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Color the vertices of a periodic graph independently red or blue with equal probabilities, and declare a vertex of its line graph open exactly when the corresponding edge has equal-colored endpoints. The resulting site percolation has marginal $1/2$ and is a finite-range dependent random field: sets at line-graph distance greater than one use disjoint color inputs. Red and blue open clusters cannot join. If the original graph has an infinite red component and an infinite blue component, its line graph consequently has at least two infinite percolation clusters. One explicit amenable graph with that property is obtained by replacing each square-lattice vertex by a complete graph on ten vertices, joining neighboring blocks completely. Blocks containing both colors have independent probability $511/512$; a Peierls argument proves that they percolate, and their infinite cluster carries both a red and a blue component. Thus finite-range dependence, even with all site marginals nondegenerate, does not substitute for finite energy in percolation uniqueness.

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.

Assume short-range ferromagnetic exchange $J>0$ and zero external field. In the one-dimensional Ising model, a domain wall costs only $2J$, independent of the lengths of the domains it separates. There are order-$N$ positions for one wall, so its positional entropy offsets any finite energy cost at nonzero temperature as the system grows. Domain walls consequently have a positive density at every $T>0$, destroying a uniform spontaneous spin magnetization. Periodic boundaries require an even number of walls, but that constraint does not remove their nonzero thermodynamic-limit density.

The One-dimensional Ising correlation function makes this argument explicit. A bond is broken with probability $p_w=e^{-2{\beta }_{\mathrm{th}}J}/(1+e^{-2{\beta }_{\mathrm{th}}J})$, and the infinite-chain correlation isThe dominant eigenvalue of the transfer matrix for the one-dimensional Ising model is $2\cosh {\beta }_{\mathrm{th}}J$, so the free energy per spin $-k_{B}T\log [2\cosh {\beta }_{\mathrm{th}}J]$ is analytic for $T>0$. There is no finite-temperature transition in one dimension; only the zero-temperature limit orders.

In two dimensions a reversed cluster is enclosed by a Peierls contour. A contour of length $\ell$ costs $2J\ell$, while the number of such contours surrounding a fixed site grows at most as a polynomial in $\ell$ times $3^{\ell }$: after the initial edge, a nonbacktracking square-lattice walk has at most three choices per step. With plus boundary conditions, flipping the interior of a specified contour yields the Boltzmann suppression $e^{-2{\beta }_{\mathrm{th}}J\ell }$. ThereforeFor sufficiently low temperature this bound is below $1/2$, giving a positive spontaneous magnetization. This is the energy-entropy mechanism of the Peierls argument.

At sufficiently high temperature, the graphical expansion bounds correlations by sums of connecting paths weighted by $(\tanh {\beta }_{\mathrm{th}}J{)}^{\ell }$. The same nonbacktracking count converges geometrically when $3\tanh {\beta }_{\mathrm{th}}J<1$, giving exponential decay and a disordered phase. Since an ordered low-temperature phase and a disordered high-temperature phase both occur, the two-dimensional model has a finite-temperature phase transition. The argument establishes its existence and the dimensional distinction, not the exact value of its critical temperature.

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.

The three-dimensional Peierls argument replaces planar contours by closed dual plaquette surfaces surrounding the negative cluster of the origin. A surface of area $m$ costs $e^{-2\beta m}$, and the number of connected surfaces of area $m$ containing a fixed nearby plaquette is at most $C^m$. HenceThis is below $1/2$ for sufficiently large $\beta$, so $⟨{\sigma }_0{⟩}_{\beta ,0}^{{\mathbb{Z}}^3,+}>0$.

It suffices to use the two-dimensional coordinate half-plane inside $H$. A Peierls argument shows that when $p$ is sufficiently close to one, the probability that a fixed open vertex is separated from distance $n$ by a closed dual contour is summable over contour lengths: the number of length-$m$ contours is at most exponential in $m$, whereas each is closed with probability $(1-p{)}^m$. Hence the origin has positive probability of belonging to an infinite open cluster. The event that some infinite cluster exists is invariant and therefore has probability zero or one because the boundary translation is an ergodic transformation; its positive probability makes it almost sure. The same cluster is also an infinite cluster of $H$ in every $d\ge 2$.