# 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*](cubic-gauss-sums-and-pattersons-conjecture.md) and [*Large sieves for cubic characters*](large-sieves-for-cubic-characters.md), with their notation. From *The Riemann zeta function*: [*The Gamma function and Stirling's formula*](course:NT-ZETA/NT-ZETA-03), 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 \(\nu\). - At a prime \(\pi\mid r\) of norm \(p\), let \(e\in\{0,1\}\) and \(k\ge0\) be its exponents in \(h\) and \(h'\). The local factor is at most \[ \begin{aligned} &\sum_{e=0}^1\sum_{k\ge0}p^{\mathbf 1_{e+k>0}}\,p^{k/2}\,p^{-(e+3k)(1+\sigma)}\\ &\qquad=1+p^{-\sigma}+O\bigl(p^{-3/2-3\sigma}\bigr). \end{aligned} \] - For fixed \(\eta>0\), the product of these factors over \(\pi\mid r\) is \(O_{\sigma,\eta}(R^\eta)\): finitely many small primes give a bounded factor, and each other factor is at most \(p^\eta\). - The sum over \(w\) converges at the exponent \(1+\sigma\). - The powers of \(\lambda\) give at most \(\sum_{j\ge-1}3^{j/3}3^{-j(1+\sigma)}<\infty\). - The sum over \(d\) converges at the exponent \(\frac52+3\sigma\). 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 \(X^{0.40}\sqrt R\,T^2\).* 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}