PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

11.2 Percolation
Percolation and Disordered Systems 253
Can C contain long paths? More concretely, can C contain a crossing from
left to right of the original unit square C0 (which is to say that C contains
a connected subset which has non-trivial intersection with the left and right
sides of the unit square)? Let LR denote the event that such a crossing exists
in C, and define the percolation probability
(11.4) θ(p) = Pp(LR).
In [91], it was proved that there is a non-trivial critical probability
pc = sup{p : θ(p) = 0}.
Theorem 11.5. We have that 0 < pc < 1, and furthermore θ(pc) > 0.
Partial Proof. This proof is taken from [91] with help from [113]. Clearly
, and we shall prove next that
pc ≥ 1
9
pc ≤ 8
9
� �
64 7
63 ≃ 0.99248 . . . .
Write C = (C0, C1, . . . ). We call C 1-good if |C1| ≥ 8. More generally, we
call C (k+1)-good if at least 8 of the squares in C1 are k-good. The following
fact is crucial for the argument: if C is k-good then Ck contains a left-right
crossing of the unit square (see Figure 11.2). Therefore (using the fact that
the limit of a decreasing sequence of compact connected sets is connected,
and a bit more 9 )
(11.6) Pp(C is k-good) ≤ Pp(Ck crosses C0) ↓ θ(p) as k → ∞,
whence it suffices to find a value of p for which πk = πk(p) = Pp(C is k-good)
satisfies
(11.7) πk(p) → π(p) > 0 as k → ∞.
We define π0 = 1. By an easy calculation,
(11.8) π1 = 9p 8 (1 − p) + p 9 = Fp(π0)
9 There are some topological details which are necessary for the limit in (11.6). Look
at the set Sk of maximal connected components of Ck which intersect the left and right
sides of C0. These are closed connected sets. We call such a component a child of a
member of Sk−1 if it is a subset of that member. On the event {C crosses C0}, the
ensuing family tree has finite vertex degrees and contains an infinite path S1, S2, . . . of
compact connected sets. The intersection S∞ = limk→∞ Sk is non-empty and connected.
By a similar argument, S∞ has non-trivial intersections with the left and right sides of
C0. It follows that {Ck crosses C0} ↓ {C crosses C0}, as required in (11.6). Part of this
argument was suggested by Alan Stacey.