TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Finite Edwards–Sokal coupling and correlation identity

Statement

Let $G = (V, E)$ be a finite simple graph, let $Q \geq 2$ be an integer, and let each edge activity $v_{e}$ be finite and nonnegative. Use the Potts weight $w_{P} (σ) = \prod_{e = {u_{e}, w_{e}}} (1 + v_{e} 1_{{σ (u_{e}) = σ (w_{e})}})$ and FK weight $w_{F K} (A) = Q^{k_{G} (A)} \prod_{e \in A} v_{e}$ from the finite definitions. Define $W (σ, A) : = \prod_{e \notin A} 1 \prod_{e \in A} v_{e} 1_{{σ (u_{e}) = σ (w_{e})}}, (σ, A) \in [Q]^{V} \times 2^{E} .$ The normalized joint law $ν (σ, A) : = W (σ, A) / \sum_{τ, B} W (τ, B)$ has Potts marginal $w_{P} / Z_{P}$ and FK marginal $w_{F K} / Z_{F K}$ , with $Z_{P} = Z_{F K} = \sum_{τ, B} W (τ, B)$ . Conditional on an edge set $A$ of positive probability, the colors of its connected components are independent and uniform on $[Q]$ . Thus for every $x, y \in V$ , $P (σ (x) = σ (y)) = \frac{1}{Q} + (1 - \frac{1}{Q}) P_{F K} (x \leftrightarrow y),$ where $x \leftrightarrow y$ means that $x$ and $y$ lie in the same component of $G_{A}$ .

Statement references

Finite Potts and Fortuin–Kasteleyn weights

Facts & Assumptions

Given: A finite simple graph $G = (V, E)$ , integer $Q \geq 2$ , and finite activities $v_{e} \geq 0$ .

[F1]

Finite Potts and Fortuin–Kasteleyn weights: the spin weight is $\prod_{e} (1 + v_{e} 1_{{σ (u_{e}) = σ (w_{e})}})$ , the FK weight of $A$ is $Q^{k_{G} (A)} \prod_{e \in A} v_{e}$ , and both finite normalizers are positive.

Proof

technique · direct

1.1givenF1algebra

For fixed $σ$ , summing $W (σ, A)$ over all $A \subseteq E$ chooses, independently for each edge, either its absent factor $1$ or its present factor $v_{e} 1_{{σ (u_{e}) = σ (w_{e})}}$ . Distributivity of the finite product therefore gives $\sum_{A} W (σ, A) = w_{P} (σ)$ . Summing this equality over the finite set $[Q]^{V}$ gives $\sum_{σ, A} W (σ, A) = Z_{P}$ .

1.2F1algebra

Fix $A \subseteq E$ . The product $W (σ, A)$ is zero unless the spins agree across every edge in $A$ . Equality propagates along each finite path, so the admissible configurations are exactly those constant on each connected component of $G_{A}$ . Assigning one of $Q$ colors independently to each of its $k_{G} (A)$ components gives exactly $Q^{k_{G} (A)}$ such configurations, each with weight $\prod_{e \in A} v_{e}$ . Hence $\sum_{σ} W (σ, A) = w_{F K} (A)$ , including when this weight is zero. Summing over $A$ gives $\sum_{σ, A} W (σ, A) = Z_{F K}$ .

2.1F1step 1.1step 1.2algebra

The common total is positive by [F1], so dividing $W$ by it defines a probability law. The finite marginal sums in steps 1.1 and 1.2 show that its spin and edge marginals are respectively $w_{P} / Z_{P}$ and $w_{F K} / Z_{F K}$ , and that $Z_{P} = Z_{F K}$ .

2.2step 1.2algebra

For a fixed $A$ with positive FK probability, every spin assignment constant on its components has the same joint weight $\prod_{e \in A} v_{e}$ . By step 1.2 there are $Q^{k_{G} (A)}$ such assignments, so the conditional law is uniform on them; this is equivalent to assigning independent uniform colors to the components.

3.1step 2.1step 2.2algebra∎

If $x, y$ are connected in $G_{A}$ , step 2.2 makes their colors equal with conditional probability $1$ . If they are in distinct components, their independent uniform colors agree with probability $1 / Q$ . Averaging over the FK marginal from step 2.1 yields $P (σ (x) = σ (y)) = P_{F K} (x \leftrightarrow y) + Q^{- 1} P_{F K} (x \leftrightarrow̸ y) = Q^{- 1} + (1 - Q^{- 1}) P_{F K} (x \leftrightarrow y)$ .

Depends on

Finite Potts and Fortuin–Kasteleyn weights