A theta expression for the coefficient sums

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson writes each coefficient sum \(B_m(Z)\) of The coefficient family and the plan of the zero-free argument as an average of the cubic theta function. The theta function is translated by a diagonal element \(a(y)\) at the places of \(S\) and by finitely many unipotent elements at the primes dividing \(m\); the unipotent translations, weighted by Fourier masks, produce the twist \(\chi_{\mathfrak A}(m)\). The average runs over a set of fixed norm modulo sixth powers of \(S\)-units, against a character with a nontrivial angular part, which removes the constant term. Unfolding the remaining Fourier series gives the coefficients of \(B_m\) multiplied by the profile \(V\) of Norm sheets and exact Gaussian smoothing, and the kernel \(\mathcal K\) of that lesson turns \(V\) into the Gaussian weight. The method is the reflection method of [QuasiRH]; Lesson 9 applies the Weyl element to the same average.

Prerequisites. S-integers and sixth-power classes, Lemma 2.1, Section 3, Lemma 4.1 and Definition 4.2; Sextic reciprocity and Kummer characters, Section 1 and Proposition 2.1; Gauss sums and the correction characters, Lemma 1.1 and Lemma 6.1; The coefficient family and the plan of the zero-free argument, (3.1), Lemma 3.1 and Section 4; Cubic theta and its Whittaker coefficients, N1, (1.2), (2.1), (2.3), Lemma 3.1, Proposition 4.1, Corollary 5.1 and Lemma 6.1; Norm sheets and exact Gaussian smoothing, Section 4, (5.1), Lemma 5.1 and Lemma 6.1; Dirichlet's unit theorem, Theorem 9.4.

1. Rows and their unit classes

Keep \(F\), \(S\), \(R=\mathcal O_S\), \(U_S=R^\times\), the class group \(\mathcal V=F_S^\times/F_S^{\times6}\) with its subgroups \(E\) and \(B\), and the allowed generators of Lesson 2. Put \[ \Gamma=U_S^6,\qquad U_6=U_S/\Gamma . \] The group \(U_6\) is finite, since \(U_S\) is finitely generated by the \(S\)-unit theorem (Theorem 9.4 of Dirichlet's unit theorem).

For a nonzero \(m\in R\) let \(\mathfrak m=mR\), choose a balanced allowed generator \(m_0\) of \(\mathfrak m\) (Definition 4.2 of Lesson 2), and write \(m=\varepsilon_mm_0\) with \(\varepsilon_m\in U_S\). Put \(\upsilon_m=\varepsilon_m\Gamma\in U_6\). For every prime \(\mathfrak p\mid\mathfrak m\) let \(j_{\mathfrak p}\in\{0,\dots,5\}\) be the residue of \(v_{\mathfrak p}(\mathfrak m)\) modulo \(6\). With the unit mask of Lesson 3, for every exterior ideal \(\mathfrak A\) with allowed generator \(A\), \[ \begin{aligned} \chi_{\mathfrak A}(m)&=\chi_{\mathfrak A}(\varepsilon_m)\prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(A)^{j_{\mathfrak p}} .\quad(1.1) \end{aligned} \] Indeed (2.2) of Lesson 3 gives \(\chi_{\mathfrak A}(m)=\chi_{\mathfrak A}(\varepsilon_m)\chi_{\mathfrak m}(A)\), and \(\chi_{\mathfrak p}(A)^{v_{\mathfrak p}(\mathfrak m)}=\chi_{\mathfrak p}(A)^{j_{\mathfrak p}}\) because \(\chi_{\mathfrak p}\) has order \(6\) on units and both sides are \(0\) when \(\mathfrak p\mid\mathfrak A\), also when \(j_{\mathfrak p}=0\).

Lemma 1.1 (rows [OpenAI-029, Section 6.1]). Let \(\mathscr K\subset F_S^\times\) be compact and let \(\mathscr B_Z(\mathscr K)\) be the row set of Section 4 of Lesson 5.

  1. The value \(B_m(Z)\) depends on \(\varepsilon_m\) only through \(\upsilon_m\).
  2. For every \(Z\) and every exterior ideal \(\mathfrak m\), the number of \(m\in\mathscr B_Z(\mathscr K)\) with \(mR=\mathfrak m\) is at most the finite number \(|U_S\cap\mathscr K\mathscr K^{-1}|\).

Proof. (1) By (2.3) of Lesson 3, \(\chi_{\mathfrak A}(\varepsilon)=H_S(\varepsilon,A)^{-1}\), which depends on \(\varepsilon\) only through its class in \(\mathcal V\); elements of \(\Gamma\) have trivial class. The other factors in (1.1) do not involve \(\varepsilon_m\).

(2) If \(m,m'\in\mathscr B_Z(\mathscr K)\) generate the same ideal, then \(m'/m\in U_S\), and \(m'/m\in(\ell_{Z^{M_0}}\mathscr K)(\ell_{Z^{M_0}}\mathscr K)^{-1}\)\(=\mathscr K\mathscr K^{-1}\), a compact subset of \(F_S^\times\), hence of \(F_S\). By Lemma 2.1 of Lesson 2, \(R\) is a lattice in \(F_S\), so it meets this compact set in finitely many points, and \(U_S\subseteq R\). Fixing one such \(m\), the map \(m'\mapsto m'/m\) is injective. \(\square\)

2. The twisting characters

For \(\varepsilon\in U_S\) define \(\lambda_\varepsilon(y)=H_S(\varepsilon,y)^{-1}\) on \(F_S^\times\), and with the character \(L_S\) of Lesson 4 put \[ L_\upsilon=L_S\,\lambda_\varepsilon\qquad(\upsilon=\varepsilon\Gamma\in U_6). \]

Lemma 2.1 ([OpenAI-029, (6.2)]). The character \(L_\upsilon\) depends only on \(\upsilon\), and the characters \(L_\upsilon\), \(\upsilon\in U_6\), form a finite family of smooth unitary characters of \(F_S^\times\). Each is trivial on \(U_S\), radially trivial at the complex places, and of angular exponent \(1\) or \(-1\) at each complex place. For every exterior ideal \(\mathfrak A\) with allowed generator \(A\), \[ L_\upsilon(A)=\vartheta(\mathfrak A)\,\chi_{\mathfrak A}(\varepsilon). \tag{2.1} \] Consequently, for every nonzero \(m\in R\), with \(A=cn^3\) the product of allowed generators \(c,n\) of \(\mathfrak c,\mathfrak n\), \[ \begin{gathered} \begin{aligned} B_m(Z)&=\sum_{\substack{\mathfrak c\ \text{squarefree}\\ \mathfrak n}}\frac{\gamma_2(\mathfrak c)}{q_{\mathfrak c}^{1/2}q_{\mathfrak n}}\\ &\quad \,L_{\upsilon_m}(A)\prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(A)^{j_{\mathfrak p}}\\ &\quad \,P_G\Bigl(\frac{q_{\mathfrak A}}Z\Bigr), \end{aligned}\\ \mathfrak A=\mathfrak c\mathfrak n^3 .\quad(2.2) \end{gathered} \]

Proof. The symbol \(H_S(\varepsilon,y)\) is bimultiplicative and depends only on the classes in \(\mathcal V\) (Section 3 of Lesson 2); so \(\lambda_\varepsilon\) is a character of finite order, locally constant, trivial at the complex places, and \(\lambda_{\varepsilon\gamma}=\lambda_\varepsilon\) for \(\gamma\in\Gamma\). For \(u\in U_S\), \(H_S(\varepsilon,u)=1\) because the image \(E\) of \(U_S\) satisfies \(E^\perp=E\) (Lemma 4.1 of Lesson 2); and \(L_S(u)=1\) by Lemma 6.1 of Lesson 4, which also gives the complex components of \(L_S\). By (6.2) of Lesson 4 and (2.3) of Lesson 3, \(L_\upsilon(A)=\vartheta(\mathfrak A)H_S(\varepsilon,A)^{-1}=\vartheta(\mathfrak A)\chi_{\mathfrak A}(\varepsilon)\). Finally, \(A=cn^3\) is an allowed generator of \(\mathfrak A\) (Definition 4.2 of Lesson 2), and (2.2) follows from (3.1) of Lesson 5, (1.1) and (2.1). \(\square\)

The right side of (2.2) does not depend on the choice of allowed generators: two of them differ by an element \(\gamma\in\Gamma\), with \(L_\upsilon(\gamma)=1\) and \(\chi_{\mathfrak p}(\gamma)=1\).

3. Fourier masks at the row primes

Let \(\mathfrak p\) be an exterior prime with allowed generator \(p\), residue field \(k_{\mathfrak p}=R/\mathfrak p\) and norm \(q_{\mathfrak p}\). The character \(\chi_{\mathfrak p}\) of Lesson 3 has order exactly \(6\) on the cyclic group \(k_{\mathfrak p}^\times\), and \(x\mapsto\psi_{\mathfrak p}(x/p)\) is a well-defined nontrivial additive character of \(k_{\mathfrak p}\), because \(\psi_{\mathfrak p}\) is trivial on \(\mathcal O_{\mathfrak p}\) and not on \(\mathfrak p^{-1}\) (Section 1 of Lesson 2). For \(1\le j\le5\) put \[ \begin{aligned} &\tau_{\mathfrak p,j}\\ &\quad =\sum_{x\in k_{\mathfrak p}^\times}\chi_{\mathfrak p}(x)^{-j}\,\psi_{\mathfrak p}(x/p),\quad(3.1) \end{aligned} \] a Gauss sum of a nontrivial character, so \(|\tau_{\mathfrak p,j}|=q_{\mathfrak p}^{1/2}\) by Lemma 1.1 of Lesson 4. Define the mask \(\alpha_{\mathfrak p,j}\) on \(k_{\mathfrak p}\), for \(1\le j\le5\) and for \(j=0\) respectively, by \[ \begin{aligned} \alpha_{\mathfrak p,j}(x)&=\begin{cases}\tau_{\mathfrak p,j}^{-1}\chi_{\mathfrak p}(x)^{-j},&x\ne0,\\ 0,&x=0,\end{cases}\\ \alpha_{\mathfrak p,0}(x)&=\begin{cases}1-q_{\mathfrak p}^{-1},&x=0,\\ -q_{\mathfrak p}^{-1},&x\ne0.\end{cases}\quad(3.2) \end{aligned} \]

Lemma 3.1 (masks [OpenAI-029, (6.4)]). For \(0\le j\le5\) and every \(\nu\in\mathcal O_{\mathfrak p}\), with the unit mask, \[ \begin{aligned} &\sum_{x\in k_{\mathfrak p}}\alpha_{\mathfrak p,j}(x)\,\psi_{\mathfrak p}(\nu x/p)\\ &\quad =\chi_{\mathfrak p}(\nu)^j .\quad(3.3) \end{aligned} \] The total mass \(\sum_x|\alpha_{\mathfrak p,j}(x)|\) is \((q_{\mathfrak p}-1)q_{\mathfrak p}^{-1/2}<q_{\mathfrak p}^{1/2}\) for \(j\ne0\) and \(2(1-q_{\mathfrak p}^{-1})<2\) for \(j=0\).

Proof. The left side depends only on \(\nu\) modulo \(\mathfrak p\). Let \(j\ne0\). For a unit \(\nu\), the substitution \(x\mapsto\nu^{-1}x\) gives \(\tau_{\mathfrak p,j}^{-1}\chi_{\mathfrak p}(\nu)^j\sum_{x\ne0}\chi_{\mathfrak p}(x)^{-j}\psi_{\mathfrak p}(x/p)\)\(=\chi_{\mathfrak p}(\nu)^j\). For \(\nu\in\mathfrak p\) the additive factor is \(1\), and \(\sum_{x\ne0}\chi_{\mathfrak p}(x)^{-j}=0\) because \(\chi_{\mathfrak p}^{-j}\) is nontrivial; the right side is \(0\) by the unit mask. For \(j=0\) the left side is \(1-q_{\mathfrak p}^{-1}\sum_{x\in k_{\mathfrak p}}\psi_{\mathfrak p}(\nu x/p)\), which is \(1\) for a unit \(\nu\) and \(0\) for \(\nu\in\mathfrak p\), the unit indicator. The masses are \((q_{\mathfrak p}-1)q_{\mathfrak p}^{-1/2}\) and \((1-q_{\mathfrak p}^{-1})+(q_{\mathfrak p}-1)q_{\mathfrak p}^{-1}\). \(\square\)

4. Translated theta functions

For each \(\upsilon\in U_6\), Section 4 of Lesson 7, applied to the finite family \(\{L_\upsilon\}\), gives one compact open subgroup \(U\) of the finite part of \(F_S^\times\), contained in the sixth powers and in the kernels of the finite components of all \(L_\upsilon\), and a vector \(\varphi_\upsilon\) of the cubic theta representation that is the distinguished spherical vector at every place outside \(S\) and whose \(S\)-component \(\mathcal W_{S,\upsilon}\) in the normalized factorization (3.1) of Lesson 6 satisfies \[ \begin{aligned} f_\upsilon(y)&:=\mathcal W_{S,\upsilon}(a(y))=\mathbf 1_U(y_f)\\ &\quad \prod_{v\mid\infty}|y_v|_v\,\omega_{m_v}(y_v)\\ &\quad \,K_{1/3}\bigl(c_v|y_v|_v^{1/2}\bigr), \end{aligned} \] with the angular exponents \(m_v\) of \(L_\upsilon\). In particular \[ \begin{aligned} &L_\upsilon(y)^{-1}f_\upsilon(y)=\mathbf 1_U(y_f)\prod_{v\mid\infty}|y_v|_v\\ &\qquad K_{1/3}\bigl(c_v|y_v|_v^{1/2}\bigr),\quad(4.1) \end{aligned} \] the integrand of the profile \(V\) of (6.1) in Lesson 7. Write \(\theta_\upsilon=\theta_{\varphi_\upsilon}\), and \(\theta_{\upsilon,N}\) for its constant term (N1(3)).

Fix a nonzero row \(m\), put \(\upsilon=\upsilon_m\), and consider digit vectors \(\mathbf x=(x_{\mathfrak p})_{\mathfrak p\mid\mathfrak m}\) with \(x_{\mathfrak p}\in k_{\mathfrak p}\), each represented by an element of \(\mathcal O_{\mathfrak p}\). For \(y\in F_S^\times\) let \(g(y,\mathbf x)\) be the adelic element with component \(a(y)\) at \(S\), component \(n(x_{\mathfrak p}/p)\) at each \(\mathfrak p\mid\mathfrak m\), and the identity at the other exterior places, all lifted by the sections of N1(1). Put \[ \begin{gathered} \boldsymbol\alpha_m(\mathbf x)=\prod_{\mathfrak p\mid\mathfrak m}\alpha_{\mathfrak p,j_{\mathfrak p}}(x_{\mathfrak p}),\\ w_+=1,\\ w_-=w, \end{gathered} \] and define the digit sums \[ \begin{gathered} \begin{aligned} \mathscr D_m^\pm(y)&=\sum_{\mathbf x}\boldsymbol\alpha_m(\mathbf x)\,\theta_\upsilon\bigl(w_\pm g(y,\mathbf x)\bigr), \end{aligned}\\ \begin{aligned} \mathscr D_{m,N}^\pm(y)&=\sum_{\mathbf x}\boldsymbol\alpha_m(\mathbf x)\\ &\quad \,\theta_{\upsilon,N}\bigl(w_\pm g(y,\mathbf x)\bigr). \end{aligned}\quad(4.2) \end{gathered} \] Here \(w\) is the rational Weyl element. The right side of (4.2) does not depend on the representatives of the digits, because \(n(\mathcal O_{\mathfrak p})\) fixes the spherical vector at \(\mathfrak p\).

Lemma 4.1 (descent [OpenAI-029, (6.6)–(6.7)]). \(\mathscr D_m^+=\mathscr D_m^-\). For every \(\gamma\in\Gamma\) and \(y\in F_S^\times\), each of the four functions in (4.2) takes the same value at \(\gamma y\) and at \(y\).

Proof. The theta function is invariant under the rational points of the cover (N1(3)), and \(w\) is rational; so \(\theta_\upsilon(wg)=\theta_\upsilon(g)\), which gives \(\mathscr D_m^+=\mathscr D_m^-\).

Let \(\gamma\in\Gamma\). For each digit \(x_{\mathfrak p}\) choose a representative \(x'_{\mathfrak p}\equiv\gamma x_{\mathfrak p}\pmod{\mathfrak p}\) and put \(u_{\mathfrak p}=(\gamma x_{\mathfrak p}-x'_{\mathfrak p})/p\in\mathcal O_{\mathfrak p}\). Let \(k\) be the adelic element with component \(1\) at \(S\), \(n(u_{\mathfrak p})a(\gamma)\) at \(\mathfrak p\mid\mathfrak m\), and \(a(\gamma)\) at the other exterior places. Since \(\gamma\) is a unit outside \(S\), every exterior component of \(k\) lies in the split compact subgroup of N1(1) and fixes the spherical vector; so right translation by \(k^{-1}\) fixes \(\varphi_\upsilon\). We claim \[ \begin{aligned} &g(\gamma y,\mathbf x')=a(\gamma)\,g(y,\mathbf x)\,k^{-1},\\ &w\,g(\gamma y,\mathbf x')\\ &\quad=a(\gamma^{-1})\,w\,g(y,\mathbf x)\,k^{-1},\quad(4.3) \end{aligned} \] with \(a(\gamma)\) the rational diagonal element. At \(S\): \(a(\gamma)a(y)=\mathcal C_S(\gamma,y)\,a(\gamma y)\) by (1.2) of Lesson 6, and \(\mathcal C_S(\gamma,y)=H_S(\gamma,y)^2=1\) by Lemma 6.1 there, since \(\gamma\) is a sixth power. At \(\mathfrak p\mid\mathfrak m\): by (2.1) of Lesson 6, \(a(\gamma)n(x_{\mathfrak p}/p)a(\gamma)^{-1}n(-u_{\mathfrak p})\)\(=n(\gamma x_{\mathfrak p}/p-u_{\mathfrak p})=n(x'_{\mathfrak p}/p)\). At the other exterior places both sides are \(1\). The second identity follows from the first and \(wa(\gamma)=a(\gamma^{-1})w\).

By rational invariance and \(\Theta(k^{-1})\varphi_\upsilon=\varphi_\upsilon\), (4.3) gives \(\theta_\upsilon(w_\pm g(\gamma y,\mathbf x'))=\theta_\upsilon(w_\pm g(y,\mathbf x))\). The constant term has the same invariance: for \(q\in F^\times\) and adelic \(h\), \[ \begin{aligned} \theta_{\upsilon,N}(a(q)h)&=\int_{F\backslash\mathbb A_F}\theta_\upsilon(n(t)a(q)h)\,dt\\ &=\int_{F\backslash\mathbb A_F}\theta_\upsilon(n(t/q)h)\,dt\\ &=\theta_{\upsilon,N}(h), \end{aligned} \] using \(n(t)a(q)=a(q)n(t/q)\), rational invariance, and the invariance of the measure on \(F\backslash\mathbb A_F\) under \(t\mapsto t/q\), which holds because \(|q|_{\mathbb A}=1\) by the product formula. Right translation by \(k^{-1}\) commutes with taking the constant term. Finally \(\chi_{\mathfrak p}(\gamma)=1\), so \(\alpha_{\mathfrak p,j}(\gamma x)=\alpha_{\mathfrak p,j}(x)\), and \(\mathbf x\mapsto\mathbf x'\) is a bijection of digit vectors. Reindexing the sums proves the lemma. \(\square\)

5. Averages over norm sheets and the constant term

Recall the norm sheets \(X_l=\ell_lF_S^1\) and their measures from (5.1) of Lesson 7. Since \(|\gamma|_S=1\) for \(S\)-units, \(\Gamma\) acts on each \(X_l\) by multiplication, preserving the measure. The quotient \(F_S^1/U_S\) is compact (proof of Lemma 4.1 of Lesson 2, balance), and \(\Gamma\) has finite index in \(U_S\); so \(X_l/\Gamma\) is compact. Choose a relatively compact measurable fundamental set \(\Omega_1\subset F_S^1\) for \(\Gamma\) and put \(\Omega_l=\ell_l\Omega_1\). An integral over \(X_l/\Gamma\) of a \(\Gamma\)-invariant function means its integral over \(\Omega_l\) with the sheet measure; it does not depend on the fundamental set.

The functions \(\mathbf 1_B([y])\), with \([y]\) the class of \(y\) in \(\mathcal V\), and \(L_\upsilon(y)^{-1}\) are \(\Gamma\)-invariant, because \([\gamma]=1\) and \(L_\upsilon(\gamma)=1\). The digit sums are continuous, and by Lemma 4.1 \(\Gamma\)-invariant.

Lemma 5.1 (angular cancellation [OpenAI-029, (6.8)]). For every \(l>0\), \[ \begin{aligned} &\int_{X_l/\Gamma}\mathbf 1_B([y])\,L_\upsilon(y)^{-1}\\ &\qquad \,\mathscr D_{m,N}^\pm(y)\,d_l^\times y=0 .\quad(5.1) \end{aligned} \]

Proof. Fix a complex place \(v_0\) and \(z\in\mathbb C\) with \(|z|=1\), and let \(z_0\in F_S^1\) have component \(z\) at \(v_0\) and \(1\) at the other places of \(S\). Multiplication by \(z_0\) maps \(X_l\) to itself, commutes with \(\Gamma\) and preserves the measure, so it preserves the integral of every integrable \(\Gamma\)-invariant function on \(X_l/\Gamma\).

The cover is split at complex places, so \(g(z_0y,\mathbf x)=a(z_0)g(y,\mathbf x)\), where \(a(z_0)\) acts on the left at \(v_0\) only, and \(wg(z_0y,\mathbf x)=a(z_0^{-1})wg(y,\mathbf x)\). By N1(3) the constant term depends radially on a complex diagonal element acting on the left; so \(\mathscr D_{m,N}^\pm(z_0y)=\mathscr D_{m,N}^\pm(y)\). Also \([z_0y]=[y]\), as \(\mathbb C^\times\) is divisible. But \(L_\upsilon(z_0y)^{-1}=z^{-m_{v_0}}L_\upsilon(y)^{-1}\) with \(m_{v_0}=\pm1\). If \(I\) is the left side of (5.1), these facts give \(I=z^{-m_{v_0}}I\); taking \(z=i\) gives \(I=0\). \(\square\)

Define the theta average \[ \begin{aligned} \mathcal T_m(l)&=\int_{X_l/\Gamma}\mathbf 1_B([y])\,L_\upsilon(y)^{-1}\\ &\quad \,\mathscr D_m^+(y)\,d_l^\times y\\ &=\int_{X_l/\Gamma}\mathbf 1_B([y])\,L_\upsilon(y)^{-1}\\ &\quad \,\mathscr D_m^-(y)\,d_l^\times y ,\quad(5.2) \end{aligned} \] the two expressions being equal by Lemma 4.1. Lesson 9 starts from the second one.

6. The theta representation

Let \(V\) and \(\mathcal K\) be the profile (6.1) and the kernel of Lemma 6.1 of Lesson 7, built with the subgroup \(U\) of Section 4.

Lemma 6.1 (theta representation [OpenAI-029, Lemma 6.1]). Let \(m\in R\) be nonzero, \(\upsilon=\upsilon_m\), and \(l,Z>0\). With \(\mathfrak A=\mathfrak c\mathfrak n^3\) and \(A=cn^3\) as in (2.2), \[ \begin{aligned} \mathcal T_m(l)&=\sum_{\substack{\mathfrak c\ \text{squarefree}\\ \mathfrak n}}\frac{\gamma_2(\mathfrak c)}{q_{\mathfrak c}^{1/2}q_{\mathfrak n}}\,L_\upsilon(A)\\ &\quad \prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(A)^{j_{\mathfrak p}}\,V(q_{\mathfrak A}l)\\ &=\sum_{\substack{\mathfrak c\ \text{squarefree}\\ \mathfrak n}}\frac{\gamma_2(\mathfrak c)}{q_{\mathfrak c}^{1/2}q_{\mathfrak n}}\,\vartheta(\mathfrak A)\\ &\quad \,\chi_{\mathfrak A}(m)\,V(q_{\mathfrak A}l),\quad(6.1) \end{aligned} \] \[ \begin{aligned} B_m(Z)&=\int_0^\infty\mathcal K(T)\,\mathcal T_m(T/Z)\\ &\quad \,\frac{dT}T .\quad(6.2) \end{aligned} \] The series converges absolutely, and for every \(\sigma>1/2\) the sum of the absolute values of its terms is at most \(C_\sigma\,l^{-\sigma}\zeta_F^S(\sigma+\frac12)\zeta_F^S(3\sigma+1)\). The integral (6.2), with the series inserted, converges absolutely.

Proof. Fourier expansion. By (2.3) and Lemma 3.1 of Lesson 6, for each digit vector, \[ \begin{aligned} \theta_\upsilon(g(y,\mathbf x))&=\theta_{\upsilon,N}(g(y,\mathbf x))\\ &\quad +\sum_{\nu\in F^\times}\mathcal W_{S,\upsilon}\bigl(a(\nu)a(y)\bigr)\\ &\quad \prod_{\mathfrak p\mid\mathfrak m}W_{\mathfrak p}\bigl(a(\nu)n(x_{\mathfrak p}/p)\bigr)\\ &\quad \prod_{\mathfrak q\notin S,\ \mathfrak q\nmid\mathfrak m}W_{\mathfrak q}(a(\nu)), \end{aligned} \] where \(a(\nu)a(y)\) is formed at \(S\). By (1.2) of Lesson 6, \(\mathcal W_{S,\upsilon}(a(\nu)a(y))=\mathcal C_S(\nu,y)f_\upsilon(\nu y)\). By (2.1) of Lesson 6, \(a(\nu)n(x)=n(\nu x)a(\nu)\), and the Whittaker functional transforms by \(\psi\) under \(n\); so \(W_{\mathfrak p}(a(\nu)n(x_{\mathfrak p}/p))\)\(=\psi_{\mathfrak p}(\nu x_{\mathfrak p}/p)W_{\mathfrak p}(a(\nu))\). By Proposition 4.1 of Lesson 6, \(W_{\mathfrak q}(a(\nu))=0\) when \(v_{\mathfrak q}(\nu)<0\); so only \(\nu\in R\setminus\{0\}\) contribute. For such \(\nu\), Lemma 3.1 evaluates the digit sum: \[ \sum_{\mathbf x}\boldsymbol\alpha_m(\mathbf x)\prod_{\mathfrak p\mid\mathfrak m}\psi_{\mathfrak p}(\nu x_{\mathfrak p}/p)=\prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(\nu)^{j_{\mathfrak p}} . \] Hence \[ \begin{aligned} \mathscr D_m^+(y)&=\mathscr D_{m,N}^+(y)\\ &\quad +\sum_{\nu\in R\setminus\{0\}}\mathcal C_S(\nu,y)\,f_\upsilon(\nu y)\\ &\quad \prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(\nu)^{j_{\mathfrak p}}\\ &\quad \prod_{\mathfrak q\notin S}W_{\mathfrak q}(a(\nu)).\quad(6.3) \end{aligned} \]

Support. Multiply by \(\mathbf 1_B([y])L_\upsilon(y)^{-1}\). If \(f_\upsilon(\nu y)\ne0\), the finite part of \(\nu y\) lies in \(U\), which consists of sixth powers, so \([\nu y]=1\). If also \([y]\in B\), then \([\nu]\in B\), and \[ \begin{aligned} \mathcal C_S(\nu,y)&=H_S(\nu,y)^2\\ &=\bigl(H_S(\nu,\nu^{-1})\,H_S(\nu,\nu y)\bigr)^2\\ &=1, \end{aligned} \] because \(H_S(\nu,\nu^{-1})=H_S(\nu,\nu)^{-1}=1\) (Lemma 3.1 of Lesson 2) and \([\nu y]=1\). An element \(\nu\in R\setminus\{0\}\) with \([\nu]\in B\) is an allowed generator of the integral ideal \(\mathfrak A=\nu R\). Fix one allowed generator \(A\) of each integral exterior ideal \(\mathfrak A\); since allowed generators are unique up to \(\Gamma\) (Lemma 4.1 of Lesson 2), the map \((\mathfrak A,\gamma)\mapsto\gamma A\) is a bijection from pairs of an integral exterior ideal and an element of \(\Gamma\) onto these \(\nu\). By Corollary 5.1 of Lesson 6 and \(\chi_{\mathfrak p}(\gamma)=1\), the exterior factors at \(\nu=\gamma A\) are those at \(A\), and \(\prod_{\mathfrak q}W_{\mathfrak q}(a(A))\) is \(\gamma_2(\mathfrak c)/(q_{\mathfrak c}^{1/2}q_{\mathfrak n})\) if \(\mathfrak A=\mathfrak c\mathfrak n^3\) with \(\mathfrak c\) squarefree and \(0\) otherwise.

Unfolding. For a fixed \(\mathfrak A\), the sets \(\gamma\Omega_l\), \(\gamma\in\Gamma\), partition \(X_l\) up to null sets. Substituting \(y\mapsto\gamma^{-1}y\) on each of them, and then \(v=Ay\), which maps \(X_l\) onto \(X_{q_{\mathfrak A}l}\) because \(|A|_S=q_{\mathfrak A}\) (Lemma 2.1 of Lesson 2) and carries one sheet measure to the other, \[ \begin{aligned} &\sum_{\gamma\in\Gamma}\int_{\Omega_l}\mathbf 1_B([y])L_\upsilon(y)^{-1}\\ &\qquad f_\upsilon(\gamma Ay)\,d_l^\times y=L_\upsilon(A)\\ &\qquad \int_{X_{q_{\mathfrak A}l}}\mathbf 1_B\bigl([A]^{-1}[v]\bigr)L_\upsilon(v)^{-1}\\ &\qquad f_\upsilon(v)\,d_{q_{\mathfrak A}l}^\times v=L_\upsilon(A)\,V(q_{\mathfrak A}l).\quad(6.4) \end{aligned} \] For the last equality, \([v]=1\) on the support of \(f_\upsilon\) and \([A]\in B\), so the indicator is \(1\); then (4.1) and (6.1) of Lesson 7 identify the integral. The same computation with absolute values, and the bound (5.2) of Lesson 7 for the sheet integral of \(|f_\upsilon|\), give \[ \begin{gathered} \begin{aligned} &\sum_{\gamma\in\Gamma}\int_{\Omega_l}|f_\upsilon(\gamma Ay)|\,d_l^\times y\\ &\quad =\int_{X_{q_{\mathfrak A}l}}|f_\upsilon(v)|\,d^\times_{q_{\mathfrak A}l}v\\ &\quad \le C_\sigma(q_{\mathfrak A}l)^{-\sigma} \end{aligned}\\ (\sigma\ge0).\quad(6.5) \end{gathered} \]

Exchanges. Integrate (6.3) against \(\mathbf 1_B([y])L_\upsilon(y)^{-1}\) over \(\Omega_l\). The constant term contributes \(0\) by Lemma 5.1. In the series, only the \(\nu\) with \([\nu]\in B\) contribute, as shown, and by (6.5) and the coefficient magnitudes the absolute series is at most \[ \begin{aligned} &\sum_{\mathfrak c,\mathfrak n}q_{\mathfrak c}^{-1/2}q_{\mathfrak n}^{-1}\,C_\sigma(q_{\mathfrak c}q_{\mathfrak n}^3l)^{-\sigma}\\ &\quad \le C_\sigma\,l^{-\sigma}\zeta_F^S(\sigma+\tfrac12)\zeta_F^S(3\sigma+1) \end{aligned} \] for \(\sigma>1/2\), as in Lemma 3.1 of Lesson 5. The digit sums are finite, and before their evaluation each digit vector carries the same majorant, since \(|\psi_{\mathfrak p}|=1\); the total digit mass is finite by Lemma 3.1. So the series may be integrated term by term, and (6.4) gives the first expression in (6.1). The second follows from (2.1) and (1.1): \(L_\upsilon(A)\prod_{\mathfrak p\mid\mathfrak m}\chi_{\mathfrak p}(A)^{j_{\mathfrak p}}\)\(=\vartheta(\mathfrak A)\chi_{\mathfrak A}(\varepsilon_m)\chi_{\mathfrak m}(A)\)\(=\vartheta(\mathfrak A)\chi_{\mathfrak A}(m)\).

The Gaussian weight. Insert (6.1) with \(l=T/Z\) into (6.2). Since \(V\ge0\) and \(V(u)\le C_\sigma u^{-\sigma}\) (Lemma 5.1 of Lesson 7), the absolute double sum and integral is at most \[ \begin{aligned} &\sum_{\mathfrak c,\mathfrak n}q_{\mathfrak c}^{-1/2}q_{\mathfrak n}^{-1}\int_0^\infty|\mathcal K(T)|\\ &\qquad \,C_\sigma\Bigl(\frac{q_{\mathfrak A}T}Z\Bigr)^{-\sigma}\frac{dT}T=C_\sigma Z^\sigma M_{-\sigma}(\mathcal K)\\ &\qquad \sum_{\mathfrak c,\mathfrak n}q_{\mathfrak c}^{-1/2-\sigma}q_{\mathfrak n}^{-1-3\sigma}<\infty , \end{aligned} \] with \(M_{-\sigma}(\mathcal K)\) finite by Lemma 6.1 of Lesson 7. Term by term, (6.3) of Lesson 7 with \(u=q_{\mathfrak A}/Z\) gives \(\int_0^\infty\mathcal K(T)V(q_{\mathfrak A}T/Z)\,dT/T=P_G(q_{\mathfrak A}/Z)\), and (2.2) gives (6.2). \(\square\)

The integral in (6.4) depends on \(A\) only through its norm, because \([A]\in B\) makes the class condition automatic; so one profile \(V\) serves every ideal \(\mathfrak A\) and every row.

7. Exercises

7.1. Show that if \(6\mid v_{\mathfrak p}(\mathfrak m)\) for a prime \(\mathfrak p\mid\mathfrak m\), the digit sum at \(\mathfrak p\) removes exactly the terms of (6.3) with \(\mathfrak p\mid\nu\), and leaves the others unchanged.

7.2. With \(\tau_{\mathfrak p,j}\) as in (3.1), show that \(\tau_{\mathfrak p,j}\,\tau_{\mathfrak p,6-j}=q_{\mathfrak p}\) for \(1\le j\le5\).

7.3. In Lemma 1.1(2), let \(\mathscr K=\{y:|y_v|_v\in[1,2]\text{ for all }v\in S\}\). Describe \(U_S\cap\mathscr K\mathscr K^{-1}\) and show that it contains the roots of unity of \(F\).

7.4. Explain why Lemma 5.1 needs only one complex place at which \(L_\upsilon\) has a nonzero angular exponent, and why the argument fails for a character that is radial at every complex place.

7.5. Let \(\mathfrak A\) be an integral exterior ideal that is not of the form \(\mathfrak c\mathfrak n^3\) with \(\mathfrak c\) squarefree. Show that no frequency \(\nu\) generating \(\mathfrak A\) contributes to (6.3).

8. Solutions

7.1. With \(j_{\mathfrak p}=0\), (3.3) gives the factor \(\chi_{\mathfrak p}(\nu)^0\), which with the unit mask is \(1\) if \(\mathfrak p\nmid\nu\) and \(0\) if \(\mathfrak p\mid\nu\).

7.2. By (3.1), \(\tau_{\mathfrak p,6-j}=\sum_{x\ne0}\chi_{\mathfrak p}(x)^{j}\psi_{\mathfrak p}(x/p)\), the Gauss sum of the inverse character. Lemma 1.1 of Lesson 4 gives the product \(\chi_{\mathfrak p}(-1)^jq_{\mathfrak p}\), and \(\chi_{\mathfrak p}(-1)=1\) by (1.3) of Lesson 3.

7.3. The set \(\mathscr K\mathscr K^{-1}\) consists of the \(y\) with \(|y_v|_v\in[\frac12,2]\) at every \(v\in S\). An \(S\)-unit has absolute value \(1\) outside \(S\), so \(U_S\cap\mathscr K\mathscr K^{-1}\) is the set of \(S\)-units all of whose absolute values at \(S\) lie in \([\frac12,2]\); it is finite by Lemma 1.1(2). A root of unity has absolute value \(1\) at every place.

7.4. The proof rotates \(y\) at one place \(v_0\); it needs \(z^{-m_{v_0}}\ne1\) for some \(z\) of modulus one, that is \(m_{v_0}\ne0\). For a character that is radial at every complex place the rotation leaves the whole integrand unchanged and gives no information. Lemma 2.1 provides exponents \(\pm1\) at every complex place.

7.5. The ideals \(\mathfrak c\mathfrak n^3\) with \(\mathfrak c\) squarefree are exactly those whose valuations are all congruent to \(0\) or \(1\) modulo \(3\). So some prime \(\mathfrak q\) has \(v_{\mathfrak q}(\mathfrak A)\equiv2\pmod3\), and by Proposition 4.1 of Lesson 6, \(W_{\mathfrak q}(a(\nu))=0\) for every generator \(\nu\) of \(\mathfrak A\), whatever its unit part. The corresponding term of (6.3) is \(0\).

Sources