PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

278 Geoffrey Grimmett
If A(q) < 1 (which holds for sufficiently large q), then
φ 1 B,p,q(0 ↔ ∂B) = φ 1 � �
B,p,q OC(Γ) occurs for no Γ
≥ 1 − A(q) > 0.
(We have used the assumption of wired boundary conditions here.) This
implies, by taking the limit n → ∞, that θ1 (p, q) > 0 when p = √ q/(1 +
√
q). Using (13.39), this implies parts (a) and (b) of the theorem, when q is
sufficiently large.
For general q > Q, we have only that A(q) < ∞. In this case, we find N
(< n) such that
�
Γ outside B(N)
φ 1 � �
1
B,p,q OC(Γ) < 2
where Γ is said to be outside B(N) if it contains B(N) in its interior. This
implies that φ1 � �
1
B,p,q B(N) ↔ ∂B ≥ 2 . Let n → ∞, and deduce that
φ1 � �
1
p,q B(N) ↔ ∞ ≥ 2 , implying that θ1 (p, q) > 0 as required.
Turning to part (c) 12 , let p = pd = √ q/(1 + √ q) and n ≤ r. Let An
be the (cylinder) event that the point ( 1 1
2 , 2 ) lies in the interior of an open
circuit of length at least n, this circuit having the property that its interior
is contained in the interior of no open circuit having length strictly less than
n. We have from (13.36) and (13.45) that
(13.46) φ 0 B(r),p,q (An) ≤
∞�
m=n
mam
q
�
q
(1 + √ q) 4
� m/4
for all large r.
[Here, we use the observation that, if An occurs in B(r), then there exists
a maximal open circuit Γ of B(r) containing ( 1 1
2 , 2 ). In the dual of B(r), Γ
constitutes an outer circuit.]
We write LRn for the event that there is an open crossing of the rectangle
Rn = [0, n] ×[0, 2n] from its left to its right side, and we set λn = φ0 p,q (LRn).
We may find a point x on the left side of Rn and a point y on the right side
such that
φ 0 p,q (x ↔ y in Rn)
λn
≥ .
(2n + 1) 2
By placing six of these rectangles side by side (as in Figure 13.4), we find by
the FKG inequality that
(13.47) φ 0 p,q
�
� �
x ↔ x + (6n, 0) in [0, 6n] × [0, 2n] ≥
λn
(2n + 1) 2
� 6
.
12 The present proof was completed following a contribution by Ken Alexander, see [36]
for related material.