# Non-vanishing of twists of $GL_{4}(\mathbb{A}_{\mathbb{Q}})$ $L$-functions

Maksym Radziwiłł  UT Austin, Department of Mathematics, PMA 8.100, 2515
Speedway, Stop C1200, Austin, TX 78712<EMAIL_ADDRESS>and  Liyang
Yang  Fine Hall, 304 Washington Rd, Princeton, NJ 08544, USA
<EMAIL_ADDRESS>
###### Abstract.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $f\geq 1$ be given. We show
that there exists infinitely many primitive even (resp. odd) Dirichlet
characters $\chi$ with conductor coprime to $f$ such that
$L(s,\pi\otimes\chi)$ is non-vanishing at the central point.

Our result has applications for the construction of $p$-adic $L$-functions for
$\text{GSp}_{4}$ following Loeffler-Pilloni-Skinner-Zerbes, the Bloch-Kato
conjecture and the Birch-Swinnerton-Dyer conjecture for abelian surfaces
following Loeffler-Zerbes, strong multiplicity one results for paramodular
cuspidal representations of $\text{GSp}_{4}(\mathbb{A}_{\mathbb{Q}})$ and the
rationality of the central values of $\text{GSp}_{4}(\mathbb{A}_{\mathbb{Q}})$
$L$-functions in the remaining non-regular weight case.

Throughout we will always assume that all unitary cuspidal automorphic
representations $\pi$ are normalized so that the central characters of $\pi$
are trivial on the diagonally embedded copy of the positive reals.

The following is our main result.

######  Theorem 1.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $f\geq 1$ be given. There
exists infinitely many primitive even (resp. odd) Dirichlet characters $\chi$
with conductor coprime to $f$ such that,

$L(\tfrac{1}{2},\pi\otimes\chi)\neq 0.$

Theorem 1 is at the limit of analytic methods. For cuspidal automorphic forms
$\pi\in\mathrm{GL}_{k}(\mathbb{A}_{K})$ with $k\in\\{2,3\\}$ and $K$ a number
field the analogue of Theorem 1 is due to Shimura [Shi77], Rohrlich [Roh89],
Barthel-Ramakrishnan [BR94] and Luo [Luo05]. For
$\pi\in\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$ Luo established in [Luo05]
the non-vanishing of $L(\sigma,\pi\otimes\chi)$ for all $0<\sigma<1$ except
for $\sigma=\tfrac{1}{2}$. We also mention that the isobaric case
$\pi_{1}\boxplus\pi_{2}$ with
$\pi_{1}\in\text{GL}_{n_{1}}(\mathbb{A}_{\mathbb{Q}})$ and
$\pi_{2}\in\text{GL}_{n_{2}}(\mathbb{A}_{\mathbb{Q}})$ and $n_{1}+n_{2}=4$ is
resolved in [FKM+23] ($n_{1}=2=n_{2}$) and in recent work of Li-Li-Lin [LLL23]
($n_{1}=3$, $n_{2}=1$).

Our approach builds and expands on Luo’s approach. The point
$\sigma=\tfrac{1}{2}$ is particularly interesting because of many arithmetic
applications, some of which we detail below. We note also that our proof can
be easily modified to give non-vanishing at $\tfrac{1}{2}+it$ for any fixed
$t\in\mathbb{R}$.

In contrast algebraic methods are able to access forms of higher rank, but
only when the form is of cohomological type and provided that there exists
critical values of the $L$-function associated with $\pi$ outside of the
critical strip $0<\Re s<1$. For example if
$\pi\in\mathrm{GL}_{2n}(\mathbb{A}_{F})$ with $F$ a totally real field is of
cohomological type, $L(\tfrac{3}{2},\pi)$ is a critical value and $\pi$
satisfies some additional minor local conditions at a prime $p$ then [DJR20]
show that for all primitive characters $\chi$ of conductor $p^{k}$ with $k$
sufficiently large $L(\tfrac{1}{2},\pi\otimes\chi)\neq 0$. To appreciate the
applications of Theorem 1 it is important for us to put this result and its
assumptions in context. The value $L(\tfrac{3}{2},\pi)$ is outside of the
critical strip and thus necessarily non-zero. By establishing a congruence
relationship between $L(\tfrac{3}{2},\pi\otimes\chi)$ and
$L(\tfrac{1}{2},\pi\otimes\chi)$ the authors of [DJR20] deduce the non-
vanishing of $L(\tfrac{1}{2},\pi\otimes\chi)$. This approach cannot be
executed when the form is not of cohomological type or when there are no
critical points outside of the critical strip. A simple concrete example of
the former is $\text{Sym}^{3}\phi$ with $\phi$ a classical Hecke-Maass form of
level 1 and eigenvalue $>\tfrac{1}{4}$, while an example of the latter is
$\text{Sym}^{3}\phi$ with $\phi$ a classical holomorphic modular form of odd
weight and non-trivial level. All of the new applications of Theorem 1 concern
the case of forms of cohomological type without critical values outside of the
critical strip. We now explain those applications in more detail below.

###  0.1. Applications

Let $\Pi$ be a cuspidal automorphic representation of
$\text{GSp}_{4}(\mathbb{A}_{\mathbb{Q}})$ which is non-CAP, globally generic
and cohomological with coefficients in the algebraic representation of highest
weight $(r_{1},r_{2})$. In [LPSZ21] and [DJR20] a $p$-adic measure
interpolating the critical values of $L(s,\Pi\otimes\chi)$ is constructed
provided that $r_{1}>r_{2}$. In addition in [LPSZ21, Theorem A] the existence
of such a measure for some $p$ in the case $r_{1}=r_{2}$ was established
conditionally on the existence of a character $\chi$ for which
$L(\tfrac{1}{2},\Pi\otimes\chi)\neq 0$. This now follows from our theorem and
the generic transfer given in [AS06]. Historically assumptions of this type go
back to the work of Ash and Ginzburg [AG94]. Note incidentally that in the
case $r_{1}=r_{2}$ the $L$-function $L(s,\pi)$ has no critical points outside
of the critical strip.

In [GR14] the rationality of the critical values
$\Lambda(\tfrac{1}{2},\Pi\otimes\chi)$ is established for $r_{1}\geq r_{2}\geq
0$. The case $r_{1}>r_{2}=-1$ is then proven unconditionally in [LZ21, Theorem
3.1.6] and the remaining case $r_{1}=r_{2}=-1$ is proven in [LZ21, Theorem
3.1.6] conditionally on the existence of a primitive character $\chi$ with
specific parity such that $L(\tfrac{1}{2},\Pi\otimes\chi)\neq 0$. This now
follows from our Theorem 1.

Similarly a case of the Bloch-Kato conjecture is established in [LZ21, Theorem
B] in the case $r_{1}=r_{2}$ conditionally on the existence of a primitive
character $\chi$ such that $L(\tfrac{1}{2},\Pi\otimes\chi)\neq 0$ and two
additional local conditions. Our Theorem 1 removes the non-vanishing
assumption. This special case of the Bloch-Kato conjecture is then used in
[LZ21] to establish a special of the Birch and Swinnerton-Dyer conjecture for
abelian varieties (still conditional on other assumptions, such as the
modularity of the abelian surface).

Finally, in conjunction with [AS06], our main theorem verifies Hypothesis 1.1
in [RW17], leading to consequences towards (strong) multiplicity one for
paramodular cuspidal representation of $\text{GSp}_{4}$. For instance, if
$\Pi$ is a paramodular cuspidal representation of
$\text{GSp}_{4}(\mathbb{A}_{\mathbb{Q}})$ with archimedean component
$\Pi_{\infty}$ in the discrete series and the local representation $\Pi_{p}$
spherical at $p=2,$ then it now follows from our Theorem 1 that $\Pi$ occurs
in the cuspidal spectrum with multiplicity one.

###  0.2. Quantitative non-vanishing

Theorem 1 follows from a stronger result computing the mean-value

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm 1)L(\tfrac{1}{2},\pi\otimes\chi)$

for a set of moduli $\mathcal{Q}\subset[Q,2Q]$ co-prime to all primes smaller
than $(\log Q)^{\varepsilon}$ for some small $\varepsilon>0$. In order to
state the result we need to introduce the notion of a Siegel-Walfisz sequence.
We will say that a sequence $\alpha(n)$ is Siegel-Walfisz of level $\kappa>0$
if for every $x\geq 10$, and $(a,q)=1$ with $q\leq(\log x)^{\kappa}$,
$|t|\leq(\log x)^{\kappa}$ we have,

$\sum_{\begin{subarray}{c}p\leq x\\\ p\equiv
a\pmod{q}\end{subarray}}\alpha(p)p^{it}\ll_{A}\frac{x}{(\log x)^{A}}$

for any given $A>10$. It is known that the coefficients $\lambda_{\pi}$ of the
$L$-funtions $L(s,\pi)$ are Siegel-Walfisz for some small $\kappa>0$ following
the work of Brumley [Bru06]. Moreover if one allows for ineffective results
(i.e results in which the implicit constant in the $\ll$ cannot be specified)
then it follows from the very recent work [TH23] that $\lambda_{\pi}$ is
Siegel-Walfisz for any $\kappa>0$. Throughout Brumley’s effective result is
all that we will ever need. In particular, with some work it is possible to
exhibit a completely explicit function $g$ such that for any given $\pi$ there
exists a primitive character $\chi$ of conductor $q\leq g(N_{\pi})$ where
$N_{\pi}$ is the conductor of $\pi$ and such that
$L(\tfrac{1}{2},\pi\otimes\chi)$ is non-zero.

We are now ready to state our main technical result. We will use the notation
$a\sim A$ to mean that $a\in[A,2A)$. This notation will be never used to mean
an asymptotic relation.

######  Theorem 2.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $\kappa>0$ be an exponent such
that $\lambda_{\pi}$ is Siegel-Walfisz of level $\kappa$. Let
$\mathcal{Q}_{\delta,\nu}\subset[Q/16,16Q]$ be the set of integers that can be
written as $p_{1}p_{2}m$ with $p_{1}\sim P_{1}:=(\log Q)^{\kappa\nu}$,
$p_{2}\sim P_{2}:=(\log Q)^{10000}$ and $m\sim Q/(P_{1}P_{2})$ square-free
with at most $\delta\log\log Q+10$ distinct prime factors, all larger than
$(\log Q)^{20000}$. Then, there exists a choice of $0<\delta<\nu<1$ with
$\kappa\nu<1000$ such that,

$\displaystyle\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\
\chi\pmod{q}\\\ \text{primitive}\end{subarray}}$ $\displaystyle\chi(\pm
1)L(\tfrac{1}{2},\pi\otimes\chi)$ $\displaystyle=Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}}\cdot\Big{(}\mathbf{1}_{\pm
1=1}\cdot(\log Q)^{o(1)}+O\Big{(}(\log Q)^{-1000\delta}\Big{)}\Big{)}$

as $Q\rightarrow\infty$.

The $(\log Q)^{o(1)}$ can be replaced by a more precise $(\log\log
Q)^{-3/2-\delta+o(1)}$ if needed. It is also possible to specify an admissible
$\kappa$, $\delta$ and $\nu$ completely explicitly. We find that doing so
makes the proof harder to parse. The theorem gives immediately the infinitude
of even or odd characters $\chi$ with $L(\tfrac{1}{2},\pi\otimes\chi)\neq 0$,
using the observation that,

$\mathbf{1}_{\chi(-1)=\pm 1}=\frac{1\pm\chi(-1)}{2}.$

Moreover for any given $f\geq 1$ once $Q$ is choosen large enough all the
exhibited characters have modulus coprime to $f$.

A nearly immediate consequence of the above Theorem and the large sieve is a
quantitative result asserting that for any given $f\geq 1$ there are at least

$\frac{Q^{2}}{\log^{4}Q}$

even (resp. odd) primitive characters $\chi\pmod{q}$ with $q\in[Q/16,16Q]$ and
$(q,f)=1$ for which $L(\tfrac{1}{2},\pi\otimes\chi)\neq 0$. We record two
final observations. We can require the set of moduli to be composed of only
three primes under additional assumptions on the form $\pi$, e.g forms of
cohomological type or forms arising from lifting of
$\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$ forms, for example
$\operatorname{Sym}^{3}\phi$ with $\phi$ being a cusp form on
$\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}}).$ We discuss some of these
modifications in the remarks in the next section. While we haven’t checked all
the details it seems plausible that Theorem 2 holds in general with a set of
moduli $q$ composed of $\leq C$ distinct prime factors with $C$ an absolute
constant. We have not considered possible extensions of our result to the case
of number fields and plan to address this in later papers.

###  0.3. Notation

Throughout all constants in $O(\cdot)$ are allowed to depend on the form $\pi$
since throughout the paper the form $\pi$ is fixed. We wil from time to time
mention this dependence, with the purpose of bringing to the reader’s
attention that uniformity in $\pi$ is not considered in our bounds. We will
write $a\sim A$ to mean $a\in[A,2A)$. We will also write $a\asymp A$ to mean
that there exists an absolute constant $C$ such that $a\in[A/C,CA]$. For
example $C=10^{9}$ is an admissible choice that works throughout the paper.

###  0.4. Acknowledgments

We would like to thank Henri Darmon, Wenzhi Luo and Chris Skinner for
discussions related to this paper. The first author acknowledges support of
NSF grant DMS-1902063.

##  1\. Organization of the paper and deduction of Theorem 2

Theorem 2 follows from three Propositions whose proof will occupy us for the
rest of the paper. In this section we state these three Propositions and we
deduce Theorem 2 from them.

The first Proposition is a variant of Luo’s work [Luo05]. Similar ideas are
also used in the paper of Blomer-Milicevic [BM15]. The main ingredient in the
Proposition below is Deligne’s bound for certain exponential sums.

######  Proposition 1.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$ of conductor $N_{\pi}$. Let
$\delta\geq 0$ be given. Let $\mathcal{R}\subset[R/4,4R]$ and
$\mathcal{S}\subset[S/4,4S]$ with $R\leq S$ be two sets of of integers such
that for all $r\in\mathcal{R}$ and $s\in\mathcal{S}$,

  1. (1)

we have $(rs,N_{\pi})=1$ and $(r,s)=1$

  2. (2)

both $r$ and $s$ are square-free,

  3. (3)

we have $\omega(r)\leq 10$ and $\omega(s)\leq\delta\log\log S+10$,

Let $\mathcal{Q}$ be the set of all integers that can be written as $rs$ with
$r\in\mathcal{R}$ and $s\in\mathcal{S}$. Let $V$ be a smooth function
compactly supported in $[1/100,100]$. Finally set $Q:=RS$ and assume that
$S>R^{10}$. Then, for any $M\geq 1$,

$\displaystyle\sum_{q\in\mathcal{Q}}$
$\displaystyle\sum_{\begin{subarray}{c}\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\varepsilon(\pi,\chi)\chi(\pm
1)\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)\overline{\chi}(m)}{\sqrt{m}}V\Big{(}\frac{m}{M}\Big{)}$
$\displaystyle\ll\|V\|_{\infty,2}\cdot\Big{(}\sqrt{RM}Q(\log
Q)^{-\frac{1}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}}+Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}}\cdot R^{-1/4}(\log
Q)^{C\delta}\Big{)}$

with $C>10$ an absolute constant and where $\varepsilon(\pi,\chi)$ is the root
number of $L(s,\pi\otimes\chi)$ and where finally

$\|V\|_{\infty,2}:=\|V\|_{\infty}+\|V^{\prime}\|_{\infty}+\|V^{\prime\prime}\|_{\infty}.$

Notice that $Q^{2}(\log Q)^{-1+\delta+\delta\log\frac{1}{\delta}+o(1)}$ is the
size of the main term. The term $(\log Q)^{C\delta}$ is the loss from
Deligne’s bound. The first term in the bound in Proposition 1 corresponds to
the diagonal while the second term corresponds to the off-diagonal. Loosing a
factor of $(\log Q)^{C\delta}$ in the diagonal would not make a difference to
our argument. We in fact carry out the proof of the Theorem with this
additional loss to make this point.

The second Proposition relies on a variant of a recent dispersion estimate of
Fouvry-Radziwiłł [FR22]. Results of this type go back to the work of Green
[Gre18] and use the bounds of Duke-Friedlander-Iwaniec [DFI97] as their main
arithmetic input. We refer the reader to [GS19] for another variation on the
work of Green and to [BC18] for a recent improvements of the bounds of Duke-
Friedlander-Iwaniec.

######  Proposition 2.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $0<\kappa<1$ be such that
$\lambda_{\pi}$ is Siegel-Walfisz of level $\kappa$. Let $0<\nu,\delta<1/1000$
be given. Let $\mathcal{Q}_{\delta,\nu}\subset[Q/16,16Q]$ be the set of
square-free integers that can be written as $p_{1}p_{2}m$ with $p_{1}\sim
P_{1}:=(\log Q)^{\kappa\nu}$, $p_{2}\sim P_{2}:=(\log Q)^{10000}$ and $m\sim
Q/(P_{1}P_{2})$ having less than $\delta\log\log Q+10$ prime factors, all
larger than $(\log Q)^{20000}$. Let $V$ be smooth and compactly supported in
$[1/100,100]$. Then, for any $N$ in the range,

$Q^{2}(\log Q)^{-10^{9}}\leq N\leq Q^{2}(\log Q)^{10^{9}}$

we have, for any $A>10$,

$\displaystyle\sum_{q\in\mathcal{Q}}$
$\displaystyle\sum_{\begin{subarray}{c}\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm
1)\sum_{n}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$
$\displaystyle\ll\|V\|_{\infty,2}\cdot
Q\sqrt{N}\cdot\Big{(}\frac{e^{C/\delta}}{(\log Q)^{3/2}}\cdot(\log
Q)^{4\nu+10\delta+\delta\log\frac{1}{\delta}}+C(A,\nu,\kappa)(\log
Q)^{-A}\Big{)}$

with $C(A,\nu,\kappa)$ a constant depending only on $A$, $\nu$ and $\kappa$.

######  Remark 1.

This bound can be improved by a small power of $\log x$ if we assume that
there exists a $\eta>0$ such that,

(1) $\sum_{p\leq x}|\lambda_{\pi}(p)|^{2+\eta}\ll\frac{x}{\log x}.$

The bound (1) is true for $\pi$ of cohomological type or $\pi$ arising from
certain functorial lifting of
$\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$-cuspidal representations, namely,
for $0<\eta<2/3,$

  * •

$\pi=\operatorname{Sym}^{3}\sigma,$ where $\sigma$ is a
$\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$-cuspidal representation which is
not of tetrahedral or octahedral type. In this case we have
$|\lambda_{\pi}(p)|^{2+\eta}\ll
1+|\lambda_{\operatorname{Sym}^{4}\sigma}(p)|^{2}$ and
$\operatorname{Sym}^{4}\sigma$ is cuspidal (cf. [KS02b] and [KS02a]).

  * •

$\pi=\sigma_{1}\boxtimes\sigma_{2},$ where $\sigma_{1}$ and $\sigma_{2}$ are
$\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$-cuspidal representations which are
not of dihedral type and $\operatorname{Sym}^{2}\sigma_{1}$ is not a twist of
$\operatorname{Sym}^{2}\sigma_{2}.$ In this case we have
$|\lambda_{\pi}(p)|^{2+\eta}\ll
1+|\lambda_{\operatorname{Sym}^{2}\sigma_{1}\boxtimes\sigma_{2}}(p)|^{2}+|\lambda_{\sigma_{1}\boxtimes\operatorname{Sym}^{2}\sigma_{2}}(p)|^{2}$
and $\operatorname{Sym}^{2}\sigma_{1}\boxtimes\sigma_{2}$ and
$\sigma_{1}\boxtimes\operatorname{Sym}^{2}\sigma_{2}$ are cuspidal
representations of $\mathrm{GL}_{6}(\mathbb{A}_{\mathbb{Q}})$ (cf. [Ram00],
[KS02a] and [RW04]).

The bound of Proposition 2 is essentially dominated by the contribution of
primes which we don’t know how to bound other than trivially.

The final Proposition is rather trivial.

######  Proposition 3.

Let $\pi$ be a unitary cuspidal automorphic representation of
$GL_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $V$ be a smooth function compactly
supported in $[N/100,100N]$. Let $\mathcal{Q}\subset[Q/16,16Q]$ be a set of
squarefree integers that can be written as $p_{1}p_{2}m$ with $p_{1}\sim
P_{1}$, $p_{2}\sim P_{2}$ and $m\sim Q/(P_{1}P_{2})$ having at most
$\delta\log\log Q+10$ prime factors, all larger than $(\log Q)^{20000}$. We
assume that $4P_{1}<2P_{2}<(\log Q)^{20000}$. Then,

$\displaystyle\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm
1)\sum_{n}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}=V\Big{(}\frac{\pm
1}{N}\Big{)}$ $\displaystyle Q^{2}(\log
Q)^{\delta-1+\delta\log\frac{1}{\delta}+o(1)}$ $\displaystyle+O(Q\sqrt{N}(\log
Q)^{2}\|V\|_{\infty})$

Notice that the main term above is zero unless $N\leq 100$ and
$\textbf{1}_{\pm 1=1}$.

We are now ready to describe how Theorem 2 follows from these two Proposition.

###### Deduction of Theorem 2 from Proposition 2 and Proposition 1.

Let

$N_{\nu}=Q^{2}(\log Q)^{1-200\nu}\text{ and }M_{\nu}=Q^{2}(\log
Q)^{-1+200\nu}.$

Let also $N_{\nu,\varepsilon}^{+}:=N_{\nu}(\log Q)^{\varepsilon}$ and
$M_{\nu,\varepsilon}^{+}:=M_{\nu}(\log Q)^{\varepsilon}$. By the approximate
functional equation,

$L(\tfrac{1}{2},\pi\otimes\chi)=\sum_{n}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}W_{1}\Big{(}\frac{n}{N_{\nu}}\Big{)}+\varepsilon(\pi,\chi)\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)\overline{\chi(m)}}{\sqrt{m}}W_{2}\Big{(}\frac{m}{M_{\nu}}\Big{)}$

with $W_{1},W_{2}$ smooth functions such that $W_{i}(x)=1+O_{A}(x^{A})$ for
$x<1$ and any given $A>10$, and $W_{i}(x)=O(x^{-A})$ for $x>1$ and any given
$A>10$. We also introduce a partition of unity on the $n$ and $m$ sum, i.e we
pick a smooth function $V$ compactly supported in $[1/4,4]$ and such that for
all $x>0$,

$1=\sum_{N}V\Big{(}\frac{x}{N}\Big{)}$

with $N$ running over powers of two. Notice that by the large sieve and
Rankin-Selberg,

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\frac{\lambda_{\pi}(n)\chi(\pm
n)}{\sqrt{n}}W_{1}\Big{(}\frac{n}{N_{\nu}}\Big{)}V\Big{(}\frac{n}{N}\Big{)}\ll_{A,\varepsilon}Q^{2}(\log
Q)^{-A}$

for any $A>10$, $V$ smooth and compactly supported away from zero and
$N>N^{+}_{\nu,\varepsilon}$. Similarly,

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\varepsilon(\pi,\chi)\sum_{n}\frac{\lambda_{\tilde{\pi}}(n)\overline{\chi(\pm
n)}}{\sqrt{n}}W_{2}\Big{(}\frac{m}{M_{\nu}}\Big{)}V\Big{(}\frac{m}{M}\Big{)}\ll_{A,\varepsilon}Q^{2}(\log
Q)^{-A}$

for any $A>10$, $V$ smooth and compactly supported away from $0$ and
$M>M^{+}_{\nu,\varepsilon}$. Therefore, we can write

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm
1)L(\tfrac{1}{2},\pi\otimes\chi)=\sum_{N\leq
N^{+}_{\nu,\varepsilon}}\mathcal{F}_{N}+\sum_{M\leq
M^{+}_{\nu,\varepsilon}}\mathcal{D}_{M}+O_{A,\varepsilon}(Q^{2}(\log Q)^{-A})$

with $N$, $M$ running over powers of two, and where,

$\mathcal{F}_{N}:=\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\
\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\sum_{n}\frac{\lambda_{\pi}(n)\chi(\pm
n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}W_{1}\Big{(}\frac{n}{N_{\nu}}\Big{)}$

and $V$ is a smooth function compactly supported in $[1/4,4]$ and similarly,

$\mathcal{D}_{M}:=\sum_{\begin{subarray}{c}q\in\mathcal{Q}_{\delta,\nu}\\\
\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\varepsilon(\pi,\chi)\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)\overline{\chi}(\pm
m)}{\sqrt{m}}V\Big{(}\frac{m}{M}\Big{)}W_{2}\Big{(}\frac{m}{M_{\nu}}\Big{)}.$

We now collect our previous bounds in each of the ranges.

###  1.1. $\mathcal{F}_{N}$ in the range $N\leq Q^{2}(\log Q)^{-10^{9}}$

Appealing to Proposition 3 we find,

$\sum_{N\leq Q^{2}(\log Q)^{-10^{9}}}\mathcal{F}_{N}=\mathbf{1}_{\pm
1=1}Q^{2}(\log Q)^{-1+\delta+\delta\log\frac{1}{\delta}+o(1)}+O(Q^{2}(\log
Q)^{-100}).$

as $Q\rightarrow\infty$.

###  1.2. $\mathcal{F}_{N}$ in the range with $N_{\nu,\varepsilon}^{+}\geq
N\geq Q^{2}(\log Q)^{-10^{9}}$

We first note that there exists a small $0<\kappa<1$ for which $\lambda_{\pi}$
is Siegel-Walfisz. This follows for example from Brumley’s result [Bru06] (see
also [KT22] where the details of the deduction are worked out or Lemma 17
later in this paper). There are $\log\log Q$ dy-adic ranges to consider in
this case. According to Proposition 2 their total contribution is,

$\displaystyle\ll
e^{O(1/\delta)}\cdot\frac{\sqrt{N_{\nu,\varepsilon}^{+}}Q}{(\log
Q)^{3/2}}(\log
Q)^{4\nu+\delta\log\frac{1}{\delta}+10\delta+2\varepsilon}+C(\nu,\kappa)Q^{2}(\log
Q)^{-10}$ $\displaystyle\ll e^{O(1/\delta)}\cdot Q^{2}(\log
Q)^{-1-100\nu+\varepsilon+\delta\log\frac{1}{\delta}+10\delta+2\varepsilon}+C(\nu,\kappa)Q^{2}(\log
Q)^{-10}$

with $C(\nu,\kappa)$ a constant depending only on $\nu$ and $\kappa$. The
above is negligible compared to the main term $Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}+o(1)}$ for any fixed
$0<\varepsilon<\delta<\nu<1/1000$.

###  1.3. $\mathcal{D}_{M}$ in the range $M\leq Q^{2}(\log Q)^{-10^{9}}$

In this range we interpret $\mathcal{Q}_{\delta,\nu}$ as a product of a set
$\mathcal{R}$ consisting of a product of two primes $p_{1}p_{2}$ with
$p_{1}\sim P_{1}:=(\log Q)^{\kappa\nu}$, $p_{2}\sim P_{2}:=(\log Q)^{10000}$
and another set $\mathcal{S}$ consisting of $m\sim Q/(P_{1}P_{2})$ an integer
with at most $\delta\log\log Q+10$ prime factors, all larger than $(\log
Q)^{20000}$. Let $M_{\text{max}}:=Q^{2}(\log Q)^{-10^{9}}$. Applying
Proposition 1 we get a bound,

$\displaystyle(\log
Q)^{C\delta+2\varepsilon}\cdot\Big{(}\sqrt{M_{\text{max}}}Q\sqrt{R}+Q^{2}R^{-1/4}\Big{)}$

If $\kappa\nu<10^{3}$ then $(\log Q)^{10000}\leq R\leq(\log Q)^{20000}$. If in
addition we assume that $C\delta<10^{3}$ and $0<\varepsilon<1$, then since
$M_{\text{max}}\leq Q^{2}(\log Q)^{-10^{9}}$ the above is

$\ll\frac{Q^{2}}{(\log Q)^{10}}.$

We record again the conditions $\kappa\nu<10^{3}$, $C\delta<10^{3}$ and
$0<\varepsilon<1$.

###  1.4. $\mathcal{D}_{M}$ in the range $M_{\nu,\varepsilon}^{+}\geq M\geq
Q^{2}(\log Q)^{-10^{9}}$

In this range we interpret $\mathcal{Q}_{\delta,\nu}$ as a product of a set
$\mathcal{R}$ consisting of one prime $p_{1}\sim P_{1}:=(\log Q)^{\kappa\nu}$
and another set $\mathcal{S}$ consisting of $p_{2}m$ with $p_{2}\sim
P_{2}:=(\log Q)^{10000}$ and $m\sim Q/(P_{1}P_{2})$ an integer with at most
$\delta\log\log Q+10$ prime factors, all larger than $(\log Q)^{20000}$.
Applying Proposition 1 we get a bound,

$\displaystyle\mathcal{D}_{M}$ $\displaystyle\ll(\log
Q)^{C\delta+2\varepsilon}\cdot\Big{(}\sqrt{MR}Q(\log
Q)^{-\frac{1}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}}+Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}}R^{-1/4}\Big{)}$
$\displaystyle\ll(\log Q)^{C\delta+2\varepsilon}\cdot\Big{(}Q^{2}(\log
Q)^{-\frac{1}{2}+100\nu+\kappa\nu/2-\frac{1}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}}+Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}-\kappa\nu/4}\Big{)}$

Notice that we could have written down a more careful bound without the loss
of $(\log Q)^{C\delta+2\varepsilon}$ in the first term. It however makes no
difference. Summing all these contributions only adds a factor of $\log\log Q$
which is negligible for our purposes. We therefore require that,

(2)
$\displaystyle\begin{cases}C\delta+2\varepsilon-1+100\nu+\frac{\kappa\nu}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}&<-1+\delta\log\frac{1}{\delta}-10^{9}\cdot\delta\\\
C\delta+2\varepsilon-1+\delta+\delta\log\frac{1}{\delta}-\frac{\kappa\nu}{4}&<-1+\delta\log\frac{1}{\delta}-10^{9}\cdot\delta\end{cases}$

Finally we have the conditions $0<C\delta<10^{3}$, $\kappa\nu<10^{3}$,
$0<\varepsilon<1$ and $0<\varepsilon<\delta<\nu<1/1000$ carrying over from the
previous ranges. For convenience we rewrite (2) as

$\displaystyle\begin{cases}2\varepsilon+(100+\kappa/2)\nu+(C+1/2+10^{9})\delta&<\frac{\delta}{2}\log\frac{1}{\delta}\\\
2\varepsilon+(1+10^{9}+C)\delta&<\frac{\kappa}{4}\cdot\nu\end{cases}$

If the conditions in (2) are met we obtain,

$\sum_{Q^{2}(\log Q)^{-10^{9}}\leq M\leq
M_{\nu,\varepsilon}^{+}}\mathcal{D}_{M}\ll(\log\log Q)\cdot Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}-10^{3}\cdot\delta}$

which is entirely acceptable. A choice of parameters that satisfies all the
conditions is not difficult to find, for example one can pick

$\varepsilon=\delta\ ,\ \nu=\delta\sqrt{\log\frac{1}{\delta}}$

and pick $\delta>0$ sufficiently small to make all the inequalities true.

Notice incidentally that the limiting case $\delta=0$ barely fails to work
(ignoring the factors $e^{C/\delta}$ which are mere technicalities). Note also
that $\delta$ shrinks when either $\kappa$ shrinks or $C$ grows, as expected.
∎

######  Remark 2.

Even though the above proof barely fails to work in the limiting case
$\delta=0$ it can be made to work in the case $\delta=0$ under the additional
assumption that there exists an exponent $\eta>0$ such that,

(3) $\sum_{p\leq x}|\lambda_{\pi}(p)|^{2+\eta}\ll\frac{x}{\log x},$

since this would provide a decisive (for the case $\delta=0$) improvement in
Proposition 2. In particular under the assumption (3) one could take
$\mathcal{Q}$ to consist of a product of three primes. In an earlier version
of the manuscript we operated under the assumption of (3) until we realized
that integers with $\leq\delta\log\log Q+20$ prime factors can be used
successfully thanks to the fact that $\delta\log\frac{1}{\delta}$ is larger
than $C\delta$ (any any fixed $C>10$) for small $\delta>0$. Thus the density
of such integers grows sufficiently quicker (but only for small $\delta$!) to
allow us to overcome the loss from Deligne’s bound.

##  2\. Complete character sums

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Denote by $\varepsilon(\pi,\chi)$
the root number of the $L$-function $L(s,\pi\otimes\chi)$. Throughout this
section let

$\mathcal{E}_{\pi}(m;q):=\sum_{\begin{subarray}{c}\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\varepsilon(\pi,\chi)\overline{\chi}(m)$

Notice that $\mathcal{E}_{\pi}(m;q)$ is $q$-periodic in $m$. We use the
notation $\mathcal{E}$ to signal that $\mathcal{E}$ is some sort of transform
of the root number $\varepsilon(\pi,\chi)$. The bulk of this section is spent
analyzing $\mathcal{E}$ and various complete exponential sums arising from
$\mathcal{E}$.

From [Luo05] we know that, for $(q,N_{\pi})=1$ where $N_{\pi}$ is the
conductor of $\pi,$

$\varepsilon(\pi,\chi)=c_{\pi}w_{\pi}(q)\chi(N_{\pi})\cdot\frac{\tau(\chi)^{4}}{q^{2}}\
,\ \tau(\chi):=\sum_{x\pmod{q}}\chi(x)e\Big{(}\frac{x}{q}\Big{)},$

where $|c_{\pi}|=1$, $w_{\pi}(q)$ is the central unitary character of $\pi$,
$N_{\pi}$ is the conductor of $\pi$ and $\tau(\chi)$ denotes the Gauss sum
associated to $\chi$. Therefore, for $(m,q)=1$,

(4)
$\frac{1}{\sqrt{q}}\cdot\mathcal{E}_{\pi}(m;q)=c_{\pi}w_{\pi}(q)\mathcal{T}_{4}(N_{\pi}\overline{m};q)$

where

$\mathcal{T}_{k}(\ell;q):=\frac{1}{\sqrt{q}}\sum_{\begin{subarray}{c}\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\frac{\tau(\chi)^{k}}{q^{k/2}}\chi(\ell).$

For $(m,q)\neq 1$ even though the inverse $\overline{m}$ is undefined we
declare that

$\mathcal{T}_{4}(N_{\pi}\overline{m};q):=0$

so that (4) holds for all $m$. Notice that for $k$ even,
$\mathcal{T}_{k}(\ell;q)$ is real-valued because

$\overline{\tau(\chi)^{k}\chi(l)}=\tau(\overline{\chi})^{k}\overline{\chi}(l)$

since $\chi(-1)^{k}=1$ when $k\in 2\mathbb{Z}$. To a first approximation
$\mathcal{T}_{k}(\ell;q)$ is a hyper-Kloosterman sum, this would be
identically the case if in the definition of $\mathcal{T}_{k}$ the sum over
primitive characters were replaced by a sum over all characters.

######  Lemma 1.

Let $r,s$ be squarefree with $(r,s)=1$. Then, for $k\geq 1$,

$\mathcal{T}_{k}(m;rs)=\mathcal{T}_{k}(mr^{k};s)\mathcal{T}_{k}(ms^{k};r).$

###### Proof.

Any primitive character $\chi\pmod{rs}$ can be written as $\psi\nu$ with
$\psi\pmod{r}$ and $\nu\pmod{s}$. Furthermore

$\tau(\psi\nu)=\tau(\psi)\tau(\nu)\nu(r)\psi(s).$

Therefore,

$\displaystyle\mathcal{T}_{k}(\ell;rs)$
$\displaystyle=\frac{1}{\sqrt{rs}}\sum_{\begin{subarray}{c}\psi\pmod{r}\\\
\nu\pmod{s}\\\ \text{all
primitive}\end{subarray}}\frac{\tau(\psi\nu)^{k}}{r^{k/2}s^{k/2}}(\psi\nu)(\ell)$
$\displaystyle=\Big{(}\frac{1}{\sqrt{r}}\sum_{\begin{subarray}{c}\psi\pmod{r}\\\
\text{primitive}\end{subarray}}\frac{\tau(\psi)^{k}}{r^{k/2}}\psi(s^{k}\ell)\Big{)}\cdot\Big{(}\frac{1}{\sqrt{s}}\sum_{\begin{subarray}{c}\nu\pmod{s}\\\
\text{primitive}\end{subarray}}\frac{\tau(\nu)^{k}}{s^{k/2}}\nu(r^{k}\ell)\Big{)}$
$\displaystyle=\mathcal{T}(s^{k}\ell;r)\mathcal{T}(r^{k}\ell;s)$

as claimed. ∎

Using this proposition we obtain the following.

######  Lemma 2.

Let $(r,s_{1}s_{2})=1$ with $r$, $s_{1}$ and $s_{2}$ squarefree. Let
$v_{1},v_{2}$ be integers with $(v_{1}v_{2},rs_{1}s_{2})=1$. Let
$d=(s_{1},s_{2})$ and $s_{i}^{\star}=s_{i}/d,$ $i=1,2$. Then, for $k\geq 1$,

$\displaystyle\mathcal{K}_{k}(v_{1},v_{2},\ell;r,s_{1},s_{2})$
$\displaystyle:=\frac{1}{\sqrt{rs_{1}s_{2}}}\sum_{\begin{subarray}{c}x\pmod{rs_{1}s_{2}}\\\
(x,rs_{1}s_{2})=1\end{subarray}}\mathcal{T}_{k}(v_{1}x;rs_{1})\mathcal{T}_{k}(v_{2}x;rs_{2})e\Big{(}\frac{\ell
x}{rs_{1}s_{2}}\Big{)}$

is zero unless $(\ell,s_{1}^{\star}s_{2}^{\star})=1$ and $d|\ell$. In the
remaining case it is equal to,

$\displaystyle\sqrt{d}$
$\displaystyle\mathcal{T}_{k+1}(v_{1}\overline{\ell}s_{2}^{\star}r^{k+1}d^{k+2};s_{1}^{\star})\mathcal{T}_{k+1}(v_{2}\overline{\ell}s_{1}^{\star}r^{k+1}d^{k+2};s_{2}^{\star})$
$\displaystyle\times\mathcal{K}_{k}(v_{1}{s_{1}^{\star}}^{k+1}s_{2}^{\star},v_{2}s_{1}^{\star}{s_{2}^{\star}}^{k+1},\ell/d;rd,1,1)$

###### Proof.

Since $(r,s_{1}s_{2})=1$ with $s_{1}$ and $s_{2}$ squarefree, $rd$ is coprime
to $s_{1}^{\star}s_{2}^{\star}$. By Lemma 1

$\mathcal{T}_{k}(v_{i}x;rs_{i})=\mathcal{T}_{k}(v_{i}x;rds_{i}^{\star})=\mathcal{T}_{k}({s_{i}^{\star}}^{k}v_{i}x;rd)\mathcal{T}_{k}(v_{i}xr^{k}d^{k};s_{i}^{\star}).$

We express $(x,rs_{1}s_{2})=1$ as $x=ard^{2}+bs_{1}^{\star}s_{2}^{\star}$ with
$(a,s_{1}^{\star}s_{2}^{\star})=1$ and $(b,rd)=1$, $b\pmod{rd^{2}}$. In this
way,

$\mathcal{T}_{k}(v_{i}x;rs_{i})=\mathcal{T}_{k}({s_{i}^{\star}}^{k}v_{i}x;rd)\mathcal{T}_{k}(v_{i}xr^{k}d^{k};s_{i}^{\star})=\mathcal{T}_{k}(s_{1}^{\star}s_{2}^{\star}{s_{i}^{\star}}^{k}v_{i}b;rd)\mathcal{T}_{k}(v_{i}ar^{k+1}d^{k+2};s_{i}^{\star})$

Thus, we find,

$\displaystyle\mathcal{K}_{k}(v_{1},v_{2},\ell;r,s_{1},s_{2})$
$\displaystyle=\Big{(}\frac{1}{\sqrt{s_{1}^{\star}s_{2}^{\star}}}\sum_{\begin{subarray}{c}a\pmod{s_{1}^{\star}s_{2}^{\star}}\\\
(a,s_{1}^{\star}s_{2}^{\star})=1\end{subarray}}\mathcal{T}_{k}(v_{1}ar^{k+1}d^{k+2};s_{1}^{\star})\mathcal{T}_{k}(v_{2}ar^{k+1}d^{k+2};s_{2}^{\star})e\Big{(}\frac{\ell
a}{s_{1}^{\star}s_{2}^{\star}}\Big{)}\Big{)}$
$\displaystyle\times\Big{(}\frac{1}{\sqrt{r}d}\sum_{\begin{subarray}{c}b\pmod{rd^{2}}\\\
(b,rd)=1\end{subarray}}\mathcal{T}_{k}({s_{1}^{\star}}^{k+1}s_{2}^{\star}v_{1}b;rd)\mathcal{T}(s_{1}^{\star}{s_{2}^{\star}}^{k+1}v_{2}b;rd)e\Big{(}\frac{\ell
b}{rd^{2}}\Big{)}\Big{)}$

In the first sum we can write $a=xs_{1}^{\star}+ys_{2}^{\star}$ with
$(x,s_{2}^{\star})=1=(y,s_{1}^{\star})$ thus factoring the sum further as

$\Big{(}\frac{1}{\sqrt{s_{2}^{\star}}}\sum_{\begin{subarray}{c}x\pmod{s_{2}^{\star}}\\\
(x,s_{2}^{\star})=1\end{subarray}}\mathcal{T}_{k}(v_{2}xs_{1}^{\star}r^{k+1}d^{k+2};s_{2}^{\star})e\Big{(}\frac{\ell
x}{s_{2}^{\star}}\Big{)}\Big{)}\cdot\Big{(}\frac{1}{\sqrt{s_{1}^{\star}}}\sum_{\begin{subarray}{c}y\pmod{s_{1}^{\star}}\\\
(y,s_{1}^{\star})=1\end{subarray}}\mathcal{T}_{k}(v_{1}ys_{2}^{\star}r^{k+1}d^{k+2};s_{1}^{\star})e\Big{(}\frac{\ell
y}{s_{1}^{\star}}\Big{)}\Big{)}.$

Using the definition of $\mathcal{T}_{k}$ we then find for $i,j\in\\{1,2\\}$
with $i\neq j$,

$\displaystyle\frac{1}{\sqrt{s_{i}^{\star}}}\sum_{\begin{subarray}{c}x\pmod{s_{i}^{\star}}\\\
(x,s_{i}^{\star})=1\end{subarray}}$
$\displaystyle\mathcal{T}_{k}(v_{i}xs_{j}^{\star}r^{k+1}d^{k+2};s_{i}^{\star})e\Big{(}\frac{\ell
x}{s_{i}^{\star}}\Big{)}$
$\displaystyle=\frac{1}{\sqrt{s_{i}^{\star}}}\sum_{\begin{subarray}{c}\chi\pmod{s_{i}^{\star}}\\\
\text{primitive}\end{subarray}}\frac{\tau(\chi)^{k}}{{s_{i}^{\star}}^{k/2}}\sum_{\begin{subarray}{c}x\pmod{s_{i}^{\star}}\\\
(x,s_{i}^{\star})=1\end{subarray}}\chi(v_{i}xs_{j}^{\star}r^{k+1}d^{k+2})e\Big{(}\frac{\ell
x}{s_{i}^{\star}}\Big{)}$

This evaluates to

$\mathbf{1}_{(\ell,s_{i}^{\star})=1}\sum_{\begin{subarray}{c}\chi\pmod{s_{i}^{\star}}\\\
\text{primitive}\end{subarray}}\frac{\tau(\chi)^{k+1}}{{s_{i}^{\star}}^{(k+1)/2}}\cdot\chi(v_{i}\overline{\ell}s_{j}^{\star}r^{k+1}d^{k+2})=\mathcal{T}_{k+1}(v_{i}\overline{\ell}s_{j}^{\star}r^{k+1}d^{k+2};s_{i}^{\star})\mathbf{1}_{(\ell,s_{i}^{\star})=1}$

by the fact that the Gauss sum

$\displaystyle\sum_{\begin{subarray}{c}x\pmod{s_{i}^{\star}}\\\
(x,s_{i}^{\star})=1\end{subarray}}\chi(x)e\Big{(}\frac{\ell
x}{s_{i}^{\star}}\Big{)}=0$

unless $(\ell,s_{i}^{\star})\neq 1.$ Finally it remains to evaluate

$\frac{1}{\sqrt{r}d}\sum_{\begin{subarray}{c}b\pmod{rd^{2}}\\\
(b,rd)=1\end{subarray}}\mathcal{T}_{k}({s_{1}^{\star}}^{k+1}s_{2}^{\star}v_{1}b;rd)\mathcal{T}(s_{1}^{\star}{s_{2}^{\star}}^{k+1}v_{2}b;rd)e\Big{(}\frac{\ell
b}{rd^{2}}\Big{)}.$

Write $b=x+yrd$ with $y\pmod{d}$ and $x\pmod{rd}$ with $(x,rd)=1$. Plugging
this inside the sum evaluates to

$\frac{1}{\sqrt{r}d}\sum_{y\pmod{d}}e\Big{(}\frac{\ell
y}{d}\Big{)}\sum_{\begin{subarray}{c}x\pmod{rd}\\\
(x,rd)=1\end{subarray}}\mathcal{T}_{k}({s_{1}^{\star}}^{k+1}s_{2}^{\star}v_{1}x;rd)\mathcal{T}(s_{1}^{\star}{s_{2}^{\star}}^{k+1}v_{2}x;rd)e\Big{(}\frac{\ell
x}{rd^{2}}\Big{)}.$

This is zero unless $d\mid\ell$. And in the case when $d|\ell$ we can rewrite
the above as

$\sqrt{d}\mathcal{K}_{k}(v_{1}{s_{1}^{\star}}^{k+1}s_{2}^{\star},v_{2}s_{1}^{\star}{s_{2}^{\star}}^{k+1},\ell/d;rd,1,1).$

∎

We now record bounds for $\mathcal{T}_{k}$ and $\mathcal{K}_{k}$.

######  Lemma 3.

Let $q$ be squarefree and $\ell\in\mathbb{Z}$. Then, for $k\geq 2$,

$|\mathcal{T}_{k}(\ell;q)|\leq C^{\omega(q)}$

for some constant $C>10$ depending only on $k$.

###### Proof.

For $(\ell,q)\neq 1$ we have $\mathcal{T}_{k}(\ell;q)=0$. We can therefore
assume $(\ell,q)=1$. Using Lemma 1 we see that it’s enough to bound
$\mathcal{T}_{k}(v;p)$ for $(v,p)=1$. Note that

$\displaystyle\mathcal{T}_{k}(v;p)$
$\displaystyle=\frac{1}{\sqrt{p}}\sum_{\chi\pmod{p}}\frac{\tau(\chi)^{k}}{p^{k/2}}\chi(v)-\frac{(-1)^{k}}{p^{(k+1)/2}}$
$\displaystyle=\frac{\varphi(p)}{p^{(k+1)/2}}\sum_{x_{1}\ldots
x_{k}\equiv\overline{v}\pmod{p}}e\Big{(}\frac{x_{1}+\ldots+x_{k}}{p}\Big{)}-\frac{(-1)^{k}}{p^{(k+1)/2}}.$

and this $\ll_{k}1$ by Deligne’s bound. Therefore by twisted multiplicativity
(i.e Lemma 1) we get a bound of $\leq C^{\omega(q)}$ for $q$ squarefree and
with a constant $C$ depending solely on $k$. ∎

We also have a similar bound for $\mathcal{K}_{k}$.

######  Lemma 4.

Let $q$ be squarefree and $\ell\in\mathbb{Z}$. Then for $k\geq 2$ with
$k\equiv 0\pmod{2}$,

$|\mathcal{K}_{k}(v_{1},v_{2},\ell;q,1,1)|\leq\sqrt{(\ell,q,v_{1}-v_{2})}\cdot
C^{\omega(q)}$

for some constant $C>10$ depending only on $k$.

######  Remark 3.

Though it’s irrelevant for our purpose $C^{\omega(q)}$ can be improved to
$C^{\omega(q/(q,\ell))}$.

###### Proof.

First note that if $(v_{1}v_{2},q)\neq 1$ then there is nothing to prove since
in that case,

$\mathcal{K}_{k}(v_{1},v_{2},\ell;q,1,1)=0.$

Therefore we can assume that $(v_{1}v_{2},q)=1$. Suppose $q=rs$ with $r,s\geq
1$ and $(r,s)=1$. Recall that,

$\mathcal{K}_{k}(v_{1},v_{2},\ell;rs,1,1)=\frac{1}{\sqrt{rs}}\sum_{\begin{subarray}{c}x\pmod{rs}\\\
(x,rs)=1\end{subarray}}\mathcal{T}_{k}(v_{1}x;rs)\mathcal{T}_{k}(v_{2}x;rs)e\Big{(}\frac{\ell
x}{rs}\Big{)}.$

Write $x=ar+bs$ with $(a,s)=1$ and $(b,r)=1$. Using Lemma 1

$\mathcal{T}_{k}(v_{i}x;rs)=\mathcal{T}_{k}(v_{i}xs^{k};r)\mathcal{T}_{k}(v_{i}xr^{k};s)=\mathcal{T}_{k}(v_{i}bs^{k+1};r)\mathcal{T}_{k}(v_{i}ar^{k+1};s).$

Therefore

$\displaystyle\mathcal{K}_{k}(v_{1},v_{2},\ell;q,1,1)$
$\displaystyle=\Big{(}\sum_{\begin{subarray}{c}a\pmod{s}\\\
(a,s)=1\end{subarray}}\mathcal{T}_{k}(v_{1}ar^{k+1};s)\mathcal{T}_{k}(v_{2}ar^{k+1};s)e\Big{(}\frac{\ell
a}{s}\Big{)}\Big{)}$
$\displaystyle\times\Big{(}\sum_{\begin{subarray}{c}b\pmod{r}\\\
(b,r)=1\end{subarray}}\mathcal{T}_{k}(v_{1}bs^{k+1};r)\mathcal{T}_{k}(v_{2}bs^{k+1};r)e\Big{(}\frac{\ell
b}{r}\Big{)}\Big{)}$
$\displaystyle=\mathcal{K}_{k}(v_{1}s^{k+1},v_{2}s^{k+1},\ell;r,1,1)\mathcal{K}_{k}(v_{1}r^{k+1},v_{2}r^{k+1},\ell;s,1,1)$

Therefore it is enough to bound $\mathcal{K}_{k}(v_{1},v_{2},\ell;p,1,1)$ with
$p$ prime. We notice that,

(5) $\displaystyle\mathcal{K}_{k}(v_{1},v_{2},\ell;p,1,1)$
$\displaystyle=\frac{1}{\sqrt{p}}\sum_{\begin{subarray}{c}x\pmod{p}\\\
(x,p)=1\end{subarray}}\mathcal{T}_{k}(v_{1}x;p)\mathcal{T}_{k}(v_{2}x;p)e\Big{(}\frac{\ell
x}{p}\Big{)}.$

By definition

$\mathcal{T}_{k}(v_{1}x;p)=\frac{p-1}{p}K_{k}(\overline{v_{1}x};p)-\frac{(-1)^{k}}{p^{(k+1)/2}}$

where $K_{k}(v;p)$ denotes a hyper-Kloosterman sum

$K_{k}(v;p):=\frac{1}{p^{(k-1)/2}}\sum_{\begin{subarray}{c}x_{1}\ldots
x_{k}\equiv
u\pmod{p}\end{subarray}}e\Big{(}\frac{x_{1}+\ldots+x_{k}}{p}\Big{)}.$

Therefore, using Deligne’s bound $K_{k}(u;p)\ll_{k}1$, we find,

$\displaystyle\mathcal{K}_{k}(v_{1},v_{2},\ell;p,1,1)=\frac{1}{\sqrt{p}}\sum_{\begin{subarray}{c}x\pmod{p}\\\
(x,p)=1\end{subarray}}K_{k}(\overline{v_{1}x};p)K_{k}(\overline{v_{2}x};p)e\Big{(}\frac{\ell
x}{p}\Big{)}+O_{k}\Big{(}\frac{1}{\sqrt{p}}\Big{)}.$

If $v_{1}\not\equiv v_{2}\pmod{p}$ then by [FKM15, Corollary 3.3] this is
$\ll_{k}1$ since $k$ is even so that $K_{k}$ is self-dual. Moreover if
$\ell\not\equiv 0\pmod{p}$ then also by [FKM15, Corollary 3.3] we have

$|\mathcal{K}_{k}(v_{1},v_{2},\ell;p,1,1)|\ll_{k}1.$

There remains the case when $p|\ell$ and $v_{1}\equiv v_{2}\pmod{p}$. In this
case, by definition of $\mathcal{T}_{k}$ the exponential sum (5) can be
rewritten as

$\displaystyle\frac{1}{\sqrt{p}}\sum_{\begin{subarray}{c}x\pmod{p}\\\
(x,p)=1\end{subarray}}\Big{|}\frac{1}{\sqrt{p}}\sum_{\begin{subarray}{c}\chi\pmod{p}\\\
\text{primitive}\end{subarray}}\frac{\tau(\chi)^{k}}{p^{k/2}}\chi(v_{1}x)\Big{|}^{2}$
$\displaystyle=\frac{\varphi(p)}{p^{3/2}}\sum_{\begin{subarray}{c}\chi\pmod{p}\\\
\text{primitive}\end{subarray}}\Big{|}\frac{\tau(\chi)^{k}}{p^{k/2}}\Big{|}^{2}\leq\sqrt{p}$

And this gives the bound
$|\mathcal{K}_{k}(v_{1},v_{2},\ell;p,1,1)|\leq\sqrt{p}$ in the case $p|\ell$
and $v_{1}\equiv v_{2}\pmod{p}$. Using twisted multiplicativity we then obtain
the claim. ∎

We are now ready to state our bounds for the exponential sums
$\mathcal{K}_{4}$.

######  Lemma 5.

Let $r,s_{1},s_{2}$ be squarefree with $(r,s_{1}s_{2})=1$. Let
$v_{1},v_{2}\in\mathbb{Z}$. Write $d=(s_{1},s_{2})$ and
$s_{i}^{\star}:=s_{i}/d$ for $i\in\\{1,2\\}$. Then,

$\mathcal{K}_{4}(v_{1},v_{2},\ell;r,s_{1},s_{2})\leq\mathbf{1}_{d|\ell}\cdot\mathbf{1}_{(\ell,s_{1}^{\star}s_{2}^{\star})=1}\cdot\sqrt{d}C^{\omega(rs_{1}s_{2})}\cdot\sqrt{(\ell/d,rd,v_{1}{s_{1}^{\star}}^{4}-v_{2}{s_{2}^{\star}}^{4})}$

with $C>10$ a constant depending only on $k$.

###### Proof.

Upon combining Lemma 2, Lemma 3 and Lemma 4 we obtain

$\mathcal{K}_{4}(v_{1},v_{2},\ell;r,s_{1},s_{2})\leq\mathbf{1}_{d|\ell}\cdot\mathbf{1}_{(\ell,s_{1}^{\star}s_{2}^{\star})=1}\cdot\sqrt{d}C^{\omega(rs_{1}s_{2})}\cdot\sqrt{(\ell/d,rd,v_{1}{s_{1}^{\star}}^{5}s_{2}^{\star}-v_{2}{s_{2}^{\star}}^{5}s_{1}^{\star})}$

We then simplify this bound by noting that since
$(rd,s_{1}^{\star}s_{2}^{\star})=1$ we have,

$(\ell/d,rd,v_{1}{s_{1}^{\star}}^{5}s_{2}^{\star}-v_{2}{s_{2}^{\star}}^{5}s_{1}^{\star})=(\ell/d,rd,v_{1}{s_{1}^{\star}}^{4}-v_{2}{s_{2}^{\star}}^{4}).$

∎

We also record the following simple (diagonal) case.

######  Lemma 6.

Let $(r,s)=1$. Then,

$|\mathcal{K}_{4}(v,v,0;r,s,s)|\leq\sqrt{r}s.$

###### Proof.

For $(v,rs)\neq 1$ there is nothing to show because
$\mathcal{K}_{4}(v,v,0;r,s,s)=0$. Hence, assume that $(v,rs)=1$. Note that,

$\mathcal{K}_{k}(v,v,0;r,s,s)=\frac{1}{\sqrt{r}s}\sum_{x\pmod{rs^{2}}}|\mathcal{T}_{4}(vx;rs)|^{2}$

Recall that

$\mathcal{T}_{4}(x;q):=\frac{1}{\sqrt{q}}\sum_{\begin{subarray}{c}\chi\pmod{q}\\\
\text{primitive}\end{subarray}}\frac{\tau(\chi)^{4}}{q^{2}}\chi(x).$

Therefore, by orthogonality of characters,

$\mathcal{K}_{k}(v,v,0;r,s,s)=\frac{\varphi(rs^{2})}{r^{3/2}s^{2}}\sum_{\begin{subarray}{c}\chi\pmod{rs}\\\
\text{primitive}\end{subarray}}\Big{|}\frac{\tau(\chi)^{k}}{q^{k/2}}\Big{|}^{2}\leq\sqrt{r}s$

as claimed. ∎

With all this information in place we are finally ready to bound exponential
sums with $\mathcal{E}_{\pi}$.

######  Lemma 7.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $v_{1},v_{2}\in\mathbb{Z}$.
Let $r,s_{1},s_{2}$ be squarefree with $(r,s_{1}s_{2})=1$,
$(rs_{1}s_{2},N_{\pi})=1$. Write $d=(s_{1},s_{2})$ and set
$s_{i}^{\star}:=s_{i}/d$ for $i\in\\{1,2\\}$. For $\ell\in\mathbb{Z}$, define

$\mathcal{K}_{\pi}(v_{1},v_{2},\ell;r,s_{1},s_{2}):=\frac{1}{\sqrt{rs_{1}s_{2}}}\sum_{\begin{subarray}{c}x\pmod{rs_{1}s_{2}}\\\
(x,rs_{1}s_{2})=1\end{subarray}}\frac{\mathcal{E}_{\pi}(v_{1}x;rs_{1})}{\sqrt{rs_{1}}}\frac{\overline{\mathcal{E}_{\pi}}(v_{2}x;rs_{2})}{\sqrt{rs_{2}}}e\Big{(}\frac{\ell
x}{rs_{1}s_{2}}\Big{)}.$

Then,

$\displaystyle\mathcal{K}_{\pi}$
$\displaystyle(v_{1},v_{2},\ell;r,s_{1},s_{2})$
$\displaystyle\ll\sqrt{N_{\pi}}\mathbf{1}_{d|\ell}\cdot\mathbf{1}_{(\ell,s_{1}^{\star}s_{2}^{\star})=1}\cdot\sqrt{d}\cdot
C^{\omega(rs_{1}s_{2})}\sqrt{(\ell/d,rd,v_{1}{s_{1}^{\star}}^{4}-v_{2}{s_{2}^{\star}}^{4})}$

Also, for $(r,s)=1$ with $(rs,N_{\pi})=1$ and $v\in\mathbb{Z}$,

$|\mathcal{K}_{\pi}(v,v,0;r,s,s)|\leq\sqrt{r}s.$

###### Proof.

Without loss of generality we can assume that $(v_{1}v_{2},rs_{1}s_{2})=1$,
otherwise $\mathcal{K}_{\pi}(v_{1},v_{2},\ell;r,s_{1},s_{2})=0$ and we are
done. The first assertion follows Lemma 5, the observation that,

$\frac{1}{\sqrt{q}}\mathcal{E}_{\pi}(v;q)=c_{\pi}w_{\pi}(q)\mathcal{T}_{4}(\overline{v}N_{\pi};q)$

with $w_{\pi}(q)$ the central character of $\pi$ and $|c_{\pi}|=1$ and the
observation that $\mathcal{T}_{4}(v\overline{N_{\pi}};q)$ is real-valued. The
second assertion follows by using Lemma 6 instead. ∎

##  3\. Proof of Proposition 1

Proposition 1 will follow quickly from the next Lemma.

######  Lemma 8.

Let $\delta>0$, $M,N\geq 1$ and $V$ be a smooth function compactly supported
in $[1/100,100]$. Let $\mathcal{S}\subset[S/4,4S]$ be a set of integers such
that all $s\in\mathcal{S}$ are square-free and have at most $\delta\log\log
S+20$ prime factors. Let $r$ be a square-free integer co-prime to all elements
of $\mathcal{S}$, composed of at most $20$ distinct prime factors and
contained in $[R/4,4R]$. Suppose that $S>R^{10}$. Then,

$\displaystyle\sum_{s_{1},s_{2}\in\mathcal{S}}$
$\displaystyle\Big{|}\sum_{(m,rs_{1}s_{2})=1}V\Big{(}\frac{m}{M}\Big{)}\frac{\mathcal{E}_{\pi}(\pm
m;rs_{1})}{\sqrt{rs_{1}}}\frac{\overline{\mathcal{E}_{\pi}}(\pm
m;rs_{2})}{\sqrt{rs_{2}}}\Big{|}$
$\displaystyle\ll\Big{(}(\log(RS))^{C\delta}\cdot S\sqrt{R}\cdot\Big{(}S(\log
S)^{-1+\delta+\delta\log\frac{1}{\delta}}\Big{)}^{2}+MS(\log
S)^{-1+\delta+\delta\log\frac{1}{\delta}}\Big{)}\|V\|_{\infty,2}$

where

$\|V\|_{p,r}:=\sum_{i=0}^{r}\|V^{(i)}\|_{p}$

######  Remark 4.

In the above $(\log(RS))^{C\delta}$ is the loss from applying Deligne’s bound
to integers having many prime factors. The term $S\mathcal{L}_{\delta}:=S(\log
S)^{-1+\delta+\delta\log\frac{1}{\delta}}$ is an upper bound for the number of
square-free integers in $[S/4,4S]$ that have $\leq\delta\log\log S+20$
distinct prime factors. The term $S\sqrt{R}\cdot(S\mathcal{L}_{\delta})^{2}$
corresponds to the bound for the off-diagonal. The term
$MS\mathcal{L}_{\delta}$ accounts for the diagonal.

###### Proof.

By Poisson summation,

$\displaystyle\sum_{(m,rs_{1}s_{2})=1}$ $\displaystyle
V\Big{(}\frac{m}{M}\Big{)}\frac{\mathcal{E}_{\pi}(\pm
m;rs_{1})}{\sqrt{rs_{1}}}\frac{\overline{\mathcal{E}_{\pi}}(\pm
m;rs_{2})}{\sqrt{rs_{2}}}$
$\displaystyle=\frac{M}{rs_{1}s_{2}}\sum_{\ell}\widehat{V}\Big{(}\frac{\ell
M}{rs_{1}s_{2}}\Big{)}\Big{(}\sum_{(x,rs_{1}s_{2})=1}\frac{\mathcal{E}_{\pi}(\pm
x;rs_{1})}{\sqrt{rs_{1}}}\frac{\overline{\mathcal{E}_{\pi}}(\pm
x;rs_{2})}{\sqrt{rs_{2}}}e\Big{(}\frac{\ell x}{rs_{1}s_{2}}\Big{)}\Big{)}.$

###  3.1. The case $\ell=0$

This corresponds to

$\widehat{V}(0)\sum_{\begin{subarray}{c}s_{1},s_{2}\in\mathcal{S}\\\
d:=(s_{1},s_{2})\\\
s_{i}^{\star}:=s_{i}/d\end{subarray}}\frac{M}{\sqrt{rs_{1}s_{2}}}\cdot\frac{1}{\sqrt{rs_{1}s_{2}}}\sum_{(x,rs_{1}s_{2})=1}\frac{\mathcal{E}_{\pi}(\pm
x;rs_{1})}{\sqrt{rs_{1}}}\frac{\overline{\mathcal{E}_{\pi}}(\pm
x;rs_{2})}{\sqrt{rs_{2}}}.$

According to Lemma 7 (with $\ell=0$), if this is non-zero then
$1=(0,s_{1}^{\star}s_{2}^{\star})=s_{1}^{\star}s_{2}^{\star}$. However since
$s_{i}^{\star}=s_{i}/d$ this implies that $s_{1}=s_{2}$. Therefore the above
simplifies to

$\widehat{V}(0)\sum_{s\in\mathcal{S}}\frac{M}{\sqrt{r}s}\cdot\Big{(}\frac{1}{\sqrt{r}s}\sum_{\begin{subarray}{c}x\pmod{rs^{2}}\\\
(x,rs)=1\end{subarray}}\Big{|}\frac{\mathcal{E}_{\pi}(\pm
x;rs)}{\sqrt{rs}}\Big{|}^{2}\Big{)}.$

Again, by Lemma 7 the exponential sum inside the brackets is $\leq\sqrt{r}s$.
Therefore, the above expression is,

$\ll\|V\|_{\infty}\cdot\frac{M}{\sqrt{R}S}\sum_{s_{1}\in\mathcal{S}}\sqrt{R}S\ll\|V\|_{\infty}\cdot
MS(\log S)^{-1+\delta+\delta\log\frac{1}{\delta}},$

as needed.

###  3.2. The case $\ell\neq 0$

In this case we aim to bound,

$\sum_{\begin{subarray}{c}s_{1},s_{2}\in\mathcal{S}\\\ d:=(s_{1},s_{2})\\\
s_{i}^{\star}:=s_{i}/d\end{subarray}}\frac{M}{\sqrt{rs_{1}s_{2}}}\sum_{\ell\neq
0}\Big{|}\widehat{V}\Big{(}\frac{\ell
M}{rs_{1}s_{2}}\Big{)}\Big{|}\cdot\Big{|}\mathcal{K}_{\pi}(\pm 1,\pm
1,\ell;r,s_{1},s_{2})|$

using the notation of Lemma 7. By Lemma 7 this is bounded above by,

$\ll\sum_{\begin{subarray}{c}s_{1},s_{2}\in\mathcal{S}\\\ d:=(s_{1},s_{2})\\\
s_{i}^{\star}:=s_{i}/d\end{subarray}}\frac{M}{\sqrt{rs_{1}s_{2}}}\sum_{\begin{subarray}{c}\ell\neq
0\\\ d|\ell\end{subarray}}\Big{|}\widehat{V}\Big{(}\frac{\ell
M}{rs_{1}s_{2}}\Big{)}\Big{|}\sqrt{d}C^{\omega(rs_{1}s_{2})}\sqrt{(\ell/d,rd,{s_{1}^{\star}}^{4}-{s_{2}^{\star}}^{4})}$

We notice that due to the assumption on the number of prime factors of
$r,s_{1},s_{2}$ we have,

$C^{\omega(rs_{1}s_{2})}\ll(\log(RS))^{C\delta}.$

We bound,

$\sqrt{d}\cdot\sqrt{(\ell/d,rd,{s_{1}^{\star}}^{4}-{s_{2}^{\star}}^{4})}\leq
d\cdot\sqrt{(r,\ell/d)}\leq d\sum_{\begin{subarray}{c}f|r\\\
df|\ell\end{subarray}}\sqrt{f}$

We then find,

$\displaystyle(\log(RS))^{C\delta}$
$\displaystyle\sum_{\begin{subarray}{c}ds_{1},ds_{2}\in\mathcal{S}\\\
(d,s_{1}s_{2})=1\end{subarray}}\frac{Md}{\sqrt{r}S}\sum_{f|r}\sqrt{f}\sum_{\begin{subarray}{c}\ell\neq
0\\\ df|\ell\end{subarray}}\Big{|}\widehat{V}\Big{(}\frac{\ell
M}{rd^{2}s_{1}s_{2}}\Big{)}\Big{|}$

The sum over $\ell\neq 0$ is then easily bounded; write $\ell=df\ell^{\prime}$
and use the bounds

$\Big{|}\widehat{V}\Big{(}\frac{\ell^{\prime}dfM}{rd^{2}s_{1}s_{2}}\Big{)}\Big{|}\ll\|V\|_{\infty,2}\cdot\begin{cases}1&\text{
if }|\ell^{\prime}|\leq RS^{2}/(dfM)\\\
\frac{R^{2}S^{4}}{d^{2}f^{2}M^{2}}\cdot\frac{1}{\ell^{\prime 2}}&\text{ if
}|\ell^{\prime}|>RS^{2}/(dfM).\end{cases}$

Therefore the previous bound becomes

$\displaystyle\ll\|V\|_{\infty,2}\cdot(\log(RS))^{C\delta}\sum_{\begin{subarray}{c}ds_{1},ds_{2}\in\mathcal{S}\\\
(d,s_{1}s_{2})=1\end{subarray}}\frac{Md}{\sqrt{r}S}\sum_{f|r}\sqrt{f}\cdot\frac{RS^{2}}{dfM}$

Since $\omega(r)\leq 20$ and $r$ is square-free, we have,

$\sum_{f|r}\frac{1}{\sqrt{f}}\leq 100.$

It follows that the above expression is bounded by

$\|V\|_{\infty,2}(\log(RS))^{C\delta}\sqrt{R}S\Big{(}\sum_{\begin{subarray}{c}ds\in\mathcal{S}\\\
(d,s)=1\end{subarray}}1\Big{)}^{2}$

It remains to notice that,

$\sum_{\begin{subarray}{c}ds\in\mathcal{S}\\\
(d,s)=1\end{subarray}}1\leq\sum_{s\in\mathcal{S}}2^{\omega(s)}\ll(\log
S)^{\delta}S(\log S)^{-1+\delta+\delta\log\frac{1}{\delta}}$

using the fact that $\omega(s)\leq\delta\log\log S+20$. ∎

We are now ready to prove Proposition 1.

###### Proof of Proposition 1.

Using the first Lemma,

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\
\chi\pmod{q}\end{subarray}}\varepsilon(\pi,\chi)\chi(\pm
1)\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)\overline{\chi}(m)}{\sqrt{m}}V\Big{(}\frac{m}{M}\Big{)}=\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)}{\sqrt{m}}V\Big{(}\frac{m}{M}\Big{)}\sum_{q\in\mathcal{Q}}\mathcal{E}_{\pi}(\pm
m;q).$

Moreover by definition of $\mathcal{Q}$,

$\sum_{q\in\mathcal{Q}}\mathcal{E}_{\pi}(\pm
m;q)=\sum_{\begin{subarray}{c}r\in\mathcal{R}\\\
s\in\mathcal{S}\end{subarray}}\mathcal{E}_{\pi}(\pm m;rs).$

We can therefore re-write the previous sum as

$\sum_{r\in\mathcal{R}}\sqrt{r}\sum_{m}\frac{\lambda_{\tilde{\pi}}(m)}{\sqrt{m}}V\Big{(}\frac{m}{M}\Big{)}\sum_{s\in\mathcal{S}}\sqrt{s}\cdot\frac{\mathcal{E}_{\pi}(\pm
m;rs)}{\sqrt{rs}}$

Ignoring the sum over $r\in\mathcal{R}$, applying Cauchy-Schwarz and the
Rankin-Selberg bound to the sum over $m$ the above is less than

$\|V\|_{\infty}^{1/2}\cdot
R^{3/2}\cdot\max_{r\in\mathcal{R}}\Big{(}\sum_{m}V\Big{(}\frac{m}{M}\Big{)}\Big{|}\sum_{s\in\mathcal{S}}\sqrt{s}\cdot\frac{\mathcal{E}_{\pi}(\pm
m;rs)}{\sqrt{rs}}\Big{|}^{2}\Big{)}^{1/2}.$

Expanding the square and applying the previous Lemma we conclude with the
following bound

$\displaystyle\ll\|V\|_{\infty,2}\cdot
R^{3/2}\cdot\Big{(}(\log(RS))^{C\delta}R^{1/4}S^{2}(\log
S)^{-1+\delta+\delta\log\frac{1}{\delta}}+\sqrt{M}S(\log
S)^{-\frac{1}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}}\Big{)}$
$\displaystyle\asymp\|V\|_{\infty,2}\cdot\Big{(}(\log Q)^{C\delta}\cdot
Q^{2}(\log Q)^{-1+\delta+\delta\log\frac{1}{\delta}}R^{-1/4}+\sqrt{MR}\cdot
Q(\log
Q)^{-\frac{1}{2}+\frac{\delta}{2}+\frac{\delta}{2}\log\frac{1}{\delta}}\Big{)}$

as claimed. ∎

##  4\. Character sums & Proof of Proposition 3

We collect in this section a number of “generic” estimates for character sums.

######  Lemma 9.

Let $\alpha(n)$ be a sequence supported in $[N/4,4N]$ with $N>4$. Let
$\mathcal{Q}\subset[Q/16,16Q]$ be a set of moduli and
$\mathcal{D}\subset[1,16Q]$ an arbitrary subset of the integer. Then,

$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
d\in\mathcal{D}\end{subarray}}\mu(c)\varphi(d)\sum_{n\equiv\pm
1\pmod{d}}\frac{\alpha(n)}{\sqrt{n}}\ll Q\sqrt{N}(\log
N)^{3/2}\Big{(}\sum_{n}\frac{|\alpha(n)|^{2}}{n}\Big{)}^{1/2}.$

######  Remark 5.

Notice that $\alpha(1)=0$. This is important since in the case when
$n\equiv\pm 1\pmod{q}$ is chosen to mean $n\equiv 1\pmod{q}$ the term $n=1$
yields a main term of size

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}1.$

###### Proof.

Replacing the condition $cd\in\mathcal{Q}$ by the condition $cd\in[Q/16,16Q]$,
and bounding trivially the above is

$\ll Q\sum_{d\leq 16Q}\frac{\varphi(d)}{d}\sum_{n\equiv\pm
1\pmod{d}}\frac{|\alpha(n)|}{\sqrt{n}}$

Bounding $\varphi(d)\leq d$ and interchanging sums the above is

$\leq Q\sum_{n}\frac{|\alpha(n)|}{\sqrt{n}}\cdot d(n\mp 1).$

The final bound now follows from applying Cauchy-Schwarz. ∎

As an immediate Corollary (with $\mathcal{D}=[1,16Q]$) we record the
following.

######  Corollary 1.

Let $\alpha(n)$ be a sequence supported in $[N/4,4N]$ with $N>10$. Let
$\mathcal{Q}\subset[Q/16,16Q]$ be a set of moduli. Then,

$\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm
1)\sum_{n}\frac{\alpha(n)\chi(n)}{\sqrt{n}}\ll Q\sqrt{N}(\log
N)^{3/2}\Big{(}\sum_{n}\frac{|\alpha(n)|^{2}}{n}\Big{)}^{1/2}$

We will also need the following estimate following from the Selberg-Delange
method.

######  Lemma 10.

Let $\mathcal{Q}$ be a set of squarefree integers $\subset[Q/16,16Q]$ that can
be written as $p_{1}p_{2}m$ with $p_{1}\sim P_{1}$, $p_{2}\sim P_{2}$ and
$m\sim Q/(P_{1}P_{2})$ having less than $\delta\log\log Q+10$ prime factors
all larger than $(\log Q)^{20000}$. We assume that $4P_{1}\leq 2P_{2}\leq(\log
Q)^{20000}$. Then,

$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)=Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}+o(1)}.$

as $Q\rightarrow\infty$.

###### Proof.

By definition of $\mathcal{Q}$ we can write,

$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)=\sum_{\begin{subarray}{c}\omega(m)\leq\delta\log\log
Q+10\\\ p|m\implies p>(\log Q)^{20000}\\\ m\sim
Q/(P_{1}P_{2})\end{subarray}}\mu^{2}(m)\sum_{\begin{subarray}{c}p_{1}\sim
P_{1}\\\ p_{2}\sim P_{2}\end{subarray}}\mathfrak{C}(p_{1}p_{2}m)$

where

$\mathfrak{C}(n):=\sum_{n=cd}\mu(c)\varphi(d).$

Since $\mathfrak{C}(n)$ is multiplicative and $p_{1},p_{2}$ and $m$ are
mutually co-prime the above re-arranges into

$\Big{(}\sum_{p_{1}\sim
P_{1}}\mathfrak{C}(p_{1})\Big{)}\cdot\Big{(}\sum_{p_{2}\sim
P_{2}}\mathfrak{C}(p_{2})\Big{)}\cdot\Big{(}\sum_{\begin{subarray}{c}\omega(m)\leq\delta\log\log
Q+10\\\ p|m\implies p>(\log Q)^{20000}\\\ m\sim
Q/(P_{1}P_{2})\end{subarray}}\mu^{2}(m)\mathfrak{C}(m)\Big{)}.$

Since $\mathfrak{C}(p)=p-2$ the sum over $P_{1}$ and $P_{2}$ is respectively
$\asymp P_{1}^{2}(\log P_{1})^{-1}$ and $\asymp P_{2}^{2}(\log P_{2})^{-1}$.
To compute the sum over $m$ we use the Selberg-Delange method. We quickly
sketch the proof without giving the full details, since those can be found in
any standard textbook e.g [Ten15]. Let $M:=Q/(P_{1}P_{2})$ and $w=(\log
Q)^{20000}$. We note that $\log\log M=\log\log Q+o(1)$. First by contour
integration, for $z$ with $\Re z=\log\frac{1}{\delta}$, we have,

$\sum_{\begin{subarray}{c}m\sim M\\\ p|m\implies
p>w\end{subarray}}\mathfrak{C}(m)e^{-z\omega(m)}=\frac{A(z)M^{2}}{\log
M}\cdot\exp\Big{(}(\log\log M-\log\log w)e^{-z}\Big{)}+O\Big{(}M^{2}(\log
M)^{\delta-2}\Big{)},$

with $A(z)$ an analytic function with $A(x)>0$ for $x\geq 0$. Subsequently we
apply the inverse Laplace transform,

$\mathbf{1}\Big{(}\omega(m)\leq\delta\log\log Q+10\Big{)}=\frac{1}{2\pi
i}\int_{\Re z=\log\frac{1}{\delta}}e^{z(\delta\log\log
Q+10)-z\omega(m)}\cdot\frac{dz}{z}$

and using the saddle-point method we obtain that the sum over $m$ is

$\asymp A\Big{(}\log\frac{1}{\delta}\Big{)}M^{2}(\log
M)^{-1+\delta+\delta\log\frac{1}{\delta}}\cdot\Big{(}\log\frac{\log M}{\log
w}\Big{)}^{-1/2}\cdot(\log w)^{-\delta}$

This gives the claim since $\log w\asymp\log\log Q$.

∎

We are now ready to prove Proposition 3.

###### Proof.

If $n=1$ falls outside of the support of $V(n/N)$ then the result follows from
Lemma 9. In the other case since $q\in\mathcal{Q}$ has all of its prime
factors larger than $1000$111we are assuming without loss of generality that
$Q$ is taken sufficiently large if $cd\in\mathcal{Q}$ then $d=1$ or $d>500$.
Therefore, if $n=1$ is in the support of $V(n/N)$,

$\displaystyle\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm
1)\sum_{n}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}$ $\displaystyle
V\Big{(}\frac{n}{N}\Big{)}=\sum_{\begin{subarray}{c}c\in\mathcal{Q}\end{subarray}}\mu(c)\sum_{n}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$
$\displaystyle+\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
d>500\end{subarray}}\mu(c)\varphi(d)\sum_{n\equiv\pm
1\pmod{d}}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$

Notice that in the case when $\pm 1=-1$ the second term is zero while the
first term is $\ll_{\pi}Q\|V\|_{\infty}$. We can therefore assume that $\pm
1=1$. Therefore the above amounts to

$\sum_{c\in\mathcal{Q}}\mu(c)\sum_{n}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}+V\Big{(}\frac{1}{N}\Big{)}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
d>500\end{subarray}}\mu(c)\varphi(d).$

This can be further re-written as

$\sum_{c\in\mathcal{Q}}\mu(c)\sum_{n\neq
1}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}+V\Big{(}\frac{1}{N}\Big{)}\sum_{cd\in\mathcal{Q}}\mu(c)\varphi(d).$

Bounding everything trivially we now find that the first term is
$\ll_{\pi}Q\|V\|_{\infty}$ while by the Selberg-Delange method,

$\sum_{cd\in\mathcal{Q}}\mu(c)\varphi(d)=Q^{2}(\log
Q)^{-1+\delta+\delta\log\frac{1}{\delta}+o(1)},$

thus giving the claim. ∎

##  5\. Roadmap and deduction of Proposition 2

The proof of Proposition 2 is a little bit more involved. We state now three
Propositions from which it follows. Their proofs occupy the remainder of the
paper. In the first Proposition we bound the contribution of integers that are
nearly primes (i.e small factor times a product of a few large primes), the
bound is essentially trivial but as accurate as possible.

######  Proposition 4.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $0<\nu,\delta<1/1000$. Let
$\mathcal{Q}\subset[Q/16,16Q]$ be a set of squarefree integers such that all
$q\in\mathcal{Q}$ have $\leq\delta\log\log Q+20$ distinct prime factors. Let
$\mathcal{E}\subset[N/16,16N]$ denote the set of integers without prime
factors in the range $[\exp(\log^{\nu}N),N^{1/1000}]$. Then, for any smooth
function $V$ compactly supported in $[1/100,100]$ and $N$ in the range
$Q^{2}(\log Q)^{-10^{9}}\leq N\leq Q^{2}(\log Q)^{10^{9}}$ we have,

$\displaystyle\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)$
$\displaystyle\sum_{\begin{subarray}{c}n\in\mathcal{E}\\\ n\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$
$\displaystyle\ll e^{C/\delta}\cdot\frac{\sqrt{N}Q}{(\log Q)^{3/2}}\cdot(\log
Q)^{4\nu+\delta\log\frac{1}{\delta}+10\delta}\cdot(\|V\|_{\infty}+\|V^{\prime}\|_{\infty}).$

with $C>10$ an absolute constant.

Once the contribution of the primes is discarded the remaining integers are
composite and therefore naturally endowed with a bilinear structure. This is
the object of the next Proposition.

######  Proposition 5.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $0<\nu<1/1000$. Let
$\mathcal{Q}\subset[Q/16,16Q]$ be a set of squarefree integers such that all
$q\in\mathcal{Q}$ have at most $\delta\log\log Q+20$ distinct prime factors.
Let $\mathcal{P}_{\nu}(N)$ denote the set of primes in
$[\exp(\log^{\nu}N),N^{1/1000}]$. Let $V$ be a smooth function compactly
supported in $[1/100,100]$. Then, for all

$\frac{Q^{2}}{(\log Q)^{10^{9}}}\leq N\leq Q^{2}(\log Q)^{10^{9}},$

we have,

$\displaystyle\sum_{\begin{subarray}{c}q\in\mathcal{Q}\\\ \chi\pmod{q}\\\
\text{primitive}\end{subarray}}\chi(\pm 1)\sum_{n}$
$\displaystyle\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}=\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}(m,p)=1\\\
m\geq 1,\ p\in\mathcal{P}_{\nu}(N)\\\ pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)\alpha(m)}{\sqrt{pm}}V\Big{(}\frac{pm}{N}\Big{)}$
$\displaystyle\ \ \ \ \ \ +O\Big{(}e^{C/\delta}\cdot\frac{Q\sqrt{N}}{(\log
N)^{3/2}}\cdot(\log
N)^{4\nu+10\delta+\delta\log\frac{1}{\delta}}\cdot(\|V\|_{\infty}+\|V^{\prime}\|_{\infty})\Big{)}.$

with $C>10$ an absolute constant and where

$\alpha(m):=\frac{\lambda_{\pi}(m)\mu_{\mathcal{P}}^{2}(m)}{1+\omega(m;\mathcal{P}_{\nu}(N))}$

where $\omega(m;\mathcal{P}_{\nu}(N))$ counts the number of distinct prime
factors of $m$ that belong to $\mathcal{P}_{\nu}(N)$ and
$\mu_{\mathcal{P}}^{2}(n)$ is the indicator function of integers $n$ not
divisible by $p^{2}$ with $p\in\mathcal{P}_{\nu}(N)$.

To bound the bilinear sum we need a dispersion estimate; this is stated below
and is a variant of the recent work of Fouvry-Radziwiłł.

######  Proposition 6.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $\kappa>0$ be such
$\lambda_{\pi}(m)$ satisfies a level $\kappa$ Siegel-Walfisz condition. Let
$\alpha(m)$ be a sequence with $|\alpha(m)|\leq|\lambda_{\pi}(m)|$ for all
$m\geq 1$. Let $0<\nu<1/1000$ be given. Let $\mathcal{P}_{\nu}(N)$ denote the
set of primes contained in $[\exp(\log^{\nu}N),N^{1/1000}]$. Let
$\mathcal{Q}\subset[Q/16,16Q]$ be a set of moduli such that all
$q\in\mathcal{Q}$ are squarefree and moreover $q\in\mathcal{Q}$ has a single
prime factor $\leq(\log Q)^{\nu\kappa}$ and all the other prime factors of $q$
are larger than $z>(\log Q)^{10000}$. Let $V$ be a smooth function compactly
supported in $[1/100,100]$. Then, for all,

$\frac{Q^{2}}{\log^{10^{9}}Q}\leq N\leq Q^{2}\log^{10^{9}}Q$

we have,

$\displaystyle\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}(m,p)=1\\\
m\geq 1,p\in\mathcal{P}_{\nu}(N)\\\ pm\equiv\pm 1\pmod{d}\end{subarray}}$
$\displaystyle\frac{\lambda_{\pi}(p)\alpha(m)}{\sqrt{pm}}V\Big{(}\frac{pm}{N}\Big{)}$
$\displaystyle\ll\|V\|_{\infty,2}\cdot
Q\sqrt{N}\cdot\Big{(}\frac{C(A,\nu,\kappa)}{(\log N)^{A}}+\frac{(\log
Q)^{4}}{z^{1/4}}\Big{)}$

for any given $A>10$, and where the constant $C(A,\nu,\kappa)$ depends only on
$A$, $\nu$ and $\kappa$.

With all these ingredients in place we can now easily prove Proposition 2.

###### Proof of Proposition 2.

Proposition 2 follows immediately from combining Proposition 5 and Proposition
6 above. ∎

##  6\. Proposition 4: Bounding the contribution of primes

We start with a trivial Lemma.

######  Lemma 11.

Let $0<\delta<1$ and $10<\alpha$. Let $n$ and $k$ be two integers with,

$|n-\alpha\log\log Q|\leq 100\text{ and }|k-\delta\log\log Q|\leq 100$

Then,

$\sum_{j\leq k}\binom{n}{j}\ll e^{C\alpha/\delta}\cdot(\log
Q)^{\delta\log\frac{1}{\delta}+(1+\log\alpha)\delta}$

with $C>10$ an absolute constant.

###### Proof.

First, notice that for $Q\leq\exp(\exp(C/\delta))$ the bound,

$\binom{n}{k}\leq 2^{n}$

is sufficient. Therefore without loss of generality we can assume that
$Q>\exp(\exp(C/\delta))$ for some sufficiently large constant $C>10$. We have
(see e.g [Gal, Theorem 3.1]222note that all logarithms in [Gal] are base 2,
while all logarithms in this paper are natural base)

$\sum_{j\leq k}\binom{n}{j}\leq\exp(nH(k/n))$

where

$H(p):=p\log\frac{1}{p}+(1-p)\log\frac{1}{1-p}.$

Notice that,

$\frac{k}{n}=\frac{\delta}{\alpha}+O\Big{(}\frac{1}{\alpha\log\log Q}\Big{)}$

with an absolute constant in the $O(\cdot)$ thanks to the conditions
$0<\delta<1$. Therefore,

$nH(k/n)=\alpha H(\delta/\alpha)\log\log Q+O(\alpha/\delta).$

Finally,

$\alpha
H(\delta/\alpha)\leq\delta\log\frac{1}{\delta}+\delta\log\alpha+(\alpha-\delta)\log\frac{1}{1-\delta/\alpha}\leq\delta\log\frac{1}{\delta}+\delta(1+\log\alpha)$

where we used the inequality,

$-\log(1-x)\leq\frac{x}{1-x}\ ,\ 0<x<1$

in the last step. ∎

We will also need the following.

######  Lemma 12.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $\eta>0$. Let
$\mathcal{B}\subset[B/4,4B]$ denote a set of integers all of whose prime
factors are greater than $B^{\eta}$. Then,

$\sum_{b\in\mathcal{B}}|\lambda_{\pi}(b)|^{2}\ll_{\eta}\frac{B}{\log B}$

###### Proof.

This follows from [ST19, Proposition 6.2] ∎

Finally, we will also appeal to the following well-known bound.

######  Lemma 13 (Shiu’s lemma).

Let $C>10$ and $\eta>0$ be given. Let $f\geq 0$ be a multiplicative function
with $f(p^{k})\leq C^{k}$ for all prime $p$ and $k\geq 1$. Suppose that for
every $\varepsilon>0$ there exists a $B(\varepsilon)>0$ such that $f(n)\leq
B(\varepsilon)n^{\varepsilon}$ for all $n\geq 1$. Then, for all $y\geq 1$,
$1\leq q\leq y^{1-\eta}$ and $(a,q)=1$,

$\sum_{\begin{subarray}{c}x\leq n\leq x+y\\\ n\equiv
a\pmod{q}\end{subarray}}f(n)\ll_{\eta}\frac{y}{\varphi(q)}\exp\Big{(}\sum_{\begin{subarray}{c}p\leq
x\\\ p\nmid q\end{subarray}}\frac{f(p)}{p}\Big{)}$

###### Proof.

See [Shi80]. ∎

For convenience we recall the statement of Proposition 4 below.

See 4

###### Proof.

Let $\mathcal{A}\subset[1,\exp(\log^{4\nu}N)]$ be a set of integers that are
$\exp(\log^{\nu}N)$ smooth, and let $\mathcal{B}\subset[1,N]$ be a set of
integers such that all prime factors of $b\in\mathcal{B}$ are larger than
$N^{1/1000}$. If $n\in\mathcal{E}$ then either $n$ can be written as $ab$ with
$a\in\mathcal{A}$ and $b\in\mathcal{B}$ or $n$ belongs to a subset
$\mathcal{U}\subset[N/4,4N]$ of integers that can be written as $uv$ with
$u>\exp(\log^{4\nu}N)$ and having all prime factors $\leq\exp(\log^{\nu}N)$
and $(u,v)=1$. We denote by $\mathcal{U}_{1}$ the set of integers that are
larger than $\exp(\log^{4\nu}N)$ and all of whose prime factors are
$\leq\exp(\log^{\nu}N)$.

First notice that, by Lemma 9

$\displaystyle\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)$
$\displaystyle\sum_{\begin{subarray}{c}n\in\mathcal{U}\\\ n\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(n)\chi(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$
$\displaystyle\ll Q\sqrt{N}(\log
Q)^{3/2}\Big{(}\sum_{n\in\mathcal{U}}\frac{|\lambda_{\pi}(n)|^{2}}{n}\Big{)}^{1/2}.$

Using the definition of $\mathcal{U}$ we can bound,

$\sum_{n\in\mathcal{U}}\frac{|\lambda_{\pi}(n)|^{2}}{n}\leq\sum_{u\in\mathcal{U}_{1}}\frac{|\lambda_{\pi}(u)|^{2}}{u}\sum_{n\leq
4N}\frac{|\lambda_{\pi}(n)|^{2}}{n}.$

By Rankin-Selberg the sum over $n$ is $\ll\log N$. If therefore remains to
bound the sum over $\mathcal{U}_{1}$ using Rankin’s trick. Indeed all terms
are larger than $\exp(\log^{4\nu}N)$ and are $\exp(\log^{\nu}N)$ smooth,
therefore we can bound the sum over $u\in\mathcal{U}_{1}$ by,

$\displaystyle\exp$
$\displaystyle\Big{(}-\iota\log^{4\nu}N\Big{)}\sum_{n\in\mathcal{U}_{1}}\frac{|\lambda_{\pi}(n)|^{2}}{n^{1-\iota}}$
$\displaystyle\ \ \
\leq\exp\Big{(}-\iota\log^{4\nu}N\Big{)}\prod_{p\leq\exp(\log^{\nu}N)}\Big{(}1+\sum_{1\leq\ell\leq
2\log^{\nu}N}\frac{|\lambda_{\pi}(p^{\ell})|^{2}}{p^{\ell(1-\iota)}}\Big{)}$

We pick $\iota=\log^{-2\nu}N$ so that,

$\displaystyle\sum_{\ell\leq\log^{\nu}N}\sum_{\log p\leq
2\log^{\nu}N}\frac{|\lambda_{\pi}(p^{\ell})|^{2}}{p^{\ell(1-\iota)}}\ll\sum_{\ell\leq\log^{\nu}N}\sum_{\log
p\leq
2\log^{\nu}N}\frac{|\lambda_{\pi}(p^{\ell})|^{2}}{p^{\ell}}\ll\log^{\nu}N.$

As a result,

$\sum_{n\in\mathcal{U}_{1}}\frac{|\lambda_{\pi}(n)|^{2}}{n}\ll\exp\Big{(}-\log^{\nu}N\Big{)},$

for all $N$ large enough.

Therefore we can now focus on the contribution of

$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}a\in\mathcal{A},b\in\mathcal{B}\\\
ab\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(ab)}{\sqrt{ab}}V\Big{(}\frac{ab}{N}\Big{)}.$

Re-arranging sums and opening $V$ into a Mellin transform, we can express this
as

$\frac{1}{2\pi i}\int_{(0)}\widetilde{V}(s)\sum_{\begin{subarray}{c}\log
A\leq\log^{4\nu}N\\\ AB\asymp
N\end{subarray}}\sum_{\begin{subarray}{c}a\in\mathcal{A},b\in\mathcal{B}\\\
a\sim A,b\sim
B\end{subarray}}\frac{\lambda_{\pi}(ab)}{(ab)^{1/2+s}}\Big{(}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
ab\equiv\pm 1\pmod{d}\end{subarray}}\mu(c)\varphi(d)\Big{)}N^{s}ds,$

where $A$ and $B$ run over powers of two. Notice that
$B\gg_{\varepsilon}Q^{2-\varepsilon}$ for any $\varepsilon>0$. Therefore we
focus now on bounding,

$\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\ b\sim
B\end{subarray}}\frac{\lambda_{\pi}(b)}{b^{1/2+s}}\Big{(}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
ab\equiv\pm 1\pmod{d}\end{subarray}}\mu(c)\varphi(d)\Big{)}$

with $B\asymp N/A$. By Cauchy-Schwarz this is less than,

$\frac{1}{\sqrt{B}}\cdot\Big{(}\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\
b\sim
B\end{subarray}}|\lambda_{\pi}(b)|^{2}\Big{)}^{1/2}\cdot\Big{(}\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\
b\sim B\end{subarray}}\Big{|}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
ab\equiv\pm 1\pmod{d}\end{subarray}}\mu(c)\varphi(d)\Big{|}^{2}\Big{)}^{1/2}.$

Because of Lemma 12 it remains to evaluate

$\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\ b\sim
B\end{subarray}}\Big{|}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\ d|ab\mp
1\end{subarray}}\mu(c)\varphi(d)\Big{|}^{2}$

For our purpose there is little gain to be made from the term $\mu(c)$ and we
bound the above by,

$\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\ b\sim
B\end{subarray}}\Big{|}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\ d|ab\mp
1\end{subarray}}\varphi(d)\Big{|}^{2}.$

First we notice that,

(6) $\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\ d|ab\mp
1\end{subarray}}\varphi(d)\leq
Q\sum_{\begin{subarray}{c}\omega(d)\leq\delta\log\log Q+20\\\ d|ab\mp
1\end{subarray}}\mu^{2}(d)\cdot\frac{\varphi(d)}{d}\ll
Q\sum_{\begin{subarray}{c}\omega(d)\leq\delta\log\log Q+20\\\ d|ab\mp
1\end{subarray}}\mu^{2}(d).$

In particular we can restrict the sum to integers $b$ with $\omega(ab\mp
1)<100\log\log Q$ by using Rankin’s trick and the divisor bound. Indeed the
complement is bounded by

$\displaystyle Q^{2}\sum_{\begin{subarray}{c}b\sim B\end{subarray}}d(ab\mp
1)^{2}e^{\omega(ab\mp 1)-100\log\log Q}$ $\displaystyle\ll\frac{Q^{2}}{(\log
Q)^{100}}\sum_{\begin{subarray}{c}n\asymp AB\\\ n\equiv\pm
1\pmod{a}\end{subarray}}d(n)^{2}e^{\omega(n)}$
$\displaystyle\ll\frac{Q^{2}B}{(\log Q)^{50}}$

using Shiu’s bound (Lemma 13, or see [Shi80]). On the remaining integers with
$\omega(ab\mp 1)<100\log\log Q$, using (6) and Lemma 11,

$\displaystyle\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\ d|ab\mp
1\end{subarray}}\varphi(d)$ $\displaystyle\ll
Q\sum_{\begin{subarray}{c}\omega(d)\leq\delta\log\log Q+20\\\ d|ab\mp
1\end{subarray}}\mu^{2}(d)$ $\displaystyle\ll Q\sum_{j\leq\delta\log\log
Q+20}\binom{\lfloor 100\log\log Q\rfloor}{j}\ll e^{C/\delta}\cdot Q(\log
Q)^{10\delta+\delta\log\frac{1}{\delta}}$

with $C>10$ an absolute constant. Therefore it remains to bound,

$e^{C/\delta}Q(\log
Q)^{10\delta+\delta\log\frac{1}{\delta}}\sum_{\begin{subarray}{c}b\sim B\\\
b\in\mathcal{B}\end{subarray}}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
d|ab\mp 1\end{subarray}}\varphi(d).$

Interchanging sums we get

$e^{C/\delta}Q(\log
Q)^{10\delta+\delta\log\frac{1}{\delta}}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\varphi(d)\sum_{\begin{subarray}{c}b\sim
B\\\ b\in\mathcal{B}\\\ d|ab\mp 1\end{subarray}}1.$

Using Brun-Titchmarsh inequality [MV73, Theorem 2] to bound the sum over $b$,
the above equation is

$\displaystyle\ll e^{C/\delta}\cdot Q(\log
Q)^{10\delta+\delta\log\frac{1}{\delta}}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}$
$\displaystyle\varphi(d)\cdot\frac{B}{\varphi(d)\log Q}$
$\displaystyle\ll\frac{e^{C/\delta}QB}{(\log Q)}(\log
Q)^{10\delta+\delta\log\frac{1}{\delta}}\sum_{n\in\mathcal{Q}}d(n)$

Any $n\in\mathcal{Q}$ is square-free and has at most $\delta\log\log Q+20$
prime factors, therefore $d(n)=2^{\omega(n)}\ll(\log Q)^{\delta}$. Therefore
the previous equation is bounded by,

$e^{C/\delta}\frac{Q^{2}B}{(\log Q)^{2}}\cdot(\log
Q)^{20\delta+2\delta\log\frac{1}{\delta}}.$

which is sufficient. We have therefore obtained,

$\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\ b\sim
B\end{subarray}}\Big{|}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\ d|ab\mp
1\end{subarray}}\varphi(d)\Big{|}^{2}\ll e^{C/\delta}\cdot\frac{Q^{2}B}{(\log
Q)^{2}}\cdot(\log Q)^{20\delta+2\delta\log\frac{1}{\delta}}.$

Combining the bounds together we get

$\sum_{\begin{subarray}{c}b\in\mathcal{B}\\\ b\sim
B\end{subarray}}\frac{\lambda_{\pi}(b)}{b^{1/2+s}}\Big{(}\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\\\
ab\equiv\pm 1\pmod{d}\end{subarray}}\mu(c)\varphi(d)\Big{)}\ll
e^{C/\delta}\cdot\frac{Q\sqrt{B}}{(\log Q)^{3/2}}\cdot(\log
Q)^{\delta\log\frac{1}{\delta}+10\delta}.$

Finally it remains to sum trivially over $a\sim A$ and over the sum remaining
powers of two $A,B$ satisfying $\log A\leq\log^{4\nu}Q$ and $AB\asymp N$. This
gives the final bound,

$\ll e^{C/\delta}\cdot\frac{Q\sqrt{N}}{(\log Q)^{3/2}}\cdot(\log
Q)^{4\nu+\delta\log\frac{1}{\delta}+10\delta}.$

We also notice that,

$\int_{|t|\leq
1}|\widetilde{V}(it)|dt\leq\widetilde{V}(0)=\int_{\mathbb{R}}V(x)x^{-1}dx\ll\|V\|_{\infty}.$

and

$\int_{|t|\geq
1}|\widetilde{V}(it)|dt\ll\Big{(}\int_{\mathbb{R}}|t\widetilde{V}(it)|^{2}dt\Big{)}^{1/2}=\Big{(}\int_{\mathbb{R}}x|V^{\prime}(x)|^{2}dx\Big{)}^{1/2}\ll\|V^{\prime}\|_{\infty}.$

Therefore,

$\int_{\mathbb{R}}|\widetilde{V}(it)|dt\ll\|V\|_{\infty}+\|V^{\prime}\|_{\infty}.$

Altogether this yields the result. ∎

##  7\. Proposition 5: Reduction to dispersion estimate

For convenience we recall here the statement of Proposition 5. The proof
crucially relies on Proposition 4 established in the previous section.

See 5

Throughout let $\mathcal{P}_{\nu}(N)$ denote the set of primes contained in
$[\exp(\log^{\nu}N),N^{1/1000}]$. Denote by $\mu^{2}_{\mathcal{P}}(n)$ the
indicator function of integers $n$ not divisible by $p^{2}$ with
$p\in\mathcal{P}_{\nu}(N)$.

We will also need the following Lemmas.

######  Lemma 14.

Let $\pi$ be a fixed unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Let $k\geq 2$ be given. Then, for
all $10\leq P\leq N$,

$\sum_{P\leq p\leq N}\frac{|\lambda_{\pi}(p^{k})|^{2}}{p^{k}}\ll P^{-1/11}\log
N,$

where the implied constant depends on $\pi.$

######  Remark 6.

The dependence on $\pi$ in the bound is irrelevant to us and will be omitted.

###### Proof.

Assume that $P$ is sufficiently large so that for all primes $p>P$ the local
representation $\pi_{p}$ is unramified. Let $\alpha_{\pi}(p,j)$ with $1\leq
j\leq 4$ denote the Satake parameters. Define,

(7) $a_{\pi}(p^{\ell}):=\sum_{j=1}^{4}\alpha_{\pi}(p,j)^{\ell}.$

By the Hecke relation

$k\lambda_{\pi}(p^{k})=\sum_{\ell=1}^{k}a_{\pi}(p^{\ell})\cdot\lambda_{\pi}(p^{k-\ell}).$

and Cauchy-Schwarz we have

$\sum_{P\leq p\leq
N}\frac{|\lambda_{\pi}(p^{k})|^{2}}{p^{k}}\ll\frac{1}{k}\sum_{\ell=1}^{k-1}\sum_{P\leq
p\leq
N}|a_{\pi}(p^{\ell})|^{2}\cdot\frac{|\lambda_{\pi}(p^{k-\ell})|^{2}}{p^{k-\ell}}+\sum_{P\leq
p\leq N}\frac{|a_{\pi}(p^{k})|^{2}}{p^{k}}.$

By the Kim-Sarnak bound [Kim03, Appendix 1], $|a_{\pi}(p^{\ell})|\ll
p^{\ell/2-\ell/11}.$ So, by Rankin-Selberg,

$\displaystyle\frac{1}{k}\sum_{\ell=1}^{k-1}\sum_{P\leq p\leq
N}|a_{\pi}(p^{\ell})|^{2}\cdot\frac{|\lambda_{\pi}(p^{k-\ell})|^{2}}{p^{k-\ell}}\ll\frac{1}{k}\sum_{\ell=1}^{k-1}\sum_{P\leq
p\leq
N}p^{-\frac{2\ell}{11}}\cdot\frac{|\lambda_{\pi}(p^{k-\ell})|^{2}}{p^{k-\ell}}\ll
P^{-\frac{2}{11}}\log N.$

The contribution from $\ell=k$ can be handled separately as follows. According
to the functoriality established by Kim [Kim03], the exterior square
$\Pi=\Lambda^{2}\pi$ of $\pi$ is an automorphic representation of
$\mathrm{GL}_{6}(\mathbb{A}_{\mathbb{Q}}).$

Since $\pi_{p}$ is unitary and unramified, we have

$\displaystyle\big{\\{}\overline{\alpha_{\pi}(p,1)},\overline{\alpha_{\pi}(p,2)},\overline{\alpha_{\pi}(p,3)},\overline{\alpha_{\pi}(p,4)}\big{\\}}=\big{\\{}\alpha_{\pi}(p,1)^{-1},\alpha_{\pi}(p,2)^{-1},\alpha_{\pi}(p,3)^{-1},\alpha_{\pi}(p,4)^{-1}\big{\\}}.$

As a consequence, the Langlands class
$\big{\\{}\alpha_{\pi}(p,1),\alpha_{\pi}(p,2),\alpha_{\pi}(p,3),\alpha_{\pi}(p,4)\big{\\}}$
is one of the following forms

  1. (a).

$\big{\\{}\alpha,\beta,\gamma,\delta\big{\\}},$ where
$|\alpha|=|\beta|=|\gamma|=|\delta|=1.$

  2. (b).

$\big{\\{}p^{t}\alpha,p^{-t}\alpha,\beta,\gamma\big{\\}},$ where $t>0,$
$|\alpha|=|\beta|=|\gamma|=1.$

  3. (c).

$\big{\\{}p^{t}\alpha,p^{-t}\alpha,p^{s}\beta,p^{-s}\beta\big{\\}},$ where
$t>0,$ $s>0,$ $|\alpha|=|\beta|=1.$

In the Case (a), we have $|\alpha_{\pi}(p,j)|\leq 4,$ $1\leq j\leq 4.$ In the
Case (b),
$|\lambda_{\pi}(p)|=|\alpha_{\pi}(p,1)+\alpha_{\pi}(p,2)+\alpha_{\pi}(p,3)+\alpha_{\pi}(p,4)|\geq
p^{t}-3.$ Hence,

$\max_{1\leq j\leq 4}|\alpha_{\pi}(p,j)|=p^{t}\leq|\lambda_{\pi}(p)|+3.$

In the Case (c), we have,
$\lambda_{\Pi}(p)=(p^{s+t}+p^{-s+t}+p^{s-t}+p^{-s-t})\alpha\beta+\alpha^{2}+\beta^{2},$
for $s>0,$ $t>0.$ Hence $|\lambda_{\Pi}(p)|\geq
p^{s+t}+p^{-s+t}+p^{s-t}+p^{-s-t}-2.$ So

$\max_{1\leq j\leq
4}|\alpha_{\pi}(p,j)|=\max\\{p^{s},p^{t}\\}\leq|\lambda_{\Pi}(p)|+2.$

Consequently, we have

$\displaystyle\max_{1\leq j\leq
4}|\alpha_{\pi}(p,j)|\leq|\lambda_{\Pi}(p)|+|\lambda_{\pi}(p)|+4.$

Substituting this estimate into (7) we then obtain

$\displaystyle\sum_{P\leq p\leq N}\frac{|a_{\pi}(p^{k})|^{2}}{p^{k}}\ll
3^{2k}\sum_{P\leq p\leq
N}\frac{|\lambda_{\Pi}(p)|^{2k}+|\lambda_{\pi}(p)|^{2k}+4^{2k}}{p^{k}},$

which, by the Kim-Sarnak bound for the automorphic representation $\Pi$ (resp.
$\pi$) of $\mathrm{GL}_{6}(\mathbb{A}_{\mathbb{Q}})$ (resp.
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$) and Rankin-Selberg, is

$\displaystyle\ll 3^{2k}P^{-(k-1)/11}\sum_{P\leq p\leq
N}\frac{|\lambda_{\Pi}(p)|^{2}+|\lambda_{\pi}(p)|^{2}}{p}+12^{2k}\cdot
P^{-(k-1)}\ll P^{-1/11}\log N,$

where the implied constant depends on $\pi.$ ∎

######  Lemma 15.

Let $\pi$ be a fixed unitary cuspidal automorphic representation of
$\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Then, for $10\leq P\leq N$,

$\sum_{P\leq p\leq N}\frac{|\lambda_{\pi}(p)|^{4}}{p^{2}}\ll P^{-2/11}\log N,$

where the implied constant depends on $\pi.$

###### Proof.

By the Kim-Sarnak bound,

$\sum_{P\leq p\leq N}\frac{|\lambda_{\pi}(p)|^{4}}{p^{2}}\leq
P^{-2/11}\sum_{P\leq p\leq N}\frac{|\lambda_{\pi}(p)|^{2}}{p}\ll P^{-2/11}\log
N,$

where the last inequality follows from Rankin-Selberg convolution. ∎

We are now ready to prove Proposition 5.

###### Proof of Proposition 5.

Executing the sum over $q$ we need to estimate

$\sum_{cd\in\mathcal{Q}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}n\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}.$

###  7.1. Integers with large square factors from $\mathcal{P}_{\nu}(N)$

First we handle the contribution of

$\displaystyle\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)$
$\displaystyle\sum_{n\equiv\pm
1\pmod{d}}\frac{\lambda_{\pi}(n)(1-\mu_{\mathcal{P}}^{2}(n))}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}$

Using Lemma 9 this is bounded by,

$Q\sqrt{N}(\log Q)^{2}\Big{(}\sum_{n\sim
N}\frac{|\lambda_{\pi}(n)|^{2}(1-\mu^{2}_{\mathcal{P}}(n))}{n}\Big{)}^{1/2}\|V\|_{\infty}.$

By the union bound, the above sum over $n$ is

$\sum_{\begin{subarray}{c}k\geq 2\\\ P^{k}\ll N\\\ (\log N)^{\nu}\leq\log
P\end{subarray}}\sum_{p\sim
P}\frac{|\lambda_{\pi}(p^{k})|^{2}}{p^{k}}\sum_{\begin{subarray}{c}n\asymp
N/P^{k}\end{subarray}}\frac{|\lambda_{\pi}(n)|^{2}}{n}.$

By Lemma 14 and Rankin-Selberg, this is,

$\ll\exp\Big{(}-\frac{1}{11}(\log N)^{\nu}\Big{)}\cdot(\log
N)^{20}\ll\exp\Big{(}-\frac{1}{20}(\log N)^{\nu}\Big{)}.$

∎

###  7.2. Integers without prime factors in $\mathcal{P}_{\nu}(N)$

Applying Proposition 4 we obtain

$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}n\in\mathcal{E}\\\
n\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(n)}{\sqrt{n}}V\Big{(}\frac{n}{N}\Big{)}\ll\frac{\sqrt{N}Q}{(\log
N)^{3/2}}\cdot(\log N)^{4\nu+10\delta+\delta\log\frac{1}{\delta}}.$

###  7.3. Integers with a single prime factor from $\mathcal{P}_{\nu}(N)$

Let $\mathcal{E}$ denote the set of integers $n$ that cannot be written as
$pm$ with $p\in\mathcal{P}_{\nu}(N)$. We can write,

$\displaystyle\lambda_{\pi}(n)\mu^{2}_{\mathcal{P}}(n)\mathbf{1}_{n\not\in\mathcal{E}}$
$\displaystyle=\sum_{\begin{subarray}{c}n=pm\\\ p\in\mathcal{P}_{\nu}(N)\\\
(m,p)=1\end{subarray}}\frac{\lambda_{\pi}(p)\lambda_{\pi}(m)\mu^{2}_{\mathcal{P}}(m)}{1+\omega(m;\mathcal{P}_{\nu}(N))}.$

Therefore the contribution of integers $n\not\in\mathcal{E}$ is exactly

(8)
$\sum_{\begin{subarray}{c}cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}(m,p)=1\\\
p\in\mathcal{P}_{\nu}(N),m\geq 1\\\ pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)\alpha(m)}{\sqrt{pm}}V\Big{(}\frac{pm}{N}\Big{)},$

where

$\alpha(m):=\frac{\lambda_{\pi}(m)\mu_{\mathcal{P}}^{2}(m)}{1+\omega(m;\mathcal{P}_{\eta}(N))}.$

We have thus reduced the problem to estimating (8).

##  8\. Proposition 6: Dispersion estimate

Our main technical input will be the following estimate for bilinear forms in
Kloosterman fractions due to Duke, Friedlander and Iwaniec [DFI97]. This
estimate was recently improved by Bettin and Chandee [BC18], but we will not
need the stronger version.

######  Lemma 16.

Let $F(x,y)$ be a smooth function with,

$\frac{\partial^{j}}{\partial x^{j}}\frac{\partial^{k}}{\partial
y^{k}}F(x,y)\ll\Delta^{j+k}x^{-j}y^{-k}$

for all $0\leq j,k\leq 2$ and $\Delta>10$. For any integer $\ell\neq 0$ and
sequence $\alpha,\beta$ supported respectively on $[M,2M)$ and $[N,2N)$ we
have,

$\displaystyle\sum_{m,n\geq 1}$
$\displaystyle\alpha(m)\beta(n)e\Big{(}\frac{\ell\overline{m}}{n}\Big{)}F(m,n)$
$\displaystyle\ll_{\varepsilon}\Delta^{2}\cdot\Big{(}\sum_{m}|\alpha(m)|^{2}\Big{)}^{1/2}\cdot\Big{(}\sum_{n}|\beta(n)|^{2}\Big{)}^{1/2}\cdot(|\ell|+MN)^{\frac{3}{8}}\cdot(M+N)^{\frac{11}{48}+\varepsilon}$

for every $\varepsilon>0$.

###### Proof.

See [DFI97, Theorem 2]. ∎

######  Lemma 17.

Let $\pi$ be a unitary cuspidal automorphic representation of
$\text{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$. Then, there exists an absolute
constant $\kappa>0,$ such that for every $x\geq 10$, and $(a,q)=1$ with
$q\leq(\log x)^{\kappa}$, $|t|\leq(\log x)^{\kappa}$ we have,

$\displaystyle\sum_{\begin{subarray}{c}p\sim x\\\ p\equiv
a\pmod{q}\end{subarray}}\lambda_{\pi}(p)p^{it}\ll_{A}\frac{x}{(\log x)^{A}},$

where the implied constant depends only on $A.$

######  Remark 7.

We will use Brumley’s [Bru06] effective zero-free region. Using the recent
Thorner-Harcos [TH23] zero-free region would have allowed us to prove Lemma 17
with $\kappa$ any fixed positive number, however the bounds would involve an
ineffective implicit constant.

###### Proof.

By the orthogonality of characters, we have

$\displaystyle\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv
a\pmod{q}\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\ll\max_{\chi\pmod{q}}\Big{|}\sum_{\begin{subarray}{c}n\sim
x\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\chi(n)\Big{|},$

where $\Lambda(\cdot)$ is the von Mangoldt function. Proceeding as [IK04,
§5.6] (in conjunction with Rankin-Selberg) we then obtain

(9) $\sum_{\begin{subarray}{c}n\sim
x\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\chi(n)=-\frac{x^{\beta-
it}}{\beta-it}+O(q^{2}(1+|t|)^{2}x\exp(-c_{\pi}\sqrt{\log x})),$

if $\beta$ is the corresponding Landau-Siegel zero of the $L$-function
$L(s,\pi\times\chi),$ and

(10) $\sum_{\begin{subarray}{c}n\sim
x\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\chi(n)=O(q^{2}(1+|t|)^{2}x\exp(-c_{\pi}\sqrt{\log
x})),$

if $L(s,\pi\times\chi)$ does not have such an exceptional zero. Here
$c_{\pi}>0$ is a constant depending only on $\pi.$

Suppose the $L$-function $L(s,\pi\times\chi)$ has a Landau-Siegel zero
$\beta\in[1/2,1).$ By [Bru06, Theorem 5] we have

(11) $L(1,\pi\times\chi)\gg_{A}q^{-A}$

for some absolute constant $A>0.$ It follows from the mean value theorem that

(12)
$L(1,\pi\times\chi)=L(1,\pi\times\chi)-L(\beta,\pi\times\chi)\ll(1-\beta)\max_{\beta\leq\sigma\leq
1}|L^{\prime}(\sigma,\pi\times\chi)|.$

Combining (11), (12) with the trivial estimate

$\max_{\beta\leq\sigma\leq 1}|L^{\prime}(\sigma,\pi\times\chi)|\ll_{\pi}q$

we obtain that $\beta\leq 1-c_{A,\pi}q^{-A-1},$ where $c_{A,\pi}>0$ is a
constant relying only on $A$ and $\pi.$ Take $\kappa=1/(2A+2).$ Then (9) and
(10) leads to

$\displaystyle\sum_{\begin{subarray}{c}n\sim
x\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\chi(n)\ll
x\exp(-c_{A,\pi}\sqrt{\log x})+x(\log x)^{4\kappa}\exp(-c_{\pi}\sqrt{\log x})$

for all $q\leq(\log x)^{\kappa},$ $|t|\leq(\log x)^{\kappa},$ and
$\chi\pmod{q}.$ Therefore,

$\displaystyle\sum_{\begin{subarray}{c}n\sim x\\\ n\equiv
a\pmod{q}\end{subarray}}\Lambda(n)\lambda_{\pi}(n)n^{it}\ll
x\exp(-c_{\pi}^{\prime}\sqrt{\log x})$

for some constant $c_{\pi}^{\prime}$ which depends only on $\pi.$ Now Lemma 17
follows from integration by parts. ∎

We first establish a dispersion estimate from which Proposition 6 readily
follows (see the end of this section). This is a variant of the work of
Fouvry-Radziwiłł. Stronger version with much larger ranges could be easily
obtained but are of no relevance for us.

######  Lemma 18.

Let $M\geq 1$, $0<\nu<1/1000$ and $\exp(\log^{\nu}M)\leq P\leq M^{1/1000}$ be
given. Let $\kappa$ be such that $\lambda_{\pi}(p)$ is Siegel-Walfisz of level
$\kappa$. Let $\mathcal{D}\subset[D/100,100D]$ be a set of square-free
integers such that all prime factors of $\mathcal{D}$ are either $\leq(\log
P)^{\kappa}$ or larger than $z\geq(\log D)^{10000}$. Let $\beta$ be arbitrary
coefficients with $|\beta|\leq 1$ and the support of $\beta$ is contained in
$\mathcal{D}$, and $\alpha$ an arbitrary sequence supported in $[M,2M]$. Let
$|t|\leq(\log P)^{\kappa}$. Then, provided that $D^{2}(\log D)^{-10^{9}}\leq
M$, we have for any given $A>10$,

$\displaystyle\sum_{d\sim D}\beta(d)\sum_{\begin{subarray}{c}(m,p)=1\\\ p\sim
P,\ m\sim M\\\ pm\equiv\pm 1\pmod{d}\end{subarray}}$
$\displaystyle\frac{\lambda_{\pi}(p)p^{it}\alpha(m)}{\sqrt{pm}}$
$\displaystyle\ll\Big{(}\sum_{m}\frac{|\alpha(m)|^{2}}{m}\Big{)}^{1/2}\sqrt{MP}\cdot\Big{(}\frac{C(A)}{(\log
P)^{A}}+\frac{(\log D)^{2}}{z^{1/4}}\Big{)}$

where $C(A)$ is a constant depending only on $A$.

###### Proof.

First we remove the condition $(m,p)=1$. Notice that the contribution of
integers $m$ divisible by $p$ is

$\ll\sum_{d\sim D}\sum_{\begin{subarray}{c}p\sim P,m\sim M/P\\\
p^{2}m\equiv\pm
1\pmod{d}\end{subarray}}\frac{|\lambda_{\pi}(p)\alpha(pm)|}{p\sqrt{m}}\ll\sum_{p\sim
P}\frac{|\lambda_{\pi}(p)|}{p}\sum_{m\sim
M/P}\frac{|\alpha(pm)|}{\sqrt{m}}d(p^{2}m\mp 1)$

To proceed further we use the Kim-Sarnak bound $|\lambda_{\pi}(p)|\ll
p^{1/2-1/11}$, getting that the above is

$\ll P^{-1/11}\sum_{\begin{subarray}{c}p\sim P,m\sim
M/p\end{subarray}}\frac{|\alpha(pm)|}{\sqrt{pm}}\cdot d(p^{2}m\mp 1)$

By Cauchy-Schwarz this is

$\ll P^{-1/11}\Big{(}\sum_{\begin{subarray}{c}p\sim P,m\sim
M/P\end{subarray}}\frac{|\alpha(pm)|}{pm}\Big{)}^{1/2}\cdot\Big{(}\sum_{\begin{subarray}{c}p\sim
P,m\sim M/P\end{subarray}}d(p^{2}m\mp 1)^{2}\Big{)}^{1/2}.$

In the first sum we group terms according to $n=pm$. Each integer $n$ has at
most $\log M$ such representations. For the second term we use Shiu’s bound,
to find that,

$\sum_{m\sim M/P}d(p^{2}m\mp 1)^{2}\ll\frac{M}{P}\cdot(\log M)^{3}.$

Therefore the entire expression is bounded by

$\ll\sqrt{MP}\cdot\Big{(}\sum_{m\asymp
M}\frac{|\alpha(m)|^{2}}{m}\Big{)}^{1/2}\cdot P^{-1/11}(\log M)^{4}.$

This is negligible compared to our target bound. We can therefore drop the
condition $(m,p)=1$.

It remains to bound,

$\sum_{d\sim D}\beta(d)\sum_{\begin{subarray}{c}p\sim P,\ m\sim M\\\
pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)p^{it}\alpha(m)}{\sqrt{pm}}.$

Interchanging sums and using Cauchy-Schwarz, we bound this by,

$\Big{(}\sum_{m\sim M}\Big{|}\sum_{\begin{subarray}{c}p\sim P,d\sim D\\\
pm\equiv\pm
1\pmod{d}\end{subarray}}\beta(d)\frac{\lambda_{\pi}(p)}{\sqrt{p}}p^{it}\Big{|}^{2}\Big{)}^{1/2}\cdot\Big{(}\sum_{m}\frac{|\alpha(m)|^{2}}{m}\Big{)}^{1/2}.$

Expanding the square we get

$\sum_{\begin{subarray}{c}p_{1},p_{2}\sim P\\\ d_{1},d_{2}\sim D\\\
(p_{1},d_{1})=1\\\
(p_{2},d_{2})=1\end{subarray}}\beta(d_{1})\overline{\beta(d_{2})}\frac{\lambda_{\pi}(p_{1})\overline{\lambda_{\pi}(p_{2})}}{\sqrt{p_{1}p_{2}}}p_{1}^{it}p_{2}^{-it}\sum_{\begin{subarray}{c}m\equiv\pm\overline{p_{1}}\pmod{d_{1}}\\\
m\equiv\pm\overline{p_{2}}\pmod{d_{2}}\end{subarray}}V\Big{(}\frac{m}{M}\Big{)}$

Notice that in the sum above $(m,d_{1}d_{2})=1$. Write $d_{1}=dd_{1}^{\prime}$
and $d_{2}=dd_{2}^{\prime}$ with $d=(d_{1},d_{2})$. Then we have
$mp_{1}\equiv\pm 1\pmod{d_{1}}$ and $mp_{2}\equiv\pm 1\pmod{d_{2}}$ and
therefore $d|m(p_{1}-p_{2})$ and hence $d|p_{1}-p_{2}$ since
$(m,d_{1}d_{2})=1$. We notice that this stage that $d$ can be restricted to be
less than $(\log P)^{\kappa}$. Indeed if that were not the case then $d>z$,
and the trivial bound yields,

$\frac{1}{P}\sum_{\begin{subarray}{c}p_{1},p_{2}\sim
P\end{subarray}}|\lambda_{\pi}(p_{1})\lambda_{\pi}(p_{2})|\sum_{\begin{subarray}{c}d|p_{1}-p_{2}\\\
d,d_{1}^{\prime},d_{2}^{\prime}\ll D\\\
d>z\end{subarray}}\frac{M}{dd_{1}^{\prime}d_{2}^{\prime}}$

We now use the inequality

$|\lambda_{\pi}(p_{1})\lambda_{\pi}(p_{2})|\leq|\lambda_{\pi}(p_{1})|^{2}+|\lambda_{\pi}(p_{2})|^{2}$

and complete the sum over $p_{1},p_{2}\sim P$ to a sum over all integers in
the range $[P,2P)$. Therefore the above expression is bounded by

$\displaystyle\frac{1}{P}\sum_{\begin{subarray}{c}m_{1},m_{2}\sim
P\end{subarray}}|\lambda_{\pi}(m_{1})|^{2}\sum_{\begin{subarray}{c}d|m_{1}-m_{2}\\\
d,d_{1}^{\prime},d_{2}^{\prime}\ll D\\\
d>z\end{subarray}}\frac{M}{dd_{1}^{\prime}d_{2}^{\prime}}$
$\displaystyle\ll\frac{1}{P}\sum_{m_{1}\sim
P}|\lambda_{\pi}(m_{1})|^{2}\sum_{\begin{subarray}{c}d,d_{1}^{\prime},d_{2}^{\prime}\ll
D\\\
d>z\end{subarray}}\frac{M}{dd_{1}^{\prime}d_{2}^{\prime}}\cdot\Big{(}\frac{P}{d}+1\Big{)}$
$\displaystyle\ll\frac{MP}{z}\cdot(\log D)^{2}+M(\log D)^{3}\ll PM(\log
D)^{4}\cdot\Big{(}\frac{1}{\sqrt{z}}+\frac{1}{P}\Big{)}.$

as needed.

Let $a$ run through congruence classes $\pmod{d}$ and write $p_{1}\equiv
p_{2}\equiv a\pmod{d}$. Thus we rewrite the above sum as,

(13) $\displaystyle\sum_{\begin{subarray}{c}d\ll(\log P)^{\kappa}\\\
a\pmod{d}\\\ (a,d)=1\end{subarray}}\sum_{\begin{subarray}{c}p_{1},p_{2}\sim
P\\\ p_{1}\equiv a\pmod{d}\\\ p_{2}\equiv
a\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p_{1})p_{1}^{it}\overline{\lambda_{\pi}(p_{2})}p_{2}^{-it}}{\sqrt{p_{1}p_{2}}}\sum_{\begin{subarray}{c}dd_{1}^{\prime},dd_{2}^{\prime}\sim
D\\\ (d_{1},p_{1})=1\\\ (d_{2},p_{2})=1\\\
(d,d_{1}^{\prime}d_{2}^{\prime})=1\end{subarray}}\beta(dd_{1}^{\prime})\overline{\beta(dd_{2}^{\prime})}\sum_{\begin{subarray}{c}m\equiv\pm\overline{p_{1}}\pmod{d_{1}^{\prime}}\\\
m\equiv\pm\overline{p_{2}}\pmod{d_{2}^{\prime}}\\\
m\equiv\pm\overline{a}\pmod{d}\end{subarray}}V\Big{(}\frac{m}{M}\Big{)}.$

Notice in (13) the congruence condition can be re-written as,

$m\equiv\pm((\overline{p_{1}})_{d_{1}^{\prime}}d_{2}(\overline{d_{2}})_{d_{1}^{\prime}}+(\overline{p_{2}})_{d_{2}^{\prime}}d_{1}(\overline{d_{1}})_{d_{2}^{\prime}}+(\overline{a})_{d}(d_{1}^{\prime}d_{2}^{\prime})(\overline{d_{1}^{\prime}d_{2}^{\prime}})_{d})\pmod{dd_{1}^{\prime}d_{2}^{\prime}}$

Therefore Poisson summation gives,

$\sum_{\begin{subarray}{c}m\equiv\pm\overline{p_{1}}\pmod{d_{1}^{\prime}}\\\
m\equiv\pm\overline{p_{2}}\pmod{d_{2}^{\prime}}\\\
m\equiv\pm\overline{a}\pmod{d}\end{subarray}}V\Big{(}\frac{m}{M}\Big{)}=\frac{M}{dd_{1}^{\prime}d_{2}^{\prime}}\sum_{\ell}\widehat{V}\Big{(}\frac{\ell
M}{dd_{1}^{\prime}d_{2}^{\prime}}\Big{)}e\Big{(}\frac{\pm\ell\overline{p_{1}d_{2}}}{d_{1}^{\prime}}+\frac{\pm\ell\overline{p_{2}d_{1}}}{d_{2}^{\prime}}+\frac{\pm\ell\overline{ad_{1}^{\prime}d_{2}^{\prime}}}{d}\Big{)}$

After this transformation (13) summed becomes,

(14) $\displaystyle M\sum_{\begin{subarray}{c}d\ll(\log P)^{\kappa}\\\
a\pmod{d}\\\
(a,d)=1\end{subarray}}\frac{1}{d}\sum_{\begin{subarray}{c}p_{1},p_{2}\sim P\\\
p_{1}\equiv a\pmod{d}\\\ p_{2}\equiv a\pmod{d}\end{subarray}}$
$\displaystyle\frac{\lambda_{\pi}(p_{1})p_{1}^{it}\overline{\lambda_{\pi}(p_{2})}p_{2}^{-it}}{\sqrt{p_{1}p_{2}}}\sum_{\begin{subarray}{c}dd_{1}^{\prime},dd_{2}^{\prime}\sim
D\\\ (d_{1},p_{1})=1\\\ (d_{2},p_{2})=1\\\
(d,d_{1}^{\prime}d_{2}^{\prime})=1\end{subarray}}\frac{\beta(dd_{1}^{\prime})}{d_{1}^{\prime}}\frac{\overline{\beta(dd_{2}^{\prime})}}{d_{2}^{\prime}}$
$\displaystyle\times\sum_{\ell}\widehat{V}\Big{(}\frac{\ell
M}{dd_{1}^{\prime}d_{2}^{\prime}}\Big{)}e\Big{(}\frac{\pm\ell\overline{p_{1}d_{2}}}{d_{1}^{\prime}}+\frac{\pm\ell\overline{p_{2}d_{1}}}{d_{2}^{\prime}}+\frac{\pm\ell\overline{ad_{1}^{\prime}d_{2}^{\prime}}}{d}\Big{)}$

We also split $d_{1}^{\prime}$ and $d_{2}^{\prime}$ into residue classes
$\pmod{d}$, so that $d_{1}^{\prime}$ and $d_{2}^{\prime}$ are respectively
congruent to $\beta_{1}$ and $\beta_{2}$ modulo $d$. To wit, we re-write the
previous equation as

(15) $\displaystyle M\sum_{\begin{subarray}{c}d\ll(\log P)^{\kappa}\\\
a,\beta_{1},\beta_{2}\pmod{d}\\\
(a\beta_{1}\beta_{2},d)=1\end{subarray}}\frac{1}{d}\sum_{\begin{subarray}{c}p_{1},p_{2}\sim
P\\\ p_{1}\equiv a\pmod{d}\\\ p_{2}\equiv a\pmod{d}\end{subarray}}$
$\displaystyle\frac{\lambda_{\pi}(p_{1})p_{1}^{it}\overline{\lambda_{\pi}(p_{2})}p_{2}^{-it}}{\sqrt{p_{1}p_{2}}}\sum_{\begin{subarray}{c}dd_{1}^{\prime},dd_{2}^{\prime}\sim
D\\\ (d_{1},p_{1})=1\\\ (d_{2},p_{2})=1\\\
d_{1}^{\prime}\equiv\beta_{1}\pmod{d}\\\
d_{2}^{\prime}\equiv\beta_{2}\pmod{d}\end{subarray}}\frac{\beta(dd_{1}^{\prime})}{d_{1}^{\prime}}\frac{\overline{\beta(dd_{2}^{\prime})}}{d_{2}^{\prime}}$
$\displaystyle\times\sum_{\ell}\widehat{V}\Big{(}\frac{\ell
M}{dd_{1}^{\prime}d_{2}^{\prime}}\Big{)}e\Big{(}\frac{\pm\ell\overline{p_{1}d_{2}}}{d_{1}^{\prime}}+\frac{\pm\ell\overline{p_{2}d_{1}}}{d_{2}^{\prime}}+\frac{\pm\ell\overline{a\beta_{1}\beta_{2}}}{d}\Big{)}$

Using the bound,

(16) $\widehat{V}\Big{(}\frac{\ell
M}{dd_{1}^{\prime}d_{2}^{\prime}}\Big{)}\ll\|V\|_{\infty,2}\cdot\frac{D^{4}}{d^{2}|\ell|^{2}M^{2}}$

we see that the total contribution of terms with $|\ell|>P^{3}$ to (15) is

$MP\cdot\frac{D^{4}}{P^{3}M^{2}}\cdot(\log M)^{10}\ll\frac{M}{P}$

using the bound $D^{2}\leq M(\log M)^{10^{10}}$. This is negligible. Therefore
we can truncate the sum in (15) at $|\ell|\leq P^{3}$

###  8.1. Diagonal terms

In (15) we isolate the contribution of the diagonal terms $\ell=0$. This
amounts to

$M\sum_{\begin{subarray}{c}d\ll(\log P)^{\kappa}\\\ a\pmod{d}\\\
(a,d)=1\end{subarray}}\frac{1}{d}\cdot\Big{|}\sum_{\begin{subarray}{c}dd^{\prime}\sim
D\\\
(d^{\prime},d)=1\end{subarray}}\frac{\beta(dd^{\prime})}{d^{\prime}}\sum_{\begin{subarray}{c}p\sim
P\\\ (p,d^{\prime})=1\\\ p\equiv
a\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)}{\sqrt{p}}p^{it}\Big{|}^{2}.$

Notice that we can bound this sum by appealing to Siegel-Walfisz for
$\pi\in\mathrm{GL}_{4}(\mathbb{A}_{\mathbb{Q}})$, namely, Lemma 17. This
yields, a bound of

$\ll_{A}MP\Big{(}(\log P)^{-A}+P^{-2/11}\log^{2}M\Big{)}\ll_{A}MP(\log
P)^{-A},$

for any given $A>10$. Note that $(\log M)P^{-1/11}$ is the contribution of
primes $p\sim P$ dividing $d^{\prime}$, and the exponent $P^{-2/11}$ arises
from an application of the Kim-Sarnak bound $|\lambda_{\pi}(p)|\ll
p^{1/2-1/11}$.

###  8.2. Off-diagonal terms with $p_{1}=p_{2}$

We bound the contribution of terms with $\ell\neq 0$ and $p_{1}=p_{2}$ in (15)
as,

$\ll M\sum_{d\ll(\log P)^{\kappa}}\frac{1}{d}\sum_{p\sim
P}\frac{|\lambda_{\pi}(p)|^{2}}{p}\sum_{dd_{1}^{\prime},dd_{2}^{\prime}\sim
D}\frac{1}{d_{1}^{\prime}d_{2}^{\prime}}\sum_{\ell\neq
0}\Big{|}\widehat{V}\Big{(}\frac{\ell
M}{dd_{1}^{\prime}d_{2}^{\prime}}\Big{)}\Big{|}$

Using (16) for $|\ell|>D^{2}/(Md)$ and the trivial bound
$\widehat{V}\ll\|V\|_{\infty}$ otherwise, we conclude that the above is

$\ll M(\log M)^{2}\cdot\frac{D^{2}}{M}\ll M(\log M)^{C}$

with $C>10$ an absolute constant since $D\ll\sqrt{M}(\log M)^{10^{9}}$.

###  8.3. Off-diagonal terms with $p_{1}\neq p_{2}$

We transform the phase of the exponential sum by noticing that, for $p_{1}\neq
p_{2}$ with $p_{1}\equiv p_{2}\equiv a\pmod{d}$,
$d_{1}^{\prime}\equiv\beta_{1}\pmod{d}$ and
$d_{2}^{\prime}\equiv\beta_{2}\pmod{d}$,

$\displaystyle\frac{\overline{p_{1}d_{2}}}{d_{1}^{\prime}}+\frac{\overline{p_{2}d_{1}}}{d_{2}^{\prime}}+\frac{\overline{a\beta_{1}\beta_{2}}}{d}$
$\displaystyle\equiv\frac{\overline{p_{1}d_{2}}}{d_{1}^{\prime}}-\frac{\overline{d_{2}^{\prime}}}{p_{2}d_{1}}+\frac{\overline{a\beta_{1}\beta_{2}}}{d}+\frac{1}{p_{2}d_{1}d_{2}^{\prime}}$
$\displaystyle\equiv\frac{\overline{p_{1}d_{2}}}{d_{1}^{\prime}}-\frac{\overline{d_{2}}}{p_{2}d_{1}^{\prime}}-\frac{\overline{p_{2}d_{1}^{\prime}d_{2}^{\prime}}}{d}+\frac{\overline{a\beta_{1}\beta_{2}}}{d}+\frac{1}{pd_{1}d_{2}^{\prime}}$
$\displaystyle\equiv\frac{(p_{2}-p_{1})\overline{p_{1}d_{2}}}{p_{2}d_{1}^{\prime}}-\frac{\overline{a\beta_{1}\beta_{2}}}{d}+\frac{\overline{a\beta_{1}\beta_{2}}}{d}+\frac{1}{p_{2}d_{1}d_{2}^{\prime}}$
$\displaystyle\equiv\frac{((p_{2}-p_{1})/d)\overline{p_{1}d_{2}^{\prime}}}{p_{2}d_{1}^{\prime}}+\frac{1}{p_{2}dd_{1}^{\prime}d_{2}^{\prime}}\pmod{1}$

It then remains to bound,

(17) $\displaystyle\|V\|_{\infty}\cdot\frac{1}{P}$
$\displaystyle\sum_{\begin{subarray}{c}d\ll(\log P)^{\kappa}\\\
a,\beta_{1},\beta_{2}\pmod{d}\\\
(a\beta_{1}\beta_{2},d)=1\end{subarray}}\sum_{\begin{subarray}{c}p_{1}\neq
p_{2}\sim P\\\ p_{1}\equiv a\pmod{d}\\\ p_{2}\equiv
a\pmod{d}\end{subarray}}\frac{dM}{D^{2}}\sum_{0\neq|\ell|\leq
P^{3}}|\lambda_{\pi}(p_{1})\lambda_{\pi}(p_{2})|$ (18) $\displaystyle\ \
\times\Big{|}\sum_{v_{1},v_{2}\sim
PD/d}\gamma_{p_{1},p_{2},d,\beta_{2}}(v_{2})\gamma_{p_{2},p_{1},d,\beta_{1}}(v_{1})e\Big{(}\frac{\pm\ell((p_{2}-p_{1})/d)\overline{v_{2}}}{v_{1}}\Big{)}\widehat{V}\Big{(}\frac{\ell
Mp_{1}p_{2}}{dv_{1}v_{2}}\Big{)}e\Big{(}\frac{\pm\ell
p_{1}}{dv_{1}v_{2}}\Big{)}\Big{|},$

where

$\gamma_{p_{1},p_{2},d,\beta}(v):=\sum_{\begin{subarray}{c}v=p_{1}\delta\\\
(\delta,p_{2})=1\\\
\delta\equiv\beta\pmod{d}\end{subarray}}\beta(d\delta)\ll\log M$

Applying Lemma 16 with

$F(v_{1},v_{2}):=\widehat{V}\Big{(}\frac{\ell
Mp_{1}p_{2}}{dv_{1}v_{2}}\Big{)}e\Big{(}\frac{\pm\ell
p_{1}}{dv_{1}v_{2}}\Big{)}$

we obtain the following bound for (17)

$\ll P^{100}\cdot(PD)^{2-\frac{1}{40}}\ll M$

since $D\leq\sqrt{M}(\log M)^{10^{9}}$ and $\exp(\log^{\nu}M)\leq P\leq
M^{1/1000}$. Combining all these estimates together it follows that (13) is

$\ll_{A}\frac{MP}{(\log P)^{A}}.$

for any given $A>10$. Collecting all the previous bounds the claim follows. ∎

For convenience we recall here the statement of Proposition 6.

See 6

###### Proof.

We localize $c\sim C$, $d\sim D$, $p\sim P$ and $m\sim M$ with $CD\asymp Q$
and $PM\asymp N$. We then open $V$ into Mellin transform, so that the sum is

$\displaystyle\ll\sum_{\begin{subarray}{c}CD\asymp Q\\\ PM\asymp
N\end{subarray}}\int_{\mathbb{R}}|\widetilde{V}(it)|\Big{|}\sum_{\begin{subarray}{c}c\sim
C,d\sim D\\\
cd\in\mathcal{Q}\end{subarray}}\mu(c)\varphi(d)\sum_{\begin{subarray}{c}(m,p)=1\\\
p\sim P,m\sim M\\\ pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)\alpha(m)}{\sqrt{pm}}(pm)^{it}\Big{|}dt$

We can further bound this by

(19) $\ll\sum_{\begin{subarray}{c}CD\asymp Q\\\ PM\asymp
N\end{subarray}}D\int_{\mathbb{R}}|\widetilde{V}(it)|\sum_{c\sim
C}\Big{|}\sum_{d\sim D}\beta_{c}(d)\sum_{\begin{subarray}{c}(m,p)=1\\\ p\sim
P,m\sim M\\\ pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)p^{it}\alpha_{t}(m)}{\sqrt{pm}}\Big{|}dt$

where $|\beta_{c}(d)|\leq 1$, uniformly in $c$, and

$|\alpha_{t}(m)|\leq|\lambda_{\pi}(m)|$

uniformly in $t\in\mathbb{R}$. The sum over $d,p$ and $m$ is trivially bounded
by

$\displaystyle\ll\frac{1}{\sqrt{PM}}\sum_{\begin{subarray}{c}(p,m)=1\\\ p\sim
P\\\ m\sim M\end{subarray}}|\lambda_{\pi}(p)\alpha_{t}(m)|d(pm\mp 1)$
$\displaystyle\ll\frac{\log M}{\sqrt{PM}}\sum_{n\asymp
MP}|\lambda_{\pi}(n)|\cdot d(n\mp 1)\ll\sqrt{MP}(\log M)^{10}$

where in the second bound we used that
$|\alpha_{t}(m)|\leq|\lambda_{\pi}(m)|$, $(p,m)=1$ and that $m$ has at most
$\log M$ prime divisors, and where in the last bound we used Cauchy-Schwarz
and Rankin-Selberg. Therefore, the contribution of $|t|>(\log P)^{\kappa}$ to
(19) is

$\ll_{A,\kappa}\sqrt{MP}(\log P)^{-A}$

for any given $A>10$. This allows us to truncate the integral in (19) to
$|t|\leq(\log P)^{\kappa}$. In the remaining range $|t|\leq(\log P)^{\kappa}$,
by Proposition 6,

$\sum_{d\sim D}\beta(d)\sum_{\begin{subarray}{c}(p,m)=1\\\ p\sim P,m\sim M\\\
pm\equiv\pm
1\pmod{d}\end{subarray}}\frac{\lambda_{\pi}(p)p^{it}\alpha(m)}{\sqrt{pm}}\ll\sqrt{MP}\cdot\Big{(}\frac{C(A)}{(\log
P)^{A}}+\frac{(\log M)^{2}}{z^{1/4}}\Big{)}.$

for any given $A>10$. Therefore, upon summing over all ranges, (19) is bounded
by

$\ll Q\sqrt{N}\cdot\Big{(}\frac{C(A)}{(\log N)^{\nu A-10}}+\frac{(\log
Q)^{4}}{z^{1/4}}\Big{)}$

and this gives the claim upon taking $A$ large enough with respect to $\nu$
(thus making the constant $C(A)$ depend on both $\nu$ and $A$). ∎

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