arxiv.org e-print archive · arxiv:math/0301160v1 [math.pr] 15 jan 2003...
TRANSCRIPT
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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
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)
3
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);
∃ path r4 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 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)
10
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);
∃ path r2 on B(n) inside S(r∗1, r∗3, e
∗1, e
∗2) from va(e1) to vc(e2);
∃ path r4 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 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);
∃ path r2 on B(n) inside S(r∗1, r∗3, e
∗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
Pp(f is pivotal for A(b, e), not for B(b, 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
Pp(f is pivotal for A(b, e), not for B(b, 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)
+∑
case 2Pp(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)
+∑
case 3Pp(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
∑
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)
≤ 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
∑
case 1Pp(f is pivotal for A(b, e), not for B(b, f),
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
∑
case 2Pp(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)
≤ 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
∑
case 2Pp(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)‖ = 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
∑
case 2Pp(f is pivotal for A(b, e), not for B(b, f),
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
∑
case 3Pp(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)
≤ 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
∑
case 3Pp(f is pivotal for A(b, e), not for B(b, f),
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
∑
case 3Pp(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)
≤ 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
(Part 2 under case 2) ≤ Cn3/2L5(p). (2.32)
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
(Part 2 under case 3) ≤ Cn3/2L5(p). (2.34)
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,
|∂eC(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∂eC(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,
|∂eC(v1(b))| ≤ ‖v1(b)− v1(e)‖4/3+4δ, |∂eC(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
[Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp1(R(b, e), ‖v1(b)− v1(e)‖ = i)
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
[Pp2(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp1(R(b, e), ‖v1(b)− v1(e)‖ = i)
p2 − p1]
+∑
b∈B(n′)
L(p4)1+ǫ∑
i=L1−200ǫ(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]
+∑
b∈B(n′)
2n∑
i=L1+ǫ(p4)
∑
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]
= I1 + II1 + III1.
Similarly,
Zn(p4)− Zn(p3)
p4 − p3=
∑
b∈B(n′)
L1−200ǫ(p1)∑
i=1
∑
e∈B(n′)e 6=b
[Pp4(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp3(R(b, e), ‖v1(b)− v1(e)‖ = i)
p4 − p3]
+∑
b∈B(n′)
L(p4)1+ǫ∑
i=L1−200ǫ(p1)
∑
e∈B(n′)e 6=b
[Pp4(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp3(R(b, e), ‖v1(b)− v1(e)‖ = i)
p4 − p3]
+∑
b∈B(n′)
2n∑
i=L1+ǫ(p4)
∑
e∈B(n′)e 6=b
[Pp4(R(b, e), ‖v1(b)− v1(e)‖ = i)− Pp3(R(b, e), ‖v1(b)− v1(e)‖ = i)
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
49