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Interacting marginals are not free Gibbs laws

Statement refuted

A finite marginal of an interacting Gibbs chain always equals the free Gibbs law obtained by retaining only interaction terms wholly inside the observed sites.

Facts & Assumptions

Given: A three-site zero-field Ising chain with spins s1,s2,s3∈{−1,+1}, open-chain Hamiltonian H=−J(s1s2+s2s3), β>0, and J≠0.

[A1]

Put K=βJ and define sinh⁡K:=(eK−e−K)/2, cosh⁡K:=(eK+e−K)/2, and tanh⁡K:=sinh⁡K/cosh⁡K=(eK−e−K)/(eK+e−K). The denominator is positive by positivity of the exponential; since K≠0, strict increase gives eK≠e−K, so tanh⁡K≠0 (The real exponential function and the number e by a power series, The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x), The exponential function is strictly increasing).

[A2]

The specification uses conditional laws given the exterior; it does not identify them with unconditional finite marginals (Lattice Gibbs kernels and DLR states).

[F1]

The finite Gibbs weights are proportional to e−βH and have positive finite normalizer (Finite Gibbs relative entropy and free-energy minimum).

Counterexample

1.1givenA1F1algebra

With K, sinh⁡K, cosh⁡K, and tanh⁡K as in [A1], summing over the middle spin gives Z=2(2cosh⁡K)2 and E(s1s3)=2(2sinh⁡K)22(2cosh⁡K)2=tanh⁡2K>0. Thus the endpoint marginal is correlated.

2.1A2F1step 1.1algebra∎

The only interaction terms are on bonds {1,2} and {2,3}, so no term is wholly supported on the retained endpoint set {1,3}. The claimed retained-term free law therefore has zero energy and assigns independent uniform endpoint spins, with correlation zero. This contradicts step 1.1, refuting the statement.

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