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Finite-range lattice interaction and conditional energy

1 · Prerequisites

2 · Summary

This page defines bounded finite-range interactions on finite-alphabet lattices and distinguishes finite-volume free energies from conditional energies. It develops the finite Gibbs entropy identity, the product measure framework (with its exact Countable Choice assumption), Gibbs specifications and DLR states, and local proofs of existence, stationarity, uniqueness bounds and low-temperature coexistence for specified models. The pressure variational result is one-way: DLR states are variational equilibria, with no converse asserted. Continuum stability terminology is defined separately and does not imply that finite range alone gives stability. Physical interpretations are explicitly adopted model assumptions; all consequences remain conditional, with no empirical measurements or critical point claims.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-range lattice interaction and conditional energy

Definition

Fix an integer dimension d≥1 and a finite nonempty alphabet A. A configuration is a function σ:Zd→A. An interaction list consists of a finite index set R and, for each r∈R, a finite nonempty shape Sr⊂Zd containing 0 and a function Φr:ASr→R. The index r is part of the data: two equal shapes with different indices represent two terms and are both counted. Every translate x+Sr contributes once, with value Φr((σx+y)y∈Sr).

The range is R=max⁡r∈Rdiam⁡∞(Sr), with R=0 when R is empty. For finite Λ⊂Zd, define the free-volume energy and the energy conditioned on an exterior configuration η∈AΛc by HΛfree(σ)=∑r∈R ∑x:x+Sr⊆ΛΦr(σx+Sr),EΛ(τΛ∣η)=∑r∈R ∑x:(x+Sr)∩Λ≠∅Φr((τΛη)x+Sr). Here τΛ∈AΛ, and τΛη∈AZd is the concatenated configuration equal to τΛ on Λ and to η on Λc. These are finite sums: for each finite shape Sr, only finitely many anchors x make x+Sr meet Λ. Terms meeting Λ but not contained in it appear in the conditional energy and not in the free-volume energy. The definitions are mathematical and assign no physical units; a physical model must state its units separately.

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Stability, superstability and regular tails

Definition

Fix d≥1. A pair potential is an even Borel function ϕ:Rd→R∪{+∞}. For a finite list of positions X=(x1,…,xN), set U0=U1=0 and UN(X)=∑1≤i<j≤Nϕ(xi−xj). The value +∞ is allowed and represents a forbidden configuration; −∞ is not allowed.

The potential is stable if some finite B≥0 satisfies UN(X)≥−BN for every N and every X. Fix a cell side ℓ>0 and the half-open cubic tiling Ck=ℓ(k+[0,1)d), k∈Zd, and let nk(X) count the positions of X in Ck. The potential is cell-superstable if there are constants A∗>0 and B∗≥0 such that UN(X)≥A∗∑k∈Zdnk(X)2−B∗N for every finite configuration; the sum is finite for finite X.

For disjoint finite cell-index sets I,J, let WI,J(X) be the sum of pair terms with one point in a cell indexed by I and the other in a cell indexed by J. The potential is lower regular for this tiling if a nonnegative envelope Ψ:Zd∖{0}→[0,∞) exists with ∑k≠0Ψ(k)<∞ and WI,J(X)≥−∑k∈I∑l∈JΨ(k−l)nk(X)nl(X) for all such I,J,X. The ordered-pair convention in this formula is fixed as written.

A decreasing-integrable tail bound means that for some R0≥0 there is a nonnegative nonincreasing function ψ:[R0,∞)→[0,∞) such that ∣ϕ(x)∣≤ψ(∣x∣) for ∣x∣≥R0 and ∫R0∞ψ(r)rd−1 dr<∞. This explicit envelope condition is stronger than bare Lebesgue integrability of an arbitrary signed tail. Stability, cell-superstability, lower regularity, and a tail bound are separate properties; none is inferred from finite range alone.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Classical Gibbs quantities and SI reference conventions

Physical scope

SI units and reference conventions for the specified classical lattice and continuum Gibbs models; no empirical calibration or material-fit claim.

Definition

Use SI units. Let kB>0 be Boltzmann's constant in joules per kelvin, let T>0 be absolute temperature in kelvin, and define β=(kBT)−1 in joules−1. For a finite-alphabet lattice model in dimension d≥1, the lattice spacing a>0 is in metres; a spin is a dimensionless label. In the nearest-neighbor Ising convention HΛ=−JI∑{i,j}⊂Λ, ∣i−j∣1=1sisj−h∑i∈Λsi, the coupling JI and field parameter h are energies in joules, so βJI and βh are dimensionless. Here h is an energy bias per spin, not by itself a magnetic-field strength.

For a finite nonempty region Λ⊂Zd, set HΛ:=HΛfree from Finite-range lattice interaction and conditional energy, with each physical interaction value Φr and total energy measured in joules. Thus the partition below uses free boundary conditions: only interaction supports contained in Λ contribute. Exterior-conditioned energies EΛ(⋅∣η) and their partitions carry a separate exterior label and are not denoted by this HΛ or ZΛ. Let ZΛ=∑σ∈AΛe−βHΛ(σ) and define dimensionless pressure per site pΛ=∣Λ∣−1log⁡ZΛ. When p=lim⁡n→∞∣Bn∣−1log⁡ZBn exists for cubes Bn=[−n,n]d∩Zd, define physical pressure P=kBT p/ad in J m−d (pascals when d=3). For a stationary state μ on a finite alphabet (a probability invariant under every translation (θjσ)x=σx+j), let Hn=−∑σ∈ABnμBn(σ)log⁡μBn(σ) with 0log⁡0=0; when the limit exists, its Shannon entropy density is hSh=lim⁡n→∞Hn/∣Bn∣ in nats per site. The mean energy per site is u=∑r∈REμ[Φr] in joules per site. Define reference spin entropy density s=kBhSh/ad and free-energy density f=u/ad−Ts; both are per physical volume. Thus f and P have units J m−d, while s has units J K−1 m−d; for d=3 these are the usual per-cubic-metre units.

For a continuum model in dimension d, let x,p∈Rd be position and momentum, let mp>0 be particle mass in kilograms, and use kinetic energy ∣p∣2/(2mp). Assume Countable Choice for the Lebesgue change-of-variables interface The Axiom of Countable Choice (ACω). Tonelli's theorem Tonelli's theorem for nonnegative measurable functions on a sigma-finite product and the one-dimensional Gaussian integral The Gaussian integral ∫−∞∞e−x2 dx=π give ∫Rde−∣q∣2 ddq=πd/2. Set c=β/(2mp)>0 and apply the nonnegative change-of-variables theorem A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions to the C1 diffeomorphism q↦cq of Rd and f(q)=e−∣q∣2. Since its Jacobian determinant is cd, ∫Rde−β∣p∣2/(2mp) ddp=c−dπd/2=(2πmpkBT)d/2. Under the phase-space reference measure ddx ddp/hrefd, with reference action href>0 in joule-seconds, the integrated momentum factor is href−d∫Rde−β∣p∣2/(2mp) ddp=(2πmpkBThref)d=λT−d,λT=href2πmpkBT. For chemical potential μ in joules, define the classical activity zc=eβμλT−d, with units m−d. Thus λT is a length and zc is inverse volume. This momentum integration explains the one-particle thermal factor; it does not derive particle indistinguishability or a factorial convention. The choice of href is a reference convention, not a consequence of the classical equations of motion. For continuum interactions, the potential hypotheses and energy meaning are as in Stability, superstability and regular tails.

The parameter status in these formulas is model-relative. T, a, JI, and h are inputs for the chosen lattice model; mp and μ are inputs for the chosen continuum model. Here h is an energy bias per dimensionless spin, not a magnetic-field strength. The reference action href>0 is a primitive normalization convention, not a derived quantity or an assertion that it equals Planck's constant. In the current SI, kB has the exact defining value 1.380649×10−23 J K−1; it is a fixed SI conversion constant, not a measured fit parameter in this model. The inverse temperature β, thermal wavelength λT, activity zc, pressure and entropy densities, energy density u, entropy density s, and free-energy density f are defined or derived from those inputs and the state as specified above; their formulas do not calibrate the inputs to a material or assert empirical accuracy.

PostulateStatement: AI-adaptedProof: Not applicableOpen item page →

Interacting classical Gibbs equilibrium adoption

Physical scope

Adopted equilibrium interpretation for the chosen classical Hamiltonian, temperature and reference measure; no claim of equilibration or empirical accuracy.

Postulate

Fix a classical model with physical energy measured in joules and a positive inverse-temperature parameter β=(kBT)−1 in J−1. Its global configuration space is a measurable space (X,F). For each finite region Λ, specify a local state measurable space (XΛ,FΛ), an admissible exterior measurable space (EΛ,EΛ), a measurable exterior restriction map rΛ:(X,F)→(EΛ,EΛ), and a measurable concatenation map cΛ:XΛ×EΛ→X. Require rΛ(cΛ(x,η))=η for every x∈XΛ and η∈EΛ, and write xη:=cΛ(x,η). Specify a jointly measurable conditional energy HΛ:XΛ×EΛ→R∪{+∞}, where +∞ marks a forbidden local state, and a measurable kernel η↦mΛη of positive σ-finite reference measures on (XΛ,FΛ); σ-finite means that XΛ is a countable union of measurable sets of finite mΛη-measure. The reference normalization and any units carried by its coordinates are part of the model data. For a thermodynamic partition function, choose the reference normalization so ZΛη is dimensionless. The product βHΛ is dimensionless.

For η∈EΛ, define ZΛη:=∫XΛe−βHΛ(x∣η) dmΛη(x),γΛ(C∣η):=(ZΛη)−1∫XΛ1C(xη)e−βHΛ(x∣η) dmΛη(x),C∈F, with e−β(+∞):=0. Adopt these conditional probability kernels only when 0<ZΛη<∞ for every finite Λ and admissible η, and when their dependence on η is measurable. A probability measure μ on (X,F) is a DLR state for this adopted model if, for every finite Λ and every C∈F, μ(C)=∫XγΛ(C∣rΛ(y)) dμ(y). A specification is a family of these probability kernels that is proper when γΛ(rΛ−1(D)∣η)=1D(η) for every D∈EΛ and η∈EΛ, and consistent when γΛγΔf=γΛf for every bounded measurable f:X→R and finite Δ⊆Λ. Here γΛ(f∣η):=∫Xf(y) γΛ(dy∣η), and composition is explicitly (γΛγΔf)(η):=∫XγΔ(f∣rΔ(y)) γΛ(dy∣η). These identities are required when a DLR specification is asserted; normalization alone does not imply them.

In the finite-alphabet lattice instance, let A be finite and nonempty, X=AZd with the cylinder-generated Borel σ-algebra, XΛ=AΛ with its power-set σ-algebra, and EΛ=AΛc. The restriction map is rΛ(σ)=σ∣Λc and concatenation is coordinatewise patching. For the finite-range interaction of Finite-range lattice interaction and conditional energy, use counting measure on AΛ and the energy EΛ(⋅∣η) in joules; the resulting kernels are exactly those in Lattice Gibbs kernels and DLR states. The consistency result Conditional Gibbs kernel consistency supplies their properness and consistency. The cylinder-measure extension in Finite-alphabet product and probability compactness uses the Countable Choice axiom The Axiom of Countable Choice (ACω) to construct an infinite-volume probability; no finite conditional sum spends that choice assumption.

When particle-number sectors are present, use the exact interface of Classical particle counting and physical observables. For species counts Na with total N=∑aNa≥1 in d spatial dimensions, the phase-space definition Classical configuration, cotangent phase and reference measures sets the labelled reference to mN=cNΛN with arbitrary cN>0. This postulate chooses cN=a0−dN, where a0>0 is a prescribed reference action in joule-seconds, and uses the species factor 1/∏aNa! in the normalized quotient reference. Equivalently, the symmetric labelled integral uses (∏aNa! a0dN)−1dq dp; the factorial is included once, and the phase-volume dimension (J s)dN makes this reference dimensionless. For the vacuum sector N=0, define the zero-fold phase space to be a singleton with unit Dirac reference measure (consistent with 0!=1 and a00=1); in a finite region with nonempty exterior, the local vacuum still has the energy HΛ(∅∣η) in its Gibbs weight. If the thermal-wavelength convention of Classical Gibbs quantities and SI reference conventions is used, set href=a0; omit the species factorial for operationally distinguishable labels. These are adopted reference and counting conventions, not consequences of Hamiltonian mechanics or evidence of equilibration. No empirical parameter, measured transition temperature, or uncertainty is asserted here.

This is a model postulate. Normalization, stationarity of a Hamiltonian flow, and finite-volume calculations do not prove equilibration or establish that the model describes nature.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite Gibbs relative entropy and free-energy minimum

Statement

Let S be a finite nonempty set, H:S→R, and β>0. Define Z=∑s∈Se−βH(s) and γ(s)=Z−1e−βH(s). For every probability vector p on S, with 0log⁡0=0, its Shannon entropy HSh(p)=−∑sp(s)log⁡p(s) and relative entropy D(p∥γ)=∑p(s)>0p(s)log⁡(p(s)/γ(s)) satisfy D(p∥γ)=−HSh(p)+β∑sp(s)H(s)+log⁡Z≥0. Equality holds exactly when p=γ. Equivalently, γ is the unique minimizer of ∑sp(s)H(s)−β−1HSh(p), and the minimum is −β−1log⁡Z.

Facts & Assumptions

Given: A finite nonempty state set S, a real-valued energy H, an inverse-temperature parameter β>0, and a probability vector p.

[A1]

Every term e−βH(s) is a positive finite real number; hence 0<Z<∞ and γ is a strictly positive probability vector.

Proof

1.1A1F1algebra

For every r>0, [F1] gives −log⁡r=∫r1dt/t≥1−r when r<1, and −log⁡r=−∫1rdt/t≥1−r when r>1; both comparisons are strict unless r=1. For p(s)>0, put rs=γ(s)/p(s)>0. Then p(s)log⁡(p(s)/γ(s))≥p(s)−γ(s). Summing over the support of p yields D(p∥γ)≥1−∑p(s)>0γ(s)≥0.

2.1A1step 1.1algebra

If D(p∥γ)=0, the two nonnegative gaps in step 1.1 both vanish: every scalar inequality on the support is equality and the omitted γ-mass is zero. The strictness just proved gives γ(s)/p(s)=1 for every p(s)>0, while positivity of γ makes zero omitted mass imply p(s)>0 for every s. Thus p=γ. Conversely p=γ makes every summand zero.

3.1step 1.1step 2.1algebra∎

Expanding log⁡γ(s)=−βH(s)−log⁡Z in the finite sum defining D gives the displayed identity. Rearranging the nonnegativity and equality statement from steps 1.1–2.1 proves the unique-minimum formulation.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite Gibbs response and covariance identities

Statement

Let S be finite nonempty, let H0,M:S→R, fix β>0, and set Ht=H0−tM and Z(t)=∑s∈Se−βHt(s) for t∈R. For the normalized finite Gibbs law γt(s)=Z(t)−1e−βHt(s), ddtlog⁡Z(t)=β γt(M),d2dt2log⁡Z(t)=β2Var⁡γt(M)≥0, where Var⁡γ(M):=γ(M2)−γ(M)2. For fixed H:S→R, the corresponding inverse-temperature identities are ddβlog⁡Z(β)=−γβ(H),d2dβ2log⁡Z(β)=Var⁡γβ(H)≥0. These are finite-state identities; they do not justify differentiating an infinite-volume limit.

Facts & Assumptions

Given: A finite nonempty state set, real-valued observables and energies, and a positive finite parameter β.

[A1]

For finite energies and β>0, the finite Gibbs weights normalize to a strictly positive probability law (Finite Gibbs relative entropy and free-energy minimum).

[F1]

The real exponential is differentiable with derivative exp⁡ (The exponential function is smooth and (exp⁡)′=exp⁡).

[F2]

The natural logarithm is differentiable on (0,∞) with derivative 1/x (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Proof

1.1givenA1F1F2F3

Since S is finite, repeated use of the sum rule and chain rule [F1, F3] differentiates each summand and gives Z′(t)=β∑sM(s)e−βHt(s). Positivity of Z(t) follows from [A1]; the logarithm chain rule and [F2, F3] give (log⁡Z)′=Z′/Z=βγt(M).

2.1A1F1F3step 1.1

Differentiate the finite-sum numerator ∑sM(s)e−βHt(s) and denominator Z(t) by [F1, F3], then apply the quotient rule from [F3]. This gives ddtγt(M)=β(γt(M2)−γt(M)2). Multiplying by the factor β in step 1.1 gives the stated second derivative, which is nonnegative because it equals β2γt((M−γt(M))2).

3.1givenA1F1F2F3∎

For the inverse-temperature family, finite-sum differentiation using [F1, F3] gives Z′(β)=−∑sH(s)e−βH(s) and Z′′(β)=∑sH(s)2e−βH(s). Applying the logarithm chain rule and quotient rule from [F2, F3] yields the stated mean and variance identities.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite-alphabet product and probability compactness

Statement

Assume the Axiom of Countable Choice. Let d≥1, let A be a finite nonempty set, and put X=AZd. Fix a bijection (in)n≥1:N>0→Zd and give X the metric dX(σ,η)=∑n≥12−n1{σin≠ηin}. For a finite F⊆Zd and B⊆AF, write [F,B]={σ∈X:σ∣F∈B} and let C={[F,B]:F⊆Zd finite, B⊆AF} be the cylinder algebra. A family (νF)F⊆Zd finite of probability measures on AF is consistent if (πEF)∗νF=νE whenever E⊆F, where πEF:AF→AE is restriction. Such a family defines μ0([F,B])=νF(B); consistency makes this a well-defined normalized finitely additive probability on C. Then X is compact, the Borel sigma-algebra is σ(C), and μ0 extends uniquely to a probability on σ(C). Moreover, every sequence of probabilities on this sigma-algebra has a subsequence whose cylinder probabilities converge to those of a probability measure; along that subsequence integrals of every continuous f:X→R converge.

Facts & Assumptions

Given: d≥1, a finite nonempty alphabet A, the fixed bijective site enumeration, a consistent family (νF) of finite-dimensional probabilities, and the Axiom of Countable Choice for the extension step.

[A1]

The Axiom of Countable Choice says every sequence of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)).

[F1]

Under Countable Choice, the outer measure induced by a premeasure on an algebra restricts to a measure on the generated sigma-algebra extending that premeasure (Assuming countable choice, a premeasure extends through its induced outer measure).

[F2]

If two probability measures agree on a pi-system, they agree on the sigma-algebra it generates (Dynkin's pi-lambda theorem).

[F3]

Every bounded real sequence has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

Proof

1.1givenconstruct

Enumerate the sites and order the finite alphabet. From any sequence in X, retain an infinite subsequence constant at i1, then a further infinite subsequence constant at i2, and so on. At each stage use the first alphabet value in the fixed order occurring infinitely often in the current sequence, and enumerate its retained indices increasingly. This is a specified recursive rule, not a sequence of arbitrary dependent choices. Taking the diagonal subsequence makes each fixed coordinate eventually constant. For ε>0, choose K with 2−K<ε. Two sufficiently late diagonal terms agree at the first K sites, so their metric distance is at most 2−K<ε and they converge to the coordinatewise limit.

1.2givenalgebra

If [F,B]=[G,D], pass to H=F∪G and use consistency to express both νF(B) and νG(D) as the νH-probability of the same subset of AH; hence μ0([F,B])=μ0([G,D]). To check finite additivity, represent any finite collection of cylinders on the common finite coordinate set H; disjointness and unions then reduce to finite additivity of νH. The cylinder X has mass one, so μ0 is a normalized finitely additive probability.

1.3F3givenconstruct

Write the given probability sequence as (μk)k≥1. The cylinder algebra is countable: there are countably many finite subsets of Zd, and each finite coordinate set has only finitely many subsets of the finite set AF. Enumerate it as C1,C2,… (allow repetitions if it is finite). Specify a convergent-subsequence selector for any sequence (uk) in [0,1] as follows. Start with the closed interval I0=[0,1] and all indices. Bisect the current interval and retain its left closed half if infinitely many current indices have values there; otherwise retain its right closed half. The retained index set is infinite because the two halves cover the previous interval. After stage r, let kr be the least retained index larger than kr−1, with k0=0. The intervals are nested with diameter 2−r, so any two selected values from stages at least r differ by at most 2−r. By [F3] the bounded selected sequence has a subsequential limit; the diameter bound then makes the whole selected sequence converge to that limit. No witness subsequence from [F3] is selected by the rule: each interval, retained set and index is uniquely specified. Starting with a0(k)=k, apply this selector to (μaj−1(k)(Cj))k≥1 and let aj be the resulting increasing subsequence of aj−1. Ordinary deterministic recursion defines all aj without dependent choice. The diagonal indices nj=aj(j) are strictly increasing, since aj+1(j+1)≥aj(j+1)>aj(j); for each fixed cylinder their probabilities converge along this diagonal. Limits of finite sums preserve normalization and finite additivity, hence all finite-coordinate consistency identities.

2.1givenstep 1.1construct

Every cylinder is clopen: if F⊆{i1,…,iK}, then the ball of radius 2−K around any σ∈[F,B] lies in [F,B], and its complement is a finite union of cylinders. Conversely, for any ε>0, choose K with 2−K<ε; the cylinder fixing the first K coordinates of σ lies in the ε-ball about σ. Thus the countable cylinder family is a basis for the metric topology and generates its Borel sigma-algebra. To prove compactness, suppose an open cover has no finite subcover. Starting with X, successively split the current cylinder by the next coordinate into finitely many subcylinders; at least one remains without a finite subcover, and choose the first such cylinder in a fixed order on A. The resulting nested cylinders determine one configuration σ. Some cover member contains σ and, being open, contains a basic cylinder fixing finitely many coordinates of σ. A sufficiently late nested cylinder is contained in that member, contradicting its selection. Thus every open cover has a finite subcover.

3.1A1F1F2step 2.1

If a cylinder C is a disjoint union of countably many cylinders Cn, compactness of C gives a finite subcover from the open cover (Cn). Disjointness forces every other Cn to be empty, so finite additivity gives countable additivity whenever the union is in C; the cylinder weight is therefore a premeasure. Apply [F1] to extend it under the stated Countable Choice assumption. The cylinder family is a pi-system generating σ(C), so [F2] gives uniqueness.

4.1A1F1step 1.1step 2.1step 3.1step 1.3algebra∎

The extension in step 3.1 gives a probability measure with those cylinder values. Local functions are finite linear combinations of cylinder indicators. Every continuous f is uniformly approximated by local functions. If uniform continuity failed, [A1] would choose, for each K, one pair agreeing on the first K coordinates whose f-values differ by at least a fixed ε>0. These are independent nonempty sets of pairs indexed by K, so countable choice suffices. Step 1.1 gives a convergent subsequence of the first components; along the same indices the second components converge to the same configuration, since their distances are at most 2−K. Continuity then contradicts the fixed separation of their f-values. Thus for large K, fixing the coordinates after the first K to any fixed reference tail gives a local approximation with uniform error at most ε. Cylinder convergence therefore gives convergence of integrals first for local functions and then for every continuous f.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Lattice Gibbs kernels and DLR states

Definition

Assume Countable Choice for the product-measure extension framework of The Axiom of Countable Choice (ACω) and Finite-alphabet product and probability compactness. Work in the finite-alphabet configuration space X=AZd with its cylinder-generated Borel sigma-algebra. Let the finite-range interaction and its conditional energy be as in Finite-range lattice interaction and conditional energy, and fix β>0. For a finite Λ⊂Zd and exterior configuration η∈AΛc, put ZΛη=∑τ∈AΛe−βEΛ(τ∣η),γΛ(f∣η)=∑τ∈AΛf(τη)e−βEΛ(τ∣η)ZΛη for bounded measurable f:X→R. Here γΛf denotes the function η↦γΛ(f∣η), composition is operator composition, and μγΛ(f)=∫γΛ(f∣η) dμ(η). A proper kernel leaves every exterior-measurable event fixed. A specification is a family (γΛ)Λ finite of proper probability kernels satisfying γΛγΔ=γΛ whenever Δ⊆Λ. A DLR state is a probability measure μ on X satisfying μγΛ=μ for every finite Λ.

For the existence and uniqueness results below, a specification has finite range when for every finite Δ and every cylinder event C the function η↦γΔ(1C∣η) depends on only finitely many coordinates of η. The displayed finite-range Gibbs formula has this property because only finitely many interaction terms meet Δ, and a cylinder indicator depends on only finitely many additional coordinates.

The formula is the candidate Gibbs specification; Conditional Gibbs kernel consistency ↗ establishes its positive finite normalization, measurability, properness and consistency under the stated hypotheses. The Countable Choice assumption is used for the product-measure construction, not for any finite conditional sum.

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Conditional Gibbs kernel consistency

Statement

Assume the finite-alphabet product-measure convention and conditional Gibbs formula of Lattice Gibbs kernels and DLR states, a finite-range interaction, and β>0. For every finite Λ and exterior η, the displayed partition ZΛη is finite and strictly positive, the formula γΛ(⋅∣η) defines a measurable proper probability kernel, and the family satisfies γΛγΔ=γΛ whenever finite Δ⊆Λ.

Facts & Assumptions

Given: The finite-range interaction, β>0, and the conditional Gibbs formula and operator convention of Lattice Gibbs kernels and DLR states.

[F1]

The finite Gibbs variational identity defines the strictly positive normalized finite-volume law whenever the finite state set is nonempty and the energies are real (Finite Gibbs relative entropy and free-energy minimum).

[F2]

The candidate kernel, specification terminology, and finite-range locality condition are defined in Lattice Gibbs kernels and DLR states.

Proof

1.1givenF1algebra

For fixed finite Λ, only finitely many translated interaction supports meet it, so every EΛ(τ∣η) is a finite real number and AΛ is finite nonempty. Thus ZΛη is a positive finite sum and [F1] supplies its normalized probability weights.

2.1F2step 1.1algebra

For Δ⊆Λ, split the terms meeting Λ into those meeting Δ and those disjoint from Δ. The second sum is independent of τΔ, so EΛ(τΛ∣η)=EΔ(τΔ∣τΛ∖Δη)+C(τΛ∖Δ,η). Consequently ZΛη=∑ξ∈AΛ∖Δe−βC(ξ,η)ZΔξη. In (γΛγΔf)(η), first sum over the γΛ spin assignment on Δ while holding ξ=τΛ∖Δ fixed. Its weights sum to ZΔξη, which cancels the denominator in γΔ(f∣ξη). The remaining sum is ZΛ−1∑ξ∈AΛ∖Δe−βC(ξ,η)∑ζ∈AΔf(ζξη)e−βEΔ(ζ∣ξη), which is exactly γΛf(η) by reindexing (ζ,ξ) as a configuration on Λ. The factor e−βC remains; it does not cancel by itself.

2.2F2step 1.1algebra

Each replacement map η↦τΛη is continuous in the product topology, and each weight depends on finitely many boundary coordinates; finite sums therefore make γΛ(f∣η) measurable for bounded measurable f. When f=1C for a cylinder event C, its value additionally depends only on the finitely many coordinates specified by C, so the resulting function of η depends on finitely many coordinates. For an exterior event D⊆AΛc measurable in the exterior cylinder sigma-algebra, let B={σ∈X:σ∣Λc∈D}. Then 1B(τΛη)=1D(η) for every local assignment, so γΛ(1B∣η)=1D(η); this is properness. Whenever a kernel is integrated against a probability on X or composed with another kernel, its exterior argument is the restriction of that global configuration to the relevant complement.

3.1step 1.1step 2.2step 2.1∎

Steps 1.1, 2.2 and 2.1 prove normalization, measurability, properness and consistency, so the candidate family is a specification.

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Finite-range lattice DLR existence and closed convex state set

Statement

Assume Countable Choice, a finite nonempty alphabet A, and a proper consistent finite-range specification γ on X=AZd, with finite range as defined in Lattice Gibbs kernels and DLR states. No translation invariance is required. For every exhaustion Λn↑Zd and every sequence of exterior configurations ηn, the measures γΛn(⋅∣ηn) extended with the fixed exterior have a cylinder-convergent subsequence. Every such subsequential limit is a DLR state. Consequently the DLR state set is nonempty, convex, and closed under cylinder convergence. No uniqueness is asserted.

Facts & Assumptions

Given: A proper consistent finite-range lattice specification, an exhausting sequence of finite volumes, exterior configurations, and Countable Choice for the product-measure extension.

[A1]

Assume Countable Choice: every countable family of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)). The measure-extension and probability-compactness conclusions used below are supplied by [F1], not by this axiom definition.

[A2]

The specification is proper, consistent, and finite range in the cylinder-local sense of Lattice Gibbs kernels and DLR states.

[F1]

Cylinder convergence is sequentially compact and determines a unique probability measure under Countable Choice (Finite-alphabet product and probability compactness).

Proof

1.1A1A2F1construct

Each proper kernel value γΛn(⋅∣ηn) is already a probability measure on X. By [F1], a subsequence converges on every cylinder to a probability measure μ.

2.1A2F1step 1.1

Fix a finite Δ and cylinder event C. By [A2], choose a finite coordinate set S on which γΔ(C∣⋅) depends. For sufficiently large n, Δ∪S⊆Λn. Consistency in [A2] then gives μnγΔ(C)=μn(C), and both integrands are cylinder functions, so cylinder convergence passes the identity to μγΔ(C)=μ(C).

3.1F1step 2.1

Cylinder events form a pi-system generating the Borel sigma-algebra. The two probability measures μγΔ and μ agree on the generating cylinders by step 2.1. Uniqueness of measures on the cylinder-generated sigma-algebra, part of [F1], extends equality to every Borel event. Thus μ is DLR for each Δ.

4.1A2F1step 3.1∎

Convex combinations preserve the linear identities μγΔ=μ. A cylinder-convergent limit of DLR laws satisfies the same local identities by the argument of step 2.1, hence is DLR. Since step 1.1 constructs at least one such law, the state set is nonempty, convex and closed.

Remarks

The cited Princeton notes make different broad claims about projective consistency in distinct model sections: the O(N) existence discussion at PDF pp. 16–17 says its finite-volume Gibbs laws do not satisfy the Kolmogorov-extension consistency hypothesis, while the Ising/DLR discussion at pp. 31–32 says the finite-volume family is consistent in that sense. Because those passages concern different model sections, they do not by themselves establish a contradiction for the finite-alphabet specification in this theorem. Neither passage is used here. This proof obtains a DLR state from a cylinder-convergent subsequence, local kernel consistency, and measure extension; it does not treat the finite-volume laws as a projectively consistent family.

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Translation-averaged stationary Gibbs state

Statement

Assume Countable Choice, a translation-invariant finite-range interaction on Zd with finite alphabet, and a nonempty DLR state set. Then there exists at least one translation-stationary DLR state. This assertion does not imply uniqueness or ergodicity.

Facts & Assumptions

Given: A translation-invariant specification with at least one DLR probability measure and the Countable Choice measure framework.

[F1]

Under Countable Choice the DLR set is nonempty, convex and closed under cylinder convergence (Finite-range lattice DLR existence and closed convex state set).

[F2]

Every sequence of probabilities on the finite-alphabet product has a cylinder-convergent subsequence whose limit is a probability; two probabilities agreeing on all cylinders are equal (Finite-alphabet product and probability compactness).

[A1]

Set X=AZd and Bn=[−n,n]d∩Zd. For j∈Zd, define the translation θj:X→X by (θjσ)x=σx+j and its action on measures by push-forward, (θjμ)(C)=μ(θj−1C).

Proof

1.1givenF1A1choose

Choose a DLR state μ. The translated interaction list gives (γΛf)∘θj=γΛ+j(f∘θj) for every bounded measurable f, by reindexing each anchor and each local spin in the finite sums. Hence (θjμ)γΛ(f)=μγΛ+j(f∘θj)=μ(f∘θj)=(θjμ)(f), so every translate is DLR; by convexity in [F1], the cube average μˉn=∣Bn∣−1∑j∈Bnθjμ is DLR.

2.1A1step 1.1algebra

For any fixed shift k=(k1,…,kd)∈Zd and bounded local f, cancellation of terms indexed by Bn∩(Bn+k) gives ∣μˉn(f)−(θkμˉn)(f)∣≤2∥f∥∞∣Bn△(Bn+k)∣/∣Bn∣. For all sufficiently large n, ∣Bn∣=(2n+1)d and ∣Bn∩(Bn+k)∣=∏q=1d(2n+1−∣kq∣), so the ratio equals 2(1−∏q=1d(1−∣kq∣/(2n+1)))→0.

3.1F1F2step 1.1step 2.1∎

By [F2], a subsequence of μˉn converges on cylinders to a probability μˉ; since each average is DLR, closedness in [F1] makes its limit DLR. Step 2.1 passes to this limit for every local f and fixed k, so μˉ and θkμˉ agree on all cylinders; uniqueness of measures on cylinders in [F2] gives θkμˉ=μˉ. Thus μˉ is stationary.

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Finite maximal coupling and heat-bath stationary law

Statement

For probability vectors p,q on a finite set A, there is a coupling (X,Y) with Pr⁡(X≠Y)=12∑a∈A∣p(a)−q(a)∣. For a finite region Λ with strictly positive single-site Gibbs conditionals, the random-scan heat-bath chain has the finite-volume Gibbs law as its unique stationary distribution. This chain is an auxiliary finite-state proof device, not a physical time evolution.

Facts & Assumptions

Given: A finite state alphabet and the normalized finite-volume conditional laws of Conditional Gibbs kernel consistency.

[F1]

The finite-volume conditional Gibbs laws are positive probability kernels consistent under nested regions (Conditional Gibbs kernel consistency).

Proof

1.1givenalgebra

Put m(a)=min⁡{p(a),q(a)} and r=1−∑am(a)=12∑a∣p(a)−q(a)∣. Couple the common mass m(a) on the diagonal. If r>0, couple the residual masses p−m and q−m by their product divided by r; their supports are disjoint, so this part always disagrees. If r=0, the diagonal coupling already has the required marginals.

1.2F1algebra

On the finite configuration space AΛ, choose a site uniformly and resample its spin from the conditional Gibbs law. The finite-volume Gibbs law is invariant under each such update because it resamples from its own conditional distribution; hence it is stationary for the random-scan mixture.

2.1F1step 1.2algebra∎

Let N=∣Λ∣>0 and fix a target configuration. Since there are finitely many strictly positive conditional probabilities, their minimum p∗>0. In N updates, the probability of selecting a fixed ordering of all sites is N−N, and conditional on that ordering the probability of setting every spin to its target is at least p∗N. Thus every starting state reaches the target with probability at least ε=N−Np∗N>0. The N-step kernel has a common component εδtarget, so it contracts total variation by at most 1−ε; two stationary laws must coincide. If Λ=∅, the configuration space is a singleton and uniqueness is immediate.

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Finite-range Dobrushin uniqueness and boundary influence

Statement

Assume Countable Choice, d≥1, a finite nonempty alphabet A, and a proper, consistent finite-range specification γ on X=AZd with its cylinder-generated Borel σ-algebra, using Lattice Gibbs kernels and DLR states. Here consistency means γΛγΔ=γΛ for finite Δ⊆Λ, properness means γΛ(rΛ−1(D)∣η)=1D(η) for each exterior-measurable event D⊆AΛc and η∈AΛc, where rΛ:X→AΛc is coordinate restriction; in compositions and integrals against measures on X, every kernel takes the appropriate exterior restriction of the global configuration as its argument, and finite range means each γΔ(1C∣⋅) for finite Δ and cylinder event C depends on only finitely many exterior coordinates. Also assume a uniform finite integer radius R≥0 such that each one-site kernel γ{i} depends only on exterior spins j with ∥i−j∥∞≤R. For each pair of sites define cij as the supremum of the half-total-variation change in the one-site conditional law at i when only exterior spin j is changed, where half-total variation for probability vectors p,q on A is 12∑a∈A∣p(a)−q(a)∣, and set cii=0. If c=sup⁡i∑jcij<1, then the DLR state is unique. Moreover, for a finite region Λ and two exterior conditions ξ,ζ, couple the random-scan finite-volume heat-bath chains by choosing the same uniformly random site and using a maximal coupling of its two single-site conditional laws. A stationary joint law of this pair chain exists. If the distance dist⁡∞(i,Λc):=inf⁡j∉Λ∥i−j∥∞ is at least r(R+1) for a nonnegative integer r, then its disagreement probability at i is at most cr/(1−c). Consequently every local conditional expectation becomes uniformly insensitive to boundary conditions as Λ↑Zd.

Facts & Assumptions

Given: The proper consistent finite-alphabet specification, the uniform single-site locality radius R, the half-total-variation influence matrix (cij), the strict row-sum bound c<1, and Countable Choice.

[A1]

Countable Choice is used for the DLR existence and product-measure extension inputs (The Axiom of Countable Choice (ACω)).

[F1]

For any two probability vectors on a finite alphabet there is a coupling whose disagreement probability equals their half-total-variation distance (Finite maximal coupling and heat-bath stationary law).

[F2]

A DLR state exists under Countable Choice for a proper consistent finite-range finite-alphabet specification (Finite-range lattice DLR existence and closed convex state set).

[F3]

By the stated uniform locality radius, cij=0 whenever ∥i−j∥∞>R.

[F4]

Every bounded real sequence has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

[F5]

For 0≤c<1 and integer r≥0, the geometric tail satisfies ∑k≥rck=cr/(1−c) (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

Proof

1.1given

Fix a finite Λ and an exterior condition ξ. On AΛ, define the random-scan heat-bath chain that chooses one site uniformly and resamples it from γ{i}(⋅∣xΛ∖{i}ξ). Properness and consistency imply that the finite-volume law γΛ(⋅∣ξ) is stationary for this chain: applying a one-site kernel to it leaves it unchanged by γΛγ{i}=γΛ.

1.2givenF1F4algebra

Any two stationary laws for the same-boundary chain have a stationary coupling: start with any coupling of them, evolve the pair using a common random site and the maximal coupling [F1], and take a convergent subsequence of Cesàro averages using [F4]. The marginals remain the given stationary laws. If qi is the stationary disagreement probability, then stationarity makes it equal to the mismatch probability on updating i. Telescoping the one-site conditional laws over the finitely many differing internal spins gives qi≤∑j∈Λcijqj. Therefore max⁡iqi≤cmax⁡iqi, so all qi=0. Thus the finite-volume Gibbs law is the unique stationary law; this proof needs no positivity assumption on the one-site conditionals.

2.1F1F4step 1.1

For two boundary conditions ξ,ζ, start the pair chain at the product of the two finite-volume Gibbs laws and use common-site maximal updates. Its marginal laws remain stationary by step 1.1; Cesàro averaging and [F4] therefore give a stationary coupling whose marginals are those Gibbs laws. For a current pair X,Y, telescoping the differing internal and exterior spins one at a time bounds the update mismatch at site i by ∑j∈Λcij1{Xj≠Yj}+bi, where bi=∑j∉Λcij1{ξj≠ζj}. Under the stationary coupling, if qi=Pr⁡(Xi≠Yi), random-scan stationarity gives qi equal to the mismatch probability on updating i, hence qi≤∑j∈Λcijqj+bi. For Λ=∅ there are no coordinates to bound.

3.1givenF3F5step 2.1algebra

Let CΛ=(cij)i,j∈Λ. Iterating the inequality in step 2.1 gives q≤CΛn1+∑k=0n−1CΛkb. Every row sum of CΛ is at most c<1, so ∥CΛn1∥∞≤cn→0. A nonzero term of (CΛkb)i is an influence path of k internal edges followed by one edge to the exterior; each edge changes ℓ∞-distance by at most R by [F3]. For R≥1 and dist⁡∞(i,Λc)≥r(R+1), such a path has at least r+1 edges, hence k≥r. Since 0≤bi≤c<1, its total weight is bounded by ∑k≥rck=cr/(1−c) by [F5]. If R=0, all off-diagonal influences vanish and the boundary term is zero, so the same bound holds.

4.1step 3.1algebra

For a bounded local f, let δi(f) be its maximum change when only spin i changes. Coordinate-by-coordinate coupling gives ∣γΛ(f∣ξ)−γΛ(f∣ζ)∣≤∑i∈supp⁡fδi(f)qi. Step 3.1 makes this bound tend to zero uniformly over the boundary conditions as Λ exhausts the lattice.

5.1A1F2step 4.1∎

Let μ,ν be DLR states, which exist by [F2]. Integrate the uniform boundary difference from step 4.1 against μ(dξ)ν(dζ); using μγΛ=μ and νγΛ=ν gives μ(f)=ν(f) for every bounded local f. Cylinder indicators are local, so the measures agree on a generating family and hence are equal.

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Ising influence bounded by hyperbolic tangent

Statement

Let d≥1, let spins be si∈{−1,+1}, let hi∈R, and let Jij=Jji∈R with Jii=0 and Jij=0 whenever ∥i−j∥∞>R for some finite integer R. No translation invariance is assumed. For finite Λ⊂Zd and exterior η∈{−1,+1}Λc, define s^i=si on Λ and s^i=ηi outside, and use EΛ(sΛ∣η)=−∑{i,j}:{i,j}∩Λ≠∅Jijs^is^j−∑i∈Λhisi, where the pair sum is over unordered distinct pairs and is finite by the range assumption. For site i and exterior spins η∈{−1,+1}Zd∖{i}, the one-site law is explicitly γi(s∣η)=exp⁡ ⁣(β[his+∑k≠iJiksηk])∑u∈{−1,+1}exp⁡ ⁣(β[hiu+∑k≠iJikuηk]),s∈{−1,+1}; the sums are finite by the range condition. For i≠j, define cij:=sup⁡η,η′: ηk=ηk′ (k≠j)12∑s∈{−1,+1}∣γi(s∣η)−γi(s∣η′)∣,cii:=0. Then cij=0 if Jij=0, and otherwise cij≤tanh⁡(β∣Jij∣). In particular, sup⁡i∑jtanh⁡(β∣Jij∣)<1 is a sufficient Dobrushin row-sum bound for every bounded field profile (hi).

Facts & Assumptions

Given: The finite-range Ising pair couplings, field profile, conditional energy, one-site Gibbs laws, and β>0 specified in the Statement.

[F1]

The hyperbolic functions are defined by sinh⁡x=(ex−e−x)/2, cosh⁡x=(ex+e−x)/2, and tanh⁡x=sinh⁡x/cosh⁡x; their addition and product identities include tanh⁡(x+b)−tanh⁡(x−b)=sinh⁡(2b)/(cosh⁡(x+b)cosh⁡(x−b)) and cosh⁡(x+b)cosh⁡(x−b)=(cosh⁡(2x)+cosh⁡(2b))/2 (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

Proof

1.1givenF1algebra

Fix i and all boundary spins except j. Write x=β(hi+∑k∉{i,j}Jikηk) and b=βJij; these sums are finite by the range condition. Direct normalization of the two weights gives γi(+1∣η)=[1+tanh⁡(x+b)]/2 and γi(+1∣η′)=[1+tanh⁡(x−b)]/2.

2.1step 1.1F1algebra

For Bernoulli laws the half-total-variation distance equals the absolute difference of the probabilities of +1. By [F1], cosh⁡(x+b)cosh⁡(x−b)=(cosh⁡(2x)+cosh⁡(2b))/2≥cosh⁡2b, so their difference is at most ∣sinh⁡(2b)∣/(2cosh⁡2b)=tanh⁡∣b∣. If Jij=0, the conditional weights are identical and cij=0.

3.1F1step 2.1∎

Summing the bound in step 2.1 over the finite interaction neighborhood proves the stated sufficient row-sum criterion; the supremum over i is the Dobrushin constant used by the uniqueness theorem.

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Unique interacting Ising state at sufficient high temperature

Statement

Assume Countable Choice, spins si∈{−1,+1}, inverse temperature β>0, a constant finite field h∈R, and symmetric pair couplings with zero diagonal, Jij=Jji and Jii=0, satisfying Ji+k,j+k=Jij for all i,j,k∈Zd, and Jij=0 when ∥i−j∥∞>R for some finite integer R≥0. For finite Λ, use the conditional energy EΛ(sΛ∣η)=−∑{i,j}:{i,j}∩Λ≠∅Jijs^is^j−h∑i∈Λsi, where s^i=si on Λ and s^i=ηi outside; let γ be the corresponding finite-alphabet Gibbs specification from Lattice Gibbs kernels and DLR states. If sup⁡i∑jtanh⁡(β∣Jij∣)<1, then there exists exactly one DLR state. For homogeneous nearest-neighbor coupling J this sufficient condition is 2dtanh⁡(β∣J∣)<1. It is not asserted to be sharp.

Facts & Assumptions

Given: The finite-range Ising pair Hamiltonian and one-site conditional Gibbs laws in the Statement, with β>0 and constant finite field h.

[A1]

Assume Countable Choice: every countable family of nonempty sets admits a choice function (The Axiom of Countable Choice (ACω)). Under that assumption, every proper consistent cylinder-local specification on this finite-alphabet product has a DLR state (Finite-range lattice DLR existence and closed convex state set).

[F1]

The one-site influence satisfies cij≤tanh⁡(β∣Jij∣) for this Hamiltonian normalization (Ising influence bounded by hyperbolic tangent).

[F2]

A finite-range specification with strict influence row sum c<1 has exactly one DLR state (Finite-range Dobrushin uniqueness and boundary influence).

[F3]

The finite-range lattice Gibbs kernels are proper, consistent and finite range under the stated hypotheses (Conditional Gibbs kernel consistency).

[F4]

To realize the displayed pair Hamiltonian in the interaction-list convention of Finite-range lattice interaction and conditional energy, choose one displacement v from each pair {v,−v} with J0v≠0, use shape Sv={0,v} and term Φv(s0,sv)=−J0vs0sv, and add the onsite shape S0={0} with Φ0(s0)=−hs0. Every unordered pair is counted once through its unique representative and translated anchor; this finite list has uniform one-site dependence radius at most R.

Proof

1.1givenF1algebra

By [F1], sup⁡i∑jcij≤sup⁡i∑jtanh⁡(β∣Jij∣)<1.

2.1A1F2F3F4step 1.1∎

The interaction list in [F4] has exactly the conditional energy displayed in the Statement: each unordered pair occurs once, and each site contributes the single field term −hsi. Thus [F3] gives its proper consistent finite-range Gibbs specification. Existence follows from [A1] for this specification. Apply [F2] with the uniform radius in [F4] and row-sum bound from step 1.1 to obtain uniqueness. This criterion is sufficient only.

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Bounded lattice pressure on van Hove sets

Statement

Let d≥1, A be finite nonempty, and let a bounded translation-invariant finite-range interaction on AZd be as in Finite-range lattice interaction and conditional energy. Fix β>0, set CL={1,…,L}d for each integer L≥1, and define ZL=∑σ∈ACLe−βHCLfree(σ) and pL=∣CL∣−1log⁡ZL. For a van Hove sequence of finite sets Λn⊂Zd, meaning ∣Λn∣→∞ and ∣{x∈Λn:dist⁡∞(x,Λnc)≤r}∣/∣Λn∣→0 for every fixed r, the free-boundary pressure pΛn=1∣Λn∣log⁡∑σ∈AΛne−βHΛnfree(σ) and the cubic sequence pL both converge to the same finite limit p. Bounded finite-range interactions are essential; this statement does not concern continuum particles or long-range models.

Facts & Assumptions

Given: A finite alphabet, a bounded translation-invariant finite-range interaction, β>0, and a van Hove sequence.

[A1]

The shape list, translated terms, range and free-volume energy are exactly those of Finite-range lattice interaction and conditional energy.

[F1]

For a finite alphabet and bounded finite-range interaction list on supports contained in a fixed coordinate box, the free cubic log-partition density has a finite limit; a uniform energy perturbation bounded by CbLd−1 preserves it (Finite-range lattice log-partition limit). The local interaction convention in Finite-range lattice interaction and conditional energy is a special case: its shapes contain 0 and have diameter at most R, hence lie in [−R,R]d, and its translated terms and free-volume energy have the same form. Thus [F1] supplies the limit p defined in the Statement.

Proof

1.1A1given

Fix an integer block side ℓ>2R and tile the lattice by disjoint ℓ-cubes. For Λn, retain only cubes fully contained in Λn and denote their number by mn; the leftover sites lie within distance at most ℓ of Λnc, so ∣Λn∖(full cubes)∣=o(∣Λn∣) and mnℓd/∣Λn∣→1.

2.1A1step 1.1algebra

Erase each interaction term whose support crosses a grid seam or touches a leftover site. At fixed ℓ, leftover-touching anchors are o(∣Λn∣) by the van Hove condition. Each full ℓ-cube has at most CRℓd−1 anchors within distance R of one of its seams, so seam-crossing anchors total at most CR∣Λn∣/ℓ+o(∣Λn∣); here CR depends only on the finite shape list, R, and dimension. Since each term is bounded, the original and cut energies differ uniformly by at most C∣Λn∣/ℓ+o(∣Λn∣).

3.1givenstep 2.1algebra

If two energies on a finite state set differ uniformly by at most D, their partition functions differ by factors between e−βD and eβD, so their log partition functions differ by at most βD. The cut partition function is Zℓmn∣A∣∣Λn∣−mnℓd, where Zℓ is the free cubic partition function. Dividing by ∣Λn∣ and taking n→∞ shows that the limsup and liminf of pΛn each differ from ℓ−dlog⁡Zℓ by at most Cβ/ℓ.

4.1F1step 3.1∎

By [F1], ℓ−dlog⁡Zℓ tends to the cubic pressure as ℓ→∞. Sending ℓ→∞ in the bound from step 3.1 makes the limsup and liminf of pΛn equal to that same value, proving convergence.

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Stationary lattice entropy and energy densities

Statement

Assume Countable Choice and let A be finite nonempty. A consistent translation-stationary family means probability measures νF on AF for every finite F⊂Zd, with νE=(πEF)∗νF whenever E⊆F and νF+j=(τj)∗νF for every j∈Zd, where (πEFx)=x∣E and (τjx)i+j=xi. This family extends to a unique translation-stationary probability μ on the cylinder sigma-algebra. Set Bn=[−n,n]d∩Zd for each integer n≥1. For any such stationary μ and any bounded translation-invariant finite-range interaction (R,(Sr,Φr)r∈R) as in Finite-range lattice interaction and conditional energy, define Hn=−∑σ∈ABnμBn(σ)log⁡μBn(σ) with 0log⁡0=0, and u(μ)=∑r∈R∑a∈ASrμ(σ∣Sr=a)Φr(a). Then h(μ)=lim⁡n→∞Hn∣Bn∣=inf⁡ℓ≥1Hℓ∣Bℓ∣ exists in [0,log⁡∣A∣], and the free-boundary interaction-energy density satisfies lim⁡n→∞Eμ[HBnfree]∣Bn∣=u(μ).

Facts & Assumptions

Given: A finite alphabet lattice, a bounded translation-invariant finite-range interaction, Countable Choice, and a consistent stationary family of finite-dimensional probabilities.

[A1]

Countable Choice and cylinder extension give a unique probability measure from a consistent family of finite-dimensional probabilities (Finite-alphabet product and probability compactness).

[F1]

The finite Gibbs variational identity implies nonnegative relative entropy on each finite state set (Finite Gibbs relative entropy and free-energy minimum).

Proof

1.1givenA1

Apply [A1] to the consistent stationary cylinder probabilities. The extension is unique; translating it leaves the same cylinder values, so uniqueness makes the resulting measure μ stationary.

1.2F1algebra

For finite random vectors X,Y, nonnegative relative entropy of their joint law relative to the product of its marginals gives H(X,Y)≤H(X)+H(Y). If the product law has zeros, restrict to its support; the joint law is supported there, and [F1] applies on that finite support. For any single-site law q on A, applying [F1] with zero energy and the uniform Gibbs law gives D(q∥unifA)=log⁡∣A∣−H(q)≥0, so H(q)≤log⁡∣A∣.

2.1givenstep 1.2algebra

Tile a large cube Bn by disjoint translates of Bℓ and leftover sites. By stationarity, each full block has entropy Hℓ; each leftover site contributes at most log⁡∣A∣. Repeated subadditivity from step 1.2 gives Hn≤mnHℓ+∣Rn∣log⁡∣A∣, with mn∣Bℓ∣/∣Bn∣→1 and ∣Rn∣/∣Bn∣→0 for fixed ℓ. Thus lim sup⁡Hn/∣Bn∣≤Hℓ/∣Bℓ∣ for every ℓ. Since the infimum is no larger than any ratio, the liminf is at least the infimum, proving the stated limit formula.

3.1A1givenalgebra∎

For each interaction type r, the number of anchors with x+Sr⊆Bn, divided by ∣Bn∣, tends to 1 because Sr is finite. Stationarity gives expectation EμΦr for each included translate, so summing the finite interaction list and taking the limit yields u(μ). Nonnegativity and the bound Hn≤∣Bn∣log⁡∣A∣ give 0≤h(μ)≤log⁡∣A∣.

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DLR states attain the bounded lattice pressure variational principle

Statement

Assume Countable Choice, a bounded translation-invariant finite-range lattice interaction on a finite alphabet, and β>0. Let p be its dimensionless pressure per site, and let μ be a translation-stationary probability with entropy density h(μ) and energy density u(μ) as in Stationary lattice entropy and energy densities. Then h(μ)−βu(μ)≤p. If μ is a DLR state for the specification, equality holds. Since stationary DLR states exist under these hypotheses, the supremum of h(μ)−βu(μ) over stationary probabilities is attained and equals p. The converse “every maximizing stationary probability is DLR” is not asserted.

Facts & Assumptions

Given: A bounded finite-range translation-invariant interaction, β>0, a stationary lattice probability, and Countable Choice for the DLR existence input.

[A1]

Countable Choice is retained in the finite-alphabet DLR measure construction (The Axiom of Countable Choice (ACω)).

[F1]

Stationary finite-alphabet laws have entropy and energy densities with the stated limits (Stationary lattice entropy and energy densities).

[F2]

The free-boundary pressure along cubes and van Hove sets has a finite limit p (Bounded lattice pressure on van Hove sets).

[F3]

For each finite state set, the Gibbs relative-entropy identity is D(ρ∥γ)=−HSh(ρ)+βρ(H)+log⁡Z≥0 (Finite Gibbs relative entropy and free-energy minimum).

[F4]

For the stated finite-range interaction and β>0, its candidate Gibbs kernels are proper, consistent and cylinder-local (Conditional Gibbs kernel consistency). Under Countable Choice, such a specification has a nonempty DLR state set (Finite-range lattice DLR existence and closed convex state set). For a translation-invariant interaction with a nonempty DLR set, translation averaging then gives a stationary DLR state (Translation-averaged stationary Gibbs state).

Proof

1.1F1F2F3givenalgebra

Under the Countable Choice assumption [A1], apply [F3] to the Bn marginal of μ and the free Gibbs law for HBnfree. Rearranging gives Hn−βEμHBnfree≤log⁡ZBn. Divide by ∣Bn∣ and use [F1] and [F2] to obtain h(μ)−βu(μ)≤p.

1.2givenF1F3algebra

Suppose μ is DLR. Its Bn marginal is a mixture of the conditional finite-volume Gibbs laws γBn(⋅∣η). Let R be the interaction range, ∂RBn={x∈Bn:dist⁡∞(x,Bnc)≤R}, and C=∑r∈R∣Sr∣∥Φr∥∞. A support meeting both Bn and its complement contains an interior site in ∂RBn; for each such site and type r there are at most ∣Sr∣ possible anchors. Thus the conditional energy differs from HBnfree by at most Dn=C∣∂RBn∣. Termwise comparison shows the likelihood ratio of each conditional law to the free Gibbs law lies in [e−2βDn,e2βDn]; the same holds for their mixture. Hence 0≤D(μBn∥γBnfree)≤2βDn.

2.1F1F2F3step 1.2algebra

Since Dn/∣Bn∣→0, the finite relative-entropy identity [F3], divided by volume and passed to the limits in [F1]–[F2], gives h(μ)−βu(μ)=p.

3.1A1F4step 1.1step 2.1∎

The kernel consistency and existence results in [F4] first give a nonempty DLR set under [A1]; translation averaging in [F4] then gives a stationary DLR state and hence attains equality from step 2.1. Combining with the upper bound of step 1.1 proves the supremum statement, without any converse implication.

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Square-lattice even graphs and parity contour separation

Statement

Let n≥1, set Bn=[−n,n]2∩Z2, let s:Bn→{−1,+1}, and extend s by +1 on Z2∖Bn. The dual lattice has vertex set Z2+(12,12) and edges between its nearest-neighbor vertices. Draw a dual edge across each primal nearest-neighbor bond whose endpoint spins disagree. The resulting finite dual graph has even degree at every vertex and decomposes into edge-disjoint simple dual cycles. For each such cycle γ, define the inside of a primal site by the parity of intersections of the horizontal ray from that site to +∞ with γ. Then the inside is finite and contained in Bn, and two neighboring primal sites have different inside parity exactly when their bond is crossed by γ. Consequently, if s(0)=−1, at least one cycle in the decomposition has the origin inside.

The schematic below places primal sites at integer-coordinate intersections and dual vertices at pairs of half-integers. Each thick dual segment crosses the perpendicular primal bond at its midpoint; the dashed ray from the origin crosses the displayed cycle once. The center spin is negative and the surrounding spins are positive. This is one local picture of the parity convention, not a proof of the separation or cycle-decomposition claims.

+1++++¡++++primalspinsandbondsdualcontour

Facts & Assumptions

Given: A square-lattice configuration with all exterior spins +1 and the nearest-neighbor mismatch graph on its dual square lattice.

[A1]

The model uses the finite-range lattice configuration convention of Finite-range lattice interaction and conditional energy.

Proof

1.1givenA1algebra

Around each dual vertex the product of the four primal nearest-neighbor spin products is 1, since each corner spin occurs twice. Hence the number of disagreeing bonds incident to that vertex is even. The mismatch graph is finite because all spins outside the finite box are +1.

2.1step 1.1construct

Start from an unused edge and extend a trail along unused edges. Since every vertex has even degree, the trail cannot stop at a vertex other than its starting vertex; finiteness therefore gives a closed trail. Taking the segment between the first repeated vertex extracts a simple cycle. Removing its edges preserves even degree at every vertex, so repeating the finite procedure decomposes every edge into edge-disjoint cycles. The square lattice has no loops, parallel edges, or triangles, so each cycle has length at least four.

3.1step 2.1construct

Dual vertices have half-integer coordinates, so a ray from an integer-height primal site meets no dual vertex. For horizontally adjacent sites, the symmetric difference of their rightward rays is exactly the horizontal unit segment crossing their joining primal bond, so their parity differs exactly when that bond is crossed by γ. For vertically adjacent sites u=(i,j) and v=(i,j+1), choose an integer M larger than every horizontal coordinate of γ and consider the rectangle [i,M]×[j,j+1]. Its boundary contains no dual vertices. As the closed cycle is traversed, the indicator of whether its current vertex lies inside this rectangle toggles at each transverse boundary crossing and returns to its starting value, so the total number of boundary crossings is even. There is no crossing on the far side x=M; the other crossings are exactly those of the two truncated horizontal rays from u,v and the dual edge across the primal bond {u,v}. Hence the two ray-intersection counts have different parity exactly when that vertical bond is crossed by γ.

4.1givenstep 2.1step 3.1∎

Every primal site outside a rectangle containing γ has parity zero, since its ray either misses γ or crosses the full cycle an even number of times; thus the inside is finite. In a plus-boundary box, any site outside the box has an outward lattice path to infinity lying entirely in the plus exterior. Along this path no bond is a mismatch, so step 3.1 keeps its parity zero; hence every cycle-inside site lies in the box. Along any primal path, the spin changes exactly on mismatch edges. A path from the origin to the all-plus exterior therefore crosses an odd number of mismatch edges if s0=−1. For each cycle in the edge-disjoint decomposition from step 2.1, step 3.1 says its parity changes along the path exactly when that path crosses one of its edges; its parity is zero at the exterior endpoint. The sum modulo two of the cycle parity changes is the total mismatch-edge count, which is odd, so at least one cycle has origin parity one, as claimed.

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Peierls contour probability bound

Statement

Let n∈Z≥0 and Bn=[−n,n]2∩Z2. For s∈{−1,+1}Bn, set s^i=si for i∈Bn and s^i=+1 for i∉Bn, and define HBn+(s)=−J∑{i,j}⊂Z2: ∣i−j∣1=1, {i,j}∩Bn≠∅s^is^j,ZBn+=∑s∈{−1,+1}Bne−βHBn+(s),γBn+(s)=(ZBn+)−1e−βHBn+(s). The energy sum is finite because Bn is finite and interactions are nearest-neighbor. For the nearest-neighbor Ising model with coupling J>0, zero field and β>0, any simple dual cycle γ that occurs in the disagreement graph of a configuration determines the event Aγ that all primal bonds crossed by γ have disagreeing spins. Then γBn+(Aγ)≤e−2βJ∣γ∣. For each such cycle, the parity-inside is defined by Square-lattice even graphs and parity contour separation and lies in the variable box.

Facts & Assumptions

Given: A finite plus-boundary square-lattice Ising system with J>0, h=0, β>0, and a disagreement cycle γ.

[F1]

For n≥1, in a disagreement graph of a plus-boundary configuration, ray parity defines a finite inside whose boundary bonds are exactly the cycle edges (Square-lattice even graphs and parity contour separation).

[F2]

A finite Gibbs probability assigns weights proportional to e−βH with a finite positive normalizer (Finite Gibbs relative entropy and free-energy minimum).

Proof

1.1givenF1construct

For n≥1, let Cγ be the parity-inside primal sites from [F1]; that lemma gives Cγ⊆Bn and identifies its boundary bonds with the edges crossed by γ. For n=0, B0={0}: the plus spin has no disagreement edges, and the minus spin has exactly the four dual edges forming the unit square centered at 0. This is the only possible simple disagreement cycle; its ray-parity inside is {0} and its crossed bonds are exactly the four bonds incident to 0. Thus the same containment and boundary identification hold also for n=0. The map that flips every spin in Cγ and leaves all other spins fixed is an involution on the finite configuration set, preserves the fixed plus exterior, and is injective.

2.1step 1.1algebra

Every bond crossing γ has opposite spins on Aγ and becomes aligned after the flip; its contribution to HBn+ changes from +J to −J, a decrease of 2J. Every other bond has both endpoints inside or both outside Cγ, so its spin product is unchanged. Therefore HBn+(σ)=HBn+(flipCγσ)+2J∣γ∣ on Aγ.

3.1F2step 2.1algebra∎

By [F2], summing e−βH(σ) over Aγ equals e−2βJ∣γ∣ times the weight sum over the image of the injective flip map. That image is a subset of all box configurations, so its total weight is at most the full partition function. Dividing by that positive partition function proves the bound.

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Two-dimensional Ising Gibbs coexistence at low temperature

Statement

Assume Countable Choice. For the nearest-neighbor ferromagnetic Ising model on Z2 with coupling J>0, zero field, and βJ≥2, there exist two distinct translation-stationary DLR states μ+ and μ− such that for every site i, μ+(si)≥m∗>0,μ−(si)≤−m∗. One may take m∗=1−2δ(βJ), where δ(K)=43 r4(5−4r)(1−r)2,r=3e−2K. This is a sufficient low-temperature bound; the states are not asserted to be extremal and K=2 is not claimed to be critical.

Facts & Assumptions

Given: A two-dimensional nearest-neighbor Ising model with J>0, h=0, β>0, plus boundary conditions on finite squares, and Countable Choice.

[A1]

Countable Choice is retained exactly for the probability extension and DLR existence steps (The Axiom of Countable Choice (ACω)).

[F1]

If the origin has spin −1 under plus boundary, at least one simple disagreement cycle has the origin in its parity-inside (Square-lattice even graphs and parity contour separation).

[F2]

For each such cycle γ, the event that all its crossed bonds disagree has Gibbs probability at most e−2βJ∣γ∣ (Peierls contour probability bound).

[F3]

Subsequential limits of finite-volume Gibbs laws are DLR, and translation averaging yields a stationary DLR law (Finite-range lattice DLR existence and closed convex state set, Translation-averaged stationary Gibbs state).

[F4]

The exponential is defined by its positive power series, so for x>0, ex=∑n=0∞xn/n! (The real exponential function and the number e by a power series).

[F7]

The nearest-neighbor model is the finite-range interaction with onsite term Φ0(s0)=0 on {0} and pair terms Φk(s0,sek)=−Js0sek on {0,ek} for k=1,2. Its finite conditional kernels are proper and consistent by Conditional Gibbs kernel consistency.

Proof

1.1givenF1construct

Write K=βJ and r=3e−2K. A cycle with the origin in its parity-inside has an odd number of crossings of the positive horizontal ray. Along the full horizontal line, each crossing changes the sign of a dual vertex's y-coordinate; after one closed traversal that sign returns to its starting value, so the total number of crossings is even. Therefore the negative ray also has a crossing. Dual vertices have half-integer coordinates, so positive-ray crossings occur on vertical dual edges at x=j+12 for integers j≥0. If such a cycle has length ℓ, its horizontal coordinate range includes a negative coordinate and this x>0; a closed lattice walk traverses that range in both directions, so it uses at least 2x horizontal edges and x≤ℓ/2. Thus there are at most ℓ+1 possible positive-ray crossing edges. Choose the lower endpoint of each such edge; from it there are at most 4⋅3ℓ−1 nonbacktracking walks of length ℓ, which overcounts the simple cycles using that crossing. Hence there are at most 4(ℓ+1)3ℓ−1 candidate simple cycles of length ℓ.

2.1F1F2F5step 1.1algebra

By [F2] and the union bound, γBn+(s0=−1)≤43∑ℓ≥4(ℓ+1)rℓ=43 r4(5−4r)/(1−r)2=δ(K), using [F5] for 0<r<1.

3.1givenF4F6step 2.1algebra

The first seven nonnegative terms of [F4] give exp⁡(4)>∑n=064n/n!=437/9>48. Since K≥2, monotonicity and reciprocal positivity in [F6] imply r=3exp⁡(−2K)≤3exp⁡(−4)<1/16. Thus δ(K)<43(1/16)4⋅5/(15/16)2<1/2. The same translated contour count applies to every site in the box, so the plus-boundary magnetization there is at least m∗=1−2δ(K)>0.

4.1A1F3F7step 3.1∎

The onsite and nearest-neighbor list in [F7] realizes the stated Hamiltonian, and its specification is proper and consistent. Take a subsequential limit of plus-boundary laws; by [F3] it is DLR and retains μ(si)≥m∗ for each fixed site. The zero-field nearest-neighbor Hamiltonian is invariant under simultaneous spin reversal, so the spin-reversed law is also DLR and has one-site expectations at most −m∗. Translation averaging as in [F3] preserves each bound and yields stationary μ+,μ−. They are distinct because their one-site expectations have opposite signs.

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Coexistence implies an Ising pressure cusp

Statement

Assume Countable Choice and fix β>0 and J>0 with βJ≥2. For Bn=[−n,n]2∩Z2, define the free-boundary field-dependent energy and partition function by HBn(h),free(s)=−J∑x∈Bn, k∈{1,2}:x+ek∈Bnsxsx+ek−h∑x∈Bnsx,Zn(h)=∑s∈{−1,+1}Bne−βHBn(h),free(s). The bounded finite-range pressure theorem gives the finite limit p(h):=lim⁡n→∞∣Bn∣−1log⁡Zn(h) for each h∈R. Here h is the field energy per spin, while hSh(μ) below denotes Shannon entropy per site. The dimensionless pressure p(h) is not differentiable at h=0. More precisely, if the stationary DLR states at zero field have magnetizations at least m∗>0 and at most −m∗, then every positive-field secant slope from zero is at least βm∗ and every negative-field secant slope is at most −βm∗. This proves a sufficient coexistence cusp; it does not identify the critical coupling.

Facts & Assumptions

Given: The finite-range two-dimensional Ising model, βJ≥2, and Countable Choice.

[A1]

Countable Choice is retained for the DLR and stationary-state construction (The Axiom of Countable Choice (ACω)).

[F1]

At βJ≥2 there are stationary DLR states with one-site magnetizations at least m∗>0 and at most −m∗ (Two-dimensional Ising Gibbs coexistence at low temperature).

[F2]

For each fixed field h, every stationary law satisfies hSh(μ)−βuh(μ)≤p(h); a DLR law attains equality. Here hSh(μ) denotes the entropy density h(μ) of DLR states attain the bounded lattice pressure variational principle, and uh is its mean energy density with the field contribution included.

Proof

1.1A1F1F2

Under Countable Choice [A1], let μ+ and μ− be the states in [F1], and let m=μ+(s0)≥m∗>0. At zero field, [F2] gives p(0)=hSh(μ+)−βu0(μ+).

2.1F2step 1.1algebra

For h>0, the energy per site of the same stationary law is uh(μ+)=u0(μ+)−h m. Applying the variational inequality [F2] to μ+ yields p(h)≥p(0)+βhm≥p(0)+βhm∗.

3.1F1F2step 2.1algebra∎

For μ− with magnetization at most −m∗, the same inequality gives p(h)≥p(0)+βh μ−(s0). If h<0, divide by h to get (p(h)−p(0))/h≤−βm∗. For h>0, step 2.1 gives the slope at least βm∗. The one-sided secant slopes are separated, so no derivative at zero exists.

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Exact convergent Ising-chain high-temperature series

Statement

For the zero-field nearest-neighbor Ising chain with spins si∈{−1,+1}, coupling J∈R, inverse temperature β>0, and K=βJ, let P(K) be the dimensionless pressure per site on periodic chains. For ∣K∣<12, P(K)=log⁡(2cosh⁡K)=log⁡2+∑m=1∞(−1)m+1m(cosh⁡K−1)m. The series converges absolutely at every such K and uniformly on each compact subinterval of (−12,12). This is a sufficient convergence range for this specific one-dimensional model, not a generic cluster-expansion result.

Facts & Assumptions

Given: A finite periodic Ising chain with zero field, J∈R, β>0, and the dimensionless coupling K=βJ.

[A1]

The chain Hamiltonian is HN(s)=−J∑i=1Nsisi+1 with sN+1=s1, and its transfer matrix is Ts,t=eKst.

[F1]

For this finite-alphabet periodic chain, the thermodynamic pressure exists and is log⁡λ+, where λ+ is the larger eigenvalue of the stated positive transfer matrix (One-dimensional periodic Ising-chain limit).

[F3]

The mean value theorem gives g(b)−g(a)=g′(c)(b−a) for continuous g on [a,b] and differentiable g on (a,b) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[F4]

The hyperbolic functions are defined from the exponential, satisfy cosh⁡K−1=2sinh⁡2(K/2), and have (sinh⁡x)′=cosh⁡x with cosh⁡ increasing on [0,∞) (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[F5]
[F6]

The exponential is positive, satisfies exp⁡(−x)=1/exp⁡x, and is strictly increasing (The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x), The exponential function is strictly increasing).

Proof

1.1A1F1F2algebra

At zero field the larger transfer-matrix eigenvalue in [F1] is λ+=eK+e−K=2cosh⁡K. Therefore the limiting pressure is P(K)=log⁡(2cosh⁡K)=log⁡2+log⁡(1+w) with w=cosh⁡K−1≥0.

1.2F2F3F4F5F6algebra

Put u=∣K∣/2<1/4. If u=0, then w=0. If u>0, [F3] applied to sinh⁡ on [0,u] gives sinh⁡u=ucosh⁡ξ for some 0<ξ<u; by [F4], cosh⁡ξ<cosh⁡u, while [F6] and the definition of cosh⁡ give cosh⁡u<eu, so sinh⁡u<ueu. The power-series definition of exp⁡ in [F2], with j!≥1, and [F5] give eu=∑j≥0uj/j!≤∑j≥0uj=(1−u)−1<4/3. Thus 0≤w=2sinh⁡2u<2(4u/3)2=8K2/9<2/9<1/4 when u>0, and w=0 when u=0.

2.1step 1.2F5algebra

For 0≤t≤w<1, the finite geometric identity gives ∑j=0n(−t)j=(1−(−t)n+1)/(1+t); the remainder is bounded uniformly by wn+1/(1−w). The pointwise sum is also (1+t)−1 by [F5], so these same remainder bounds show uniform convergence on [0,w].

3.1F2F5step 1.2step 2.1algebra∎

Integrate the finite geometric identity from 0 to w. The integral of its remainder has absolute value at most wn+2, since 1+t≥1 on [0,w]; hence these finite integrals converge to ∫0w(1+t)−1dt=log⁡(1+w) by [F2]. Integrating each finite sum gives ∑m=1n+1(−1)m+1wm/m, so passage to the limit yields log⁡(1+w)=∑m≥1(−1)m+1wm/m. If ∣K∣≤r<1/2 with r≥0, step 1.2 gives 0≤w≤8r2/9<2/9=:wr<1. The tail of the series of absolute values after index M is at most ∑m>Mwrm=wrM+1/(1−wr)→0, uniformly in K, by [F5]. This proves absolute convergence and uniform convergence on every such compact interval, including r=0.

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Positive Ising transfer matrix and stationary Markov law

Statement

Assume Countable Choice. For finite β>0 and J,h∈R, set K=βJ, b=βh, and define the positive symmetric transfer matrix on {−1,+1} by Ts,t=eKst+b(s+t)/2. Let λ+ be its larger eigenvalue and let r be its strictly positive unit eigenvector, with coordinates rs. Then Qs,t=Ts,trtλ+rs,πs=rs2 define a stochastic matrix Q and invariant probability vector π. Their consistent consecutive-block laws πsℓ∏i=ℓr−1Qsi,si+1 extend uniquely to a stationary probability measure on {−1,+1}Z.

Facts & Assumptions

Given: A one-dimensional finite-alphabet lattice and the zero-field-or-field nearest-neighbor Ising transfer matrix specified above, with β>0 and finite real J,h.

[A1]

Adopt Countable Choice for the consistent-cylinder extension step.

[F1]

The finite-alphabet product-measure lemma extends a consistent family of finite-dimensional probabilities uniquely to a probability on the cylinder sigma-algebra, under Countable Choice (Finite-alphabet product and probability compactness).

[F2]

The one-dimensional Ising transfer matrix has eigenvalues λ±=eKcosh⁡b±e2Ksinh⁡2b+e−2K with λ+>∣λ−∣ (One-dimensional periodic Ising-chain limit).

Proof

1.1A1F2algebra

The off-diagonal entries of T are positive. For a positive symmetric 2×2 matrix, its larger eigenvalue exceeds both diagonal entries; explicitly, if T=(accd) with c>0, then λ+−a>0 and the vector (c,λ+−a) is strictly positive and satisfies Tr=λ+r by the characteristic equation. Normalize it to length one.

2.1step 1.1algebra

Since Tr=λ+r, ∑tQs,t=1 for each s. Symmetry of T gives ∑sπsQs,t=rt∑sTt,srs/λ+=rt2=πt. Thus π is invariant.

3.1step 2.1algebra

The finite block probabilities in the Statement are normalized. Summing over the right endpoint uses the row sums of Q; summing over the left endpoint uses invariance of π. Summing over omitted sites therefore gives a consistent probability family on all finite coordinate sets.

4.1A1F1step 3.1∎

Apply [F1] to that consistent family. The resulting measure is stationary because its cylinder probabilities are shift invariant, and uniqueness on cylinders makes it the unique measure with the stated block laws.

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All-positive-temperature one-dimensional Ising DLR uniqueness

Statement

Assume Countable Choice, finite β>0, finite J,h∈R, and the one-dimensional nearest-neighbor Ising specification from Lattice Gibbs kernels and DLR states. Then there is exactly one DLR state. It is the stationary Markov law constructed from the positive transfer matrix in Positive Ising transfer matrix and stationary Markov law. For every fixed cylinder, finite-interval Gibbs probabilities converge to that law uniformly over the two endpoint boundary spins.

Facts & Assumptions

Given: The one-dimensional Ising chain, finite real parameters, and Countable Choice for the product-measure construction.

[A1]

Countable Choice is assumed only for the DLR measure extension and uniqueness framework (The Axiom of Countable Choice (ACω)).

[F1]

The positive transfer matrix has a stationary Markov law with transition probabilities Qs,t=Ts,trt/(λ+rs) and stationary weights πs=rs2 (Positive Ising transfer matrix and stationary Markov law).

[F2]

For every exhaustion and boundary sequence, subsequential finite-volume limits are DLR, and DLR laws are determined by their cylinder probabilities (Finite-range lattice DLR existence and closed convex state set).

[F3]

In d=1, the transfer-matrix model is realized by the interaction list with onsite shape {0} and Φ0(s0)=−hs0, and pair shape {0,1} and Φ1(s0,s1)=−Js0s1; its finite conditional kernels are proper and consistent (Finite-range lattice interaction and conditional energy, Conditional Gibbs kernel consistency).

Proof

1.1A1F1algebra

Let r be the positive unit eigenvector of T and w an orthogonal unit eigenvector. By [F1] and the eigenvalue formula, Tn/λ+n=rrT+(λ−/λ+)nwwT, with ∣λ−/λ+∣<1. Since both entries of r are positive, the denominator entries have a uniform positive lower bound after division by λ+n.

2.1F1step 1.1algebra

Fix an interval [−n,n], endpoint spins a,b, and a consecutive observed block sℓ,…,sr. Its conditional Gibbs probability is (Tn+ℓ+1)a,sℓ(∏i=ℓr−1Tsi,si+1)(Tn−r+1)sr,b(T2n+2)a,b. Using the uniform spectral expansion in step 1.1, this converges uniformly in a,b to πsℓ∏i=ℓr−1Qsi,si+1. Finite summation over missing intermediate spins gives the same convergence for every cylinder.

3.1A1F1F2F3step 2.1

The limiting cylinder law is the Markov law μ in [F1]. The specification in [F3] has the exact finite conditional energy used by the interval laws. By [F2], a subsequential limit of those finite-volume laws is DLR; step 2.1 forces every such limit to equal μ, so μ is DLR.

4.1F2step 2.1step 3.1∎

If ν is any DLR state, then for every cylinder C, ν(C)=νγ[−n,n](C). The conditional probability inside this integral differs uniformly in the exterior endpoints from μ(C) by a quantity tending to zero by step 2.1. Thus ν(C)=μ(C); [F2] gives ν=μ.

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Conditional Ising equilibrium phase predictions

Physical scope

Classical translation-invariant finite-range Ising pair models on Z^d, d>=1, under the stated sufficient uniqueness condition; finite-temperature nearest-neighbor chains; and zero-field two-dimensional nearest-neighbor ferromagnets with J>0 and beta J>=2 for coexistence and the pressure cusp. Conditional model predictions only; no measured critical point or universal phase diagram.

Statement

Assume Countable Choice and adopt Interacting classical Gibbs equilibrium adoption for the specified classical Ising model. In the SI interpretation of Classical Gibbs quantities and SI reference conventions, β=(kBT)−1 is in J−1, J is the coupling energy JI in joules, and h is an energy bias per dimensionless spin in joules, not a magnetic-field strength; hence βJ and βh are dimensionless. The conditional model predictions are: (i) for a one-dimensional nearest-neighbor chain with finite β>0, J,h∈R, the DLR equilibrium state is unique; (ii) for a translation-invariant finite-range Ising pair interaction with symmetric zero-diagonal couplings Jij=Jji, Jii=0 and constant finite field, the sufficient condition sup⁡i∑jtanh⁡(β∣Jij∣)<1 gives a unique DLR equilibrium state; and (iii) for the two-dimensional nearest-neighbor ferromagnet at h=0 and βJ≥2, there are distinct stationary DLR equilibrium states with opposite nonzero magnetization and the pressure is nondifferentiable in h at zero. These deductions are restricted to the stated classical model; they do not report a measured transition or infer a phase diagram between the sufficient regimes.

Facts & Assumptions

Given: A specified classical Ising Hamiltonian, β>0, finite-range couplings and fields as stated, Countable Choice, and the adopted Gibbs-equilibrium postulate.

[A1]

The equilibrium interpretation is an adopted model assumption, not a theorem about equilibration or nature (Interacting classical Gibbs equilibrium adoption).

[A2]

Countable Choice is used in the lattice probability construction (The Axiom of Countable Choice (ACω)).

[F1]

Under Countable Choice, the one-dimensional nearest-neighbor Ising chain with finite β>0 and finite J,h∈R has exactly one DLR state (All-positive-temperature one-dimensional Ising DLR uniqueness).

[F2]

Under a strict Ising influence row-sum bound, the finite-range DLR state is unique (Unique interacting Ising state at sufficient high temperature).

[F3]

Under Countable Choice, for the zero-field nearest-neighbor Ising ferromagnet on Z2 with J>0 and β>0 satisfying βJ≥2, there exist stationary DLR states μ+ and μ− whose one-site expectations satisfy μ+(si)≥m∗>0 and μ−(si)≤−m∗ at every site (Two-dimensional Ising Gibbs coexistence at low temperature).

[F4]

Those coexistence states force nondifferentiability of the field-dependent pressure at h=0 (Coexistence implies an Ising pressure cusp).

Proof

1.1A1A2F1

Apply [F1] to the one-dimensional model to obtain uniqueness of its mathematical DLR equilibrium state; by the adoption [A1], this is the model's unique equilibrium-state prediction under the stated conditions.

1.2A1A2F2

Under [A2], apply [F2] whenever its strict influence sum is below one. This is a sufficient uniqueness regime and makes no assertion about parameters where the sum is at least one.

1.3A1A2F3

For the stated zero-field two-dimensional nearest-neighbor ferromagnet with J>0 and βJ≥2, [F3] supplies two stationary DLR states with magnetizations separated from zero.

2.1A1F4step 1.3∎

By [F4], their variational inequalities force a nonzero gap between the left and right pressure secant slopes. Thus coexistence and the cusp follow conditionally from the adopted postulate and the stated parameters; no empirical accuracy or exact critical value is inferred.

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Gibbs pressure and entropy with physical reference units

Physical scope

Bounded finite-alphabet translation-invariant finite-range lattice models under the stated DLR and SI reference conventions; conditional model deductions only.

Statement

Assume the adopted interacting Gibbs-equilibrium postulate for a bounded finite-range translation-invariant lattice model in dimension d≥1, Countable Choice, lattice spacing a>0 metres, and temperature T>0 kelvin. For any stationary DLR equilibrium state with dimensionless pressure per site p, Shannon entropy density hSh in nats per site, and mean energy u in joules per site, the conditional model prediction is P=kBTadp,s=kBadhSh,f=uad−Ts=−kBTadp. The generalized pressure density P and free-energy density f have units J m−d; for d=3, P is in pascals (Pa=J m−3). The reference spin entropy density s has units J K−1m−d. This is a consequence of the bounded-lattice DLR variational identity; it makes no claim that the model describes a material.

Facts & Assumptions

Given: The adopted equilibrium model, a stationary DLR state for a bounded finite-range lattice interaction, SI parameters as in the local definition, and Countable Choice.

[F1]

The equilibrium meaning is adopted as a postulate for a chosen Hamiltonian and reference; it does not imply empirical accuracy or equilibration (Interacting classical Gibbs equilibrium adoption).

[F2]

The dimensionless pressure and entropy/energy densities satisfy p=hSh−βu for stationary DLR states under Countable Choice (DLR states attain the bounded lattice pressure variational principle).

[F3]

β=(kBT)−1, P=kBTp/ad, and s=kBhSh/ad in the declared SI convention (Classical Gibbs quantities and SI reference conventions).

[A1]

Countable Choice is the measure-extension assumption carried by [F2] (The Axiom of Countable Choice (ACω)).

Proof

1.1A1F2algebra

Under Countable Choice [A1], [F2] gives p=hSh−βu. Multiplication by −kBT/ad gives −kBTp/ad=−kBThSh/ad+(kBTβ)u/ad.

2.1F1F3step 1.1algebra∎

By [F3], kBTβ=1, s=kBhSh/ad, and P=kBTp/ad. Substitution in step 1.1 gives f=u/ad−Ts=−P. This exact conclusion is conditional on [F1] and the declared reference convention, not an empirical claim.

5 · Examples, experiments and counterexamples

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