College of Computer & Information Science Spring 2012 Northeastern University 28 February 2012 CS 7880: Network Algorithms and Analysis Scribe: Bochao Shen

Percolation on ℤ^{2}

The following notes are largely based on the excellent coverage of percolation by Bollobas and Riordan [BO06].

1 Critical Probability of Percolation on ℤ^{2}

In the last lecture, we showed that the critical probability p_{c} of percolation on ℤ^{2} satisfies p_{c} ≥ \frac{1}{3}. This time we will prove an upper bound of p_{c} such that p_{c} ≤ \frac{2}{3}. First, we need to give out some definitions we will work with.

Definition 1. Let θ(p) denote the probability that there is an infinite connected component in ℤ^{2} with percolation probability p”.

Definition 2. Let θ_{x}(p) denote the probability that there is an infinite connected component in ℤ^{2} containing the point x ∈ ℤ^{2} with percolation probability p. Specially, θ_{0}(p) denote the probability that there is an infinite connected component in ℤ^{2} containing the origin with percolation probability p.

To prove the above upper bound, we first show that θ_{0}(p) > 0, (p > \frac{2}{3}). By the fact that θ(p) ≥ θ_{0}(p), we can say that θ(p) > 0, (p > \frac{2}{3}). Then, by Kolmogrov 0-1 Law, we can claim that θ(p) = 1, (p > \frac{2}{3}). Thus, \frac{2}{3} is an upper bound for the critical probability p_{c} of percolation on ℤ^{2}.

Now, we prove that if p > \frac{2}{3}, then θ_{0}(p) > 0.

First, we construct ℤ^{2∗} the dual of ℤ^{2} by putting a vertex in every face and placing an edge between two vertices corresponding to two faces which share an edge. Each edge in ℤ^{2∗} will intersect with a corresponding edge in ℤ^{2}. As we can see, each edge in ℤ^{2∗} has a one-to-one mapping with an edge in ℤ^{2}. Furthermore, we can say that each edge in ℤ^{2∗} is open (or selected) if and only if that corresponding edge in ℤ^{2} is closed (or not selected).

Second, we introduce a special subgraph of ℤ^{2}, B(m) a square box with side length 2m containing the vertices in {−m, …, 0, …, m} × {−m, …, 0, …, m}. Now, we define two events G_{m} and F_{m} as following.

Definition 3. G_{m}: Event that every edge in B(m) is open.

Definition 4. F_{m}: Event that there is an open circuit in ℤ^{2∗} containing B(m) in its interior.

Now, we let M(n) denote the number of circuits of length n surrounding the origin. Since the number of paths of length n is no greater than 4 · 3^{n−1}, M(n) satisfies,

M(n) ≤ # of starting points × 4 · 3^{n−1}
≤ Poly(n) · 4 · 3^{n−1}.

(1)

Then, the probability that there exists a circuit of length n around the origin that is open in ℤ^{2∗}. is no greater than Poly(n) · 4 · 3^{n−1}(1 − p)^{n} ≤ Poly(n) · 4 · (3(1 − p))^{n−1}.

As we can see, G_{m} ⋀ \overline{F}_{m} ⇒; there is a shortest path of length ≥ m from the origin to the outside of the box B(m). So, the probability that there is a shortest path of length ≥ m from the origin to the outside of the box B(m) is greater than Pr(G_{m}) · Pr(\overline{F}_{m}).

Pr(G_{m}) · Pr(\overline{F}_{m}) ≥ Pr(G_{m}) · [1 − ∞ ; ∑_{n≥8m}4 · Poly(n)(3(1 − p))^{n−1}]. (2)

By Eq. 2, when p > \frac{2}{3}, for sufficiently large m, we will have,

∞ ; ∑_{n≥8m}4 · Poly(n)(3(1 − p))^{n−1} < 1.  (3)

Meanwhile, Pr(G_{m}) = (p)^{4m^{2}} > 0. Then, we can claim when m → ∞, the probability that there is a shortest path of length ≥ m from the origin to the outside of the box B(m) is greater than zero. By the fact that this event implies that there is an infinite connected component in ℤ^{2}. Thus, the probability that there is an infinite connected component in ℤ^{2}. is greater than zero. By Kolmogrov 0-1 Law, we can say that when p > \frac{2}{3}, the probability that there is an infinite connected component in ℤ^{2} equals one. Hence, \frac{2}{3} is an upper bound for the critical probability of percolation in ℤ^{2}.

2 Harris’s Theorem: θ(\frac{1}{2}) = 0

Definition 5. Let H(R) denote the event that there exists an open path crossing R horizontally. Similarly, let V (R) denote the event that there exists an open path crossing R vertically.

Definition 6. Let R^{h} denote the horizontal dual of R. Similarly, let R^{v} denote the vertical dual of R.

Lemma 1. Exactly one of the events of H(R) and V (R^{h}) holds.

Corollary 1. Given R is a k × (l − 1) rectangle, R′ is a (k − 1) × l rectangle, then we will have the following hold.

i) P_{p}{H(R)}+P_{1−p}{V (R′)}=1;

ii) If R is an (n + 1) × n rectangle, then P_{\frac{1}{2}} \{H(R)\} = \frac{1}{2};

iii) If S is n × n square, then P_{\frac{1}{2}} \{H(S)\} = P_{\frac{1}{2}} \{V (S)\} ≥ \frac{1}{2}.

Proof:

i) By Lemma 1, we will have P_{p}{H(R)} + P_{p}{V (R^{h})} = 1. Observe that R^{h} in ℤ^{2∗} is isomorphic to R′ in ℤ^{2}, and by the fact that the edge in ℤ^{2∗} is open if and only if the corresponding edge in ℤ^{2} is closed, we will have P_{p}{V (R^{h})} = P_{1−p}{V (R′)} hold. Hence, P_{p}{H(R)} + P_{1−p}{V (R′)} = 1.

ii) By i), this can be immediately proved by setting p = \frac{1}{2}, k = l = n + 1.

iii) By ii), given a (n + 1) × n rectangle R, we will have P_{\frac{1}{2}} \{H(R)\} = \frac{1}{2}. Suppose S is a n×n square, we can see that H(R) implies H(S). Hence, we will have P_{\frac{1}{2}} H(S) ≥ P_{\frac{1}{2}} H(R) = \frac{1}{2}. Similarly, we can get P_{\frac{1}{2}} \{V (S)\} ≥ \frac{1}{2}.

Definition 7. Let h_{p}(m, n) denote the probability that there exists a horizontal open crossing in a m × n rectangle with percolation probability p; Specifically, we let h(m, n) = h_{\frac{1}{2}} (m, n).

Lemma 2. h(6n, 2n) ≥ 2^{−25}.

Proof: We will cover this proof in the next lecture.

Theorem 1 (Harris’s Theorem). θ(1/2) = 0.

Proof: First, construct four overlapping 6n by 2n rectangles in ℤ^{2∗} in the way in Fig. 1 (taken from [BO06].

By Lemma 2, the probability for an open crossing in each 6n by 2n rectangle is at least 2^{−25}. By FKG inequality (which will be covered in the next lecture), the probability for all four crossing in these four 6n by 2n rectangles is at least γ = 2^{−100}.

For k ≥ 1, let A_{k} be the small square surrounded by those four 6n by 2n rectangles in ℤ^{2∗} which is centered on (1/2, 1/2). Meanwhile, the small centered square and the outer big square have radii 3^{k} and 3^{k+1}, respectively.

Now, let E_{k} be the event that A_{k} contains an open dual cycle surrounding the interior of A_{k}. Then P(E_{k}) ≥ γ. Since those crossings surrounding A_{k} varying by k are disjoint, the events E_{k} are independent. If E_{k} holds, then no

Figure 1: Four 6n by 2n rectangles overlapping

point inside A_{k} in ℤ^{2} can be reached from any point in ℤ^{2} outside the cycle formed by those four open crossing in ℤ^{2∗}. Thus, r_{0} the radius of A_{k} is bounded by 3^{k+1}. Then, we will have

P\{r_{0} ≥ 3^{l+1}\} ≤ P\{ l ; ⋂_{k=1}\overline{E}_{k}\} = l ; ∏_{k=1}P(\overline{E}_{k}) ≤ (1 − γ)^{l}.  (4)

By Eq. 4, we have

P{r_{0} = ∞} ≤ P{r_{0} ≥ n} ≤ (1 − γ)^{log_{3} n−1} = n^{log_{3}(1−γ)}/(1 − γ) ≤ n^{−c}. (5)

where, c is an constant for the above inequality’s holding.

Now, we can see that θ_{0}(\frac{1}{2}) = 0. Since it is an infinite lattice, by symmetry, for any vertex x = (i, j) ∈ ℤ^{2}, we will have θ_{x}(1/2) = θ_{0}(1/2) = 0. And, there are countable infinite vertices in ℤ^{2}. Thus, θ(1/2) satisfies following,

θ(1/2) = ∑_{∀x=(i,j)∈ℤ^{2}}θ_{x}(1/2) = 0.  (6)

References

[BO06] B. Bollobás and Riordan O. Percolation. Cambridge University Press, 2006.