Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 38 / 4 / Solution

On the coordinatewise ordered configuration space, ${\mu }_1$ stochastically dominates ${\mu }_2$ when $\int fd{\mu }_1\ge \int fd{\mu }_2$ for every increasing real function $f$. Equivalently, ${\mu }_1(A)\ge {\mu }_2(A)$ for every increasing event. The Holley condition is the sufficient cross-lattice inequalityHere join and meet are coordinatewise maximum and minimum. This condition implies ${\mu }_1{\ge }_{\mathrm{st}}{\mu }_2$ by the Holley inequality; it is not asserted to be necessary. It can be checked on unnormalized weights, since their normalization constants cancel.

Use the weightsFor the upper bound, take ${\mu }_1={\varphi }_{p,1}$ and ${\mu }_2={\varphi }_{p,q}$. Since open-edge count is modular, the Bernoulli factors cancel in the Holley ratio, leavingThe last inequality uses $q\ge 1$ and the fact that removing edges cannot reduce the number of components. Thus ${\varphi }_{p,q}{\le }_{\mathrm{st}}{\varphi }_{p,1}$.

For the lower bound, let $p^{\prime }=p/[p+q(1-p)]$. Its odds satisfySet $a=o(\omega \vee \eta )-o(\omega )=o(\eta )-o(\omega \wedge \eta )$, the number of edges added to $\omega$ in the union. For ${\mu }_1={\varphi }_{p,q}$ and ${\mu }_2={\varphi }_{p^{\prime },1}$, the corresponding ratio isAdding one edge either merges two components or leaves the count unchanged. Therefore adding $a$ edges reduces $k$ by at most $a$, proving the exponent nonnegative. This also proves the hinted monotonicity of $k(\omega )+o(\omega )$ rather than assuming it. The Holley condition now proves both Bernoulli bounds for the random-cluster model:

The same argument holds with wired boundary conditions: identify the wired boundary vertices first and count components on that quotient graph. Taking finite-volume wired limits preserves every increasing cylinder-event inequality. Approximating $\{0\leftrightarrow \infty \}$ by the decreasing events $\{0\leftrightarrow \partial [-n,n{]}^d\}$ consequently givesLet $a=p_c(1)$. For $p<a$, the upper bound makes the wired percolation probability zero, so $p_c(q)\ge a$. If $a<1$ and $p^{\prime }>a$, the lower bound makes it positive. Solving this strict inequality,Taking infima provesFor $a=1$, the lower bound and the trivial upper bound $p_c(q)\le 1$ give the same conclusion. No assumption about percolation at the critical parameter is required.