bibkey: liu2026quasirhresidue
authors: Baiying Liu
year: 2026
title: Joint mixed-Mobius estimates and the actual exceptional residue interface
doi: null
url: https://arxiv.org/abs/2610.12234v2
claim: Repo-derived conditional joint Mellin bounds pay the fifteen radial mixed-Mobius terms; fixed-cutoff polygonal localization pays the all-left and outer-face remainders of the original all-K scalar, retaining its fifteen coupled residue terms. A separately redesigned BPY Mellin cutoff and the actual bare-J4 finite-ray table pay the conditional local nonzero exceptional branch with one common budget; the original scalar, generalized angular factors and full signed J40 contraction remain unresolved.
strata_touched: []
license: citation-only
triage: anchor

Joint mixed-Mobius estimates and the actual exceptional residue

This is a repo-derived companion to the four-factor source reduction. The deductions below are not attributed to Liu, Grafakos–Teschl or Xu–Sang. They reuse C37, C48 and C51–C76, without certifying the upstream main proof. In particular, $R_{7/8}$ is an explicit source-conditional premise, not a newly proved zero-free strip. The statements are paper mathematics, without Lean or CI certification. Labels R1–R46 belong only to this companion; C1–C76 retain their scope.

Keep $K=\mathbb{Q}(\sqrt{-3})$, its original finite bad set $S$, ideals counted once, $D=U^{1/2}$, $A=U^{\alpha }$, $0<\alpha <1/4$, and the original real smooth cutoff $W$ of C37. On precisely the original inverse annulus, $b_A=\mu +K_A$, where $K_A(d)={\sum }_{e\mid d}\mu (e)2^{\omega (e)}W(q_e/A)$. The four-way squarefree product, every coprimality puncture, raw sextic/cubic zero extension, profiles, signed coefficients and common beta remain those of C59. Here $P$ is a squarefree puncture outside $S$; in the frequency application it is $P_a$ of C58, not the product-squarepart parameter of J40. R1–R42 introduce no nonprincipal inverse-Hecke premise; R43–R46 separately state the fixed finite-order suppliers for the bare-J4 residual. No new column coefficient is introduced.

The joint estimate on each original nonempty subset

For $\varnothing \ne M\subseteq \{1,2,3,4\}$, put $m=|M|$, $T=\{1,2,3,4\}\setminus M$ and $r=|T|$. Define

$$Q_M(P;\Psi )=\sum _{\begin{array}{c}{d}_1{d}_2{d}_3{d}_4\mathrm{sf}\\ (d_1{d}_2{d}_3{d}_4,SP)=1\end{array}}\prod _{i\in M}\mu (d_i)\prod _{k\in T}{K}_A(d_k)\Psi (q_{d_1}/D,\dots ,q_{d_4}/D).\text{\hspace{1em}(R1)}$$

The profile is any smooth complex function supported on the same fixed compact rectangle in $(0,\infty {)}^4$ as C52. Let $N_J(\Psi )$ bound its partial derivatives up to order $J$ in each variable there. Assume $1/2\le \beta <1$ and exactly $R_{\beta }$ of C48, including polynomial vertical growth on each smaller closed strip. For every fixed $C$, $q_P\le U^C$, and $\epsilon >0$, one finite $J$ gives, uniformly in all fifteen subsets,

$$|Q_M(P;\Psi )|{\ll }_{\epsilon }{N}_J(\Psi )(U{q}_P{)}^{\epsilon }{D}^{m\beta }{\left[D{A}^{-(1-\beta )}+D^{1/2}{A}^{\beta -1/2}\right]}^r.\text{\hspace{1em}(R2)}$$

Constants depend on $K,S,W,\alpha ,C$, the fixed rectangle, the strip margins and the constants of $R_{\beta }$. For $r=0$ the bracket is $1$. This is a joint estimate for the actual mixed coefficients, rather than an estimate obtained by fixing three columns and taking their masses.

Literal Euler product. Expand only the $r$ actual divisor sums in R1 and Mellin-invert their cutoffs. Write $x_i=q^{-s_i}$ and $y_k=q^{-z_k}$. At $p\nmid SP$, squarefreeness of the full product allows the prime in at most one $d_i$. A Mobius slot contributes $-x_i$; a $K_A$ slot contributes $x_k(1-2y_k)$. Thus the joint series is

$$\begin{array}{rl}{F}_P(\mathbf{s},\mathbf{z})& =\prod _{p\nmid SP}\left(1-\sum _{i\in M}{x}_i+\sum _{k\in T}{x}_k(1-2y_k)\right)\\ & =H_P(\mathbf{s},\mathbf{z})\prod _{i\in M}{\zeta }_K(s_i{)}^{-1}\prod _{k\in T}{\zeta }_K(s_k){\zeta }_K(s_k+z_k{)}^{-2}.\end{array}\text{\hspace{1em}(R3)}$$

The first product is literal for $\Re s_i>1$, $\Re z_k>0$. It encodes the same forbidden cross-column prime collisions as C53; there are no independent pairwise products. At an unexcluded prime, with $L_p$ the parenthesis in R3,

$$H_{P,p}=L_p\prod _{i\in M}(1-x_i{)}^{-1}\prod _{k\in T}(1-x_k)(1-x_k{y}_k{)}^{-2}.\text{\hspace{1em}(R4)}$$

At $p\mid SP$, the factor is the same expression with $L_p=1$. All linear monomials cancel at unexcluded primes. On every closed substrip of

$$\Re s_i>1/2(1\le i\le 4),\Re (s_k+z_k)>1/2(k\in T),\text{\hspace{1em}(R5)}$$

let $a>1/2$ be a common lower bound for these real parts. Expansion of the displayed denominators, whose moduli are bounded away from zero, gives $H_{P,p}=1+O(q^{-2a})$, uniformly in all heights. The constant uses only the closed real substrip. The unexcluded product converges normally and is bounded there. No $L_p$ is inverted, so its zeros create no polar divisor.

For an excluded prime the modulus is at most $\exp (C_a{q}^{-a})$ after enlarging $C_a$ to cover finitely many small primes. For each $\eta >0$, $C_a{q}^{-a}\le \eta \log q$ for all sufficiently large $q$. The product over the remaining small primes is bounded independently of $P$. Consequently $H_P{\ll }_{\eta }{q}_P^{\eta }$, uniformly in heights on these substrips. This proves polynomial-puncture uniformity without assuming $q_P\le A^2$ or dividing by a local numerator.

Contour calculation. Use $\widetilde\Psi(\mathbf s)=\int\Psi(\mathbf x) \prod_i x_i^{s_i-1}d\mathbf x$ and $\hat{W}(z)={\int }_0^{\infty }W(t)t^{z-1}dt$ initially for $\Re z>0$. The exact $4+r$ dimensional inversion has integrand

$$\tilde{\Psi }(\mathbf{s})D^{{\sum }_i{s}_i}\prod _{k\in T}[\hat{W}(z_k)A^{z_k}]F_P(\mathbf{s},\mathbf{z}),(2\pi i{)}^{-(4+r)}d\mathbf{s}d\mathbf{z}.\text{\hspace{1em}(R6)}$$

Choose $0<\delta <(1-\beta )/10$, later small enough for the stated loss, and start with every $\Re s_i=2$ and $\Re z_k=c=\beta -1/2+2\delta >0$. Absolute convergence justifies this inversion and all initial interchanges. First move the $m$ Mobius lines to $\beta +\delta$. Next move each $K_A$ line to $\Re s_k=1/2+\delta$, keeping its $z_k$ line at $c$, and take its single $s_k=1$ residue. Throughout, $\Re (s_k+z_k)\ge \beta +3\delta$. The only poles crossed are the simple poles of the ordinary ${\zeta }_K(s_k)$. Inverse-zeta factors are holomorphic by $R_{\beta }$; R5 holds during each move. Intersections of ordinary simple poles are handled by these successive one-variable residues.

For a term with residue set $E\subseteq T$, now shift just its $z_k$, $k\in E$, to $-h=\beta -1+\delta$. In those slots $s_k=1$, so the reciprocal arguments have real part at least $\beta +\delta$. All other slots retain $\Re s_k=1/2+\delta$, $\Re z_k=c$. Hence R5 and $R_{\beta }$ still apply to the whole joint correction, including all mixed prime monomials. The cutoff continuation is

$$\hat{W}(z)=-\frac{1}{z}{\int }_1^2{W}^{\prime }(t)t^{z}dt.\text{\hspace{1em}(R7)}$$

Its sole pole is simple at zero. In a residue slot the double zero of ${\zeta }_K(1+z{)}^{-2}$ cancels that pole. Thus no cutoff residue is crossed. The same cancellation is not claimed at zeros of ${\zeta }_K$; all paths stay strictly to the right of those allowed by $R_{\beta }$.

A term with $\ell =r-|E|$ boundary slots has modulus bounded by a constant times the common seminorm and

$$q_P^{\eta }{D}^{m(\beta +\delta )}(D{A}^{\beta -1+\delta }{)}^{r-\ell }(D^{1/2+\delta }{A}^{\beta -1/2+2\delta }{)}^{\ell }.\text{\hspace{1em}(R8)}$$

Each ordinary residue contributes the fixed field residue $\kappa$, absorbed in the constant. Summing the $2^r$ choices and allocating $\delta ,\eta$ inside the final loss proves R2. In particular this proof uses $A^{\beta -1/2}$ in each boundary slot. No stronger boundary estimate is needed for the conclusion.

One derivative and height envelope. Here is explicit sufficient control for the contour argument. Let $B$ be the reciprocal growth degree from C48 for margin $\delta$, and let $L$ bound the ordinary zeta growth on the fixed $s$ strips, off fixed small neighborhoods of its pole. Such a finite $L$ is supplied by fixed-field continuation and the functional equation. Smooth compact support and integration by parts in each log-coordinate give $|\widetilde\Psi(\mathbf s)|\ll_J N_J(\Psi) \prod_i(1+|\Im s_i|)^{-J}$ uniformly on the real strips used. R7 gives the analogous arbitrary decay of $\hat{W}$ on each vertical line. Near its crossed pole, use the holomorphic canceled product instead. Every reciprocal factor in a $K_A$ slot has growth at most $(1+|\Im s_k+\Im z_k|{)}^{2B}$, bounded by the product of the two separate weights of degree $2B$. A Mobius slot has degree $B$. Thus $J>L+4B+20$ suffices for absolute integration in all at most seven variables and for the vanishing of horizontal faces. Use truncated rectangles, detour the ordinary poles at bounded height, apply the successive residue theorem, and let all heights tend to infinity under these integrable bounds. This also justifies interchanges in every residual integral, without a moving-height reciprocal premise.

Enlarge this same finite $J$ once for the original derivatives and profile separation; $W$ and its required derivatives are fixed. For C59 use, for one fixed $B_0\ge 4$, a common bound on $(1+z{)}^{B_0}$ times all needed $x$-derivatives of ${\Psi }_z$, including any originally required profile derivatives. Schwartz seminorms of the original row Fourier transform and annular column seminorms supply it. Their orders are enlarged once to pay $B_0$ and $J$. For a component/height measure $dc$, the assumption is precisely $\sum \int |dc|N_{J,B_0}(\Psi )\le {\mathcal{N}}_J$, with ${\mathcal{N}}_J$ fixed or polylogarithmic in $U$. Include the original $|\eta |$ and any external signed coefficient costs in this measure. For the four norm heights, a polynomial seminorm cost of degree $g$ is integrable against log-Fourier decay of order greater than $g+5$; choose that order once within the same envelope. Angular and other original heights must satisfy their original absolute-measure contract too. This is a retained hypothesis, not a claim that an unspecified full-J40 table or height measure has been checked. No derivative acts on $b_A$; no envelope is restarted per subset, frequency, class, or component.

Full-frequency consequence and positive reserve

Apply R2 to exactly ${\Psi }_z$ of C59 and $P=P_a$ of C58. Define ${\mathcal{R}}_{\mu }$ by the original signed sum of all fifteen R1 terms, including their individual $\eta$, component coefficients and height integrals. The finite six-unit and $S$-exponent sums are unchanged. There is no cancellation assumption between subsets.

Here are explicit full-frequency bounds. On the ball $z\le 1$ there are $O(U^{1/6})$ ideals $a$; on a dyadic shell $z\asymp R\ge 1$ there are $O((RU{)}^{1/6})$. On $1\le R\le U$, $q_{P_a}\le q_a\ll U^{1/3}$. Use R2 with one fixed polynomial range, choosing its internal loss small enough to pay this puncture factor and all $U$ losses. Its seminorm is $O({\mathcal{N}}_J{R}^{-B_0})$, and ${\sum }_{R\ge 1}{R}^{1/6-B_0}<\infty$ over dyadic shells. For $R>U$ do not apply a nonuniform puncture estimate: use $|K_A(d)|{\ll }_W{3}^{\omega (d)}$ on squarefree ideals, the original annulus, and the fixed-field divisor bound. Each four-column absolute sum is $O_{\epsilon }(D^{4+\epsilon })$. Consequently the entire tail is

$$\ll {\mathcal{N}}_J{U}^{13/6+\epsilon }\sum _{R>U}{R}^{1/6-B_0}\ll {\mathcal{N}}_J{U}^{7/3-B_0+\epsilon }.\text{\hspace{1em}(R9)}$$

With $B_0\ge 4$ this is negligible. All shell bounds are integrated against the same absolute component measure. This proves summability of the complete original frequency identity, without discarding large punctures or altering its signed coefficients.

At $A=U^{1/12}$ and source-conditional $R_{7/8}$, a Mobius slot in R8 costs $U^{42/96}$, a residue slot $U^{47/96}$, and a boundary slot $U^{27/96}$, apart from the allocated loss. After the frequency count, a subset with $m$ Mobius slots and $\ell$ boundary slots costs $U^{(204-5m-20\ell )/96+\epsilon }$. Therefore

$$|{\mathcal{R}}_{\mu }|{\ll }_{\epsilon }{\mathcal{N}}_J{U}^{199/96+\epsilon },\rho =1+B_{\ast }-199/96=\frac{393-160\sqrt{6}}{96}>0,B_{\ast }=(31-10\sqrt{6})/6.\text{\hspace{1em}(R10)}$$

Indeed $393^2-6\cdot 160^2=849>0$; numerically $\rho =0.01126709536137\dots$. Choose, for example, $\nu =\rho /4$ and $0<\epsilon <\rho /4$, absorbing the stipulated polylogarithms with a smaller internal loss. This pays this radial fifteen-term obligation at $U^{1+B_{\ast }-\nu }$. It improves its C60 certificate by $101/48-199/96=1/32$. The other fourteen terms and every boundary term are covered by the same estimate, rather than assigned a fresh budget.

These are the unnormalized local Gauss powers of C59. Restore the original $D^{-1}$ on each side of J4: their product is $D^{-2}=U^{-1}$, giving $U^{103/96+\epsilon }$ for this local normalized remainder, below $U^{B_{\ast }-\nu }$. Keep every additional physical-row, core, gcd, bad-dual and detector/rho prefactor in its original place. The $U^2$ multiplying the all-$K_A$ residue in C60 is already $D^4$; it is not an additional multiplier on R10. C56 fixes the planar measure; there is no extra area factor or unit division. This local calculation does not bound those other parameters or identify the local Gauss object with the full J40 contraction.

The actual profile moment and finite scalar

This section is a reuse application of the existing radial identities, not a new Fourier theorem. Use Grafakos–Teschl, Section 2 for the two-dimensional Fourier–Bessel normalization, and the source Xu–Sang v2, Section 3.4 for its radial Mellin identity. The standard Bessel Mellin value is DLMF 10.22.43. Parameter correspondence: $c=2/\sqrt{3}$, $w(z)=W_{\mathrm{row}}(|z{|}^2)$, Euclidean Fourier frequency radius $c\sqrt{t}$, with measure multiplier $c$ from C56; Bessel order $0$ and power $2v-1$. At $v=1/6$ that power is $-2/3$, satisfying $\Re (\mu +0)>-1$ and $\Re \mu <1/2$ in the cited Bessel identity. It is absolutely integrable even at infinity at this parameter. Equivalently these are dimension $2$, Riesz potential order $5/3$, and spatial kernel $|z{|}^{-1/3}$. No unrestricted inverse-$L^p$ assertion is used.

For the original smooth compact annular row and column profiles set

$$\begin{array}{rl}{M}_i& ={\int }_0^{\infty }{V}_i(x)x^{-1/3}dx=\mathcal{M}{V}_i(2/3),\\ C_w& ={\int }_0^{\infty }{\hat{w}}_{\mathrm{rad}}(t)t^{-5/6}dt={\left(\frac{2\pi }{\sqrt{3}}\right)}^{2/3}\frac{\Gamma (1/6)}{\Gamma (5/6)}{\int }_0^{\infty }{W}_{\mathrm{row}}(u)u^{-1/6}du,\\ {\hat{I}}_4(1/6)& =C_w{M}_1{M}_2\overline{M_3{M}_4}.\end{array}\text{\hspace{1em}(R11)}$$

The row constant is the cited radial identity with the stated measure and frequency scale; its gamma ratio is the standard Bessel Mellin value, not a new source arithmetic constant. The last line follows directly by substituting $t=z/(x_1{x}_2{x}_3{x}_4)$ in C59 and using Fubini; compact annular column support, smoothness at zero and rapid Fourier decay ensure absolute integrability. The resulting power of each $x_i$ is $1/6-1/2=-1/3$. Thus these factor Mellin values are at $2/3$. They must not be confused with either ${\hat{I}}_4(1/6)$ or the source total-column $1/6$ moment: the total column has already absorbed the $t^{1/2}$ Gauss normalization in C57. The row profile $W_{\mathrm{row}}$ is not the divisor cutoff $W$ of C37/C69.

For a nonnegative nonzero original row, $C_w>0$. A diagonal pairing with $V_3=V_1$, $V_4=V_2$ has profile functional $C_w|M_1{M}_2{|}^2$, positive if both Mellin values are nonzero. Nonnegative nonzero column profiles have that property. This states positivity of a profile functional, not of a selected Poisson component or of an arithmetic residue. Complex actual profiles can have zero moments; their values must be substituted, rather than inferred from smoothness.

For the documented J4 diagonal with no additional side multiplier, J28 has ${\zeta }_i=1$. On a common ray refinement of order $h$, its remaining finite table is $1_{g_1{g}_2=\sigma }{1}_{g_3{g}_4=\sigma }$: the $\xi$ and $\chi (-1)$ phases pair with their conjugates on the same total class, and the exceptional numerator characters cancel as in C58. Each relation admits $h$ pairs, so normalized counting on four factors gives $\eta =h^2/h^4=h^{-2}$. Angular Haar measures are normalized; this diagonal has no extra angular multiplier. The six units and $S$ exponents then give $(6/h^2){\prod }_{p\in S}{\sum }_{j=0}^5{q}_p^{-j/6}>0$. Four specified individual classes would instead give $h^{-4}$. This finite Fourier calculation assumes no statistical independence of the actual ideal classes or angles.

Additional original side multipliers must be averaged as their actual finite/angular table. For a side multiplier $B$ the constant pair coefficient is $L_{\sigma }(B)=h^{-2}{\sum }_{g_1{g}_2=\sigma }\int Bd{\theta }_{1}d{\theta }_2$; a polarized table has coefficient $L_{\sigma }(B)\overline{L_{\sigma }(B^{\prime })}$, not an average of $|B{|}^2$. The full J40 component and height aggregate is unspecified here; the J4 diagonal value cannot substitute for that aggregate.

Writing $d_{\mathbf{r}}=q_{e_S(\mathbf{r})}$, retain the exact interface

$$\begin{array}{rl}{S}_{\mathrm{actual}}& =\sum _{ϵ,\mathbf{r}}\int {\eta }_{ϵ,\mathbf{r},c}{d}_{\mathbf{r}}^{-1/6}{C}_{w_c}{M}_{1,c}{M}_{2,c}\overline{M_{3,c}{M}_{4,c}}dc,\\ {\mathcal{E}}_{\mathrm{rad}}& =\frac{\bar{C}}{6}{U}^{13/6}{\bar{Q}}_A{S}_{\mathrm{actual}}+O_{\epsilon }\left({\mathcal{N}}_J[U^{199/96+\epsilon }+U^{47/24+\epsilon }+U^{95/48+\epsilon }]\right).\end{array}\text{\hspace{1em}(R12)}$$

Here $dc$ is the actual signed component/height measure; for a single component omit the integral. The individual row constants stay inside if rows differ. Absolute integrability follows from the same envelope, and C74–C75 supply exactly $\bar{C}$, ${\bar{Q}}_A$ and the other two errors. The common cutoff, field and $S$ are those of the local C59 branch; R12 does not combine branches with different constants by silently identifying them. All original physical prefactors still multiply their own components. This retains the source total-column identity as an application and does not claim a new Fourier result.

Unpaid mathematical interfaces

R10 pays only the original constant-ray radial fifteen-Mobius sum. The all-$K_A$ principal bound remains $U^{17/8+\epsilon }$ under $R_{7/8}$. Neither $S_{\mathrm{actual}}=0$ nor ${\bar{Q}}_A=0$ is proved. A nonzero finite/profile factor would not give a lower bound on ${\bar{Q}}_A$, the principal term or the signed physical contraction. C76, actual finite/profile cancellation, or another sufficient estimate for this principal scalar remains necessary for this route.

A nonnegative nonzero row cannot annihilate R11’s positive row functional. Replacing it by a signed zero-moment kernel requires an unproved transfer inequality for the same original positive row energy, actual coefficient space and paid Type-I consumers. No such redesign or automatic full-$b$ cancellation is used here. R10 does not turn upper bounds into a same-realization cancellation identity.

Outside the bare-J4 finite-table application R43–R46, the twisted residual ${\mathcal{R}}_{\mathrm{tw}}$ remains unpaid. The nonexceptional C68 terms, other cores/gcds/bad duals, original common-e signed detector/rho recombination at the same beta, physical-row restoration and full J40 height/table aggregation remain unpaid. The original Type-I coefficient and consumers are retained without new claims. No selected-Poisson positivity, principal lower bound, stronger zero-free strip, upstream main-proof certification, signed Robin restoration or RH conclusion follows. The full research goal remains RH.

Original-product zeros and finite rectangular residues

Retain exactly C73’s ${\bar{Q}}_A,\bar{H},{\bar{D}}_q$ and the original $W,S$; the results here concern that local principal scalar. They add finite polar information to R12, without evaluating $S_{\mathrm{actual}}$ or changing R1–R12. Put $L=\log A$ and

$$N_q(\mathbf{z})=1-\frac{2{\sum }_{i=1}^4{q}^{-z_i}}{{\bar{D}}_q},D_q(\mathbf{z})=\prod _{i=1}^4(1-q^{-1-z_i}{)}^{-2},{\bar{D}}_q=q+5+\frac{1}{q-1}.\text{\hspace{1em}(R13)}$$

Thus ${\bar{H}}_p=N_{q_p}{D}_{q_p}$ outside $S$, and ${\bar{H}}_p=D_{q_p}$ inside $S$. These are C73’s original factors, with the four-way exclusive-prime rule; no product of independent two-slot factors is substituted.

The finite norm-7 factor. The fixed arithmetic datum in the pinned primary manuscript, notation preceding its additive character puts every prime above $6$ in $S$, together with the original fixed modulus supports. In $K=\mathbb{Q}(\sqrt{-3})$, $3$ ramifies, a rational prime $p\equiv 1(\mathrm{mod}3)$ splits into two primes of norm $p$, and $p\equiv 2(\mathrm{mod}3)$ is inert with norm $p^2$. For example these alternatives follow by factoring $X^2+X+1$ modulo $p\ne 3$: its roots are precisely the nontrivial cube roots of unity. The primes over $2,3$ are excluded; the next possible good norm is $7$, followed by $13$ (the inert prime $5$ has norm $25$). In particular no good norm lies below $13$ except $7$. Define, using the unchanged $S$,

$$n_7=\#\{p\notin S:q_p=7\}\in \{0,1,2\},P_7(\mathbf{z})=1-\frac{12}{73}\sum _{i=1}^4{7}^{-z_i}.\text{\hspace{1em}(R14)}$$

No norm-7 prime is newly excluded. Since ${\bar{D}}_7=73/6$, construct $H_{\ast }$ directly as follows: at every unexcluded norm-7 prime use $D_7$; at other unexcluded primes use $N_q{D}_q$; at primes in $S$ use $D_q$. This deletes only those finitely many numerators and never divides by a numerator. Then

$$\bar{H}=P_7^{n_7}{H}_{\ast },H_{\ast }\text{ is holomorphic and nowhere zero near every point of }\{\Re z_i\ge -1/4(1\le i\le 4)\}.\text{\hspace{1em}(R15)}$$

Indeed for $q\ge 13$ throughout that polytube,

$$|1-N_q|\le \frac{8q^{1/4}}{{\bar{D}}_q}<1.$$

The strict inequality follows from $q+5>8q^{1/4}$: it holds at $13$, as $18^4>8^4\cdot 13$, and the difference has derivative $1-2q^{-3/4}>0$ thereafter. All denominators are nonzero since $\Re (1+z_i)\ge 3/4$. C73’s normal convergence, unchanged by finitely many factors, and the locally summable deviations from $1$ imply nonvanishing of the product of these nonzero factors. The strict bounds and C73’s open domain $\Re z_i>-1/2$ also give the asserted local neighborhoods. Thus R15 identifies the complete zero locus of $\bar{H}$ in this polytube: it is empty when $n_7=0$, and is exactly $P_7=0$ when $n_7>0$.

There are two useful consequences with different scopes. First, $\bar{H}$ itself is nowhere zero on $\Re z_i\ge -1/5$: the only additional check is $8\cdot 7^{1/5}<73/6$, equivalently $7\cdot 48^5<73^5$. Second, on a conjugate-balanced tuple,

$$H_{\ast }(a,a,\bar{a},\bar{a})>0(\Re a\ge -1/4).\text{\hspace{1em}(R16)}$$

For each remaining good numerator its value is $1-8q^{-\Re a}\cos (\Im a\log q)/{\bar{D}}_q>0$. The denominator at every prime is $|1-q^{-1-a}{|}^{-8}>0$; removed norm-7 numerators leave only this denominator. Normal convergence makes the positive product strictly positive. This is a statement about the Euler correction, not positivity of the physical contraction or a selected Poisson component.

A genuine one-coordinate polar slice. Suppose $\varrho$ is a zero of this same ${\zeta }_K$, of multiplicity $m\ge 1$, with $3/4\le \Re \varrho <1$, and set $a=\varrho -1$. Fix arbitrary $z_2,z_3,z_4$ with $\Re z_j\ge 1$. In this slice $\bar{H}(a,z_2,z_3,z_4)\ne 0$: at every good prime $q\ge 7$,

$$|1-N_q|\le \frac{2(q^{1/4}+3/q)}{{\bar{D}}_q}<1,$$

because $2q^{1/4}<q$ and $6/q<5$. The same nonzero-product argument applies, including norm $7$. The other three reciprocal zeta factors are nonzero since their arguments have real part at least $2$. Consequently C73’s $\bar{F}$ has a pole of exact order $2m$ in $z_1$ at $a$ on this slice. For its cutoff integrand below, the order is $2m-v$ if $\hat{W}$ has vanishing order $v<2m$ at $a$ and the other three cutoff values are nonzero; if $v\ge 2m$ the singularity is removable. In particular the original cutoff’s nonvanishing at $a$ is an explicit sufficient hypothesis for a genuine order-$2m$ pole. No such zero or cutoff nonvanishing is asserted to exist. This rules out an automatic Euler-numerator cancellation of all these zero divisors, without asserting noncancellation at every four-zero intersection.

All finite faces and residues. Reuse the finite-contour principle of SV5 and its retained remainder for the present original four-variable integrand

$$G_A(\mathbf{z})=\bar{H}(\mathbf{z})\prod _{i=1}^4\left[\hat{W}(z_i)A^{z_i}{\zeta }_K(1+z_i{)}^{-2}\right],{\bar{Q}}_A=\frac{1}{(2\pi i{)}^4}{\int }_{(c{)}^4}{G}_A(\mathbf{z})d\mathbf{z},c>0.\text{\hspace{1em}(R17)}$$

The inversion is initially absolute, using C73. Fix $0<h<1/2$, write $b=-h$, and take a finite $T>0$ for which the boundary of $[b,c]+i[-T,T]$ meets no zero of ${\zeta }_K(1+z)$. Let $\mathcal{Z}$ be its finite set of distinct interior zero positions $a=\varrho -1$, recording each zero’s multiplicity separately. The only possible polar divisors of $G_A$ in this rectangle product are these coordinate divisors. At $z_i=0$ the simple pole of $\hat{W}$ is canceled by the double zero of the reciprocal zeta square, as in C74; there is no zero-position residue at $0$ and no mixed polar divisor.

For a meromorphic function $f$, let the following operators act only in coordinate $i$:

$$\begin{array}{rlrl}{I}_{b,i}f& =\frac{1}{2\pi i}{\int }_{b-iT}^{b+iT}fd{z}_i,& Z_{i}f& =\sum _{a\in \mathcal{Z}}{\mathrm{Res}}_{z_i=a}f,\\ H_{+,i}f& =\frac{1}{2\pi i}{\int }_{b+iT}^{c+iT}fd{z}_i,& H_{-,i}f& =\frac{1}{2\pi i}{\int }_{b-iT}^{c-iT}fd{z}_i.\end{array}$$

Both horizontal integrals are oriented left to right. The full exact finite-height identity is

$$\begin{array}{rl}{\bar{Q}}_A& =\sum _{E\bigsqcup B\bigsqcup C_{+}\bigsqcup C_{-}=[4]}(-1{)}^{|C_{-}|}\left(\prod _{i\in E}{Z}_i\right)\left(\prod _{i\in B}{I}_{b,i}\right)\left(\prod _{i\in C_{+}}{H}_{+,i}\right)\left(\prod _{i\in C_{-}}{H}_{-,i}\right)G_A+E_T,\\ E_T& =\frac{1}{(2\pi {)}^4}{\int }_{{\mathbb{R}}^4\setminus [-T,T{]}^4}{G}_A(c+i{t}_1,\dots ,c+i{t}_4)d\mathbf{t},\\ |E_T|& {\ll }_{N,c,K,S,W}{A}^{4c}(1+T{)}^{-N}(A\ge 1)\text{for each fixed }N.\end{array}\text{\hspace{1em}(R18)}$$

The sum runs over all $4^4$ ordered partitions; empty products are identities. Each $Z_i$ is itself the full finite sum over zero positions. Thus a term with $k$ residues retains the factor $(2\pi i{)}^{-(4-k)}$ from its remaining paths. This includes every mixed residue/left-line/horizontal contribution, not just the full corners. To prove it, truncate the original four right lines, apply $I_{c,i}=I_{b,i}+Z_i+H_{+,i}-H_{-,i}$ successively, and expand. The one-variable residue theorem gives that sign by the positively oriented rectangle starting up its right edge. Compact boundary paths, finitely many interior zeros and absence of mixed polar divisors justify these finite interchanges and commuting residues. The tail is only on the original right lines. There $\bar{H}$ and the reciprocal zetas are bounded, and R7 gives arbitrary Mellin decay of the fixed smooth cutoff, proving its displayed bound. No estimate on a left-strip reciprocal, no infinite zero sum and no disappearance of horizontal connectors is used. Substitution into R12 retains its exact multiplier $(\bar{C}/6)U^{13/6}{S}_{\mathrm{actual}}$ and all component-specific physical prefactors, with no new field residue or unit factor.

Multiplicity and the coupled derivative. For a selected residue coordinate $i$, let ${\varrho }_i$ have multiplicity $m_i$, put $a_i={\varrho }_i-1$ and define the holomorphic local function

$$b_i(z)=\hat{W}(z){\left(\frac{(z-a_i{)}^{m_i}}{{\zeta }_K(1+z)}\right)}^2,b_i(a_i)=\hat{W}(a_i){\left(\frac{m_i!}{{\zeta }_K^{(m_i)}({\varrho }_i)}\right)}^2.\text{\hspace{1em}(R19)}$$

For any subset $E$ the residue at its chosen zero positions is exactly

$$A^{{\sum }_{i\in E}{a}_i}{\left[\prod _{i\in E}\frac{(L+{\partial }_i{)}^{2m_i-1}}{(2m_i-1)!}\left\{\bar{H}(\mathbf{z})\prod _{i\in E}{b}_i(z_i)\right\}\right]}_{z_i=a_i(i\in E)}\prod _{j\notin E}\left[\hat{W}(z_j)A^{z_j}{\zeta }_K(1+z_j{)}^{-2}\right].\text{\hspace{1em}(R20)}$$

This follows by extracting the coefficient of ${\prod }_{i\in E}(z_i-a_i{)}^{2m_i-1}$ after factoring out the poles and $A^{\sum a_i}$. Vanishing cutoffs or numerator factors are already included; $2m_i$ is the possible order, not an assertion that every intersection has that order.

For four simple zeros, R20 is $A^{\sum_i a_i}[\prod_i(L+\partial_i) {\bar H\prod_i b_i}]_{\mathbf a}$. If every original $\hat{W}(a_i)\ne 0$, an equivalent explicit coupled formula is

$$\begin{array}{rl}{\mathrm{Res}}_{\mathbf{z}=\mathbf{a}}{G}_A& =A^{{\sum }_i{a}_i}\left(\prod _i{B}_i\right){\left[\prod _i(L+d_i+{\partial }_i)\bar{H}\right]}_{\mathbf{a}},\\ B_i& =\frac{\hat{W}(a_i)}{{\zeta }_K^{\prime }({\varrho }_i{)}^2},d_i=\frac{{\hat{W}}^{\prime }(a_i)}{\hat{W}(a_i)}-\frac{{\zeta }_K^{\prime \prime }({\varrho }_i)}{{\zeta }_K^{\prime }({\varrho }_i)}.\end{array}\text{\hspace{1em}(R21)}$$

Each $d_i$ in the operator is its fixed displayed value; derivatives act on the full $\bar{H}$. In particular this is not a product of four independent one-variable residue constants.

Nonzero simple corners, including norm-7 resonance. Suppose $\varrho =\sigma +i\gamma$ is a simple nonreal zero of ${\zeta }_K$, $3/4\le \sigma <1$, and the actual $\hat{W}(a)\ne 0$ at $a=\varrho -1$. Such corners enter R18 when $1-h<\sigma$ and $|\gamma |<T$; the boundary is still required to avoid all zeros. Since $W$ is real, the conjugate values agree under conjugation. Set $u=7^{-a}=x+iy$, $k_7=12/73$, $c_7=k_7\log 7$, and $B_{\varrho }=\hat{W}(a)/{\zeta }_K^{\prime }(\varrho {)}^2$. For the ordered balanced corner $\mathbf{a}=(a,a,\bar{a},\bar{a})$ write $P=P_7(\mathbf{a})=1-4k_{7}x$ and $V=H_{\ast }(\mathbf{a})|B_{\varrho }{|}^4>0$. Its residue is $A^{4(\sigma -1)}$ times a real polynomial in $L$. Its degree and leading coefficient are

$$\begin{array}{ccc}\text{condition}& \text{degree}& \text{leading coefficient}\\ n_7=0\text{ or }P\ne 0& 4& P^{n_7}V\\ P=0,n_7=1& 3& (\log 7)V\\ P=0,n_7=2& 2& 4c_7^2(3x^2+y^2)V.\end{array}\text{\hspace{1em}(R22)}$$

In the first row $P^0=1$, even at $P=0$. All listed coefficients are nonzero. To see the resonant rows directly, ${\partial }_i{P}_7=c_7{u}_i$ and ${\partial }_i{\partial }_j{P}_7=0$ for $i\ne j$. When $P=0$, ${\sum }_i{\partial }_i{P}_7=\log 7$. For $P_7^2$, the first nonzero squarefree derivative sum is

$$\sum _{i<j}{\partial }_i{\partial }_j(P_7^2)=2c_7^2\sum _{i<j}{u}_i{u}_j=4c_7^2(3x^2+y^2),(u_1,u_2,u_3,u_4)=(u,u,\bar{u},\bar{u}).$$

Every smaller derivative of $P_7^{n_7}$ vanishes at resonance, so the highest surviving power of $L$ places exactly $n_7$ derivatives on that factor. Derivatives of $H_{\ast }$ and $b_i$ then affect only lower powers. Resonance forces $x=73/48>0$, proving the last coefficient is nonzero. Conjugation followed by slot permutation fixes the balanced residue, proving reality. The six distinct permutations have equal residues by symmetry of the original product and cutoff, so their sum is six times this same polynomial.

For the repeated simple-zero corner $(a,a,a,a)$ the residue is $A^{4a}$ times a complex polynomial. Put $P=1-4k_{7}u$ and $V^{\prime }=H_{\ast }(a,a,a,a)B_{\varrho }^4\ne 0$. Away from resonance (or when $n_7=0$), its degree is $4$ with leading coefficient $P^{n_7}{V}^{\prime }$. At $P=0$, $u=73/48$; its degree and leading coefficient are $3,(\log 7)V^{\prime }$ for $n_7=1$, and $2,\frac{3}{4}(\log 7{)}^2{V}^{\prime }$ for $n_7=2$. The latter follows from $2{\sum }_{i<j}{c}_7^2{u}^2=12c_7^2{u}^2=\frac{3}{4}(\log 7{)}^2$. No extra factorial arises from repeating the zero in distinct coordinates. Multiplicity of a zero of ${\zeta }_K$ itself is instead handled by R19–R20. These corner statements are conditional on simplicity and the original cutoff value; neither is inferred from $R_{7/8}$. They do not show that the sum over other zeros, mixed paths and all common-scale signed components is nonzero or large.

Domain at the requested depth. R15 covers a rectangle product of depth $h=249/1000<1/4$. In C76’s formal exponent arithmetic, $12\Delta =20\sqrt{6}-48$, and, for the positive margin $\nu =1/3000$,

$$4h-12(\Delta +\nu )=\frac{6124}{125}-20\sqrt{6}=0.002205144336438\dots >0,6124^2-6\cdot 2500^2=3376>0.\text{\hspace{1em}(R23)}$$

Thus the original-product factorization reaches a depth that would be sufficient if the entire R18 expression admitted the corresponding power bound. R23 alone is domain applicability only. The $\bar{H}$ nonvanishing tube of depth $1/5$ reaches only the formal all-line exponent $4/5$; it gives no reciprocal-zeta strip or contour bound. The polygonal application below pays the regular boundaries while retaining the complete finite residue aggregate. It does not deduce a deeper zero-free strip from $R_{7/8}$ or take an infinite-height residue limit.

Fixed-cutoff polygonal localization of the original scalar

This is a source-specific repo-derived paper application of the published Buck/Jensen supplier already recorded at LC168–LC182: Matthew Buck, Integer points of meromorphic functions, DOI 10.1017/S0013091513000436, Lemma 2.4, equation (2.2), Remark 2.3, and the Jensen inequality used there. No new exceptional-disc theorem is proposed. Keep every coefficient and the ideal mask of C73, the same $K,S$, and any fixed original admissible smooth $W$. Constants may use finitely many of its seminorms for each requested error power. No Gevrey condition, zero simplicity, zero spacing, derivative floor, RH or ordinary-zeta reciprocal-height supplier is used.

Dedekind normalization and the local supplier. Write $g(s)=(s-1){\zeta }_K(s)$, an entire function, and for every integer $n$ put ${\tau }_n=n+1/2$, $Q_n=3+|{\tau }_n|$ and

$$p_n(w)=\frac{g(2+i{\tau }_n+w)}{g(2+i{\tau }_n)},p_n(0)=1,\underset{|w|\le 16}{\max }|p_n(w)|\le V_n=C{Q}_n^{B_0},V_n\ge 2.\text{\hspace{1em}(R24)}$$

Here $C,B_0$ are fixed, independent of $n,A,W$. The upper-growth input is precisely the unconditional fixed-field strip growth used in R6–R8 and C74, now on $-14\le \Re s\le 18$ for the regularized $g$. For this degree-two field the completed function is $(\sqrt{3}/(2\pi ){)}^s\Gamma (s){\zeta }_K(s)$ with functional equation $s\leftrightarrow 1-s$; its strip upper bounds and the absolutely convergent right-half-plane Euler product supply a finite polynomial degree on that fixed strip. The pole at $1$ has been removed before taking the maximum, including the bounded-height windows. On the normalizing line the same Dedekind Euler product gives ${\zeta }_K(2{)}^{-1}\le |{\zeta }_K(2+i{\tau }_n)|\le {\zeta }_K(2)$, so $|g(2+i{\tau }_n)|\asymp Q_n$. Thus this is a Dedekind-specific normalization; it imports no lower bound inside the zero strip.

In Buck’s notation take $u=1/p_n$, scale $s=2$, zero-count parameter $m=0$, and $ϵ=r/48$, where $r=10^{-6}$ is fixed. Jensen’s identity, with zero additive constant because $p_n(0)=1$, gives $T(16,1/p_n)=T(16,p_n)\le \log V_n$. Lemma 2.4 consequently supplies a finite union $U_n$ of open discs in the $w$ plane, of total radius at most $48ϵ=r$, such that

$$|p_n(w)|\ge V_n^{-k},|w|\le 4,w\notin U_n,k=6+\frac{\log (48/r)}{\log 2}.\text{\hspace{1em}(R25)}$$

The accompanying Jensen count is $n(4,1/p_n)\le \log V_n/\log 4$, with multiplicities; poles of $1/p_n$ count zeros of $g$, not the removed zeta pole. In particular every such zero in $|w|\le 4$ belongs to $U_n$. Translate these discs to the $z$ plane by $z=1+i{\tau }_n+w$, and call their union $E_n$. Its real and imaginary projections each have total interval length at most $2r$. For $-249/1000\le \Re z\le 1$ and $|\Im z-{\tau }_n|\le 1$, the corresponding $|w|<2<4$. Outside $E_n$, R24–R25 and ${\zeta }_K(1+z{)}^{-1}=z/g(1+z)$ therefore give, for one fixed $B$,

$$|{\zeta }_K(1+z{)}^{-1}|\le C{Q}_n^B\le C^{\prime }(3+|\Im z|{)}^B.\text{\hspace{1em}(R26)}$$

The formula at $z=0$ is interpreted by continuation and equals zero. Indeed $|z|\ll Q_n$ and $|g(2+i{\tau }_n)|\gg Q_n$, so one may take $B=B_{0}k$ after adjusting the fixed constant. This polynomial bound holds on the chosen complement, including low heights. It is not a polynomial bound on a complete zero-bearing vertical line. The arbitrarily close zeros and their full multiplicities remain inside the exceptional discs.

An actual finite polygon, with bounded local length. Set

$$h_0=\frac{497}{2000},h_1=\frac{249}{1000},I=(-h_1,-h_0),H=\lceil A\rceil +4(A\ge 2).$$

Use the finite collection $E_n$, $-H-1\le n\le H$. For each $-H\le n<H$ choose $x_n\in I\setminus {\pi }_{\Re }(E_n)$; this is possible since $|I|=1/2000>2r$. For $-H<n<H$ choose $y_n\in (n-1/8,n+1/8)$ outside the imaginary projections of $E_{n-1}\cup E_n$. Their total length is at most $4r<1/4$. Choose a single $T\in (H-1/8,H+1/8)$ outside the imaginary projections of $E_{H-1}\cup E_H$ and the negatives of the imaginary projections of $E_{-H-1}\cup E_{-H}$. The excluded length is at most $8r<1/4$. Put $y_{-H}=-T$, $y_H=T$. All choices can be made outside the closed projection intervals as well, since their finite total lengths have these strict margins.

Define ${\Gamma }_A$ upwards: in slab $n$ take the vertical segment $[x_n+i{y}_n,x_n+i{y}_{n+1}]$; at each internal $y_n$ join $x_{n-1}+i{y}_n$ to $x_n+i{y}_n$ horizontally. Each vertical segment avoids $E_n$ by its real coordinate, and each join avoids both adjacent unions by its height. All these segments have $|\Im z-{\tau }_n|\le 5/8$ in at least one of their stated windows. R26 therefore holds on the entire path. The two outer horizontal segments from its endpoints to $1\pm iT$ also avoid the relevant unions, so R26 holds on them as well. The right side is $[1-iT,1+iT]$, where the Euler-product reciprocal bound already holds without disc deletion.

The heights $y_n$ strictly increase, by at least $3/4$. Distinct vertical interiors occupy disjoint height intervals; horizontal joins meet only their adjacent vertical endpoints. Removing a zero-length join if necessary, ${\Gamma }_A$ is thus simple and meets neither outer segment except at its own endpoint. Together with the right side and outer segments it bounds a simple polygon ${\Omega }_A$, oriented positively by going up the right side. There are no holes or additional boundary components. Every boundary point has $-h_1<\Re z\le 1$; the entire closed polygon is in $\Re z>-1/2$. In any height interval of length one there are at most three vertical pieces, with total vertical length at most one, and at most three joins, each of length less than $1/2000$. In particular its local arclength is bounded by $2$, uniformly in $A,H$, the discs and the choices. Its total length is $O(T+1)$, and $T\asymp A$ is finite for every $A\ge 2$.

Let ${\mathcal{Z}}_A$ be the distinct positions $a=\varrho -1$ with $a\in {\Omega }_A$ and ${\zeta }_K(\varrho )=0$, retaining each multiplicity $m_{\varrho }$. This is a finite set: $g$ is nonzero as an entire function, the closed polygon is compact, and R26 excludes boundary zeros. Membership is in this polygon’s actual interior, equivalently winding number one for its boundary, not the conditions $\Re \varrho >1-h_0$ or $>1-h_1$ in a replacement rectangle. Zeros in the narrow intervening band are selected by the actual vertical segments and joins. All four coordinates use this same polygon, zero set and top/bottom heights.

Finite identity, with all fifteen terms. For coordinate $i$ let $J_i=(2\pi i{)}^{-1}{\int }_{{\Gamma }_A}d{z}_i$, let $Z_i={\sum }_{a\in {\mathcal{Z}}_A}{\mathrm{Res}}_{z_i=a}$, and let ${\ell }_{\pm ,A}$ be the segment from the corresponding endpoint of ${\Gamma }_A$ to $1\pm iT$, left to right. Let $H_{\pm ,i}$ integrate over that segment with factor $(2\pi i{)}^{-1}$. Let $I_{1,i}$ be the truncated upward right integral. The polygonal residue theorem gives exactly $I_{1,i}=J_i+Z_i+H_{+,i}-H_{-,i}$. Define

$$\begin{array}{rl}{P}_A& =\sum _{\varnothing \ne E\subseteq [4]}\left(\prod _{i\in E}{Z}_i\right)\left(\prod _{j\notin E}{J}_j\right)G_A,\\ L_A& =\left(\prod _{i=1}^4{J}_i\right)G_A,\\ C_A& =\sum _{\begin{array}{c}E\bigsqcup B\bigsqcup C_{+}\bigsqcup C_{-}=[4]\\ C_{+}\cup C_{-}\ne \varnothing \end{array}}(-1{)}^{|C_{-}|}\left(\prod _{i\in E}{Z}_i\right)\left(\prod _{i\in B}{J}_i\right)\left(\prod _{i\in C_{+}}{H}_{+,i}\right)\left(\prod _{i\in C_{-}}{H}_{-,i}\right)G_A,\\ {\bar{Q}}_A& =P_A+L_A+C_A+E_T.\end{array}\text{\hspace{1em}(R27)}$$

Here $E_T$ is exactly R18’s original right-line tail with $c=1$. The fifteen $P_A$ terms have sizes $4,6,4,1$ for residue-subset cardinalities $1,2,3,4$, all with positive operator sign. A term of cardinality $k$ has exactly $(2\pi i{)}^{-(4-k)}$ on its remaining paths. Each coordinate residue operator includes every zero position in ${\mathcal{Z}}_A$. Formula R20 applies without alteration, with arbitrary multiplicities and derivatives of the full coupled $\bar{H}$. It allows vanishing cutoff values and Euler numerators; no individual residue or inverse derivative is estimated here.

The cancelled product at $z_i=0$ remains holomorphic, as in R17–R18. C73 puts $\bar{H}$ holomorphic on a neighborhood of the entire closed polygon product, with no mixed polar divisors. Successive one-coordinate Cauchy deformations on compact paths consequently prove R27, and the finite residues commute. There is no omitted cutoff residue, no change of mask, and no infinite zero sum.

Payment of the all-left term and every outer face. On ${\Gamma }_A$, both outer segments and the right side put $v_A(z)=|\hat{W}(z)A^z{\zeta }_K(1+z{)}^{-2}|$. R7 and R26, the local-length bound, and the fixed distance of the left band from zero give

$$\begin{array}{rl}{\int }_{{\Gamma }_A}{v}_A(z)|dz|& \ll A^{-h_0},\\ {\int }_{[1-iT,1+iT]}{v}_A(z)|dz|& \ll A,\\ {\int }_{{\ell }_{\pm ,A}}{v}_A(z)|dz|& {\ll }_{M}A(1+T{)}^{-M}\text{for every fixed }M>0.\end{array}\text{\hspace{1em}(R28)}$$

For example on the left choose a Mellin decay order larger than $2B+2$ and sum the weights over unit-height intervals. On an outer segment use the same uniform strip decay at height $T$ with order larger than $2B+M$; its length is at most $1+h_1$ and $A^{\Re z}\le A$. The original right side uses its Euler-product bound and R7. These arguments use the fixed smooth cutoff’s arbitrary polynomial decay, without an exponential-decay or varying-cutoff assertion. C73 gives a uniform bound for $\bar{H}$ throughout the fixed real polytube $[-h_1,1{]}^4$, independently of all heights. It follows at once that $|L_A|\ll A^{-4h_0}=A^{-497/500}$.

To pay a face containing residues, keep each $Z_i$ as its whole finite operator and replace it by $I_{1,i}-J_i-H_{+,i}+H_{-,i}$ on the closed boundary. Its weighted contour mass is $O(A)$ by R28, uniformly in the number, closeness and multiplicities of the enclosed zeros. After this replacement in any term of $C_A$, at least one outer segment remains and all other factors have weighted mass $O(A)$. The uniform bound for the coupled $\bar{H}$ then proves

$$|C_A|+|E_T|{\ll }_M{A}^4(1+T{)}^{-M},{\bar{Q}}_A=P_A+O_{K,S,W}(A^{-497/500}),A\ge 2.\text{\hspace{1em}(R29)}$$

More precisely R27 retains $L_A$ exactly, and for every prescribed $D>0$ the combined outer-face and original right-tail error is $O_{D,K,S,W}(A^{-D})$: use $T\asymp A$ and $M>D+4$. The contour and $P_A$ need not change with $D$; only the fixed cutoff seminorms used in the estimate increase. This bounds the complete finite faces, including those with several residue coordinates, without summing absolute values of individual residues or assuming a lower bound on any ${\zeta }_K^{(m)}(\varrho )$.

The exact signed physical consumer. R29 is unconditional for the scalar C73 with fixed $K,S,W$. For R12 retain its source-conditional $R_{7/8}$, original profiles, coefficient tables and common absolute component/height envelope; those are premises of its existing evaluation and other errors. With the same $A=U^{1/12}$, R27 gives the exact substitution of $P_A+L_A+C_A+E_T$ for ${\bar{Q}}_A$ in R12. In particular

$$\begin{array}{rl}{\mathcal{E}}_{\mathrm{rad}}& =\frac{\bar{C}}{6}{U}^{13/6}{P}_A{S}_{\mathrm{actual}}\\ & +O_{K,S,W}\left(\left|\frac{\bar{C}}{6}\right|U^{13/6}|S_{\mathrm{actual}}|A^{-497/500}\right)\\ & +O_{\epsilon }\left({\mathcal{N}}_J[U^{199/96+\epsilon }+U^{47/24+\epsilon }+U^{95/48+\epsilon }]\right).\end{array}\text{\hspace{1em}(R30)}$$