# A cubic mean-square bound for angular sextic Gauss sums ## 1. The coefficient and the result The mixed inverse/plain detector arising in the quasi-Riemann-hypothesis framework of OpenAI [1] leads, after a cubic-theta reflection, to the coefficient $$ c(b)=\overline{\alpha(b)}\gamma_{-1}(b) $$ against a fourth power of a sextic residue character. This is a specific arithmetic coefficient, and replacing it by an arbitrary bounded sequence loses useful information. We prove a mean-square estimate for this coefficient and a companion angular mode. The estimate does not by itself improve a zero-free region. Let $\mathcal O=\mathbb Z[\omega]$, where $\omega=e^{2\pi i/3}$, and put $\lambda=1+2\omega$. Write $q_a=Na$. Fix a finite set $S$ of prime ideals containing the primes above $6$. A good ideal is an integral ideal prime to $S$. We identify it with its unique primary generator congruent to $1$ modulo $3$, and set $\alpha(a)=a/\sqrt{q_a}$. These generators are multiplicative. For a good prime $p$, let $\chi_p(x)$ be the sixth root of unity congruent to $x^{(q_p-1)/6}$ modulo $p$, and extend it by zero at nonunits. Extend multiplicatively in the denominator. Every power of a character, including its zeroth or sixth power, retains the nonunit zero. With the additive character $$ e(z)=\exp\!\left(2\pi i\frac{z-\bar z}{\lambda}\right), $$ define on good squarefree ideals $$ \gamma_r(b)=q_b^{-1/2}\sum_{x\bmod b}\chi_b(x)^r e(x/b). \tag{1.1} $$ The unit ideal has Gauss sum one. These are the normalizations of [1, Section 4]. Let $\nu$ be a fixed multiplicative finite ray character, with its entire defining modulus supported in $S$. Let $W\in C_c^\infty(0,\infty)$ have support in a fixed annulus. Let $\Phi(q_z)$ be a nonnegative radial Schwartz function on $\mathbb C$. For $j\in\{-2,2\}$ put $$ D_j(b)=\mu(b)\alpha(b)^j\gamma_2(b)\nu(b). \tag{1.2} $$ For a fixed squarefree good ideal $Q$, and $w(b)$ equal either to $1$ or to $\varphi(b)/q_b$, define $$ \mathcal H_j(N,B;Q,w)= \sum_{z\in\mathcal O}\Phi(q_z/N) \left|\sum_{\substack{b\ { m good, squarefree}\\(b,Q)=1}} \frac{D_j(b)w(b)}{\sqrt{q_b}}\chi_b(z)^4W(q_b/B)\right|^2. \tag{1.3} $$ The puncture $Q$ is fixed during this row average; it may vary between invocations of the theorem. **Theorem 1.** For $N,B\geq1$ and every $\varepsilon>0$, $$ \mathcal H_j(N,B;Q,w) \ll (NBq_Q)^\varepsilon \left(N+\frac{B^3}{N}+N^{2/3}B^{1/2}\right). \tag{1.4} $$ The implicit constant depends on $\varepsilon$, the fixed arithmetic data and finitely many seminorms of $W,\Phi$. In particular, $$ N\geq B^{3/2}\quad\Longrightarrow\quad \mathcal H_j(N,B;Q,w)\ll (NBq_Q)^\varepsilon N. \tag{1.5} $$ The same assertions hold for a fixed finite linear combination of the characters $\nu$. The finite Gauss identities of [1, Lemmas 4.2–4.4] give $$ \gamma_{-1}(b)=\mu(b)\overline{\alpha(b)}\gamma_2(b)K(b), \tag{1.6} $$ where $K$ is a function on a fixed finite ray group. Expanding $K$ into genuine group characters shows that Theorem 1 applies to $c(b)$, with both allowed weights and the stated puncture loss. The positive outer sum may be restricted to squarefree rows. On that subset, the generic cubic large sieve of Heath-Brown [2] has factor $$ N+B+(NB)^{2/3} \tag{1.7} $$ for a column vector with bounded squared norm. Its $O(N)$ range is $N\geq B^2$. At $N=B^{3/2}$, (1.4) gives $B^{3/2+\varepsilon}$ in place of the generic $B^{5/3+\varepsilon}$. We use (1.7) only on its squarefree support; it is not an estimate for unrestricted rows as written in (1.3). The proof has two parts. First, we extend the unmarked inverse-moment argument of [1] to the fixed angular coefficients $\mu\alpha^{\pm2}$. Second, cubic Poisson summation converts (1.3) to those inverse coefficients. In the latter step, a common column divisor becomes part of the new row, with divisor multiplicity instead of a separate length count. ## 2. Related work and the imported framework Heath-Brown [2] proved the cubic large sieve used in the comparison. De Faveri [3, Theorem 1.1] proves the corresponding optimal upper bounds for fixed-order character families, including power-free indices. Those bounds permit arbitrary coefficients. Our estimate uses (1.2) essentially and makes no improvement to their operator norms. Dunn and Radziwiłł [4, Propositions 5.2–5.3] give angular cubic-theta transformations as part of their work on Gauss-sum bias. The pure-derivative mechanism in Section 3 is consistent with those formulas. What is needed here is the simultaneous finite-mask reflection of [1, Proposition 5.1], including its exponent-zero branches and moving punctures, followed by its coefficient-sensitive recursive inverse moment. A twisted transformation formula alone is insufficient for that transfer. The first-moment theorem for cubic Gauss sums in [5] concerns primes and supplies fixed angular cancellation at a different scale. The cubic/quartic nonorthogonality results of de Faveri, Dunn and Hoffstein [6], and the sextic lower bound of Xu and Sang [7], concern large-sieve operator norms and Gauss averages in different row families. In particular [7, Lemma 4.1] is a sextic Gauss-sum second moment with a sextic row character. Our row character is cubic, and its Gauss root cancels the $\gamma_2$ in (1.2). The conclusions are compatible. The analytic dependency in this paper is the cubic-theta and recursive moment framework of [1], at the version pinned in the references. We prove the modifications required for the angular coefficients, and give the full new gcd calculation. We do not use the final zero-free theorem of [1] as a hypothesis, claim a complete independent audit of that manuscript, or claim a new formal verification. All cited analytic inputs are stated unconditionally in that source. ## 3. Higher angular reflection with the same conductor power For a fixed nonzero integer $\ell$, set $$ C_\ell(n)=\alpha(n)^\ell\gamma_2(n). $$ In the notation and complete zero extensions of [1, Proposition 5.1], define $$ \mathcal T_\ell(s,\Psi)= \left(\sum_{n\ { m sf}}\frac{C_\ell(n)\Psi(n)}{q_n^s}\right) \left(\sum_b\frac{\alpha(b)^{3\ell}\Psi(b)^3}{q_b^{3s-1/2}}\right). \tag{3.1} $$ Both primal variables avoid the fixed excluded primes; they may share primes. The second factor is the cube completion. **Lemma 2.** The reflection of [1, Proposition 5.1] holds for (3.1), with the following replacements and no others: * the dual coefficient $d(\mu)\alpha(\mu)$ becomes $d(\mu)\alpha(\mu)^{-\ell}$; * the angular part $-i\overline{\alpha(c)}^2/81$ of its scalar becomes $i^\ell\alpha(c)^{2\ell}/81$; * its gamma quotient becomes $$ R_{|\ell|}(t)=\prod_{\pm} \frac{\Gamma((|\ell|+1)/2+t\pm1/6)} {\Gamma((|\ell|+1)/2-t\pm1/6)}. \tag{3.2} $$ The finite local factors, active and inactive prime sets, cusp coefficient functions and their supports, additive factors, and branch compatibility are unchanged. The kernel still has the argument $q_\mu X/q_c^2$. Its Euler derivatives have the same bounds $$ |(x\partial_x)^m V_\ell^\sharp(x)| \ll_{\ell,A,m}\min\{x^{1/4},(1+x)^{-A}\}\,\mathcal S_{A,m,\ell}(V), \tag{3.3} $$ where $\mathcal S$ is a fixed finite smooth seminorm. No uniformity as $|\ell|\to\infty$ is asserted. **Proof.** We describe the changed analytic calculation explicitly. The source uses a finite sum $F(z,v)$ of translated conjugate cubic theta functions. At a rational cusp $a/c$, its inverse hyperbolic coordinate is $$ z'=-\frac{\delta'}c-\frac{\bar z}{c^2(v^2+z\bar z)}, \qquad v'=\frac{v}{q_c(v^2+z\bar z)}. \tag{3.4} $$ For a pure Wirtinger derivative at $z=0$, hold $z=0$ while differentiating with respect to $\bar z$. In this Taylor jet, $z'$ is affine in $\bar z$, while its conjugate coordinate and $v'$ are constant. Therefore, for $h\geq1$, the $h$th derivative of the transformed theta function is exactly $$ \bar\kappa(G)(-c^{-2}v^{-2})^h \left[\partial_z^h\bar\theta\bigl(H(z,(q_cv)^{-1})\bigr)\right]_{z=-\delta'/c}. \tag{3.5} $$ There are no lower derivative terms. Every cusp constant, being independent of the horizontal coordinate, is killed. For $\ell=-h$, form $$ J_h(s)=\int_0^\infty [\partial_{\bar z}^hF(z,v)]_{z=0}\,v^{2s+h-2}\,dv. $$ A Fourier term $d(\mu)vK_{1/3}(4\pi|\mu|v)\check e(\mu z)$ contributes $$ \frac{i^h}{4}(2\pi)^{-2s} \Gamma(s+h/2-1/6)\Gamma(s+h/2+1/6) d(\mu)\overline{\alpha(\mu)}^h q_\mu^{-s}. \tag{3.6} $$ At the isolated infinity indices $\mu=\lambda^{-3}nb^3$, the coefficient formula [1, (5.9)–(5.10)] and $\overline{\alpha(\lambda^{-3})}=-i$ give $$ J_h(s)=\frac{3^{5/2}}4 \left(\frac{27}{(2\pi)^2}\right)^s \Gamma(s+h/2-1/6)\Gamma(s+h/2+1/6) \mathcal T_{-h}(s,\Psi). \tag{3.7} $$ Using (3.5) and substituting $v\mapsto(q_cv)^{-1}$, the reflected integral has the factor $$ \frac{(-i)^h}{4}(2\pi)^{2s-2} \overline{\alpha(c)}^{2h}q_c^{1-2s}\bar\kappa(G) \prod_\pm\Gamma(1-s+h/2\pm1/6), \tag{3.8} $$ and dual summands $d_H(\mu)\alpha(\mu)^h\check e(-\delta'\mu/c)q_\mu^{-(1-s)}$. The $q_c^h$ from the change of variable cancels the norm part of $c^{-2h}$. Thus the conductor exponent is exactly $1-2s$, independent of $h$. For $\ell=h>0$, use $\partial_z^h$ instead. The direct side has the extra scalar $(-1)^h$, and division by it changes the reflected angular scalar to $i^h\alpha(c)^{2h}$. This proves the two signs in the statement. The local finite Fourier calculations in [1, (5.11)–(5.17)] are untouched: differentiation introduces only a dual angular factor and the displayed conductor scalar. Differentiated cusp expansions decay exponentially at both ends of the integral. The source's continuation and strip argument consequently applies, and Mellin inversion gives (3.2). The first numerator pole of (3.2) is at $-|\ell|/2-1/3\leq-5/6$. Its fixed-strip Stirling bound is $|R_{|\ell|}(\sigma+it)|\ll (1+|t|)^{4\sigma}$. The source shifts to $\Re t=-1/4$ and to positive lines are therefore still permitted. The same integration by parts proves (3.3) and its common logarithmic Fourier-profile version. This also preserves compatibility of inactive local branches. $\square$ ## 4. The fixed angular inverse moment For $j\in\{-2,2\}$, $\epsilon_\chi\in\{-1,1\}$ and a smooth annular $W$, put $$ M_{j,u}(D;W)=D^{-1/2}\sum_n \mu(n)\alpha(n)^j\nu(n)\chi_n(u)^{\epsilon_\chi}W(q_n/D). \tag{4.1} $$ **Lemma 3.** If $H,D\geq1$ and $H\geq D^{1+c}$ for fixed $c>0$, then $$ \sum_{0