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Take a directed graph with vertices $v_0,v_1,\dots$ and $n+1$ parallel bonds from $v_n$ to $v_{n+1}$. This is a locally finite graph, although the incoming degrees are unbounded. For bond percolation, ${\theta }_b(p)={\prod }_{n\ge 0}(1-(1-p{)}^{n+1})>0$ whenever $p>0$: the failure probabilities are summable, so the infinite product is positive. For site percolation, survival requires every vertex of the ray to be open, which has probability zero for $p<1$. Thus the survival and susceptibility thresholds for directed percolation satisfy $p_H^b=0$ and $p_H^s=1$.

In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability $p$, and state zero, meaning closed, with probability $1-p$. Thus the configuration law is the product measure ${\mathbb{P}}_p={⨂}_{e\in E}\mathrm{Bernoulli}(p)$. An open path in bond percolation uses only open edges. An increasing event $A$ satisfies $\omega \in A,{\omega }^{\prime }\ge \omega \Rightarrow {\omega }^{\prime }\in A$: opening further edges cannot destroy the event.

The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write $A\square B$ when disjoint sets of open edges separately witness $A$ and $B$, so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.

For $1\le m<n$, take a simple open path from the origin to the radius-$n$ boundary and let $x$ be its first vertex on the radius-$m$ boundary. Its initial segment witnesses $0\leftrightarrow x$. If $z$ is its final vertex, then $\Vert z-x{\Vert }_{\infty }\ge n-m$. Consequently the remaining segment reaches $x+\partial {\Lambda }_{n-m}$, and its initial portion until that visit witnesses $x\leftrightarrow x+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor $m=n$ the same bound follows directly from the union bound, with $g_0=1$. With the conventional $\partial {\Lambda }_0=\{0\}$, $m=0$ gives equality.

Write $C(0)$ for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then $a_m\to 0$. For $0<p<1$, choose $m$ with $0<a_m=a<1$. Iterating the boundary-splitting estimate with this fixed $m$ givesFor $k\ge 2m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To handle all smaller positive radii without an unspecified prefactor, observe that $g_k\le g_1=1-(1-p{)}^4<1$. ThereforeFor $1\le k<2m$, the second chosen rate gives $e^{-\gamma k}\ge g_1$, and for larger $k$ the first rate applies. At $k=0$ both sides equal one. The endpoint $p=0$ has $g_k=0$ for every $k\ge 1$, so any positive rate works; $p=1$ is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.

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 $p_c$:

In bond percolation on ${\mathbb{Z}}^d$, edges are independently open with probability $p$ and closed otherwise. An open percolation cluster is a connected component using only open edges. In site percolation it is the graph vertices that are independently declared open. The percolation probability $\theta (p)=P_p(0\leftrightarrow \infty )$ measures the chance that the origin lies in an infinite percolation cluster, and the percolation critical probability is $p_c=\inf \{p:\theta (p)>0\}$. Monotone coupling of Bernoulli percolation by independent uniform distribution makes $\theta$ nondecreasing.

The sharpness of the percolation transition says that below $p_c$ connection probabilities on ${\mathbb{Z}}^d$ decay exponentially with distance, while above $p_c$ the percolation probability is positive. There is almost surely a unique infinite percolation cluster in the supercritical regime. On the square lattice, the Harris-Kesten theorem gives bond threshold $1/2$, and there is no infinite percolation cluster at percolation critical probability. The emergence of connections on arbitrarily large scales is the geometric meaning of the phase transition.

Several observables describe this singular behavior. The percolation susceptibility is $\chi (p)=E_p|C(0)|$ for $p<p_c$. A directional correlation length can be defined through exponential two-point decay, $\xi (p{)}^{-1}=-{\lim }_n{n}^{-1}\log P_p(0\leftrightarrow n{e}_1)$. Percolation critical exponents describe leading powers, conveniently defined logarithmically rather than requiring exact asymptotic amplitudes:The first limit of a sequence approaches from above, the other two from below. At percolation critical probability, percolation two-point connection probability is expected to have leading power $|x{|}^{-(d-2+\eta )}$, and the number per site of percolation clusters of size $s$ has leading power $s^{-\tau }$. This cluster-count probability distribution differs from the size-biased probability distribution seen by choosing a site. Scaling relations for critical exponents such as $\gamma =(2-\eta )\nu$ and $2\beta +\gamma =d\nu$ connect the exponents in their applicable scaling regime; hyperscaling relation is not valid above the upper critical dimension of percolation in ordinary mean-field regimes.

The percolation universality hypothesis predicts that dimension and large-scale symmetry, rather than microscopic graph details, determine these exponents and scaling functions. Thus square-lattice bond and triangular-lattice site models are expected to share the planar graph universality class despite different microscopic definitions. Universality does not mean identical critical probabilities, amplitudes or finite-lattice distributions. Exact values and existence of exponents require model-specific theorems, not merely this prediction.

Two dimensions have an additional structure: conformal maps preserve angles, and critical crossing probabilities are expected to be unchanged under conformal transport of the domain and its marked boundary arcs. A crossing probability then depends only on the conformal shape of that marked quadrilateral. The Cardy boundary crossing formula gives this dependence in terms of its boundary cross ratio. For critical triangular-lattice site percolation, conformal invariance and the resulting exponents are established; examples are $\beta =5/36$, $\gamma =43/18$, $\nu =4/3$, with the rigorous statements in leading-power form. These values are summarized in Smirnov and Werner's original paper. Their universality extension to arbitrary planar lattices should not be presented as the same theorem.

An exploration path separating the two colors, with opposite boundary colors on two arcs, has a domain Markov property of a percolation exploration: conditional on the explored portion, the unexplored colors retain their independent laws with the newly exposed boundary colors. Conformal invariance, this domain Markov property and suitable regularity lead to Schramm–Loewner evolution. In the upper half-plane its chordal equation iswhere $B_t$ is standard Brownian motion and time is normalized by half-plane capacity. Critical percolation corresponds to $\kappa =6$. Thus the complicated discrete interface becomes a conformally natural random curve driven by one real Brownian motion. Its crossing and arm probabilities supply critical exponents and, through scaling relations for critical exponents, near-critical singularities. This makes conformal methods predictive tools for the transition, rather than only descriptions of an interface's appearance.

We use the Van den Berg-Kesten inequality: for increasing events $A,B$ under independent bond percolation, their disjoint occurrence $A\square B$, meaning that they have disjoint sets of open edges witnessing the two events, satisfies ${\mathbb{P}}_p(A\square B)\le {\mathbb{P}}_p(A){\mathbb{P}}_p(B)$. The finite-edge statement applies here; connection to a finite-box boundary can always be witnessed before the first exit from that box.

On the event $0\leftrightarrow \partial {\Lambda }_{m+n}$ choose a simple open path to that boundary, and let $y$ be its first point on $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow \partial {\Lambda }_m$ through $y$, inside ${\Lambda }_m$. The remaining segment eventually reaches maximum-norm distance at least $n$ from $y$, because its endpoint has norm $m+n$ while $\Vert y{\Vert }_{\infty }=m$. Stop that segment when it first reaches $y+\partial {\Lambda }_n$. These two segments use disjoint edges.

For fixed $y$, let $A_y$ be connection from zero to $y$ inside ${\Lambda }_m$, and $B_y$ be connection from $y$ to $y+\partial {\Lambda }_n$ inside $y+{\Lambda }_n$. Then ${\mathbb{P}}_p(A_y)\le {\beta }_m$ and, by translation invariance, ${\mathbb{P}}_p(B_y)={\beta }_n$. The union bound and the stated disjoint-occurrence inequality give

Set $x_n=\log {\beta }_n$ and ${\alpha }_n=\log |\partial {\Lambda }_n|$. The previous part applies, sinceMoreover a straight path of $n$ open bonds has probability $p^n$, so $p^n\le {\beta }_n\le 1$. Thus the limit is finite andEquivalently, for the nonnegative decay rate $\kappa =-\gamma$, the bound is ${\beta }_n\ge |\partial {\Lambda }_n{|}^{-1}{e}^{-n\kappa }$.

The printed final exponent has the opposite sign while retaining the definition of $\gamma$ as the limit of $n^{-1}\log {\beta }_n$. With that definition it cannot be correct in general. Indeed, if $(2d-1)p<1$, count self-avoiding paths of length $\ell \ge n$ to getHence $\gamma \le \log ((2d-1)p)<0$. A lower bound with $e^{-n\gamma }$ would eventually exceed one, because the boundary factor grows only polynomially. The valid bound is the boxed expression with $e^{n\gamma }$, or the equivalent expression using $\kappa =-\gamma$.

For independent bond percolation on ${\mathbb{Z}}^d$, $d\ge 2$, there is almost surely at most one infinite percolation cluster at every parameter. This is the Burton-Keane theorem. It does not require an assumption about the behavior at the critical probability. If the probability that the origin lies in an infinite percolation cluster is positive, translation ergodicity of Bernoulli percolation also gives almost-sure existence and hence exactly one.

The two structural ingredients are translation invariance and finite-energy property of Bernoulli percolation. For $0<p<1$, every prescribed configuration of finitely many bonds has positive conditional probability given all the other bonds. In particular, opening or closing a fixed finite set of bonds transforms events of positive probability into events of positive probability. The product measure is ergodic under translations. One way to see this is to approximate a translation-invariant event by a finite-cylinder event, translate the cylinder sufficiently far to make its coordinates disjoint, and use independence. Sending the approximation error to zero gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.

Let $K$ be the number of infinite percolation clusters. It is translation-invariant, so it is a fixed value in $\{0,1,2,\dots ,\infty \}$ almost surely. First exclude a finite value $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Force every internal bond in that box to be open. This joins those clusters and can only merge other clusters; it cannot split an infinite percolation cluster. It also cannot create a new infinite percolation cluster out of finite ones, since only finitely many clusters touch the modified region. Thus the number of infinite percolation clusters decreases. finite-energy property of Bernoulli percolation makes the modified event have positive probability, contradicting its almost-sure value $k$.

Suppose next that $K=\infty$. Some finite box meets three different infinite percolation clusters with positive probability. Each such cluster has an infinite component outside the box: deletion of a finite set in a locally finite graph leaves only finitely many components adjacent to that set, so an infinite percolation cluster cannot have all of them finite. Choose one infinite exterior component from each of the three clusters, and one bond connecting it to the box.

Inside the box, connect the three selected exterior endpoints by a finite tree, preserving just their three selected entrance bonds. Take the minimal subtree connecting these endpoints; they are leaves, since their only retained connection into the box is their chosen entrance bond. This tree has a branch graph vertex of degree three. Close every other internal or boundary bond of the box. Its three exterior components remain distinct outside, since they belonged to different original clusters. The branch graph vertex now has exactly three open incident bonds, and deleting it separates its cluster into three infinite components. Such a graph vertex is a trifurcation vertex in percolation. There are finitely many choices for the endpoints and tree within the chosen region, so at least one fixed finite modification of Bernoulli percolation has positive probability. finite-energy property of Bernoulli percolation consequently gives positive probability of a trifurcation at some fixed graph vertex. Translation invariance makes this probability the same number $q>0$ at every graph vertex.

The decisive fact is that a finite box cannot contain a positive-volume density of these branching points. Here is the boundary count, including its combinatorial justification. For a finite graph vertex set $W$, let ${\partial }^{+}W$ be the exterior graph vertices adjacent to it. In each infinite percolation cluster meeting $W$, contract each of its components outside $W$ that is adjacent to $W$ to a terminal graph vertex. The quotient graph consisting of the cluster's graph vertices in $W$ and these terminals is finite and connected. A trifurcation graph vertex in $W$ separates its terminals into at least three nonempty groups, since each of its infinite branches must leave $W$.

Take a minimal subtree of the quotient graph connecting all the terminals. Every such trifurcation graph vertex must belong to that subtree and have degree at least three: otherwise the terminals in one of its separated groups could not connect to the other groups. All leaves of the minimal subtree are terminals. A cluster with only one terminal has no counted trifurcation. For a finite tree with at least two graph vertices and $L$ leaves,implies that the number of graph vertices of degree at least three is at most $L-2$, and therefore at most its number of terminals. Distinct exterior components can be assigned distinct graph vertices of ${\partial }^{+}W$. Terminals from different clusters are disjoint as well. Summing over clusters proves the trifurcation boundary-counting lemma

Take $W=[-n,n{]}^d\cap {\mathbb{Z}}^d$. The expected left side is $q|W|$, of order $q{n}^d$, while the right side is $O_d(n^{d-1})$. Division by $|W|$ and passage to infinity force $q=0$, contradicting the positive probability obtained by finite modification of Bernoulli percolation. Thus infinitely many infinite percolation clusters are impossible. The surface-to-volume comparison is the geometric reason for uniqueness on the lattice; the same argument is tied to amenability on more general graphs.

At $p=0$ there is no infinite percolation cluster, and at $p=1$ the whole lattice is the unique infinite percolation cluster. This completes all parameter cases. Finally, if the origin's infinite-cluster probability is zero, the countable union over all graph vertices shows there are no infinite percolation clusters almost surely. If it is positive, the translation-invariant existence event has positive probability and hence probability one. Together with uniqueness this distinguishes absence from the case of a single infinite percolation cluster, without making an unsupported assertion that every graph vertex belongs to it.

The Harris theorem for square-lattice bond percolation asserts $\theta (1/2)=0$, and hence $p_c\ge 1/2$. Work in the dual square lattice, declaring a dual bond open precisely when its primal bond is closed. At $p=1/2$ this is again independent bond percolation of density one half.

Around a square centered at the origin place four dual rectangles of dimensions $6r$ by $2r$, one on each side, with overlap squares at the corners. Long crossings of all four must meet in the corner squares and their union contains a dual circuit surrounding the central square. The fixed-aspect crossing bound and the Harris lemma give this dual barrier a probability at least $\eta :=c_3^4>0$, independent of $r$.

Take geometrically increasing integer radii, for example with a factor of eight, and shift by half a lattice spacing to lie on the dual lattice. The four-rectangle annuli then have disjoint bond sets. Their barrier events are independent. An infinite primal open path from the origin would have to avoid every closed dual circuit, so the independent annular barriers for percolation giveLetting $j\to \infty$ provesThe extension to smaller $p$ follows by the monotone coupling of Bernoulli percolation. Translation and a countable union also show that no infinite open cluster exists anywhere at $p=1/2$.

View the accessible rooms as the open cluster of bond percolation on a rooted ternary tree. Each accessible room has three forward corridors, independently open with probability $2/3$, so its number of accessible children has the binomial distribution $\mathrm{Bin}(3,2/3)$. The counts at successive levels form a binomial branching process with offspring probability generating functionLet $q_N$ be the probability that the root has no open path to level $N$. Then $q_0=0$, and independence of the three descendant subtrees givesThe sequence increases to the smallest fixed point $q=\varphi (q)$ in $[0,1]$, the branching extinction probability. Thus $q_N$ is close to that solution for large $N$.

Connectivity is the relevant escape event: if no window is reachable, escape is impossible; if one is reachable, ordinary random wandering reaches a window eventually with probability one. Before reaching level $N$ the guest moves within a finite connected graph, so an accessible exit cannot be avoided forever. This interpretation does not require the guest to move monotonically outward.

No. The parallel-bond ray with different site and bond thresholds supplies a locally finite graph counterexample. Take vertices $v_0,v_1,\dots$ and $n+1$ parallel directed edges from $v_n$ to $v_{n+1}$. Its underlying multigraph is connected and every individual degree is finite, but there is no uniform incoming-degree bound.

For bond percolation rooted at $v_0$, survival means that every successive group of parallel bonds contains at least one open bond. Independence givesFor $p>0$, the sum of the failure probabilities is ${\sum }_{n\ge 0}(1-p{)}^{n+1}=(1-p)/p<\infty$. Eventually these terms are at most $1/2$, and $\log (1-t)\ge -2t$ for $0\le t\le 1/2$. The tail of the infinite product is therefore bounded below by a positive exponential; its finitely many initial factors are positive as well. Thus ${\theta }_b(p)>0$ for every $p>0$.

For site percolation, however, every vertex of the unique underlying ray must be open. For $p<1$, the probability that its first $N$ vertices are all open is $p^N\to 0$. At $p=1$ the ray is open. Consequently

Use the incoming-edge coupling of site and bond percolation. Start with independent random variables indicating open bonds, each with probability $q$. For a vertex $v$ of incoming degree $d_v$, let $J_v$ indicate that at least one incoming bond is open. ThenThe indicators $J_v$ are independent: the sets of bonds entering different vertices are disjoint. Using an additional independent coin at each vertex, turn some zeros into ones so that the resulting indicators $S_v$ are independent with common probability $r$. Explicitly, when $J_v=0$, set $S_v=1$ with conditional probability $(r-\mathbb{P}(J_v=1))/(1-\mathbb{P}(J_v=1))$.

Every vertex other than $x$ in an open bond percolation out-cluster has an open incoming bond, and is consequently an open site. On $\{S_x=1\}$, the whole bond out-cluster is therefore contained in the site percolation out-cluster. The event of infinite bond reachability does not depend on bonds entering $x$: delete cycles from any finite open directed path starting at $x$. It is thus independent of $S_x$, including its additional coin. HenceFor every $q>p_H^b$ below $1$, monotonicity and the definition of the survival threshold give ${\theta }_b(q)>0$. Letting $q\downarrow p_H^b$ yields the quantitative boundIn particular, if $p_H^b<1$, the right-hand side is strictly less than $1$. If $p_H^b=1$, no separation from $1$ is implied or generally possible.

Let $C_x^b$ and $C_x^s$ be the out-clusters in directed percolation, with the site percolation cluster empty when $x$ is closed. Write ${\theta }_b(p)={\mathbb{P}}_p(|C_x^b|=\infty )$ and ${\chi }_b(p)={\mathbb{E}}_p|C_x^b|$, and similarly for site percolation. The survival and susceptibility thresholds for directed percolation are the infima of the parameters at which these quantities become positive and infinite, respectively.

We construct a coupling of probability distributions that compares the whole clusters. First condition the site percolation root to be open. Explore its out-cluster, using a fixed order of the finitely many outgoing bonds at each vertex. When a previously untested vertex $v$ is first encountered along a bond $u\to v$ from an already reached vertex, use a fresh Bernoulli random variable of parameter $p$ both for the state of $v$ and for the state of this particular bond. If $v$ is open, add it to the exploration; if it is closed, mark it as tested and never add it. Assign fresh independent random variables of the same parameter to every other bond, including bonds leading to vertices already tested. The choice of the next unassigned bond depends only on previously revealed information. Its assigned coin is therefore independent of that information, so the resulting bond configuration has exactly the product bond percolation law. Parallel bonds cause no difficulty; loops may be discarded.

Every reached site has an open parent bond leading from a previously reached site. Consequently the conditional site percolation cluster is contained in the bond percolation cluster. The connected graph is countable because it is a locally finite graph; taking increasing finite stages of the exploration proves the same containment for infinite clusters. Removing the conditioning on the root givesThus positive site percolation survival implies positive bond percolation survival at the same parameter, and infinite percolation susceptibility for site percolation implies infinite percolation susceptibility for bond percolation. Taking the infima in the definitions proves both comparisons:

We use three precise inputs. The Harris-Kesten theorem says that independent bond percolation on the square lattice has percolation critical probability $p_c=1/2$. Sharpness of the percolation transition says that, for every $p<p_c$, there are $c,C>0$ such that ${\mathbb{P}}_p(0\leftrightarrow \partial [-R,R{]}^2)\le C{e}^{-cR}$ for all positive integers $R$. Finally, the BK inequality says that increasing connection events with disjoint bond witnesses have joint probability at most the product of their individual probabilities. All three statements concern the independent product law used here.

The sharpness bound implies finite percolation susceptibility. Indeed, with $\tau (u,v)={\mathbb{P}}_p(u\leftrightarrow v)$, summing over square shells givesA radius bound alone would give only a square-root exponential volume bound. To obtain the requested exponential volume bound, we prove the tree-graph moment bound for a percolation cluster:where $(-1)!!=1$.

Expand the left-hand side as a sum over ordered marked vertices $x_1,\dots ,x_k$ of the probability that $0,x_1,\dots ,x_k$ belong to one open cluster. Whenever that event occurs, a finite open tree connects the marked vertices. Delete unmarked leaves and suppress unmarked degree-two vertices. Treat a marked internal vertex as a leaf attached by a zero-length connection, and split any higher-degree branch into degree-three branch points, again with zero-length connections. This gives an abstract binary tree with $k+1$ labelled leaves, $k-1$ branch points and $2k-1$ connections. Its nonempty connecting paths use disjoint sets of open bonds. Coincident marked vertices are allowed: their zero-length connections have empty witnesses.

There are $(2k-3)!!$ such binary tree topologies. For example, the topology with two leaves is unique; inserting the next labelled leaf subdivides one of the $2k-1$ edges of a topology with $k+1$ leaves, giving the usual double-factorial recursion. For a fixed topology and fixed positions of all its vertices, the BK inequality bounds the disjoint connection event by the product of the corresponding $\tau$ factors. Sum over all $x_i$ and all branch-point positions. Keep the root fixed, and successively sum a leaf away. Translation invariance gives ${\sum }_y\tau (z,y)=\chi$ at every step. The resulting sum is ${\chi }^{2k-1}$. Summing the topologies proves the bound.

Since $(2k-3)!!\le 2^{k-1}(k-1)!$, expansion of the exponential, justified by monotone convergence, givesTake $t=1/(4{\chi }^2)$ and write $M=1+\log 2/(2\chi )$. Markov's inequality yields ${\mathbb{P}}_p(|C_0|\ge n)\le M{e}^{-tn}$. To remove the prefactor, choose an integer $N\ge 2$ with $tN\ge 2\log M$. For $n\ge N$ the bound is at most $e^{-tn/2}$. For $2\le n\le N$, useFor $0<p<1/2$, setting $a=\min \{t/2,-\log u/N\}>0$ therefore proves the bound with unit prefactor:At $p=0$, $|C_0|=1$ and any positive $a$ 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

For bond percolation, let $C(0)$ be the percolation cluster of the origin. Its percolation probability isWe prove $\theta (1/2)=0$ by alternating arms at the self-dual parameter, using the uniqueness of the infinite percolation cluster proved in Question 4. Suppose instead that $\theta (1/2)>0$.

At $p=1/2$, declaring a dual edge open exactly when its crossing primal edge is closed produces independent dual bond percolation with the same parameter on ${\mathbb{Z}}^2+(1/2,1/2)$. It therefore also has an infinite percolation cluster almost surely, and at most one. The two preliminary arguments below imply that, for a sufficiently large square, the probability of two primal infinite exterior arms starting through its top and bottom sides exceeds $3/4$. Applying the same arguments to the dual lattice gives probability greater than $3/4$ for dual infinite exterior arms starting through the left and right sides. Thus all four arms occur together with positive probability, by the union bound.

Here are precise boundary conventions for the alternating exterior-arm separation lemma. Use the primal graph vertex box $[-N,N{]}^2$, the dual graph vertex box $[-N+1/2,N-1/2{]}^2$, and the common continuous square with sides at $\pm (N+1/4)$. The primal arms first cross its top and bottom; the dual arms first cross its left and right. Subsequent graph vertices of every arm lie outside this continuous square. Large $N$ gives the required probabilities for both boxes, by the same rotational symmetry and boundary-side estimate. Primal open paths and dual open paths cannot cross, since the two states of a crossing edge are complementary.

The two primal arms cannot meet in the exterior. If they did, a finite simple primal crosscut in the exterior would join the top and bottom entry points. Together with one of the two arcs of the square boundary it would enclose a bounded region containing one of the dual entry points. That dual exterior arm could not reach infinity without crossing the primal crosscut or the forbidden square interior. This is the alternating exterior-arm separation lemma. Thus the two primal arms are disjoint.

Now modify only the primal edges with both endpoints in $[-N,N{]}^2$, making them open. Their grid supplies a simple interior path joining the two primal arm roots. The unchanged exterior arms and this connector form a proper doubly infinite simple primal path. Such a path separates the plane into two sides, by the Jordan curve theorem applied to its one-point compactification in the sphere. The left and right dual arms lie on opposite sides. Preserve their tails from their first exterior dual graph vertices; none of those tails crosses a modified edge. Their original boundary attachment edges may be destroyed, but the infinite exterior tails themselves remain. They now belong to two distinct infinite dual clusters, since any dual connection between them would have to cross the doubly infinite primal path.

This modification has positive probability by the finite-energy property of Bernoulli percolation: condition on the exterior configuration and open the finitely many interior edges. The exterior projections of the four-arm event have positive probability. The resulting positive-probability event with two infinite dual clusters contradicts uniqueness of the infinite percolation cluster on the dual lattice. Therefore

Only the selected paths are drawn. The black connector lies in the finitely modified box; the two blue infinite dual tails remain outside and on opposite sides.

The proof uses self-duality, symmetry, positive association, finite-energy property of Bernoulli percolation and uniqueness; it does not assume the Harris-Kesten theorem as a black box. The boundary language means an infinite ray whose first graph vertex is on the side and all later graph vertices are outside the graph vertex box. It cannot mean literal intersection of two disjoint sets, a closed box boundary and its strict complement.

At $p=1/2$, explore the directed out-cluster of the origin by a deterministic breadth-first procedure. Maintain the set of reached graph vertices and a queue of untested edges with one reached and one unreached endpoint. For each such edge, reveal its orientation. Add the unreached endpoint if the arrow points from the reached endpoint to it; otherwise discard that edge. edges whose endpoints are both already reached need no further test.

An untested edge has an independent fair orientation even though the choice of the next edge depends on earlier discoveries. Its probability of pointing toward the unreached endpoint is $1/2$, whichever geometric direction that represents. Thus the discovery procedure has exactly the same transition probabilities as the ordinary bond percolation cluster exploration at density $1/2$. In particular, for every $m$, the probability of reaching at least $m$ graph vertices agrees in the two models. The uniform random orientation out-cluster comparison impliesA locally finite graph has an infinite directed simple ray from the origin exactly when its directed out-cluster is infinite: the exploration predecessor edges form a finitely branching rooted tree, to which the König infinity lemma applies. Conversely, such a ray visits infinitely many reachable graph vertices. HenceFairness of orientations is essential to this comparison; for general $p$ the success probability depends on the geometric direction of exploration. As usual, an infinite directed path means a ray with distinct graph vertices. Allowing a finite directed cycle to be traversed repeatedly would define an infinite walk instead and would make the claimed conclusion false.

The Burton-Keane theorem says that for independent bond percolation on the hypercubic lattice ${\mathbb{Z}}^d$, $d\ge 2$, at every fixed $p\in [0,1]$, there is almost surely at most one infinite percolation cluster. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.

Let $N_{\infty }$ be the number of infinite percolation clusters. This random variable is translation invariant. By translation ergodicity of Bernoulli percolation, it is almost surely a constant $k\in \{0,1,2,\dots ,\infty \}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of edges, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its probability equals its square, hence is zero or one.

The other probabilistic input is the finite-energy property of Bernoulli percolation. Conditionally on all edges outside a finite set, every prescribed inside configuration has positive probability, namely a product of powers of $p$ and $1-p$. Thus an event of positive probability can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.

First exclude a finite $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Open all edges in and incident to that box. These clusters merge. Opening finitely many edges cannot create an infinite percolation cluster solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of infinite percolation clusters decreases with positive probability, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.

Now suppose $k=\infty$. With positive probability a finite box meets at least three distinct infinite percolation clusters. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary edges by a minimal tree joining their three attachment graph vertices, and close all unused edges there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch graph vertex $v$ with exactly three incident open edges, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a graph vertex is a trifurcation vertex in percolation. A finite box has only finitely many candidate locations and modifications; therefore some fixed graph vertex has positive trifurcation probability. By translation invariance, that probability is a common number $\tau >0$ at every graph vertex.

The decisive deterministic input is the trifurcation boundary-counting lemma. Let $T$ be all these degree-three trifurcation vertices in percolation in a finite box $B_n$. In each infinite percolation cluster meeting $T$, delete the graph vertices of $T$, contract each remaining connected component of a graph adjacent to $T$ to a node, and retain the incidences with the deleted graph vertices, including direct edges between graph vertices of $T$. This is a finite tree: a cycle would connect two neighbours of an original trifurcation vertex in percolation without passing through it, contradicting its three-way separation. Every deleted graph vertex has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent trifurcation vertex in percolation. The elementary tree degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.

Each such infinite component reaches the outer graph vertex boundary ${\partial }^{+}{B}_n$, and different components reach different boundary graph vertices. Summing over the clusters yields, in particular,For a nearest-neighbour lattice box,Taking expectations and using the common trifurcation probability givesThis contradicts $\tau >0$ and excludes infinitely many infinite percolation clusters. Thus only $k=0$ or $k=1$ remains:The proof explains the role of the amenable boundary growth of lattice boxes: finite-energy property of Bernoulli percolation would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.

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