2 Statement of results

We begin by fixing notation, which we mostly borrow from [Reference Iwaniec7]. Let $\unicode[STIX]{x1D6E4}=\unicode[STIX]{x1D6E4}_{0}(N)$ for some squarefree integer $N$ and let $k\geqslant 0$ be an even integer. The weight $k$ Petersson inner product is defined as

$$\begin{eqnarray}\langle h_{1},h_{2}\rangle =\iint _{\unicode[STIX]{x1D6E4}\backslash {\mathcal{H}}}h_{1}(z)\overline{h_{2}(z)}y^{k}\,\frac{dx\,dy}{y^{2}}.\end{eqnarray}$$

Here $z=x+\text{i}y$ and $h_{1},h_{2}$ are holomorphic cusp forms of weight $k$ or Maass cusp forms (in the latter case $k=0$ ).

Let $S_{k}(N)$ denote the space of holomorphic cusp forms of weight $k$ on $\unicode[STIX]{x1D6E4}_{0}(N)$ , and let ${\mathcal{B}}_{k}(N)$ denote an orthonormal basis of $S_{k}(N)$ . We will always use $f$ or $g$ to denote an element of ${\mathcal{B}}_{k}(N)$ . The Fourier expansion of such an $f$ at a cusp $\mathfrak{a}$ of $\unicode[STIX]{x1D6E4}$ is given by

(2.1) $$\begin{eqnarray}j(\unicode[STIX]{x1D70E}_{\mathfrak{a}},z)^{-k}f(\unicode[STIX]{x1D70E}_{\mathfrak{ a}}z)=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D70C}_{\mathfrak{a}f}(n)e(nz),\quad j\left(\left(\begin{array}{@{}cc@{}}a & b\\ c & d\end{array}\right),z\right)=cz+d,\end{eqnarray}$$

where $\unicode[STIX]{x1D70E}_{\mathfrak{a}}$ is a scaling matrix (see §3). When $\mathfrak{a}=\infty$ we will often drop the dependence on $\mathfrak{a}$ from the notation. We adopt the standard notation $e(x):=\text{e}^{2\unicode[STIX]{x1D70B}\text{i}x}$ . We normalize the coefficients $\unicode[STIX]{x1D70C}_{\mathfrak{a}f}(n)$ by setting

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{\mathfrak{a}f}(n)=\biggl(\frac{\unicode[STIX]{x1D70B}^{-k}\unicode[STIX]{x1D6E4}(k)}{(4n)^{k-1}}\biggr)^{1/2}\unicode[STIX]{x1D70C}_{\mathfrak{a}f}(n).\end{eqnarray}$$

Throughout this paper we work with a fixed orthonormal basis ${\mathcal{B}}_{k}(N)$ whose elements are eigenforms of the Hecke operators $T_{n}$ for $(n,N)=1$ with eigenvalues $\unicode[STIX]{x1D706}_{f}(n)$ , and which satisfy

(2.2) $$\begin{eqnarray}N^{-k/2}z^{-k}f(-1/Nz)=\unicode[STIX]{x1D714}_{f}f(z),\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{f}=\pm 1$ is the eigenvalue of the Fricke involution. With $\unicode[STIX]{x1D70E}_{0}$ as in (3.1) below, we find from (2.1) and (2.2) that

(2.3) $$\begin{eqnarray}\unicode[STIX]{x1D70C}_{0f}(n)=\unicode[STIX]{x1D714}_{f}\,\unicode[STIX]{x1D70C}_{\infty f}(n).\end{eqnarray}$$

Similarly, let ${\mathcal{U}}(N)=\{u_{j}\}$ denote a complete orthonormal system of Maass cusp forms with Fourier expansions

$$\begin{eqnarray}u_{j}(\unicode[STIX]{x1D70E}_{\mathfrak{a}}z)=\sqrt{y}\mathop{\sum }_{n\neq 0}\unicode[STIX]{x1D70C}_{\mathfrak{a}j}(n)K_{\text{i}t_{j}}(2\unicode[STIX]{x1D70B}|n|y)e(nx),\end{eqnarray}$$

where $\frac{1}{4}+\text{i}t_{j}$ is the Laplace eigenvalue and $K_{\unicode[STIX]{x1D708}}(x)$ is the $K$ -Bessel function. We normalize the coefficients $\unicode[STIX]{x1D70C}_{\mathfrak{a}j}(n)$ by

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{\mathfrak{a}j}(n)=\biggl(\frac{\unicode[STIX]{x1D70B}}{\cosh (\unicode[STIX]{x1D70B}t_{j})}\biggr)^{1/2}\unicode[STIX]{x1D70C}_{\mathfrak{a}j}(n).\end{eqnarray}$$

We may also assume, as above, that the $u_{j}$ are eigenforms for the Hecke operators $T_{n}$ for $(n,N)=1$ and for the Fricke involution (i.e. that $u_{j}$ satisfies (2.2) with $k=0$ ). We write their eigenvalues $\unicode[STIX]{x1D706}_{j}(n)$ and $\unicode[STIX]{x1D714}_{j}$ , respectively.

Let $(\unicode[STIX]{x1D708},\unicode[STIX]{x1D706})$ denote one of the pairs $(\unicode[STIX]{x1D708}_{f},\unicode[STIX]{x1D706}_{f})$ or $(\unicode[STIX]{x1D708}_{j},\unicode[STIX]{x1D706}_{j})$ . As long as $(n,N)=1$ we have the relation

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D708}(m)\unicode[STIX]{x1D706}(n)=\mathop{\sum }_{d|(m,n)}\unicode[STIX]{x1D708}\biggl(\frac{mn}{d^{2}}\biggr)\end{eqnarray}$$

see [Reference Iwaniec7, (8.37)]. This implies that

$$\begin{eqnarray}\unicode[STIX]{x1D708}(n)=\unicode[STIX]{x1D708}(1)\unicode[STIX]{x1D706}(n)\end{eqnarray}$$

as long as $(n,N)=1$ .

For each cusp $\mathfrak{c}$ of $\unicode[STIX]{x1D6E4}$ (see §3) and for $\operatorname{Re}(u)>1$ let

$$\begin{eqnarray}E_{\mathfrak{c}}(z,u):=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{\infty }\backslash \unicode[STIX]{x1D6E4}}\operatorname{Im}(\unicode[STIX]{x1D70E}_{\mathfrak{c}}^{-1}\unicode[STIX]{x1D6FE}z)^{u}\end{eqnarray}$$

denote the Eisenstein series associated to $\mathfrak{c}$ . This has Fourier expansion

$$\begin{eqnarray}E_{\mathfrak{c}}(\unicode[STIX]{x1D70E}_{\mathfrak{a}}z,u)=\unicode[STIX]{x1D6FF}_{\mathfrak{a}\mathfrak{c}}y^{u}+\unicode[STIX]{x1D70C}_{\mathfrak{ a}\mathfrak{c}}(0,u)y^{1-u}+\sqrt{y}\mathop{\sum }_{n\neq 0}\unicode[STIX]{x1D70C}_{\mathfrak{a}\mathfrak{c}}(n,u)K_{u-1/2}(2\unicode[STIX]{x1D70B}|n|y)e(nx)\end{eqnarray}$$

which has a meromorphic continuation to $u\in \mathbb{C}$ . On the line $\operatorname{Re}(u)=\frac{1}{2}$ we normalize the coefficients by

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{\mathfrak{a}\mathfrak{c}}(n,t)=\biggl(\frac{\unicode[STIX]{x1D70B}}{\cosh \unicode[STIX]{x1D70B}t}\biggr)^{1/2}\unicode[STIX]{x1D70C}_{\mathfrak{a}\mathfrak{c}}\biggl(n,\frac{1}{2}+\text{i}t\biggr).\end{eqnarray}$$

For the remainder of the paper, fix a holomorphic newform $g\in S_{\unicode[STIX]{x1D705}}(1)$ . For $s=\unicode[STIX]{x1D70E}+\text{i}t$ with $\unicode[STIX]{x1D70E}$ sufficiently large, let us call

$$\begin{eqnarray}\widetilde{L}(s,h)=\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D708}_{h}(n)}{n^{s}}\quad \text{and}\quad \widetilde{L}(s,h\times g)=\unicode[STIX]{x1D701}_{N}(2s)\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{g}(n)\unicode[STIX]{x1D708}_{h}(n)}{n^{s}},\end{eqnarray}$$

where $\unicode[STIX]{x1D701}_{N}(s)=\prod _{p\nmid N}(1-p^{-s})^{-1}$ . These are the “raw $L$ -functions” involving $\unicode[STIX]{x1D708}_{h}$ as opposed to $\unicode[STIX]{x1D706}_{h}$ . We choose this notation in order to simultaneously cover oldforms. If $h$ is a newform (holomorphic or Maass), then we simply have $\widetilde{L}(s,h)=\unicode[STIX]{x1D708}_{h}(1)L(s,h)$ and $\widetilde{L}(s,g\times h)=\unicode[STIX]{x1D708}_{h}(1)L(s,g\times h)$ , where $L$ without a tilde denotes the usual $L$ -function. If $f$ is a newform of level $M$ with $M|N$ and $h=f(dz)$ for some $d|N/M$ , then one can see that $\widetilde{L}(s,h)=\unicode[STIX]{x1D708}_{f}(1)d^{-s}L(s,f)$ . For the Rankin–Selberg $L$ -function $\widetilde{L}(s,g\times h)$ the relationship is more complicated, but by applying Hecke relations one can see that it has the same analytic behavior as $L(s,g\times f)$ in the region $\operatorname{Re}(s)\geqslant 0$ . The coefficients $\unicode[STIX]{x1D708}_{h}(n)$ come up in applications of the Bruggeman–Kuznetsov trace formula and hence the Dirichlet series $\widetilde{L}(s,g\times h)$ naturally appears.

We specialize now to the case $N=q$ is prime. Let $\unicode[STIX]{x1D711}$ be a smooth function defined on the non-negative reals such that

$$\begin{eqnarray}\unicode[STIX]{x1D711}(0)=0\quad \text{and}\quad \unicode[STIX]{x1D711}^{(j)}(x)\ll (1+x)^{-2-\unicode[STIX]{x1D716}}\quad \text{for }j=0,1,2,\end{eqnarray}$$

and let $\unicode[STIX]{x1D711}_{h}$ and $\unicode[STIX]{x1D711}_{+}$ denote the integral transforms in (3.6) below. Define

$$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}^{d}(p,q;s;\unicode[STIX]{x1D711}) & = & \displaystyle \mathop{\sum }_{u_{j}\in {\mathcal{U}}(q)}\unicode[STIX]{x1D714}_{j}\,\unicode[STIX]{x1D711}_{+}(t_{j})\widetilde{L}(s,g\times u_{j})^{2}\unicode[STIX]{x1D706}_{j}(p),\nonumber\\ \displaystyle {\mathcal{N}}_{g}^{h}(p,q;s;\unicode[STIX]{x1D711}) & = & \displaystyle \mathop{\sum }_{\substack{ \ell \text{ even}}}i^{\ell }\unicode[STIX]{x1D711}_{h}(\ell )\mathop{\sum }_{f\in {\mathcal{B}}_{\ell }(q)}\unicode[STIX]{x1D714}_{f}\,\widetilde{L}(s,g\times f)^{2}\unicode[STIX]{x1D706}_{f}(p),\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}^{c}(p,q;s;\unicode[STIX]{x1D711}) & = & \displaystyle \frac{1}{4\unicode[STIX]{x1D70B}}\mathop{\sum }_{\mathfrak{c}}\int _{-\infty }^{\infty }\unicode[STIX]{x1D711}_{+}(t)\widetilde{L}\biggl(s,g\times E_{\mathfrak{c}}\biggl(\unicode[STIX]{x1D70E}_{0}\ast ,\frac{1}{2}+\text{i}t\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\widetilde{L}\biggl(s,g\times E_{\mathfrak{c}}\biggl(\ast ,\frac{1}{2}-\text{i}t\biggr)\biggr)\unicode[STIX]{x1D70F}_{\text{i}t}(p)\,dt,\nonumber\end{eqnarray}$$

where $\sum _{\mathfrak{c}}$ is over a set of inequivalent cusps of $\unicode[STIX]{x1D6E4}$ . Call

$$\begin{eqnarray}{\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D711})={\mathcal{N}}_{g}^{d}(p,q;s;\unicode[STIX]{x1D711})+{\mathcal{N}}_{g}^{c}(p,q;s;\unicode[STIX]{x1D711})+{\mathcal{N}}_{g}^{h}(p,q;s;\unicode[STIX]{x1D711}).\end{eqnarray}$$

With $\unicode[STIX]{x1D711}$ as above, the function ${\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D711})$ is holomorphic at $s=\frac{1}{2}$ , which is the point we are most interested in, and we let ${\mathcal{N}}_{g}(p,q;\unicode[STIX]{x1D711}):={\mathcal{N}}_{g}(p,q;\frac{1}{2};\unicode[STIX]{x1D711})$ .

We are now ready to state our main theorem.

Theorem 2.1. Let $p$ and $q$ be distinct primes and let $\unicode[STIX]{x1D719}$ be a function on $[0,\infty )$ satisfying the conditions in the beginning of Theorem 5.1. With the notation above, define

$$\begin{eqnarray}\displaystyle {\mathcal{M}}_{g}(p,q;\unicode[STIX]{x1D711}) & := & \displaystyle (1-p^{-2})\,{\mathcal{N}}_{g}(p,q;\unicode[STIX]{x1D711})-\frac{2\unicode[STIX]{x1D706}_{g}(p)}{\sqrt{p}}(1-p^{-1})\,{\mathcal{N}}_{g}(1,q;\unicode[STIX]{x1D711})\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{\sqrt{p}}\,{\mathcal{N}}_{g}(1,pq;\unicode[STIX]{x1D711}).\nonumber\end{eqnarray}$$

Then

$$\begin{eqnarray}\sqrt{q}\,{\mathcal{M}}_{g}(p,q;\unicode[STIX]{x1D719})=\sqrt{p}\,{\mathcal{M}}_{g}(q,p;\unicode[STIX]{x1D6F7}),\end{eqnarray}$$

where $\unicode[STIX]{x1D6F7}$ is an integral transform of $\unicode[STIX]{x1D719}$ given in (5.1).

The various steps of the proof are detailed in the remaining sections, but we give a high-level outline here.

Proof of Theorem 2.1.

Using equation (4.1) with ${\mathcal{S}}$ defined as in (4.6) below we have that

$$\begin{eqnarray}\displaystyle {\mathcal{S}}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \frac{1-p^{-4s}}{\unicode[STIX]{x1D701}_{pq}(2s)^{2}}{\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle -\,\frac{2\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\frac{(1-p^{-2s})}{\unicode[STIX]{x1D701}_{pq}(2s)^{2}}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719})+\frac{{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})}{\sqrt{p}\unicode[STIX]{x1D701}_{pq}(2s)^{2}}.\nonumber\end{eqnarray}$$

Specialization of $\unicode[STIX]{x1D701}_{pq}(2s)^{2}{\mathcal{S}}(p,q;s;\unicode[STIX]{x1D719})$ at $s={\textstyle \frac{1}{2}}$ yields ${\mathcal{M}}_{g}(p,q;\unicode[STIX]{x1D719})$ . Now we apply the reciprocity relation for ${\mathcal{S}}$ in Theorem 5.1 below, which gives

$$\begin{eqnarray}\unicode[STIX]{x1D701}_{pq}(2s)^{2}{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})=\unicode[STIX]{x1D701}_{pq}(2s)^{2}\biggl(\frac{p}{q}\biggr)^{2s-1}\frac{\sqrt{p}}{\sqrt{q}}{\mathcal{S}}(q,p;s;\unicode[STIX]{x1D6F7})\end{eqnarray}$$

with $\unicode[STIX]{x1D6F7}$ as in (5.1). Specializing to $s={\textstyle \frac{1}{2}}$ yields the result.◻

3 The Bruggeman–Kuznetsov trace formula

The purpose of this section is to state the Bruggeman–Kuznetsov trace formula associated to the $0\infty$ cusp pair, which will allow us to relate the moments in Theorem 2.1 to sums of Kloosterman sums.

Let $\unicode[STIX]{x1D6E4}=\unicode[STIX]{x1D6E4}_{0}(N)$ and let $\mathfrak{a},\mathfrak{b}\in \mathbb{Q}\cup \{\infty \}$ denote two cusps. A scaling matrix $\unicode[STIX]{x1D70E}_{\mathfrak{a}}\in \operatorname{SL}_{2}(\mathbb{R})$ for the cusp $\mathfrak{a}$ satisfies the properties

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{\mathfrak{a}}\infty =\mathfrak{a}\quad \text{and}\quad \unicode[STIX]{x1D70E}_{\mathfrak{a}}^{-1}\unicode[STIX]{x1D6E4}_{\mathfrak{a}}\unicode[STIX]{x1D70E}_{\mathfrak{a}}=\left\{\pm \left(\begin{array}{@{}cc@{}}1 & n\\ 0 & 1\end{array}\right):n\in \mathbb{Z}\right\},\end{eqnarray}$$

where $\unicode[STIX]{x1D6E4}_{\mathfrak{a}}$ is the stabilizer of $\mathfrak{a}$ in $\unicode[STIX]{x1D6E4}$ . We are primarily interested in the cases $\mathfrak{a}=\infty$ and $\mathfrak{a}=0$ , for which we have

(3.1) $$\begin{eqnarray}\unicode[STIX]{x1D70E}_{\infty }=\left(\begin{array}{@{}cc@{}}1 & 0\\ 0 & 1\end{array}\right)\quad \text{and}\quad \unicode[STIX]{x1D70E}_{0}=\left(\begin{array}{@{}cc@{}}0 & -1/\sqrt{N}\\ \sqrt{N} & 0\end{array}\right).\end{eqnarray}$$

Given a pair of cusps $\mathfrak{a},\mathfrak{b}$ and associated scaling matrices $\unicode[STIX]{x1D70E}_{\mathfrak{a}},\unicode[STIX]{x1D70E}_{\mathfrak{b}}$ a set of allowed moduli for the Kloosterman sums is given by

$$\begin{eqnarray}{\mathcal{C}}_{\mathfrak{a},\mathfrak{b}}=\left\{\unicode[STIX]{x1D6FE}>0:\left(\begin{array}{@{}cc@{}}\ast & \ast \\ \unicode[STIX]{x1D6FE} & \ast \end{array}\right)\in \unicode[STIX]{x1D70E}_{\mathfrak{a}}^{-1}\unicode[STIX]{x1D6E4}\unicode[STIX]{x1D70E}_{\mathfrak{b}}\right\}.\end{eqnarray}$$

The Kloosterman sum for a modulus $\unicode[STIX]{x1D6FE}\in {\mathcal{C}}_{\mathfrak{a},\mathfrak{b}}$ is defined as

(3.2) $$\begin{eqnarray}S_{\mathfrak{a}\mathfrak{b}}^{\unicode[STIX]{x1D6E4}}(m,n;\unicode[STIX]{x1D6FE})=\mathop{\sum }_{\left(\!\begin{smallmatrix}a & b\\ \unicode[STIX]{x1D6FE} & d\end{smallmatrix}\!\right)\in \unicode[STIX]{x1D6E4}_{\infty }\backslash \unicode[STIX]{x1D70E}_{\mathfrak{a}}^{-1}\unicode[STIX]{x1D6E4}\unicode[STIX]{x1D70E}_{\mathfrak{b}}/\unicode[STIX]{x1D6E4}_{\infty }}e\biggl(\frac{am+dn}{\unicode[STIX]{x1D6FE}}\biggr).\end{eqnarray}$$

When $\mathfrak{a}=\mathfrak{b}=\infty$ we obtain the usual Kloosterman sum (for $c\equiv 0\hspace{0.6em}({\rm mod}\hspace{0.2em}N)$ )

$$\begin{eqnarray}S_{\infty \infty }^{\unicode[STIX]{x1D6E4}}(m,n;c)=S(m,n;c):=\mathop{\sum }_{\substack{ d\hspace{0.2em}{\rm mod}\hspace{0.2em}c \\ (d,c)=1}}e\biggl(\frac{\overline{d}m+dn}{c}\biggr).\end{eqnarray}$$

For these sums we have Weil’s bound

(3.3) $$\begin{eqnarray}|S(m,n;c)|\leqslant \unicode[STIX]{x1D70F}_{0}(c)(m,n,c)^{1/2}c^{1/2},\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{0}(c)$ is the number of divisors of $c$ . For the $\infty 0$ cusp-pair, and for the choices above, the set of allowed moduli is ${\mathcal{C}}_{\infty ,0}=\{\unicode[STIX]{x1D6FE}=c\sqrt{N}:(c,N)=1\}$ (see [Reference Iwaniec6, §4.2]). Then (3.2) can be expressed in terms of $S(m,n;c)$ via

$$\begin{eqnarray}S_{\infty 0}^{\unicode[STIX]{x1D6E4}}(m,n;c\sqrt{N})=S(\overline{N}m,n;c).\end{eqnarray}$$

Kloosterman sums appear in the Fourier expansion of the Eisenstein series. For $n\neq 0$ and $\operatorname{Re}(u)>1$ we have (see [Reference Iwaniec7, Theorem 3.4])

$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{\mathfrak{a}\mathfrak{c}}(n,u)=\frac{2\unicode[STIX]{x1D70B}^{u}}{\unicode[STIX]{x1D6E4}(u)}|n|^{u-1/2}\mathop{\sum }_{c\in {\mathcal{C}}_{\mathfrak{c},\mathfrak{a}}}\frac{S_{\mathfrak{c}\mathfrak{a}}(0,n;c)}{c^{2u}}.\end{eqnarray}$$

So for $q$ prime we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{\infty 0}(n,u) & = & \displaystyle \frac{2\unicode[STIX]{x1D70B}^{u}}{\unicode[STIX]{x1D6E4}(u)}|n|^{u-1/2}\mathop{\sum }_{(c,q)=1}\frac{S(0,n;c)}{q^{u}c^{2u}},\nonumber\\ \displaystyle \unicode[STIX]{x1D70C}_{\infty \infty }(n,u) & = & \displaystyle \frac{2\unicode[STIX]{x1D70B}^{u}}{\unicode[STIX]{x1D6E4}(u)}|n|^{u-1/2}\mathop{\sum }_{q\mid c>0}\frac{S(0,n;c)}{c^{2u}}.\nonumber\end{eqnarray}$$

The Kloosterman sum $S(0,n;c)$ is equal to the Ramanujan sum

$$\begin{eqnarray}\mathop{\sum }_{d\mid (n,c)}\unicode[STIX]{x1D707}(c/d)d,\end{eqnarray}$$

whose Dirichlet series is given by

$$\begin{eqnarray}\mathop{\sum }_{c=1}^{\infty }\frac{S(0,n;c)}{c^{2u}}=\frac{\unicode[STIX]{x1D70F}_{u-1/2}(n)}{\unicode[STIX]{x1D701}(2u)},\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{w}(n)=\sum _{ab=|n|}(a/b)^{w}$ (cf. [Reference Iwaniec7, (2.27)]). It follows that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{\infty 0}(n,u)=\frac{2\unicode[STIX]{x1D70B}^{u}\unicode[STIX]{x1D70F}_{u-1/2}^{(q)}(n)}{q^{u}\unicode[STIX]{x1D6E4}(u)\unicode[STIX]{x1D701}_{q}(2u)} & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{\infty \infty }(n,u)=\frac{2\unicode[STIX]{x1D70B}^{u}}{\unicode[STIX]{x1D6E4}(u)}\biggl(\frac{\unicode[STIX]{x1D70F}_{u-1/2}(n)}{\unicode[STIX]{x1D701}(2u)}-\frac{\unicode[STIX]{x1D70F}_{u-1/2}^{(q)}(n)}{\unicode[STIX]{x1D701}_{q}(2u)}\biggr), & & \displaystyle \nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{w}^{(q)}(n)=\unicode[STIX]{x1D70F}_{w}(n/(n,q))$ . From this we can derive a Hecke-type relation for the pair $(\unicode[STIX]{x1D708}_{\mathfrak{a}\mathfrak{c}}(n,t),\unicode[STIX]{x1D70F}_{\text{i}t}(p))$ , where $p\neq q$ is a prime. Suppose that $(n,q)=1$ and that $(\mathfrak{a},\mathfrak{c})=(\infty ,\infty )$ or $(\infty ,0)$ . Then since

$$\begin{eqnarray}\unicode[STIX]{x1D70F}_{w}^{(q)}(m)\unicode[STIX]{x1D70F}_{w}^{(q)}(n)=\mathop{\sum }_{d\mid (m,n)}\unicode[STIX]{x1D70F}_{w}^{(q)}\biggl(\frac{mn}{d^{2}}\biggr)\end{eqnarray}$$

we have the relation

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D708}_{\mathfrak{a}\mathfrak{c}}(n,t)\unicode[STIX]{x1D70F}_{\text{i}t}(p)=\unicode[STIX]{x1D708}_{\mathfrak{a}\mathfrak{c}}(np,t)+\unicode[STIX]{x1D6FF}_{p\mid n}\unicode[STIX]{x1D708}_{\mathfrak{a}\mathfrak{c}}(n/p,t).\end{eqnarray}$$

The Bruggeman–Kuznetsov trace formula relates sums of Kloosterman sums to coefficients of cusp forms and Eisenstein series. Let $\unicode[STIX]{x1D711}$ be a smooth function defined on the non-negative reals satisfying

(3.5) $$\begin{eqnarray}\unicode[STIX]{x1D711}(0)=0\quad \text{and}\quad \unicode[STIX]{x1D711}^{(j)}(x)\ll (1+x)^{-2-\unicode[STIX]{x1D716}}\quad \text{for }j=0,1,2,\end{eqnarray}$$

and define

(3.6) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{h}(\ell ) & = & \displaystyle \int _{0}^{\infty }J_{\ell -1}(x)\unicode[STIX]{x1D711}(x)\,\frac{dx}{x},\nonumber\\ \displaystyle \unicode[STIX]{x1D711}_{+}(t) & = & \displaystyle \frac{\text{i}}{2\sinh \unicode[STIX]{x1D70B}t}\int _{0}^{\infty }(J_{2\text{i}t}(x)-J_{-2\text{i}t}(x))\unicode[STIX]{x1D711}(x)\,\frac{dx}{x}.\end{eqnarray}$$

The following can be found in [Reference Iwaniec7, Theorem 9.8].

Proposition 3.1 (Bruggeman–Kuznetsov).

Suppose that $m,n\geqslant 1$ . Let $\unicode[STIX]{x1D711}$ be a smooth function defined on $[0,\infty )$ which satisfies (3.5). Define

$$\begin{eqnarray}{\mathcal{K}}(m,n,N;\unicode[STIX]{x1D711})=\mathop{\sum }_{(c,N)=1}\frac{S(\overline{N}m,n;c)}{c\sqrt{N}}\unicode[STIX]{x1D711}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{N}}\biggr).\end{eqnarray}$$

Then

$$\begin{eqnarray}{\mathcal{K}}={\mathcal{K}}^{d}+{\mathcal{K}}^{c}+{\mathcal{K}}^{h},\end{eqnarray}$$

where

4 Calculations with the Hecke relations and Kloosterman sums

Let $\unicode[STIX]{x1D719}$ be as in Theorem 2.1. The purpose of this section is to prove the identity

(4.1) $$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \frac{2\unicode[STIX]{x1D706}_{g}(p)}{p^{s}(1+p^{-2s})}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{\sqrt{p}}\frac{{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})}{(1-p^{-4s})}+\unicode[STIX]{x1D701}_{pq}(2s)^{2}\frac{{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})}{(1-p^{-4s})},\qquad\end{eqnarray}$$

where ${\mathcal{S}}$ is defined in (4.6) below. The following section shows that ${\mathcal{S}}$ is the quantity that satisfies a natural reciprocity relation.

We begin with explicit computations in the case of the holomorphic spectrum ${\mathcal{N}}_{g}^{h}(p,q;s;\unicode[STIX]{x1D719})$ . Suppose that $\unicode[STIX]{x1D70E}>1$ . Opening the $L$ -functions as Dirichlet series we find that

(4.2) $$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}^{h}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \unicode[STIX]{x1D701}_{q}^{2}(2s)\mathop{\sum }_{m\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)}{m^{s}}\mathop{\sum }_{\ell \equiv 0(2)}i^{\ell }\unicode[STIX]{x1D719}_{h}(\ell )\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{f\in {\mathcal{H}}_{\ell }(q)}\unicode[STIX]{x1D714}_{f}\unicode[STIX]{x1D708}_{f}(m)\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}_{f}(n)\unicode[STIX]{x1D706}_{f}(p).\end{eqnarray}$$

We apply the Hecke relations (2.4) in order to absorb the $\unicode[STIX]{x1D706}_{f}(p)$ factor.

Lemma 4.1. Let $p$ be a prime number and let $g\in S_{\unicode[STIX]{x1D705}}(1)$ . Suppose that $(\unicode[STIX]{x1D708},\unicode[STIX]{x1D706})$ is a pair of arithmetic functions satisfying the relation

(4.3) $$\begin{eqnarray}\unicode[STIX]{x1D708}(n)\unicode[STIX]{x1D706}(p)=\unicode[STIX]{x1D708}(np)+\unicode[STIX]{x1D6FF}_{p\mid n}\unicode[STIX]{x1D708}(n/p)\end{eqnarray}$$

and the bound $\unicode[STIX]{x1D708}(n)\ll n^{\unicode[STIX]{x1D6FC}}$ for some $\unicode[STIX]{x1D6FC}>0$ . Then for $\unicode[STIX]{x1D70E}=\operatorname{Re}(s)>1+\unicode[STIX]{x1D6FC}$ we have

$$\begin{eqnarray}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(n)\unicode[STIX]{x1D706}(p)=\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(n)+(1-p^{-2s})\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(np).\end{eqnarray}$$

Proof. Using the relation (4.3) we find that

$$\begin{eqnarray}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(n)\unicode[STIX]{x1D706}(p)=\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(np)+\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(np)}{(np)^{s}}\unicode[STIX]{x1D708}(n).\end{eqnarray}$$

Since $\unicode[STIX]{x1D706}_{g}$ also satisfies the Hecke relations (4.3) we see that

$$\begin{eqnarray}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(np)}{(np)^{s}}\unicode[STIX]{x1D708}(n)=\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(n)-\frac{1}{p^{2s}}\mathop{\sum }_{n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(n)}{n^{s}}\unicode[STIX]{x1D708}(np).\end{eqnarray}$$

The lemma follows. ◻

Applying Lemma 4.1 and the relation (2.3) to (4.2), we find that

(4.4) $$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}^{h}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}{\mathcal{N}}_{g}^{h}(1,q;s;\unicode[STIX]{x1D719})+(1-p^{-2s})\unicode[STIX]{x1D701}_{q}(2s)^{2}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}{\mathcal{K}}^{h}(m,pn,q;\unicode[STIX]{x1D719}).\end{eqnarray}$$

A similar computation confirms that (4.4) holds for the discrete and continuous spectra (using (3.4) in the latter case), simply replacing $h$ by $d$ or $c$ above. After applying Proposition 3.1, it follows that

(4.5) $$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719})+(1-p^{-2s})\unicode[STIX]{x1D701}_{q}(2s)^{2}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}{\mathcal{K}}(m,pn,q;\unicode[STIX]{x1D719}).\end{eqnarray}$$

Our next aim is to relate the sum on the right-hand side of (4.5) to the sum

(4.6) $$\begin{eqnarray}{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719}):=\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}\mathop{\sum }_{(c,pq)=1}\frac{S(m\overline{q},np;c)}{c\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mnp}}{c\sqrt{q}}\biggr).\end{eqnarray}$$

This involves a sieving process which leaves us with a sum over $c$ relatively prime to $pq$ .

Proposition 4.2. Let $p$ be a prime number and let $g\in S_{\unicode[STIX]{x1D705}}(1)$ . For $\unicode[STIX]{x1D70E}>\frac{5}{4}$ we have

$$\begin{eqnarray}\displaystyle & & \displaystyle (1+p^{-2s})\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}{\mathcal{K}}(m,pn,q;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle \quad ={\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})-\frac{1}{\sqrt{p}}\unicode[STIX]{x1D701}_{pq}(2s)^{-2}{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\unicode[STIX]{x1D701}_{q}(2s)^{-2}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719}).\nonumber\end{eqnarray}$$

In the proof of Proposition 4.2 we will need the following facts about Kloosterman sums, all of which follow from standard exponential sum manipulations.

Proof of Proposition 4.2.

Let

$$\begin{eqnarray}\displaystyle {\mathcal{R}} & := & \displaystyle \mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}{\mathcal{K}}(m,pn,q;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}\mathop{\sum }_{(c,q)=1}\frac{S(\overline{q}m,pn;c)}{c\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mnp}}{c\sqrt{q}}\biggr),\nonumber\end{eqnarray}$$

and write ${\mathcal{R}}={\mathcal{S}}+{\mathcal{T}}$ , where ${\mathcal{S}}={\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})$ is defined in (4.6) and ${\mathcal{T}}$ comprises those terms of ${\mathcal{R}}$ with $p\mid c$ . If furthermore $p^{2}\mid c$ , we apply (4.8), and if $p\mid \mid c$ , we apply (4.7), obtaining

$$\begin{eqnarray}\displaystyle {\mathcal{T}} & = & \displaystyle \mathop{\sum }_{\substack{ m,n\geqslant 1}}\frac{\unicode[STIX]{x1D706}_{g}(pm)\unicode[STIX]{x1D706}_{g}(n)}{(pm)^{s}n^{s}}\mathop{\sum }_{\substack{ (c,q)=1 \\ p\mid c}}\frac{S(\overline{q}pm,pn;pc)}{pc\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{q}}\biggr)\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{\substack{ m,n\geqslant 1 \\ p\nmid m}}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{m^{s}n^{s}}\mathop{\sum }_{\substack{ (c,pq)=1}}\frac{S(\overline{qp}m,n;c)}{pc\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{pq}}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,(p-1)\mathop{\sum }_{\substack{ m,n\geqslant 1 \\ p\mid m}}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{m^{s}n^{s}}\mathop{\sum }_{\substack{ (c,pq)=1}}\frac{S(\overline{qp}m,n;c)}{pc\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{pq}}\biggr).\nonumber\end{eqnarray}$$

Now we apply (4.9) to the terms of the first sum. In the last line we separate $p-1$ , combining the $p$ term with the first line, and the $-1$ term with the second. We find that

$$\begin{eqnarray}\displaystyle {\mathcal{T}} & = & \displaystyle \mathop{\sum }_{\substack{ m,n\geqslant 1}}\frac{\unicode[STIX]{x1D706}_{g}(pm)\unicode[STIX]{x1D706}_{g}(n)}{(pm)^{s}n^{s}}\mathop{\sum }_{\substack{ (c,q)=1}}\frac{S(\overline{q}m,n;c)}{c\sqrt{q}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{q}}\biggr)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{\sqrt{p}}\mathop{\sum }_{\substack{ m,n\geqslant 1}}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{m^{s}n^{s}}\mathop{\sum }_{\substack{ (c,pq)=1}}\frac{S(\overline{qp}m,n;c)}{c\sqrt{pq}}\unicode[STIX]{x1D719}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mn}}{c\sqrt{pq}}\biggr).\nonumber\end{eqnarray}$$

Finally, we apply the Hecke relation again to $\unicode[STIX]{x1D706}_{g}(pm)$ in the first term, and conclude that

$$\begin{eqnarray}{\mathcal{T}}=\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\unicode[STIX]{x1D701}_{q}(2s)^{-2}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719})-\frac{1}{p^{2s}}{\mathcal{R}}-\frac{1}{\sqrt{p}}\unicode[STIX]{x1D701}_{pq}(2s)^{-2}{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})\end{eqnarray}$$

since

$$\begin{eqnarray}S(\overline{q}pm,n;c)=S(pm,\overline{q}n;c)=S(\overline{q}n,pm;c).\end{eqnarray}$$

On the other hand, ${\mathcal{T}}={\mathcal{R}}-{\mathcal{S}}$ , so we have

$$\begin{eqnarray}\displaystyle (1+p^{-2s}){\mathcal{R}} & = & \displaystyle {\mathcal{S}}-\frac{1}{\sqrt{p}}\unicode[STIX]{x1D701}_{pq}(2s)^{-2}{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\unicode[STIX]{x1D701}_{q}(2s)^{-2}{\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719}),\nonumber\end{eqnarray}$$

as desired. ◻

Now let us combine the last two results. Combining (4.5) and Proposition 4.2 we obtain

$$\begin{eqnarray}\displaystyle {\mathcal{N}}_{g}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}{\mathcal{N}}(1,q,s,\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x1D701}_{pq}(2s)^{2}(1+p^{-2s})}{(1-p^{-2s})(1+p^{-2s})}\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}{\mathcal{K}}(m,pn;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & = & \displaystyle \biggl(\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}+\frac{\unicode[STIX]{x1D706}_{g}(p)}{p^{s}}\frac{(1-p^{-2s})}{(1+p^{-2s})}\biggr){\mathcal{N}}_{g}(1,q;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{\sqrt{p}}\frac{{\mathcal{N}}_{g}(1,pq;s;\unicode[STIX]{x1D719})}{(1-p^{-4s})}+\unicode[STIX]{x1D701}_{pq}(2s)^{2}\frac{{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})}{(1-p^{-4s})}.\nonumber\end{eqnarray}$$

Rearranging the terms in the first parentheses we obtain (4.1).

5 Reciprocity for ${\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})$

In this section we will prove the following reciprocity relation for ${\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})$ .

Theorem 5.1. Let $\unicode[STIX]{x1D719}$ be a smooth test function satisfying $\unicode[STIX]{x1D719}^{(j)}(0)=0$ and $\unicode[STIX]{x1D719}^{(j)}(x)\ll (1+x)^{-A}$ for $0\leqslant j\leqslant 12$ and for some $A>12$ . Suppose that $\operatorname{Re}(s)=\unicode[STIX]{x1D70E}>\frac{5}{4}$ and that $p,q$ are distinct primes. Then

$$\begin{eqnarray}\sqrt{q}\,{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719})=\biggl(\frac{p}{q}\biggr)^{2s-1}\sqrt{p}\,{\mathcal{S}}(q,p;s,\unicode[STIX]{x1D6F7}),\end{eqnarray}$$

where, for any $\unicode[STIX]{x1D709}$ satisfying $0<\unicode[STIX]{x1D709}+12<A$ , $\unicode[STIX]{x1D6F7}(x)$ is defined as

(5.1) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(x) & = & \displaystyle \unicode[STIX]{x1D6F7}_{\unicode[STIX]{x1D705},s}(x)\nonumber\\ \displaystyle & := & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}\widetilde{\unicode[STIX]{x1D719}}(u)2^{-u}\biggl[\frac{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}+1)/2-s-u/2)}{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}-1)/2+s+u/2)}\biggr]^{2}\biggl(\frac{x}{2}\biggr)^{u+4s-2}\,du\quad\end{eqnarray}$$

and $\widetilde{\unicode[STIX]{x1D719}}$ is the Mellin transform

$$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D719}}(u)=\int _{0}^{\infty }t^{u}\unicode[STIX]{x1D719}(t)\,\frac{dt}{t}.\end{eqnarray}$$

Furthermore, the function $\unicode[STIX]{x1D6F7}_{\unicode[STIX]{x1D705},s}$ can be analytically continued to $\frac{1}{2}\leqslant \unicode[STIX]{x1D70E}\leqslant \frac{5}{4}+\unicode[STIX]{x1D716}$ and it satisfies (3.5) in that region.

Suppose that $\unicode[STIX]{x1D719}^{(j)}(0)=0$ and $\unicode[STIX]{x1D719}^{(j)}(x)\ll (1+x)^{-A}$ for all $j\in \{0,1,2,\ldots ,J\}$ , for some integer $J$ . We will later see that we may take $J=12$ . By the decay of $\unicode[STIX]{x1D719}$ and its derivatives at $0$ and $\infty$ , we may apply integration by parts $j$ times and obtain that

(5.2) $$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D719}}(u)=(-1)^{j}\int _{0}^{\infty }\unicode[STIX]{x1D719}^{(j)}(x)\frac{x^{u+j}}{u(u+1)\cdots (u+j-1)}\,\frac{dx}{x}\ll (1+|u|)^{-j}\end{eqnarray}$$

as long as the integral converges. Near $0$ we have $\unicode[STIX]{x1D719}^{(j)}(x)\ll x^{J-j}$ , so the integrand is majorized by $x^{J+\unicode[STIX]{x1D709}-1}$ (with $\unicode[STIX]{x1D709}=\operatorname{Re}(u)$ ) as $x\rightarrow 0$ . For large $x$ the integrand is majorized by $x^{-A+\unicode[STIX]{x1D709}+j-1}$ . It follows that the integral is absolutely convergent (and thus (5.2) holds with $j=J$ ) as long as $-J<\unicode[STIX]{x1D709}<A-J$ .

Starting with (4.6), we write $\unicode[STIX]{x1D719}$ as the inverse Mellin transform of $\widetilde{\unicode[STIX]{x1D719}}$ via

$$\begin{eqnarray}\unicode[STIX]{x1D719}(x)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}\widetilde{\unicode[STIX]{x1D719}}(u)x^{-u}\,du.\end{eqnarray}$$

We then interchange integral and summation to obtain

(5.3) $$\begin{eqnarray}\displaystyle \sqrt{q}\,{\mathcal{S}}(p,q;s,\unicode[STIX]{x1D719}) & = & \displaystyle \mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}\mathop{\sum }_{(c,pq)=1}\frac{S(m\overline{q},np;c)}{c}\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}\widetilde{\unicode[STIX]{x1D719}}(u)\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mnp}}{c\sqrt{q}}\biggr)^{-u}\,du\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}\widetilde{\unicode[STIX]{x1D719}}(u)\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{p}}{\sqrt{q}}\biggr)^{-u}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{(c,pq)=1}\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)S(\overline{q}m,pn;c)}{c^{1-u}(mn)^{s+u/2}}\,du.\end{eqnarray}$$

As the following proposition shows, the innermost sum above satisfies a functional equation which exchanges the roles of $p$ and $q$ .

Proposition 5.2. Let $g\in S_{k}(1)$ be a newform. Suppose that $a$ , $b$ , and $c$ are integers with $c$ positive and $(a,c)=(b,c)=1$ . For $\unicode[STIX]{x1D70E}>1$ , define

$$\begin{eqnarray}D_{g}(a,b,c;s):=\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)S(am,bn;c)}{(mn)^{s}}.\end{eqnarray}$$

Then $D_{g}(a,b,c;s)$ extends to an entire function of $s$ which satisfies the functional equation

(5.4) $$\begin{eqnarray}D_{g}(a,b,c;s)=\biggl(\frac{2\unicode[STIX]{x1D70B}}{c}\biggr)^{4s-2}\biggl[\frac{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}+1)/2-s)}{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}-1)/2+s)}\biggr]^{2}D_{g}(\overline{a},\overline{b},c;1-s),\end{eqnarray}$$

where $a\overline{a}\equiv b\overline{b}\equiv 1\hspace{0.6em}({\rm mod}\hspace{0.2em}c)$ . Furthermore, for any $\unicode[STIX]{x1D716}>0$ , as $|t|\rightarrow \infty$ with $\unicode[STIX]{x1D70E}$ bounded we have

(5.5) $$\begin{eqnarray}D_{g}(a,b,c;s)\ll \left\{\begin{array}{@{}ll@{}}c^{1/2+\unicode[STIX]{x1D716}}\quad & \text{if }\unicode[STIX]{x1D70E}\geqslant 1+\unicode[STIX]{x1D716},\\ c^{5/2-2\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D716}}|t|^{2-2\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D716}}\quad & \text{if }-\unicode[STIX]{x1D716}\leqslant \unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D716},\\ c^{5/2-4\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D716}}|t|^{2-4\unicode[STIX]{x1D70E}}\quad & \text{if }\unicode[STIX]{x1D70E}\leqslant -\unicode[STIX]{x1D716}.\end{array}\right.\end{eqnarray}$$

Proof. Suppose that $\unicode[STIX]{x1D70E}>1$ . Opening up the Kloosterman sum, we find that

(5.6) $$\begin{eqnarray}D_{g}(a,b,c;s)=\sum ^{\ast }_{d\hspace{0.2em}{\rm mod}\hspace{0.2em}c}L\biggl(s,g,\frac{a\overline{d}}{c}\biggr)L\biggl(s,g,\frac{b\overline{d}}{c}\biggr),\end{eqnarray}$$

where

$$\begin{eqnarray}L(s,g,x):=\mathop{\sum }_{m=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{g}(m)e(mx)}{m^{s}}.\end{eqnarray}$$

The $L$ -series $L(s,g,x)$ extends to an entire function of $s$ which satisfies, for $x=d/c$ with $(d,c)=1$ , the functional equation

$$\begin{eqnarray}L\biggl(s,g,\frac{d}{c}\biggr)=(-1)^{\unicode[STIX]{x1D705}/2}\biggl(\frac{2\unicode[STIX]{x1D70B}}{c}\biggr)^{2s-1}\frac{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}+1)/2-s)}{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}-1)/2+s)}L\biggl(1-s,g,-\frac{\overline{d}}{c}\biggr).\end{eqnarray}$$

Thus $D_{g}(a,b,c;s)$ inherits the functional equation (5.4) after using that $S(-a,-b;c)=S(a,b;c)$ . To prove the estimates (5.5) we apply the Phragmen–Lindelöf principle (see, e.g., [Reference Iwaniec and Kowalski8, Theorem 5.53]).

The estimate in the range $\unicode[STIX]{x1D70E}\geqslant 1+\unicode[STIX]{x1D716}$ follows from a standard computation involving the Weil bound (3.3). In the range $\unicode[STIX]{x1D70E}\leqslant -\unicode[STIX]{x1D716}$ we apply the functional equation (5.4), together with Stirling’s formula and the bound in the $\unicode[STIX]{x1D70E}\geqslant 1+\unicode[STIX]{x1D716}$ range. Then Phragmen–Lindelöf yields the bound in the middle range $-\unicode[STIX]{x1D716}\leqslant \unicode[STIX]{x1D70E}\leqslant 1+\unicode[STIX]{x1D716}$ .◻

The innermost sum in (5.3) equals $c^{u-1}D_{g}(\overline{q},p,c;s+u/2)$ . We would like to apply the functional equation (5.4) and write the result as a sum of Kloosterman sums. To do this, we will need to work in the region of absolute convergence of the function $D_{g}(q,\overline{p},c;1-u/2-s)$ , so we move the line of integration in $u$ to $\unicode[STIX]{x1D709}=-2\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D716}$ . By the estimates (5.5) and (5.2), the resulting integral converges as long as $J\geqslant 4$ . Note that $\unicode[STIX]{x1D70E}>{\textstyle \frac{5}{4}}$ and hence the sum over $c$ converges. The function is entire, as can be seen from (5.6) and we do not pick up any poles in the processFootnote ¹ .

Proof of Theorem 5.1.

Starting with the representation (5.3) of $\sqrt{q}\,{\mathcal{S}}(p,q;s)$ with $\unicode[STIX]{x1D709}$ satisfying $\unicode[STIX]{x1D709}=-2\unicode[STIX]{x1D70E}-\unicode[STIX]{x1D716}$ , we apply the functional equation (5.4) to obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \sqrt{q/p}\,{\mathcal{S}}(p,q;s;\unicode[STIX]{x1D719})\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl(\frac{p}{q}\biggr)^{2s-1}\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}\widetilde{\unicode[STIX]{x1D719}}(u)\biggl[\frac{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}+1)/2-s-u/2)}{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}-1)/2+s+u/2)}\biggr]^{2}2^{-4s-2u+2}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}\mathop{\sum }_{(c,pq)=1}\frac{S(qm,\overline{p}n;c)}{c\sqrt{p}}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mnq}}{c\sqrt{p}}\biggr)^{4s+u-2}\,du.\nonumber\end{eqnarray}$$

The integral and sums are absolutely convergent for $\unicode[STIX]{x1D70E}>\frac{5}{4}$ , so we can interchange them to obtain

$$\begin{eqnarray}\displaystyle \sqrt{q/p}\,{\mathcal{S}}(p,q;s;\unicode[STIX]{x1D719}) & = & \displaystyle \biggl(\frac{p}{q}\biggr)^{2s-1}\mathop{\sum }_{m,n\geqslant 1}\frac{\unicode[STIX]{x1D706}_{g}(m)\unicode[STIX]{x1D706}_{g}(n)}{(mn)^{s}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{(c,pq)=1}\frac{S(qm,\overline{p}n;c)}{c\sqrt{p}}\unicode[STIX]{x1D6F7}\biggl(\frac{4\unicode[STIX]{x1D70B}\sqrt{mnq}}{c\sqrt{p}}\biggr).\nonumber\end{eqnarray}$$

The sums on the right-hand side are exactly ${\mathcal{S}}(q,p;s;\unicode[STIX]{x1D6F7})$ .

Let us finish the proof by specifying the conditions for $\unicode[STIX]{x1D719}$ so that $\unicode[STIX]{x1D6F7}$ satisfies (3.5). This is necessary in order for us to apply the Kuznetsov formula to the sum of Kloosterman sums with the weight function $\unicode[STIX]{x1D6F7}$ .

So far we have required that $J\geqslant 4$ and that $-J<\unicode[STIX]{x1D709}<A-J$ . By Stirling’s formula the integrand in the definition of $\unicode[STIX]{x1D6F7}$ is majorized by $(1+|\Im (u)|)^{-J+2-4\unicode[STIX]{x1D70E}-2\unicode[STIX]{x1D709}}$ , so the integral converges as long as $2\unicode[STIX]{x1D709}+4\unicode[STIX]{x1D70E}>3-J$ . Suppose that $\unicode[STIX]{x1D70E}\geqslant \frac{1}{2}$ . Then $\unicode[STIX]{x1D6F7}(0)=0$ as long as we can take $\unicode[STIX]{x1D709}>0$ (note that we can easily stay away from the poles of the gamma factor since $\unicode[STIX]{x1D705}\geqslant 12$ ). This requires $A>J$ .

On the other hand, to estimate $\unicode[STIX]{x1D6F7}^{(\ell )}(x)$ for large $x$ and for $\ell =0,1,2$ , we want to move $\unicode[STIX]{x1D709}$ as far to the left as possible. Differentiating under the integral, we find that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}^{(\ell )}(x) & \ll & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(\unicode[STIX]{x1D709})}|\widetilde{\unicode[STIX]{x1D719}}(u)|\biggl|\frac{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}+1)/2-s-u/2)}{\unicode[STIX]{x1D6E4}((\unicode[STIX]{x1D705}-1)/2+s+u/2)}\biggr|^{2}x^{4\unicode[STIX]{x1D70E}-2+\unicode[STIX]{x1D709}-\ell }(1+|u|)^{\ell }\,du\nonumber\\ \displaystyle & \ll & \displaystyle (1+x)^{4\unicode[STIX]{x1D70E}-2+\unicode[STIX]{x1D709}-\ell },\nonumber\end{eqnarray}$$

assuming the integral converges. To ensure convergence, we need that $2\unicode[STIX]{x1D709}+4\unicode[STIX]{x1D70E}>\ell +3-J$ , and for $\unicode[STIX]{x1D6F7}^{(\ell )}(1+x)\ll x^{-2-\unicode[STIX]{x1D716}}$ to hold we need that $4\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D709}<\ell$ . Suppose that $\frac{1}{2}\leqslant \unicode[STIX]{x1D70E}\leqslant \frac{5}{4}+\unicode[STIX]{x1D716}$ . Then we need

(5.7) $$\begin{eqnarray}\ell -J+1<2\unicode[STIX]{x1D709}<2\ell -10\quad \text{for }\ell =0,1,2.\end{eqnarray}$$

For this interval to be non-empty, it is enough to take $J\geqslant 12$ .◻