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Planar random clusters, sharpness and the exact square-lattice critical point — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Square-lattice Ising critical coupling from the two-state Potts model

Example

On Λn=[−n,n]2∩Z2, let En be the nearest-neighbor edges with both endpoints in Λn and fix plus boundary spins σx=+1 whenever ∥x∥∞=n. For σx∈{−1,+1}, define the finite-volume Ising Hamiltonian and Gibbs law by HI,n(σ)=−JI∑{x,y}∈Enσxσy,Pβ,nI,+(σ)∝e−βHI,n(σ),JI>0. Define mI+(β):=lim⁡n→∞Pβ,nI,+(σ0=+1)−1/2 when the limit exists, and βcI:=inf⁡{β>0:mI+(β)>0}. The limit exists and βcIJI=12log⁡(1+2).

Facts & Assumptions

Given: The Ising interaction above, JI>0, and the square-lattice Q=2 Potts critical-temperature theorem.

[F1]

For σx,σy∈{−1,+1}, 1{σx=σy}=(1+σxσy)/2 by checking the two possible products +1 and −1.

[F2]

For the two-state Potts model with energy −JP∑{x,y}1{τx=τy}, its critical inverse temperature βcP satisfies βcPJP=log⁡(1+2) (Square-lattice Potts critical inverse temperature).

Verification

technique · direct coupling conversion
1.1F1algebra

For each finite box and edge, [F1] gives −JP1{σx=σy}=−(JP/2)σxσy−JP/2. Setting JP=2JI makes the finite Potts Hamiltonian equal to HI,n(σ)−JI∣En∣. The difference is independent of σ, so it cancels from the normalized finite Gibbs probabilities. The boundary color 1 in the Potts model corresponds to the Ising plus spin +1. Thus the two models have the same plus-boundary probabilities and the same critical inverse temperature.

2.1F2step 1.1algebra∎

Let βcP and βcI denote the critical inverse temperatures in the two definitions. Step 1.1 gives βcI=βcP. By [F2] with JP=2JI, βcP(2JI)=log⁡(1+2). Divide by 2 to obtain βcIJI=12log⁡(1+2). This is an exact coupling conversion; the constant energy shift does not affect probabilities.

Remarks

The factor of two comes from 1{σx=σy}=(1+σxσy)/2; no critical exponent or transition order is inferred.

Sources