Before starting, we have to notice that the terminology of percolation is adopted in different situations and generally in the contexts of graphs - it could be various graphs - like lattice Zd, random graphs, random maps etc. But the phenomena is a little universal that a path starts from 0 and goes to infinite faraway. In this series of talk, Hugo focuses on the situation of 2D Bernoulli percolation.
p≤pc,θ(p)=P[0↔∞]=0p>pc,θ(p)=P[0↔∞]>0
In 2D model, the critical point pc=1/2. Some simple argument supports this point. For example, if we draw the dual percolation between the face whose frontier is closed, we get a dual graph with probability 1−p. If we suppose that the critical point is unique, then the transition of face happens at the same time in both the primal graph and the dual graph. So pc=1−pc and pc=1/2.
Definition and critical point
We give some definitions precisely. In the lattice graph Z2, each site has 4 neighbors and every edge has independently a probability p to be present (open) or a probability 1−p to be absent (closed). Then there exits a critical point pc : when p≤pc, with probability 0 there exists a path from 0 to ∞, while when p>pc, with probability θ(p) there exists a path from 0 to ∞.p≤pc,θ(p)=P[0↔∞]=0p>pc,θ(p)=P[0↔∞]>0
In 2D model, the critical point pc=1/2. Some simple argument supports this point. For example, if we draw the dual percolation between the face whose frontier is closed, we get a dual graph with probability 1−p. If we suppose that the critical point is unique, then the transition of face happens at the same time in both the primal graph and the dual graph. So pc=1−pc and pc=1/2.
Quantitative analysis
We hope to get some stronger result, The following theorem is first obtained by Menshikov, Aizeman and Michael ∀p<pc,∃Cp>0 s.t Pp[0↔∂Bn]≤exp(−Cpn)∀p>pc,∃C>0 s.t Pp[0↔∞]≥C(p−pc)
Here, we denote the ball of radius n by ∂Bn and this theorem indeed, gives some numerical estimation of the speed of decrements.
Proof 1 by (Menshikov, Aizeman, Michael)
We note θn(p)=Pp[0↔∂Bn] and ϕp(S)=∑x∈S,y∉S,x∼ypP[0↔Sx]. We can prove it by 5 steps.
- We admit at first this important inequalityddpθn(p)≥1p(1−p)[inf0∈S⊂Bnϕp(S)](1−θn(p))then we define ˜pc=sup{p:∃S∋0,s.tϕp(S)<1}we can prove θ(P)≥1p(1−˜pc)(p−˜pc)
- We choose S⊂Bk−1 such that ϕp(S)<1 and proveθnk(p)≤(ϕp(S))n
- We conclude that ˜pc=pc
- Verify the identity ddpPp(X)=∑e∈E1p(1−p)Cov(we,X)
- We put X=−I0!↔∂Bn and we prove the important inequality.
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