Proof. Inclusion–exclusion followed by \(k=d\ell\) gives \[\sum_k\chi(k)\mathbf 1_{(k,\mathfrak r)=1}\Phi(\mathrm N_{K/\mathbb Q}(k)/\mathcal{H}) =\sum_{d\mid\operatorname{rad}\mathfrak r}\mu(d)\chi(d) \sum_\ell\chi(\ell)\Phi\!\Bigl(\frac{\mathrm N_{K/\mathbb Q}(\ell)}{\mathcal{H}/\mathrm N_{K/\mathbb Q}(d)}\Bigr).\] Apply lattice Poisson summation [22] in \(\ell\) on residue classes modulo \(\mathfrak m\), at scale \(\mathcal H/\mathrm N_{K/\mathbb Q}(d)\). The primitive Gauss-sum identity [22] evaluates the finite Fourier transform as \(\sqrt{\mathrm N_{K/\mathbb Q}(\mathfrak m)}\gamma(\chi)\overline{\chi(h)}\) for every \(h\), giving (36). For a nonprincipal primitive character, \(\overline{\chi(0)}=0\), so the zero-frequency term vanishes. For the principal character, sum \(\mu(d)/\mathrm N_{K/\mathbb Q}(d)\) over \(d\mid\operatorname{rad}\mathfrak r\) to obtain the displayed zero-frequency contribution. ◻

Proof. Let \(C_{\mathfrak r}(X;k,f)\) denote the inner column sum defining \(\mathcal E_{\mathfrak r}\), with \(\xi,W\) fixed. Inclusion–exclusion, \(n=dm\), and (34) give \[\begin{aligned} C_{\mathfrak r}(X;k,f) &=\sum_{d\mid r_0}\mu(d) \sum_{\substack{(n,S)=1\\d\mid n}}^* a_\xi(n)\chi_n(k)\chi_n(f)^4W(\mathrm N_{K/\mathbb Q}(n)/X)\\ &=\sum_{\substack{d\mid r_0\\(d,f)=1}} \mu(d)a_\xi(d)\chi_d(k)\chi_d(f)^4 C_1(X/\mathrm N_{K/\mathbb Q}(d);k,df). \end{aligned}\] Indeed, \(\chi_m(d)^4\) enforces \((m,d)=1\) and combines with \(\chi_m(f)^4\); terms with \((d,f)\ne1\) vanish. Each exterior coefficient has modulus at most one. Apply Cauchy–Schwarz in \(d\), then enlarge the injective image \(f\mapsto df\) to the squarefree range \(F\mathrm N_{K/\mathbb Q}(d)\le\mathrm N_{K/\mathbb Q}(df)<2F\mathrm N_{K/\mathbb Q}(d)\). The normalizing product is unchanged: \((X/\mathrm N_{K/\mathbb Q}(d))(F\mathrm N_{K/\mathbb Q}(d))=XF\). ◻

Proof. Write \(I=[a_0,b_0]\). Choose a nonnegative radial Schwartz weight \(\Phi\) such that \(\Phi(x)\ge1\) on \([0,1]\) and \(\operatorname{supp}\widehat\Phi\subset[0,C_\Phi]\) for some fixed \(C_\Phi>0\). It suffices to prove \(\mathcal M_D\ll_{\nu,S,I,\vartheta,\varepsilon} \|W\|_{C^{J'}(I)}^2HD^\varepsilon\) for some \(J'=J'(\vartheta,\varepsilon)\), where \[ \mathcal M_D:=\frac1D\sum_{u\in\mathcal O}\Phi(\mathrm N_{K/\mathbb Q}(u)/H) \Bigl|\sum_{(n,S)=1}\mu(n)\nu(n)\chi_n(u)W(\mathrm N_{K/\mathbb Q}(n)/D)\Bigr|^2. \tag{41}\] Conjugate the inner sum, expand the square, and put \(g=(n_1,n_2)\) and \(n_j=gz_j\). The squarefree indices satisfy \((z_1,z_2)=(z_1z_2,g)=1\), and \[\overline{\chi_{n_1}(u)}\chi_{n_2}(u) =\mathbf 1_{(u,g)=1}\overline{\chi_{z_1}(u)}\chi_{z_2}(u).\] Apply Lemma 5 with \(\chi=\overline{\chi_{z_1}}\chi_{z_2}\) and exclusion \(g\): \[ \begin{aligned} &\sum_u\Phi(\mathrm N_{K/\mathbb Q}(u)/H)\mathbf 1_{(u,g)=1}\chi(u)\\ &\quad=\sum_{e\mid g} \frac{H\mu(e)\chi(e)\gamma(\chi)} {\mathrm N_{K/\mathbb Q}(e)\sqrt{\mathrm N_{K/\mathbb Q}(z_1)\mathrm N_{K/\mathbb Q}(z_2)}} \sum_h\overline{\chi(h)} \widehat\Phi\!\Bigl(\frac{H\mathrm N_{K/\mathbb Q}(h)}{\mathrm N_{K/\mathbb Q}(e)\mathrm N_{K/\mathbb Q}(z_1)\mathrm N_{K/\mathbb Q}(z_2)}\Bigr). \end{aligned} \tag{42}\] The zero frequency occurs only when \(z_1=z_2=1\) and contributes \[Z=\frac HD\widehat\Phi(0) \sum_{(g,S)=1}^*|W(\mathrm N_{K/\mathbb Q}(g)/D)|^2 \prod_{p\mid g}\Bigl(1-\frac1{\mathrm N_{K/\mathbb Q}(p)}\Bigr) \ll_{\Phi,I}H\|W\|_\infty^2.\] Retain all \(h\ne0\), including those with \(z_1=z_2=1\).

Expand the fixed ray class function \(\overline{\nu(t)}G(t^{-1})=\sum_\xi c_\xi\xi(t)\). The Chinese remainder theorem and (33)–(35) give \[ \mu(z_1)\mu(z_2)\overline{\nu(z_1)}\nu(z_2) \gamma(\overline{\chi_{z_1}}\chi_{z_2}) =\sum_\xi c_\xi a_\xi(z_1)\overline{a_\xi(z_2)}. \tag{43}\] Write \(N=\mathrm N_{K/\mathbb Q}\) for the remainder of this proof, and put \(W_0(x)=x^{-1/2}\overline{W(x)}\). Substituting (43) into (42), including the normalization \(1/D\) in (41), gives \[\mathcal M_D-Z=\sum_\xi c_\xi\mathcal S_\xi, \qquad |\mathcal M_D-Z|\ll\max_\xi|\mathcal S_\xi|,\] since the character sum is fixed and finite. Fix \(\xi\). Using \(\overline{\chi_z(e)}=\chi_z(e^5)\), its contribution is \[\begin{aligned} \mathcal S_\xi ={}&\frac H{D^2} \sum_g^*\sum_{e\mid g}\frac{\mu(e)N(g)}{N(e)} \sum_{h\ne0} \sum_{\substack{z_1,z_2\\(z_1,z_2)=1\\(z_1z_2,g)=1}}^* a_\xi(z_1)\overline{a_\xi(z_2)} \chi_{z_1}(he^5)\overline{\chi_{z_2}(he^5)} \\ &\quad{}\times W_0\!\Bigl(\frac{N(gz_1)}D\Bigr) \overline{W_0\!\Bigl(\frac{N(gz_2)}D\Bigr)} \widehat\Phi\!\Bigl( \frac{HN(h)}{N(e)N(z_1)N(z_2)} \Bigr). \end{aligned}\] Here and below every starred variable is squarefree, primary, and prime to \(S\), while \(h,k\) range over nonzero elements of \(\mathcal O\). We must show \(|\mathcal S_\xi|\ll_{\nu,S,I,\vartheta,\varepsilon} \|W\|_{C^{J'}(I)}^2HD^\varepsilon\).

Insert the coprimality identity and change variables: \[\mathbf 1_{(z_1,z_2)=1} =\sum_{v\mid z_1,\ v\mid z_2}\mu(v), \qquad z_j=vm_j.\] By (34), the common factor from the two columns satisfies \[a_\xi(vm)=a_\xi(v)a_\xi(m)\chi_m(v)^4, \qquad |a_\xi(v)\chi_v(he^5)|^2=\mathbf 1_{(v,h)=1},\] where \((v,e)=1\). Thus the same sum becomes \[\begin{aligned} \mathcal S_\xi ={}&\frac H{D^2} \sum_{\substack{g,v\\(g,v)=1}}^* \sum_{e\mid g}\frac{\mu(e)\mu(v)N(g)}{N(e)} \sum_{\substack{h\ne0\\(h,v)=1}} \sum_{\substack{m_1,m_2\\(m_1m_2,gv)=1}}^* a_\xi(m_1)\overline{a_\xi(m_2)} \\ &\quad{}\times \chi_{m_1}(he^5v^4)\overline{\chi_{m_2}(he^5v^4)} W_0\!\Bigl(\frac{N(gvm_1)}D\Bigr) \overline{W_0\!\Bigl(\frac{N(gvm_2)}D\Bigr)} \\ &\quad{}\times \widehat\Phi\!\Bigl( \frac{HN(h)}{N(e)N(v)^2N(m_1)N(m_2)} \Bigr). \end{aligned}\] There is now no restriction \((m_1,m_2)=1\). Make the bijective change of variables \[\begin{gathered} b=g/e,\qquad f=ev,\qquad k=eh,\\ e=(f,k),\qquad v=f/e,\qquad g=be,\qquad h=k/e. \end{gathered}\] The resulting \(b,f\) are coprime and squarefree, \(k\ne0\) is arbitrary, and \[gv=bf,\qquad he^5v^4=kf^4,\qquad \mu(e)\mu(v)=\mu(f),\qquad \frac{N(g)}{N(e)}=N(b).\] Consequently, \[ \begin{aligned} \mathcal S_\xi ={}&\frac H{D^2} \sum_{\substack{b,f\\(b,f)=1}}^*\mu(f)N(b) \sum_{k\ne0} \sum_{\substack{m_1,m_2\\(m_1m_2,b)=1}}^* a_\xi(m_1)\overline{a_\xi(m_2)} \chi_{m_1}(kf^4)\overline{\chi_{m_2}(kf^4)} \\ &\quad{}\times W_0\!\Bigl(\frac{N(bfm_1)}D\Bigr) \overline{W_0\!\Bigl(\frac{N(bfm_2)}D\Bigr)} \widehat\Phi\!\Bigl( \frac{HN(k)}{N(f)^2N(m_1)N(m_2)} \Bigr). \end{aligned} \tag{44}\] The characters enforce \((m_j,f)=1\), leaving only the displayed exclusion \((m_j,b)=1\). The supports of \(W_0\) and \(\widehat\Phi\) give \[N(b)N(f)\le b_0D,\qquad 0<N(k)\le\frac{C_\Phi b_0^2D^2}{HN(b)^2}.\]

Partition \(B\le N(b)<2B\) and \(F\le N(f)<2F\) into dyadic ranges. Denote the corresponding contribution to (44) by \(\mathcal S_{\xi;B,F}\), and put \[X=\frac D{BF},\qquad \mathcal H=\frac{C_{I,\Phi}D^2}{HB^2},\qquad C_{I,\Phi}=\max(2,C_\Phi b_0^2).\] These satisfy (39). In the variables \(x_j=N(m_j)/X\), the coupled smooth weight is \[\begin{gathered} \mathcal K_{b,f,k}(x_1,x_2) =W_0(r x_1)\overline{W_0(r x_2)} \widehat\Phi\!\Bigl(\frac{a}{x_1x_2}\Bigr),\\ r=\frac{N(b)N(f)}{BF}\in[1,4),\qquad a=\frac{HN(k)}{N(f)^2X^2}\le C_{I,\Phi}. \end{gathered}\] For every integer \(q\ge0\), these kernels satisfy \[\sup_{b,f,k}\|\mathcal K_{b,f,k}\|_{C^q([a_0/4,b_0]^2)} \ll_{I,\Phi,q}\|W\|_{C^q(I)}^2.\] For \(U\in C_c^\infty(I_*)\), where \(I_*=[a_0/8,2b_0]\), put \[S_U(b,f,k)=\sum_{\substack{(n,S)=1\\(n,b)=1}}^*a_\xi(n) \chi_n(k)\chi_n(f)^4U(N(n)/X).\] The shifts \((X,F)\mapsto(X/N(d),FN(d))\) preserve (39). Thus Lemma 7 and (40), with \(J=J(\vartheta,\varepsilon/4)\), give \[\begin{aligned} \sum_{f,k}|S_U(b,f,k)|^2 &=XF\,\mathcal E_{(b)}(\mathcal H,X,F;\xi,U)\\ &\le XF\,\tau_{\mathrm{div}}(b) \sum_{d\mid b}\mathcal E(\mathcal H,X/N(d),FN(d);\xi,U)\\ &\ll_{\nu,S,I,\vartheta,\varepsilon} D^{\varepsilon/2}(XF)^2\|U\|_{C^J(I_*)}^2. \end{aligned}\] Here \(N(b)\ll_I D\), so the divisor factors are absorbed in \(D^{\varepsilon/4}\). We use the following smooth-weight principle, stated and proved in Lemma 23: a mean-square bound valid for every common test function, with a \(C^J\) norm, also bounds the corresponding quadratic sum with a kernel depending on the row, at a cost given by the kernel’s \(C^{2J+4}\) norm. For each fixed \(b\), apply Lemma 23 with row index \((f,k)\), coefficient \(\mu(f)\mathbf 1_{(f,b)=1}\), and kernel \(\mathcal K_{b,f,k}\). The larger row range \(0<N(k)\le\mathcal H\) adds only terms whose original kernel vanishes. With \(q=2J+4\), this gives \[ \begin{aligned} |\mathcal S_{\xi;B,F}| &\ll_{\nu,S,I,\vartheta,\varepsilon} D^{\varepsilon/2}\|W\|_{C^q(I)}^2 \sum_{B\le N(b)<2B}^*\frac{HN(b)}{D^2}(XF)^2\\ &\ll D^{\varepsilon/2}\|W\|_{C^q(I)}^2 \frac{HB}{D^2}\,B\,(XF)^2 =D^{\varepsilon/2}\|W\|_{C^q(I)}^2H, \end{aligned} \tag{45}\] where \(XF=D/B\) and ideal counting gives \(O(B)\) choices of \(b\). Summing the \(O((\log D)^2)\) nonempty dyadic ranges bounds \(\mathcal S_\xi\) as required. Sum over the fixed finite set of \(\xi\), include \(Z\), and restore the factor \(D\) to obtain (26). The required derivative order depends only on \(\vartheta,\varepsilon\). ◻

Proof. Write \(N=\mathrm N_{K/\mathbb Q}\) and \(I=[u,v]\). Complete multiplicativity in (48), including at the zeros of \(\Psi_k\), gives \[ \begin{aligned} &X^{-1/2}\sum_{(n,S)=1}^*a_\xi(n)\chi_n(k)\chi_n(f)^4 W(N(n)/X)\\ &\qquad=\sum_{(h,S)=1} \frac{\mu(h)\overline{\alpha(h)}^{\,3}\Psi_k(h)^3}{N(h)} T(X/N(h)^3;k,f). \end{aligned} \tag{53}\] Indeed, after substitution of (48), the coefficient of the total cube index \(b\) contains \(\sum_{h\mid b}\mu(h)=\mathbf 1_{b=1}\). This is Möbius inversion; compare [22] and [7].

Split the right-hand side of (53) into \(P_{\rm short}(k,f)+P_{\rm long}(k,f)\), where \[\begin{aligned} P_{\rm short}(k,f)&= \sum_{\substack{(h,S)=1\\N(h)\le H_c}} \frac{\mu(h)\overline{\alpha(h)}^{\,3}\Psi_k(h)^3}{N(h)} T(X/N(h)^3;k,f),\\ \mathcal E_{\rm short}&=\frac1F\sum_{\substack{F\le N(f)<2F\\(f,S)=1}}^* \sum_{0<N(k)\le\mathcal H}|P_{\rm short}(k,f)|^2. \end{aligned}\] Define \(\mathcal E_{\rm long}\) in the same way using \(P_{\rm long}\). Weighted Cauchy–Schwarz for each \(k,f\), followed by Proposition 10, gives \[ \begin{aligned} \mathcal E_{\rm short} &\le\Bigl(\sum_{N(h)\le H_c}\frac1{N(h)}\Bigr) \sum_{N(h)\le H_c}\frac1{N(h)} \frac1F\sum_{\substack{F\le N(f)<2F\\(f,S)=1}}^* \sum_{0<N(k)\le\mathcal H}|T(X/N(h)^3;k,f)|^2\\ &\ll_{I,\nu,S,C_0,\varepsilon} D^{\varepsilon/2}\|W\|_{C^J(I)}^2 \Bigl(\mathcal H+\frac{\mathcal H^2FH_c^3}{X}\Bigr) \ll D^{\varepsilon/2}\Sigma\|W\|_{C^J(I)}^2. \end{aligned} \tag{54}\] Here \(X/N(h)^3\ge1\), and the reciprocal-norm sum contributes only a logarithm. We use Proposition 10 with exponent \(\varepsilon/4\) and range \(C_0+1\), since \(N(f)<2D^{C_0}\). If \(H_c<1\), the sum is empty.

Expand \(T\) using (48) in the remaining terms, with cube index \(c\): \[\begin{aligned} P_{\rm long}(k,f) ={}&\frac1{\sqrt X} \sum_{\substack{(h,S)=1\\N(h)>H_c}}\sum_{(c,S)=1} \mu(h)\sqrt{N(hc)}\,\overline{\alpha(hc)}^{\,3}\Psi_k(hc)^3\\ &\quad\times\sum_{(n,S)=1}^*a_\xi(n)\chi_n(k)\chi_n(f)^4 W\!\Bigl(\frac{N(n)N(hc)^3}{X}\Bigr). \end{aligned}\] Put \(b=hc\); the pairs giving \(b\) are exactly \((h,b/h)\) with \(h\mid b\) and \(N(h)>H_c\). Since \(\Psi_k(b)^3=\xi(b)^3\chi_b(k)^3\mathbf 1_{(b,f)=1}\), this gives \[\begin{aligned} P_{\rm long}(k,f) ={}&\sum_{\substack{(b,S)=1\\N(b)>H_c}} \frac{\beta_0(b,f)\chi_b(k)^3}{N(b)} \frac1{\sqrt{L_b}} \sum_{(n,S)=1}^*a_\xi(n)\chi_n(k)\chi_n(f)^4 W(N(n)/L_b),\\ \beta_0(b,f):={}&\overline{\alpha(b)}^{\,3}\xi(b)^3 \mathbf 1_{(b,f)=1}\sum_{\substack{h\mid b\\N(h)>H_c}}\mu(h), \qquad |\beta_0(b,f)|\le\tau_{\mathrm{div}}(b). \end{aligned}\] The divisor function \(\tau_{\mathrm{div}}(b)\) counts ideal divisors. All these indices are primary and prime to \(S\); \(b\) need not be squarefree. The support of \(W\) restricts the sum to \(N(b)^3\le vX\). Weighted Cauchy–Schwarz now gives \[\begin{aligned} \mathcal E_{\rm long} &\le\Bigl(\sum_{\substack{H_c<N(b)\le(vX)^{1/3}\\(b,S)=1}} \frac{\tau_{\mathrm{div}}(b)}{N(b)}\Bigr) \sum_{\substack{H_c<N(b)\le(vX)^{1/3}\\(b,S)=1}} \frac{\tau_{\mathrm{div}}(b)}{N(b)} \mathcal E(\mathcal H,L_b,F;\xi,W). \end{aligned}\] The reciprocal divisor sum is \(\ll_I\log^2(2+X)\), since \[\sum_{N(b)\le R}\frac{\tau_{\mathrm{div}}(b)}{N(b)} =\sum_{N(cd)\le R}\frac1{N(c)N(d)} \le\Bigl(\sum_{N(c)\le R}\frac1{N(c)}\Bigr)^2 \ll\log^2(2R)\qquad(R\ge1).\] If \(L_b\le1\), the column sum is empty unless \(L_b\ge1/v\); otherwise it has \(O_I(1)\) terms, so \[\mathcal E(\mathcal H,L_b,F;\xi,W) \ll_I\frac{\mathcal H}{L_b}\|W\|_\infty^2 \ll_I\Sigma\|W\|_\infty^2.\] Absorb the logarithms in \(D^\varepsilon\) and combine the two parts using \(\mathcal E\le2\mathcal E_{\rm short}+2\mathcal E_{\rm long}\). This proves (52). If \(\mathcal H^2\le X\), then \(H_c=X^{1/3}\), so \(N(b)>H_c\) implies \(L_b<1\). ◻

Proof of Proposition 9. Fix \(\kappa>0\) and \(C_0\ge1\). We prove by induction on \(j\ge0\) that the proposition holds under the additional restriction \(\mathcal H\le D^{j\kappa}\), with the derivative order and implied constant allowed to depend on \(j\). More precisely, for every \(\varepsilon>0\) there is an integer \(J=J(j,\kappa,C_0,\varepsilon)\ge1\) such that \[\mathcal E(\mathcal H,X,F;\xi,W) \ll_{I,\nu,S,j,\kappa,C_0,\varepsilon} D^\varepsilon\Sigma\|W\|_{C^J(I)}^2\] for every compact interval \(I=[u,v]\subset(0,\infty)\) and \(W\in C_c^\infty(I)\), under the hypotheses of the proposition. The derivative order is independent of \(I\); this allows us to apply the induction hypothesis on the enlarged interval in the transfer estimate. For the base case \(j=0\), we have \(\mathcal H\le1\). In (37), the support of \(W\) restricts \(n\) to the \(O_I(X)\) ideals with \(\mathrm N_{K/\mathbb Q}(n)\in[uX,vX]\). The factors \(a_\xi(n)\), \(\chi_n(k)\), and \(\chi_n(f)^4\) have absolute value at most one, so the triangle inequality bounds each inner sum by \(O_I(X\|W\|_\infty)\). There are \(O(F)\) choices of \(f\) and \(O(\mathcal H)\) choices of \(k\). Consequently, \[\mathcal E(\mathcal H,X,F;\xi,W) \ll_I\frac{F\mathcal H}{XF} \bigl(X\|W\|_\infty\bigr)^2 =\mathcal H X\|W\|_\infty^2.\] Dividing by \(\Sigma=XF\) and using \(\mathcal H\le1\) and \(F\ge1\) proves the assertion for \(j=0\), with \(J=1\).

Suppose the assertion holds for \(j\), and fix \(\varepsilon>0\). Let \(m\) be the derivative order supplied by the induction hypothesis with exponent \(\varepsilon/3\). For \(\mathcal H\le D^{(j+1)\kappa}\), apply Lemma 11 with exponent \(\varepsilon/3\). We must bound the mean squares in its supremum uniformly in \(b\). If this supremum is nonempty, then \(\mathcal H^2>X\) and \(H_c^3=X^2/\mathcal H^2\). For each such \(b\), recall that \(L_b=X/\mathrm N_{K/\mathbb Q}(b)^3>1\). Since \(L_b\le X\) and \(\Sigma\ge\max\{\mathcal H,L_bF\}\), Proposition 12 applies at column scale \(L_b\), with derivative order \(m\), and exponent \(\varepsilon/3\). Its estimate also bounds \(\mathcal E(\mathcal H,L_b,F;\xi,W)\) because \(\mathcal E(\mathcal H,L_b,F;\xi,W)\le\mathcal A(W)\).

Let \(\mathcal H',X',F'\) be any parameters in the supremum in (56), with \(\Sigma'=X'F'\) as in that proposition. By (57) and \(\mathrm N_{K/\mathbb Q}(b)>H_c\), \[\begin{aligned} \mathcal H' &\le\frac{\mathcal H L_b}{\Sigma F} =\frac{\mathcal H}{\mathrm N_{K/\mathbb Q}(b)^3F^2} <\mathcal H\Bigl(\frac{\mathcal H}{\Sigma}\Bigr)^2 \le D^{-2\kappa}\mathcal H\le D^{j\kappa},\\ \frac{\mathcal H'}{\Sigma'} &\le\frac{\mathcal H}{\Sigma}\le D^{-\kappa}, \qquad \Sigma'\le L_b\le\Sigma\le D^{C_0}. \end{aligned}\] Thus each of these mean squares is covered by the induction hypothesis. Its weight \(U\) is supported in \([u/16,4v]\) and has \(C^m\) norm at most one. Applying the induction hypothesis on this interval gives \[\frac{\mathcal E(\mathcal H',X',F';\xi',U)}{\Sigma'} \ll_{I,\nu,S,j,\kappa,C_0,\varepsilon}D^{\varepsilon/3}.\] The enlarged interval is determined by \(I\), so the implied constant has only the permitted dependence on the support. Substituting into (56) yields, uniformly in \(b\), \[\mathcal E(\mathcal H,L_b,F;\xi,W) \ll_{I,\nu,S,j,\kappa,C_0,\varepsilon} D^{2\varepsilon/3}\Sigma\|W\|_{C^{4m+12}(I)}^2.\] Substitution into (52) contributes the remaining factor \(D^{\varepsilon/3}\). Choose \(J\) at least \(4m+12\) and at least the derivative order required by Lemma 11. We obtain \[\mathcal E(\mathcal H,X,F;\xi,W) \ll_{I,\nu,S,j,\kappa,C_0,\varepsilon} D^\varepsilon\Sigma\|W\|_{C^J(I)}^2.\] The choice of \(J\) depends only on \(j,\kappa,C_0,\varepsilon\). If the supremum in (52) is empty, the same bound follows directly from that lemma. This completes the induction.

Finally, take \(j=\lceil C_0/\kappa\rceil\). The hypothesis \(\mathcal H\le\Sigma\le D^{C_0}\) ensures \(\mathcal H\le D^{j\kappa}\), so the induction gives the proposition. ◻

Proof of Proposition 3. Fix \(0<\vartheta\le1/10\) and put \(H=D^{1+\vartheta}\). The ranges in (39) give \[ \mathcal H\le\frac{CD^{1-\vartheta}}{B^2},\qquad \Sigma=XF=\frac DB,\qquad \frac{\mathcal H}{\Sigma}\ll D^{-\vartheta}. \tag{58}\] For large \(D\), Proposition 9 applies with \(\kappa=\vartheta/2\) and \(C_0=2\). If \(X<1\), nonempty support forces \(X\gg_I1\), and counting gives \(\mathcal E\ll_I\mathcal H\|W\|_\infty^2 \ll\Sigma\|W\|_\infty^2\); if \(\mathcal H<1\), the sum is empty. Thus (40) holds, with derivative order depending only on \(\vartheta,\varepsilon\), and Proposition 8 proves (26). Bounded \(D\) is again covered by counting. ◻

Proof. The coefficient formula (12) gives \[c_\theta(nb^3)\phi_k(nb^3) =3^{5/2}|b|\gamma_2(n)\Psi_k(n)\Psi_k(b)^3.\] Substitution into (48), using \(V_*(y)=\sqrt y\,W(y)\), proves (61). Finite Fourier inversion gives \[\phi_k(n)=\sum_{h\in\mathcal O/(q)}\widehat\phi_k(h)e(nh/q), \qquad \sum_{h\in\mathcal O/(q)}\widehat\phi_k(h)=\phi_k(0)=0.\] Multiplication of the \(\ell\)th Fourier mode by \(e(\lambda^3\ell h/q)\) translates \(z\) to \(z+\lambda^2h/q\). The second identity cancels the constant term, proving (62). ◻

Proof. This is Goldmakher and Louvel’s quadratic large sieve [13], after fixing the product of the prime factors of \(n\) lying in \(S\), and finitely many ray classes. For completeness, let \(e_k\in\{0,1\}\) according as \(\mathrm N_{K/\mathbb Q}(k)\equiv1,3\pmod4\), and let \(\kappa_\lambda\) be the nontrivial character modulo \(\lambda\). The character \(x\mapsto(x/k)_2\kappa_\lambda(x)^{e_k}\) is trivial on units and has primitive conductor \(k\lambda^{e_k}\). When \(e_k=0\), the factor \(\kappa_\lambda(x)^{e_k}\) is omitted. Classes modulo \(24\mathcal O\) fix the supplementary characters and reciprocity factors; for coprime \(k_1,k_2\) in the same class, the product character has conductor \(k_1k_2\). These are the hypotheses in [13]. ◻

Proof. By homogeneity assume \(\|W\|_{C^J(I)}\le1\), with \(J\) chosen below. Write \[s=t k_0,\qquad t=\prod_{\substack{p\mid s\\p\mid g\ \text{or}\ p\in S}}p,\qquad N(k_0)\le\mathcal H_0:=\frac{\mathcal H}{N(t)}.\] Fix \(t\); then \(k_0\) is squarefree and primary, with \((k_0,g)=1\) and \((k_0,S)=1\). Discard empty ranges, so \(\mathcal H_0\ge1\), and retain these restrictions below. For a coefficient function \(A(n,b)\), a positive scale \(Y\), and an integer \(m\ge-4\), put \[ \mathscr S_m[A,Y](k_0):= \sum_{\substack{n,b\equiv1\ (3)\\n\ {\rm squarefree}}} \frac{A(n,b)\chi_{k_0}(nb)^3}{3^{m/3}\sqrt{N(n)}\,N(b)} V_*^\sharp\!\Bigl(\frac{3^mN(n)N(b)^3\mathcal H_0^2} {YN(k_0)^2}\Bigr). \tag{73}\]

We shall obtain, after partitioning \(k_0\) into fixed ray classes, \[T(X;u_0tk_0,g) =\sum_{\iota\in\mathcal I}\sum_{m\ge-4} c_{\iota,m}(k_0)\mathscr S_m[A_{\iota,m},Y_\iota](k_0),\] where \(\mathcal I\) is independent of \(k_0\), \(|\mathcal I|\preccurlyeq 1\), \(|c_{\iota,m}(k_0)|\ll1\), and \(A_{\iota,m}\) is independent of \(k_0\). For some \(a_\iota,Y_\iota>0\), the coefficients and lengths satisfy \[ |A_{\iota,m}(n,b)|\ll a_\iota,\qquad a_\iota^2\le1,\qquad a_\iota^2Y_\iota\ll \frac{\mathcal H_0^2}{X}N(t)^2N(g). \tag{74}\] All column restrictions are included by extending \(A_{\iota,m}\) by zero. Put \(k=u_0tk_0\). For \(\mathcal P=\{p\notin S:p\mid kg\}\), sextic reciprocity (29) gives \[ \Psi_k(n)=\Psi_0(n)\prod_{p\in\mathcal P}\chi_p^{j_p}(n), \qquad j_p\equiv v_p(k)+4v_p(g)\pmod6, \qquad 0\le j_p\le5. \tag{75}\] Even when \(j_p=0\), the factor \(\chi_p^0(n)=\mathbf 1_{p\nmid n}\) retains the zero extension at \(p\mid kg\). The fixed factor \(\Psi_0\) contains \(\xi\), the factors at \(S\), the unit factors, and \(n\mapsto\mathcal R(n,\prod_{p\notin S}p^{v_p(k)})\). The fourth power at \(g\) contributes no reciprocity sign, and the factors at \(S\) depend only on exponents modulo six. Thus \(\Psi_0\) ranges over a fixed finite family. In the notation of Lemma 15, take \(\mathcal P_{\rm fix}=\{p\notin S:p\mid tg\}\), so that \(\mathcal P=\mathcal P_{\rm fix}\sqcup\{p:p\mid k_0\}\); the exponents on \(\mathcal P_{\rm fix}\) are fixed with \(t,g\). Partitioning \(k_0\) into fixed ray classes fixes \(\Psi_0\) and, by Lemma 15, the data \(d,\psi,c_0\) in each transformed term.

Fix one transformed term, with active primes \(\mathcal A_{\rm fix}\) away from \(k_0\), and put \(c_*=c_0\prod_{p\in\mathcal A_{\rm fix}}p\). Thus \(c=c_*k_0\) in (67). By (68), write \(\ell=u\lambda^mnb^3\), with \(u,m,n,b\) as there. Every prime dividing \(k_0\) has \(j_p=1\), and hence \[ B_{p,1}(u\lambda^{m+4}nb^3) =\chi_p(u\lambda^{m+4})^3\chi_p(nb)^3. \tag{76}\] Indeed, the transformed exponent is \(-1-2\equiv3\pmod6\) and \(\chi_p(b)^9=\chi_p(b)^3\), also when \(p\mid b\) by zero extension. The coefficient bound and rapid decay in Lemma 16 give absolute convergence of the transformed series, so we may regroup its terms below. Set \[A_{u,m}(n,b)= \frac{d(u\lambda^mnb^3)\alpha(u\lambda^mnb^3) \psi(u\lambda^{m+4}nb^3)}{3^{m/6}\sqrt{N(b)}}, \qquad |A_{u,m}(n,b)|\le27.\] The transformed term is therefore \[\begin{aligned} &C(k_0)\sum_{u\in\mathcal O^\times}\sum_{m\ge-4} \chi_{k_0}(u\lambda^{m+4})^3 \sum_{\substack{n,b\equiv1\ (3)\\n\ {\rm squarefree}}} \frac{A_{u,m}(n,b)\chi_{k_0}(nb)^3} {3^{m/3}\sqrt{N(n)}\,N(b)}\\ &\qquad\times\prod_{p\in\mathcal A_{\rm fix}} B_{p,j_p}(u\lambda^{m+4}nb^3) V_*^\sharp\!\Bigl(\frac{3^mN(n)N(b)^3X} {N(c_*)^2N(k_0)^2}\Bigr),\qquad |C(k_0)|\ll1. \end{aligned}\] The representation is unique: away from \(\lambda\), the prime exponents of \(\ell\) are \(3v_p(b)\) or \(1+3v_p(b)\). Apart from the quadratic character and weight, the \(k_0\)-dependence is a bounded scalar.

Fix \(u,m\) and write \(q=N(p)\) for an active prime \(p\nmid k_0\). For \(j_p=4\), the local identity is \[B_{p,4}(u\lambda^{m+4}nb^3) =-q^{-1/2}+q^{1/2}\mathbf 1_{p\mid n} +q^{1/2}\mathbf 1_{p\nmid n,\ p\mid b}.\] To display the reindexing in (73), let \(A^{(p)}(n,b)\) include all the other local factors and set \[A^{(p,n)}(n,b)=\mathbf 1_{p\nmid n}A^{(p)}(pn,b),\qquad A^{(p,b)}(n,b)=\mathbf 1_{p\nmid n}A^{(p)}(n,pb).\] Then the changes \(n=pn'\) and \(b=pb'\) give the exact identity \[\begin{aligned} &\mathscr S_m[A^{(p)}(n,b)B_{p,4}(u\lambda^{m+4}nb^3),q^2Y](k_0)\\ &\quad=-q^{-1/2}\mathscr S_m[A^{(p)},q^2Y](k_0) +\chi_{k_0}(p)^3\mathscr S_m[A^{(p,n)},qY](k_0)\\ &\qquad\quad+q^{-1/2}\chi_{k_0}(p)^3 \mathscr S_m[A^{(p,b)},Y/q](k_0). \end{aligned}\] In the second term \(p\nmid n'\) preserves squarefreeness; in the third, \(p\nmid n\) is retained and \(b'\) is unrestricted at \(p\). Both extracted phases have modulus one. The factors \(q^{1/2}\) in the local identity cancel against \(\sqrt{N(pn')}\) or leave \(q^{-1/2}\) after division by \(N(pb')\).

Let \(a\) track a bound \(|A(n,b)|\le27a\) for the coefficient in (73), and let \(Y\) be its scale. Before inserting the active primes, these are \(a=1\) and \(Y=N(c_0)^2\mathcal H_0^2/X\). Each active prime contributes \(q^2\) to the squared conductor norm. Including this factor, the updates are \[(a^2,Y)\longmapsto \begin{cases} (a^2,q^2Y),&j_p\notin\{0,4\},\\ (a^2/q,q^2Y),&j_p=0,\\ (a^2/q,q^2Y),\ (a^2,qY),\ (a^2/q,Y/q),&j_p=4. \end{cases}\] Thus \(a^2\) never increases. If \(p\mid t\) and \(p\notin S\), then \(j_p\) is odd, so \(a^2Y\) costs at most \(N(p)^2\). If \(p\mid g\) and \(p\nmid t\), the cost is \(N(p)\) for \(j_p=0,4\) and \(N(p)^2\) for \(j_p=2\); the latter case requires \(v_p(g)\ge2\). Hence \[a^2Y\ll\frac{\mathcal H_0^2}{X} \prod_{\substack{p\mid t\\p\notin S}}N(p)^2 \prod_{\substack{p\mid g\\p\nmid t,\ p\notin S}}N(p)^{v_p(g)} \le\frac{\mathcal H_0^2N(t)^2N(g)}X.\] Reindexing at distinct primes preserves the form and zero extensions. By Lemma 15, the transformed terms, units, and at most three choices per prime \(p\in\mathcal A_{\rm fix}\) with \(j_p=4\) form an index set of divisor-bounded size in \(tg\). Here and below, divisor-bounded in \(a\) means bounded by \(C\tau_{\mathrm{div}}(a)^A\) for fixed constants \(C,A\); in particular, this is \(\ll_\varepsilon\mathrm N_{K/\mathbb Q}(a)^\varepsilon\) for every \(\varepsilon>0\). For one index \(\iota\), abbreviate \(a=a_\iota\), \(Y=Y_\iota\), and \(A_m=A_{\iota,m}\). The local updates give \(Y\ll D^{C_1}\) for some fixed \(C_1=C_1(C_0)\). Split \(b\) into \(B\le N(b)<2B\), and \(n\) by a smooth dyadic partition \(V(N(n)/U)\), with \(U,B\ge1\) and uniformly bounded cutoffs. For this part of (73), write \[\begin{aligned} \mathscr S_{m;U,B}(k_0) &=3^{-m/3}\sum_{B\le N(b)<2B}\frac{\chi_{k_0}(b)^3}{N(b)}F_b(k_0),\\ F_b(k_0) &=\sum_{n\ {\rm squarefree}} \frac{A_m(n,b)\chi_{k_0}(n)^3}{\sqrt{N(n)}}V(N(n)/U) V_*^\sharp\!\Bigl(\frac{3^mN(n)N(b)^3\mathcal H_0^2} {YN(k_0)^2}\Bigr). \end{aligned}\] Here \(n,b\) remain primary. Put \(z_*=3^mUB^3/Y\) and fix a decay exponent \(A>0\). With \(m,U,B,Y\) fixed, Lemma 22, applied only in \(N(n)/U\), gives the following representation in a real Mellin variable \(s\): \[ \begin{aligned} F_b(k_0)&=\int_{\mathbb R}c_{b,k_0}(s)G_{b,s}(k_0)\,ds,\\ G_{b,s}(k_0)&=\sum_{\substack{n\ {\rm squarefree}\\N(n)\asymp U}} \frac{A_m(n,b)\chi_{k_0}(n)^3}{\sqrt{N(n)}} (N(n)/U)^{\mathrm i s},\\ |c_{b,k_0}(s)|&\ll_{I,A}(1+z_*)^{-A}(1+|s|)^{-2}. \end{aligned} \tag{77}\] Indeed, the scale \(R\) in (107) is \(3^mUN(b)^3\mathcal H_0^2/(YN(k_0)^2)\ge z_*\). Lemma 16 supplies the required derivative bounds, independently of \(b,k_0,U,B,m\). Since \(\sum_{N(n)\asymp U}N(n)^{-1}\ll1\), Lemma 17 gives, for any \(\sigma>0\), \[\sum_{N(k_0)\le\mathcal H_0}^*|G_{b,s}(k_0)|^2 \ll_{S,\sigma}a^2(\mathcal H_0U)^\sigma(\mathcal H_0+U).\] Apply weighted Cauchy–Schwarz to the Mellin integral, as recorded in (111), with the common majorant in (77), then weighted Cauchy–Schwarz in \(b\), using \(\sum_{B\le N(b)<2B}N(b)^{-1}\ll1\): \[ \begin{aligned} \sum_{N(k_0)\le\mathcal H_0}^*|\mathscr S_{m;U,B}(k_0)|^2 &\le3^{-2m/3}\Bigl(\sum_{B\le N(b)<2B}\frac1{N(b)}\Bigr) \sum_{B\le N(b)<2B}\frac1{N(b)} \sum_{N(k_0)\le\mathcal H_0}^*|F_b(k_0)|^2\\ &\ll_{I,S,A,\sigma}a^2 3^{-2m/3} (\mathcal H_0U)^\sigma(\mathcal H_0+U)(1+z_*)^{-2A}. \end{aligned} \tag{78}\] Since \(m\ge-4\), we have \(U\le81Yz_*\) and hence \(\mathcal H_0U\ll D^{C_0+C_1}(1+z_*)\). Choose \(0<\sigma\le1\) small in terms of \(\varepsilon_1>0,C_0\) and take \(A=3\). Taking square roots in (78) gives \[\Bigl(\sum_{N(k_0)\le\mathcal H_0}^*|\mathscr S_{m;U,B}(k_0)|^2\Bigr)^{1/2} \ll aD^{\varepsilon_1}3^{-m/3} (\sqrt{\mathcal H_0}+\sqrt U)(1+3^mUB^3/Y)^{-2}.\] The infinite dyadic sums converge: for dyadic \(U,B\ge1\) and every \(Z>0\), \[\sum_{U,B\ {\rm dyadic}}(1+UB^3/Z)^{-2}\ll\log^2(2+Z), \qquad \sum_{U,B\ {\rm dyadic}}\sqrt U(1+UB^3/Z)^{-2}\ll\sqrt Z.\] These estimates make the sum of \(\ell^2(k_0)\) norms finite. The triangle inequality and the geometric sum over \(m\ge-4\) therefore give \[\Bigl(\sum_{N(k_0)\le\mathcal H_0}^* \Bigl|\sum_{m\ge-4}c_{\iota,m}(k_0) \mathscr S_m[A_m,Y](k_0)\Bigr|^2\Bigr)^{1/2} \ll aD^{\varepsilon_1} \bigl(\sqrt{\mathcal H_0}\log^2(2+Y)+\sqrt Y\bigr).\] Combine the divisor-bounded choices of \(\iota\) and use (74) to obtain \[\sum_{N(k_0)\le\mathcal H_0}^*|T(X;u_0tk_0,g)|^2 \preccurlyeq\max_\iota a_\iota^2(\mathcal H_0+Y_\iota) \ll\mathcal H_0+\frac{\mathcal H_0^2N(t)^2N(g)}X \le\mathcal H+\frac{\mathcal H^2N(g)}X.\] Sum the fixed ray classes and the divisor-bounded choices \(t\mid\operatorname{rad}(g\prod_{\mathfrak p\in S}\mathfrak p)\), choosing \(\varepsilon_1\) and the other small-power losses in terms of \(\varepsilon\). The weight estimates above require a fixed number \(J=J(\varepsilon,C_0)\) of derivatives. Homogeneity restores \(\|W\|_{C^J(I)}^2\) and proves (72). ◻

Proof of Proposition 10. Write uniquely \(k=u_0sv^2\), with \(s\) squarefree and \(s,v\) among the chosen ideal generators. No condition \((s,v)=1\) is imposed. Since \(8\equiv2\pmod6\), including the zero extensions, \[\chi_n(u_0sv^2)\chi_n(f)^4 =\chi_n(u_0s)\chi_n(fv^2)^4,\qquad T(X;u_0sv^2,f)=T(X;u_0s,fv^2).\] Apply Lemma 18 with row bound \(\mathcal H/N(v)^2\) and auxiliary twist \(g=fv^2\). Here \(N(g)\le\mathcal H N(f)\le D^{2C_0}\), so use that lemma with \(2C_0\). Summing its bounds gives \[\begin{aligned} \sum_{0<N(k)\le\mathcal H}|T(X;k,f)|^2 &=\sum_{u_0\in\mathcal O^\times}\sum_{N(v)^2\le\mathcal H} \sum_{\substack{N(s)\le\mathcal H/N(v)^2\\s\ {\rm squarefree}}} |T(X;u_0s,fv^2)|^2\\ &\ll D^\varepsilon\|W\|_{C^J(I)}^2 \Bigl(\mathcal H+\frac{\mathcal H^2N(f)}X\Bigr) \sum_v\frac1{N(v)^2}. \end{aligned}\] The last sum is \(\zeta_K(2)<\infty\), where \(\zeta_K(s)=L_K(s,\mathbf1)\) is the Dedekind zeta function of \(K\). This proves (50). ◻

Proof. Expand the square defining \(\mathcal A(W)\), and write \(n_i=Cu_i\), where \(C=(n_1,n_2)\) and \((u_1,u_2)=(u_1u_2,C)=1\). The row character is \(\chi_{u_1}\overline{\chi_{u_2}}\), primitive modulo \(u_1u_2\), with the extra restriction \((k,C)=1\). Lemma 5 introduces \(d\mid C\) and a frequency \(h\). Let \(Z\) denote the zero-frequency contribution. It requires \(u_1=u_2=1\), so \[|Z|\ll\frac{\mathcal H}{LF} \sum_{F\le N(f)<2F}^*\sum_C^*|W(N(C)/L)|^2 \ll_I\mathcal H\|W\|_\infty^2.\] For the other frequencies, the Chinese remainder theorem, (32), and (35) give the paired identity \[ a_\xi(u_1)\overline{a_\xi(u_2)} \gamma(\chi_{u_1}\overline{\chi_{u_2}}) =\mu(u_1)\mu(u_2)(\xi G)(u_1u_2^{-1}). \tag{81}\] Expand \((\xi G)(z)=\sum_{\xi_1}\ell_{\xi_1}\xi_1(z)\) on the fixed ray class group. Insert \(\mathbf 1_{(u_1,u_2)=1}=\sum_{t\mid u_1,\ t\mid u_2}\mu(t)\) and put \(u_i=tx_i\). The inverse is \(x_i=u_i/t\); squarefreeness imposes \((x_i,t)=1\), but no condition \((x_1,x_2)=1\) remains. The factors at \(C,t\) cancel against their conjugates, leaving \(\mu(d)\mu(t)\) and the restrictions \((f,Ct)=(h,t)=1\). Thus \[\begin{aligned} &\mathcal A(W)-Z =\sum_{\xi_1}\ell_{\xi_1} \sum_{\substack{C,t\text{ squarefree}\\(Ct,S)=1, (C,t)=1\\ N(C)N(t)\le2vL}} \sum_{d\mid C}\mu(d)\mu(t)w_{C,d}\\ &\quad\times\sum_{\substack{F\le N(f)<2F\\(f,Ct)=1}}^* \sum_{\substack{h\ne0, (h,t)=1\\ N(h)\le C_\Phi v^2L^2N(d)/(\mathcal H N(C)^2)}} \sum_{x_1,x_2}^* p_{hf^2}(x_1)\overline{p_{hf^2}(x_2)} \mathscr K_h(N(x_1)/\ell,N(x_2)/\ell), \end{aligned}\] where, for each integer \(j_0\ge0\), \[\begin{aligned} \mathscr K_h(z_1,z_2) &=(z_1z_2)^{-1/2}W(z_1)\overline{W(z_2)} \widehat\Phi\!\Bigl( \frac{\mathcal H N(h)N(C)^2}{L^2N(d)z_1z_2}\Bigr),\\ \sup_{C,d,h}\|\mathscr K_h\|_{C^{j_0}(I^2)} &\ll_{I,\Phi,j_0}\|W\|_{C^{j_0}(I)}^2. \end{aligned}\] Here \(\chi_{x_i}(C)^4\) supplies \((x_i,C)=1\); the other original zeros are supplied by \(p_{hf^2}(x_i)\). The displayed frequency range follows from the support of \(\widehat\Phi\).

For fixed \(C,d,t\), the map \((f,h)\mapsto y=hf^2\) has divisor-bounded multiplicity in \(y\). Since \(\Sigma/(LF)\ge1\), its nonzero image satisfies \(N(y)\le4C_\Phi v^2L^2F^2N(d)/(\mathcal H N(C)^2)\le Y_{C,d}\). Since \(\Phi\ge1\) on \([0,1]\) and is nonnegative, \[\sum_{C,t}\sum_{d\mid C}w_{C,d}\sum_{f,h}|P_{C,d,t}(hf^2;U)|^2 \preccurlyeq\mathcal Q_{\xi_1}(U) \le M\|U\|_{C^j(I_*)}^2,\] with the preceding ranges on the left. Apply Lemma 23 to the full displayed expression for \(\mathcal A(W)-Z\). The kernel bound gives \[|\mathcal A(W)-Z| \ll D^{\varepsilon_0/2}M\|W\|_{C^{2j+4}(I)}^2.\] Together with \(\mathcal H\le\Sigma\), this proves the result after allocating the divisor losses within \(\varepsilon_0\). ◻

Proof. Expand the square, put \(g=(n_1,n_2)\) and \(n_i=gz_i\). The nonzero terms satisfy \[(z_1,z_2)=(z_1z_2,gCt)=(g,Ct)=1,\qquad \overline{\chi_{n_1}(y)}\chi_{n_2}(y) =\mathbf 1_{(y,g)=1}\overline{\chi_{z_1}(y)}\chi_{z_2}(y).\] Apply Lemma 5 to this last character, with exclusion \(g\) and divisor \(e\mid g\). Its zero frequency occurs exactly when \(z_1=z_2=1\) and gives the displayed \(Z\); counting \(g\) gives its bound. For the nonzero frequencies, (33)–(35) and the Chinese remainder theorem give \[\mu(z_1)\mu(z_2)\xi_1(z_1)\overline{\xi_1(z_2)} \gamma(\overline{\chi_{z_1}}\chi_{z_2}) =\sum_{\xi'}c_{\xi'}a_{\xi'}(z_1)\overline{a_{\xi'}(z_2)}.\] Use also \[\begin{aligned} \chi_z(dh)\overline{\chi_z(e)}\chi_z(C)^4 &=\chi_z(deh)\chi_z(Ce)^4,\\ \frac{U(N(gz_1)/\ell)\overline{U(N(gz_2)/\ell)}} {\sqrt{N(z_1)N(z_2)}} &=\frac{N(g)}\ell U_0(N(gz_1)/\ell)\overline{U_0(N(gz_2)/\ell)}. \end{aligned}\] Thus the full nonzero contribution is \[\begin{aligned} \mathcal M-Z={}&\sum_{\xi'}c_{\xi'} \sum_{\substack{g\\(g,Ct)=1}}^*\sum_{e\mid g} \frac{Y\mu(e)N(g)}{N(e)\ell}\sum_{h\ne0}\\ &\quad\times\sum_{\substack{z_1,z_2\\(z_1,z_2)=1\\(z_1z_2,gCt)=1}}^* a_{\xi'}(z_1)\overline{a_{\xi'}(z_2)} \chi_{z_1}(deh)\overline{\chi_{z_2}(deh)} \chi_{z_1}(Ce)^4\overline{\chi_{z_2}(Ce)^4}\\ &\quad\times U_0\!\Bigl(\frac{N(gz_1)}\ell\Bigr) \overline{U_0\!\Bigl(\frac{N(gz_2)}\ell\Bigr)} \widehat\Phi\!\Bigl(\frac{YN(h)}{N(e)N(z_1)N(z_2)}\Bigr). \end{aligned}\] All starred ideal indices are prime to \(S\). Insert \[\mathbf 1_{(z_1,z_2)=1}=\sum_{w\mid z_1,\ w\mid z_2}\mu(w), \qquad z_i=wn_i.\] Here \((w,gCt)=1\). Equation (34) and the zero-extended characters give \[\begin{gathered} a_{\xi'}(wn)=a_{\xi'}(w)a_{\xi'}(n)\chi_n(w)^4 \quad((w,n)=1),\\ |a_{\xi'}(w)\chi_w(deh)\chi_w(Ce)^4|^2=\mathbf 1_{(w,h)=1},\\ \mathbf 1_{(n,gCtw)=1}\chi_n(deh)\chi_n(Cew)^4 =\mathbf 1_{(n,tg/e)=1}\chi_n(deh)\chi_n(Cew)^4. \end{gathered}\] Consequently the preceding full sum becomes \[\begin{aligned} &\mathcal M-Z=\sum_{\xi'}c_{\xi'} \sum_{\substack{g,w\\(g,Ct)=(w,gCt)=1}}^* \sum_{e\mid g}\frac{Y\mu(e)\mu(w)N(g)}{N(e)\ell} \sum_{\substack{h\ne0\\(h,w)=1}}\\ &\quad\times\sum_{(n_1n_2,S)=1}^*a^\sharp(n_1)\overline{a^\sharp(n_2)} U_0\!\Bigl(\frac{N(gwn_1)}\ell\Bigr) \overline{U_0\!\Bigl(\frac{N(gwn_2)}\ell\Bigr)} \widehat\Phi\!\Bigl( \frac{YN(h)}{N(e)N(w)^2N(n_1)N(n_2)}\Bigr). \end{aligned}\] There is no remaining condition \((n_1,n_2)=1\). The column support gives \(N(gw)\le b\ell\) and \(N(z_1z_2)\le b^2\ell^2/N(g)^2\); the support of \(\widehat\Phi\) therefore imposes the stated bound on \(N(h)\). Substituting (83) gives (84), with only finitely many terms. ◻

Proof. Write \(N=\mathrm N_{K/\mathbb Q}\). Apply Lemma 20 with input length \(\ell\), row scale \(Y_{C,d}\), and the same \(C,d,t\), using the character \(\xi_1\). Here \(h\) is the new Fourier variable for the sum over \(y\). Its nonzero output has squarefree \(g,w\), \(e\mid g\), and \[(g,Ct)=(w,gCth)=1,\qquad r=tg/e,\quad f'=Cew,\quad k'=deh.\] Thus \(r,f'\) are coprime and squarefree, and the coefficient \(a^\sharp\) from (83) becomes \[a^\sharp(n) =a_{\xi'}(n)\mathbf 1_{(n,r)=1}\chi_n(k')\chi_n(f')^4.\] Write \(X'_0\) for the individual column scale \(X'\) in (83); below, \(X'\) will denote a common dyadic scale. The length, coefficient, and Fourier parameter simplify to \[\begin{aligned} X'_0&=\frac{\ell}{N(g)N(w)}=\frac L{N(r)N(f')},\\ w_{C,d}\frac{Y_{C,d}N(g)}{N(e)\ell} &=\frac{c_I\Sigma N(r)}{L^2},\\ \frac{Y_{C,d}N(h)N(g)^2}{N(e)\ell^2} &=\frac{c_I\Sigma F N(k')N(r)^2}{\mathcal H L}. \end{aligned}\] Put \(U_0(x)=x^{-1/2}U(x)\). The coupled kernel is therefore \[\mathscr K_{r,f',k'}(x_1,x_2) =U_0(x_1)\overline{U_0(x_2)} \widehat\Phi\!\Bigl(\frac{c_I\Sigma F N(k')N(r)^2}{\mathcal H Lx_1x_2}\Bigr),\] with both columns evaluated at \(x_j=N(n_j)/X'_0\). In particular its nonzero support requires \[N(r)N(f')\le 2vL,\qquad 0<N(k')\le\frac{4C_\Phi v^2}{c_I} \frac{\mathcal H L}{\Sigma F N(r)^2} \le\frac{\mathcal H L}{\Sigma F N(r)^2},\] since \(C_\Phi(2v)^2/c_I\le1\).

For fixed \(r,f',k'\), all preimages are obtained by choosing \[t\mid r,\qquad f'=Cew,\qquad e\mid k',\qquad (w,k')=1,\qquad d\mid(C,k'),\] and setting \(g=e(r/t)\), \(h=k'/(de)\). The factorization \(f'=Cew\) is disjoint and squarefree. All these preimages have the same kernel: \(Ctgw=rf'\), and the Fourier parameter displayed above depends only on \(r,k'\). Thus the support restrictions discard only zero kernels. The choices \(t\mid r\) contribute \(\tau_{\mathrm{div}}(r)\). At a prime \(p\mid f'\), the choices \(p\mid C\), \(p\mid e\), and \(p\mid w\), respectively, contribute \[\bigl(1+\mathbf 1_{p\mid k'}\bigr) -\mathbf 1_{p\mid k'}-\mathbf 1_{p\nmid k'}=\mathbf 1_{p\mid k'}\] to the sum of \(\mu(e)\mu(w)\) over the preimages; the two choices inside the first term are \(p\nmid d\) and \(p\mid d\). Let \(\mathcal Z\) be the total zero-frequency contribution: the sum of the terms \(Z\) from Lemma 20, with the outer weights \(w_{C,d}\) and ranges in (80). Consequently, regrouping before taking absolute values gives \[\begin{aligned} \mathcal Q_{\xi_1}(U)-\mathcal Z ={}&\frac{c_I\Sigma}{L^2} \sum_{\xi'}c_{\xi'} \sum_{\substack{r,f'\ \mathrm{squarefree}\\(r,f')=1}} N(r)\tau_{\mathrm{div}}(r)\sum_{\substack{k'\ne0\\f'\mid k'}}\\ &\quad\times \sum_{n_1,n_2}^*a^\sharp(n_1)\overline{a^\sharp(n_2)} \mathscr K_{r,f',k'}\!\Bigl(\frac{N(n_1)}{X'_0}, \frac{N(n_2)}{X'_0}\Bigr). \end{aligned}\]

Fix \(R\le N(r)<2R\), \(F'\le N(f')<2F'\) dyadically, and put \[X'=\frac L{RF'},\qquad \Sigma'=X'F'=\frac LR,\qquad \mathcal H'=\frac{\mathcal H L}{\Sigma F R^2}.\] Retain only those \(r\) for which a nonzero term occurs in these ranges. Since \(f'\mid k'\), each such \(r\) satisfies \[F'\le N(f')\le N(k')\le \frac{\mathcal H L}{\Sigma F N(r)^2}.\] In particular \(\mathcal H'\ge1\). For every \(j\mid r\), using \(\mathcal H\le\Sigma\), we also have \[ \frac{X'}{N(j)} \ge\frac{\Sigma F N(r)^2}{\mathcal H R N(j)} \ge\frac{F\Sigma}{\mathcal H}\ge1. \tag{85}\] Moreover \(1\le X'/X'_0<4\). The rescaled kernels \[\mathcal K_{r,f',k'}(x_1,x_2) =\mathscr K_{r,f',k'}\!\Bigl(\frac{X'}{X'_0}x_1, \frac{X'}{X'_0}x_2\Bigr)\] have support in \([u/8,2v]^2\subset\operatorname{int}(I'^2)\) and satisfy \[\sup_{r,f',k'}\|\mathcal K_{r,f',k'}\|_{C^{2m+4}(I'^2)} \ll_{I_*,m,\Phi}\|U\|_{C^{2m+4}(I_*)}^2,\] where the supremum is over the retained indices in the dyadic block. This follows from the Schwartz bounds for \(\widehat\Phi\). The common parameters satisfy exactly \[\frac{\mathcal H'}{\Sigma'}= \frac{\mathcal H}{\Sigma FR}\le\frac{\mathcal H}{\Sigma},\qquad \Sigma'\le L,\qquad \mathcal H'\le\frac{\mathcal H L}{\Sigma F}.\]

Fix \(\delta>0\), to be chosen in terms of \(\varepsilon_0\) at the end. For each retained \(r\), enlarge the \(f',k'\) ranges by positivity, keeping this set of \(r\) fixed. Lemma 7 then gives, for \(V\in C_c^\infty(I')\), \[\begin{aligned} \sum_{F'\le N(f')<2F'}^*\sum_{0<N(k')\le\mathcal H'} \Bigl|\sum_n^*a^\sharp(n)V(N(n)/X')\Bigr|^2 &=\Sigma'\mathcal E_{(r)}(\mathcal H',X',F';\xi',V)\\ &\ll_\delta D^\delta(\Sigma')^2S_m \|V\|_{C^m(I')}^2. \end{aligned}\] Indeed exclusion removal replaces \((X',F')\) by \((X'/N(j),F'N(j))\), \(j\mid r\), preserving \(\mathcal H'\) and \(\Sigma'\). Equation (85) gives \(X'/N(j)\ge1\), and \(F'N(j)\ge1\) as well. Thus every resulting mean square lies in the defining supremum.

There are \(O(R)\) ideals \(r\) in the dyadic range. Apply Lemma 23 with row index \((r,f',k')\), the preceding mean-square bound, and the uniform kernel bound. The absolute contribution of this dyadic block is at most \[D^{2\delta}\frac{\Sigma R}{L^2}\, R(\Sigma')^2S_m\|U\|_{C^{2m+4}(I_*)}^2 =D^{2\delta}\Sigma S_m\|U\|_{C^{2m+4}(I_*)}^2,\] because \((\Sigma R/L^2)R(L/R)^2=\Sigma\).

Finally, the zero-frequency bound in Lemma 20 gives \[\begin{aligned} |\mathcal Z| &\ll_{I_*,\Phi}\|U\|_\infty^2 \sum_{N(Ct)\le 2vL}\sum_{d\mid C}w_{C,d}Y_{C,d}\ell\\ &\ll_{I_*}\Sigma\|U\|_\infty^2 \sum_C\frac{\tau_{\mathrm{div}}(C)}{N(C)^2} \sum_{N(t)\le 2vL}\frac1{N(t)} \ll_{I_*,\delta}D^\delta\Sigma\|U\|_\infty^2. \end{aligned}\] This uses only \(Y_{C,d}>0\), not \(Y_{C,d}\ge1\). Sum the \(O_{I_*,C_0}((\log(2D))^2)\) dyadic blocks and the fixed finite character set, and take \(\delta=\varepsilon_0/4\). The stated bound follows, with implied constant depending only on \(I_*,m,C_0,\varepsilon_0\), the fixed cutoffs and arithmetic data. ◻

Proof of Proposition 12. Use \(S_m\) as in Lemma 21, with \(m\) as in the proposition. Lemma 21, with loss \(D^{\varepsilon/4}\), supplies the hypothesis of Lemma 19 with \(j=2m+4\) and \(M\ll D^{\varepsilon/4}\Sigma(1+S_m)\). Applying that lemma with the same loss gives \[\mathcal A(W)\ll D^{\varepsilon/2}\Sigma(1+S_m) \|W\|_{C^{4m+12}(I)}^2,\] which proves (56). ◻

Proof of Lemma 4. Proof of (30). For squarefree \(n\), set \(G(n)=\overline{\chi_n(4)}\gamma_3(n)\). Its dependence on a fixed ray class will be verified below. Fix a primary prime \(p\equiv1\pmod3\) outside \(S\). For multiplicative characters \(A,B\) of \((\mathcal O/(p))^\times\), extended by zero at \(0\), write \[J(A,B)=\sum_{x\bmod p}A(x)B(1-x).\] For each \(y\in\mathcal O/(p)\), the substitution \(u=2x-1\) gives \[\#\{x\bmod p:4x(1-x)=y\} =\#\{u\bmod p:u^2=1-y\} =1+\chi_p^3(1-y),\] since \(p\nmid2\) and \(\chi_p^3\) is the quadratic character, extended by zero at \(0\). Hence \[\chi_p(4)J(\chi_p,\chi_p) =\sum_{x\bmod p}\chi_p\bigl(4x(1-x)\bigr) =\sum_{y\bmod p}\chi_p(y)\bigl(1+\chi_p^3(1-y)\bigr) =J(\chi_p,\chi_p^3).\] The Gauss–Jacobi identity (cf. [22]) gives \[\frac{\gamma_1(p)\gamma_3(p)}{\gamma_4(p)} =\frac{J(\chi_p,\chi_p^3)}{\sqrt{\mathrm N_{K/\mathbb Q}(p)}} =\chi_p(4)\frac{J(\chi_p,\chi_p)}{\sqrt{\mathrm N_{K/\mathbb Q}(p)}} =\chi_p(4)\frac{\gamma_1(p)^2}{\gamma_2(p)}.\] For \(j\not\equiv0\pmod6\), the normalized Gauss sums satisfy \[\gamma_j(p)\gamma_{-j}(p)=\chi_p(-1)^j.\] Taking \(j=2\) gives \(\gamma_2(p)\gamma_4(p)=\chi_p^2(-1)=1\), since \(\chi_p^2\) is cubic. Combining this with the Gauss–Jacobi relation above gives, after cancelling \(\gamma_1(p)\) and rearranging, \[\gamma_1(p)\gamma_2(p)=\overline{\chi_p(4)}\gamma_3(p)\gamma_2(p)^3.\]

To finish the prime case of (30), it remains to evaluate \(\gamma_2(p)^3\). We do so by computing \(J(\chi_p^2,\chi_p^2)\). Put \(q=\mathrm N_{K/\mathbb Q}(p)\). We have \(J(\chi_p^2,\chi_p^2)\in\mathcal O\) and \(\mathrm N_{K/\mathbb Q}\bigl(J(\chi_p^2,\chi_p^2)\bigr)=q\). For \(0\le\ell\le(q-1)/3\), the exponent \((q-1)/3+\ell\) lies strictly between \(0\) and \(q-1\), so \(\sum_{x\bmod p}x^{(q-1)/3+\ell}=0\) in \(\mathcal O/(p)\). Reducing modulo \(p\) and expanding the binomial therefore gives \[\begin{aligned} J(\chi_p^2,\chi_p^2) &\equiv\sum_{x\bmod p}x^{(q-1)/3}(1-x)^{(q-1)/3}\\ &=\sum_{\ell=0}^{(q-1)/3}(-1)^\ell\binom{(q-1)/3}{\ell} \sum_{x\bmod p}x^{(q-1)/3+\ell} =0\pmod p. \end{aligned}\] Thus \(J(\chi_p^2,\chi_p^2)/p\in\mathcal O\) has norm one, so \(J(\chi_p^2,\chi_p^2)\) is a unit multiple of \(p\). To determine the unit, write \(\chi_p^2(x(1-x))=\omega^{j_x}\) with \(j_x\in\{0,1,2\}\) for \(x\in\mathcal O/(p)\setminus\{0,1\}\). Then \[\prod_{x\ne0,1}x(1-x)=1 \quad\Longrightarrow\quad \sum_{x\ne0,1}j_x\equiv0\pmod3.\] Since \((\omega-1)^2\) generates \((3)\), expansion modulo this ideal gives \[J(\chi_p^2,\chi_p^2) =\sum_{x\ne0,1}\omega^{j_x} \equiv(q-2)+(\omega-1)\sum_{x\ne0,1}j_x \equiv-1\pmod3.\] Together with \(p\equiv1\pmod3\), this fixes the unit: \[J(\chi_p^2,\chi_p^2)=-p.\] Consequently \[\gamma_2(p)^3 =\frac{\gamma_2(p)^2}{\gamma_4(p)} =\frac{J(\chi_p^2,\chi_p^2)}{\sqrt{\mathrm N_{K/\mathbb Q}(p)}} =-\alpha(p).\] The Chinese remainder theorem extends these prime-modulus identities to (30) for squarefree \(n\), with one minus sign for each prime factor accounting for \(\mu(n)\).

Proof of (34). For coprime squarefree primary \(a,b\) prime to \(S\), the Chinese remainder theorem gives \[\gamma_2(ab)=\gamma_2(a)\gamma_2(b) \chi_a(b)^2\chi_b(a)^2,\] and cubic reciprocity gives \(\chi_a(b)^2=\chi_b(a)^2\). Multiplying by \(\overline{\alpha(ab)}\xi(ab)\) therefore proves (34).

Proof of (29). We now prove (29) and the formula for \(G(n)\) in (31), including their dependence on a fixed ray class. We begin by evaluating the normalized quadratic Gauss sum. For \(c\in\mathcal O\) coprime to \(2\), we will show that \[ \Gamma_{\rm quad}(c):=|c|^{-1}\sum_{x\bmod c}e(x^2/c) =\frac12\sum_{y\bmod2\mathcal O}e(-cy^2/4). \tag{86}\] To justify (86), apply Poisson summation over \(z\in\mathcal O\) to \(e(z^2/c)e^{-\pi\eta|z|^2}\), with \(\eta>0\). The Gaussian makes the sum convergent. The Fourier transform of the product is \[\frac{|c|}{2\sqrt{r_\eta}} e^{-\pi\eta \mathrm N_{K/\mathbb Q}(c)|y|^2/(4r_\eta)}e(-cy^2/(4r_\eta)), \qquad r_\eta=1+3\mathrm N_{K/\mathbb Q}(c)\eta^2/16.\] Divide both sides by the Gaussian mass \[\sum_{z\in\mathcal O}e^{-\pi\eta|z|^2}\asymp\eta^{-1}.\] On the Fourier side, replacing \(r_\eta\) by \(1\) has total error \[O_c\!\biggl(\eta^2\sum_{y\in\mathcal O}|y|^2e^{-C_c\eta|y|^2}\biggr) =O_c(1),\] where \(C_c>0\) is fixed. The normalized error is therefore \(O_c(\eta)\to0\). Grouping the original sum modulo \(c\) and the Fourier sum modulo \(2\mathcal O\), then letting \(\eta\downarrow0\), gives (86).

For \(c=a+b\omega\), formula (86) becomes \[\Gamma_{\rm quad}(a+b\omega)=\frac{1+\mathrm i^{-b}+\mathrm i^a+\mathrm i^{b-a}}2.\] Thus \(\Gamma_{\rm quad}(c)\) depends only on \(c\bmod4\mathcal O\) and is invariant under multiplication by a square in \((\mathcal O/4\mathcal O)^\times\). Representatives for the four square classes and their values are \[\begin{array}{c|rrrr} c&1&-1&\lambda&-\lambda\\ \hline \Gamma_{\rm quad}(c)&1&1&\mathrm i&-\mathrm i \end{array}.\] Consequently the function \[\mathcal R(c_1,c_2):=\frac{\Gamma_{\rm quad}(c_1c_2)}{\Gamma_{\rm quad}(c_1)\Gamma_{\rm quad}(c_2)}\] on these square classes satisfies \[\mathcal R((-1)^e\lambda^f,(-1)^g\lambda^h)=(-1)^{eh+fg+fh}, \qquad e,f,g,h\in\{0,1\},\] and is a symmetric bicharacter. For a prime \(p\), each \(y\in\mathcal O/(p)\) has \(1+\chi_p^3(y)\) square roots. Grouping by \(y=x^2\) gives \[\Gamma_{\rm quad}(p) =\frac1{|p|}\sum_{y\bmod p}\bigl(1+\chi_p^3(y)\bigr)e(y/p) =\gamma_3(p),\] since \(\sum_{y\bmod p}e(y/p)=0\). The Chinese remainder theorem extends this equality to squarefree \(n\). Together with cubic reciprocity, it identifies the above \(\mathcal R\) with the quotient \(\chi_b(a)/\chi_a(b)\) for coprime primary \(a,b\). Furthermore \(\chi_c(4)=(-2/c)_3=(c/(-2))_3\) is a character modulo \(2\). Consequently \(G(c)=\overline{\chi_c(4)}\Gamma_{\rm quad}(c)\) factors through a fixed ray class group and satisfies (29), with \(\mathcal R(c,c)=\chi_c(-1)\).

Proof of (35). In the fixed ray class group, the multiplicative relation for \(G\) gives \[G(a^{-1})=\chi_a(-1)\overline{G(a)}.\] Using the symmetry and multiplicativity of \(\mathcal R\), we obtain \[G(ba^{-1}) =G(b)G(a^{-1})\mathcal R(b,a^{-1}) =\chi_a(-1)\overline{G(a)}G(b)\mathcal R(a,b),\] which is (35).

Proof of (32)–(33). The second identity in (31) follows from the inverse-character Gauss identity: \(\gamma_1(n)\gamma_{-1}(n)=\chi_n(-1)\). It remains to prove (32)–(33). Multiplying the second identity in (30) by \(\overline{\alpha(n)}\) gives (32), and combining it with this inverse-character identity gives (33). ◻

Proof of Proposition 14 and Lemmas 15 and 16. Finite Fourier expansion. Define \(\phi:\mathcal O\to\mathbb C\) by \[\phi(n)= \begin{cases} \chi_n(\lambda)^2\Psi_0(n),&n\equiv1\pmod3,\quad(n,S)=1,\\ 0,&\text{otherwise}. \end{cases}\] For \(\Psi(n)=\Psi_0(n)\prod_{p\in\mathcal P}\chi_p^{j_p}(n)\), define \[ \Theta_\Psi(z,v)= \sum_{\substack{\ell\in\lambda^{-3}\mathcal O\\\ell\ne0}} \overline{\tau(-\ell)}\phi(\lambda^3\ell) \Bigl(\prod_{p\in\mathcal P}\chi_p^{j_p}(\lambda^3\ell)\Bigr) vK_{1/3}(4\pi|\ell|v)\breve e(\ell z). \tag{87}\] When \(\Psi=\Psi_k\), the multiplier of \(\overline{\tau(-\ell)}\) is \(\phi_k(\lambda^3\ell)\), so this definition recovers \(\Theta_k\) in (60). We first write \(\Theta_\Psi\) as a finite sum of translates of \(\overline\theta\) and determine their reduced denominators. Choose once and for all a nonzero \(L\in\mathcal O\), with prime divisors in \(S\), divisible by the conductor of \(\Psi_0\), every prime in \(S\), and sufficiently high powers of the primes above \(2\) and \(3\). The supplementary law of cubic reciprocity for \(\lambda\) [7], which evaluates \(\chi_n(\lambda)^2=(\lambda/n)_3\), makes \(\phi\) periodic modulo \(L\). As before (62), define \[\widehat\phi(h_0)=\frac1{\mathrm N_{K/\mathbb Q}(L)}\sum_{x\bmod L}\phi(x)e(-h_0x/L), \qquad h_0\in\mathcal O/(L).\] Since \(|\phi|\le1\), we have \(|\widehat\phi(h_0)|\le1\).

For \(p\in\mathcal P\) and \(0\le j\le5\), we expand \(\chi_p^j\) on \(\mathcal O/(p)\) in additive characters. Its Fourier coefficients are defined, for \(h_p\in\mathcal O/(p)\), by \[\begin{align*} C_{p,j}(h_p)&:=\frac1{\mathrm N_{K/\mathbb Q}(p)}\sum_{y\bmod p}\chi_p^j(y)e(-h_py/p). \\ \intertext{Finite Fourier inversion then gives, for $x\in\mathcal O$,} \chi_p^j(x)&=\sum_{h_p\in\mathcal O/(p)}C_{p,j}(h_p)e(h_px/p). \tag{88}\end{align*}\]

Multiplying (88) over the primes in \(\mathcal P\) gives, for \(x\in\mathcal O\), \[\prod_{p\in\mathcal P}\chi_p^{j_p}(x) =\sum_{\substack{h_p\in\mathcal O/(p)\\p\in\mathcal P}} \Bigl(\prod_{p\in\mathcal P}C_{p,j_p}(h_p)\Bigr) e\!\Bigl(x\sum_{p\in\mathcal P}\frac{h_p}{p}\Bigr).\] Each summand is indexed by a tuple \((h_p)_{p\in\mathcal P}\). At \(x=\lambda^3\ell\), its additive character gives the shift \(\lambda^2\sum_{p\in\mathcal P}h_p/p\) of \(\overline\theta\). Writing \(\boldsymbol h=(h_0,(h_p)_{p\in\mathcal P})\) with \(h_0\in\mathcal O/(L)\), we claim that \[\Theta_\Psi(z,v)=\sum_{\boldsymbol h} c_{\mathrm F}(\boldsymbol h)\overline{\theta(z+z_{\boldsymbol h},v)},\] where \[ z_{\boldsymbol h}=\lambda^2\Bigl(h_0/L+\sum_{p\in\mathcal P}h_p/p\Bigr), \qquad c_{\mathrm F}(\boldsymbol h)=\widehat\phi(h_0)\prod_{p\in\mathcal P} C_{p,j_p}(h_p). \tag{89}\]

To verify this identity, expand the fixed factor \(\phi\) as well. The finite Fourier expansions identify all nonzero Fourier coefficients, and the constant terms of the translates cancel because \[\sum_{\boldsymbol h} c_{\mathrm F}(\boldsymbol h)=\phi(0)\prod_{p\in\mathcal P}\chi_p^{j_p}(0)=0.\]

For \(h_p\ne0\), substitute \(y=h_p^{-1}u\) in the definition of \(C_{p,j}(h_p)\). For \(h_p=0\) and \(j\ne0\), use character orthogonality; for \(j=0\), sum the additive character over nonzero residues directly. These calculations give \[ C_{p,j}(h_p)= \begin{cases} \mathrm N_{K/\mathbb Q}(p)^{-1/2}\chi_p(-1)^j\gamma_j(p)\chi_p(h_p)^{-j}, &j\ne0,\ h_p\ne0,\\ 0, &j\ne0,\ h_p=0,\\ -\mathrm N_{K/\mathbb Q}(p)^{-1},&j=0,\ h_p\ne0,\\ 1-\mathrm N_{K/\mathbb Q}(p)^{-1},&j=0,\ h_p=0. \end{cases} \tag{90}\]

For the tuple \(\boldsymbol h\), define its set of active primes by \(\mathcal A(\boldsymbol h):=\{p\in\mathcal P:h_p\ne0\}\). By (90), \(c_{\mathrm F}(\boldsymbol h)\ne0\) implies \(\{p\in\mathcal P:j_p\ne0\}\subseteq\mathcal A(\boldsymbol h)\). Thus only primes with \(j_p=0\) can be inactive, as asserted in the proposition. Put \(r=\prod_{p\in\mathcal A(\boldsymbol h)}p\), and let \(c_0\) be the reduced denominator of \(\lambda^2h_0/L\). At each active prime \(p\), \[v_p(\lambda^2h_p/p)=-1, \qquad v_p(z_{\boldsymbol h}-\lambda^2h_p/p)\ge0,\] so the factor \(p\) cannot cancel from the denominator. At primes dividing \(L\), all terms \(\lambda^2h_p/p\) are integral. Thus the reduced denominator is \(c=c_0r\) up to a unit, including when \((h_0,L)\ne1\). The finite Fourier expansion has therefore expressed \(\Theta_\Psi\) as \(O_L(2^{|\mathcal P|})\) groups of translates \(\overline{\theta(z+z_{\boldsymbol h},v)}\), with shifts \(z_{\boldsymbol h}\) and coefficients \(c_{\mathrm F}(\boldsymbol h)\) given by (89). The groups are indexed by \(h_0\in\mathcal O/(L)\) and the active set \(\mathcal A\). Each group has a common reduced denominator \(c=c_0\prod_{p\in\mathcal A}p\) up to a unit. For each translate \(\overline{\theta(z+z_{\boldsymbol h},v)}\), we next identify which of the three Fourier expansions in (66) will be used after changing coordinates at \(z_{\boldsymbol h}\).

Reduced denominators and cusp coefficients. We next express each translate \(\overline{\theta(z+z_{\boldsymbol h},v)}\) using one of the three cusp expansions identified above. Our representatives \(\gamma_0,\gamma_+,\gamma_-\) correspond to \(\gamma_1,\gamma_{10},\gamma_{19}\), respectively, in the numbering of [7], which follows [38]. We allow \((h_0,L)\ne1\) and keep the reduced denominator \(c_0\) of \(\lambda^2h_0/L\) in the calculation; this is the extension beyond [7].

Set \(M=\lambda^{12}L^4\). Since \((p,M)=1\) for \(p\in\mathcal A(\boldsymbol h)\), the Chinese remainder theorem lets us choose representatives \(h_p\in\mathcal O\) with \(h_p\equiv0\pmod{M^2}\). Changing representatives modulo \(p\) changes \(z_{\boldsymbol h}\) by an element of \(\lambda^2\mathcal O=3\mathcal O\), under which \(\theta\) is periodic. To track dependence on the primes in \(r\), restrict \(r\) to a residue class modulo \(M^2\). For fixed \(h_0\), active set, and this residue class, write \(z_{\boldsymbol h}=a/c\) in lowest terms, normalizing \(a\equiv1\pmod3\) if \(\lambda\mid c\), and \(c\equiv1\pmod3\) otherwise. Since \[a=\lambda^2c\Bigl(h_0/L+\sum_{p\mid r}h_p/p\Bigr),\] these choices fix the residue of \(a\) modulo \(Mc_0\). Since \((a,c)=1\) and \((r,M)=1\), the Chinese remainder theorem gives \(\delta'\in\mathcal O\) satisfying the following congruences, where \(q\) ranges over prime divisors of the indicated elements: \[\begin{align*} a\delta'&\equiv1\pmod{q^{v_q(Mc_0)}}&& (q\mid c_0),\\ \delta'&\equiv0\pmod{q^{v_q(M)}}&& (q\mid M,\ q\nmid c_0),\\ a\delta'&\equiv1\pmod r. \end{align*}\] Put \[b_g=\frac{a\delta'-1}{c}, \qquad g=\begin{pmatrix}a&b_g\\c&\delta'\end{pmatrix}\in\mathrm{SL}_2(\mathcal O).\] The congruences for \(\delta'\) give \[b_g\equiv \begin{cases} 0,&q\mid c_0,\\ -c^{-1},&q\mid M,\ q\nmid c_0, \end{cases} \pmod{q^{v_q(M)}}.\]

To choose the cusp expansion at \(a/c\), put \[H= \begin{cases} I,&3\mid c,\\[2pt] \bigl(\begin{smallmatrix}1&0\\u_0&1\end{smallmatrix}\bigr), &v_\lambda(c)=1,\quad u_0\in\{\lambda,-\lambda\},\quad u_0\equiv c\pmod3,\\[2pt] \bigl(\begin{smallmatrix}u_0&-1\\1&0\end{smallmatrix}\bigr), &(c,\lambda)=1,\quad u_0\equiv a\pmod3. \end{cases}\] In the last case choose \(u_0\in\mathcal O\) from a fixed set of representatives modulo \(3\). In each case, put \(g_1=gH^{-1}\). Then \(g=g_1H\) and \(g_1\equiv I\pmod3\), as in [7]. The invariances \[\theta(\gamma w)=\theta(w)\quad(\gamma\in\mathrm{SL}_2(\mathbb Z)), \qquad \theta(z+t,v)=\theta(z,v)\quad(t\in\mathbb Z+3\mathcal O)\] reduce \(\overline\theta(Hw)\) to one of the three functions \(\overline{\theta(\gamma_\sigma w)}\), \(\sigma\in\{0,-,+\}\), with the matrices \(\gamma_\sigma\) from (65). These representatives give the infinity expansion and the two additional cusp expansions needed here: translating by \(\omega\) or \(-\omega\), then applying inversion, gives the representatives labeled \(-\) and \(+\), respectively [7]. Write \(t_\sigma(\ell)\) for the coefficient of \(vK_{1/3}(4\pi|\ell|v)\breve e(\ell z)\) in \(\theta(\gamma_\sigma w)\). Let \(\tau_1,\tau_2:\lambda^{-4}\mathcal O\setminus\{0\}\to\mathbb C\) be the coefficient sequences defined in [7]. Patterson’s cusp calculation [38] then gives \[ t_0(\ell)=\tau(\ell),\qquad t_-(\ell)=\omega^2\tau_1(\omega^2\ell)\breve e(\ell), \qquad t_+(\ell)=\omega\tau_2(\omega\ell)\breve e(\ell). \tag{91}\] Complex conjugation changes the Fourier frequency from \(\ell\) to \(-\ell\). Thus the coefficients in the normalization (66) are \[ d_\sigma(\ell)=\overline{t_\sigma(-\ell)},\qquad \sigma\in\{0,+,-\}. \tag{92}\] For indices \(\ell=u\lambda^mnb^3\) as in (68), the coefficient magnitudes satisfy \[|t_0(\ell)|\le \begin{cases} 3^{k/2+2}|b|,&m=3k-4,\ k\ge1,\\ 3^{k/2+5/2}|b|,&m=3k-3,\ k\ge0. \end{cases}\] The \(m=-4\) coefficients of \(t_-\) and \(t_+\) have magnitude at most \(9|b|\). For the chosen \(H\), let \(\sigma\in\{0,+,-\}\) index the corresponding expansion in (66). For each translate \(\overline{\theta(z+a/c,v)}\), we have identified the Fourier expansion in \(z\) of \(\overline{\theta(H(z,v))}\) as one of the three expansions in (66): \[\overline{\theta(H(z,v))} =\frac{3^{5/2}}2\mathbf 1_{\sigma=0}v^{2/3} +\sum_{0\ne\ell\in\lambda^{-4}\mathcal O} d_\sigma(\ell)vK_{1/3}(4\pi|\ell|v)\breve e(\ell z).\] These coefficients satisfy the support restriction (68) and bound (69), proving the coefficient assertions of Lemma 16. The next step uses \(g=g_1H\) to express the original translate through this Fourier series evaluated at \(g^{-1}(z+a/c,v)\), and computes the accompanying multiplier and additive phase. The matrix \(g\) maps \(\infty\) to \(a/c\), which is why this is called an expansion at the cusp \(a/c\).

The local character transformation. The matrix factorization \(g=g_1H\) lets us apply the automorphy law \[\theta(g_1w)=\kappa(g_1)\theta(w),\qquad \kappa(g_1)=(c_1/a_1)_3,\qquad g_1=\Bigl(\begin{matrix}a_1&b_1\\c_1&d_1\end{matrix}\Bigr)\equiv I\pmod3,\] where \(\kappa\) is Kubota’s cubic character [7]. We will combine its conjugate with the finite Fourier coefficients \(C_{p,j}(h_p)\) in (90) to obtain the factors \(B_{p,j}\) of (63). For the translated theta function, the automorphy law gives \[ \overline{\theta(z+a/c,v)} =\overline{\kappa(g_1)}\,\overline{\theta\bigl(Hg^{-1}(z+a/c,v)\bigr)}, \tag{93}\] where \[ g^{-1}(z+a/c,v) =\Bigl(-\frac{\delta'}c-\frac{\bar z}{c^2(v^2+|z|^2)}, \frac{v}{\mathrm N_{K/\mathbb Q}(c)(v^2+|z|^2)}\Bigr). \tag{94}\] At \(z=0\), the term with Fourier index \(\ell\) in the expansion of \(\overline\theta(Hw)\) therefore acquires the phase \(\breve e(-\delta'\ell/c)\). We claim that the multiplier is given by \[ \kappa(g_1)= \begin{cases} (c_0/a)_3(a/r)_3,&3\mid c,\\ (-u_0/(a-u_0b_g))_3\,((c_0/u_0)/a)_3(a/r)_3, &v_\lambda(c)=1,\\ (a/c_0)_3(a/r)_3,&(c,\lambda)=1. \end{cases} \tag{95}\] To verify (95), use the determinant equation and cubic reciprocity. For the case \(v_\lambda(c)=1\), the required congruences are \[a(c-u_0\delta')\equiv-u_0\pmod{a-u_0b_g}, \qquad b_gc\equiv-1\pmod a.\] For \((c,\lambda)=1\), the congruence \[-b_gc=1-a\delta'\equiv1\pmod9\] removes the supplementary factors; reciprocity at the remaining primes then gives \((\delta'/(-b_gc))_3=1\). The factors other than \((a/r)_3\) depend only on the fixed residues at primes in \(S\). We may therefore write \[\kappa(g_1)=\kappa_0(a/r)_3 =\kappa_0\prod_{p\mid r}\chi_p(a)^2, \qquad |\kappa_0|=1,\] where \(\kappa_0\) is fixed once \(h_0\), the active set, and \(r\bmod M^2\) are fixed.

Fix \(h_0\) and the active set, so that the denominator \(c=c_0r\) is fixed while the nonzero residues \(h_p\) vary. Put \(D_0=\lambda^3c_0\). For each active prime \(p\), define in \(\mathcal O/(p)\) \[\sigma_p=\lambda^2c/p,\qquad \epsilon_p=-\bigl((\lambda^3c/p)\sigma_p\bigr)^{-1}.\] The inverse in \(\epsilon_p\) is taken in the field \(\mathcal O/(p)\); it exists because \(r\) is squarefree and \(p\nmid\lambda c_0\). The expression for \(a\) gives \(a\equiv\sigma_ph_p\pmod p\). For a dual Fourier index \(\ell\in\lambda^{-4}\mathcal O\), the additive form of the Chinese remainder theorem yields \[ \breve e(-\delta'\ell/c) =\psi(\lambda^4\ell)\prod_{p\mid r}e(\epsilon_ph_p^{-1}\lambda^4\ell/p), \tag{96}\] where \[\psi(\lambda^4\ell):=e(-\delta'_0r^{-1}\lambda^4\ell/D_0), \qquad \delta'_0=\delta'\bmod D_0.\] The residues \(\delta'_0\) and \(r^{-1}\bmod D_0\) are fixed by the choices above, so \(\psi\) ranges over a fixed finite family of additive characters of \(\mathcal O\). We claim the local transformation identity \[ \sum_{h_p\ne0} C_{p,j}(h_p)\chi_p(a)^{-2} e(\epsilon_ph_p^{-1}\lambda^4\ell/p) =\chi_p(\sigma_p)^{-2}\omega_{p,j}B_{p,j}(\lambda^4\ell), \tag{97}\] where \(B_{p,j}\) is defined in (63), and \[\omega_{p,j}= \begin{cases} \chi_p(-1)^j\gamma_j(p)\gamma_{j+2}(p)\chi_p(\epsilon_p)^{-j-2},&j\ne0,4,\\ \gamma_4(p),&j=4,\\ -\gamma_2(p)\chi_p(\epsilon_p)^{-2},&j=0\ {\rm active}. \end{cases}\] The indices of \(\gamma_j(p)\) are read modulo six; in the second case, \(\chi_p(-1)^4=1\). In each case \(|\omega_{p,j}|=1\). To prove (97), combine (90), the conjugate of (95), and (96). Setting \(y=h_p^{-1}\) changes the character exponent as follows: \[\chi_p(h_p)^{-j}\chi_p(a)^{-2} =\chi_p(\sigma_p)^{-2}\chi_p(h_p)^{-j-2} =\chi_p(\sigma_p)^{-2}\chi_p(y)^{j+2}, \qquad y=h_p^{-1}.\] For \(j\ne0,4\), the character \(\chi_p^{j+2}\) is nontrivial, and its Gauss sum gives the first case of \(B_{p,j}\). For \(j=4\), it is trivial on nonzero residues; after rescaling by \(\epsilon_p\ne0\), the sum is \[\sum_{y\ne0}e(\lambda^4\ell y/p) =-1+\mathrm N_{K/\mathbb Q}(p)\mathbf 1_{p\mid\lambda^4\ell}.\] For active \(j=0\), the coefficient \(-\mathrm N_{K/\mathbb Q}(p)^{-1}\) combines with a cubic Gauss sum to give \[B_{p,0}(\lambda^4\ell) =\mathrm N_{K/\mathbb Q}(p)^{-1/2}\chi_p(\lambda^4\ell)^{-2},\] with the unit factors included in \(\omega_{p,0}\). This proves (97) in every case. An inactive prime has \(j=0\) and contributes only the scalar \(1-\mathrm N_{K/\mathbb Q}(p)^{-1}\). Consequently, for each fixed \(h_0\) and active set \(\mathcal A\), combining the Fourier coefficients \(d_\sigma(\ell)\) of \(\overline{\theta(H(z,v))}\) with the sums over \(h_p\ne0\) gives \[d_\sigma(\ell)\psi(\lambda^4\ell) \prod_{p\in\mathcal A}B_{p,j_p}(\lambda^4\ell),\] up to a scalar independent of \(\ell\). These are the arithmetic factors in the dual sum (67). In particular, \(B_{p,1}(\lambda^4\ell)=\chi_p(\lambda^4\ell)^3\) is the quadratic factor used in (76).

The archimedean transform. We now derive the transformed weight \(V_*^\sharp\) in (64) and the scalar multiplying the dual sum. For \(\operatorname{Re}s>1\), introduce the Dirichlet series associated with the completed sum (48): \[\mathcal T(s,\Psi)= \Bigl(\sum_{\substack{n\in\mathcal O\\n\equiv1\ (3)\\(n,S)=1}}^* \overline{\alpha(n)}\gamma_2(n)\Psi(n)\mathrm N_{K/\mathbb Q}(n)^{-s}\Bigr) \Bigl(\sum_{\substack{b\in\mathcal O\\b\equiv1\ (3)\\(b,S)=1}}\overline{\alpha(b)}^{\,3} \Psi(b)^3\mathrm N_{K/\mathbb Q}(b)^{-3s+1/2}\Bigr).\] With \(V_*\) from (48) and its Mellin transform defined before (64), Mellin inversion gives \[ T(X;\Psi)=\frac1{2\pi \mathrm i}\int_{(\sigma)} \widehat V_*(s-\tfrac12)\mathcal T(s,\Psi)X^{s-1/2}\,ds, \qquad \sigma>1. \tag{98}\] The identity (61), with \(\Psi\) in place of \(\Psi_k\), identifies these coefficients with those of \(\Theta_\Psi\) after inserting the factor \(\overline{\alpha(nb^3)}\). We now compute the normalization relating \(\mathcal T(s,\Psi)\) to the Mellin transform of the derivative of \(\Theta_\Psi\).

For \(z=x+\mathrm i y\), use \(\partial_{\bar z}=(\partial_x+\mathrm i\partial_y)/2\) and \(\partial_z=(\partial_x-\mathrm i\partial_y)/2\). Differentiation in \(\bar z\) supplies a factor \(\bar\ell\) in each Fourier mode. Combined with the factor \(|\ell|^{-1}\) from the Bessel integral below, this produces the required angular factor \(\overline{\alpha(\ell)}\). We therefore define, initially for \(\operatorname{Re}s>1\), the Mellin transform \[\mathcal J(s)=\int_0^\infty \partial_{\bar z}\Theta_\Psi(z,v)\Bigr|_{z=0} v^{2s-1}\,dv.\] For each nonzero Fourier mode, differentiation and the Mellin integral for \(K_{1/3}\) (see [35]) give the following formulas: \[\begin{align*} \partial_{\bar z}\breve e(\ell z) &=2\pi \mathrm i\bar\ell\,\breve e(\ell z),\tag{99}\\ \int_0^\infty v^{2s}K_{1/3}(4\pi|\ell|v)\,dv &=\frac{2^{2s-1}\Gamma(s+1/3)\Gamma(s+2/3)} {(4\pi|\ell|)^{2s+1}} \qquad \operatorname{Re}s>0. \tag{100}\end{align*}\]

For \(\operatorname{Re}s>1\), the coefficient formula (12) and (100) show that the sum of the integrals of the absolute values is finite. We may therefore integrate the differentiated Fourier series term by term.

For \(\ell=\lambda^{-3}nb^3\), the phase and scale simplify to \[\mathrm i\overline{\alpha(\ell)}|\ell|^{-2s} =27^s\overline{\alpha(n)}\, \overline{\alpha(b)}^{\,3}\mathrm N_{K/\mathbb Q}(n)^{-s}\mathrm N_{K/\mathbb Q}(b)^{-3s}.\] Combining this with (12) gives \[ \mathcal J(s)=\frac{3^{5/2}}4 \Bigl(\frac{27}{(2\pi)^2}\Bigr)^s \Gamma(s+1/3)\Gamma(s+2/3)\mathcal T(s,\Psi). \tag{101}\] We next show that \(\mathcal J(s)\) is entire and use (101) to continue \(\mathcal T(s,\Psi)\).

Write \((z',v')=g^{-1}(z+a/c,v)\). Differentiating (94) at \(z=0\) gives \[\frac{\partial z'}{\partial\bar z}\bigg|_{z=0} =-\frac1{c^2v^2}, \qquad \frac{\partial\overline{z'}}{\partial\bar z}\bigg|_{z=0}=0, \qquad \frac{\partial v'}{\partial\bar z}\bigg|_{z=0}=0.\] Thus the derivative in cusp coordinates is \(-(cv)^{-2}\partial_{z'}\), with no height-derivative term. The defining Fourier series (87) for \(\Theta_\Psi\) controls \(v\to\infty\). For \(v\to0\), use (93) and the cusp expansions (66). In each expansion the horizontal derivative removes the constant mode. The remaining modes decay exponentially as \(v\to\infty\); at \(v\to0\), the transformed height \(1/(\mathrm N_{K/\mathbb Q}(c)v)\) tends to infinity, giving exponential decay in \(1/v\). Thus the integral defining \(\mathcal J(s)\) converges for every \(s\) and defines an entire function. Equation (101), after division by its gamma factors, also continues \(\mathcal T(s,\Psi)\) to an entire function.

For the term indexed by \(\boldsymbol h\) in (89), write \(c_{\boldsymbol h},\delta'_{\boldsymbol h},g_{1,\boldsymbol h},H_{\boldsymbol h}\) for the corresponding choices above. Let \(\sigma_{\boldsymbol h}\) be the cusp index determined by \(H_{\boldsymbol h}\). Define its cusp Mellin transform by \[ \mathcal J_{\boldsymbol h}^\vee(s)=\int_0^\infty \partial_z\bigl\{\overline{\theta(H_{\boldsymbol h}(z,v))}\bigr\} \Bigr|_{z=-\delta'_{\boldsymbol h}/c_{\boldsymbol h}}v^{2s-1}\,dv. \tag{102}\] Substituting \(v\mapsto(\mathrm N_{K/\mathbb Q}(c_{\boldsymbol h})v)^{-1}\) in each translated term of \(\mathcal J(s)\), using (93), gives \[ \mathcal J(s)=-\sum_{\boldsymbol h} c_{\mathrm F}(\boldsymbol h)\overline{\kappa(g_{1,\boldsymbol h})}\, \overline{\alpha(c_{\boldsymbol h})}^{\,2}\mathrm N_{K/\mathbb Q}(c_{\boldsymbol h})^{1-2s}\mathcal J_{\boldsymbol h}^\vee(1-s). \tag{103}\] Here \(c_{\mathrm F}(\boldsymbol h)\) is the coefficient in (89). The cusp coefficients \(d_{\sigma_{\boldsymbol h}}(\ell)\) and (100) give, for \(\operatorname{Re}s>1\), \[ \mathcal J_{\boldsymbol h}^\vee(s)=\frac{\mathrm i\,\Gamma(s+1/3)\Gamma(s+2/3)}{4(2\pi)^{2s}} \sum_{0\ne\ell\in\lambda^{-4}\mathcal O} \frac{d_{\sigma_{\boldsymbol h}}(\ell)\alpha(\ell)}{\mathrm N_{K/\mathbb Q}(\ell)^s} \breve e(-\delta'_{\boldsymbol h}\ell/c_{\boldsymbol h}). \tag{104}\] The Dirichlet series \(\mathcal T(s,\Psi)\) converges absolutely for \(\operatorname{Re}s>1\), while the series in (104) at \(1-s\) converges absolutely for \(\operatorname{Re}s<0\). The functional equation (103) and Stirling’s formula [22] therefore give polynomial bounds in \(|\operatorname{Im}s|\) for \(\mathcal T(s,\Psi)\) on both sides of the strip \(0\le\operatorname{Re}s\le1\). Splitting the integral defining \(\mathcal J(s)\) at \(v=1\) gives finite order; Phragmén–Lindelöf [22] then gives the same type of bound inside the strip (compare the arguments of Dunn and Radziwiłł in [7]). Together with the rapid decay of \(\widehat V_*\) on vertical lines, these bounds justify moving the \(s\)-contour in (98) to \(\operatorname{Re}s<0\). No poles are crossed, since \(\mathcal T(s,\Psi)\) is entire. Setting \(t=\tfrac12-s\) then gives a line \(\operatorname{Re}t>1/2\), where the dual coefficient series converges absolutely.

For fixed \(h_0\) and active set, insert (104) into (103) and use (97) to sum over the nonzero \(h_p\). After the normalization in (101), the scalar \(C\) in (67) for this group of translates is \[C=-\frac{\mathrm i}{81}\overline{\alpha(c)}^{\,2} \widehat\phi(h_0)\overline{\kappa_0} \prod_{p\ {\rm inactive}}(1-\mathrm N_{K/\mathbb Q}(p)^{-1}) \prod_{p\ {\rm active}}\chi_p(\sigma_p)^{-2}\omega_{p,j_p}, \qquad |C|\le1/81 .\] Dividing (101) by its gamma factors and setting \(t=\tfrac12-s\) produces the gamma quotient in (64). Its numerator gamma factors have their first pole at \(t=-5/6\), and its reciprocal denominator gamma factors are entire. Thus no poles lie between the current contour \(\operatorname{Re}t>1/2\) and \(\operatorname{Re}t=0\). The rapid decay of \(\widehat V_*\), together with Stirling’s formula, allows us to shift each kernel contour to \(\operatorname{Re}t=0\). This gives the weight \(V_*^\sharp\) defined in (64), evaluated at \(\mathrm N_{K/\mathbb Q}(\ell)X/\mathrm N_{K/\mathbb Q}(c)^2\). Together with (97), this gives the dual sum (67).

For \(k_0\) as in Lemma 15, fix \(h_0\), the factors of \(\Psi\) other than \(\chi_{k_0}\), and the active/inactive choices at their primes. Every prime dividing \(k_0\) has exponent \(j_p=1\) and is therefore active, so \[r=k_0\prod_{p\in\mathcal A_{\rm fix}}p.\] Thus \(r/k_0\) is fixed, and fixing \(k_0\bmod M^2\) fixes \(r\bmod M^2\). The choices of \(H,c_0\) and the residues in (96) are therefore fixed in each such class. Thus \(d=d_\sigma\), \(\psi\), and \(c_0\) have precisely the asserted dependence on \(k_0\).

It remains to bound the transformed weight. The first numerator pole of the gamma quotient in (64) is at \(t=-5/6\). We may therefore shift the kernel contour to \(\operatorname{Re}t=-1/4\) for \(0<x\le1\), obtaining the factor \(x^{1/4}\), and to \(\operatorname{Re}t=A\) for \(x\ge1\), obtaining \(x^{-A}\). Each application of \(x\partial_x\) introduces a factor \(-t\). Stirling’s formula on \(-1/4\le\operatorname{Re}t\le A\) then gives, for every \(A>0\) and \(j\ge0\), \[ \begin{aligned} |(x\partial_x)^jV_*^\sharp(x)| &\ll_{A,j}\min\{x^{1/4},(1+x)^{-A}\}\\ &\quad\times\sup_{-A\le\eta\le1/4}\int_{\mathbb R} (1+|u|)^{\lceil4A\rceil+j+2}|\widehat V_*(\eta+\mathrm i u)|\,du. \end{aligned} \tag{105}\] When \(V_*\) is supported in a fixed compact interval \(I\subset(0,\infty)\), repeated integration by parts in its Mellin transform gives rapid decay in \(|u|\), uniformly for \(-A\le\eta\le1/4\). Thus, for a sufficiently large \(J=J(A,j)\), the supremum of integrals in (105) is \(\ll_{A,j,I}\|V_*\|_{C^J(I)}\). This proves (70). Thus (67) expresses \(T(X;\Psi)\) as \(O_{\Psi_0,S}(2^{|\mathcal P|})\) dual sums with \(|C|\le1/81\), the asserted ray class dependence of \((d,\psi,c_0)\), and the weight decay just established. Together with the support and coefficient bounds proved above, this completes the proof of Proposition 14 and Lemmas 15 and 16. ◻

Proof. Take the Fourier transform in logarithmic coordinates: \[b(\mathbf t)=\frac1{(2\pi)^d}\int_{\mathbb R^d} \mathcal K(e^{y_1},\ldots,e^{y_d}) e^{-\mathrm i\mathbf t\cdot\mathbf y}\,d\mathbf y.\] Applying Fourier inversion to this gives (106). Since the intervals \(I_j\) are fixed and bounded away from zero, the \(C^q\) norm in logarithmic coordinates is bounded by a constant times \(\|\mathcal K\|_{C^q(\mathcal I)}\). Applying \((1-\Delta_{\mathbf y})^{q/2}\) under the integral therefore gives \[|b(\mathbf t)|\ll_{\mathcal I,q,d} (1+|\mathbf t|)^{-q}\|\mathcal K\|_{C^q(\mathcal I)} \qquad(q\ge0\text{ even}).\] Multiplication by \((1+|\mathbf t|)^J\) and integration proves the bound when \(q>J+d\). See also the multivariable Mellin formulation in [39]. ◻

Proof. Choose a real \(V\in C_c^\infty(I_*)\) equal to one on \(I\), and set \(U_t(x)=V(x)x^{\mathrm i t}\). Apply (106) of Lemma 22 to each \(\mathcal K_r\) with \(d=2\) and \(\mathcal I=I^2\). Replacing the second Mellin variable by its negative and multiplying by \(V(x)V(y)\), which equals one on the support of \(\mathcal K_r\), gives \[\mathcal K_r(x,y)=\int_{\mathbb R^2}b_r(s,t) U_s(x)\overline{U_t(y)}\,ds\,dt,\] with the following uniform bound, obtained from the pointwise coefficient estimate in the proof of Lemma 22 with \(d=2\) and \(\mathcal I=I^2\): \[\sup_r|b_r(s,t)|\ll_{I,q} \frac{\sup_r\|\mathcal K_r\|_{C^q(I^2)}}{(1+|s|+|t|)^q} \qquad(q\ge0\text{ even}).\] The hypothesis applies to each \(U_t\), and \(\|U_t\|_{C^m(I_*)}\ll_{I,I_*,m}(1+|t|)^m\). Taking the common bound for \(b_r\) before integrating and applying Cauchy–Schwarz in \(r\) therefore bounds the absolute value in (109) by a constant depending on \(I,I_*,m,q\) times \[ \sqrt{M_1M_2}\sup_r\|\mathcal K_r\|_{C^q(I^2)} \int_{\mathbb R^2} \frac{(1+|s|)^m(1+|t|)^m}{(1+|s|+|t|)^q}\,ds\,dt. \tag{110}\] Taking \(q=2m+4\) makes the integral converge and proves the claim. ◻

Proof. For each fixed \(r\), weighted Cauchy–Schwarz gives \[\Bigl|\sum_\iota\int c_{\iota,r}(\mathbf t) F_{\iota,\mathbf t}(r)\,d\mathbf t\Bigr|^2 \le M\sum_\iota\int b_\iota(\mathbf t) |F_{\iota,\mathbf t}(r)|^2\,d\mathbf t.\] Multiply by \(w_r\) and sum over \(r\) to obtain (111). The same argument with sums in place of integrals proves the finite version. ◻