Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 15 / 2

We use three precise inputs. The Harris-Kesten theorem says that independent bond percolation on the square lattice has percolation critical probability $p_c=1/2$. Sharpness of the percolation transition says that, for every $p<p_c$, there are $c,C>0$ such that ${\mathbb{P}}_p(0\leftrightarrow \partial [-R,R{]}^2)\le C{e}^{-cR}$ for all positive integers $R$. Finally, the BK inequality says that increasing connection events with disjoint bond witnesses have joint probability at most the product of their individual probabilities. All three statements concern the independent product law used here.

The sharpness bound implies finite percolation susceptibility. Indeed, with $\tau (u,v)={\mathbb{P}}_p(u\leftrightarrow v)$, summing over square shells givesA radius bound alone would give only a square-root exponential volume bound. To obtain the requested exponential volume bound, we prove the tree-graph moment bound for a percolation cluster:where $(-1)!!=1$.

Expand the left-hand side as a sum over ordered marked vertices $x_1,\dots ,x_k$ of the probability that $0,x_1,\dots ,x_k$ belong to one open cluster. Whenever that event occurs, a finite open tree connects the marked vertices. Delete unmarked leaves and suppress unmarked degree-two vertices. Treat a marked internal vertex as a leaf attached by a zero-length connection, and split any higher-degree branch into degree-three branch points, again with zero-length connections. This gives an abstract binary tree with $k+1$ labelled leaves, $k-1$ branch points and $2k-1$ connections. Its nonempty connecting paths use disjoint sets of open bonds. Coincident marked vertices are allowed: their zero-length connections have empty witnesses.

There are $(2k-3)!!$ such binary tree topologies. For example, the topology with two leaves is unique; inserting the next labelled leaf subdivides one of the $2k-1$ edges of a topology with $k+1$ leaves, giving the usual double-factorial recursion. For a fixed topology and fixed positions of all its vertices, the BK inequality bounds the disjoint connection event by the product of the corresponding $\tau$ factors. Sum over all $x_i$ and all branch-point positions. Keep the root fixed, and successively sum a leaf away. Translation invariance gives ${\sum }_y\tau (z,y)=\chi$ at every step. The resulting sum is ${\chi }^{2k-1}$. Summing the topologies proves the bound.

Since $(2k-3)!!\le 2^{k-1}(k-1)!$, expansion of the exponential, justified by monotone convergence, givesTake $t=1/(4{\chi }^2)$ and write $M=1+\log 2/(2\chi )$. Markov's inequality yields ${\mathbb{P}}_p(|C_0|\ge n)\le M{e}^{-tn}$. To remove the prefactor, choose an integer $N\ge 2$ with $tN\ge 2\log M$. For $n\ge N$ the bound is at most $e^{-tn/2}$. For $2\le n\le N$, useFor $0<p<1/2$, setting $a=\min \{t/2,-\log u/N\}>0$ therefore proves the bound with unit prefactor:At $p=0$, $|C_0|=1$ and any positive $a$ works.

Set $q=1-p<1/2$. Declare a bond of the planar dual graph open when its crossing primal bond is closed. These states have the independent dual bond percolation law with parameter $q$; the dual square lattice is a translate of the primal square lattice. Part (i) supplies an $a=a(q)>0$ such thatIn particular every dual cluster is finite almost surely.

Suppose the primal percolation cluster $C_0$ is finite. Fill its finite complementary holes. Every bond on the resulting outer boundary is closed, since an open bond leaving the cluster would reach another cluster vertex. The crossing dual bonds form a connected enclosing contour, possibly touching itself, contained in one dual cluster $K$. This is the dual contour bound for a finite planar cluster. If $m=|K|$, a connected set of $m$ square-lattice dual vertices has each coordinate width at most $m-1$. All vertices of $C_0$ lie inside the enclosing contour, soMoreover the contour surrounds $0$ and the diameter of $K$ is at most $m-1$ in the graph distance. Thus every vertex of $K$ lies in the square of radius $m+1$ about $0$. There are at most $(2m+3{)}^2$ possible dual vertices in that square, with a harmless enlargement if a boundary convention requires it.

On $\{n\le |C_0|<\infty \}$ there is therefore an integer $m\ge \lceil \sqrt{n}\rceil$ and a dual vertex in that square whose dual cluster has size exactly $m$. The enclosing contour has at least four vertices. A union bound and part (i) givefor constants $C_1,c_1>0$ depending only on $q$. The last estimate follows by bounding the polynomial factor by a constant times $e^{am/2}$ and summing a geometric series. This explains the square root: an enclosing dual cluster must have size at least the square root of the primal volume.

For $p<1$, the Harris-Kesten theorem and $p>p_c$ imply $\theta (p)>0$, so $v:={\mathbb{P}}_p(|C_0|<\infty )=1-\theta (p)<1$. Choose $N$ with $\log C_1\le (c_1/2)\sqrt{N}$. For $n\ge N$ use the preceding bound; for $1\le n\le N$ use the bound $v$. The positive constant $b=\min \{c_1/2,-\log v/\sqrt{N}\}$ gives the required conclusion:For $p=1$ the event is empty, so any $b>0$ works.