When Kloosterman sums meet Hecke eigenvalues

Abstract

By elaborating a two-dimensional Selberg sieve with asymptotics and equidistributions of Kloosterman sums from $\ell$-adic cohomology, as well as a Bombieri–Vinogradov type mean value theorem for Kloosterman sums in arithmetic progressions, it is proved that for any given primitive Hecke–Maass cusp form of trivial nebentypus, the eigenvalue of the n-th Hecke operator does not coincide with the Kloosterman sum $\mathrm {Kl}(1,n)$ for infinitely many squarefree n with at most 100 prime factors. This provides a partial negative answer to a problem of Katz on modular structures of Kloosterman sums.

Appendices

Appendix A: Multiplicative functions against Möbius

We would like to evaluate a weighted average of general multiplicative functions against Möbius function. This will be employed in the evaluation of Selberg sieve weights essentially given by (2.2).

Let g be a non-negative multiplicative function with $0\leqslant g(p)<1$ for each $p\in {\mathcal {P}}$. Suppose the Dirichlet series

$$\begin{aligned} {\mathcal {G}}(s):=\sum _{n\geqslant 1}\mu ^2(n)g(n)n^{-s} \end{aligned}$$

(A.1)

converges absolutely for $\mathfrak {R}s>1.$ Assume there exist a positive integer $\kappa$ and some constants $L,c_0>0,$ such that

$$\begin{aligned} {\mathcal {G}}(s)=\zeta (s+1)^\kappa {\mathcal {F}}(s), \end{aligned}$$

(A.2)

where ${\mathcal {F}}(s)$ is holomorphic for $\mathfrak {R}s\geqslant -c_0$ and does not vanish in the region

$$\begin{aligned} {\mathcal {D}}:=\Big \{\sigma +it: t\in {\mathbf {R}},\sigma \geqslant -\frac{1}{L\cdot \log (|t|+2)}\Big \}, \end{aligned}$$

(A.3)

and $|1/{\mathcal {F}}(s)|\leqslant L$ for all $s\in {\mathcal {D}}.$ We also assume

$$\begin{aligned} \left| \sum _{p\leqslant x}g(p)\log p-\kappa \log x\right| \leqslant L \end{aligned}$$

(A.4)

holds for all $x\geqslant 3$ and

$$\begin{aligned} \sum _{p}g(p)^2p^{2c_0}<+\infty . \end{aligned}$$

(A.5)

We are interested in the asymptotic behaviour of the sum

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&=\sum _{\begin{array}{c} n\leqslant x\\ n\mid P(z)\\ (n,q)=1 \end{array}}\mu (n)g(n)\Big (\log \frac{x}{n}\Big )^\kappa , \end{aligned}$$

where q is a positive integer and $x,z\geqslant 3$.

Lemma A.1

Let $q\geqslant 1.$ Under the assumption as above, we have

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&=H\cdot \prod _{p\mid q}(1-g(p))^{-1}\cdot m_\kappa (s)+O(\kappa ^{\omega (q)}(\log z)^{-A}) \end{aligned}$$

for all $A>0,x\geqslant 2,z\geqslant 2$ with $x\leqslant z^{O(1)}$, where $s=\log x/\log z,$

$$\begin{aligned} H=\prod _{p}(1-g(p))\Big (1-\frac{1}{p}\Big )^{-\kappa }, \end{aligned}$$

and $m_\kappa (s)$ is a continuous solution to the differential-difference equation

$$\begin{aligned} {\left\{ \begin{array}{ll} m_\kappa (s)=\kappa !,\ \ &{}s\in ~]0,1],\\ sm_\kappa '(s)=\kappa m_\kappa (s-1),&{}s\in ~]1,+\infty [.\end{array}\right. } \end{aligned}$$

(A.6)

The implied constant depends on $A,\kappa ,L$ and $c_0.$

Proof

We are inspired by [15, Appendix A.3]. Write ${\mathcal {M}}_\kappa (x,x;q)={\mathcal {M}}_\kappa (x;q)$. By Mellin inversion, we have

$$\begin{aligned} {\mathcal {M}}_\kappa (x;q)&=\sum _{\begin{array}{c} n\leqslant x\\ (n,q)=1 \end{array}}\mu (n)g(n)\Big (\log \frac{x}{n}\Big )^\kappa =\frac{\kappa !}{2\pi i}\int _{2-i\infty }^{2+i\infty }{\mathcal {G}}(t,q)\frac{x^t}{t^{\kappa +1}}\mathrm {d}t, \end{aligned}$$

where

$$\begin{aligned} {\mathcal {G}}(t,q)=\sum _{\begin{array}{c} n\geqslant 1\\ (n,q)=1 \end{array}}\frac{\mu (n)g(n)}{n^t},\ \ \mathfrak {R}t>1. \end{aligned}$$

Note that

$$\begin{aligned} {\mathcal {G}}(t,q)=\prod _{p\not \mid q}\Big (1-\frac{g(p)}{p^t}\Big )=\prod _{p\mid q}\Big (1-\frac{g(p)}{p^t}\Big )^{-1}\frac{{\mathcal {G}}^*(t)}{\zeta (t+1)^\kappa }, \end{aligned}$$

where

$$\begin{aligned} {\mathcal {G}}^*(t)=\prod _{p}\Big (1-\frac{g(p)}{p^t}\Big ) \Big (1-\frac{1}{p^{t+1}}\Big )^{-\kappa }=\prod _{p} \Big (1-\frac{g(p)^2}{p^{2t}}\Big )\frac{1}{{\mathcal {F}}(t)}, \end{aligned}$$

which is absolutely convergent and holomorphic for $t\in {\mathcal {C}}$ by (A.2), (A.4) and (A.5). Hence we find

$$\begin{aligned} {\mathcal {M}}_\kappa (x;q)&=\frac{\kappa !}{2\pi i}\int _{2-i\infty }^{2+i\infty }\prod _{p\mid q}\Big (1-\frac{g(p)}{p^t}\Big )^{-1}\frac{{\mathcal {G}}^*(t)x^t}{\zeta (t+1)^\kappa t^{\kappa +1}}\mathrm {d}t. \end{aligned}$$

Shifting the t-contour to the left boundary of ${\mathcal {C}}$ and passing one simple pole at $t=0$, we get

$$\begin{aligned} {\mathcal {M}}_\kappa (x;q)&=\kappa !{\mathcal {G}}^*(0)\prod _{p\mid q}(1-g(p))^{-1}+O(\kappa ^{\omega (q)}(\log 2x)^{-A}) \end{aligned}$$

for any fixed $A>0$.

For $s=\log x/\log z,$ we expect that

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&=c(q)m_\kappa (s)+O(\kappa ^{\omega (q)}(\log z)^{-A}) \end{aligned}$$

(A.7)

for all $A>0,x\geqslant 2,z\geqslant 2$ and $q\geqslant 1$, where c(q) is some constant defined in terms of g and depending also on q, and $m_\kappa (s)$ is a suitable continuous function in $s>0.$ As mentioned above, this expected asymptotic formula holds for $0<s\leqslant 1,$ in which case we may take

$$\begin{aligned} c(q)={\mathcal {G}}^*(0)\prod _{p\mid q}(1-g(p))^{-1},\ \ \ m_\kappa (s)=\kappa !. \end{aligned}$$

We now move to the case $s>1$ and prove the asymptotic formula (A.7) by induction. Since $x\leqslant z^{O(1)},$ this induction will have a bounded number of steps. We first consider the difference ${\mathcal {M}}_\kappa (x,z;q)-{\mathcal {M}}_\kappa (x;q)$. In fact, each n that contributes to this difference has a prime factor at least z, and we may decompose $n=mp$ uniquely up to the restriction $z\leqslant p<x,$$m\mid P(p).$ Hence

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&={\mathcal {M}}_\kappa (x;q)+\sum _{\begin{array}{c} z\leqslant p<x\\ (p,q)=1 \end{array}}g(p)\sum _{\begin{array}{c} m\leqslant x/p\\ m\mid P(p)\\ (m,q)=1 \end{array}}\mu (m)g(m)\Big (\log \frac{x}{mp}\Big )^\kappa \nonumber \\&={\mathcal {M}}_\kappa (x;q)+\sum _{\begin{array}{c} z\leqslant p<x\\ (p,q)=1 \end{array}}g(p){\mathcal {M}}_\kappa (x/p,p;q). \end{aligned}$$

(A.8)

Substituting (A.7) to (A.8), we get

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&=c(q)\kappa !+c(q)\sum _{\begin{array}{c} z\leqslant p<x\\ (p,q)=1 \end{array}}g(p)m_\kappa \Big (\frac{\log (x/p)}{\log p}\Big )\\&\quad +O(\kappa ^{\omega (q)}(\log x)^{-A}) +O\Big (\kappa ^{\omega (q)}\sum _{\begin{array}{c} z\leqslant p<x\\ (p,q)=1 \end{array}}g(p)(\log (2x/p))^{-A}\Big ). \end{aligned}$$

By partial summation, we find

$$\begin{aligned} {\mathcal {M}}_\kappa (x,z;q)&=c(q)\Big \{\kappa !+\kappa \int _1^sm_\kappa \Big (\frac{s}{u}-1\Big )\frac{\mathrm {d}u}{u}\Big \}+O(\kappa ^{\omega (q)}(\log z)^{-A}). \end{aligned}$$

Hence, by (A.7), $m_\kappa (s)$ should satisfy the equation

$$\begin{aligned} m_\kappa (s)=\kappa !+\kappa \int _1^sm_\kappa \Big (\frac{s}{u}-1\Big )\frac{\mathrm {d}u}{u}=\kappa !+\kappa \int _1^sm_\kappa (u-1)\frac{\mathrm {d}u}{u} \end{aligned}$$

for $s>1$. Taking the derivative with respect to s gives (A.6). $\square$

Remark 6

To extend $m_\kappa (s)$ to be defined on ${\mathbf {R}},$ we may put $m_\kappa (s)=0$ for $s\leqslant 0$.

Appendix B: A two-dimensional Selberg sieve with asymptotics

This section devotes to present a two-dimensional Selberg sieve that plays an essential role in proving Proposition 2.2.

Let h be a non-negative multiplicative function. Suppose the Dirichlet series

$$\begin{aligned} {\mathcal {H}}(s):=\sum _{n\geqslant 1}\mu ^2(n)h(n)n^{-s} \end{aligned}$$

(B.1)

converges absolutely for $\mathfrak {R}s>1.$ Assume there exist some constants $L,c_0>0,$ such that

$$\begin{aligned} {\mathcal {H}}(s)=\zeta (s)^2{\mathcal {H}}^*(s), \end{aligned}$$

(B.2)

where ${\mathcal {H}}^*(s)$ is holomorphic for $\mathfrak {R}s\geqslant 1-c_0,$ and does not vanish in the region ${\mathcal {D}}$ as given by (A.3) and $|1/{\mathcal {H}}^*(s)|\leqslant L$ for all $s\in {\mathcal {D}}$. We also assume

$$\begin{aligned} \left| \sum _{p\leqslant x}\frac{h(p)\log p}{p}-2\log x\right| \leqslant L \end{aligned}$$

(B.3)

holds for all $x\geqslant 3$ and

$$\begin{aligned} \sum _{p}h(p)^2p^{2c_0-2}<+\infty . \end{aligned}$$

(B.4)

Define

$$\begin{aligned} S(X,z;h,\varvec{\varrho })=\sum _{n\geqslant 1}\varPsi \Big (\frac{n}{X}\Big )\mu ^2(n)h(n)\Big (\sum _{d|(n,P(z))}\varrho _d\Big )^2, \end{aligned}$$

where $\varvec{\varrho }=(\varrho _d)$ is given as in (2.2) and $\varPsi$ is a fixed non-negative smooth function supported in [1, 2] with normalization (2.3).

Theorem B.1

Let $X,D,z\geqslant 3$ with $X\leqslant D^{O(1)}$ and $X\leqslant z^{O(1)}.$ Put $\tau =\log D/\log z$ and $\sqrt{D}=X^\vartheta \exp (-\sqrt{{\mathcal {L}}}),\vartheta \in ~]0,\frac{1}{2}[.$ Under the above assumptions, we have

$$\begin{aligned} S(X,z;h,\varvec{\varrho })&=(1+o(1)){\mathfrak {S}}(\vartheta ,\tau )X{\mathcal {L}}^{-1}, \end{aligned}$$

where ${\mathfrak {S}}(\vartheta ,\tau )$ is defined by

$$\begin{aligned} {\mathfrak {S}}(\vartheta ,\tau )&=16\mathrm {e}^{2\gamma }\Big (\frac{c_1(\tau )}{4\tau \vartheta ^2}+\frac{c_2(\tau )}{\tau ^2\vartheta }\Big ), \end{aligned}$$

(B.5)

where

$$\begin{aligned} c_1(\tau )&=\int _0^1\sigma '((1-u)\tau ){\mathfrak {f}}(u\tau )^2\mathrm {d}u,\\ c_2(\tau )&=\int _0^1\int _0^1\sigma '((1-u)\tau ) {\mathfrak {f}}(u\tau -2v)\{2{\mathfrak {f}}(u\tau )-{\mathfrak {f}}(u\tau -2v)\}\mathrm {d}u\mathrm {d}v. \end{aligned}$$

Here $\sigma (s)$ is the continuous solution to the differential-difference equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \sigma (s)=\dfrac{s^2}{8\mathrm {e}^{2\gamma }},\ \ &{}s\in ~]0,2],\\ {} (s^{-2}\sigma (s))'=-\,2s^{-3}\sigma (s-2),&{}s\in ~]2,+\infty [, \end{array}\right. } \end{aligned}$$

(B.6)

and ${\mathfrak {f}}(s)=m_2(s/2)$ as given by (A.6), i.e., ${\mathfrak {f}}(s)$ is the continuous solution to the differential-difference equation

$$\begin{aligned} {\left\{ \begin{array}{ll} {\mathfrak {f}}(s)=2,\ \ &{}s\in ~]0,2],\\ s{\mathfrak {f}}'(s)=2{\mathfrak {f}}(s-2),&{}s\in ~]2,+\infty [.\end{array}\right. } \end{aligned}$$

(B.7)

Remark 7

Theorem B.1 is a generalization of [42, Proposition 4.1] with a general multiplicative function h and the extra restriction $d\mid P(z)$, but specializing $k=2$ therein. It would be rather interesting to extend the case to a general $k\in {\mathbf {Z}}^+$ and we would like to concentrate this problem in the near future.

We now choose $z=\sqrt{D}$, so that the restriction $d\mid P(z)$ is redundant, in which case one has $\tau =2.$ Note that

$$\begin{aligned} c_1(2)&=4\int _0^1\sigma '(2u)\mathrm {d}u=\frac{1}{\mathrm {e}^{2\gamma }},\\ c_2(2)&=4\int _0^1\sigma '(2(1-u))u\mathrm {d}u=\frac{1}{3\mathrm {e}^{2\gamma }}. \end{aligned}$$

For $\vartheta =1/4,$ we find ${\mathfrak {S}}(\vartheta ,\tau )={\mathfrak {S}}(1/4,2)=112/3$, which coincides with $4{\mathfrak {c}}(2,F)$ in [42, Proposition 4.1] by taking $F(x)=x^2$ therein.

We now give the proof of Theorem B.1. To begin with, we write by (8.4) that

$$\begin{aligned} S(X,z;h,\varvec{\varrho })&=\sum _{d\mid P(z)}\xi (d)\sum _{n\equiv 0\,({{\,\mathrm{mod}\,}}{ d})}\varPsi \Big (\frac{n}{X}\Big )\mu ^2(n)h(n)\\&=\sum _{d\mid P(z)}\xi (d)h(d)\sum _{(n,d)=1}\varPsi \Big (\frac{nd}{X}\Big )\mu ^2(n)h(n). \end{aligned}$$

By Mellin inversion,

$$\begin{aligned} \sum _{(n,d)=1}\varPsi \Big (\frac{nd}{X}\Big )\mu ^2(n)h(n)&=\frac{1}{2\pi i}\int _{(2)}{\widetilde{\varPsi }}(s)(X/d)^s{\mathcal {H}}^\flat (s,d)\mathrm {d}s, \end{aligned}$$

where, for $\mathfrak {R}s>1,$

$$\begin{aligned} {\mathcal {H}}^\flat (s,d)=\sum _{\begin{array}{c} n\geqslant 1\\ (n,d)=1 \end{array}}\frac{\mu ^2(n)h(n)}{n^s}. \end{aligned}$$

For $\mathfrak {R}s>1,$ we first write

$$\begin{aligned} {\mathcal {H}}^\flat (s,d)&=\prod _{p\not \mid d}\Big (1+\frac{h(p)}{p^s}\Big )=\prod _{p\mid d}\Big (1+\frac{h(p)}{p^s}\Big )^{-1}{\mathcal {H}}(s)\\&=\prod _{p\mid d}\Big (1+\frac{h(p)}{p^s}\Big )^{-1}\zeta (s)^2{\mathcal {G}}(s). \end{aligned}$$

Note that

$$\begin{aligned} {\mathcal {G}}(1)=\lim _{s\rightarrow 1}\frac{{\mathcal {H}}(s)}{\zeta (s)^2}=\prod _p \Big (1+\frac{h(p)}{p}\Big )\Big (1-\frac{1}{p}\Big )^2. \end{aligned}$$

By (B.2), ${\mathcal {H}}^\flat (s,d)$ admits a meromorphic continuation to $\mathfrak {R}s\geqslant 1-c_0.$ Shifting the s-contour to the left beyond $\mathfrak {R}s=1,$ we may obtain

$$\begin{aligned}&\sum _{(n,d)=1}\varPsi \Big (\frac{nd}{X}\Big )\mu ^2(n)h(n)\\&\quad ={{\,\mathrm{Res}\,}}_{s=1}{\widetilde{g}}(s){\mathcal {G}}(s)(X/d)^s\prod _{p\mid d}\Big (1+\frac{h(p)}{p^s}\Big )^{-1}\zeta (s)^2+O((X/d){\mathcal {L}}^{-100}). \end{aligned}$$

We compute the residue as

$$\begin{aligned}&{{\,\mathrm{Res}\,}}_{s=1}[\cdots ]\\ =&\,\frac{\mathrm {d}}{\mathrm {d}s}{\widetilde{\varPsi }}(s){\mathcal {G}}(s)(X/d)^s\prod _{p\mid d}\Big (1+\frac{h(p)}{p^s}\Big )^{-1}\zeta (s)^2(s-1)^2\Big |_{s=1}\\ =&\,{\widetilde{\varPsi }}(1){\mathcal {G}}(1)\prod _{p\mid d}\Big (1+\frac{h(p)}{p}\Big )^{-1}\frac{X}{d}\Big (\log (X/d)+\sum _{p\mid d}\frac{h(p)\log p}{p+h(p)}+c\Big )\\ =&\,{\mathcal {G}}(1)\prod _{p\mid d}\Big (1+\frac{h(p)}{p}\Big )^{-1}\frac{X}{d}\Big (\log X-\sum _{p\mid d}\frac{p\log p}{p+h(p)}+c\Big ), \end{aligned}$$

where c is some constant independent of d.

Define $\beta$ and $\beta ^*$ to be multiplicative functions supported on squarefree numbers via

$$\begin{aligned} \beta (p)=\frac{p}{h(p)}+1,\ \ \ \ \beta ^*(p)=\beta (p)-1=\frac{p}{h(p)}. \end{aligned}$$

Define L to be an additive function supported on squarefree numbers via

$$\begin{aligned} L(p)=\frac{\beta ^*(p)\log p}{\beta (p)}. \end{aligned}$$

Therefore, for each squarefree number d, we have

$$\begin{aligned} \beta (d)=\prod _{p\mid d}\Big (\frac{p}{h(p)}+1\Big ),\ \ \ \ \beta ^*(d)=\frac{d}{h(d)},\ \ \ \ L(d)= \sum _{p\mid d}\frac{\beta ^*(p)\log p}{\beta (p)}. \end{aligned}$$

In this way, we may obtain

$$\begin{aligned} S(X;h,\varvec{\varrho })&={\mathcal {G}}(1)X\{S_1(X)\cdot (\log X+c)-S_2(X)\}+O(X{\mathcal {L}}^{-2}), \end{aligned}$$

where

$$\begin{aligned} S_1(X)&=\sum _{d\mid P(z)}\frac{\xi (d)}{\beta (d)},\\ S_2(X)&=\sum _{d\mid P(z)}\frac{\xi (d)}{\beta (d)}L(d). \end{aligned}$$

Note that

$$\begin{aligned} S_1(X)&=\mathop {\sum \sum }_{d_1,d_2\mid P(z)}\frac{\varrho _{d_1}\varrho _{d_2}}{\beta ([d_1,d_2])}\\&=\mathop {\sum \sum }_{d_1,d_2\mid P(z)}\frac{\varrho _{d_1}\varrho _{d_2}}{\beta (d_1)\beta (d_2)}\beta ((d_1,d_2))\\&=\mathop {\sum \sum }_{d_1,d_2\mid P(z)}\frac{\varrho _{d_1}\varrho _{d_2}}{\beta (d_1)\beta (d_2)} \sum _{l\mid (d_1,d_2)}\beta ^*(l). \end{aligned}$$

Hence we may diagonalize $S_1(X)$ by

$$\begin{aligned} S_1(X)&=\sum _{\begin{array}{c} l\leqslant \sqrt{D}\\ l\mid P(z) \end{array}}\beta ^*(l)y_l^2, \end{aligned}$$

(B.8)

where, for each $l\mid P(z)$ and $l\leqslant \sqrt{D},$

$$\begin{aligned} y_l=\sum _{\begin{array}{c} d\mid P(z)\\ d\equiv 0\,({{\,\mathrm{mod}\,}}{ l}) \end{array}}\frac{\varrho _d}{\beta (d)}. \end{aligned}$$

From the definition of sieve weights (2.2), we find

$$\begin{aligned} y_l&=\frac{4\mu (l)}{\beta (l)(\log D)^2}\sum _{\begin{array}{c} d\leqslant \sqrt{D}/l\\ dl\mid P(z) \end{array}}\frac{\mu (d)}{\beta (d)}\Big (\log \frac{\sqrt{D}/l}{d}\Big )^2. \end{aligned}$$

Applying Lemma A.1 with $g(p)=1/\beta (p)$ and $q=l$, we have

$$\begin{aligned} y_l&=\frac{4\mu (l)}{{\mathcal {G}}(1)\beta ^*(l)(\log D)^2}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )+O\Big (\frac{\tau (l)}{\beta (l)}(\log z)^{-A}\Big ). \end{aligned}$$

(B.9)

Inserting this expression to (B.8), we have

$$\begin{aligned} S_1(X)&=\frac{16(1+o(1))}{{\mathcal {G}}(1)^2(\log D)^4}\sum _{\begin{array}{c} l\leqslant \sqrt{D}\\ l\mid P(z) \end{array}}\frac{1}{\beta ^*(l)}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )^2. \end{aligned}$$

Following [17, Lemma 6.1], we have

$$\begin{aligned} \sum _{\begin{array}{c} l\leqslant x\\ l\mid P(z) \end{array}}\frac{1}{\beta ^*(l)}=\frac{1}{W(z)}\Big \{\sigma (2\log x/\log z)+O\Big (\frac{(\log x/\log z)^5}{\log z}\Big )\Big \} \end{aligned}$$

(B.10)

with

$$\begin{aligned} W(z)&=\prod _{p<z}\Big (1-\frac{1}{\beta (p)}\Big ), \end{aligned}$$

from which and partial summation, we find

$$\begin{aligned} S_1(X)&=\frac{16\tau c_1(\tau )}{{\mathcal {G}}(1)^2W(z)(\log D)^4}\cdot (1+o(1)) \end{aligned}$$

with $\tau =\log D/\log z$ and

$$\begin{aligned} c_1(\tau )=\int _0^1\sigma '((1-u)\tau ){\mathfrak {f}}(u\tau )^2\mathrm {d}u. \end{aligned}$$

(B.11)

We now turn to consider $S_2(X)$. Note that L(d) is an additive function supported on squarefree numbers. We then have

$$\begin{aligned} S_2(X)&=\mathop {\sum \sum }_{d_1,d_2\mid P(z)}\frac{\varrho _{d_1}\varrho _{d_2}}{\beta ([d_1,d_2])}L([d_1,d_2])\\&=\mathop {\sum \sum }_{dd_1d_2\mid P(z)}\frac{\varrho _{dd_1}\varrho _{dd_2}}{\beta (dd_1d_2)} \{L(d)+L(d_1)+L(d_2)\}, \end{aligned}$$

where there is an implicit restriction that $d,d_1,d_2$ are pairwise coprime. By Möbius formula, we have

$$\begin{aligned} S_2(X)&=\mathop {\sum \sum \sum }_{dd_1,dd_2\mid P(z)}\frac{\varrho _{dd_1}\varrho _{dd_2}}{\beta (d)\beta (d_1) \beta (d_2)}\{L(d)+L(d_1)+L(d_2)\}\sum _{l\mid (d_1,d_2)}\mu (l)\\&=\mathop {\sum \sum \sum \sum }_{ldd_1,ldd_2\mid P(z)}\frac{\mu (l)\varrho _{ldd_1}\varrho _{ldd_2}}{\beta (l)^2\beta (d) \beta (d_1)\beta (d_2)}\{L(ldd_1)+L(ldd_2)-L(d)\}\\&=2S_{21}(X)-S_{22}(X) \end{aligned}$$

with

$$\begin{aligned} S_{21}(X)&=\sum _{l\mid P(z)}\beta ^*(l)y_ly_l',\\ S_{22}(X)&=\sum _{l\mid P(z)}v(l)y_l^2, \end{aligned}$$

where for each $l\mid P(z),l\leqslant \sqrt{D},$

$$\begin{aligned} y_l'=\sum _{\begin{array}{c} d\mid P(z)\\ d\equiv 0\,({{\,\mathrm{mod}\,}}{ l}) \end{array}}\frac{\varrho _dL(d)}{\beta (d)}. \end{aligned}$$

and

$$\begin{aligned} v(l)&=\beta (l)\sum _{uv=l}\frac{\mu (u)L(v)}{\beta (u)}. \end{aligned}$$

(B.12)

Moreover, we have

$$\begin{aligned} y_l'&=\sum _{dl\mid P(z)}\frac{\varrho _{dl}L(dl)}{\beta (dl)}=\sum _{d\mid P(z)}\frac{\varrho _{dl}L(d)}{\beta (dl)}+L(l)y_l\\&=\sum _{p<z}\frac{\beta ^*(p)\log p}{\beta (p)}\sum _{d\mid P(z)}\frac{\varrho _{pdl}}{\beta (pdl)}+L(l)y_l\\&=\sum _{p<z}\frac{y_{pl}\beta ^*(p)\log p}{\beta (p)}+L(l)y_l. \end{aligned}$$

It then follows that

$$\begin{aligned} S_{21}(X)&=\sum _{p<z}\frac{\beta ^*(p)\log p}{\beta (p)}\sum _{l\mid P(z)}\beta ^*(l)y_ly_{pl}+\sum _{l\mid P(z)}L(l)\beta ^*(l)y_l^2\\&=\sum _{p<z}\frac{\beta ^*(p)\log p}{\beta (p)}\sum _{pl\mid P(z)}\beta ^*(l)y_ly_{pl}\\&\quad +\sum _{p<z}\frac{\beta ^*(p)^2\log p}{\beta (p)}\sum _{pl\mid P(z)}\beta ^*(l)y_{pl}^2\\&=S_{21}'(X)+S_{21}''(X), \end{aligned}$$

say.

From (B.9), it follows, by partial summation, that

$$\begin{aligned} S_{21}'(X)&=-\frac{16(1+o(1))}{{\mathcal {G}}(1)^2(\log D)^4}\sum _{l\mid P(z)}\frac{1}{\beta ^*(l)}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )\\&\quad \times \sum _{\begin{array}{c} p<z\\ p\not \mid l \end{array}}\frac{\log p}{\beta (p)}m_2\Big (\frac{\log (\sqrt{D}/(pl))}{\log z}\Big ). \end{aligned}$$

Up to a minor contribution, the inner sum over p can be relaxed to all primes $p\leqslant z.$ In fact, the terms with $p\mid \ell$ contribute at most

$$\begin{aligned}&\ll \frac{1}{(\log D)^4}\sum _{l\mid P(z)}\frac{1}{\beta ^*(l)}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )\sum _{p\mid l}\frac{\log p}{p}\\&\ll \frac{1}{(\log D)^3\log \log D}\sum _{l\mid P(z)}\frac{1}{\beta ^*(l)}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )\\&\ll \frac{1}{W(z)(\log D)^3\log \log D}. \end{aligned}$$

We then derive that

$$\begin{aligned} S_{21}'(X)&=-\frac{16(1+o(1))}{{\mathcal {G}}(1)^2(\log D)^4}\sum _{l\mid P(z)}\frac{1}{\beta ^*(l)}m_2\Big (\frac{\log (\sqrt{D}/l)}{\log z}\Big )\\&\quad \times \sum _{p<z}\frac{\log p}{\beta (p)}m_2\Big (\frac{\log (\sqrt{D}/(pl))}{\log z}\Big )\\&\qquad +O\Big (\frac{\log z}{\log \log z}\frac{1}{W(z)(\log D)^4}\Big )\\&=-\frac{32\tau c_{21}'(\tau )\log z}{{\mathcal {G}}(1)^2W(z)(\log D)^4}\cdot (1+o(1)), \end{aligned}$$

where

$$\begin{aligned} c_{21}'(\tau )=\int _0^1\int _0^1\sigma '((1-u)\tau ){\mathfrak {f}}(u\tau ){\mathfrak {f}}(u\tau -2v)\mathrm {d}u\mathrm {d}v. \end{aligned}$$

In a similar manner, we can also show that

$$\begin{aligned} S_{21}''(X)&=\frac{32\tau c_{21}''(\tau )\log z}{{\mathcal {G}}(1)^2W(z)(\log D)^4}\cdot (1+o(1)),\end{aligned}$$

where

$$\begin{aligned} c_{21}''(\tau )=\int _0^1\int _0^1\sigma '((1-u)\tau ){\mathfrak {f}}(u\tau -2v)^2\mathrm {d}u\mathrm {d}v. \end{aligned}$$

In conclusion, we obtain

$$\begin{aligned} S_{21}(X)=S_{21}'(X)+S_{21}''(X)&=\frac{32\tau (c_{21}''(\tau )-c_{21}'(\tau ))\log z}{{\mathcal {G}}(1)^2W(z)(\log D)^4}\cdot (1+o(1)), \end{aligned}$$

We now evaluate $S_{22}(X)$. For each squarefree $l\geqslant 1$, we have

$$\begin{aligned} v(l)&=\beta (l)\sum _{u\mid l}\frac{\mu (u)}{\beta (u)}\sum _{p\mid l/u}\frac{\beta ^*(p)\log p}{\beta (p)}\\&=\beta (l)\sum _{p\mid l}\frac{\beta ^*(p)\log p}{\beta (p)}\sum _{u\mid l/p}\frac{\mu (u)}{\beta (u)}\\&=\beta (l)\sum _{p\mid l}\frac{\beta ^*(l/p)\beta ^*(p)\log p}{\beta (l/p)\beta (p)}\\&=\beta ^*(l)\log l. \end{aligned}$$

Hence

$$\begin{aligned} S_{22}(X)&=\sum _{p<z}\beta ^*(p)\log p\sum _{pl\mid P(z)}\beta ^*(l)y_{pl}^2\\&=\frac{16(1+o(1))}{{\mathcal {G}}(1)^2(\log D)^4}\sum _{l\mid P(z)}\frac{1}{\beta ^*(l)}\sum _{\begin{array}{c} p<z\\ p\not \mid l \end{array}}\frac{\log p}{\beta ^*(p)}m_2\Big (\frac{\log (\sqrt{D}/(pl))}{\log z}\Big )^2 \end{aligned}$$

by (B.9). From partial summation, it follows that

$$\begin{aligned} S_{22}(X)&=\frac{32\tau c_{21}''(\tau )\log z}{{\mathcal {G}}(1)^2W(z)(\log D)^4}\cdot (1+o(1)). \end{aligned}$$

Combining all above evaluations, we find

$$\begin{aligned}&S(X,z;h,\varvec{\varrho })\\ =&\,{\mathcal {G}}(1)X\{S_1(X)\cdot (\log X+c)-2S_{21}(X)+S_{22}(X)\}+O(X{\mathcal {L}}^{-2})\\ =&\,(1+o(1))\frac{16\tau X\log z}{{\mathcal {G}}(1)W(z)(\log D)^4} \Big \{c_1(\tau )\frac{\log X}{\log z}+4c_{21}'(\tau )-2c_{21}''(\tau ))\Big \}. \end{aligned}$$

Hence Theorem B.1 follows by observing that $c_2(\tau )=2c_{21}'(\tau )-c_{21}''(\tau )$ and

$$\begin{aligned} {\mathcal {G}}(1)W(z)&=\prod _{p<z}\Big (1-\frac{1}{p}\Big )^2\cdot \prod _{p\geqslant z}\Big (1+\frac{h(p)}{p}\Big )\Big (1-\frac{1}{p}\Big )^2\\&=(1+o(1))\frac{\mathrm {e}^{-2\gamma }}{(\log z)^2} \end{aligned}$$

by Mertens’ formula.

Appendix C: Chebyshev approximation

A lot of statistical analysis of $GL_2$ objects relies heavily on the properties of Chebychev polynomials $\{U_k(x)\}_{k\geqslant 0}$ with $x\in [-\,1,1],$ which can be defined recursively by

$$\begin{aligned}&U_0(x)=1,\ \ U_1(x)=2x,\\&U_{k+1}(x)=2xU_k(x)-U_{k-1}(x),\ \ k\geqslant 1. \end{aligned}$$

It is well-known that Chebychev polynomials form an orthonormal basis of $L^2([-\,1, 1])$ with respect to the measure $\frac{2}{\pi }\sqrt{1-x^2}\mathrm {d}x$. In fact, for any $f\in {\mathcal {C}}([-\,1,1])$, the expansion

$$\begin{aligned} f(x)=\sum _{k\geqslant 0}\beta _k(f)U_k(x) \end{aligned}$$

(C.1)

holds with

$$\begin{aligned} \beta _k(f)=\frac{2}{\pi } \int _{-1}^1f(t)U_k(t)\sqrt{1-t^2}\mathrm {d}t. \end{aligned}$$

In practice, the following truncated approximation is usually more effective and useful, which has its prototype in [28, Theorem 5.14].

Lemma C.1

Suppose $f:[-\,1,1]\rightarrow {\mathbf {R}}$ has $C+1$ continuous derivatives on $[-\,1,1]$ with $C\geqslant 2$. Then for each positive integer $K>C,$ there holds the approximation

$$\begin{aligned} f(x)=\sum _{0\leqslant k\leqslant K}\beta _k(f)U_k(x)+O\Big (K^{1-C}\Vert f^{(C+1)}\Vert _1\Big ) \end{aligned}$$

uniformly in $x\in [-\,1,1]$, where the implied constant depends only on C.

Proof

For each $K>C$, we introduce the operator $\vartheta _{K}$ mapping $f \in {\mathcal {C}}^{C+1}([-\,1,1])$ via

$$\begin{aligned} (\vartheta _{K}f)(x):=\sum _{0\leqslant k\leqslant K}\beta _k(f)U_k(x)-f(x). \end{aligned}$$

This gives the remainder of approximation by Chebychev polynomials up to degree K. Obviously, $(\vartheta _{K}f)(\cdot )\in {\mathcal {C}}^{C+1}([-\,1,1])$ and in fact, $\vartheta _{K}$ is a bounded linear functional on ${\mathcal {C}}^{C+1}([-\,1,1]),$ which vanishes on polynomials of degree $\leqslant K$.

Using a theorem of Peano ([7, Theorem 3.7.1]), we find that

$$\begin{aligned} (\vartheta _{K}f)(x)=\frac{1}{C!}\int _{-1}^1f^{(C+1)}(t)H_K(x,t)\mathrm {d}t, \end{aligned}$$

(C.2)

where

$$\begin{aligned} H_K(x,t)=-\sum _{k>K}\lambda _k(t)U_k(x) \end{aligned}$$

with

$$\begin{aligned} \lambda _k(t)=\frac{2}{\pi }\int _t^1\sqrt{1-x^2}(x-t)^CU_k(x)\mathrm {d}x. \end{aligned}$$

Put $x=\cos \theta ,t=\cos \phi$, so that

$$\begin{aligned} \lambda _k(t) = \lambda _k(\cos \phi )=\frac{2}{\pi }\int _0^\phi (\cos \theta -\cos \phi )^C\sin \theta \sin ((k+1)\theta )\mathrm {d}\theta . \end{aligned}$$

We deduce from integration by parts that

$$\begin{aligned} \Vert \lambda _k\Vert _\infty \ll \frac{1}{k\genfrac(){0.0pt}0{k-1}{C}}, \end{aligned}$$

where the implied constant is absolute. For any $x,t\in [-\,1,1]$, the Stirling’s formula $\log \Gamma (k) = (k-1/2)\log k - k +\log \sqrt{2\pi } + O(1/k)$ gives

$$\begin{aligned}H_K(x,t)&\ll \sum _{k>K}\frac{1}{\genfrac(){0.0pt}0{k-1}{C}}=C!\sum _{k>K}\frac{\Gamma (k-C)}{\Gamma (k)}\\&\ll \sum _{k>K} \Big (\frac{\mathrm {e}}{k-C}\Big )^C \Big (1-\frac{C}{k}\Big )^{k-1/2}\\&\ll K^{1-C}, \end{aligned}$$

from which and (C.2) we conclude that

$$\begin{aligned} \Vert \vartheta _{K}f\Vert _\infty \ll K^{1-C}\Vert f^{(C+1)}\Vert _1. \end{aligned}$$

This completes the proof of the lemma. $\square$

We now turn to derive a truncated approximation for |x| on average.

Lemma C.2

Let k, J be two positive integers and $K>1.$ Suppose $\{x_j\}_{1\leqslant j\leqslant J}\in [-\,1,1]$ and ${\mathbf {y}}:=\{y_j\}_{1\leqslant j\leqslant J}\in {\mathbf {C}}$ are two sequences satisfying

$$\begin{aligned} \max _{1\leqslant j\leqslant J}|y_j|\leqslant 1,\ \ \ \Bigg |\sum _{1\leqslant j\leqslant J}y_jU_k(x_j)\Bigg |\leqslant k^BU \end{aligned}$$

(C.3)

with some $B\geqslant 1$ and $U>0$. Then we have

$$\begin{aligned} \sum _{1\leqslant j\leqslant J}y_j|x_j| =\frac{4}{3\pi }\sum _{1\leqslant j\leqslant J}y_j+O\Big (UK^{B-1}(\log K)^{\delta (B)}+\frac{\Vert {\mathbf {y}}\Vert _1^2}{UK^B}\Big ). \end{aligned}$$

where $\delta (B)$ vanishes unless $B=1$, in which case it is equal to 1, and the O-constant depends only on B.

Proof

In order to apply Lemma C.1, we would like to introduce a smooth function $R:[-\,1,1]\rightarrow [0,1]$ with $R(x)=R(-\,x)$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} R(x)=0,\ \ &{}x\in [-\,\varDelta ,\varDelta ],\\ R(x)=1,&{}x\in [-\,1,-2\varDelta ]\cup [2\varDelta ,1],\end{array}\right. } \end{aligned}$$

where $\varDelta \in ~]0,1[$ be a positive number to be fixed later. We also assume the derivatives satisfy

$$\begin{aligned} R^{(j)}(x)\ll _j \varDelta ^{-j} \end{aligned}$$

for each $j\geqslant 0$ with an implied constant depending only on j.

Put $f(x):=R(x)|x|.$ Due to smooth decay of R at $x=0,$ we may apply Lemma C.1 to f(x) with $C=2$, getting

$$\begin{aligned} f(x)&=\sum _{0\leqslant k\leqslant K}\beta _k(f)U_k(x)+O(K^{-1}\Vert f'''\Vert _1). \end{aligned}$$

Note that $f'''(x)$ vanishes unless $x\in [-\,2\varDelta ,-\varDelta ]\cup [\varDelta ,2\varDelta ]$, in which case we have $f'''(x)\ll \varDelta ^{-2}.$ It then follows that

$$\begin{aligned} f(x)&=\sum _{0\leqslant k\leqslant K}\beta _k(f)U_k(x)+O\Big (\frac{1}{K\varDelta }\Big ). \end{aligned}$$

Moreover, $f(x)-|x|$ vanishes unless $x\in [-\,2\varDelta ,2\varDelta ]$. This implies that $f(x)=|x|+O(\varDelta )$. In addition, $\beta _0(f)=\frac{4}{3\pi }+O(\varDelta )$. Therefore,

$$\begin{aligned} |x|&=\frac{4}{3\pi }+\sum _{1\leqslant k\leqslant K}\beta _k(f)U_k(x)+O\Big (\varDelta +\frac{1}{K\varDelta }\Big ). \end{aligned}$$

We claim that

$$\begin{aligned} \beta _k(f)\ll k^{-2} \end{aligned}$$

(C.4)

for all $k\geqslant 1$ with an absolute implied constant. It then follows that

$$\begin{aligned}&\sum _{1\leqslant j\leqslant J}y_j|x_j|-\frac{4}{3\pi }\sum _{1\leqslant j\leqslant J}y_j\\ =&\,\sum _{1\leqslant k\leqslant K}\beta _{k}(f)\sum _{1\leqslant j\leqslant J}y_jU_{k}(x_j)+O\Big (\Vert {\mathbf {y}}\Vert _1\varDelta +\frac{\Vert {\mathbf {y}}\Vert _1}{K\varDelta }\Big )\\ \ll&\, U\sum _{1\leqslant k\leqslant K}k^{B-2}+\Vert {\mathbf {y}}\Vert _1\varDelta +\frac{\Vert {\mathbf {y}}\Vert _1}{K\varDelta }\\ \ll&\, UK^{B-1}(\log K)^{\delta (B)}+\Vert {\mathbf {y}}\Vert _1\varDelta +\frac{\Vert {\mathbf {y}}\Vert _1}{K\varDelta }, \end{aligned}$$

where the implied constant depends only on B. To balance the first and last terms, we take $\varDelta =\Vert {\mathbf {y}}\Vert _1/(UK^B)$, which yields

$$\begin{aligned} \sum _{1\leqslant j\leqslant J}y_j|x_j|-\frac{4}{3\pi }\sum _{1\leqslant j\leqslant J}y_j&\ll UK^{B-1}(\log K)^{\delta (B)}+\frac{\Vert {\mathbf {y}}\Vert _1^2}{UK^B} \end{aligned}$$

as expected.

It remains to prove the upper bound (C.4). Since $U_k(\cos \theta ){=}\sin ((k{+}1)\theta )/\sin \theta$, it suffices to show that

$$\begin{aligned} \beta _k:=\int _0^{\frac{\pi }{2}}R(\cos \theta )(\sin 2\theta ) \sin ((k+1)\theta )\mathrm {d}\theta \ll k^{-2} \end{aligned}$$

(C.5)

for all $k\geqslant 3$ with an absolute implied constant. From the elementary identity $2\sin \alpha \sin \beta =\cos (\alpha -\beta )-\cos (\alpha +\beta )$, it follows that

$$\begin{aligned} \beta _k&=\int _0^{\arccos \varDelta }R(\cos \theta )(\sin 2\theta ) \sin ((k+1)\theta )\mathrm {d}\theta \\&=\frac{\alpha (k-1,R)-\alpha (k+3,R)}{2}, \end{aligned}$$

where, for $\ell \geqslant 2$ and a function $g\in {\mathcal {C}}^2([-\,1,1])$,

$$\begin{aligned} \alpha (\ell ,g):=\int _0^{\arccos \varDelta }g(\cos \theta )\cos (\ell \theta )\mathrm {d}\theta . \end{aligned}$$

From integration by parts, we derive that

$$\begin{aligned} \alpha (\ell ,g)&=\frac{1}{\ell }\int _0^{\arccos \varDelta } g'(\cos \theta )(\sin \theta )\sin (\ell \theta )\mathrm {d}\theta \\&=\frac{\alpha (\ell -1,g')-\alpha (\ell +1,g')}{2\ell }, \end{aligned}$$

and also

$$\begin{aligned} \alpha (\ell ,g')&=\frac{\alpha (\ell -1,g'')-\alpha (\ell +1,g'')}{2\ell }. \end{aligned}$$

It then follows that

$$\begin{aligned} \alpha (\ell ,g)&=\frac{\alpha (\ell -2,g'')-\alpha (\ell ,g'')}{4\ell (\ell -1)}-\frac{\alpha (\ell ,g'')-\alpha (\ell +2,g'')}{4\ell (\ell +1)}. \end{aligned}$$

We then further have

$$\begin{aligned} \beta _k&=\frac{1}{8}(\beta _{k,1}-\beta _{k,2}) \end{aligned}$$

with

$$\begin{aligned} \beta _{k,1}&=\frac{\alpha (k-3,R'')-\alpha (k-1,R'')}{(k-1)(k-2)} -\frac{\alpha (k-1,R'')-\alpha (k+1,R'')}{k(k-1)},\\ \beta _{k,2}&=\frac{\alpha (k-1,R'')-\alpha (k+1,R'')}{k(k+1)} -\frac{\alpha (k+1,R'')-\alpha (k+3,R'')}{(k+1)(k+2)}. \end{aligned}$$

Note that

$$\begin{aligned}&\alpha (k-3,R'')-\alpha (k-1,R'')\\ =&\,\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta )\{\cos ((k-3)\theta )-\cos ((k-1)\theta )\}\mathrm {d}\theta \\ =&\,2\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) (\sin (k-2)\theta )(\sin \theta )\mathrm {d}\theta \end{aligned}$$

and

$$\begin{aligned} \alpha (k-1,R'')-\alpha (k+1,R'')&=2\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) (\sin k\theta )(\sin \theta )\mathrm {d}\theta . \end{aligned}$$

Hence

$$\begin{aligned} \beta _{k,1}&=\frac{2}{(k-1)(k-2)}\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) (\sin (k-2)\theta )(\sin \theta )\mathrm {d}\theta \\&\ \ \ \ \ -\frac{2}{k(k-1)}\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) (\sin k\theta )(\sin \theta )\mathrm {d}\theta \\&=\frac{2}{(k-1)(k-2)}\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) \{\sin (k-2)\theta -\sin k\theta \}(\sin \theta )\mathrm {d}\theta \\&\ \ \ \ \ +\frac{4}{k(k-1)(k-2)}\int _{\arccos 2\varDelta }^{\arccos \varDelta } R''(\cos \theta )(\sin k\theta )(\sin \theta )\mathrm {d}\theta . \end{aligned}$$

The first term can be evaluated as

$$\begin{aligned}&=\frac{2}{(k-1)(k-2)}\int _{\arccos 2\varDelta }^{\arccos \varDelta }R''(\cos \theta ) (\sin (k-1)\theta )(\cos \theta )(\sin \theta )\mathrm {d}\theta \\&\ll \frac{1}{k^2}\int _{\arccos 2\varDelta }^{\arccos \varDelta }\varDelta ^{-2} \cos \theta \mathrm {d}\theta \ll \frac{1}{k^2}. \end{aligned}$$

Again from the integration by parts, the second term is

$$\begin{aligned}&\frac{4}{(k-1)(k-2)}\int _{\arccos 2\varDelta }^{\arccos \varDelta }R'(\cos \theta ) (\cos k\theta )\mathrm {d}\theta \\&\quad \ll \frac{1}{k^2}\int _{\arccos 2\varDelta }^{\arccos \varDelta }\varDelta ^{-1} \mathrm {d}\theta \ll \frac{1}{k^2}. \end{aligned}$$

Hence $\beta _{k,1}\ll k^{-2}$, and similarly $\beta _{k,2}\ll k^{-2}.$ These yield (C.5), and thus (C.4), which completes the proof of the lemma. $\square$

Note that $U_k(\cos \theta )=\mathrm {sym}_k(\theta ).$ Taking $x_j=\cos \theta _j$ in Lemma C.2, we obtain the following truncated approximation for $|\cos |$.

Lemma C.3

Let k, J be two positive integers and $K>1.$ Suppose $\{\theta _j\}_{1\leqslant j\leqslant J}\in [0,\pi ]$ and ${\mathbf {y}}:=\{y_j\}_{1\leqslant j\leqslant J}\in {\mathbf {C}}$ are two sequences satisfying

$$\begin{aligned} \max _{1\leqslant j\leqslant J}|y_j|\leqslant 1,\ \ \ \Bigg |\sum _{1\leqslant j\leqslant J}y_j\mathrm {sym}_k(\theta _j)\Bigg |\leqslant k^BU \end{aligned}$$

with some $B\geqslant 1$ and $U>0.$ Then we have

$$\begin{aligned} \sum _{1\leqslant j\leqslant J}y_j|\cos \theta _j| =\frac{4}{3\pi }\sum _{1\leqslant j\leqslant J}y_j+O\Big (UK^{B-1}(\log K)^{\delta (B)}+\frac{\Vert {\mathbf {y}}\Vert _1^2}{UK^B}\Big ). \end{aligned}$$

where $\delta (B)$ is defined as in Lemma C.2.