Arithmetic smoothing and a complete short-inverse quartic Gauss mean

Author: bramble_trace. Author proof; independent review pending.

1. The result and its scope

Let $\mathcal O=\mathbb Z[i]$, $\lambda=1+i$, and write $q_a=Na$. All ideal indices are nonzero, integral and odd. We identify an odd ideal with its unique generator congruent to $1$ modulo $\lambda^3$. Write $\chi_n(h)=(h/n)_4$, with its literal zero at a common prime, and use the standard additive character $e(z)=\exp(2\pi i\operatorname{Re}z)$. Put

$$ G(n,h)=q_n^{-1/2}\sum_{z\bmod n}\chi_n(z)e(hz/n). \tag{1} $$

A dyadic ray below is a fixed class for which both $\chi_n(i)$ and $\chi_n(\lambda)$ are constant. Fix such a nonempty ray $\mathcal C$ and a real smooth profile $W$ supported in a compact subinterval of $(1,2)$.

Choose a fixed real $\phi\in C_c^\infty((1,2))$ with $\int_0^\infty\phi(u)\,du/u=1$. Define

$$ V_{\rm ar}(x)=\frac8\pi\sum_{a\ {\rm odd}} \frac{\phi(q_a x)}{q_a},\qquad c_X(n)=\sum_{d\mid n}\mu(d)V_{\rm ar}(q_d/X). \tag{2} $$

The cutoff is bounded, smooth, supported through two, and flat with value one at zero. It need not be monotone or bounded by one. Consider the full ordinary Gaussian-row mean

$$ \mathcal E(H,L,X)=\frac1L\sum_{h\in\mathcal O}e^{-\pi q_h/H} \left|\sum_{n\in\mathcal C\atop n\ {\rm squarefree}} c_X(n)W(q_n/L)G(n,h)\right|^2. \tag{3} $$

Theorem 1. For $0\le x<27/40$ and every $\epsilon>0$,

$$ \mathcal E(L,L,L^x)\ll_{\epsilon,\phi,W}L^{e(x)+\epsilon},\qquad e(x)=\begin{cases} 5/4-2x,&0\le x\le1/8,\\ 1,&1/8\le x\le1/3,\\ 1/2+3x/2,&1/3\le x\le3/8,\\ 7/8+x/2,&3/8\le x\le1/2,\\ 3/4+3x/4,&1/2\le x\le3/5,\\ 4/5+2x/3,&3/5\le x<27/40. \end{cases} \tag{4} $$

The assertion includes every physical row and Fourier frequency. It also holds for a fixed finite union of rays by Cauchy. In particular, the exponent is one throughout $L^{1/8}\le X\le L^{1/3}$, is $9/8$ at $X=L^{1/16}$ and $X=L^{1/2}$, and is $6/5$ at $X=L^{3/5}$. The arbitrary-coefficient quartic sieve gives $L^{5/4+\epsilon}$ for this same mean; the saving is strict for each fixed $0<x<27/40$.

The proof gives a rectangular statement. All implicit constants may depend on a fixed number of profile derivatives, the ray, and the displayed fixed exponents.

Theorem 2. Let $1\le H\le L^2$, $1\le X\le L$ and $0<\eta<1$. Then $\mathcal E(H,L,X)$ is at most $(HL)^\epsilon$ times a constant multiple of

$$ \begin{aligned} H+L &+\min\left\{H^{1/2}X+H^{1/8}L^{3/4}X^{1/2} +H^{1/4}L^{1/2}X^{3/4},\ H^{1/2}X^{3/2}\right\}\\ &+H^{3/4}L^{1/2}X^{-2\eta}+H^{3/4}L^{-1/2}X. \end{aligned} \tag{5} $$

There are two ingredients specific to the prescribed coefficient. An arithmetic Mellin factor in (2) cancels the principal reciprocal zeta in the entire exterior vector, not only in one coordinate. For the other frequencies, the actual tilted inverse splits into ordinary Möbius polynomials with an absolutely summable powerful part. A maximum over the inverse length lets us aggregate the forced divisors at each natural frequency, where their column exclusions are already character zeros.

The moment inputs are de Faveri's quartic sieve [1] and the natural fourth moment proved in [2]. Neither result alone estimates the present coefficient. The lower bounds of de Faveri, Dunn and Hoffstein [3] explain why an arbitrary-coefficient diagonal sieve cannot replace this argument. The short mollified central moments of Castillo, de Faveri and Dunn [4] concern a different coefficient and family range. We make no novelty claim for lattice Poisson, the Euler expansion, the elementary lcm identity, or these published inputs. A second polynomial bound follows from the quadratic large sieve in David, Dunn, Hamieh and Lin [5]. Its short-product consequence is proved in Appendix A, with the common-profile and natural-zero conventions needed here.

The last balanced branch uses the actual reciprocal property. Buffered zero-density bins from [2] and [6], together with the natural fourth moment, give a third maximal mixed estimate. Its proof verifies the order of the actual dyadic twists and restores small conductors and frequencies. It is an additional author deduction, not a signed covariance theorem.

This theorem concerns a Gauss-weighted source with squarefree columns. The ordinary zero detector has a different column phase. Moreover, $X=L^{3/4}$ lies outside (4). Section 8 states the precise remaining consumer issues; no new zero-density, zero-free or RH theorem is claimed.

2. The arithmetic cutoff

Let $f(z)=\phi(|z|^2)/|z|^2$ on $\mathbb C$, extended by zero near zero. The primary generators form the lattice coset $1+\lambda^3\mathcal O$, of Euclidean covolume eight. Since $\int_{\mathbb C} f=\pi$, Poisson gives

$$ V_{\rm ar}(x)=1+O_{A,j}(x^A),\qquad (x\partial_x)^j(V_{\rm ar}(x)-1)=O_{A,j}(x^A) \quad(0<x\le1). \tag{6} $$

Indeed (2) is $(8/\pi)x\sum_{n\in1+\lambda^3\mathcal O}f(\sqrt x n)$; every nonzero dual-lattice frequency has length bounded below, and the Fourier transform and all required derivatives are Schwartz. This proves the asserted smooth extension at zero and uniform boundedness.

Write $\zeta_o(s)=(1-2^{-s})\zeta_{\mathbb Q(i)}(s)$; its residue at one is $\pi/8$. Absolute interchange for $\operatorname{Re}z>0$ gives

$$ \widehat V_{\rm ar}(z)=\frac8\pi\zeta_o(1+z)\widehat\phi(z), \qquad \widehat\phi(z)=\int_0^\infty\phi(u)u^{z-1}\,du. \tag{7} $$

Opening the sum also gives, for squarefree $n$,

$$ c_X(n)=\frac8\pi\sum_b\frac{\phi(q_b/X)}{q_b} \prod_{p\mid n}(1-q_p1_{p\mid b}). \tag{8} $$

The local factors have zero complete residue mean. The proof below does not infer cancellation in the incomplete $b$ annulus from that fact. In particular, for a prime $p$ with $q_p>2X$, $c_X(p)=V_{\rm ar}(1/X)$, so the prime tail is retained. Always $|c_X(n)|\ll\tau(n)$.

3. The principal fourth-power packet

Put $a_n=c_X(n)W(q_n/L)$. For squarefree $n,m$ in the ray, define the full finite correlation

$$ F(n,m;j)=\sum_{mx-ny\equiv j\bmod nm} \chi_n(x)\overline{\chi_m(y)}. \tag{9} $$

Both residue variables run over their complete quotients. Additive orthogonality and Poisson give exactly

$$ \mathcal E=\frac HL\sum_{n,m}\frac{a_n\overline{a_m}}{\sqrt{q_nq_m}} \sum_{j\in\mathcal O}F(n,m;j) e^{-\pi Hq_j/(q_nq_m)}. \tag{10} $$

At $j=0$, $F(n,m;0)=1_{n=m}\varphi(n)$, so the origin costs $O(HL^\epsilon)$. Let $P_1$ be the contribution of every nonzero $j=\omega\lambda^k t^4$, where $\omega\in\{1,i,-1,-i\}$, $k\ge0$, and $t$ is odd primary. This representation is unique.

Write $n=ga,m=gb$ with $g,a,b$ pairwise coprime and squarefree. The complete local correlations give

$$ F(n,m;\omega\lambda^k t^4) =\varphi(g)1_{g\mid t}1_{(t,ab)=1}. \tag{11} $$

To check the phase, put $e(v)=(q_v-1)/4\pmod2$. Quartic reciprocity has defect $(-1)^{e(u)e(v)}$ and $\chi_v(-1)=(-1)^{e(v)}$. The common ray gives $e(a)=e(b)$; the reciprocity defect and the residual $-1$ factor cancel. The $\omega\lambda^k$ phase cancels between the two columns on the same ray. At a shared prime the local correlation is $q_p-1$ precisely when $p\mid t$; otherwise it vanishes. These are complete correlations, including all shared-prime zeros.

Set $\delta(n)=\varphi(n)/q_n$ on squarefree ideals and

$$ Q(a)=\sum_{n,m}\frac{a_n\overline{a_m}}{(q_nq_m)^{1/4}} \delta(\operatorname{lcm}(n,m)). \tag{12} $$

We need the packet estimate with its total error:

$$ P_1=\kappa\frac{H^{3/4}}LQ(a)+O((HL)^\epsilon H),\qquad \kappa=\frac{\pi^{3/4}\Gamma(1/4)}{8(1-2^{-1/4})}. \tag{13} $$

Here and subsequently errors allow the uniform divisor bound for $a_n$. For a proof, let

$$ S(y)=\sum_{t\ {\rm odd\ primary}}e^{-\pi(q_t/y)^4} =\rho I y+E(y),\quad \rho=\pi/8,\quad I=\Gamma(1/4)/(4\pi^{1/4}). $$

Poisson on the primary coset, using $e^{-\pi|z|^8}$, gives $E(y)=O_A(y^{-A})$ for $y\ge1$ and $E(y)=O(y)$ for $y\le1$. The coset contains no origin. For each pair in (11), expand coprimality with $ab$ to obtain $\varphi(g)\sum_{\ell\mid ab}\mu(\ell)S(T/(q_gq_\ell))$, where $T=(q_nq_m/(H2^k))^{1/4}$. The main density is $\delta(\operatorname{lcm}(n,m))$.

The number of pairs with fixed $g,\ell$, $\ell\mid ab$ and $q_n,q_m\asymp L$ is at most $O(L^2 2^{\omega(\ell)}/(q_g^2q_\ell))$, by allocating each prime of $\ell$ to $a$ or $b$ and counting their shortened ideal ranges. The normalized error is consequently at most a small power times

$$ H\sum_{z\ {\rm squarefree}}\frac{4^{\omega(z)}}{q_z} \sup_{c\asymp1}|E(cT_0/q_z)|, \qquad T_0=(L^2/(H2^k))^{1/4}. $$

Splitting at $q_z=T_0$ bounds this sum by $H(1+T_0)^\epsilon\min(1,T_0)$. Its sum in $k$ is logarithmic and then geometric. The four units and the geometric main term give exactly (13). Thus the error is summed over the whole packet, not an error repeated without control at every frequency.

The elementary identity

$$ \delta(\operatorname{lcm}(n,m))= \delta(n)\delta(m)\sum_{e\mid n,m}\frac1{\varphi(e)} $$

gives the positive exterior norm

$$ Q(a)=\sum_{e\ {\rm squarefree}}\frac{|A_e|^2}{\varphi(e)},\qquad A_e=\sum_{n\in\mathcal C\atop e\mid n} c_X(n)W(q_n/L)\delta(n)q_n^{-1/4}. \tag{14} $$

Define $\alpha_p=(q_p-1)/(q_p^2+q_p-1)$ and $\alpha_e=\prod_{p\mid e}\alpha_p$. Smooth squarefree counting gives

$$ \sum_{n\in\mathcal C\atop\ell\mid n}\delta(n)q_n^{-1/4}W(q_n/L) =C_W L^{3/4}\alpha_\ell +O_\epsilon(L^{1/4+\epsilon}q_\ell^{-1/2+\epsilon}). \tag{15} $$

For completeness, expand the ray into finitely many fixed dyadic characters $\xi$. The unforced Dirichlet series is $L_o(s,\xi)B_\xi(s)$, where $B_{\xi,p}(s)=(1+\delta(p)\xi(p)q_p^{-s})(1-\xi(p)q_p^{-s})$ converges normally for $\operatorname{Re}s>1/2$. Forcing $\ell$ inserts $\prod_{p\mid\ell}\delta(p)\xi(p)q_p^{-s}/(1+\delta(p)\xi(p)q_p^{-s})$, of size $O(q_\ell^{-1/2+\epsilon})$ on a line strictly to the right of $1/2$. Shift the smooth Mellin integral only to that line. Only the principal $\xi$ has a residue, and its local ratio is $\alpha_p$. Finite deletion products are subpower. This proves (15), including empty forced ranges, with polynomial height bounds absorbed by the profile.

Opening the original $c_X$ now yields

$$ A_e=C_W L^{3/4}\alpha_e M_e(X)+E_e,\qquad M_e(X)=\sum_d\mu(d)V_{\rm ar}(q_d/X) \frac{\alpha_{\operatorname{lcm}(e,d)}}{\alpha_e}, \tag{16} $$

and

$$ \sum_e\frac{|E_e|^2}{\varphi(e)} \ll L^{1/2+\epsilon}X. \tag{17} $$

Indeed write $d=hb$, $h\mid e$, $(b,e)=1$. The range is $q_b\le2X/q_h$, and $q_{\operatorname{lcm}(e,d)}=q_eq_b$. Thus the forced error sum is $O(X^{1/2+\epsilon}q_e^{-1/2+\epsilon})$. Squaring and summing uses $\sum_e q_e^{-1+\epsilon}/\varphi(e)<\infty$ for sufficiently small $\epsilon$. The sum in (17) can extend over all squarefree $e$; (15) and the cancellation in (16) also hold for $q_e>2L$.

The main vector is where the chosen cutoff matters. Define

$$ H(z)=\prod_p\frac{1-\alpha_pq_p^{-z}}{1-q_p^{-1-z}},\qquad g_e(z)=\frac{\alpha_e}{\sqrt{\varphi(e)}} \prod_{p\mid e}\frac{1-q_p^{-z}}{1-\alpha_pq_p^{-z}}. \tag{18} $$

On an absolute right line, the Dirichlet series in (16) contains $H(z)/\zeta_o(1+z)$. Equation (7) cancels this reciprocal there, giving the identity in $\ell^2$,

$$ \left(\frac{\alpha_e M_e(X)}{\sqrt{\varphi(e)}}\right)_e =\frac8\pi\frac1{2\pi i} \int_{(c)}\widehat\phi(z)X^zH(z)g(z)\,dz. \tag{19} $$

For every fixed $\eta<1$, both $H$ and $g$ are holomorphic and bounded on $\operatorname{Re}z\ge-\eta$. In fact the local deviations for $H$ are $O(q_p^{-2+\eta})$, and the local squared nontrivial coordinates of $g$ are $O(q_p^{-3+2\eta})$. The denominators cannot vanish there, since $\alpha_p<1/q_p$. These convergent majorants on a slightly wider half-plane also justify strong $\ell^2$ holomorphy.

Shift (19) to $\operatorname{Re}z=-\eta$ as a Bochner integral. Rapid vertical decay of $\widehat\phi$ controls tails and horizontals; there are no poles or reciprocal-zero residues. Minkowski gives

$$ \sum_e\frac{\alpha_e^2|M_e(X)|^2}{\varphi(e)} \ll_{\phi,\eta} X^{-2\eta}. \tag{20} $$

Combining (13)--(20), including the cross term by the Hilbert norm, proves

$$ |P_1|\ll(HL)^\epsilon \left(H^{3/4}L^{1/2}X^{-2\eta} +H^{3/4}L^{-1/2}X+H\right). \tag{21} $$

The exact packet need not be positive; only its averaged main form is.

4. A maximal tilted-inverse mean on natural frequencies

Fix a dyadic character $\nu$ and set $\psi_j(v)=\nu(v)\chi_v(j)$, retaining its literal zeros. Throughout this section $j$ is odd primary; units and powers of $\lambda$ will be restored in Section 6. We use the natural fourth moment of [2]: for $R\ge1$, $1/2\le\sigma\le2$,

$$ \sum_{q_j\le R\atop\psi_j\ {\rm nonprincipal}} |L_o(\sigma+i\tau,\psi_j)|^4 \ll_\epsilon R^{1+\epsilon}(1+|\tau|)^C. \tag{22} $$

The principal/nonprincipal distinction refers to the inducing character. Reciprocity changes orientation using finitely many fixed dyadic twists. The natural fourth moment and its central-value transfer in [2, Sections 8--9] keep the deleted Euler factors: primitive conductor times redundant natural radical is $O(q_j)$. The common truncation there can be taken as $(2R)^{10}$ for all these rows. On $\sigma>1/2$, each annulus gains a decreasing real norm weight; the same Minkowski argument proves (22). All fixed dyadic twists belong to a finite list. No varying odd twist or radical occurs in (22).

Fix $\delta>0$, $s=1/2+\delta+i\tau$, and a common odd mask $e$. For a smooth bounded cutoff $V$ supported through two and flat at zero, define

$$ M_{T,\psi,e}(s)=\sum_{a\ {\rm squarefree}\atop(a,e)=1} \mu(a)V(q_a/T)\prod_{p\mid a}\frac{z_p}{1+z_p}, \qquad z_p=\psi(p)q_p^{-s}. \tag{23} $$

Expand $-z/(1+z)=\sum_{k\ge1}(-z)^k$. With $\lambda_0(m)=(-1)^{\Omega(m)}$, the exact regrouping is

$$ M_{T,\psi,e}(s)=\sum_{b\ {\rm powerful}\atop(b,e)=1} \frac{\lambda_0(b)\psi(b)}{q_b^s} \sum_{a\ {\rm squarefree}\atop(a,be)=1} \frac{\mu(a)\psi(a)}{q_a^s} V\left(\frac{q_aq_{\operatorname{rad}(b)}}T\right). \tag{24} $$

The decomposition has $(a,b)=1$ and is unique; powerful means every positive prime exponent is at least two. The cutoff involves the radical of $b$, not $b$. The original radical support is finite, so the local expansions are absolutely legitimate. Also

$$ \sum_{b\ {\rm powerful}}q_b^{-1/2-\delta} =\prod_p\left(1+\frac{q_p^{-1-2\delta}} {1-q_p^{-1/2-\delta}}\right)<\infty. \tag{25} $$

Write $j=rf^4$ uniquely, with $r$ fourth-power-free. We do not require $(r,f)=1$. Put $R_f=R/q_f^4$. Exactly,

$$ \psi_{rf^4}(v)=\psi_r(v)1_{(v,f)=1},\qquad L_o(s,\psi_{rf^4})=L_o(s,\psi_r) \prod_{p\mid f}(1-\psi_r(p)q_p^{-s}). \tag{26a} $$

The inducing character is unchanged. A factor already deleted at $r$ is one. Uniformly for $\operatorname{Re}s\ge1/2$, this product has modulus $\ll_\epsilon q_f^\epsilon$: treat finitely many small primes separately, and use $1+q^{-1/2}\le q^\epsilon$ for the others. The new polynomial zeros are the common mask $f$.

For fixed $b,f$, any hard-capped inner polynomial of (24), with an extra factor $q_a^{-z}$, $\operatorname{Re}z\ge0$, has the following two fourth-moment bounds on $q_r\le R_f$:

$$ \begin{aligned} \sum_r|P_r|^4 &\ll(R_fX)^\epsilon (R_f+X^2+R_f^{3/4}X+R_f^{1/2}X^{3/2}),\\ \sum_r|P_r|^4&\ll(R_fX)^\epsilon(R_f+X^3). \end{aligned} \tag{26b} $$

The first is [1, Theorem 1.1] applied to the polynomial square, whose columns are cubefree, have length $O(X^2)$ and coefficients bounded by $\tau(m)q_m^{-1/2-\delta}$. Its squared coefficient mass is bounded. The second follows from Appendix A with two equal factors, after splitting the hard cap into common norm annuli and using Minkowski. On an annulus of length $T$, the central normalization gives $R_f+T^3$; the extra real norm weights decrease. The logarithmic number of annuli is absorbed in the small power. Fixed twists and masks are common contractions in both arguments, with no dependence on their moduli.

We need a maximum over $0<T\le X$, not a separate uniform bound for each $T$. For fixed $b$, set $Y_b=2X/q_{\operatorname{rad}(b)}$ and $U=T/q_{\operatorname{rad}(b)}$. If $Y_b<1$, the inner polynomial vanishes. Otherwise insert the same cap $q_a\le Y_b$ for every $T$. For a small fixed $\kappa>0$, Mellin inversion gives exactly

$$ \begin{aligned} &\sum_{a\ {\rm squarefree}\atop(a,be f)=1} \mu(a)\psi_r(a)q_a^{-s}V(q_a/U)\\ &=\frac1{2\pi i}\int_{(\kappa)}\widehat V(z)U^z \sum_{a\ {\rm squarefree},\ q_a\le Y_b\atop(a,be f)=1} \mu(a)\psi_r(a)q_a^{-s-z}\,dz. \end{aligned} \tag{26c} $$

The possible Mellin pole at zero lies to the left. On this positive line $\widehat V$ decays faster than any fixed height power. Taking the supremum on the left costs at most $X^\kappa$ times the common integral of polynomial moduli on the right. In particular the coefficients inside each integral are common to every row; the maximum does not authorize row-dependent choices in a sieve.

Apply Hölder to (22) and each alternative of (26b) on the fixed-$f$ family. Minkowski in row $\ell^2$ then applies to the common integral (26c) and the powerful sum (24). The factors $\psi_r(b)$ are row contractions; (25) pays their sum without changing the $L$ factor. The first resulting mixed bound has monomials $R_f^\alpha X^\beta$ with $(\alpha,\beta)=(1,0),(1/2,1),(7/8,1/2),(3/4,3/4)$; the second has $(1,0),(1/2,3/2)$. Finally sum the positive squared moments over $f$. Every sum converges since $4\alpha\ge2>1$. One can first truncate the powerful expansion and pass to the limit using its summable norm majorant. All constants in (26b) are uniform in $b,e,f$ and the Mellin height. Choosing the subsidiary losses after the requested final $\epsilon$ proves

$$ \begin{aligned} &\sum_{q_j\le R\atop\psi_j\ {\rm nonprincipal}} |L_o(s,\psi_j)|^2\sup_{0<T\le X}|M_{T,\psi_j,e}(s)|^2\\ &\qquad\ll(RX)^\epsilon(1+|\tau|)^C \min\{A(R,X),C(R,X)\},\\ A(R,X)&=R+R^{1/2}X+R^{7/8}X^{1/2}+R^{3/4}X^{3/4},\\ C(R,X)&=R+R^{1/2}X^{3/2}. \end{aligned} \tag{27} $$

Here $\delta,\kappa$ are chosen sufficiently small in terms of the final loss and then fixed. We prove both bounds for the entire family and take their minimum afterward. Hölder precedes the fourth-power tower sum; reversing that order gives a weaker estimate.

4.1. A buffered reciprocal alternative

For the actual dyadic twists used by the ray projections, fix $B$ and suppose $R^{3/5}\le X\le R^B$. Then the left side of (27) also satisfies

$$ \ll(RXq_e)^\epsilon(1+|\tau|)^C B_*(R,X),\qquad B_*(R,X)=R^{4/5}X^{2/3}+R^{1/2}X. \tag{27a} $$

Here $\delta>0$ is fixed. The constants may depend on $B,\delta$ and the final loss. We prove this alternative before applying it to (3).

The supplementary quartic symbols at $i,\lambda$, and the reciprocity corrections, factor through the units modulo $\lambda^7$. The local logarithm identifies $U_3/U_7$ with $\lambda^3\mathcal O_\lambda/\lambda^7\mathcal O_\lambda$; multiplication by four kills this quotient. Since $\mathcal O_\lambda^\times=\mu_4\times U_3$, the unit group modulo $\lambda^7$ has exponent four. Thus every character in the actual finite ray projection, including the unit and valuation twists, has order dividing four. This statement does not concern higher-order dyadic twists. For a fourth-power-free primary parameter $r$, the odd local exponents $v_p(r)\in\{1,2,3\}$ recover its primitive odd character, up to the fixed reciprocity orientation. A dyadic twist changes none of these exponents. Its primitive conductor has norm $O(q_r)$, with bounded dyadic multiplicity. Consequently the full conductor density of [7, Theorem 2] restricts to each actual ray family. This supplies the order hypothesis left explicit in the comparison at the end of [2, Section 10].

First work on $q_r\le P$ with fourth-power-free $r$. Choose a small mesh $u>0$ and use the buffered maximum construction of [7, Section 6], with floor $1/2+u$ and $O(1/u)$ successive height windows. Include the conjugate row; its zeros are the conjugates of those of the original row. The maximum abscissae increase by at most $1/2$ in total, so some consecutive pair increases by at most $u$. A nonfloor bin of lower edge $a$ is certified by an actual zero with abscissa at least $a$, while the enlarged window has no zero right of $a+2u$. Only finitely many bins and window indices occur. Write $d=2a-1$.

We recall the reciprocal step to specify its uniformity. Continue $\log L$ from the Euler product to a disk centered at $2+it$ inside that zero-free rectangle. Borel--Carathéodory gives $O_u(\log P)$ on an outer disk; the Euler logarithm is bounded on the inner disk of radius $49/100$. Three-circles on $\log L$ gives $O_u((\log P)^\theta)$, $\theta<1$, on a fixed intermediate disk. Thus $L$ and $1/L$ are bounded by any prescribed small power of $P$ on the line $a+6u$ within a reserved height window. This is a bound on each row's certified rectangle, not a common zero-free assertion.

Deleting a common mask $E$ multiplies the reciprocal by

$$ \prod_{p\mid E}(1-\psi_r(p)q_p^{-w})^{-1} \ll_{u,\epsilon}q_E^\epsilon \quad(\operatorname{Re}w\ge1/2+u). $$

Already ramified primes give one. Fixed derivatives have the same bound after reserving a further subsidiary loss. Shifting inverse Mellin inversion for a common smooth annular profile at scale $Y\ge1$ therefore gives

$$ \left|\sum_{a\ {\rm squarefree}\atop(a,E)=1} \mu(a)\psi_r(a)q_a^{-s}W_0(q_a/Y)\right|^2 \ll(Pq_E)^\epsilon Y^{d+O(u)}. \tag{27b} $$

The sharper exponent subtracts $2\delta$, which we may discard. Contour portions outside the buffer stay on an absolute line; arbitrary smooth Mellin decay pays their fixed polynomial scale cost. All length exponents are in a fixed range. Choose the finite mesh first, the minimum positive height allowance in the density statements next, and the tail order last. Reduce the height allowance to pay (22)'s height polynomial and all constant multiples of the consecutive windows.

Partition $V(q_a/U)$ into norm annuli. Their profiles are $\eta(t)V(Yt/U)$, with $Y/U$ bounded above whenever the annulus is nonempty, so all required seminorms are uniform. The same buffered rectangle gives (27b) simultaneously for every $U\le X$. Summing $O(\log X)$ annuli bounds the square of the maximum by $(PXq_E)^\epsilon X^{d+O(u)}$. Values $U<1$ with nonempty support have only boundedly many terms. Insert the powerful expansion (24), keeping $E=e\operatorname{rad}(b)$ and length $U=T/q_{\operatorname{rad}(b)}$. Its support has $q_{\operatorname{rad}(b)}\le2X$ and its absolute mass is (25). Choose the subsidiary mask loss smaller than $\delta$. It follows that the tilted inverse itself obeys

$$ \sup_{0<T\le X}|M_{T,\psi_r,e}(s)|^2 \ll(PXq_e)^\epsilon X^{d+O(u)}. \tag{27c} $$

The maximum uses pointwise reciprocal bounds for a uniform profile class, not a sieve with independently chosen row coefficients.

For these fourth-free rows, [2, Corollary 3] gives density exponent $g(d)=3(1-d)/(3-2d)$ up to $d=2/3$. For $d\ge2/3$, the four relevant branches of [7, Theorem 2] give

$$ D(d)\le\frac95(1-d). \tag{27d} $$

Indeed their ratios to $1-d$ are respectively $12/(2+7d),8/(15d-6),4/(1+2d),3/2$, on the intervals with endpoints $2/3,22/31,8/11,5/6,1$. Each ratio is at most $9/5$; equality holds at $2/3$. Cauchy using (22) on the selected bin and (27c) gives its contribution, up to small powers, as

$$ P^{(1+D(d))/2}X^d. \tag{27e} $$

On $d\le2/3$ and $x=\log_P X\ge3/5$, the derivative of its exponent is $x-3/(2(3-2d)^2)\ge x-27/50>0$. On $d\ge2/3$, (27d) bounds the exponent by $7/5+(x-9/10)d$. The endpoints $d=2/3,1$ therefore give precisely $B_*(P,X)$. The floor bin costs $PX^{O(u)}$ and fits this bound. For the last bins within $O(u)$ of one, one may use the direct inverse bound $O(X^{1/2+\epsilon})$ and density $P^{O(u)}$; this gives $P^{1/2+O(u)}X$ without invoking density at $\sigma=1$. Choose $u$ after the fixed upper length exponent and final loss.

Restore $j=rf^4$ only now. Put $P=R/q_f^4$. For $P\ge R^{1/10}$, the length has fixed upper exponent $10B$ in $P$, and $X\ge P^{3/5}$. Equation (26a) and the common mask $ef$ introduce only small powers. The two tower sums have norm exponents $16/5$ and $2$ and converge. For $P<R^{1/10}$, direct summation of the inverse and (22) give the crude mixed bound $PX$, with small powers and a fixed height polynomial. Its tower sum is $O(R^{13/40+\epsilon}X)$, below $R^{1/2}X$. This includes bounded cores and shared support $(r,f)>1$. Hence (27a) holds at sufficiently small height power. For larger $|\tau|$, use (27): its ratio to $B_*(R,X)$ is at most a fixed power of $R$ in the stated length range, absorbed by enlarging the fixed height polynomial. This proves (27a).

5. The exact nonprincipal source and its forced divisors

For $j\ne0$, put $J=\operatorname{rad}_o(j)$. At a prime $p\mid J$ set $a_p(j)=-1$ if $v_p(j)=1$, and $a_p(j)=q_p-1$ otherwise; write $A_j(d)=\prod_{p\mid d}a_p(j)$. The complete local correlations give zero unless $\gcd(n,J)=\gcd(m,J)=d$, $n=da,m=db$, $(ab,j)=1$, and $(a,b)=1$. On that locus

$$ F(n,m;j)=A_j(d)\chi_a(j)\overline{\chi_b(j)}. \tag{28} $$

At a shared prime the correlation is $-1$ at valuation one and $q_p-1$ at higher valuation; at a nonshared prime the residual character enforces the zero. The reciprocity calculation of Section 3 fixes the phase. Möbius inversion of $(a,b)=1$ therefore gives, for finite column coefficients $b_n$ on the ray,

$$ \sum_{n,m}b_n\overline{b_m}F(n,m;j) =\sum_{d\mid J}A_j(d) \sum_{e\ {\rm squarefree}\atop(e,j)=1}\mu(e) \left|\sum_{v\ {\rm squarefree}\atop(v,ej)=1,\,dev\in\mathcal C} b_{dev}\chi_v(j)\right|^2. \tag{29} $$

Both exterior signs remain in this identity; the kernel need not be positive. Its polarized version allows independent column profiles.

With $b_n=a_n/\sqrt{q_n}$, the exterior weight is $A_j(d)/(q_dq_e)$. For $(v,de)=1$, the actual convolution satisfies

$$ c_X(dev)=\sum_{u\mid de}\mu(u)c_{X/q_u}(v). \tag{30} $$

Every prime of $d$ divides the actual frequency $j$. Thus the natural character $\psi_j$ already vanishes on the omitted $d$ columns, even when $v_p(j)=4$ or $8$ and its primitive inducing character is unramified at $p$. Only $e$ is an additional common mask. For each daughter in (30), its squarefree Dirichlet series, initially on an absolute line, is

$$ \frac{L_o(s,\psi_j)}{L_o(2s,\psi_j^2)} \prod_{p\mid e}(1+\psi_j(p)q_p^{-s})^{-1} M_{X/q_u,\psi_j,e}(s). \tag{31} $$

A factor with $p\mid d$ would equal one. Deleting the natural zeros before making this observation would invalidate (31).

Project $dev\in\mathcal C$ onto a fixed finite list of dyadic characters. The coefficients of the projection can depend on $de$, but have bounded modulus. Outside the packet of Section 3, every resulting $\psi_j$ is nonprincipal: a dyadic twist cannot cancel a nontrivial odd local character. For each such term move only the counting line to $\operatorname{Re}s=1/2+\delta$. The inverse $L_o(2s,\psi_j^2)$ has an absolute Euler bound, the finite $e$ product costs $q_e^\epsilon$, and the tilted inverse is holomorphic for $\operatorname{Re}s>0$. No reciprocal contour crosses unknown zeros. At column length $L/(q_dq_e)$, the central normalization costs $(L/(q_dq_e))^\delta\ll L^\delta$ on occupied ranges.

Cauchy in the finite $u$ sum replaces all its inverse lengths by the maximum in (27), at a divisor loss. More explicitly, let $S_{j,d,e}$ be a separated inner sum in (29) after removing its exterior $1/\sqrt{q_dq_e}$. The counting integral and Cauchy give

$$ \begin{aligned} |S_{j,d,e}|^2\ll(q_jq_eL)^\epsilon\sum_\nu\int_{\mathbb R} w(\tau)|L_o(1/2+\delta+i\tau,\psi_j)|^2 \sup_{0<T\le X}|M_{T,\psi_j,e}(1/2+\delta+i\tau)|^2\,d\tau. \end{aligned} \tag{31a} $$

For each pair of original column heights, $w$ is a common nonnegative rapidly decaying weight, independent of $j,d,e$; its weighted integrals cost only a fixed polynomial in those column heights. The finite sum ranges over the fixed ray characters. Both original heights stay in the common outer Fourier integral. We now aggregate $d$ pointwise at $j$:

$$ \sum_{d\mid J}\frac{|A_j(d)|}{q_d} =\prod_{p\mid J}\left(1+\frac{|a_p(j)|}{q_p}\right) \le2^{\omega(J)}\ll_\epsilon q_j^\epsilon. \tag{32} $$

The preceding divisor losses cost only another subsidiary power of $q_jq_e$. We have not applied a uniform row bound separately for every $d$. The original $e$ support has $q_e\ll L$, and its weight $q_e^{-1+\epsilon}$ has a subpower ideal sum. For each fixed $e$, drop $(e,j)=1$ by positivity while keeping the common coefficient mask $e$. After summing the fixed finite list of ray characters, (27) now applies to the whole natural-frequency family. Consequently, on $q_j\asymp K$, all the nonprincipal terms of (29), after absolute exterior signs and common profile separation, are bounded by

$$ (KLX)^\epsilon\min\{A(K,X),C(K,X)\}. \tag{33} $$

There is no exterior-divisor cutoff or discarded tail.

6. Restoration of the Gaussian profile

The Gaussian in (10) is not a product of two column profiles. On fixed dyads in $j,n,m$, separate its three log-norm variables by one common Fourier integral. The two column heights are independent. Apply Cauchy to the polarized version of (29) under this integral. All its Mellin and fourth-moment height costs are integrable after sufficiently many profile derivatives. The seminorms are uniform for $HK/L^2\le1$ and decay faster than every fixed power for $HK/L^2>1$. The frequency norm phase is a scalar of modulus one.

For $j=\omega\lambda^a j_o$, the odd row length is $R\asymp K/2^a$. The unit and $a$ modulo four give a fixed finite list of twists. All powers of $R$ in $A,C$ are positive, so their valuation sums converge. The same positive powers pay the lower frequency dyads; Schwartz decay pays the upper ones. Prove the entire restoration once using $A$ and once using $C$, then take the smaller bound. Restoring $H/L$ in (10) and putting $K=L^2/H$ gives

$$ \begin{aligned} \frac HL\min\{A(K,X),C(K,X)\} =L+\min\{&H^{1/2}X+H^{1/8}L^{3/4}X^{1/2} +H^{1/4}L^{1/2}X^{3/4},\\ &H^{1/2}X^{3/2}\}. \end{aligned} \tag{34} $$

Add the principal estimate (21) and the origin to prove Theorem 2.

For Theorem 1 choose $\eta<1$ sufficiently close to one in terms of the final $\epsilon$. Since $X\le L$, the factor $X^{2(1-\eta)}$ can be absorbed in that loss, uniformly also at $x=1/8$. Thus

$$ \begin{aligned} \mathcal E(L,L,X)\ll L^\epsilon\bigl[L &+\min\{L^{1/2}X+L^{7/8}X^{1/2}+L^{3/4}X^{3/4}, L^{1/2}X^{3/2}\}\\ &+L^{5/4}X^{-2}+L^{1/4}X\bigr]. \end{aligned} \tag{35} $$

Writing $X=L^x$, its exponent is

$$ \max\left\{1,\frac54-2x,\frac14+x, \min\left(\max\left(\frac12+x,\frac78+\frac x2, \frac34+\frac{3x}4\right), \frac12+\frac{3x}2\right)\right\}. \tag{36} $$

This proves the first five branches of (4), restricting the last of them to $x\le3/5$. To use (27a) for $3/5\le x\le9/10$, an additional small-frequency split is needed. At odd frequency length $R\ge L^{1/10}$ and $R\le X^{5/3}$, its upper length exponent is at most nine, so (27a) applies uniformly. At $R<L^{1/10}$ use the crude mixed bound $RX$ from the proof above. Summing these dyads gives $L^{1/10+\epsilon}X$, below $B_*(L,X)$. High $\lambda$ valuations are summed by their odd length $R$ in the same way; the number of physical dyads contributes only logarithms.

Lower frequency dyads with $R\ge L^{1/10}$ sum by the positive powers in $B_*$. For upper dyads up to $R=X^{5/3}$, $B_*(R,X)$ grows at most linearly in $R/L$. At the transition $X=R^{3/5}$, every term of $A(R,X)$ is at most $B_*(R,X)$. All its $R$ powers are at most one, so beyond that transition it remains bounded by a constant times $(R/L)B_*(L,X)$. The Gaussian profile's rapid decay pays that growth. The height polynomials in (27a) remain inside the same common integrals as (31a); choose the profile derivative orders after those polynomials. The exterior $e$ cost is still a subpower, since $q_e\ll L$ and its weight is $1/q_e$.

Thus for $L^{3/5}\le X\le L^{9/10}$ the nonprincipal contribution is $O(L^\epsilon B_*(L,X))$. The principal terms in (35) and the second term of $B_*$ are at most $L^{4/5}X^{2/3}$ on this interval. Consequently

$$ \mathcal E(L,L,L^x)\ll L^{4/5+2x/3+\epsilon} \quad(3/5\le x\le9/10). \tag{36a} $$

The exponent agrees with the preceding branch at $3/5$, and is below $5/4$ precisely for $x<27/40$. Combining the two estimates proves (4). At $x=0$ the generic exponent is recovered; throughout $0<x<27/40$ it is strictly improved.

7. One exact detector cutoff with logarithmic gap loss

Take the specific Gevrey-2 bump $\phi(u)=C\exp[-1/((u-1)(2-u))]$ on $(1,2)$, extended by zero. Its planar function $f$ in Section 2 satisfies $\|\partial_v^k f\|_1\le C_0C_1^k(k!)^2$ for unit directions $v$. This follows from the standard derivative estimate for $e^{-1/t}$, Leibniz, and analytic composition on the fixed annulus. Integrating by parts $k$ times and optimizing $k\asymp\sqrt{|\xi|}$ gives $|\widehat f(\xi)|\ll e^{-c\sqrt{|\xi|}}$, also after any fixed differential-polynomial transform. Poisson thus refines (6) to

$$ |\partial_x^j(V_{\rm ar}(x)-1)| \ll_j e^{-c_jx^{-1/4}}\quad(0<x\le1). \tag{37} $$

Polynomial factors in $x^{-1}$ are absorbed by decreasing $c_j$.

Fix a global detector parameter $Q\ge3$, a precision $M$, and a finite profile order $J$. Choose a fixed sufficiently large $B_0$, put $\tau=(B_0\log Q)^{-4}$, and let

$$ V_Q(x)=V_{\rm ar}(x)+\rho(x/\tau)(1-V_{\rm ar}(x)), \tag{38} $$

where $\rho$ is fixed, equals one through one and zero from two onward. Equation (37) gives $\|V_Q-V_{\rm ar}\|_{C^J}\ll Q^{-M}$. This single cutoff has uniformly bounded fixed seminorms, is one for $x\le\tau$, and is zero for $x\ge2$. Whenever $X\tau\ge1$, its convolution has unit coefficient one and vanishes for $1<q_n\le X\tau$; the inverse has length at most $2X$.

For every $n$, the change in the coefficient is $O(Q^{-M}\tau(n))$. Since $|G(n,h)|\le1$ on squarefree columns, the direct physical-row bound for the perturbation of (3) is $O(HL^{1+\epsilon}Q^{-2M})$. Choose $M$ after a fixed polynomial range for $H,L,X$; Minkowski restores the cross term. Thus (4), (5) persist through every annulus of that range using the same $V_Q$. The gap is $X/(B_0\log Q)^4$, not literally $X$. The constant $B_0$ is fixed after the precision and profile requirements.

8. Evidence, consumer and remaining problem

The supplement includes exact finite controls and the analytic audits. The powerful regrouping passes 1,250 Gaussian-rational cases, including 610 with literal zeros; omitting powerful factors fails 1,232 negative controls. The global-frequency control checks 27,648 natural-character and tower identities, 6,912 redundant-mask identities, 216 divisor-weight identities and 324 shortened convolutions. Dropping fourth-power zeros fails 2,704 negative controls. An exact rational ledger checks the five branches before the buffered refinement, every affine crossing and the rectangular normalization. A common-cap Mellin experiment checks four lengths with a toy cutoff of known transform, with error below $3.23\times10^{-11}$ and the stated elementary integral-tail allowances. It is a floating check, not an interval certificate. Earlier controls check complete finite correlations, principal residue ratios and the full exterior vector. Appendix A also has exact ordinary-column restoration and floating full Poisson controls. These verify algebra and conventions; none proves an infinite estimate.

For the buffered alternative, a separate exact check enumerates the 128 residue classes modulo $\lambda^7$, its 64 units and 16 primary units, and verifies the exponent-four decomposition. Twelve complete-interval certificates check its density majorants, exponent comparisons and frequency restoration. The changed ordinary consumer below is verified by 120 exact semialgebraic cases, covering every possible endpoint and intersection minimum, including the unbounded last interval.

There is a direct scoped consumer. A bounded nonnegative row weight on $q_h\asymp L$ is bounded by a fixed multiple of the Gaussian weight in (3). Its restriction to squarefree rows therefore inherits (4). Its exact column diagonal costs $O(L^{1+\epsilon})$, so the absolute off-diagonal is bounded by the total plus that diagonal. This supplies a saving for the full prescribed Gauss source, not merely for its principal packet.

For polynomially related scales, the strict saving in (4) requires $0<x<27/40$, or $L>X^{40/27}$ with $X>1$. A detector tail starts at $X/(B_0\log Q)^4$ and includes earlier annuli. These are still not covered by the saving. At $X=L^{3/4}$, (27a) improves the mixed exponent from $21/16$ to $13/10$, a gain of $1/80$. It remains $1/20$ above the generic exponent $5/4$, so a saving for this long source remains open.

There is also an essential phase distinction. For squarefree $n$, $G(n,h)=\gamma(n)\overline{\chi_n(h)}$ with $|\gamma(n)|=1$. The ordinary zero detector [6, Section 5] uses $c_X(n)\psi_h(n)$, without this column Gauss phase. A diagonal unitary change preserves an arbitrary-coefficient operator norm, but does not transfer an estimate for one prescribed coefficient to another. No such transfer is established here. Ordinary-column restoration, additional quadratic parameters in product detectors, and selected-height covariance are also not supplied by (4). The partial result is a complete source mean and a controlled arithmetic cutoff; the all-zero Riemann hypothesis remains unresolved.

The direct ordinary convolution admits a different estimate: its Dirichlet series is exactly $L(s,\psi)M_X(s,\psi)$, so its own smooth counting contour uses (27) without a Gauss-phase transfer. The full consumer calculation in the supplement retains both selected central alternatives and the finite upper endpoint of the tail. Use the common smooth exact-plateau cutoff (38); the buffered reciprocal argument does not assert the same bound for a sharp inverse. Its optimized density exponent remains $g(d)=3(1-d)/(3-2d)$ from [2], where $d=2\sigma-1$ and $0<d\le3/4$. For $0<d\le1/4$, the best infinite-tail exponent is $f(d)=1-d(1-d)/(3-4d)$, and

$$ f(d)-g(d)=\frac{d^2(1-2d)}{(3-4d)(3-2d)}\ge0. $$

For $1/4\le d\le1/2$ it is $1-d/4-d^2/2$, again at least $g(d)$. The buffered bound changes the longer range. For $1/2\le d\le2/3$, the minimizing inverse exponent and difference are

$$ x_* = \frac{9(8d-1)}{10(3+2d)},\qquad f(d)-g(d)=\frac{3(2d-1)(2-3d)(3-4d)}{5(3-2d)(3+2d)}\ge0. $$

For $2/3\le d\le3/4$, the minimum is exactly $g(d)$, at $x_*=3/(6-4d)$. To include finite endpoints, put $q(\ell)=\max(1,\ell,3/4+\ell/2,1/2+3\ell/4)$, let $a(x)$ be the best mixed exponent, and define $z(x)\ge x$ by $q(z)=a(x)$. The infinite tail is $F_d(x)=\max(q(x)-dx,a(x)-dz(x))$. If the finite upper endpoint $y\ge z$, it already incurs this cost. If $y<z$, its combination with the new central alternative $a(x)-dy$ still costs at least $F_d$. With the original selected fourth-moment central alternative the cap has not acted, and the positive affine comparison in the supplement gives $g(d)$. Hence neither option improves the input density. For $d>2/3$, the transferred conductor bound [6] is already stronger for the actual rays. These are comparisons of upper-bound envelopes, not lower bounds for the true arithmetic mean.

The new use of reciprocal dependence is therefore limited to the stated source estimate. The remaining problem is a stronger joint arithmetic estimate for $L$ and its inverse, or cancellation in the signed source before positive separation.

Appendix A. The short-product estimate

We supply the author argument behind the second bound in (26b). The quadratic-sieve input is [5, Theorem 5.1 and (7.15)]. The short-product estimate is now published as [8, Lemma 6]. We retain its proof here to make the common profiles, ordinary supports and natural zeros explicit in the present application. It is an input, not a new contribution of this paper.

Lemma A.1. For common coefficients on ordinary primary ideals $q_u\asymp U$, $q_v\asymp V$, with $|a_u|\le\tau(u)^D$, $|b_v|\le\tau(v)^D$ and fixed $D$, let $A_r=\sum_u a_u\chi_r(u)$, $B_r=\sum_v b_v\chi_r(v)$. For a fixed nonnegative smooth compact row profile $W_0$ and $R,U,V\ge1$,

$$ \sum_{r\ {\rm ordinary}}W_0(q_r/R)|A_rB_r|^2 \ll(RUV)^\epsilon\{RUV+(UV)^2(U+V)\}. \tag{A1} $$

Fixed ray restrictions, common multiplicative twists, norm phases and coprimality masks are permitted, uniformly in their moduli. By dyadic summation the same bound holds for a cumulative row cap.

Proof. For squarefree primary $c$ write $\gamma(c)=q_c^{-1/2}g_4(1,c)$. Gauss multiplication gives, for coprime squarefree $u,v$,

$$ \gamma(uv)=s(u,v)\gamma(u)\gamma(v)(u/v)_2,\qquad s(u,v)=(-1)^{((q_u-1)/4)((q_v-1)/4)}. \tag{A2} $$

Partition into finitely many primary ray classes to separate the sign. For squarefree coefficient supports, Cauchy and the quadratic sieve give

$$ \left|\sum_{u\asymp U}\sum_{v\asymp V} \alpha_u\beta_v(u/v)_2\right|^2 \ll(UV)^\epsilon UV(U+V). \tag{A3} $$

Indeed $\sum|\alpha_u|^2\ll U^{1+\epsilon}$ and the quadratic sieve bounds the remaining square mean by $(UV)^\epsilon(U+V)V$. The quadratic symbol supplies zero at a shared prime. All extra characters of one factor, including their zeros and arbitrary row parameters, remain in its coefficient vector. This is the uniform Type-II bound [5, (7.15)].

First restrict the whole product $c=uv$ to be squarefree and put $\lambda_c=\sum_{uv=c}a_ub_v$, $Z=UV$. Reciprocity and a finite ray partition turn the row character into $\chi_c(r)$; the resulting fixed multiplicative twist of $c$ respects the convolution. Enlarge each positive row packet to the whole Gaussian lattice with a radial majorant. The equal-column contribution is $O(RZ^{1+\epsilon})$. For unequal columns put $c_i=gb_i$, $G=q_g$, with $g,b_1,b_2$ pairwise coprime and squarefree. Expand the remaining row mask $1_{(r,g)=1}$ over $m\mid g$, and put $M=q_m$. The ratio character on $b_1b_2$ is primitive and nonprincipal. Lattice Poisson after $r=mt$ has zero principal frequency, amplitude and dual norm scale

$$ \frac{RG}{MZ},\qquad K_0=\frac{MZ^2}{RG^2}. \tag{A4} $$

The factors $q_{b_i}/(Z/G)$ are retained in the smooth profile. The primitive Gauss ratio splits exactly as

$$ \tau(\chi_{b_1}\overline{\chi_{b_2}}) =\chi_{b_1}(b_2)\overline{\chi_{b_2}(b_1)}\chi_{b_2}(-1) \tau(\chi_{b_1})\overline{\tau(\chi_{b_2})}. \tag{A5} $$

Its cross factor is a finite reciprocity sign. The two remaining Gauss roots and all characters at $m$ and the dual row separate between the left and right column vectors. Thus each vector has the form $\lambda_{gb}\gamma(b)$ times common multiplicative twists.

For a nonzero dual-row dyad $q_h\asymp J$, use Fourier inversion on the compact logarithmic variables $q_{b_1}/(Z/G)$, $q_{b_2}/(Z/G)$, $q_h/J$. Radiality removes any conductor angle. Schwartz decay gives a common weighted $L^1$ norm $O_A((1+J/K_0)^{-A})$. The two column heights are independent norm phases, which factor inside $\lambda$. This separates the actual profile without replacing moving norms by constants or choosing coefficients separately for each physical row.

Only now remove $(b_1,b_2)=1$ by Möbius inversion. Write $b_i=en_i$, $E=q_e$. Each $n_i$ is squarefree and prime to $ge$. This extends an algebraic expression already restricted to nonzero dual rows: it does not assign a primitive Poisson formula or a zero frequency to the new noncoprime pairs. Extract $\gamma(e)$ by (A2); it adds only fixed quadratic characters of $n_i$. The factor $|\chi_e(h)|^2$ is a row mask.

In each vector allocate the primes of $ge$ to the two original factors, $ge=d_ud_v$, $u=d_uu'$, $v=d_vv'$. The daughter coefficients are the actual $a_{d_uu'}$, $b_{d_vv'}$, with common exclusions and lengths $U'=U/q_{d_u}$, $V'=V/q_{d_v}$, $U'V'=Z/(GE)$. Gauss multiplication (A2) now supplies a quadratic cross symbol between $u',v'$. Formula (A3) yields, pointwise in the nonzero dual row,

$$ |\text{daughter polynomial}|^2 \ll(RUV)^\epsilon U'V'(U'+V') \le(RUV)^\epsilon\frac{Z(U+V)}{GE}. \tag{A6} $$

Prime allocations and finite Cauchy cost a subpower. Different allocations and the two independent column heights are handled by Cauchy in the same dual-row sum. Its weighted nonzero row count is $O(K_0)$; if $K_0<1$, it is $O(K_0^A)\le O(K_0)$ for $A>1$. The product of amplitude, row count and (A6) is exactly

$$ \frac{RG}{MZ}\frac{MZ^2}{RG^2}\frac{Z(U+V)}{GE} =\frac{Z^2(U+V)}{G^2E}. \tag{A7} $$

The ideal sum in $g$ converges; the $e$ sum on $GE\ll Z$ is harmonic; the sum over $m\mid g$ and all allocations cost subsidiary powers. Before using the primitive ratio, separate the bounded residual annuli $G\asymp Z$ directly in the original square. There are $O(Z)$ possible $g$, boundedly many residual pairs and $O(R)$ rows. They contribute $O(RZ^{1+\epsilon})$, including the residual-unit endpoint. This proves (A1) for squarefree whole products.

Finally every ordinary product has a unique form $c=q^2df_0$, $d\mid\operatorname{rad}(q)$, with $f_0$ squarefree and $(f_0,q)=1$. The exponent of $q$ at $p$ is $\lfloor v_p(c)/2\rfloor$; $d$ records the odd exponents at primes already in $q$. Fix $q,d$ and allocate $q^2d=u_0v_0$ to the original factors before removing its row scalar $\chi_r(q^2d)$ of modulus at most one. The residual product $u_1v_1=f_0$ is squarefree, with lengths $U'=U/q_{u_0}$, $V'=V/q_{v_0}$ and $Z'=Z/(q_q^2q_d)$. The squarefree argument applies to the common actual daughter coefficients. Nonempty scales below one are in a fixed bounded interval and cause only a constant. Minkowski bounds the root moment by a small power times

$$ \sum_{q_q\ll\sqrt Z}\tau(q)^{O_D(1)} \left(\frac{\sqrt{RZ}}{q_q} +\frac{Z\sqrt{U+V}}{q_q^2}\right). \tag{A8} $$

The first sum is harmonic up to subsidiary powers and the second converges. Squaring proves (A1). Taking $U=V=T$ and dividing by $T^2$ gives the normalized fourth moment $R+T^3$ used in (26b). $\square$

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