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In bond percolation on the square lattice, the outer boundary of a finite percolation cluster, after filling its holes, consists of closed bonds. Their crossing edges in the planar dual graph form a connected enclosing contour, and hence belong to a single dual bond percolation cluster $K$. If $|K|=m$, both its coordinate widths are at most $m-1$, so the enclosed primal cluster contains at most $m^2$ vertices. A contour enclosing the origin also places every vertex of $K$ within distance $m+1$ of the origin. An exponential volume tail for the subcritical dual bond percolation clusters thus gives a bound $C{e}^{-c\sqrt{n}}$ for a finite primal cluster of size at least $n$.

For a graph with a cellular embedding on a surface, the dual has one vertex for each face and an edge crossing each original edge between its adjacent faces. Each edge is counted, including a loop when the two incident faces coincide. The planar dual graph is the case of a planar embedding.

The honeycomb lattice is the periodic degree-three planar graph consisting of the vertices and edges of a regular hexagonal tiling. It has two vertex classes; every edge joins opposite classes. Its planar dual graph is the triangular lattice.

A self-avoiding walk of length $n$ is a sequence of $n+1$ distinct graph vertices joined consecutively by edges. Split a rooted self-avoiding walk of length $m+n$ after $m$ steps. There are ${\kappa }_m$ possible prefixes; each remaining segment is a self-avoiding $n$-step walk from its current graph vertex, and forbidding intersections with the prefix can only reduce its count below ${\kappa }_n$. Translation invariance therefore givesApply the Fekete lemma, proved in Question 4(a), to $\log {\kappa }_n$. It gives $\log \mu ={\inf }_n{n}^{-1}\log {\kappa }_n$ and hence existence of the connective constant. Walks using only the $d$ positive coordinate directions never revisit a graph vertex, so there are at least $d^n$. A walk has $2d$ choices for its first step and at most $2d-1$ subsequently because immediate reversal is forbidden. Thus

For independent bond percolation, let $C(0)$ be the percolation cluster of the origin, $\theta (p)=P_p(|C(0)|=\infty )$, and $p_c=\inf \{p:\theta (p)>0\}$. An infinite percolation cluster in a locally finite graph contains an open self-avoiding path of every length. The union bound givesWhen $p\mu <1$, the right-hand side tends to zero, since its $n$th root tends to $p\mu$. Hence the connective-constant lower bound for percolation is

For the upper bound in two dimensions, use the square planar dual graph. Call a dual edge available when its crossing primal edge is closed. Each dual edge is available with probability $q=1-p$, independently. The planar graph boundary lemma says that a finite primal percolation cluster is enclosed by a graph cycle consisting of such edges: its external edge boundary gives this graph cycle. We show that $q\mu <1$ guarantees positive percolation probability.

Let $N_n$ count enclosing graph cycles of length $n$. Because the origin lies inside their coordinate bounding box and their diameter is at most $n$, every such graph cycle has a graph vertex within a box of side $2n+3$. From each possible root, the first $n-1$ edges give a self-avoiding walk; the final closing edge is then determined. Dropping the closure restriction gives the convenient boundChoose $\rho$ with $q\mu <\rho <1$. The connective-constant limit of a sequence implies ${\kappa }_{n-1}{q}^n\le C{\rho }^n$. Consequently the large-circuit tailis summable.

Condition all primal edges inside a large square $[-L,L{]}^2$ to be open; this event has positive probability. The origin percolation cluster then contains that square. If it is finite, its enclosing graph cycle has length at least $L$. A proposed graph cycle using a pinned-open edge now has conditional probability zero; every other proposed graph cycle still has probability $q^n$ by independence of random variables. For $L$ large enough the union bound for the remaining graph cycles is less than one. There is therefore positive conditional probability that the origin percolation cluster is infinite. This proves the connective-constant Peierls bound without needing the exact value of $p_c$:

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.

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:

Independent bond percolation turns a deterministic graph into a random network: each edge is open with probability $p$, independently, and a percolation cluster is a connected component of its open subgraph. Site percolation instead randomizes vertices. On the cubic lattice the percolation probability $\theta (p)=P_p(0\leftrightarrow \infty )$ is the order parameter and $p_c=\inf \{p:\theta (p)>0\}$ is the critical probability. A monotone coupling, using independent uniform edge labels and opening labels below $p$, makes the growth of connectivity with $p$ transparent.

For $d\ge 2$, $0<p_c<1$: self-avoiding walk counting gives $p_c\ge 1/\mu (d)$, and the planar bound proved above gives an upper bound below one. Below $p_c$, clusters are finite and connection probabilities decay exponentially, a percolation sharpness theorem rather than a consequence of the threshold definition alone. Above $p_c$ there is almost surely a unique infinite percolation cluster. At criticality, the presence of an infinite percolation cluster must be addressed for the particular model. In two dimensions there is none. The transition replaces a finite characteristic length by scale-free geometry, followed by a macroscopic connected component.

Percolation critical exponents quantify different aspects of this change. Use power-law notation for leading exponents; it may suppress amplitudes and logarithmic corrections. From the subcritical side, define the percolation susceptibility $\chi (p)={\mathbb{E}}_p|C(0)|$ and an exponential correlation length $\xi (p)=1/\lambda (p)$, using a fixed norm for the one-arm rate. Above criticality the unrestricted expectation is infinite, so a finite-cluster percolation susceptibility must exclude the infinite percolation cluster. The principal observables are:

The last distribution is not the distribution seen from a uniformly chosen site: the latter is weighted by cluster size. In the scaling description this accounts for $1/\delta =\tau -2$. Under the usual below-upper-critical-dimension scaling hypotheses, further scaling relations for critical exponents includeThey organize the exponents into a small number of independent quantities; they are not automatic identities following only from their definitions. A percolation cluster fractal dimension $D_f$ describes the mass of a large critical cluster of radius $r$ as roughly $r^{D_f}$, with the scaling relation for critical exponents $D_f=d-\beta /\nu$ in this regime.

The percolation universality hypothesis is that ordinary short-range independent models in the same spatial dimension share these exponents and their continuum behavior despite differing microscopic lattices or using bonds rather than sites. The threshold itself is not universal. Long-range connections, correlations or changes of geometry can change the universality class, so the hypothesis does not include every random graph called percolation.

Two dimensions have a particularly strong geometric structure. Closed planar dual graph circuits obstruct primal open paths. The Harris-Kesten theorem fixes the square lattice bond percolation threshold at $p_c=1/2$, using planar duality and crossing estimates. Self-duality alone is not a proof; uniform rectangle-crossing control is a crucial extra ingredient. The Russo-Seymour-Welsh theorem keeps critical crossing probabilities of rectangles of fixed aspect ratio bounded away from zero and one at all scales. Closed dual circuits on disjoint annular scales then rule out an infinite critical cluster and reveal why arbitrarily large finite structures persist.

Critical site percolation on the triangular lattice has rigorously established conformal invariance of planar percolation for crossing limits and an SLE description of interfaces. These results are specific to this model. The exact planar valuesare rigorously established there; its one-arm probability has exponent $5/48$. In the corresponding scaling description these values give $\eta =5/24$, $\delta =91/5$, $\tau =187/91$ and $D_f=91/48$. The percolation universality hypothesis predicts the same planar exponents for square lattice bond percolation, but the triangular-lattice theorem by itself does not prove that transfer. This distinction separates exact model-specific mathematics from a broader physical prediction.

Dimension changes the importance of loops and correlations between growing branches. The mean-field percolation exponents come from approximating critical clusters by branching processes. Their values are $\beta =1$, $\gamma =1$, $\delta =2$, $\nu =1/2$, $\eta =0$ and $\tau =5/2$. The predicted upper critical dimension of percolation is six. One way to see its role is the triangle diagram appearing in the percolation triangle condition: if the long-distance Fourier connectivity behaves like $|k{|}^{-2}$, its infrared contribution is proportional to ${\int }_0^{\epsilon }{r}^{d-7}dr$. It is finite above six and logarithmically divergent at six. This motivates mean-field behavior above six and logarithmic corrections at six, without turning that heuristic into a proof for every lattice.

Lace expansion makes the mean-field picture rigorous in suitable high-dimensional regimes. For nearest-neighbor percolation, sufficiently high dimensions are covered; sufficiently spread-out models have mean-field results for every $d>6$. Between two and six the exponents depend nontrivially on dimension, and exact three-dimensional values are not supplied by the planar theory. Hyperscaling relation illustrates the change: mean-field values give $2\beta +\gamma =3$, while $d\nu =d/2$, so the naive equality fails above six. In $d=1$, $p_c=1$ and no infinite percolation cluster exists for $p<1$; this is a degenerate endpoint case rather than the ordinary interior transition.

For a finite graph $G=(V,E)$, let $o(\omega )={\sum }_{e\in E}\omega (e)$, and let $k_G(\omega )$ count connected components of the spanning open subgraph, including isolated graph vertices. The random-cluster model with $p\in [0,1]$ and $q>0$ isThe partition function $Z_{G,p,q}$ is the sum of these weights over all configurations, making the expression a probability measure.

For the lattice box, take all nearest-neighbor edges with both endpoints in ${\Lambda }_n$. A random-cluster boundary condition $\xi$ is a partition of the boundary graph vertices. Vertices in one block are identified, or wired together, before counting components; the identifications do not add random edges. Let $k_{\xi }(\omega )$ be the number of components of the resulting quotient open graph. ThenThe free random-cluster boundary condition has singleton blocks, while the wired random-cluster boundary condition has one boundary block. Arbitrary partitions are allowed; no assumption that the partition itself has a planar realization is needed for the monotonicity statement.

Write $\xi ⪯\zeta$ if every block of $\xi$ is contained in a block of $\zeta$, so $\zeta$ makes at least as many identifications. The precise boundary monotonicity of the random-cluster measure isEquivalently, the expectation of every real-valued order-preserving function is larger under the more wired measure. This is stochastic domination of probability measures on the coordinatewise configuration order.

To prove it, condition on every edge except $e=\{u,v\}$. If $u,v$ are already connected using those open edges and the boundary wiring, opening $e$ does not change $k_{\xi }$, so its conditional open probability is $p$. If they are not connected, opening it reduces $k_{\xi }$ by one. The ratio of the open weight to the closed weight is then $p/[q(1-p)]$. Thus the random-cluster single-edge conditional probability isFor $q\ge 1$ the second number is at most the first. Adding open edges or making the boundary partition coarser can only turn a disconnected pair into a connected one. Therefore these conditional open probabilities are increasing in both the exterior configuration and the amount of wiring.

Run a heat-bath Markov chain for each boundary partition. At every step choose the same uniformly sampled edge in both chains, sample the same independent uniform variable $U$, and set that edge open if $U$ is below its conditional open probability. Start both chains at the all-closed configuration. By the preceding inequality, the two configurations remain ordered at every update. Each marginal chain has its corresponding random-cluster measure as its stationary distribution: resampling one coordinate from its conditional distribution preserves that law. Because $0<p<1$ and $q>0$, every update gives both possible states positive probability; the finite chain is an irreducible Markov chain with aperiodicity, and hence converges to its unique stationary distribution. Taking expectations of any order-preserving function and passing to the limit proves the claimed stochastic domination of probability measures. At $q=1$ the two conditional probabilities agree, and the boundary condition has no effect on the independent bond percolation law.

For the final identity fix a plane embedding of the finite planar graph, and include the unbounded face in its planar dual graph. In the dual configuration ${\bar{\omega }}^{\ast }$, a dual edge is open exactly when its crossed primal edge is closed. Let $m=o(\omega )$. The spanning open primal subgraph has $|V|$ graph vertices, $m$ edges, and $k(\omega )$ components. The Euler formula for a connected planar graph, applied componentwise with the common exterior face accounted for, gives its number of faces asDeleting a closed primal edge merges its incident original faces precisely when its dual edge joins distinct dual components. Consequently the faces of the open primal subgraph correspond exactly to components of the open complementary dual subgraph. This remains true for bridges and loops, with their dual loops and bridges, and for a disconnected primal graph. Thus the planar cluster-count identity isAt the self-dual parameter of the random-cluster model, put $s=\sqrt{q}$ andFor fixed $G$ the factor $(1-p_{\mathrm{sd}}{)}^{|E|}$ is independent of $\omega$. The weight is therefore proportional toAbsorbing the remaining graph-dependent factor into the normalization gives the symmetric cluster weight at the self-dual parameter:Counting the outer dual face and all isolated primal graph vertices is essential for the constant in the Euler identity. No assumption that $G$ is self-dual is required for this proportionality.

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.

First obtain a polynomial critical one-arm upper bound from part (a). Fix a sufficiently large integer radius $r$. Use rotated-coordinate boxes $B_r=[-r,r{]}^2$ for this geometric construction, in the axes of part (a). Surround $B_r$ by a square ring inside $B_{3r}$. It can be assembled from the four stripsThe first two require horizontal dual crossings; the last two require vertical dual crossings. Their corner overlaps are squares, so the adjacent long crossings intersect there and their union contains a dual graph cycle surrounding the inner box. Translate the strips by the half-lattice displacement of the planar dual graph and adjust the endpoints by one lattice unit if needed. All aspect ratios remain bounded. Part (a) and the same overlap gluing give a uniform lower bound $c_1>0$ for each dual crossing at $1/2$. Applying the Harris-FKG inequality to these decreasing events in the primal configuration gives a closed dual circuit with probability at least $\delta =c_1^4>0$. We may replace $\delta$ by a smaller number in $(0,1)$.

Use the independent annular barriers for percolation at radii $r_j=r_{0}10^j$. Their supporting edge sets are disjoint, so the circuit events are independent. In the original unrotated coordinates, $B_{3r}$ lies inside ${\Lambda }_{5r}$ and the ring still separates the origin from infinity. These fixed geometric comparisons let us bound the original $g_N$. An open graph path from $0$ to distance $N$ must avoid every such closed dual circuit inside ${\Lambda }_N$. Thus, for some constants $C<\infty$ and $\alpha >0$,where the first bound is used only when at least one of the indicated rings fits; increasing $C$ handles smaller $N$.

Now let $\epsilon =p-1/2>0$ and use the monotone coupling of Bernoulli percolation on edges. The event $\{0\leftrightarrow \partial {\Lambda }_N\}$ only needs edges whose endpoints lie in ${\Lambda }_N$, because a first boundary hit gives an internal graph path. There are exactly $4N(2N+1)$ such edges. Each changes state between parameters $1/2$ and $p$ with probability $\epsilon$. The union bound therefore gives the finite-box comparison for percolation parametersFor sufficiently small $\epsilon$, choose $N=\lfloor {\epsilon }^{-1/(\alpha +2)}\rfloor$, so that $N$ is at least half the unrounded value. Both terms then have order ${\epsilon }^{\alpha /(\alpha +2)}$. In particular,for a finite $A$. Increase $A$ to cover the remaining compact range of parameters by $\theta (p)\le 1$. This near-critical percolation power upper bound requires only a positive crossing constant, not the exact value of a critical exponent. It also implies $\theta (1/2)=0$ through the critical one-arm probability bound.

We use planar duality for rectangle crossings, Harris-FKG inequality and the permitted exponential decay of subcritical percolation. A critical-point detail matters: uniqueness stated only for $p>p_c$ cannot by itself be applied at $p=p_c$. We prove the Burton-Keane theorem by the boundary counting proof of percolation uniqueness, establishing at most one infinite percolation cluster at every parameter and avoiding that gap.

For $0<p<1$, the number $N$ of infinite percolation clusters is almost surely constant by translation ergodicity of Bernoulli percolation. It cannot be a finite constant greater than one. A sufficiently large box meets two different infinite percolation clusters with positive probability; opening its finitely many interior edges joins them without creating a new infinite percolation cluster, decreasing $N$. Finite modification of Bernoulli percolation gives positive probability to this modification, contradicting constancy.

Nor can $N=\infty$. A finite box then meets three different infinite percolation clusters with positive probability. Select one infinite exterior branch from each. In the box and a finite collar retain just an embedded three-armed tree joining those branches and close all other incident edges there. Its branching graph vertex separates three infinite components when removed, so is a trifurcation vertex in percolation. All selections concern finitely many possible entrances and edge patterns; finite-energy property of Bernoulli percolation gives positive probability for at least one such pattern. Translation invariance then gives a common positive trifurcation probability $q$.

Here is the counting contradiction in detail. For a finite box $W$, contract each open component outside $W$ that touches $W$ to a boundary terminal. In each cluster, the resulting incidence graph is finite and connected. Every trifurcation in $W$ separates at least three sets of terminals, since each infinite branch must leave $W$. A minimal subtree joining all terminals must therefore contain that graph vertex with degree of a vertex at least three. Its leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of those branching graph vertices by the terminal count. Different terminals are represented by different exterior graph neighbours of $W$. Thus the trifurcation boundary-counting lemma givesFor $W=[-r,r{]}^2\cap {\mathbb{Z}}^2$, the two sizes are $(2r+1{)}^2$ and $8r+4$. Letting $r\to \infty$ contradicts $q>0$. Therefore $N\in \{0,1\}$ almost surely, with the same conclusion for the translated planar dual graph.

Now prove absence of percolation at $p=1/2$ by Zhang's argument. Suppose instead that $\theta (1/2)>0$. Almost surely an infinite primal cluster exists, and self-duality gives an infinite dual cluster as well. An infinite connected subgraph of this locally finite lattice contains a graph ray, by the König infinity lemma. Expanding square boxes meet the infinite percolation clusters with probability tending to one.

For $B_r=[-r,r{]}^2\cap {\mathbb{Z}}^2$, let $A_N,A_E,A_S,A_W$ mean that an open graph ray takes an outward edge through the indicated side and subsequently uses graph vertices outside $B_r$. The union occurs whenever the box meets an infinite percolation cluster: take the last exit of an infinite graph ray from the finite box. Conversely an exterior arm supplies an infinite percolation cluster meeting its boundary. Quarter-turn symmetry makes the four probabilities equal. Their complements are decreasing events, so the square-root trick for positively associated events givesFor the planar dual graph use the box with boundary coordinates $\pm (r+1/2)$ and define its exterior side-arm events in the same way. This box also has quarter-turn symmetry and meets the dual infinite percolation cluster with probability tending to one. Thus each dual side has an infinite exterior arm with probability tending to one. Primal outward edges cross this dual-box contour on the corresponding sides; their remaining graph vertices stay outside it. A union bound makes the simultaneous event of primal north/south arms and dual east/west arms have positive probability for a sufficiently large box.

The four arms alternate around the contour. Open all edges with both endpoints in $B_r$, leaving all outward and exterior edges intact. This joins the two primal entrance points and preserves all four exterior arms. Finite-energy property of Bernoulli percolation keeps the event's probability positive. The primal north and south arms are now connected through the box. The two dual arms must lie in different infinite dual clusters. To see the separation, a hypothetical finite dual graph path joining the east and west arms cannot enter the dual-box interior: its boundary-crossing and interior edges cross primal edges that were just opened. Erase its loops and cut it at successive hits of the contour. A resulting exterior dual crosscut, together with the corresponding contour arc, encloses one of the two intervening primal entrance points. The infinite primal graph ray from that point would have to cross a dual-open edge, which is impossible. This is the planar separation used in the alternating arms argument at the self-dual percolation parameter. It contradicts uniqueness of the infinite dual cluster. HenceFinally suppose $p_c(2)>1/2$. At $p=1/2$ the permitted exponential decay of subcritical percolation would supply $C,c>0$ with ${\mathbb{P}}_{1/2}(v\leftrightarrow \partial B_n(v))\le C{e}^{-cn}$. Use the rectangle with graph vertices $\{0,\dots ,n\}\times \{0,\dots ,n-1\}$, and let $H_n$ be its left-to-right open crossing. Planar duality for rectangle crossings identifies its complement with a dual top-to-bottom crossing of a rectangle of width $n-1$ and height $n$. Rotation and translation give the same crossing law; side edges at the entrance and exit boundaries are irrelevant. Thus the exact self-dual rectangle crossing probability isBut a crossing has some starting graph vertex on its $n$-vertex left side connected to distance $n$. The union bound and exponential decay of subcritical percolation imply ${\mathbb{P}}_{1/2}(H_n)\le nC{e}^{-cn}\to 0$, a contradiction. ThereforeNo Russo-Seymour-Welsh theorem or continuity assertion for $\theta$ is being used, and critical uniqueness was proved rather than inferred from the supercritical hypothesis.

Declare an edge of the planar dual graph open exactly when its crossed primal edge is closed. This dual bond percolation has parameter $q=1-p<1/2=p_c$, using the Harris-Kesten theorem. We use the standard exponential tail of subcritical cluster size: for some $K,a>0$, uniformly in the dual vertex $v$,This is decay of the number of vertices in the percolation cluster, not merely its radius.

On $\{|C|=n\}$, the geometric fact allowed in the paper supplies a simple dual cycle in a graph surrounding $C$, with length $\ell \ge \alpha \sqrt{n}$. Every crossed edge is in the edge boundary of $C$ and is therefore closed, so the dual cycle is open. Its $\ell$ distinct vertices lie in one dual percolation cluster of size at least $\ell$.

A dual cycle in a graph of length $\ell$ surrounding the origin has all its vertices within sup-norm distance $\ell +1$ of the origin: its coordinate spans are at most $\ell$, and the origin lies between each pair of extreme coordinates. There are at most $K_0(\ell +1{)}^2$ possible dual vertices there. The union bound over lengths and possible vertices yieldsfor some $K_1,b>0$, since a polynomial factor can be absorbed into a slower exponential decay.

To remove the constant prefactor for all $n\ge 1$, first choose $n_0$ so that the bound is at most $e^{-(b/2)\sqrt{n}}$ for $n\ge n_0$. For the remaining finitely many $n$, use ${\mathbb{P}}_p(|C|=n)\le 1-\theta (p)<1$ and chooseFor $1\le n<n_0$, the chosen $c_2$ ensures $e^{-c_2\sqrt{n}}\ge 1-\theta (p)$.

In bond percolation on the square lattice, exactly one of the following occurs in the rectangle: an open left-to-right primal crossing, or a closed top-to-bottom crossing in the planar dual graph. These alternatives are disjoint and exhaustive by planar duality for rectangle crossings.

At $p=1/2$, the closed dual edges have the same law as open primal edges. Rotating the dual rectangle through a right angle identifies its top-to-bottom crossing with the original left-to-right crossing; the slight difference between the side lengths $\ell +1$ and $\ell$ is exactly the boundary shift introduced by dualization. Thus the event and its complement have equal probability, so

A planar finite network with three boundary terminals induces one of five connection partitions: all three connected, all three separate, or one connected pair and an isolated third terminal. Put three dual terminals on the intervening boundary arcs. Passing to the planar dual graph interchanges the all-connected and all-separate partitions and permutes the three pair partitions. If the network is symmetric under all terminal permutations, equality $P_3=P_0$ of the first two probabilities makes its whole partition law dual-invariant. Independent cells on a self-dual triangular terminal arrangement then give identical primal and dual coarse connection laws. Infinite connectivity is preserved because cells are uniformly finite. For a martini lattice cell with six independent bonds of common parameter $p$, $P_3=3p^5-2p^6$ and $P_3-P_0=(2p^2-1)(p^4-3p^3+2p^2+1)$. The unique zero in $[0,1]$ is $1/\sqrt{2}$.