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Euclidean positivity, controlled limits and reconstruction boundaries — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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

Ordinary positive measure need not be reflection positive

Statement refuted

A positive probability measure with a positive-definite ordinary covariance matrix must be reflection positive for the involution exchanging two sites.

Facts & Assumptions

Given: Fix a real parameter $r$ with $0 < r < 1$ . On the four-point set ${(- 1, - 1), (- 1, 1), (1, - 1), (1, 1)}$ assign probabilities

P (X = 1, Y = 1) = P (X = - 1, Y = - 1) = \frac{1 - r}{4}, P (X = 1, Y = - 1) = P (X = - 1, Y = 1) = \frac{1 + r}{4} .

On the complex algebra of functions on this four-point space, generated by the coordinate functions $X, Y$ with $X^{2} = Y^{2} = 1$ , define the conjugate-linear map by $Θ (c 1) = \overset{ˉ}{c} 1$ , $Θ X = Y$ , $Θ Y = X$ , and $Θ (a b) = Θ (b) Θ (a)$ . These rules give an involution because they preserve the relations and square to the identity on scalars and generators. Define the positive-site polynomial algebra to be the unital subalgebra generated by $Y$ .

[F1]

Euclidean Schwinger hierarchies and controlled limit data: ordinary positivity tests $E [\overset{ˉ}{F} F] \geq 0$ , while reflection positivity tests $E [(Θ F) F]$ for positive-site polynomials.

Counterexample

1.1givenalgebra

All four weights are positive and sum to $(1 - r) / 2 + (1 + r) / 2 = 1$ . Symmetry of the support weights gives $E [X] = E [Y] = 0$ , and $X^{2} = Y^{2} = 1$ at every support point, so the covariance diagonal is $(1, 1)$ . The equal-sign outcomes have total probability $(1 - r) / 2$ and product $X Y = 1$ ; the opposite-sign outcomes have total probability $(1 + r) / 2$ and product $X Y = - 1$ . Therefore $E [X Y] = (1 - r) / 2 - (1 + r) / 2 = - r$ .

2.1F1givenstep 1.1algebra∎

The covariance matrix is $C = (\begin{array}{cc} 1 & - r \\ - r & 1 \end{array})$ . For real $u, v$ , $(u, v) C (u, v)^{T} = (u - r v)^{2} + (1 - r^{2}) v^{2},$ which is strictly positive unless $(u, v) = (0, 0)$ because $0 < r < 1$ . Thus $C$ is positive definite and the probability law is ordinarily positive. Since $Y$ belongs to the declared positive-site algebra, take $F = Y$ . Reflection positivity would require $E [(Θ F) F] = E [X Y] \geq 0$ , but this is $- r < 0$ . Hence the statement is false.