arXiv.org e-Print archive · arXiv:math/0301160v1 [math.PR] 15 Jan 2003...

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A power law for the free energy in two dimensional

percolation

Yu Zhang

April 17, 2019

Abstract

Consider bond percolation on the square lattice and site percolation on the trian-gular lattice. Let κ(p) be the free energy at the zero field. If we assume the existenceof the critical exponents for the three arm and four arm paths and these critical expo-nents are −2/3 and −5/4, respectively, then we can show the following power law forthe free energy function κ(p):

κ′′′(p) = +(1/2− p)−1/3+δ(|1/2−p|) for p < 1/2

κ′′′(p) = −(1/2− p)−1/3+δ(|1/2−p|) for p > 1/2,

where δ(x) goes to zero as x → 0. Note that the critical exponents for four arm andthree arm paths indeed are proven to exist and equal −5/4 and −2/3 on the triangularlattice and the above power law for κ(p) therefore holds for the triangular lattice. Notethat the above power law for κ(p) implies κ(p) is not third differentiable at the criticalpoint of the triangular lattice. This answers a long time conjecture that κ(p) has asingularity at 1/2 since 1964 affirmatively.

Key words and phrases: percolation, free energy, power laws, scaling relations.

Mathematics subject classification: 60K 35.

1 Introduction and statement of results.

Consider bond percolation on the square lattice or site percolation on the triangular lattice

in which bonds or sites are independently occupied with probability p and vacant with prob-

ability 1−p. The triangular lattice may be viewed as being obtained from the square lattice

by adding all the northwest-southeast diagonals. The corresponding probability measure on

the configurations of occupied and vacant bonds or sites is denoted by Pp. We also denote

by Ep the expectation with respect to Pp. The cluster of the vertex x, C(x), consists of all

vertices which are connected to x by an occupied path. For bond percolation on the square

lattice, C(x) always contains vertex x. For site percolation on the triangular lattice, C(x)

1

http://arxiv.org/abs/math/0301160v1

is an empty set if x is vacant. Here a path from u to v is a sequence (v0, ..., vi, vi+1, ..., vn)

with distinct vertices vi (0 ≤ i ≤ n) such that v0 = u and vn = v. A circuit is a path with

distinct vertices vi (1 ≤ i ≤ n− 1) and v0 = vn. For any collection A of vertices, |A| denotesthe cardinality of A. We choose 0 as the origin. The percolation probability is

θ(p) = Pp(|C(0)| = ∞),

and the critical probability is

pc = sup{p : θ(p) = 0}.It has been proved by Kesten, H. (see Chapter 3 in Kesten (1982)) that for bond percolation

on the square lattice and site percolation on the triangular lattice

pc =1

2.

We denote the cluster distribution by

θn(p) = Pp(|C(0)| = n).

By analogy with the Ising model, we introduce the magnetization function as

M(p, h) = 1−∞∑

n=0

θn(p)e−nh for h ≥ 0.

By setting h = 0 in the magnetization function,

M(p, 0) = θ(p).

Using the term by term differentiation, we also have

limh→0+

∂M(p, h)

∂h= Ep(|C(0)|; |C(0)| < ∞) = χf(p).

χf(p) is called the mean cluster size. The free energy F (p, h) is defined by

F (p, h) = h(1− θ0(p)) +∞∑

n=1

1

nθn(p)e

−hn for h > 0.

If we differentiate with respect to h, we find

∂F (p, h)

∂h= M(p, h).

For h > 0, the free energy is infinitely differentiable with respect to p. The zero-field free

energy F (p, 0) is a more interesting object of study. By our definition,

F (p, 0) = E(|C(0)|−1; |C(0)| > 0).

2

Grimmett G. (1981) discovered that the zero-field free energy also coincides with the number

of clusters per vertex. Let us define the number of clusters per vertex as follows. Note that

any two vertices x, y ∈ B(n) = [−n, n]2 are said to be connected in B(n) if either x = y

or there exists a path γ consisting of occupied bonds such that γ connects x and y and

γ ⊆ B(n). Connectedness in B(n) defines an equivalence relation in B(n) and it decomposes

B(n) into connected components. Each connected component is called a cluster in B(n).

Let Mn be the number of clusters in B(n). By a standard ergodic theorem (see Theorem 4.2

in Grimmett, G. (1999)) the limit

limn→∞

1

|B(n)|Mn = κ(p) a.s. and L1

exists for all 0 ≤ p ≤ 1. Let Kn = E(Mn). Then

limn→∞

1

|B(n)|Kn(p) = κ(p).

κ(p) is called the number of clusters per vertex. Grimmett (1981) proved that

κ(p) = F (p, 0). (1.1)

Sykes and Essam were perhaps the first to introduce the number of clusters per vertex in

1964, and they tried to use it to compute pc. They explored a beautiful geometric argument

in their paper to show

κ(p)− κ(1− p) = 2p− 1 (1.2)

for bond percolation on the Z2 lattice and site percolation on the triangular lattice. Sykes

and Essam argued that phase transition in percolation must be manifested by a singularity

at the critical value pc. If pc is indeed the only singularity of κ(p), then (1.2) implies that

pc = 1/2 for both bond percolation on Z2 and site percolation on the triangular lattice. For

many years, the singularity criterion of Sykes and Essam has offered a tantalizing approach

to the famous problem that pc = 1/2. Kesten, H. gave another method in 1980 to show

pc = 1/2. However, until the present paper, there was no proof of a singularity at pc. We

would like to mention some progress for κ(p) throughout the years. It has been ruled out

that κ(p) has another singularity on p for p 6= pc. Furthermore, κ(p) is analytic for p 6= pc(see Chapter 9 in Kesten 1982). On the other hand, it has also been proved (see Chapter 9

in Kesten (1982)) that κ(p) is twice differentiable at pc. This tells us κ(p) is a very smooth

function. Indeed, the smoothness of κ(pc) might tell why the singularity at pc is difficult

to prove. The main result obtained here is to understand the behavior of κ at the critical

point. If pc is indeed a singularity of κ(p), then it is natural to ask the behavior of the

singularity. Physicists believe that the zero-field free energy is not three times differentiable.

It is believed that the behavior of percolation functions can be described in terms of critical

exponents as p approaches to pc. For κ(p), it is conjectured that there exists a constant α

such that

κ′′′(p) ≈ |p− pc|−1−α. (1.3)

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It is not known even among physicists how strong they expect such an asymptotic “ ≈ ”

relation to be, and it is for this reason that we shall use the logarithmic relation. More

precisely, f(p) ≈ g(p) or fn ≈ gn means

log g(p)/ log f(p) → 1 as p → pc or log fn/ log gn → 1 as n → ∞.

The exponent α is called the heat exponent and (1.3) is called the power law for the free

energy. Numerical computations indicate α = −2/3. In addition to this power law, it is also

widely believed that the exponents satisfy the following so called scaling laws. To be more

specific we need to introduce all the other critical exponents and power laws. We denote the

correlation length by

ξ−1(p) = limn→∞

{−1

nlogPp(0 → ∂B(n), |C(0)| < ∞)} if p 6= pc,

the probability on the tail of |C(0)| at pc

π(n) = Ppc(n ≤ |C| < ∞),

where ∂B(n) is the surface of the box B(n) and A → B means that there exists an occupied

path from some vertex of A to some vertex of B for any sets A and B.

The power laws are introduced as follows:

θ(p) ≈ (p− pc)β for p > pc,

χf(p) ≈ (pc − p)−γ for p 6= pc,

ξ(p) ≈ |pc − p|−ν for p 6= pc, (1.4)

π(n) ≈ n−1/ρ for n ≥ 1, (1.5)

κ′′′(p) ≈ |p− pc|−1−α for p 6= pc. (1.6)

Numerical computations indicate that

β =5

36, γ =

43

18, ν =

4

3, ρ =

48

5.

In addition to the power laws, it is also widely believed that the exponents satisfy the

following so called scaling laws:

α = 2− 2ν (1.7)

β =2ν

ρ+ 1, (1.8)

γ = 2νρ− 1

ρ+ 1. (1.9)

4

In particular, (1.7) is called a hyper scaling relation. Moreover, let us introduce k arm paths.

Consider the annulus

A(m,n) = {B(n) \B(m)} ∪ {∂B(m)} for m < n.

Let Qk(m,n) be the event that there exist i disjoint occupied paths and j disjoint vacant

paths with i+ j = k for all i, j ≥ 1 from ∂B(m) to ∂B(n) inside A(m,n). It is believed that

P1/2(Qk(m,n)) = n−(k2−1)/12+o(1) (1.10)

for fixed m as n → ∞. For the triangular lattice, (1.10) has indeed proved by Smirnov and

Werner (see Theorem 4 in Smirnov, S. and Werner,. W 2002).

We know that (1.10) together with an argument of Kesten (see Corollary 1 and 2 in

Kesten 1987) imply that all the power laws and scaling relations hold except possibly for

(1.7). Here we will give the following theorem to discuss the power laws concerning for the

free energy function.

Theorem 1. Consider either bond percolation on the square lattice or site percolation

on the triangular lattice. If (1.10) holds for k = 3 and 4, then

κ′′′(p) = (1/2− p)−1/3+δ(|1/2−p|) for p < 1/2

κ′′′(p) = −(p− 1/2)−1/3+δ(|1/2−p|) for p > 1/2,

where δ(x) → 0 as x → 0.

Remarks. 1. If (1.10) holds for k = 4, we know that ν = 4/3 (see Cor 2 and (4.5) in

Kesten (1987)). On the other hand, by our Theorem 1 we know α = −2/3 if (1.10) holds.

Therefore, if (1.10) holds for k = 3 and 4, then a hyper scaling relation in two dimensions:

α = 2− 2ν. (1.11)

2. The proof of Theorem 1 depends on pc = 1/2 for percolation on the square and trian-

gular lattices. Our method encounters difficulties when applied to percolation models with

pc 6= 1/2.

On the triangular lattice it is known that (1.10) holds and therefore on the triangular

lattice we have the following Corollary.

Corollary 2. For the site percolation on the triangular lattice,

|κ′′′(p)| ≈ |p− 1/2|−1/3+δ(|1/2−p|) for p 6= 1/2, (1.12)

where δ(x) → 0 as x → 0.

5

Indeed, Corollary 2 completes the proof of the conjecture made by Sykes and Essam that

1/2 is the unique singularity of κ(p) on the triangular lattice.

The rest of this paper is organized in the following manner. Since the proofs of the The-

orems are very similar for both the square and triangular lattices, we would rather deal with

only one lattice. Since the proofs for the triangular lattice are easier to handle, we would

rather show the details of the proofs for the square lattice, but outline the proofs for the

triangular lattice in the end of section three.

In section two we introduce a few basic results in percolation on the square lattice. We

will use Russo’s formula to estimate the first, second and third derivatives of κ(p). After the

third derivatives, there are both positive and negative terms, and we give upper bounds for

the positive terms and negative terms separately. To handle the sum of positive and negative

terms, we show that the second derivative of κ(p) is decreasing when p ↓ 1/2 and increasing

when p ↑ 1/2. This is also independently interesting, since many functions such as θ(p) and

ξ(p) have been investigated for their concavity for many years. However, as far as we know,

it seems that no one can solve these problems. The methods in Theorem 3 might offer a way

to solve these problems when p is near pc.

Acknowledgments. The author would like to thank Harry Kesten for pointing out a

serious mistake in the first version and for many fruitful conversations and he would also like

to thank Geoffrey Grimmett for his many comments.

2 Preliminaries.

In this section we introduce a few basic properties of percolation in the square lattice. We

first introduce the definition of planar duality. Define Z∗ as the dual graph with vertex set

{v+(12, 12) : v ∈ Z2} and edges joining all pairs of vertices which are one unit apart. For any

bond set A, we write A∗ for the corresponding dual bonds of A. For each bond b∗ ∈ Z∗, we

declare that b∗ is occupied or vacant if b is occupied or vacant, respectively. In other words,

each occupied (vacant) b∗ crosses a corresponding occupied (vacant) bond in Z2. With this

duality, we have (see Prop. 11.2 in Grimmett, G. 1999) the following Proposition.

Proposition 1. Let G be a finite connected subgraph of Z2. There exists a unique cir-

cuit σ(G) on Z∗ containing G in its interior and with the property that every edge of σ(G)

crosses an edge of ∆G. In other words, σ(G) is the smallest circuit containing G in its interior.

Here ∆G is the outer edge boundary of G defined to be the set of edges e of Z2 such

that: e does not lie in G but e is incident to at least one vertex of G, and there is no circuit

in G∗ enclosing any vertex of ∆G. In other words, each vertex of (∆G)∗ can be connected to

6

∞ by a dual path without using any edge of G∗. If G is occupied and finite, then it follows

from Proposition 1 that there exists a vacant unique circuit on Z∗ containing G in its interior.

Now we define an occupied and a vacant crossings in a box. A left-right (respectively

top-bottom) occupied crossing of B(n) is an occupied path in B(n) that joins some vertex

on the left (respectively upper) side of B(n) to some vertex on the right (respectively lower)

side of B(n) but which uses no edges joining two vertices in the boundary of B(n). We

denote the occupied and the vacant crossing probabilities of B(n) and B∗(n) by

σ(p, n) = Pp(∃ a left-right occupied crossing of B(n));

σ∗(p, n) = Pp(∃ a top-bottom vacant dual crossing of B∗(n)).

We need to show that the vacant crossing and occupied crossing probabilities of squares are

bounded away from zero when p is near pc. To make this precise, we first define (see (1.21)

in Kesten (1987))

L(p) = min{n : σ(p, n) ≥ 1− ǫ0} for p > 1/2 and L(p) = min{n : σ∗(p, n) ≥ ǫ0} for p < 1/2,

where ǫ0 is some small, but strictly positive number whose precise value is not important.

The important property is that ǫ0 can be chosen such that there exists a constant δ for which

σ(p, n) ≥ δ, and σ∗(p, n) ≥ δ

uniformly in n ≤ L(p). L(p) is also called the correlation length and it is proved (see

Corollary 2 in Kesten, H. (1987)) that

L(p) ≍ ξ(p). (2.0)

Here f(p) ≍ g(p) means f(p)/g(p) is bounded away from 0 and ∞ for an interval containing

1/2. Note that if (1.10) holds for k = 4, as we mentioned in section one for any δ > 0 there

exist constants C1(δ) and C2(δ) such that

C1|1/2− p|−4/3+δ ≤ L(p) ≍ ξ(p) ≤ C2|1/2− p|−4/3−δ. (2.1)

On the other hand, if p = 1/2, it is known (see Chapter 11 in Gremmitt, G. (1999)) that for

any n there exists C > 0 such that

σ(1/2, n) ≍ σ∗(1/2, n) ≍ C.

In this paper, all C or Cq, for q = 1, 2, 3, 4, 5, represents a positive constant bounded away

from zero whose value does not depend on p, n, k,m, i and j but may depend on ǫ and δ.

On the other hand, C or Ci may change from appearance to another appearance.

Next we will define four arm paths, two occupied on Z2 and two vacant on Z∗ starting

from one edge. Let e0 be the bond connecting (0, 0) and (1, 0) and let (see Fig. 1)

7

[−n, n]2

[−n, n]2

[−n, n]2

[−n, n]2

e0

e0

b

b

e

e

Figure 1: The left-top figure is event Q4(n). The left-bottom figure is event ∆4(n). EventsQ3(n) and ∆3(n) can use the same graphs but ignore r4 and A(3, n). The right-top figure isevents R(b, e). Together the right-top and right-bottom figures are event R(b, e). Here thesolid paths are occupied and the dot paths are dual vacant.

8

Q4(n) = {∃ paths r2 and r4 on B(n) from (0, 0) and (1, 0) to ∂B(n),

respectively; paths r∗1 and r∗3 on B∗(n) from (1/2, 1/2) and (1/2,−1/2)

to ∂B∗(n), respectively; any two these paths

only have the bond e0 in common; r∗i is vacant and ri+1 is occupied,

i = 1, 3, except possible at e0}.

Similarly, we can define three arm paths

Q3(n) = {∃ paths r∗1 and r∗3 on B∗(n) from (1/2, 1/2) and (1/2,−1/2) to ∂B(n),

respectively, ∃ path r2 from (0, 0) to ∂B(n)

all ri only have e0 in common; all edges of r∗1 ∪ r∗3 other than e0

are occupied, and all edges of r2 other than e0 are vacant}.

If (1.10) holds for k = 3 and 4, we know that

P1/2(Q3(n)) = n−2/3+δ(1/n) and P1/2(Q4(n)) = n−5/4+δ(1/n), (2.2)

where δ(1/n) → 0 as n → ∞.

If p 6= 1/2 and n ≤ L(p), Kesten’s Lemma 8 (1987) shows Pp(Q4(n)) has the same decay

rate as P1/2(Q(n)). The same proof in his Lemma 8 can be carried out for the three arm

paths. Here we summarize his results as the following Proposition.

Propostion 2 (Kesten) If n ≤ L(p),

Pp(Q4(n)) ≍ P1/2(Q4(n)) and Pp(Q3(n)) ≍ P1/2(Q3(n)).

For any bond e, let v1(e) and v2(e) be the two vertices of e. In fact, we can define v1(e)

and v2(e) in a unique manner as follows. Suppose that e is a horizontal edge. Then we

denote by v1(e) and v2(e) the left and right vertices of e. Suppose e is a vertical edge. We

denote by v1(e) and v2(e) the lower and upper vertices of e.

Given two edges e1 and e2 in Z2, if the two vertices of e∗1 and the two vertices of e∗2 are

connected by two disjoint vacant paths r∗1 and r∗3 on Z∗, respectively, then e∗1 ∪ r∗1 ∪ e∗2 ∪ r∗3is a circuit. Let S(r∗1, r

∗3, e

∗1, e

∗2) be the region in R2 enclosed by the circuit. Here we assume

that S(r∗1, r∗3, e

∗1, e

∗2) does not contain any bonds in the circuit (see Fig. 1). In other words,

S(r∗1, r∗3, e

∗1, e

∗2) is an open set in R2 enclosed by the circuit e∗1 ∪ r∗1 ∪ e∗2 ∪ r∗3. Let (see Fig. 1).

R(e1, e2) =

{∃ disjoint paths r∗1 and r∗3 in B∗(n) from va(e∗1) and vb(e

∗1) to vc(e

∗2) and vf(e

∗2);

9

∃ a path r2 on B(n) inside S(r∗1, r∗3, e

∗1, e

∗2) from va(e1) to vc(e2); ∃ disjoint paths

r4 and r5 from vb(e1) and vf (e2) to ∂B(n), but outside the closure of S(r∗1, r∗3, e

∗1, e

∗2) or

∃ path r6 inside B(n) from vb(e1) to vf (e2) but outside the closure of S(r∗1, r∗3, e

∗1, e

∗2);

r∗1 and r∗3 are vacant and rl is occupied for l = 2, 4, 5 or 2, 6},

where a 6= b and c 6= f are either 1 or 2 such that va(e∗1), vc(e

∗2) ∈ S(r∗1, r

∗3, e

∗1, e

∗2). When we

estimate the derivatives of Kn, we only need to deal with the following simpler event (see

Fig. 1).

R(e1, e2) =

{∃ disjoint paths r∗1 and r∗3 in B∗(n) from va(e∗1) and vb(e

∗1) to vc(e

∗2) and vf(e

∗2);

∃ path r2 on B(n) inside S(r∗1, r∗3, e

∗1, e

∗2) from va(e1) to vc(e2);


∗1, e

∗2);

r∗1 and r∗3 are vacant and r2 and r4 are occupied},

With these definitions, we will show the following lemmas.

Lemma 1. If n < L(p), for any e1, e2 ∈ B(κn) and for some 0 < κ < 1 there exist two

positive constants C1 and C2 such that

C1P2p (Q4(‖v1(e1)− v1(e2)‖)) ≤ Pp(R(e1, e2)) ≤ C2P

2p (Q4(‖v1(e1)− v1(e2)‖)), (2.3)

where ‖y‖ = |y1|+ |y2| for y = (y1, y2).

Before the proof of Lemma 1 we would like to introduce the four arm extension argument

by H. Kesten (see section 2 Kesten 1987). Let ∆4(n) be the subevent of Q4(n) as follows (see

Fig. 1): ∆4(n) occurs with the four paths r∗1, r∗3, r2, r4 satisfying four additional requirements

(see Fig. 1)

r1 ∩ [−n, n]2 \ (−n/2, n/2)2 ⊂ A(1, n) =: [−n,−n/2]× [−n/2, n/2],

r3 ∩ [−n, n]2 \ (−n/2, n/2)2 ⊂ A(3, n) =: [n/2, n]× [−n/2, n/2],

r2 ∩ [−n, n]2 \ (−n/2, n/2)2 ⊂ B(2, n) =: [−n/2, n/2]× [−n,−n/2],

r4 ∩ [−n, n]2 \ (−n/2, n/2)2 ⊂ B(4, n) =: [−n/2, n/2]× [n/2, n].

In addition, we require that there exist occupied vertical crossings on Z2 of A(i, n), i = 1, 3

and vacant horizontal crossings on Z∗ ofB∗(i, n), i = 2, 4. With these definitions, if n ≤ L(p),

it is proved by Kesten, H. (see Lemma 4 in Kesten 1987) that there exists C > 0 such that

Pp(Q4(n)) ≤ CPp(∆4(n)). (2.4)

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We may also deal with ∆3(n) instead of ∆4(n) by only considering r∗1, r3∗ and r2 in A(1, n),

B(2, n) and B(4, n). Then the same proof Kesten (1987) did in his Lemma 4 can be adopted

to show if n ≤ L(p),

Pp(Q3(n)) ≤ CPp(∆3(n)). (2.5)

Furthermore, as we defined in section one Q4(n,m) and Q3(n,m) are the events that

four arm paths and three arm path are from ∂B(n) to ∂B(m) in the annulus area {B(m) \B(n)} ∪ {∂B(n)}. Similarly, for 2n ≤ m we may define ∆4(n,m) as the event that the four

arm paths in Q4(n,m) satisfies the following four additional requirments.

r1 ∩ [−2n, 2n]2 \ (−n, n)2 ⊂ A(1, n) =: [−2n,−n]× [−n, n],

r3 ∩ [−2n, 2n]2 \ (−n, n)2 ⊂ A(3, n) =: [n, 2n]× [−2n, n],

r2 ∩ [−2n, 2n]2 \ (−n, n)2 ⊂ B(2, n) =: [−n, n]× [−2n,−n],

r4 ∩ [−2n, 2n]2 \ (−n, n)2 ⊂ B(4, n) =: [−n, n]× [n, 2n].

In addition, we require that there exist occupied vertical crossings on Z2 of A(i, n), i = 1, 3

and vacant horizontal crossings on Z∗ of B∗(i, n), i = 2, 4. Similarly, we can define ∆3(n,m).

It is also proved by H. Kesten (see Lemma 5 in Kesten 1987) that if n ≤ L(p),

Pp(Q4(n,m)) ≤ CPp(∆4(n,m)) for all m > n (2.6)

Also, the same proof can be adopted to show if n ≤ L(p),

Pp(Q3(n,m)) ≤ CPp(∆3(n,m)) for all m > n (2.7)

With extra vertical and horizontal crossings in events ∆4(n,m) and ∆3(n,m), it is easy to

extend these four or three arm paths twice long from inside or outside of an annulus. More-

over, if both ∆l(n) and ∆l(n,m) occur for l = 3 or 4, we can also use these vertical and

horizontal crossings together with extra occupied and vacant crossings to connect occupied

and vacant arms such that Ql(m) occurs. We call these extensions as Kesten’s extension

method. The key point is that by using (2.4)-(2.7) and the RSW lemma the probability will

not change but with a constant correction after these extensions. With Kesten’s extension

method let us show Lemma 1.

Proof of Lemma 1. For any bond e ∈ Z2, let Q4(v1(e), n) be the event by replacing

v1(e0) by v1(e) and ri by v1(e) + ri for i = 1, 2, 3, 4 in the event Q4(n). In other words,

we just consider the same event Q4(n) in v1(e) + B(n) rather than in B(n). We can define

∆4(v1(e), n) by the same way. Clearly, it follows from the definiton of R and Q that

Pp(R(e1, e2)) ≤ CPp(Q4(v1(e1), ‖v1(e1)− v1(e2)‖/3),Q4(v1(e2), ‖v1(e1)− v1(e2)‖/3)), (2.8)

where we assume that ‖v1(e1)− v1(e2)‖/3 is an positive interger without loss of generality.

Note that Q4(v1(e1), ‖v1(e1)−v1(e2)‖/3) and Q4(v1(e2), ‖v1(e1)−v1(e2)‖/3) only depend on

11

occupied or vacant of bonds in v1(e1) +B(‖v1(e1)− v1(e2)‖/3) and in v1(e2) +B(‖v1(e1)−v1(e2)‖/3), respectively and

{v1(e1) +B(‖v1(e1)− v1(e2)‖/3)} ∩ {v1(e2) +B(‖v1(e1)− v1(e2)‖/3)} = ∅

so that by Kesten’s extension method there exist positive C,C1 such that

Pp(R(e1, e2))

≤ Pp(Q4(v1(e1), ‖v1(e1)− v1(e2)‖/3),Q4(v1(e2), ‖v1(e1)− v1(e2)‖/2)≤ Pp(Q4(v1(e1), ‖v1(e1)− v1(e2)‖/3))Pp(Q4(v1(e2), ‖v1(e1)− v1(e2)‖/2)≤ CP 2

p (Q4(v1(e1), ‖v1(e1)− v1(e2)‖/3))≤ C1P

2p (Q4(v1(e1), ‖v1(e1)− v1(e2)‖)).

The second inequality in (2.3) is now proven. To show the first inequality in (2.3), let us

assume that both Q4(v1(e1),13‖v1(e1)−v1(e2)‖) and Q4(v1(e2),

13‖v1(e1)−v1(e2)‖) occur. By

the independence discussed before and (2.3) we have

P 2p (Q4(

1

3‖v1(e1)−v1(e2)‖)) ≤ CPp(∆4(v1(e1),

1

3‖v1(e1)−v1(e2)‖)Pp(∆4(v1(e2),

1

3‖v1(e1)−v1(e2)‖)

(2.9)

for some constant C > 0. To show the first inequality in (2.3) from (2.9) we would rather

use the following graph instead of using words to make the proof easier to understand. ✷

Given p 6= pc, we start by taking the first derivative of Kn, where n is some integer much

larger than L(p). Note that

Kn = EMn =∞∑

l=1

Pp(Mn ≥ l)

and the event {Mn ≥ l} is decreasing. Let {Mn ≥ l}(b) be the event that b is a pivotal

bond (see the definition for pivotal bonds in Grimmett, G. (1999)) for Mn ≥ l. By Russo’s

formula

K ′n(p) = −

∞∑

l=1

∑

b∈B(n)

Pp({Mn ≥ l}(b))

= −∑

b∈B(n)

∞∑

l=1

Pp(Mn = l − 1or l if b is occupied or vacant)

= −∑

b∈B(n)

Pp(b is a pivotal bond for the connection of v1(b) and v2(b) in B(n)).

Let

E(b) = {b is a pivotal band for the connection of v1(b) and v2(b) in B(n)}.It is easy to check that E(b) is the event that there does not exist an occupied path connecting

v1(b) to v2(b) inside B(n) \ b. In fact, if there was such a path, then v1(b) and v2(b) would

12

[−n, n]2

e1

e2

Figure 2: In this graph we only show how to connect two occupied paths from two squares.Here η is the area formed by three rectangles with width and length larger than 1

2‖v1(e1)−

v1(e2)‖} and less than 2‖v1(e1)−v2(e2)‖. By the FKG inequality and the RSW lemma, thereare occupied vertical or horizontal crossings in these rectangles with a probability larger thanC > 0. On the other hand, if ∆4(e1,

13‖v1(e1) − v2‖) and ∆4(e2,

13‖v1(e1) − v2‖) occur, and

there exist such occupied crossings, this would imply that R(e1, e2) occurs, where the solidpaths are occupied and the dot paths are dual vacant. Therefore, the first inequality in (2.3)follows from these observations, (2.9) and Lemma 3 in Kesten, H. (1987).

13

always be connected by an occupied path whenever b is occupied or vacant. Let Cn(v1(b)) bethe connected component containing v1(b) inside B(n) if we delete all vacant edges of B(n)

and b. It follows from Proposition 1 there exists a dual circuit D∗ with D∗ ⊂ ∆Cn(v1(b))surrounding v1(b) such that

edges in D∗ ∩ {B∗(n) \ (∂B∗(n) ∪ b∗)} are vacant.

On the other hand, v2(b) 6∈ Cn(v1(b)), otherwise there exists an occupied path inside B(n)\ bconnecting v1(b) and v2(b). Note also that v2(b) is connected to v1(b) by the edge b and

C(v1(b)) does not contain v2(b) so that b ∈ ∆Cn(v1(b)). Moreover, D∗ does not exist if edge

b is removed. Therefore, b ∈ D.

Now we divide D∗ to the following two cases:

1. D∗ does not contain any edge of ∂B∗(n). In this case, there exists a vacant dual path in

B∗(n) from one vertex of b∗ to the other one which separates v1(b) from v2(b).

2. D∗ contains an edge of ∂B∗(n). In this case there exists an occupied path from v1(b) to

∂B(n).

Similarly, we can discuss the situation for v2(b). With the observation, we could express

E(b) as the event that there exists a vacant dual path in B∗(n) from one vertex of b∗ to the

other one without using b, or E(b) occurs and there exist disjoint two occupied paths from

v1(b) and v2(b) to ∂B(n), respectively. We denote by D(b) the first event that there exists a

vacant dual path in B∗(n) from one vertex of b∗ to the other one without using b. Then

K ′n(p) = −

∑

b∈B(n)

Pp(E(b))

= −∑

b∈B(n)

Pp(D(b))−∑

b∈B(n)

Pp(E(b) \ D(b)).

Lemma 2. For p ≤ 1/2

limn→∞

1

|B(n)|∑

b∈B(n)

Pp(E(b) \ D(b)) = 0

uniformly on [0, 1/2].

Proof. For p ≤ pc if D(b) does not occur, then there is no vacant dual path from v1(b∗)

to v2(b∗) inside B∗(n). In other words, there exists an occupied path from either v1(b) or

v2(b) to ∂B(n). Therefore, it follows from an appropriate ergodic theorem (see Dunford and

Schwarz (1988)) that

lim supn→∞

1

n2

∑

b∈B(n)

Pp(E(b) \ D(b)) ≤ lim supn→∞

2

n2

∑

v∈B(n)

Pp(v → ∂B(n)) = 2θ(p) ≤ Cθ(pc) = 0

14

for some C > 0. This implies that

limn→∞

1

|B(n)|∑

b∈B(n)

Pp(E(b) \ D(b)) = 0 uniformly on [0, pc].

Therefore, Lemma 2 follows. ✷

Let

Nn(p) = −∑

b∈B(n)

Pp(D(b)).

We summarize the above arguments as the following Lemma.

Lemma 3. For all p ≤ 1/2,

limn→∞

Nn(p)

|B(n)| = −κ′(p).

Applying Russo’s formula again,

N ′n(p) = −[−

∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(e is pivotal for D(b))] =∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(e is pivotal for D(b)).

Now we show the following lemma.

Lemma 4.

{e is pivotal for D(b)} = R(b, e).

Proof. Suppose that R(b, e) occurs. There exists a vacant path from v1(b∗) to v2(b

∗) if

e∗ is vacant. Then D(b) occurs if e∗ is vacant. On the other hand, if e is occupied, there is

either an occupied circuit (except for b) from v1(b) to v2(b) in B(n) \ b and then from v1(b)

to v2(b) using b, or there exist two disjoint occupied paths one from v1(b) to ∂B(n) and the

other one from v2(b) to ∂B(n). We focus on the first case, that is there is an occupied circuit

(except for b) from v1(b) to v2(b) in B(n) \ b. Note that either v1(b∗) or v2(b∗) is inside of thecircuit or outside of the circuit so that without loss generality we assume that v1(b

∗) is inside

of the circuit, but v2(b∗) is not. We denote the circuit by F . We consider C∗

n(v1(b∗)) to be

the connected component inside of B(n) and containing v1(b∗) if we delete all occupied edges

of B∗(n) and b∗. Now we show C∗n(v1(b

∗)) does not contain v2(b∗). By Proposition 1 there

exists a unique circuit, denoted by F1, inside ∆C∗n(v1(b

∗)). It also follows from Proposition

1 that the edges enclosed by F1 are the edges enclosed by F . Therefore, v2(b∗) is outside of

F1 so that C∗n(v1(b)) does not contain v2(b

∗). This implies that D(b) does not occur if e is

occupied. In the second case, we consider the circuit formed by these two occupied paths

and the boundary of B(n). Then the same argument implies v1(b∗) and v2(b

∗) are also not

connected by any vacant dual paths inside of B∗(n) if e is occupied. In either case, D(b) will

not occur if e is occupied. Therefore, e is pivotal for D(b) so that

R(b, e) ⊂ {e is pivotal for D(b)}.

15

Now we will show the other direction in Lemma 4. Suppose that e is a pivotal for D(b).

If e is vacant, D(b) occurs. It implies that there exists a vacant path inside B∗(n) from v1(b∗)

to v2(b∗) if e is vacant. We denote the path by r∗. On the other hand, if e is occupied, D(b)

does not occur so that r∗ has to use e∗. Therefore, there exist r∗1 and r∗3 and S(r∗1, r∗3, b

∗, e∗)

defined in the event R(b, e). We shall find the other paths r2, r4 and r5, or r2 and r6. Note

that the boundary of S(r∗1, r∗3, b

∗, e∗) is a Jorden’s curve so that it divides Z∗ into two parts:

the inside part and the outside part. Without loss of generality, we assume that v1(b) belongs

to the inside part. Let us consider Cn(v1(b)) which is the connected component inside of B(n)

containing v1(b) if we delete all vacant edges of B(n), e and b. It follows from Proposition 1

there exists a circuit D∗ such that D∗ ⊂ ∆Cn(v1(b)). By Proposition 1 again D∗ has to be

enclosed by the boundary of the inside part. On the other hand, note that D∗ \ {b∗ ∪ e∗}is vacant and any such circuit has to use e since e∗ is a pivotal for D(b) so that e∗ ∈ D∗.

Therefore, there exists an occupied r2 from one of vertex of e to v1(b) inside S(r∗1, r∗3, b

∗, e∗).

We suppose without loss of generality that v1(e) and v1(b) are connected by r2.

Next we need to find r4 and r5 or r6 in R(b, e). Now we consider Cn(v2(b)) which is the

connected component inside of B(n) and containing v2(b) if we delete all vacant edges of

B∗(n) and b. Clearly, Cn(v2(b)) belongs to the outside part. Suppose that Cn(v2(b)) containsv2(e). Then r6 exists. Now we suppose that Cn(v2(b)) does not contain v2(e). By Proposition

1, there exists D∗3 ⊂ ∆Cn(v2(b)). If Cn(v2(b)) does not contain a vertex of ∂B(n), by the

same argument above, D∗3 \ b connects v1(b∗) to v2(b

∗) without using e. Since all edges of D∗3

inside of B(n) (not in its boundary) are vacant, it will contradict that e is a pivotal for D(b).

Therefore, r4 exists. Similarly, we can consider Cn(v2(e)) which is the component inside of

B(n) and containing v2(e) if we delete all vacant edges of B∗(n), b and e. With the same

argument, we can find r5. This implies R(b, e) occurs. ✷

By Lemma 4 we know

N ′n(p) =

∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(R(b, e)).

As we showed in Lemma 3, we see there are no long occupied paths so that the case of

existence r4 and r5 unlikely occurs. In other words, we can use R(b, e) to replace R(b, e)

without losing too much. More precisely,

Lemma 5. If p ≤ pc,

N ′n

|B(n)| =1

|B(n)|∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(R(b, e)) +o(n2)

n2.

Proof. Since the proof is the same as the proof in Lemma 3, we shall omit it. ✷

16

Now we shall focus on some p ∈ (0, p0] for some p0 < 1/2 and try to differentiate N ′n(p)

for p < p0. Note that R(e, b) is neither increasing nor decreasing so that we have to introduce

the following more general Russo’s formula: if B is increasing and A is decreasing, then (see

Lemma 1 in Kesten, H. (1987))

dP (A∩ B)dp

= −∑

f

P (f is pivotal for A, not for B, B occurs)

+∑

f

P (f is pivotal for B, not for A, A occurs).

To use the formular for R(b, e), let us recall (see Fig. 1)

R(e1, e2)

= {∃ disjoint paths r∗1 and r∗3 in B∗(n) from va(e∗1) and vb(e

∗1) to vc(e

∗2) and vf(e

∗2);


∗1, e

∗2) from va(e1) to vc(e2);


∗1, e

∗2);

r∗1 and r∗3 are vacant and r2 and r4 are occupied,},

where a 6= b and c 6= f are either 1 or 2 such that va(e∗1), vc(e

∗2) ∈ S(r∗1, r

∗3, e

∗1, e

∗2). We may

deal with only one case for R(e1, e2) without loss of generality when we pick a = 1, b = 2,

c = 1 and f = 2. We denote the case by

R(e1, e2)

= {∃ disjoint paths r∗1 and r∗3 in B∗(n) from v1(e∗1) and v2(e

∗1) to v1(e

∗2) and v2(e

∗2);


∗1, e

∗2) from v2(e1) to v1(e2);

∃ path r4 inside B(n) from v1(e1) to v2(e2) but outside the closure of S(r∗1, r∗3, e

∗1, e

∗2);

r∗1 and r∗3 are vacant and r2 and r4 are occupied}.

Now we divide R(b, e) into the intersection of a decreasing and an increasing events, that

is

R(b, e) = A(b, e) ∪ B(b, e),where

A(e1, e2) = {∃ two disjoint vacant paths r∗1 and r∗3 in B∗(n) from

v1(e∗1) and v2(e

∗1) to v1(e

∗2) and v2(e

∗2), respectively},

B(e1, e2) = {∃ two disjoint occupied paths r2 and r4 on B(n) from

v2(e1) and v1(e1) to v1(e2) and v2(e2), respectively}.

Furthermore, let

Mn(p) =∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(A(b, e) ∩ B(b, e)).

17

[−n, n]2 [−n, n]2

e

b

f

b

f

e

Figure 3: The left figure shows that f is pivotal for A(e, b), not for B(e, b), B(b, e) occurs,and the right figure shows that f is pivotal for B(e, b), not for A(e, b), A(b, e) occurs, wherethe solid paths are occupied and the dot paths are dual vacant.

18

Therefore, by the new Russo formula and the symmetry (see Fig.3)

dMn(p)

dp

= −∑

b

∑

e 6=b

∑

f 6=b,f 6=e

Pp(f is pivotal for A(b, e), not for B(b, e),

B(b, e) occurs)+

∑

b

∑

e 6=b

∑

f 6=b,f 6=e

Pp(f is pivotal for B(b, e), not for A(b, e),

A(b, e) occurs)

= −I + II,

where each sum above is taken over all bonds on B(n). We would like to give upper bounds

for I and II.

Let us focus on I. We divide the sum I into two parts

(a) (Part 1): e, b, f ∈ B(n′);

(b) (Part 2): One of e, b, f in B(n) \B(n′),

where n′ = n−√n for

√n ≫ L(p). Furthermore, we suppose that

‖v1(b)− v1(e)‖ = j,

‖v1(b)− v1(f)‖ = i.

We also suppose that

i ≤ j and ‖v1(b)− v1(f)‖ ≤ ‖v1(e)− v1(f)‖.We now construct

B1 = v1(b) +B(j

2− 1), B2 = v1(e) +B(

j

2− 1) and B3 = v1(f) +B(

i

2− 1).

The first two square boxes are not overlapped and B3 does not contain b and e. We first

estimate part 1. Now we also divide the following cases:

Case 1. j ≤ L(p).

Case 2. L(p) ≤ i.

Case 3. i ≤ L(p) ≤ j.

Then for a fixed b ∈ B(n′) by the symmetry∑

e 6=b

∑

f 6=b,f 6=e

Pp(f is pivotal for A(b, e), not for B(b, e), B(b, e) occurs)

≤ 3j∑

i=1

2n∑

j=1

∑

e 6=b

∑

f 6=b,f 6=e


B(b, e) occurs ‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = i),

where ‖v1(b)− v1(f)‖ ≤ ‖v1(e)− v1(f)‖ in the above right side sums. Then

j∑

i=1

2n∑

j=1

∑

e 6=b

∑

f 6=b,f 6=e


19

B(b, e) occurs ‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = i)

=∑

case 1Pp(f is pivotal for A(b, e), not for B(b, f),


+∑



+∑


B(b, e) occurs ‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = i),

where the sum for case l (l=1,2,3) is the sum taking over all possible choices in case l.

In case 1 (see Fig.3), note that if f is vacant, then A(b, e)∩B(b, e) occurs so that R(b, e)

occurs. In other words, if f is vacant, there are two disjoint vacant dual paths r∗1 and r∗3from the two vertices b∗ to the two vertices of e∗, and there are two disjoint occupied paths

from the two vertices of e to the two vertices of b. Therefore,

Q4(v1(b),j − 1

2) ∩ Q4(v1(e),

j − 1

2) occurs .

Here without loss of generality, we assume that (j − 1)/2 is a positive integer. Similarly,

(i−1)/2 or L(p)/2 are all integers in the later cases. Note also that f is a pivotal for A(b, e)

so that any vacant path from v1(b∗) to v1(e

∗) or any vacant path from v2(b∗) to v2(e

∗) has to

use f ∗. In either case, there exist two disjoint vacant dual paths without using f ∗ from the

two vertices of f ∗ to ∂B∗3 (See Fig. 3). Now we show that there exist two disjoint occupied

paths from the two vertices of f to ∂B3.

To show this, let Cn(v1(f)) and Cn(v2(f)) be the components inside B3 which contain

v1(f) and v2(f), respectively if we delete all vacant edges inside B3 and f . First we show

Cn(v1(f)) and Cn(v2(f)) are different. To show that, if they are the same, then it implies

that there exists an occupied path inside B3 from v1(f) to v2(f) without using f . In other

words, there exists an occupied circuit inside B3 that encloses either v1(f∗) or v2(f

∗) if f is

occupied. However, by Proposition 1 either vacant cluster v1(f∗) or v2(f

∗) cannot reach to

∂B3 if f is removed. It will contradict the event that there exist two disjoint dual vacant

paths from the two vertices of f ∗ to ∂B3.

Secondly, we show that Cn(vi(f)) contains a vertex of ∂B3 for both i = 1, 2. To show

this, if we suppose they did not, then by Proposition 1 there exists a vacant dual path γ

inside B3 from v1(f∗) or v2(f

∗) without using f ∗. We know as we mentioned before there

exists a vacant dual path from a vertex of e to a vertex of b and any such vacant dual path

has to use f ∗. However, on the other side, we can always find a vacant dual path from e to f

without using f since we can use the part of γ to avoid using f . This will contradict that f

20

is a pivotal bond for A(b, e). Therefore, Q4(v1(f), (i− 1)/2) occurs. We know B1 ∩B2 = ∅,but B3∩B1 may not be empty. To eliminate this problem, we can use the rerouting method

in Fig. 4 (see the graph proof for this argument in Fig. 4) to obtain the upper bound for a

fixed e, b and f , that is

Pp(f is pivotal for A(b, e), not for B(b, f),B(b, e) occurs ‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = i)

≤ CPp(Q4(v1(e),j − 1

2)Pp(Q4(v1(b),

j − 1

2)Pp(Q4(v1(f),

i− 1

2). (2.10)

Note that there are at most 8n2 choices for b and there are at most 8i choices for v1(f) if

‖v1(b)− v1(f)‖ = i

so that

∑

b

∑


B(b, e) occurs‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = i)

≤ C64n2L(p)∑

j=1

j∑

i=1

iPp(Q4(i/2))jP2p (Q4(j/2)) (C is the correction constant in Fig. 4)

≤ C2n2L(p)∑

j=1

j∑

i=1

iPp(Q4(i))jP2p (Q4(j)) (by Kesten’s extension method).

By Proposition 2

L(p)∑

j=1

j∑

i=1

iPp(Q4(i))jP2p (Q4(j)) ≤ C

L(p)∑

j=1

j∑

i=1

iP1/2(Q4(i))jP21/2(Q4(j)). (2.11)

For ǫ > 0, by (2.2) there exists C(ǫ) such that

P1/2(Q4(i)) ≤ Ci−5/4+ǫ for all i. (2.12)

Thenj∑

i=1

iP1/2(Q4(i)) ≤ Cj∑

i=1

i−1/4+ǫ ≤ C1j3/4+ǫ, (2.14)

If we substitute this into the right side of (2.11) and use (2.12) again,

L(p)∑

j=1

j∑

i=1

iP1/2Q4(i))jP21/2(Q(j)) ≤ C1

L(p)∑

j=1

jj3/4+ǫP 21/2(Q4(L(j)) ≤ C2

L(p)∑

j=1

j1+3/4−5/2−3ǫ.

21

[−n, n]2

b f

e

Figure 4: In this Fig. we shall show how to bound case 1 by CP 2p (Q4(j))Pp(Q4(i)) for a fixed

b, e and f . Here the box enclosing f is B3 and the largest box containing b and f , but not eis B1. We can see that Q4(v1(f), (i− 1)/2) occurs. We divide two situations, that is i < j/2and j ≥ i ≥ j/2. Here our graph is the first situation. We can see that Q4(v1(e), (j − 1)/2)also occurs independently fromQ4(v1(f), (i−1)/2). Besides these two events, we can also seethat there are four paths, two occupied and two vacant from b to the boundary of the smallestbox containing b, and two occupied and two vacant crossings in the annulus enclosing b andf . Since these four events use different bonds, they are independent. Furthermore, we canuse Kesten’s extension method to connect these four paths in the smallest box enclosing b,and the four crossings in the annulus. Note that the smallest box is v1(b)+B((i−1)/2) andthe annulus is v1(b) + {B(j/2) \B(i)} so the connection has the same probability but witha constant correction. Therefore, we have the right upper bound in (2.10) for case 1. In thesecond situation, we have an upper bound CP 2

p (Q4(j))Pp(Q4(i)) directly.

22

By (2.1) there exists C(ǫ) such that for all p

L(p) ≤ C(1/2− p)−4/3−ǫ. (2.16)

With these observations,

L(p)∑

j=1

j∑

i=1

iPp(Q4(i))jP2p (Q4(j)) ≤ CL(p)1/4+3ǫ ≤ C1(1/2− p)−1/3−4ǫ. (2.18)

By (2.18)L(p)∑

j=1

j∑

i=1

iPp(Q4(i))jP2p (Q4(j)) ≤ C(1/2− p)−1/3+δ(|1/2−p|). (2.20)

Therefore, there exists C > 0 such that∑

b

∑


B(b, e) occurs ‖v1(e)− v1(b)‖ = i, ‖v1(f)− v1(b)‖ = j)

≤ Cn2(1/2− p)−1/3+δ(|1/2−p|).

The case 1 has the right upper bound.

In case 2, let

B1 = v1(b) +B(L(p)− 1

2), B2 = v1(e) +B(

L(p)− 1

2) and B3 = v1(f) +B(

L(p)− 1

2).

Note that B1, B2 and B3 are disjoint. It follows from the same argument as we did in case

1 therefore we can show

Q4(v1(b),L(p)− 1

2) ∩Q4(v1(e),

L(p)− 1

2) ∩ Q4(v1(f),

L(P )− 1

2) occurs .

Also note that there are at least two disjoint occupied paths, one from B2 to the boundary

of v1(e) +B(j/2) and another one from B3 to the boundary of v1(f) +B(i/2). Therefore,∑

b

∑



≤ 64n2[Pp(Q4(L(p)/2))]3

2n∑

i=L(p)/2

2n∑

j=L(p)/2

iPp(B(L(p)/2) → ∂B(i/2)))jPp(B(L(p)/2) → ∂B(j/2)))

It follows from (2.24) in Kesten, H. (1987) that there exists C3 > 0 which is independent

of L(p) such that

∑

i≥L(p)

iPp(B(L(p)/2 → ∂B(i)) ≤∞∑

k=1

∑

kL(p)≤i≤(k+1)L(p)

iPp(B(L(p)/2)) → ∂B(i))

≤ 4C2L2(p)

∞∑

k=1

(k + 1)2e−C3k

≤ C3L2(p). (2.22)

23

Therefore, by Proposition 2, (2.2) and (2.22)

∑

b

∑


B(b, e) occurs ‖v1(e)− v1(b)‖ = j, ‖v1(f)− v1(b)‖ = j)

≤ C1n2L(p)4P 3

p (Q4(L(p)/2)

≤ Cn2L4−15/4+δ(|1/2−p|)(p) ≤ Cn2(1/2− p)−1/3+δ(|1/2−p|). (2.24)

In conclusion,

∑

b

∑


B(b, e) occurs‖v1(e)− v1(b)‖ = i, ‖v1(f)− v1(b)‖ = j)

≤ Cn2(pc − p)−1/3+δ(|1/2−p|).

In case 3, we can also use the the same arguments as in case 1, but with j ≥ L(p), to show

∑

b

∑



≤ Cn2∑

j≥L(p)/2

jP 2p (Q4(j))

L(p)∑

i=1

iPp(Q4(i))). (2.25)

By Proposition 2, (2.2) and (2.24) in Kesten (1987) we can show for all j

Pp(Q4(j)) ≤ Cj−5/4+δ(|1/2−p|). (2.26)

Therefore, by using (2.12) and (2.26) in (2.25)

∑

b

∑


B(b, e) occurs‖v1(e)− v1(b)‖ = i, ‖v1(f)− v1(b)‖ = j)

≤ C1n2L−1/2+3/4+δ(|1/2−p|)(p) ≤ (1/2− p)−1/3+δ(|1/2−p|). (2.27)

In conclusion,

∑

b

∑



≤ C2(1/2− p)−1/3+δ(|p−1/2|).

In summary of cases 1-3, under the condition in part 1 there exists a constant C > 0 such

that

I ≤ Cn2(1/2− p)−1/3+δ(|p−1/2|) (2.28)

24

Under the condition in part 2 we assume without loss of generality that b ∈ B(n)\B(n′).

Then there are at most Cn3/2 choices for b. On the other hand, we can also divide f and e

in the following three cases.

Case 1. i, j ≤ L(p).

Case 2. Both i, j > L(p).

Case 3. i ≤ L(p) < j.

Here i, j, are defined as the same as before. For case 1, it is simple to derive

(Part 2 under case 1) ≤ Cn3/2L4(p). (2.30)

For case 2, note that there exist occupied paths from v1(e) + B(L(p)/2) to v1(e) + B(i/2)

and v1(f) + B(L(p)/2) to v1(f) + B(j/2) . Therefore, the same estimate in Part 1, case 2

shows that there exists C > 0


Similarly, for case 3 there are at most L2(p) choices for f and we may use the same argument

of part 1, case 3 and (2.24) in Kesten (1987) to treat e to get


Together with part 1 and 2 this implies that

I ≤ Cn2(1/2− p)−1/3+δ(|1/2−p|) + Cn3/2L5(p). (2.36)

To estimate II we can use the same method as we used in I by simply interchanging the

roles of occupied and vacant. Recall that we only estimate a special case for R(b, e) when

a = 1, b = 2, c = 1, f = 2. The same estimate can be copied word for word to show the same

upper bound for the other cases. In other words,

M ′n(p) ≍ N ′′

n(p) + o(n2)/n2.

Therefore,

|N′′n(p)

|B(n)| | ≤ C(1/2− p)−1/3+δ(|1/2−p|) +o(n2)

n2. (2.38)

Now we try to find N ′′′p (n). We shall show that

|N′′′n (p)

|B(n)| | ≤ C(1/2− p)−4/3+δ(|1/2−p|) +o(n2)

n2. (2.40)

25

Since the estimate is the same as we did for N ′′n(p), we just give an outline for the third

derivative of N ′′′n (p). To estimate N ′′′

n (p) we only need to estimate M ′′n(p) so we only need to

estimate

−∑

b

∑

e 6=b

∑

f 6=e,f 6=b

dPp(R1(b, e, f))

dp+

∑

b

∑

e 6=b

∑

f 6=e,f 6=b

dPp(R2(b, e, f))

dp,

where

R1(b, e, f) = {f is pivotal for A(b, e) not for B(b, e) and B(b, e) occurs}

and

R2(b, e, f) = {f is pivotal for B(b, e) not for A(b, e) A(b, e) occurs}.Let us work on the first sum. We divide it into an intersection of an increasing event and a

decresing event. More precisely, we collect all necessary occupied paths connecting b, e and

f as an increasing event and all necessary vacant paths connecting b∗, e∗ and f ∗, where these

occupupied paths and vacant paths make R1(b, e, f) occur. We then use the new Russo’s

formula for positive and negative events. Besides b, e and f we will need to handle another

pivotal bond g. We denote by B4 = v1(g)+B(m) the box containing g and with side length

m, where m = ‖v1(g) − v1(b)‖. By the same argument as we did for M ′n(p), we can show

Q4(g, (m− 1)/2) occurs. We can use the rerouting method in Fig. 4 to separate this event

from the other events. Note that the other events in the three boxes centered at b, e and f

will contribute as the same as before, that is (1/2 − p)−1/3+δ(|1/2−p|). The new pivotal will

contribute at most∑2n

m=1 mPp(Q4((m− 1)/2). Then the upper bound of both the incresing

and decresing events are less than

[(1/2− p)−1/3+δ(|1/2−p|)][2n∑

m=1

mPp(Q4(m)] +o(n2)

n2.

By (2.22) and (2.14),

2n∑

m=1

mPp(Q(m) ≤ C(1/2− p)−1+δ(|p−1/2|).

Therefore, the first sum is less than C(1/2 − p)−4/3+δ(|1/2−p)|). Similarly, we can treat the

second sum. Therefore, we can show

|M′′n(p)

|B(n)| | ≤ C(1/2− p)−4/3+δ(|1/2−p|) +o(n2)

n2.

Note that M ′′′n (p) is a special case of N ′′′

n (p). But the other cases can be use the same way

to show the same upper bound so we have

|N′′′n (p)

|B(n)| | ≤ C1(1/2− p)−4/3+δ(|1/2−p|) +o(n2)

n2. (2.42)

26

We summarize (2.38) and (2.42) as the following Lemma.

Lemma 6. If p < 1/2, there exists C that does not depend on p such that

|N(l)n (p)

|B(n)| | ≤ C1(1/2− p)−1/3−(l−2)+δ(|1/2−p|) +o(n2)

n2

for l = 2, 3.

Now we would like to focus on the lower bound of N ′′n(p). We know N ′′

n consists of the

positive and the negative parts. In fact, it is possible to give a lower bound for the positive

or negative part separately. But it seems to be diffucult to give a lower bound of the sum,

since we do not know whether the negative terms and the positive terms will cancel each

other. Therefore, we have to go back to check N ′n(p). We know by Lemma 5

N ′n(p)

|B(n)| =1

|B(n)|∑

b∈B(n)

∑

e∈B(n)e 6=b

Pp(R(b, e)) +o(n2)

n2

=1

|B(n)| [∑

b∈B(n)\B(n′)

∑

e∈B(n)\B(n′);e 6=b

Pp(R(b, e)) +∑

b∈B(n′)

∑

e∈B(n′)e 6=b

Pp(R(b, e))

+2∑

b∈B(n)\B(n′)

∑

e∈B(n′)e 6=b

Pp(R(b, e))] +o(n2)

n2

= (Part 1)+ (Part 2) +(Part 3) +o(n2)

n2,

where n′ = n − √n ≫ L(p). Note that there are less than Cn3/2 choices for b if b ∈

B(n) \ B(n′). Note also that there is an occupied path from vi(e) to vj(b) if R(b, e) occurs

for some i = 1, 2 and j = 1, 2 so that

|Part 1| ≤ Cn−1/2∑

e∈B(n)

Pp(R(b, e)) ≤ Cn−1/2Ep(|C(0)|). (2.44)

We know that E(|C(0)|) < ∞ if p < 1/2. It also follows from the same estimate for Part 1

that

(Part 3) ≤ Cn−1/2E|C(0)|. (2.46)

Let us work on part 2. Let

Zn(p) =∑

b

∑

e∈B(n′)e 6=b

Pp(R(b, e))

=∑

b

∑

0≤i≤2n

∑

e∈B(n′)e 6=b

Pp(R(b, e), ‖v1(b)− v1(e)‖ = i).

27

Therefore,N ′

n(p)

|B(n)| =Zn(p)

|B(n)| + part 1 + part 3 +o(n2)

n2.

It follows from Lemma 6 for p0 < 1/2 and the Ascoli-Arzella theorem that there exists a

subsequence {ni} such that for p0 < 1/2

N ′ni(p)/|B(ni)| → κ′′(p) uniformly on [0, p0].

Therefore, by (2.44) and (2.46)

Zni(p)/|B(ni)| → κ′′(p) uniformly on [0, p0]. (2.48)

It follows from (2.48) that we only need to work on the bounds for Zn(p).

To show the lower bound, the key step is to compare Pp2(R(b, e)) with Pp1(R(b, e)) for

p1 < p2 and any two edges b and e. We know that if p + q = 1, then the largest value of

pm · qm is (1/4)2m when both p = 1/2 and q = 1/2. This seems to give us an intuitive feeling

that Pp(R(b, e)) is increasing as p ↑ 1/2 since R(b, e) is the event that there are occupied

and vacant paths. However, the lengths of occupied paths and vacant paths may not be the

same. In other words, we may deal with pnqm for n 6= m. Forturnately, when p is near 1/2,

n and m are not too distant, we have this increasing property but with an error correction.

More precisely, we can show the following theorem that is sufficient for our purpose.

Proposition 3. For any 0 < p1 < p2 ≤ 1/2 and two edges e and b with

L71/72(p1) ≤ ‖v1(b)− v1(e)‖ ≤ L73/72(p1)

there exists C that does not depend on p such that

Pp1(R(b, e)) ≤ Pp2(R(b, e)) + C(1/2− p1)1/72P1/2(R(b, e)).

Before the proof of Proposition 3, we would like to introduce a few definitions and lemmas.

We know if R(b, e) occurs, then there are two occupied paths, and two vacant dual paths

from b and b∗ to e and e∗, where two occupied paths are separated by two vacant paths. We

name the path sets by {r∗1}, {r∗3}, {r2} and {r4}, respectively, where r2 and r4 are occupied

and the others are vacant. We may consider the vacant cluster C∗(v1(b∗)) and the occupied

cluster C(v1(b)) by removing b and e. If R(b, e) occurs, as we discussed in Lemma 4, b∗

and e∗ belong to the unique circuit σ(C(v1(b))) and b and e belong to the unique circuit

σ(C∗(v1(b∗))), where σ(G), for a cluster G, is defined in Proposition 1. This implies that

σ(C∗(v1(b∗))) will contain r2 ∈ {r2} and r4 ∈ {r4}, and σ(C(v1(b))) will contain r∗1 ∈ {r∗1} and

r∗3 ∈ {r∗3}. Conversely, if σ(C(v1(b))) contains r∗1 and r∗3 and σ(C∗(v1(b∗))) contains r2 and

r4, then R(b, e) occurs. With the observation, we may use clusters C(v1(b)) and C∗(v1(b∗))

28

to decompose event R(b, e). We now want to fix these ∆C∗(v1(b∗)) and ∆C(v1(b)). We know

the pair

{ all boundary edges of C(v1(b)), C(v1(b))}is unique for each configuration. But if we work on this pair, it seems that we need to count

too many edges. Instead, we will work on pairs

{∂eC∗(v1(b∗)),∆C∗(v1(b

∗))} and {∂eC(v1(b)),∆C(v1(b))},

where

∂eC(v1(b)) = {e : e is an edge in C(v1(b)) and adjacent to at least one vertex of ∆C(v1(b))}.

Given a cluster C(v1(b)) we know it is a random animal. We may think ∂eC(v1(b)) as its skinand ∆C(v1(b)) as its fur. With these definitions we claim that

Pp(R(b, e)) =∑

s,f

as,f(b, e)ps(1− p)f , (2.49)

where the sum takes over all possible s and f . Here as,f(b, e) is the number of animals

{As,f(b, e)} such that As,f(b, e) consists of two disjoint clusters G1 and G∗2 inside B(n) by

removing two bonds b and e that satisfy:

Condition 1: v1(b) ∈ G1 and v1(b∗) ∈ G∗

2;

v1(b∗), v2(b

∗), v1(e∗), v2(e

∗) ∈ σ(G1) and v1(b), v2(b), v1(e), v2(e) ∈ σ(G∗2);

Note that b and e are removed so both ∆G1 and ∆G∗2 are divided into two separate compo-

nents, respectively. We require the component of (∆G1)∗ containing v1(b

∗) belongs to G∗2,

and the component of (∆G2)∗ containing v1(b) belongs to G1;

Condition 2. |∆G∗2 ∪ ∂eG1|e = s and |∆G1 ∪ ∂eG

∗2|e = f , where |G|e is the number of edges

in G and G1 and G∗2 do not have a crossing edge if they are disjoint.

To see (2.49) as we discussed above,

R(b, e) =⋃

G1,G∗

2

{∆C(v1(b)) = ∆G1,∆C∗(v1(b∗)) = ∆G∗

2},

where G1 and G∗2 take all edge sets satisfying condition 1 above. However, for different pairs

{G1, G∗2} and {G1, G2},

{∆C(v1(b)) = ∆G1,∆C∗(v1(b∗)) = ∆G∗

2} and {∆C(v1(b)) = ∆G1,∆C∗(v1(b∗)) = ∆G2

∗}

may not be disjoint. To elimate the problem, we treat {G1, G∗2} and {G1, G2

∗} as the same

pair if

∆G1 = ∆G1 and ∆G∗2 = ∆G∗

2.

If they are not the same, we label the two pairs as different pairs. With this new decompo-

sition,

R(b, e) =⋃{∆C(v1(b)) = ∆G1,∆C∗(v1(b

∗)) = ∆G∗2},

29

where the union takes all the different pairs {G1, G∗2} that satisfy condition 1. For each

configuration, the outer boundary ∆C(v1(b)) defines a unique bond set. On the other hand,

if ∆(G1) is vacant, then ∂G1 is also uniquely determined and occupied. With these obser-

vations, we can decompose R(b, e) into the following union of disjoint sets:

R(b, e) =⋃{∆C(v1(b)) = ∆G1, ∂eC(v1(b)) = ∂eG1,∆C∗(v1(b

∗)) = ∆G∗2, ∂eC∗(v1(b

∗)) = ∂eG∗2},

(2.50)

where the union takes all the different pairs {G1, G∗2} that satisfy condition 1. By (2.50)

Pp(R(b, e))

=∑

Pp(∆C(v1(b)) = ∆G1, ∂eC(v1(b)) = ∂eG1,∆C∗(v1(b∗)) = ∆G∗

2, ∂eC∗(v1(b∗)) = ∂eG

∗2),

where the sum takes all different pairs that satisfy condition 1. Note that

{∆C(v1(b)) = ∆G1, ∂eC(v1(b)) = ∂eG1,∆C∗(v1(b∗)) = ∆G∗

2, ∂eC∗(v1(b∗)) = ∂eG

∗2}

only involves in the edges in ∆G1, ∂eG1, ∆G∗2 and ∂eG

∗2. Note also that ∂eG1 and ∆G∗

2 are

occupied and ∂eG∗2 and ∆G1 are vacant so we have

∑

s,f

as,f(b, e)ps(1− p)f .

Therefore, (2.49) follows.

It is well known (see Theorem 4.20 in Grimmett (1999)) that the proportion of an occu-

pied cluster and its edge boundary is about p/(1− p). In the following lemma we show the

proportion of the skin and fur of an occupied cluster is also about p/(1− p).

Lemma 7. For fixed b and e there exists ǫ > 0 such that for 0 < x < ǫ

∑

f :|fp−(1−p)s|>2xs

as,f(b, e)ps(1− p)f ≤ 4s exp[−sx2].

Proof. The proof follows from the idea of Kunz and Souillard (see Theorem 4.20 in

Grimmett 1999). From (2.49)

Pp(R(b, e)) =∑

s,f

as,f(b, e)ps(1− p)f ≤ 1.

We then know for all of s, f and all of p

as,f(b, e) ≤ p−s(1− p)−f .

We choose p = s/(s+ f) and 1− p = f/(s+ f) to obtain

∑

f :|fp−(1−p)s|>2xs

as,f(b, e)ps(1− p)f ≤

∑

f :|fp−(1−p)s|>2xs

[(s + f)p/s]s[(s+ f)(1− p)/f ]f . (2.51)

30

Note that each vertex in the skin of G1 has to be adjacent to its fur. Similarly, each vertex

in the skin of G∗2 also has to be adjacent to its fur. Then we have

1 ≤ f ≤ 4s. (2.52)

By (2.51) and (2.52) we have

∑

f :|fp−(1−p)s|>2xs

as,f(b, e)ps(1− p)f ≤ 4s max

f :|fp−(1−p)s|>2xs{g(f)}, (2.53)

where

g(f) = [(s+ f)p/s]s[(s+ f)(1− p)/f ]f .

By the exact same proof of Theorem 4.20 (see page 83 in Grimmett 1999) we know that for

fixed s and for f satisfing f ≤ 4s

maxf

{g(f) : |s/p− f/(1− p)| > 2xs} ≤ exp(−1/3sx2p2(1− p))} (2.54)

for all positive value x. Lemma 7 follows from (2.53) and (2.54). ✷

With Lemma 7 we know that the proportion of s and f is about p/(1 − p). Next, we

want to study the sizes of the skin and fur in a cluster that connects to both b and e. Note

that C(v1(b)), and C∗(v1(b∗)) are denoted to be the occupied cluster and the vacant cluster

by removing b and e.

Lemma 8. For all p < 1/2 and all 1/6 ≥ δ > 0 if ‖v1(b)− v1(e)‖ ≥ L1−δ(p), there exists

a positive constant C1(δ) such that

Pp(|∆C(v1(b))| ≥ ‖v1(b)− v1(e)‖4/3+4δ,R(b, e)) ≤ C1((1/2− p)δPp(R(b, e)), (2.55)

Pp(|∂eC(v1(b))| ≥ ‖v1(b)− v1(e)‖4/3+4δ,R(b, e)) ≤ C1((1/2− p)δPp(R(b, e)), (2.56)

Pp(|∆C∗(v1(b∗1)| ≥ ‖v1(b)− v1(e)‖4/3+4δ,R(b, e)) ≤ C1((1/2− p)δPp(R(b, e)), (2.57)

Pp(|∂eC∗(v1(b∗1))| ≥ ‖v1(b)− v1(e)‖4/3+4δ,R(b, e)) ≤ C1(1/2− p)δPp(R(b, e)). (2.58)

Proof. We only prove (2.55) since the same proof can be carried out to show the other

three inequalities in Lemma 8. We first estimate

Ep(|∆C(v1(b))|I(R(b, e))) =∑

x

Pp(x ∈ ∆C(v1(b)),R(b, e)),

where I(R(b, e)) is the indicator of R(b, e). If x ∈ ∆C(v1(b)), as we did in Lemma 6, we can

construct three boxes centered at v1(b), x and v1(e), respectively:

B1 = v1(b) +B(j

2− 1), B2 = x+B(

j

2− 1) and B3 = v1(e) +B(

i

2− 1),

31

y

γ∗1

γ∗2

x

e

b

Figure 5: In this Fig. we assume that x 6∈ γ∗1 ∪ γ∗

2 , but x ∈ ∆C(v1(b)). Here y ∈ Z∗ ∩ γ∗1 is

the only common vertex such that any vacant dual path from x to b∗ has to use it. Then wecan check from our graph in the small box containing x and the annulus containing x andy to see Q3(x, l/2) and Q3(y, l, j) occur. Besides, Q4(v1(b), j/2) occurs and there are fourarm paths from e and e∗ so we may use the method in Fig. 4 to reconnect them such thatR(b, e) occurs.

where

‖v1(b)− x‖ = j and ‖v1(e)− x‖ = i.

We suppose that

j ≤ i.

On R(b, e) we know by Proposition 1 that there exist two vacant path γ∗1 and γ∗

2 staying

σ(C(v1(b))) such that together with b and e they form a closed circuit enclosing C(v1(b)) inits interior.

If x ∈ γ∗1 ∪ γ∗

2 , there are two disjoint vacant paths either using γ∗1 or γ∗

2 from x and its

neighbor to v1(b∗) and v1(e

∗), respectively and there is another occupied path from x′s dual

neighbor to v1(b) by using C(v1(b)). Therefore, Q3(x, j/3) occurs. Besides, we can use the

method in Fig. 4 to reconnect the four arm paths from bonds b and b∗ to the other four arm

paths from e and e∗ such that R(v, u) occurs. These reconnections cost a constant. Here we

32

assume that j ≤ i. By the symmetry, we can also treat the case if j ≥ i. Therefore, by (2.2)

and (2.24) in Kesten (1987) there exists a constant C1(δ) such that for all j

Pp(x ∈ ∆C(v1(b));R(b, e))) ≤ CPp(R(b, e))Pp(Q3(j/3)) ≤ C1Pp(R(b, e))j−2/3+δ/3. (2.60)

If x 6∈ γ∗1 ∪ γ∗

2 (see Fig. 5), then there exists a vacant dual path connecting x to either

γ∗1 or γ∗

2 (see Fig. 5) since x ∈ ∆C(v1(b)). Without loss of generality we assume that the

path comes to γ∗1 . We denote by y the common vertex of the path and γ∗

1 . In fact, by

the uniqueness in Proposition 1, there is only one vertex, our y, in γ∗1 such that any vacant

path from x to γ∗1 has to use it. We denote by f ∗ the bond that is adjacent to y and any

path from x to γ1 has to use it. By the Proposition 1 again, if we change f from vacant

to occupied, then there exists an occupied circuit enclosing x and separating x from γ1 (see

Fig. 5). Since the occupied circuit has to be a part of ∂eC(v1(b)), x is adjacent to the dual of

the circuit (see Fig. 5). Therefore, there exist one vacant path from x to y and two disjoint

occupied paths from neighbors of x also to the vertices of f , respectively. We still denote by

Q3(x, ‖x− y‖/2) the event. Now we assume that

‖x− y‖ = l

and divide the following two cases:

(1) l < j/4 and (2) l ≥ j/4. For the first case, we construct boxes

x+B(l/2) and y +B(j/2) and y +B(2l).

Note that box x + B(l/2) is contained in y + B(2l) and y + B(2l) is also contained in

y+B(j/2). By the discussion above we know that Q3(x, l/2) occurs. Since y ∈ γ1, we know

that there exist two occupied paths and a vacant path from ∂{y +B(2l)} to ∂{y +B(j/2)}inside {y + B(j/2)} \ {y + B(2l)}. We denote the event by Q3(y, l, j). By Proposition 2,

(2.2), Kesten’s extension method and (2.24) in Kesten (1987) we have for all j,

Pp(Q3(x, l/2)Pp(Q3(y, l, j)) ≤ C1l−2/3+δ/6(

j

l)−2/3+δ/6 ≤ C1j

−2/3+δ/3.

On the other hand, we can treat the four arm paths from b and b∗ and the other four arm

paths from e and e∗ by the same way as we did for x ∈ γ1 ∪ γ2. Therefore, in case (1)

Pp(x ∈ ∆C(v1(b));R(b, e))) ≤ C1Pp(R(b, e))j−2/3+δ/3. (2.61)

In case (2) by the same discussion above we know that Q3(x, j/3) occurs so that after

we reroute these four arm paths as we did before we still have

Pp(x ∈ ∆C(v1(b));R(b, e))) ≤ C1Pp(R(b, e))j−2/3+δ/3.

In summary, we always have

Pp(x ∈ ∆C(v1(b));R(b, e))) ≤ C1Pp(R(b, e))j−2/3+δ/3. (2.62)

33

Now we divide the mean into two parts.

Ep(|∆C(v1(b))|I(R(b, e)))

=∑

j≤L1+δ

jPp(x ∈ ∆C(v1(b)),R(b, e), ‖x− v1(b)‖ = j)

+∑

j≥L1+δ(p)

Pp(x ∈ ∆C(b, e),R(b, e), ‖x− v1(b)‖ = j)

= I + II.

With these observations and (2.62) there exists C which may depend on δ but not p such

that

I ≤ CPp(R(b, e))L1+δ(p)∑

j=1

jj−2/3+δ/3 ≤ CPp(R(b, e))L4/3+4δ/3+δ/3+δ2/3(p). (2.63)

By (2.62) and the same estimate in (2.22) (by replaying L(p) in (2.22) with L1+δ(p))

II ≤ CPp(R(b, e))L(p)2(1+δ) exp(−C1Lδ(p)). (2.64)

Together with (2.63) and (2.64) for δ < 1/6 and for all p < 1/2 there exists a constant C(δ)

such that

Ep(|∆C(v1(b))|I(R(b, e))) ≤ CPp(R(b, e))L4/3+2δ(p). (2.65)

By Markov’s inequality and (2.65)

Pp(|∆C(b, e)| ≥ ‖v1(b)− v1(e)‖4/3+4δ,R(b, e))

= Pp(|∆C(b, e)|I(R(b, e)) ≥ ‖v1(b)− v1(e)‖4/3+4δ)

≤ Ep(|∆C(v, u)|I(R(b, e)))

L(1−δ)(4/3+4δ)(p)

≤ CPp(R(v, u))L4/3+2δ(p)

L(1−δ)(4/3+4δ)(p)

≤ C1(1/2− p)δPp(R(b, e)). (2.66)

(2.55) follows ✷.

Proof of Proposition 3. Let us start at Pp1(R(b, e)). On R(b, e) we know that if

∆C(v1(b)) and ∆C(v1(b∗)) are fixed, then r∗1, r∗3, and r2, r4 exist and are contained in σC(v1(b))

and σC(v1(b∗)), respectively. By lemma 8 for δ = 1/72

Pp1(R(b, e)) ≤ Pp(R(b, e), |∆C(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∆C(v1(b∗))| ≤ ‖v1(b)− v1(e)‖4/3+4δ,

|∂eC(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∂eC(v1(b∗)| ≤ ‖v1(b)− v1(e)‖4/3+4δ)

+ C1Pp1(R(b, e))(1/2− p)δ. (2.67)

By Lemma 7, on R(b, e)

s = |∂eC(v1(b)) ∪∆C(v1(b∗))| ≥ ‖v1(b)− v1(e)‖ ≥ L1−δ(p1)

34

so we know that

Pp1(R(b, e), |∆C(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∆C(v1(b∗)| ≤ ‖v1(b)− v1(e)‖4/3+4δ,


≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f + C1L3(1+δ)(p1) exp(−CL(1−δ)δ(p1)). (2.68)

We know that

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f =∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f [p1(1− p1)]s(1− p1)

f−s.

Note that if 0 < p1 < p2 ≤ 1/2

p1(1− p1) ≤ p2(1− p2)

so that

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+4δ

as,fps1(1−p1)

f ≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(p2)s(1−p2)

f(1− p11− p2

)f−s. (2.69)

By taking ǫ = (1/2− p1)

maxs≤4‖v1(b)−v1(e)‖4/3+4δ ,

f≤s(1−p1)/p1+s1/2+δ

[1− p11− p2

]f−s

≤ maxs≤4‖v1(b)−v1(e)‖4/3+4δ ,

f≤s(1−p1)/p1+s1/2+δ

[2(1− p1)]f−s

≤ maxs≤4‖v1(b)−v1(e)‖4/3+4δ ,

f≤s(1−p1)/p1+s1/2+δ

(1 + 2ǫ)s(1+2δ)/2

(1 + 2ǫ)s[(1−p1)/p1−1]

≤ (1 + 2ǫ)4‖v1(b)−v1(e)‖(4/3+4δ)(1+2δ)/2

(1 + 2ǫ)2ǫp−11 ‖v1(b)−v1(e)‖4/3+4δ

. (2.70)

By our assumption that ‖v1(b)−v1(e)‖ ≤ L1+δ(p1) note that by (2.1) there exists a constant

C(δ) such that

L(p1) ≤ Cǫ(−4/3−δ)

so that the right of (2.70) is less than

(1 + 2ǫ)4(L(p1))(1+δ)(4/3+4δ)(1+2δ)/2

(1 + 2ǫ)2ǫp−11 (L(p1))(1+δ)(4/3+4δ)

≤ (1 + 2ǫ)4ǫ−8/9−7δ

(1 + 2ǫ)2ǫ−7/9−7δp−1

1

≤ (1 + C3ǫ1/9−7δ) = (1 + C3(1/2− p1)

1/9−7δ), (2.71)

35

where C3 does not depend on p1. By using (2.69)-(2.71) in (2.68)

Pp1(R(b, e), |∆C(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∆C(v1(b∗))| ≤ ‖v1(b)− v1(e)‖4/3+4δ,


≤ [∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps2(1− p2)

f ][1 + C3ǫ1/9−7δ] + C1L

2(1+δ)(p1) exp(−CL(1−δ)δ(p1))

≤ Pp2(R(b, e)[1 + C3(1/2− p1)1/9−7δ] + C1L

3(1+δ)(p1) exp(−CL(1−δ)δ(p1)).

Therefore, together with (2.67)

Pp1(R(b, e)) ≤ Pp2(R(b, e)[1 + C3(1/2− p1)1/9−7δ]

+C1L3(1+δ)(p1) exp(−CL(1−δ)δ(p1)) + C1Pp1(R(b, e))(1/2− p1)

δ (2.72)

Note that there exists C(δ) such that

L3(1+δ)(p1) exp(−CL(1−δ)δ(p)) ≤ CPp1(R(b, e))(1/2− p1)δ

for all p1. By Lemma 1 and Proposition 2 for all p and b and e there exists C such that

Pp(R(b, e)) ≤ CP1/2(R(b, e)).

Therefore, if we take δ = 1/72, then for all p we have

Pp1(R(b, e)) ≤ Pp2(R(b, e)) + C(1/2− p)1/72P1/2(R(b, e)). (2.73)

Proposition 3 follows. ✷

We use the idea of the proof in Proposition 3 to show the following Theorem.

Proposition 4. For all 0 < p1 < p2 < p3 < p4 < 1/2 with

L71/72(p4) < L73/72(p1),

then for any two edges b and e with

L71/72(p4) ≤ ‖v1(b)− v1(e)‖ ≤ L73/72(p1)

there exists C1 such that

Pp2(R(b, e))− Pp1(R(b, e))

p2 − p1≤ Pp4(R(b, e))− Pp3(R(b, e))

p4 − p3+ C1

(1/2− p1)1/73P1/2(R(b, e))

min{p4 − p3, p2 − p1}.

Proof. Note that

L71/72(p2) ≤ L71/72(p4) ≤ ‖v1(b)− v1(e)‖ ≤ L73/72(p1) ≤ L73/72(p2).

36

Note also that

(1− p2)/p2 ≤ (1− p1)/p1

so by (2.67) and (2.68) in Proposition 3 if δ = 1/72, we know that

Pp2(R(v, u))− Pp1(R(b, e))

≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps2(1− p2)

f −∑

s,f

as,fps1(1− p1)

f + C(1/2− p2)1/72P1/2(R(b, e)) (2.75)

≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps2(1− p2)

f −∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f + C(1/2− p1)1/72P1/2(R(b, e)).

Let us focus on the first sum in (2.75). We know that∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(ps2(1−p2)

f) =∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(ps2(1−p2)

s)2−(f−s)[2(1−p2)]f−s.

By the same estimates as (2.69)-(2.71) there exists a constant C such that

maxs≤4‖v1(b)−v1(e)‖4/3+4δ ,

f≤s(1−p1)/p1+s1/2+δ

[2(1− p2)]f−s

≤ maxs≤4‖v1(b)−v1(e)‖4/3+4δ ,

f≤s(1−p1)/p1+s1/2+δ

[2(1− p1)]f−s

≤ [1 + C(1/2− p1)1/72].

Note also that∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps2(1− p2)

s2−(f−s) ≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(1/2)s(1/2)f .

Then∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(ps2(1− p2)

f)

≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps2(1− p2)

s2−(f−s) + C(1/2− p1)1/72P1/2(R(b, e)). (2.76)

Let us focus on the second sum in (2.75). We know that∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)[2(1− p1)]f−s. (2.77)

37

If we set ǫ = (1/2− p1),

mins≤4‖v1(b)−v1(e)‖4/3+δ ,

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

[2(1− p1)]f−s

≥ mins≤4‖v1(b)−v1(e)‖4/3+δ ,

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

(1 + 2ǫ)−s1/2+δ

≥ (1 + 2ǫ)−4‖v1(b)−v1(e)‖(4/3+4δ)(1/2+δ)

≥ (1 + 2ǫ)−4L(p1)(1+δ)(4/3+4δ)(1/2+δ)

≥ (1 + 2ǫ)−4ǫ−8/9−7δ

(by the same estimate in (2.71))

≥ (1 + 2ǫ)−4ǫ−71/72

(note that δ = 1/72)

≥ 1− C3(1/2− p1)1/72. (2.78)

Together with (2.77) and (2.78) we have

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)[1− C3(1/2− p1)1/72]

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)

−∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,f(1/2)s(1/2)f [C3(1/2− p1)

1/72]

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s) − C3(1/2− p1)1/72P1/2(R(b, e)). (2.79)

By Lemma 7

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)

=∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)

−∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s−s1/2+δ

as,fps1(1− p1)

s2−(f−s)

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s)

38

−∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s−s1/2+δ

as,f(1/2)s(1/2)f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s) − L(p1)3(1+δ) exp(−Lδ(p1)). (2.80)

If we substitute (2.80) into the first sum in the right side of (2.79), note that the exponential

term in the right side of (2.79) is extremely small so there exists a constant C(δ) such that

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps1(1− p1)

s2−(f−s) − C(1/2− p1)1/72P1/2(R(b, e)).

Therefore, together with the estimate of the first sum

Pp2(R(b, e))− Pp1(R(b, e))

≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f [(p2(1− p2))s − (p1(1− p1))

s]2−(f−s)

+ C(1/2− p1)1/72P1/2(R(b, e)). (2.81)

Similarly,

Pp4(R(v, u)) ≥∑

s≤4‖v1(b)−v1(e)‖4/3+δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,f(p4(1− p4))s2−(f−s)[2(1− p4)]

f−s

Replace L(p1) and ǫ = (1/2− p1) in (2.78) by L(p4) and ǫ = 1/2− p4 to obtain

mins≤4‖v1(b)−v1(e)‖4/3+δ ,

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

[2(1− p4)]f−s ≥ 1−C(1/2− p4)

1/72 ≥ 1−C(1/2− p1)1/72. (2.82)

Therefore,

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

s2−(f−s)[1− C(1/2− p1)1/72]

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

s2−(f−s)

39

−∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,f(1/2)s(1/2)f [C(1/2− p1)

1/72]

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δ≤f≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

s2−(f−s) − C(1/2− p1)1/72P1/2(R(b, e)). (2.83)

By the same estimate of (2.80) we have

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

s−s1/2+δf≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

f

≥∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps4(1− p4)

s2−(f−s) − CL(p1)3(1+δ) exp(−Lδ(p1)). (2.84)

Now we need to deal with Pp3(R(b, e)). By the same estimate of (2.76) we have

∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,f(ps3(1− p3)

f)

≤∑

s≤4‖v1(b)−v1(e)‖4/3+4δ

f≤s(1−p1)/p1+s1/2+δ

as,fps3(1− p3)

s2−(f−s) + C(1/2− p1)1/72P1/2(R(b, e)). (2.85)

Together with (2.83)-(2.85) we have

Pp4(R(b, e))− Pp3(R(b, e))

≥∑

s≤4‖v1(b)−v1(e)‖4/3+δ

f≤s(1−p1)/p1+s1/2+δ

as,f [(p4(1− p4))s − (p3(1− p3))

s]2−(f−s)

− C(1/2− p1)1/72P1/2(R(b, e)). (2.86)

Note that if p < 1/2 and s(1− 2p) ≥ 1, then the second derivative of [p(1− p)]s is positive.

Note also that if 1/2− p1 is small, by (2.1)

s ≥ ‖v1(b)− v1(e)‖ ≥ L71/72(p4) ≫ (1/2− p4)−1.

This implies that s(1− 2p) ≥ 1 for all p1 ≤ p ≤ p4 so

(p2(1− p2))s − (p1(1− p1))

s

p2 − p1≤ (p4(1− p4))

s − (p3(1− p3))s

p4 − p3. (2.87)

Therefore, Proposition 4 follows from (2.81), (2.86) and (2.87) when 1/2− p1 is small so it

holds for all of p with a constant correction. ✷

40

3 Proof of Theorems

Proof of Theorem 1 for the Z2 lattice. By Lemma 3, Lemma 6 and Ascoli and Arzella’s

theorem we know that for p < 1/2

κ′′′(p) ≤ (1/2− p)−1/3+δ(1/2−p). (3.1)

Therefore, the upper bound in Theorem 1 for bond percolation in Z2 follows from (3.1) when

p < 1/2.

Now we focus on the lower bound of Theorem 1. Recall that

Zn(p) =∑

b

∑

0≤i≤2n

∑

e∈B(n′)e 6=b

Pp(R(b, e), ‖v1(b)− v1(e)‖ = i).

It also follows from (2.48) that there exists {ni} such that

limi→∞

Zni(p)/|B(n2

i )| = κ′′(p) (3.2)

For 1/6 > δ > 0 by (2.2) there exists C that may depend on δ, but not p such that

P1/2(Q4(i)) ≤ Ci−5/4+δ for all i. (3.3a)

For the δ we select p0 so that

(1/2− p)−4/3+δ2 ≤ L(p) ≤ (1/2− p)−4/3−δ2 for all p0 ≤ p ≤ 1/2. (3.3b)

For the δ we require

100δ ≤ 1/144. (3.3c)

For the δ we pick p0 ≤ p1 < p2 < 1/2 such that

L1+δ(p1) ≤ L(p2) ≤ L1+2δ(p1). (3.3d)

Since

(1/2− p2)−4/3(1−δ) ≤ L(p2) ≤ L1+2δ(p1) = (1/2− p1)

(−4/3−δ)(1+2δ) ,

(1/2− p2)−1 ≤ (1/2− p1)

−(1+3δ)/(1−δ). (3.3e)

By (2.1) and the first inequality in (3.3d),

(1/2− p1)(−4/3+δ2)(1+δ) ≤ L1+δ(p1) ≤ L(p2) ≤ (1/2− p2)

−4/3−δ2 .

Therefore, the distance from p2 to 1/2 is much smaller than the distance from p2 to p1 so

(1/2− p1) ≤ 2(p2 − p1). (3.3f)

41

For p1 < p2 < 1/2 satisfying (3.3d) we divide

Zn(p2)− Zn(p1)

p2 − p1=

∑

b∈B(n′)

L1−100δ(p1)∑

i=1

∑

e∈B(n′)e 6=b

[Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp1(R(b, e), ‖v1(b)− v1(e)‖ = i)

p2 − p1]

+∑

b∈B(n′)

L1+δ(p1)∑

i=L1−100δ(p1)

∑

e∈B(n′)e 6=b


p2 − p1]

+1

p2 − p1

∑

b∈B(n′)

2n∑

i=L1+δ(p1)

∑

e∈B(n′)e 6=b

Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)

− 1

p2 − p1

∑

b∈B(n′)

2n∑

i=L1+δ(p1)

∑

e∈B(n′)e 6=b

Pp1(R(b, e), ‖v1(b)− v1(e)‖ = i). (3.4)

We estimate the first term in (3.4). By using the estimates in case 1 of I and case 1 of II in

(2.18) and the mean value theorem we know that for the δ there exists u ∈ (p1, p2)

|∑

b∈B(n′)

L1−100δ(p1)∑

i=0

∑

e∈B(n′)e 6=b

[Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp1(R(b, e)), ‖v1(b)− v1(e)‖ = i)

p2 − p1]|

= |∑

b∈B(n′)

∑

0≤i≤L1−100δ(p1)

∑

e∈B(n′)e 6=b

dPu(R(b, e), ‖v1(b)− v1(e)‖ = i)

du|

≤ Cn2L1−100δ(p1)∑

j=1

j∑

i=1

iPu(Q4(i)jP2u (Q4(j))

≤ C1n2(1/2− p2)

(−1/3−3δ)(1−100δ) (by the assumption in (3.3b) and the estimate in (2.18))

≤ C1n2(1/2− p2)

−1/3+30δ

≤ C1n2(1/2− p1)

(−1/3+30δ)(1+3δ)/(1−δ) (by (3.3e))

≤ C1n2(1/2− p1)

(−1/3+29δ)/(1−δ)

≤ C1n2(1/2− p1)

−1/3+28δ, (3.5)

where C1 may depend on δ but not p1, p2 and u. Now let us estimate the second sum in

(3.4). By using Proposition 3 in the second sum we know that

∑

b∈B(n′)

L1+δ(p1)∑

i=L1−100δ(p1)

∑

e∈B(n′)e 6=b

[Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp1(R(b, e)), ‖v1(b)− v1(e)‖ = i)

p2 − p1]

≥ −Cn2(1/2− p1)1/72 1

p2 − p1

L1+δ(p1)∑

i=L1−100δ(p1)

∑

e∈B(n′)e 6=b

P1/2(R(b, e)), ‖v1(b)− v1(e)‖ = i)

42

≥ −C1n2(1/2− p1)

1/72 1

p2 − p1

L1+δ(p1)∑

i=L1−100δ(p1)

iP 21/2(Q4(i)) ( by Lemma 1). (3.6)

Then we know that

1

p2 − p1

L1+δ(p1)∑

i=L1−100δ(p1)

iP 21/2(Q4(i))

≤ C3(1/2− p1)−1

∞∑

i=(1/2−p1)(−4/3+δ)(1−100δ)

i−3/2+δ (by (3.3a) and (3.3b) and (3.3f))

≤ C4(1/2− p1)−1(1/2− p1)

2/3−67δ

≤ C4(1/2− p1)−1/3−67δ. (3.7)

By (3.6) and (3.7) the second sum in (3.4) is larger than

−C5n2(1/2− p1)

−1/3+1/72−67δ . (3.8)

We focus on the third sum in (3.4). By the same estimate in (3.7) the third sum in (3.4) is

∑

b

2n∑

i=L(p1)1+δ

∑

e∈B(n′)e 6=b

Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)

p2 − p1

≥ Cn2

p2 − p1

2L(p1)1+δ∑

i=L(p1)1+δ

iP 2p2(Q4(i)) (by (3.3d) and Lemma 1)

≥ Cn2

p2 − p1

2L(p1)1+δ∑

i=L(p1)1+δ

i−3/2−δ(by (3.3d), Proposition 2 and (2.2))

≥ C(1/2− p1)−1L(p1)

1+δL(p1)(1+δ)(−3/2−δ))

≥ C1n2(1/2− p1)

−1(1/2− p1)(−4/3+δ)(1+δ)(1/2− p1)

(−4/3−δ)(1+δ)(−3/2−δ) (by (3.3b))

≥ C1n2(1/2− p1)

−1−4/3+2−4δ/3+δ+δ2+3δ/2+2δ+3/2δ2+4δ/3+δ2+4δ2/3+δ3

≥ C1n2(1/2− p1)

−1/3+11δ. (3.9)

Now we work on the fourth sum. Note that there is at least one occupied path from b to e

so by the same estimate of (2.22) (by replacing L(p) in (2.22) with L1+δ(p) here)

|∑

b

∑

2n≥i≥L1+δ(p1)

∑

e∈B(n′)e 6=b

Pp1(R(b, e), ‖v1(b)− v1(e)‖ = i)| ≤ C2n2[L(p1)]

2(1+δ)e−C3Lδ(p1).

(3.10).

If we compare our sums, we can find that the third sum dominates in these sums for all

p0 ≤ p1 < p2. Indeed, if we put these four sums together

Zn(p2)− Zn(p1)

p2 − p1

≥ Cn2[−(1/2− p1)−1/3+28δ − (1/2− p1)

−1/3+1/72−67δ

+(1/2− p1)−1/3+11δ − (1/2− p1)

−8 exp−(1/2− p1)δ].

43

In summary, if we take 0 < δ < 1/14400 satisfying (3.3a)-(3.3c) and take p1 and p2 satisfying

(3.3d), then there exist C(δ) and p0 such that for all p0 ≤ p1 < p2 < 1/2

Zn(p2)− Zn(p1)

p2 − p1≥ Cn2(1/2− p1)

−1/3+11δ. (3.11)

By (3.2) and the mean value theorem we know that there is a subsequence pj → 1/2 such

that

κ′′′(pj) ≥ C(1/2− pj)−1/3+δ(1/2−pj ).

Now we need to show this holds for each sequence that goes to 1/2. In order to do that we

denote intervals I1, ..., Ik, ... by

Ik = [1/2− 1/2k−1, 1/2− 1/2k].

We only need to show for each x ∈ Ik

κ′′′(x) ≥ (1/2− x)−1/3+δ.

Let us focus on Ik. If we pick any two points from the interval, (3.3d) will not hold so

we cannot get (3.11). Therefore, we enlarge the interval to [1/2 − 1/2k(1−ǫ), 1/2 − 1/2k] for

1/6 > ǫ > 0. We pick p1 = 1/2− 1/2k(1−ǫ) and p2 = 1/2− 1/2k. For the ǫ > 0 note that by

(2.1) if we take k large, then

L(1+ǫ)(p1) = L1+ǫ(1/2−1/2k(1−ǫ)) ≤ 24k(1−ǫ)(1+ǫ)(1+ǫ2/4)/3 and L(p2) = L(1/2−1/2k) ≥ 24k(1−ǫ2/4)/3.

On the other hand, we take k large such that

L(p2) ≤ 24k(1+ǫ2)/3 and L1+2ǫ(p1) ≥ 24k(1−ǫ2)(1+2ǫ)(1−ǫ)/3.

Therefore, there exists M0 such that for all k ≥ M0,

L1+ǫ(p1) ≤ L(p2) ≤ L1+2ǫ(p1).

We also pick ǫ > 0 such that

300ǫ < 1/146.

Then for the ǫ (3.3a)-(3.3f) hold. By (3.11)

Zn(p2)− Zn(p1)

p2 − p1≥ Cn2(1/2− p1)

−1/3+11ǫ ≥ Cn22k(1/3−12ǫ). (3.12)

Let us compare with Zn(p2)−Zn(p1)n2(p2−p1)

and Zn(p4)−Zn(p3)n2(p4−p3)

for p3 = 1/2−1/2k and p4 = 1/2− [1/2k−1/2k(1+100ǫ)]. Note that

p4 − p3 ≤ 1/2k(1+100ǫ) and (1/2− p4) = 1/2k − 1/2k(1+100ǫ) ≥ 1/2k+1. (3.13)

44

Note also that by (2.1) there exists M1 such that for k ≥ M1

L71/72(p4) ≤ 2(4(k+1)/3+ǫ)71/72 ≤ 2k(4/3−ǫ)(1−ǫ)73/72 ≤ L73/72(p1). (3.14)

Now we can use Proposition 4 for pi (i = 1, 2, 3, 4) because of (3.14). Let us divideZn(p2)−Zn(p1)

n2(p2−p1)into

Zn(p2)− Zn(p1)

p2 − p1=

∑

b∈B(n′)

L1−200ǫ(p1)∑

i=1

∑

e∈B(n′)e 6=b


p2 − p1]

+∑

b∈B(n′)

L(p4)1+ǫ∑

i=L1−200ǫ(p1)

∑

e∈B(n′)e 6=b


p2 − p1]

+∑

b∈B(n′)

2n∑

i=L1+ǫ(p4)

∑

e∈B(n′)e 6=b


p2 − p1]

= I1 + II1 + III1.

Similarly,

Zn(p4)− Zn(p3)

p4 − p3=

∑

b∈B(n′)

L1−200ǫ(p1)∑

i=1

∑

e∈B(n′)e 6=b


p4 − p3]

+∑

b∈B(n′)

L(p4)1+ǫ∑

i=L1−200ǫ(p1)

∑

e∈B(n′)e 6=b


p4 − p3]

+∑

b∈B(n′)

2n∑

i=L1+ǫ(p4)

∑

e∈B(n′)e 6=b


p4 − p3]

= I2 + II2 + III2.

It follows from the estimate in (3.5) that for the ǫ

|I1| ≤ Cn2(1/2− p2)−1/3+64ǫ ≤ Cn22(1/3−64ǫ)k (3.15)

and

|I2| ≤ Cn2(1/2− p4)−1/3+64ǫ ≤ Cn22(k+1)(1/3−64ǫ) (3.16)

To compare II1 with II2 we need to use Proposition 4. With (3.14) to use Proposition 4 it

remains to check that each pair b, e in the sum II satisfies

L71/72(p4) ≤ ‖v1(b)− v1(e)‖ ≤ L73/72(p1).

45

Note that for each such pair

L1−200ǫ(p4) ≤ ‖v1(b)− v1(e)‖ ≤ L1+ǫ(p1) and 300ǫ ≤ 1/146

so that the above inequality follows for all large k. By Proposition 4 we know that

II1 ≤ II2 + C(1/2− p1)

1/73

p4 − p3n2

L1+ǫ(p4)∑

i=L1−200ǫ(p1)

∑

e∈B(n′)e 6=b

P1/2(R(b, e), ‖v1(b)− v1(e)‖ = i).

By the same estimate as (3.7)

1

p4 − p3

L1+ǫ(p4)∑

i=L1−200ǫ(p1)

∑

e∈B(n′)e 6=b

P1/2(R(b, e), ‖v1(b)− v1(e)‖ = i) ≤ C2k(1+100ǫ)2[(−2/3+134ǫ)]k.

Therefore,

II1 ≤ II2 + C12k(1/3−1/73+234ǫ). (3.17)

By the same estimate of (3.10)

|III1| ≤ Cn2L(p4)2(1+ǫ) exp(−L(p1)

ǫ) (3.18)

and

|III2| ≤ Cn2L(p4)2(1+ǫ) exp(−L(p1)

ǫ). (3.19)

Therefore, by (3.12), (3.15)-(3.19)

Zn(p4)− Zn(p3)

n2(p4 − p3)

≥ Zn(p2)− Zn(p1)

n2(p2 − p1)− C2k(1/3−64ǫ) − C2k(1/3−1/73+234ǫ) − C24k exp(−2ǫk)

≥ C12k(1/3−12ǫ) − C2k(1/3−64ǫ) − C2k(1/3−1/73+234ǫ) − C24k exp(−2ǫk).

Therefore, there exist M1 and C(ǫ) such that for all k ≥ M1 and for the subsequence in (3.2)

Zni(p4)− Zni

(p3)

n2i (p4 − p3)

≥ C2k(1/3−12ǫ). (3.20)

By (3.20) for the subsequence and the mean value theorem again we know that there exists

x ∈ [1/2− 1/2k, 1/2− (1/2k − 1/2k(1+100ǫ))] such that for all k > M1

κ′′′(x) ≥ C2(1/3−12ǫ)k. (3.21)

We know that for any u ∈ [1/2− 1/2k, 1/2− (1/2k − 1/2k(1+100ǫ))]

κ′′′(u) = κ′′′(x) + κ′′′′(l)|(u− x)|

46

for some l ∈ (x, u). By Lemma 6 there exist M2 and C(ǫ) such that for all k ≥ M2

|κ′′′′(l)| ≤ C(1/2− l)−4/3−ǫ ≤ 4C2k(4/3+ǫ) (3.22)

By (3.22) and (3.21) we know that for all u ∈ [1/2− 1/2k, 1/2− (1/2k − 1/21+100ǫ)] and for

all k ≥ M4 = max(M1,M2)

κ′′′(u) ≥ κ′′′(x)− C2k(4/3+ǫ−1−100ǫ) ≥ C12(1/3−12ǫ)k − C22

(1/3−99ǫ)k ≥ C32(1/3−12ǫ)k. (3.23)

If we check every step from (3.13)-(3.23), for all k ≥ M4 and for any p4, p3 ∈ [1/2−1/2k, 1/2−1/2k+1] with p4 − p3 ≤ 1/2k(1+100ǫ), then

κ′′′(u) ≥ C32(1/3−12ǫ)k (3.24)

for all u ∈ [p3, p4]. Therefore, there exists M3, fixed and large, such that for all k ≥ M3 and

for any point u ∈ [1/2− 1/2k, 1/2− 1/2k+1]

κ′′′(u) ≥ C32(1/3−11ǫ)k ≥ C4(1/2− u)−1/3+12ǫ, (3.25)

where C may depend on ǫ, but not u and k. The lower bound in Theorem 1 for p < 1/2

follows from (3.25) for bond percolation in Z2. Also, Theorem 1 follows from Sykes and

Essam’s identity together with the lower and upper bounds when p > 1/2.

Proof of Theorem 1 for site percolation in the triangular lattice.

We denote as we did in section one by

κ(p) = Ep(C(0)−1I(|C(0)| > 0)).

Similar to bond percolation if we denote by Mn the number of occupied clusters in B(n)

and denote by Kn the mean of Mn, then (see Page 526 in Aizenman, Kesten and Newman

(1987))

limn→∞

Kn

|B(n)| = κ(p) a.s. and in L1.

What changes is the formula for dKn/dp. Let us define for x ∈ B(n), Nn(x) as the number

of distinct occupied cluster, obtained after setting x to be vacant, which contains a neighbor

of x. It follows from (5.5) in Aizenman, Kesten and Newman (1987) or Theorem 4.3 in

Grimmett (1981) thatdKn

dp=

∑

x∈B(n)

Ep(1−Nn(x)) (3.26)

Note that unlike bond percolation dKn/dp is neither increasing nor decreasing in p. We use

the method in Aizenman, Kesten and Newman (1987) to add one term to make a monotone

property for dKn/dp. To do it let us define

Vn(x) = number of neighbors of x that are vacant.

47

After adding this new term, Nn(x) + Vn(x) cannot increasing when a site is changed from

vacant to occupied for each x ∈ B(n). Hence

Ep(Nn(x) + Vn(x)) is decreasing in p.

For each fixed x, by Russo’s formula and the same discussion as we did in Lemma 4

dEp(Nn(x) + Vn(x))

dp= −

∑

y 6=x

Pp(R(x, y)), (3.27)

where

R(x, y) =

{∃ disjoint paths r1 and r3 in B(n) from two neighbors of x to two neighbors of y;

∃ path r2 on B(n) inside S(r1, r3, x, y) from a neighbor of x to a neighbor of y;

∃ disjoint paths r4 and r5 from a neighbor of x and a neighbor of y to ∂B(n)

but outside the closure of S(r1, r3, x, y) or ∃ path r6 inside B(n)

from a neighbor of x to a neighbor of y but outside the closure of S(r1, r3, x, y);

r1 and r3 are occupied and and rl is vacant for l = 2, 4, 5 or 2, 6},

where S(r1, r3, x, y) is the open set enclosed by the circuit r1 ∪ x ∪ r3 ∪ y. Note that there

are six neighbors for each vertex of x

EpVn(x) = 6(1− p) (3.28)

By (3.26) and (3.28) we have

dKn

|B(n)|dp = 1− 1

|B(n)|∑

x∈B(n)

Ep(Nn(x) + Vn(x)) + 6(1− p). (3.29)

By (3.27) and (3.29)

K ′′n

|B(n)| =1

|B(n)|∑

x∈B(n)

∑

y 6=x

Pp(R(x, y))− 6 (3.30)

After (3.30) we can follow the exact estimates in bond percolation to find the upper bond

and lower bound in Theorem 1 for the third derivative for p < 1/2. Then we use the Sykes

and Essam’s identity for p > 1/2 to find the upper and lower bound in Theorem 1.

References

[1] Aizenman, M., Kesten, H. and Newman, C. (1987), Uniqueness of the infinite cluster

and continuity of connectivity functions for short and long range percolation, Comm. Math.

Phys. 111, 503-532.

48

Dunford, N. and Schwartz, T. (1958) Linear operators, Vol 1, Wiley-Intrscience, New York.

Grimmett, G. (1981) On the differentiability of the number of clusters per vertex in perco-

lation model. J Lond Math Soc (2) 23, 372-384.

Grimmett, G. (1999) Percolation. Springer-Verlag, New York.

Kesten, H. (1982). Percolation Theory for Mathematicians, Birkhauser, Boston.

Kesten, H. (1987). Scaling relations for 2D-percolation. Comm. Math. Phys. 109, 109-156.

Smirnov, S. and Wendelin,. W. (2001) Critical exponent for two dimensional percolation.

Math. Res. Letter 8 729-744.

Sykes, M. F., and Essam, J. W. (1964) Exact critical percolation probabilities for site and

bond problems in two dimensions, J. Math. Phys. 5 1117-1127.

Yu Zhang

Department of mathematics

University of Colorado

Colorado Springs, CO 80933

[email protected]

49

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