Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 204 / 2 / c / Solution

Let $c_n$ count $n$-step self-avoiding walks on the square lattice from one fixed graph vertex. Splitting a self-avoiding walk after $m$ steps and dropping the avoidance constraint on its suffix gives $c_{m+n}\le c_m{c}_n$. The Fekete lemma therefore gives the connective constant $\mu =\lim c_n^{1/n}$. The planar dual graph of the square lattice is a translated copy of it, so its connective constant is also $\mu$.

Put $q=1-p$ and assume $q\mu <1$. Choose $a>\mu$ with $qa<1$. The definition of the connective constant provides a finite $K$ such that $c_j\le K{a}^j$ for every $j\ge 0$. A simple dual graph cycle of length $\ell$ surrounding $0$ has a graph vertex in a box of radius $\ell +1$: its diameter is at most $\ell$, and its coordinate ranges straddle the origin. Choose such a graph vertex as the starting point, orient the graph cycle, and omit its closing edge. The remaining self-avoiding walk has length $\ell -1$. Consequently the number $N_{\ell }$ of these graph cycles obeysfor fixed finite constants. A specified graph cycle is open in dual bond percolation, equivalently all its crossed primal edges are closed, with probability $q^{\ell }$. HenceTo make this tail argument valid for every $q\mu <1$, rather than only extremely small $q$, use a finite modification of Bernoulli percolation. Condition all primal edges within ${\Lambda }_N$ to be open. This event has positive probability for $p>0$. If the resulting percolation cluster of $0$ is finite, its outer boundary contains a simple closed dual graph cycle surrounding the whole box. Such a graph cycle has length tending to infinity with $N$ and crosses no forced-open edge. Under the conditioning its remaining edge states retain the original independent law, so its probability is still $q^{\ell }$. Choose $N$ so that the above tail is less than $1/2$. The conditional probability that $0$ belongs to an infinite percolation cluster is then at least $1/2$, and thus $\theta (p)>0$.

This connective-constant Peierls bound proves that every $p>1-{\mu }^{-1}$ is above or at the onset of positive percolation probability. Taking the infimum givesThe case $p=1$ is immediate. This proof uses the Peierls argument, without assuming the exact Harris-Kesten theorem.