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International Mathematics Research Notices
, Volume 2018 (2) – Jan 1, 2018

79 pages

/lp/ou_press/kloosterman-sums-and-maass-cusp-forms-of-half-integral-weight-for-the-1ELYRPLzVq

- Publisher
- Oxford University Press
- Copyright
- © The Author(s) 2016. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permission@oup.com.
- ISSN
- 1073-7928
- eISSN
- 1687-0247
- D.O.I.
- 10.1093/imrn/rnw234
- Publisher site
- See Article on Publisher Site

Abstract We estimate the sums ∑c≤xS(m,n,c,χ)c, where the $$S(m,n,c,\chi)$$ are Kloosterman sums associated with a multiplier system $$\chi$$ of half-integral weight on the modular group. Our estimates are uniform in $$m,n,$$ and $$x$$ in analogy with Sarnak and Tsimerman’s improvement of Kuznetsov’s bound for the ordinary Kloosterman sums. Among other things this requires us to develop mean value estimates for coefficients of Maass cusp forms of weight $$1/2$$ and uniform estimates for $$K$$-Bessel integral transforms. As an application, we obtain an improved estimate for the classical problem of estimating the size of the error term in Rademacher’s formula for the partition function $$p(n)$$. 1 Introduction and Statement of Results The Kloosterman sum S(m,n,c)=∑dmodc(d,c)=1e(md¯+ndc),e(x):=e2πix plays a leading part in analytic number theory. Indeed, many problems can be reduced to estimates for sums of Kloosterman sums (see, e.g., [18] or [46]). In this paper we study sums of generalized Kloosterman sums attached to a multiplier system of weight $$1/2$$ (the sums $$S(m,n,c)$$ are attached to the trivial integral weight multiplier system). Kloosterman sums with general multipliers have been studied by Bruggeman [10], Goldfeld–Sarnak [16], and Pribitkin [37], among others. We focus on the multiplier system $$\chi$$ for the Dedekind eta-function and the associated Kloosterman sums S(m,n,c,χ)=∑0≤a,d<c (abcd)∈SL2(Z)χ¯((abcd))e(m~a+n~dc),m~:=m− 2324. (1.1) These sums are intimately connected to the partition function $$p(n)$$, and we begin by discussing an application of our main theorem to a classical problem. In the first of countless important applications of the circle method, Hardy and Ramanujan [17] proved the asymptotic formula p(n)∼143neπ2n/3 (and in fact developed an asymptotic series for $$p(n)$$). Perfecting their method, Rademacher [40, 41] proved that p(n)=2π(24n−1)34∑c=1∞Ac(n)cI32(π24n−16c), (1.2) where $$I_{\frac32}$$ is the $$I$$-Bessel function, Ac(n):=∑dmodc(d,c)=1eπis(d,c)e(−dnc), and $$s(d,c)$$ is a Dedekind sum (see (2.8)). The Kloosterman sum $$A_c(n)$$ is a special case of (1.1); in particular we have (see Section 2.8 below) Ac(n)=−iS(1,1−n,c,χ). (1.3) The series (1.2) converges very rapidly. For example, when $$n=100$$, the first four terms give p(100)≈190,568,944.783+348.872−2.598+0.685=190,569,291.742, while the actual value is $$p(100) = 190,569,292$$. A natural problem is to estimate the error which results from truncating the series (1.2) after the $$N$$th term, or in other words to estimate the quantity $$R(n, N)$$ defined by p(n)=2π(24n−1)34∑c=1NAc(n)cI32(π24n−16c)+R(n,N). Since $$I_\frac32(x)$$ grows exponentially as $$x\to\infty$$, one must assume that $$N\gg\sqrt{n}$$ in order to obtain reasonable estimates. For $$\alpha>0$$, Rademacher [40, (8.1)] showed that R(n,αn)≪αn−14. Lehmer [29, Theorem 8] proved the sharp Weil-type bound |Ac(n)|<2ωo(c)c, (1.4) where $$\omega_o(c)$$ is the number of distinct odd primes dividing $$c$$. Shortly thereafter [30] he used this bound to prove that R(n,αn)≪αn−12logn. (1.5) In 1938, values of $$p(n)$$ had been tabulated only for $$n\leq 600$$. Using (1.4), Lehmer showed [29, Theorem 13] that $$p(n)$$ is the nearest integer to the series (1.2) truncated at $$\frac 23\sqrt{n}$$ for all $$n>600$$. Building on work of Selberg and Whiteman [54], Rademacher [42] later simplified Lehmer’s treatment of the sums $$A_c(n)$$. Rademacher’s book [43, Chapter IV] gives a relatively simple derivation of the error bound $$n^{-\frac38}$$ using these ideas. Using equidistribution results for Heegner points on the modular curve $$X_0(6)$$, Folsom and Masri [14, Corollary 1.2] improved Lehmer’s estimate. They proved that if $$24n-1$$ is squarefree, then there exists a constant $$c>0$$ such that for any $$\epsilon>0$$ and any $$0<b<\frac 1{12c}$$ we have R(n,n6)=Oϵ(n−712+bc+ϵ)+Oϵ(n−12−b+ϵ). (1.6) Here the value of $$c$$ and the implied constant can in principle be made explicit. Assuming GRH, Folsom and Masri obtain R(n,n6)=Oϵ(n−34+bc′+ϵ)+Oϵ(n−12−b+ϵ), where $$c'>0$$ is another constant and $$0<b<\frac 1{4c'}$$ (see [14, Remark 1.3]). The authors comment [14, Remark 1.4] that it should be possible to remove the assumption on $$24n-1$$ with some additional technical work. Using the estimates for sums of Kloosterman sums in Theorem 1.4, we obtain an improvement to (1.6), and we (basically) remove the assumption that $$24n-1$$ is squarefree. For simplicity in stating the results, we will assume that 24n−1is not divisible by 54 or 74. (1.7) As the proof will show (cf. Section 8) the exponent $$4$$ in (1.7) can be replaced by any positive integer $$m$$; such a change would have the effect of changing the implied constants in (1.8) and (1.9) Theorem 1.1. Suppose that $$\alpha>0$$. For any $$\epsilon>0$$ and any $$n\geq1$$ satisfying (1.7) we have R(n,αn12)≪α,ϵn−12−1168+ϵ. (1.8) □ In fact, our method optimizes when $$N$$ is slightly larger with respect to $$n$$. Theorem 1.2. Suppose that $$\alpha>0$$. For any $$\epsilon>0$$ and any $$n\geq1$$ satisfying (1.7) we have R(n,αn12+5252)≪α,ϵn−12−128+ϵ. (1.9) □ Remark In principle, the implied constants in these results (and the results below) could be made explicit. □ Remark Suppose, for example, that $$24n-1$$ is divisible by $$5^4$$. Then $$n$$ has the form $$n = 625 m + 599$$, so that $$p(n)\equiv 0\pmod{625}$$ by work of Ramanujan [5, Section 22]. A sharp estimate for $$R(n,N)$$ for such $$n$$ is less important since $$p(n)$$ can be determined by showing that $$|R(n,N)| < 312.5$$. The situation is similar for any power of $$5$$ or $$7$$. □ Remark The estimates in Theorems 1.1 and 1.2 depend on progress toward the Ramanujan–Lindelöf conjecture for coefficients of certain Maass cusp forms of weight 1/2. Assuming the conjecture, we can use the present methods to prove R(n,αn12)≪α,ϵn−12−116+ϵ. The analogue of Linnik and Selberg’s conjecture (1.11) would give R(n,αn12)≪α,ϵn−34+ϵ. Computations suggest that this bound might be optimal. For example, see Figure 1. □ Fig. 1. View largeDownload slide $$|R(n,\sqrt n)|$$ versus $$n^{-\frac 34}$$. Fig. 1. View largeDownload slide $$|R(n,\sqrt n)|$$ versus $$n^{-\frac 34}$$. Theorems 1.1 and 1.2 follow from estimates for weighted sums of the Kloosterman sums (1.1). For the ordinary Kloosterman sum $$S(m,n,c)$$, Linnik [31] and Selberg [49] conjectured that there should be considerable cancellation in the sums ∑c≤xS(m,n,c)c. (1.10) Sarnak and Tsimerman [47] studied these weighted sums for varying $$m,n$$ and put forth the following modification of Linnik and Selberg’s conjecture with an “$$\epsilon$$-safety valve” in $$m,n$$: ∑c≤xS(m,n,c)c≪ϵ(|mn|x)ϵ. (1.11) Using the Weil bound [53] |S(m,n,c)|≤τ(c)(m,n,c)12c, (1.12) where $$\tau(c)$$ is the number of divisors of $$c$$, one obtains ∑c≤xS(m,n,c)c≪τ((m,n))x12logx. Thus the conjecture (1.11) represents full square-root cancellation. The best current bound in the $$x$$-aspect for the sums (1.10) was obtained by Kuznetsov [28], who proved for $$m,n>0$$ that ∑c≤xS(m,n,c)c≪m,nx16(logx)13. Recently, Sarnak and Tsimerman [47] refined Kuznetsov’s method, making the dependence on $$m$$ and $$n$$ explicit. They proved that for $$m$$, $$n>0$$ we have ∑c≤xS(m,n,c)c≪ϵ(x16+(mn)16+(m+n)18(mn)θ2)(mnx)ϵ, where $$\theta$$ is an admissible exponent toward the Ramanujan–Petersson conjecture for coefficients of weight 0 Maass cusp forms. By work of Kim and Sarnak [25, Appendix 2], the exponent $$\theta=7/64$$ is available. A generalization of their results to sums taken over $$c$$ which are divisible by a fixed integer $$q$$ is given by Ganguly and Sengupta [15]. We study sums of the Kloosterman sums $$S(m,n,c,\chi)$$ defined in (1.1). In view of (1.3) we consider the case when $$m$$ and $$n$$ have mixed sign (as will be seen in the proof, this brings up a number of difficulties which are not present in the case when $$m$$, $$n>0$$). Theorem 1.3. For $$m>0$$, $$n<0$$ and for any $$\epsilon>0$$ we have ∑c≤xS(m,n,c,χ)c≪ϵ(x16+|mn|14)|mn|ϵlogx. □ While Theorem 1.3 matches [47] in the $$x$$ aspect, it falls short in the $$mn$$ aspect. As we discuss in more detail below, this is due to the unsatisfactory Hecke theory in half-integral weight. This bound is also insufficient to improve the estimate (1.5) for $$R(n,N)$$; for this we sacrifice the bound in the $$x$$-aspect for an improvement in the $$n$$-aspect. The resulting theorem, which for convenience we state in terms of $$A_c(n)$$, leads to the estimates of Theorems 1.1 and 1.2. Theorem 1.4. Fix $$0<\delta<1/2$$. Suppose that $$\epsilon>0$$ and that $$n\geq 1$$ satisfies (1.7). Then ∑c≤xAc(n)c≪δ,ϵn14−156+ϵx34δ+(n14−1168+ϵ+x12−δ)logx. □ As in [28] and [47], the basic tool is a version of Kuznetsov’s trace formula. This relates sums of the Kloosterman sums $$S(m,n,c,\chi)$$ weighted by a suitable test function to sums involving Fourier coefficients of Maass cusp forms of weight $$1/2$$ and multiplier $$\chi$$. Proskurin [39] proved such a formula for general weight and multiplier when $$mn>0$$; in Section 4 we give a proof in the case $$mn<0$$ (Blomer [6] has recorded this formula for twists of the theta-multiplier by a Dirichlet character). The Maass cusp forms of interest are functions which transform like $$\text{Im}(\tau)^{\frac 14}\eta(\tau)$$, where η(τ)=e(τ24)∏n=1∞(1−e(nτ)),Im(τ)>0, (1.13) is the Dedekind eta-function, and which are eigenfunctions of the weight $$1/2$$ Laplacian $$\Delta_{1/2}$$ (see Section 2 for details). To each Maass cusp form $$F$$ we attach an eigenvalue $$\lambda$$ and a spectral parameter $$r$$ which are defined via Δ12F+λF=0 and λ=14+r2. We denote the space spanned by these Maass cusp forms by $$\mathcal S_\frac12(1,\chi)$$. Denote by $$\{u_j(\tau)\}$$ an orthonormal basis of Maass cusp forms for this space, and let $$r_j$$ denote the spectral parameter attached to each $$u_j$$. Then (recalling the notation $$\tilde n=n-\frac{23}{24}$$), $$u_j$$ has a Fourier expansion of the form uj(τ)=∑n≠0ρj(n)Wsgn(n)4,irj(4π|n~|y)e(n~x), where $$W_{\kappa,\mu}(y)$$ is the $$W$$-Whittaker function (see Section 2.3) and $$\text{sgn}(n):=\frac{n}{|n|}$$ (we never require $$\text{sgn}(0)$$). In Section 4 we will prove that ∑c>0S(m,n,c,χ)cϕ(4πm~|n~|c)=8im~|n~|∑rjρj(m)¯ρj(n)chπrjϕ(rj), (1.14) where $$\phi$$ is a suitable test function and $$\check\phi$$ is the $$K$$-Bessel transform ϕˇ(r):=chπr∫0∞K2ir(u)ϕ(u)duu. (1.15) (For notational convenience we use $$\text{ch}$$, $$\text{sh}$$, $$\text{th}$$ for the hyperbolic trigonometric functions.) In Section 5 we introduce theta lifts (as in [36], [51]) which give a Shimura-type correspondence S12(N,ψχ,r)→S0(6N,ψ2,2r). (1.16) Here $$\mathcal S_k(N,\nu,r)$$ denotes the space of Maass cusp forms of weight $$k$$ on $$\Gamma_0(N)$$ with multiplier $$\nu$$ and spectral parameter $$r$$. These lifts commute with the action of the Hecke operators on the respective spaces, and in particular allow us to control the size of the weight $$1/2$$ Hecke eigenvalues, which becomes important in Section 8. The existence of these lifts shows that there are no small eigenvalues; this fact is used in Sections 6, 7, and 9. In particular, from (1.16) it will follow that if $$\mathcal S_{1/2}(1,\chi,r)\neq 0$$ then either $$r={\rm i}/4$$ or $$r>1.9$$. To estimate the right-hand side of (1.14), we first obtain a mean value estimate for the sums |n~|∑0<rj≤x|ρj(n)|2chπrj. In analogy with Kuznetsov’s result [28, Theorem 6] in weight $$0$$, we prove in Section 3 a general mean value result for weight $$\pm 1/2$$ Maass cusp forms which has the following as a corollary. Theorem 1.5. Let notation be as above. Then for any $$\epsilon>0$$ we have |n~|∑0<rj≤x|ρj(n)|2chπrj={x525π2+Oϵ(x32logx+|n|12+ϵx12) if n<0,x323π2+Oϵ(x12logx+n12+ϵ) if n>0. □ We then require uniform estimates for the Bessel transform (1.15), which are made subtle by the oscillatory nature of $$K_{{\rm ir}}(x)$$ for small $$x$$ and by the transitional range of the $$K$$-Bessel function. In Section 6 we obtain estimates for $$\check\phi(r)$$ which, together with Theorem 1.5, suffice to prove Theorem 1.3. To prove Theorem 1.4 we require a second estimate for the Fourier coefficients $$\rho_j(n)$$. In [47] such an estimate is obtained via the simple relationship a(n)=λ(n)a(1) satisfied by the coefficients $$a(n)$$ of a Hecke eigenform with eigenvalue $$\lambda(n)$$. This relationship is not available in half-integral weight (the best which one can do is to relate coefficients of index $$m^2n$$ to those of index $$n$$). As a substitute, we employ an average version of a theorem of Duke [12]. Duke proved that if $$a(n)$$ is the $$n$$th coefficient of a normalized Maass cusp form in $$\mathcal S_{1/2}(N,\left(\frac{{D}}{\bullet}\right) \nu_\theta,r)$$ ($$\nu_\theta$$ is the theta-multiplier defined in Section 2.2) then for squarefree $$n$$ we have |a(n)|≪ϵ|n|−27+ϵ|r|52−sgn(n)4ch(πr2). This is not strong enough in the $$r$$-aspect for our purposes (Baruch and Mao [4] have obtained a bound which is stronger in the $$n$$-aspect, but even weaker in the $$r$$-aspect). Here we modify Duke’s argument to obtain an average version of his result. We prove in Theorem 8.1 that if the $$u_j$$ as above are eigenforms of the Hecke operators $$T(p^2)$$, $$p\nmid 6$$, then we have ∑0<rj≤x|ρj(n)|2chπrj≪ϵ|n|−47+ϵx5−sgnn2, for any collection of values of $$n$$ such that $$24n-23$$ is not divisible by arbitrarily large powers of $$5$$ or $$7$$. In Section 9 we use Theorem 8.1 and a modification of the argument of [47] to prove Theorem 1.4. Section 10 contains the proofs of Theorems 1.1 and 1.2. 1.1 Notation Throughout, $$\epsilon$$ denotes an arbitrarily small positive number whose value is allowed to change with each occurrence. Implied constants in any equation which contains $$\epsilon$$ are allowed to depend on $$\epsilon$$. For all other parameters, we will use a subscript (e.g., $$\ll_{a,b,c}$$) to signify dependencies in the implied constant. Since the proofs are quite involved, we pause to provide an overview of the structure of the paper. Theorem 1.3 is proved in Section 7. The proof relies on Theorem 1.5 (which is a special case of Theorem 3.1), Theorem 4.1, Corollary 5.3 (although a weaker result would suffice) and Theorem 6.1. Theorem 1.4 (which is the same as Theorem 9.1) is proved in Section 9. Its proof relies in addition on Theorems 5.1 and 8.1. Finally, Theorems 1.1 and 1.2 are proved in Section 10. These follow in a relatively straightforward manner from Theorem 1.4. 2. Background We give some background on Maass forms and Kloosterman sums of general weight $$k$$ and multiplier $$\nu$$ (references for this material are [13, 39, 46] along with the original sources [33, 34, 45, 48, 49]). In the body of the paper we will work primarily in integral or half-integral weight and with the multiplier attached to a Dirichlet character or to the Dedekind eta-function, but for much of the background it is no more complicated to describe the general case. For the benefit of the reader we have attempted to provide a relatively self-contained account by providing details at various points. 2.1 Eigenfunctions of the Laplacian Let $$k$$ be a real number and let $$\mathbb{H}$$ denote the upper half-plane. The group $${\rm SL}_2(\mathbb{R})$$ acts on $$\tau=x+{\rm i}y\in\mathbb{H}$$ via linear fractional transformations γτ:=aτ+bcτ+d,γ=(abcd). For $$\gamma\in {\rm SL}_2(\mathbb{R})$$ we define the weight $$k$$ slash operator by f|kγ:=j(γ,τ)−kf(γτ),j(γ,τ):=cτ+d|cτ+d|=eiarg(cτ+d), where we always choose the argument in $$(-\pi,\pi]$$. The weight $$k$$ Laplacian Δk:=y2(∂2∂x2+∂2∂y2)−iky∂∂x commutes with the weight $$k$$ slash operator for every $$\gamma\in {\rm SL}_2(\mathbb{R})$$. The operator $$\Delta_k$$ can be written as Δk=−Rk−2Lk−k2(1−k2), (2.1) Δk=−Lk+2Rk+k2(1+k2), (2.2) where $$R_k$$ is the Maass raising operator Rk:=k2+2iy∂∂τ=k2+iy(∂∂x−i∂∂y), and $$L_k$$ is the Maass lowering operator Lk:=k2+2iy∂∂τ¯=k2+iy(∂∂x+i∂∂y). These raise and lower the weight by $$2$$: we have Rk(f|kγ)=(Rkf)|k+2γandLk(f|kγ)=(Lkf)|k−2γ. From (2.1) and (2.2) we obtain the relations RkΔk=Δk+2RkandLkΔk=Δk−2Lk. (2.3) A real analytic function $$f:\mathbb{H}\to\mathbb{C}$$ is an eigenfunction of $$\Delta_k$$ with eigenvalue $$\lambda$$ if Δkf+λf=0. (2.4) If $$f$$ satisfies (2.4) then for notational convenience we write λ=14+r2, and we refer to $$r$$ as the spectral parameter of $$f$$. From (2.3) it follows that if $$f$$ is an eigenfunction of $$\Delta_k$$ with eigenvalue $$\lambda$$ then $$R_k f$$ (resp. $$L_k f$$) is an eigenfunction of $$\Delta_{k+2}$$ (resp. $$\Delta_{k-2}$$) with eigenvalue $$\lambda$$. 2.2 Multiplier systems Let $$\Gamma = \Gamma_0(N)$$ for some $$N\geq 1$$. We say that $$\nu:\Gamma\to \mathbb{C}^\times$$ is a multiplier system of weight $$k$$ if (1) $$|\nu|=1$$, (2) $$\nu(-I)={\rm e}^{-\pi {\rm i} k}$$, and (3) $$\nu(\gamma_1 \gamma_2) \, j(\gamma_1\gamma_2,\tau)^k=\nu(\gamma_1)\nu(\gamma_2) \, j(\gamma_2,\tau)^k j(\gamma_1,\gamma_2\tau)^k$$ for all $$\gamma_1,\gamma_2\in \Gamma$$. If $$\nu$$ is a multiplier system of weight $$k$$, then it is also a multiplier system of weight $$k'$$ for any $$k'\equiv k\pmod 2$$, and the conjugate $$\bar\nu$$ is a multiplier system of weight $$-k$$. If $$\nu_1$$ and $$\nu_2$$ are multiplier systems of weights $$k_1$$ and $$k_2$$ for the same group $$\Gamma$$, then their product $$\nu_1\nu_2$$ is a multiplier system of weight $$k_1+k_2$$ for $$\Gamma$$. When $$k$$ is an integer, a multiplier system of weight $$k$$ for $$\Gamma$$ is simply a character of $$\Gamma$$ which satisfies $$\nu(-I)={\rm e}^{-\pi {\rm i} k}$$. If $$\psi$$ is an even (resp. odd) Dirichlet character modulo $$N$$, we can extend $$\psi$$ to a multiplier system of even (resp. odd) integral weight for $$\Gamma$$ by setting $$\psi(\gamma):=\psi(d)$$ for $$\gamma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right) $$. Given a cusp $$\mathfrak{a}$$ let $$\Gamma_\mathfrak{a}:=\{\gamma\in\Gamma:\gamma\mathfrak{a}=\mathfrak{a}\}$$ denote its stabilizer in $$\Gamma$$, and let $$\sigma_\mathfrak{a}$$ denote the unique (up to translation by $$\pm\left({{1}\atop{0}} \quad {{*}\atop {1}}\right)$$ on the right) matrix in $${\rm SL}_2(\mathbb{R})$$ satisfying $$\sigma_\mathfrak{a} \infty = \mathfrak{a}$$ and $$\sigma_\mathfrak{a}^{-1}\Gamma_\mathfrak{a}\sigma_\mathfrak{a}=\Gamma_\infty$$. Define $$\alpha_{\nu,\mathfrak{a}} \in [0,1)$$ by the condition ν(σa(1101)σa−1)=e(−αν,a). We say that $$\mathfrak{a}$$ is singular with respect to $$\nu$$ if $$\nu$$ is trivial on $$\Gamma_\mathfrak{a}$$, that is, if $$\alpha_{\nu,\mathfrak{a}}=0$$. For convenience we write $$\alpha_{\nu}$$ for $$\alpha_{\nu,\mathfrak{a}}$$ when $$\mathfrak{a}=\infty$$. If $$\alpha_\nu>0$$ then αν¯=1−αν. (2.5) We are primarily interested in the multiplier system $$\chi$$ of weight $$1/2$$ on $${\rm SL}_2(\mathbb{Z})$$ (and its conjugate $$\bar\chi$$ of weight $$-1/2$$) where η(γτ)=χ(γ)cτ+dη(τ),γ=(abcd)∈SL2(Z), (2.6) and $$\eta$$ is the Dedekind eta-function defined in (1.13). From condition (ii) above we have $$\chi(-I)=-{\rm i}$$. From the definition of $$\eta(\tau)$$ we have $$\chi(\left({{1} \atop {0}} \quad {{1} \atop {1}}\right))=e(1/24)$$; it follows that αχ=2324andαχ¯=124, so the unique cusp $$\infty$$ of $${\rm SL}_2(\mathbb{Z})$$ is singular neither with respect to $$\chi$$ nor with respect to $$\bar\chi$$. For $$\gamma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right)$$ with $$c>0$$ there are two useful formulas for $$\chi$$. Rademacher (see, e.g., (74.11), (74.12), and (71.21) of [43]) showed that χ(γ)=−ie−πis(d,c)e(a+d24c), (2.7) where $$s(d,c)$$ is the Dedekind sum s(d,c)=∑r=1c−1rc(drc−⌊drc⌋−12). (2.8) On the other hand, Petersson (see, e.g., Section 4.1 of [26]) showed that χ(γ)={(dc)e(124[(a+d)c−bd(c2−1)−3c]) if c is odd,(cd)e(124[(a+d)c−bd(c2−1)+3d−3−3cd]) if c is even. (2.9) We will also use the multiplier system $$\nu_\theta$$ of weight $$1/2$$ on $$\Gamma_0(4N)$$ defined by θ(γτ)=νθ(γ)cτ+dθ(τ),γ∈Γ0(4N), where θ(τ):=∑n∈Ze(n2τ). Explicitly, we have νθ((abcd))= (cd)εd−1, (2.10) where $$\left(\tfrac{\cdot}{\cdot}\right)$$ is the extension of the Kronecker symbol given, for example, in [50] and εd=(−1d)12={1 if d≡1(mod4),i if d≡3(mod4). Let $$\Gamma_0(N,M)$$ denote the subgroup of $$\Gamma_0(N)$$ consisting of matrices whose upper-right entry is divisible by $$M$$. Equation (2.9) shows that χ(γ)= (cd)e(d−18)for γ=(abcd)∈Γ0(24,24). This implies that χ((a24bc/24d))= (12d)(cd)εd−1for (abcd)∈Γ0(576), (2.11) which allows one to relate automorphic functions with multiplier $$\chi$$ on $${\rm SL}_2(\mathbb{Z})$$ to those with multiplier $$\left(\frac{{12}}{{\bullet}}\right)\nu_\theta$$ on $$\Gamma_0(576)$$. 2.3 Maass forms A function $$f:\mathbb{H}\to\mathbb{C}$$ is automorphic of weight $$k$$ and multiplier $$\nu$$ for $$\Gamma=\Gamma_0(N)$$ if f|kγ=ν(γ)f for all γ∈Γ. Let $$\mathcal{A}_k(N,\nu)$$ denote the space of all such functions. A smooth, automorphic function which is also an eigenfunction of $$\Delta_k$$ and which satisfies the growth condition f(τ)≪yσ+y1−σfor some σ, for all τ∈H (2.12) is called a Maass form. We let $$\mathcal{A}_k(N,\nu,r)$$ denote the vector space of Maass forms with spectral parameter $$r$$. From the preceding discussion, the raising (resp. lowering) operator maps $$\mathcal{A}_k(N,\nu,r)$$ into $$\mathcal{A}_{k+2}(N,\nu,r)$$ (resp. $$\mathcal{A}_{k-2}(N,\nu,r)$$). Also, complex conjugation $$f\to \bar f$$ gives a bijection $$\mathcal{A}_k(N,\nu,r) \longleftrightarrow \mathcal{A}_{-k}(N,\bar\nu,r)$$. If $$f \in \mathcal{A}_k(n,\nu,r)$$, then it satisfies $$f(\tau+1)=e(-\alpha_\nu)f(\tau)$$. For $$n\in\mathbb{Z}$$ define nν:=n−αν. (2.13) Then $$f$$ has a Fourier expansion of the form f(τ)=∑n=−∞∞a(n,y)e(nνx). By imposing condition (2.4) on the Fourier expansion we find that for $$n_\nu\neq0$$, the function $$a(n,y)$$ satisfies a″(n,y)(4π|nν|y)2+(1/4+r2(4π|nν|y)2+ksgn(nν)8π|nν|y−14)a(n,y)=0. The Whittaker functions $$M_{\kappa,\mu}(y)$$ and $$W_{\kappa,\mu}(y)$$ are the two linearly independent solutions to Whittaker’s equation [35, Section 13.14] W″+(1/4−μ2y2+κy−14)W=0. As $$y\to\infty$$, the former solution grows exponentially, while the latter decays exponentially. Since $$f$$ satisfies the growth condition (2.12), we must have a(n,y)=a(n)Wk2sgn nν,ir(4π|nν|y) (2.14) for some constant $$a(n)$$. For $$n_\nu=0$$ we have a(n,y)=a(0)y12+ir+a′(0)y12−ir. We call the numbers $$a(n)$$ (and $$a'(0)$$) the Fourier coefficients of $$f$$. For $$\text{Re}(\mu-\kappa+1/2)>0$$ we have the integral representation [35, (13.14.3), (13.4.4)] Wκ,μ(y)=e−y/2yμ+1/2Γ(μ−κ+12)∫0∞e−yttμ−κ−12(1+t)μ+κ−12dt. (2.15) When $$\kappa=0$$ we have [35, (13.18.9)] W0,μ(y)=yπKμ(y/2), where $$K_{\mu}$$ is the $$K$$-Bessel function. For Maass forms of weight $$0$$, many authors normalize the Fourier coefficients in (2.14) so that $$a(n)$$ is the coefficient of $$\sqrt y \, K_{{\rm ir}}(2\pi|n_\nu|y)$$, which has the effect of multiplying $$a(n)$$ by $$2|n_\nu|^{1/2}$$. By [35, (13.15.26)], we have yddyWκ,μ(y)=(y2−κ)Wκ,μ(y)−Wκ+1,μ(y)=(κ−y2)Wκ,μ(y)+(κ+μ−12)(κ−μ−12)Wκ−1,μ(y). This, together with [35, (13.15.11)] shows that Lk(Wk2sgn(nν),ir(4π|nν|y)e(nνx))=Wk−22sgn(nν),ir(4π|nν|y)e(nνx)×{−(r2+(k−1)24) if nν>0,1 if nν<0, from which it follows that if $$f_L:=L_k f$$ has coefficients $$a_L(n)$$, then aL(n)={−(r2+(k−1)24)a(n) if nν>0,a(n) if nν<0. (2.16) Similarly, for the coefficients $$a_R(n)$$ of $$f_R:=R_k f$$, we have aR(n)={−a(n) if nν>0,(r2+(k+1)24)a(n) if nν<0. From the integral representation (2.15) we find that $$\bar{W_{\kappa,\mu}(y)}=W_{\kappa,\bar\mu}(y)$$ when $$y,\kappa\in\mathbb{R}$$, so $$W_{\kappa,\mu}(y)$$ is real when $$\mu \in \mathbb{R}$$ or (by [35, (13.14.31)]) when $$\mu$$ is purely imaginary. Suppose that $$f$$ has multiplier $$\nu$$ and that $$\alpha_\nu\neq 0$$. Then by (2.5) we have $$-n_\nu=(1-n)_{\overline\nu}$$, so the coefficients $$a_c(n)$$ of $$f_c:= \bar f$$ satisfy ac(n)=a(1−n)¯. (2.17) 2.4 The spectrum of $$\boldsymbol\Delta_\textit{k}$$ Let $$\mathcal{L}_k(N,\nu)$$ denote the $$L^2$$-space of automorphic functions with respect to the Petersson inner product ⟨f,g⟩:=∫Γ∖Hf(τ)g(τ)¯dμ,dμ:=dxdyy2. (2.18) Let $$\mathcal{B}_k(N,\nu)$$ denote the subspace of $$\mathcal{L}_k(N,\nu)$$ consisting of smooth functions $$f$$ such that $$f$$ and $$\Delta_k f$$ are bounded on $$\mathbb{H}$$. By (2.1), (2.2), and Green’s formula, we have the relations (see also Section 3 of [44]) ⟨f,−Δkg⟩=⟨Rkf,Rkg⟩−k2(1+k2)⟨f,g⟩=⟨Lkf,Lkg⟩+k2(1−k2)⟨f,g⟩ (2.19) for any $$f,g\in \mathcal{B}_k(N,\nu)$$. It follows that $$\Delta_k$$ is symmetric and that ⟨f,−Δkf⟩≥λ0(k)⟨f,f⟩,λ0(k):=|k|2(1−|k|2). (2.20) By Friedrichs’ extension theorem (see, e.g., [23, Appendix A.1]) the operator $$\Delta_k$$ has a unique self-adjoint extension to $$\mathcal{L}_k(N,\nu)$$ (denoted also by $$\Delta_k$$). From (2.20) we see that the spectrum of $$\Delta_k$$ is real and contained in $$[\lambda_0(k),\infty)$$. The holomorphic forms correspond to the bottom of the spectrum. To be precise, if $$f_0\in \mathcal{L}_k(N,\nu)$$ has eigenvalue $$\lambda_0(k)$$ then the equations above show that f0(τ)={yk2F(τ) if k≥0,y−k2F¯(τ) if k<0, where $$F:\mathbb{H}\to\mathbb{C}$$ is a holomorphic cusp form of weight $$k$$; that is, $$F$$ vanishes at the cusps and satisfies F(γτ)=ν(γ)(cτ+d)kF(τ)for all γ=(a)bcd∈Γ. In particular, $$\lambda_0(k)$$ is not an eigenvalue when the space of cusp forms is trivial. Note also that if $$a_0(n)$$ are the coefficients of $$f_0$$, then a0(n)=0 when sgn(nν)=−sgn(k). The spectrum of $$\Delta_k$$ on $$\mathcal{L}_k(N,\nu)$$ consists of an absolutely continuous spectrum of multiplicity equal to the number of singular cusps, and a discrete spectrum of finite multiplicity. The Eisenstein series, of which there is one for each singular cusp $$\mathfrak{a}$$, give rise to the continuous spectrum, which is bounded below by $$1/4$$. When $$(k,\nu)=(1/2,\chi)$$ or $$(-1/2,\bar\chi)$$ and $$\Gamma={\rm SL}_2(\mathbb{Z})$$ there are no Eisenstein series since the only cusp is not singular. Let $$\mathcal{S}_k(N,\nu)$$ denote the orthogonal complement in $$\mathcal{L}_k(N,\nu)$$ of the space generated by Eisenstein series. The spectrum of $$\Delta_k$$ on $$\mathcal{S}_k(N,\nu)$$ is countable and of finite multiplicity with no limit points except $$\infty$$. The exceptional eigenvalues are those which lie in $$(\lambda_0(k),1/4)$$. Let $$\lambda_1(N,\nu,k)$$ denote the smallest eigenvalue larger than $$\lambda_0(k)$$ in the spectrum of $$\Delta_k$$ on $$\mathcal{S}_k(N,\nu)$$. Selberg’s eigenvalue conjecture states that λ1(N,1,0)≥14, that is, there are no exceptional eigenvalues. Selberg [49] showed that λ1(N,1,0)≥316. The best progress toward this conjecture was made by Kim and Sarnak [25, Appendix 2], who proved that λ1(N,1,0)≥14−(764)2=9754096, as a consequence of Langlands functoriality for the symmetric fourth power of an automorphic representation on $${\rm GL}_2$$. 2.5 Maass cusp forms The subspace $$\mathcal{S}_k(N,\nu)$$ consists of functions $$f$$ whose zeroth Fourier coefficient at each singular cusp vanishes. Eigenfunctions of $$\Delta_k$$ in $$\mathcal{S}_k(N,\nu)$$ are called Maass cusp forms. Let $$\{f_j\}$$ be an orthonormal basis of $$\mathcal{S}_k(N,\nu)$$, and for each $$j$$ let λj=14+rj2 denote the Laplace eigenvalue and $$\{a_j(n)\}$$ the Fourier coefficients. Then we have the Fourier expansion fj(τ)=∑nν≠0aj(n)Wk2sgnnν,irj(4π|nν|y)e(nνx). Suppose that $$f$$ has spectral parameter $$r$$. Then by (2.19) we have ‖Lkf‖2=(r2+(k−1)24)‖f‖2. (2.21) Weyl’s law describes the distribution of the spectral parameters $$r_j$$. Theorem 2.28 of [19] shows that ∑0≤rj≤T1−14π∫−TTφ′φ(12+it)dt=vol(Γ∖H)4πT2−K0πTlogT+O(T), where $$\varphi(s)$$ and $$K_0$$ are the determinant (see [19, p. 298]) and dimension (see [19, p. 281]), respectively, of the scattering matrix $$\Phi(s)$$ whose entries are given in terms of constant terms of Eisenstein series. In particular we have ∑0≤rj≤T1=112T2+O(T)for (k,ν)=(12,χ). (2.22) We will be working mostly in the space $$\mathcal S_\frac12(1,\chi)$$. Throughout the paper we will denote an orthonormal basis for this space by $$\{u_j\}$$, and we will write the Fourier expansions as uj(τ)=∑n≠0ρj(n)Wsgn n4,irj(4π|n~|y)e(n~x),n~:=n−2324. (2.23) 2.6 Hecke operators We introduce Hecke operators on the spaces $$\mathcal S_0(N,\boldsymbol 1)$$ and $$\mathcal S_\frac12(1,\chi)$$. For each prime $$p\nmid N$$ the Hecke operator $$T_p$$ acts on a Maass form $$f\in \mathcal{A}_0(N,\boldsymbol 1)$$ as Tpf(τ)=p−12(∑jmodpf(τ+jp)+f(pτ)). The $$T_p$$ commute with each other and with the Laplacian $$\Delta_0$$, so $$T_p$$ acts on each $$\mathcal{S}_0(N,\boldsymbol 1, r)$$. The action of $$T_p$$ on the Fourier expansion of $$f\in\mathcal{S}_0(N,\boldsymbol 1, r)$$ is given by Tpf(τ)=∑n≠0(p12a(pn)+p−12a(n/p))W0,ir(4π|n|y)e(nx). Hecke operators on half-integral weight spaces have been defined for the theta multiplier (see, e.g., [24]). We define Hecke operators $$T_{p^2}$$ for $$p\geq 5$$ on $$\mathcal A_{1/2}(1,\chi)$$ directly by Tp2f=1p[∑bmodp2e(−b24)f|12 (1pbp0p)+e(p−18)∑h=1p−1e(−hp24) (hp)f|12 (1hp01)+f|12 (p001p)]. One can show that (Tp2f)|12 (1101)=e(124)fand(Tp2f)|12 (0−110)=−if and that $$T_{p^2}$$ commutes with $$\Delta_{\frac 12}$$, so $$T_{p^2}$$ is an endomorphism of $$\mathcal{S}_{\frac 12}(1,\chi,r)$$. To describe the action of $$T_{p^2}$$ on Fourier expansions, it is convenient to write $$f\in \mathcal{S}_{\frac 12}(1,\chi,r)$$ in the form f(τ)=∑n≠0a(n)Wsgn(n)4,ir(π|n|y6)e(nx24). Then a standard computation gives Tp2f(τ)=∑n≠0(pa(p2n)+p−12 (12np)a(n)+p−1a(n/p2))Wsgn n4,ir(π|n|y6)e(nx24). (2.24) 2.7 Further operators The reflection operator f(x+iy)↦f(−x+iy) defines an involution on $$\mathcal S_0(N,\boldsymbol 1)$$ which commutes with $$\Delta_0$$ and the Hecke operators. We say that a Maass cusp form is even if it is fixed by reflection; such a form has Fourier expansion f(τ)=∑n≠0a(n)W0,ir(4π|n|y)cos(2πnx). We also have the operator f(τ)↦f(dτ) which in general raises the level and changes the multiplier. We will only need this operator in the case $$d=24$$ on $$\mathcal S_{1/2}(1,\chi)$$, where (2.11) shows that f(τ)↦f(24τ):S12(1,χ,r)→S12(576,(12∙)νθ,r). 2.8 Generalized Kloosterman sums Our main objects of study are the generalized Kloosterman sums given by S(m,n,c,ν):=∑0≤a,d<cγ=(abcc)∈Γν¯(γ)e(mνa+nνdv)=∑γ∈Γ∞∖Γ/Γ∞γ=(abcc)ν¯(γ)e(mνa+nνdc) (2.25) (to see that these are equal, one checks that each summand is invariant under the substitutions $$\gamma\to \left({{1} \atop {0}} \quad {{1} \atop {1}}\right)\gamma$$ and $$\gamma\to \gamma\left({{1} \atop {0}} \quad {{1} \atop {1}}\right)$$ using axiom (iii) for multiplier systems and (2.13)). When $$\Gamma={\rm SL}_2(\mathbb{Z})$$ and $$\nu=\boldsymbol 1$$ we recover the ordinary Kloosterman sums S(m,n,c)=S(m,n,c,1)=∑dmodc(d,c)=1e(md¯+ndc), where $$d\bar d\equiv 1\pmod c$$. For the eta-multiplier, (2.7) gives S(m,n,c,χ)=i∑dmodc(d,c)=1eπis(d,c)e((m−1)d¯+(n−1)dc), so the sums $$A_c(n)$$ appearing in Rademacher’s formula are given by Ac(n)=−iS(1,1−n,c,χ). We recall the bounds (1.4) and (1.12) from the introduction. A general Weil-type bound for $$S(m,n,c,\chi)$$ which holds for all $$m,n$$ follows from work of the second author [1], and is given in the following proposition. Proposition 2.1. Let $$m,n\in \mathbb{Z}$$ and write $$24n-23=N$$ and $$24m-23=v^2 M$$, with $$M$$ squarefree. For $$c\geq 1$$ we have |S(m,n,c,χ)|≪τ((v,c))τ(c) (MN,c)12c. (2.26) □ Proof We apply Theorem 3 of [1]. In the notation of that paper, we have $$S(m,n,c,\chi)=\sqrt{{\rm i}} \, K(m-1,n-1;c)$$. Estimating the right-hand side of [1, (1.20)] trivially, we find that |S(m,n,c,χ)|≤43τ((v,c))c12R(MN,24c), (2.27) where R(y,ℓ)=#{xmodℓ:x2≡ymodℓ}. The function $$R(y,\ell)$$ is multiplicative in $$\ell$$. If $$p$$ is prime and $$p\nmid y$$, then R(y,pe)≤{2 if p is odd,4 if p=2. (2.28) If $$y=p^{d}y'$$ with $$p\nmid y'$$ then $$R(y,p^e)\leq p^{d/2} R(y',p^{e-d})$$. It follows that R(y,ℓ)≤2⋅2ω(ℓ)(y,ℓ)12, (2.29) where $$\omega(\ell)$$ is the number of primes dividing $$\ell$$. By (2.27)–(2.29), and the fact that $$2^{\omega(\ell)}\leq \tau(\ell)$$, we obtain (2.26). ■ One useful consequence of Proposition 2.1 is the “trivial” estimate ∑c≤X|S(m,n,c,χ)|c≪X12logX|mn|ϵ, (2.30) which follows from a standard argument involving the mean value bound for $$\tau(c)$$. 3. A Mean Value Estimate for Coefficients of Maass Cusp Forms In this section we prove a general version of Theorem 1.5 which applies to the Fourier coefficients of weight $$\pm 1/2$$ Maass cusp forms with multiplier $$\nu$$ for $$\Gamma_0(N)$$, where $$\nu$$ satisfies the following assumptions: (1) There exists $$\beta=\beta_\nu \in (1/2,1)$$ such that ∑c>0|S(n,n,c,ν)|c1+β≪νnϵ. (3.1) (1) None of the cusps of $$\Gamma_0(N)$$ is singular for $$\nu$$. By (2.30) the multipliers $$\chi$$ and $$\bar\chi$$ on $${\rm SL}_2(\mathbb{Z})$$ satisfy these assumptions with $$\beta=1/2+\epsilon$$. Fix an orthonormal basis of cusp forms $$\{f_j\}$$ for $$\mathcal{S}_k(\Gamma_0(N),\nu)$$. For each $$j$$, let $$a_j(n)$$ denote the $$n$$th Fourier coefficient of $$f_j$$ and let $$r_j$$ denote the spectral parameter. Theorem 3.1. Suppose that $$k=\pm 1/2$$ and that $$\nu$$ satisfies conditions (1) and (2) above. Then for all $$x\geq 2$$ and $$n\geq 1$$ we have nν∑0<rj≤x|aj(n)|2chπrj={x323π2+Oν(x12logx+nβ+ϵ) if k=1/2,x525π2+Oν(x32logx+nβ+ϵx12) if k=−1/2. (3.2) □ The $$n>0$$ case of Theorem 1.5 follows from a direct application of Theorem 3.1 when $$(k,\nu)=(1/2,\chi)$$. For $$n<0$$, we apply Theorem 3.1 to the case $$(k,\nu)=(-1/2,\bar\chi)$$ and use the relation (2.17). 3.1 An auxiliary version of the Kuznetsov trace formula We begin with an auxiliary version of Kuznetsov’s formula ([28, Section 5]) which is Lemma 3 of [39] with $$m=n$$. By assumption (2) there are no Eisenstein series for the multiplier system $$\nu$$. While Proskurin assumes that $$k>0$$ throughout his paper, this lemma is still valid for $$k<0$$ by the same proof. We include the term $$\Gamma(2\sigma-1)$$ which is omitted on the right-hand side of [39, Lemma 3]. Lemma 3.2. Suppose that $$k=\pm 1/2$$ and that $$\nu$$ satisfies conditions (1) and (2) above. For $$n>0$$, $$t\in\mathbb{R}$$, and $$\sigma>1$$ we have −1ik+1∑c>0S(n,n,c,ν)c2σ∫LKit(4πnνcq)(q+1q)2σ−2qk−1dq=Γ(2σ−1)4(2πnν)2σ−1−21−2σπ2−2σnν2−2σΓ(σ−k2+it2)Γ(σ−k2−it2)∑rj|aj(n)|2Λ(σ+it2,σ−it2,rj), (3.3) where $$L$$ is the contour $$|q|=1$$, $$\text{Re}(q)>0$$, from $$-{\rm i}$$ to $${\rm i}$$ and Λ(s1,s2,r)=Γ(s1−12−ir)Γ(s1−12+ir)Γ(s2−12−ir)Γ(s2−12+ir). □ We justify substituting $$\sigma=1$$ in this lemma. Suppose that $$t\in \mathbb{R}$$ and write $$a = 4\pi n_\nu/c$$ and $$q={\rm e}^{{\rm i}\theta}$$ with $$-\pi/2<\theta<\pi/2$$. By [35, (10.27.4), (10.25.2), and (5.6.7)] we have, for $$c$$ sufficiently large, Kit(aeiθ)≪e−π|t|2(eθt+e−θt). So $$K_{\rm it}(a q) \ll 1$$ on $$L$$ and the integral over $$L$$ is bounded uniformly in $$a$$. By this and the assumption (3.1), the left-hand side of (3.3) is absolutely uniformly convergent for $$\sigma \in [1,2]$$. In this range we have |Γ(σ−12+iy)|=|Γ(σ+12+iy)||σ−12+iy|≤2|Γ(σ+12+iy)|. Using this together with the fact that $$\Lambda\left(\sigma+\tfrac{{\rm i}t}2,\sigma-\tfrac{{\rm i}t}2,r_j\right)$$ is positive, we see that the convergence of the right-hand side of (3.3) for $$\sigma>1$$ implies the uniform convergence of the right-hand side for $$\sigma \in [1,2]$$. We may therefore take $$\sigma=1$$. To evaluate the integral on the left-hand side of (3.3) we require the following lemma. Lemma 3.3. Suppose that $$a>0$$. Then ∫LKit(aq)qk−1dq=i2×{1a∫0au−12∫0∞cos(tξ)(cos(uchξ)−sin(uchξ))dξdu if k=12,a∫a∞u−32∫0∞cos(tξ)(cos(uchξ)+sin(uchξ))dξdu if k=−12. □ Proof The case $$k=1/2$$ follows from equation (44) in [39] (correcting a typo in the upper limit of integration) and the integral representation [35, (10.9.8)]. For $$k=-1/2$$, we write ∫LKit(aq)qk−1dq=−(∫−iR−i+∫iiR+∫LR)Kit(aq)qk−1dq, where the first two integrals are along the imaginary axis, and $$L_R$$ is the semicircle $$|q|=R$$, $$\text{Re}(q)>0$$, traversed clockwise. By [35, (10.40.10)], the integral over $$L_R$$ approaches zero as $$R\to\infty$$. For the first and second integrals we change variables to $$q=-{\rm iu}/a$$ and $$q={\rm iu}/a$$, respectively, to obtain ∫LKit(aq)qk−1dq=−a∫a∞(−iKit(iu)−iKit(−iu))duu3/2. The lemma follows from writing $$K_{\rm it}(\pm {\rm iu})$$ as a linear combination of $$J_{\rm it}(u)$$ and $$J_{-{\rm it}}(u)$$ using [35, (10.27.9), (10.27.3)] and applying the integral representation [35, (10.9.8)]. ■ Starting with (3.3) with $$\sigma=1$$ and using the fact that Λ(1+it,1−it,r)=π2ch(πr+πt)ch(πr−πt), (3.4) we replace $$t$$ by $$2t$$, multiply by 4nνπ2g(t), where g(t):=|Γ(1−k2+it)|2shπt, and integrate on $$t$$ from $$0$$ to $$x$$. Applying Lemma 3.3 to the result, we obtain nν∑rj|aj(n)|2chπrjhx(rj)=12π3∫0xg(t)dt+(4nν)1−k2ikπ2+k∑c>0S(n,n,c,ν)c2−kI(x,4πnνc), (3.5) where hx(r):=∫0x2shπtchπrch(πr+πt)ch(πr−πt)dt (3.6) and I(x,a):={∫0xg(t)∫0au−12∫0∞cos(2tξ)(cos(uchξ)−sin(uchξ))dξdudt if k=12,∫0xg(t)∫a∞u−32∫0∞cos(2tξ)(cos(uchξ)+sin(uchξ))dξdudt if k=−12. (3.7) 3.2 The function $$g(t)$$ and the main term A computation involving [35, (5.11.3), (5.11.11)] shows that for fixed $$x$$ and $$y\geq 1$$ we have Γ(x+iy)Γ(x−iy)=2πy2x−1e−πy(1+O(1y)), from which it follows that g(t)=πt1−k+O(t−k)as t→∞]. (3.8) We have g(t)=πΓ(1−k2)2t+O(t2)as t→0. (3.9) This gives the main term 12π3∫0xg(t)dt={x323π2+O(x12) if k=12,x525π2+O(x32) if k=−12. (3.10) 3.3 Preparing to estimate the error term The error term in (3.2) is obtained by uniformly estimating the integral $$I(x,a)$$ in the ranges $$a\leq 1$$ and $$a\geq 1$$. The analysis is similar in spirit to [28, Section 5], but the nature of the function $$g(t)$$ introduces substantial difficulties. In this section we collect a number of facts which we will require for these estimates. Given an interval $$[a,b]$$, let $$\text{var} f$$ denote the total variation of $$f$$ on $$[a,b]$$ which, for $$f$$ differentiable, is given by varf=∫ab|f′(x)|dx. (3.11) We begin by collecting some facts about the functions $$g'(t)$$, $$(g(t)/t)'$$, and $$t(g(t)/t)'$$. Lemma 3.4. Let $$k=\pm 1/2$$. If $$x\geq 1$$ then on the interval $$[0,x]$$ we have var(g(t)/t)′≪1+x−1−k, (3.12) vart(g(t)/t)′≪1+x−k. (3.13) For $$t>0$$ we have g′(t)>0,sgn(g(t)/t)′=−sgnk. (3.14) □ Proof Let $$\sigma=1-k/2$$. Taking the logarithmic derivative of $$g(t)/t$$, we find that (g(t)t)=g(t)h(t), (3.15) where h(t):=1t[i(ψ(σ+it)−ψ(σ−it))−1t+πcothπt],ψ:=Γ′Γ. We have g′(t)=g(t)[i(ψ(σ+it)−ψ(σ−it))+πcothπt], and h′(t)=−h(t)t−1t[ψ′(σ+it)+ψ′(σ−it)−1t2+π2(cschπt)2]. At $$t=0$$ we have the Taylor series g(t)=πΓ(σ)2t+πΓ(σ)2(π26−ψ′(σ))t3+…,h(t)=π23−2ψ′(σ)+(13ψ‴(σ)−π445)t2+…. Thus we have the estimates g(t)≪t,g′(t)≪1,h(t)≪1,andh′(t)≪t as t→0. For large $$t$$ we use [35, (5.11.2) and (5.15.8)] to obtain the asymptotic expansions i(ψ(σ+it)−ψ(σ−it))=−2arctan(tσ)−tσ2+t2+O(1t2), (3.16) ψ′(σ+it)+ψ′(σ−it)=2σσ2+t2+σ2−t2(σ2+t2)2+O(1t3). (3.17) By (3.8), (3.16), and (3.17) we have the estimates g(t)≪t1−k,g′(t)≪t−k,h(t)≪1t2,andh′(t)≪1t3as t→∞. (3.18) From (3.18) and (3.11) it follows that var(g(t)/t)′=∫0x|g(t)h′(t)+g′(t)h(t)|dt≪1+x−1−k and vart(g(t)/t)′=∫0x|tg(t)h′(t)+tg′(t)h(t)+g(t)h(t)|dt≪1+x−k. Using [35, (4.36.3), (5.7.6)] we have h(t)=2∑n=1∞(1t2+n2−1t2+(n−k2)2). The second claim in (3.14) follows from (3.15). The first is simpler. ■ Suppose that $$f\geq 0$$ and $$\phi$$ are of bounded variation on $$[a,b]$$ with no common points of discontinuity. By Corollary 2 of [27], there exists $$[\alpha,\beta]\subseteq[a,b]$$ such that ∫abfdϕ=(inff+varf)∫αβdϕ, where the integrals are of Riemann–Stieltjes type. We obtain the following by taking $$\phi(x)=\int_0^x G(t)\, {\rm d}t$$. Lemma 3.5. Suppose that $$F\geq 0$$ and $$G$$ are of bounded variation on $$[a,b]$$ and that $$G$$ is continuous. Then |∫abF(x)G(x)dx|≤(infF+varF)sup[α,β]⊆[a,b]|∫αβG(x)dx|. □ We will frequently make use of the following well-known estimates for oscillatory integrals (see, e.g., [52, Chapter IV]). Lemma 3.6. (First-derivative estimate). Suppose that $$F$$ and $$G$$ are real-valued functions on $$[a,b]$$ with $$F$$ differentiable, such that $$G(x)/F'(x)$$ is monotonic. If $$|F'(x)/G(x)|\geq m>0$$ then ∫abG(x)e(F(x))dx≪1m. □ Lemma 3.7. (Second-derivative estimate). Suppose that $$F$$ and $$G$$ are real-valued functions on $$[a,b]$$ with $$F$$ twice differentiable, such that $$G(x)/F'(x)$$ is monotonic. If $$|G(x)|\leq M$$ and $$|F''(x)|\geq m>0$$ then ∫abG(x)e(F(x))dx≪Mm. □ 3.4 The first error estimate We estimate the error arising from $$I(x,a)$$ (recall (3.7)) when $$k=1/2$$. Proposition 3.8. Suppose that $$k=1/2$$. For $$a>0$$ we have I(x,a)≪{a(log(1/a)+1) if a≤1,1 if a≥1. □ Proof We write $$I(x,a)$$ as a difference of integrals, one involving $$\cos(u\text{ch}\xi)$$, the other involving $$\sin(u\text{ch} \xi)$$. We will estimate the first integral, the second being similar. We have ∫0∞cos(2tξ)cos(uchξ)dξ=∫0Tcos(2tξ)cos(uchξ)dξ+RT(u,t), where, by the second-derivative estimate, RT(u,t)≪u−12e−T2. For $$0<\varepsilon < a$$ we have, using (3.8), ∫0xg(t)∫εau−12RT(u,t)dudt≪e−T2x32(|loga|+|logε|)→0 as T→∞. So our goal is to estimate limε→0+limT→∞I(x,ε,a,T), where I(x,ε,a,T)=∫0xg(t)∫εau−12∫0Tcos(2tξ)cos(uchξ)dξdudt. Integrating the innermost integral by parts, we find that I(x,ε,a,T)=12∫0xg(t)tsin(2tT)dt∫εau−12cos(uchT)du+12∫0Tshξ∫0xg(t)tsin(2tξ)dt∫εausin(uchξ)dudξ. (3.19) The first-derivative estimate (recalling (3.9) and (3.14)) gives ∫0xg(t)tsin(2tT)dt≪1T and ∫εau−12cos(u ch T)du≪1εchT, so the first term on the right-hand side of (3.19) approaches zero as $$T\to\infty$$. Turning to the second term, set Gx(ξ):=∫0xg(t)tsin(2tξ)dt. Integrating by parts gives Gx(ξ)=πΓ(34)22ξ−g(x)xcos(2xξ)2ξ+12ξ∫0x(g(t)t)′cos(2tξ)dt. (3.20) Applying Lemma 3.5 and (3.12) to the integral in (3.20) we find that 1ξ∫0x(g(t)t)′cos(2tξ)dt≪1ξsup0≤α<β≤x|∫αβcos(2tξ)dt|≪1ξ2. On the other hand, (3.14) gives |1ξ∫0x(g(t)t)′cos(2tξ)dt|≤1ξ∫0x−(g(t)t)′dt≪1ξ. (3.21) It follows that Gx(ξ)=πΓ(34)22ξ−g(x)xcos(2xξ)2ξ+O(min(ξ−1,ξ−2)). (3.22) The integral ∫εausin(uchξ)du evaluates to $$H(a,\xi) - H(\varepsilon, \xi)$$, where H(u,ξ):=−ucos(uchξ)chξ+π/2(chξ)32C(2uchξπ), (3.23) and $$C(x)$$ denotes the Fresnel integral C(x):=∫0xcos(π2t2)dt. For $$u>0$$, write J(x,u):=∫0∞sh ξGx(ξ)H(u,ξ)dξ. (3.24) To show that $$J(x, u)$$ converges, suppose that $$B\geq A\geq T$$. By (3.22), (3.23), and the bound $$C(x)\leq 1$$ we have ∫ABsh ξGx(ξ)H(u,ξ)dξ∫=−u∫ABth ξcos(uchξ)(πΓ(34)22ξ−g(x)xcos(2xξ)2ξ+O(1ξ2))dξ+O(∫ABsh ξξ(chξ)32dξ)∫≪u−12e−A/2+u12A−1+e−A/2, where we have used the first-derivative estimate in the last inequality. We have limε→0+limT→∞I(x,ε,a,T)=12J(x,a)−12limε→0+J(x,ε), (3.25) so it remains only to estimate $$J(x,u)$$ in the ranges $$u\leq 1$$ and $$u\geq 1$$. First, suppose that $$u\leq 1$$. We estimate each of the six terms obtained by multiplying (3.22) and (3.23) in (3.24). Starting with the terms involving $$\cos(u\text{ch}\xi)$$, we have u∫0∞th ξξcos(uchξ)dξ=u(∫0log1u+∫log1u∞)th ξξcos(u ch ξ)dξ≪u(log(1/u)+1) by applying the second-derivative estimate to the second integral. By the same method, ug(x)x∫0∞thξξcos(2xξ)cos(u ch ξ)dξ≪u(log(1/u)+1), since $$g(x)/x\ll 1$$. For the third term we have u∫0∞thξcos(u ch ξ)min(ξ−1,ξ−2)dξ≪u∫01thξξdξ+u∫1∞thξξ2dξ≪u. For the terms involving the Fresnel integral, we use the trivial estimate $$C(x)\leq \min(x,1)$$. For the first of these terms we have ∫0∞shξξ(chξ)32C(2uchξπ)dξ≪u∫0log1uthξξdξ+∫log1u∞dξchξ≪u(log(1/u)+1). By the same method, the remaining two terms are $$\ll \sqrt{u} \,(\log(1/u)+1)$$. It follows that J(x,u)≪u(log(1/u)+1) for u≤1. (3.26) For $$u\geq 1$$ we show that $$J(x, u)\ll 1$$. Write J(x,u)=J1(x,u)+J2(x,u):=(∫01/u+∫1/u∞)shξGx(ξ)H(u,ξ)dξ. Since $$G_x(\xi)\ll \xi^{-1}$$ and $$H(u,\xi)\ll \sqrt{u}/\text{ch}\xi$$, we have J1(x,u)≪u∫01/uthξξdξ≪1. We break $$J_2(x,u)$$ into six terms, using (3.20) in place of (3.22). We start with the terms involving $$\pi\Gamma(3/4)^2/2\xi$$. By the first-derivative estimate we have u∫1/u∞thξξcos(uchξ)dξ≪1. Estimating the Fresnel integral trivially, we have ∫1/u∞shξξ(chξ)32C(2uchξπ)dξ≪∫0∞dξchξ≪1. The two terms involving $$\cos(2x\xi)/\xi$$ are treated similarly, and are $$\ll 1$$. The term involving the integral in (3.20) and the Fresnel integral can be estimated trivially using (3.21); it is also $$\ll $$1. The final term, u∫1/u∞shξcos(uchξ)ξchξ∫0x(g(t)t)′cos(2tξ)dtdξ, is more delicate. We write the integral as u∫1/u∞{1ξ ch ξ∫0x(g(t)t)′cos(2tξ)dt}⋅{shξcos(uchξ)dξ}=u∫1/u∞UdV and integrate by parts. By (3.21) we have uUV|1/u∞≪u⋅uch(1/u)⋅1u≪1. Write $$U'=R+S$$, where R=(1ξ ch ξ)′∫0x(g(t)t)′cos(2tξ)dt,S=2ξ ch ξ∫0xt(g(t)t)′sin(2tξ)dt. By (3.21) we have R≪1ξ ch ξ(1ξ+thξ)≪{(ξ2chξ)−1 if ξ≤1,(ξ ch ξ)−1 if ξ≥1. Applying Lemma 3.5 and the estimate (3.13) we find that S≪1ξ2 ch ξ. Thus we have u∫1/u∞VdU≪1u∫1/u1dξξ2+1u∫1∞dξξ ch ξ≪1. We conclude that J(x,u)≪1 for u≥1. (3.27) The proposition follows from (3.25)–(3.27). ■ 3.5 The second error estimate In the case when $$k=-1/2$$ we must keep track of the dependence on $$x$$. Proposition 3.9. Suppose that $$k=-1/2$$. For $$a>0$$ we have I(x,a)≪x×{a−12(log(1/a)+1) if a≤1,a−1 if a≥1. □ Proof As before, we estimate the term involving $$\cos(u\text{ch}\xi)$$ and we write ∫0∞cos(2tξ)cos(u ch ξ)dξ=12tsin(2tT)cos(uchT)+u2t∫0Tsh ξsin(2tξ)sin(uchξ)dξ+RT(u,t), with $$R_T(u,t)\ll u^{-1/2}{\rm e}^{-T/2}$$. Recalling (3.8) and (3.9), we have ∫0xg(t)∫a∞u−32RT(u,t)dudt≪a−1x52e−T/2→0 as T→∞ and, by the first-derivative estimate, ∫0xg(t)tsin(2tT)dt∫a∞u−32cos(uchT)du≪a−32x32e−T→0 as T→∞. Our goal is to estimate J(x,a):=∫0∞shξGx(ξ)H(a,ξ)dξ, where Gx(ξ):=∫0xg(t)tsin(2tξ)dt (3.28) and H(a,ξ):=∫a∞u−12sin(uchξ)du. (3.29) The integral defining $$J(x,a)$$ converges by an argument as in the proof of Proposition 3.8. Recalling (3.8) and (3.14), the first-derivative estimate gives Gx(ξ)≪xξ. (3.30) For a better estimate, we integrate by parts in (3.28) to get Gx(ξ)=πΓ(54)22ξ−g(x)xcos(2xξ)2ξ+12ξ∫0x(g(t)t)′cos(2tξ)dt. (3.31) We estimate the third term in two ways. Taking absolute values and using (3.14), it is $$\ll \sqrt x/\xi$$; by (3.12) and Lemma 3.5 it is $$\ll 1/\xi^2$$. Thus 1ξ∫0x(g(t)t)′cos(2tξ)dt≪min(xξ,1ξ2). (3.32) We also need two estimates for $$H(a,\xi)$$. The first-derivative estimate applied to (3.29) gives H(a,ξ)≪1achξ, (3.33) while integrating (3.29) by parts and applying the first-derivative estimate gives H(a,ξ)=cos(a ch ξ)a ch ξ+O(1a32(chξ)2). (3.34) First suppose that $$a\leq 1$$. Write J(x,a)=J1(x,a)+J2(x,a)=(∫01/a+∫1/a∞)shξGx(ξ)H(a,ξ)dξ. By (3.30) and (3.33) we have J1(x,a)≪xa(∫01+∫11/a)th ξξdξ≪xa(1+log(1/a)). By (3.34), (3.30), (3.31), and the second bound in (3.32) we have J2(x,a)=1a∫1/a∞th ξGx(ξ)cos(achξ)dξ+O(xa32∫1/a∞shξξ(chξ)2dξ)≪1a∫1/a∞th ξξcos(achξ)dξ+g(x)xa∫1/a∞thξξcos(2xξ)cos(achξ)dξ+x/a. Applying the first-derivative estimate to each integral above, we find that $$J_2(x,a)\ll \sqrt{x/a}$$, from which it follows that J(x,a)≪xa(log(1/a)+1) for a≤1. (3.35) Now suppose that $$a\geq 1$$. Using (3.30), the contribution to $$J(x,a)$$ from the second term in (3.34) is a−32∫0∞sh ξ(ch ξ)2Gx(ξ)dξ≪a−32x. We break the integral involving $$\cos(a\text{ch} \xi)$$ at $$1/\sqrt a$$. By (3.30) the contribution from the initial segment is $$\ll a^{-1}\sqrt x$$. It remains to estimate 1a∫1/a∞thξGx(ξ)cos(achξ)dξ, which we break into three terms using (3.31). For the first two terms the first-derivative estimate gives 1a∫1/a∞thξξcos(achξ)dξ≪1a,g(x)xa∫1/a∞thξξcos(2xξ)cos(achξ)dξ≪xa. The remaining term, 1a∫1/a∞thξξcos(achξ)∫0x(g(t)t)′sin(2tξ)dtdξ, requires more care, but we follow the estimate for the corresponding term in the proof of Proposition 3.8 using Lemma 3.5 and (3.13). The details are similar, and the contribution is $$ \ll \frac{\sqrt x}{a}$$, which, together with (3.35), completes the proof. ■ 3.6 Proof of Theorem 3.1 We give the proof for the case $$k=1/2$$, as the other case is analogous. By Proposition 3.8 we have I(x,a)≪min(1,a1/2log(1/a))≪δaδfor any 0≤δ<1/2. So by (3.5) and (3.10) we have nν∑rj|aj(n)|2chπrjhx(rj)=x323π2+Oδ(x12+n12+δ∑c>0|S(n,n,c,ν)|c3/2+δ). Setting $$\delta=\beta-1/2$$, with $$\beta$$ as in (3.1), this becomes nν∑rj|aj(n)|2chπrjhx(rj)=x323π2+O(x12+nβ+ϵ). (3.36) The function $$h_x(r)$$ (recall (3.6)) is a smooth approximation to the characteristic function of $$[0,x]$$. We recall some properties from [28, (5.4–5.7)] for $$x\geq 1$$: hx(r)=2πarctan(eπx−πr)+O(e−πr)for r≥1, hx(r)=1+O(x−3+e−πr)for 1≤r≤x−logx, (3.37) hx(r)≪e−π(r−x)for r≥x+logx, (3.38) 0<hx(r)<1for r≥0. (3.39) For $$x\geq 1$$ and $$0\leq r\leq x$$ we bound $$h_x(r)$$ uniformly from below as follows. If $$r\geq 1$$ we have hx(r)=4chπr∫0xshπtch2πr+ch2πtdt≥2chπrch2πr∫0rshπtdt>14, while if $$r\leq 1$$ we have hx(r)≥4chπr∫01shπtch2πr+ch2πtdt≥4∫01shπtch2π+ch2πtdt>3100. Similarly, we have $$h_x(r)>2/5$$ for $$r\in {\rm i}[0,1/4]$$. Set A(x):=nν∑rj≤xorrj∈iR|aj(n)|2chπrj. (3.40) Then A(x)≪nν∑rj≤xor rj∈iR|aj(n)|2chπrjhx(rj)≪nν∑rj|aj(n)|2chπrjhx(rj) which, together with (3.36), gives A(x)≪x32+nβ+ϵ. (3.41) Let $$A^*(x)$$ denote the sum (3.40) restricted to $$1\leq r_j\leq x$$. By (3.41) with $$x=1$$ we have $$A(x)=A^*(x)+O(n^{\beta+\epsilon})$$, so in what follows we work with $$A^*(x)$$. Assume that $$x\geq 2$$ and set $$X=x+2\log x$$ so that $$X-\log X\geq x$$. Then by (3.37) there exist $$c_1,c_2>0$$ such that nν∑1≤rj|aj(n)|2chπrjhX(rj)≥A∗(x)(1−c1x3)−c2nν∑1≤rj≤x|aj(n)|2chπrje−πrj. On the other hand, (3.36) gives nν∑1≤rj|aj(n)|2chπrjhX(rj)=x323π2+O(x12logx+nβ+ϵ). By (3.41) we have nν∑1≤rj≤x|aj(n)|2ch πrje−πrj=nν∑ℓ=1⌊x⌋∑ℓ≤rj≤ℓ+1|aj(n)|2chπrje−πrj≪∑ℓ=1∞e−πℓ(ℓ32+nβ+ϵ)≪nβ+ϵ, (3.42) so that A∗(x)≤x323π2+O(x12logx+nβ+ϵ). (3.43) Similarly, replacing $$x$$ by $$x-\log x$$ in (3.36) and using (3.38), (3.39) give A∗(x)≥x323π2+O(x12logx+nβ+ϵ)+O(nν∑rj>x|aj(n)|2chπrje−π(rj−x+logx)). Arguing as in (3.42), the error term is $$\ll x^{-\frac 32} + n^{\beta+\epsilon} x^{-3}$$, from which we obtain A∗(x)≥x323π2+O(x12logx+nβ+ϵ). (3.44) Equations (3.43) and (3.44)together give Theorem 3.1. ■ 4. The Kuznetsov Trace Formula We require a variant of Kuznetsov’s formula [28, Theorem 1], which relates weighted sums of the Kloosterman sums $$S(m,n,c,\chi)$$ on one side to spectral data on the other side (in this case, weighted sums of coefficients of Maass cusp forms). In [39], Proskurin generalized Kuznetsov’s theorem to any weight $$k$$ and multiplier system $$\nu$$, but only for the case $$m,n>0$$. Here we treat the mixed sign case when $$k=1/2$$ and $$\nu=\chi$$; the proof follows the same lines. Blomer [6] has recorded this formula for twists of the theta-multiplier by a Dirichlet character; we provide a sketch of the proof in the present case since there are some details which are unique to this situation. Suppose that $$\phi: [0, \infty)\rightarrow \mathbb{C}$$ is four times continuously differentiable and satisfies ϕ(0)=ϕ′(0)=0,ϕ(j)(x)≪ϵx−2−ϵ(j=0,…,4)as x→∞ (4.1) for some $$\epsilon>0$$. Define ϕˇ(r):=chπr∫0∞K2ir(u)ϕ(u)duu. (4.2) We state the analogue of the main result of [39] in this case. Theorem 4.1. Suppose that $$\phi$$ satisfies the conditions (4.1). As in (2.23) let $$\rho_j(n)$$ denote the coefficients of an orthonormal basis $$\{u_j\}$$ for $$\mathcal{S}_{1/2}(1,\chi)$$ with spectral parameters $$r_j$$. If $$m>0$$ and $$n<0$$ then ∑c>0S(m,n,c,χ)cϕ(4πm~|n~|c)=8im~|n~|∑rjρj(m)¯ρj(n)chπrjϕˇ(rj). □ The proof involves evaluating the inner product of Poincaré series in two ways. Let $$\tau=x+{\rm i}y \in \mathbb{H}$$ and $$s=\sigma+{\rm i}t\in \mathbb{C}$$. Let $$k\in \mathbb{R}$$ and let $$\nu$$ be a multiplier system for $$\Gamma$$ in weight $$k$$. For $$m>0$$ the Poincaré series $$\mathcal U_m(\tau,s,k,\nu)$$ is defined by Um(τ,s,k,ν):=∑γ∈Γ∞∖Γν(γ)¯j(γ,τ)−kIm(γτ)se(mνγτ),σ>1. (4.3) This function satisfies the transformation law Um(⋅,s,k,ν)|kγ=ν(γ)Um(⋅,s,k,ν)for all γ∈Γ and has Fourier expansion (see equation (15) of [39]) Um(τ,s,k,ν)=yse(mντ)+ys∑ℓ∈Z∑c>0S(m,ℓ,c,ν)c2sB(c,mν,ℓν,y,s,k)e(ℓνx), (4.4) where B(c,mν,ℓν,y,s,k)=y∫−∞∞e(−mνc2y(u+i))(u+i|u+i|)−ke(−ℓνyu)duy2s(u2+1)s. (4.5) In the next two lemmas we will obtain expressions for the inner product ⟨Um(⋅,s1,12,χ), U1−n(⋅,s2,−12,χ¯)¯⟩. The first lemma is stated in such a way that symmetry may be exploited in its proof. Lemma 4.2. Suppose that $$(k, \nu)$$ is one of the pairs $$(1/2, \chi)$$ or $$(-1/2, \bar\chi)$$. Suppose that $$m>0$$, $$n<0$$. Then for $$\text{Re} s_1>1$$, $$\text{Re} s_2>1$$ we have ⟨Um(⋅,s1,k,ν), U1−n(⋅,s2,−k,ν¯)¯⟩=i−k23−s1−s2π (mν|nν|)s2−s12Γ(s1+s2−1)Γ(s1−k/2)Γ(s2+k/2)∑c>0S(m,n,c,ν)cs1+s2Ks1−s2(4πmν|nν|c). (4.6) □ Proof of Lemma 4.2 We will prove (4.6) under the assumption $$\text{Re} s_2>\text{Re} s_1$$. Recall the definitions (2.25), (2.5), and (2.13), and recall that (1−n)ν¯=−nν. Replacing $$\gamma$$ by $$-\gamma^{-1}$$ in (2.25) and using $$\nu(-I)={\rm e}^{-\pi {\rm i} k}$$, we find that S(m,n,c,ν)=eπikS(1−n,1−m,c,ν¯). The case $$\text{Re} s_1>\text{Re} s_2$$ then follows since both sides of the putative identity are invariant under the changes $$s_1\leftrightarrow s_2$$, $$k\leftrightarrow -k$$, $$\nu\leftrightarrow \overline\nu$$, $$m\leftrightarrow 1-n$$ (see [35, (10.27.3)]). The case $$\text{Re} s_1=\text{Re} s_2$$ follows by continuity. Let $$I_{m,n}(s_1,s_2)$$ denote the inner product on the left-hand side of (4.6). Unfolding using (4.4) and (4.3), we find that Im,n(s1,s2)=∫Γ∞∖HUm(τ,s1,k,ν)(Imτ)s2e(−nν)dμ=∫0∞ys1+s2−2∑c>0S(m,n,c,ν)c2s1B(c,mν,nν,y,s1,k)e2πnνydy. (4.7) By [39, (17)] we have B(c,mν,nν,y,s1,k)≪s1y1−2Res1e−π|nν|y. Therefore the integral in (4.7) is majorized by the convergent (since $$\text{Re} s_2>\text{Re} s_1$$) integral ∫0∞yRe s2−Re s1−1e3πnνydy. Interchanging the integral and sum in (4.7) and using (4.5), we find that Im,n(s1,s2)=∑c>0S(m,n,c,ν)c2s1∫−∞∞(u+i|u+i|)−k(u2+1)−s1×(∫0∞ys2−s1−1e(−mνc2y(u+i)−nνyu)e2πnνydy)du. Setting $$w=2\pi n_\nu({\rm iu}-1)y$$ and $$a=4\pi\sqrt{ m_\nu| n_\nu|}/c$$, the inner integral becomes (2πnν(iu−1))s1−s2∫0(1−iu)∞exp(−w−a24w)dwws1−s2+1. We may shift the path of integration to the positive real axis since ∫T(1−iu)Texp(−w−a24w)dwws1+s2+1≪T−Re(s1−s2)e−Tas T→∞. Using the integral representation [35, (10.32.10)] for the $$K$$-Bessel function, we find that the inner integral is equal to 2(mν|nν|)s2−s12(1−iu)s1−s2cs1−s2Ks1−s2(4πmν|nν|c). It follows that Im,n(s1,s2)=2(mν|nν|)s2−s12∑c>0S(m,n,c,ν)cs1+s2Ks1−s2(4πmν|nν|cc)×∫−∞∞(u+i|u+i|)−k(u2+1)−s1(1−iu)s1−s2du. Lemma 4.2 follows after using [35, (5.12.8)] to evaluate the integral. ■ Recall the definition Λ(s1,s2,r)=Γ(s1−12−ir)Γ(s1−12+ir)Γ(s2−12−ir)Γ(s2−12+ir). Lemma 4.3. Suppose that $$\text{Re} s_1>1, \text{Re} s_2>1$$. As in (2.23) let $$\rho_j(n)$$ and $$r_j$$ denote the coefficients and spectral parameters of an orthonormal basis $$\{u_j\}$$ for $$\mathcal{S}_{1/2}(1,\chi)$$. If $$m>0$$ and $$n<0$$ then ⟨Um(⋅,s1,12,χ), U1−n(⋅,s2,−12,χ¯)¯⟩=(4πmχ)1−s1(4π|nχ|)1−s2Γ(s1−1/4)Γ(s2+1/4)∑rjρj(m)¯ρj(n)Λ(s1,s2,rj). □ Proof of Lemma 4.3 In this situation there are no Eisenstein series. So for any $$f_1,f_2 \in \mathcal{L}_{\frac 12}(1,\chi)$$ we have the Parseval identity [39, (27)] ⟨f1,f2⟩=∑rj⟨f1,uj⟩⟨f2,uj⟩¯. (4.8) By [39, (32)] we have ⟨Um(z,s1,12,χ),uj⟩=ρj(m)¯(4πmχ)1−s1Γ(s1−12−irj)Γ(s1−12+irj)Γ(s1−14). (4.9) Unfolding as in the last lemma using the definition (4.3) for $$\mathcal U_{1-n}$$ and the Fourier expansion (2.23) for $$u_j$$ gives ⟨U1−n(⋅,s2,−12,χ¯)¯,uj⟩¯=⟨uj,U1−n(⋅,s2,−12,χ¯)¯⟩=ρj(n)∫0∞ys2−2W−14,irj(4π|nχ|y)e2πnχydy. Using [35, (13.23.4) and (16.2.5)] the latter expression becomes ρj(n)(4π|nχ|)1−s2Γ(s2−12−irj)Γ(s2−12+irj)Γ(s2+14). The lemma follows from this together with (4.8) and (4.9). ■ We are now ready to prove Theorem 4.1. Recall the notation $$\tilde n=n_\chi$$. Proof of Theorem 4.1 Equating the right-hand sides of Lemmas 4.2 (for $$k=1/2$$) and 4.3 and setting s1=σ+it2,s2=σ−it2 we obtain (for $$\sigma>1$$) i−1223−2σπΓ(2σ−1)∑c>0S(m,n,c,χ)c2σKit(4πm~|n~|c) =(4πm~)1−σ(4π|n~|)1−σ∑rjρj(m)¯ρj(n)Λ(σ+it2,σ−it2,rj). (4.10) We justify the substitution of $$\sigma=1$$ in (4.10). By [35, (10.45.7)] we have Kit(x)≪(t shπt)−1/2as x→0. Using (2.30) we see that the left side of (4.10) converges absolutely uniformly for $$\sigma\in[1, 2]$$. For the right-hand side we use the inequality $$|\rho_j(m)\rho_j(n)|\leq |\rho_j(m)|^2+|\rho_j(n)|^2$$. The argument which follows Lemma 3.2 shows that the right side converges absolutely uniformly for $$\sigma\in[1, 2]$$. Using (3.4), we find that i−12∑c>0S(m,n,c,χ)c2Kit(4πm~|n~|c)=π2∑rjρj(m)¯ρj(n)chπ(t2−rj)chπ(t2+rj). (4.11) Letting $$\phi$$ be a function satisfying the conditions (4.1), multiply both sides of (4.11) by 2π2tshπt∫0∞Kit(u)ϕ(u)duu2 and integrate from $$0$$ to $$\infty$$. We apply the Kontorovich–Lebedev transform ([38, (35)] or [35,(10.43.30–31)]) 2π2∫0∞Kit(x)tshπt(∫0∞Kit(u)ϕ(u)duu2)dt=ϕ(x)x to the left-hand side of (4.11) and the transform [38, (39)] ∫0∞tshπtchπ(t2+r)chπ(t2−r)(∫0∞Kit(u)ϕ(u)duu2)dt=2chπrϕˇ(r) (recalling the definition (4.2)) to the right-hand side. Then (4.11) becomes i−1/24πm~|n~|∑c>0S(m,n,c,χ)cϕ(4πm~|n~|c)=2π∑rjρj(m)¯ρj(n)chπrjϕˇ(rj), and Theorem 4.1 follows. ■ 5. A Theta Lift for Maass Cusp Forms In this section we construct a version of the Shimura correspondence as in [46, Section 3] and [24, Section 4] for Maass forms of weight $$1/2$$ on $$\Gamma_0(N)$$ with the eta multiplier twisted by a Dirichlet character. Throughout this section, $$N\equiv 1\bmod {24}$$ is a positive integer and $$\psi$$ is an even Dirichlet character modulo $$N$$. When working with the Shimura correspondence and the Hecke operators, it is convenient to write the Fourier expansion of $$G\in \mathcal{S}_{\frac 12}(N,\psi\chi,r)$$ as G(τ)=∑n≡1(24)a(n)Wsgn(n)4,ir(π|n|y6)e(nx24). (5.1) When $$G=u_j$$ as in (2.23) we have the relation ρj(n+2324)=a(n). For each $$t\equiv 1\bmod {24}$$, the following theorem gives a map St:S12(N,ψχ,r)→S0(6N,ψ2,2r). We will only apply this theorem in the case $$N=1$$. The proof in the case $$N=t=1$$ is simpler than the general proof given below; we must work in this generality since the proof in the case $$t>1$$ requires the map $$S_1$$ on $$\Gamma_0(t)$$ forms. Theorem 5.1. Let $$N\equiv 1\bmod{24}$$ be squarefree. Suppose that $$G\in \mathcal{S}_{\frac 12}(N,\psi\chi,r)$$ with $$r\neq {\rm i}/4$$ has Fourier expansion (5.1). Let $$t\equiv 1\bmod{24}$$ be a squarefree positive integer and define coefficients $$b_t(n)$$ by the relation ∑n=1∞bt(n)ns=L(s+1,ψ(t∙))∑n=1∞ (12n)a(tn2)ns−12, (5.2) where $$L(s,\psi)$$ is the usual Dirichlet $$L$$-function. Then the function $$S_t(G)$$ defined by (StG)(τ):=∑n=1∞bt(n)W0,2ir(4πny)cos(2πnx) is an even Maass cusp form in $$\mathcal{S}_0(6N,\psi^2,2r)$$. □ An important property of the Shimura correspondence is Hecke equivariance. Since we only need this fact for $$N=1$$, we state it only in that case. Corollary 5.2. Suppose that $$G\in \mathcal{S}_{\frac 12}\left(1,\chi,r\right)$$ with $$r\neq {\rm i}/4$$ and that $$t\equiv 1\bmod {24}$$ is a squarefree positive integer. Then for any prime $$p\geq 5$$ we have TpSt(G)=(12p)St(Tp2G). □ Proof We prove this in the case $$p\nmid t$$; the other case is similar, and easier. Let $$G$$ have coefficients $$a(n)$$ as in (5.1). The coefficients $$b_t(n)$$ of $$S_t(G)$$ are given by bt(n)=∑jk=n (tj) (12k)kja(tk2), and the coefficients $$A(n)$$ of $$T_{p^2} G$$ are given by A(n)=pa(p2n)+p−12 (12np)a(n)+p−1a(n/p2). We must show that p12bt(pn)+p−12bt(n/p)=(12p)∑jk=n(tj)(12k)kjA(tk2). (5.3) Writing $$n=p^\alpha n'$$ with $$p\nmid n'$$, the left-hand side of (5.3) equals p−12∑ℓ=0α+1Sℓ+p12∑ℓ=0α−1Sℓ,whereSℓ=p−α(tp)α+1−ℓ(12p)ℓp3ℓ2∑jk=n′ (tj)(12k)kja(tp2ℓk2). A computation shows that the right-hand side of (5.3) equals p−12∑ℓ=0αSℓ+1+p−12S0+p12∑ℓ=1αSℓ−1, and the corollary follows. ■ Using Theorem 5.1 we can rule out the existence of exceptional eigenvalues in $$\mathcal S_{1/2}(1,\chi)$$ and obtain a lower bound on the second smallest eigenvalue $$\lambda_1=\frac14+r_1^2$$ (we note that Bruggeman [9, Theorem 2.15] obtained $$\lambda_1>\frac14$$ using different methods). Theorem 5.1 shows that $$2r_1$$ is bounded below by the smallest spectral parameter for $$\mathcal S_0(6,\boldsymbol 1)$$. Huxley [20, 21] studied the problem of exceptional eigenvalues in weight $$0$$ for subgroups of $${\rm SL}_2(\mathbb{Z})$$ whose fundamental domains are “hedgehog” shaped. On page 247 of [20] we find a lower bound which, for $$N=6$$, gives $$2r_1>0.4$$. Computations of Strömberg [32] suggest (since $$2r_1$$ corresponds to an even Maass cusp form) that $$2r_1 \approx 3.84467$$. Using work of Booker and Strömbergsson [7] on Selberg’s eigenvalue conjecture we can prove the following lower bound. Corollary 5.3. Let $$\lambda_1=\frac 14+r_1^2$$ be as above. Then $$r_1>1.9$$. □ Proof Let $$\{\tilde r_j\}$$ denote the set of spectral parameters corresponding to even forms in $$\mathcal S_0(6,\boldsymbol 1)$$. We show that each $$\tilde r_j>3.8$$ using the method described in [7, Section 4] in the case of level $$6$$ and trivial character. We are thankful to the authors of that paper for providing the numerical details of this computation. Given a suitable test function $$h$$ such that $$h(t)\geq 0$$ on $$\mathbb{R}$$ and $$h(t)\geq 1$$ on $$[-3.8,3.8]$$, we compute via an explicit version of the Selberg trace formula [7, Theorem 4] that ∑r~jh(r~j)<1, (5.4) so we cannot have $$\tilde r_j \leq 3.8$$ for any $$j$$. Let f(t)=(sint/6t/6)2∑n=02xncos (nt3), where x0=1.167099688946872,x1=1.735437017086616,x2=0.660025094420283. With $$h=f^2$$, we compute that the sum in (5.4) is approximately $$0.976$$. ■ The proof of Theorem 5.1 occupies the remainder of this section. We first modify the theta-functions introduced by Shintani [51] and Niwa [36]. Next, we construct the Shimura lift for $$t=1$$ and derive the relation (5.2) in that case. The function $$S_t(G)$$ is obtained by applying the lift $$S_1$$ to the form $$G(t\tau)$$. We conclude the section by showing that $$S_t(G)$$ has the desired level. 5.1 The theta-functions of Shintani and Niwa In this subsection we adopt the notation of [36] for easier comparison with that paper. Let $$Q=\frac{1}{12N} \left(\begin{smallmatrix} &&-2 \\ &1& \\ -2&& \end{smallmatrix}\right)$$, and for $$x,y\in \mathbb{R}^3$$ define ⟨x,y⟩=xTQy=112N(x2y2−2x1y3−2x3y1). The signature of $$Q$$ is $$(2,1)$$. Let $$L\subset L'\subset L^*$$ denote the lattices L=NZ⊕12NZ⊕6NZ,L′=Z⊕NZ⊕6NZ,L∗=Z⊕Z⊕6Z. Then $$L^*$$ is the dual lattice of $$L$$, and for $$x,y\in L$$ we have $$\langle x,y \rangle \in \mathbb{Z}$$ and $$\langle x,x \rangle \in 2\mathbb{Z}$$. Let $$f:\mathbb{R}^3\to \mathbb{C}$$ be a Schwarz function satisfying the conditions of [36, Corollary 0] for $$\kappa=1$$. If $$\psi$$ is a character mod $$N$$ and $$h=(h_1,Nh_2,0)\in L'/L$$, define ψ1(h):=ψ(h1)(12h2). For $$z=u+{\rm i}v$$ we define θ(z,f,h):=v−14∑x∈L[r0(σz)f](x+h) and θ(z,f):=∑h∈L′/Lψ¯1(h)θ(z,f,h), (5.5) where $$\sigma\mapsto r_0(\sigma)$$ is the Weil representation (see [36, p. 149]) and σz:= (v12uv−120v−12),z=u+iv∈ H. The following lemma gives the transformation law for $$\theta(z,f)$$. Lemma 5.4. Suppose that $$N\equiv 1 \pmod {24}$$ and that $$\sigma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right)\in \Gamma_0(N)$$. With $$\theta(z, f)$$ defined as in (5.5) we have θ(σz,f)=ψ¯(d) (Nd)χ(σ)(cz+d)12θ(z,f), where $$\chi$$ is the eta multiplier (2.6). □ In the proof of Lemma 5.4 we will encounter the quadratic Gauss sum G(a,b,c)=∑xmodce(ax2+bxc). With $$g=(a,c)$$, we have G(a,b,c)={gG(a/g,b/g,c/g) if g∣b,0 otherwise. (5.6) We require the following identity relating the Dedekind sums $$s(d,c)$$ to these Gauss sums. This generalizes the results of Lemma 1 and Section 4 of [54], and of Lemma 5 of [1]. Lemma 5.5. Suppose that $$(c,d)=1$$ and that $$k\in\mathbb{Z}$$. Let $$\bar d$$ satisfy $$d\bar d\equiv 1\pmod c$$. Then 12c(12k)e(d¯(k2−1)24c)eπis(d,c)=∑hmod12(12h)e(−d(h2−124)c−hk12c)G(−6d,dh+k,c). (5.7) □ Proof First suppose that $$(k,6)=1$$. If $$c$$ is even and $$d\bar d\equiv 1\pmod {2c}$$, or if $$c$$ is odd, then by [1, Lemma 5 and (4.1)] we have 12c(12k)e(d¯(k2−1)24c)eπis(d,c)=e(k12c)∑jmod2ce(−3dj2−dj+j(c+k)2c)+e(−k12c)∑jmod2ce(−3dj2−dj+j(c−k)2c). (5.8) If $$c$$ is even and $$d\bar d\not\equiv 1\pmod{2c}$$ then, applying (5.8) with $$\bar d$$ replaced by $$\bar d+c$$, we see that (5.8) is true in this case as well. Splitting each of the sums in (5.8) into two sums by writing $$j=2x$$ or $$j=2x+1$$ shows that (5.8) equals the right-hand side of (5.7). It remains to show that the right-hand side of $$(5.7)$$ is zero when $$\delta:=(k,6)>1$$. If $$(\delta, c)>1$$ then (5.6) shows that $$\left(\frac{{12}}{{h}}\right)G(-6d,dh+k,c)=0$$. If $$(\delta,c)=1$$, write A(h)=e(−d(h2−124)c−hk12c)G(−6d,dh+k,c). If $$\delta=2$$ or $$6$$ then a computation shows that, replacing $$x$$ by $$x-\bar 2$$ in $$G(-6d,dh+k,c)$$, we have $$A(h)=A(h-6)$$ for $$(h,6)=1$$. Similarly, if $$\delta=3$$ and $$c$$ is odd then, replacing $$x$$ by $$x+\bar 6 h$$, we find that $$A(h)=-A(-h)$$. Finally, if $$\delta=3$$ and $$c$$ is even then, replacing $$x$$ by $$x - \bar 3$$, we find that $$A(1)=A(5)$$ and $$A(7)=A(11)$$. In each case we find that ∑hmod12 (12h)A(h)=0. ■ Proof of Lemma 5.4 Since $$\sigma_{z+1}=\left({{1} \atop {0}} \quad {{1} \atop {1}}\right)\sigma_z$$, Proposition 0 of [36] gives θ(z+1,f)=∑h∈L′/Le(Nh2224)ψ¯1(h)θ(z,f,h)=e(124)θ(z,f). For $$\sigma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right) \in \Gamma_0(N)$$ with $$c>0$$ we have, by Corollary 0 of [36], θ(σz,f)=(cz+d)12∑k∈L∗/Lθ(z,f,k)⋅−i∑h∈L′/Lψ¯1(h)c(h,k)σ, where c(h,k)σ=1(Nc)3212e(d⟨k,k⟩2c)∑r∈L/cLe(a⟨h+r,h+r⟩2c−⟨k,h+r⟩c). (5.9) For $$h\in L'/L$$, $$k\in L^*/L$$, and $$r\in L/cL$$, we write $$h=(h_1,Nh_2,0)$$, $$k=(k_1,k_2,6k_3)$$, and $$r=(Nr_1,12Nr_2,6Nr_3)$$. Then the sum on $$r$$ in (5.9) equals e(aNh22−2h2k224c)G(6aN,aNh2−k2,c)S(c), (5.10) where (recalling that $$N \mid c$$), S(c)=e(h1k3Nc)∑r1,r3(c)e(−aNr1r3−ah1r3+r1k3+r3k1c)={Nce(dk1k3′c) if k3=Nk3′ and k1≡ah1(modN),0 otherwise. (5.11) So by (5.10), (5.11), and (5.6) we have $$c(h,k)_\sigma=0$$ unless $$k=(k_1,Nk_2',6Nk_3') \in L'$$, in which case $$k_1\equiv ah_1\pmod N$$. For such $$k$$ we have ∑h∈L′/Lψ¯1(h)c(h,k)σ=ψ¯(dk1)12c′e(d(k2′)224c′)∑h2mod12(12h2)e(ah22−2h2k2′24c′)G(6a,ah2−k2′,c′). Applying Lemma 5.5 we have −i∑h∈L′/Lψ¯1(h)c(h,k)σ=ψ¯(d)ψ¯1(k)−ie(a+d24c′)eπis(−a,c′). By (68.4) and (68.5) of [43] we have $$s(-a,c')=-s(a,c')=-s(d,c')$$. Therefore θ(σz,f)=ψ¯(d)χ (aNbc′d)(cz+d)12θ(z,f). By (2.9) and the assumption $$N\equiv 1\pmod{24}$$, we have $$\chi \left( \left( {{a}\atop{{c}'}}\quad {{Nb} \atop {d}} \right) \right)=\left( \frac{N}{d} \right)\chi (\left( {a \atop c} \quad {b \atop d} \right))$$, from which the lemma follows. ■ For $$w=\xi+{\rm i}\eta\in \mathbb{H}$$ and $$0\leq \theta<2\pi$$, let g=g(w,θ)= (1ξ01)(η1200η−12) (cosθsinθ−sinθcosθ)∈SL2(R). The action of $${\rm SL}_2(\mathbb{R})$$ on $$\mathbb{R}^3$$ is given by $$g.x=x'$$, where (x1′x2′/2x2′/2x3′)=g(x1x2/2x2/2x3)gT. (5.12) Since $$\langle x,x \rangle=-\frac 1{3N} \det\left({{x_1} \atop {x_2/2}} \quad {{x_2/2} \atop {x_3}}\right)$$ we have $$\langle gx,gx \rangle = \langle x,x \rangle$$. The action of $${\rm SL}_2(\mathbb{R})$$ on functions $$f:\mathbb{R}^3\to \mathbb{C}$$ is given by $$gf(x):=f(g^{-1}x)$$. We specialize to $$f=f_3$$, with f3(x)=exp(−π12N(2x12+x22+2x32)) as in [36, Example 3], and we consider the function $$\theta(z,gf_3)$$. Let $$k(\theta)=\left({{\cos \theta}\atop {-\sin\theta}}\quad {{\sin\theta} \atop {\cos\theta}}\right)$$. Since $$f_3(k(\theta)x)=f_3(x)$$, the function $$\theta(z,gf_3)$$ is independent of the variable $$\theta$$. Therefore it makes sense to define ϑ(z,w):=θ(z,g(w,0)f3)=v12∑x∈L′ψ¯1(x)e(u2⟨x,x⟩)f3(vσw−1x), (5.13) where the second equality follows from [r0(σz)f](x)=v34e(u2⟨x,x⟩)f(vx). To determine the transformation of $$\vartheta(z,w)$$ in the variable $$w$$, we use the relation σγw=γσwk(arg(cw+d)),γ=(a)bcd∈SL2(R). Suppose that $$\gamma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right)\in \Gamma_0(6N)$$. If $$x'=\gamma x$$ then by (5.12) we have x1′=a2x1+abx2+b2x3,x2′=2acx1+(1+2bc)x2+2bdx3, (5.14) x3′=c2x1+cdx2+d2x3. (5.15) It follows that $$\gamma L'=L'$$ and that $$\psi_1(x')=\psi^2(a)\psi_1(x)$$. Thus ϑ(z,γw)=ψ2(d)ϑ(z,w)for all γ∈Γ0(6N). (5.16) A computation shows that replacing $$w=\xi+{\rm i}\eta$$ by $$w'=-\xi+{\rm i}\eta$$ in $$f_3(\sqrt{v}\sigma_w^{-1}x)$$ has the same effect as replacing $$x_2$$ by $$-x_2$$, so ϑ(z,−ξ+iη)=ϑ(z,ξ+iη). (5.17) 5.2 Proof of Theorem 5.1 in the case $$t=1$$ Suppose that $$F\in \mathcal{S}_{\frac 12}(N,\bar\psi\left(\tfrac{N}{\bullet}\right) \chi, r)$$ with $$r\neq i/4$$. For $$z=u+{\rm i}v$$ and $$w=\xi+{\rm i}\eta$$ we define ΨF(w):=∫Dv14ϑ(z,w)¯F(z)dμ, where $$\mathcal D=\Gamma_0(N)\backslash\mathbb{H}$$ and $${\rm d}\mu=\frac{{\rm d}u\,dv}{v^2}$$. The integral is well-defined by Lemma 5.4. There is no issue with convergence since $$\vartheta$$ and $$F$$ are both rapidly decreasing at the cusps. Let $$D^{(w)}$$ denote the Casimir operator for $${\rm SL}_2(\mathbb{R})$$ defined by D(w)=η2(∂2∂ξ2+∂2∂η2)−η∂2∂ξ∂θ. To show that $$\Psi_F$$ is an eigenform of $$\Delta_0^{(w)}$$, we use Lemmas 1.4 and 1.5 of [51], which give v−12r0(σz)D(w)f=Δ12(z)[v−12r0(σz)f]. Since $$\Psi_F$$ is constant with respect to $$\theta$$, we have $$\Delta_0^{(w)}\Psi_F = D^{(w)}\Psi_F$$. By the lemma on p. 304 of [46], it follows that Δ0(w)ΨF+(14+(2r)2)ΨF=0. This, together with the transformation law (5.16) shows that $$\Psi_F$$ is a Maass form. The following lemma shows that $$\Psi_F$$ is a cusp form. This is the only point where we use the assumption that $$N$$ is squarefree. It would be possible to remove this assumption with added complications by arguing as in [11]. For the remainder of this section, to avoid confusion with the eta-function, we write $$w=\xi+{\rm i}y$$. Lemma 5.6. Suppose that $$N\equiv 1\pmod{24}$$ is squarefree and that $$F\in \mathcal{S}_{\frac 12}(N,\bar\psi\left(\tfrac{N}{\bullet}\right) \chi, r)$$. Let $$\eta_N(w):=y^{\frac 14}\eta(Nw)$$. For each cusp $$\mathfrak a=\gamma_{\mathfrak a}\infty$$ there exists $$c_{\mathfrak a}\in \mathbb{C}$$ such that as $$y\to\infty$$ we have (ΨF|0γa)(iy)={ca⟨ηN,F⟩y+O(1) if ψ is principal,O(1) otherwise. (5.18) □ Proof For each $$d\mid 6N$$, let $$f\mapsto f\big|_0 W_d$$ be the Atkin–Lehner involution [2, Section 2] given by any matrix $$W_d \in {\rm SL}_2(\mathbb{R})$$ of the form Wd= (dαβ/d6Nγ/ddδ),α,β,γ,δ∈Z. Since $$N\equiv 1\pmod{24}$$ is squarefree, every cusp of $$\Gamma_0(6N)$$ is of the form $$W_d \infty$$ for some $$d$$. Thus it suffices to establish (5.18) for $$\gamma_{\mathfrak a}=W_d$$. By (5.14) and (5.15) we have $$W_d L' = L'$$, so ϑd(z,w):=ϑ(z,Wdw)=v12∑x∈L′ψ¯1(Wdx)e(u2⟨x,x⟩)f3(vσw−1x). To determine the asymptotic behavior of $$\vartheta_d(z,{\rm i}y)$$ as $$y\to\infty$$, we follow the method of Cipra [11, Section 6.1]. We write ϑd(z,iy)=v12∑x∈L′x3=0ψ¯1(Wdx)e(z24Nx22)exp(−πv6Ny2x12)+ε(z,y), (5.19) where |ε(z,y)|≤v12∑x1∈Zexp(−πv6Ny2x12)∑x2∈Zexp(−πv12Nx22)∑x3≠0exp(−πvy26Nx32)≪(v+1v)ye−c1vy2 for some $$c_1>0$$ (see [11, Appendix B]). As in [11, Proposition B.3], the contribution of $$\varepsilon(z,y)$$ to $$(\Psi_F\big|_0 W_d)({\rm i}y)$$ is $$o(1)$$ as $$y\to\infty$$. For $$x_3 = 0$$ and $$W_d x=(x_1',x_2',x_3')$$ we have x1′≡dα2x1(modN),x2′≡(1+12Ndγβ)x2(mod12) (in particular, $$\bar\psi_1(W_d x)=0$$ unless $$d\in \{1,2,3,6\}$$). Thus for some $$c_2$$ the main term of (5.19) equals c2v12∑x1∈Zψ¯(x1)exp(−πv6Ny2x12)∑x2∈NZ(12x2)e(z24Nx22)=2c2η(Nz)v12∑hmodNψ¯(h)θ1(iv12Ny2,h,N), where $$\theta_1(\cdot,h,N)$$ is as in [11, Theorem 1.10(i)]. By the first two assertions of that theorem we have v12∑hmodNψ¯(h)θ1(iv12Ny2,h,N)=6Ny∑hmodNψ¯(h)+ON(v). The latter sum is zero unless $$\psi$$ is principal. In that case we have (ΨF|0Wd)(iy)=c3y∫Dv14η(Nz)¯F(z)dμ+O(1) for some $$c_3\in \mathbb{C}$$; otherwise $$\Psi_F$$ is $$O(1)$$ at the cusps. ■ If the spectral parameter of $$F$$ is not $${\rm i}/4$$ then $$F$$ is orthogonal to $$\eta_N$$. Since a Maass form that is bounded at the cusps is a cusp form, we obtain the following proposition (recall (5.17)). Proposition 5.7. Suppose that $$F\in \mathcal{S}_{\frac 12}(N,\bar\psi\left(\tfrac{N}{\bullet}\right) \chi, r)$$, where $$N\equiv 1\pmod{24}$$ is squarefree and $$r \neq {\rm i}/4$$. Then $$\Psi_F$$ is an even Maass cusp form in $$\mathcal{S}_0(6N,\bar\psi^2,2r)$$. □ With $$G$$ as in Theorem 5.1 let $$F:=G\big|_{\frac 12}w_N$$, where wN= (0−1/NN0) is the Fricke involution. To obtain the $$t=1$$ case of Theorem 5.1 we will compute the Fourier expansion of ΦF(w):=ΨF(−1/6Nw). Once we show that the coefficients of $$\Phi_F$$ satisfy the relation (5.2), Proposition 5.7 completes the proof. We first show that $$F$$ defined in this way satisfies the hypotheses of Proposition 5.7. Lemma 5.8. Suppose that $$N\equiv 1\pmod{24}$$. If $$G\in \mathcal{S}_{\frac 12}(N,\psi\chi,r)$$ and $$F=G\big|_{\frac 12}w_N$$, then $$F\in \mathcal{S}_{\frac 12}\big(N,\bar\psi \left(\tfrac{N}{\bullet}\right) \chi, r\big)$$. □ Proof Since the slash operator commutes with $$\Delta_k$$, it suffices to show that $$F$$ transforms with multiplier $$\bar\psi(\tfrac N\bullet)\chi$$. Writing $$\gamma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right)\in \Gamma_0(N)$$ and $$\gamma'=w_N\gamma w_N^{-1}=\left({{d} \atop {-bN}} \quad {{-c/N}\atop {a}}\right)$$ we have F|12γ=ψ¯(d)χ(γ′)F. Thus it suffices to show that $$\chi(\gamma')=\left(\frac{N}{d}\right) \chi(\gamma)$$. For this we argue as in [50, Proposition 1.4]. Write $$w_N=SV$$, where S=(0−110),V= (N001/N). Let $$\tilde \eta(w):= y^{1/4}\eta(w)$$ and let $$g:=\tilde\eta\big|_{\frac 12}V$$. Then by (2.9), $$g$$ transforms on $$\Gamma_0(N)$$ with multiplier $$\left(\frac{N}{\bullet}\right) \chi$$. We compute χ(wNγwN−1)η~=η~|12wNγwN−1=−ig|12γwN−1=−i (Nd)χ(γ)g|12wN−1= (Nd)χ(γ)η~. ■ We will determine the Fourier coefficients of $$\Phi_F$$ by computing its Mellin transform $$\Omega(s)$$ in two ways. We have Ω(s):=∫0∞ΦF(iy)ysdyy=(6N)−s∫0∞ΨF(i/y)ysdyy. (5.20) For $$w$$ purely imaginary, the sum in (5.13) simplifies to ϑ(z,iy)=ϑ1(z)ϑ2(z,y), where $$\vartheta_1(z)=2\eta(Nz)$$ and ϑ2(z,y)=v12∑m,n∈Zψ¯(m)e(−umn)exp(−πv(m26Ny2+6Ny2n2))=1y6N∑m,n∈Zψ¯(m)exp(−π6Ny2v|mz+n|2). The latter equality follows from Poisson summation on $$n$$. Setting $$A:=\frac\pi{6Nv}|mz+n|^2$$ for the moment, we find that Ω(s)=2(6N)−s−12∑m,nψ(m)∫Dv14η(Nz)¯F(z)∫0∞yse−Ay2dydμ. For $$\text{Re}(s)>1$$ the inner integral evaluates to $$\frac 12 A^{\frac{-s-1}{2}} \Gamma\left(\frac{{s+1}}{2}\right)$$, so we have Ω(s)=Γ(s+12)(6N)s2πs+12∑m,nψ(m)∫Dv14η(Nz)¯F(z)(v|mz+n|2)s+12dμ. Replacing $$z$$ by $$w_N z$$, recalling that $$G=F|_\frac12w_N$$ and using the fact that $$\psi$$ is even, we obtain Ω(s)=iN146s2πs+12Γ(s+12)∑m,nψ(m)∫Dv14η(z)¯G(z) (v|Nnz+m|2)s+12dμ. For $$(m,N)=1$$, write $$m=gd$$ and $$Nn=gc$$ with $$g=(m,n)=(m,Nn)$$ so that ∑m,nψ(m)(v|Nnz+m|2)s+12=L(s+1,ψ)∑γ∈Γ∞∖Γ0(N)ψ(γ)Im(γz)s+12, where $$\psi(\gamma)=\psi(d)$$ for $$\gamma=\left({{a} \atop {c}} \quad {{b} \atop {d}}\right)$$. Thus Ω(s)=c(s)∑γ∈Γ∞∖Γ0(N)ψ(γ)∫Dv14η(z)¯G(z)Im(γz)s+12dμ, where c(s)=iN146s2πs+12Γ(s+12)L(s+1,ψ). Since $$\psi(\gamma)v^{1/4}\bar{\eta(z)}G(z) = \text{Im}(\gamma z)^{1/4}\bar{\eta(\gamma z)}G(\gamma z)$$ for $$\gamma\in \Gamma$$, we have Ω(s)=c(s)∫Γ∞∖Hvs+12+14η(z)¯G(z)dμ. Recall that $$G$$ has Fourier expansion G(z)=∑n≡1(24)a(n)Wsgn(n)4,ir(π|n|v66)e(nu24). Thus we have Ω(s)=c(s)∑n=1∞(12n)a(n2)∫0∞vs2−54W14,ir(πn2v6)e−πn212vdv. By [35, (13.23.4) and (16.2.5)], the integral evaluates to (6πn2)s2−14Γ(s2+14+ir)Γ(s2+14−ir)Γ(s+12), so we conclude that Ω(s)=i(πN6)14π−s−12Γ(s2+14+ir)Γ(s2+14−ir)L(s+1,ψ)∑n=1∞(12n)a(n2)ns−12. On the other hand, since $$\Psi_F$$ is an even Maass cusp form, it follows that $$\Phi_F$$ is also even. Since $$\Phi_F$$ has eigenvalue $$\frac 14+(2r)^2$$, it has a Fourier expansion of the form ΦF(w)=2∑n=1∞b(n)W0,2ir(4πny)cos(2πnx). By the definition (5.20) of $$\Omega(s)$$ and [35, (13.18.9) and (10.43.19)] we have Ω(s)=4∑n=1∞nb(n)∫0∞K2ir(2πny)ys−12dy=π−s−12Γ(s2+14+ir)Γ(s2+14−ir)∑n=1∞b(n)ns. So the coefficients $$b(n)$$ are given by the relation ∑n=1∞b(n)ns=i(πN6)14L(s+1,ψ)∑n=1∞(12n)a(n2)ns−12. This proves Theorem 5.1 in the case $$t=1$$ since $$b(n)$$ is a constant multiple of the function $$b_1(n)$$ in (5.2). 5.3 The case $$t>1$$ For squarefree $$t\equiv 1\pmod{24}$$ with $$t>1$$ we argue as in Section 3 of [36]. For a function $$f$$ on $$\mathbb{H}$$ define $$f_t(\tau):=f(t\tau)$$. We apply Theorem 5.1 to Gt(τ)=∑n≡1(24)a(n/t)Wsgn(n)4,ir(π|n|y6)e(nx24)∈S12(Nt,ψ(t∙)χ,r). The coefficients $$c(n)$$ of $$S_1(G_t) \in \mathcal{S}_0(6Nt,\psi^2,2r)$$ are given by ∑n=1∞c(n)ns=L(s+1,ψ(t∙))∑n=1∞(12n)a(n2/t)ns−12=L(s+1,ψ(t∙))t−s+12∑n=1∞(12n)a(tn2)ns−12. Thus $$c(n)=0$$ unless $$t\mid n$$, in which case $$c(n)=\sqrt t \, b_t(n/t)$$, where $$b_t(n)$$ are the coefficients of $$S_t(G)$$. We conclude that $$S_1(G_t) = \sqrt t \, [S_t(G)]_t$$. By a standard argument (see, e.g., [36, Section 3]) we have St(G)∈S0(6N,ψ2,2r). This completes the proof of Theorem 5.1. 6. Estimates for a $$K$$-Bessel Transform This section contains uniform estimates for the $$K$$-Bessel transform $$\check\phi(r)$$ which are required in Sections 7 and 9. Recall that ϕˇ(r)=chπr∫0∞K2ir(u)ϕ(u)duu, where $$\phi$$ is a suitable test function (see (4.1)). Given $$a,x,T>0$$ with T≤x3 and T≍x1−δ,0<δ<12, we choose $$\phi=\phi_{a,x,T}:[0,\infty)\to[0,1]$$ to be a smooth function satisfying (i) $$\phi(t)=1$$ for $$\frac{a}{2x} \leq t \leq \frac ax$$, (ii) $$\phi(t)=0$$ for $$t\leq \frac{a}{2x+2T}$$ and $$t\geq \frac{a}{x-T}$$, (iii) $$\phi'(t) \ll \left(\frac{a}{x-T} - \frac ax\right)^{-1} \ll \frac{x^2}{aT}$$, and (iv) $$\phi$$ and $$\phi'$$ are piecewise monotone on a fixed number of intervals (whose number is independent of $$a$$, $$x$$, and $$T$$). In Theorem 4.1, the function $$\check\phi(r)$$ is evaluated at the spectral parameters $$r_j$$ corresponding to the eigenfunctions $$u_j$$ as in (2.23). In view of Corollary 5.3, we require estimates only for $$r \geq 1$$. We will prove the following theorem. Theorem 6.1. Suppose that $$a,x,T$$, and $$\phi=\phi_{a,x,T}$$ are as above. Then ϕˇ(r)≪{r−32e−r/2 if 1≤r≤a8x,r−1 if max(a8x,1)≤r≤ax,min(r−32,r−52xT) if r≥max(ax,1). □ To prove Theorem 6.1 we require estimates for $$K_{{\rm i}v}(vz)$$ which are uniform for $$z\in (0,\infty)$$ and $$v\in [1,\infty)$$. 6.1 Uniform estimates for the $$K$$-Bessel function We estimate $$K_{{\rm i}v}(vz)$$ in the following ranges as $$v\to \infty$$: (A) the oscillatory range $$0<z \leq 1-O(v^{-\frac23})$$, (B) the transitional range $$1-O(v^{-\frac 23}) \leq z \leq 1+O(v^{-\frac23})$$, and (C) the decaying range $$z \geq 1+O(v^{-\frac 23})$$. Suppose that $$c$$ is a positive constant. In the transitional range there is a significant “bump” in the $$K$$-Bessel function. By [8, (14) and (21)] we have eπv2Kiv(vz)≪cv−13 for z≥1−cv−23. (6.1) In the decaying range the $$K$$-Bessel function is positive and decreasing. By [8, (14)] we have eπv2Kiv(vz)≪ce−vμ(z)v12(z2−1)14 for z≥1+cv−23, (6.2) where μ(z):=z2−1−arccosh(1z). The oscillatory range is much more delicate. Balogh [3] gives a uniform asymptotic expansion for $$K_{{\rm i}v}(vz)$$ in terms of the Airy function $$\text{Ai}$$ and its derivative $$\text{Ai}'$$. For $$z\in (0,1)$$ define w(z):=arccosh(1z)−1−z2 and define $$\zeta$$ and $$\xi$$ by 23ζ32=−iw(z),ξ=v23ζ. Taking $$m=1$$ in equation (2) of [3] we have eπv2Kiv(vz)=π2v13(−ζ1−z2)14{Ai(ξ)[1+A1(ζ)v2]+Ai′(ξ)B0(ζ)v43+eivw(z)1+|ξ|14O(v−3)}, uniformly for $$v\in [1,\infty)$$, where A1(ζ):=45510368w(z)2−7(3z2+2)1728(1−z2)32w(z)−81z4+300z2+41152(1−z2)3,B0(ζ):=(23)13e2πi/3w(z)13(3z2+224(1−z2)32−572w(z)). (6.3) A computation shows that $$A_1(\zeta)$$ and $$B_0(\zeta)$$ have finite limits as $$z\to 0^+$$ and as $$z\to 1^-$$, so both functions are $$O(1)$$ for $$z\in(0,1)$$. Note that $$\arg \xi=-\frac {\pi}3$$. In order to work on the real line, we apply [35, (9.6.2–3)] to obtain Ai(ξ)=(23)1/6i23/2(vw(z))13H13(1)(vw(z)),Ai′(ξ)=−(32)1/6i23/2(vw(z))23H23(1)(vw(z)), where $$H_{\frac 13}^{(1)}$$ and $$H_{\frac 23}^{(1)}$$ are Hankel functions of the first kind. So we have eπv2Kiv(vz)=π2e2πi/3w(z)12(1−z2)14H13(1)(vw(z))[1+A1(ζ)v2]+(32)1/3π2e−πi/3w(z)56(1−z2)14H23(1)(vw(z))B0(ζ)v+O(v−72(1−z2)14). (6.4) Since $$w(z)\to\infty$$ as $$z\to0$$ and $$(1-z^2)^{-\frac14}\to\infty$$ as $$z\to 1$$, we derive more convenient expressions for (6.4) for $$z$$ in the intervals $$(0,3/4]$$ and $$[3/16,1-cv^{-\frac 23})$$. Proposition 6.2. Suppose that $$z\in (0, 3/4]$$ and that $$v\geq 1$$. Then eπv2Kiv(vz)=eπi/4(π2)12eivw(z)v12(1−z2)14[1−i3z2+224v(1−z2)32]+O(v−52). (6.5) □ Proof Since $$(1-z^2) \gg 1$$ for $$z\leq 3/4$$, the error term in (6.4) is $$\ll v^{-\frac 72}$$. By [35, (10.17.5) and Section 10.17(iii)] we have H13(1)(vw(z))=(2π)12e−5πi/12eivw(z)(vw(z))12(1−5i72vw(z))+O((vw(z))−52),H23(1)(vw(z))=(2π)12e−7πi/12eivw(z)(vw(z))12+O((vw(z))−32). In particular, this implies that w(z)12(1−z2)14H13(1)(vw(z))A1(ζ)v2≪v−52, so by (6.4) we have eπv2Kiv(vz)=eπi/4(π2)12eivw(z)v12(1−z2)14[1−5i72vw(z)+e−7πi/6(32)13w(z)13B0(ζ)v]+O(v−52). Using (6.3), we obtain (6.5). ■ We require some notation for the next proposition. Let $$J_\nu(x)$$ and $$Y_\nu(x)$$ denote the $$J$$ and $$Y$$-Bessel functions, and define Mν(x)=Jν2(x)+Yν2(x). Proposition 6.3. Suppose that $$c>0$$. Suppose that $$v\geq 1$$ and that $$\frac 3{16}\leq z\leq 1-cv^{-\frac 23}$$. Then eπv2Kiv(vz)=π2e2πi/3w(z)12(1−z2)14M13(vw(z))eiθ13(vw(z))+Oc(v−4/3), (6.6) where $$\theta_{\frac 13}(x)$$ is a real-valued continuous function satisfying θ13′(x)=2πxM132(x). (6.7) □ Proof Since $$(1-z^2) \gg_c v^{-\frac 23}$$ for $$z \leq 1-cv^{-\frac 23}$$, the error term in (6.4) is $$\ll_c v^{-\frac {10}3}$$. The modulus and phase of $$H_\alpha^{(1)}(x)$$ are given by [35, Section 10.18] Hα(1)(x)=Mα(x)eiθα(x), where $$M_\alpha^2(x)\theta_\alpha'(x)=2/\pi x$$. A straightforward computation shows that $$w(z)\gg(1-z)^{\frac 32}$$. It follows that $$vw(z) \gg_c 1$$ for $$z\leq 1-cv^{-\frac 23}$$, so by [35, (10.18.17)] we obtain Mα(vw(z))≪α,c1(vw(z))12. This, together with (6.4) and the fact that $$A_1(\zeta)$$ and $$B_0(\zeta)$$ are $$O(1)$$, gives (6.6). ■ 6.2 Estimates for $$\check\phi(r)$$ We treat each of the three ranges considered in Theorem 6.1 separately in the following propositions. We will make frequent use of an integral estimate which is an immediate corollary of the second mean value theorem for integrals. Lemma 6.4. Suppose that $$f$$ and $$g$$ are continuous functions on $$[a,b]$$ and that $$g$$ is piecewise monotonic on $$M$$ intervals. Then |∫abf(x)g(x)dx|≤2Msupx∈[a,b]|g(x)|sup[α,β]⊆[a,b]|∫αβf(x)dx|. □ Proposition 6.5. With the notation of Theorem 6.1, suppose that $$1\leq r\leq a/8x$$. Then ϕˇ(r)≪r−32e−r/2. □ Proof Since $$r\leq a/8x$$ and $$T\leq x/3$$, we have $$\frac{a}{4(x+T)r} \geq \frac32$$. Taking $$c=2^{-1/3}$$ in (6.4) gives ϕˇ(r)=chπr∫a4(x+T)ra2(x−T)rK2ir(2ry)ϕ(2ry)dyy≪r−12∫32∞e−2rμ(y)dyy(y2−1)14. Since $$\mu'(y) = \sqrt{y^2-1}/y$$ we have ϕˇ(r)≪r−32∫32∞(−e−2rμ(y))′dy(y2−1)34≪r−32e−2rμ(3/2), which, together with $$\mu(3/2)\approx 0.277$$, proves the proposition. ■ Proposition 6.6. With the notation of Theorem 6.1, suppose that $$\max(a/8x,1)\leq r\leq a/x$$. Then ϕˇ(r)≪r−1. □ Before proving Proposition 6.6, we require a lemma describing the behavior of the function $$M_{\frac 13}(x)$$ in Proposition 6.3. Lemma 6.7. The function $$\tilde M(x) := x M_{\frac 13}^2(x)$$ is increasing on $$[0,\infty)$$ with $$\lim\limits_{x\to\infty} \tilde M(x)\,{=}\, \frac 2\pi$$. □ Proof We will prove that $$w(x):=\sqrt{\smash[b]{\tilde M(x)}}$$ is increasing. From [35, (10.7.3–4), (10.18.17)] we have M~(0)=0,0≤M~(x)≤2π, and limx→∞M~(x)=2π. It is therefore enough to show that $$w''(x)<0$$ for all $$x>0$$. In view of the second-order differential equation [35, (10.18.14)] satisfied by $$w$$ it will suffice to prove that M~(x)>12xπ36x2+5. (6.8) The inequality (6.8) can be proved numerically, using the expansions of $$\tilde M(x)$$ at $$0$$ and $$\infty$$. By [35, Section 10.18(iii)] we have M~(x)≥2π(1+∑k=1n1⋅3⋯(2k−1)2⋅4⋯(2k)(49−1)(49−9)⋯(49−(2k−1)2)(2x)2k) (6.9) for odd $$n\geq 1$$. Taking $$n=3$$ in (6.9), we verify numerically that (6.8) holds for $$x>2.34$$. From [35, (10.2.3)] we have M~(x)=4x3(J132(x)−J13(x)J−13(x)+J−132(x)). So by [35, (10.8.3)] it follows that the series for $$\tilde M(x)$$ at $$0$$ is alternating, and that $$\tilde M(x)$$ is larger than the truncation of this series after the term with exponent $$47/3$$. Moreover, this truncation is larger than $$\frac{12x}{\pi\sqrt{36x^2+5}}$$ for $$x<2.45$$. The claim (6.8) follows. ■ Proof of Proposition 6.6 Fix $$c=\frac 12$$ and write ϕˇ(r)=ϕˇ1(r)+ϕˇ2(r)+ϕˇ3(r)=(∫a4(x+T)r1−cr−23+∫1−cr−231+cr−23+∫1+cr−23∞)chπrK2ir(2ry)ϕ(2ry)dyy. We will show that $$\check\phi_i(r)\ll r^{-1}$$ for $${\rm i}=1,2,3$$. Note that $$\frac{a}{4(x+T)r}\geq \frac 3{16}$$. By Proposition 6.3 we have ϕˇ1(r)≪|∫a4(x+T)r1−cr−23eiθ13(2rw(y))M13(2rw(y))w(y)12ϕ(2ry)y(1−y2)14dy|+r−43∫3161ϕ(2ry)dyy. (6.10) The error term is $$O(r^{-{4/3}})$$. Using (6.7) and the fact that $$w'(y)=-\sqrt{1-y^2}/y$$, the first term in (6.10) equals π2(2r)32|∫a4(x+T)r1−cr−23(eiθ13(2rw(y)))′ (2rw(y)M132(2rw(y)))32(1−y2)34ϕ(2ry)dy|. Note that $$w(y)$$ is decreasing, so by Lemma 6.7 the function $$2rw(y)M_{\frac 13}^2(2rw(y))$$ is decreasing. We apply Lemma 6.4 three times; first with the decreasing function g(y)=(2rw(y)M13(2rw(y)))32≪1, next with the increasing function g(y)=(1−y2)−34≪r12, and then with $$g(y)=\phi(2ry)$$. We conclude that $$\check\phi_1(r)\ll r^{-1}$$. For $$\check\phi_2(r)$$ we apply (6.1) to obtain ϕˇ2(r)≪r−13∫1−cr−231+cr−23dy≪r−1. For $$\check\phi_3(r)$$ we argue as in the proof of Proposition 6.5 to obtain ϕˇ3(r)≪r−32∫1+cr−23∞(−e−2rμ(y))′dy(y2−1)34. For $$y\geq 1+cr^{-\frac 23}$$ we have $$(y^2-1)^{\frac 34}\gg r^{-\frac 12}$$, so $$\check\phi_3(r)\ll r^{-1}$$. ■ Proposition 6.8. Suppose that $$r\geq \max(a/x,1)$$. Then ϕˇ(r)≪min(r−32,r−52xT). □ Proof Since $$r\geq a/x$$ and $$T\leq x/3$$, the interval on which $$\phi(2ry)\neq 0$$ is contained in $$(0, 3/4]$$. So by Proposition 6.2 we have ϕˇ(r)≪r−12|I0|+r−32|I1|+r−52∫a4(x+T)ra2(x−T)rdyy, (6.11) where Iα=∫a4(x+T)ra2(x−T)re2irw(y)ϕ(2ry)(3y2+2)αy(1−y2)14+32αdy. The third term in (6.11) equals r−52log2(x+T)x−T≪r−52. For the other terms we use $$w'(y)=-\sqrt{1-y^2}/y$$ to write Iα≪r−1|∫a4(x+T)ra2(x−T)r(e2irw(y))′ϕ(2ry)(3y2+2)α(1−y2)34+32αdy|. (6.12) Since $$y\leq 3/4$$ we have (3y2+2)α(1−y2)34+32α≪1, so by Lemma 6.4 we obtain $$I_\alpha \ll r^{-1}$$ and therefore $$\check\phi(r)\ll r^{-\frac 32}$$. For large $$r$$, we obtain a better estimate by integrating by parts in (6.12) for $$\alpha=0$$. Since $$\phi(2ry)=0$$ at the limits of integration, we find that I0≪r−1|∫a4(x+T)ra2(x−T)re2irw(y)yϕ(2ry)(1−y2)74dy|+r−1|∫a4(x+T)ra2(x−T)re2irw(y)rϕ′(2ry)(1−y2)34dy|=:J1+J2. As above, Lemma 6.4 gives J1≪r−2|∫a4(x+T)ra2(x−T)r(e2irw(y))′y2ϕ(2ry)(1−y2)94dy|≪r−2. For $$J_2$$, we apply Lemma 6.4 with the estimates $$\phi'(2ry)\ll x^2/aT$$ and $$y \ll a/rx$$ to obtain J2=r−1|∫a4(x+T)ra2(x−T)r(e2irw(y))′yϕ′(2ry)(1−y2)54dy|≪r−2xT. So $$I_0 \ll r^{-2}x/T$$. This, together with (6.11) and our estimate for $$I_1$$ above, gives (6.8). ■ 7. Application to Sums of Kloosterman Sums We are in a position to prove Theorem 1.3, which asserts that for $$m>0$$, $$n<0$$ we have ∑c≤XS(m,n,c,χ)c≪(X16+|mn|14)|mn|ϵlogX. (7.1) This will follow from an estimate for dyadic sums. Proposition 7.1. If $$m>0$$, $$n<0$$, and $$x>4\pi\sqrt{\tilde m|\tilde n|}$$ then ∑x≤c≤2xS(m,n,c,χ)c≪(x16logx+|mn|14)|mn|ϵ. □ To obtain Theorem 1.3 from the proposition, we estimate using (2.30) to see that the initial segment $$c\leq 4\pi\sqrt{\tilde m|\tilde n|}$$ of (7.1) contributes $$O(|mn|^{\frac 14+\epsilon})$$. We break the rest of the sum into dyadic pieces $$x\leq c\leq 2x$$ with $$4\pi\sqrt{\tilde m|\tilde n|} < x\leq X/2$$. Estimating each of these with Proposition 7.1 and summing their contributions give (7.1). Proof of Proposition 7.1 Let a:=4πm~|n~|,T:=x23, and suppose that $$x>a$$. We apply Theorem 4.1 using a test function $$\phi$$ which satisfies conditions (i)–(iv) of Section 6. Proposition 2.1 and the mean value bound for the divisor function give |∑c=1∞S(m,n,c,χ)cϕ(ac)−∑x≤c≤2xS(m,n,c,χ)c|≤∑x−T≤c≤x2x≤c≤2x+2T|S(m,n,c,χ)|c≪Tlogxx|mn|ϵ≪|mn|ϵx16logx. (7.2) Using Theorem 4.1, it will therefore suffice to obtain the stated estimate for the quantity ∑c=1∞S(m,n,c,χ)cϕ(ac)=8im~|n~|∑1<rjρj(m)¯ρj(n)chπrjϕˇ(rj), (7.3) where we have used Corollary 5.3 and the fact that the eigenvalue $$r_0={\rm i}/4$$ makes no contribution since $$n$$ is negative. We break the sum (7.3) into dyadic intervals $$A\leq r_j\leq 2A$$. Using the Cauchy–Schwarz inequality together with Theorems 1.5 and 6.1 (recall that $$a/x<1 < r_j$$) we obtain m~|n~|∑A≤rj≤2A|ρj(m)¯ρj(n)chπrjϕˇ(rj)|≪min(A−32,A−52x13)(m~∑A≤rj≤2A|ρj(m)|2chπrj)12(|n~|∑A≤rj≤2A|ρj(n)|2chπrj)12≪min(A−32,A−52x13)(A32+m12+ϵ)12(A52+|n|12+ϵA12)12≪min(A12,A−12x13)(1+m14+ϵA−34+|n|14+ϵA−1+|mn|14+ϵA−74). Summing the contribution from the dyadic intervals gives m~|n~|∑1<rjρj(m)¯ρj(n)chπrjϕˇ(rj)≪x16+|mn|14+ϵ, and Proposition 7.1 follows. ■ 8. A Second Estimate for Coefficients of Maass Cusp Forms In the case $$m=1$$ we can improve the estimate of Proposition 7.1 by using a second estimate for the sum of the Fourier coefficients of Maass cusp forms in $$\mathcal{S}_\frac12(1,\chi)$$. The next theorem is an improvement on Theorem 1.5 only when $$n$$ is much larger than $$x$$. Theorem 8.1. Suppose that $$\{u_j\}$$ is an orthonormal basis of $$\mathcal{S}_\frac12(1,\chi)$$ with spectral parameters $$r_j$$ and coefficients $$\rho_j(n)$$ as in (2.23). Suppose that the $$u_j$$ are eigenforms of the Hecke operators $$T_{p^2}$$ for $$p\nmid 6$$. If $$24n-23$$ is not divisible by $$5^4$$ or $$7^4$$ then ∑0<rj≤x|ρj(n)|2chπrj≪|n|−47+ϵx5−sgnn2. (8.1) □ Remark As the proof will show, the exponent $$4$$ in the assumption on $$n$$ in Theorem 8.1 can be replaced by any positive integer $$m$$. Increasing $$m$$ has the effect of increasing the implied constant in (8.1). □ We require an average version of an estimate of Duke [12, Theorem 5]. Let $$\nu_\theta$$ denote the weight $$1/2$$ multiplier for $$\Gamma_0(4)$$ defined in (2.10). Proposition 8.2. Let $$D$$ be an even fundamental discriminant, let $$N$$ be a positive integer with $$D\mid N$$, and let (k,ν)=(12,νθ(|D|∙))or(32,ν¯θ(|D|∙)). Suppose that $$\{v_j\}$$ is an orthonormal basis of $$\mathcal{S}_k(N,\nu)$$ with spectral parameters $$r_j$$ and coefficients $$b_j(n)$$. If $$n>0$$ is squarefree then ∑0≤rj≤x|bj(n)|2chπrj≪N,Dn−47+ϵx5−k. (8.2) □ Sketch of proof Let $$\phi,\hat\phi$$ be as in the proof of Theorem 5 of [12]. Then $$\hat\phi(r)>0$$ and ϕ^(r)∼12π2|r|k−5as |r|→∞. (8.3) Set V1(n,n):=n∑rj|bj(n)|2chπrjϕ^(rj). By Theorem 2 of [12] we have $$|V_1(n,n) +V_2(n,n)|\ll |S_N| + |V_3(n,n)|$$, where $$S_N$$ is defined in the proof of Theorem 5 and $$V_2,V_3$$ are defined in Section 3 of that paper. Since $$V_1(n,n)$$ and $$V_2(n,n)$$ are visibly non-negative, we have V1(n,n)≪|SN|+|V3(n,n)|. (8.4) The terms $$S_N$$ and $$V_3(n,n)$$ are estimated by averaging over the level. For $$P>(4\log 2n)^2$$ let Q¯={pN:p prime, P<p≤2P,p∤2n}. Summing (8.4) gives ∑Q∈Q¯V1(Q)(n,n)≪∑Q∈Q¯|SQ|+∑Q∈Q¯|V3(Q)(n,n)|, (8.5) where $$V_1^{(Q)}$$, $$S_Q$$, and $$V_3^{(Q)}$$ are the analogues of $$V_1$$, $$S_N$$, and $$V_3$$ for $$\Gamma_0(Q)$$. For each $$Q\in \bar Q$$, the functions $$\{[\Gamma_0(N):\Gamma_0(Q)]^{-1/2}u_j(\tau)\}$$ form an orthonormal subset of $$\mathcal{S}_k(Q,\nu)$$. It follows that V1(Q)(n,n)≥n[Γ0(N):Γ0(Q)]∑rj|bj(n)|2chπrjϕ^(rj)=V1(n,n)[Γ0(N):Γ0(Q)]. Since $$[\Gamma_0(N):\Gamma_0(Q)]\leq p+1\ll P$$ we find that V1(n,n)≪PV1(Q)(n,n) for all Q∈Q¯. Since $$|\bar Q|\asymp P/\log P$$ we conclude that V1(n,n)≪logP∑Q∈Q¯V1(Q)(n,n). (8.6) In the proof of Theorem 5 of [12], Duke gives the estimates ∑Q∈Q¯|V3(Q)(n,n)|≪N,Dn37+ϵ and ∑Q∈Q¯|SQ|≪N,D((n/P)12+(nP)38)nϵ (8.7) which follow from work of Iwaniec [22]. By (8.5)–(8.7) we conclude that V1(n,n)≪N,DlogP((n/P)12+(nP)38+n37)nϵ. Choosing $$P=n^{1/7}$$, we find that $$V_1(n,n)\ll_{N,D} n^{3/7+\epsilon}$$. By (8.3) we have n∑0≤rj≤x|bj(n)|2chπrj≪n∑0≤rj≤xrj5−k|bj(n)|2chπrjϕ^(rj)≪x5−kV1(n,n), from which (8.2) follows. ■ We turn to the proof of Theorem 8.1. Proof of Theorem 8.1 Set $$\alpha := [\Gamma_0(1):\Gamma_0(24,24)]^{1/2}$$. With $$\{u_j\}$$ as in the hypotheses, and recalling (2.21), define vj(τ):=1αuj(24τ)=∑n≡1(24)bj(n)W14sgnn,irj(4π|n|y)e(nx),vj′(τ):=1α(rj2+116)−12L12uj¯(24τ)=∑n≡23(24)bj′(n)W34sgnn,irj(4π|n|y)e(nx). By (2.11) we have vj∈S12(576,νθ(12∙)),vj′∈S32(576,νθ¯(12∙)). From (2.18) and (2.21) we see that $$\{v_j\}$$, $$\{v_j'\}$$ are orthonormal sets which can be extended to orthonormal bases for these spaces. Using (2.16) and (2.17) and comparing Fourier expansions, we find for positive $$n$$ that bj(n)=1αρj(n+2324),bj′(n)=1α(rj2+116)−12ρj(−n+2324)¯. It follows from Proposition 8.2 that for squarefree $$n$$ we have ∑0<rj≤x|ρj(n+2324)|2chπrj≪|n|−47+ϵx5−sgnn2. (8.8) To establish (8.8) for non-squarefree $$n$$ we use the assumption that the $$\{u_j\}$$ are Hecke eigenforms. Recall the definition of the lift $$S_t$$ from Theorem 5.1. For each $$j$$ there exists some squarefree positive $$t\equiv 1\pmod{24}$$ for which $$S_t(u_j)\neq 0$$. Denote by $$\lambda_j(p)$$ the eigenvalue of $$u_j$$ under $$T_{p^2}$$. From Corollary 5.2 it follows that $$\left(\tfrac{12}{p}\right)\lambda_j(p)$$ is a Hecke eigenvalue of the weight zero Maass cusp form $$S_t(u_j)$$. For these eigenvalues we have the estimate |λj(p)|≤p764+p−764 due to Kim and Sarnak [25, Appendix 2]. The Hecke action (2.24) gives pρj(p2n+2324)=λj(p)ρj(n+2324)−p−12(12npρj(n+2324))−p−1ρj(n/p2+2324). (8.9) Suppose that $$p^2 \nmid n$$. Then |ρj(p2n+2324)|≤(p−5764+p−7164+p−32)|ρj(n+2324)|≤{1.25p−47 if p=5 or 7,p−47 if p≥11. Thus (8.8) holds whenever $$n$$ is not divisible by $$p^4$$ for any $$p$$. To treat the remaining cases, assume that $$p\geq 11$$ and that for $$p^2\nmid n$$ we have |ρj(p2ℓn+2324)|≤p−4ℓ7|ρj(n+2324)|,ℓ≤m−1. Then for $$m\geq 2$$, the relation (8.9) gives |ρj(p2mn+2324)|≤(p−5764+p−7164)|ρj(p2m−2n+2324)|+p−2|ρj(p2m−4n+2324)|≤[(p−5764+p−7164)p47+p−2+87]p−4m7|ρj(n+2324)|. For $$p\geq 11$$ the quantity in brackets is $$<1$$. It follows that (8.8) holds for all $$n$$ which are not divisible by $$5^4$$ or $$7^4$$. ■ 9. Sums of Kloosterman Sums in the Case $$m=1$$ Throughout this section, $$n$$ will denote a negative integer. In our application to the error term in Rademacher’s formula we need an estimate for sums of the Kloosterman sums $$S(1, n, c, \chi)$$ which improves the bound of Theorem 1.3 with respect to $$n$$. We will also make the assumption throughout that 24n−23 is not divisible by 54 or 74 (9.1) so that we may apply Theorem 8.1 (see the remark after that theorem). Theorem 9.1. For $$0<\delta<1/2$$ and $$n<0$$ satisfying (9.1) we have ∑c≤XS(1,n,c,χ)c≪δ|n|1356+ϵX34δ+(|n|41168+ϵ+X12−δ)logX. (9.2) □ Theorem 9.1 will follow from an estimate for dyadic sums. Proposition 9.2. Suppose that $$0<\delta<1/2$$ and that $$n<0$$ satisfies (9.1). Then ∑x≤c≤2xS(1,n,c,χ)c≪δ|n|4128+ϵx−52+|n|1356+ϵx34δ+x12−δlogx. (9.3) □ Deduction of Theorem 9.1 from Proposition 9.2 We break the sum (7.3) into an initial segment corresponding to $$c\leq |n|^\alpha$$ and dyadic intervals of the form (9.3) with $$x\geq |n|^\alpha$$. Estimating the initial segment using Lehmer’s bound (1.4) and applying Proposition 9.2 to each of the dyadic intervals, we find that ∑c≤XS(1,n,c,χ)c≪δ,α|n|α2+ϵ+|n|1356+ϵX34δ+(|n|4128−52α+ϵ+X12−δ)logX. Theorem 9.1 follows upon setting $$\alpha=41/84$$. ■ Proof of Proposition 9.2 Let $$T$$ satisfy T≍x1−δ and T≤x3, where $$0<\delta<1/2$$ is a parameter to be chosen later, and set a:=4π|n~|. Arguing as in (7.2) (using Lehmer’s bound (1.4) to remove the dependence on $$n$$) we have |∑c=1∞S(1,n,c,χ)cϕ(ac)−∑x≤c≤2xS(1,n,c,χ)c|≪δx12−δlogx. (9.4) Let $$\{u_j\}$$ be an orthonormal basis for $$\mathcal S_\frac12(1,\chi)$$ as in (2.23) which consists of eigenforms for the Hecke operators $$T_{p^2}$$ with $$p\nmid 6$$. Theorem 4.1 gives ∑c=1∞S(1,n,c,χ)cϕ(ac)=8i|n~|∑rjρj(1)¯ρj(n)chπrjϕˇ(rj). (9.5) We break the sum on $$r_j$$ into ranges corresponding to the three ranges of the $$K$$-Bessel function: (i) $$r_j\leq \frac a{8x}$$, (ii) $$\frac a{8x}<r_j<\frac ax$$, (iii) $$r_j\geq \frac ax$$. We may restrict our attention to $$r_j>1$$ by Corollary 5.3 (the eigenvalue $$r_0={\rm i}/4$$ does not contribute). For the first range, Theorem 6.1 gives $$\check\phi(r_j) \ll r_j^{-3/2} {\rm e}^{-r_j/2}$$. By Theorem 8.1 we have ρj(1)chπrj≪rj94,ρj(n)chπrj≪|n|−27+ϵrj114. Combining these estimates gives |n~|∑1<rj≤a8x|ρj(1)¯ρj(n)chπrjϕˇ(rj)|≪|n|314+ϵ∑1<rj≤a8xrj72e−rj/2≪|n|314+ϵ, (9.6) where we have used (2.22) to conclude that the latter sum over $$r_j$$ is $$O(1)$$. For the other ranges we need the mean value estimates of Theorem 1.5: ∑1<rj≤x|ρj(1)|2chπrj≪x32, (9.7) |n~|∑1<rj≤x|ρj(n)|2chπrj≪x52+|n|12+ϵx12, (9.8) as well as the average version of Duke’s estimate (Theorem 8.1): |n~|∑rj≤x|ρj(n)|2chπrj≪|n|37+ϵx112. (9.9) Using (9.7) and (9.8) with the Cauchy–Schwarz inequality, we find that |n~|∑rj≤x|ρj(1)¯ρj(n)|chπrj≪x34(x52+|n|12+ϵx12)12≪x2+|n|14+ϵx. Using (9.7) and (9.9) we obtain |n~|∑rj≤x|ρj(1)¯ρj(n)|chπrj≪x34(|n|37+ϵx112)12≪|n|314+ϵx72. (9.10) In the second range Theorem 6.1 gives $$\check\phi(r_j)\ll r_j^{-1}$$. It follows from (9.10) (assuming as we may that $$a/x\geq 1$$) that |n~|∑a8x<rj<ax|ρj(1)¯ρj(n)chπrjϕˇ(rj)|≪|n|314+ϵ(ax)52≪|n|4128+ϵx−52. (9.11) In the third range Theorem 6.1 gives ϕˇ(rj)≪min(rj−32, rj−52xT). We use dyadic sums corresponding to intervals $$A\leq r_j\leq 2A$$ with $$A\geq \max\left(\frac ax,1\right)$$. For such a sum we have |n~|∑A≤rj≤2A|ρj(1)¯ρj(n)chπrjϕˇ(rj)|≪min(A−32, A−52xT)min(A2+|n|14+ϵA, |n|314+ϵA72)≪min(A12,A−12xT)(1+|n|1356+ϵA14)≪|n|1356+ϵmin(A34,A−14xT), where we have used the fact that for positive $$B$$, $$C$$, and $$D$$ we have min(B,C+D)≤min(B,C)+min(C,D)andmin(B,C)≤BC. Combining the dyadic sums, we find that |n~|∑rj≥ax|ρj(1)¯ρj(n)chπrjϕˇ(rj)|≪|n|1356+ϵ(xT)34≪δ|n|1356+ϵx34δ. (9.12) Using (9.5) with (9.6), (9.11), and (9.12) we obtain ∑c=1∞S(1,n,c,χ)cϕˇ(ac)≪δ|n|4128+ϵx−52+|n|1356+ϵx34δ. Proposition 9.2 follows from this estimate together with (9.4). ■ 10. Application to the Remainder Term in Rademacher’s Formula In the final section, we use Theorem 9.1 to prove Theorems 1.1 and 1.2. Recall that we wish to bound $$R(n, N)$$, where p(n)=2π(24n−1)34∑c=1NAc(n)cI32(π24n−16c)+R(n,N), and Ac(n)=−iS(1,1−n,c,χ). To match the notation of the previous sections, we assume that $$n< 0$$ satisfies (9.1) and we provide a bound for R(1−n,N)=2π−i2434|n~|−34∑c>NS(1,n,c,χ)cI32(ac), (10.1) where a:=23π|n~|. (10.2) We will apply partial summation to (10.1). For fixed $$\nu,M>0$$ and $$0\leq z\leq M$$, the asymptotic formula [35, (10.30.1)] gives Iν(z)≪ν,Mzν. (10.3) We have a straightforward lemma. Lemma 10.1. Suppose that $$a>0$$ and that $$\alpha>0$$. Then for $$\frac ta \geq \alpha$$ we have (I32(a/t))′≪αa32t−52. □ Proof By [35, (10.29.1)] we have I32′(t)=12(I12(t)+I52(t)). It follows that (I32(a/t))′=−a2t2(I12(a/t)+I52(a/t)). (10.4) For $$\frac ta \geq \alpha$$, equations (10.3) and (10.4) give (I32(a/t))′≪αa32t−52+a72t−92, and the lemma follows. ■ Proof of Theorems 1.1 and 1.2 Let $$a$$ be as in (10.2). Then R(1−n,N)≪|n~|−34|∑c>NS(1,n,c,χ)cI32(ac)|. Let S(n,X):=∑c≤XS(1,n,c,χ)c, so that Theorem 9.1 (for $$0<\delta<1/2$$) gives S(n,X)≪δ|n|1356+ϵX34δ+(|n|41168+ϵ+X12−δ)logX. By partial summation using (10.3) we have ∑c>NS(1,n,c,χ)cI32(ac)=−S(n,N)I32(a/N)−∫N∞S(n,t)(I32(a/t))′dt. Let $$\alpha>0$$ be fixed and take N=α|n|12+β, where $$\beta\in [0, 1/2]$$ is to be chosen. By (10.3) we have I32(a/N)≪α|n|34N−32 and by Lemma 10.1 we have (I32(a/t))′≪α|n|34t−52for t≥N. Thus S(n,N)I32(a/N)≪δ,α|n|34N−32(|n|1356N34δ+|n|41168+N12−δ)|n|ϵ, and the same bound holds for the integral term. Therefore R(1−n,α|n|12+β)≪δ,α|n|(12+β)(34δ−32)+1356+ϵ+|n|−32(12+β)+41168+ϵ+|n|−(12+β)(1+δ)+ϵ. (10.5) When $$\beta=0$$, the estimate (10.5) becomes R(1−n,α|n|12)≪δ,α|n|38δ−2956+ϵ+|n|−85168+ϵ+|n|−12−12δ+ϵ. For any choice of $$\delta\in [\frac1{84}, \frac2{63}]$$ this gives R(1−n,α|n|12)≪α|n|−12−1168+ϵ. Theorem 1.1 follows after replacing $$n$$ by $$1-n$$. To optimize, we choose $$\beta=\frac{5}{252}$$ and $$\delta=\frac{4}{131}$$, which makes all three exponents in (10.5) equal to $$-\frac12-\frac{1}{28}$$. 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