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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

We first specify the standard hypotheses of percolation uniqueness on an amenable quasi-transitive graph. A locally finite graph has finite degree at every vertex; a quasi-transitive graph, or graph of finite type, has finitely many vertex orbits under the acting group $\Gamma$ of graph automorphisms. Here the site percolation law is invariant under $\Gamma$ and independent, with one parameter $p_j$ for each orbit. Initially suppose $0<p_j<1$. This gives the finite-energy property of Bernoulli percolation: conditional on all states outside any finite set, every assignment inside that set has positive probability. An amenable graph has finite sets $W_n$ with $|{\partial }^{+}{W}_n|/|W_n|\to 0$, where ${\partial }^{+}{W}_n$ denotes the external vertex boundary. Finite type bounds the degrees uniformly, so internal and external boundary conventions are equivalent here.

Let $N$ be the number of infinite percolation clusters, taking values in $\{0,1,2,\dots ,\infty \}$. Each event $\{N=j\}$ is invariant. Part (i), together with countability of these possibilities, shows that $N$ is an almost-sure constant.

Suppose first that this constant is a finite integer $j\ge 2$. Some finite connected set $F$ meets all $j$ infinite percolation clusters with positive probability, since increasing finite sets exhaust the graph. Make every vertex of $F$ open. This joins all the infinite clusters. Opening a finite set cannot produce a new infinite cluster from finite clusters: only finitely many clusters meet its finite boundary, and their finite union is finite. By the finite-energy property the modified event has positive probability, contradicting the almost-sure value $j$. Thus a finite $N$ must be $0$ or $1$.

Suppose instead that $N=\infty$ almost surely. Some finite connected set $F$ meets three distinct infinite percolation clusters with positive probability. Open all of $F$. Its vertices belong to one open cluster, and deleting $F$ from that cluster leaves at least three infinite components. This is an encounter box in percolation. We use boxes rather than assume that the local geometry permits a single trifurcation vertex in percolation.

Here is the boundary-counting contradiction in detail. Translates of $F$ have the same positive encounter probability. Finite type and bounded degree allow a deterministic packing of mutually disjoint such translates whose centers have positive lower density in the interiors of the Følner sets $W_n$: every vertex is within a fixed distance of one chosen orbit, and a bounded-radius greedy packing discards only a bounded number of possible centers per chosen center. Boundary layers of any fixed thickness have size $o(|W_n|)$. The expected number of encounter boxes wholly inside $W_n$ is consequently at least $c|W_n|$ for a fixed $c>0$ and all large $n$.

On the other hand, contract the open encounter boxes in each cluster to vertices and retain the part inside $W_n$. Each contracted encounter box separates at least three branches that run to infinity, and each branch must leave $W_n$. Choose a minimal forest connecting all these boundary exits. Every encounter junction has degree at least three in that forest: its three infinite branches remain in different components after deleting the junction, so none can be omitted. The leaves are boundary exits. For each nontrivial tree, the degree-sum identity gives at most the number of leaves minus two vertices of degree at least three. Thus the number of encounter boxes is at most a constant times $|{\partial }^{+}{W}_n|$, where the constant accounts for the bounded boundary degrees. Taking expected values contradicts amenability. This excludes $N=\infty$ and provesThere is a genuine endpoint qualification if different orbit parameters are allowed to be $0$ or $1$. Subdivide each horizontal edge of the square lattice once. Open every original vertex deterministically and close every new midpoint deterministically. This is an invariant independent law on an amenable graph that is also a quasi-transitive graph, but its infinite open clusters are the distinct vertical columns. Thus the finite-energy property must be included when interpreting a general vector $\mathbf{p}$. The homogeneous endpoint laws themselves cause no problem: they give zero or one infinite cluster.

The conclusion is false for $k$-independent laws, even with all site marginals equal to $1/2$. Here is a fully specified monochromatic line-graph percolation counterexample. Form a graph $B$ by replacing each vertex of ${\mathbb{Z}}^2$ by a complete graph on ten vertices and putting every possible edge between the blocks of two neighboring lattice vertices. Every vertex of $B$ has degree $9+4\cdot 10=49$. The graph is connected, periodic and amenable: a square of blocks has volume of order $n^2$ and boundary of order $n$. Assign independent fair red/blue colors to the vertices of $B$.

Call a block good when it contains both colors. Good-block indicators are independent, withThey percolate. To verify this without presupposing a site threshold, put $q=1-g=1/512$. A finite nearest-neighbor good-block cluster around a good origin is surrounded by a circuit of bad blocks in the matching square graph, where horizontal, vertical and diagonal neighbors are allowed. A surrounding circuit of length $\ell$ has an anchor within $[-\ell ,\ell {]}^2$. Overcounting its choices by $9{\ell }^2{8}^{\ell }$ gives the Peierls argumentThe sum bound follows directly from ${\sum }_{\ell \ge 1}{\ell }^2{r}^{\ell }=r(1+r)/(1-r{)}^3$, with $r=1/64$. An infinite good-block cluster therefore exists with positive probability, and by part (i) it exists almost surely. Inside each of its blocks the red vertices are connected to one another, as are the blue vertices; between adjacent good blocks the complete bipartite joining edges connect each color. Hence $B$ has both an infinite red component and an infinite blue component almost surely.

Now use site percolation on $\Gamma =L(B)$, the line graph of $B$: its vertices are the edges $\{u,v\}$ of $B$, two adjacent when they share an endpoint. Declare such a site open exactly when $u,v$ have the same color. Each site has open probability $1/2$. Sets of sites at graph distance greater than one involve disjoint sets of underlying color variables, so this is a translation-invariant $1$-independent law. The line graph is again connected, locally finite, amenable and of finite type. The infinite red and blue components of $B$ yield infinite open clusters in $L(B)$. An open path in $L(B)$ cannot change color at a shared endpoint, so those two clusters are distinct. Thus $N\ge 2$ almost surely. Local color constraints destroy the finite-energy property, precisely the ingredient missing from the uniqueness proof.

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

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.

Give each nearest-neighbor edge of the cubic lattice an independent Bernoulli distribution state, open with probability $p$. The resulting product measure is denoted ${\mathbb{P}}_p$. In this bond percolation model, $C(0)$ is the percolation cluster of the origin, andThe uniform-label monotone coupling of Bernoulli percolation shows that ${\theta }_d$ is increasing.

Let $c_n(d)$ count $n$-step self-avoiding walks starting at the origin, with $c_0(d)=1$. Splitting a walk after $n$ steps, and discarding the avoidance constraint between the two pieces, gives $c_{n+m}(d)\le c_n(d)c_m(d)$. The Fekete lemma therefore gives the connective constantThe locally finite graph structure means that an infinite percolation cluster at the origin supplies an open self-avoiding walk of every length. Each specified walk has $n$ distinct edges and is open with probability $p^n$. The union bound givesFor $p\mu (d)<1$ the right side tends to zero. Hence the connective-constant lower bound for percolation is $p_c(d)\ge \mu (d{)}^{-1}$.

For the upper bound first work on the square lattice. A finite open percolation cluster has an outer boundary containing a simple closed graph cycle of dual edges, all crossing closed primal edges. Such a dual bond percolation circuit separates that cluster from infinity. Write $N_{\ell }$ for the number of simple dual circuits of length $\ell$ surrounding the origin. Each circuit meets the positive horizontal ray at distance at most $\ell$: its bounding box contains the origin and its diameter is bounded by its length. Choose such an intersection as an anchor and orient the circuit. Removing its last edge leaves a rooted self-avoiding walk of length $\ell -1$ in the translated square lattice. Consequently, for an absolute constant $K$,The exact constant and this possible overcount do not matter. If $(1-p)\mu (2)<1$, the root test givesA summable circuit count alone need not give a total sum below one. To use its tail correctly, let $B_R=[-R,R{]}^2\cap {\mathbb{Z}}^2$ and condition every edge internal to $B_R$ to be open. This finite event has positive probability. A closed dual circuit surrounding all of $B_R$ crosses no internal edge of $B_R$, and so its closed-edge probability remains $(1-p{)}^{\ell }$ under this conditioning. Its length tends to infinity with $R$. Choose $R$ so large that the union bound for all such circuits is below one. With positive conditional probability, none occurs.

On that event, $C(0)$ contains $B_R$ and cannot be finite: a finite cluster containing $B_R$ would have an enclosing closed dual circuit. Thus ${\theta }_2(p)>0$ whenever $p>1-\mu (2{)}^{-1}$. This is the connective-constant Peierls bound, proved by excluding short circuits through the open-box conditioning. An embedded coordinate plane in the cubic lattice has exactly the same bond percolation law as the square lattice, so $p_c(d)\le p_c(2)$. Together,There are $2d$ choices for the first step of a self-avoiding walk and at most $2d-1$ thereafter, because immediate reversal is forbidden. Hence $c_n(d)\le 2d(2d-1{)}^{n-1}$ and $\mu (d)\le 2d-1$. In particular $\mu (2)\le 3$. Substituting with the correct directions of the inequalities gives

The purpose of the RSW lemma is to turn local crossing information into control at every fixed shape and scale. Work with independent bond percolation on the square lattice. Let $H(a,b)$ be the event of an open left-to-right graph path in $[0,a]\times [0,b]\cap {\mathbb{Z}}^2$, and write $h_p(a,b)={\mathbb{P}}_p(H(a,b))$.

A useful precise uniform version of the RSW lemma is the following uniform RSW crossing estimate. If for a fixed parameter $p$ and some $\delta >0$,then for every fixed aspect ratio $\rho >1$ there is $c(\delta ,\rho )>0$, independent of $n$, such thatIn particular this bounds length $2n$ in terms of the uniform square bound at length $n$. The constants need not be sharp. The hypotheses used in the proof are planarity, translation and reflection symmetries, the Harris-FKG inequality and independence of unexplored edges. At a self-dual parameter the same argument for dual crossings gives an upper bound strictly below one as well. Neither an exact value of $p_c$ nor the existence of an infinite cluster is an assumption of this estimate.

Here is the gluing picture behind the RSW lemma. For increasing crossing or attachment events, the Harris-FKG inequality provides a lower bound on their joint probability. If an event of probability at least $\delta$ is a union of two reflection-related increasing alternatives $A_1,A_2$, the square-root trick for positively associated events givesIndeed their decreasing complements are positively associated too, so ${\mathbb{P}}_p(A_1^c\cap A_2^c)\ge {\mathbb{P}}_p(A_1^c{)}^2$. This prevents a square crossing from concentrating all its useful attachment locations on just one side.

The nontrivial first gluing step enlarges a square to aspect ratio $3/2$. Explore an extremal square crossing, revealing the edges on the explored side but leaving its other side unexamined. In that unexplored region the conditional law is still independent bond percolation. Compare possible attachments to the crossing with their reflected alternatives. Reflection symmetry and the preceding square-root estimate give a positive bound, depending only on $\delta$, for the required attachment after averaging over the explored crossing. Carry out the reflected construction at the other end and use the Harris-FKG inequality to combine the increasing attachment events with the original crossing. Planarity ensures that the relevant transverse paths actually meet. This RSW reflection extension lemma yields a lower bound $u(\delta )>0$ for the $3n/2$-by-$n$ crossing. The exploration is important: a reflection compares the laws of fresh configurations; reflecting the picture of an open path does not make the reflected edges open.

From that first extension, longer rectangles are obtained by a genuinely transverse gluing. Two $3n/2$-by-$n$ rectangles shifted by $n/2$ overlap in an $n$-by-$n$ square. Require a horizontal crossing in each long rectangle and a vertical crossing of the overlap square. Each horizontal crossing crosses that overlap from left to right, and therefore meets its vertical crossing. Their union crosses the $2n$-by-$n$ rectangle. The Harris-FKG inequality gives a lower bound $u(\delta {)}^2\delta$. Repeat a bounded number of times for any fixed $\rho$. Integer rounding uses neighboring lattice rectangles and bounded additional gluing steps; finitely many smallest scales can be absorbed into the constant. The constants deteriorate with $\rho$ but not with $n$. Merely requiring horizontal crossings in adjacent squares would not suffice, since their endpoints need not coincide; the overlap crossing solves that problem.

I apply the RSW lemma to the exact threshold for bond percolation on the square lattice. The planar duality for rectangle crossings says that an open horizontal crossing and a closed dual vertical crossing are complementary. At $p=1/2$, the dual edge states have the same law as the primal ones. For the balanced lattice rectangle with side lengths $n+1,n$, the rotated dual crossing rectangle has those same side lengths: its transverse lengths before rotation are $n,n+1$. Thus the exact self-dual rectangle crossing probability is $1/2$. Restricting a crossing of this rectangle to its first visit to the shorter vertical side givesThe one-unit balance avoids assuming an exact one-half probability for every finite vertex-square convention. The RSW lemma now gives scale-independent positive bounds for all fixed-aspect-ratio primal and closed dual rectangle crossings.

Arrange four appropriately overlapping long rectangles around a square ring. Closed dual crossings along the four sides, with transverse overlap crossings if needed, join to a closed dual graph cycle surrounding the inner square. The Harris-FKG inequality, applied to the closed dual states, and the RSW lemma give a constant $b>0$ for this circuit event, uniformly over ring size. Choose disjoint rings with radii increasing, for example, by a factor of four. Their circuit events depend on disjoint edge sets, so they are independent. An open graph path from the origin to infinity would have to avoid every one of these dual barriers. Its probability is at most $(1-b{)}^j$ after $j$ rings and therefore zero. HenceThis is the independent annular barriers for percolation argument. It is worth separating it from the other inequality: the absence of an infinite cluster at one parameter alone does not prove that every larger parameter percolates.

For the reverse inequality use sharpness of the percolation transition: below $p_c$, independent bond percolation on the cubic lattice has exponentially decaying connection probabilities. One can see why this is the relevant general ingredient through the finite-set criterion for percolation sharpness. For a finite set $S$ containing the origin putIf ${\phi }_p(S)<1$, split a long open self-avoiding walk at its first exit from $S$. Its internal connection, exit edge and subsequent connection have disjoint witnesses. The BK inequality gives a contraction by ${\phi }_p(S)$ each time distance decreases by the diameter of $S$ plus a fixed step. Iteration proves exponential decay of subcritical percolation at such $p$.

To identify this finite-set threshold with $p_c$, the Margulis–Russo formula expresses the derivative of $g_n(p)$ as the sum of pivotal-edge probabilities. Explore the cluster attached to the box boundary, and let $S$ be its complement. On failure of the origin-to-boundary event, $0\in S$; all edges from $S$ to the boundary cluster are closed, while the internal edges of $S$ remain fresh. A boundary edge is pivotal precisely when its endpoint in $S$ is connected to the origin inside $S$. Removing the factor $1-p$ for a closed pivotal edge yields the general differential inequalityFor $p>{\tilde{p}}_c$ every finite-set quantity on the right is at least one. Integrating from any $p_0\in ({\tilde{p}}_c,p)$ gives a positive lower bound for $g_n(p)$ independent of $n$, and taking $n\to \infty$ gives $\theta (p)>0$. Combined with the contraction below ${\tilde{p}}_c$, this proves $p_c={\tilde{p}}_c$ and the stated sharpness conclusion. This outline supplies the extra threshold argument rather than assuming the desired critical value.

If $p_c(2)>1/2$, sharpness would give $g_n(1/2)\le C{e}^{-cn}$. But a square crossing starting somewhere on the left side entails an open connection from that starting graph vertex to distance $n$. A union bound over the $n+1$ possible starting graph vertices givescontradicting exponential decay. Therefore $p_c(2)\le 1/2$, and the two directions establishThis is the Harris-Kesten theorem. The overall mechanism is that self-duality supplies a square crossing, the RSW lemma transports it between shapes and creates barriers at every scale, and sharpness converts the finite-scale crossing information into the exact threshold.

The union bound gives $g_k\le {\sum }_{x\in \partial {\Lambda }_k}{\mathbb{P}}_p(0\leftrightarrow x)$. Hence some boundary graph vertex has percolation two-point connection probability at least $g_k/(8k)$. The rotations and reflections of the square lattice let us choose such a graph vertex as $x=(k,j)$, $-k\le j\le k$. Reflection in the vertical line through $x$ sends $0$ to $e_{2k}=(2k,0)$ and fixes $x$. ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from $0$ to $e_{2k}$ reaches $\partial {\Lambda }_{2k}$, so $h_{2k}\le g_{2k}$. Taking roots givesBoth outside expressions have limit of a sequence $\gamma$, proving the even case. For $p>0$, the Harris-FKG inequality with the last horizontal edge gives $h_{2k+1}\ge p{h}_{2k}$, and also $h_{2k+1}\le g_{2k+1}$. The lower bound has rootThe upper bound has the same limit of a sequence. At $p=0$ all positive-distance connection probabilities vanish. Thus ${\lim }_{n\to \infty }{h}_n^{1/n}=\gamma$ for every $p\in [0,1]$.

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.

Write $H(R)$ and $V(R)$ for the horizontal and vertical open-crossing increasing events of a rectangle in the rotated square lattice. At parameter $1/2$, planar duality for rectangle crossings says that failure of $H$ is exactly a closed dual vertical crossing. Quarter-turn symmetry and the boundary convention of the rotated rectangles identify the two alternatives on a square. Thus its crossing probability is $1/2$; a rectangle obtained by a harmless one-step boundary adjustment has crossing probability at least $1/2$. This is the finite-domain exact self-dual rectangle crossing probability, not an assumption about infinite percolation clusters.

We first prove a RSW reflection extension lemma. In coordinate units adapted to the rotated drawing, take $S=[0,n{]}^2$, $T=[-n,n{]}^2$, and $R=[-n,2n]\times [-n,n]$. Reveal the rightmost open vertical crossing $\Gamma$ of $S$, when one exists, using only its edges and the region to its right. Reflect $\Gamma$ across the horizontal axis. The part $D$ of $T$ to the left of these two paths remains unexposed and has the independent parameter-$1/2$ law. A horizontal crossing of $T$ must reach their union. Its probability is at least $1/2$. Reflection symmetry and the union bound give a probability at least $1/4$ of reaching the upper path $\Gamma$ through $D$. Conditional opening of the edges on $\Gamma$ can only help. Since $V(S)$ has probability at least $1/2$, the increasing event $B$ that a vertical crossing of $S$ is connected inside $T$ to its left side has probability at least $1/8$.

Reflect this construction in the vertical midline of $S$ to obtain an increasing event $B^{\prime }$ of the same probability, connecting a vertical crossing of $S$ to the right side of $[0,2n]\times [-n,n]$. On $H(S)\cap B\cap B^{\prime }$, the horizontal crossing of $S$ meets both vertical crossings, so all three graph paths join and give $H(R)$. The Harris-FKG inequality now yieldsThe exploration only conditions the exposed side of the extremal graph path; reflecting a drawn path does not assert that its reflected edges are open. This distinction is what makes the extension argument valid.

To pass from aspect ratio $3/2$ to $3$, put four rectangles of width $3h/2$ and height $h$ next to each other, with successive left boundaries separated by $h/2$. Adjacent rectangles overlap in a square of side $h$. Require horizontal crossings of the four rectangles and vertical crossings of the three overlap squares. Every consecutive pair of horizontal crossings meets the vertical crossing in its overlap; together they form a horizontal crossing of width $3h$. The Harris-FKG inequality gives a strictly positive bound, for these compatible scales, ofThe construction works on the rotated square lattice with the boundary sites used in the source drawing. Integer roundings do not create a scale restriction: for an odd height use the largest smaller compatible even height, and use eight horizontal rectangles and seven overlap squares, whose union has auxiliary aspect ratio $5$, so that its width exceeds the desired $3m$ width for all sufficiently large $m$. Truncate its horizontal crossing on first reaching the desired right side. The number of extra rectangles and overlap-square crossings is bounded independently of $m$, and every factor is bounded below by the same positive constants. The finitely many smallest rectangles each have positive crossing probability, since a specified finite open graph path suffices. Therefore a constant $c>0$ works at every integer scale. Taking $\tau =c/2$ ensures the requested strict inequality:This proves the required instance of the Russo-Seymour-Welsh theorem through reflection, exploration and gluing. The parameter in this calculation is $1/2$.

Use dimensionless interaction couplings in the exponent of the Boltzmann weight. In the following formulas $K,L$ denote the physical interaction energies divided by $k_{B}T$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.

Split the original square lattice into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained Ising spins by $s_e$ and eliminated Ising spins by ${\eta }_o$. For fixed retained Ising spins defineThe blocked Boltzmann weight in the checkerboard decimation of the square-lattice Ising model isUse the independent uniform product measure on Ising spins on $O$ to perform a cumulant expansion. Its one-spin mean vanishes, and $⟨{\eta }_o{\eta }_{o^{\prime }}{⟩}_0={\delta }_{o{o}^{\prime }}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.

Each eliminated site has four retained neighbours, soIts six distinct neighbour pairs comprise four corner pairs at distance $\sqrt{2}a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives spin inversion symmetry, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.

The primitive vectors of the retained square lattice are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt{2}$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt{2}$, the leading checkerboard Ising decimation recursion is thereforeThe constant part of the weight is also determined: $4(K^2/2)(N/2)=N{K}^2$, so the original partition function equalsto the same order in the logarithm of the weights. That normalization changes free energy but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed ${\beta }_T$ the relations read $K_{\mathrm{energy}}^{\prime }=L_{\mathrm{energy}}+2{\beta }_T{K}_{\mathrm{energy}}^2$ and $L_{\mathrm{energy}}^{\prime }={\beta }_T{K}_{\mathrm{energy}}^2$.

The fixed-point equations are $L_{\ast }=K_{\ast }^2$ and $K_{\ast }=3K_{\ast }^2$. Their only finite real solutions areThe origin is the high-temperature fixed point; its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete renormalization-group transformation there:The characteristic polynomial is ${\lambda }^2-(4/3)\lambda -2/3$, givingEigenvectors may be chosen as $(1,2/(3{\lambda }_{\pm }){)}^T$. The positive eigenvalue greater than one is the relevant direction of a fixed point; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function eigenvalues.

Let $t$ be the relevant thermal scaling coordinate, so $t^{\prime }={\lambda }_{+}t$ to first order. A correlation length measured in restored lattice units satisfies $\xi (t^{\prime })=\xi (t)/b$. If $\xi (t)\sim |t{|}^{-\nu }$, then ${\lambda }_{+}^{-\nu }=b^{-1}$. Hence the thermal exponent from a discrete renormalization map isThis is the value predicted by the stated low-order real-space renormalization group, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice Ising model.

The strips are nested. Every connection permitted inside $T_k$ is also permitted inside $T_{k+1}$, so $q_k(n)\le q_{k+1}(n)$ for every fixed separation. ConsequentlyThe preceding nonnegative bound makes this a decreasing sequence bounded below. By monotone convergence of real sequences,There is no assertion that this limit of a sequence is strictly positive: positivity at each fixed width need not survive an increasing-width limit of a sequence.

One can also identify the limit of a sequence. Let $q(n)$ be the percolation two-point connection probability in the whole square lattice. Every finite connecting graph path has a bounded vertical extent, so $q_k(n)\uparrow q(n)$. Commuting infima gives the strip approximation to the planar connection decay rate:The same positive-association argument identifies the final infimum with the whole-plane normalized logarithmic limit of a sequence. This is an interchange of infima justified by monotonicity at fixed $n$. For example, when $p>1/2$, the threshold proved in question 2 and uniqueness give an infinite percolation cluster with root probability $\theta (p)>0$. The Harris-FKG inequality makes the probability that both endpoints belong to it at least $\theta (p{)}^2$, so $q(n)\ge \theta (p{)}^2$. The whole-plane rate, and hence ${\lim }_k{f}_k(p)$, is then zero, despite the strict positivity of every finite-strip rate.