工程紀錄 · 第五弧線 v3.12 · 2026-09-03 · UMP4_COUNTEREXAMPLE · CONSTRAINT_ADAPTED_CENTERING · DOUBLE_AXIS_REMOVAL · RH_CLAIM_FALSE # Modulated Pairing Uniformity：UMP4 反例、Parallelogram Constraint-Adapted Centering 與 Double-Axis Remainder **RH-ModulatedPairingUniformity v3.12** 本節點承接： - `RH-Parallelogram-AddingFractionsRestriction v3.11` - `RH-Parallelogram-SobolevTailTransfer v3.10` - Montgomery–Soundararajan fourth-moment pairing method - Kuperberg smooth weighted pairing theorem v3.11 提出： $$ \operatorname{UMP4}(\delta) $$ 作為 candidate： > 將 smooth fourth-order pairing theorem 對 modulation sign pattern > $$ > (+\lambda,-\lambda,-\lambda,+\lambda) > $$ > uniformize，並保留： > $$ > H^{2-\delta} > $$ > pairing-free error。 v3.12 完成 proof-portability audit。 結果不是： ```text UMP4 still open ``` 而是： $$ \boxed{ \operatorname{UMP4}(\delta) \text{ is false for every }\delta>0. } $$ 甚至 normalized finite-character averaging後，標準 three-pairing subtraction仍留下： $$ \boxed{ \Theta(H^3) } $$ resonant remainder。 核心 exact counterexample只需要 finite conductor： $$ \boxed{q=2} $$ 與： $$ \boxed{\lambda=\frac12}. $$ 這個 failure 精確對應 Montgomery–Soundararajan proof 中原本被視為 lower-order 的： $$ \boxed{ \text{three-or-more equal denominators} } $$ stratum。 在 unshifted setting，frequency： $$ \frac12 $$ 遠離 integer，smooth exponential sum有 cancellation。 modulation： $$ \lambda=\frac12 $$ 將它平移至： $$ 0, $$ 使同一 stratum resonance 到： $$ H^4. $$ 因此 local shift-uniform two-point lemmas不能推出 global modulation-uniform pairing theorem。 --- 更深一層，v3.12 發現： $$ \boxed{ \text{global Wick centering} \neq \text{parallelogram constraint-adapted centering}. } $$ 對： $$ \mu_x=\mathfrak S(x)-1, $$ full Wick-connected fourth cumulant可寫成： $$ \boxed{ \mathfrak C_4(h,d) = \mathfrak S_0(0,h,d,h+d) - \mu_h^2 - \mu_d^2 - \mu_{h+d}\mu_{h-d}. } $$ 但 v3.6 canonical covariance是： $$ \boxed{ \mathfrak K_4(h,d) = \mathfrak S_0(0,h,d,h+d) - \mu_h^2. } $$ 在 finite squarefree modulus $q$ 上，令： $$ A(q) = \prod_{p\mid q} \left( 1+\frac1{(p-1)^3} \right). $$ 則： $$ \boxed{ \mathbb E_{h,d} \mathfrak S_{0,q}(0,h,d,h+d) = A(q)-1, } $$ $$ \boxed{ \mathbb E_h\mu_q(h)^2 = A(q)-1, } $$ 而： $$ \boxed{ \mathbb E_{h,d} \mu_q(h+d)\mu_q(h-d) = \mathbf1_{2\mid q}. } $$ 所以： $$ \boxed{ \mathbb E\mathfrak C_{4,q} = - [A(q)-1] - \mathbf1_{2\mid q}, } $$ 通常不是零。 反而： $$ \boxed{ \mathbb E\mathfrak K_{4,q}=0 } $$ exactly。 所以在 parallelogram slice 上，第二與第三 pairing channels不是可獨立丟掉的 nuisance terms。 它們參與 constraint-induced cancellation。 --- v3.12 因此提出新的 exact decomposition。 v3.8 已知： $$ \widehat{\mathfrak K}_{4,q}(\alpha,0)=0. $$ 又有： $$ \boxed{ \widehat{\mathfrak K}_{4,q}(0,\beta) = \widehat{\mu_q^2}(\beta) \qquad \beta\ne0. } $$ Define： $$ \boxed{ \mathfrak A_q(d) = \mu_q(d)^2 - [A(q)-1]. } $$ and： $$ \boxed{ \mathfrak K_{4,q}^{\perp}(h,d) = \mathfrak K_{4,q}(h,d) - \mathfrak A_q(d). } $$ Then： ## Double-Axis Theorem $$ \boxed{ \widehat{ \mathfrak K_{4,q}^{\perp} }(\alpha,0) = 0 \quad \forall\alpha, } $$ and： $$ \boxed{ \widehat{ \mathfrak K_{4,q}^{\perp} }(0,\beta) = 0 \quad \forall\beta. } $$ Moreover： $$ \boxed{ \mathfrak K_{4,q}^{\perp}(h,d) = \mathfrak K_{4,q}^{\perp}(d,h). } $$ Thus the deterministic parallelogram model now splits into： ```text one-dimensional centered pair-square axis + genuinely two-dimensional axis-free remainder. ``` This is the correct constraint-adapted replacement for UMP4。 No RH claim is made。 **RH_CLAIM = False.** --- # 0. Claim register ```text RH_PROVED = FALSE RH_DISPROVED = FALSE UMP4(delta) = FALSE FOR EVERY delta > 0 AVERAGED_UMP4(delta) = FALSE FOR EVERY delta > 0 Q2_POINTWISE_RESONANCE = CLOSED Q2_AVERAGED_RESONANCE = CLOSED MODULATION_PORTABILITY_BLOCKER = HIGHER_DENOMINATOR_MULTIPLICITY_RESONANCE LOCAL_SHIFT_UNIFORM_LEMMAS = TRUE_EXTERNALLY GLOBAL_SHIFT_UNIFORM_PAIRING_THEOREM = FALSE GLOBAL_WICK_CENTERING_IS_PARALLELOGRAM_CENTERING = FALSE FINITE_Q_CONNECTED_CUMULANT_MEAN = CLOSED FINITE_Q_K4_MEAN_ZERO = CLOSED_FROM_V3_7 DOUBLE_CENTERING_AXIS_DECOMPOSITION = CLOSED K4_PERP_BETA_ZERO_AXIS = ZERO K4_PERP_ALPHA_ZERO_AXIS = ZERO K4_PERP_H_D_SYMMETRY = CLOSED MIXED_TWO_DIRECTION_SOBOLEV_UNIFORMITY = OPEN ONE_DIMENSIONAL_AXIS_TAIL = OPEN FULL_DETERMINISTIC_ETA_M_POSITIVE = NOT_PROVED ACTUAL_PRIME_DEVIATION = OPEN GLOBAL_RH_CERTIFICATE = FALSE ``` --- # 1. Smooth exponential-sum notation Let： $$ f:\mathbb R\to\mathbb R $$ be nonzero, smooth, compactly supported。 Define： $$ \boxed{ E_{f,H}(\alpha) = \sum_{n\in\mathbb Z} f(n/H)e(n\alpha). } $$ If： $$ \int f\ne0, $$ then： $$ \boxed{ E_{f,H}(0) = H\int_{\mathbb R}f(x)dx + O_f(1). } $$ Thus： $$ E_{f,H}(0)\asymp_f H. $$ --- # 2. Finite conductor $q=2$ In the finite refined singular-series moment： $$ V_4(q,H;f_1,\ldots,f_4), $$ take： $$ q=2. $$ The only allowed divisor： $$ q_i>1,\quad q_i\mid2 $$ is： $$ q_i=2. $$ The only reduced numerator is： $$ a_i=1. $$ Also： $$ \frac12+\frac12+\frac12+\frac12=2\in\mathbb Z. $$ Since： $$ \frac{\mu(2)}{\phi(2)}=-1, $$ the four coefficient product is： $$ 1. $$ Therefore： $$ \boxed{ V_4(2,H;f_1,\ldots,f_4) = \prod_{i=1}^4 E_{f_i,H}\left(\frac12\right). } $$ --- # 3. Modulation sign pattern Introduce the v3.11 modulation： $$ (\lambda_1,\lambda_2,\lambda_3,\lambda_4) = (\lambda,-\lambda,-\lambda,\lambda). $$ Then： $$ \boxed{ V_4^{(\lambda)}(2,H) = E_1\left(\frac12+\lambda\right) E_2\left(\frac12-\lambda\right) E_3\left(\frac12-\lambda\right) E_4\left(\frac12+\lambda\right). } $$ Here： $$ E_i=E_{f_i,H}. $$ --- # 4. Every pair term at $q=2$ For any pair： $$ (i,j), $$ the finite： $$ V_2^{(\lambda_i,\lambda_j)}(2,H) $$ contains exactly the same denominator： $$ 2 $$ and residue： $$ 1. $$ Hence the product over either member of any perfect matching gives： $$ \boxed{ \prod_{i=1}^4 E_i\left( \frac12+\lambda_i \right). } $$ There are： $$ \boxed{3} $$ perfect matchings of four labeled slots。 Therefore the standard pairing sum is exactly： $$ \boxed{ 3 \prod_{i=1}^4 E_i\left( \frac12+\lambda_i \right). } $$ --- # 5. Exact $q=2$ pairing-free remainder Define： $$ \mathcal R_{4,2}^{(\lambda)} = V_4^{(\lambda)} - \sum_{\sigma\in B_4} \prod_{(i,j)\in\sigma} V_2^{(\lambda_i,\lambda_j)}(i,j). $$ Sections 3–4 give： ## Theorem 5.1 $$ \boxed{ \mathcal R_{4,2}^{(\lambda)} = -2 E_1\left(\frac12+\lambda\right) E_2\left(\frac12-\lambda\right) E_3\left(\frac12-\lambda\right) E_4\left(\frac12+\lambda\right). } $$ This is exact。 --- # 6. Resonant modulation $\lambda=1/2$ Take： $$ \boxed{ \lambda=\frac12. } $$ Modulo one： $$ \frac12+\lambda=0, $$ and： $$ \frac12-\lambda=0. $$ Therefore： ## Theorem 6.1 · Resonant UMP Counterexample $$ \boxed{ \mathcal R_{4,2}^{(1/2)} = -2 \prod_{i=1}^4 E_{f_i,H}(0). } $$ If： $$ \int f_i\ne0, $$ then： $$ \boxed{ \mathcal R_{4,2}^{(1/2)} = -2H^4 \prod_{i=1}^4 \left( \int f_i \right) + O(H^3). } $$ Thus： $$ \boxed{ | \mathcal R_{4,2}^{(1/2)} | \asymp H^4. } $$ --- # 7. UMP4 is false Any： $$ \operatorname{UMP4}(\delta) $$ candidate with： $$ \delta>0 $$ would demand： $$ \sup_\lambda | \mathcal R_{4}^{(\lambda)} | \ll H^{2-\delta+\varepsilon}. $$ But Section 6 gives： $$ H^4. $$ Therefore： ## Corollary 7.1 $$ \boxed{ \operatorname{UMP4}(\delta) \text{ is false for every }\delta>0. } $$ Indeed even： $$ O(H^{4-\varepsilon}) $$ uniformity fails at this resonance。 --- # 8. Why this does not contradict the unshifted theorem At： $$ \lambda=0, $$ the problematic factor is： $$ E_{f,H}\left(\frac12\right). $$ For smooth compactly supported $f$，Poisson summation gives rapid decay when the frequency remains a fixed positive distance from the integers。 So the all-equal-denominator configuration is small in the unshifted setting。 At： $$ \lambda=\frac12, $$ the same frequency is translated to： $$ 0, $$ and： $$ E_{f,H}(0)\asymp H. $$ Therefore modulation changes the strength class of a denominator-multiplicity stratum。 --- # 9. Exact proof-audit blocker The Montgomery–Soundararajan fourth-moment proof isolates as main terms denominator tuples： $$ \boxed{ q_i\text{ equal in pairs with no further equalities}. } $$ Configurations in which： $$ \boxed{ \text{three or more }q_i\text{ are equal} } $$ are placed inside the lower-order error。 The $q=2$ counterexample lies exactly in this excluded higher-multiplicity stratum。 Hence the proof-portability blocker is： $$ \boxed{ \text{modulation can resonate a higher-multiplicity denominator collision}. } $$ --- # 10. Local shift-uniform estimates are not sufficient Smooth two-point lemmas can be stated for arbitrary additive shifts。 However a shifted second-moment estimate itself contains resonant spikes。 For example, even modulus and shift near： $$ \frac12 $$ can retain an： $$ H^2 $$ main contribution。 Therefore： $$ \boxed{ \text{local shift-uniform statement} \not\Longrightarrow \text{uniform global pairing decomposition}. } $$ v3.11's inference direction is rejected。 --- # 11. Averaged modulation does not repair UMP4 Now take： $$ f_1=f_2=f_3=f_4=f $$ real。 From Theorem 5.1 and： $$ E_f(-x)=\overline{E_f(x)}, $$ we get： $$ \boxed{ \mathcal R_{4,2}^{(\lambda)} = -2 \left| E_{f,H} \left( \frac12+\lambda \right) \right|^4. } $$ Let： $$ \lambda_j=\frac jM, $$ and choose： $$ M $$ larger than the complete additive diameter of the lattice support of： $$ f(n/H). $$ Then finite character orthogonality gives： ## Theorem 11.1 $$ \boxed{ \frac1M \sum_{j=0}^{M-1} \left| E_{f,H} \left( \frac12+\frac jM \right) \right|^4 = \sum_{ n_1+n_2=n_3+n_4 } c_{n_1}c_{n_2}c_{n_3}c_{n_4}, } $$ where： $$ c_n=f(n/H). $$ Equivalently： $$ \boxed{ = \sum_s \left| (c*c)(s) \right|^2. } $$ --- # 12. The averaged resonance is $H^3$ For nonzero compactly supported continuous： $$ f, $$ Riemann-sum scaling gives： $$ \boxed{ \sum_s | (c*c)(s) |^2 = H^3 \int_{\mathbb R} | (f*f)(x) |^2dx + o(H^3). } $$ The convolution is not identically zero when： $$ f\ne0. $$ Therefore： ## Theorem 12.1 · Averaged UMP Counterexample $$ \boxed{ \frac1M \sum_{j=0}^{M-1} \mathcal R_{4,2}^{(j/M)} = - 2C_fH^3 + o(H^3), } $$ where： $$ \boxed{ C_f = \int |(f*f)(x)|^2dx >0. } $$ So normalized exact-character averaging still leaves： $$ \boxed{ \Theta(H^3). } $$ --- # 13. Averaged UMP is false A hoped-for averaged estimate： $$ \frac1M \sum_j \mathcal R_4^{(j/M)} = O( H^{2-\delta} ) $$ cannot hold for any： $$ \delta>0. $$ Thus： $$ \boxed{ \text{finite Fourier averaging does not rescue standard Wick subtraction}. } $$ The problem is structural centering, not loss from triangle inequality。 --- # 14. Parallelogram finite-$q$ means Let： $$ q $$ be squarefree。 Define： $$ \boxed{ A(q) = \prod_{p\mid q} \left( 1+\frac1{(p-1)^3} \right). } $$ v3.7 gives： $$ \boxed{ \mathbb E_{h,d} \mathfrak S_{0,q}(0,h,d,h+d) = A(q)-1. } $$ Also： $$ \boxed{ \mathbb E_h\mu_q(h)^2 = A(q)-1. } $$ --- # 15. The third pairing mean Use the Fourier expansion： $$ \boxed{ \mu_q(x) = \sum_{\theta\in G_q} \rho_q(\theta)^2e(x\theta). } $$ Then： $$ \begin{aligned} &\mathbb E_{h,d} \mu_q(h+d) \mu_q(h-d) \\ &= \sum_{\theta,\varphi} \rho_q(\theta)^2 \rho_q(\varphi)^2 \mathbf1_{\theta+\varphi=0} \mathbf1_{\theta-\varphi=0}. \end{aligned} $$ Thus： $$ 2\theta=0. $$ The zero frequency has： $$ \rho_q(0)=0. $$ A nonzero order-two frequency exists iff： $$ 2\mid q, $$ namely： $$ \theta=\frac12. $$ Its reduced denominator is： $$ 2, $$ so： $$ \rho_q(1/2)^2=1. $$ Hence： ## Theorem 15.1 $$ \boxed{ \mathbb E_{h,d} \mu_q(h+d) \mu_q(h-d) = \mathbf1_{2\mid q}. } $$ --- # 16. Global Wick-connected mean on the slice Define： $$ \boxed{ \begin{aligned} \mathfrak C_{4,q}(h,d) &= \mathfrak S_{0,q}(0,h,d,h+d) \\ &\quad - \mu_q(h)^2 - \mu_q(d)^2 \\ &\quad - \mu_q(h+d)\mu_q(h-d). \end{aligned} } $$ Sections 14–15 give： ## Theorem 16.1 $$ \boxed{ \mathbb E\mathfrak C_{4,q} = - [A(q)-1] - \mathbf1_{2\mid q}. } $$ So the globally Wick-connected fourth cumulant is not centered on the parallelogram slice。 --- # 17. Why v3.7 $\mathfrak K_4$ stays centered Recall： $$ \boxed{ \mathfrak K_{4,q} = \mathfrak S_{0,q} - \mu_q(h)^2. } $$ Then： $$ \boxed{ \mathfrak K_{4,q} = \mathfrak C_{4,q} + \mu_q(d)^2 + \mu_q(h+d)\mu_q(h-d). } $$ The two retained pairing channels have mean： $$ [A(q)-1] + \mathbf1_{2\mid q}, $$ which exactly cancels Theorem 16.1。 Therefore： $$ \boxed{ \mathbb E\mathfrak K_{4,q}=0. } $$ This is the constraint-adapted cancellation identified in v3.7。 --- # 18. Constraint-adapted centering principle A conditional slice can change which lower-order channels must remain grouped。 Therefore： $$ \boxed{ \text{unconstrained cumulant decomposition} } $$ need not be compatible with： $$ \boxed{ \text{conditional / constrained centering}. } $$ For the parallelogram： $$ d_1+d_4=d_2+d_3, $$ separating all three Wick channels destroys an exact deterministic cancellation。 This is the main conceptual correction of v3.12。 --- # 19. The surviving one-dimensional spectral axis v3.8 proved： $$ \widehat{\mathfrak K}_{4,q}(\alpha,0)=0. $$ It also identified： $$ \boxed{ \widehat{\mathfrak K}_{4,q}(0,\beta) = \widehat{\mu_q^2}(\beta) } $$ for： $$ \beta\ne0. $$ Therefore the complete： $$ \alpha=0 $$ nonconstant axis is exactly the second pair-square channel。 --- # 20. Centered pair-square axis Define： $$ \boxed{ \mathfrak A_q(d) = \mu_q(d)^2 - [A(q)-1]. } $$ Then： $$ \mathbb E_d\mathfrak A_q(d)=0. $$ Its Fourier support is： $$ \boxed{ \alpha=0, \qquad \beta\ne0. } $$ --- # 21. Double-centered parallelogram remainder Define： ## Definition 21.1 $$ \boxed{ \mathfrak K_{4,q}^{\perp}(h,d) = \mathfrak K_{4,q}(h,d) - \mathfrak A_q(d). } $$ Equivalently： $$ \boxed{ \begin{aligned} \mathfrak K_{4,q}^{\perp}(h,d) &= \mathfrak S_{0,q}(0,h,d,h+d) \\ &\quad - \mu_q(h)^2 - \mu_q(d)^2 + [A(q)-1]. \end{aligned} } $$ This expression is manifestly symmetric in： $$ h,d. $$ --- # 22. Double-axis removal theorem The： $$ \beta=0 $$ axis of： $$ \mathfrak K_{4,q} $$ is already zero。 The subtraction： $$ \mathfrak A_q(d) $$ has zero： $$ \beta=0 $$ coefficient。 Therefore： $$ \boxed{ \widehat{ \mathfrak K_{4,q}^{\perp} }(\alpha,0)=0. } $$ For： $$ \alpha=0,\quad\beta\ne0, $$ Section 19 and the definition of： $$ \mathfrak A_q $$ give exact cancellation。 At： $$ (0,0), $$ both objects have mean zero。 Hence： ## Theorem 22.1 · Double-Axis Removal $$ \boxed{ \widehat{ \mathfrak K_{4,q}^{\perp} }(\alpha,0)=0 \quad \forall\alpha, } $$ and： $$ \boxed{ \widehat{ \mathfrak K_{4,q}^{\perp} }(0,\beta)=0 \quad \forall\beta. } $$ --- # 23. Symmetry From Definition 21.1： ## Corollary 23.1 $$ \boxed{ \mathfrak K_{4,q}^{\perp}(h,d) = \mathfrak K_{4,q}^{\perp}(d,h). } $$ So the genuinely two-dimensional remainder treats the two parallelogram gaps symmetrically。 --- # 24. New deterministic decomposition The finite covariance is now： $$ \boxed{ \mathfrak K_{4,q}(h,d) = \mathfrak A_q(d) + \mathfrak K_{4,q}^{\perp}(h,d). } $$ Interpretation： ### Axis component $$ \boxed{ \mathfrak A_q(d) } $$ is a one-dimensional centered pair-square problem。 ### Axis-free component $$ \boxed{ \mathfrak K_{4,q}^{\perp}(h,d) } $$ has no one-dimensional Fourier axes。 This is more adapted to v3.10 restriction analysis than the full Wick cumulant。 --- # 25. Potential two-direction Sobolev quantity Since： $$ \alpha\ne0, \qquad \beta\ne0 $$ for every Fourier mode of： $$ \mathfrak K_{4,q}^{\perp}, $$ define the mixed candidate： $$ \boxed{ \mathfrak H_{hd}^{\perp}(q) = \sum_{\alpha\ne0,\beta\ne0} \frac{ | \widehat{ \mathfrak K_{4,q}^{\perp} }(\alpha,\beta) |^2 }{ |1-e(\alpha)|^2 |1-e(\beta)|^2 }. } $$ A uniform bound would provide a two-direction periodic antiderivative。 This is **not proved** in v3.12。 --- # 26. Why a mixed Sobolev theorem would matter If： $$ \Delta_h\Delta_d \mathfrak G_q^{(2)} = \mathfrak K_{4,q}^{\perp}, $$ with uniformly bounded normalized： $$ L^2 $$ energy, then two summations by parts would act on： $$ \Omega_N. $$ The mixed second finite difference of the piecewise-cubic Cesàro weight is smaller than the raw weight。 This may reduce the finite-box restriction burden before any high-conductor large-sieve step。 The axis term： $$ \mathfrak A_q(d) $$ would be handled separately with one-dimensional Ramanujan-tail methods。 --- # 27. Reference finite-conductor diagnostics The package checks： $$ q=2,3,6,15,30,210. $$ For each： - $\mathbb E\mathfrak K_{4,q}=0$； - $\mathbb E\mathfrak C_{4,q}$ matches Theorem 16.1； - $\mathbb E\mu(h+d)\mu(h-d)$ matches $\mathbf1_{2\mid q}$； - both Fourier axes of $\mathfrak K_{4,q}^{\perp}$ vanish； - $\mathfrak K_{4,q}^{\perp}$ is symmetric。 These are exact-structure normalization checks。 --- # 28. $q=2$ scaling diagnostics Using a fixed smooth bump： $$ f(x) = \begin{cases} \exp[-1/(x(1-x))],&00 $$ ```text OPEN ``` ### G10. Actual prime deviation ```text OPEN ``` ### G11. Complete quartic： $$ \eta_Q>0 $$ ```text OPEN ``` ### G12. RH ```text OPEN ``` --- # 33. Trust boundary ```text RH_PROVED = FALSE RH_DISPROVED = FALSE UMP4 FALSE = EXACT FINITE-CONDUCTOR COUNTEREXAMPLE AVERAGED UMP FALSE = EXACT CHARACTER-ORTHOGONALITY ARGUMENT LITERATURE PAIRING THEOREM REMAINS CORRECT IN ITS STATED SETTING FAILURE IS IN THE PROPOSED MODULATED TRANSPLANT CONSTRAINT-ADAPTED DOUBLE CENTERING = EXACT FINITE-Q ALGEBRA NO MIXED SOBOLEV THEOREM PROVED NO ETA_M > 0 PROVED NO ETA_Q > 0 PROVED GLOBAL_RH_CERTIFICATE = FALSE ``` Forbidden： $$ \text{UMP4 counterexample} \Longrightarrow \text{Kuperberg theorem false}. $$ The published theorem is not being contradicted。 Forbidden： $$ \text{connected fourth cumulant} \Longrightarrow \text{correct constrained residual}. $$ Forbidden： $$ \text{both axes removed} \Longrightarrow \text{automatic two-dimensional power saving}. $$ --- # 34. One-line status > v3.12 disproves the proposed UMP4 route by an exact finite-conductor resonance. At $q=2$, the only residue frequency is $1/2$. Under modulation $(+\lambda,-\lambda,-\lambda,+\lambda)$ with $\lambda=1/2$, all four exponential sums move to zero frequency. The fourth moment is therefore the product of four zero-frequency sums, while each of the three Wick pairings gives the same product; the pairing-free remainder is exactly minus twice that product and has size $\Theta(H^4)$. Even normalized finite-character averaging does not rescue the route: for equal real smooth weights the averaged remainder is minus twice the additive energy of the sampled weight sequence, hence $\Theta(H^3)$. This resonance occurs precisely in the higher-denominator-multiplicity stratum that is lower-order in the original unshifted Montgomery–Soundararajan proof. The deeper correction is centering: the fully Wick-connected fourth cumulant has nonzero mean on the parallelogram slice, whereas the AMRAL covariance $\mathfrak K_4=\mathfrak S_0-\mu_h^2$ remains mean-zero because the other two pairing channels cancel the constraint-induced connected mean. The correct next decomposition is constraint-adapted double centering: subtract the centered $\mu_d^2$ axis from $\mathfrak K_4$, leaving a symmetric remainder whose entire $\alpha=0$ and $\beta=0$ Fourier axes vanish. The deterministic branch should now analyze the one-dimensional axis and the genuinely two-dimensional axis-free remainder separately rather than attempt any global uniform Wick theorem. --- # 35. References 1. Hugh L. Montgomery, K. Soundararajan, **Primes in short intervals**, *Communications in Mathematical Physics* 252 (2004), 589–617. arXiv: https://arxiv.org/abs/math/0409258 2. Vivian Kuperberg, **Sums of singular series along arithmetic progressions and with smooth weights**, *International Journal of Number Theory* 21 (2025), 53–74. DOI: https://doi.org/10.1142/S1793042125500046 arXiv: https://arxiv.org/abs/2301.06095 3. AMRAL, **RH-Parallelogram-AddingFractionsRestriction v3.11**. 4. AMRAL, **RH-Parallelogram-RamanujanTail v3.8**. --- # 36. Provenance 研究主導：Neo.K v3.12 exact $q=2$ modulation counterexample、character-averaged resonance、proof-portability blocker、constraint-adapted centering theorem、double-axis decomposition、reference implementation 與 canonical source 整理：ChatGPT / GPT-5.6 Sol 日期：2026-09-03 研究定位：AMRAL 黎曼猜想半自主研究線，第五弧線 modulated pairing counterexample / constraint-adapted centering 節點。 本文件是 research source artifact，不是 peer-reviewed publication。 所有 claim 必須依 Claim register、GAP ledger 與 Trust boundary 解讀。