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Classical Potts random clusters and planar loop transfer — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Two-vertex Potts and FK weights

Example

Let $G = (V, E)$ have vertex set $V = {x, y}$ with $x \neq y$ and edge set $E = {e}$ , where $e = {x, y}$ has finite activity $v \geq 0$ , and let $Q \geq 2$ be an integer. Its FK normalizer is $Z_{F K} = Q^{2} + Q v$ , and under the normalized FK law $P_{F K} (x \leftrightarrow y) = v / (Q + v)$ . The finite Potts normalizer is the same and the same-color probability is $P (σ (x) = σ (y)) = \frac{1 + v}{Q + v} .$

Example references

Finite Edwards–Sokal coupling and correlation identity

Facts & Assumptions

Given: The graph $G = (V, E)$ with $V = {x, y}$ , $x \neq y$ , and $E = {e}$ for $e = {x, y}$ , integer $Q \geq 2$ , and finite activity $v \geq 0$ .

[F1]

Finite Potts and Fortuin–Kasteleyn weights: for a subset $A \subseteq E$ , its FK weight is $Q^{k_{G} (A)} \prod_{e \in A} v_{e}$ .

[F2]

Finite Edwards–Sokal coupling and correlation identity: $P (σ (x) = σ (y)) = Q^{- 1} + (1 - Q^{- 1}) P_{F K} (x \leftrightarrow y)$ .

Verification

technique · direct finite enumeration

1.1givenF1algebra

There are two edge subsets. The empty subset has two isolated components and weight $Q^{2}$ ; the occupied subset has one component and weight $Q v$ . Thus $Z_{F K} = Q^{2} + Q v$ and $P_{F K} (x \leftrightarrow y) = Q v / (Q^{2} + Q v) = v / (Q + v)$ , also when $v = 0$ .

2.1F2step 1.1algebra

Substitution in [F2] gives $P (σ (x) = σ (y)) = Q^{- 1} + (1 - Q^{- 1}) v / (Q + v) = (1 + v) / (Q + v)$ .

3.1givenF1step 2.1algebra∎

Direct spin enumeration checks the same result: $Q$ equal-color assignments each have weight $1 + v$ , and $Q (Q - 1)$ unequal-color assignments each have weight $1$ , so $Z_{P} = Q (1 + v) + Q (Q - 1) = Q^{2} + Q v$ and the same-color probability is $Q (1 + v) / (Q^{2} + Q v) = (1 + v) / (Q + v)$ .