D-011 — Task 1.3: third moment tr G̃³ — what survives unconditionally?

Date: 2026-08-17 | Session: 001 | Task: 1.3 | Status: analysis complete; decisive numerics (E-005) designed Sources: D-004 (prime-side machinery), D-005 (chain), D-006 (§7.2 walls), D-010 (T1 moments). Tags: [VERIFIED] derivations / [ANALYSIS] conclusions.

1. Reduction of tr G̃³ to a triple zero correlation

G̃ = (aL²)⁻¹ Σ_ρ m_ρ v_ρv_ρᵀ ⟹ tr G̃³ = (aL²)⁻³ Σ_{ρ₁,ρ₂,ρ₃} m₁m₂m₃ (v₁·v₂)(v₂·v₃)(v₃·v₁). By Lemma 2.1, v_ρ·v_ρ' = LΦ(γ_ρ − γ_ρ') + edge, Φ = ϕ²̂: tr G̃³ ≈ (a²L³)⁻¹ Σ_{ρ₁,ρ₂,ρ₃} m₁m₂m₃ Φ(γ₁−γ₂)Φ(γ₂−γ₃)Φ(γ₃−γ₁). [VERIFIED algebra] — a triple zero-correlation with band-limited kernel.

2. Prime-side cost (why k=3 is different from k=2)

Fourier-expand each Φ(γᵢ−γⱼ) = ∫ϕ²(u)e^{−iu(γᵢ−γⱼ)}du (supp ϕ² ⊂ [−L/2,L/2]): the triple sum factorizes into integrals of products of THREE single-zero sums Σ_ρ e^{−iγt}, each |t| ≤ L. Via the explicit formula each becomes a prime sum P_X(t); the diagonal method must then evaluate shift-correlations ∫ P_X(τ)P_X(τ+x)P_X(τ+y)dτ.

• Diagonal: multiplicative relations n₁n₂ = n₃ (sign pattern ++−) — EVALUABLE (von Mangoldt convolution identities, Λ∗Λ etc.).
• Off-diagonal: frequencies log(n₁n₂) for n₁n₂ ≤ X² — about X² log X distinct frequencies; Montgomery–Vaughan separation δ⁻¹ ~ X²; the MV bound on the off-diagonal costs a factor X²/T against the main term. ⟹ NEEDS X² ≲ T, i.e. X³ ≲ T^{3/2}... the Rudnick–Sarnak condition is Xᵏ ≤ T^{2−ε}: for k=3, X ≤ T^{2/3−ε} [RS96]. At X = T/2π: FAILS. (k=2: X² = T²/(2π)² ≤ T^{2−ε} ✔ — the 1/(2π)² margin is why Thm 5.7 works at full bandwidth. Verified consistent with D-004.)

3. The bandwidth–moment tradeoff (the naive wall)

Work at reduced bandwidth: L′ = λℓ, X = (T/2π)^λ (Remark 6.1). Then:

• tr G̃³ available unconditionally ⟺ X³ ≤ T^{2−ε} ⟺ λ ≤ 2/3 − o(1).
• Grid is critical-density for the window: h′ = 2π/L′, so d = TL′/2π = λ·(Tℓ/2π) = (λ + o(1))N.
• The certificate satisfies N^s_0 + o(N) ≥ rank P₁ with rank P₁ ≤ d: any certificate ≤ λ + o(1). ⟹ third moment usable only at λ ≤ 2/3, where the dimension cap is 2/3. The moment becomes available EXACTLY where it stops helping: the RS wall for this method is the coincidence λ_RS(k=3) = 2/3 = current certificate. [This is the precise mechanistic form of the paper's "higher moments add nothing" — sharper: they add nothing BECAUSE availability and usefulness thresholds coincide at 2/3.]

4. The oversampling escape hatch (OPEN — needs numerics)

The naive wall assumes the critical-density grid h = 2π/L′. But Lemma 2.1's no-aliasing only needs supp(ϕ∗ϕ) ⊂ (−2π/h, 2π/h): an OVERSAMPLED grid h = 2π/(qL′), q ≥ 1, keeps the Poisson–Gabor identity EXACT (m ≠ 0 terms evaluate H at multiples of qL′ outside supp H = [−L′,L′]). [VERIFIED against Lemma 2.1's proof — H support argument.] With q > 1: d = qλN and the d-cap moves to qλ — at q = 2, λ = 2/3: d = 4N/3 > N, cap gone. THE NAIVE WALL HAS A HOLE. But: G̃ = VᵀV with V having N(I′) rows ⟹ rank G̃ ≤ N(I′) regardless of d; oversampling pads the spectrum with correlated/near-zero directions. Open questions (decisive for task 1.3): Q1. Does the oversampled (q=2) Gram matrix carry MORE usable positive index than the critical one, or is the extra rank spectrally empty? Q2. With 3 moments at λ = 2/3, q = 2: does the CMS/Christoffel bound for n₊(G̃)/d × (d/N) exceed 2/3? Q3. What does the renormalized trace bookkeeping do to the certificate constants (tr scales by q; certificate must be q-invariant to be meaningful — check). These are EXACTLY numerically decidable on the sine process (the method's own model): compute m₁,m₂,m₃(λ,q) for the λ-bandwidth, q-oversampled sinc-Gram, run the 3-moment one-sided (Markov) bound, scan λ ∈ [1/2,1], q ∈ {1,2}. → E-005 (next sub-stage). NOTE: E-005 decides the METHOD-level question (what the data class can certify). If E-005 shows > 2/3 is achievable at λ ≤ 2/3, q = 2, the remaining analytic work is making tr G̃³ rigorous at that λ (RS-range, so plausible) — a genuine theorem path. If E-005 shows ≤ 2/3 everywhere: G-002 with the exact obstruction.

5. Sanity-gate preview (adversarial)

The sine-process model CANNOT see the D–H/Epstein obstruction (it models zeta's actual statistics). A positive E-005 result would still require the full prime-side evaluation — the model only falsifies or motivates. Also note: at λ < 1 the certificate's prime side changes (R_λ), so the comparison must use the λ-dependent 2-moment baseline R_λ(ψ₀) = 1/λ + λ/3 (Remark 6.1, D-004 §9) — built into E-005 design.

6. Status

Task 1.3 analysis COMPLETE: the wall is precisely located (λ_RS = 2/3 coincidence), and a concrete escape hatch (oversampled frames at reduced bandwidth) is formulated with decidable questions. E-005 next.