# Non-Linear Additive Twists of $\mathrm{GL}_{3}$ Hecke Eigenvalues

Ikuya Kaneko  The Division of Physics, Mathematics and Astronomy, California
Institute of Technology, 1200 E. California Blvd., Pasadena, CA 91125, USA
<EMAIL_ADDRESS>https://sites.google.com/view/ikuyakaneko/ and  Wing Hong
Leung  Department of Mathematics, Mailstop 3368, Texas A&M University, College
Station, TX 77843-3368, USA<EMAIL_ADDRESS>https://sites.google.com/view/winghongleung/

###### Abstract.

We bound non-linear additive twists of $\mathrm{GL}_{3}$ Hecke eigenvalues,
improving upon the work of Kumar–Mallesham–Singh (2022). The proof employs the
DFI circle method with standard manipulations (Voronoĭ, Cauchy–Schwarz,
lengthening, and additive reciprocity). The main novelty includes the
conductor lowering mechanism, albeit sacrificing some savings to remove an
analytic oscillation, followed by the iteration ad infinitum of Cauchy–Schwarz
and Poisson. The resulting character sums are estimated via the work of
Adolphson–Sperber (1993). As an application, we prove nontrivial bounds for
the first moment of $\mathrm{GL}_{3}$ Hardy’s function, which corresponds to
the cubic moment of Hardy’s function studied by Ivić (2012).

###### Key words and phrases:

Additive twists, $\mathrm{GL}_{3}$ Hecke eigenvalues, circle method, Voronoĭ,
iteration

###### 2020 Mathematics Subject Classification:

11F66, 11M41 (primary); 11F55 (secondary)

IK acknowledges the support of the Masason Foundation.

###### Contents

  1. 1 Introduction
  2. 2 Conventions
  3. 3 Applications
  4. 4 Arithmetic Preliminaries
  5. 5 Automorphic Preliminaries
  6. 6 Proof of Theorem 1.1
  7. A Computations of Character Sums

##  1\. Introduction

###  1.1. Statement of Main Results

Let $\mathbb{A}_{\mathbb{Q}}$ be the ring of adèles over $\mathbb{Q}$, and let
$\mathfrak{F}_{k}$ be the universal family of all cuspidal automorphic
representations of $\mathrm{GL}_{k}(\mathbb{A}_{\mathbb{Q}})$ with unitary
central character, which we normalise to be trivial on the diagonally embedded
copy of the positive reals so that $\mathfrak{F}_{k}$ is discrete; see, for
example, Brumley [Bru06, Corollary 9]. If $(a_{n})$ represents a sequence of
arbitrary complex numbers, then a well-known problem in number theory is to
estimate nontrivially its correlations against _non-linear additive twists_ of
the shape111One often considers a dyadically smoothed version of (1.1) by
inserting a smooth weight function therein, but it is aesthetically convenient
in the present paper to deal with the unsmoothed version (1.1).

$\sum_{n\leq T}a_{n}e(\alpha n^{\beta}),$ (1.1)

where $T\geq 1$ and $\alpha,\beta\in\mathbb{R}$. The summand is called an
additive twist of $(a_{n})$ when $\beta=1$. Of particular interest is when
$(a_{n})$ has an automorphic origin due to its relevance to moments of
families of $L$-functions and subconvex bounds thereof. For example, if
$(a_{n})$ arises from the Hecke eigenvalues of $\varphi\in\mathfrak{F}_{2}$,
then the study of (1.1) has a distinguished history. In particular, Jutila
[Jut87] improves upon a _Wilton-type bound_ and shows uniformly in
$\alpha\in\mathbb{R}$ that

$\sum_{n\leq T}\lambda_{\varphi}(n)e(\alpha n)\ll\sqrt{T},$ (1.2)

where $\lambda_{\varphi}(n)$ is the $n$-th Hecke eigenvalue of $\varphi$.

Given $\pi\in\mathfrak{F}_{k}$, emblematic generalisations of (1.2) include

$\sum_{n\leq T}\lambda_{\pi}(n)e(\alpha n),$ (1.3)

where $\lambda_{\pi}(n)\coloneqq\lambda_{\pi}(1,n)$ is the $n$-th Hecke
eigenvalue of $\pi$. For $k=3$, the best known bound of Miller [Mil06, Theorem
1.1] goes halfway between the trivial and optimal bound, namely

$\sum_{n\leq T}\lambda_{\pi}(n)e(\alpha
n)\ll_{\pi,\varepsilon}T^{\frac{3}{4}+\varepsilon}$ (1.4)

uniformly in $\alpha\in\mathbb{R}$. For $k\geq 4$, nontrivial bounds for (1.3)
exhibit a hindrance of progress at the current state of knowledge; see [RY15a,
RY15b] for further details.

If $\pi\in\mathfrak{F}_{3}$ and $\alpha,\beta>0$, then the result of Ren–Ye
[RY15b, Theorem 2] states222There is a typographical error in the exponent of
$\alpha$ in [KMS22, Equation (3)]. that

$\mathcal{A}_{\pi}(T,\alpha,\beta)\coloneqq\sum_{n\leq
T}\lambda_{\pi}(n)e(\alpha n^{\beta})\ll_{\pi,\beta}(T^{\frac{2}{3}}+(\alpha
T^{\beta})^{\frac{3}{2}})\log T$

for $\beta\neq\frac{1}{3}$, and

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll_{\pi,\varepsilon}(1+\alpha^{\frac{1}{14}+\varepsilon})(T^{\frac{2}{3}}\log
T+\alpha^{2}T^{\frac{1}{3}+\varepsilon})$

for $\beta=\frac{1}{3}$. Furthermore, Kumar–Mallesham–Singh [KMS22, Theorem 1]
prove

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll_{\pi,\varepsilon}\alpha\sqrt{\beta}T^{\frac{3}{4}+\frac{9\beta}{28}+\varepsilon}$
(1.5)

for $\frac{1}{3}<\beta<\frac{7}{9}$. In particular, for fixed
$\alpha\in\mathbb{R}$, (1.5) implies

$\mathcal{A}_{\pi}(T,\alpha)\coloneqq\mathcal{A}_{\pi}\left(T,\alpha,\frac{2}{3}\right)=\sum_{n\leq
T}\lambda_{\pi}(n)e(\alpha
n^{\frac{2}{3}})\ll_{\pi,\varepsilon}T^{\frac{27}{28}+\varepsilon}.$

This work provides a substantial improvement over (1.5). We regard $\alpha$
and $\beta$ as fixed, but our method generalises without any problems to let
them vary in their appropriate regimes.

######  Theorem 1.1.

Let $\pi\in\mathfrak{F}_{3}$, and let $\alpha\in\mathbb{R}$,
$\frac{5}{14}<\beta<\frac{4}{5}$. Then we have for any $\varepsilon>0$ that

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll_{\pi,\alpha,\beta,\varepsilon}T^{\frac{2}{3}+\frac{5\beta}{12}+\varepsilon}+T^{\frac{7}{8}+\varepsilon}+T^{\frac{19}{14}-\beta+\varepsilon}.$

Theorem 1.1 improves upon [KMS22, Theorem 1] for
$\frac{17}{37}<\beta<\frac{4}{5}$. The proof is predicated upon the
Duke–Friedlander–Iwaniec circle method together with the fairly standard
toolbox including Voronoĭ, Poisson, Cauchy–Schwarz, lengthening, and additive
reciprocity. We use conductor lowering, but sacrifice some savings to
eliminate a cumbersome analytic oscillation after Voronoĭ. Another key input
is an iteration ad infinitum of Cauchy–Schwarz and Poisson summation to attain
additional savings. We highlight that this iteration gives rise to certain
intricate character sums that are viewed as a simpler analogue of what is
handled in [ALM22]. This is where a result of Adolphson–Sperber [AS93] comes
into play. They extended Dwork’s cohomology theory from smooth projective
hypersurfaces in characteristic $p$ to a sufficiently wide class of
exponential sums.

######  Corollary 1.2.

Let $\pi\in\mathfrak{F}_{3}$, and let $\alpha\in\mathbb{R}$. Then we have for
any $\varepsilon>0$ that

$\mathcal{A}_{\pi}(T,\alpha)\ll_{\pi,\alpha,\varepsilon}T^{\frac{17}{18}+\varepsilon}.$

Corollary 1.2 has an application to the first moment of $\mathrm{GL}_{3}$
Hardy’s function; see Section 3.

We now formulate a smoothed version of Theorem 1.1. Let $1\leq Y\ll
T^{\beta-\varepsilon}$ be a parameter at our disposal, and let $W$ be a smooth
function supported on $[1,2]$ such that $x^{j}W^{(j)}(x)\ll_{j}Y^{j}$ for all
fixed $j\in\mathbb{N}_{0}$.

######  Theorem 1.3.

Keep the notation and assumptions as above. Then we have for any
$\varepsilon>0$ that

$\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta):=\sum_{n=1}^{\infty}\lambda_{\pi}(n)e(\alpha
n^{\beta})W\left(\frac{n}{T}\right)\ll_{\pi,\alpha,\beta,\varepsilon}T^{\frac{2}{3}+\frac{5\beta}{12}+\varepsilon}+T^{\frac{7}{8}+\varepsilon}.$

###  1.2. Heuristic Sketch

This section overviews the method of the proof of Theorem 1.1 in a back-of-
the-envelope fashion, presenting a high-level sketch geared to experts. This
section is not intended to entail rigour but might serve as a roadmap. In
particular, we ignore some subtleties such as complicated smooth weights,
coprimality conditions, and common divisors.

####  1.2.1. Trivial Bound

By Cauchy–Schwarz and the Rankin–Selberg bound (also known as the Ramanujan
conjecture on average), the trivial bound amounts to

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll_{\pi}T$

uniformly in $\alpha,\beta\in\mathbb{R}$. In what follows, we shall implicitly
employ the Rankin–Selberg bound to visualise how much we ought to save from
the trivial bound at each step.

####  1.2.2. Smoothing

Via a standard smoothing argument, the proof of Theorem 1.1 reduces to the
proof of Theorem 1.3, because roughly

$\mathcal{A}_{\pi}(T,\alpha,\beta)=\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta)+O\left(\frac{T^{\frac{19}{14}+\varepsilon}}{Y}\right).$
(1.6)

####  1.2.3. Applying the $\delta$-Symbol

The Duke–Friedlander–Iwaniec circle method now implies333Here and henceforth,
we employ the symbol $\approx$ to indicate that the left-hand side may roughly
be written as an expression resembling the right-hand side with an admissible
error term. Furthermore, we replace the condition $n\leq T$ with $n\sim T$ for
brevity, and this tentative simplification applies to other summations.

$\displaystyle\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta)$
$\displaystyle\approx\sum_{m\sim T}\lambda_{\pi}(m)\sum_{n\sim T}e(\alpha
n^{\beta})W\left(\frac{n}{T}\right)\delta(m=n)$
$\displaystyle\approx\frac{1}{Q^{2}}\sum_{q\sim Q}\
\sideset{}{{}^{\ast}}{\sum}_{a{\@displayfalse\pmod{q}}}\sum_{m\sim
T}\lambda_{\pi}(m)e\left(\frac{am}{q}\right)\sum_{n\sim T}e(\alpha
n^{\beta})e\left(-\frac{an}{q}\right)W\left(\frac{n}{T}\right)U\left(\frac{m-n}{qQ}\right)$

for any parameter $Q\geq 1$ and some fixed function $U\in
C_{c}^{\infty}([-1,1])$. The trivial bound at this stage is $O_{\pi}(T^{2})$,
and we need to save $T$ and a little more.

####  1.2.4. Voronoĭ Summation in $m$

As an initial manoeuvre, we use $\mathrm{GL}_{3}$ Voronoĭ summation in the sum
over $m$. A sophisticated analysis of the integral transform on the dual side
yields

$\sum_{m\sim
T}\lambda_{\pi}(m)e\left(\frac{am}{q}\right)U\left(\frac{m-n}{qQ}\right)\approx\frac{Q^{2}}{T}\sum_{m\sim\frac{T^{2}}{Q^{3}}}\lambda_{\pi}(m)S(\overline{a},m;q)e\left(\frac{3(mn)^{\frac{1}{3}}}{q}\right),$

where $S(a,b;c)$ denotes the standard Kloosterman sum.

####  1.2.5. Poisson Summation in $n$

We apply Poisson summation to the sum over $n$. Choosing $Q\gg
T^{\frac{1-\beta}{2}+\varepsilon}$ yields something like

$\displaystyle\sum_{n\sim T}e\left(-\frac{an}{q}+\alpha
n^{\beta}+\frac{3(mn)^{\frac{1}{3}}}{q}\right)W\left(\frac{n}{T}\right)\approx
T\sum_{\begin{subarray}{c}n\sim\frac{Q}{T^{1-\beta}}\\\ n\equiv
a{\@displayfalse\pmod{q}}\end{subarray}}\mathcal{I}(m,n,q),$

where

$\displaystyle\mathcal{I}(m,n,q)\coloneqq\int_{0}^{\infty}W(y)e\left(\alpha(Ty)^{\beta}+\frac{3(mTy)^{\frac{1}{3}}}{q}-\frac{nTy}{q}\right)dy.$

Hence, there holds

$\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta)\approx\sum_{m\sim\frac{T^{2}}{Q^{3}}}\sum_{n\sim\frac{Q}{T^{1-\beta}}}\sum_{q\sim
Q}\lambda_{\pi}(m)S(m,\overline{n};q)\mathcal{I}(m,n,q).$

A stationary phase analysis yields
$\mathcal{I}(m,n,q)\ll_{\varepsilon}T^{-\frac{\beta}{2}+\varepsilon}$, thus
the trivial bound at this stage is
$O_{\pi,\alpha,\beta}(Q^{-\frac{1}{2}}T^{1+\frac{\beta}{2}})$.

####  1.2.6. Rewriting the Integral

It is now convenient to rewrite the oscillatory factor
$e(\frac{3(mTy)^{\frac{1}{3}}}{q})$ via Mellin inversion. In fact, one
approximates

$e(x)\approx\frac{1}{\sqrt{x}}\int_{t\sim x}f(t)x^{it}dt$

for some appropriate function $f\ll 1$. This allows one to separate $m$ from
$q$ in the ensuing Cauchy–Schwarz step at the sacrifice of losing a square-
root saving from the above integral.

####  1.2.7. Cauchy–Schwarz and Lengthening in $m$

Cauchy–Schwarz implies

$\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta)\ll_{\pi,\alpha,\beta}\frac{T^{\frac{3}{2}}}{Q^{\frac{5}{2}}}\sup_{t}\left(\sum_{m\sim\frac{T^{2}}{Q^{3}}}\left|\sum_{n\sim\frac{Q}{T^{1-\beta}}}\sum_{q\sim
Q}S(m,\overline{n};q)\mathcal{J}(n,q)\right|^{2}\right)^{\frac{1}{2}},$ (1.7)

where

$\displaystyle\mathcal{J}(n,q)\coloneqq
q^{-it}\int_{0}^{\infty}W(y)e\left(\alpha(Ty)^{\beta}-\frac{nTy}{q}\right)dy\ll
T^{-\beta/2+\varepsilon}.$

The square-root cancellation heuristic now demonstrates that the best possible
bound for the expression (1.7) is
$O_{\pi,\alpha,\beta,\varepsilon}(Q^{-\frac{5}{2}}T^{2+\varepsilon})$. The
expression inside the square root in (1.7) becomes

$\sum_{m\sim\frac{T^{2}}{Q^{3}}}\mathop{\sum\sum}_{n_{1},n_{2}\sim\frac{Q}{T^{1-\beta}}}\mathop{\sum\sum}_{q_{1},q_{2}\sim
Q}S(m,\overline{n_{1}};q_{1})S(m,\overline{n_{2}};q_{2})\mathcal{J}(n_{1},q_{1})\overline{\mathcal{J}(n_{2},q_{2})}.$

Rigorously, we introduce an lengthening parameter $L\geq 1$ and make the sum
over $m$ of length $\frac{LT^{2}}{Q^{3}}$ whose purpose is to rebalance the
contributions of zero frequency and nonzero frequencies in Poisson summation
in some non-generic cases. Since we choose $L\asymp 1$ in the generic case, we
here refrain from discussing this step in detail.

####  1.2.8. Poisson Summation in $m$

We are in a position to execute Poisson summation in $m$, obtaining something
like

$\sum_{m\sim\frac{T^{2}}{Q^{3}}}S(m,\overline{n_{1}};q_{1})S(m,\overline{n_{2}};q_{2})\approx\frac{T^{2}}{Q^{2}}\cdot\delta(n_{1}=n_{2})\delta(q_{1}=q_{2})\\\
+\frac{T^{2}}{Q^{5}}\sum_{m\sim\frac{Q^{5}}{T^{2}}}\sum_{\gamma{\@displayfalse\pmod{q_{1}q_{2}}}}S(\gamma,\overline{n_{1}};q_{1})S(\gamma,\overline{n_{2}};q_{2})e\left(\frac{\gamma
m}{q_{1}q_{2}}\right).$ (1.8)

####  1.2.9. Additive Reciprocity

By twisted multiplicativity for Kloosterman sums and additive reciprocity, the
sum over $\gamma$ in (1.8) transforms into (ignoring an archimedean
exponential)

$q_{1}q_{2}\cdot
e\left(\frac{\overline{mn_{1}}q_{2}}{q_{1}}+\frac{\overline{mn_{2}}q_{1}}{q_{2}}\right)\approx
q_{1}q_{2}\cdot
e\left(\frac{\overline{q_{1}}q_{2}}{mn_{1}}+\frac{q_{1}\overline{q_{2}}}{mn_{2}}\right).$

Hence, there holds

$\mathcal{A}_{\pi}^{Y}(T,\alpha,\beta)\ll_{\pi,\alpha,\beta,\varepsilon}Q^{-\frac{5}{2}}T^{2+\varepsilon}+\Omega,$
(1.9)

where the contribution of nonzero frequencies is defined by

$\Omega\coloneqq\frac{T^{\frac{5}{2}}}{Q^{4}}\left(\sum_{m\sim\frac{Q^{5}}{T^{2}}}\mathop{\sum\sum}_{n_{1},n_{2}\sim\frac{Q}{T^{1-\beta}}}\mathop{\sum\sum}_{q_{1},q_{2}\sim
Q}e\left(\frac{\overline{q_{1}}q_{2}}{mn_{1}}+\frac{q_{1}\overline{q_{2}}}{mn_{2}}\right)\mathcal{J}(n_{1},q_{1})\overline{\mathcal{J}(n_{2},q_{2})}\right)^{\frac{1}{2}}.$

The trivial bound for $\Omega$ is
$O(\sqrt{Q}T^{\frac{1}{2}+\frac{\beta}{2}})$.

####  1.2.10. Cauchy–Schwarz and Poisson Summation in $q_{1}$

We apply Cauchy–Schwarz to take out everything except the $q_{2}$-sum, and
introduce a lengthening parameter $R=1+Q^{6}T^{-4+2\beta}$ to the $q_{1}$-sum.
Upon applying Poisson summation in $q_{1}$, the lengthening forces the dual
side to degenerate to frequency zero. While this step does not theoretically
affect the final bound, it substantially facilitates the rest of the proof. To
summarise, we are led to the expression

$\Omega\ll\frac{T^{\frac{5}{2}}}{Q^{4}}\left(\frac{RQ^{4}}{T^{2-\beta}}\right)^{\frac{1}{4}}\Bigg{(}\sum_{m\sim\frac{Q^{5}}{T^{2}}}\mathop{\sum\sum}_{n_{1},n_{2}\sim\frac{Q}{T^{1-\beta}}}\mathop{\sum\sum}_{\begin{subarray}{c}q_{2},q_{3}\sim
Q\\\ q_{2}\equiv q_{3}{\@displayfalse\pmod{n_{1}n_{2}}}\end{subarray}}\\\
\times
S(\overline{q_{2}}-\overline{q_{3}},\overline{n_{1}n_{2}}(q_{2}-q_{3});m)\mathcal{J}(n_{2},q_{2})\overline{\mathcal{J}(n_{2},q_{3})}\Bigg{)}^{\frac{1}{4}}.$

The trivial bound for $\Omega$ at this stage is
$O(Q^{-\frac{5}{8}}R^{\frac{1}{4}}T^{\frac{5}{4}})$.

####  1.2.11. Iteration Ad Infinitum

One observes that the dual side preserves the structure

$\displaystyle\mathop{\sum\sum}_{q_{2},q_{3}\sim Q}\mathcal{C}(q_{2},q_{3})$

for a more intricate character sum $\mathcal{C}(q_{2},q_{3})$ defined modulo
$mn_{1}n_{2}$. This enables the iteration of the above procedure. Iterating
$k$ times for sufficiently large $k$, the total saving amounts to

$\displaystyle\left(\frac{Q^{\frac{5}{2}}}{RT}\right)^{\frac{1}{8}}\times\left(\frac{Q^{\frac{5}{2}}}{RT}\right)^{\frac{1}{16}}\times\cdots\times\left(\frac{Q^{\frac{5}{2}}}{RT}\right)^{\frac{1}{2^{k}}}\approx\left(\frac{Q^{\frac{5}{2}}}{RT}\right)^{\frac{1}{4}}T^{-\varepsilon}.$

This implies

$\displaystyle\Omega\ll\frac{R^{\frac{1}{4}}T^{\frac{5}{4}+\varepsilon}}{Q^{\frac{5}{8}}}\left(\frac{RT}{Q^{\frac{5}{2}}}\right)^{\frac{1}{4}}\ll\frac{T^{\frac{3}{2}+\varepsilon}}{Q^{\frac{5}{4}}}\left(1+\frac{Q^{3}}{T^{2-\beta}}\right).$
(1.10)

Inserting (1.10) into (1.9) along with the optimisation
$Q=\min\\{T^{\frac{2-\beta}{3}},\sqrt{T}\\}$ justifies the claim of Theorem
1.3. Theorem 1.1 now follows by choosing $Y=T^{\beta-\varepsilon}$ in (1.6).

### Acknowledgements

The authors thank Keshav Aggarwal for helpful discussions that led to this
paper being written.

##  2\. Conventions

Throughout, we use the shorthand $e(x)=e^{2\pi ix}$. We make use of the letter
$\varepsilon>0$ to denote an arbitrarily small positive quantity that is
possibly different in each instance. The notation $f\ll_{\nu}g$ or
$f=O_{\nu}(g)$ indicates that there exists an effectively computable constant
$c_{\nu}>0$, depending at most on $\nu$, such that $|f(z)|\leq c_{\nu}|g(z)|$
for all $z$ in a range that is apparent from context. If no parameter $\nu$ is
present, then $c$ is absolute. If $a$ and $b$ are either both positive or both
negative, then the notation $a\sim b$ means $k_{1}<\frac{a}{b}<k_{2}$ for some
absolute and effectively computable constants $k_{1},k_{2}>0$. If they depend
on $\nu$, then we attach a subscript and write $a\sim_{\nu}b$. The Kronecker
symbol $\delta(\mathrm{S})$ detects $1$ if the statement $\mathrm{S}$ is true,
and $0$ otherwise. For any positive integer $n$ with its prime factorisation
$n=\prod_{i=1}^{k}p_{i}^{a_{i}}$, we write
$n_{\square}=\prod_{\begin{subarray}{c}1\leq i\leq k\\\ a_{i}\geq
2\end{subarray}}p_{i}^{a_{i}}$.

##  3\. Applications

This section describes an application of Theorem 1.1 to moments of
$\mathrm{GL}_{3}$ Hardy’s function.

###  3.1. Classical Triumphs

Hardy’s function plays a crucial rôle in the study of the zeros of the Riemann
zeta function $\zeta(s)$ on the critical line $\mathrm{Re}(s)=\frac{1}{2}$.
Given $t\in\mathbb{R}$, it is defined by

$Z(t)\coloneqq\zeta\left(\frac{1}{2}+it\right)\chi\left(\frac{1}{2}+it\right)^{-\frac{1}{2}},\qquad\zeta(s)=\chi(s)\zeta(1-s).$

As Ivić discusses in [Ivi13], Hardy’s function $Z(t)$ is real-valued, even,
and smooth. Moreover, $|Z(t)|=|\zeta(\frac{1}{2}+it)|$ implies that its zeros
match the zeros of $\zeta(s)$ on the critical line. Hardy’s original
application of $Z(t)$ is to prove that $\zeta(s)$ has infinitely many
nontrivial zeros; see the book of Titchmarsh [Tit51]. The proof begins with an
assumption that $Z(t)$ does not change sign for sufficiently large $t>0$ and
then yields a contradiction. Hardy [Har14] later refines on his reasoning to
show that $\zeta(s)$ has $\gg T$ zeros $\beta+i\gamma$ satisfying
$|\gamma|\leq T$. Selberg [Sel89] then improves upon this estimate to $\gg
T\log T$, followed by its further refinements by successors.

We are interested in the moments of $Z(t)$, but only odd moments represent a
novelty since $|Z(t)|=|\zeta(\frac{1}{2}+it)|$. Given $k\in\mathbb{N}$ fixed
and $T\geq 1$, consider the $k$-th moment of $Z(t)$, namely

$\mathcal{F}_{k}(T)\coloneqq\int_{0}^{T}Z(t)^{k}dt.$

When $k=1$, Korolev [Kor08] and Jutila [Jut09, Jut11] prove independently the
true order of the first moment

$\mathcal{F}_{1}(T)=O(T^{\frac{1}{4}}),\qquad\mathcal{F}_{1}(T)=\Omega_{\pm}(T^{\frac{1}{4}}),$

Nonetheless, higher odd moments, where one also expects some nontrivial
cancellations, still remain unamenable. In particular, the following problem
has been unsolved since being posed as a question of study during the
conference “Elementary and Analytic Number Theory” at the Mathematisches
Forschungsinstitut Oberwolfach.

######  Problem 3.1 (Ivić 2003).

Does there exist a constant $0<c<1$ such that

$\mathcal{F}_{3}(T)=O(T^{c})?$ (3.1)

The approximate functional equation shows [Ivi13, Equation (7.5)] that

$\mathcal{F}_{3}(T)=2\pi\sqrt{\frac{2}{3}}\sum_{(\frac{T}{2\pi})^{\frac{3}{2}}\leq
n\leq(\frac{T}{\pi})^{\frac{3}{2}}}d_{3}(n)n^{-\frac{1}{6}}\cos\left(3\pi
n^{\frac{2}{3}}+\frac{\pi}{8}\right)+O_{\varepsilon}(T^{\frac{3}{4}+\varepsilon}),$
(3.2)

where $d_{3}(n)$ is the $3$-fold divisor function generated by $\zeta(s)^{3}$.
The problem of obtaining (3.1) for $\frac{3}{4}<c<1$ boils down to the
estimation of the right-hand side of (3.2). It is known that

$\sum_{n\leq T}d_{3}(n)e(\alpha
n^{\frac{2}{3}})\ll_{\varepsilon}T^{\frac{5}{6}+\varepsilon}$ (3.3)

uniformly in $\alpha\in\mathbb{R}$. Nevertheless, this corresponds to the
exponent $c=1+\varepsilon$ in Problem 3.1, and is in fact weaker than the
trivial bound because Cauchy–Schwarz gives

$\mathcal{F}_{3}(T)\ll\left(\int_{0}^{T}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{2}dt\right)^{\frac{1}{2}}\left(\int_{0}^{T}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{4}dt\right)^{\frac{1}{2}}\ll
T(\log T)^{\frac{5}{2}},$ (3.4)

where we use the classical bounds due to Hardy–Littlewood and Ingham [Ing27],
namely

$\int_{0}^{T}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{2}dt\ll T\log
T,\qquad\int_{0}^{T}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{4}dt\ll
T(\log T)^{4}.$

For the cubic moment of Hardy’s function, there seems to exist no better
estimate than (3.4), and any improvement thereof would necessitate a delicate
analysis of the additive twist (3.3). The difficulty in the estimation of
(3.3) lies in the presence of $d_{3}(n)$, which, despite its simple
appearance, is quite difficult to handle. In fact, a variant of Voronoĭ
summation is essentially involutory and is of no great use in this context.
Moreover, writing $n=k\ell m$ transforms (3.3) into a three-dimensional
exponential sum over the variables $k,\ell,m\in\mathbb{N}$, but the theory of
exponent pairs is not capable of yielding any nontrivial bounds. We also refer
the reader to the pioneering papers of Ivić [Ivi10, Ivi13, Ivi17c, Ivi17b,
Ivi17a].

###  3.2. $\mathrm{GL}_{3}$ Hardy’s Function

Given $t\in\mathbb{R}$, $\mathrm{GL}_{3}$ Hardy’s function is defined by

$Z(t,\pi)\coloneqq
L\left(\frac{1}{2}+it,\pi\right)\chi\left(\frac{1}{2}+it,\pi\right)^{-\frac{1}{2}},\qquad
L(s,\pi)=\chi(s,\pi)L(1-s,\widetilde{\pi}).$

We then define the first moment of $\mathrm{GL}_{3}$ Hardy’s function by

$\mathcal{F}_{1}(T,\pi)\coloneqq\int_{0}^{T}Z(t,\pi)dt.$

The key difference from the cubic moment of Hardy’s function for the Riemann
zeta function is that general $\mathrm{GL}_{3}$ $L$-functions do not factorise
into lower degree $L$-functions. This precludes us from using Cauchy–Schwarz
as in (3.4). Instead, if one assumes the second moment bound

$\int_{0}^{T}\left|L\left(\frac{1}{2}+it,\pi\right)\right|^{2}dt\ll_{\pi,\varepsilon}T^{\frac{3}{2}-\delta+\varepsilon}$

for some $\delta>0$, then

$\mathcal{F}_{1}(T,\pi)\ll_{\pi,\varepsilon}T^{\frac{5}{4}-\frac{\delta}{2}+\varepsilon}.$
(3.5)

Conforming to [Ivi13, Chapter 7] _mutatis mutandis_ , one can establish the
following reduction.

######  Theorem 3.2.

Let $\pi\in\mathfrak{F}_{3}$. Assume that

$\sum_{n\leq T}\lambda_{\pi}(n)e(\alpha
n^{\frac{2}{3}})\ll_{\pi,\alpha,\varepsilon}T^{1-\eta+\varepsilon}$

for some $\eta>0$. Then we have for any $\varepsilon>0$ that

$\mathcal{F}_{1}(T,\pi)\ll_{\pi,\varepsilon}T^{\frac{5}{4}-\frac{3\eta}{2}+\varepsilon}.$
(3.6)

Substituting $\eta=\frac{1}{18}$ from Corollary 1.2 into (3.6) yields a
refined exponent $\frac{7}{6}=1.16666\cdots$.

######  Corollary 3.3.

Let $\pi\in\mathfrak{F}_{3}$. Then we have for any $\varepsilon>0$ that

$\mathcal{F}_{1}(T,\pi)\ll_{\pi,\varepsilon}T^{\frac{7}{6}+\varepsilon}.$

If $\eta>\frac{1}{6}$, then one can prove the infinitude of zeros of
$\mathrm{GL}_{3}$ $L$-functions on the critical line; see the Ph.D. thesis of
Rajkumar [Raj12]. It is commensurate with Weyl-strength subconvex bounds for
$\mathrm{GL}_{3}$ $L$-functions (at least) in the $t$-aspect.

##  4\. Arithmetic Preliminaries

This section reviews arithmetic tools that we shall adopt in the proof of
Theorem 1.1.

###  4.1. Circle Method

There are two oscillatory components in the non-linear additive twist that we
address. The idea is to separate these oscillations via the circle method. One
seeks for a Fourier expansion that matches the Kronecker symbol $\delta(n=0)$.

######  Lemma 4.1 (Leung [Leu22]).

Let $n\in\mathbb{Z}$ be such that $|n|\ll N$, $q\in\mathbb{N}$, and let
$C>N^{\varepsilon}$. Let $U\in C_{c}^{\infty}(\mathbb{R})$ and $W\in
C_{c}^{\infty}([-2,-1]\cup[1,2])$ be nonnegative even functions such that
$U(x)=1$ for $x\in[-2,2]$. Then we have that

$\delta(n=0)=\frac{1}{\mathcal{C}}\sum_{c=1}^{\infty}\frac{1}{cq}\sum_{a{\@displayfalse\pmod{cq}}}e\left(\frac{an}{cq}\right)V_{0}\left(\frac{c}{C},\frac{n}{cCq}\right),$

where

$\mathcal{C}\coloneqq\sum_{c=1}^{\infty}W\left(\frac{c}{C}\right)\sim C,$

and

$V_{0}(x,y)\coloneqq W(x)U(x)U(y)-W(y)U(x)U(y)$

is a smooth function localising to $\delta(|x|,|y|\ll 1)$.

In view of the result of Heath-Brown [HB96], Lemma 4.1 is the
Duke–Friedlander–Iwaniec circle method [DFI94] with a simpler weight function
that restricts $|n|\ll cCQ$. This feature facilitates the forthcoming integral
analysis once we make use of the $\mathrm{GL}_{3}$ Voronoĭ summation formula
(Lemma 5.4). We direct the reader to [KL23, Leu22] for further discussions.

###  4.2. Poisson Summation

For $n\in\mathbb{N}$ and an integrable function
$w\colon\mathbb{R}^{n}\to\mathbb{C}$, denote its Fourier transform by

$\widehat{w}(y)\coloneqq\int_{\mathbb{R}^{n}}w(x)e(-\langle x,y\rangle)dx,$

where $\langle\cdot,\cdot\rangle$ denotes the standard inner product on
$\mathbb{R}^{n}$. Furthermore, if $c\in\mathbb{N}$ and
$K\colon\mathbb{Z}\to\mathbb{C}$ is a periodic function of period $c$, then
its Fourier transform $\widehat{K}$ is again the periodic function of period
$c$, defined on $\mathbb{Z}$ by

$\widehat{K}(n)\coloneqq\
\sum_{a{\@displayfalse\pmod{c}}}K(a)e\left(-\frac{an}{c}\right).$

There is a minor inconsistency in sign choices, and
$\widehat{\widehat{K}}(n)=K(-n)$ for all $n\in\mathbb{Z}$.

######  Lemma 4.2 (Fouvry–Kowalski–Michel [FKM15, Lemma 2.1]).

For any $c\in\mathbb{N}$, any $c$-periodic function $K$, and any even smooth
function $V$ compactly supported on $\mathbb{R}$, we have that

$\sum_{n=1}^{\infty}K(n)V(n)=\frac{1}{c}\sum_{n\in\mathbb{Z}}\widehat{K}(n)\widehat{V}\left(\frac{n}{c}\right).$

###  4.3. The Gamma Function

For fixed $\sigma\in\mathbb{R}$, real $|\tau|\geq 3$, and any $M>0$, we often
make use of Stirling’s formula

$\Gamma(\sigma+i\tau)=e^{-\frac{\pi|\tau|}{2}}|\tau|^{\sigma-\frac{1}{2}}\exp\left(i\tau\log\frac{|\tau|}{e}\right)g_{\sigma,M}(\tau)+O_{\sigma,M}(|\tau|^{-M}),$
(4.1)

where

$g_{\sigma,M}(\tau)=\sqrt{2\pi}\exp\left(\frac{\pi}{4}(2\sigma-1)i\operatorname{sgn}(\tau)\right)+O_{\sigma,M}(|\tau|^{-1})$

and

$|\tau|^{j}g_{\sigma,M}^{(j)}(\tau)\ll_{j,\sigma,M}1$

for all fixed $j\in\mathbb{N}_{0}$.

##  5\. Automorphic Preliminaries

This section compiles automorphic tools that we shall adopt in the proof of
Theorem 1.1; see the book of Goldfeld [Gol06] for a thorough account of the
theoretical background.

###  5.1. Hecke Eigenvalues

Denote by $\pi$ a Hecke–Maaß cusp form of type $(\nu_{1},\nu_{2})$ for
$\mathrm{SL}_{3}(\mathbb{Z})$ with the Fourier–Whittaker coefficients
$\lambda_{\pi}(m,n)$ normalised so that $\lambda_{\pi}(1,1)=1$. As usual, one
abbreviates $\lambda_{\pi}(1,n)\coloneqq\lambda_{\pi}(n)$. The corresponding
Langlands parameters are given by

$\mu_{1}=-\nu_{1}-2\nu_{2}+1,\qquad\mu_{2}=-\nu_{1}+\nu_{2},\qquad\mu_{3}=2\nu_{1}+\nu_{2}-1.$

The Ramanujan conjecture then states that $\mathrm{Re}(\mu_{i})=0$. The work
of Jacquet–Shalika [JS90] shows $|\mathrm{Re}(\mu_{i})|<\frac{1}{2}$. The
Fourier–Whittaker coefficients obey the Hecke relation

$\lambda_{\pi}(n_{1},1)\lambda_{\pi}(1,n_{2})=\sum_{d\mid(n_{1},n_{2})}\lambda_{\pi}\left(\frac{n_{1}}{d},\frac{n_{2}}{d}\right).$
(5.1)

The Rankin–Selberg bound is known in the following form.

######  Lemma 5.1.

We have for any $\varepsilon>0$ that

$\underset{n_{1}^{2}n_{2}\leq
N}{\sum\sum}|\lambda_{\pi}(n_{2},n_{1})|^{2}\ll_{\pi,\varepsilon}N^{1+\varepsilon}.$

As a corollary, we deduce the following estimate.

######  Corollary 5.2.

We have for any $\varepsilon>0$ that

$\sum_{n_{1}\leq N_{1}}\sum_{n_{2}\leq
N_{2}}|\lambda_{\pi}(n_{1},n_{2})|^{2}\ll(N_{1}N_{2})^{1+\varepsilon}.$

###### Proof.

Applying the Hecke relation (5.1) in tandem with Möbius inversion yields

$\displaystyle\sum_{n_{1}\leq N_{1}}\sum_{n_{2}\leq
N_{2}}|\lambda_{\pi}(n_{1},n_{2})|^{2}=$ $\displaystyle\sum_{n_{1}\leq
N_{1}}\sum_{n_{2}\leq
N_{2}}\left|\sum_{d\mid(n_{1},n_{2})}\mu(d)\lambda_{\pi}\left(\frac{n_{1}}{d},1\right)\lambda_{\pi}\left(1,\frac{n_{2}}{d}\right)\right|^{2}.$

By the Cauchy–Schwarz inequality to take out the $d$-sum, the right-hand side
is bounded by

$X^{\varepsilon}\sum_{n_{1}\leq N_{1}}\sum_{n_{2}\leq
N_{2}}\sum_{d\mid(n_{1},n_{2})}\left|\lambda_{\pi}\left(\frac{n_{1}}{d},1\right)\lambda_{\pi}\left(1,\frac{n_{2}}{d}\right)\right|^{2}\\\
=X^{\varepsilon}\sum_{d}\sum_{n_{1}\leq\frac{N_{1}}{d}}\sum_{n_{2}\leq\frac{N_{2}}{d}}|\lambda_{\pi}(n_{1},1)\lambda_{\pi}(1,n_{2})|^{2}.$

Corollary 5.2 now follows from Lemma 5.1. ∎

We record the best known bound for individual Fourier–Whittaker coefficients
$\lambda_{\pi}(n_{1},n_{2})$.

######  Lemma 5.3 (Kim–Sarnak [Kim03]).

We have for any $\varepsilon>0$ that

$|\lambda_{\pi}(n_{1},n_{2})|\ll_{\pi,\varepsilon}(n_{1}n_{2})^{\frac{5}{14}+\varepsilon}.$

###  5.2. $\mathrm{GL}_{3}$ Voronoĭ Summation

In conjunction with Poisson summation in Section 4.2, the proof of Theorem 1.1
requires the Voronoĭ summation formula attached to Hecke–Maaß cusp forms on
$\mathrm{Z}(\mathbb{R})\mathrm{SL}_{3}(\mathbb{Z})\backslash\mathrm{GL}_{3}(\mathbb{R})/\mathrm{O}(3)$.
Let $h$ be a compactly supported smooth function on $\mathbb{R}_{+}$ and
denote the corresponding Mellin transform by

$\widetilde{h}(s)\coloneqq\int_{0}^{\infty}h(x)x^{s-1}dx.$

Given
$\sigma>-1+\max\\{-\mathrm{Re}(\alpha_{1}),-\mathrm{Re}(\alpha_{2}),-\mathrm{Re}(\alpha_{3})\\}$
and $a\in\\{0,1\\}$, define

$\gamma_{a}(s)\coloneqq\frac{\pi^{-3s-\frac{3}{2}}}{2}\prod_{i=1}^{3}\frac{\Gamma(\frac{1+s+\alpha_{i}+a}{2})}{\Gamma(\frac{-s-\alpha_{i}+a}{2})}.$

Let $\gamma_{\pm}(s)\coloneqq\gamma_{0}(s)\mp i\gamma_{1}(s)$, and define

$\mathcal{H}_{\pm}(x)\coloneqq\frac{1}{2\pi
i}\int_{(\sigma)}x^{-s}\gamma_{\pm}(s)\widetilde{h}(-s)ds.$ (5.2)

We now formulate a version of the $\mathrm{GL}_{3}$ Voronoĭ summation formula;
cf. [Blo12, Li11, MS06].

######  Lemma 5.4 ($\mathrm{GL}_{3}$ Voronoĭ summation).

Let $a,\overline{a}\in\mathbb{Z}$, $q\in\mathbb{N}$ with $(a,q)=1$,
$a\overline{a}\equiv 1{\@displayfalse\pmod{q}}$. Let $h$ be a compactly
supported smooth function on $\mathbb{R}_{+}$. Then we have that

$\sum_{n=1}^{\infty}\lambda_{\pi}(n)e\left(\frac{an}{q}\right)h(n)=q\sum_{\pm}\sum_{n_{0}\mid
q}\sum_{n=1}^{\infty}\frac{\lambda_{\pi}(n,n_{0})}{nn_{0}}S\left(\overline{a},\pm
n;\frac{q}{n_{0}}\right)\mathcal{H}_{\pm}\left(\frac{nn_{0}^{2}}{q^{3}}\right).$

The behaviour of the integral transform $\mathcal{H}_{\pm}$ on the dual side
is determined in the literature.

######  Lemma 5.5 (Blomer [Blo12, Lemma 6]).

Let $h$ be a compactly supported smooth function on $[a,b]\subset(0,\infty)$,
and let $\mathcal{H}_{\pm}$ be defined by (5.2). Then there exist constants
$\gamma_{\ell}$ depending only on the Langlands parameters
$(\alpha_{1},\alpha_{2},\alpha_{3})$ such that

$\mathcal{H}_{\pm}(x)=x\int_{0}^{\infty}h(y)\sum_{\ell=1}^{L}\frac{\gamma_{\ell}}{(xy)^{\frac{\ell}{3}}}e(\pm
3(xy)^{\frac{1}{3}})dy+\mathcal{R}(x),$

where the error term $\mathcal{R}(x)$ satisfies

$x^{k}\frac{d^{k}}{dx^{k}}\mathcal{R}(x)\ll\left\lVert
h\right\rVert_{\infty}^{1+\frac{k-L}{3}}$

with an implicit constant depending at most on
$a,b,\alpha_{1},\alpha_{2},\alpha_{3},k$, and $L$.

##  6\. Proof of Theorem 1.1

In this section, we embark on the proof of Theorem 1.1. Recall that our goal
is to estimate

$\mathcal{A}_{\pi}(T,\alpha,\beta)\coloneqq\sum_{n\leq
T}\lambda_{\pi}(n)e(\alpha
n^{\beta})\ll_{\pi,\alpha,\beta,\varepsilon}T^{\frac{9+10\beta}{17}+\varepsilon}+T^{\frac{19}{14}-\beta+\varepsilon}.$

We will consider $\pi$, $\alpha$, $\beta$ fixed, and employ the usual
convention that $\varepsilon$ is an arbitrarily small positive quantity, which
may change from line to line. We will then drop the dependence on $\pi$,
$\alpha$, $\beta$, $\varepsilon$, but each inequality is allowed to possess an
implicit constant depending on them. Without loss of generality, we shall
assume that $\alpha>0$.

###  6.1. Smoothing

A dyadic subdivision yields

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll T^{\varepsilon}\sup_{1\leq N\leq
T}\sum_{\frac{N}{2}<n\leq N}\lambda_{\pi}(n)e(\alpha n^{\beta}).$

Let $Y\geq 1$ be a parameter at our disposal. Approximating the indicator
function $\mathbf{1}_{(\frac{N}{2},N]}$ by a $Y$-inert function
$W(\frac{\cdot}{N})$ that takes $1$ on
$[\frac{1}{2}+\frac{1}{Y},1-\frac{1}{Y}]$ and $0$ outside $[\frac{1}{2},1]$,
and bounding the error term with Lemma 5.3, we deduce

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll T^{\varepsilon}\sup_{1\leq N\leq
T}\left(|\mathcal{S}(N)|+\frac{N^{\frac{19}{14}+\varepsilon}}{Y}\right),$
(6.1)

where

$\mathcal{S}(N)\coloneqq\sum_{n}\lambda_{\pi}(n)e(\alpha
n^{\beta})W\left(\frac{n}{N}\right).$

###  6.2. Applying the $\delta$-Symbol

We now use Lemma 4.1 to separate the oscillation of $\lambda_{\pi}(n)$ from
$e(\alpha n^{\beta})W(\frac{n}{N})$. Let $C\geq 1$ be a parameter at our
disposal, and fix a smooth function $U$ that takes $1$ on $[1,2]$ and $0$
outside $[\frac{1}{2},\frac{5}{2}]$. Then

$\displaystyle\mathcal{S}(N)$
$\displaystyle=\sum_{m}\lambda_{\pi}(m)U\left(\frac{m}{N}\right)\sum_{n}e(\alpha
n^{\beta})W\left(\frac{n}{N}\right)\delta(n=m)$
$\displaystyle=\frac{1}{\mathcal{C}}\sum_{c}\frac{1}{c}\sum_{m}\lambda_{\pi}(m)U\left(\frac{m}{N}\right)\sum_{n}e(\alpha
n^{\beta})W\left(\frac{n}{N}\right)\sum_{a{\@displayfalse\pmod{c}}}e\left(\frac{a(m-n)}{c}\right)V_{0}\left(\frac{c}{C},\frac{m-n}{cC}\right)$

for some $\mathcal{C}\sim C$ and a fixed smooth function $V_{0}$ satisfying
$V_{0}(x,y)\ll\delta(|x|,|y|\ll 1)$. Pulling out the divisor $(a,c)$ implies

$\mathcal{S}(N)\ll\frac{1}{C}\sum_{b}\sum_{c}\frac{1}{bc}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{a{\@displayfalse\pmod{c}}}\sum_{m}\lambda_{\pi}(m)e\left(\frac{am}{c}\right)U\left(\frac{m}{N}\right)\\\
\times\sum_{n}e\left(\alpha
n^{\beta}-\frac{an}{c}\right)W\left(\frac{n}{N}\right)V_{0}\left(\frac{bc}{C},\frac{m-n}{bcC}\right).$

###  6.3. Voronoĭ Summation in $m$

Applying $\mathrm{GL}_{3}$ Voronoĭ summation (Lemma 5.4 together with Lemma
5.5) to the $m$-sum, we deduce

$\sum_{m}\lambda_{\pi}(m)e\left(\frac{am}{c}\right)U\left(\frac{m}{N}\right)V_{0}\left(\frac{bc}{C},\frac{m-n}{bcC}\right)\\\
=\frac{N^{\frac{2}{3}}}{c}\sum_{\pm}\sum_{m_{0}\mid
c}\sum_{m}\lambda_{\pi}(m,m_{0})\left(\frac{m_{0}}{m}\right)^{\frac{1}{3}}S\left(\overline{a},\pm
m;\frac{c}{m_{0}}\right)\mathcal{J}_{1}(m_{0}^{2}m,n,c),$

where444We suppress less important variables from the notation, and it applies
to the ensuing integral transforms.

$\mathcal{J}_{1}(m_{0}^{2}m,n,c)\coloneqq\int_{0}^{\infty}\Phi\left(y,\frac{m_{0}^{2}m}{c^{3}}\right)V_{0}\left(\frac{bc}{C},\frac{Ny-n}{bcC}\right)e\left(\pm\frac{3(m_{0}^{2}mNy)^{\frac{1}{3}}}{c}\right)dy$
(6.2)

for some flat function $\Phi$ supported on
$[\frac{1}{2},\frac{5}{2}]\times\mathbb{R}$. This leads to the expression

$\mathcal{S}(N)\ll\frac{N^{\frac{2}{3}}}{C}\sum_{\pm}\sum_{b}\sum_{c}\frac{1}{bc^{2}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{a{\@displayfalse\pmod{c}}}\sum_{m_{0}\mid
c}\sum_{m}\lambda_{\pi}(m,m_{0})\left(\frac{m_{0}}{m}\right)^{\frac{1}{3}}S\left(\overline{a},\pm
m;\frac{c}{m_{0}}\right)\\\ \times\sum_{n}e\left(\alpha
n^{\beta}-\frac{an}{c}\right)W\left(\frac{n}{N}\right)\mathcal{J}_{1}(m_{0}^{2}m,n,c).$

###  6.4. Poisson Summation in $n$

Applying Poisson summation (Lemma 4.2) to the $n$-sum, we deduce

$\sum_{n}e\left(\alpha
n^{\beta}-\frac{an}{c}\right)W\left(\frac{n}{N}\right)\mathcal{J}_{1}(m_{0}^{2}m,n,c)=\frac{N}{c}\sum_{n\in\mathbb{Z}}\sum_{\gamma{\@displayfalse\pmod{c}}}e\left(\frac{(n-a)\gamma}{c}\right)\mathcal{J}_{2}(m_{0}^{2}m,n,c),$

where

$\displaystyle\begin{split}\mathcal{J}_{2}(m_{0}^{2}m,n,c)&\coloneqq\int_{0}^{\infty}W(x)\mathcal{J}_{1}(m_{0}^{2}m,Nx,c)e\left(\alpha(Nx)^{\beta}-\frac{nNx}{c}\right)dx\\\
&=\int_{0}^{\infty}\int_{0}^{\infty}W(x)\Phi\left(y,\frac{m_{0}^{2}m}{c^{3}}\right)V_{0}\left(\frac{bc}{C},\frac{N(y-x)}{bcC}\right)\\\
&\quad\times
e\left(\pm\frac{3(m_{0}^{2}mNy)^{\frac{1}{3}}}{c}+\alpha(Nx)^{\beta}-\frac{nNx}{c}\right)dxdy.\end{split}$
(6.3)

The sum over $\gamma$ is equal to $c\cdot\delta(n\equiv
a{\@displayfalse\pmod{c}})$. Making a change of variables $c\mapsto cm_{0}$
gives

$\mathcal{S}(N)\ll\frac{N^{\frac{5}{3}}}{C}\sum_{\pm}\sum_{b}\sum_{c}\frac{1}{bc^{2}}\sum_{m_{0}}\sum_{m}\frac{\lambda_{\pi}(m,m_{0})}{m_{0}^{\frac{5}{3}}m^{\frac{1}{3}}}\sum_{\begin{subarray}{c}n\in\mathbb{Z}\\\
(n,cm_{0})=1\end{subarray}}S(\pm
m,\overline{n};c)\mathcal{J}_{2}(m_{0}^{2}m,n,cm_{0}).$

If we choose $Y\leq N^{\beta}T^{-\varepsilon}$ and $C\leq\sqrt{N}$, then
repeated integration by parts in $x$ and $y$ in (6.3) ensures an arbitrary
saving unless

$m_{0}^{2}m\ll\frac{N^{2}T^{\varepsilon}}{b^{3}C^{3}},\qquad
n>0,\qquad\frac{cm_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n\ll\frac{cm_{0}T^{\varepsilon}}{N^{1-\beta}}.$

Therefore, we arrive at

$\mathcal{S}(N)\ll\frac{N^{\frac{5}{3}}}{C}\sum_{\pm}\sum_{b}\sum_{c}\frac{1}{bc^{2}}\mathop{\sum\sum}_{m_{0}^{2}m\ll\frac{N^{2}T^{\varepsilon}}{b^{3}C^{3}}}\frac{\lambda_{\pi}(m,m_{0})}{m_{0}^{\frac{5}{3}}m^{\frac{1}{3}}}\\\
\times\sum_{\begin{subarray}{c}\frac{cm_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n\ll\frac{cm_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n,cm_{0})=1\end{subarray}}S(\pm
m,\overline{n};c)\mathcal{J}_{2}(m_{0}^{2}m,n,cm_{0})+T^{-2023}.$

###  6.5. Rewriting the Integral Transform

In anticipation of the Cauchy–Schwarz step, it is convenient to rewrite
slightly the integral $\mathcal{J}_{2}$ and extract some of its oscillations.
Making a change of variables

$Ny\mapsto Nx+bcm_{0}Cu,$

we obtain

$\mathcal{J}_{2}(m_{0}^{2}m,n,cm_{0})=\frac{bcm_{0}C}{N}\int_{\mathbb{R}}\int_{0}^{\infty}W(x)\Phi\left(x+\frac{bcm_{0}Cu}{N},\frac{m}{c^{3}m_{0}}\right)V_{0}\left(\frac{bcm_{0}}{C},u\right)\\\
\times
e\left(\pm\frac{3(m(Nx+bcm_{0}Cu))^{\frac{1}{3}}}{cm_{0}^{\frac{1}{3}}}+\alpha(Nx)^{\beta}-\frac{nNx}{cm_{0}}\right)dxdu.$

######  Remark 6.1.

This is one of the strengths of the circle method that we choose. The bump
function $V_{0}$ being fixed allows one to extract the oscillations via a
simple change of variables.

Note that for any $x\in\mathbb{R}$ and $-1<\sigma<0$,

$e^{ix}-1=\int_{(\sigma)}\Gamma(s)e\left(\frac{s}{4}\right)x^{-s}\frac{ds}{2\pi
i}.$ (6.4)

This is immediate from shifting the contour to the left and then comparing it
with the Taylor expansion of $e^{ix}-1$. The identity (6.4) gives

$\displaystyle\mathcal{J}_{2}(m_{0}^{2}m,n,cm_{0})$
$\displaystyle=\frac{bcm_{0}C}{N}\int_{\mathbb{R}}\int_{0}^{\infty}W(x)\Phi\left(x+\frac{bcm_{0}Cu}{N},\frac{m}{c^{3}m_{0}}\right)$
$\displaystyle\quad\times
V_{0}\left(\frac{bcm_{0}}{C},u\right)e\left(\alpha(Nx)^{\beta}-\frac{nNx}{cm_{0}}\right)$
$\displaystyle\quad\times\left(1+\int_{(-\frac{1}{2}-\varepsilon)}\Gamma(s)e\left(\frac{s}{4}\right)\left(\pm\frac{3(m(Nx+bcm_{0}Cu))^{\frac{1}{3}}}{cm_{0}^{\frac{1}{3}}}\right)^{-s}\frac{ds}{2\pi
i}\right)dxdu.$

It follows from Stirling’s formula (4.1) that the $s$-integral converges
absolutely. Furthermore, repeated integration by parts in the $s$-integral
ensures an arbitrary saving unless

$|\mathrm{Im}(s)|\ll\left(\frac{mN}{c^{3}m_{0}}\right)^{\frac{1}{3}}T^{\varepsilon}.$

###  6.6. Cauchy–Schwarz in $m$

Applying dyadic subdivisions on $m_{0}^{2}m\sim M$ and $cm_{0}\sim C_{0}$, and
the Cauchy–Schwarz inequality with Corollary 5.2 to take out the
$(b,m_{0},m)$-sums and $(u,s)$-integrals with the truncation
$|\mathrm{Im}(s)|\ll\frac{(MN)^{\frac{1}{3}}T^{\varepsilon}}{C_{0}}$, we
derive

$\mathcal{S}(N)\ll\sup_{\pm}\sup_{\eta\in\\{0,1\\}}\sum_{b}\sup_{m_{0}\sim
M_{0}}\sup_{|u|\ll
1}\sup_{C_{0}\ll\frac{C}{b}}\sup_{M\ll\frac{N^{2}T^{\varepsilon}}{b^{3}C^{3}}}\int_{|t|\ll\frac{(MN)^{\frac{1}{3}}T^{\varepsilon}}{C_{0}}}\frac{M^{\frac{1}{6}}N^{\frac{2}{3}}T^{\varepsilon}}{\sqrt{1+|t|^{2+\varepsilon}}}\sqrt{\mathcal{R}(N)}dt+T^{-2023},$
(6.5)

where

$\mathcal{R}(N)\coloneqq\sum_{M\leq
m_{0}^{2}m<2M}\left|\sum_{\frac{C_{0}}{m_{0}}\leq
c<\frac{2C_{0}}{m_{0}}}\frac{1}{c}\sum_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n,cm_{0})=1\end{subarray}}S(\pm
m,\overline{n};c)\mathcal{J}_{3}^{\eta}(m,n,c)\right|^{2}$

with

$\mathcal{J}_{3}^{\eta}(m,n,c)\coloneqq
V_{0}\left(\frac{bcm_{0}}{C},u\right)\int_{0}^{\infty}W(x)\Phi\left(x+\frac{bcm_{0}Cu}{N},\frac{m}{c^{3}m_{0}}\right)\\\
\times
e\left(\alpha(Nx)^{\beta}-\frac{nNx}{cm_{0}}\right)\left(\frac{(MN)^{\frac{1}{6}}}{\sqrt{m_{0}}}\left(\frac{x+\frac{bcm_{0}Cu}{N}}{c^{3}}\right)^{\frac{1}{6}+\frac{\varepsilon}{3}-\frac{it}{3}}\right)^{\eta}dx.$
(6.6)

Before we open the square, we lengthen the $m$-sum in anticipation of Poisson
summation in the ensuing analysis. Let $L\geq 1$ be a parameter at our
disposal, and fix a smooth function $U_{0}$ supported on $[-3,3]$ such that
$U_{0}(x)=1$ for $x\in[-2,2]$. Then

$\mathcal{R}(N)\leq\sum_{m}U_{0}\left(\frac{m_{0}^{2}m}{LM}\right)\left|\sum_{\frac{C_{0}}{m_{0}}\leq
c<\frac{2C_{0}}{m_{0}}}\frac{1}{c}\sum_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n,cm_{0})=1\end{subarray}}S(\pm
m,\overline{n};c)\mathcal{J}_{3}^{\eta}(m,n,c)\right|^{2}.$ (6.7)

This step counterweights zero frequency and nonzero frequencies in Poisson
summation, and the optimisation of the parameter $L$ thus governs the quality
of the final bound.

###  6.7. Poisson Summation in $m$

Opening the square, the $m$-sum boils down to

$\sum_{m}U_{0}\left(\frac{m_{0}^{2}m}{LM}\right)S(\pm
m,\overline{n_{1}};c_{1})S(\pm
m,\overline{n_{2}};c_{2})\mathcal{J}_{3}^{\eta}(m,n_{1},c_{1})\overline{\mathcal{J}_{3}^{\eta}(m,n_{2},c_{2})}.$

Poisson summation (Lemma 4.2) then yields

$\frac{LM}{c_{1}c_{2}}\sum_{m}\mathcal{C}^{\pm}(m,n_{1},n_{2};c_{1},c_{2})\mathcal{J}_{4}^{\eta}(m,n_{1},n_{2},c_{1},c_{2}),$

where

$\mathcal{C}^{\pm}(m,n_{1},n_{2};c_{1},c_{2})\coloneqq\sum_{\gamma{\@displayfalse\pmod{c_{2}c_{2}}}}S(\pm\gamma,\overline{n_{1}};c_{1})S(\pm\gamma,\overline{n_{2}};c_{2})e\left(\frac{m\gamma}{c_{1}c_{2}}\right)$

and

$\mathcal{J}_{4}^{\eta}(m,n_{1},n_{2},c_{1},c_{2})\coloneqq\int_{\mathbb{R}}U_{0}(y)\mathcal{J}_{3}^{\eta}(LMy,n_{1},c_{1})\overline{\mathcal{J}_{3}^{\eta}(LMy,n_{2},c_{2})}e\left(-\frac{mLMy}{c_{1}c_{2}}\right)dy.$
(6.8)

Repeated integration by parts in the $y$-integral ensures an arbitrary saving
unless

$|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}.$

Inserting the above manipulations back into (6.7), we obtain

$\mathcal{R}(N)\ll LM\mathop{\sum\sum}_{\frac{C_{0}}{m_{0}}\leq
c_{1},c_{2}<\frac{2C_{0}}{m_{0}}}\frac{1}{(c_{1}c_{2})^{2}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1},c_{1}m_{0})=(n_{2},c_{2}m_{0})=1\end{subarray}}\\\
\times\sum_{|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\mathcal{C}^{\pm}(m,n_{1},n_{2};c_{1},c_{2})\mathcal{J}_{4}^{\eta}(m,n_{1},n_{2},c_{1},c_{2})+T^{-2023}.$
(6.9)

For the character sum $\mathcal{C}^{\pm}$, summing over
$\gamma{\@displayfalse\pmod{c_{1}c_{2}}}$ yields

$\mathcal{C}^{\pm}(m,n_{1},n_{2};c_{1},c_{2})=c_{1}c_{2}\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{1}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{2}}}}}_{\alpha_{1}c_{2}+\alpha_{2}c_{1}\equiv\mp
m{\@displayfalse\pmod{c_{1}c_{2}}}}e\left(\frac{\overline{\alpha_{1}n_{1}}}{c_{1}}+\frac{\overline{\alpha_{2}n_{2}}}{c_{2}}\right).$

Write $c_{0}=(c_{1},c_{2})$, $c_{1,0}=(\frac{c_{1}}{c_{0}},c_{0}^{\infty})$,
$c_{2,0}=(\frac{c_{2}}{c_{0}},c_{0}^{\infty})$,
$c_{1}=c_{0}c_{1,0}c_{1}^{\prime}$, $c_{2}=c_{0}c_{2,0}c_{2}^{\prime}$, and
$(c_{1}^{\prime},c_{2}^{\prime})=(c_{1}^{\prime}c_{2}^{\prime},c_{0})=1$. The
congruence condition $\alpha_{1}c_{2}+\alpha_{2}c_{1}\equiv\mp
m{\@displayfalse\pmod{c_{1}c_{2}}}$ then decomposes as

$\displaystyle c_{0}\mid
m,\qquad\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}},$
$\displaystyle\alpha_{1}\equiv\mp
m\overline{c_{2}}{\@displayfalse\pmod{c_{1}^{\prime}}},\qquad\alpha_{2}\equiv\mp
<EMAIL_ADDRESS>
By the Chinese Remainder Theorem, we observe

$\mathcal{C}^{\pm}(m,n_{1},n_{2};c_{1},c_{2})=c_{1}c_{2}e\left(\mp\frac{c_{2}\overline{c_{0}c_{1,0}mn_{1}}}{c_{1}^{\prime}}\mp\frac{c_{1}\overline{c_{0}c_{2,0}mn_{2}}}{c_{2}^{\prime}}\right)\\\
\times\delta(c_{0}\mid
m)\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}_{\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}}e\left(\frac{\overline{\alpha_{1}c_{1}^{\prime}n_{1}}}{c_{0}c_{1,0}}+\frac{\overline{\alpha_{2}c_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\right).$

In particular, if $m=0$, then

$\displaystyle\mathcal{C}^{\pm}(0,n_{1},n_{2};c_{1},c_{2})$
$\displaystyle=c_{1}^{2}\delta(c_{1}=c_{2})\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha{\@displayfalse\pmod{c_{1}}}}e\left(\frac{\alpha(\overline{n_{1}}-\overline{n_{2}})}{c_{1}}\right)$
$\displaystyle=c_{1}^{2}\sum_{ac^{\prime}=c_{1}}\mu(a)c^{\prime}\delta(c_{1}=c_{2},n_{1}\equiv
n_{2}{\@displayfalse\pmod{c^{\prime}}}).$

Hence, the contribution of $m=0$ to (6.9) is given by

$LM\sum_{\frac{C_{0}}{m_{0}}\leq
c<\frac{2C_{0}}{m_{0}}}\frac{1}{c^{2}}\sum_{ac^{\prime}=c}\mu(a)c^{\prime}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\ n_{1}\equiv
n_{2}{\@displayfalse\pmod{c^{\prime}}}\\\
(n_{1}n_{2},cm_{0})=1\end{subarray}}\mathcal{J}_{4}^{\eta}(0,n_{1},n_{2},c_{1},c_{2}).$

Applying the second derivative test on the integral transform
$\mathcal{J}_{3}^{\eta}$ in (6.6) yields

$\mathcal{J}_{4}^{\eta}(0,n_{1},n_{2},c_{1},c_{2})=\int_{\mathbb{R}}U_{0}(y)\mathcal{J}_{3}^{\eta}(LMy,n_{1},c_{1})\overline{\mathcal{J}_{3}^{\eta}(LMy,n_{2},c_{2})}dy\ll\frac{(MN)^{\frac{1}{3}}T^{\varepsilon}}{C_{0}N^{\beta}},$

where we use $Y\leq N^{\beta}T^{-\varepsilon}$. As a result, the contribution
of $m=0$ to (6.9) is bounded by

$LM\sum_{\frac{C_{0}}{m_{0}}\leq
c<\frac{2C_{0}}{m_{0}}}\frac{1}{c^{2}}\sum_{ac^{\prime}=c}\mu(a)c^{\prime}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\ n_{1}\equiv
n_{2}{\@displayfalse\pmod{c^{\prime}}}\end{subarray}}\frac{(MN)^{\frac{1}{3}}T^{\varepsilon}}{C_{0}N^{\beta}}\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}.$

It now transpires that

$\displaystyle\begin{split}\mathcal{R}(N)&\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+LM\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{c_{0}m_{0}}\leq
c_{1,0}c_{1}^{\prime},c_{2,0}c_{2}^{\prime}<\frac{2C_{0}}{c_{0}m_{0}}\\\
(c_{1}^{\prime},c_{2}^{\prime})=(c_{1}^{\prime}c_{2}^{\prime},m)=1\end{subarray}}\frac{1}{c_{0}^{2}c_{1,0}c_{2,0}c_{1}^{\prime}c_{2}^{\prime}}\\\
&\quad\times\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1},c_{0}c_{1}^{\prime}m_{0})=(n_{2},c_{0}c_{2}^{\prime}m_{0})=1\end{subarray}}\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}_{\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}}e\left(\frac{\overline{\alpha_{1}c_{1}^{\prime}n_{1}}}{c_{0}c_{1,0}}+\frac{\overline{\alpha_{2}c_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\right)\\\
&\quad\times
e\left(\mp\frac{c_{2,0}c_{2}^{\prime}\overline{c_{1,0}mn_{1}}}{c_{1}^{\prime}}\mp\frac{c_{1,0}c_{1}^{\prime}\overline{c_{2,0}mn_{2}}}{c_{2}^{\prime}}\right)\mathcal{J}_{4}^{\eta}(m,n_{1},n_{2},c_{0}c_{1,0}c_{1}^{\prime},c_{0}c_{2,0}c_{2}^{\prime}).\end{split}$

###  6.8. Additive Reciprocity

Additive reciprocity implies

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+LM\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{c_{0}m_{0}}\leq
c_{1,0}c_{1}^{\prime},c_{2,0}c_{2}^{\prime}<\frac{2C_{0}}{c_{0}m_{0}}\\\
(c_{1}^{\prime},c_{2}^{\prime})=(c_{1}^{\prime}c_{2}^{\prime},m)=1\end{subarray}}\frac{1}{c_{0}^{2}c_{1,0}c_{2,0}c_{1}^{\prime}c_{2}^{\prime}}$
$\displaystyle\quad\times\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1},c_{0}c_{1}^{\prime}m_{0})=(n_{2},c_{0}c_{2}^{\prime}m_{0})=1\end{subarray}}\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}_{\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}}e\left(\frac{\overline{\alpha_{1}c_{1}^{\prime}n_{1}}}{c_{0}c_{1,0}}+\frac{\overline{\alpha_{2}c_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\right)$
$\displaystyle\quad\times
e\left(\pm\frac{c_{2,0}c_{2}^{\prime}\overline{c_{1}^{\prime}}}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}c_{1}^{\prime}\overline{c_{2}^{\prime}}}{c_{2,0}mn_{2}}\right)\mathcal{J}_{5}^{\eta}(m,n_{1},n_{2},c_{0},c_{1,0},c_{2,0},c_{1}^{\prime},c_{2}^{\prime}),$

where

$\mathcal{J}_{5}^{\eta}(m,n_{1},n_{2},c_{0},c_{1,0},c_{2,0},c_{1}^{\prime},c_{2}^{\prime})\\\
\coloneqq
e\left(\mp\frac{c_{2,0}c_{2}^{\prime}}{c_{1,0}c_{1}^{\prime}mn_{1}}\mp\frac{c_{1,0}c_{1}^{\prime}}{c_{2,0}c_{2}^{\prime}mn_{2}}\right)\mathcal{J}_{4}^{\eta}(m,n_{1},n_{2},c_{0}c_{1,0}c_{1}^{\prime},c_{0}c_{2,0}c_{2}^{\prime}).$

Tracing back to the definitions (6.2), (6.3), (6.6), and (6.8) shows

$\mathcal{J}_{5}^{\eta}(m,n_{1},n_{2},c_{0},c_{1,0},c_{2,0},c_{1}^{\prime},c_{2}^{\prime})\\\
=T^{2\varepsilon}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\int_{\mathbb{R}}\int_{0}^{\infty}\int_{0}^{\infty}W(x_{1})W(x_{2})U_{0}(y)\Phi_{m,n_{1},n_{2}}^{c_{0},c_{1,0},c_{2,0}}(c_{1}^{\prime},c_{2}^{\prime},x_{1},x_{2},y)\\\
\times
e\left(\alpha(Nx_{1})^{\beta}-\frac{n_{1}Nx_{1}}{c_{0}c_{1,0}c_{1}^{\prime}}-\alpha(Nx_{2})^{\beta}+\frac{n_{2}Nx_{2}}{c_{0}c_{2,0}c_{2}^{\prime}}\right)dx_{1}dx_{2}dy,$

where

$\displaystyle\Phi_{m,n_{1},n_{2}}^{c_{0},c_{1,0},c_{2,0}}(c_{1}^{\prime},c_{2}^{\prime},x_{1},x_{2})$
$\displaystyle\coloneqq
T^{-2\varepsilon}\left(\frac{m_{0}}{C_{0}}\right)^{2\varepsilon\eta}V_{0}\left(\frac{bc_{1}m_{0}}{C},u\right)V_{0}\left(\frac{bc_{2}m_{0}}{C},u\right)$
$\displaystyle\quad\times
e\left(\mp\frac{c_{2,0}c_{2}^{\prime}}{c_{1,0}c_{1}^{\prime}mn_{1}}\mp\frac{c_{1,0}c_{1}^{\prime}}{c_{2,0}c_{2}^{\prime}mn_{2}}-\frac{mLMy}{c_{1}c_{2}}\right)$
$\displaystyle\quad\times\Phi\left(x_{1}+\frac{bc_{1}m_{0}Cu}{N},\frac{LMy}{c_{1}^{3}m_{0}}\right)\overline{\Phi\left(x_{2}+\frac{bc_{2}m_{0}Cu}{N},\frac{LMy}{c_{2}^{3}m_{0}}\right)}$
$\displaystyle\quad\times\left(\frac{C_{0}^{3}\left(x_{1}+\frac{bc_{1}m_{0}Cu}{N}\right)}{c_{1}^{3}m_{0}^{3}}\right)^{(\frac{1}{6}+\frac{\varepsilon}{3}-\frac{it}{3})\eta}\left(\frac{C_{0}^{3}\left(x_{2}+\frac{bc_{2}m_{0}Cu}{N}\right)}{c_{2}^{3}m_{0}^{3}}\right)^{(\frac{1}{6}+\frac{\varepsilon}{3}+\frac{it}{3})\eta}$

is a $T^{\varepsilon}$-inert function satisfying
$\Phi_{m,n_{1},n_{2}}^{c_{0},c_{1,0},c_{2,0}}(c_{1}^{\prime},c_{2}^{\prime},x_{1},x_{2})\ll
1$.

It is also convenient to remove the coprimality condition
$(c_{1}^{\prime},c_{2}^{\prime})=1$ via Möbius inversion:

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+LM\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}\mathop{\sum\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{c_{0}m_{0}}\leq
ac_{1,0}c_{1}^{\prime},ac_{2,0}c_{2}^{\prime}<\frac{2C_{0}}{c_{0}m_{0}}\\\
(ac_{1}^{\prime}c_{2}^{\prime},m)=1\end{subarray}}\frac{\mu(a)}{a^{2}c_{0}^{2}c_{1,0}c_{2,0}c_{1}^{\prime}c_{2}^{\prime}}$
$\displaystyle\quad\times\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1},n_{2}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1},ac_{0}c_{1}^{\prime}m_{0})=(n_{2},ac_{0}c_{2}^{\prime}m_{0})=1\end{subarray}}\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}_{\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}}e\left(\frac{\overline{\alpha_{1}ac_{1}^{\prime}n_{1}}}{c_{0}c_{1,0}}+\frac{\overline{\alpha_{2}ac_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\right)$
$\displaystyle\quad\times
e\left(\pm\frac{c_{2,0}c_{2}^{\prime}\overline{c_{1}^{\prime}}}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}c_{1}^{\prime}\overline{c_{2}^{\prime}}}{c_{2,0}mn_{2}}\right)\mathcal{J}_{5}^{\eta}(m,n_{1},n_{2},c_{0},c_{1,0},c_{2,0},ac_{1}^{\prime},ac_{2}^{\prime}).$

###  6.9. Cauchy–Schwarz in $c_{1}^{\prime}$

We now take advantage of the $(c_{1}^{\prime},c_{2}^{\prime})$-sums. To
simplify the forthcoming analysis, we pull out common divisors of $n_{1}$ and
$n_{2}$. Write $n_{0}=(n_{1},n_{2})$, $n_{1,0}=(n_{1},n_{0}^{\infty})$,
$n_{2,0}=(n_{2},n_{0}^{\infty})$, $(n_{1}/(n_{0}n_{1,0},m^{\infty}))=m_{1}$,
$(n_{2}/(n_{0}n_{2,0},m^{\infty}))=m_{2}$,
$n_{1}=m_{1}n_{0}n_{1,0}n_{1}^{\prime}$ and
$n_{2}=m_{2}n_{0}n_{2,0}n_{2}^{\prime}$. Under this notation, our estimate
transforms into

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+LM\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}\mathop{\sum\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{c_{0}m_{0}}\leq
ac_{1,0}c_{1}^{\prime},ac_{2,0}c_{2}^{\prime}<\frac{2C_{0}}{c_{0}m_{0}}\\\
(ac_{1}^{\prime}c_{2}^{\prime},m)=1\end{subarray}}\frac{\mu(a)}{a^{2}c_{0}^{2}c_{1,0}c_{2,0}c_{1}^{\prime}c_{2}^{\prime}}$
$\displaystyle\quad\times\sum_{(n_{0},ac_{0}c_{1}^{\prime}c_{2}^{\prime}m_{0})=1}\mathop{\sum\sum}_{\begin{subarray}{c}n_{1,0},n_{2,0}\mid
n_{0}^{\infty}\\\
(n_{1,0},n_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}m_{1},m_{2}\mid
m^{\infty}\\\ (m_{1},m_{2})=(m_{1}m_{2},ac_{0}m_{0}n_{0})=1\\\
(m_{1},c_{1}^{\prime})=(m_{2},c_{2}^{\prime})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1}=m_{1}n_{0}n_{1,0}n_{1}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{2}=m_{2}n_{0}n_{2,0}n_{2}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1}^{\prime},n_{2}^{\prime})=(n_{1}^{\prime}n_{2}^{\prime},ac_{0}m_{0}mn_{0})=1\\\
(n_{1}^{\prime},c_{1}^{\prime})=(n_{2}^{\prime},c_{2}^{\prime})=1\end{subarray}}$
$\displaystyle\quad\times\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}_{\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}}e\left(\frac{\overline{\alpha_{1}ac_{1}^{\prime}n_{1}}}{c_{0}c_{1,0}}+\frac{\overline{\alpha_{2}ac_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\pm\frac{c_{2,0}c_{2}^{\prime}\overline{c_{1}^{\prime}}}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}c_{1}^{\prime}\overline{c_{2}^{\prime}}}{c_{2,0}mn_{2}}\right)$
$\displaystyle\quad\times\mathcal{J}_{5}^{\eta}(m,n_{1},n_{2},c_{0},c_{1,0},c_{2,0},ac_{1}^{\prime},ac_{2}^{\prime}).$

Note that the second derivative test with the choice $Y\leq
N^{\beta}T^{-\varepsilon}$ implies that for any $T^{\varepsilon}$-inert
function $F$,

$\int_{0}^{\infty}W(x)F(x)e\left(\alpha(Nx_{1})^{\beta}-\frac{n_{1}Nx}{ac_{0}c_{1,0}c_{1}^{\prime}}\right)dx\ll
N^{-\frac{\beta}{2}}T^{\varepsilon}.$

By the Cauchy–Schwarz inequality to take out the
$(a,c_{0},c_{1,0},c_{2,0},c_{1}^{\prime},\alpha_{1},\alpha_{2},m,n_{0},n_{1,0},n_{2,0},n_{1}^{\prime},n_{2}^{\prime})$-sums
and the $y$-integral, along with a lengthening parameter $R_{1}\geq 1$, we
deduce

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+\frac{LMT^{\varepsilon}}{N^{\frac{\beta}{2}}}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\Bigg{\\{}\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{(a,m)=1}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}$
$\displaystyle\quad\times\sum_{(n_{0},ac_{0}c_{1}^{\prime}c_{2}^{\prime}m_{0})=1}\mathop{\sum\sum}_{\begin{subarray}{c}n_{1,0},n_{2,0}\mid
n_{0}^{\infty}\\\
(n_{1,0},n_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}m_{1},m_{2}\mid
m^{\infty}\\\
(m_{1},m_{2})=(m_{1}m_{2},ac_{0}m_{0}n_{0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1}=m_{1}n_{0}n_{1,0}n_{1}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{2}=m_{2}n_{0}n_{2,0}n_{2}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1}^{\prime},n_{2}^{\prime})=(n_{1}^{\prime}n_{2}^{\prime},ac_{0}m_{0}mn_{0})=1\end{subarray}}$
$\displaystyle\quad\times\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}\sum_{(c_{1}^{\prime},mn_{0}n_{1}^{\prime})=1}\frac{m_{0}mn_{0}n_{1}^{\prime}n_{2}^{\prime}}{a^{2}c_{0}^{2}c_{1,0}c_{2,0}^{2}C_{0}\sqrt{(mn_{0})_{\square}}}U_{0}\left(\frac{ac_{0}c_{1,0}c_{1}^{\prime}m_{0}}{C_{0}R_{1}}\right)$
$\displaystyle\quad\times\Bigg{|}\sum_{\begin{subarray}{c}\frac{C_{0}}{c_{0}m_{0}}\leq
ac_{2,0}c_{2}^{\prime}<\frac{2C_{0}}{c_{0}m_{0}}\\\
(c_{2}^{\prime},mn_{0}n_{2}^{\prime})=1\\\
\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}\end{subarray}}\frac{1}{c_{2}^{\prime}}e\left(\frac{\overline{\alpha_{2}ac_{2}^{\prime}n_{2}}}{c_{0}c_{2,0}}\pm\frac{c_{2,0}c_{2}^{\prime}\overline{c_{1}^{\prime}}}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}c_{1}^{\prime}\overline{c_{2}^{\prime}}}{c_{2,0}mn_{2}}\right)$
$\displaystyle\quad\times\int_{0}^{\infty}W(x)\varphi(c_{1}^{\prime},c_{2}^{\prime},x)e\left(-\alpha(Nx)^{\beta}+\frac{n_{2}Nx}{ac_{0}c_{2,0}c_{2}^{\prime}}\right)dx\Bigg{|}^{2}\Bigg{\\}}^{\frac{1}{2}}$

for some $T^{\varepsilon}$-inert function
$\varphi(c_{1}^{\prime},c_{2}^{\prime},x)\ll 1$.

###  6.10. Poisson Summation in $c_{1}^{\prime}$

Opening the square, the $c_{1}^{\prime}$-sum is now given by

$\sum_{(c_{1}^{\prime},mn_{0}n_{1}^{\prime})=1}U_{0}\left(\frac{ac_{0}c_{1,0}c_{1}^{\prime}m_{0}}{C_{0}R_{1}}\right)\varphi(c_{1}^{\prime},c_{2}^{\prime},x_{2})\overline{\varphi(c_{1}^{\prime},c_{3}^{\prime},x_{3})}e\left(\pm\frac{c_{2,0}\overline{c_{1}^{\prime}}(c_{2}^{\prime}-c_{3}^{\prime})}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}c_{1}^{\prime}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}mn_{2}}\right).$

By Poisson summation (Lemma 4.2), this is equal to

$\frac{n_{0}C_{0}R_{1}}{ac_{0}c_{1,0}^{2}c_{2,0}m_{0}mn_{1}n_{2}}\sum_{c_{1}^{\prime}}\mathcal{C}^{\prime}(c_{1}^{\prime},c_{2}^{\prime},c_{3}^{\prime})\int_{\mathbb{R}}U_{0}(y)\varphi\left(\frac{C_{0}R_{1}y}{ac_{0}c_{1,0}m_{0}},c_{2}^{\prime},x_{2}\right)\\\
\times\overline{\varphi\left(\frac{C_{0}R_{1}y}{ac_{0}c_{1,0}m_{0}},c_{3}^{\prime},x_{3}\right)}e\left(-\frac{c_{1}^{\prime}n_{0}C_{0}R_{1}y}{ac_{0}c_{1,0}^{2}c_{2,0}m_{0}mn_{1}n_{2}}\right)dy,$

where

$\mathcal{C}^{\prime}(c_{1}^{\prime},c_{2}^{\prime},c_{3}^{\prime})\coloneqq\sum_{\begin{subarray}{c}\gamma{\@displayfalse\pmod{c_{1,0}c_{2,0}mn_{1}n_{2}/n_{0}}}\\\
(\gamma,mn_{0}n_{1}^{\prime})=1\end{subarray}}e\left(\pm\frac{c_{2,0}\overline{\gamma}(c_{2}^{\prime}-c_{3}^{\prime})}{c_{1,0}mn_{1}}\pm\frac{c_{1,0}\gamma(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}mn_{2}}+\frac{c_{1}^{\prime}n_{0}\gamma}{c_{1,0}c_{2,0}mn_{1}n_{2}}\right).$

Repeated integration by parts in the $y$-integral ensures an arbitrary saving
unless

$|c_{1}^{\prime}|\ll\frac{ac_{0}c_{1,0}^{2}c_{2,0}m_{0}mn_{1}n_{2}}{n_{0}C_{0}R_{1}}T^{\varepsilon}\ll\frac{ac_{0}c_{1,0}^{2}c_{2,0}m_{0}mC_{0}}{n_{0}N^{2-2\beta}R_{1}}T^{\varepsilon}.$

Choosing
$R_{1}=1+\frac{ac_{0}c_{1,0}^{2}c_{2,0}m_{0}mC_{0}}{n_{0}N^{2-2\beta}}T^{2\varepsilon}$
ensures an arbitrary saving unless $c_{1}^{\prime}=0$, in which case

$\displaystyle\mathcal{C}^{\prime}(0,c_{2}^{\prime},c_{3}^{\prime})$
$\displaystyle=\sum_{\begin{subarray}{c}\gamma{\@displayfalse\pmod{c_{1,0}c_{2,0}mn_{1}n_{2}/n_{0}}}\\\
(\gamma,mn_{0}n_{1}^{\prime})=1\end{subarray}}e\left(\frac{c_{2,0}\overline{\gamma}(c_{2}^{\prime}-c_{3}^{\prime})}{c_{1,0}mn_{1}}+\frac{c_{1,0}\gamma(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}mn_{2}}\right)$
$\displaystyle=n_{2}^{\prime}\delta(c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{n_{2}^{\prime}}})$
$\displaystyle\quad\times
S(c_{1,0}^{2}m_{1}n_{1,0}n_{1}^{\prime}\overline{n_{2}^{\prime}}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}}),c_{2,0}^{2}m_{2}n_{2,0}(c_{2}^{\prime}-c_{3}^{\prime});c_{1,0}c_{2,0}m_{1}m_{2}mn_{0}n_{1,0}n_{2,0}n_{1}^{\prime}).$

Because $c_{1,0}\mid c_{0}^{\infty}\mid m^{\infty}$, we observe that for any
$p\mid c_{1,0}$, the Kloosterman sum evaluates to $0$ unless the $p$-adic
valuation obeys

$v_{p}(c_{1,0}^{2}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}}))=v_{p}(c_{2}^{\prime}-c_{3}^{\prime}),\qquad
c_{1,0}(c_{1,0}^{\infty},m)\mid c_{2}^{\prime}-c_{3}^{\prime}.$

By $p^{k}\mid c_{2}^{\prime}-c_{3}^{\prime}\Leftrightarrow
p^{k}\mid\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}}$, the former
condition cannot hold unless $c_{2}^{\prime}=c_{3}^{\prime}$. Therefore, the
Kloosterman sum evaluates to $0$ unless $c_{1,0}(c_{1,0}^{\infty},m)\mid
c_{2}^{\prime}-c_{3}^{\prime}$. Similarly, we obtain the condition
$c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0}(mn_{0},(c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0})^{\infty})\mid
c_{2}^{\prime}-c_{3}^{\prime}$. Writing
$d=(mn_{0},(c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0})^{\infty})$, one derives

$\mathcal{C}^{\prime}(0,c_{2}^{\prime},c_{3}^{\prime})=\varphi(d)c_{1,0}c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0}n_{2}^{\prime}\delta(c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{c_{1,0}c_{2,0}dm_{1}m_{2}n_{1,0}n_{2,0}n_{2}^{\prime}}})\\\
\times
S\left(\frac{c_{1,0}n_{1}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}dm_{2}n_{2,0}n_{2}^{\prime}},\frac{c_{2,0}(c_{2}^{\prime}-c_{3}^{\prime})}{c_{1,0}dm_{1}n_{1,0}};\frac{mn_{0}n_{1}^{\prime}}{d}\right),$

where $\varphi(d)$ denotes Euler’s totient function. Since $(mn_{0},n_{1})=1$,
we arrive at

$\mathcal{C}^{\prime}(0,c_{2}^{\prime},c_{3}^{\prime})=\varphi(d)c_{1,0}c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0}n_{2}^{\prime}\\\
\times\sum_{b_{0}b_{1}=n_{1}^{\prime}}\mu(b_{0})b_{1}\delta(c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{b_{1}c_{1,0}c_{2,0}dm_{1}m_{2}n_{1,0}n_{2,0}n_{2}^{\prime}}})\\\
\times
S\left(\frac{c_{1,0}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}dm_{2}n_{2,0}n_{2}^{\prime}},\frac{c_{2,0}\overline{b_{0}}(c_{2}^{\prime}-c_{3}^{\prime})}{b_{1}c_{1,0}dm_{1}n_{1,0}};\frac{mn_{0}}{d}\right).$

On the other hand, there holds

$\begin{dcases}\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}},\\\
\alpha_{1}c_{2,0}c_{3}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}\end{dcases}\\\
\Longleftrightarrow\quad\begin{dcases}\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}},\\\
c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{c_{0}c_{1,0}}}.\end{dcases}$

Because $c_{0}\mid m$ and $(m,c_{2,0}^{\infty})\mid d$, we have that
$c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}$.

Gathering the above observations together yields

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+\frac{LMT^{\varepsilon}}{N^{\frac{\beta}{2}}}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\Bigg{\\{}\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{(a,m)=1}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}$
$\displaystyle\quad\times\sum_{(n_{0},ac_{0}c_{1}^{\prime}c_{2}^{\prime}m_{0})=1}\mathop{\sum\sum}_{\begin{subarray}{c}n_{1,0},n_{2,0}\mid
n_{0}^{\infty}\\\
(n_{1,0},n_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}m_{1},m_{2}\mid
m^{\infty}\\\
(m_{1},m_{2})=(m_{1}m_{2},ac_{0}m_{0}n_{0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1}=m_{1}n_{0}n_{1,0}n_{1}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{2}=m_{2}n_{0}n_{2,0}n_{2}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1}^{\prime},n_{2}^{\prime})=(n_{1}^{\prime}n_{2}^{\prime},ac_{0}m_{0}mn_{0})=1\end{subarray}}$
$\displaystyle\quad\times\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{1}{\@displayfalse\pmod{c_{0}c_{1,0}}}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\alpha_{2}{\@displayfalse\pmod{c_{0}c_{2,0}}}}}\frac{\varphi(d)m_{1}m_{2}n_{2}^{\prime}R_{1}}{a^{3}c_{0}^{3}c_{1,0}^{2}c_{2,0}^{2}\sqrt{(mn_{0})_{\square}}}\sum_{b_{0}b_{1}=n_{1}^{\prime}}\mu(b_{0})b_{1}$
$\displaystyle\quad\times\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{ac_{0}c_{2,0}m_{0}}\leq
c_{2}^{\prime},c_{3}^{\prime}<\frac{2C_{0}}{ac_{0}c_{2,0}m_{0}}\\\
(c_{2}^{\prime}c_{3}^{\prime},mn_{0}n_{2}^{\prime})=1\\\
\alpha_{1}c_{2,0}c_{2}^{\prime}+\alpha_{2}c_{1,0}c_{1}^{\prime}\equiv\mp\frac{m}{c_{0}}\overline{a}{\@displayfalse\pmod{c_{0}c_{1,0}c_{2,0}}}\end{subarray}}\frac{1}{c_{2}^{\prime}c_{3}^{\prime}}\mathcal{C}_{1}(c_{2}^{\prime},c_{3}^{\prime})\mathcal{I}_{1}(c_{2}^{\prime},\ell)\Bigg{\\}}^{\frac{1}{2}},$

where

$\mathcal{C}_{1}(c_{2}^{\prime},c_{3}^{\prime})=\delta(c_{2}^{\prime}\equiv
c_{3}^{\prime}{\@displayfalse\pmod{q}})S\left(\frac{c_{1,0}(\overline{c_{2}^{\prime}}-\overline{c_{3}^{\prime}})}{c_{2,0}dm_{2}n_{2,0}n_{2}^{\prime}},\frac{c_{2,0}\overline{b_{0}}(c_{2}^{\prime}-c_{3}^{\prime})}{b_{1}c_{1,0}dm_{1}n_{1,0}};\frac{mn_{0}}{d}\right)$

with the notation

$d=(mn_{0},(c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0})^{\infty}),\qquad
q=b_{1}[c_{0},d]c_{1,0}c_{2,0}m_{1}m_{2}n_{1,0}n_{2,0}n_{2}^{\prime},$

and

$\mathcal{I}_{1}(c_{2}^{\prime},\ell)\coloneqq\int_{\mathbb{R}}\int_{0}^{\infty}\int_{0}^{\infty}W(x_{2})W(x_{3})U_{0}(y)\varphi\left(\frac{C_{0}R_{1}y}{ac_{0}c_{1,0}m_{0}},b_{2}c_{2}^{\prime},x_{2}\right)\overline{\varphi\left(\frac{C_{0}R_{1}y}{ac_{0}c_{1,0}m_{0}},b_{2}c_{3}^{\prime},x_{3}\right)}\\\
\times
e\left(-\alpha(Nx_{2})^{\beta}+\frac{n_{2}Nx_{2}}{ac_{0}c_{2,0}c_{2}^{\prime}}+\alpha(Nx_{3})^{\beta}-\frac{n_{2}Nx_{3}}{ac_{0}c_{2,0}c_{3}^{\prime}}\right)dx_{2}dx_{3}dy.$

###  6.11. Iteration Ad Infinitum

For any $k\in\mathbb{N}$, we iterate the above process $k$ times with a
different lengthening parameter

$R_{2}=1+\frac{ac_{0}c_{1,0}c_{2,0}^{2}m_{0}mC_{0}}{n_{0}N^{2-2\beta}}T^{2\varepsilon}.$

This implies

$\displaystyle\mathcal{R}(N)$
$\displaystyle\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+\frac{LMT^{\varepsilon}}{N^{\beta(1-2^{-k-1})}}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\Bigg{\\{}\sum_{0\neq|m|\ll\frac{C_{0}^{2}T^{\varepsilon}}{m_{0}^{2}LM}}\sum_{(a,m)=1}\sum_{c_{0}\mid
m}\mathop{\sum\sum}_{\begin{subarray}{c}c_{1,0},c_{2,0}\mid c_{0}^{\infty}\\\
(c_{1,0},c_{2,0})=1\end{subarray}}$
$\displaystyle\quad\times\sum_{(n_{0},ac_{0}c_{1}^{\prime}c_{2}^{\prime}m_{0})=1}\mathop{\sum\sum}_{\begin{subarray}{c}n_{1,0},n_{2,0}\mid
n_{0}^{\infty}\\\
(n_{1,0},n_{2,0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}m_{1},m_{2}\mid
m^{\infty}\\\
(m_{1},m_{2})=(m_{1}m_{2},ac_{0}m_{0}n_{0})=1\end{subarray}}\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{1}=m_{1}n_{0}n_{1,0}n_{1}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
\frac{C_{0}T^{-\varepsilon}}{N^{1-\beta}}\ll
n_{2}=m_{2}n_{0}n_{2,0}n_{2}^{\prime}\ll\frac{C_{0}T^{\varepsilon}}{N^{1-\beta}}\\\
(n_{1}^{\prime},n_{2}^{\prime})=(n_{1}^{\prime}n_{2}^{\prime},ac_{0}m_{0}mn_{0})=1\end{subarray}}$
$\displaystyle\quad\times\sum_{b_{0}b_{1}=n_{1}^{\prime}}\frac{(b_{1}\varphi(d)m_{1}m_{2}n_{2}^{\prime}R_{1})^{2^{k}}R_{2}^{2^{k}-1}}{a^{2^{k+1}+1}c_{0}^{2^{k+1}}(c_{1,0}c_{2,0})^{3\cdot
2^{k}-1}(n_{1,0}n_{2,0})^{2^{k}-1}(mn_{0})_{\square}^{2^{k}-\frac{1}{2}}}$
$\displaystyle\quad\times\mathop{\sum\sum}_{\begin{subarray}{c}\frac{C_{0}}{ac_{0}c_{2,0}m_{0}}\leq
c_{k+2}^{\prime},c_{k+3}^{\prime}<\frac{2C_{0}}{ac_{0}c_{2,0}m_{0}}\\\
(c_{k+2}^{\prime}c_{k+3}^{\prime},mn_{0}n_{2}^{\prime})=1\end{subarray}}\frac{1}{c_{k+2}^{\prime}c_{k+3}^{\prime}}\mathcal{C}_{k+1}(c_{k+2}^{\prime},c_{k+3}^{\prime})\mathcal{I}_{k+1}(c_{k+2}^{\prime},c_{k+3}^{\prime})\Bigg{\\}}^{\frac{1}{2^{k+1}}},$

where $\mathcal{C}_{k+1}(A,B)$ is defined recursively by

$\displaystyle\mathcal{C}_{k+1}(A,B)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}\gamma{\@displayfalse\pmod{c_{1,0}c_{2,0}mn_{1}n_{2}/n_{0}}}\\\
(\gamma,mn_{0}n_{1}^{\prime})=1\end{subarray}}\mathcal{C}_{k}(\gamma,A)\mathcal{C}_{k}(\gamma,B),$
$\displaystyle\mathcal{C}_{1}(A,B)$ $\displaystyle\coloneqq\delta(A\equiv
B{\@displayfalse\pmod{q}})S\left(\frac{c_{1,0}(\overline{A}-\overline{B})}{c_{2,0}dm_{2}n_{2,0}n_{2}^{\prime}},\frac{c_{2,0}\overline{b_{0}}(A-B)}{b_{1}c_{1,0}dm_{1}n_{1,0}};\frac{mn_{0}}{d}\right),$

and for $k\in\mathbb{N}$,

$\mathcal{I}_{k+1}(A,B)\coloneqq\int_{\mathbb{R}}\int_{0}^{\infty}\int_{0}^{\infty}W(x_{1})W(x_{2})U_{0}(y)\varphi_{k+1}\left(\frac{C_{0}Ry}{ac_{0}c_{1,0}m_{0}},A,x_{1}\right)\overline{\varphi_{k+1}\left(\frac{C_{0}Ry}{ac_{0}c_{1,0}m_{0}},B,x_{2}\right)}\\\
\times
e\left((-1)^{k}\left(\alpha(Nx_{1})^{\beta}-\frac{n_{2}Nx_{1}}{ac_{0}c_{2,0}A}-\alpha(Nx_{2})^{\beta}+\frac{n_{2}Nx_{2}}{ac_{0}c_{2,0}B}\right)\right)dx_{1}dx_{2}dy$

with $\varphi_{k+1}=\varphi$ if $k$ is even and $\overline{\varphi}$ if $k$ is
odd.

###  6.12. Endgame

By Lemma A.2, we obtain

$\mathcal{C}_{k+1}(c_{k+2}^{\prime},c_{k+3}^{\prime})\ll\left(\frac{b_{0}mn_{0}}{[c_{0},d]}C_{1}(C_{2},c_{1,0}c_{2,0})\left(\frac{mn_{0}}{d},q^{\infty}\right)\right)^{2^{k}},$

where

$C_{1}=\left(\frac{mn_{0}}{(mn_{0},q^{\infty})}\right)_{\square},\qquad
C_{2}=\frac{mn_{0}}{\left(mn_{0},q^{\infty}\right)C_{1}}.$

Taking $k$ sufficiently large and bounding everything else trivially yields

$\displaystyle\begin{split}\mathcal{R}(N)&\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+LMT^{\varepsilon}\frac{R}{N^{\beta}}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\\\
&\quad\times\sup_{\cdots}\cdots\sup_{\cdots}\frac{\sqrt{b_{1}dm_{1}m_{2}n_{2}^{\prime}R_{1}R_{2}}}{ac_{0}(c_{1,0}c_{2,0})^{\frac{3}{2}}\sqrt{n_{1,0}n_{2,0}(mn_{0})_{\square}}}\sqrt{\frac{b_{0}mn_{0}}{[c_{0},d]}C_{1}(C_{2},c_{1,0}c_{2,0})\left(\frac{mn_{0}}{d},q^{\infty}\right)}\\\
&\ll\frac{LM^{\frac{4}{3}}T^{\varepsilon}}{N^{\frac{2}{3}}}+\frac{C_{0}^{2}\sqrt{LM}T^{\varepsilon}}{m_{0}N}\left(\frac{(MN)^{\frac{1}{3}}}{C_{0}}\right)^{\eta}\left(1+\frac{C_{0}^{3}}{m_{0}LMN^{2-2\beta}}\right).\end{split}$
(6.10)

It now follows from (6.5) and (6.10) that

$\displaystyle\mathcal{S}(N)\ll
T^{\varepsilon}\sup_{M\ll\frac{N^{2}T^{\varepsilon}}{C^{3}}}\left(\sqrt{L}M^{\frac{5}{6}}N^{\frac{1}{3}}+CL^{\frac{1}{4}}M^{\frac{5}{12}}N^{\frac{1}{6}}\left(1+\frac{(MN)^{\frac{1}{6}}}{\sqrt{C}}\right)\left(1+\frac{C^{\frac{3}{2}}}{\sqrt{LM}N^{1-\beta}}\right)\right),$

where $L\geq 1$, $Y\leq N^{\beta}T^{-\varepsilon}$, and $C\leq\sqrt{N}$. If we
choose

$L=\left(\frac{N^{2}T^{\varepsilon}}{C^{3}M}\right)^{\frac{1}{3}},$

then the above estimate simplifies to

$\mathcal{S}(N)\ll\left(\frac{N^{2}}{C^{\frac{5}{2}}}+\frac{N^{\frac{3}{2}}}{C^{\frac{5}{4}}}+C^{\frac{7}{4}}N^{\beta-\frac{1}{2}}\right)T^{\varepsilon}.$

The optimal choice of $C$ is given by

$C=\min\\{N^{\frac{2-\beta}{3}},\sqrt{N}\\}.$

Altogether, we arrive at

$\mathcal{S}(N)\ll(N^{\frac{2}{3}+\frac{5\beta}{12}}+N^{\frac{7}{8}})T^{\varepsilon},$
(6.11)

where $\beta<\frac{4}{5}$. This completes the proof of Theorem 1.3. Moreover,
inserting (6.11) into (6.1) and choosing $Y=N^{\beta}T^{-\varepsilon}$, we
conclude that

$\mathcal{A}_{\pi}(T,\alpha,\beta)\ll
T^{\frac{2}{3}+\frac{5\beta}{12}+\varepsilon}+T^{\frac{7}{8}+\varepsilon}+T^{\frac{19}{14}-\beta+\varepsilon}$

where $\beta<\frac{4}{5}$. The proof of Theorem 1.1 is now complete. ∎

##  Appendix A Computations of Character Sums

To handle the character sum that we encounter, we need the following lemmata.

######  Lemma A.1.

Let $A,B,C,k,u,v\in\mathbb{N}$ with $C$ squarefree. Consider the character sum
defined recursively by

$\displaystyle\mathcal{C}_{1}(A,B)$
$\displaystyle=S(u(\overline{A}-\overline{B}),v(A-B);C),$
$\displaystyle\mathcal{C}_{k+1}(A,B)$
$\displaystyle=\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\gamma{\@displayfalse\pmod{C}}}\mathcal{C}_{k}(\gamma,A)\mathcal{C}_{k}(\gamma,B).$

Then we have that

$\displaystyle\mathcal{C}_{k}(A,B)\ll((C,uv)C)^{2^{k-1}-\frac{1}{2}}\sqrt{(C,A-B)}.$

###### Proof.

We first assume that $C=p$ is prime. If $p\mid uv$, then trivial bound
$\mathcal{C}_{k}\ll p^{2^{k}-1}$ follows. On the other hand, if $p\nmid
uv(A-B)$, then we apply the method of Adolphson–Sperber [AS93]. For $k=1$, the
Weil bound for Kloosterman sums implies $\mathcal{C}_{1}(A,B)\ll\sqrt{p}$. For
$k=2$,

$\displaystyle\mathcal{C}_{2}(A,B)=\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}\
\mathop{\sum\nolimits^{\mathrlap{\ast}}}\
\mathop{\sum\nolimits^{\mathrlap{\ast}}}}_{\alpha_{1},\alpha_{2},\beta{\@displayfalse\pmod{p}}}e\left(\frac{f_{2}(\alpha_{1},\alpha_{2},\beta)}{p}\right),$

where

$f_{2}(\alpha_{1},\alpha_{2},\beta)=u\alpha_{1}\overline{\beta}+v\overline{\alpha_{1}}\beta-u\alpha_{1}\overline{A}-v\overline{\alpha_{1}}A+u\alpha_{2}\overline{\beta}+v\overline{\alpha_{2}}\beta-u\alpha_{2}\overline{B}-v\overline{\alpha_{2}}B.$

Considering $f_{2}(x,y,z)\in\mathbb{F}_{p}^{\times}[x,y,z,(xyz)^{-1}]$, we see
that the locus of $\frac{\partial f_{2,\sigma}}{\partial x}=\frac{\partial
f_{2,\sigma}}{\partial y}=\frac{\partial f_{2,\sigma}}{\partial z}=0$ is empty
in $(\mathbb{F}_{p}^{\times})^{3}$ for any face $\sigma$ of the Newton
polyhedron $\Delta(f)$ since $p\nmid uv(A-B)$. Hence the work of
Adolphson–Sperber [AS93] yields $\mathcal{C}_{2}(A,B)\ll p^{\frac{3}{2}}$.

For $k\geq 3$, we observe that the character sum $\mathcal{C}_{k}(A,B)$ can be
written in the form

$\displaystyle\mathcal{C}_{k}(A,B)=\mathop{\mathop{\sum\nolimits^{\mathrlap{\ast}}}\cdots\mathop{\sum\nolimits^{\mathrlap{\ast}}}}_{\begin{subarray}{c}\alpha_{1},\ldots,\alpha_{2^{k-1}}{\@displayfalse\pmod{p}}\\\
\beta_{1},\ldots,\beta_{2^{k-2}}{\@displayfalse\pmod{p}}\\\
\gamma_{1},\ldots,\gamma_{2^{k-2}-1}{\@displayfalse\pmod{p}}\end{subarray}}e\left(\frac{f_{k}(\alpha_{1},\ldots,\alpha_{2^{k-1}},\beta_{1},\ldots,\beta_{2^{k-2}},\gamma_{1},\ldots,\gamma_{2^{k-2}-1})}{p}\right),$

where

$f_{k}(\alpha_{1},\ldots,\alpha_{2^{k-1}},\beta_{1},\ldots,\beta_{2^{k-2}},\gamma_{1},\ldots,\gamma_{2^{k-2}-1})\\\
=\sum_{j=1}^{2^{k-1}}\left(u\alpha_{j}\overline{\beta_{\lceil\frac{j}{2}\rceil}}+v\overline{\alpha_{j}}\beta_{\lceil\frac{j}{2}\rceil}-u\alpha_{j}\overline{\gamma_{g(j)}}-v\overline{\alpha_{j}}\gamma_{g(j)}\right)-u\alpha_{2^{k-2}}\overline{A}-u\overline{\alpha_{2^{k-2}}}A-v\alpha_{2^{k-1}}\overline{B}-v\overline{\alpha_{2^{k-1}}}B.$

For $j=2^{\ell}+2^{\ell+2}h,\,2^{\ell}+2^{\ell+1}(2h+1)$ with $\ell,h\geq 0$,
the function $g(j)$ above is given by

$\displaystyle g(j)=2^{k-2}-2^{k-\ell-2}+h.$

Considering
$f_{k}(x_{1},\ldots,x_{2^{k}-1})\in\mathbb{F}_{p}^{\times}[x_{1},\ldots,x_{2^{k}-1},(x_{1}\cdots
x_{2^{k}-1})^{-1}]$, we see that locus of $\frac{\partial
f_{k,\sigma}}{\partial x_{j}}=0$ for all $1\leq j\leq 2^{k}-1$ is empty in
$(\mathbb{F}_{p}^{\times})^{2^{k}-1}$ for any face $\sigma$ of the Newton
polyhedron $\Delta(f_{k})$ since $p\nmid uv(A-B)$. Hence the work of
Adolphson–Sperber [AS93] yields $\mathcal{C}_{k}(A,B)\ll
p^{2^{k-1}-\frac{1}{2}}$.

Finally, we perform induction to prove that

$\displaystyle\mathcal{C}_{k}(A,B)\ll p^{2^{k-1}-\frac{1}{2}}\sqrt{(p,A-B)}$

when $p\nmid uv$. We are left with the case where $p\mid(A-B)$. Note that when
$k=1$, the Weil bound for Kloosterman sums justifies the desired claim. For
$k\geq 2$, we have

$\displaystyle\mathcal{C}_{k}(A,B)=\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\gamma{\@displayfalse\pmod{p}}}\mathcal{C}_{k-1}(\gamma,A)\mathcal{C}_{k-1}(\gamma,B).$

By induction hypothesis,

$\displaystyle\mathcal{C}_{k-1}(\gamma,A)\ll
p^{2^{k-2}-\frac{1}{2}}\sqrt{(p,\gamma-A)}.$

Therefore, we obtain

$\displaystyle\mathcal{C}_{k}(A,B)\ll
p^{2^{k-1}}=p^{2^{k-1}-\frac{1}{2}}\sqrt{(p,A-B)}.$

The Chinese Remainder Theorem completes the proof of Lemma A.1 for squarefree
$C$. ∎

######  Lemma A.2.

Let $A,B,C,k,u,v,q_{1},q_{2},q_{3},q,Q\in\mathbb{N}$ be such that $C\mid Q$,
$q_{1},q_{2},q_{3}\mid q\mid Q$, and $(q_{3},\frac{CQ}{q_{3}})=1$. Consider
the character sum defined recursively by

$\displaystyle\mathcal{C}_{1}(A,B)$ $\displaystyle=\delta(A\equiv
B{\@displayfalse\pmod{q}})S\left(\frac{u(\overline{A}-\overline{B})}{q_{1}},\frac{v(A-B)}{q_{2}};C\right),$
$\displaystyle\mathcal{C}_{k+1}(A,B)$
$\displaystyle=\sum_{\begin{subarray}{c}\gamma{\@displayfalse\pmod{Q}}\\\
(\gamma,\frac{Q}{q_{3}})=1\end{subarray}}\mathcal{C}_{k}(\gamma,A)\mathcal{C}_{k}(\gamma,B).$

Then we have that

$\displaystyle\mathcal{C}_{k}(A,B)\ll\left(\frac{Q}{q}C_{1}(C_{2},uv)(C,q^{\infty})\right)^{2^{k-1}},$

where $C_{1}C_{2}=\frac{C}{(C,q^{\infty})}$ with $C_{1}$ squarefull, $C_{2}$
squarefree, and $(C_{1},C_{2})=1$.

###### Proof.

Write $C_{0}=(C,q^{\infty})$, $C=C_{0}C^{\prime}$, $Q_{0}=(Q,q^{\infty})$ and
$Q=Q_{0}Q^{\prime}$. Then

$\displaystyle\mathcal{C}_{k}(A,B)=\mathcal{C}_{k,1}(A,B)\mathcal{C}_{k,2}(A,B),$

where

$\displaystyle\mathcal{C}_{1,1}(A,B)$ $\displaystyle=\delta(A\equiv
B{\@displayfalse\pmod{q}})S\left(\overline{C^{\prime
2}}\frac{u(\overline{A}-\overline{B})}{q_{1}},\frac{v(A-B)}{q_{2}};C_{0}\right),$
$\displaystyle\mathcal{C}_{1,2}(A,B)$
$\displaystyle=S(\overline{C_{0}^{2}q_{1}q_{2}}u(\overline{A}-\overline{B}),v(A-B);C^{\prime}),$
$\displaystyle\mathcal{C}_{k+1,1}(A,B)$
$\displaystyle=\sum_{\begin{subarray}{c}\gamma{\@displayfalse\pmod{Q_{0}}}\\\
(\gamma,\frac{Q_{0}}{q_{3}})=1\end{subarray}}\mathcal{C}_{k,1}(\gamma,A)\mathcal{C}_{k,1}(\gamma,B),$
$\displaystyle\mathcal{C}_{k+1,2}(A,B)$
$\displaystyle=\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\gamma{\@displayfalse\pmod{Q^{\prime}}}}\mathcal{C}_{k,2}(\gamma,A)\mathcal{C}_{k,2}(\gamma,B)$
$\displaystyle=\varphi\left(\frac{Q^{\prime}}{(Q^{\prime},C^{\infty})}\right)\frac{(Q^{\prime},C^{\infty})}{C^{\prime}}\mathop{\sum\nolimits^{\mathrlap{\ast}}}_{\gamma{\@displayfalse\pmod{C^{\prime}}}}\mathcal{C}_{k,2}(\gamma,A)\mathcal{C}_{k,2}(\gamma,B).$

For $\mathcal{C}_{k,1}(A,B)$, bounding the Kloosterman sum trivially implies

$\displaystyle\mathcal{C}_{k,1}(A,B)\ll\left(\frac{Q_{0}}{q}\right)^{2^{k-1}-1}C_{0}^{2^{k-1}}.$

For $\mathcal{C}_{k,2}(A,B)$, note that the character sum is an iteration of
Kloosterman sums. Factorising $C^{\prime}=C_{1}C_{2}$ uniquely with $C_{1}$
squarefull, $C_{2}$ squarefree, and $(C_{1},C_{2})=1$, and then applying the
Chinese Remainder Theorem accordingly, Lemma A.1 yields

$\displaystyle\mathcal{C}_{k,2}(A,B)\ll\left(\varphi\left(\frac{Q^{\prime}}{(Q^{\prime},C^{\infty})}\right)\frac{(Q^{\prime},C^{\infty})}{C^{\prime}}\right)^{2^{k-1}-1}C_{1}^{2^{k}-1}\left(\left(C_{2},uv\right)C_{2}\right)^{2^{k-1}-\frac{1}{2}}\sqrt{(C_{2},A-B)}.$

Combining the above estimates completes the proof of Lemma A.2. ∎

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