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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)
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Trivial reflection-positive Schwinger hierarchy

Example

For each n≥1, let Sn be the zero functional on S((R2)n), and set S0=1. This is a degenerate but reflection-positive Euclidean Schwinger hierarchy in the two-dimensional conventions of this page.

Facts & Assumptions

Given: Use the definitions of a Euclidean Schwinger hierarchy and its reflection-positive test algebra from this pair.

[F1]

A Schwinger hierarchy fixes S0=1 and continuous tempered functionals Sn on ((R2)n) (Euclidean Schwinger hierarchy in two dimensions).

[F2]

Euclidean covariance and block symmetry mean Sn(Tg,nf)=Sn(f) and Sn(Tπ,nf)=Sn(f) (Euclidean Schwinger hierarchy in two dimensions).

[F3]

The reflection-positive quadratic form is QS(F)=∑j,kaj‾akSj+k(Θfj⊗fk) (Reflection-positive Euclidean field data).

[A1]

The example chooses Sn=0 for every n≥1 and S0=1.

Verification

1.1givenA1F1F2algebra

The zero functional is complex-linear and continuous on every Schwartz space. Its pullback by every Euclidean isometry or block permutation is still zero, so the normalization in [F1] and covariance and symmetry equalities in [F2] hold; the chosen degree-zero value is the required normalization.

2.1givenA1F3step 1.1algebra

In [F3], every summand with j+k≥1 vanishes by [A1]. The only possible surviving term is j=k=0, and it equals ∣a0∣2∣f0∣2S0=∣a0f0∣2. Hence QS(F)=∣a0f0∣2≥0 is real for every finite family, proving reflection positivity.

3.1givenA1step 2.1construct∎

Let η(s)=exp⁡(−1/(1−s2)) for ∣s∣<1 and η(s)=0 for ∣s∣≥1, and choose f1(t,x)=η(t−2)η(x), which is a nonzero smooth compactly supported test function in O+; set a1=1 and a0=0, with every other coefficient zero. Then the polynomial is nonzero while step 2.1 gives QS(F)=0, so the positive form is degenerate. The example asserts neither a nontrivial field nor OS reconstruction.

Depends on

Dependency tree · 2 levels

Sources