PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

196 Geoffrey Grimmett
4. The open copy of B(m), constructed above, may be used as a ‘seed’ for
iterating the above construction. When doing this, we shall need some
control over where the seed is placed. It may be shown that every face
of ∂B(n) contains (with large probability) a point adjacent to some
seed, and indeed many such points.
Above is the scheme for constructing long finite paths, and we turn to the
second step.
5. This construction is now iterated. At each stage there is a certain
(small) probability of failure. In order that there be a strictly positive
probability of an infinite sequence of successes, we iterate ‘in two independent
directions’. With care, one may show that the construction
dominates a certain supercritical site percolation process on L 2 .
6. We wish to deduce that an infinite sequence of successes entails an
infinite open path of L d within the corresponding slab. There are
two difficulties with this. First, since there is not total control of the
positions of the seeds, the actual path in L d may leave every slab. This
may be overcome by a process of ‘steering’, in which, at each stage,
we choose a seed in such a position as to compensate for any earlier
deviation in space.
7. A larger problem is that, in iterating the construction, we carry with
us a mixture of ‘positive’ and ‘negative’ information (of the form that
‘certain paths exist’ and ‘others do not’). In combining events we cannot
use the FKG inequality. The practical difficulty is that, although
we may have an infinite sequence of successes, there will generally
be breaks in any corresponding open route to ∞. This is overcome by
sprinkling down a few more open edges, i.e., by working at edge-density
p + δ where δ > 0, rather than at p.
In conclusion, we show that, if θ(p) > 0 and δ > 0, then there is (with
large probability) an infinite (p + δ)-open path in a slice of the form
Tk = {x ∈ Z d : 0 ≤ xj ≤ k for j ≥ 3}
where k is sufficiently large. This implies that p+δ > pc(T) = limk→∞ pc(Tk)
if p > pc, i.e., that pc ≥ pc(T). Since pc(T) ≥ pc by virtue of the fact that
Tk ⊆ Z d for all k, we may conclude that pc = pc(T), implying also that
pc = pc(S).
Henceforth we suppose that d = 3; similar arguments are valid when
d > 3. We begin with some notation and two key lemmas. As usual, B(n) =
[−n, n] 3 , and we shall concentrate on a special face of B(n),
F(n) = {x ∈ ∂B(n) : x1 = n},
and indeed on a special ‘quadrant’ of F(n),
T(n) = {x ∈ ∂B(n) : x1 = n, xj ≥ 0 for j ≥ 2}.