Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 30 / 3

Order configurations coordinatewise: $\omega \le \eta$ means $\omega (e)\le \eta (e)$ for every edge. Then stochastic domination is the conditionEquivalently, ${\mathbb{E}}_{{\mu }_1}f\ge {\mathbb{E}}_{{\mu }_2}f$ for every real-valued order-preserving function $f$. The equivalence follows on this finite space by writing $f$, up to an additive constant, as a nonnegative linear combination of indicators of its increasing upper level sets. A coupling of probability distributions supported on pairs $(X_1,X_2)$ with $X_1\ge X_2$ proves this domination; conversely such an ordered coupling of probability distributions exists on a finite partially ordered space.

For the coordinatewise lattice operations $\omega \vee \eta$ and $\omega \wedge \eta$, Holley's condition isThis sufficient condition implies ${\mu }_1\ge {\mu }_2$ by the Holley inequality. It is often stated first for strictly positive probability weights, as will hold for the random-cluster application, but the displayed condition is also sufficient for nonnegative weights.

For clarity about the zero-weight case, apply the four functions theorem to an increasing event $A$, with the first two functions ${\mu }_1{1}_{A^c}$ and ${\mu }_2{1}_A$, and the last two ${\mu }_1{1}_A$ and ${\mu }_2{1}_{A^c}$. If $\omega \notin A$ and $\eta \in A$, then $\omega \vee \eta \in A$ and $\omega \wedge \eta \notin A$. The pointwise hypothesis is exactly the displayed cross-lattice inequality. Its summed conclusion giveswhich rearranges to ${\mu }_1(A)\ge {\mu }_2(A)$. This establishes the same statement without a positivity-of-support assumption.

Identify a configuration with its set $S$ of open edges. Let $k(S)$ count the connected components of a graph on the full graph vertex set, including isolated graph vertices. The random-cluster measure isFor $p\in (0,1)$ and finite $q\ge 1$ all its weights are positive. We compare $1\le q_1\le q_2$, taking ${\mu }_1={\mu }_{p,q_1}$ and ${\mu }_2={\mu }_{p,q_2}$.

The key fact is supermodularity of graph component count:To prove it, orient the graph edges arbitrarily and let $U_S$ be the real span of their incidence vectors for edges in $S$. Its dimension of a vector space is $r(S)=|V|-k(S)$: on each connected component of a graph these vectors span the vectors whose coordinates sum to zero. Since $U_{S\cup T}=U_S+U_T$ and $U_{S\cap T}\subset U_S\cap U_T$, the dimension formula for a sum of subspaces giveswhich is exactly the component-count inequality.

Put $a=k(S)-k(S\cup T)\ge 0$ and $b=k(S\cap T)-k(T)\ge 0$. The preceding inequality gives $b\ge a$. In the ratio of the two sides of Holley's condition, the normalization constants and Bernoulli edge factors cancel, because $|S\cup T|+|S\cap T|=|S|+|T|$. The ratio is thereforeBy the Holley inequality, ${\mu }_{p,q_1}$ stochastically dominates ${\mu }_{p,q_2}$. Thus increasing cluster weight suppresses every increasing event:The random-cluster single-edge conditional probability confirms the mechanism: it is $p$ when the endpoints are already connected, and $p/(p+q(1-p))$ otherwise. Larger $q$ favours configurations with more separate components. The condition $q\ge 1$ enters explicitly in the last factor $q_2^{b-a}$.

Use consistently wired boundary conditions in increasing finite boxes and let ${\varphi }_{p,q}^{\mathrm{w}}$ be the usual wired infinite-volume random-cluster measure. The finite-volume comparison just proved applies with boundary graph vertices identified, since it is still a comparison on a finite graph. Passing to the infinite-volume limit preserves the inequality for every increasing event determined by finitely many edges.

For fixed $m$, let $A_m$ be the event that the origin connects by an open path in $[-m,m{]}^2$ to its boundary. It is such a finite increasing event. Therefore, for $q_1\le q_2$,The events decrease to the event that the origin has an infinite percolation cluster, by local finiteness. Continuity from above of a measure givesDefine the random-cluster critical probability by $p_c(q)=\inf \{p:{\theta }^{\mathrm{w}}(p,q)>0\}$. The set in which percolation occurs at $q_2$ is a subset of that at $q_1$, whenceThe same proof works if free boundary conditions are used consistently throughout. It compares critical probabilities; it does not presume whether an infinite percolation cluster exists at the critical parameter itself.