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PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

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234 <strong>Geoffrey</strong> Grimmett<br />

the volume of ∂B(β k−1 ) has order β (k−1)(d−1) . The above construction can<br />

therefore only succeed when 4 k < β (k−1)(d−1) for all large k, which is to say<br />

that β > 4 1/(d−1) .<br />

This crude picture suggests the necessary inequalities<br />

(10.3) 4 1/(d−1) < β < 4,<br />

which can be satisfied if and only if d ≥ 3.<br />

Assume now that d ≥ 3. Our target is to show that the infinite cluster I<br />

contains sufficiently many disjoint paths to enable a comparison of its effective<br />

resistance with that of the tree in Figure 10.1, and with some value of β<br />

satisfying (10.3). In presenting a full proof of this, we shall use the following<br />

two percolation estimates, which are consequences of the results of Chapter<br />

7.<br />

Lemma 10.4. Assume that p > pc.<br />

(a) There exists a strictly positive constant γ = γ(p) such that<br />

� � −γn<br />

(10.5) Pp B(n) ↔ ∞ ≥ 1 − e for all n.<br />

(b) Let σ > 1, and let A(n, σ) be the event that there exist two vertices inside<br />

B(n) with the property that each is joined by an open path to ∂B(σn) but<br />

that there is no open path of B(σn) joining these two vertices. There exists<br />

a strictly positive constant δ = δ(p) such that<br />

� � −δn(σ−1)<br />

(10.6) Pp A(n, σ) ≤ e<br />

for all n.<br />

We restrict ourselves here to the case d = 3; the general case d ≥ 3 is<br />

similar. The surface of B(n) is the union of six faces, and we concentrate on<br />

the face<br />

F(n) = {x ∈ Z 3 : x1 = n, |x2|, |x3| ≤ n}.<br />

We write Bk = B(3 k ) and Fk = F(3 k ). On Fk, we distinguish 4 k points,<br />

namely<br />

xk(i, j) = (idk, jdk), −2 k−1 < i, j ≤ 2 k−1<br />

where dk = ⌊(4/3) k ⌋. The xk(i, j) are distributed on Fk in the manner of a<br />

rectangular grid, and they form the ‘centres of attraction’ corresponding to<br />

the kth generation of the tree discussed above.<br />

With each xk(i, j) we associate four points on Fk+1, namely those in the<br />

set<br />

Ik(i, j) = {xk+1(r, s) : r = 2i − 1, 2i, s = 2j − 1, 2j}.<br />

These four points are called children of xk(i, j). The centroid of Ik(i, j) is<br />

denoted Ik(i, j). We shall attempt to construct open paths from points near<br />

xk(i, j) to points near to each member of Ik(i, j), and this will be achieved<br />

with high probability. In order to control the geometry of such paths, we<br />

shall build them within certain ‘tubes’ to be defined next.

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