PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

Percolation and Disordered Systems 241
Next, we construct an electrical network E(ρ) on L 2 by, for each edge
e of L 2 , placing an electrical resistor of size ρ(e) at e. Let R = R(Z) be
the effective resistance of this network between 0 and ∞ (which is to say
that R = limn→∞ Rn, where Rn is the resistance between 0 and a composite
vertex obtained by identifying all vertices in ∂B(n)).
Lemma 10.12. We have that P � R(Z) = ∞ � �0 is a rw point � = 1.
Proof. We define the ‘edge-boundary’ ∆eB(n) of B(n) to be the set of edges
e = 〈x, y〉 with x ∈ ∂B(n) and y ∈ ∂B(n + 1). We claim that there exists a
positive constant c and a random integer M such that
(10.13) ρ(e) ≥ c
log n for all e ∈ ∆eB(n) and n ≥ M.
To show this, we argue as follows. Assume that e = 〈x, y〉 is normal, and let
λ1 be the number of edges in the path π(e) on one side of e (this side being
chosen in an arbitrary way), and λ2 for the number on the other side. Since
each new vertex visited by the path is a rw point with probability prw, and
since no vertex appears more than twice in π(e), we have that
P � l(e) > 2k, e is normal � ≤ 2P � λ1 ≥ k, e is normal � ≤ 2(1 − prw) 1
2 (k−1) .
Therefore, for c > 0 and all n ≥ 2,
�
P ρ(e) < c
log n
�
for some e ∈ ∆eB(n)
≤ 4(2n + 1)P
≤ βn 1−α
�
l(e) >
log n
, e is normal
c
where α = α(c) = −(4c) −1 log(1 − prw) and β = β(c, prw) < ∞. We choose c
such that α > 5
2 , whence (10.13) follows by the Borel–Cantelli lemma.
The conclusion of the lemma is a fairly immediate consequence of (10.13),
using the usual argument which follows. From the electrical network E(ρ)
we construct another network with no larger resistance. This we do by identifying
all vertices contained in each ∂B(n). In this new system there are
|∆eB(n)| parallel connections between ∂B(n) and ∂B(n + 1), each of which
has (for n ≥ M) a resistance at least c/ log n. The effective resistance from
the origin to infinity is therefore at least
∞�
n=M
c
|∆eB(n)| log n =
∞�
n=M
c
= ∞,
4(2n + 1)log n
and the proof of the lemma is complete. �
�