工程紀錄 · 第四弧線 v3.2 · 2026-09-03 · COMPLETED_REWARD_STRENGTH_AUDIT · STRICT_MODE_SENSITIVITY · BRANCH_STOP · RH_CLAIM_FALSE # Completed Reward Correlation Kernel：Zero-Mode Transfer、Strict Positivity 與 Strength Audit **RH-CompletedReward-CorrelationKernel v3.2** 本節點承接： - `RH-ExternalWork-InventoryBound v3.1` - `RH-ArithmeticInnovation-WorkIdentity v3.0` - `RH-CauchyPoisson-TwistScalarization v2.5` v3.1 已得到 exact renewal–reward decomposition： $$ \boxed{ \int_T^{T+L} \mathcal C_h(t)dt = \Delta\mathcal R_h[T,T+L] + \Delta\mathcal U_h[T,T+L]. } $$ 其中： - $\mathcal R_h$ 是 completed lifecycle bulk reward； - $\mathcal U_h$ 是 finite-memory unfinished boundary inventory。 v3.2 問： > one-sided completed reward $\Delta\mathcal R_h$ 是否比原本 positive Cauchy block energy更弱，因此可能更容易證明？ 本輪結論： $$ \boxed{ \text{沒有找到 theorem-strength advantage。} } $$ 更精確地說： 1. completed reward kernel對任何 admissible off-axis conjugate zero mode都保留相同 horizontal exponent； 2. its transfer coefficient is strictly positive for every vertical ordinate； 3. 因此 causal ordering沒有 vertical blind spot，也沒有 fixed-exponent discount； 4. 但 $\Delta\mathcal R_h$ 是 signed correlation，而原 Cauchy energy是 positive quadratic form； 5. 所以 causal ordering在 proof engineering 上反而失去 positivity； 6. v3.1 的 boundary inventory仍然不能忽略； 7. completed-reward branch適合作為： - diagnostic； - event tracker； - correlation decomposition； 但目前不適合作為 RH 主 closure target。 Canonical branch decision： ```text COMPLETED_REWARD_BRANCH REPRESENTATION VALUE = KEEP CERTIFICATE VALUE = KEEP MAIN RH STRENGTH TARGET = STOP / DEPRIORITIZE RETURN MAIN ATTACK TO POSITIVE CAUCHY / INNOVATION ENERGY. ``` **RH_CLAIM = False.** --- # 0. Claim register ```text RH_PROVED = FALSE RH_DISPROVED = FALSE ONE_SIDED_COMPLETION_KERNEL = CLOSED CONJUGATE_ZERO_TRANSFER = CLOSED TRANSFER_COEFFICIENT_STRICT_POSITIVITY = CLOSED VERTICAL_BLIND_SPOTS = NONE FIXED_EXPONENT_DISCOUNT = NONE RH_IMPLIES_POLYNOMIAL_COMPLETED_REWARD = CLOSED_FROM_RH_PNT_INPUT POLYNOMIAL_COMPLETED_REWARD_IMPLIES_RH = CLOSED_AS_ZERO_MODE_REDUCTION FORMAL_GLOBAL_QUADRATIC_EXPLICIT_FORMULA = OPEN_FORMALIZATION COMPLETED_REWARD_POSITIVITY = FALSE ORIGINAL_CAUCHY_BLOCK_POSITIVITY = TRUE THEOREM_STRENGTH_ADVANTAGE = NOT_ESTABLISHED BRANCH_DECISION = DEPRIORITIZE_AS_MAIN_CLOSURE_ROUTE GLOBAL_RH_CERTIFICATE = FALSE ``` --- # 1. Full lifecycle reward kernel From v3.1: $$ \boxed{ \mathcal H_h(d) = e^{-|d|} C_h(d), } $$ where: $$ C_h = T_h\ast\widetilde T_h. $$ Writing: $$ r=|d|, $$ $$ \boxed{ C_h(d) = \begin{cases} \displaystyle \frac{2h^3}{3} - hr^2 + \frac{r^3}{2}, & 0\le r\le h, \\ \displaystyle \frac{(2h-r)^3}{6}, & h\le r\le2h, \\ 0, & r\ge2h. \end{cases} } $$ Thus: $$ \operatorname{supp}\mathcal H_h \subset[-2h,2h]. $$ --- # 2. One-sided completion rate For a smooth real source density: $$ f(u), $$ completed pair lifecycles are registered at the EXIT time of the earlier source. Let: $$ x=t-h. $$ Then the off-diagonal completion rate is: $$ \boxed{ \dot{\mathcal R}_h(t) = 2f(x) \int_0^{2h} \mathcal H_h(r) f(x+r)dr. } $$ For atomic sources, the same formula becomes a Stieltjes completion measure with explicit diagonal rewards. --- # 3. One hypothetical off-axis zero mode Consider a real conjugate pair mode: $$ \boxed{ f(u) = A e^{\lambda u} + \overline A e^{\overline\lambda u}, } $$ with: $$ \lambda = \delta+i\gamma, $$ $$ 0<\delta<\frac12. $$ This is the generic horizontal/vertical structure generated by one hypothetical off-critical zeta zero after critical normalization. --- # 4. Completed-reward zero-mode expansion Insert Section 3 into Section 2. There are: - an $A^2e^{2\lambda x}$ oscillatory term; - a $\overline A^2e^{2\overline\lambda x}$ oscillatory term; - a conjugate-pair nonoscillatory term at exponent: $$ 2\delta. $$ The nonoscillatory contribution is: $$ \boxed{ \dot{\mathcal R}^{(0)}_h(t) = 2|A|^2 e^{2\delta(t-h)} \Phi_{h,\delta}(\gamma), } $$ where: $$ \boxed{ \Phi_{h,\delta}(\gamma) = 2 \int_0^{2h} e^{-(1-\delta)r} C_h(r) \cos(\gamma r)dr. } $$ Therefore the causal completion observable retains the candidate off-axis exponent: $$ \boxed{ 2\delta. } $$ --- # 5. Even-kernel interpretation Define: $$ \boxed{ F_{h,\delta}(r) = e^{-(1-\delta)|r|} C_h(r). } $$ Since both factors are even: $$ \boxed{ \Phi_{h,\delta}(\gamma) = \widehat F_{h,\delta}(\gamma) } $$ under the Fourier convention: $$ \widehat f(\xi) = \int_{\mathbb R} f(r)e^{-i\xi r}dr. $$ --- # 6. Positive-definite factors First: $$ \boxed{ C_h = T_h\ast\widetilde T_h. } $$ Therefore: $$ \boxed{ \widehat C_h(\xi) = |\widehat T_h(\xi)|^2 \ge0. } $$ Explicitly: $$ \boxed{ \widehat T_h(\xi) = \frac{ 2(1-\cos(h\xi)) }{ \xi^2 }, } $$ with removable value: $$ \widehat T_h(0)=h^2. $$ Second, set: $$ a=1-\delta. $$ Because: $$ 0<\delta<\frac12, $$ we have: $$ a>0. $$ The exponential kernel: $$ E_a(r)=e^{-a|r|} $$ has Fourier transform: $$ \boxed{ \widehat E_a(\xi) = \frac{ 2a }{ a^2+\xi^2 } >0 } $$ for every real $\xi$. --- # 7. Strict transfer positivity Since: $$ F_{h,\delta} = E_aC_h, $$ the product-to-convolution rule gives: $$ \boxed{ \widehat F_{h,\delta}(\gamma) = \frac1{2\pi} \int_{\mathbb R} \frac{ 2a }{ a^2+\xi^2 } \left| \widehat T_h(\gamma-\xi) \right|^2 d\xi. } $$ The integrand is: - nonnegative everywhere; - strictly positive on a set of positive measure. Therefore: ## Theorem 7.1 · Strict completion transfer For every: $$ h>0, $$ $$ 0<\delta<\frac12, $$ and every: $$ \gamma\in\mathbb R, $$ $$ \boxed{ \Phi_{h,\delta}(\gamma)>0. } $$ Thus the completed-reward observable has **no vertical blind spots** for off-axis conjugate modes. --- # 8. Unit-block completed reward contribution Integrate the nonoscillatory term over: $$ [T,T+1]. $$ For: $$ \delta>0, $$ $$ \boxed{ \begin{aligned} \Delta\mathcal R_h^{(0)}[T,T+1] &= |A|^2 e^{-2\delta h} \Phi_{h,\delta}(\gamma) \\ &\quad\times \frac{ e^{2\delta}-1 }{ \delta } e^{2\delta T}. \end{aligned} } $$ The coefficient is strictly positive. Therefore a hypothetical zero at horizontal displacement $\delta$ creates a nonoscillatory completion-reward component of fixed exponential type: $$ \boxed{ 2\delta. } $$ Causal ordering does not lower the rightmost-zero exponent. --- # 9. Multiple zeros at the same rightmost exponent Suppose several modes have the same maximal horizontal displacement: $$ \delta. $$ Pairing each mode with its own conjugate produces a zero-frequency coefficient proportional to: $$ |A_j|^2 \Phi_{h,\delta}(\gamma_j)>0. $$ Cross-pairings between distinct ordinates have nonzero vertical frequencies: $$ \gamma_j-\gamma_k. $$ Thus the grouped zero-frequency component is positive. A Fejér / long smooth time average isolates this nonoscillatory contribution. Hence cross-oscillations cannot erase the fixed exponent detected by the self-conjugate pairs. --- # 10. Fixed-exponent detector Define the completed-reward block exponent schematically by: $$ \boxed{ \eta_{\mathcal R} = \frac12 \limsup_{T\to\infty} \frac{ \log^+ |\Delta\mathcal R_h[T,T+1]| }{ T }, } $$ with the usual caveat that a signed observable may require a local/smoothed absolute-value version to avoid isolated zeros. Theorem 7.1 and Section 9 give at zero-mode level: $$ \boxed{ \eta_{\mathcal R} = \Delta_\zeta. } $$ No exponent gain is obtained from the causal ordering. The fully formal global version requires the standard grouped explicit-formula / smoothing justification and remains a formalization task. --- # 11. RH-side polynomial bound Under RH, prior AMRAL work and the explicit-formula literature give polynomial control of the weighted cumulative innovation: $$ \boxed{ \mathfrak b(t) = O(t^A) } $$ for some finite $A$. For example the weighted Chebyshev estimate: $$ \sum_{n\le x} \frac{\Lambda(n)}{\sqrt n} = 2\sqrt x + O(\log^3x) $$ gives: $$ \mathfrak b(t)=O(t^3). $$ A completed-reward block is a double Stieltjes integral of: $$ d\mathfrak b(u)d\mathfrak b(v) $$ against a fixed compact piecewise-smooth kernel. Integrating by parts in both variables over a fixed-width translated region therefore gives: $$ \boxed{ RH \Longrightarrow \Delta\mathcal R_h[T,T+1] = \operatorname{poly}(T). } $$ No boundedness claim is required. --- # 12. Polynomial completed-reward converse Conversely, if a rigorous global theorem gave: $$ \boxed{ \Delta\mathcal R_h[T,T+1] = \operatorname{poly}(T) } $$ in a form stable under fixed smooth time averaging, then an off-axis zero would contradict Section 8 because it creates a strictly positive: $$ e^{2\delta T} $$ nonoscillatory component. Thus: ## Reduction 12.1 At the standard explicit-formula mode level: $$ \boxed{ \text{polynomial completed-reward blocks} \Longrightarrow RH. } $$ This is not yet packaged as a standalone formal theorem in Lean/Coq; the infinite zero-series grouping / smoothing step remains to be formalized. --- # 13. Comparison with the positive Cauchy block scalar Original v2.5 object: $$ \boxed{ \mathscr W_h(T) = \int_T^{T+1} \mathcal C_h(t)dt \ge0. } $$ Completed-reward object: $$ \boxed{ \Delta\mathcal R_h[T,T+1] } $$ is signed. Both retain the same off-axis fixed exponent. So: ```text CAUCHY BLOCK positive RH-complete fixed exponent 2 Delta_zeta COMPLETED REWARD signed same off-axis exponent causal/event-friendly no demonstrated strength discount ``` --- # 14. Renewal identity does not create a weaker theorem Recall: $$ \boxed{ \mathscr W_h(T) = \Delta\mathcal R_h[T,T+1] + \Delta\mathcal U_h[T,T+1]. } $$ The identity is exact. But neither term is known to be lower-order unconditionally. Therefore one cannot infer: ```text completed reward is easier ``` merely because the other term has finite temporal memory. Finite memory is a representation property, not an asymptotic bound. --- # 15. Positivity loss The full Cauchy block is a positive quadratic form: $$ \mathscr W_h(T)\ge0. $$ Completed reward increments can have either sign. Finite actual-prime samples in v3.1 already show: $$ \Delta\mathcal R_q $$ switching sign. Therefore completion ordering removes a useful structural constraint. Any future proof using $\Delta\mathcal R_h$ must reconstruct enough signed cancellation control to compensate for the loss of positivity. --- # 16. Is one-sided ordering analytically weaker? At present: $$ \boxed{ \text{NO}. } $$ What v3.2 establishes is: - same fixed horizontal exponent; - no vertical annihilation; - signed rather than positive; - boundary term not known negligible. So no theorem-strength advantage has been demonstrated. The one-sided representation may still improve: - streaming computation; - certificates; - causal decomposition; - diagnostics of which prime events complete correlations. Those are engineering advantages, not a weaker RH theorem. --- # 17. External literature alignment Weighted Selberg integrals are quadratic means and are linked through dispersion to arithmetic correlation averages. Coppola–Laporta explicitly develop this correlation-average viewpoint and Cesàro weighting. The AMRAL completed reward: $$ \Delta\mathcal R_h $$ is another compact correlation average, with a causal ordering attached to the same finite-range two-point arithmetic content. This supports the v3.2 conclusion: > causal ordering changes bookkeeping more clearly than it changes arithmetic strength. --- # 18. Cauchy mean alignment Weber's Cauchy means of Dirichlet polynomials use the same fundamental exponential ratio kernel: $$ e^{-|\log(m/n)|}. $$ The v3.2 reward kernel adds the compact tent autocorrelation: $$ C_h(\log(m/n)). $$ So the remaining correlation is still within the same broad Cauchy / Dirichlet-polynomial operator family. No known external theorem located in this audit gives a fixed-exponent improvement for the centered von-Mangoldt completion reward. --- # 19. Branch decision The completed-reward machinery remains valuable. Keep: 1. 23-state renewal tracker; 2. completion-field moments; 3. actual-prime completion diagnostics; 4. lifecycle decomposition; 5. correlation-kernel audit. But deprioritize: ```text PROVE RH BY BOUNDING DELTA R FIRST ``` as the main route. Reason: $$ \boxed{ \text{same exponent sensitivity} + \text{loss of positivity} + \text{non-negligible boundary term}. } $$ --- # 20. Return target The preferred main object returns to: $$ \boxed{ \mathscr W_h(T) } $$ or its causal equivalent: $$ \boxed{ 2 \int_T^{T+1} \|s_t\|_2^2dt. } $$ This retains positivity. The next useful theorem should exploit arithmetic innovation before taking absolute values, but preserve the positive energy output. --- # 21. Candidate next mechanisms Promising directions after the branch audit: ### A. Innovation spectral measure Study whether the centered innovation has a positive spectral measure / density constraint under fixed local filtering. ### B. Positive-kernel comparison Search for a kernel: $$ 0\preceq K_{\rm easy}\preceq K_{\rm Cauchy} $$ or reverse comparison whose arithmetic quadratic form has known estimates. ### C. Innovation square-function Build a Littlewood–Paley / multiscale positive decomposition of: $$ \mathscr W_h. $$ ### D. Causal martingale-like decomposition Use the v2.6 causal state but keep the final square norm positive. These retain positivity while seeking weaker arithmetic input. --- # 22. Suggested v3.3 direction Recommended: `RH-PositiveKernel-InnovationSpectrum v3.3` Tasks: 1. derive the exact Fourier symbol of the positive Cauchy/tent energy; 2. factor it into positive spectral multipliers; 3. compare with classical Selberg/Gallagher kernels; 4. search for monotone positive-kernel inequalities; 5. determine whether any known mean-square theorem controls a nontrivial positive portion of the RH-complete energy; 6. construct a scale ladder where partial kernel control yields quantitative zero-strip progress; 7. reject signed causal decompositions unless they restore a clear positivity advantage. --- # 23. GAP ledger ## CLOSED / AUDITED ### G1. One-sided completed-reward kernel ```text CLOSED ``` ### G2. Off-axis conjugate transfer ```text CLOSED ``` ### G3. Strict vertical positivity ```text CLOSED ``` $$ \Phi_{h,\delta}(\gamma)>0. $$ ### G4. Fixed-exponent discount ```text NONE ``` ### G5. Completed-reward main-route strength advantage ```text NOT ESTABLISHED ``` ### G6. Branch decision ```text DEPRIORITIZE ``` --- ## OPEN ### G7. Formal global quadratic explicit-formula theorem ```text OPEN_FORMALIZATION ``` ### G8. Positive-kernel arithmetic inequality ```text OPEN ``` ### G9. Any fixed exponent $\sigma<1/2$ ```text OPEN ``` ### G10. Polynomial positive Cauchy energy ```text OPEN_RH_COMPLETE ``` ### G11. RH ```text OPEN ``` --- # 24. Trust boundary ```text RH_PROVED = FALSE RH_DISPROVED = FALSE STRICT ZERO-MODE TRANSFER POSITIVITY = EXACT COMPLETED-REWARD EXPONENT ALIGNMENT = ZERO-MODE REDUCTION FULL GLOBAL EXPLICIT-FORMULA QUADRATIC GROUPING = NOT FORMALLY VERIFIED COMPLETED REWARD SIGNEDNESS = REAL OBSTACLE NO STRENGTH ADVANTAGE CLAIM GLOBAL_RH_CERTIFICATE = FALSE ``` Forbidden: $$ \Phi_{h,\delta}(\gamma)>0 \Longrightarrow RH. $$ Forbidden: $$ \text{same exponent detector} \Longrightarrow \text{same proof difficulty exactly}. $$ Forbidden: $$ \text{causal representation} \Longrightarrow \text{weaker theorem}. $$ --- # 25. One-line status > v3.2 completes the strength audit of the completed-lifecycle branch. For a conjugate off-axis mode $A e^{(\delta+i\gamma)u}+\overline A e^{(\delta-i\gamma)u}$, the completed-reward flux contains a nonoscillatory term $2|A|^2e^{2\delta(t-h)}\Phi_{h,\delta}(\gamma)$. The transfer coefficient is the Fourier transform of the even kernel $e^{-(1-\delta)|r|}C_h(r)$ and is strictly positive for every real $\gamma$, because it is the convolution of the everywhere-positive Fourier transform of $e^{-(1-\delta)|r|}$ with the nonnegative nonzero spectrum $|\widehat T_h|^2$. Thus causal completion ordering has no vertical blind spots and no fixed-exponent discount: every off-axis zero retains the same $e^{2\delta t}$ quadratic growth class. Under RH, known polynomial control of the weighted cumulative innovation makes completed-reward blocks polynomial after compact-kernel integration by parts; conversely the positive nonoscillatory mode shows why a polynomial completed-reward theorem would exclude off-axis zeros, modulo formalization of the grouped quadratic explicit formula. However completed reward is signed, while the original Cauchy block energy is positive, and the finite-memory boundary term is not known negligible. Therefore no theorem-strength advantage has been established. The causal branch should remain as a diagnostic/certificate decomposition, while the main RH attack returns to positive Cauchy/innovation energy and positive-kernel inequalities. --- # 26. References 1. Giovanni Coppola, Maurizio Laporta, **Generations of Correlation Averages**, *Journal of Numbers* (2014), Article ID 140840. DOI: https://doi.org/10.1155/2014/140840 2. Giovanni Coppola, Maurizio Laporta, **A generalization of Gallagher's lemma for exponential sums**, arXiv:1411.1739. 3. Michel J. G. Weber, **Cauchy Means of Dirichlet Polynomials**, *Journal of Approximation Theory* 204 (2016), 61–79. arXiv: https://arxiv.org/abs/1412.7812 4. AMRAL, **RH-ExternalWork-InventoryBound v3.1**. 5. AMRAL, **RH-CauchyPoisson-TwistScalarization v2.5**. --- # 27. Provenance 研究主導：Neo.K v3.2 completed-reward zero-mode transfer、strict positivity theorem、strength audit、branch-stop decision、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，第四弧線 causal-completion branch strength audit / positive-kernel return 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。