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The level formula and Type I sums

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

A Type I sum has a short outer variable \(r\) with arbitrary coefficients and a long, smoothly weighted inner variable \(u\): \[ \sum_ra_r\sum_uG(ru)\,W_r\Bigl(\frac{N(u)}U\Bigr). \] For a single \(r\) the inner sum is a smoothed partial sum of the Dirichlet series \(\sum_uG(ru)N(u)^{-s}\), which has a pole at \(s=5/6\). Dunn and Radziwiłł proved a summation formula for it, a Voronoi formula in the level \(r\), whose main term is that pole. This lesson shows that the main term is exactly the contribution of the model \(c_*\mu^2(ru)N(ru)^{-1/6}\). It then bounds the Type I sums with the model subtracted, and also after twisting by \(N(u)^{it}\) at large heights \(t\). Everything is done for every fixed angular order \(\ell\), as the last lesson needs the twisted sums.

Prerequisites. The previous lessons, Cubic Gauss sums and Patterson's conjecture and Large sieves for cubic characters, with their notation. From The Riemann zeta function: The Gamma function and Stirling's formula, Theorem 2.2 (reflection) and Theorem 3.1 (Stirling).

1. Notation

For \(\ell\in\mathbb Z\) and \(z\ne0\) put \(\theta_\ell(z)=(z/|z|)^\ell\), and for primary \(n\) \[ \begin{gathered} G_\ell(n)=\theta_\ell(n)G(n),\\ \mathcal D_\ell(n)=G_\ell(n)-\frac{c_*\,\theta_\ell(n)\,\mu^2(n)}{N(n)^{1/6}}. \end{gathered} \] The function \(\theta_\ell\) is multiplicative, and \(\mathcal D_\ell(n)=0\) unless \(n\) is squarefree. Put \[ A_0=\frac{(2\pi)^{5/3}}{3^{7/2}\,\Gamma(2/3)} . \] The primary elements form the coset \(1+3\mathcal O\), of index \(9\) in \(\mathcal O\), so they have density \(\frac2{9\sqrt3}\) per unit of Euclidean area. In the variable \(N(z)=|z|^2\) this becomes \(\frac{2\pi}{9\sqrt3}\) per unit length. One checks directly that \[ c_*\cdot\frac{2\pi}{9\sqrt3}=A_0 . \tag{1.1} \] Indeed both sides equal \((2\pi)^{5/3}/(3^{3}\sqrt3\,\Gamma(2/3))\).

2. The level Voronoi formula

For primary squarefree \(r\), \(R=N(r)\), a smooth function \(W\) with compact support in \((0,\infty)\), \(V>0\) and \(\ell\in\mathbb Z\), put \[ \begin{aligned} &\mathcal C_{r,\ell}(V;W)\\ &=\sum_{\substack{d,u\\ (d,r)=1}}|d|\,G(ru)\,\theta_\ell(rud^3)\\ &\qquad\cdot W\Bigl(\frac{N(ud^3)}V\Bigr). \end{aligned} \] The variable \(d\) completes the sum over \(u\) by cubes; the completion is removed in Section 4.

Theorem 2.1 (the level Voronoi formula, N2). For \(0<\sigma<10^{-4}\) and \(k=|\ell|/2\) put \[ \begin{gathered} Q_\ell(z)=\frac{\Gamma(\frac56+k-z)\Gamma(\frac76+k-z)}{\Gamma(z+k-\frac16)\Gamma(z+k+\frac16)},\\ \check W_\ell(v)=\frac1{2\pi i}\int_{(-\sigma)}v^zQ_\ell(z)\,\mathcal MW(z)\,dz . \end{gathered} \] Then \[ \begin{aligned} &\mathcal C_{r,\ell}(V;W)\\ &=\mathbf 1_{\ell=0}\,A_0\,\mathcal MW(\tfrac56)\,V^{5/6}\frac{\varphi(r)}{R^{7/6}}\\ &\quad+\frac{R^{1/2}\,\theta_{3\ell}(r)}{3^{7/2}(2\pi)^2}\,\Sigma_{r,\ell}(V;W), \end{aligned} \] with the dual sum \[ \begin{aligned} \Sigma_{r,\ell}&=\sum_{\substack{\nu,\,d\\ (d,r)=1}}\frac{a_r(\nu)\,b_r(\nu)\,\theta_{-\ell}(d^3\nu)}{N(\nu)\,N(d)^{5/2}}\\ &\qquad\cdot\check W_\ell\Bigl(\frac{(2\pi)^4N(d^3\nu)V}{R^2}\Bigr), \end{aligned} \] over \(0\ne\nu\in\lambda^{-1}\mathcal O\) and primary \(d\). The coefficients \(a_r(\nu),b_r(\nu)\) depend only on \(r\) and \(\nu\). They vanish unless \[ \nu=\lambda^j\zeta\,h\,w\,h'^3,\qquad j\ge-1, \] with a unit \(\zeta\), primary \(h,h',w\), \(h\) and \(h'\) composed of primes dividing \(r\), \(w\) prime to \(r\), and \(hw\) squarefree. On this support, \[ \begin{gathered} |a_r(\nu)|\ll3^{\max(j,0)/3}\,|h'|,\\ |b_r(\nu)|\le N\bigl((\lambda\nu,r)\bigr). \end{gathered} \]

Theorem 2.1 is Proposition 5.3 of [DR], multiplied by \(G(r)\theta_\ell(r)\), and it is used without proof in the programme. Its left side becomes \(\mathcal C_{r,\ell}\) because \(G(ru)=\overline{\chi_r(u)}G(r)G(u)\) for \(u\) prime to \(r\), and both sides vanish otherwise. In the main term and the dual factor, \(G(r)\overline{g(r)}=R^{1/2}\) and \(\theta_\ell(r)(\bar r/r)^{-\ell}=\theta_{3\ell}(r)\). The coefficients are \(a_r=a^\dagger\) and \(b_r=b_r^\dagger\) of Proposition 5.2 of [DR], with the support stated there. The bound for \(b_r\) is immediate from its formula \(b_r(\nu)=\mu(r/(\lambda\nu,r))\varphi(r)/\varphi(r/(\lambda\nu,r))\). For \(j\ge0\), \(a_r(\nu)\) is the conjugate of a Fourier coefficient \(\tau(\mu)\) of Patterson's cubic theta function, whose formula is recorded in Section 5.1 of [DR]. These coefficients are Gauss sums of absolute value \(|c|\) times \(|d/c|\,3^{n/2+O(1)}\) at arguments \(\lambda^{3n-3}cd^3\) or \(\lambda^{3n-4}cd^3\), up to units. At \(\nu\) as above the argument has \(d=h'\) and \(n\le\frac23j+O(1)\), which gives the bound. For \(j=-1\) the two other cusp coefficients \(\tau_1,\tau_2\) are bounded by \(9|d|\).

Lemma 2.2 (the truncated formula). Let \(\ell\) be fixed, and suppose that \(V,V^{-1},R\le X^{C_0}\), that \(W\) is supported in a fixed compact subinterval of \((0,\infty)\), and that its derivatives of each fixed order are bounded by a fixed power of \(L\). Then \[ \begin{aligned} &\mathcal C_{r,\ell}(V;W)\\ &=\mathbf 1_{\ell=0}A_0\,\mathcal MW(\tfrac56)\,V^{5/6}\frac{\varphi(r)}{R^{7/6}}\\ &\quad+O_{\ell,\epsilon}\bigl(X^\epsilon R^{1/2}\bigr). \end{aligned} \]

Proof. On the line \(\operatorname{Re}z=-\sigma\), Stirling's formula gives \(|Q_\ell(z)|\ll_{\ell,\sigma}(1+|\operatorname{Im}z|)^{2+4\sigma}\). Indeed each of the two quotients \(\Gamma(a-z)/\Gamma(z+b)\) has size \(|\operatorname{Im}z|^{1+2\sigma}\). With the decay of \(\mathcal MW\) from Lemma 1.2 of the previous lesson, \[ \begin{aligned} &\Bigl|\check W_\ell\Bigl(\frac{(2\pi)^4N(d^3\nu)V}{R^2}\Bigr)\Bigr|\\ &\qquad\ll X^{O(\sigma)}L^{O(1)}N(d^3\nu)^{-\sigma}. \end{aligned} \] It remains to bound \(\sum|a_rb_r|N(\nu)^{-1-\sigma}N(d)^{-5/2-3\sigma}\). It factors over the parts of \(\nu\).

Choosing \(\sigma\) and \(\eta\) small in terms of \(\epsilon\) gives the bound \(X^\epsilon R^{1/2}\) for the dual term. \(\square\)

3. Lattice sums over squarefree elements

Lemma 3.1. Let \(r\) be primary and squarefree, \(U\ge1\) bounded by a power of \(X\), and \(W\) smooth with fixed compact support in \((0,\infty)\), with derivatives of fixed order bounded by fixed powers of \(L\).

(a) \[ \begin{aligned} &\sum_{(u,r)=1}\mu^2(u)\,W\Bigl(\frac{N(u)}U\Bigr)\\ &=\frac{2\pi U}{9\sqrt3}\,\mathcal MW(1)\,\frac{\varphi(r)}{N(r)}\sum_{(c,r)=1}\frac{\mu(c)}{N(c)^2}\\ &\quad+O_\epsilon(X^\epsilon\sqrt U). \end{aligned} \]

(b) For every fixed integer \(\ell\ne0\), \[ \begin{aligned} &\sum_{(u,r)=1}\mu^2(u)\,\theta_\ell(u)\,W\Bigl(\frac{N(u)}U\Bigr)\\ &\qquad\ll_{\ell,\epsilon}X^\epsilon\sqrt U. \end{aligned} \]

Proof. For primary \(u\), the indicator that \(u\) is squarefree and prime to \(r\) equals \[ \sum_{\substack{d\mid r\\d\mid u}}\mu(d)\sum_{\substack{c^2\mid u\\ (c,r)=1}}\mu(c), \] and the conditions say \(u=c^2dv\) with primary \(v\). For a smooth function \(f\) on \(\mathbb C\) supported in an annulus of radii comparable to \(\sqrt Y\), with \(|\nabla f|\ll L^C/\sqrt Y\), comparing each fundamental cell of \(1+3\mathcal O\) with its integral gives \[ \begin{aligned} \sum_{v\in1+3\mathcal O}f(v)&=\frac2{9\sqrt3}\int_{\mathbb C}f\,dA\\ &\quad+O\bigl(L^C(1+\sqrt Y)\bigr). \end{aligned} \] (a) For \(f(z)=W(|z|^2/Y)\) the integral is \(\pi Y\mathcal MW(1)\). With \(Y=U/N(c^2d)\), and \(N(c)\ll\sqrt U\) for nonempty sums, \[ \sum_{d\mid r}\frac{\mu(d)}{N(d)}=\frac{\varphi(r)}{N(r)} , \] and completing the sum over \(c\) costs \(O(\sqrt U)\). The errors add up to \(O(\tau(r)L^C\sqrt U\log U)\).

(b) Here \(\theta_\ell(u)=\theta_\ell(c^2d)\theta_\ell(v)\), and \(f(z)=\theta_\ell(z)W(|z|^2/Y)\) has integral zero, because \(\int_0^{2\pi}e^{i\ell\phi}d\phi=0\). Its gradient is \(O_\ell(L^C/\sqrt Y)\) on its support, so only the error terms remain. \(\square\)

4. Type I sums at bounded heights

Proposition 4.1 (Type I comparison). Let \(\ell\) be fixed, \(RU\asymp X\) and \(1\le R\ll X^{0.51}\). Let \((a_r)\) be divisor-bounded and supported on \(r\) of length \(R\). Let \(W_r\) be smooth functions supported in a common fixed compact subinterval of \((0,\infty)\), whose derivatives of each fixed order are bounded by a fixed power of \(L\), uniformly in \(r\). Then \[ \begin{aligned} &\sum_ra_r\sum_u\mathcal D_\ell(ru)\,W_r\Bigl(\frac{N(u)}U\Bigr)\\ &\qquad\ll_\ell X^{5/6-1/100}. \end{aligned} \]

Proof. Only squarefree \(r\) contribute. For such \(r\) we have the exact identity \[ \begin{aligned} &\sum_uG_\ell(ru)\,W_r\Bigl(\frac{N(u)}U\Bigr)\\ &=\sum_{(c,r)=1}\mu(c)\,|c|\,\theta_{3\ell}(c)\\ &\qquad\cdot\mathcal C_{r,\ell}\Bigl(\frac U{N(c)^3};W_r\Bigr).\quad(4.1) \end{aligned} \] Indeed, expanding the right side and putting \(e=cd\), the term with \(u\) and \(e\) carries \(|e|\,\theta_\ell(rue^3)\sum_{c\mid e}\mu(c)\), which vanishes unless \(e=1\). All sums are finite. Let \[ \begin{gathered} C_*=X^{0.13}\ \text{ if }R\le X^{0.40},\\ C_*=X^{0.02}\ \text{ if }R>X^{0.40}. \end{gathered} \]

Short completions. For \(N(c)\le C_*\) apply Lemma 2.2. Its error terms, summed over \(c\) with \(|c|=N(c)^{1/2}\) and over \(r\), give \[ X^\epsilon R^{3/2}C_*^{3/2}\ll X^{0.795+\epsilon} \] in both ranges of \(R\).

Main terms and the model, \(\ell=0\). The main terms of Lemma 2.2, completed to all \(c\), give \[ \begin{aligned} &A_0\,\mathcal MW_r(\tfrac56)\,U^{5/6}\frac{\varphi(r)}{R^{7/6}}\\ &\qquad\cdot\sum_{(c,r)=1}\frac{\mu(c)}{N(c)^2},\quad(4.2) \end{aligned} \] with a completion error \(O(X^{5/6+\epsilon}C_*^{-1})\) after summing over \(r\). On the other side, the model is \[ \begin{aligned} &\sum_uc_*\frac{\mu^2(ru)}{N(ru)^{1/6}}W_r\Bigl(\frac{N(u)}U\Bigr)\\ &=\frac{c_*}{(RU)^{1/6}}\sum_{(u,r)=1}\mu^2(u)\,\widetilde W_r\Bigl(\frac{N(u)}U\Bigr), \end{aligned} \] with \(\widetilde W_r(x)=x^{-1/6}W_r(x)\), up to the factor \((N(r)/R)^{-1/6}\), which is put into \(\widetilde W_r\). Lemma 3.1(a), with \(\mathcal M\widetilde W_r(1)=\mathcal MW_r(\frac56)\), and the identity (1.1) show that the model equals (4.2) up to \(O(X^\epsilon R^{-1/6}U^{1/3})\). Summed over \(r\), this is \[ X^\epsilon R^{5/6}U^{1/3}\asymp X^{1/3+\epsilon}R^{1/2}\ll X^{0.589+\epsilon}. \]

Main terms and the model, \(\ell\ne0\). Lemma 2.2 has no main term, and Lemma 3.1(b) bounds the model by the same \(X^{0.589+\epsilon}\).

Long completions. It remains to bound the terms of (4.1) with \(N(c)>C_*\). Combine \(c\) and \(d\) into \(e=cd\) and restrict \(N(e)\) to a dyad of length \(E\gg C_*\). The coefficient of \(e\) is \[ |e|\,\theta_{3\ell}(e)\sum_{\substack{c\mid e\\N(c)>C_*}}\mu(c), \] divisor-bounded times \(E^{1/2}\). The inner variable \(u\) then has length \(U/E^3\). Bounding \(|G|\le1\) gives \[ \ll X^{1+\epsilon}E^{-3/2}, \] which is \(O(X^{0.805+\epsilon})\) when \(R\le X^{0.40}\). When \(R>X^{0.40}\), fix \(e\) and write \(G(ru)=\overline{\chi_u(r)}G(r)G(u)\). Separate the weights \(W_r\) by Mellin inversion: the dependence on \(r\) passes into the outer coefficients, with a common integrable majorant. By Cauchy's inequality in \(r\) and Theorem 4.1 of the previous lesson, both variables being squarefree, the sum over \(r,u\) is \[ \begin{gathered} \ll X^\epsilon\sqrt{RU_e}\,\bigl(R+U_e+(RU_e)^{2/3}\bigr)^{1/2},\\ U_e=U/E^3 . \end{gathered} \] Summing the coefficients of the \(e\) in the dyad costs \(X^\epsilon E^{3/2}\), so the dyad contributes \[ \ll X^{1/2+\epsilon}\Bigl(R+\frac U{E^3}+\Bigl(\frac X{E^3}\Bigr)^{2/3}\Bigr)^{1/2}. \] For \(X^{0.40}<R\ll X^{0.51}\) and \(E\gg X^{0.02}\), the three terms have exponents at most \[ \begin{gathered} 0.755,\qquad 0.77,\\ \frac12+\frac{1-0.06}3=\frac56-\frac1{50}. \end{gathered} \] The completion error of the main term, \(X^{5/6+\epsilon}C_*^{-1}\), has the same last exponent in this range. All the exponents are below \(\frac56-\frac1{100}\) after the small losses, and there are \(O(L)\) dyads. \(\square\)

5. Type I sums at large heights

At heights \(T\) up to \(X^{1/6+\rho}\), the model term is treated separately (in the lesson Bilinear sums and the sharp cutoff), and the Gauss sums are bounded alone. The tool is a mean value over \(t\) of the completed sums twisted by \(N(\cdot)^{it}\): \[ \begin{aligned} &\mathcal C_{r,\ell,t}(V;W)\\ &=\sum_{\substack{d,u\\ (d,r)=1}}|d|\,G(ru)\,\theta_\ell(rud^3)\\ &\qquad\cdot N(ud^3)^{it}\,W\Bigl(\frac{N(ud^3)}V\Bigr), \end{aligned} \] \[ \begin{aligned} &M_{r,\ell,t}(V;W)\\ &=\mathbf 1_{\ell=0}\,A_0\,\mathcal MW(\tfrac56+it)\\ &\qquad\cdot V^{5/6+it}\,\frac{\varphi(r)}{R^{7/6}} . \end{aligned} \]

Lemma 5.1 (completed sums at large heights). Let \(\ell\) be fixed, \(r\) primary and squarefree, \(R=N(r)\), and let \(W\) be supported in a fixed compact subinterval of \((0,\infty)\), with derivatives bounded independently of \(X\). For \(T\ge2\), \(V\gg1\) and \(R,V,T\le X^{O(1)}\), \[ \begin{aligned} &\int_{T\le|t|\le2T}\bigl|\mathcal C_{r,\ell,t}(V;W)\\ &\qquad-M_{r,\ell,t}(V;W)\bigr|\,\frac{dt}T\\ &\ll_{\ell,\epsilon}X^\epsilon\sqrt V\,R^{1/4}\sqrt T . \end{aligned} \]

Proof. Short sums: \(V\le\sqrt R\,T^2\). Collect the terms with equal \(m=N(ud^3)\). By Lemma 1.1 of the previous lesson the squared coefficients sum to \[ \begin{aligned} &\ll X^\epsilon\sum_dN(d)\,\#\{u:N(ud^3)\asymp V\}\\ &\ll X^\epsilon V\sum_dN(d)^{-2}\ll X^\epsilon V . \end{aligned} \] The mean value inequality (Lemma 5.1 of the previous lesson) and Cauchy's inequality bound the mean of \(|\mathcal C_{r,\ell,t}|\) by \(X^\epsilon\sqrt V(1+V/T)^{1/2}\le X^\epsilon\sqrt VR^{1/4}\sqrt T\). For \(\ell=0\), the subtracted term is \(O_D(V^{5/6}R^{-1/6}T^{-D})\) by the decay of \(\mathcal MW\). Its ratio to the claimed bound is at most \(V^{1/3}R^{-5/12}T^{-D-1/2}\le R^{-1/4}T^{1/6-D}\le1\) for \(D\ge1\).

Long sums: \(V>\sqrt R\,T^2\). Apply Theorem 2.1 to \(W_t(x)=W(x)x^{it}\), whose Mellin transform is \(\mathcal MW(z+it)\), and multiply by \(V^{it}\). The main term is \(M_{r,\ell,t}\). In \(\check W_\ell\), move the line of integration to \(\operatorname{Re}z=-A\). There \(|Q_\ell(z)|\ll(1+|\operatorname{Im}z|)^{2+4A}\), and the decay of \(\mathcal MW(z+it)\) in \(\operatorname{Im}z+t\) bounds the transform at \(v=(2\pi)^4N(d^3\nu)V/R^2\) by \[ \ll v^{-A}T^{2+4A}=T^2\Bigl(\frac{R^2T^4}{(2\pi)^4N(d^3\nu)V}\Bigr)^{A}. \] For a fixed small \(\delta>0\) and large \(A\), the terms with \(N(d^3\nu)>I_{\max}=X^\delta R^2T^4/V\) are therefore smaller than every fixed power of \(X^{-1}\). This holds after summing the absolutely convergent coefficients of Lemma 2.2. The cutoff does not depend on \(t\). The retained frequencies satisfy \(N(d^3\nu)\le I_{\max}\le X^\delta R^{3/2}T^2\).

For the retained terms, move the line back to \(\operatorname{Re}z=\frac12\). No pole is crossed, since the poles of \(\Gamma(\frac56+\frac{|\ell|}2-z)\) lie at \(\operatorname{Re}z\ge\frac56\). On this line \(|Q_\ell(\frac12+iy)|=1\), because each quotient \(\Gamma(a-iy)/\Gamma(a+iy)\) with real \(a\) has absolute value one. After the substitution \(y\mapsto y-t\), the transform is an integral of \(\mathcal MW(\frac12+iy)\) against \(N(d^3\nu)^{i(y-t)}\) times factors of absolute value one. Minkowski's inequality in \(y\) reduces the claim to the mean over \(t\) of Dirichlet polynomials in \(N(d^3\nu)\). Their coefficients have absolute value a constant times \[ \sqrt V\,R^{-1/2}\,\frac{|a_r(\nu)b_r(\nu)|}{N(d)\sqrt{N(\nu)}} . \]

Fix \(d,j,\zeta,h,h'\) in \(\nu=\lambda^j\zeta hwh'^3\), and put \(D_d=N(d)\), \(E=N(h)\), \(P'=N(h')\) and \(g=N((\lambda\nu,r))\le\min(R,EP')\). The free squarefree variable \(w\) has length at most \(Y'=I_{\max}/(D_d^3\,3^jEP'^3)\). By the bounds of Theorem 2.1 its coefficients are \(N(w)^{-1/2}\) times \[ \sqrt V\,R^{-1/2}\,\frac{3^{-j/6}\,g}{D_d\,P'\sqrt E}. \] By the mean value inequality on dyads of \(N(w)\), the mean over \(t\) is at most \(X^\epsilon\) times this factor times \(1+\sqrt{Y'/T}\).

For the term \(1\), use \(g\le\sqrt{REP'}\). The sum over \(j\) is geometric and the sum over \(d\) is \(O(L)\). There are at most \(\tau(r)\) choices of \(h\), since \(h\) is squarefree and divides \(r\), and \(\sum_{h'}P'^{-1/2}\ll R^\epsilon\). The total is \(O(X^\epsilon\sqrt V)\).

For the term \(\sqrt{Y'/T}\), use \(g\le EP'\). The total is at most \[ \begin{aligned} &X^\epsilon\sqrt V\sqrt{\frac{I_{\max}}{RT}}\sum_{d,j,h,h'}\frac{3^{-2j/3}}{D_d^{5/2}P'^{3/2}}\\ &\qquad\ll X^{\epsilon+\delta}\sqrt V\,R^{1/4}\sqrt T . \end{aligned} \] Choosing \(\delta\) small proves the lemma. \(\square\)

Proposition 5.2 (Type I at large heights). Let \(\ell\) be fixed, \(RU\asymp X\), \((a_r)\) divisor-bounded on \(r\) of length \(R\), and let \(W_r\) be smooth with a common fixed compact support in \((0,\infty)\) and derivatives bounded independently of \(X\). For \(2\le T\le X^{O(1)}\) and every fixed \(D>0\), \[ \begin{aligned} &\int_{T\le|t|\le2T}\sum_r|a_r|\\ &\quad\cdot\Bigl|\sum_uG_\ell(ru)N(u)^{it}W_r\Bigl(\frac{N(u)}U\Bigr)\Bigr|\frac{dt}T\\ &\ll_{\ell,\epsilon}X^{1/2+\epsilon}R^{3/4}T^{1/2}\\ &\quad+\mathbf 1_{\ell=0}O_D\bigl(X^{5/6}L^CT^{-D}\bigr), \end{aligned} \] where \(C\) depends only on the coefficient bounds.

Proof. Only squarefree \(r\) contribute. Twisting (4.1) by \(N(u)^{it}\) gives the same identity with \(\mathcal C_{r,\ell,t}\) and the extra factor \(N(c)^{3it}\). If \(b_0\) bounds the supports of the \(W_r\), only \(N(c)\le(b_0U)^{1/3}\) contribute, and then \(V=U/N(c)^3\ge1/b_0\). By Lemma 5.1 the error terms contribute \[ \begin{aligned} &\sum_r|a_r|\sum_{N(c)\ll U^{1/3}}|c|\,X^\epsilon\sqrt{\frac U{N(c)^3}}\\ &\qquad\cdot R^{1/4}\sqrt T\ll X^\epsilon RL^C\sqrt UR^{1/4}\sqrt T, \end{aligned} \] which is \(X^{1/2+\epsilon}R^{3/4}T^{1/2}\). For \(\ell=0\), the terms \(M_{r,0,t}\) are \(O_D(U^{5/6}N(c)^{-5/2}R^{-1/6}T^{-D})\). The sum over \(c\) converges, and summing over \(r\) gives \(X^{5/6}L^CT^{-D}\). \(\square\)

Remark 5.3. For \(\ell=0\) the same bound can be obtained from Heath-Brown's mean-square estimate for the Dirichlet series \(\sum_uG(ru)N(u)^{-s}\) on the line \(\operatorname{Re}s=\frac12+\epsilon\); this is the route of [OpenAI-023], Proposition 3.2. The proof above uses only the level formula.

Two ranges of Proposition 5.2 are used later. With \(R\le X^{0.40}\) and \(T\ll X^{0.01}\), the exponent of the first term is at most \(\frac12+\frac34(0.40)+\frac12(0.01)=0.805\). With \(R\le X^{1/3-\kappa/2}\) and \(T\ll X^{1/6+\rho}\), it is \[ \begin{aligned} &\frac12+\frac34\Bigl(\frac13-\frac\kappa2\Bigr)+\frac12\Bigl(\frac16+\rho\Bigr)\\ &\qquad=\frac56-\frac{3\kappa}8+\frac\rho2 , \end{aligned} \] which is below \(\frac56\) when \(\rho<\frac34\kappa\).

6. Exercises

6.1. Verify the identity (1.1).

6.2. Show that \(|Q_\ell(\frac12+iy)|=1\) for all real \(y\), and that on the line \(\operatorname{Re}z=-\sigma\) the function \(Q_\ell\) grows like \(|\operatorname{Im}z|^{2+4\sigma}\).

6.3. In the proof of Proposition 4.1, check the claim that the short-completion error \(X^\epsilon R^{3/2}C_*^{3/2}\) is \(O(X^{0.795+\epsilon})\) in both ranges of \(R\). Check also that the trivial bound for the long completions is \(O(X^{0.805+\epsilon})\) when \(R\le X^{0.40}\).

6.4. Explain why the identity (4.1) would fail if the coefficient \(\theta_{3\ell}(c)\) were replaced by \(\theta_{\ell}(c)\).

6.5. Let \(\ell\ne0\) and let \(W\) be a fixed smooth function with compact support in \((0,\infty)\). Use Poisson summation (Lemma 2.1 of the previous lesson) to show that \[ \begin{aligned} &\sum_{v\in1+3\mathcal O}\theta_\ell(v)W(N(v)/Y)\\ &\qquad=O_{\ell,W,A}(Y^{-A}) \end{aligned} \] for every \(A>0\). Why does the proof of Lemma 3.1 use only the weaker bound \(O(1+\sqrt Y)\)?

7. Solutions

6.1. \(c_*\frac{2\pi}{9\sqrt3}=\frac{(2\pi)^{2/3}}{3\Gamma(2/3)}\cdot\frac{2\pi}{9\sqrt3}=\frac{(2\pi)^{5/3}}{27\sqrt3\,\Gamma(2/3)}\), and \(27\sqrt3=3^{7/2}\).

6.2. At \(z=\frac12+iy\) the two quotients are \(\Gamma(\frac13+k-iy)/\Gamma(\frac13+k+iy)\) and \(\Gamma(\frac23+k-iy)/\Gamma(\frac23+k+iy)\), with \(k=|\ell|/2\). Since \(\Gamma(\bar w)=\overline{\Gamma(w)}\), each has absolute value one. On \(\operatorname{Re}z=-\sigma\), Stirling's formula gives \(|\Gamma(a+iy)|\asymp|y|^{a-1/2}e^{-\pi|y|/2}\) for fixed real \(a\) and large \(|y|\). The first quotient therefore has size \(|y|^{(\frac56+k+\sigma-\frac12)-(k-\sigma-\frac16-\frac12)}=|y|^{1+2\sigma}\), and the second has the same size.

6.3. For \(R\le X^{0.40}\): \(\frac32(0.40)+\frac32(0.13)=0.795\). For \(X^{0.40}<R\ll X^{0.51}\): \(\frac32(0.51)+\frac32(0.02)=0.795\). The trivial bound is \(X^{1+\epsilon}E^{-3/2}\) with \(E\gg X^{0.13}\), and \(1-\frac32(0.13)=0.805\).

6.4. After \(e=cd\) the angular factor of a term is \(\theta_{3\ell}(c)\theta_\ell(rud^3)=\theta_\ell(ru)\theta_{3\ell}(cd)\)\(=\theta_\ell(rue^3)\). Only this combination depends on \(c\) and \(d\) through \(e\) alone, which is what makes \(\sum_{c\mid e}\mu(c)\) appear. With \(\theta_\ell(c)\) the coefficient of a fixed \(e\) would be \(\sum_{c\mid e}\mu(c)\theta_{-2\ell}(c)\), which does not vanish for \(e\ne1\) in general.

6.5. Put \(f(z)=\theta_\ell(z)W(|z|^2/Y)\). Lemma 2.1 of the previous lesson with \(b=3\), \(c=1\) gives \[ \sum_{v\in1+3\mathcal O}f(v)=\frac19\sum_{y\in\lambda^{-1}\mathcal O}e\Bigl(\operatorname{Tr}\frac y3\Bigr)\widehat f\Bigl(\frac y3\Bigr). \] The term \(y=0\) is \(\frac19\int f\,d\mathfrak m=0\), since the angular integral of \(e^{i\ell\phi}\) vanishes. The function \(f\) is supported where \(|z|\asymp\sqrt Y\), and each derivative of order \(k\) is \(O_{\ell,W,k}(Y^{-k/2})\), since both \(\theta_\ell\) and \(W(|z|^2/Y)\) vary on the scale \(\sqrt Y\). Integrating by parts \(k\) times gives \(|\widehat f(\eta)|\ll Y\,(\sqrt Y|\eta|)^{-k}\). For \(y\ne0\) we have \(|y/3|\ge1/(3\sqrt3)\), and summing over \(y\) with \(k\ge3\) gives \(O(Y^{1-k/2})\). In Lemma 3.1 the weights have derivatives bounded only by powers of \(L\), and the sums occur with \(Y=U/N(c^2d)\) ranging down to \(1\). The cell comparison gives one bound valid uniformly in all these cases, and it is sufficient.

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