (oriented, as always) path Pr horn x to yr Let Pr, he m shortest open path how a site a on P i to /1 2 : such a path exists as /12 Then Pr and Pr are vertex disjoint except at a (otherwise. Pr could be shortened) Let us split PI into two paths. a path P; horn a: to and a path Pi/ horn a to yr: see Figure 15 ()witting the initial vertices horn the paths P(. P(' and P, gives disjoint witnesses for the events IT + --, — I/1 { -- !P I }* respectively.Tills ' if 1./2 E Cr. Olen the event -

0 {a + IL}

Ica some tt E A (Recall that 0 is an associative operation) Using the can den Bet g- IKest en inequality, it follows that holds

FOP fi g )
0 we have L (exp(tIC, I)) = T

E (t. I k

Pr:!)

(2k - 3)H

, ( (A )-) A k

i

116

Erponentiol decay and el Then/ pinbalnlihes

As (2k -a) /1,7! < 2 k . this expectation is finite lot 0 < t < (‘')- 2 /2 As E(exp(ird)) -

exp(tn)PaC r i = n),

it follows that 1.11 (r i d = 11) < exp(-/o) fin all sufficiently huge 11, and the result follows,. SO fat. we assumed that all vertices ate out-like As plot/Used, extending the result from this case to the general case is sttaightfonvatd Indeed, the pool is the same, except that we 'enlace by \" = SUP



P UP . —

In= SU P

\if (v)I p

In place of 22), we then have 11) (lif

(b)
0 be given Since E is measurable, there is a finite set F of sites of A and a cylindrical event Er depending only on the states of the sites in F. such that P(ELEF) < E

(1)

Let AI = max{d(xo, q) : y E F} Since A is locally finite, the ball = {z : d(x0 ,z) < 2A1} is finite Thus there is a site x with xo and d(x,,r0 ) > 2.111 Let (p be an autonnuphism of A mapping xo to a. For y E F we have Bom(xo)

d(te,y)(0)

d(au, Y0'0)- d(V( to), (PM = der° , 37) - (1.070 ,:y) > 251 - AI =

so cp(y) F. Thus the sets of sites F and y(F) are disjoint It follows that the event y(EF ), defined in the natural way, is independent of Er SO

NEL

n v(EI )) = F ( Er) F (y (E r)) =INET )2

Thus, 1P(E) F(Er)2I = 1.P(E n E) -P(Er n ))I < P((E n E)A(Ef, n y(EF ))) For any sets A, B, C. D we have (Anr)A(c Thus, as E is automm phisrn-inwniant,

I P ( E ) - P(Er )2

1


0 is arbitr ^



9E

1 + i

r (EF) 2— P(E)21

45.

it follows that P(E) = P(E) 2 , so F(E) is 0 or 1 0

ii tji

120 Uniqueness of the infinite open cluster and critical pivhabilities In the proof above, we did not need E to be invariant uncle/ all antomot phisms of A. just under a set of autornmphisms large enough that any finite set can be separated horn itself by such an autornmphisin In the context of lattices in R d . for example, invariance under a single automm phism y of A conesponding to a translation of IF'' thiough any vector (a l , .0) is enough. In particular. Lemma 1 ati) (0,0, is often stated for translation- invariant events In the terminology of ergodic theory, we have shown that the induced automorphism : R Q is et godic. NAre now turn to the particular event lk that there are exactly k open clusters, where (1 < k < x We star t with a lemma of Newman and Schulman 119811, showing that, with probability 1, there 1 or infinitely many open clusters are

0,

Lemma 2, Let. A be run infinite locally finite. finite-type graph. and le E (0.0 Then

= ft

PA , \: 0 fon some 2 < k < c Let To be any fixed site of A, and let be the event that lk holds, and each infinite cluster contains a site in B„(x0 ) As the balls B„(ry) cover A. we have Lk = u n T, k, ,/ P (11,), and Oleic is air 11 such P(Ty that P(T„ , r) > 0 Changing the state of every site in B„(,ry) to open, we see that P(./ 1 ) > 0 Indeed, spelling everything out in great detail, the event T„ A is the disjoint union of the events P100.1

so

=T„

.:n{5' = s}.

where s (s„);,,Bu(,,„,) E (13 " 1 " ;) , and 5' = (5,.):,..EBu(,,„) is the (to) vector giving the states of all sites in Thus there is an s which 0. Now if w E T„ k s and w' is the configuration obtained P (-rs k s) horn w by changing the state of each of the closed sites in f3„(ro) h om closed to open, then w' E In Thus k

for

>

P(1 1 )

La ( {if

: E T„

= (p1(1.

p))1(TT„ s ) > O.

5 I The Airiemnan-Aesten-Neuunav Theorem

121

where c is the number of sites in .13„(:ro) that are closed when S =s We have shown that if P(//,,) > 0 for some 2 < k < cc. , then P(I,) > 0 But then P(4) = P(1 1 ) = 1 by Lentela 1 As 1k fl it = 0, this is ❑ impossible To prove the main result of this section.we shall need a simple deterinistic lemma concerning finite graphs L emma 3, Let C; he a finite graph with k components and let L and , es } he disjoint sets of ITT bees of C. with at least one et C = in each coinponent of Let . be integers each at least 3 Suppose that fin each i deleting the reflex m disconnects the component containin f« . 1 info mullet components. in; of which contain mudd i es of L Then

ILI > 2k 4- E(In i — 2) Proof By considering each component: of G separately, we 11/111/ assume that k = 1, i e that C is connected Flemming an edge from Ce can only increase tire number of components of — e i containing elements of L.. so we may assume that C is minimal subject to CI being connected and containing C'U L Thus C is a tree all of whose leaxes are in CUL As > 3, no vertex E C' can be a leaf of C. so all leases are in L (there ma ■ also he internal vertices in 14 Since C — has in; components, the vertex has degree in; But for any tree with at least one edge, the number of leaves is exactly 2} (d(e)— 2), where the sum HMS over internal vertices. As d(v)> 2 for each internal vet tex, the sum is at least. 5-7= ,(nc —2), and the result follows. ❑

5

Let us say that an infinite graph A is untenable if 1.5' ,,(r)iABflOpi —' as n for each site 3, i e . if large balls contain man y more sites than their boundary spheres. There me several notions of amenability for graphs; in the present context this \In hint seems to be the most useful. In fact, the concept of amenability originated in group theory, where it is defined somewhat differently. Note that. if A is amenable am( of finite type, then the limit above is automatically uniform in r. The following result is dire to Aizemnan, Nester' and Newman [1984 the proof we shall present is that of Bruton and Keane 119891 Theorem 4. Let A he a connoted. locally finite. finite type. amenable

122 Uniqueness of the infinite open cluster and critical probabilities infinite graph, and let p E (0,1) Then either 1FX p (I0 ) = 1 or rA p(B) 1, where Ik i s the event that there are exactly k infinite open clusters in the site percolation WI A Proof As usual, let us write P for P;\ p In the light of Lemma 2, this result is equivalent to the assertion that P(/,) = 0. In fact, we shall show that the probability that there are at least three (possibly infinitely many) infinite open clusters is zero Suppose for a contradiction that this is not the case.. Let to be any site of A, and let X 0 be the set of sites equivalent to zo As the balls Br (ra), r > 1, cover A, there is an p such that, with positive probability, Br (x0 ) contains sites From (at least) three infinite open clusters. For the rest of the argument, we fix such an I Let Tr (_y) be the event that every site in B, (x) is open, and there is au infinite open cluster C) such that when the states of all the sites in B, (a) arc changed horn open to closed, 0 is disconnected into at least three infinite open clusters; see Figure 1 If w is a configuration in which

Figure I A site a for which Tr (c) holds, Ever y site in B, (r) is open; the rest of the open subgraph is shown by solid lines If the sites in Br (x) ale deleted, or their states changed to closed, then the infinite open cluster C) falls into four pieces, three of which are infinite

B, (to) meets at least three infinite open clusters, and w' is obtained from w by changing the states of all the sites in B, (to) to open, then E Ti (a0 ). Hence, IP(Tr (to)) > Thus, for all sites x E X0 we have

P(T, (a)) = a. for some constant a > 0 Our next aim is to sinus that, if

/I

(2)

is much larger than 1, then we can

5 I The kicennion-ICesten-Arcoonon Theorem

123

find marry disjoint balls B, (:r), :r 6 X0 , inside the ball B„(:to) In fact, to make the picture clearer, we shall find balls B, (.r) that are far from each other. To do this, let; If c X 0 n B„_, (x0 ) be maximal subject to the ba lls (c), w E W. being disjoint If lot E Xo n B„_, (x0), then (1.04/, w) < 4t for some w E otherwise. w' could have been added to If . As A is connected and of finite type, there is a constant C such that ever y site is within distance C of a site in X 0 Thus, every y e B„_,—a:ro) is within distance t of some w' E X0 n B„_,.(3,0), and hence within distance 4t+t of some w E W In other words, for a > +11, the balls BID-4w), w E H. cover B„_, _ ( Gro) Thus, IH1

1-812-1-(edi/P-1,-H:(34

ver y crudely. 1B„,n(x0 )1 ,5_ IB„_,_ 1 0.0 )10 + A + A 2 +

A'±t+/),

where A < x is t he maximum degree of A Thus. since is a c > 0 such that

I

is fixed, there

(:co)i for all n > t

C. As A is amenable, it follows that 111 7 1 > a -1 1,9„ + ("di

if n is large enough, where a is the constant in (2) Let us fix such an a Let us call a ball B, (w) a cut-boll if w E 117 C B,,-, (::0 ) and T, (w) holds. Note that if Br (a.,) is a cut-ball. then B, (iv) C B„ (co), and every site in B,(w) is open Since w ti ,r0 for every w E by linearity of expectation ' the expected number of cut-balls is

E P(T,(w)) = 011171

IS„.H(370)1

„'Em Hence, as P(Z > IE(Z)) > 0 for any random variable Z, there is a configuration w such that, in this configuration, we have s 18n-m(tie)1,

(3)

where s is the number of cut-balls As we shall soon see, iris contradicts Lemma 3 For the rest of the argument, we consider one particular configuration w for which (3) holds: in the rest of the argument Uncle is no randomness. Let denote the union of all infinite open clusters of the configuration meeting B„ (4), considered as a subgraph of A. In the configuration

EL)

124 Uniqueness of the infinite open cluster rind ci Weal probabilities w' obtained from w b y changing the states of all sites in cut-balls to closed. the (perhaps already discotmected) clustet 0 is disconnected into several open clustels, some infinite and some finite Let the infinite ones Fu Each L i contains 1 be L 1 , L 2 . ,L t , and the finite ones site in 5'„+1(:ro)„ so TIM/]

(-I)

Let the cut-balls be CIL , C. We define a graph H from 0 by coneach F4 to a single vet tex trading each cut-ball Ci to a single vertex and each L i to a single vet tex en see Figure 2 In the graph H. there is an edge how Ci to ci , for example, if and only if some site of fq is adjacent to sonic site of CI; Infinite components of (9 cot espond to components of H containing at least one vertex in L = Tints, the condition that Ci is a cut-ball says exactly that deleting c i Flom H disconnects a component into at least (Mee components containing vet tices of L. Thus we /nay „ s, to conclude that apply Lemma 3 with in; > 3, i = 1.2, t=

2±

E ( 3 – 2) = s + 2.

This cont /adios (3) and (- ). completing the wool

❑

Simpl y put, Theorem 4 tells us that, above the critical probability /4 1 . almost smelt- theme is a unique infinite open clustet TO conclude this section we remark that, by a simple argument of van den Beta and Keane (198. 1], Theorem 4 implies that 0(1 .,p) is a continuous function of p, except possibly at p =14[(A)

5.2 The Harris–Kesten Theorem revisited Combined with Menshilrov's Theotem, Theorem -t leads to vet another proof of the thuris-Kesten result that pl'i (Z2 ) = (22 ) = 1/2 This moot will adapt easily to give the exact values of the critical ptobabilities tot certain tithe/ planar lattices We start by improving the 'ease' inequality that p li ),(Z2 ) > 1/2 Mote precisely, we shall deduce Mattis's 'McGloin, restated below, nom Theorem -1 The mg:tin/eat: we give is due to Zhang; see Ctimmett, (1999, p 2891 Theorem 5. Fm bone! percolation cm E 2 l i am 0(1/2) = 0

5 :2 The Harris taster Theorem revisited



125

Figure 2 The upper figure shows the union 0 of all infinite open clusters meeting B„dro) The shaded halls, in which all sites are O pen, are the cutballs. The lower figure shows the corresponding graph H The filled circles ar e the vertices ei corresponding to the cut: balls, the hollow dines the vertices t i corresponding to the infinite clusters L t , and the crosses the vertices di cm esponding to t he finite clusters

Hive( Suppose not. Them applying Theorem -1 to site percolation on the line graph of Z2 we see that iP itq l ) = 1 . where I I is the event that

126 Uniqueness of the infinite open eluslel and critical pmbalatities

there is exactly one infinite open cluster, mid It follows that there is an no such that, if n > no, then the probability that an infinite open cluster meets a given by a square is at least 1 - 10-1 Let n = no + 1, and let 5' be an by square in 22 . Suppose that some site x in S is in an infinite open cluster Then there is an infinite open path P starting at Let y be the last site on .P that is in S, and let P' be the sub-path of P starting at in then P' is an open path from S to infinity, using only bonds outside 5'. Let L i be the event that there is an infinite open path P' as above leaving S upwards, i.e with the initial site y on the upper side of 5, the initial bond vertical, and all bonds outside S. as in Figure 3 Let LO, L 3 and be defined analogously,

Figure 3 An infinite open path P starting at a site a in a square S sub-path P' [torn the last site y of P in S leaves S upwards.

rotating 8 though 90 degrees each time Thus P 1/2 (L 1 ) = P i/2 (L i ) kn all i. We lime Pip (LI U L9 UL3UL 4) > 1 - 10-1

(5)

As the L i are increasing events, it follows from H a rris's Lemma by the 'nth-loot trick' (equation (8) of Chapter 2) that P(L 1 ) > 1 - 1/10 for each i: other wise, INV!) > 1/10 for each and, as the L ate decreasing and hence positively correlated, L?) 10-1, contradicting (5) Recall that the planar dual of the square lattice 2 2 is the lattice Z2 + (1/2,1/2), and that we take a dual bond e* crossing a bond c of 22 to be open if and only if e is closed Let 5" be an 11 - 1 by 11- 1

5.2 The Harris-Kesten Theorem revisited

127

Figure 4 Infinite open paths P, and P4 in the lattice V, leaving a equate S to the right and to the left The infinite open paths P. PI in the dual lattice leave 8 1 upwards and downwards. If P( and may be connected by open dual bonds in S' t hen there ate at least two infinite open clusters, one containing A, and one containing Pi

square in the dual lattice inside 8, as in Figure As the bonds of the dual lattice ate also open independently with probability 1/2, and as n — 1 = ne > no, the argument above shows that P i r(L;) > 9/10, where is the event that an infinite open dual path leaves the ith side of Si . Thus the event E = fl L2 fl L43 1-1 LI illustrated in Figure I has probability at least 1 x (1 — 9/10), 6/10 > 0. The event E depends only on the states of bonds outside 8'. Thus, with positive probability, E holds and every dual bond in 5' is open But then the paths P. and .P:13 may be joined to form a doubly infinite path P' that separates the plane into two pieces. As P' consists of open dual edges, and an open edge of Z2 cannot cross an open dual edge, the open paths ./?, and lie on opposite sides of P', and thus in separate open clusters. Hence, there are at least two infinite open clusters

In short, starting horn the assumption IP, /2(1 1 ) = 1, we have shown that with positive probability there me at least two infinite open clusters: a contradiction It follows that P i pe(11 ) = 0, and hence that 0(1/2) = 0

El

128 Uniqueness of the infinite open cluster, and critical plobabilitics Using Menshikov S l /WU/ it is veil, eas y to complete l et another proof of the fla t r is [Kristen Theorem. Them cm 13 of Chapter 3. restated below Theorem 6. Pot bond p rcolation on the square lattice pm ! = pr = 1/2 e

Proof As misted in Chapter 4. Theorems 7 and 9 of that chapter. stated

for oriented site percolation. appl y also to tu t or iented bond percolation and in particular. to bond percolation on E 2 (Formalit y one can apply the theorems to the graph obtained front the line gr aph of 2:2 by replacing each bond Ir y two ofmositd y oriented bonds ) In particular, lenshikoyis result, 'Theorem 7 of Chapin 4. gives ',I li C.2;2 1 = p r (7, 2 ) Since 0(1/2) (2V) < 1/2 implies ph (;Y,') > 1/2 it tints suffices to prove that This is immediate flour Theorem 9 of Chapter -I and Lemma 1 of Chapter :3 Indeed. suppose that p.11(272 ) > 1/2 Then. In the first of these results. as 1/2 is less than the critical probabilit y, there is an o > (1 such that LP it i(f 21() < exp(—mt) for all sites ,r and integers n. where Tr 21 1 denotes the own that .r is joined by an open path to some site at graph distance n front r Taking n large enough. we have j

----JJ) C 1/(10(M) 113 1/2 (ft "H Let S be an n by n square in 2T2 As before. let 1-1(8) be the event that thane is an open horizontal crossing of S II 1-1(S) holds. then one of the n sites on the left of 5 is joined by art open path ill S to a site on the right of S. tit distance at least n — I Hence. (H(S))
I/2, one ma y prove a Russo-Seymour-Welsh (RS \V) type theorem as in Chapter 3. 'elating crossings of rectangles o crossings of squares Alternatively. one can deduce the result horn the Aizeurnan- Kristen- NtiNV/IMIt ' filet)/ C/11, Theorem 4 lb prove that i ii(G 2 ) C 1/2, ha y ing moved an RS NV type theorem. one can apply one of tat ions sharp-threshold results as in Chapter 3 Alternatively

58

Site percolation on the Niangidat and square lattices

129

one can deduce that ph(Z-) = (Z2 ) > 1/2 directly Flom Monshikov's Theorem (Fm the last pint: we do not need exponential decay of the radius as used above: the almost exponential decay given directly by Theorem 6 of Chapter 4 is more than enough ) The approach used in Chapter 3 is perhaps mote down-to-earth, and simpler it/ any one given case The advantage of the Afenshikow Aizennnt- Kest:en-Newman approach illustrated in this chapter is that the tools rue very general, so genetalizing the moth to other settings is easier. Of course. one still needs a suitable star ting point, given by sortie kind of self-duality Let us note that, in this latter approach, moving that percolation does occur undo suitable conditions. which was historically much the l u odei part of the Flattis-liesten result, is vent easy: the deduction hour Alenshikov's Them em is very simple. and can be applied in a peat: variet y of settings. In contrast, showing that percolation does not occur. which was histoticallt the easier part of the result, is more difficult: the deduction Man the Aizen n ian Kesten Newman Them ent is not quite so simple, and requi t es write assumptions. We shall see this phenomenon again when we consider percolation on (whet lattices

5.3 Site percolation on the triangular and square lattices We next to p side! the (equilateral) tit iangulat lattice T C IF:2 Pot definiteness, let US imam and scaler so that (0.0) and (1, 0) ale sites of T. and all bowls hate length 1 Portions of T me illustrated in Figures 5 and 7 below Out aim is to show that psil (1) = = 1/2. As a stinting point, we need a suitable sell-dualit y property In bond percolation on V. the outer boundary of a finite open cluster can be viewed as tin open cycle in the dual lattice V ± (1/2,1/2) For site percolation Oil T. it is east to see that any finite open cluster is bounded by a closed cycle in the 5anie lattice T. Also, an open path in T cannot stair inside and end outside a closed cycle in T: indeed, the latter statement holds for site percolation on any plane graph, as a cycle in a plane graph separates the plane into two components These observations give a sufficient still ting point to enable us to prove that Mr(T) = p)(T) = 1/2, using the results of Menshiko y and of Aizenman, Kesten mid Newman. In fact, as in Chapter 3, it is easy to prove a striking ; huge-scale' consequence of the self-duality As usual, we write t he percolation nwastue undet considelat ion, in this case, for P)1/ P„ Lemma 7. Let 1?„ he the rhombus in

until 11 sites mi a ,side s"loam

130 Uniqueness of the infinite open cluster and critical probabilities

in Figure 5, and let fl(R,,) be the event that there is an open path in T consisting of sites in R.„, starting at a site on the left-hand side of R„, and ending at a. site on the right-hand side. 'Then P 1/2 (11(1?„)) = 1/2 for every rr >

• • • •• .•A. • • ••

Figure 5 closed sit

V

to rhombus nu: solid citcles represent open sites, and hollow circles

Plug. Let IiI (R„) be the event that there is a closed path in R„ joining the top of R,, to the bottom. Reflecting R„ in its long diagonal, and exchanging closed and open, we see drat

1E„(11(R„)) = Pr_glit'(/?„)) for any p and any n. In particular, Pit,(H(R„)) = 1P it (17 *(R„)). It thus suffices to prove that 17[72(H(R„)) +FP1/2(r(R„)) = 1 As in Chapter 3, we shall prove the much more detailed result that, whatever the states of t h e sites in R„ exactly one of the events 11(1?„) and 17 *(1?„) holds.. The proof is essentially the same as that of Lemma 1 of Chapter 3, although the picture is somewhat simpler One can replace each site of T with a regular hexagon to obtain a tiling of the plane. Thus, what we have to show is that in the game of Hex, no draw is possible: if all t i re hexagons corresponding to R„ ate coloured black m white, then either there is a black path from left to right, of a white path from top to bottom, but not both (On a symmetric board, it follows easily that the first player has a winning strategy. )

5 .9 Site percolation on the triangular and square lattices

131

To see this, we shall consider face percolation on the hexagonal lattice, which, as noted in Chapter 1, is equivalent: to site percolation on T More precisely, let; us replace each open site of R„ by a black hexagon, and each closed site of R„ by a white hexagon, and consider additional black hexagons to the left and right of R„, and white hexagons above and below R„, as in Figure 6

Figure 6. A partial tiling of the plane corresponding to Figure 5, obtained by replacing each open site in Ro by a black hexagon and each closed site by a white hexagon, with additional black and white hexagons around the outside The thick line is a path separating black and white hexagons, starting at :r, with black hexagons on the right This path must end at y (as shown) or at re The rest of the proof is exactly as for Lennart 1 of Chapter 3: let I be the interface graph formed by those edges of hexagons that separate a black region horn a white region, with the endpoints of these edges as the vertices. Then every vertex of I has degree exactly 2, except for the four vertices ,r, y, z and w of degree 1 shown in Figure 6 The component of I containing a is thus a path. Following the path star ting at there is always a black hexagon on the right and a white one on the left, so the path ends either at y or at ro In the for men case, the black hexagons on the right contain a path iu T witnessing H(R„), in the latter case, the white hexagons on the left witness V"(.11.,,) As before, 11- (1?„) and V"(R.„) cannot both hold as other wise Kr, could be drawn in the plane Using the results of Menslrikov and of Aizeinn, Kesten and Newman, it is easy to deduce that the critical probability for site percolation on T is 1/2, a result due to Kesten [19821 Menshikov's Theorem (Theorem 7

132 Unpteney; of the infinite open Haslet and et Hirai ptvbabilitics of Chapter -1) cells us that m = p H in this context: horn now on, we wine Th. for their common yable Theorem 8. Let I be the equilateral trianottla, lattice in the pla ns Then ms.(T) = 1/2 Proof By Theorem 7 of Chapter 4 we have M i (T) = Suppose first that /VT). it; (T) > 1/2 Then, by Theorem 9 of Chapter 1, we have exponential decay of the radius of an open cluster at p = 1/2, i e., there is an a > 0 stud/ that 111 1/ 9(0 < exp(—or/).. DefinR„ as in Lemma 7, any of the sites on the right-hand side of is ing at distance at least it — t hour any of the a sites on the left-hand side. so Pit2

(11(R„)) < Itifi l/2 (0

:)

n exp(-0(n — 1 ))

As n — oc the thud humid tends to zero, contradicting Le11/1118 7 Suppose next that p;'(T) = M i (T) < 1/2, so 8(1/2) > 0 Then. MI heorem 4, in the p = 1/2 site percolation on T there is with probability I a unique infinite open cluster Let /1„ be the hexagon centred at the or igin h " sit es on " eh side dm)" iu Figure 7 As U„ = T, if n is large enough then the ihr-p t obability that some site in H„ is in an infinite open cluster is at least 1 — sap. Numbering the six sides of 11„ in cyclic order let L 1 be the event that an infinite open path leaves horn side i Mote precisely, L i is the event that there is an infinite open path in T with initial site on the nth side of fl„ (we may include both cot nets), and all other sites outside ll„. Then j i L 1 is exact') the event that there is an infinite open cluster meeting > 1 — 10 1 Then we may chose p E ((l,1) with p < .1)!!(T) and 1 — p < p!)(11) Let us take the bonds c E E (T) to be open independently with probability p, and each dual bond c' E E(H) to be open if and onl y if c is closed. Then la Menshikov's Theorem we have exponential decay of the t adius of open clusters both itr T and hi H.. Hence, taking a huge enough 'rectangle' R as in Figure 9, with probability 99% t here is neither au open path in 71 massing R how left to tight, not au open path in H c t ossing I? limn top to bottom But by planar duality, them is always a path of one of these two types: this is a special case of Le11/111i1 2 of Chaplet 3, w hose proof is the same as that of 1,0111111i1 t 01 that chaplet In this case. the figure obtained la replacing each degree d site of H ot its dual by a 2d-gon, and each bond-dual bond pan by a squate, is the (3,12') lattice shown irr Figure 1$ To complete the moor of (8), it suffices to show that lot any p, at most one of the petcolation probabilities 0(T; p) and 0(1-1; 1 — is strictly positive 'This follows from Theorem -I as above: if both 0(1;p) and — p) ate strictly positive then, taking bonds of T open independently with probability p, with probability 1 them is a unique infinite open cluster in T, and a unique infinite open cluster in II Considering a huge enough hexagon in T. it follows as berate that with positive probability thew ate infinite open paths in 71 leaving the hexagon flout the 1st and Std sides, and infinite open paths in H leaving from the 2nd and dth sides. If the bonds of H inside the hexagon me also open, we find a doubly infinite open path iu H separating two infinite open components in It a cmaradict ion ❑

Having proved (7) in two special cases. fin A = V, and rot A = 21, we bun to a considerably more genet al result The arguments we have given so lar used the fact, that A had a suitable rotational symmetry, of onto 4 in the M i st case. and order 6 in the second In fact, the weaker assumption of onto 2 rotational symmetry is enough, although

138 Uniqueness of the infinile open (larder and critical probabilities one has to work a little harder to obtain (7) in this case Also, there is a natural generalization of (T) to certain settings in which bonds of different 'types' may be open with different probabilities. In this context, it is convenient to wad; with a weighted graph (A, p), i.e , a graph A together with an assignment of a weight p c E (0,1] to every bond e of A Fin each weighted graph there is a corresponding independent bond percolation model NI ARA, p), in which the bonds of A are open independently, and each bond e is open with probability pc To state a formal result, by a planar lattice we mean a connected, locally finite plane graph A (i e a planar graph with a given drawing' in the plane), with V(A) a discrete subset of 1R 2 , such that therm are translations 21,, and of IR 2 through two independent vectors tt i and rto each of which acts on A as a graph isomorphism In particular. all the Archimedean lattices are planar lattices Recall that two sites and a', or two bonds c and e', are equivalent ill a graph A if there is an automorphism y of A mapping r' to c to c' Note that any lattice is a finite-type graph, in the sense that there ale finitely many equivalence classes of sites and of bonds under this 'elation To allow for models in which edges have different probabilities of being open.. we define a weighted planar lattice (A, p) as above: A is a planar lattice, and there are two translations T„, and T. as above acting as antonnophisnis of (A,p) as a weighted graph, i e preserving the edge weights Perhaps the simplest non-trivial example is the square lattice, with = p for every horizontal bond and p r = q lot ever y ye t Heal bond, where 0 < p, q < 1 kesten [1982] showed that in this case, percolation occurs it and only it p+ q > 1: see Themern 15 below Another simple example is shown in Figure 10. Note that in a weighted planar lattice, there can only be finitely many distinct edge weights We sus that a graph A drawn in the plane is centrally symmetric, or simple symmeta lc, if the map a t--Y from P. 2 to itself acts on A as a graph isomorphism For a weighted graph, this map should also preserve the weights For example, the (weighted) planar lattice shown in in Figure 10 is str ut/nettle if one takes the origin to be the centre of an appropriate face If A is a planar lattice then, taking the vertices of the planar dual A* to be the centroids of the faces of A, sav, one can draw A' as a planar lattice, as in Figure 10 We assume 1:111°110/out that the bonds of both A and A* are drawn with piecewise linear craves in the plane If A is symmetric, then we may draw A* so that it is also symmetric The dual of a weighted planar lattice (A, p) is the weighted planar lattice (A*,q) in which the dual e' of a bond e

5.4 Bond percolation on a lattice and its dual

139

Figure 10 The planar lattice A (solid lines and filled circles) obtained by adding diagonals to every fourth face of Z 2 If the horizontal. vertical and diagonal bonds are assigned weights p, q, and r respectively, then A becomes a weighted planar lattice. The dual A' of A is drawn with hollow circles, at the centroids of the faces of A, and dashed lines

has weight qc . = 1 — As shown by Bollobtis and Riordan (2006d1, percolation cannot occur simultaneously on a symmetr is planar weighted lattice and on its dual Theorem 12. Let (A. p) be a squattete ic weighted planar lattice. with 0 < p„ < 1 fat every bond c Then either 0(A: p) = 0 al OW: where (A", q) is the dual weighted lattice Proof Suppose lin a court adiction that 0(A; p) > 0 and 0(„V;q) > 0. In the proof of Theorem 4 it was not relevant that all bonds were open with equal probability. Thus. writing Al and AP for the independent bond percolation models associated to (A, p) and to (A*, q), we see that in each of Al and AI* there is a unique infinite open cluster with probability 1 As usual, we realize the bond percolation models 11I and Al* simultaneously on the same probability space, by taking the dual e* of a bond e E A to be open if turd only if e is closed Throughout the proof we write I? for the probability measure on this probability space The basic idea of the proof is as follows: as before, any large square S is very likely to meet the unique infinite open clusters in A and in A". If we had four-fold symmetry then, using the 'nth-root trick', we could deduce that for each side of S. with high probability there are infinite open paths in A and ili A* leaving S from that side With only central symmetry, all we can conclude immediately is that there is some pair

110 Uniqucee s of the nrfiurlu open cluster and cr rhea/ probabililics of opposite sides of 5 how which infinite open pat hs in A leave S with high pr obalitlik■ The key idea is to move the 'corners' of S while keeping S the Sante Mote precisely. iliStetuf of a square, We take S to he a circle whose 24,/, and cot/sides infinite boundary is divided into four arcs A l , open paths leaving S hunt each A; If we move the dividing point; between two arcs, then pa t hs leaving one become more likely, and paths leaving tin other less Mi ch If we move the dividing point grachially, then the ptobabilities will change in a roughly continuous umnner, so at some point they will be roughly equal 13v moving two opposite division points while preserving symmetry. we can find a symmetr ic decomposition of the boundar y of 5' into four ales so that open paths of A leaving the tom arcs a t e roughly equall y likely Now, using the lout th-root for even/ ate A 1 it is sett likel y that tittle is an infinite open path in A leaving S front this ate We cannot say that the same applies to AL as we have chosen the ales for A and not for A' We observe. however, that among out lour arcs there is sonic pair of opposite at cs of S horn which infinite open paths of A' leave with high probability Indeed. this Follows from symmetr y and I he squaw-root hick This gives us infinite pat hs in A. AL. A and A' leaving the aces of 5' in miler and, as before, we can deduce a contradiction lit showing that the two paths in A' may be joined within .5. giving two infinite open clusters in A. We shall now make this aigument precise. Let .5 = S r be the dicky centred at t he or igin with nadirs r Let E(S) bet he event that some site of A inside S is in an initiate open clustet iu A, and let P(S) he t he men( that some site of A' inside .5' lies in an infinite open cluster in A" \\ T ilting D, for the disc bounded In 5, , we have IR', so the union U , E(S, ) is simply the event that there is U, D, infinite open cluster somewhere in A. and we have lin t ,— NE(5,.)) = S(A. p) he ;r positive constant that we and similar Iv for E' (5', ) Let shall specify later Choosing r large enough, we have NE(S,)) l —s 00d P(E' (S,))? I—

(9)

For simplicit y, we shall assume throughout this ptoof that no site of A ot A' lies on 5, and that tic bond of A ot A' is tangent to ,S, (Mote pletiselv, recalling that bonds of A and A' me drawn as sequences of line segments, we asSinne that none of these segments is tangent to S, ) This assumption is satisfied ha all but a countable set of values of r. Lin the test of the wool we fix such an r large enough that (9) holds, and mite tot S,

5 4 130nd petcolaijon on 0 lattice and its dual



141

Let e 1 1 < i < 4, he four distinct points on the boundfu y of S. 'nunbered in anticlockwise order We write c for the quad/ uple (c,,/!,,c3 We shall always choose these points so that no c; is on a bond of A of. A* We write .4 1 = A t (c) lur the boundary 01 r of S front cr to ciir t taking e tr r If MC is a bond of A with o inside S and to outside. then ate leaves S front the arc A; iL travelling along the (piecewise linear) bond ca! from V to 0', the last point of S lies on the ate .4 1 Let E 1 = E 1 (c) be the event that there is an infinite open path in A leaping 51 from the arc outside S with co inside S and i.e an open path P = roo t co. for all j > such that non leaves S from the arc .4 1 ; see Figure 11

Figure I I Possible open paths Pt and Pa witnessing the events E t = (c) and Ea = E:dc) Usuallt, the bonds toe straight line segments, as in Pi but they need not be

The precise details of the definition are not that important: the -soft' arguments we shall present go through with many minor variants For example, we could consider the last time the whole path leaves S. even if this is MI a hood soul with vo S Fot 'nice' drawings of 'nice' lattices, a bond typically crosses S at most once, so the condition is essentially that vo vf crosses Ai Set fdc) = INEi(c)) and odc)

— f dc) = P(Ei(c)"),

y find define f and 91 similarly, using the dual lattice As aninfinite

142 Uniqueness of' the infinite open dusted' and critical probabilities open path starting inside S must leave S somewhere, U 1 E1 (5') is exactly the event E(S) The events Ea(S) are increasing, so then complements .E1 (S)" are decreasing Thus. by Harris's Lemma (Lennna 3 of Chap. ter 2), for any c we have P(E(8)1

P (Ei (c) c )=

E;(c)c) i=i

H

(c)

er_r

Front (9) it follows that gi(e)

1.17( c )


1 is immediate from Theorem 12: if this inequality does not hold, then there (Al But then 0(A; p) and is a p E (0, 1) with p > p ck.'(A) and 1 - p > O(A'; 1 - p) are strictly positive, contradicting Theorem 12 ❑ Theorem 13 includes the Harris-Kesten Theorem, Theorem 13 of Chapter 3, and Theorem 11 as special cases. It also applies to many other lattices, for example, all the Archimedean lattices shown in Figure 18 Turning to site percolation, Kesten [1982] pointed out that the site percolation models on certain pairs of graphs are related in a way that is analogous to the connection between bond percolation on a planar graph and its dual; he called such graphs matching pairs, and noted that any planar lattice matches some graph To see this, let. A be a planar lattice, and let A x be the graph on the same vertex set obtained from A by adding all diagonals to all faces. For example, if A = A 0 , then A'' = Ao The wool of Lemma 9 extends immediately to show that, for a suitably chosen 'rectangle' in A, whatever the states of the sites of 11 (A) = 1,7 (A" ), either there is an open A-crossing from the left to the t ight. or a closed A x -crossing front the top to the bottom: to obtain the picture corresponding to Figure 8, replace each site v of A with degree d by a 2d-gon that is black if v is open mid white if v is closed, and each f-sided face of A by a white f-gon If A is symmetric, then trivial modifications of the proof of Them ern 12 show that 01,(A) and 0 1 _ 0 (A X ) cannot both be strictly positive, while p-,5,(A) + p s,„(A ) < 1 is again immediate from 'A lenshiltov's Theorem, giving the following analogue of Theorem 13. Let A be a symmetric planar lattice. and let A' he the graph obtained from A by adding all diagonals to all faces of A. Then

Theorem 14.

K.(A) + p,s,(A x ) = 1

❑

As noted above, = A E , so Theorem 14 implies Theorem 10 Since every face of the tr hingular lattice T is a triangle, T X = T. so Theorem 1.1 implies Them ern 8 as well

5.4 Bond percolation

On

a lattice and ids dual

147

We conclude this section with an application of Theorem 12 to a weighted graph. Let (2 2 , p,., pr ) be the graph 22 , in which each hot izontal bond has weight p, and each vet deal bond weight pm Kesten [1982, R 82] showed that the 'cr itical line' for this model is given by px ±pg = 1. Theorem 15. Let 0(Th„ p„) denote llw probability that the origin is in an infinite open cluster in the independent bond percolation on (2 2 ,p,, p„) Fm 0 < p,,, pr < 1, we have 0(p„,,py )> 0 if and only if >1

Proof The planar dual of t he weighted graph A = (2 2 , p„, p„) is A' = (22 + (1/2,1/2), 1-6, t -p„), the usual dual of 2 2 with weight 1 -py on each horizontal bond, and weight 1- on each vertical bond Rotating and translating, A' is isomorphic to (Z 2 ,1 - p,, 1 - pil). Suppose first that O(p,,,p„) = 0 for sonic p„, p,, with p„, > 1, and fix 0 < < p„. and 0 < jig < pr with 11, > 1. By the weighted version of Menshiko y 's Theorem, the radius of the open cluster containing a given site of A' = (22,[4,p'9) decays exponentially. As 1 - p'y < Tit and I - < p'„, the same is tr ne in the dual, (22 + (1/2,1/2), 1 - 1 -1/J..) But then the probability that a large square has either an open horizontal crossing or an open thud vertical crossing tends to zero, contradicting Lemma 1 of Chapter :3 We have shown that the condition > 1 is sufficient for 0(p„,,p„) to be lion-zero To show that it is necessary, it suffices to show that p, 1 - p) = 0 for every 0 < p < 1 Since (22 , p. 1 - p) is symmetric and self-dual, this follows from Theorem 12. ❑

As it happens, one does not need Them ern 12 to prove Theorem 13; the proof of the Harris-Kesten Theorem given in this chapter adapts immediately Indeed, suppose that O(p, 1 - p) > 0 Considering a large square S, it follows from the 'four th-root' trick that there is some side of S from which an infinite open path leaves with high probability Of course, the same holds for the opposite side The dual weighted lattice is isomorphic to the original lattice rotated through 90 degrees, so infinite open dual paths leave the remaining two sides of 5' with high probability, and one can complete the proof as before. In the next section we shall apply Theo/ em 12 to prove a mot e difficult result, that a cer talc analogue of Theorem 15 holds for the triangular lattice

118 Uniqueness of the infinite! open (A ide, and e l itical prohabiltlics 5.5 The star-delta transformation Sykes and Essam 119631 noticed a second connection between bond petcolation on the hexagonal and ttiangulat lattices H and T. other than that given by duality This connection involves the star-thavilln formation 01 star-delta transformation, a basic transformation in the theory O f electrical networks To describe this, let G I and Gi be the two graphs shown in Figure lt Suppose that the bonds of G I ate open eit

y

Figure Id A triangle, G I and a star, and

02,

with the same :at lac:lune ' sites

independently with probability p i , and those of GO with p t obabilit■ ya. In either glaph, there rue five possibilities fot which sites among 1,r, y, :1 me connected to each other by open paths: all thtee may be connected, none 111M , he connected and some pair may be connected to each other but not to the third In other wo t ds, the par tition of fr. y, cl induced {{,r}, f lb ca one In tety subgiaph 01 0 1 Of of 02 is {{:r. {c}} These cases have the of the three pat titions isomorphic to {{,r, probabilities shown below: pairs connected probability in C I ;ill none

+ 3/4(1 — pi) (l — / 03 Pr ( 1 — Pt )2

probability in at

— p2/ 3 ± 3/12( 1 — P2 p4(1 — Ai)

Serendipitously, there is a solution to the duce equations suggested by the table above, i e., to Pi

+ 3/4(1 — p i ) = ( 1 —

/ 3 =

Pi — P0 2 =

P2,

— p2Fi IIp2 (1

p ) 2 . and

(17)

P_( 1— P2)

i i Substituting p = 1—p 1Indeed, the last equation is satisfied whenever into either of the first two equations gives t he stare equation. iii

—$pi=1=(l.

5 5 The star-delta tionsjhrination



1-19

This e quation has a unique solution in (0.1 ), nameh pp = 2 sin(7/18) = 0 3172 Let a, i 1,2, be the random (open) subigraph of 65 obtained by selecting each bond independently with probability pi . whew p i = pc, and pc = 1 – pa As all Hum equations in (17) me satisfied, the random graphs 0 1 and a, are equivalent with respect to the sites 11, and z: We may couple 0 1 and a so that exactly the same pairs of sites front { Ti m ate connected in 0 as in 0 In the context of independent bond percolation, each bond of a (usually infinite) graph is open independently with a certain probability, which we may think of as a weight The obser ations above mean that it a (finite ca infinite) weighted graph A has G as a subgraph, with bond weights pu, then We may replace G I by 6'0, with weights 1–pp. Ignoring the Internal' site of this operation does not change the distribution of open clusters This is a simple example of the 'substitution method' that we shall return to in the next chapter Using the star-triangle transformation, it is easy to deduce bow TheOFC111 11 that p(1'(T) = pp This result was derived by Sykes and Essan [1963; 1961] without rigorous moof. \Vicuna' [1981] gave the first rigorous moot, based on the star-triangle transformation and a RussoSeyniour-AVelsh type theorem

Theorem 16. Let T be the triangular lattice in the plane.and FI the hexagonal oi honeycomb lattice Then p(1 .'(T)= 2 sin(–/18) and

p{I VI) = 1 – 2sin(rr/18) Proof As before. it follows nom Nlenshikov's Themem that the two critical probabilities associated to each lattice are equal, so it is legitimate to write pr:' for their common value. Let H' he the graph obtained by replacing every second triangle T by a star with the same attachment sites; see Figure 15 Then H' is isomorphic to /1; we shall keep the notation separate to indicate the different relationships to 71, reserving H lot the planar dual of T. Infonnidly, bond percolation on I with parameter po is equivalent to bond percolation on H' with parameter 1 – po, so both ate supercritical. both me

CSN

150 Unigaeness of Ike infinite open. cluslet and critical probabilities

WAWA WA res WA' a A SS WA TA Tata Figure 15 The triangular lattice 7, and the graph H' 01) each downward pointing face of 7' by a star. The sites o (solid and hollow); its bonds ate the dashed lines.

red by replacing are the circles

01 both are critical By Theorem 11, it Follows that both are critical, giving the result More formally, by a (10M(1411 in or in H', we shall mean one of the triangles to which we applied the star-triangle transformation, or the resulting star in .11 ' Consider the probability measures 111 ,1 mi in which bonds of T me open independently with probability pa, and nu, in which bonds of H' are open independentl y with probability 1 — Po. We have shown above that the restrictions of these measures to a single domain D may be coupled so that the same attachment sites of D are joined by open paths within D in the two measures We may extend the coupling to all domains simultaneously by independence. Any path in T . or, in H' between two sites of T may be split into a sequence of paths P1 within domains D i , with the ends of Pi being attachment sites of Di. It follows that, under our coupling, two sites y of T are joined by an open path in T if and only if they are joined by an open path in H'. Recalling that 0 is a site of T, let Co be the open cluster of T containing 0, and let C') be the open cluster of H' containing 0 Under our coupling, we have (2() n V(T) = Co. As Cf, is a connected subgraph of H' and sites of V(H') \ V(T) me joined only to sites in V(T), which have degree three, we have 'Cid < 41C1-, n 1 ,1 (T)I Thus

subevitic

C 10) 1

I

411C01

holds always in the given coupling, so 111. 0co

l

n) POC(d

POCH ^



151

5.5 The situ-delta leeinsfortnatimi

for every a Letting n — we see that 0(H';1 - po) = 0(T; po) In other words, as H' is isomor phic to the hexagonal lattice H. we have 0(11; 1 -po) = 0(T; po) In proving Theorem 11, we showed that, for any p, at most one of 0(11; 1 - p) and 0(Thp) can be strictly positive Thus, = 0,

0 ( 11 ; 1 — Po) = O CT ;

which gives p,i (T)> po and pcl.'(1.)> 1 - po Since p (1 )(T)+ plc)(11) = 1 ❑ by Theorem 11, it follows that; Al' (T) = po and 144.11) = 1 - po. Let us summarize what the results above, the Harr is-Kesten Theo/ern. Theorem 8 and Theorem 16, tell us about the critical probabilities associated to the three regular planar lattices. Theorem 17. Poi the aware lattice V. art have mi ./ (E 2) = 1 /2. fin the trianyalai lattice T. K(T)

= 1/2 wad 1 . (T) = 9 n(d/18),

and lot Mc he:rayon& of honeycomb lattice TT ple'(H) = 1 - 2 sine,T/18)

In a sense, the summary above is a little misleading: for these three lattices, four of the six critical probabilities are known exactly. but there are very few other natural lattices km which even one critical prohabilit\ is known exactly Tire observation of Sykes and Essar [1903 .1 concerning the star-delta transformation is a little more general: let G I be a triangle in which the bonds have probabilities p„, ph and pc, of being open, and G, a star in which the corresponding (i.e., opposite) bonds have probabilities r,,, and ra of being open Then G i and Go are equivalent if and only if 1'1)0

- r n

=

k



(1.8)

lot fi. j.1.1 = {a. 0, el Pa

pap,: + (1 - P)PbPc ±

pa ( 1 POP



p„ph(i — )

/ a/ /

and (1 - p)(1 -pb )(1 -pc ) = (1 - /„)(1 -10(1- le)* l„(1-ts)(1 -t„)+(1-1„),/,(1---/J ±(1 -t„)(1-

152 Unignenciis of the infinite open elimle, and critical probabilities The equations (18) ate satisfied by taking then froth the remaining equations reduce to

= 1 —

p„pem — pa — ye — pc + I = 0

for each i and (19)

This more general ,tai-triangle transformation was used by Sykes and Essarn 119041 to study percolation on a ti iangular lattice in which the states of the bonds ate independent., but the probability that a bond is Open depends on its orientation They derive (normigoirmsl y ) equation (19) lot the critical surface' in this three-parameter model Using Theorem 13 in place of [Theorem 11, the proof Of Theorem In given above adapts immediately to this weighted model, to give the Following result = fir . p„. y z ) be the weighted le iangulal lattice in which bonds in the three directions have weights Th. IL, and fi, sprelively when.: 0 < pr . p„. fi, < I Let 0(),„, Th) be the mailability that the oe igin is in an infinite open citadel in the independent bond maculation model corresponding /o A Then 0(p,. y y . p„)> 0 if and otilg

Theorem 18. Let

+ P y P P ;PO) z > 1

❑

This result g as Lamed by Grim/nett [1999[.. using ideas of Kristen [1982; 1988] In the light of VIenslnkm i s Thement, the hind pint is to show that percolation does not ()Celli when ys p„ T p, — pr y„p, < Kristen [1982] deduced this result Flout a theorem that he stated without proof A N'elSiOn of this liniment that is in litany was !note general was later proved by Gandolfi, Keane and Russo L19881, but then result assumes synunetiv untie/ reflections in the coordinate axes, which this model does not have. A star-delta tt ansfot mat ion 'elated to that discussed above is important in the them, of electrical networks. where it has a much longer p iston The operation on the graph is the same. but the weights (resistances) t i anslo t aiireleutiv. to satisfy the different notion of equivalence (that the ' espouse i . net cadent at each attachment vertex. to each input. i e. set of potentials at the attachment vertices. is the same) 6n electrical networks. it hums out that even sla t is etpthalent to kt triangle. and vice versa: see Bollobris [1998. pp .13 -H] Rental kablv. Al t man 119811 w as able to use the star—triangle tianslotmat ion with unequal edge weights to obtain t he exact critical probability

5 5 111P, si tu-della Emusformation

153

Cot a certain lattice, where each bond is open with the same probability Let ,5 1e be the square lattice with one diagonal added to every second lace, shown (rotated) on the left of Figure 16, together with its dual D

Figure lb 1 he lattice S obtained bow the swami Ire t ice by ridding a diagonal to ever} other face, shown on the left (solid circles and lines) toget hen wit its dual D (dashed lines and hollow cir cies) Fru clarity is draw n sepalateIN on the light Let S' he the lattice shown on the Hi in Figure IT below, obtained Inuit S by replacing each of the diagonal bonds b t a double bond Then SI-. with tinily bond open with probabilit y p. is equivalent to 5' ' . with the otiginal bonds of S t open with probnbilitq p, and the new bonds open with probability =1 — (I — p)' /.1 (As usual. the states of different bonds are independent ) Appl y ing the star—triangle t t anstot motion. one obtains t he lattice I)' bu nted hour D by subdividing ce t lain howls: see Figure 17

Figure IT certain bonds of 5'' arc replaced by double bonds, and the tliangle-star t t ansibi mat ion is applied as shown. the resulting lattice is I) %vit h certain bonds subdivided Taking the undivided bonds to be open with probability q = I — p. and the divided bonds with probability q = q = I — p'. we see

151 Uniqueness of the infinite open cluster and critical probabilities that D' is equivalent to D, with every bond open with probability q. The conditions for equivalence in the star-triangle transformation are satisfied provided (19) holds with p„ = = p, P c= p' This condition reduces to — p — 6p2 + 6p3 — ps (20) Using these transformations and arguing as for the hexagonal and triangular lattices above,\Vierman deduces that p!),(S ÷ ) = 0.101518 a loot of (20). An Architnedean Ionia is a tiling of the plane by regular polygons which all vertices are equivalent, i e the autornorphism group of the tiling acts transitively on the vertices. The square, triangular and hexagonal lattices are all Archimedean, the lattice 5' and its dual are not The complete set of Archimedean lattices is shown in Figure 18, The notation, which is that of Griinbaurn and Shep pard [19871, is selfexplanatory: it gives US the orders of the faces when we go round a vertex At this point we have essentially exhausted the list of Archimedean haft:CS Pi exact critical probability is known; there are two further examples that may be easily derived from those above. Let K be the kagona'r lattice. shown in Figure 18 Then K is the lice-graph of the hone y comb II, so we have iu

gdIC), p(1,(H),----

1 — 2 sin( 7/18)

Also, let K' be the (3, 12 2 ) or extended Kagoind lattice shown in Figure 18. Then IC + is the line graph of the lattice H2 obtained by subdividing each bond of II exactly once As noted in Chapter 1, the relation fki (11(i), p!(H) I12 is immediate: an open bond in the subdivided graph is only 'useful' if its partner bond is also open. Thus, as noted by Sliding and Ziff [19991, among others, fise (K), pcIVI2 ) = p A l H) 1/2

(1 — 2 sin(rT/18)) L/2

(21)

In the next chapter we shall review some of the upper and lower bounds for the critical probabilities of Archimedean lattices.

5.5 The stun-delta hamsformahmt

•• •• •• •• •• •• •■■ ■ • ■ • SUOMI

•■■ ■ •■■ •

•■■ ■ •■■ • Sq nu ):



tome: (3,

(a,6, 12)

4•1■•■•• a • •^ • • • • a • • • • 0 11 0 (3' .1-

•A 114 •A ............A .41 y

Triangulm: (36)

3,6)

155

extended Kagon

Hexagonal: (63)

(I. 82)

(3, -1, 6, -1)

1

ally A A w IS • • A•sffa Alwarar A SAY AT • A

A A 111a • FaVa •r a•Al • rA A Va• A ValrAwA VANS • ATAYAT VA

,6)

-1)

(3 6)

Figure 18 The 11 Archimedeatt lattices, i e filings of the plane with regular convex polygons in which all vertices ate equivalent The notation for the unnamed lattices is that of Griinbainn and Shep pard (1987) 10 of the lattices me equivalent under 'oration and translation to then 'Milo' images The final lattice. (31 ,6), is not, and is shown in hot h forms

6 Estimating critical probabilities

In genet al. there is no hope of dele l mining the exact critical probabilities /OA) and pl,;(A) for a general graph A. even if A is a planar Iant ice Neverthel ess, there a l e mane ways of proving rigorous bounds on these critical probabilities In this chapter we shall describe sever al of these. soil ing with the substitution inclhod, a special case of which we saw in the previous chaPlel

6.1 The substitution method Tu describe the substitution method, we shall use the fenninolug i, of wcightcd graphs: all our graphs will haw a weight p, associated to each bond e. with 0 < < t We shall consider independent bond percolation Olt a weighted graph (A. p), whew each bond e of A is open with probability p, independent Iv of the ot het bonds. We often suppress t he weights in the notation. A weighted graph (G. p) with a specified set A of attachment sites generates a 'widow partition n of A: two sites in A are in the save class of El if they rue joined ln an open path in G As in the precious chapter. we SUN" that ( WO weighted graphs G I . C, with the saute set .1 of attachment sites ate equivalent if the associated I andoin partitions Il i II, have the same distribution. i e if the col lespondiug pet colat ion measures can be coupled so that IT = 11 2 always holds In general. exact equivalence is too much to hope for Let us saw that a partition 7ir. , of A is counsel than "Th. and mite 71 . ) > if anNr two sites of A that ale in the saute class in 7 1 ar e also in the same class in79 In other words. r, is (ionise' than r i if and in this context. a coarser partition is onl y if Ti is a 'ohne/new of 'better'. as it will correspond to more connections in the percolation model: this is the reason for our notation 'We say that a weighted

G.1 The substitution method

157

gr aph CI, is shott ipo than Cr, and WI > il rte DIM couple the

co l I esponding percolation measures so that II is alwa y s coarser than H I In this case Flo stochastically dominates FT,. Note that C I and Ga t ue equivalent it and onl ■ if each is stronger than the other Let A, and Aa he two infinite weighted graphs Suppose that A l may be decomposed into edge-disjoint domains Dr j . i = 1.2, . each having a specified set A j of attachment sites We assume that each D I is a suligmpli of Ac that (l i cit union is A t and that t. WO dOtliltillti 1118V meet onl y in sites that are at tachment sites of both Typically the graphs Du me all isomorphic Suppose that Aa has a decomposition into domains hew each Du has the same attachment sites as D i \\: e have seen an example of such a decomposition ;heady. ill connection with the sum-triangle transformation Indeed wi le/ i ing to Figure 15 of Chaplet 5. we IWIV take A l to be the tiitungular lattice. A, the hexagonal lattice fr. the domains D I j to he ever y second triangle in A l and each Du to he the cot tesponding star in A . , Aiguing as in the pool of Theorem 16 of Chaptet 5. since am, path bout one domain to ;mottle:: must pass Huough an attachment site. the onl y inopei It of D I ; that is teleiant tot percolation on A i is the induced partition 01 the set TI; Hence. if is equivalent to 171 j fin eve/ then petcolat ion occurs 011 A l if and (t h if it Deems on Am fin any fixed attachment site the percolation probabilities O(A j : 3 ) ;o ld 0(Aa: .r) ate exactly equal then 0(Aam ) > 0(A 1 : ./.): one Similarly. if Du > tot every call couple the percolation ;leashes on A, and A, so that any pair of at tachment sites that a t e joined in A i ale also joined ill A, Tins tact allows its to let iVe i elat ionsltips between the critical probabilities of two lattices: if the ethical mobahility of one is known then we can hound the critical p l ohabilit ■ of the other This technique is known as Hie substitution method. and is due to Vieunan 119901 At frost sight. it is not cleat hots one can tell whether a given weighted graph is stronger than another, but Hume is a simple algorithm An upset ill the partition lattice on a set A is simpl y a set U of partitions of E U and 71 > 71. then ira E Ll It is not too A such that whenever haul to show that > (7, if and onl y if. fo i even up-set. we have E > P ( H, E U): in fact.. this is an eas y consequence of Halls p 77j) hus, the 'Matching Theorem [1935] (see also Bollolnis (1995. condition CD > G' t is equivalent to a finite set of pol nomial inequalities on the weights of the bowls Let its illustrate this with a simple CNillipie l (it ) = = p!"(10 giving hounds On the critical probability p T the whew It is the Anytime lattice shown on the light of Figure 1

Eqimatilly dim( probabilities

158

Figtue 1 A two-step transformation 11'0111 the hexagonal lattice /I to the liagomet latt ice K: first subdivide the bonds of to obtain the lattice ifs shown on the left. Then apply the star-triangle transformation to the original sites of /1 (middle figure) The result is the kagotn6 lattice (right-hand figure). notation of atiinbaum and Shephatd [19871 for Ai chimedean lattices, is the (3,6,3,6) lattice. Ottavi [19791 noticed that the Kagoin6 lattice may be obtained horn the honeycomb or hexagonal lattice H by fist subdividing every bond, and then applying the stat-thangle transformation to the (non edgedisjoint) stay s centred at the original sites of H see Figure 1 (A mote inimitite sequence of stair ' equivalent ' graphs was shown in Figure 6 of Chaptet 1.) Let Ss denote the sla t with attachment sites {,r, y, :di in which each bond has height s, and Tr the triangle on {:v,:y,s} in tvhich each bond has weight t We have seen that 8, is equivalent to T, if and only if t po and s 1 — po, where po = 2 sin(x/18) -We would like to know lot which pails (.s, t) we have > Th and lot which pans we have Ss U and T, > then 0(K:1)> U As o(it,;s) > 0 fin any s > so.. we thus ha‘ e mi,)(K) < int

: T, >

for sonic s > so

inl ft :

Ss

p(1.)(k) > slug/ : 8, 0 > Tel

Solving the simple polynomial inequalities (1) fort with s = so, method gives the bounds U 51822

this

< pcl ."(K) < 0 5-1128

These inequalities and the proof we have just given are due to Wierman [1990] In the same paper, lie obtains the stronger upper bound pl:(K) < 0.5:335 by considering a larger substitution - replacing the union of two adjacent: stars in HO by the union of the co/responding triangles in IC In this case there are lour attachment sites, so the partition

9

1(i9

Estimating t tlical 1)101)001,1u s

lattice is now complicated 13% using huge, and lilt get substitutions. hot ten and bet k g bounds m i t\ l i e ()With/v(1 1-lowc) \ el. the calculations quickly become Wiwi/Oita' ii C111110(1 0111 ill a 11111\ C \la \ Using yxtions methods of simplifying the eill(ititat and it sttl/stitittion \\ it h six attachment sites. \Vic/in/an 120031)] shat wined these 'mitt/cis considrilabl Theorem 1, Let Ii ix the ha q Dm( o, (3.6.3.6) lattice Then 5299




is fin some PluP o s es. anueeessaliin strong. Suppose that we Wee a \\ eight ed aph A l with as a subs' aph (joined out \ at the attachment sites). mud we replace S, bt 1, to Obtain A9 Nle would like condition's

on s and t that allow us 10 deduce that 00/C01/1110/1 is mo l e likel y in Al than in A.1 Mote wriciseh we should like conditions that cosine that the went {0 — that a particular site (the might) is in an infinite open (lusty ' . is at least as likel y in A l as ill A . ) lVe write Ss :7-- 7 / it this holds Ion all /Wits of weighted graphs A.)) /elated in the win we have described. Oita\ i 119791 round the set of pai l s (s. ) lie which 5, 1– .Ej To present this t esult let us Ws( decide the states of the bowls in E the set of all bonds of A I outside 5. Note that E is also the set of bonds ()I A . ) outside T, Eon each attachment site p E u there 111/0 1/1 Wilt not be 1111 open path limn 0 to 1' in A I \ S s . and there nia \ 01 /100,' not he an infinite open (inst il ()I A i \ meeting ° diet "°° 111 111C ( " 1/18 {{) } and { } 11111% of m i n not hold \ Chen the states of the bonds in E. the conditional probability that — x depends owl \\ Melt of the events and — hold in A i \ 5. Indeed. tl — x in A t if awl only it thew ale sites el. E dial are connected within .5 – with 0 — and x in A i \ Hew. e l it.) is allowed Much of the time, this conditional wobabilit \ is 0 (it 1: it 0— v and u rot some e E z} then 0 — x even if all bonds in S„ me closed Similarl y if 0 f r fia all 1' e f.r C }. on r f x for all then 0 — x cannot hold. whatwei the states of the hoods in 5, This leaves only One() non-t thin! cases: we must have an attachment site di sat, with — :0 and a/101w/ site tr. tin) with 11)101 the Hind site, we roan 1000 0 — x. 01 0 71-^ c 7 x In the last case. — x in A t if and only it ,r and it me connected in .5, II 0 —c and x. then 0 — x A l if and 0111V if one of and : is connected to

inl

1 The substitution method

y Slunk)/Iv if d sc., then 0 — sic in A i if turd only if one of irf and c is connected to :J . in S, The relevant events in S s have respective (s) ± ( fa( ± 2 f (s) and .f:Ws) + 2 fi(s) Pwilmbililies lane shown that PA, ,X) is a weighted awlage of the quantities 1. 0, 1 3 (s) + h(s), and nr(s) + 2[ 1 (s), where the weights depend sc) is a weighted average of 0, on A t \ S. Fut Him more s P A2 (0 weights, determined by y3 (t) -I- in (1). and g3 (t)-1- 2g i (f) with the sonic \IT/ A i \S, Tints S s bolds mecisciv when the two inequalities

E(s) ± [ 1 (s) ^ 00= !EU) and / 3 (s) + 2f ( )

113(0

20 1 (1)

(2)

hold. i . when (I) holds tot = I and j = 2 This is a much tveaket condition than S s > Tt flu i trivetse relation Tr S. holds pun Wed the revel se inequalities to (2) hold so is the (*initial probabilits for [1,, and I = Ottayi showed that il s 0 52803. then E >- so the sultglaph S s of run weighted gin/tit A may be teplaced and the probabilit y that a given site is in an infinite cluster ( 0 1 that a given pair of sites ale connected by an open path) will not decrease It might seem that the inequalit y p(1 9 fi) < 0 52893 follows easily: un101 I mutt els. this is not the case As the Wit III al at gement is so close to working let us ex tunine it in detail, to see where it fails One would hope that. if T .5.. then percolation is at least as likely in a graph Bunted In gluing together copies of 1st at tlwit illlath/nem sites. as in the col responding graph obtained from copies of 5, However. the relation 8- S s allows its to replace (me cop%of Tj lw a copy of S, in an 'outside wank w [licit is the same beton . and after the substitution. but it does not allow its to conti nu e and !vitiate a second copy After enlacing the first cop y. the outside gl aphs E l and Eg ate different: in Eg Nu , have ahead -■ replaced a cop y of S s by 7r, One might hope that. as IT, is 'bet lei' than E l them is no teal problem But what does 'better' mean? ft could mean that an y connection in the outside graph E i between an at tachment site of mu second substitution and ft or isc is also present in Es Until we have looked at the second substitution. we do not know which connections we will tric l inic., so we should impose the condition that the first substitution preset yes all such connections (while pet Imps adding new ones) ilhe condition Tr S. does not allow us to do this only to p i esti' NC a connection chosen in advance The arguments above illustrate the power of the notion of stochastic domination: if I) > S s then it is ye t y cans v to prove that we nun replace as main copies of S s by copiris of Tt as we like I In ke y to the application

Estimating critical probabilities

162

of the substitution method is to find suitable weighted graphs an with Gr > ay. \Viet/Ilan 12002] used the substitution method to obtain bounds on pit i(A) for other Archimedean lattices A, obtaining the following result. Theorem 2, Antony the Atchimedean lattices A, the extended I< againá, 0013i70/2CS p!?(A), with 07 (3.12 2 ). lattice 0.7385 < pci Vii+ ) < 0 7-119 Although we have described the substitution method only lot bond percolation, essentially the same method can be used to studs site percolation. For example, \Vietnm [1095[ adapted Ids method to obtain an upper bound on p-cs,(7.9) Theorem 3. The et itical probability pi,(22 ) for site pet ()lotion on the S01107V lattice satisfies pis.(22 ) < 0 679+192 Po" a list of other rigorous bounds on the ccritical probabilities for lattices. see Wiennan and Pm viainen [2003]

6,2 Comparison with dependent percolation Another method of bounding critical probabilities is implicit in the use of dependent percolation in Chapter 3. Let us Wrist] ate thus by giving an upper bound fill icii(Z 2 ) = M i (22 ) = 14(22 ); we shall mention a much more sophistica.ted version' of this idea later. As usual, we write = P_:. t,for the independent site percolation measure on 2 2 , where each site is open with probability p; we denote this model by M. Let be a parameter to be chosen later, and partition the yet tex set of Z2 into C by ( squares S,„. e E Z2 Thus, ha v = (a b), S„ =

= Ur, y) E 2,72 : at 5 < (a +1)1', Itt < y < (b +

The set of squares St has the structure of Z2 in a natural sense To make use of this, for each bond e = u e of Z2 , let R = S„ US,.. so let is a 2t by (or t by 20 rectangle. If e and are vertex-disjoint, then the rectangles R.„ and R./. are vertex-disjoint Thus, if we define a bond percolation model M on 22 in which the state of a bond e E E(22 ) is determined 1w the states of the sites in Rt . then the associated probability measure on 2 E(:2) will be 1-independent. The idea is to define IC/ so that

6.2 Comparison with dependent percolation

163

an infinite path in Al guarantees an infinite path in tare original site percolation Al We have seen a way of doing this in Chapter 3, using 3 by 1 rectangles for clarity in the figures We use the same idea here Recall that 11(R) denotes the events that a given rectangle R is crossed horizontally by an open path, and 17 (R) the event that I? is crossed vertically by an open path. For a horizontal bond e = ((a, b). (a + 1, b)) of Z2 . let (AR() be the event Win b)) illustrated in Figure 2. For a vet tical bond

Figure 2 The figure on the left shows open paths guaranteWng that t he event

0(10 holds, where c = ((a. b), (a + l.b)) is a horizontal bond of 2: 2 ; the sites shown are open 'The corresponding event lot a vet deal bond c is shown schematically on the right.

(0, b + 1)) O r 22 , let 0(11%) = 1 .7 (R,.) n H(SR„ 0 ) in either case, let e be open in Al d and only if 0(RJ holds Since hor izontal and vertical crossings of the same square must meet, if there is an infinite open path in A/ then there is an infinite open cluster in Al Note that the probability Pp (G(Rt .)) is the same for all bonds c Let p i = p i (Z2 ) be the infimum of the set of p such that, in any 1independent bond percolation measure on 7G2 in which each bond is open with probability at least p, the origin is in an infinite open cluster with positive probability. As we saw in Chapter 3, it is very easy to show that p i < 1 Here, the value of p i is important; we shall use the result of Balistet Bollobris and \Valte i s [2005] that a t < (1 8639; see Lemma 15 of Chapter 3 Suppose that for some parameters s and p we can show that = ((a, 6),

lt(P)=Pp(Cl(R.,)) > Then lTr

is a 1-independent measure on 72 2 in which each bond is open

Estimating .

161

'cal probabilities

so with positive Limitability there is au with some probabilit y p > and hence in II/ 'Thus p> p11(Z11) Al infinite open path in Suppose that p > p11(Z 2 ) is fixed Then, by 'Theorem 10 of Chapter 5, we have 1 — p < t ,11(Am), where A D is the square lattice with both diagonals added to every face Exponential decay of closed clustets A D follows by Menshiliovs Trhecneum so the probability that a 2( by t rectangle R is crossed the shott way by a closed A D -path tends to zero Hence, by Lemma 9 of Chapter 5, the probability that R as ( is crossed the long YVilV by au open path in An tends to 1 It follows that i f (p)--(• 1, so thew is sonic 1 Stith f hat li (p)> CI 8639 > Thus,

in plinciple. Hie method above gives arbittarilr good upper bounds: for each (', find the minimal rot an almost mini/nap value of p, of p such that heir) > 0 8639. Then each p, is an tippet bound on p11(2, 2 )„ and the sequence p, converges to pi1(Z 2 ) boo, above. ft is easy to check that, 16p6 q2 +8p7 q 1 p s , \\dune q = 1 — p, giving the bourn! ',f p ) = 6p 5 M 1 p11(Z) < 0 8798 With a computer, it is eas y to show that fa(p)

117/tq l ° ±1399p9 q9 1737p lugs -1027p 11 q 7 = 5-166p12q°

+ 1527p 1 "q 1.'

2335/ J il t/ I - I - 7571, 15 q3 153 /) 16 ,12 -F 18p ri + pis

giving p1(21: 2 ) < 0. 815 An exact (ruination of 1,(p) gives /.)`,(II32) < 0 817 Linlin unat el ■ t he st r night il/Wald met hod of evaluating rpr mnd count inglia which of t he 2' 2 configut at ions in a 2l In 1 rectangle 11), the event 0(11, 1 holds, quickh becomes imm act lent. at mound I = 5 \\Ten the subst it Mimi method can be applied. it tends to give beton bounds for the same computational shirt Note, hotveier. Hurt the method described hew is much more robust Fin example. fi l eg1 1 1e/Iiii( IS in die graph Me riot a ploblem plodded we can show hat 12'1,(G(1, )") > i > pr ha every bond c one can perform different substitutions in diffetent pails of the gtaph the substitutions lune to fit together exact) to give the stt ucHne of a graph with known critical probability This will often not be possible Another very imputeant adiant age of the 1-independent approach is that, if one is p l epa l ed to accept an ewer probabilit y of 1 in a million. say. or 1 in a billion. then good bounds can easil y he obtained b y this method Indeed. there is a yeti- easy wa y to estimate the numerical configtuations in a 2( by value of E(p) very plecisek: generate rectangle I?, at random using the ['wasn't , and count the number .I/ of configurations 1hr which 0(1?,) holds. The random number Al has p) so if A is huge. a binomial dishHa t tion with parameters A . and

n',2 Commi t Cion with dependent percolation

165

then with lel \ high probability 11//A will he close to /r(p) The y IreV point is that one can boned the error probabilit y. ercu without knowing ir (Pr Given any > 0, one Call give a simple procedure for producing upper. bounds on /OE') ry ltich provably has probability' at most E 0[ giving an incorrect bound: foist (by trial and 0110 1 . or guesswork) decide on parameters I Will p for which )r(p) seems likely to he huge enough Then calculate numbers N and Ma such that TdA. > J1,(0 ) < whew Bi(N. 0 8639) Then generate A samples as ;drove. m i d if Al > 1110 o f t hese have the propert y Ci(g). assert pe r r....- 1 p as a bound IL in fact 0,(23 2 ) > p. then h(p) < 0 8639. and t he probability that the sampling procedure generates at least .3/0 successful trials is at most This method works ver y well in 0010 and the dependence ()I the I mining time on is ver y modest For eXa/111A0. using the parameters = 10 000. p = 0 591, N = 1000 and .110 = 935 with a moderate computational efhat we obtain the hound i t,".( 2 ) < 0 591 with all 19 lot plObabilit \ Of E < 10-12 Similar Iv, 161 the matching lattice AE with vet I CX set out bonds between each pair of vertices at Euclidean distance I or v' 71. using the pat ;mulcts l = 20 000. p 408. N = 1000 and A' = 935 we obtain K( A 3) < -1(18 w" a c""lidellue of1 1g-12 Together these bounds give /):1( ) E 10 592.0 5911 with extremely- high confidence. A little 'note computational effort (1= 80.(100. N = 1000. Itltt = 915) gives the result K(Z 2 ) E [(1.592 0593] with 99 9999cA confidence Note that we cannot: sat- that the probability that pr(r 2 ) lies in the inlet val (0.592,0 593] is at least 99 9999%: the critical probability is not a random < / mutt:it-v. so this statement either holds cm it does not In the language of statistics. we have given a confidence interval for the lillk110111/ deftlillillititie quantity ir,'.(Z2 ) As described, the procedure produces a confidence interval whose upper limit is either the bound 0 593. sm. that we aim for. or infinit y, if Ill < 21.4) Narrow confidence intervals for mars or her critical probabilities have been obtained by .Rionlan and \Valters [2006]; Ica example, p;1(11) E [0 6965..0 6975] with 99 9999 1X confidence. where is the hexagonal lattice As usual in statistics. we must be a little careful in generating confidence intervals: Ito example it is not legitimate to per fornt various runs

166

Ethmaling critical probabilities

of the sampling procedure with various parameters, and only use one result In practice, this is not a problem for two reasons: finitly we can perform as many • durnuty' Inns as we like to get an idea of parameters that are very likely to work, and then one teal run with these parameters Also, it is easy to get a failure probability of 10- 9 , say, for each run It is then legitimate to perform 1000 runs with different parameters, and take the best bounds obtained: as long as the probability that each run gives an incorrect bound is at most 10 -9 , the final bounds obtained still give a. 99.9999% confidence inter val for the critical probability. There is another pitfall to bear in mind when implementing the probabilistic procedure described above: we have implicitly assumed that a source of random numbers is available.. In practice, for this kind of sinndation one usualli uses a pseudo-random number generator Not all the widel y used ones are sufficiently good lb/ this pm pose Indeed, the current standard generator I t anclourfl used with the programming language C is not For example, using this generator we obtained estimates for 1' 10 (0 731) of 0 8661 ±0.000 2 and 0 8631+0 0002, depending on the order in which the random states of the sites were assigned (The turc:er tainties given ire t wo standar d deviations ) Using the much better I Merserme Twister' generator MT19337 WI ititen by Matsumoto and Nishimura, we obtain li o(0 131) = 0/8630 ± 0 0002 Note that this (presumably) true value is smaller that 0/8639, /i hile one of t he values obtained using I andom() is larger. in generating t he confidence intervals given above, we used the lersenne Twister. Of course, one can re-run the same procedure tr itlr a different generator. or with I n rte' candour numbers obtained horn, for example.. quantum noise in a diode The methods described above can be easily adapted to give good upper and lower bounds on p4!(.(1) and pII(A) for any of the Archimedean lattices A. 1701 lower bounds. to prove that percolation does not occur at a particular value of p one can use 1-independence as in the proof of exponential decay of the volume in Chapter 3. For another approach to lower bounds, let S„(0) be the set of sites at distance n 11010 the origin, and let N„ be the number of sites in 5„ (0) that may be reached from 0 by open paths using only sites within distance 11 of 0 Lemma 8 of Chapter 4 (or its equivalent, for bond percolation) states that if, for some n, we have 1/1/ ,(N„) < 1, then there is exponential decay of the radius of the open cluster Co, which certainly implies that p < pc . In am- lattice, for any p < Menshikov's 'Theorem implies that: a p (N„) — 0 as n ^ ob , so arbitrarily good bounds may in

6.3 Oriented percolation on 22 167 pr inciple be obtained in this way As before, exact calculations of Ep(Ar„) are impractical except for vet y small but estimates with rigorous error bounds may be obtained for larger re This observation is applicable to any finite-type graph, including any lattice in any dimension For bond percolation on a planar lattice A with inversion symmetry, to find a lower bound on plc' (A) = 14.11 (A) = pi (A), one may find an upper bound on pl„)(A*) and use Them ern 13 of Chapter 5, which states that p ,l (A)-t-p1JJA*) = I. Similarly, for site percolation, we may find an upper bound on p4(AX) and apply Theorem 14 of Chapter 5 This approach seems to give better results in practice The reason is that it is easier to estimate the probability of an event by sampling than to estimate the expectation of a random variable that, might in principle take quite large values occasionally

6.3 Oriented percolation on Z2 The study of oriented percolation, and, in panic/dar t bond percolation on the or iented graph 23-, is a major topic in its own right.; see Durrett [1984] for a stir vey of the early results in this area. iented percolation is in general much harder to work with than unoriented percolation, a fact that is reflected in the difficulty of obtaining good bounds on Z 2 ) Indeed, Durrett described the question' of finding a sequence of rigorous upper bounds on 1.21!1 ( Z 2 ) that decrease to the true value as an important open problem His bound, pi'( Z 2 ) < 0 84 , Was very fat from the tower bound of 0.6298 obtained by Mar [1982]. In contrast, for lower bounds, it is easy to produce hounds that do tend up to the true value Indeed, let N„ be the number of sites on the line x y = that may by reached from the origin. If Eih,(N„) < 1 for some n, then Lemma 8 of Chapter 4 shows that p < pfi ( Z 2 ) < Indeed, Hanumnsley [1957b] deduced that pl I (Z 2 ) > 0 5176 from the fact that Eb (No) - = 4p2 — 11 1 . Of course. by Nlenshikov's Theorem,Z -) = p!iFE 2 ) =1.4,'(72), say

There have been many Monte Car lo estimates of pN Z 2 ): such results are not our main focus, so we shall give only a few examples, rather than attempt a complete list: Kertósz and Vicsek [1980] gave the estimate pcb ( 2.,Y ) = 0 632 ± 0 004. Dinar and Bar ma [1981] reported p[ (G2) 0 6445 ± 0.0005, and Essam, Gultruann and De'Bell [1988] 0 644701 ± 0.000001, for example

(Z2) =

168

Eqinading et Theo! probabilities

Bali:der. Bollolgis and Stace y [1993: 199-1] gave Indict tippet bounds

ou p :l ( Z 2 ) using the basic sir ategv of comparison with l-independent percolation. Init ill a much mow complicated WaV than that described iu the previous section. Their approach does give a sequence of rigorous upper bounds tending down to the true value As Hie alguntent is lather involved. NiAi shall give onl y an outline \Alien consideting oriented percolation on

OW aim is to decide

tot which p Ha' percolation probabilit y O(p) 00 (p) is m e n-zero In doing so. we was of cows() restrict mu attention to that prat of 2)2 that Mak C011telVabh : LC l eached hoot the (nigh). namel y the positive

(radian (2 The basic idea is to use independent bond percolation on Q to define a le-Sealed I-independent bond percolation on Q To (I() this, We choose parameters b and h. and at 'fume rhombi with II+ I sites along the bottom-lett and top-tight shies as in Figure 3 Mow precisely

Pigmy: 3. Rhombi with sites along the base (bottom left)) and the top (tipper light) II each t hondtns is replaced b y all oriented edge and I he top and Lot too t of each rhombus be a \et tux the tesulting or iented pupil is isomm to 2 In realize the rhombi ate lagged : each contains exact IN :I sites ham each of t la: 9 la% epf of 772 that it meets

f

for : .r. y > ± q = for bagel wlft iug = Ei each site r iu L 1 we choose a set 5,, of consccutke sites in L h(t+. 1) . so that the sets Se ate disjoint Also. fin each bond 7 E Q we choose a (toughl y rhombic) subgtaph of Q in such a wry that

R R-

and P7

ate disjoint whenever 7 and f me bonds of Q. that do not sha l e a site \\ : e legnite that it e = UP then the bale of R T , defined in the nattu al 5„ , turd the top of 117 is 5, wr y( is exact

3 Oriented percolation On

V

169

Cit for the open out-clunk] of the i e., As before . WI it e open (oriented) the set of sites that may be reached fron t the might ( -) < 1 such that, paths. It is easy to show that there is a pu = for any l-independent bond percolation measure F on - iu which each > bond is open with probability at least p H we have P(rof the ingument is vets similto to the moot of Lemma in Chapter 3 Indeed, if Co is finite, then there is a dual cycle S surrounding the origin.. such that every oriented bond star Ling inside S and ending outside is closed As shown in Chapter 3 there me, ve tt- ctuclely, at most :te321-2 dual cycles S of length 2( surrounding the or igin For each, there is a set E of exactl y f oriented bonds starting inside S and ending outside Two horizontal bonds in E cannot share a site, and nor can two vertical C E consisting of fl/121vertex-disjoint bonds in E. so there is a set bonds The fill-probability that all bonds in ate closed is at most (1 — ) 1/2 so the expected number. of cy cles with the required property is at most

which is less than I if p i = 997. sa t Note that the expression above is exactl y the same as that appearing i l l the mool of Lemma ll of Chaplet 3, not b y coincidence By counting more car Mont the cycles .5' that ma t actually arise as the boundary of Co. as in Balister , 11n1 Stacey [19991, one can obtain a bet ter bound on pi If we define an event C:( U T- ) depending on the states of the bonds in fly so that an infinite or iented path of bonds 7 for which C(R 7 ) holds guar antees an infinite open oriented path in Q. and if. (or some p. we can show that P p (G(R T )) > pr for me t y(-T then it Follows easily that TT, i ( -) < p Howevet , it is not eas y to see how to define G(R 7 ): piecing together paths in this context is much harder than in the unot iented case. The actual atgm/rent of Balist et. Bollobfis and Stacey is notch more subtle Choosing the regions fly so that they are isomorphic to one another, the y seek a Wont—trivial up-set A C 'P(11/± II) with the Following proper ty: let i = 17 1 t, and consider the random set U of sites in S„ that may he leached nom the otigin b y open oriented paths. This set may be naturall y. identified with a subset of [b-f- I] Let V be the set or sites in S, that may be reached horn U br open paths in again this set may be identified with a subset of [b+ II The condition [conked is that.

Estimating el die& pmbabilitie5

1 70 for some p p

E (0,

l),

whenevet U E A, then 171,(1 .7

E A U) > p

(3)

There is a further restriction, that A be symmetric under the operation b + 1– I. so that the identifications described above may be made i consistently. Bollobas and Stacey [1993] show that, under these assump–'' lions, one can construct a re-scaled oriented bond per colation on Lin which each bond is open with probability at least p, and which is effectively 1-independent (Mole precisely, given the states of the bonds below sonic layer, the bonds in that layer ate 1-indepertdent..) They show that, if p > (3 – 07) )/2 and p > 1 – (1 – p) 2 , then such a process dominates the original independent process Using Menshikov's Theorem, -). one can then deduce that p> In summar y, if one can find a suitable region R., au up-set A, and a p > (3 – V17) )/2 such that (3) holds with p = 1 – (1 – p) 2 I'm all U E A, then p is ' in tippet bound fin p l!( Z 2 ) Note that this does not con espond to a direct construction of a 1-independent percolation model as in the unotiented case Such a direcl construction could be achieved by defining 0(1?7 ) so that if G(Rn" ) holds and U E A then V E A; indeed, one could take this as the definition of 0(11 7 ) But then the condition that 0(1 7-) ha t e probability at least p i amounts to Pp(U E

A

V E- A) >

which is infinitelt stronger than (3) with p In order to apply the method above, it seems that, we have to make a large rininiter of choices: lust, we have to choose a suitable legion P. E-4'.fP/t/n and a probability p Then, we must also choose one of the 2 possible up-sets to test Fmtunately, there is an ;lige/Ulm' given by Balistet, Bollob 'is and Stacey [19931 that, for a given I? and p, tests whether there is am] up-set A with the required property (and finds the maximal one if there is) Using this algorithm, together with the guluent s outlined above, they moved that p:( L 9-) < 0 6863. They also showed that. as in the case of man iented percolation, the method gives arbitraril y good bounds with sufficient computational effort This is a much harder result than the (almost trivial) equivalent for fluor iented percolation described in the previous section Using a more sophisticated version of the argument, involving a more complicated 'eduction how the dependent percolation to independent

6 3 at iented percolation Oil Z2 171 percolation. Balistei Bollobas and Stacey 1199-II obtained tare following hounds. Theorem 4. The critical piobabilitics for oriented bond and site percolation on the square lattice satisfy ( 72 %; ) pc1;'(27: 2 ) < 0 6735 and pj"(1

0, a' -F y = t.} The be reached from the origin by open paths set TIt of sites in L i that and depends only on and on the states of the bonds between L t Obi oriented bond percolation). 01 the states of the sites in L i (for site percolation) More explicitly. let us code R t by t he r-coordinates of the points it contains, setting A i =

: (r t t — :r) may be reached from 0 by an open path}

Then Ao = Fat bond percolation, given ./14 , the probability that :I! E = 2p—p2 according to whether lA t ofx-1, j is 0, = A,+1 is 0, 1 01 2. Furthermore, given A t , the events E A t+ , } are independent for different c;. For site percolation, the jiMarkov chain is the same, except that p i = p2 = p. Thus, it is natural to extend the definition of < 1 and 0 < pa < 1 the Maticov chain (A t ) to any pan/meters 0 < Note that this Markov chain has stationary transition probabilities: the distribution of 1 t t , given that = A is the same for any t. Of course, this Nlarkov chain underlies any analysis of ot hinted percolation; lot the approach described above, we did not need to define it explicitly Liggett j1995] !moved the following result. Theorem 5. If the parameters 1/2 <

< 1 and

mid ha satisfy the inequalities (1 — ) < p 2 2/3 For site percolation. whew p, the relations (-I) hold fin art's p > 3/1 Thus Liggett's pu = Ps mediatels implies the following bounds result in t Theorem 6. The (atheist probabilities lot minuted bond and site percolation on the square lattice satisfy

< 2/3 and p.( 2 ) G Given a finite set A C E. let denote the landont set obtained by minting the M iukov chain lot k steps stinting with the set A Thus, if A = {01, then .4" has the clistlibut ion (if A„ The basic idea of Ligget t's pool is as follows: we hope to define a Function on the finite subsets A of Z so that II(0) = I awl M I all A 0 lye have fl(A) < 1 and (5)

EgH( A I )) < H(A)

Of course whether such int I7 exists depends on the parameters p i and of the :\ farkov chain If such an 11 does exist, then from the Nla i km" opei tv awl (5) we have

po

E(B(. 0, by SA we mean the lattice obtained by scaling A by 6 about the origin Thus, for example, OZ 2 is the graph with vertex set {(So, : b E Z} in which two vertices at distance rS are joined by an edge Suppose that we have an assignment of states, open or closed, to the bonds or sites of SA If = (D; Po, P3, PI) is a 4-mar iced domain, then by an open crossing of D from. A i to A3 in SA, we mean an open path non r., t. in OA



Conpain& invaDaaer Smanov s Theorem

180

such that /7 1 , cr_i lie inside D 1 ,0 and id ale outside D. and t he line segments eon and et _ i n meet the arcs = (Dt land A 3 = A30).0, respectively When the context is cheat, we IllitY omit the references to the arcs .1 1 and A3 and to the lattice SA The quintessential example of a -1-mat Iced domain is a rectangle with the comers marked Let D = (a,b) x (e..d.), and let be the vertices of the rectangle. D, labelled as in Figure 2 If A is the square lattice P2

P1

J

1

I I --m- -

' i , ,

, ,

, , 11 1

R - ,,, 11 I II [

Figure 2 A path P in the lattice 0::1 The path P is an open (dossing open ciossing of D. 1 in site (a bond Inn iZOIdal n ossing of t he act angular

DI

11111-11 11: I 1

I [ I ,

I r I

, , , ,

I I I I 1: Ill 11111 111111'

II

Ir

!

crossing a reel angular •I-marked domain it all sites or bonds ()I P are open An percolation on 6: 2 is exactly an open sulnpaph l? of pc. fixed This begs the question of what happens when p = Flom now on, we take p to lie the critical probability = ThjA), and

mite /),;(D.1 , A) for .Ps(Dj . A. pc ) To be pedantic, we should indicate whether we consider site or bond percolation. but we shall not do so. \1'c have seen in Chapte i 3 that it D. 1 = (a.b) x (c, (/) is a rectangle with the corners marked, then the Russo-Seymour-Welsh Theorem implies that there is a constant: o(D.1 ) > 0 such that (l} instead of the open unit disc. We shall not consider unbounded domains hen± but the definitions and results extend easily to such domains under suitable conditions M a rking four points 2 1 , z± 011 the !cal axis to obtain a T. /narked domain U, = (CT; (z i )), the cross-ratio t(l.i ) is also given by (2) For 0 < k < 1. let U l (k) he the domain U with the foul points —14- 1 , —1,1, marked Then (1.1(k) 11/10' be mapped by a Schwartz Christoffel transformation (If \/(1 —1-)(1 — k-/-) to the interi o r of the rectangle I? w it II corners iihk(k 2 ) k2 ) \,/T. whine A

00=

di I V( — U)(1 —

A:(k2)±

186

Cortformal invariance - Spirt-t i tre's Theorem

is the complete elliptic integral of the first kind (Our notation here follows Abramowitz and Stegun [1966. p 590) The same notation 1(1,7) is sometimes used for K(u2 ) ) The aspect ratio of r is r 2K(1c2 )//C(1 — k 2 ), so, to find 7r(D.di )), one can invert (minim jenny): 7r(q), where ri= this formula to find k; then 7r(D.I (r)) = 7r((.1.1(/,:)) tkU.0)) = (1 — k) 2 /(1 + k) 2 (In particular, this shows that tkr) lAD4 (1)) is monotone decreasing in r ) Starting from Cardv's formula, Ziff [1995a; 1995b] gave a relatively simple formula not for r(D.1 (1)) itself, but for its derivative with respect to : 20ITI (2/3) (fi k(D, 1 (0) — 1(1/3)2

(n±lit

The calculations outlined above illustrate an unfortunate propert) of conformal invariance: given two families of 4-marked domains, each: parametrized by a real parameter, even if both families ate 'nice' the conformal transformation from one to the other may II ansform the N. rameter in a rather ugh. way. Thus, one can n ot expect Cand y 's for muk to take a simple for in for a given Mice' family of domains. There is, however, one family for which Cm dy's formula can he written very simply, an observation made by Lennart Carleson in connection with joint work with Peter Jones Let .D be the equilateral triangle in (0. 0), nnc (1/2, 4/2) and .p, IR2 C with vertices P i = (1,0), P, let Pi = (x, 0), 0 < 1 be constant Them is a COP shin/ e(p) > Il such that, if n > 2 and R pn by n rectangle In IR' with any orientation, then the probability that the critical site percolation On T contogys an open crossing ofR joining the two short sides is at least e(p). Here. the notion of an open crossing is that km tmarlced discrete domains. i e au open path uor,r v i in iL with m i . insicle such that the line segments von t and e t _ i V I meet opposite short sides )1 R. For a detailed proof of Theorem 3 based on the strategy used for bond percolation on 22 in Section 1 of Chapter 3, see BolloNis and Riordan (2006b]

190

Conformal irraarian CC Smintatis Theorem

We shall use the following immediate consequence of Theorem 3 several times Let A be an annulus, i.e the region between two concentric circles C 1 and 0). We say that A has an open crossing in the lattice ST if there is an open path from a site inside the inner circle C 1 to a site outside the outer circle C.)). Lemma 4. Let. A be an annulus with ginner radius r_ and onto, radius ± If 7 +11 > 2 WO r_ > 10008, then the probability that A has an open crossing in ST is at most (r_11 + )", where (11 > 0 is an absolute constant_ Proof. R eplacing a by a/2, we may assume that / 4 . = 2 k r_ for some integer k > 1. In any annulus with inner and outer radii I and 21 . , we can find six rectangles as shown in Figure 6. If 6/7 is small enough (and

Figure 6 Six rectangles inside an ru t /mins A of inner radius r and outer radius 2r If each rectangle is crossed the long way by a closed path in OT, then the annulus A cannot be crossed by an open path

< 1/1000 will certainly do), then the shorter side of each rectangle is at least 26, and the 6-neighbourhood of each rectangle is contained in the annulus, so the event that it has a closed crossing depends only on the states of sites in the annulus As K(T) = 1/2, Theorem 3 applies equally well to closed crossings Hence, for each of mu six rectangles, the probability that it contains a closed path joining the two short sides is at least c, where c > 0 is an absolute constant By Harris's Lemma

7 2 Stair non's Theorem

191

(Lemma 3 of Chapter 2), the probability that all six rectangles contain such paths is at least c 6 Hence, with probability at least d i , there is a closed cycle in ST separating the inner and outer circles of the annulus. Recall that = 2 k t Thus, inside the given annulus A, we can log(r+//_)/ log2 disjoint annuli A j C A with inner and outer find k radii and 2/ 7 , respectively Let E1 be the event that A; contains a closed cycle separating its inside from its outside. If J_ > 10008, then 1P(E 1 ) > di for each i. If A has an open crossing, then none of the events c an hold. But the events E1 are independent. so Pfil has an open crossing) < Pfno E i holds) =

(E1) < (1 — ,_r

and the result follows with a = — log(1 — c 6 )/(log 2)

❑

Thinking of open sites as black and closed sites as white, a path P is monochromatic if all sites in P have the same colour. Lemma 4 applies equally well to closed crossings, and hence to monochromatic crossings Roughly speaking. Lemma 4 says that if we have a 'small' legion of the plane then, when our mesh a is fine enough, this region is unlikely to he joined either by an open path or by a closed path to any part of the plane 'far away . To a certain extent, the form of the bound does not matter: any upper bound of the form f(r_/ r +) with 1(a) 0 as 0 suffices for the proof of Smitnov's Theorem. Note that even this weaker form of the lemma is much stronger than the fact that there is no percolation at the critical point:, which implies only that the pr oba.hility above tends to zero as I 4./5 oc with r_/8 fixed.

7.2.2 Discrete domains The heart t of Sr MAT'S proof is a lemma stating that certain probabilities associated to per colation on ST within a domain D are exactly equal (Of course, this is just a statement concerning a subgraph of the triangular lattice T ) In presenting and proving this statement, we shall often consider the (re-scaled) triangular lattice ST together with its dual, the hexagonal lattice 511 obtained by associating to each site v C ST a regular hexagon PL. in tire natural way, to obtain a tiling of the plane By a discrete domain arab mesh S we mean a finite induced subgraph Cs of ST such that the union of the (closed) hexagons 11„, v E is simply connected When considering C,1 as a graph, the mesh 8 is irrelevant, so we may take 1 and view C5 as a subgraph C of T

192

Conlin nul l thew lance - Swirrow s Theorem

In terms of the lattice T On OTT the condition that C Gs) be simpl y connected is equivalent to requiting that both G and its outer bountho y T, 0 ÷ (C), ale connected subgraphs of T. w lane 0-(G) consists of the set of sites of T\C adjacent to some site in G In fact, we shall impose the Following additional restriction on out discrete domains G: we shall assume that neither C not U+ (C) has a cat-vertex. This restriction is it televant to the mechanics of the arguments that follow but simplifies the presentation slightly The inner boundary 1)-(0) of a discrete domain C C T is just the set of sites of Cf that ate adjacent to some site of 7' \ C Om additional assumptions ensure that both 0- (C) and 0 4-(C) are the vertex sets of simple cycles in T: indeed. viewing G as a union of hexagons, its topological boundar y OG is a simple cycle in the hexagonal lattice If Following this cycle in an anticlockwise ditection, sm. the hexagons 0 e , sites of T) seen on the left hum 0- (C)„ and those on the tight form The condition that neithet G not if + (CI) has a O c (G); see Figure

Figure 7 A discrete domain G (filled uncles and the lines joining thew) drawn with the corresponding hexagons shaded The two thick lines are t he (des in T corresponding to t he Juliet and inner boundaries 0 + (G) and 3 (C) 01.61 The topological boundaly OC of (the set of hexagons corresponding to) G is I he cycle in /I separating shaded and unshaded hexagons

cut-ye t tex enstues that we do not visit the same ver tex of 0-(0) or or eTf. (G) more than once 110111 now on we shall view 0-(G) and O H- (0) as cycles in the graph 1

7

Billie1101,'S Theorem

193

In preptuation lot S i nn nov l s key lemma, we need a hu Him definition A k-ranked discrete domain is a dismete domain G (ca Gs) together with k distinct sites ,m; of its boundaly 0 - (G), appealing in this ()Mel as 0-(G) is crave/sec! anticlockwise. \Ve shall abuse notation by writing simply G lot a hr-marked discrete domain (G;v 1 „ vb.) Palely for convenience, we inmose the additional condition that each marked site v i is adjacent to at least two sites of T \G (This condition is very mild: any site n of 0-(G) that does not have two neighbours in T \G is adjacent to one that does.) = sT(G) to be Given a k-marked discrete domai t we define the the set of sites of it- (G) appearing between n i and in + , (with o k+ , = as 0-(G) is traversed anticlockwise We include both in and in.s. I into In this discrete context, an open missing of C from A; to Ai is simply an open path in G star ting at a site of A; and ending at a site of Aj Given a kunat ked disci ete domain G = (G: (n; )). as ever y tutu ked site i n has two neighbours outside G, we can partition the outer boundary i)± (G) of G into vertex-disjoint paths Ajij = Aj/j(0), 1 < i < k. so that each AY- starts at a site adjacent to vi and ends at a site adjacent to cis., Indeed. traversing id + (G ) anticlockwise we may take Ajijj to run horn the second neighbour of c i to the first neighbour of v i+j ; see Figure 8 Note that a site r E C is adjacent to some site of Air if and only if v E Recall horn Lemma 7 of Chapter 5 that a rhombus in the hit/rigida!' lattice alwa ys contains either a horizontal open crossing, or a 501tical closed crossing. but not both this statement,. and its moo!, extends immediately to -l-marked discrete domains Although the proof lo t the general case contains nothing new. we give it in lull, since we shall soon use similar ideas in a //101e complicated way Lemma 5. Lel C be a lumoked discrete domain, and let = .11(G) 1 i -1 11'haterer the stoles (t/ the sites in G. this graph contains either an open crossing from A I to A 3 . Of u (lased ree`Otithe from to As. but not Goth Ill pal (imam. the probability that C has on open crossing Pow A, to .1 and the probability that C has tell open crosYng from stt to As sane tO Proof As above. let CY (CO denote the outer boundary of C. which

is partitioned into arcs Ajt Consider the partial tiling of the plane by hexagons, consisting of one hexagon for each site of C U (G). Colour the hexagon ./1,. cot responding to c E C black if n is open, and white

Confunnel invariance chnirnoo's Theorem

191

•

•

ALVA A A A A A V ♦ V VA VA A A V •

•

•

•

•

•

•

•

Figure 8. A 1-marked discrete domain (68. e 2 . ea! et) (filled circles and lines joining thew) The outer boundary eP(G), a cycle in T, consists of the hollow circles and the lines joining them. The thick lines show the inner boundary shining endpoints and the 0-(G) of C. a writer of arcs A. I < i sjj. corresponding disjoint arcs C a c (CI)

if c is closed Colour the hexagons H,. corresponding to u E frt U At white, as in Figure 9 black, and those corresponding to E AT U Let I be the interface graph for nwd by those edges of the hexagonal lattice separating black and white hexagons, together with their endpoints as the vertices Then every vertex of I has degree two except for loin vertices Th , 1 < i shown in Figure 9, which have degree 1 Orienting each edge of 1 so that the hexagon on its right is black, the component of I starting at y i is thus a path P ending either at go or at Suppose that .P ends at !Li , as tit Figure 9. Then the black hexagons on the right of P form a connected subgraph of G U U At joining :At to Al- . Any such subgraph contains a path within CI joining a site adjacent to At- to a site tv adjacent to But then e E A, and ar E A:3, so G has an open crossing horn A i to A:3. Similarly, if .P ends at rp, then the white hexagons on the left of P give a closed (*.Tossing of GI from 4, to A i Crossings of both kinds cannot exist simultaneously, as otherwise K5 could be drawn in the plane The second statement follows immediately: as each site is open independently with probability 1/2, the probability of an open crossing

'7 2 Binh nav's Theorem

195

Figure 9 The hexagons Ii corresponding to G U &' (C). with those corresponding to the marked vertices / 4 , vo, ra, N labelled. A hexagon EL., v E is coloured black for i = 3 and white for i 2,1 A hexagon H i ., v E C. is black if t, is open and white if t ■ is closed There is an open path in GI nom A i to A 3 if and only if there is a black path in the figure horn A lt to .-11- and hence if and only if the interface between black and white hexagons joins y, to g.1

from A 2 to A., is I he same as the probability of a closed crossing from Au to A., This lemma completes mu brief review of the basic proper ties of critical site percolation on the triangular lattice T In the next subsection we present the first step in Sninnov's proof of conformal invariance

7.2.3 Colour switching We are now ready to present, the key lemma in Smiruov's proof: this 'colour switching' lemma states that the probabilities of certain events involving paths in a discrete domain are exactly equal. We shall shall often identify a site of T or 8T with the corresponding hexagon In particular, in the figures that follow, rather than drawing site percolation on the triangular lattice, we draw 'face percolation' on the hexagonal lattice, since it is easier to shade hexagons than points Let C be a 3-mar ked discrete domain, and let; x i , :to. .e, be three sites in (3 forming the vertices of a triangle in C, labelled in anticlockwise



196

Confonnal im yal lance Stnirnov s Theorem

order around this ttiangle Thinking of open sites as black and closed sites as white. we write 131 for the event that there is an open path .joining x i to Ac = ATP). i e., an open path from x i to a site in .4i, and 11'1 for the event that there is a closed path horn x i to Let 8 1 13,113 denote the event that there ate vertex disjoint paths Pi from x i to with P i mid 172 open, and P3 closed; see Figure 10. Note that

Figure 10 1 Ire hexagons {do responding to a :S-marked discrete domain II the Nei tices correc3). this time ‘‘ilhout its outer boundary (G: sponding to I he heavily shaded hexagons al e open and those corresponding to the unshaded hexagons are closed, t hen /3, /3.11 .3 holds

8 1 13Th;, = 13, ❑ 13,} J IL, is just the box product of the events 13 1 . 13, and II -3 . as defined in Chaplet 2 Define 8 1 11 .43 3 = L3 1 ❑ II -Q C /33 sinalai b. and so MI [he probability (HMI ibutimi associated to critical percolation on t he iangultu lattice is of commit 'nese' \ Cd if we change the state of every open site to closed. and vice versa Huts Ft/3 1 11 -2 11 3 )=P(11,8 2 Ba),

(5)

and so on. so there ate font potentiall y distinct probabilities associated E {13.111 SlIkh nov's Colon y to events of the lot In X i 3",2 3 . X. ). Switching Lemma states tlmt three of these are equal Lemma 6. Lela G be o 3-m a t/eed discrete domain and let c 1 . :c xa be sites of G Jointing a triangle in G. labelled in anticlockwise rode) around

7:'

Stith 1101' s

Thew ein

197

this triangle. Then P(/3 1 /32 11 .3 )

1.11(D I

-2 83 ) = POL L /32 B3 )

(6)

h t contrast to (5), there is no symmetry of the metal( setup that implies (6). Lemma 6 does not say that when looking for disjoint paths, the colours me it relevant as it makes no statement about IP(B, Bo B3) To prov e Lemma 6. we shall show that P(B, Ilji/33 ) = P(8 1 Ijillja)

(7)

Applying (5). or the relation P(B i 1 ,11'3 ) = 111-1 (8 1 which is essent ially equivalent to (7), one of the equalities in (6) Follows immediately The older inequality follows similarly, or by relabelling In tutu. (7) is equivalent to IP(/3 1 11',D3 =

(8)

The idea of the proof is as follows.. Whenevet B 1 FI holds, one can find l i n t el/mist' open and closed paths Q i and Q. , witnessing Bi condition not only on B I ji. but also on the pietise values of the paths cart find these paths without examining the states of Q i and Qo sites -outside them Next we shall show that if Fi l l ri B3 holds, then thole is nu open path p, front 3i3 to .4 3 outside the inunermost paths Q j and Qo Ilms, the conditional probabilih that 8 1 11:)/3: i holds is just the probability that the -outside' contains an open path horn ,r 3 to A 3 . As we have not yet examined the states of alp sites 'outside' Q i and Q(i. this is the same as the probabilit y that the domain 'outside' (2 1 and Q contains a closed path from to A: 1 , which is the conditional probability that B111'41:3 holds At the risk of scorning too pedantic, we shall present a detailed pool of Lemma 6 using the ideas above Note that a little caution is needed: on the level of the vague outline just given, it might seem that the same argument shows that 1P(B 1 BoB3 B I M) = P(13 1 13,11 .1 / .8 1 13,0 In fact, P(13, Vii ) and P(E3 i BoB3 ) ate not in genetal equal To find the intim most paths witnessing B 1 111i, we shall fellow a return/ interlace between hexagons. Consider the /initial tiling of the plane by hexagons. with one hexagon ha each site of U (G) As holjae, we colont a hexagon H, cm responding to a site v of CI black if e is open, and white if v is closed This time. we colon y the hexagons corresponding to Ali black, those corresponding to At white. and those to At grey As before, let I be the subg l aph of the hexagonal lattice consisting of

198

Coriformal

Smernov's Theorem

all edges between black and white hexagons, with then endpoints as the vertices This time, every vertex of I has degree 2 except for a vertex y where .41' meets At, and one or more vertices y i incident with grey hexagons; see Figure 11. Let P he the component of I containing y. Then P is a path starting at y and ending at a vertex :p i where hexagons of all three colours meet.

Figure A 3-marked discrete domain with its outer boundary Hexagons are black, those corresponding to At white, and those corresponding to to A. grey Internal hexagons are black if the cot tesponding site is open and white if it is dosed The interface path P starting at y ends at a grey hexagon

4

Let iv he the centioid of n,r0;r 3 , so a' is a vertex of the hexagonal lattice Let Tr =emu; be the edge of the hexagonal lattice separating xr from tn, oriented towards tv; see Figure 12 We shall prove Lemma 6 via a sequence of three simple claims. In the statements of these, the assumptions of Lemma 6 are to be understood. Claim 7. If Br W., holds, then the Oiler face path P sla t Hag at p travetses the edge e in the positive direction P1'00.1 Suppose that the event B I IV, holds, and let P1 be any open path horn 3$ 1 to A l , and P, any closed path from wo to ;19 Note that PI and P, ate necessmils disjoint. As any site in

is adjacent to a

7.2 Swinton . Theorem

199

Figure 12 The centroid w or a triangle ruts:es in 7', and the oriented edge -7 =71 of the hexagonal lattice H that separates x i from r

site in AI, there is a cycle C in the triangular lattice form ed as follows: follow PI from x i to A i Then follow (part of) At to its end Then follow par t of At, and, finally, follow P, from A., to :1,,} The cycle C' is shown by dotted lines in Figure 11 We may view C.', which is a cycle in the triangular lattice, as a simple closed curve in the plane in the natural way As we go around the cycle C' starting at: 1, 1 , we first visit black hexagons, and then white hexagons Thus, exactly two edges of the interface graph I cross C', the edge 7 shown in Figure 12, and an edge yy'. These me the two edges of H shown with arrows in Figure 11. The path P starts with the edge gy', which takes it inside C' As all grey hexagons are outside C, the path P must leave C at some point, which it can only do along the edge 7, proving the claim, ❑ Let P' be the path in the inter face graph I defined as follows: starting at y, continue along edges of I until either we traverse 7 in the positive direction, or we reach a grey hexagon If P does traverse the edge 7 in the positive direction, then P' is an initial segment of P Otherwise, P' is all of P. Let N(P'), the neighbourhood of P', be the set of sites of G corresponding to hexagons one or more of whose edges appears in P' Claim 8. If P' ends unlit the edge then the set N(P') C G contains (necessarily disjoint) paths Q i from. :f t to A i and (29 from, r, to Ait, with Q i open and Qs closed. Proof. The argument is as in the proof of Lemma 5: if P' ends with the edge 7, then the sites corresponding to the black hexagons on the left of

C'onforrnal Mem inner Stn0 non s Theorem

2(10

meeting A and containing P' 161111 a e()Itilecticd subglaph S of At Any such subglaph .5' includes a suhgtaph 5'' of G containing both and a site of C adjacent to At Every site u E .S'l is open: the corresponding hexagon is black (as 11 E 5), and o E C. so u is open. Finally. any a, E G adjacent to At is in A l , so there is an open path C S C N(P) how c i to A, Similarly thew is a closed path (2, C to A, ❑ Qo C .1V(P) Flom Together, the claims above show that BULL holds if and only if P' ends with the edge "(7 The proofs of Claims 7 and S also given little mote Claim 0. The event 33 I II I,23 holds if and only if 13' ands with and then, is on open pith P3 C C from 3 .3 to A$ using no site of the /31 I I)IY3 holds if and only if P' twig hbou t hood N(P') of P' and them k a closed path P3 C 0 from .r3 10 .4 using no conk with Sill!

N(P')

It static:es to move the first statement Suppose that B 1 11 433 Proof holds, so then: are disjoint: paths P nom to A, with PI open Po closed. atal fi , open Then the pool of Claim 7 shows that. tw ili t horn its initial and final edges. P' lies entirel y within a cycle C' funned by PI, Pi and parts of AI and (i e the dotted c y cle in Figure 11) Thus, even site of AT/3”) is on of inside C' But 133 cannot moss C'. so P3 lies entirely outside L. awl is disjoint from N(P') The I everse implication is immediate hum Claim 8. It is now pas) to deduce Lemma (i Proof of Lemma 0 As noted eat Het . it suffices to show that (7) holds, i e that P(BI II ,Bt) =P(Bi Let 9. he the state space consisting of all 2 1(I1 possible assignments of states (open (a closed) to the sites of C3, and note that the probability measure: P induces the nonnalized counting measure on U For w E R. let P i (w) be the path I. ' defined as above. with lespeet, to the confignt ation w Let 2 he obtained nom w by flipping (changing Croat open to closed of vice versa) the states of all sites in CI\ AI ( 331(w)). The path latly be found stepdw-step. at each step examining the colour of a hexagon adjacent to the cut tent path Hence. the event that

7.2 Stith not, s Theorem

201

p' takes a particular value is independent 0 1 the states of the sites of P'(w) so w" = w for any w E Q In patricidal P' (I) w' is a Injection Thus the map

G \

Suppose that w E B 1 11 -0.133 Then, by Claim 9, the configuration Le contains au open path in G \ N(P(w)) from ,r 3 to A3 Hence ‘21 contains a closed path G \ N (1.3' (w)) = G \ tV(TIr2)) hot u :c33 a to A Thus, by Claim 9, co' E B i lV)11 -3 Similarly, if w' E .T3 1 11011 ,-3 , then ./3,11 1 ,B 3 . As w w' is a measure-preserving injection, ❑ p(B 1 11 .08 3 ) = P(B, ll irlr3 ), completing the proof Let us remark that, while we could have used the Mtn/lace path P' to refine 'howl most' open and closed paths Q 1 limn x i to A i and Q., front .1 2 to A ‘ r, there was no need: it was simpler to work directly frith properties Of the interface itself

7.2,4

Separating probabilities

The next step is to give a kninal definition of the 'separating piobabilities: the limits of these probabilities will be harmonic functions on D From this point On we shall view the discrete domains Go that we shall consider as subgraphs of ST rather than T: this makes no difference to the properties of Gs as a gtaph, but will be convenient for taking Bunts later Let Cs = (Gs: r:i) C ST he a 3-mat lord discrete domain .r‘ith mesh 8, and let c E C be the centre of a triangle in ST As heline, we may think of z as a vertex of the hexagonal lattice SH dual to (11 A 'fey idea in Sndinov's proof of Theorem 2 is to consider the probability of the event E?:(c) = { Cs contains an open A r- A, path separating c from At }, and the events E (z) and g(z) defined similarly, where, as before. Ar i(C; s) is the path in the inner boundary of Cs starting at ri and ending at or +1 , and At = Aj (Cs) C 0 + (C; (5 ) is the corresponding path in the outer boundary of Cs Usually. we shall take z to be the centre of a triangle in Gs, although the definition makes sense for points z nearer to (or even outside) the boundary of Cs The subscript fi in out notation is shorthand to indicate the dependence both On S and on the discrete domain Gs. Needless to say, by an open Ar -Ao path Pin C7,5 sve mean a path P in the graph Gs C ST all of whose sites are open starting with a site

202

Conformal

11)0(11

Smini.ov'■; Theorem

in A i and ending with a site in An Such a path P separates z hom At if, when we complete P to a cycle C in (Yr using the arcs All and At, the point z lies in the interior of the cycle C when C is viewed as a piecewise linear closed curve; see Figure 13 Equivalently, the path P

Figure 13 A discrete domain G's (circles and the lines joinil g thew) %vith the associated coloured hexagons: open sites ate shown by Ii led circles and shaded hexagons, closed sites by empty circles and unshaded hexagons. 'The outer hexagons (those not containing circles) correspond to the u t ter boundary 0 11 (0s) of C,5 , divided into truce ales .4.1 1 , At and A4 at the thick lines Art open A 1 --.19 path P and the associated cycle C separating z, the centre of a triangle in C, from At are shoran a thick lines

At if any path in 511 consisting of edges dual to bonds of starting at z arid ending on (a dual site adjacent to a site on) At, crosses a bond in P. nor to revisit a vertex. Tints, Note that P is requited to be a path, the event .Er (z) does not in the case shown on the tight in Figure hold As before, let w E C be the centre of a triangle xr:ror t a in C. with the vertices labelled in anticlockwise order, and let z be the site of 511 adjacent to a that is farthest from x 3 ; see Figure 12 Suppose that the separates z front

72

8711:177101t ' s

'Theorem

203

Figure 11 The figure on the left shows a schematic drawing cif a 3-marked discrete domain a5 together with an A i -Aii pat h separating z hour A . If the sites on this path are open, then Ei,t(z) holds. The figure on the right shows a connected set S meeting and Ao and separating z from A4 that does not contain an Ar-Ao path separating z fron t At. If only the sites of S are open. theu Es(z) does not hold

event EPz)\E,(a,) holds, and let; P be any open path in C6 separating horn At Completing P to a cycle C as above. the cycle C winds ar ound z but not around w. Therefore, C' must use the edge t i rk2 . If we trace C by following P from Ag to A i , returning anticlockwise along At and At outside Co, then C winds i1/01/11(1 c ill tire positive sense, so we trace the edge m i x, from fo to ,r 1 . Hence, P is the disjoint 1111i0/1 of open paths Pr from m to A i and A hem r2 to A,. As we shall now see, there is also a closed path P3 horn ,r3 to .43 Claim 10. Suppose that 3,,,,r.),:ra,lir and z ate as above (sec FiffittCS 12 and 15), and that the event g(z)\Ej(w) holds. Then B I M Va holds. e.., C, contains disjoint paths P, Mining ;c 1 to Ar = A(ac), with P, and A open and Pa dosed Proof As above. let P he an open A t –A, path in Cis separating z from . We have already shown that P may be split into open paths Pi horn 3,1 to A i and A from to Ag ft remains W find a. closed path P3 joining ,r 3 to any such path is necessarily disjoint horn the open paths PI and Pg Note that x:r itself is certainly closed: otherwise, the Open path PI :r 3 A separates 11., from Once again, we follow an interface As usual, we consider the partial tiling of the plane by hexagons H,, corresponding to vertices v of Go U

Conformal 'now; iance Surirvoil5 Thcairtn

204

U a + (0 A l. where 0 + /Go) = U is the onto bounda t v of 05 We C ohan a hexagon H,. c E 61 ,r, black if l' is open and white if V closed We colour H, black if e E At U and white if u E At; sc.; Figute 15.. Let. I be the oriented interface graph whose edges a t e the

Figure IS Ilse discrete domain (circles and the lines j0 n og them), with t he associated coloured hexagcms 'The inter lace I between black (dark and lightly shaded) and tc kite hexagons has exactly two vertices of degree I, namely yr and y2 An Open À - path P and the associated cycle C' separating z from At are shown by thick lines If :rr i is not connected to :L i by a closed pat I I , then t he while component containing ,r 2 is surrounded by a connected set S of black hexagons: those not on 1" or At" U AT are shown lightly shaded The union of and P contains an A l -A 2 path sepal ating from Alt consisting of the clashed lines together with part of P The °limited edge of the hexagonal lattice crossing Tow:, is indicated b y an arrow

7

edges of the hexagonal lattice wit h a black hexagon on the t ight and a white one on the left This graph has exactly two vet tices of degree the vett:ices .111 and rki shown in Figute 15 As x/i is open and

is closed, the oriented int et lace I tont tuns the

7 2 Solo [mei,: Theorem

205

edge J o ftin hexagonal lattice (tossing .10.1 3 with 1 on the left: this edge is indicated with an arrow in Figure IS Suppose first that the component of I containing f is a path. Then it ends at a vertex of I with degree 1. which must be it, But then the chine hexagons on the HI of this path Mira a connected set containing 3). and meeting At As before„ it follows that there is a dosed path ram ,e 3 to .21 3 , as required. Suppose next that the component of I containing f is a cycle, as in igme 15. Our aim is to deduce a contradiction. Let hr. 1m... h, be the sequence of hexagons seen on the right of this ci cle as we trace it once star ting hom f Then each // i is black. h i corresponds to 3.' 2 , h, /responds to x i . and h i is adjacent to h i* , lot each i Ide n tifying a hexagon with the corresponding vertex of the lattice, we do not have Mr an y i: otherwise. the interface cycle would visit w = twice Recall that we lime open paths P I and P. , joining to A t and respectively. and that these paths can be closed to a rude C sunounding c but not w by adding parts of At and A:. No edge of the interface I can cross C. so ever v center h i lies on or outside C Let he maximal subject to h, E P.) u At, and let 5 he minimal subject to s > t and 11,, E Pr U At Note that 1 < r < s < as h i = :ro E Po and to

EPn Let S = .11,„.I} Then S is a connected set of vertices of the graph C ri — for rued from Cs by deleting the edge ..rTro: the hexagons corresponding to S ate lightl y shaded in Figtue 15 As S contains a neighbour of Pn U At and a neighborn of „Pi U At, it follows that .PI U S contains a path F' in (1,5 — r i :r9 joining A i to A, (The part of tins path off P 1 UP) is shown by dashed hues in Figthe 15 ) The corresponding hexagons are black (drawn with dark or light shading in Figure 15). so the path I" is open Fu Iv lies on or outside C. so it separates z from At As P' does not use the edge x i it follows that 13' separates w from At This contradicts mu assumption that g(w) does not hold ❑ h

The proof above is longer than one might: like, and can probable be expressed more simply Note, howeve L that one cannot give a 'purely topological' proof: in order to show that E, (m) holds, it is not enough to find a connected black set meeting A i and A, and separating to front At Indeed, one might expect that the centre c of a triangle in ST is separated fron t At by an open Am An path if and onl y if z lies 'above'

206

Conformal 11111(1rion.ye - Smirnov's

ThCOlern

the unique path component of the black/white-interface in Figure 15, However, this is not the case, due to the possibility of the configuration on the right in Figure 1-1 It is easy to see that the converse of Claim 10 also holds Claim 11. Let w E OFI be the centre of a triangle x i x 2 x3 in Gs, labelled in anticlockwise order If z E bf1 is the neighbour of w opposite x3 , then E,(2)\.E,?(w) holds if and only if B 1 Hd173 holds Proof The forward implication is exactly Claim 10. For the converse, suppose that /3 1 B911 13 holds, as in Figure 16 Then the disjoint open

Figure 16 A schematic picture of the event B 1 B211:3 horn 370 to An together give an open paths Pi from and to path P flow A 1 to As P uses the edge x i :1H the path P separates exactly one of z and w from At As there is a closed path from :r 3 to .43 , the point tv cannot be separated horn At by an open path Thus ❑ ES(_) holds and Er, (v) does not. As before, given a 3-walked discrete domain (3,5 and a point E 611, let Es(s) be the event that G5 contains an open A 1+1 --A i + 2 path separating z how At, where the subscripts are taken modulo 3 In the light of Claim 11, Lemma 6 has the following immediate consequence



7.2 Swinton's Theorem

207

Lemma 12, Let ii:roxa be a triangle in a 3-malked discrete domain

Go, with its vertices labelled M anticlockwise outer. Let w

E 811 be the centre of the triangle, and let. zr, C 2, Z3 E 611 be the neighbours of ay in 673. with z i and x i opposite for each i. Then

P(Ecilt (z i ) \ E (Iv)) = P(ERz 2 ) \ ERiv)) = P(E,Nz 3 ) \ E:J(w)). z 3 . Claim 11 states that the events g(c 3 ) \ Es(w) Proof. Setting z and 13 5 B2 W3 coincide Permuting all subscr ipts cyclically, we have E!(cr) \ = B 2 B 3 111 [ and El(z2 )\ El(w) = 133 8 1 1 , 1 7o The conclusion thus follows horn Lemma 6 ❑ For the centre z E alI of a triangle in C,s, set; 1'n(z) = P(E,i5(z))

(9)

and, if to and , me the centres of adjacent triangles in C6, set:

P(E:5 (z) \ .E7dtv))

h si (w, c) Note that, trivially,

n(z)-n(rv)=iov,z)-

(10)

As noted by Stith nov, there is a surmising amount of cancellation in (10) It turns out that for z and Iv far from the boundary of L.), the quantity ki5 (w,z) is of order 5 2/3 as 6 —4 0; the exponent 2/3 is the '3arm exponent' appear ing in relation (51) in the next section. In contrast. fA(z) — is presumably of order 6. We shall not be concerned with proving these statements, as they me not needed For the proof of Srniinov's Theorem. Roughly speaking, Lemma 12 implies that the discrete derivatives of the a ate related to each other by rotation For a precise statement, it is easier to work with integrals around contours. We consider only discrete trianguleu contour s in G1,5, i.e , equilateral triangular contours C whose corners are sites of ST and whose sides are parallel to edges of ST, such that all sites on C me vertices of C. as in Figure 17. We orient C anticlockwise Givers such a contour, for 1 < < 3 we define the discrete coolant integral •D

I

) M C ) Clz

as the usual contour integral of the piecewise constant function f(z) whose value at a point z of an edge 51, of ST on C is the value of

208

Conformal Moat lance - Stith not, s Theorem

A discrete triangular contour C The d iscrete contour integral of Figure fi r wound C' is defined using the values et .111 at points of the hexagonal lattice Along :rig. fen example. we integrate j',(ai)

IA(w) on the nearest vertex w of OH inside C, immediately to the left of

aril;

i e „ on tlw vertex of (if

see Figme 17 For example, if C is a

triangle with side length 6, mid a ' is the centte of the triangle C", then

f (z)

. 0u) identically on C. so 1■(-13 Inz)dz = 0

of unity Let w = (—I + \/-3)/2 denote one of the cube toots

Lemma 13. Let CI; be a discrete 3-hooked domain such that no point of C is within disloace a of all three arcs of OGo. where a > 20006 If C' is a discrete triangulat canton; in. Cs of length L. then (1)

(:)(1: —

f,f(z) (lc < AL(61a)"

( 1)

1.2.3 where Mc sumo set ipl is taken modulo 3, o is the constant in Lemma 4. and A is an absolute constant

fm i

Proof Ri x shall prove (11) lo t i = 1; (lir corresponding equations for

i = 2,3 follow by relabelling the domain Let a' E OH lie the centre of a triangle in Cs, and let z be a neighbour of iv in OH From Claim 10 above. if E,C(::)\E's(w) holds for some then them are monochromatic (entirely open (a entirely closed) paths from the three sites of

Or adjacent to iv to the duce boundary arcs of

Cs The point w is at distance at least a (rout one boundary ale. .-1i, say, so a monochromatic path horn a point adjacent to i v to As gives an open or closed crossing of the annulus with centre ut and how l and outer radii 6 and a. 13v Lemma

the probability that such a crossing

7 2 Swirnov s Thcomm

exists is at most 2(10006/a)° =

0(010")



209

Hence.

/a)").

hs i (w..z) =

(12)

uniforml y over C. w with a' the celiac of a triangle iu Cs and z st, denotes adjacency in the graph here and below, Let

w:

tv)11,15(

wct c.- systs where C' is the set of yet tices of OH

interim of C. and z nuts

over the duce neighbours of w in OH To evaluate — w we view w and

as complex munbets Fix w E C' , and let z i Ca be the neighbours of a ' in anticloclavise cadet, so — w =

w) =

j

(13!

From Lemma 12 we have 16(w, )= M(w..m) = Ow. za)

(,A)

Also, applying the same lemma with the toles of :I , z Ca) =

c . )) =

pet nutted,

Z)

and h(15(a)7 c 3) =

Setting z 4

(

CI , we thus have (zi

tv)q(ut, zj)

— tv)1;;(v

1±i)

j=1

w(zi -w)h,v(m.=i i i) .i=1 3

(v)16(w I=i

where the first equality is simpI N a relabelling Of the same sum. the second is Eton, (13), and the third (Li) and the following similar equalities Stumning over w E C"', it follows that S2 =

(13)

210

Conformal invariance - Snail noM ii Theorem

call write

— u01011, z) as 5'; +.57. where

=

5,, =

(z- tv)11 61 (vi, z) +

— z)h's(z, u,)

it,,z€C"

and (z — w)11:5 (ev, z).

The sum S7 has CALM) terms each bounded by Ssup„,,, , h[s(u, Using (12), it follows that 87

0(L(S I o)°)

(16)

for each For

(

z E C`) with iv ti z from (10) we have

z — a)his(lut

(a'

Z

)1/1,1(Z, w)

=

=

h ■S

WX/116(111,

(z — w)( —

(It ))

f4(w))

To obtain the must stun this last term over all unordered pairs {m, z} with m + z and E C' Collecting all terms of the Ram :17101"), E C. we thus have :17

=

E E - Jig ?, ) E ('rv-z)f,R4v)- E E (a' — z)1,(10). wE C" z—w, ze4c,

In the first sum in the final line, the coefficient of each 1, ((tv) is exactly 0 There is one term in the final sum for every edge viz of SH crossing C from the inside to the outside. The edge ruz of 511 may be obtained from the dual edge 1,"// E C of ST by a clockwise rotation through w/2 followed by scaling by a factor 1/ \/[3 (For x, y and w, see Figure 17 ) Thus, z — w = ( - 14/fr3)(y — :r) Hence, h our OM definition of the discrete contour integral, •

=

(z — . 0.1A(10 = —



.11(Z) (1Z;

(17)

VF3 c

As Si =

+

, the result follows from (15), (16) and (17)

Later, we shall show that if we have a sequence of 3-marked discrete domains Co with mesh S 0 that give finer and finer approximations



7 2 5'71th-two's Theorem

211

to a Jordan domain D3, then the functions Mz) converge to continuous functions f' on D Lemma 13 will imply that the (usual) contour integrals of these functions around a given contour are related by nultiplication by co.

7,2,5 Approximating a continuous domain In this subsection we show that one can approximate a Jordan domain D by suitable discrete domains without changing the crossing probability significantly, proving the technical Lenuna 14 below This result is immediate for sufficiently 'nice' domains, such as rectangles. The reader Amested only in such domains may wish to skip the proof. Let us recall some of the definitions involved in Theorem 2. By a classing of a 4-marked Jordan domain D4 in the lattice ST, we mean a jath in OT whose first and last edges cross the arcs A i (/).1 ) and Aa (M), vith all vertices except the first and last inside D. If the sites of a crossing are open, then it is an open Glossing of D.1 . As before, we write Po(D.,) = (D4 T) for the probability that D4 I as an open crossing. The corresponding definitions for discrete domains are simpler: a crossing of a -1-marked discrete domain Gs is simply a path in the graph C:s joining A l (C;(5 ) to A3 (Cs) We write P5 (G. 6) for the probability that a discrete domain C,, C ST has an open crossing Let us write dist(x, y) = - :y1 for the Euclidean distance between two points :ET, y E R2 C. and (list (x, A) and dist (A, B) for the distance between a point and a compact set A, or between two compact sets A and B We avoid the more usual notation d(t, y) due to potential confusion with graph distance For two compact sets A, B C C, their Hausdoijd distance is du (A, B)

sup Udist(a , B) : 0 E A} u {dist(b, A) : b e = int : A C

B C A(=)},

where A V) denotes the (closed) e-neighbour hood of A. If = (P1)) is a 4-marked „Jordan domain and (75 C 6T is a 4-marked discrete domain, then Cs is E-close to Dt if the corresponding boundm y arcs of D.1 and of G5 are within Hausdorti distance e, i.e if dri Ai(G,5)) 0, there arc (5 1 , 11> 0 such that fur all r5 < S i , any two points n: and of Cif with dist (w, z) < q may be joined by a path in GI lying in the boll 13.,(w)

Iloughl ■ speaking, L.enuna 1-1 shows that. to in (WC ThCOlent 2, it suffices to Ivor k with discrete domains Mote specifically, using the fact that (Itis close to .D. 1 , we shall show that Pi (Cn) — FID4), w hich.

together with (19), implies 'Theorem 2 The last condition in Lemma If is a technicalit y that we shall need to ensure that certain functions go we shall define ate tutifin tid y equicontinuous as (5 varies The rest of this subsection is devoted to the proof of Lemma The construction of the domains (.7;t will be broken down into a se t ies of steps, and the proof that they have the required properties into a series of claims Before getting started, we present two simple facts about lot dart curves that we shall use Lemma 15. Let D be a fixed Ionian domain with boundar y such that if w, y E f and dist (.r. y) s> them is an. = ) > and q divide F lies entirely then one of the two arcs into which within the ball B (s') Proof

This is standard.

immediate Iron the fact that 1 = (IT),

72 &nil noMs Theorem

213

v iarre i is a 1-to-1 continuous Mal) horn the circle T into C Such a nap 7 is a homeomorphisur, so both 7 and its inverse ate continuous 'unctions on a compact, set, and hence uniformly continuous Alter natively, the fact that 7 is uniformly continuous implies that, given s > 0, there exist to = 0 < tr C • • • < < = 1 such that len , 7([T, t. i .r. 1 1) lies in some ball of radius s./4 Any twosetsI F with ] — i 0. ±1 modulo n are disjoint, and so separated by a positive y of F distance Hence, there is an r > 0 such that any two points or an adjacent arcs F 1 . F ) . In either within distance t lie on the same rise, them is an rue of F joining a' to y and lying within 8c(r). ❑ emma 16. Lc/ r 6c a Ionian chive boandinga domain D. and let. E D be freed Given 5 > 0, there is an = g(F.C. ^ ) > o with the following p i ppin ty far every point .r of 11, there is an .r' E D with dist (r. .11 ) < c that may be joined to C by a piecewise intent path P with dist(P, F) > Proof Let as p}(.r i ) be a (minimal) finite set of balls covering r, and Pick one point c t E 13,p)(x t ) fl D for each I, so every E F is within distance 5 of some .-t t As D is a connected open set, each z i ma y be connected to C by a piecewise linear path P1 in 1) The minimal distance between the disjoint compact. sets P Ui and C is strictly positive, so there is an q > 0 such that the 2q-neighbour hood of P is disjoint

❑

horn F

One can show, that. given a ri-nun lied domain D., and an > 0, lot sufficiently small S there are 1-/ / ta/ ked discrete domains G:5 C ST that am s-close to D.:, such that any crossing of Gs in ST contains a crossing of D4, and any crossing of f).1 in ST contains a crossing of G,"riThis implies that P6(0,7 ) P6(D4 < Ps(Cr{ )

(20)

In fact, it will be cleaner to move a somewhat weaker statement, involving only crossings that do not pass to close to the /milked points Pi. This weaker statement does not imply (20) Howevel, we shall show. using LC1/11 / 1a 4, that it does imply Lemma 1-1, which is strong enough fon the proof of Theorem 2 The basic idea of the construction is as follows. If is a rectangle with the cm ners uni tized, then we nay take (CT to be a slightl y !tinge' and thinner 'rectangle' in the lattice, and Crq to be a slighth shorter and

214

Conformal Invariance -

no v's Theorem

fatter rectangle The case where D is a polygon is similarh easy The general case is not mate so easy: when we 'narrow' D in one direction we may end up with a disconnected graph, for example. Also, when we extend it in the opposite direction, we may bump into ourselves, and end up with a domain that is not simply connected. These are not 'real' problems, but, nevertheless, it takes a fair amount of work to overcome these difficulties. For the rest of the section, let D. 1 = (Pi)) be a fixed 4-marked Jordan domain, with boundary F. Given El > 0, in the constructio that follows we shall choose other small quantities >

>

> 5.i > Er, =

where fi is a. function of so that 6; 4, 1 is much smaller than fot each i. In particular, we shall assume that 10005 i + 1 < say Let s t > 0 be given Our first step is to simplify the crave F = OD mac. the nmrked points Pi As before, A i = A i (D) is the arc of the Jordan: curve F running from P. 1 to Pi± i . The arcs A t and A 3 are disjoint, and so separated by a positive distance The same applies to A 2 and Ai,: Let Zo e D be fixed throughout, and note that dist(zo, A i ) > 0 for each Reducing if necessary, we may assume that. dist(zo, > 106 t , and that, ri.

E1,

dist(A .4 3 ), dist(A 2 , A. 1 ) > 106 1 .

(21)

In particular. dist(Pi, Pi ) > 10s, for 1 < i < ) < 4 Choose 5;9 > 0 with E2

5 7-(r,

Ei

where T(', s) is the function' in Lemma 15 Thus, if to and z are two points of F with dist (iv, c) < then they are joined by an arc of F that remains inside M, (w) Let Bi be the open ball of radius am around Pi: and let C1 be its boundary Taking the indices modulo 1, the al c A i of F joins Bi to Bi .m, and so contains an arc F i disjoint from B i U Bi+ joining Ci to Ci + 1 ; see Figure 18 Note that for j = i + 2, i +3, we have clist(F i , pi ) > dist(A i ,A 1 + 2 ) > 105 1 . Hence, each arc F.1 is disjoint from every B1 . For each the set .4 1 \ F i consists of two arcs of F each of which joins a point P1 to a point at distance 69 from Pj ; this arc is thus contained in B_, (P 1 ) Thus, (F i , A i ) < Let be the 'simplified' curve obtained by joining P i to Pi arid P1+1 by straight line segments of length 0, as in Figure 19, and let D' be the interior of I" Let he the arc of F' starting at Pi and circling at Pi-Fr, 69,

E,

7.2 Smilwov's Theorem

Figure 18 Th e

joining



215

to

Figure 19 'The 'simplified' curve r (thick lines), and its interim D'.. The points P1 arc the cent res of the circles C; with radii am (drawn tvith solid lines) The dashed lines ate the arcs of F \ F', each of which remains distance of some Pi , as indicated by the clotted circles..

and note that di-dri , Ai) <

(22)

for each i We shall modify r in a way that is analogous to replacing a rectangle by a slightly longer and thinner one The disjoint closed sets 1 < i < 4, ate separ ated by positive distances, so there is an Ea > 0 such that dist(r i ,F;) > 10E3 (23) for i



210

Cotd611nal en001 inure Smitnoe •; Theorem

Let s;3/3)/2.

(- I)

Where. as before, t is the function in Lemma 15 Let N 1 = ft' ) he the closed s A meighboruhood of P i , and let acc (N1 ) denote its external boomlarw, i.e tire boundary of the infinite component, of C \ N 1 The Jordan curve er inds around F 1 . Flom (23), it does not meet: any other I .) , so it can cross only hr the line segments inside Bi and joining In tact, it can cross only the line segments L i and to RI and Pi+t respectively Furthermore, as Fe meets Ci and (7.';+1 at one point each. and lies outside Bi the curve (N1) meets each of L i and exactly once, at the points Q; E and (2 1; E see Figure 20 Hence, With dist((2e,R) = dist(C4,P j+i ) = E g D'(Nd consists of two arcs joining (2, and Cj i . one inside r and the other outside

r

r

Pignut 20 The curve 1 11 , (thick line) together 1A-ith its £ 4 -neighbourhood N1 (shaded) 'The extol nal boundary ir(Nd of N., crosses r at two points. the points Q, CI, on the line segments L, and 14 The ca l ve 51 is one of the (N,) joining Q, and Qf we take S, outside r for i 1,3. and two arcs of inside for i= 2 4

For i = 1.3. let .55 he the arc of D=c (AC) outside F', and, a little dissonantly. for 2,4. let Si be the arc of D'(Ari ) inside r (The external boundary of IV; = ft' ) is defined without reference to the rest of F'. so Si is part of the external boundary of N 1 even when Si is inside F'.) These choices for the arcs Si correspond to the operation of replacing a rectangle by a longer and thinner one, by moving the first and third sides of the rectangle outwards a little. and the second and fourth sides inwards a little We shall write R i for the region bounded Iry the closed (nuke formed

7 2 SinirtunCs Theorem

by Si . . and the two line segments of length of these curves

S.1

217 joining

the endpoints

Claim 17. 117e can paramelf Ltd: the open Jordan carves Sm and continuous injections sm. gm : (0.1] sm(1) C with !Mt)) alt t < fin. E Ca + , and dist( sm i (0) E C h

I;

Proof. Let tim(a) and i i (a) be palm/utilizations of the .lindan turves Sm I] and I'm traced in the appropriate directions. Define a map a : let Y„ be a point of I'm at distance [0,1] as follows: fin each a E TOO E Such a point exists as Si is writ of exactl y srr., hom X„ Note that the boundamy of Ni = j" tDefine o(u) be !Ijm(o(u)), to each 1. as there is a unique closest point of i (0) = 0 and o(1) of Q i and (X The straight line segment L„ = X„Y„ meets U only at its endpoints Also. the interior of L„ lies in A'm and hence in R i (As we Ca avetse S. on one side we have the unbounded component of C \ ) Thus, L„ separates Pi On the tithe/ side we have Ni and hence into two pieces The boundaty of one of these contains all points of Si appearing X„. and all points of L i appearing berme Y„: the boundary of the other all point's appearing after these points. As the line segments L. L„, cannot moss, it follmts that o(ad) > o(u) lot > a. So hut, we have found a win to hare Sm continuousl y so that a nearby point traces 1m monotonically. but not necessarily continuously We can trace both curves continuously by 'waiting' on Si whenever the nearby < point on I i jumps Mate fin i nally, wilting 0_(.) = sup{o(44): and 0 + 40 = inf {o(P) : > .r}. as o : [0,1] [0,1] is increasing. we can find conti n uous [Unctions a,(0 and a,(1.) horn [0.1] to [0.1] such that each a; is weakly increasing. amid

0.-(11/(0) ^ //2(0 5_ (4+(11(1)) lot even- a example, one can take u t (1) = sup{ : a + (A(d) < and a 2 (1) = ni(t). Let = ;i(a2(1)) and g i(I) ai(w2(1)) At points whew cr is continuous. we have la,(1)= 0(1 [ (I)), so, by definition of o, the points ,s;(1) and ni (t) ate at distance exactly EA As both T.Cim and in ate continuous. and each discontinuit y of a may be approached Flom both sides by points at which o is contitotous, it fellows that the points ij i (o4(u l (t.))) are also at distance exactly 5.1 from si(t). these points are within distance 2E,, of each Millet B y our choice (24) of :7: 4 , it follows that the ate of 111 joining these two points lies within

Conlomat Mom-lance Surrnoo',4 Theorem

9 18

a ball of radius E 3 /3 centred at either point:, and hence within a ball of, radius g 3 /3 centred at s i (t). As in (t) is a point of this at we have (list (il i (t), -s,(f 177.1 s3/2

So fa t , the functions v .) are only weakly increasing, so gi and are not injections We can modify the tt ) slightly so that they are shied., increasing, tot example by adding a small multiple of un to tt, and vice versa Using uniform continuity, this does not shift the points 1/1(1) s i (t) significantly, and the result; follows. 0 Let; P- be the closed curve formed by Sg, 8: 3 and S. together witl the line segments joining the endpoints of these curves to the points p and let D" be t h e interior of 1---; see Figure 21

Figure 21 The curve F - (solid lines) and its interim IF The dashed li ale the curves Each Si is palt of the external boundary of N i tr. It ll : ha i = 1,3. S i is outside i 81 is inside r

Let IT denote the arc of P starting at P1 and ending at P1+1, so FT consists of Si together with two straight hue segments From Claim 17, can find corresponding parametrizaticms 1'; and FT that remain within distance particular, we

S3 fit

for 1 < i

4. Also, piecing together these pain/netrizationsrye obtain



219

7 9 Son; non's Theorem

an t ett izations of the Jordan cm yes 17- and such that c otresponding points ate within distance 5.3 Roughly speaking, we shall take CC to consist of all sites v of ST vhose corresponding hexagons H„ a t e contained UnfoitunatelY, this set; need not be connected, so we shall have to be mot e careful Suppose that 0 < 35 <

(26)

Et/10)

wLcre q is the (Unction appearing in Lemma 16, and let D (7 be the set of sites v E ST such that H„ C D - Reducing 5, if necessary, we may some lu that < 05 4 /100,

(27)

vhele 0 is the smallest of the angles at the cot nets Pi of 171 Recall that. ca is a (fixed) point of D. with dist(co,r) > 10n, i.e with (z 0 ) C D Thom the construction of 11 ', we thus have Biie ,(zo) C winds around zo, it follows that V also winds mound :9 , so D' As zit) E in fact. B 8 _, (cry) C D Hence, the hexagon containing cu is contained in DBegat cling DS as an induced submaph of ST let Cloi be the component. of D:1 containing (the site corresponding to the hexagon containing) cu; see Figtne 22 We shall take 67 as one of our discrete apptoxitnations

r

Figure 22 A pint of C;;;" (viewed as a union of hexagons). together \in.( ' auof hexagons contained in D - 0:5 is other (small) component of the set the component of containing the point; zu

to I/ To make G s- into a 1-marked discrete domain (Gis : we take vi to be the VCI ex of 0,1 closest to Pi

1.



220

Confotwal

)11-141 I )(Wet

Sitthlt017 ti The orem

Claim 18. Ire have d ji (OCC„F`) < si/10. Proqf Suppose first that :r E 00 ,7 Then ,r is incident with hexagons H e EC6: and H„, The hexagons and H„, ate adjacent:, and, H, E DI. so by definition of 0 611 , we must have H„. 0 D:.;" Thus, 11„ meets and dist (it, < 25 < £4/10 Suppose next that :r E By (26) and Lemma 16, these is a point'. x' E D- with dist(e,,r) < £ 4 /10 and a path P joining to co with dist(P,1 1 ) > 361 But then P Os) C D- contains a path of hexagons joining zy to 3.!. so x' E G7 As :c 0 D D 07. the line segment ;ri/' ❑ meets 067, so dist.Cr, 0067 < .s.. / /10 Recall that ri. 1 < r < ate the ales into which the points P1 divide so [7- stints and ends with shot t line segments slatting and ending at P1 and Pi _e j : the test of 17 is the onve Si As any point of Si is w ithin / distance/ E. 1 of U. it follows how (23) that if .r E mid y E 17 with i ) ate at distance dist(r.y) < 106, then, swapping .r and rj if necessa t y, we have j = the point :r is on the line segment of 17 ending at Pi.+1 , and y is on the line segment of 17+ scatting at P1,1 As these line segments meet at au angle of at least 0. how (27) both e, and y ale within distance 105/0 < se/10 of Pi . i e well within the ball 131

As is a component of the union of the set of hexagons contained iu a siumb connected domain. it is simply connected Let us Imo/ the bounda t v of 0 7: anticlockwise Each boundary vet tex v is within distance 2S of a point of F. which must lie on some If o' is the boundary vet tex alto c, then a' is within distance 25 of a point of some IT from the remade above, if i j. then = {/.k± I}. and .1. y E B.,;,(124.), say hi ti the/ wonls. the closest boundary me r7 can only switch when we ate very- close to some Pt . But in this legion. we know exactly what r- looks like: it meets 13,„,„(Pk ) in two line segments From Claim 18, the boundaty of 0:11 comes within distance £ 4 /10 < £3 of a Hence 00W meets B,„(Pk ) in a single ant, and, as we trace the bomicht l y of OCC, the closest curve r7 switches horn to r I when we trace this ate Recall that c i is the closest vertex of 0 s11 to the point P1 0 CIA: Thus. v, is a boundar y vertex of 07 lying i t / /3, 00 (Pi ) Let .17 = 1 ,2 , v3 , ) be the bouudanv tuts into which the c 1 divide the boundan of QC.: Clain ' 18 and the comments Ame imply that d it (Ar. [7) 0 be given Their is an q = q(D4 , 1 ) > 0 with th following property. If E l is chosen small enough, then any two points w and z of C -5- with dist B., (w).

Z) < n OIC

:joined by a path. iu G7 lying hr

Proof We shall assume that ti < 7/2

We may assume without loss of generality that m and z lie on the boundary of Gs (viewed as a union of closed hexagons) To see this, consider the line segment wz If this lies in 67, we me done Othet wise, let x and g be the fist and last points of this segment on the boundary of so 'on'. gz C G q Then disiff, y) < y < 772, so it suffices to joins and y E OCC by a path in 67 lying within 13_ /2 (y). saw Thus, Claim 21 fat tv, z E (16- follows front the same claim fit iv', E with 7 replaced by -I/2 We may also assume that the line segment tvz does not meet OGri again: other wise.. listing the points c1 in which this segment meets 00,7 in order, it suffices to connect each to the next by a path in Thi n(ri ) C B-, (w) Then the union of these paths contains a path in B-,(w) joining ay to We shall choose 11(.0.1 ,7) .50 that g < 7/100 and 5r/ < r (F. 7/3),

(30)

where r(r ,E) is the function appeming iu Lemma 15 We shall choose ai < If the line segment iv: lies inside Gs, then we ate done, so we may assume that it lies outside As tvz meets the simple closed curve 8(77 only at its endpoints, it divides \ GIs into two components, one of which is bounded: see Figtue 23 Let R denote the bounded component. Let B be the at c. of 06T which. together with iv:, bounds R If B C L32(w),

7 2 Sunni

Theorem

99:3

Figure 23 Pmt of CC (hexagons). including two points at and z on its boundary with dist(w, z) < 'Hie shaded legion R. is bounded by B C OCC, a piecewise linen/ curve joining w to z„ and the line segment The curved line is part of F - ; the point :r' E is within distance 25 of the point x E DC", and B- C I? is the arc of from le ' to z'

then we ate clone, since B C joins ar to c Thus we may assume that d ime is a point ,r E B wit h dist(t, > 7. One of the hexagons meeting at say H1 , lies in U. while anothet, flo, does not, and so lies in 17; the hexagon tr■ is drawn with dashed hues in Figure 23 As Hi and Ho are adjacent, it follows from the definition of G ,7 that 17 .- meets Ho. at a point x', say Let w' and z' be the points at which we leave 1? when we trace the cm ye F- front a' in the two possible directions, and let B- C 1? be the ate of E- horn Iv' to z' containing x' As r- cannot cross 80:5-, the points t ri and z' lie or/ the line segment to:: Iu particulat , dist(uV , c') < As noted above, it follows front Claim 17 that we may parametrize F and so that co t responding points ale within distance 5. . 3 < E t Rom the construction of f" (see Figure 19), it follows that we may parametrize F- and r so that cot responding points ale within distance 2-7 1 < 2q Let w" and z" be the points of F cottesponding to the points ed and c' of r- Then dist(w". z") < 5q, so. IA' (30) and Lemma 15, one of the two arcs of E joining w" and z" lies in 13,)/3 (17" ) The points of I"-

Confot mal in vat iamt Suth6ov s 'Thcorem conesponding te,/ this ate Mini an am B' joining er/' and lonmining within 1.3_,,,3±,:,(6 ") C d o(w) This ate cannot be 13 - , which emit /tins the point: 2 r 13. 1 ,(r) Thus. B - U = [The calve B - U /IC:/ / is contained in B. and so does not wind mound curve 13' U tv'z' is contained in B-,(w). and so does not wind atoned r Hence. I3 U B' = P c does not wind mound x, cunt /Mid ing E

Cri T

CD

❑

The pool of Lemma 1/1 is now easily completed It remains to cuisine that CI ,7! and (.7-i" satisfy all the conditions in the definition of a discrete domain The conditions that the domains and theit outer Imundmies have no cut v e hex 11110 he et/Fenced by fit st teplacing tS bti et/10, and then I mincing each hexagon by a 'hexagon of hexagons" As noted eat Mu the condition that each marked vertex has at least two neighbours outside the domain can be ensmed by moving each walked vertex at most one step nronnd t he boundal v Proof of Lent Illa .14 \Ve have shown that given a small enough Ei > 0, = E5(EI). the construction tot described abm, e is valid lot all it < Considering a sequence of Values of m O. we may thus pick domains Cit lo t all rt smaller than some no in such a way that Cs is defined using h m(rr) — 0 as el — 0. a value s- 1 (6) The tcyuir etn ent that CST and /7 1 ale o(1}-close as — 0 Inflows nom (29) Condition (19) is given In' Chihli 20, and the final condition of Lemma 14 is gum/it/O'er! by Claim 21 ❑ Let Os I emai 1r that the proof of Lenuna 1-1 shows that when defining Pi ( /7/, TL it does not inattel exactly how we neat the houndmv proof of SmirnoWs Them cm will be valid tot any definition of a c t ossing of D., lin \vhich a flossing of the 'longer. thinner' domain C-:;" horn 01 1 (LC ) to ../1/ ! (CC) t hat stays tar ham t he cot nets guru antees a crossing of D., how .ffl to 21 3 . and prevents a mussing of DA 110111 1 2 to .4 I . FOr example, we could define a ("tossing of DA 111 ST from A i to .4 ) to be open path in D Ma/ling and ending at points within distance 1051 of Ar and .!1!

7.2,6 Completing the Proof of Smirnov's Theorem Let D4 = (D: P). P./ . P i ) he a -1-ntai keel hot dan domain. rind as berme let 7(D 1 ) denote the limiting ctossing ptolabilit S lot D I predicted by Ca t ilv l s fonnula Thus 7(/). 1 ) is a cottfbnual MN/a/iron of D I which.

7.2 Smirnov's Theorem



225

following; Ca t lescm„ may he defined as follows: let be the unique conto the equilateral triangle that maps k/rut/II map from , P2 and 8 to the vertices (1.0). (1/2. fil/2) and (0.0), respectively, so /naps p, into a pout (r, 0). 0 < ar < 1 Then rc(Ds) = Let C ST be the discrete domains approximating Ds whose existence is guaranteed by Lemma 14 Note that there is a function E = 0 as r5 -r 0, such that with s

0:sl. is ai-close to Ds

(31)

Of coutse, the same holds for Cl,t, but, as we shall see.. it suffices to consider G's-. To prove Ibeotent we shall show that

Ps(G)

rr(D.1 )

(32)

as /5 0. where Ps(CCir ) is the pH/It/Wilily that the 4-marked discrete domain contains an open crossing joining its first and Hind boundary at es. Indeed, (33) P,s(Gt ) r(D-1) can be lamed in the same way as (32) and then P i (Dr) 7i(D4) follows front (19). In fact. WI iting = P2 . P3, P I , PI ) lot the dual' marked domain. the consti net ions of the apptoxiination CI" to D4 and the apt/it/xi/nation 61,1 to DI given in the previous section ate identical Hence. (33) follows hum (32). the /elation m(D. 1 )-+ 7,-(D1) = 1 (see (4)).. and Lemma 5 ha this leason we consider only C;" Tot the moment, we shall regard the domains D., = (.D: Pt , P2. PI) and 0,T (67: e t .r u:t vs) as 3-mallred, by fmgetting the fourth mailed point.. Pi of v., We write = .4 i (D// ), 1 < i < 3, fig the boundary mcs of the 3-tai Iced domain .D 3 = (D; P}, P4 ) \\'e use corresponding notation 16/ the brt/Id/ay arcs of 67 Let f,l(z) he the 'et ossing-under' probabilities defined by (9), for the domain Gs = = , 173 ) Thus, lot z the centre. of a triangle in CI . th e have C5(z) = P(E75(.:)).

where Es(z) is the event that these is an open path in 67 born = A 1+1 (67) to .:4; 4.2 separating z horn Ai (an arc of the outer bor/ditty of 67) Hoe, the subsoipts ale taken modulo 3 Let us extend to a continuous function /115 on the closure D of D as follows Recall that is defined at/ yin ions points of i e..

226

Conformal Maw lance S t mirnov'a Theorem

the centres of various triangles (faces) of the lattice 81/' At the centre E 61-1 of any face of 45T meeting D. set: g,C(z) equal to the value of JA at the closest point ay where /,( is defined. (H there is mote than one such iv, choose one in an arbitrary way ) Note that, from (31), we have dist (II'

< E +

26.

(3

say \\1e extend hour the centres of the triangles meeting .D to the entire' triangles as follows: first, take the value of fio at the corner :r of such a triangle to be (say) the average of the values at the centres of all triangles incident with a and meeting D. Then divide each face of ST into t lace triangles meeting at the centre, and define chi on each of these triangles by linear interpolation Finally, 'forget' the values of 9 i5 outside D to obtain a function with domain of definition D \Ve shall Use two proper ties of the interpolating functions f./15. each !Ali is a continuous function on D Second, for any and c E D. there are points le t E 811 at which fA is defined. with 115 (w) < gfc(z)< fRui f ). and dist (net„). dist (iv', z) < 2E.. where = 5(5) is as in (31) The first property is inunediate from the definition of q The second follows Prow (34) Claim 22 The f (Indians

MY'

uniformly equiconlinnowi

Proof It suffices to show that t he functions fL ii ate tudIcainly equicout iummns Given ii > U, we roust show t hat t here is an q > 0 such that. for all to EDCC With dist (c. u . ) < q Mid all 8, Ale !Mee (c)— (415 001 0 and restrict our attention to the functions for S < 11 1 (0): whatever the values of (z) tSr (it) the linear inter polation in the definition of id ensures that the > functions gA. e > 0r ( i3). are uuiforruhy ecpticont imams (indeed. indica rub, Lipschitz) As no point of the 3-marked J o rdan domain 3 D3 lies on all three arcs = (D3 ). there is a constant e > 0 such that

max dist (II', A i (D3 )) > fi n

(35)

aro. rc E 1) Let us choose 'y > 0 so that < c/ltl and (63/c)'< d/2. where o > is the constant in Lemma -I By Lcuuua 1-1, if we choose Sr small enough. I Iwn t here is an I/ > tl such that any two 'mints g of GT. 8 < S I . at

7 :2 Sminuols Theorem

227

distance at most q are joined by a (geometric) path in G il that stays within tire ball 13s, CO Reducing 6 1 , if necessary. we may assume that for all 6 < 6 1 , where 6(5) is as iu (31) c(6) < Suppose that 6 < 6 1 . and ar z E D with dist(w,z) < ti/-f. It suffices to show that g,15 ( z ) < 0, 15 (w) +13 From the second property of the interpolating fitnctions q listed above, there are ///, z' E 67 with ( to') and dist,(tu,a0 < ti/4 and dist(z, z') < 0/4 such that gA(tv) > z' may be < 0(z)). Note that dist(a",zn < q The points 9,i(z) joined by a path in CI; lying within 13-,(u') As (715 is 2-connected, it follows that there is a path P' E 811 from w' to z' all of whose vertices are the centres of triangles iu CT. with P' C B22(1/P1) Suppose that EA(z') \ EA(iii') holds Then there is an edge :ry of P' such that EA (0\ EA (x) holds. But then, by Claim 10, the three sites of 07 immediately next to :c are joined by monochromatic (i.e all sites open, or all sites closed) pat hs to the three boundary arcs of CT One of these paths must: end at a distance at least c – – > c/2 front ni', say, where e is as in (35) Note that dist(r, w') < We have shown that if EA (.-2')\ EA (tv') holds, then there is a monochromatic path crossing the anutilits centred at a / with inner and outer radii 3 7 and c/2 By Lemma -1 and our choice of 2.. this event: has probability at most

2 Hi It follows that

–

' ') < ci Thus. (a

< /Pc ' ) C ( 0'1 ) + <

+

❑

as required

The functions ifs take values in [0,..11 Hence, these Functions are uniformly equicontinuous and uniformly bounded Thus, by the ArzelftAston Theorem (see, for example, Bollobtis [1999, p. 90]), any subsequence of (g-, A, q) contains a subsequence that converges uniformly continuous to a limit IP), with each The last two (rather substantial) pieces of the jigsaw puzzle needed to complete the proof of Theorem are collected in the next two claims The first describes some crucial proper ties of the possible limits (g 1 . 0 2 . ,q 3 )

Claim 23.

Suppose, fa, sonic sequence

-- 0, the G inks (11

228



C'onfot mat mew lance

converge toilfu l oily to (g 1 // l1 ) any contom C in D. we have

nov 5 l'hemmti

avtk each

oE

5C1

Then

or

(36)

A i we ham

11'(o)=

who du 'opal

coalMoon

g(c)do.

g i+I (o) do = w an foi (my point

f

0

and gi+1(z)-1-

(0)

= 1.

(37),

ipts are taken modals 3.

Proof Thioughout the proof

WC consided onl y values of in the sequence /5 lot d„ to /noid nimbi/is/Atte notation We stmt with (3(t): since the g i ale continuous functions on a compact set. they ate unikandv continuous Thus. it: suffices to consider equilateral triangular contours C with sides parallel to the bonds of T: an arbitrate contour C' in 11 call be approximated by a sum of such

SS„. wilting

CO/MOMS

Let C be an equilateral t tiangulin contour ill D with sides parallel to t he bonds of T Poi each J there is a discl et e triangular contain Cs in ST tc id lin distance ci of C. Using (31). as C is contained ill t he open set, D. lot ri sufficient-1Y 5116111 we Iliac C C7 ,-c- Recall that the discrete contom chi? [ 15 (i:)d:: is defined IA int grating a function uhosc /dues al given IA values of at lattice points wit hin dist mice and, at these lattice points. = rhii Since t he fittict ions ti,C a t e unifin nthequicontinuous. it follo/‘ s hat /)

dz

!Edo)

o(I.)

as S — whete the second it/leg/id is t he usual contom ioteguai Since the ri1c = q converge indlo t l i th to we have ( lc as 61 = d„ — deduce that

-r-

=

(I) f:/(:)(1

o(1)

Combining the relations /Write with Lemma 13. we

0

g'(1)

d o

0(1)

Since licit het integral depends on S. this proves (3(1) FO 1 (37). it suffices to prove the case i = 3. sad, flu a fixed o

E

7 2 Smil noWs Theorem Since the g of .43,

at e



229

continuous We mar assume that is not an endpoint disc(_. .*1r). (list (c.. A 2 ) > a

(38)

for sotuw (I> 1) 0. the 3-marked domain (3iis .s-close to D 3 with E = E(6) — 0 As 6 Shifting GITT with co e D and co — Hence. thew a t e points E 8H is the centre of a ttiangle 1w at most 26. we mac assume that uniformly, coo fro in C1sThus. f,((cs) is defined. and. as !LI — latch)

fhis( c ) = g (c ) + o(1)

(39)

Is — 0 Flom (31) we may choose a porn ft's On the 'mutably of CT,T (viewed and hence as a union of hexagons) so that, its (5 —+ lb we have ws first (ms, c o) 0 Let :I's E C',.,1" lie a site of the inner houndm v of Coll in whose hexagon tux lies, so (list (,ro. ws) < 6 Note that it c = P l . then we may take rs to be e 3 The :Hausdoi II distance between the discrete boundary arc A;(67) and the emit:imams air A i (Da) tends to zero 0 Thus. from (38).. lot 6 sufficiently small. the site lies on the ate A3 of C1,7 (viewed as it 3-marked domain) and the point: ws on the cottesporaling ate of the bow/du t y 30W of the union 0,7 of hexagons Suppose that D:51 (.7. 3 ) holds Thc‘n time is an open path limn Al(0,7) limn At (Gill ), and, in pal t Rada', sepal ating to .'10(03l ) sepat ating Such a path must moss the line segment: cows It follows that horn ws some site of within distance (list(:,, ws) + 26 = o(1) of c 3 is joined by au open path to snow site at distance at least a — o(i) hum c3 this event has mobabilitv o(11), so Lemma =,i) = 111'(El i3 (c 3 )) = o(1) Using (39). it follows that cia (:) = Recall that 3! 3 is a lumndat V site of A 3 (0) within distance 6 of ms Let Es be the event that some site in Baistfrc ?co+ t oc(zo) is joined by an open path to A i (CI3- )U Aii,(CC) As above, LP(E3 ) = obi ) Let 0,15 be the 4-marked domain obtained front the 3-marked domain 03 l by taking to as the 'Muth marled point Note that if c = so ,r 3 = el , then we recover the original -I-marked domain If E3 does not hold. then no open path horn A t (gli ) to A;1 (0) can tome within distance 26 of the line segment /tows. Hence E,i(z:s) holds if and only if Cfs has an open (dossing h om A [ (Cs ) = (0;C: ) to .13(6"3): see figure 2-1

Conformal

230

ar lance - Mirwou's Theorem rs-- A!,

Figure 24 A 3-marked discrete domain 61 1,1 with boundary ales A i , 1 < i < 3, and the corresponding 4-marked domain 0,1j with boundary 8104 < i < 4, obtained by marking a point ass on .4 3 . If no open path joins A I U An to the line segment ,rozo, then an open path from .1 ' to A3 separates z,) from :At if and only if it ends at a site of AC,

Hence, P(Gs has an open classing limn A', to A'a ) = d(zs) + o(1) (-10) Similarly. if E6 does not hold, then EA (As) holds if and only if C's has an open crossing from ,41, to .4 ,',, so has an open (flossing from A1, to A',1 ) =

(zs) + o(1)

Using Lemma 5. it follows that fd (z6) + ffi (-7. 6) = — o(1).

Appealing to (39) again. g 1 (z)+ 92 (z) = 1 follows, completing the proof of the claim ❑ As before,

+ fr3)/2 be a. cube root of unity

Claim 24. Let A he the equilateral triangle with vertices n i = no = w and va = w2 . There is a unique tnide (2 1 ,62 , y 3 ) of continuous functions and satisfying (36) and (37). MU theunore on.D taking values in ( 0, y i (z) = (:z(z)), where h i is the lineal function On A with // 1 (z) equal to two-thirds of the distance from z to the ith side of A, and y is the unique conformal map from D to A whose continuous extension to F maps Pi to ye for 1 < i < 3 Proof Let (0', r?, q 3 ) be continuous functions on D satisfying (3(i) and



7 2 Smirh an's Theorem (37), and set = t +0_,92 contour C in D we have g =-

9 +

g2 + w 2

w293

g3

231

Then g is continuous on D. For any 91 +w2

gl (a2.1

g'= 0.

ice, by Morera's Theorem (see, e.g Bear don 11979, p 1661), the g is analytic in D. unction f On the b011./IdatY of D, condition (37) ensures that, fors E A i , the value of g(z) is a convex combination of co 1+ r and44) 1+2 , i.e., that g maps into the line segment ig 4. 1 e i+ 0 Since g is continuous, it follows that g maps Pi into Furthermore, as z traces the boundary of D in the anticlockwise direction, g(z) remains within DA and winds exactly mice around this boundary As noted by Bettina (2005], it follows that g is a conformal map from D to A Indeed, applying the argument principle (see, e.g., Beardon (1979, p 1270, the equation g(2) = w has a unique solution for each w E A, and no solution for w outside A Since g maps the craps g and y are identical Pi to Arguing similarly, the function + g2 -1-:g3 is analytic in D. continuous on the boundary, and takes the constant value 1 on the boundary Thus g2 + ga is identically 1 This means that the real-valued functions g' are determined by g: for example, (201 — g2 ga ) / 3 + (g 1 + 9 2 + ifs)/3 = 211.e(g)/3 + 1/3 As g y, this shows that the triple (g I ,g 2 ,g3 ) is uniquely determined. The triple given by g i (z) = hi(c,..-..:(z)) satisfies conditions (36) and (37), so the claim follows ❑ We now have all the pieces iu place to prove Sinn noy 's Theorem Proof of Theorem 2 Let P3, P4 ) be a 4-mmked Jordan domain Let (.7:5- and G7 be defined as in Lemma 14, for all 0 < S < So, 7r(D4 ), since where So > 0 is constant It suffices to show that Ps(C7) the same argument shows that Ps(G,I) ir(D4 ), and then .P,5(D4,T) g (D.1 ) follows from (19). Let f‘ and g,is be defined as above, using the discrete domains 67 We claim that /4; o y, where h i and y are defined as in (Hahn 24 Suppose, for a contradiction, that this is not the case, i.e that there is there are 1 < e < 3 an E. > 0 and a sequence (50 such that for each and 2 E D with ^

(41)

Co Ilia ! Mal 1111 , 01 )(HUT S1011101 s Theorem

232

B y Claim 22. the functions g are unifounk eqificont Moons on the compam set IT Since they ate also unifinink bounded the sequence ,, , has )1 1111ilin ink COMV/ gent subsequence Li t Claims 23 and 2[1, this subsequence CO/Wetgem to (h i 0 y.11 2 0 ;. // 1 o y) contradicting (A ) The proof of Claim 23 slams that these are points c, [ n with — Pr such that f

,14

)

P,;(67 )

(cs)

o(1) =

o( 1) 1 ) = 11 2 (y:(P.1 ))

o( )

Indeed. the first equation is exact Iv (-10) For c oo- PI : as noted Mat\ e. the domain Cr[:, appeal ing in (-10) is exact IN t he -1-m a nked domain 0:[[ in this case As 11-' (y(1). 0) = 7(1 ) 1 ) by definition, t he proof is complete ❑ Site poi colatimi on the tiiiiiiLmtm lattice is t he mils standard pc[ i colat ion model For g hick toilful/nal invariance is kiwi\ iv. tittle MC SOUR' 110/1standaid models to \\ inch either Sinn nov's Themem. or its pub! has been adapted Indeed. Canria. Nelk111811 and SidthilVitillti [2002: 2004] haVe ', 1 St ithiitiii0(1 t ith i lismal invariance certain dependent site peu blat ion models nu the tiiaugular Lattice, obtained l, l imning a patIicnlur deterministic cellular automaton horn an initial state given 11\ independmo site percolation The\ establish continual invarance by showing t hat the dependent model is hi a cethriu sense a small per tin bat ion' of t he independent ittoth . 1 and t hen apply ing Themem Chaves Lei 120061 defined a Lather unusual class of percolation mids based on dependent bond percolation on t he I iangulal lattice. mid muted the equiirdent of SmitnoCs theoreur t o t these models by t I anslat ing Stint Smituoc s pool to this context ft remains an blip)! ant challenge to move conEumal invariance for am other standard model. lo t example. lo t site or bond pet colation on the square lattice. or lot Gilbert 's model or random \Mir : Hun percolation in the plane (see Chapter SI lot

7.3 Critical exponents and Sell/ amin—Loewner evolution Thee is a widel y held belief t hat the behaviour of various phenomena'. inducting critical percolation. should be chat arm/ ized by certain `ethical exponents" This opinion originated among theoretical physicists hilt by WA\ mnuy mat hematicians hale been come t red Pot percolation. these et itical exponents should (10 1 )0 1 1(1 on the dimension. but nut on the details 01 the particular nrodel considered. T[his is a eels

7 3 C ritical exponents and Schramnistocuinel evolution

233

substantial topic in its own right; a detailed discussion is beyond the scope of this book Hew. we shall briefly state the main rigorous results for percolation: the existence and values of these critical exponents for site percolation on the triangular lattice follow from Smitnov's Theorem and the work of ',alder. Schramm and Weiner. Our presentation is based on that of Smirno y and Werner [2001]: we refer to the reader to their paper for further details and full references To describe the exponents associated to critical percolation we need a few definitions. \Ve shall consider only site percol a tion on the triangular lattice, although the definitions make sense in a much broader context We write F1, 1, for the probability measure in Whiell each site of the triangular lattice I' is open with probability p, and the states of the sites are independent As before. we write Co for the open erratic; of the origin, i the largest connected subgnaph of T containing (I. all of whose sites are open Fin the critical exponents. we use standard notation: see, e g Reston [1987c] Recall that the percolation probability 0(p) is defined as = fril i ,(C la is infinite)

= 1/2. while 0(p) > 0 if p > 1/2 and that 0(p) = 0 it p < general set ting. the behaviour of 0(p) rrs p It is believed that. ill a that i tends to m hom above follmis a power 0(p) = (p

p()4*()(1)

p

p, limn above.

(42)

where ./.1 > (I is a constant that depends on the dimension but not the details of the petcolat ion model Tinning to the size of the open cluster Co when it is finite. as earlier, we write V (p) = Ep (lCal) for the expected size of (number of sites in) Co As k(p) = x for p > mr . it is rather mote informative to modify the definition slightly.. and consider \ f (P) = E a

so that

\ f (p) = A(p)

( lCallacokic)

tP,,(IC01

if p < pr Again, power-law behaviour is expected:

\ I ( p ),

_

as p— pc,

(43)

where is a positive constant. and p p, bran either side. Fru theimme, at the critical probability. the tail of the distribution

23-1

Conformal hrouriunce Smintov'S Theorem

leol is expected to follow a power law: (it < ro l < oc) = a-1/64-am as

11

as is the tail of the distribution of theradius 1-(C0): (11. S

r(Co) < oc.) = n- 1 / 6 , +00) as

a

In the last definition, it makes no difference whether we measure the radius r(Co) of Co in the graph-theoretic or the geometric sense, since these two metrics are equivalent More precisely, we may take r (C0 ) to be the maximum graph distance of a site at E C0 from 0, as before. (a we may take r (Co) to he the maximum Euclidean distance of a site in from the origin For site percolation on the triangular lattice, we have 0(pr ) = 0. so the condition roj C Do can be omitted in (bl) and (15) The ear relation length describes the 'typical radius' of art open cluster: writing for the Euclidean distance between rr E T C C and the origin, lei Co

1/2

5(M- -,01)),E7

1111 2 11',({0 —

rol < x})

n

It is conjectured that = IP — Pc1 -1 '

for

P

(-1G)

The reason for the term 'correlation length' is that ti(p) is expected to be closely related to the probability that two sites at a given distance 1. arc in the same finite open cluster: roughly speaking, this probability should decry exponentially with WE(p) Finally, it is expected that, at the critical probability, we have P P- ( 0 -

(I) = i41 2-d-11+"(1) as



(-IT)

where (/ is the dimension (so (I. = 2 for percolation on the triangular lattice ) The constants (.3„ 7, 6, (b., v and ri defined above are called critical exponents, provided they exist We have used standard notation for these exponents (though 6, is also written as p); the !bun of (47). example. shows that this notation is not the most natural for percolation. In two dhnensions, it is not hard to deduce horn the Russo -SevinottrWelsh Theorem that it one of r l and 6, exists, then so does the other, and d-2-hq=216,

(48)

235

Y3 Critical exponents and .Schramm.-Loeumer evolution

' Aced, let x i , ,1], be two points at sonic (large) distance r II there is an open path joining II and to, then each,r; must be joined by an open path to the boundar v of the ball B, i3 Ur 0, say But with probability bounded awa y from zero there arc open cycles in these two balls surrounding the centres that ate joined to each other; relation (-IS) their follows using Ian is's Lemma (Lemma 3 of Chapter 2). Mester' [19871; 1987c] established highly non-trivial relationships bemen the various exponents for two-dimensional percolation. Firstly, building on Iris work on the 'incipient infinite cluster (Kesten (19861), exists), then so does 6', he showed that if ti exists (or, equivalently. it Fur then/lore, Ire showed that if the exponents ri and 11 (5 ± 1) = with S exist, then so do the other exponents, and Ij In pa/

2 1/ 6-I- 1 '

=

6— I + 1

sind "

±I

for two-ditnensim al percolation the 'scaling relatio + 2,3 = 0(5

)

and

= ti(2

1/ ),

ns'

(J19)

hold, as do the '111 T elsealing relations. dO, = +

I

and 2 q

I

— S+

(50)

It. is believed that (-19) holds in all dimensions, and that (SO) holds fo r d < 6: see Chimmett [1999] A yet y° large number of papers have been written about the ethical exponents associated to percolation aud the relationships between them: see, for example, Rudd and Frisch [1970], Wu [1978], Kesten [1981], Aizemnan and Newman [198-1], Durrett [1985], Chayes and Chayes [1987], lasaki 11984 Mester and Zhang [1987], IKesten [1987a: 1988] and Hammond [2005] Burgs, Ch ryes, Kestim and Spencer [1999] proved the deep result that the hyperscaling relations (50) hold as long as two assumptions are satisfied: 5, exists, and, at p = N. the crossing probabilities for cuboids with fixed aspect ratios are bounded away from I. (Their results are stated for bowl percolation in Za , but proved in a more general setting.) The latter assumption is expected to hold for cl 6 Bettuning to two dimensions, Selo anun [2000] studied a certain scaling limit of 'loop-erased random walks' in the plane, which we shall riot define. He defined a family of random curves iu a domain in the plane, whose distribution depends on a real parameter which he called the stocli astic Locivrod evolution with parameter h, and denoted SLE„.. The

230

Confirm& i mam iance 5'mirnotc.s Theorem

random cunt S L E„ is often known b y the name Schwalm Locione i con: lotion. SatWI W I showed that if. as conjectured. the scaling limit of the loop-erased random walk is confounall y invariant. then it roust be SLE), Fun thermore. he showed that SLE6 is the only possible cordon wally in_ variant 'scaling limit: of el itical percolation on a lattice in a sense that we shall not untke precise Stith nov [2001a] proved that critical site percolation on T does indeed have a scaling limit. and that this limit is conformalle invarinrut and thus equal to „SLEn Considering the face percolation on the hexagonal lattice corresponding to site percolation on the triangular lattice. it liAlows that. for p 1/2. the long-range behaviour (i.e . limiting helm ion ic as the lattice spacing tends to zero) of interfaces between open and closed (black and white) regions converges to SLE1 In a series of papers Lawler. Schramm r i nd \Ve t net [2001c: 2001d; 20026; 200221] studied the behaviour of SLE: ire particular. the y deter ironed Nal •cl it ical exponents' associated to SLEi ; Combining these results whin those of Slid/ /10V [200!a]. SLIM mixt and \Vet nett [2001] established the existence and values of the various critical exponents fin sit e percolation on T Theorem 25. Poi site parrolation on the biangulat lattice the critical opponents 3. -; ti and q di/toed implicitly by(42). (-13). (.16) and (.17) and flu' inners 5

-13

=

15;

4

5 and= TT-

0

These values for the critical exponents coincide with the predictions of theoretical physicists: see Kesten (1987c]. Sion 110V and \Victim' [20011 and the references therein As noted above. 6, = 1112 = 5/48 follows relatively easily, and d = 91/5 161 IOWS ftun a the results of [(estctu [1984 The proof of Theorem 25 is based on crossings of annuli lb sa y a few words about this proof. denote tu ALt the event that in the site percolation on r restricted to the disc /3 R (0), there are (at least) distinct open clusters each of which connects a site in B, (0) to a site near the boundary of B E (0) For fixed i. the asymptotic behaviour of P IT:, (A; ??) does not depend on as long as n is large enough for tins probability to be positive \\:n it ing A llt for rrü, ??. say, S i nn no\ and \\hiller noted that. b y the results of Kesten 119874 Theorem 2,5 fellows horn the tur, relations P1 / 2 ( A IR) =

(51)

.1

ideal cdpoilelds and Schramm doewnci cuolution

237

wd (52)

P l/ q (-1 ;?) = R-5/

exists and takes Clearl y . the [list of these total ions states siniph that, the value 5/iIS Lawler, Schlanun and \ Vet net (20021)] showed that relation (51) Eelhaws flow SItlit 410V ' S COltrOl inal 'mini lance tesults and then results on SLE6 Smiinov and Wei net (20011 then punted (52) and thus 'Theoem 25 In fact_ Smir uov and \Venni' moved a none genital statement about ct ossings of the tumulus A); 1?) emitted tit the origin. with inner and ante! tadii / and R respectively As helmet we think of open Paths as black and closed paths as white Given a sequence c = (c;)1=! E ,If of colones. let H,( 1 , I?) he the event that A(t. I?) contains j vertex-disjoint monochr intuit ic paths P i whew P1 has cutout et, each 1); starts at a site of .4(7 1?) adjacent to the inside of the annulus ar id ends tit a site adjacent to the outside. and the initial sites c i al the c((, , v 1 around the cliche of tachus P1 appeal in the cyclic oi del (As the paths do not (loss.. then final vertices apnea/ in the sane cyclic order al ound the outer tit cle ) Let Ge (1. I?) he the emus corr esponding to 1-1(1 17). hut defined in the hull'-annulus =

E C : t

<

< R lite(z) >

It is not haul to sett that. For crit ical site percolation on It the probabilit y of the event C,(1 .1?) does not depend on the terms of the sequence c, only on its length Indeed.. the woof of Lemma 5 allows us to deli ie a I?) From the inner tilde 'lowest' (clockwise-most) open missing of e."(;.1?) to the ante! citele. 'whi t /revel such a crossing exists Having Found such a classing. Pi , we may then look lot a lowest open at closed crossing above PI in the same wa y. The lowest classing P i may be found without examining the states of sites above Pi . so when we look fin the next clossing, the inobabilitv of success does not depend on whetIon it is tin open cat a closed crossing that we seek It follows sin/HaiIv that ^us(C',(; R)) =

R)

lot sonic (ti (t, lit ) that depends on the length of c but not the actual sequence Sinn ion and Wet nei (2(101 showed that. , it j > 1 is fixed and d is huge enough. then (1 ) (t .1?), 1?

(53)

238

Conic)! Mal inner iance

Theolvin

on 's

x■ This exponent JO + 1)/6 is known as the j-arnt exponez in the half-plane

as R

Returning to the full annulus, one can show (see Aizenman, Duplantiét (fIc ( t, R)) is independent of the colours and Allat011y 119991) that P i r, in the sequence c, provided C contains at least one B and at least One H' This is related to Lemma 6 above: if the sequence contains two terrns of opposite colours, then it contains two consecutive terrns, and one can start by searching for an Innermost' pair of paths of opposite colours Then, working outwards, each remaining path may be found as the lowest crossing of a certain region, and the probability of finding the next path does not depend on the colour \VI iting bj (t, H.) for P i/2 (He (t,R)), wirer e c is airy sequence of length j containing at least one B and at least one H', Snriu nov and Werner [2001] showed that if ) > 2 is fixed and r is large enough, then Di (r,B)

RH12- 0/12+00)



(54)

as R ce This exponent (12 — 1)/12 is tire (inallichromati i) 'harm exponent in the plane The values of these exponents, and of the halfplane exponents, were pr edicted correctly by ph y sicists: see, for example, Salem and Duplantior [1981], Aizennum, Duplantier and Akaror/y[4999] and the references therein. If one is careful with the exact definitions at the boundary (rather than glossing over them as we do here in our brief description of the results), then the events .4 /;:i? and fle(r,R) coincide, where c is the alternating sequence of length 2k: the k black paths correspond to the k open clusters joining the inner and outer circles of A(r,/?), and tire white paths witness the fact that: these clusters are disjoint. Hence, the case 4 of (54) is exactly (52) \\T ifton': going into the details, we shall sty a few words about the proofs of (53) and (54), and hence of Theorem 25 As noted ear lie, t he key elements are the result of SMitlIOV12001a1 that the (suitably defined) scaling limit of an interface in critical site percolation on T exists and is equal to SLEG , and the results of Lawler, Schramm and Werner on the behaviour of SLE6 Putting these ingredients together, one can show that a d ( i . R)

(R 11)--.1(i+1)/ti+o(I)



(55)

as R, r — x with .1711 fixed. (In fact, to obtain (55), one first needs an 'a priori' bound of the Ram a 3 (1 R) = 0(17- I —I ) as R cc, for constants I

and > 0; see Smirnov and \Verner 120011 and the references therein )

7 3 Critical exponents and Seltrantra-Locumei evolution

239

To obtain (53), one also needs 'approximate multiplicativity' that, for < 1 < 13, (1,1(1 1, 1 -2) 65(T2, 1 3)

e(a10r,13));

see Kesten [1987c] and Kesten, SidO/aViellIS and Zliang [1998] When Considering paths of the same colour, this relation is fairly easy to derive front the Russo-Seyrnom-Welsh Theorem Fortunately, for half-annuli, One can assume that all paths have the same colour: The corresponding relation for hi is much harder; see SatittION" and Werner [2001] Finally, let us note that, in the annulus, the restriction that not all paths be the same colour really does seem to matter It is likely that 1Pr/2(lf,(7,R)) a, cc with t and j fixed, where c is a sequence of length j with = .13 for every i However, these 'monochromatic' exponents 7; are very likely different horn the multichromatic exponents (j2 - 1)/12 above " rm. ticulm, Grassberger [1999] reported numerical evidence that = 0.3568 ± 0.0008. The numerical value, or even the existence, of -;; for j > 2 is not known, although Lawler, Schramm and Werner [20026] showed that 7,, exists and is equal to the maximum eigenvalue of a certain differential operator Critical percolation is not the only discrete object known to have a cargo, mally invariant scaling limit described by SLE: Lawler, SCl/181M/1 and Werner [2004] have shown that one may define certain natural scaling limits of loop-erased random walks in the plane and of uniformly chosen spanning trees in a planar lattice; these are related to SLE2 and to SLE8 , respectively. The brief remarks hr this section hardly scratch the surface of the the theory that has grown out of conformal invariance and the study of SLE For a selection of related results see, for example, Schramm 120014 Lawler Schramm and Werner [2001a; 2001b; 2002a; 2002c; 2003], Kleban and Zagier [2003], Beflara [2004], Dubedat [2004], Than [2001], Morrow and Zhang [2005], and Rohde and Schramm [2005], Informative surv eys of the field have been written by Schramm ' [2001b], Lawler [2001; 2005], Werner [2004; 2005], Kager and Nienhuis [2004] and Canty [2005].

8 Continliuni percolation

Shortly after Broadbent and Hanttnetsley slatted percolation theory rind Eras and Bányi 11960; 1961a1, together with Gilbert 119591, founded, the theory of random graphs. Gilbert. 119611 started a closely related area that is now known as continuum percolation The basic objects of study are IV/1(101U g cometric graphs, both finite and infinite Such graphs model, for example a network of transceivers scattered at random in the plane or a planar domain, each of which can communicate with those, others within a fixed distance Although this field has attracted considerably less attention than percolation theory, its Minot tante is undeniable: in this single chapter, we cannot. do justice to these topics. Indeed, this area has been treated in hundreds of impels and several monographs, including Hall [1988] on coverage processes Monts 1199-11 on random Vorolmi tessellations, {tester and ROY [1996] on continuum percolation, and Penrose 120031 on 1. rtmlotn geometric graphs These topics are also touched upon in the books by Mathison [1975]. Santal6 119701, Stoyan, Kendall and Mecke 11987: 19951. And kutzutnian [1990] and Molchano y [2005] In the first section we present the most basic model of continuum percolation, the Gilbert disc model or Boolean model, and give sonic fundamental results on it, including bounds on the critical area In the second section we take a brief look at finite random geometric graphs, with emphasis on their connectedness The most important part of the chapter is the third section, in which we shall sketch a proof of the analogue of the Hatiis Kesten result for continuum percolation: the critical probability for random Volonoi percolation in the plane is 1/2. lyre shall frequently circuit/1ns sequences of events (A„) with E(.4„) as n :xy, using standard shorthand, we say that A„ holds why or with high probability in this case. As usual, an event holds almost snafu, or as . if it has mobabilibt

8

The Gilbert disc model

2-11

8.1 The Gilbert disc model for r > 0. the yerfes set of the standmd Gilbert diNC model, or the Boolean model. G,, is a set of points distributed 'uniformly' in the plane, with density 1 To obtain G,, join two points by an edge if the distance between them is at most i The trouble with this 'definition' is that it is not clear how we can choose points unifinntly, with a certain density n fact, it is easy to turn this hopelessly loose idea into a definition of a Poisson process in the plane, the 'proper' way of selecting points uniformly Let A be a positive real number.. and let PA C be a random countably infinite set of points in the plane Let us write bi A (U) fin the number of points of PA in a bounded Bore! set U: note that i t A (U) is a random variable \Nje call PA a homogeneous Poisson process of intensity (density) A it the following two conditions hold (i) If U,. ..U„ are pair wise disjoint bounded Borel sets, then the random variables p A (Ujj ), pA(U„) are independent; (ii) For every bounded Borel set U, the random variable / A (U) is a Poisson random variable with mean AlUi, where IUD is the standard (Lebesgne) measure of U It is easily seen that there is at most one random point process (a random countably infinite subset of the plane) satisfying these conditions; in fact, as pointed out In fiónyi in the 1050s, condition (ii) alone defines the Poisson process P A In the other direction it is not hard to show that !hele is a point process satisfying conditions (i) and (ii): for example, one can use the following concrete construction For A > 0. let {Ne i : j) E 2'} be independent Poisson random variables, each with mean A Thus,

for k = (I, )) E

0. 1

Let

P ( X i ,i = h ) = (CA\k /,•! be the unit square with bottom left vertex

Q;; = {(i.!/): I ti .< i

1. j < y < p± 1

every j) E E.2 , select Nu points independently and unifinudy from then the union of all these sets has proper ties (i) and (ii), so we may take this as the definition of PA Although we shall not study it here, let us remar k in passing that a Poisson process • P f of intensit y satisfies (i) and (ii), except that the

2-12

Continuum prErrolation

is a lion-negative _el: mean of µj(U) is given by 11 , /, where integrable function RenitIllilg to Pa, note that if Z is a Bore! set of measure 0, then the probability that PA lets any point in Z is 0; hence, in what follows, we shall assume (lint PA fl Z = th lot all measure 0 sets Z that we consider. For example, we shall assume that no point of P A is on a given (fixed) polygon; a little wore generally, given a plane lattice, every point of PA will be assumed to be an interior point of a face For A > (1 and 7' > 0, let a, a be tine random geometric graph whose vertex set is P. with an edge joining two points of Pa if they are at distance at most 1; see Figure I We call C, A the Chlber t model with parameter s r and A, of the Boolean inadel with parameter 5 r and A. With this notation, the standard Gilbert (or Boolean) model is C. = C

Figure I. Pail of Hie graph C a (dots and lines) Two vertices ale joined they are within distance i e , if each lies in the shaded circle centred on the other The model GC , A has numerous variants and extensions First, let us rescale by writing H, ,A for C.), A. This resealing is not entirely pointless, since it allows us to define a random subset of R2 in a natural way, by writing D = .D, ,A (P A ) for the union of the discs of radius 7' about the points of Pa; see Figure 2 Note that the probability that there are two points of out Poisson process at distance exactly r is (l, so it makes no difference whether we take open discs or closed discs There is a. one-to-one cot respondence between the compor ter its of C2,. a = (1,,, ',CPA ) and those of D, A = D, a(PA ); in particular, as P A has (a s )

8.1 The Gilbert disc model

243

Figure 2 Pail of the jit atilt G2, A (dots and lines) The shaded region is D,. A. of (.32,. A toe adjacent if and only if the conesponding shaded discs meet Two VW tices

no accumulation points, Go t A has an infinite component: if and only it 1),. A C R2 has an unbounded component Note that although D, A is veto close to H, A. the two models are not isornol pine, since the disc of radius I about a point may be covered by other discs making up D„ A The advantage of consider ing rather than Go, A is that the complement: of .D, A, the 'empty (or vacant) space' 1:3, ,A = 1R 2 D, A not coveted by the discs, is just as interesting as the original set D,.A particular, we can study the the component: structure Of E„ A as well: we can look for an unbounded vacant component, i e , an unbounded component of E,,A We may define variants of the graphs (1, ,A and 11, A, and sets D, ,A and E, A, by replacing the circular disc by an arbitrary centrally symmetric subset of R2 . Thus, given au open symmetric set A C R 2 for 11 C p2 we write G A (11 7 ) for the graph with vertex set 11 1 hi which two vertices x and y ate joined if x—y E A. Equivalently, for x E R2 , set Br B+x, where B = and join two points x and y of IV if 13, n B = Again, the union jrciv B„ reflects the structure of G .4 (11') If A is the disc of radius r, then we may write G, for GA (IF) In two of the most natural variants A is taken to be a square and an annulus Of course, the model also generalizes to d dimensions in a natural way. Taking for W the point set of a homogeneous Poisson process PA of intensity A, and letting A be the disc of radius r centred at the origin, we see that G A (PA ) = G, ,y More genet ally, in G A (1r) both A and 11/

211

Continuum percalat iaaa

may be chosen to lm random: one of t he simplest cases is when we assign independent identicall y dish United non-negative random variables r(x): to the points .c of a homogeneous Poisson process 'P A , and take the unit of the discs 13„,x)(3.), E PA Clearly. all the models above extend trivially to higher dimensions There me many other 'natur al ways to define random geometric graphs, some of which we shall mention' in Section' 2 If we condition on a particular point c: E 112 being in PA , then the degree of c in G, ,A has a Poisson distribution with mean riT 2 A lit rad the sir tutu e of G, ,A depends on the parameters 1 and A only through the expected degree: for Aurg A t 11): , the graphs A„ and C„ A, have the same dish ibution as abstract ,midair' graphs Thus, when studying fin exannile, the various critical phenonwna concerning these giaphs, are free r1) change either / on A, provided we keep a = / 2 A COnSrallr In view of t his, ice shall \\ /ite (7(a) fu, any of the random graphs C, A with:: o = -at 2 A, the canonical neptesentative being C, we call a the degree of 0, A The quantih a is also known as the connection arca. OF ShIlply, area: G, is ((ethicl bi joining each point x E P, to all othei points of P, in a disc with area hi an obvious sense. the random graph CI, A models an infinite corn mimic:at:ion ! Rawer k in which two transceivers can cornmunicate if their', distance is at most 1 As in the discrete case (when studying percolation On lattices or hit (ice-like infinite graphs), we sa y that (.7, A percolates if it has an infinite component N'Ve unite 0(t A) = 0(a) for the probability that the component of t he or igin is infinite To intake sense of this definition_ we shall condition C, ,\ on the or igin being one of the points; equivalently, we shall assume that the origin is in 'P A This assumption does not change the distribution of the remaining points of C, A The first question concur Mug I his ; continuum percolation' is when the 'percolation mobabiliti s 0(t, A) = 0(a) is strictl y ppositive. It is trivial. that 0(a) = 0 if a is sufficientl y small and 0(a) 1 as a Since 0(a) is monotone ha:teasing, there is a critical degree or ci Meal area a,: if a < 0, then the probability t hat C7(a) maculates is 0, and if a > at„ then this probability is strictl y positive Also, as inn the discrete case. Kohnogolo y 's 0-1 law (Theorem I of Chapter 2) implies that if a < 0„ then every component of G(a) is finite a s and if a > a, then C.;(0) has an infinite component a s Note that I he critical degree a t cot responds For percolation As we shall see later in to the critical probability Theorem 3, the natural analogue an of :/ci is equal to a,.

S I The Gitlie,t disc model

245

Our main aim in the test of this section is to give hounds on the n itica t degree a c . An easy way of bounding a, is by e0111pat ing C, A to a discrete percolation model with good bounds on its critical probability Perhaps the most natural model in the face immolation On the hexagonal lattice in which each face is open with the Sallie probability indeipen(entry of the states of the other faces, and closed odic' wise; two faces ate iwighbouts if they share au edge (As every vertex has degree 3, someshat misleadingly, this is equivalent to sliming a vertex ) This amdel is t ome usually desetibed as site petcolation on the triangular lattice Theorem 8 of Chapter 5 tells us that if p < 1/2. then a s there is no face petcolation, and if p > 1/2, then a s we have face percolation To obtain a face percolation model F l om PA. let A be a hexagonal lice, with each face a regular hexagon of side-length s. Define a face percolation model on A b y setting a face to be open if it contains i t least one point of PA. and closed ()diet wise. so each face is closed with limitabilit y c -AA and open with probability t - C AA , \viten, A = 3452 /2 is the area of a hexagonal face (see Figure 3) Front Theorem 8

010 Figure 3 Comparison between Gilberts model and face percolation on the hexagonal lattice, i e site pwcolation on the hi:log[11w lattice: a hexagon is shaded if it contains one or more points of the Poisson process The co p esponding site in the triangular lattice is then taken to be open If the hexagons have side-length s. then AB = 2 ‘,/s. so AC = of Chaplet 5, if 1 - c -A1 < 1/2 then we do tot have face pet col ion, and if 1 - c -AA > 1/2 then we do All that remains is to thaw the appropriate conclusions alma the disc model G, A Note that any two points in neighbouring faces ate at distance at most ((2A 2 1 2 ) i/2 s s ■/171, and any two paints iu non-neighbouting faces are at distance at least H Consequently, if

246

Continuum percolation

1 — e- A•1 < 1/2 and r < s then we do not have percolation in C,,A, and if 1 — c- AA > 1/2 and r > sVT .3 then we do, Taking A = 1 and substituting A = (3/:73/2) 5 2 , we find that if (3 ■41/2)s 2 > log 2 then a„ > 7r5 2 , and if (342)52 < log' 2 then a c < 137,s2 , i c ,

log 2 Mil-1o°'c, 9< a, < 34 34

We have given these inequalities to show what one can read out of the approximation of a Poisson process by face percolation on the hexagonal lattice, although the lower bound above is worse than the trivial boon 1 implied by the simplest branching process argument. On the other hand. the t i pper bound, given by Gilbert [1961] under the assumption (unproved at the time, but proved now) that the critical probability for site per colation on the triangular lattice was 1/2, has not been improved much in over four decades The slightly better upper bound in the theorem below is due to Hall [1985b], while the lower bound is the bound from the seminal paper of Gilbert [1961]. (In fact, the numerical value for the formula given there was 1 75. !) Theorem 1. Let a„ be the critical degree (area) Pi the (tither (Boolean ,lice model G,. Then 671

271-

612

< (lc < 10 588

Proof Let us start with the lower bound. Let Co he the vet tex set of the component of the origin in G, = G, , r, briefly, the component of the:: or igin We shall use a very simple process to find the points in Co one by one, but in order to achieve a concise formulation of this pr ocess, we describe it in a rasher formal way. We shall constr uct, a sequence of pairs of disjoint (finite) sets of points of the plane, (Do, Lo),(D 1 , say The points in D, are the points of the component Co that ale dead at time t: they belong to the component Co, and so do all their neighbours; the points in L 1 are hue at time t: they belong to Co, but we have made no attempt to find their neighbours. To star t t he sequence, we set Do = 0 and Lo = {X 0 }, where X 0 = 0 is the origin Next, let No be the set of neighbours of Ko, and set Dr = {X 0 } and L i = No Having found (D,, L i ), if L, = 0 then we terminate the sequence; otherwise, we pick a point K, horn L I , and define Dt+. = Dt U{Xt } = {X 0 ., X, } and L 1+1 = u L,\ {NJ, where Nt is tire set of neighbours of Kt that are not neighbours of any of

8 1 The Gi(bell (list model



247

is locally finite (no disc of radius l' contains the points in D, Since infinitel y many points of P i ), the sets D, and L, al e disjoint finite sets By construction. fun then mole, if =0 then Ca = U C Since \

I.V

0 1

At

CU

Ar

i=0

laVe

ID / I =

t—1
1 vet (ices then it has at least km/2 and at most km edges The random geometric graph H„ k is again a model of an ad hoc:: network of transceivers and, as such, it has been studied by many people,: including Kleimock and Silvestet [1978], Silvester [1980], Flajek [1983] Takagi and kleimock [1984], Hon and Li [1986], Ni and Chandler [1994] Gonzales-Banjos and Qniroz [2003], and Xue and Kumar [20041. We should like to know for which Functions k = k(n) the graph . More precisely, we should like is likely to be connected as n find a function ko(n) such that, if F.- > 0 is constant, then lim P( ,A-

ro

if k 5 (1 — 04.0(4

is connecte ) =

if k

(1 + E)ko(o)

Such a function ko(n) is considerably harder to deter mine than the ctiti-: cal deg/ ee d for connectedness in the Boolean model given by Theorem 7; since the trivial obstruction to connectedness, the existence of an isolated vet tex, is ruled out by the definition of H„ As we shall now See, simple back-of-an-envelope calculations give us the order of /,.[) (o) Nevertheless, we ate very far from determining the asymptotic value of 1;0 (0 In the arguments that fellov,, the inequalities are claimed to hold only if o is sufficiently huge. Let us see first that if r > e then k = k(n) = [clog is an upper bound fix ko(n). If every disc of area a = 7r1 2 centred at a point a: E V„ contains at most k other points of S. then ti„ e contains „ as a subgraph, whew „ is the graph on V,, obtained by joining two points if they are within distance r The variant of Theorem 7 for the square (rather than the torus) tells us that the graph C„„ is connected whp (with high probability, i e., with probability tending to 1 as n x) if a = log n + log log o, say. (In the application of the theorem, s = 1, and ev(n), log logn ) Fm this a, the probability that a fixed disc of area a. contains at least A' + I points is at most

ah±1 (k + 1)!

<

1

([/(!)/
0 nom basic proper ties of Poisson processes, it Follows that the probability that the disc of area a about seine point of 1,7„ contains more than A. other points of V„ is at most 1E' Consequently. H„ , p is connected hp, as claimed

8 2 Finite modont geometric graphs

259

1(1 — s)logn/81 — 1 is a lower Next, we show that if z. > 0, then k bound for To this end, define > 0 hi = + 1, and consider a family 'D of three concentric discs D I , D3 and D5 contained in S„, where D; has centre x and radius it: see Figure 7 We say that the family D is

WO

Figure '1 A family D = {D i , D. Dr,} of three concentric discs, and two possible discs D, touching D I For k 2. the fandiv D is had., so contains no edges from the vertices in D I to the test of the graph bad for the graph

hold: 1, (or set V„) if the following three conditions

(i) D I contains at least k + 1 points of 11„, (ii) D3 \ D I contains no point of V„, and \ at , the region D, fl (D 5 \ D3 ) contains at least (iii) for every c E + I points of 1 1„. where D, is the disc with centre c and radius dist(x..i) — 1 k is disconClearly, if some family D is bad lot H„ t , then the graph nected. A family is good if it is not bad Let us show that the probability that a fixed famil y D is bad fin H„ k is not too small First, the probability that D i contains at least k +

Continuum percolation

260

points is approximatel y 1/2; an\ at least 1/3 Next soin thin (ii) holds with ptobalrilit

Finally. condition (iii)) holds if it holds tin the points c with dist(i)..r), tar For such a point c, the area of D, n (Dr, \ Da) is en 2 for sow > 2 1111/S. one call cover D 5 \ D3 by a constant number C of regions contaitrs sonic The probability that. R., of area 27/ 2 so that any D, con) a given R, does not contain at least k+1 points oft„ is Y(1) Hence, the probability that (iii) holds is 1-0(1). and in pm Perrin/ at least 1/2 lot y large Since the events (i), (ii) and (iii) are independent, the probability that a family D is bad tor [I n k is at least fr-H-5/(i The square .5„ contains at least

>)- 101(( — x) log n)/(87v) )0 -11



// 5 log n

disjoint squirts of side-length lot so it contains at least this Walk" disjoint discs of radius St Therefore. the probability that even- familt D is good For 11„ p is at roost (

+16)

lug!,

ConsequentIt the plObabilli V that 11„ k is connected is at most cii of I) The mg uncut above can he consirkriabli simplified if all we want is that connectedness happens around k = Sflog 0: there is no need to use Penrose s Theorem. Firemen ' 7, which is quite a big gun lot such a S/11)111 sparrow Be that as it mat, the tenni/1(s above tell us that if c 1/S then /1„ r„g „ ) is disconnected whp. and if r then H„ Li l „; ,, i is connected whir :Kite and Kumar (200-11 were the first to publish bounds on li tt (n): they Firmed that 0 07. i logy is a lower hound and 5 1771logn is an tippet hound. with the upper hound firllowing from Pemose's Theorem (In fact, the upper bound 3 8597 log y is implicit in Cionzriles-Barrios and Quiroz [20031 ) Bollobris. Sailun and \Villiers 120051 considerabl y Unmoved the constants 78 and r in the Pi\ ird bounds above: in pm t they disproved the nal trial coujectuc t hat kay) is asymptotically log y which is the analogue of Theorem 6 fin H„ k Tlieolern 8. If < 3103 Own II„ , 1 „„„i I s ( I l yto li nerlfrl wh i t. I 11( 1 if „ H kfitlfleCh (I why c> 1f log I 0 51:19 fin?, „

S Punic irutdoin geometric gtretlIP

261

rather

involved proof and a nundail of related distills \VC itelet the reader to the ot iginal impel Pot I he

Instead of asking lot the connectedness of certain million' geotnet tic graphs. \ye ma y ask ha the mound domain to he cou r ted by the t andont sets we used to define the graph. These enticing() questions have a long history. and ate prominent in three boolts: Stu\ an. Nendall and Matte [1987]. Hall [1988] and Meestet awl Roy 119961 Here we shall say a few avoids about some of the man y results lust before the dawn of percolation. D ymetzEN 119561 laised a beimHiltl question collect ning cove t ing a citric la /tuition) 11(5 Let 0 < independent IN at iondont IQ say.. and then 9 < p < L close enough to I that (I — p)An d < I/10. we have EA p( )

= (1004

—

> 0 8639

'Thus. in the I-independent bond petcolation model .11 on 271 , each bond is open with plobabilitL at least 0 8639 Hence. hom the result of Ballstet BolloMis and Walters [20(15). Lemma IS of Chapter 3. with positive limb/Minty the origin is in au infinite open path P in III lViten this event happens. tlw get/met/it path 11 E 6 ( C P/ i consists entirely of black points. so the black component Ca of the origin is unbounded Hence lift the chosen valves of p and A. 1l(` have O(p) = showing that p H = m i (d)

it (C0 is unbounded) > (1 p 2 we ham p it (if)> Mont

rid C

0..

II, be the In peicrube

and. lif t t > 0 let 10 denote the /-neighbourhood of Let S, he the event that son c ball of (adios 1/6 with trentte 1-1, contains no point of P = U' PT and let U, he the event that sonar black component. (union of open Votor toi cells) meets hot H,, and a point of the boundat ■ Off/L I "1 of 11[ I ' ' ° Note that if Co is unbounded then U, holds tot every r such that Co meets

i

8 ,f Random Voconoi percolation

273

Clearl y . the event 5, depends oak on the restriction of ('P". P - ) to H;, I/6) Also. U1 . is the event that Hal e is ti black path in 11;, I/6) joining 0(11,(1/1 1 Hence U, depends onl y on the colon's of the points x (of P:) ) with a E Hil l/(1) If S„ does not hold. then the closest point E H;. 1/6) is within distance 1/3. so the colom of .v is of P to any determined by the testi iction of (PUP - ) to /4. 1/2) Thus, U,,\ Sy. U (U,,\ S e ) is determined b y the restriction of and hence U, US,. = (P + . p-) to /1,'/2) > 2 lot smile then the interiors If ti E :S U ate such that ler - the sets HI, 1/2) and 11;,.112) are disjoint. Let (7 be the graph with ver tex in which e and a' ate adjacent if nu - icT < I for even- i. so set U S, holds. U is a (3` ) -f)-regular graph faking (' to he open if we obtain a 2-independent site percolation model elf on G. As A — and p — 0 with Ap — 0, we have PA 1 ,(U, U S,) — By Lenuna II of Chapter 3. it follows that 'bete ate A > 0 and (I < p < I such that, with probability I. there is no infinite open cluster in Al 13ut then Co is filth (' with probability I t so fiii( d ) >

0

❑

The arguments above ale vet'. crude: we Mute included them only to give an indication of the basic techniques needed to handle the longr ange dependence in random Nroionoi percolation Pot huge fr. home\ el. it turns out that the method of Lemma 1(1 gives a bound that is not that tat from the bulk: the towe l bound in the result of Brdister, 13ollokis and Quas [2005] below was obtained in this way The upper bound is much mote difficult Theorem 11. If 4 et sufficiently huge, then the el Meal prob ab ility pH(d)

fin random Voronoi percolation on R d ,tatetfics 2 - (1 (9il log d)" i < p H (d) < ( 1 2 - (/ ‘71/ log& mhate C Is an absolute constant.

We now tu t u to the main topic of this section, the critical probability lot miion' Voronoi percolation in the plane If P is a Poisson process in the plane. then with probabilit y I the associated Voronor tessellation has three cells meeting at even' ve t tea: we shall assume this alwa ys holds how now on Thus the graph U-p is a triangulation of the plane.. and Eidel s lot implies that the ayetage (tepee of a vet tex of Up is 6 Thus. J on avetage', ap looks like the triangular lattice This suggests that random Voionoi percolation in the plane mat- be similar to site

27,1

Continuum percolation

percolation on the triangular lattice Of course, critical probability is not just a function of average degree For example, it is easy to see that fm any (small) e > 0 and any (large) d, h > 3, there is a k- connected lattice A with all degrees at least d such that p7;1(A) > 1 — ctThe numerical experiments of Winter Feld, &liven and Davis 119811 mentioned earlier suggested that p H (2) = 1/2 There is also a compelling mathematical reason for believing this: random Voronoi percolation in the plane has a self-duality proper ty that implies analogues of Lemma 1 and Corollary 3 of Chapter 3, the basic starting point for tire proof of the Hart Theorem, and of Kesten's result that p7.1 (T) = pfr (T) Tr, 1/2, where T is the triangular lattice (Theorem 8 of Chapter 5). Let I?= [a, /4 [c, (II be a rectangle in t he plane By a black horizontal crossing of R. ■ve mean a piecewise linear path P C 1? starting on the left-hand side of H. and ending on the the right-Laud side, such that every point :r E Pis black, i e lies in the Voronoi cell of some E P with t: open; see Figure 11 We write H(R) 111,(1?) for the event that

Figure 11 In the figure err the left, 11(R) holds: the thick line is a black horizontal crossing of R. In the figure on the right, /1(R) does not hold

1? has a black hot izontal crossing Equivalently. 11(1?) is the event that the set of black points in I? has a component, meet ing both the left- and right-hand sides of I? Ignoring probability zero events, the event .11(R) holds if and only if there is an open path P =n rr2 e t in Cp such that meets the left-hand side of R, 1:;., meets the right-hand side, and, for 1 < i < t the cells V„, and 1 7,„. , meet inside I? write 11„(1?) for the event that I? has a white horizontal crossing,

8.3 Random Voranoi percolation

275

and Vi,(/?), VAR) for the events that I? has a black or white ve rtical crossing, respectively Lemma 12 *

Let I? be a 'rectangle in R2 events Ilb (R) and V (B) holds

Then precisely one of the

Proof The proof is essentially the saute as that of the corresponding result for bond percolation on Z2 (or site percolation on the triangular lattice) Defining the colours of the points in 1? from the Poisson processes (P + ,P – ), colour the points outside I? as in Figure 12

Figure 12. the shading inside the (square) rectangle R. is flour the Poisson process 111 ,(R) holds if and only if the outer black regions rue joined by a black path, and V. (I) holds if and only if the outer white regions are joined by a white path. Tracing the in/Efface between black and while regions shows that one of these events ❑ 11;st hold Let I be the inter face punk i e the set of all points that are both black and white (lie in the boundaries of both black and white regions) With probability 1„ the set I is a drawing in the plane of a graph in which every vertex has degree exactly 3, apart front Four vertices of degree 1 As in previous proofs of this type, considering a path component of the

276

Continuum pereolai ion

interface shows t hat one Of HI V?) awl (I?) must Ind If boll; hold, A5 can be drawn in the plane Cotollaxy 13, If S 1 4, anti 9quarc /2( 11 ( 3 )) = 1/2

x [g. Ii (

>

O.

Mc

Proof Dv Le1 1 1111a 1 2 . lb/ U < p < 1 we have Pp(llt,(5))+P i,(Vi v(S)) = Flipping the states of all sites now open to closed or vice V(11S11. se that lil y ( Vw(S)) Flt-/)(11,(8)) Fioni the sin/ i n/env of the Poisson pro cuss , = Ifi.„),(III,(8)) Thus. P),(//b(8))+IPI_,,(1-1-1,(5))=. 1 Taking p = 1/2 the result follows 0 I t ;eedinau 11997j plover' ;01 analogue of Co t 011 8 1 N' 13 rtniceining 'essential loops in 'widow \lo t onoi percolation on the projective plane lie untuoks that p 8 (2) = 1/2 In' notnett v consideurtions floweret, although Co/011m v 13 is t he 'uison wh y. p 8 (2) = 1/2. I his Elkin' ohserration is even further how a woof than in t he case of bond percolation on Z2 Fin the Val ions lattices w hose et it ical ptobabilitv is known exactly, ant of the ninn y plaids of the trainer 1Kesten Theorem can be adapted with out too n i nth work 1 7 o/ /widow Votonoi percolation . the situation is different Some of the tools used to stud y percolation on lattices do cal v met t o t his cont ext , but ot het s do not In pal ticitlat e is no ohs 101111 was to wink .Nlenshikov's I hern CM. and no direct equi snleut of t he Rosso -Se% mow AVelsh Thiess is known. although, as we shall see late!. ;t weak loin; of t he hatterresult has been plover! Let us tarn w our main /11111 in t his section, a sketch of the proof t hat /111 (2) = 1/2 We star t w ith Hat rim's Lemma, t he first of the basic / esults for ordin a ry percolation Burt do extend to I he \ Tot onoi setting slate its analogue for Vor onoi per colat ion, we need t o define 'increasing (1 \ cuts' in t his context We call an event, E defined in toms of t he Poisson processes (Pt P-) block-inctrusing, or simpl y increacing, if. fin ever } configuration = (Xt. XI') E and ere; configut at ion = (X.1). ) ivith C X1E and Ail AT, we have E In other words. E is preserved lw the addit ion of open points, t he deletion of closed points 1 he definition of an increasing function j (P + P-) is analogous: / is hue/easing if adding open points or deleting closed points cannot decrease the value of .1 Since a point of 1R 2 is Hack if t he (a) newest point of p is open adding open points and deleting closed points can change fhe colon/ of a point



S Random Voinnui ' ,ovulation

277

2 how vylire to black. but not vice versa any event defined by the existence of cer brill black sets is incwasing: an example is the MVO event It is easy to see flow Hat is's Lemma that, with respect to P p . incwasing events rue positively correlated, at least if they are sufficiently well behaved In fact. all increasing events are positively correlated To see this, let us first ()onsider the case of a single Poisson process of intensity I on 1?2 Let us wtite (Q 1 iF ) lot the corresponding probability space, so Str is the set of all dist:tete subsets of F 2 . P i is the pr obabilit y Wen:it/In defining the Poisson process We shall write E i for the associated expectation. Let I. 2 he two bounded measurable in I lie sense functions on 0.1 1 , P i 1. and suppose that each is that ir) C u/ implies ,b (w) < (La') Following Roy [1991) let E k he the a-field gene/rued hi the rolloir ing iulou ttation: set g = 2 k , divide Hu. a] 2 into do squares of side-length and decide fru each whether or not it contains points of Y E Thus Er. pm (Rim 's Q., into 2 4 " pmts. with each part consisting of all E 1 that have at least one point in each of cei lain small squaws. and no points in certain other small squates Any two discrete sets lie in differ ent pat Is of E i k lot large enough so t lie sequence Er. is a lint at ion of M t , Pi) As E l (pr 1. ma y be viewed as an increasing flint( i011 On a discrete product space. bout Lemma -I of Chapter 2 (a simple coiolltu t of Hauis's Lemma) we have E R

1-1(R)=

i =

l/n,

C-2

I.

EI(EW/i :0EM2 1:0)

(EMI E k)) E 01(1/4 I Ek))(*)) Ei(m)Ei(f/2)

Since i is hounded mid Er. is a lint at ion, lir, the Mat ti Theorem (see, [or example. 'Williams I1991D g

E t ( g i I L k) 22'

le Convergence

C/a

almost er el \ here. Hence la dominated comet gence. the left-hand side of (5) converges to (g i go). proving t hat Et

( g i q2)

POE' 02)

(6)

whenever th and are nwasumble hounded and Met easing The cot tesponding result flu the colour ed process follows immediately \Vt. ) state this 418 a lemma giving onl y the character istic function case - although the proof Fin increasing functions is the smile we shall not gu

Conti truant percola lion need the result. Hem. then is the equivalent of Harris's Lemma for taudom Voronoi percolation in the plane Of course. a corresponding result holds ha random Vol own percolation in Ea

Lemma 14. Let Ei = E i (P + .2 - ). = 1.2. be two black-Me events. Then lo t any 0 < p < 1 use hone IPy (E,

n

IP,,(E1)Pp(E2),

where iti the probability measure minor:toted to fandom Volonoi percolation in the plane Pl oy/ -We shall w i ne E

i e )Cei at ion with respect. e to EL p Let, J; be the elm/atter istic function /if E h so h is black-increasing Fixing F lot (guilt h. I (P r .P - ) is increasing in P l.-. so IA O. Eli 12 I P

P- HE(/ I P-4

^

fat every possible 2 - Taking the expectation of both sides ((wet Ps), GUif2)

ri( E t (r I P - )E( ig 2))

But as 1' 1 . Ei ale dee/easing in P , so are the functions g i = E(1; I P-). Applying (6) to 1 —y 1 sad 1 -go. which ate incteasing functions ° LPthat ( thl Y. we see th a t E g ig2) > E(gr )E(q) EP4 Tr P .-p a/H.21 P - 1) ^ E(E(li P-))1E(E(./2 P-)) =

11': (

)E( /2)

Combining I he last. two inequalities, we see that Et J i Li) > as el/Mucci

f i iEt (p),

❑

The uniqueness theorem of Aizumnan. Kesler ' and Newman [1087], Theorem 4 of Chaplet 5. also carries over to the Waonor setting (in any dimension). Just as fur the Gilbert model, the proof given by Button and Keane [1989] goes through. although a little cu t e is needed with the details: we shall not spell these out

Theorem 15. Fm any A > and any 0 < p < 1 thene is alumni ninety at most one infinite open cluster in Op

❑

In Chaple t 5. we presented a proof of Hai / is's Theorem due to Zhang This moo( retied on the basic crossing lemma (Lemma 1 Of Chapter 3).

8.d Random Voronoi percolatton

279

Harris's Lemma, and the symmetr y of the square lattice In his rutpublished MSc thesis. Zva y itch [1996] pointed out that Zhang's proof carries over to the random Voronoi setting, giving the following result. Theorem 16. Fm random Voronoi percolation in the plane, 0(1/2) = 0 Hence, pn(2) > 1/2. ❑ Unfortunately, none of the many proofs of Kesten's Theorem seems to adapt to the random Voronoi setting: there seems to be no easy way of proving the analogue of Kesten's Theorem Nevertheless, using an argument which is considerably more involved that any of the proofs of Kesten's Theorem, Bollolgis and Riordan [2006al did prove this analogue

Theorem 17. Fm nindom Voronoi i noculation h i the plane, par = 1/2. As we shall see at the end of this section, the proof of this result also gives an analogue of Kesten's exponential decay result, 'Theorem 1.2 of Chapter 3, and so establishes that the natural analogue of pr is also equal to 1/2 As /Mira/ Iced above, the wool of Theorem 17 is rather involved, so we shall describe only its key ideas: for full details we refer the reader to the or iginal paper Perhaps surprisingly, the lack of independence in the random Voronoi model turns out not to be the main problem (or even one of the main problems) Indeed, it is easy to see that fig rectangles I? and say, separated by a moderate distance, any events E. E' depending only on the colours of the points in B. and B.' respectively are almost independent In particular, there are independent events E. E' that almost coincide with E and E': this property turns out to be just as good as independence We make this idea more precise in the lemma below Hem, we think of s as the scale of the rectangles I? and 1?!: we shall take s cc with all other parameters fixed From now on, we fix the irrelevant scaling of the Poisson process 'P = TA by setting A =1 Given p > 1, s > 1 and a ps by s rectangle B, C R2 with any orientation, we write F(Its ) = F,(R.,;) for the event that every ball E B, (a:), contains at least one point of P = P + U where = 2y/log s (It does not matter whether we consider open or dosed balls; up to a probability zero event, which we shall ignore, this makes no clifference )



Continuant Navarreton

28(1

Lemma 18. Let p> I be mistant let C IR 2 he a / pi ha s rectangle, > 1 god set r 2v/logs Theo (R,) = 1",(R,,) holds with probability rilso if E(R s ) is any event defined 0 0 1 0 in terms — o(1) as s of the colours of the points in R s . then E(R,)11F(R 1/2 ) depends only on the rest/idiot) of IP -1 .P — ) to the -rag a hbourhood R, Progl The firs t statement is immediate horn the properties of it Poisson process Indeed. we 1111W covet Rs with 0( s2 / I2 ) = s2 ) disjoin t (half open) scrum ps S, of side-length r [ ■122 The area of each 9, is /2/2 2 hogs. so then number of points of P in has a Poisson dist) Haitian with mean 21 0g s Hence. (he probability that a particular 8, contains II() points of P is exact Is C 2 Thus. 1eitlr in obabilit \ I ) [ A vi v S. contains a point of P. and FL/11,1 holds The second statement is immediate Itonn the definition of 17(13.4:11nr argument is as in the p i vot of Lemma 10 above Indeed E(R...) chanty depends oils on the restrict ion of (Pt P — ) to the -neighbour hood ./ef) of R, [(Ili's) holds then the closest point of P to ^ 111V .r E R, is r point of B, (r), and hence a point of ri in Mc') Thus the colour of ,r is det e i mined by the closest point of the restriction of (P I . P — ) to [((;)

Of course thew is nothing special about rectangles in Lemma 18; choosing t he slim let site of the rectangle am 010 'scale parameter s will be convenient lac er Also Lemma 18 has an analogue in am dimension, concei ving cuboids say mi ning now to the specific stud y of dimension t \so. the lundiunental quantit y that wee Awn!: %vitt; is the probabilik of a black crossing of a rectangle. Let /„(p, s) =

(HO. ps] x [0, si))

he the P/3 -piobabilits that a ps bs s Ieelangle has a black Innizontal crossing start with two easy observations about the behaviour of ji ( p s) Fi t st /p(m)s) 5 1,(112. s) whenever p l < pg, as the cor i esponding events ale nested 'The second obsei sat ion is that / /,(01

—

)

/ ),(m s )hAt)2. s )1),(

s )

(7)

foi all m > 1. The argument is exactl y as in Chapter 3: Iet R 1 and R 2 he m s ha s and / srs he s rectangles intersect lug in an s by s squaw

S Bondi n

own porcolotion

98

Figure 13 1 vo rectangles Pr and lip intersecting in a square S. 11 P and P_ are Hack f t .arzontal crossings of fir arc P2 , respectively, an 1 Pr is a black vertical crossing of then P t U U Pr contains a black horizontal crossing of Pt U Po

S. as in Figure 13 The events H(R I ). 11(R ” ) and F(S) ate increasing Thus. la Ha t is s Lemma. in the Form of Lemma H.

PgH(R

ni-R re 2 )nits1)

> Pp(H(RWPI(H(R2))11'1,(1.(3))

If 1-1(R I )na(R 2 )n r (5) holds, then so does II (R U As re,(1"(8))= P(H(8)) =

(see 1:',.igitte



1

s). I he telatimi (7) follows

The itatin al analogue of the Russo Seymour Welsh (RSW) 'Thewern would state that if fp (1, s) > s > a then p (p. s) > ird > for some function q(9, E) not depending on s (so. in particulin fr it .(2. ^ ) is bounded awa y flout zero) Such a statement: has not been proved lot ardour Voionoi nett:oration ft might seem that the proof of the RS \V-type thethent te e presented in Chapter 3 would early me t to the random Voionoi setting. The strut is indeed promising: thew is no problem defining the left-most black vertical crossing' Lt (P) of R whenever 1 1 (R) holds: we simply follow a black-white interface as befote. The problem is the next step: the event LT(R) = P is independent of the stales of the points of

P to the light of P. but not of II/eh positions: we can find LP(R) without looking at the colours of the cells to the light of 1,1"(R), hut knowing LI ; (1?) tells tts whew tlw centres of these cells are, and that thew ate no points of P in certain discs, puts of which are to the light of .LII (R) Thus, there is no simple way of showing that the limitability that LI: (R) is joined by a black path to the tight side of R is at. least P(1-1(R)) Of course, a general problem wit h iandoin Voronoi percolation is that no two regions ate independent, although well-separated legions ate asymptotically independent The first step towards the proof of Theotent 17 is the following result:. although this is considerably weaker than a direct analogue of the RSW Theorem. it tutus out to be sufficient lb/ the determination of pa (2)

Continirilin percolation

282

Theorem 19. Let 0 < p < 1 and p > 1 be fixed If lint inf, 0, then lint sup,_, f,,(p, ^ )> 0

(

s)

Lemma 12 states that the hypothesis of this themem is satisfied for p= 1/2: thus Theorem 19 has the following consequence

Corollary 20. Let p > 1 be fixed. There is a constant co = to(P) 0 such that . for every so them is an s > sa with f il s(p, ^ )> Co. The proof of Theorem 19 is rather lengthy, so we shall give only sketch Proof We proceed in two stages, throughout assuming for a contradie2 tion that the result does not hold Thus, there is a constant cm > 0 such that 40,

(8)

for all large enough s, but, for some fixed p > 1, Ip(IL, s)

(9)

0

Here, and throughout, the limit is taken as s pc with all other parameters, for example p. fixed. In this context, an event holds with high probabititw or why), if it holds with probability 1 — o(1) as s cc. 'The assumptions above imply that. Mr any fixed E > 0, fp (l +5,5) as s

0

(10)

xi Indeed. taking > (p —1)/5 and using (7) k times,

fp (p, ^ )

.n,(1

/,76, s)

fff ,(1

s)fp(1, (b,(1 +

fp (1

(k — 1) 5 , s) >

s) fp (1, s)) k f„(1, s) 71-i. 0,

contradicting (9) Condition (10) imposes very severe restrictions on the possible black paths crossing a square, fOr example: roughly speaking, no segment of such a path can cross horizontally a rectangle that is even slightl y longer than high Thus, fot any E > 0, wisp all black horizontal crossings of a given s by 5 squat e 5' pass within distance E s of the top and bottom of 5': otherwise, with positive probability we could find a black horizontal crossing of on e of two 3 by (1 — E)s rectangles, contradicting (10) The next observation is that wh i p any black horizontal crossing of air s by s square 5' starts ar i d ends near the midpoints of the vertical

S 3 Random I lmonoi percolation

283

sides. Indeed, if there is a positive pr obability that some black crossing P starts at least es/2 above the midpoint of the left-hand side, then, reflecting S in a horizontal line and shifting it vertically by es to obtain a square there is also a positive probability that some black horizontal crossing P' of 5' star ts at least 65 12 below the middle height of 8', and hence below P. By Harris's Lemma, crossings P and P' with the stated proper ties then exist simultaneously with positive probability But, whip, P must pass within distance es/3, say, of the bottom of 8, and hence, below S' This forces P and P' to meet; see Figure 14 As P' passes

8'

Figure 14 Two squares S and S'; their horizontal axes of symmetry are shown bye clotted lines Each of the horizontal crossings P. P' of S and 5' passes near the top and bottom of the square it crosses. If P starts above P' then t he paths cross: P ' must leave the region bouncied by partt of the bounds ' v of S' and the initial segment of P up to the point it

within distance Es/3 of the top of we find a black vertical crossing of a rectangle taller than it is wide Using rotational symmetry of the model., this contradicts (10) In a moment:, we shall state precisely a consequence of (10).. For now, we continue with our sketch of the argument Suppose that P is a black horizontal crossing of the square S = [O, x [0, s] Let Pr be the initial segment of P, starting on the line x = 0 and stopping the first time we reach s = 0.99s Similarly, let I2 be the final segment of P, obtained by tracing P backwards horn the = s until the first time we reach x = 0 01.s; see Figure 15. Arguments similar to those above show that each of Pr and P, star is and ends yen \' close to the line = s/2. Fur thermore, one can show that, whip, since each Pi crosses a

28-1

Continuum petrol& an

Figure 15 A squa t t S (outer lines) and a black path P crossing it (solid and dot ted curves) 'The path P i is the initial segment of P s:opr ing at Hie inner vertical line The he al segment P. of P is defined similarh Ii ;uglily speaking, each Pi crosses a square smaller than 5', and so cannot a vroach too close to the top and hot ton of .9 Thus ;24. P: her wise a slightly flattened rectangle in S would have a black horizontal crossing T he dotted p i nt at in fact passes very close to the top mid botto m of S

rectangle of width II 99s, it remains within height ±(0 -195 4- 5)s of its stinting points. It follows Hat Pr and A are wh p disjoint: otherwise, P = Pr U is a Hack crossing of 5' that does not conic within height 0 004s, say, of the top and hottom of S: the probability that such a crossing exists is o(1) by (10) But then each of Pr and P, similarly contains two disjoint crossings of rectangles of width 0 98, and so on

i

ft is not hard to make the vague rugument just outlined pwcise, although it, turns out to be bet ter to work in a r (retail*, [0. x [—Cs. Ci], say, of bounded aspect ratio 2C > I. In Bollobris and Riordan [2006a1, Hie following consequence of (10) is proved Claim 21. Let C > 0 he ficed and lel 1? =

be the y by 2Cs rectangle [0.s] x [—Cs.Csj Pot < j < 4. set [js1100.(1 90) ,4/1001 x [—Cs.Cs] Assam:My Mat (10) holds why eve, y black path. P (71),;sing i < aqierc I? ha/C.:on/oily contains 16 disjoint black paths P. 1 each Pi crosses ,/onte 1?) Ito/ icantally. In other \t'olds, the path P contains 10 disjoint paths 1 i winding

S3 Random 1 /01'0110i percolation

285

backwards and to wards in a manner similar to the paths in Figure 15 Of course, in proving Claim 21, one must: be a lit tle mo t e careful titan in the vague outlirie above Never tireless, the proof is fair ly straightfor ward: it uses only- Harris's Lemma, certain symmetr ies of the nuclei, and (H)) The last condition is applied to show that \\lip none of a fixed number N = N(C.e) of rectangles with aspect ratio 1 + s is crossed the long vu}' In a black path The conclusion of Claim 21 is clearly absurd: 1? has a black horizontal crossing with probability at least f,(1, s), which is hounded away from zero Taking the shortest black horizontal crossing of R. this somehow winds backwards and forwards almost all the way across R. containing segments that star t and end in almost t he save place lint somehow never meet each other Also, the shot test crossing of R is almost, certainly at least 16 times longer than the shortest crossing of a slightl y narrower rectangle As the length of a crossing of R, cannot scale as more than 52 , this is impossible All this sounds convincing: nevertheless. it is not so easy to deduce a contradiction The problem is t hat, while the constant 16 in Claim 21 can he replaced by any l a rger constant it cannot be replaced by a function of s tending to infinity: we can only appl y (10) directl y ton bounded number of rectangles. Otherwise.. t he o(I) error probabilities il(t1/11111late En get around this, we use alutost-independence of disjoint regions to .seurre the error probability' Tire idea is to consider the length (as a piecewise linear path) 1,(IR) of the shortest black horizontal crossing P of certain .s by 2.9 rectangles R. when such a P exists We take L(1?,) = x it I?, has no black horizontal crossing. Even fen widely separated rectangles. L(R s ) and L(131,) are not independent, so we modif y the definition slightly- to achieve independence As in Lemma 18. let P(R s ) = (R s ) be the event that e ye/ y ball of radius r = 2Vlog s centred at a point in R., contains at least one point. of P, and set 7(1?,) =

L(R)F(Rs) holds 0 of her wise

13v Lemma 18. 1,(1?,) = 1,(R.) whp Furthermore. 1.(R,) depends only on the restriction of (P + ,P-) to the t meighbom hood of 17, Hence. if R s and 1r, me separated by a distance of at least 2r o(a). tire random variables 1,(17,) and Z (TC) are independent Roughl y speaking, we wish to relate t he distribution of i,(R J. which

286

Continuum pc/colahop

depends only an to the distribution of L(I? s /2 ), using Claim 21[ fact to leave a lit t le elbow room, we relate L(I?„) to L( Bo ). say For 71> 0 a (very small) constant, define g(s) by g(s) = sup{ :

CL- (R s ) <


> cy > 0, horn (8) As L(R.,,) L,(1?,; ) Min, it follows that 0 g(s) < cc if q is chosen small enough. and then s large enough, which we shall assume from now on Also, the sumer/m i n above is attained. so Pii(L(Rs) < 1/( s ))

( 11)

Claim 21 tells us that, whp, ever y black horizontal crossing of R.,[[ including the shortest. contains 16 disjoint crossings P, of slightly rower rectangles. Using the fact that whp every crossing of a 0 96 9 by 2.5 rectangle is actually almost, a crossing of a square, we can place a bounded number of 0 47s by 0 94s rectangles 1 < i < N, such that; w hp, each Pi includes crossings of some pair (R, R,;) of these rectangles, with and Bk, separated by a distance of at least 0 01.s Here N is an absolute constant.. For each such pair (16,Ra), the variables L(B' ) and L(R k ) are independent, so (1.1 (11 ) )

7(&) < g(0 17s)) 5 Pi, (11(Rj ),11(Rk ) < q(0.47s))

= F n (2:(R') < g ( 0117s )) illp ( Z ( R k) < g(0 47s)) 5

112

from (11) Hence, the probability that i(IL) ) + 2,;(1?k ) < g(0 .17s) holds for saute pair (Hi , Bk ) separated by distance 0 Ols is at roost N9 and hence at roost 11/2 if we choose ri small enough The proof of !lemur 19 is essentially complete: horn Me remarks above, whp every black horizontal ossing of B. has length at least 16 times the minimum of L(R 1 ) + .L(Rk ) over separated pails (B1. Rk). Thus, L(R,) < 16g(0.47s))

g/2 o(1) < q,

so g(s) > 160 47s), and g(s) grows faster than s [1 , say As g(so) > 0 for some so, it follows drat there are arbitrarily huge s with (33 < g(s) < cc. But then. with probability bounded away horn 1, the shortest black hor izontal crossing of IL, = [0, ,s1 x [0,2s1 has length at least [[; 3 'This is

$ 3 Random Voronoi percolation

287

impossible, since, will). R. meets only 0(s2 ) Vorouoi cells, and each has diameter at most 0(log s) ❑ As noted in the original paper, the proof of Theorem 19 outlined above uses rather few properties of random Voronoi percolation: certain symmetries, the fact that horizontal and vertical crossings of a rectangle must meet, and an asymptotic independence property. For this reason, the proof carries over to many other contexts; see, for example, Bollobris and Riordan [2006b] \Tan den Berg, Brouwer and VrigvOlgyi [2006] proved a variant of this result for 'self-destructive percolation', a model of forest growth taking into account forest fires, which they used to prove results about the continuity of the percolation probability in this model Using Corollary 20 and Harris's Lemma, Theorem 19 easily implies that 0(1/2) = 0, giving an alternative proof of Theorem 16 Of course, this result can be proved more cleanly by adapting Zhang's proof of Kestert's Theorem; see Zvavitch [1996] However, Corollary 20 is a good star ting point for the proof of the analogue of Kesten's Theorem, namely Theorem 17 The main idea of the proof is to use some kind of sharp-threshold result to deduce that, for any p > 1/2 and any so, there is an > so with (12)

f„(3,5) > 0.99,

say Before sketching the (rather lengthy) proof of (12), let us see how Theorem 17 follows The argument:, based on 1-independent bond percolation on 2Li 2 . is very close to an al gument given in Chapter 3 Given a 3s by s rectangle I?, with the long axis horizontal, let G(R.„,) = II(R s ) n V(5 1 ) n V(S2), where S I and 5, are the

two

s by send squares' of R see Figure 16

g le Figure 16 A 3s by s rectan

such that

G(R s ) holds

B y translational and rotational symmetr y. Pp ( I ( .52)) =

(Sr =

p( ER S1

)) = (

ji,(3, ^ ) > 0 99.

288

Continuum purolation

so P 1,(0(.11„)) > 0,97 Let

G(Rh), G(R.,) n F(1?„), and define elkfil,C) similarly for a 33 by 5 rectangle RI, with the long side vertical From Lemma IS, Pp (F(Rs )) 1 as s x Thus, if s is huge enough, P, (G(R, )) > 09 (1;3) for every 3s by s rectangle Rs . By Lemma 18, the event 61 (R s ) depends only on the restriction of the Poissiu processes (P lP ,P-) to the Tmeighbewa hood of B. Choosing s larg , enough. we may assume that = 2Vlog < s/2 For each horizontal bond 0 = ((a, b). (a + 1, IR) of V let = [g as, 2as + 3s] x

2b9+ s)

be a 3 3 by s rectangl e in 1E 2 Similxtly for e = ((a. b). (a,b 1)). set ,R, = 12as, 2as + x [21m. 2bs + 33] Let us define a bond percolation model AI on 1 2 by taking a bond c to be open if and only if G(R„) holds. If e and I' ale vertex-disjoint bonds of 1 2 , then the rectangles and RI are separated by a distance of at least s. Hence. if S and T me sets of bonds of V at graph distance at least 1, then the families {C(/?) : 0 E and {6(R„ ) : e E T} depend on the restrictions of the Poisson processes (P-4),P-) to the disjoint regions MI) and U rE7 RI B t , and are thus independent Hence, the probability mensme associated to AI is 1-independent From (13), each bond is open in 21 with probability at least 0 9 so, from Lemma 15 of Chapter 3. with positive probability the origin is in an infinite open path in A7 Such a path guarantees an infinite black path meeting [0. SP (see Figure 11 of Chapter 3), so 1F),([0, s] = meets an unbounded black component) > 0. The increasing event that even- point in [0. SP is black has positive probability, If this event holds and (0, meets an unbounded black component, then Co is unbounded From Harris's Lemma, this happens with positive probability so 0( /)) > 0 As p > 1/2 was arbitrary, and 0(1/2) = Theorem 17 Follow s Our aim for the rest of the section is to sketch the proof of (12). that lot any p > 1/2 and an) so thew is an s > 3 0 with f),(3.. 3 ) > 0.99: as we



Random l'oronot percolation

289

have seen, this implies 'Theorem IT As a starting point. Corollary 20 is strong enough: one o l d\ needs / 1/ 0(3, s) > err for ma l e sufficiently huge s In a discrete setting, we could easily use the Ftiedgut-kalai sharp-threshold result (Theorem 13 of Chapter . 2) to deduce that, for this s, we have fp (3, s) > 0.99 It is not unreasonable to expect it to be a simple matter to adapt this argument to random Voronoi percolation, by choosing a suitable discrete approximation, as in the proof of Theorem 3. say However, as we shall see, there is a problem In Chapter if, we presented various methods of deducing the statement, fin bond percolation on V that corresponds to l i,(3, ^ ) 1 for p > 1/2 fixed One of these. the method based on the study of symmetric events.. originated in the context of random Voronoi percolation the others do not seem to adapt to this setting As in Chapter :3, to define symmetric events we shall work in the torus, i.e.. the quotient of Pf 2 by a lattice Mote precisely. we shall work irr the torus 72 = L o = R2/(10sZ)2 i.e the sox face obtained from the square (0. l(Is) x (0,10,s1 Ire identifying opposite sides Here s will be out scale parameter; we shall consider rectangles R in 1'2 with side-lengths Afs„ where 1 < k < 9 The notion of a random Volonoi tessellation makes perfect sense on the torus: we star t with Poisson pr ocesses P + and P - of intensity p and I -p on I'm which we may take to he the restriction of our processes on Pi2 to (0. 10sI2 • The definitions of the \Unman cells and of the graph Op. UP- ate as berme. We shall always consider s huge It is easy P = y disc in of radius \/log S. sa y, contains a point to check that whp ever of P, so the diameters of all Vinonoi cells are at most 2 Ylog s = o(s) In particular., no cell f ivr aps mound' the torus The rectangles R we consider also do not come close to wrapping around the tot us Thus, events such as .1-1 (1?) have almost the same probability whether we regard I? as a rectangle in R2 ca in T2: 7:2

11= , (H(R)) =

(I-I(R))+, o(I)

where Pi, and Pr;' = F,, ar e the probability measures associas s — ated to Poisson processes (P lil .P-) on '12 = Trotand on R 2 respectively There is ji natural notion of a's y mmetric' event with respect to the measure P2„ namely, an event that is preserved by translations of the torus The is one of t he two ingredients required for the application of the Ft iedgut-Kalai shat p-t lueshold result: the other is a discrete pr oduct space



290

Continuum percolation

Let 6 = 6(s) > 0 be such that 109/6 is an integer, and divide T 2 = Tios up into (10s/6) 2 squares 51 of side-length S in the natural way We shall approximate (Pt P-) (defined on the torus) with a finite product measure as follows: a square S i is bad if [Si n > 0, neutral if i n 0 and n P 4 > 0; see Pfl P + = 0, and good if 18 1 n Figure 17 Thus, open points (points of 7) ± ) ate good, and closed points 1S

1

= 1Si

1

1

a

Figure 17 A small part of the decomposition of the torus into squares Sr. Points of P" are shown by dots those of P.- by crosses A square is good (heavily shaded) if it meets P -1 but not r , neutral (lightly shaded) if it meets neither P 1- nor P -, and bad (unshaded) if it meets PM The 'crude state1. of the torus is given by the shading of the squares Usually, almost all squares are neutral

are bad, and the presence of a bad point outweighs that of a good one; we shall see the reason for this choice later For each 4, the probabilities that 51 is bad, neutral and good ate it had Pnemrrat Prood

- exp (-62 (1 - p))

82 ( 1 — p),

exp(-82) = 1 - 0(62 ), and eXP (

( 11)

- 62 (i - p)) (1 - exp(-52p))

respectively, and these events are independent for different; squares By the trade stale of Si we mean whether Si is had, neutral, or good; the erode slate CS = C5,5 of the torus is given by the crude states of all (10s/6) 2 squares Si It is cleat that if S is sufficiently small as a function of s, then CS essentially determines (I) + , P-), and thus Op (defined on the torus).

8.3 Random 1/monai percolation.

291

Indeed, for s fixed, the a-fields generated by CS =CSo are a. filtration 72 of that associated to PI, In fact, fen the discrete approximation to (essentially) encode the entire graph Gp, it is enough to take 6 = a(811). Indeed, it is easy to check that if S is this small, then two things happen wisp First, no Si contains more than one point of P, so CS essentially determines (p' -1 ,P – ), up to shifting each point by a distance Second, there is no point of T2 such that the distances to the four closest points of P lie in an inter val 2 0.,(5] Thus, shifting points of P by at most \FM does not change which Voronoi cells meet, and so does riot affect the graph structure of Op As we shall see. we shall not be able to take S this small, so the graph structure of Op will be affected by the discrete approximation, although only 'slightly' The possible crude states CS of the torus correspond to the elements of Q" = (-1,0,11", where a = (10s/S) 2 is the number of squares Si, and for w (w 1 ) E Q" we take (+4 to be –1, 0 or 1 if Si is bad, neutral or good, respectively. Let us write Pp_ r,. for the product probability measure on Q" in which each w i is –1, 0 or 1 with respective probabilities 1,–, 1 – p_ –p+ , anti pi. Then the discrete approximation to P. given by C'S corresponds to the measure on Q" More precisely, if E is any event that depends only on the crude state of the torus. then P„ (E)=

(F),

(15)

where F C Q." is the corresponding event. and pg„„d and p are related by (11) It will clearly be convenient to use a form of the Fr iedgut–Kalai sharpthreshold result for powers of a 3-element probability space (-1,0,11 In this context, an event E C .C2" = (-1, 0, 1}" is increasing whenever w for every then w' = (c.,1) E E An event (w i ) E .E and Ji > E C Q" is symmetric if it is invariant under the action on 4" of some group acting transitively on [rd,-- {1,2, . . From (1-1), both p kid and pg,„Ki will be very small, so the sharpthreshold result we shall need is an equivalent: of Theorem LI of Chapter 2 As noted by Bollobiis and Riordan [2006a], the proof given by iedgut and Kalai [1996] extends immediately to give the following result: Theorem 22– There is fai absolute constant Ca such that, '11' < q_ < p_ < 1/c, 0 < p+ < q+ < 1/e, E C 0, 11" is symmetric and

Contiaaoat ',cicalaI ion

292

increasing and Ft : win { ri+ — where p„„,

(E) >

p_ — q - }

then

(E) > 1 —e whenever

tid log( 1/E)p„,„, log( 1

)/ log tt.

( to)

nrx{g+.p_}.

When we come to apply this result, the hunt of (16) will /natter; this is in sharp contrast to the situation lot bond percolation on 2: 2 , Indeed, when we used Thement 13 of Chapter 2 to (move Kestells 'Theorem in Chaplet 3, the hun t of the era responding bound was not it/tiro/taut All that, mattered was that, with s: fixed and II tending to infinity, the inctease in p required to taise P,,(E) front E to 1 — E tends to zero Mete even though we are using a strange/ tesult, (with the extra factor p,„„, log(.I/p,„;,„) working in out la yout). flame is a limit as to how' small we can take (5 while still getting a useful result front Theorem 22 Vie shall eta/pant (E) with Pi t,(E) fot a suitable event E. when/ p > 1/2 is fixed. Row Corollary 20„. the hate/ probability will be at least some small constant E. and to prove (12) it will suffice to show that (E) 1 — E. In doing this, we shall make a discrete approximation as above. Et ow (1+1). when we apply Theorem 22, we shall have —

p_ — a_ ,= itS 2 — 6 2 /2 = e(a-)

Also, a will he constant. p„„, = 8(e5 2 ), and n = (10s/(5) 2 = e(r2/8.2) Substihdiug these quantities into (16)„ we see t hat this condition w ill be satisfied if and out- if a > lot some constant > 0 depending on p As p ant/ °aches 1/2 front above. the constant ^1 tends to zero Thus, we cannot aff o rd to take a lei ibly fine discrete approximation; recall that = 0(S - ) vonld be needed for the alma oxiination (access not to affect, Op at all With for - a 5 0 1 8 11 positive constant, out discretization pottiest; will inhoduce ! defects' Mime, given the 'rounded' positions of t he points of P. we do not know which Voionoi cells actually meet The density of these defects will be small (a negative power of s)„ so one might expect them not to matte!. 'r Ids tutus out to he the case, but n eeds a lot of wo rk lo prove Benjamin' and Seta anon 11998] env/untie/ ed the same p i oblent in ram ing a certain 'confoi mal oval fiance' pi upc/ tv of random Vet/m ini per colation in two and since dimensions This is not conformal inwat lance in the sense of the conjecture of skizentrunt and Langlands Pouliot and Saint-Aubin [199-f] discussed in (Maple' 7 Instead, in two dimensions., what Idenjamini and Scht alma proved is essentially the following Let

8 d Random Voronoi percolation

293

C IP1.2 he a (nice) domain. and let S i and 5"g be t g o segments of its boundm y Consider t he Vol onoi percolation associated to a Poisson po t cess O f intensity A on 1?. using a cot tarn metric ds to form the Voronoi cells, rather than tine usual Euclidean metric. Then. as A — a fixed confor 'nal (locally angle-preset ving) change in the met:tic ds does not change the probability that there is a black path horn S i to So by tame than o(1) Let us note that this is also a statement about detects: each \Towing cell is very small. and, locally, the change in the metric is multiplication he a constant factor, so there \\ill be very few 'defects' whore different. Vor (anti cells meet for the two metrics: in two dimensions the expected in three dimensions there number of defects is bounded as A — are more but the density is still vei l: low the result of Benjamini turd Schr arum is that these defects do not affect the crossing probability significantly: even though there ate very few defects the proof is far nom simple In proving Theorem 17 we have a large] densit y of defects: on the of het hand. we have the freedom to vary the probability p, replacing an err bitril/ V p> 1/2 by (p+ 1/2)/2, say. Roughl ■ speaking. we shall show that the effect of the defects is smaller than the effect of decreasing p in ti l ls way Let us nog make some of the above ideas precise Let 1? he a rectangle , in 2 NVe wish to define an event E depending on the Poisson processes (P c . P-) on 72 :Mtn that, whenever E holds. then 11(1?) holds even if we move the points of P a little. Given rt > 0. rte say that a point A t E I" E P + with dist (.r. z') > dist(1 . , z) is 1i -robustly blink if them is for all o f E P-; in other words, the nearest open point of P is at least a distance closer than Ow neatest closed point If rte move all points of Pat most a distance if/2. then any .c E 1 2 that was if-robustly black (and hence black) will still he black We SUN that a path P C 12 is it-robustly Huck if every point of P is tfa ()busily black. It tin US out that if 0 < pr < po < 1 and > 0, then we can couple the probability measures Ft.), and so that, whp, for event' path PI that is black with respect to the first measure, there is a 'nearby' path Pr that is if -robustly black with respect to the second measure, whew = < Theorem 23, Let 0 < u(s) be any function with q(s) <

and > 0 be given. Let ill = 111c way cough nil in the same

294

Con tinituill percolation

probability space Poisson processes P r* PT and o re/ = nos and 1 – po respectively, such that P;.4" with intensities p i! 1 – and pi are independent foi each and Mt; following global event Eg x fo i every piecewisefinew path P 1 that is black with holds whp as s respect to (Pt , p i ) there is a pieceivise-lineal path P that is 4h-robustly black with respect to (PT ), with (In (Pr, Po) < (logs )2

The proof of Theorem 23 is perhaps the hardest: part of the proof of Theorem 17, and we shall say very little about it Essentially, we roust deal with certain 'potential defects', where the four closest points of Pr = U' I– to some point E 11 .2 are at almost the same distance (the distances differ In at roost s –e ( I )) We call such points of Pr bad, Roughh speaking, a bad cluster is a component in the graph on the had points in which two bad points are adjacent if they are close enough to interact. The heart of the proof of Theorem 23 is a proof that, tinder the conditions of the theorem, whp the largest bad cluster has size ° (log s). As the probability that a given point is bad is 5 –e(0 , one might expect the largest bad cluster to have size 0(1); it turns out. however, that there will be bad clitsters of size e(log s/ log log .^ ) Hence, the o(logs) bound needed for the proof of Theorem 23 is not far from best possible. Using Theorems 22 and 23. it is not too hard to deduce (12) from Col olltuv 20. and so prove 'Theorem 17.. Proof of Theo:T.7n 17 Let p > 1/2 be fixed As shown above, to prove that 0(p) > 0, it suffices to show that, given so, there is an s > so for

which (12) holds Let 7 be a positive constant to be chosen later In fact, we shall set = (p – 1/2)/C, where C is absolute constant Let er > so be a large

constant. such that all statements ' if .9 is large enough' in the rest of the proof hold kw all s > sr 13), Corollary 20, there is an s > such that / 1/ .,(9, > co, where co > 0 is an absolute constant; we fix such an s throughout the proof. Let Rr be a fixed Os by 9 rectangle in P., = , so P i p(11(R 1 )) > co. Regarding R i as a rectangle in the torus II" = let E 1 be the event that R i has a black horizontal crossing in the random Vcnonoi tessellation on the torus As noted earlier, since I?, does not come close to (within distance ()(log s) of, say) wrapping mound T 2 , we have a( I% i) = Pi';2(H(Ri))

o(1)

,

8 ,9 Random Volonoi percolation

as s

295

oc. . In particular, if s is sufficiently large, we have

r;9(Er) .^ co/2,

(17)

say.

Let 6 be chosen so that 10.9/6 is an integer, and S -2-1 < 6 < s14; is possible if s is huge enough, which we /nay enforce by our choice of sr. We shall first apply Theorem 23, and then 'cliscretize' by dividing the torus ¶ 2 into (108/5) 2 small squares S i of side-length 6, as above Set p' = (p + 1/2)/2, so 1/2 < p < p, and consider the coupled Poisson processes (Pif , P and (Pt PT ) whose existence is guaranteed by Theorem 23. applied with m = 1/2, po = p', and q =16 < c1 Let R. , be an Ss by 2s rectangle obtained by moving the vertical sides of outwards by a distance 8/2, and the horizontal sides inwards by the same distance; see Figure 18 Let E2 be the event that there is a horizontal crossing of Ro that is IS-robustly black. this

i)

E

s,

Figure 18 The 9s by s and Ss by 2s rectangles A and (riot to scale), together with a black path PI (solid curve) crossing R I horizontally and a nearby robustly black path P2 (dashed curve), part of which crosses ki horizontally

Suppose that Er holds with respect to (Pi", Pr); flour (17), this event has probability at least co /2 Let Pr C R 1 C T2 be a black path that crosses R I horizontally If the global event Eg described in Theorem 23 holds, as it does whim then there is a path A with (In (Pr P2 ) < (logs)' such that P, is IS - robustly black With respect to (Pt. If s is sufficiently large that (log 8) 2 < s/2, then any such path 12, contains a sub-path A crossing horizontally, so E, holds with respect to (P:t,P.r) If s is large enough, then Eg holds with probability at least 1 — e0 /4, so (E2 ) >

(E, ) — P(Eg ") >

co/2 — co/4 >

co/I

be the event that. some Ss by 2s rectangle it/ ¶2 has a 46-robustly

296

Continuum imirolati OP

black horizontal crossing Then (E.)>

P-, (Eo) > col

Let us divide T2 into n = (10s/6) squares Si as before, and define the crude state of each Si and the etude state CS of the whole torus as above. Let :p; be the event that the crude state CS of the torus is compatible with E3. Since P(E 3 ) = E(P(E3 CS)) < IF(ED, we lune P;,, (EEO > co/4 Let ',brit' and proud be defined by (14), and let similarly, but with p' in place of p. Thus.

and p'gt.

be defined

= 1 - exp (-62 (1 - p ')) ti S2 (.I - if). and //gond =

exp (-82 (1 - p'))(1 - exp (- 62 1/ ))

(52p

As the event E:63 depends only on CS, it may be viewed as an event E3 in the finite product space Q" = {-1, 0,1}" In particular, nom (15) = P7; (E753 ) > co/4.

(19)

The event F3 is symmetric, as E3 and hence Elsi is preserved by a translation of the torus through (18, Pi. ), ) E Z. ft is not hard to check that 1:3 is also increasing From (11) and (18) we have pit „ i

(1-p1 )52 , pig ,„„ 1 t-pe, /Thad ti (1 -p)82 . and it,

p82 (2(1)

shall apply Theorem 22 to -1, 0, 11", n = (10s/8) 2 , with p_ = = p800, 1 , and with e = min{c0 /4, ill- In, say Note that z > 0 is an absolute constant Let A be the quantity appearing on the light-hand side of (16) with this choice of parameters From (20), all fern of the quantities p_ and are at most if s is large enough Hence, p„,„, < 62 , and WC

= peso" , a_ = pima and

A = 0(62 log(1/62 )/ loge). where the implicit constant is absolute By om choice of 6, we have 1/6 < s 2-f On the other hand, = (10s/8) 2 > s2 Thus, log(1/5 2 )/ log a < 27. Hence. A < Chi ,i 2 for some absolute constant C As C is an absolute constant. we may go back and choose .7 small enough that CT, < (p - p')/2. so A 5_ (P - P/)62/2

8 (( Random Vownoi percolation

997

hot (20). we have q+ - v+, P-

^" (1)- 91)52

Thus, if s is large enough, q+ -p+ and p_ - q_ ale larger. than A, i.e., the hypotheses of Theorem 22 are satisfied Theorem 22 and (19) then imply I — 5 > 1 — 10 — WO (PII) Retruning to the continuous setting, but considering t h e event Er), which depends only on CS, we have p (E) = p

F ) > -

(")

Suppose that CS is such that LI holds, and let (P P-) be aqy configuration of the Poisson process consistent with this particular elude state of the tot us By the definition of E7)') . there is another realization consistent with the same crude state such that E. holds. i e such that there is a 4S-robustly black path P Grossing some Ss be 2s rectangle in 72 Prom the definition of bad, neutral and good squares, it follows that in the realization (Pt P - ). the path P is still black Hence. -)\ riling E4 for the event that some Ss by 2s rectangle in T 2 has a black horizontal Cl owing. we have PI, (Er) ^ 1Pi, (Etr ) > 1 -

I"

The rest of t he argument is as in Chapter. 3: we can cover 7 2 with (is be -Is rectangles /?, , ,R9 5 so that, whenever Er holds, one of the R i has a black Mu izontal crossing Using the 'rah-root trick', it Follows that (H(R ) ))> 1 -

Returning to the plane, fir (3/2,45) =

(1-1(R1))+ o(1)> 0 999

s is large enough Appealing to (7), it is eas y to deduce that fp (3, -Is) > 0 99. As noted earlier, 0(p) > (1 then follows easily by considering 1independent percolation Since p > 1/2 was at bitranr, and 0(1/2) = 0 from Theorem 16, the result follows

❑

The argument: above was just a sketch of the proof of Theotem 17 hi particular. we said almost nothing about the moor of Theorem 23 Although purely 'technical', such a result seems rather difficult to plover this is actually a huge part of the work in the or iginal paper

298

Continuum percolation

As shown by Bollobiis and Riordan [2006a], for p < 1/2, the proof of 'Thoth ern 17 also gives exponential decay of the size of the open cluster Co or cot responding black cluster Crg containing the origin; this result follows from (12) by considering a suitable k-independent site percolation model on Z2 Theorem 24. Let leol denote the diameter of Co, the area of Co, or the 711110101 ICT of I/010710i cells in Co Then, for any p < 1/2 > there is a constant c(p) > 0 such that F

Pp (FCol hi every n > 1

n)

exp(—e(p)n) ❑

As in Chapters 3 and 4, this implies that the Hanunersley and Temper ley critical probabilities coincide Theorem 25. Let p i he the Tempe/ley cr itical probability for random Ijoronoi percolation in the plane above which Ej,(Kup diverges, where roi denotes the diatnetei of Co, the area of Co, or the number of Voronai = p i( = 1/2 crffs in Co Then. ❑

We have seen that :random Voronoi percolation can be very difficult to work with. Nevertheless, there is some hope that the conformal invariance conjecture of Ai-tern/mu and Langlands, Pouliot and SaintAubin [1994] discussed in Chapter 7 may be approachable for this model: the model has much [note built-in syninteti y than site or bond percolation on ain lattice. Also, the result of Ben] amini and Schramm [1998] provides a possible alternative method of attack: roughly speaking, t hey showed that for :random Voionoi percolation, conformal invariance is equivalent to 'density invariance', i e , the statement that, with suitable scaling, the crossing probabilities associated to two different inhornogeneous Poisson processes converge

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