PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
178 <strong>Geoffrey</strong> Grimmett<br />
Now φ(p) = 0 when p > pc (since Pp(An) ≥ θ(p) > 0). Therefore the<br />
above discussion needs more thought in this case. In defining the supercritical<br />
correlation length, it is normal to work with the ‘truncated’ probabilities<br />
Pp(An, |C| < ∞). It may be shown ([94, 164]) that the limit<br />
�<br />
(6.8) φ(p) = lim −<br />
n→∞<br />
1 � �<br />
log Pp 0 ↔ ∂B(n), |C| < ∞<br />
n �<br />
exists for all p, and satisfies φ(p) > 0 if and only if p �= pc. We now define<br />
the correlation length ξ(p) by<br />
(6.9) ξ(p) = 1/φ(p) for 0 < p < 1.<br />
Proof of Theorem 6.2. Define the (two-point) connectivity function τp(x, y) =<br />
Pp(x ↔ y). Using the FKG inequality,<br />
�<br />
�<br />
τp(x, y) ≥ Pp {x ↔ z} ∩ {z ↔ y} ≥ τp(x, z)τp(z, y)<br />
for any z ∈ Zd . Set x = 0, z = me1, y = (m + n)e1, to obtain that<br />
τp(r) = Pp(0 ↔ re1) satisfies τp(m + n) ≥ τp(m)τp(n). Therefore the limit<br />
φ2(p) exists by the subadditive inequality.<br />
The existence of φ1(p) may be shown similarly, using the BK inequality<br />
as follows. Note that<br />
{0 ↔ ∂B(m + n)} ⊆ � � �<br />
{0 ↔ x} ◦ {x ↔ x + ∂B(n)}<br />
x∈∂B(m)<br />
� �<br />
(this is geometry). Therefore βp(r) = Pp 0 ↔ ∂B(r) satisfies<br />
βp(m + n) ≤ �<br />
τp(0, x)βp(n).<br />
x∈∂B(m)<br />
Now τp(0, x) ≤ βp(m) for x ∈ ∂B(m), so that<br />
βp(m + n) ≤ |∂B(m)|βp(m)βp(n).<br />
With a little ingenuity, and the subadditive inequality, we deduce the existence<br />
of φ1(p) in (6.3). That φ2(p) ≥ φ1(p) follows from the fact that<br />
τp(0, ne1) ≤ βp(n). For the converse inequality, pick x ∈ ∂B(n) such that<br />
τp(0, x) ≥<br />
1<br />
|∂B(n)| βp(n),<br />
and assume that x1 = +n. Now<br />
�<br />
�<br />
τp(0, 2ne1) ≥ Pp {0 ↔ x} ∩ {x ↔ 2ne1} ≥ τp(0, x) 2<br />
by the FKG inequality. �