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PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

PERCOLATION AND DISORDERED SYSTEMS Geoffrey GRIMMETT

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256 <strong>Geoffrey</strong> Grimmett<br />

II. P(C �= ∅) > 0, dim(πC) = dimC a.s.<br />

III. dim(πC) < dimC a.s. on {C �= ∅}, but λ(πC) = 0 a.s.<br />

IV. 0 < λ(πC) < 1 a.s. on {C �= ∅}.<br />

V. P � λ(πC) = 1 � > 0 but C does not percolate a.s.<br />

VI. P(C percolates) > 0.<br />

In many cases of interest, there is a parameter p, and the ensuing fractal<br />

moves through the phases, from I to VI, as p increases from 0 to 1. There<br />

may be critical values pM,N at which the model moves from Phase M to<br />

Phase N. In a variety of cases, the critical values pI,II, pII,III, pIII,IV can be<br />

determined exactly, whereas pIV,V and pV,VI can be much harder to find.<br />

Here is a reasonably large family of random fractals. As before, they<br />

are constructed by dividing a square into 9 equal subsquares. In this more<br />

general system, we are provided with a probability measure µ, and we replace<br />

a square by the union of a random collection of subsquares sampled according<br />

to µ. This process is iterated on all relevant scales. Certain parameters are<br />

especially relevant. Let σl be the number of subsquares retained from the lth<br />

column, and let ml = E(σl) be its mean.<br />

Theorem 11.11. We have that<br />

(a) C = ∅ if and only if �3 l=1 ml ≤ 1 (unless some σl is a.s. equal to 1),<br />

(b) dim(πC) = dim(C) a.s. if and only if<br />

3�<br />

ml log ml ≤ 0,<br />

l=1<br />

(c) λ(πC) = 0 a.s. if and only if<br />

3�<br />

log ml ≤ 0.<br />

l=1<br />

For the proofs, see [112, 133]. Consequently, one may check the Phases I,<br />

II, III by a knowledge of the ml only.<br />

Next we apply Theorem 11.11 to the random Cantor set of Section 11.1,<br />

to obtain that, for this model, pI,II = 1<br />

9 , and we depart Phase II as p increases<br />

through the value 1<br />

3 . The system is never in Phase III (by Theorem 11.1(c))<br />

or in Phase IV (by Theorem 1 of [133]). It turns out that pII,V = 1<br />

3 and<br />

1<br />

2 < pV,VI < 1.<br />

For the ‘random Sierpinski carpet’ (RSC) the picture is rather different.<br />

This model is constructed as the above but with one crucial difference: at<br />

each iteration, the central square is removed with probability one, and the<br />

others with probability 1 − p (see Figure 11.4). Applying Theorem 11.11 we<br />

find that<br />

pI,II = 1<br />

8 , pII,III = 54 −1 4, pIII,IV = 18 −1 3,

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