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This lattice condition on configuration masses is the hypothesis of the FKG inequality. A product measure on a finite Boolean lattice satisfies it with equality, since in each coordinate the minimum and maximum retain the original two coordinate values. It implies positive association of random variables for increasing functions.

A fixed Ising boundary condition pins the Ising spins on a specified boundary, leaving the interior distributed with the original interaction weights. The boundary can be the inner graph vertex boundary of a finite box, or an exterior neighboring set if that convention is explicitly chosen. Conditioning on pinned Ising spins preserves the FKG lattice condition of a ferromagnetic Ising model. The FKG inequality then makes interior expectations of increasing observables nondecreasing in the pinned boundary Ising spins.

Order configurations coordinatewise: $\omega \le \eta$ when $\omega (e)\le \eta (e)$ for every coordinate. An increasing event $A$ is an upward-closed set: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

The finite-lattice FKG inequality states that a probability mass function satisfying the FKG lattice conditionis positively associated. Here meet in a lattice and join in a lattice are coordinatewise minimum and maximum. Thus for increasing observables $f,g$,Taking indicator functions gives $\mu (A\cap B)\ge \mu (A)\mu (B)$ for all increasing events. The strictly positive version suffices for the Ising weights here; the theorem also holds for nonnegative log-supermodular weights of positive total mass.

We use Zhang's argument, obtaining the critical nonpercolation conclusion from uniqueness rather than assuming it as part of the Harris-Kesten theorem. For independent bond percolation on ${\mathbb{Z}}^2$, apply part (ii) to independent site percolation on its line graph; open sites correspond to open bonds. The line graph is amenable and of finite type. Therefore an infinite bond cluster, if it exists, is almost surely unique. The same holds for the dual bond percolation model.

Assume for contradiction that $\theta (1/2)>0$. The planar dual graph is another square lattice, and its open bonds correspond to closed primal bonds, also with parameter $1/2$. Both models then have an infinite percolation cluster almost surely, by part (i). As a disc $D_R$ expands, the probability that it meets an infinite cluster tends to one, for each model. Choose $R$ avoiding all primal and dual vertices and all intersections of either graph with the four endpoints of the following arcs. Divide the boundary circle into four equal arcs $A_1,A_2,A_3,A_4$ in cyclic order. Let $L_i$ be the event of an infinite primal open ray whose last exit from $D_R$ is in $A_i$; define $L_i^{\ast }$ similarly for an infinite dual open ray. After that exit the entire ray stays outside the disc. The existence of a ray follows by finite branching in an infinite locally finite graph.

Each $L_i$ is increasing in the primal bond states, and their union has probability tending to one. Four-fold rotation symmetry makes their probabilities equal. The FKG inequality for their decreasing complements gives the square-root trick for positively associated eventsThe left side tends to zero, so $\mathbb{P}(L_i)\to 1$. Apply the same argument to the dual states to get $\mathbb{P}(L_i^{\ast })\to 1$. A union bound, requiring no independence between the models, now shows that for sufficiently large $R$ the alternating eventhas positive probability.

The alternating exterior-arm separation lemma says that the two primal rays on this event cannot be joined by a primal open path lying outside $D_R$. Indeed, such a joining path contains a simple crosscut of the exterior between the two primal exit points. Join that crosscut to a chord inside the disc. The Jordan curve theorem puts one of the alternating dual exit points on the bounded side. Its dual ray, which is restricted to the exterior of the disc and runs to infinity, would have to cross the open primal crosscut. A dual open bond cannot cross a primal open bond. This is impossible.

Let $F$ be the finite set of primal bonds whose drawn edges meet $D_R$, and let $E$ be the projection of the alternating-arm event onto the states outside $F$. Because $F$ is finite, this projection is a finite union of measurable events; $\mathbb{P}(E)>0$. Every configuration in $E$ admits an assignment on $F$ witnessing the original four arms. Now close all bonds of $F$. The infinite tails of the two primal rays beyond their first exterior vertices survive; any path joining the surviving tails would now avoid the disc. Appending their original exterior segments back to their exit points would give the forbidden exterior connection. Thus the modified configuration contains at least two infinite percolation clusters. By independence, or the finite-energy property of Bernoulli percolation, $\mathbb{P}(E\cap \{F\text{ all closed}\})=2^{-|F|}\mathbb{P}(E)>0$. This contradicts the uniqueness obtained from part (ii). Therefore

Let $G=\Lambda$ be the martini lattice and $G^{\ast }$ its planar dual graph. Write $c=p_c^b(G)$ and $c^{\ast }=p_c^b(G^{\ast })$. Both are periodic, connected, bounded-degree plane graphs with bounded face sizes. Dual bond percolation declares a dual bond open exactly when its crossing primal bond is closed, giving independent dual parameter $1-p$.

First, $c+c^{\ast }\le 1$. Suppose instead that the sum exceeds $1$, and choose $1-c^{\ast }<p<c$. Both the primal model at $p$ and the dual model at $1-p$ are strictly subcritical. Sharpness of the percolation transition gives exponential connection-radius decay on each periodic graph, uniformly over its finitely many vertex types. A crossing of a rectangle of side lengths proportional to $R$ starts at one of $O(R)$ boundary vertices and travels distance of order $R$. A union bound therefore shows that each crossing probability tends to zero. But planar duality for rectangle crossings says that exactly one of an open primal left-to-right crossing and an open dual top-to-bottom crossing occurs, using the complementary boundary convention. Their probabilities sum to $1$. This contradiction proves the first inequality.

For the opposite inequality, we need noncoexistence of primal and dual infinite percolation clusters. The four-fold version in Q3(iii) cannot be assumed without change: this decorated honeycomb lattice has three-fold rotation symmetry. Here is the necessary three-fold symmetric primal-dual noncoexistence argument.

Assume both models percolate at complementary parameters $p,1-p$. For a large disc, the probability of having no exterior infinite arm is at most $\epsilon$ in each model. Divide its boundary into six arcs by splitting each $120$-degree sector at the same relative angle. In the primal model let $a,b$ be the failure probabilities for the two alternating arc types. Rotation symmetry and the FKG inequality giveSlide the splitting point across one sector. Initially the first arc is empty; finally the second is empty. A jump occurs only when an arc gains or loses an edge-crossing point. An arm using that point requires its crossing bond to be open. The FKG inequality therefore ensures that a failure probability changes by at most a factor $(1-p{)}^{-1}$ at a single jump: for a newly added arm event $L_x$, $\mathbb{P}(L_x^c)\ge 1-p$ and $\mathbb{P}(L^c\cap L_x^c)\ge (1-p)\mathbb{P}(L^c)$. At the first reversal of the order of $a,b$, their ratio is consequently at most $(1-p{)}^{-2}$. At that splitting point,Thus every one of the six primal arcs has an exterior arm with probability tending to $1$. On the same six arcs the dual failure probabilities $a^{\ast },b^{\ast }$ satisfy $a^{\ast }{b}^{\ast }\le {\epsilon }^{1/3}$, so at least one dual arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Take two of those dual arcs separated by one arc, and the two intervening primal arcs. A union bound gives four alternating primal/dual infinite exterior arms with positive probability when $\epsilon$ is sufficiently small. The separation and finite-closure argument of Q3(iii), together with percolation uniqueness on an amenable quasi-transitive graph, gives a contradiction. This proves noncoexistence. The arc balancing is the extra step needed for three-fold symmetry; it is also developed in Bollobás and Riordan's original paper on dual lattices.

If $c+c^{\ast }<1$, choose $c<p<1-c^{\ast }$. Both complementary models would then percolate, contradicting noncoexistence. Hence $c+c^{\ast }\ge 1$. Combining the two inequalities proves the dual threshold relation:

Order configurations coordinatewise. The FKG lattice condition on the masses of a finite Boolean lattice isThe FKG inequality states that under this condition, increasing real-valued functions $f,g$ satisfy $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$. In particular increasing events satisfy $P(A\cap B)\ge P(A)P(B)$. For a product measure with coordinate weights $w_e(0),w_e(1)$, the pair $(\min ({\omega }_e,{\eta }_e),\max ({\omega }_e,{\eta }_e))$ has the same two entries as $({\omega }_e,{\eta }_e)$. Multiplying over coordinates gives equality in the lattice condition, including degenerate Bernoulli parameters without division by zero.

For independent bond percolation on the nearest-neighbor cubic lattice, write ${\theta }_d(p)={\mathbb{P}}_p(|C(0)|=\infty )$. Define the percolation critical probability and the connective constant bywhere $c_n(d)$ counts rooted $n$-step self-avoiding walks. Translation invariance makes the root irrelevant, and part (i) proves existence of the latter limit.

First work on the square lattice, whose dual is another translated square lattice. Put $q=1-p$ and assume $q\mu (2)<1$. If the open cluster at the origin is finite, its exterior edge boundary contains a closed dual simple circuit surrounding the origin. To see this planar fact, surround its finitely many vertices by their unit square cells and follow the exterior boundary: its crossing primal edges are closed. Resolving repeated boundary vertices into simple circuits leaves a circuit separating the origin from infinity.

Let $a_n$ count such dual circuits of length $n$. Every one crosses the positive horizontal ray at distance at most $n$, because it surrounds the origin and its horizontal span is at most its length. Choosing a ray-crossing edge, an orientation and all but the last edge encodes it by one of at most $Cn$ rooted length-$n-1$ self-avoiding walks. Thus $a_n\le Cn{c}_{n-1}(2)$. Choose $b>\mu (2)$ with $qb<1$. The root-count limit gives $c_{n-1}(2)\le C_b{b}^n$, soThis summable tail alone need not make the probability of every enclosing circuit less than one. Choose a large $R$ and let $A_R$ be the absence of enclosing closed dual circuits of length at least $R$, with $P(A_R)>1/2$. Let $B_R$ require every primal edge inside $[-R,R{]}^2$ to be open. It has positive probability. Both events are increasing. The finite-measure Harris-FKG inequality extends to $A_R$ by decreasing limits over finitely many circuit exclusions, so $P(A_R\cap B_R)\ge P(A_R)P(B_R)>0$.

On $B_R$ the origin is connected to every graph vertex of that box. Any closed dual circuit enclosing the origin must then enclose the whole box, hence have length at least $R$. On $A_R\cap B_R$ no such circuit exists, so the origin cluster is infinite. This proves the connective-constant Peierls bound, $p_c(2)\le 1-1/\mu (2)$. Finally the lattice in dimension $d\ge 2$ contains a coordinate copy of the square lattice, whose edge law is unchanged. Percolation in that subgraph implies percolation in the full graph, and hence

Order configurations coordinatewise: $\omega \le {\omega }^{\prime }$ means ${\omega }_e\le {\omega }_e^{\prime }$ for every edge. An increasing event $A$ is a measurable set such that $\omega \in A$ and $\omega \le {\omega }^{\prime }$ imply ${\omega }^{\prime }\in A$. Opening additional edges cannot destroy an increasing event. The Harris-FKG inequality, the product measure version of the FKG inequality, statesfor two increasing events. Equivalently, bounded coordinatewise increasing random variables $f,g$ satisfy ${\mathbb{E}}_p(fg)\ge {\mathbb{E}}_{p}f{\mathbb{E}}_{p}g$. The event inequality also holds for two decreasing events, by taking complements.

For disjoint occurrence of increasing events, a finite open edge set $K$ witnesses $A$ in $\omega$ if every configuration agreeing with $\omega$ on $K$ belongs to $A$. Since $A$ is an increasing event and all edges of $K$ are open, this says that the configuration with exactly $K$ open already forces $A$. Define $A\square B$ to consist of configurations admitting disjoint finite witnesses $K,L$ for $A,B$. For events depending on finitely many edges, this is the usual disjoint-occurrence definition; it also applies to finite-connection events on the infinite graph. The Van den Berg-Kesten inequality statesOn a countable graph, its finite-witness version follows by taking increasing unions over finite edge sets. The Harris-FKG inequality concerns simultaneous occurrence, whereas the Van den Berg-Kesten inequality requires separate certificates using disjoint coordinates.

WriteSince $e_n\in \partial B(n)$, $q_n\ge a_n$, and thereforeFor the opposite inequality, choose $x=(n,x_2,\dots ,x_d)$ as in part (ii). Reflection about $x$ sends $0$ to $2n{e}_1$ and preserves the lattice. Thus the increasing events $\{0\leftrightarrow x\}$ and $\{x\leftrightarrow 2n{e}_1\}$ have the same probability. The FKG inequality yieldsConsequentlyThe polynomial boundary-size estimate makes the last term tend to zero, while the first tends to $\varphi (p)$. Hence

LetThe events $\{0\leftrightarrow e_n\}$ and $\{e_n\leftrightarrow e_{n+m}\}$ are increasing events of bond percolation. The FKG inequality and translation invariance giveThus $(a_n)$ is a subadditive sequence. Since every probability is positive for $p>0$, Fekete lemma applies and gives

For $0\le p\le 1$ and $q\ge 1$, the FKG inequality implies that increasing functions of a random-cluster configuration have nonnegative covariance. The conclusion can fail for $0<q<1$.

For independent bond percolation on a periodic plane lattice with three-fold rotation symmetry, the primal model at $p$ and the dual bond percolation model at $1-p$ cannot both percolate. To adapt Zhang's argument, divide a large boundary circle into six arcs, repeating the division in each $120$-degree sector. Vary the division point until the failure probabilities $a,b$ of the two alternating primal arc types are comparable within a factor $(1-p{)}^{-2}$. Moving past a single crossing edge changes a failure probability by at most a factor $(1-p{)}^{-1}$, by the FKG inequality. If the probability of no primal exterior arm is at most $\epsilon$, then $a^3{b}^3\le \epsilon$ and both failures are at most $(1-p{)}^{-1}{\epsilon }^{1/6}$. For the dual model, at least one arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Four alternating primal/dual arms now occur with positive probability, contradicting the alternating exterior-arm separation lemma and uniqueness after a finite modification. This extends the four-fold argument to the martini lattice.