PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
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224 <strong>Geoffrey</strong> Grimmett<br />
9. <strong>PERCOLATION</strong> IN TWO DIMENSIONS<br />
9.1 The Critical Probability is 1<br />
2<br />
The famous exact calculation for bond percolation on L 2 is the following,<br />
proved originally by Kesten [201]. The proof given here is taken from [G].<br />
Theorem 9.1. The critical probability of bond percolation on Z2 equals 1<br />
2 .<br />
Furthermore, θ( 1<br />
2 ) = 0.<br />
Proof. Zhang discovered a beautiful proof that θ( 1<br />
2 ) = 0, using only the<br />
uniqueness of the infinite cluster. Set p = 1<br />
2 . Let T(n) be the box T(n) =<br />
[0, n] 2 , and find N sufficiently large that<br />
� � 1<br />
P1 ∂T(n) ↔ ∞ > 1 −<br />
2<br />
84 for n ≥ N.<br />
We set n = N + 1. Writing A l , A r , A t , A b for the (respective) events that the<br />
left, right, top, bottom sides of T(n) are joined to ∞ off T(n), we have by<br />
the FKG inequality that<br />
� �<br />
P1 T(n) � ∞ = P1<br />
2<br />
2 (Al ∩ Ar ∩ At ∩ Ab )<br />
by symmetry, for g = l,r,t,b. Therefore<br />
≥ P1<br />
2 (Al )P(A r )P(A t )P(A b )<br />
= P1<br />
2 (Ag ) 4<br />
P1<br />
2 (Ag � � �<br />
) ≥ 1 − 1 − P1 T(n) ↔ ∞<br />
2<br />
�1/4 > 7<br />
8 .<br />
) : 0 ≤<br />
x1, x2 < n}. Let Al d , Ar d , At d , Ab d denote the (respective) events that the left,<br />
right, top, bottom sides of T(n)d are joined to ∞ by a closed dual path off<br />
T(n)d. Since each edge of the dual is closed with probability 1<br />
2 , we have that<br />
Now we move to the dual box, with vertex set T(n)d = {x + ( 1<br />
2<br />
2 (Ag 7<br />
d ) > 8<br />
P1<br />
for g = l,r,t,b.<br />
Consider the event A = Al ∩ Ar ∩ At d ∩ Ab d , and see Figure 9.1. Clearly<br />
1<br />
1<br />
P1 (A) ≤<br />
2 2 , so that P1 (A) ≥<br />
2 2 . However, on A, either L2 has two infinite<br />
open clusters, or its dual has two infinite closed clusters. Each event has<br />
probability 0, a contradiction. We deduce that θ( 1<br />
2 ) = 0, implying that<br />
pc ≥ 1<br />
2 .<br />
Next we prove that pc ≤ 1<br />
2 . Suppose instead that pc > 1<br />
2 , so that<br />
� � −γn<br />
(9.2) P1 0 ↔ ∂B(n) ≤ e for all n,<br />
2<br />
, 1<br />
2