Mean squares of quadratic twists of the Möbius function

For all odd integers $k$ and all integers $m$, define the Gauss-type sums $G_{m}(k)$, as in [sound1, Sect. 2.2],

(2.1)

$\displaystyle G_{m}(k)=\left(\frac{1-i}{2}+\left(\frac{-1}{k}\right)\frac{1+i}{2}\right)\sum_{a\negthickspace\negthickspace\negthickspace\pmod{k}}\left(\frac{a}{k}\right)e\left(\frac{am}{k}\right),\quad\mbox{where}\quad e(x)=\exp(2\pi ix).$

Let $\varphi(m)$ be the Euler totient of $m$. Our next result is taken from [sound1, Lemma 2.3] and evaluates $G_{m}(k)$.

For any smooth function $F$, we write $\hat{F}$ for the Fourier transform of $F$ and we define

(2.2)

$$\widetilde{F}(\xi)=\frac{1+i}{2}\hat{F}(\xi)+\frac{1-i}{2}\hat{F}(-\xi)=\int\limits^{\infty}_{-\infty}\left(\cos(2\pi\xi x)+\sin(2\pi\xi x)\right)F(x)\mathrm{d}x.$$

We note the following Poisson summation formula from [sound1, Lemma 2.6].

From [iwakow, Theorem 5.19], we deduce the following.

We define for any square-free $k_{1}$,

(2.3)

$$Z(\alpha,\beta,\gamma;q,k_{1})=\sum_{k_{2}=1}^{\infty}\sum_{(n_{1},2q)=1}\sum_{(n_{2},2q)=1}\frac{\mu(n_{1})\mu(n_{2})}{n_{1}^{\alpha}n_{2}^{\beta}k_{2}^{2\gamma}}\frac{G_{k_{1}k_{2}^{2}}(n_{1}n_{2})}{n_{1}n_{2}},$$

where $G_{m}(k)$ be defined as in (2.1). Note first that Lemma 2.2 implies that $Z(\alpha,\beta,\gamma;q,k_{1})$ converges absolutely when $\Re(\alpha)$, $\Re(\beta)$, and $\Re(\gamma)$ are all strictly greater than $\tfrac{1}{2}$. We write $L_{c}(s,\chi)$ for the Euler product of $L(s,\chi)$ with the factors from $p|c$ removed. Our next lemma describes the analytical behavior of $Z$.

Expanding the square in (1.4) allows us to recast ${\mathcal{S}}(X,Y;\Phi,W)$ as

$$S(h):=\sideset{}{{}^{*}}{\sum}_{d}\sum_{n_{1}}\sum_{n_{2}}\chi_{8d}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})h(d,n_{1},n_{2}),$$

where $h(x,y,z)=W(\frac{x}{X})\Phi\left(\frac{y}{Y}\right)\Phi\left(\frac{z}{Y}\right)$ is a smooth function on ${\mathbb{R}}_{+}^{3}$. We apply the Möbius inversion to remove the square-free condition on $d$ to obtain that, for an appropriate parameter $Z$ to be chosen later,

$\displaystyle S(h)=\Big{(}\sum_{\begin{subarray}{c}a\leq Z\\ (a,2)=1\end{subarray}}+\sum_{\begin{subarray}{c}a>Z\\ (a,2)=1\end{subarray}}\Big{)}\mu(a)\sum_{(d,2)=1}\sum_{(n_{1},a)=1}\sum_{(n_{2},a)=1}\chi_{8d}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})h(da^{2},n_{1},n_{2})=:S_{1}(h)+S_{2}(h),\quad\mbox{say}.$

We first estimate $S_{2}(h)$. To this end, writing $d=b^{2}\ell$ with $\ell$ square-free, and grouping terms according to $c=ab$, we deduce

(3.1)

$$S_{2}(h)=\sum_{(c,2)=1}\sum_{\begin{subarray}{c}a>Z\\ a|c\end{subarray}}\mu(a)\sideset{}{{}^{*}}{\sum}_{\ell}\sum_{(n_{1},c)=1}\sum_{(n_{2},c)=1}\chi_{8\ell}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})h(c^{2}\ell,n_{1},n_{2}).$$

Applying Mellin transforms in the variables $n_{1}$ and $n_{2}$ yeilds that the inner triple sum over $\ell$, $n_{1}$, and $n_{2}$ in (3.1) is

(3.2)

$$\frac{1}{(2\pi i)^{2}}\int\limits_{(1+\varepsilon)}\int\limits_{(1+\varepsilon)}\sideset{}{{}^{*}}{\sum}_{\ell}{\check{h}}(c^{2}\ell;u,v)\sum_{\begin{subarray}{c}n_{1},n_{2}\\ (n_{1}n_{2},c)=1\end{subarray}}\frac{\chi_{8\ell}(n_{1})\chi_{8\ell}(n_{2})\mu(n_{1})\mu(n_{2})}{n_{1}^{u}n_{2}^{v}}\mathrm{d}u\,\mathrm{d}v,$$

where

$${\check{h}}(x;u,v)=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}h(x,y,z)y^{u}z^{v}\frac{\mathrm{d}y}{y}\frac{\mathrm{d}z}{z}.$$

Now integration by parts gives that for $\Re(u)$, $\Re(v)>0$ and any positive integers $A_{j}$, $1\leq j\leq 3$,

(3.3)

$${\check{h}}(x;u,v)\ll\left(1+\frac{x}{X}\right)^{-A_{1}}\frac{Y^{\Re(u)+\Re(v)}}{|uv|(1+|u|)^{A_{2}}(1+|v|)^{A_{3}}}.$$

Note that the sum over $n_{1}$ and $n_{2}$ in (3.2) equals $L^{-1}_{c}(u,\chi_{8\ell})L^{-1}_{c}(v,\chi_{8\ell})$ and we can thus move the lines of integration in (3.2) to $\Re(u)=\Re(v)=1/2+\varepsilon$ without encountering any poles under GRH. Moreover,

(3.4)

$\displaystyle|L^{-1}_{c}(u,\chi_{8\ell})L^{-1}_{c}(v,\chi_{8\ell})|\leq d(c)^{2}(|L^{-1}(u,\chi_{8\ell})|^{2}+|L^{-1}(v,\chi_{8\ell})|^{2}),$

where $d(c)$ denotes the value of the divisor function at $c$.

We now apply (3.3) with $A_{2}=A_{3}=1$ and $A_{1}$ sufficiently large and Lemma 2.6 to get that the expression in (3.2) is

$$\ll d(c)^{2}Y^{1+2\varepsilon}\int\limits_{-\infty}^{\infty}(1+|t|)^{-2}\sideset{}{{}^{*}}{\sum}_{\ell}\left(1+\frac{c^{2}\ell}{X}\right)^{-A_{1}}\Big{|}L(\tfrac{1}{2}+\varepsilon+it,\chi_{8\ell})\Big{|}^{-2}\ \mathrm{d}t\ll d(c)^{2}(XY)^{1+\varepsilon}/c^{2}.$$

We conclude from the above estimation and (3.1) that

(3.5)

$\displaystyle S_{2}(h)\ll(XY)^{1+\varepsilon}Z^{-1+\varepsilon}.$

We evaluate $S_{1}(h)$ now. Write for brevity $C=\cos$ and $S=\sin$. We then apply the Poisson summation formula, Lemma 2.4, to deduce that

(3.6)

$\displaystyle S_{1}(h)=\frac{X}{2}\sum_{\begin{subarray}{c}a\leq Z\\ (a,2)=1\end{subarray}}\frac{\mu(a)}{a^{2}}\sum_{k\in\mathbb{Z}}\sum_{(n_{1},2a)=1}\sum_{(n_{2},2a)=1}\frac{(-1)^{k}G_{k}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})}{n_{1}n_{2}}\int\limits_{0}^{\infty}h(xX,n_{1},n_{2})(C+S)\left(\frac{2\pi kxX}{2n_{1}n_{2}a^{2}}\right)\mathrm{d}x.$

Let $S_{1,0}(h)$ for the terms in (3.6) with $k=0$. Note that

$$\sum_{\begin{subarray}{c}a\leq Z\\ (a,2n_{1}n_{2})=1\end{subarray}}\frac{\mu(a)}{a^{2}}=\frac{1}{\zeta(2)}\prod_{p|2n_{1}n_{2}}\left(1-\frac{1}{p^{2}}\right)^{-1}+O(Z^{-1})=\frac{8}{\pi^{2}}\prod_{p|n_{1}n_{2}}\left(1-\frac{1}{p^{2}}\right)^{-1}+O(Z^{-1}).$$

Moreover, with $\square$ denoting a perfect square, Lemma 2.2 implies that $G_{0}(m)=\varphi(m)$ if $m=\square$, and is zero otherwise. Thus, upon setting $h_{1}(y,z)=\int_{\mathbb{R}_{+}}h(xX,y,z)\ \mathrm{d}x$, we infer that

(3.7)

$\displaystyle S_{1,0}(h)=\frac{4X}{\pi^{2}}\sum_{\begin{subarray}{c}(n_{1}n_{2},2)=1\\ n_{1}n_{2}=\square\end{subarray}}\mu(n_{1})\mu(n_{2})\prod_{p|n_{1}n_{2}}\left(\frac{p}{p+1}\right)h_{1}\left(n_{1},n_{2}\right)+O\Big{(}\frac{X}{Z}\sum_{\begin{subarray}{c}(n_{1}n_{2},2)=1\\ n_{1}n_{2}=\square\end{subarray}}\Big{|}\mu(n_{1})\mu(n_{2})h_{1}(n_{1},n_{2})\Big{|}\Big{)}.$

Mark that the definition of $h$ implies that $h_{1}\ll 1$ and $h_{1}=0$ unless both $n_{1}$ and $n_{2}$ are $\ll Y$. Furthermore, if $n_{1}$, $n_{2}$ are square-free, then $n_{1}n_{2}=\square$ implies that $n_{1}=n_{2}$. Consequently, the sum in the $O$-term in (3.7) is $\ll Y$ and

$\displaystyle S_{1,0}(h)=\frac{4X}{\pi^{2}}\sum_{\begin{subarray}{c}(n,2)=1\end{subarray}}\prod_{p|n}\left(\frac{p}{p+1}\right)\mu^{2}(n)h_{1}\left(n,n\right)+O\left(\frac{XY}{Z}\right).$

We now apply the Mellin transform to recast $h_{1}(n,n)$ as

$$h_{1}(n,n)=\frac{1}{2\pi i}\int\limits_{(2)}\frac{Y^{u}}{n^{u}}\widetilde{h}_{1}(u,u)\mathrm{d}u,\quad\mbox{where}\quad\widetilde{h}_{1}(u,u)=\int\limits_{\mathbb{R}_{+}}h_{1}(yY,yY)y^{u}\frac{\mathrm{d}y}{y}.$$

Similar to (3.3), we have that for $\Re(u)>0$ and any integer $B\geq 0$,

(3.8)

$$\widetilde{h}_{1}(u,u)\ll\frac{1}{|u|(1+|u|)^{B}}.$$

Now we can rewrite $S_{1,0}$ as

(3.9)

$$S_{1,0}(h)=\frac{4X}{\pi^{2}}\frac{1}{2\pi i}\int\limits_{(2)}Y^{u}\widetilde{h}_{1}(u,u)Z(u)\mathrm{d}u+O\left(\frac{XY}{Z}\right),\quad\mbox{where}\quad Z(u)=\sum_{\begin{subarray}{c}(n,2)=1\end{subarray}}\frac{\mu^{2}(n)}{n^{u}}\prod_{p|n}\left(\frac{p}{p+1}\right).$$

We compute the Euler factors of $Z(u)$ to see that

(3.10)

$\displaystyle Z(u)=\prod_{p>2}\left(1+\frac{p}{p+1}\cdot\frac{1}{p^{u}}\right)=:\zeta(u)Z_{2}(u),$

where $Z_{2}(u)$ converges absolutely in the region $\Re(u)\geq\frac{1}{2}+\varepsilon$ for any $\varepsilon>0$.

Moving the line of integration in (3.9) to $\Re(u)=\frac{1}{2}+\varepsilon$, we encounter a simple pole at $u=1$ whose residue gives rise to the main term

$$\frac{4}{\pi^{2}}XY\widetilde{h}_{1}(1,1)Z_{2}(1).$$

Now to estimate the integral on the $\frac{1}{2}+\varepsilon$ line, we apply the functional equation for $\zeta(s)$ (see [Da, §8]) and Stirling’s formula, together with the convexity bound for $\zeta(s)$, rendering

$\displaystyle\zeta(s)\ll\begin{cases}1\qquad&\Re(s)>1,\\ (1+|s|)^{(1-\Re(s))/2}\qquad&0<\Re(s)<1,\\ (1+|s|)^{1/2-\Re(s)}\qquad&\Re(s)\leq 0.\end{cases}$

The above and (3.8) with $B=1$ enable us to gather that the integral on the $\frac{1}{2}+\varepsilon$ line contributes $\ll XY^{1/2+\varepsilon}$. One can easily check here that the Lindelöf hypothesis, a consequence of GRH whose truth we assume, does not lead to a better bound. Now the above discussion, together with (3.9), implies that

(3.11)

$\displaystyle S_{1,0}(h)=\frac{4}{\pi^{2}}XY\widetilde{h}_{1}(1,1)Z_{2}(1)+O\left(\frac{XY}{Z}+XY^{1/2+\varepsilon}\right).$

Here we note that

(3.12)

$\displaystyle\widetilde{h}_{1}(1,1)=\int\limits_{\mathbb{R}}W(x)\mathrm{d}x\left(\int\limits_{\mathbb{R}}\Phi(y)\mathrm{d}y\right)^{2}.$

Let $S_{3}(h)$ denote the contribution to $S_{1}(h)$ from the terms with $k\neq 0$ in (3.6). Let $f$ be a smooth function on $\mathbb{R}_{+}$ with rapid decay at infinity and $f$ itself and all its derivatives have finite limits as $x\to 0^{+}$. We consider the transform given by

$$\widehat{f}_{CS}(y):=\int\limits_{0}^{\infty}f(x)CS(2\pi xy)\mathrm{d}x,$$

where $CS$ stands for either the cosine or the sine function. It is shown in [S&Y, Sec. 3.3] that

$$\widehat{f}_{CS}(y)=\frac{1}{2\pi i}\int\limits_{(1/2)}\widetilde{f}(1-s)\Gamma(s)CS\left(\frac{\text{sgn}(y)\pi s}{2}\right)(2\pi|y|)^{-s}\mathrm{d}s.$$

Applying the above transform, we deduce that

(3.13)

$\displaystyle\begin{split}\int\limits_{0}^{\infty}h\left(Xx,n_{1},n_{2}\right)(C+S)\left(\frac{2\pi kxX}{2n_{1}n_{2}a^{2}}\right)\mathrm{d}x=\frac{1}{2\pi iX}\int\limits_{(\varepsilon)}\check{h}\left(1-s;n_{1},n_{2}\right)\left(\frac{n_{1}n_{2}a^{2}}{\pi|k|}\right)^{s}\Gamma(s)(C+\text{sgn}(k)S)\left(\frac{\pi s}{2}\right)\mathrm{d}s,\end{split}$

where

$$\check{h}(s;y,z)=\int\limits_{0}^{\infty}h(x,y,z)x^{s}\frac{\mathrm{d}x}{x}.$$

Taking the Mellin transforms in the variables $n_{1}$ and $n_{2}$, the right-hand side of (3.13) equals

$$\frac{1}{X}\left(\frac{1}{2\pi i}\right)^{3}\int\limits_{(1)}\int\limits_{(1)}\int\limits_{(\varepsilon)}\widetilde{h}\left(1-s,u,v\right)\frac{1}{n_{1}^{u}n_{2}^{v}}\left(\frac{n_{1}n_{2}a^{2}}{\pi|k|}\right)^{s}\Gamma(s)(C+\text{sgn}(k)S)\left(\frac{\pi s}{2}\right)\mathrm{d}s\,\mathrm{d}u\,\mathrm{d}v,$$

where

$$\widetilde{h}(s,u,v)=\int_{\mathbb{R}_{+}^{3}}h(x,y,z)x^{s}y^{u}z^{v}\frac{\mathrm{d}x}{x}\frac{\mathrm{d}y}{y}\frac{\mathrm{d}z}{z}.$$

Integrating by parts implies that for $\Re(s)$, $\Re(u)$, $\Re(v)>0$ and any integers $E_{j}\geq 0$, $1\leq j\leq 3$,

(3.14)

$$|\widetilde{h}(s,u,v)|\ll\frac{X^{\Re(s)}Y^{\Re(u)+\Re(v)}}{|uvs|(1+|s|)^{E_{1}}(1+|u|)^{E_{2}}(1+|v|)^{E_{3}}}.$$

Applying the above bound in (3.6) leads to

(3.15)

$$\begin{split}S_{3}(h)=\frac{1}{2}\sum_{\begin{subarray}{c}a\leq Z\\ (a,2)=1\end{subarray}}&\frac{\mu(a)}{a^{2}}\sum_{k\neq 0}\sum_{(n_{1},2a)=1}\sum_{(n_{2},2a)=1}\frac{(-1)^{k}G_{k}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})}{n_{1}n_{2}}\\ &\times\left(\frac{1}{2\pi i}\right)^{3}\int\limits_{(\varepsilon)}\int\limits_{(\varepsilon)}\int\limits_{(\varepsilon)}\widetilde{h}\left(1-s,u,v\right)\frac{1}{n_{1}^{u}n_{2}^{v}}\left(\frac{n_{1}n_{2}a^{2}}{\pi|k|}\right)^{s}\Gamma(s)(C+\text{sgn}(k)S)\left(\frac{\pi s}{2}\right)\mathrm{d}s\,\mathrm{d}u\,\mathrm{d}v.\end{split}$$

Note that by (3.14) and the estimation (see [S&Y, p. 1107]),

(3.16)

$$\Big{|}\Gamma(s)(C\pm S)\Big{(}\frac{\pi s}{2}\Big{)}\Big{|}\ll|s|^{\Re(s)-1/2},$$

the integral over $s$ in (3.15) may be taken over any vertical lines between $0$ and $1$ and the integrals over $u,v$ in (3.15) may be taken over any vertical lines between $0$ and $2$.

Hence we arrive at

(3.17)

$$\begin{split}S_{3}(h)=\frac{1}{2}\sum_{\begin{subarray}{c}a\leq Z\\ (a,2)=1\end{subarray}}&\frac{\mu(a)}{a^{2}}\left(\frac{1}{2\pi i}\right)^{3}\int\limits_{(\varepsilon)}\int\limits_{(\varepsilon)}\int\limits_{(\varepsilon)}\widetilde{h}\left(1-s,u,v\right)\sum_{k\neq 0}\sum_{(n_{1},2a)=1}\\ &\times\sum_{(n_{2},2a)=1}\frac{(-1)^{k}G_{k}(n_{1}n_{2})\mu(n_{1})\mu(n_{2})}{n_{1}n_{2}}\frac{1}{n_{1}^{u}n_{2}^{v}}\left(\frac{n_{1}n_{2}a^{2}}{\pi|k|}\right)^{s}\Gamma(s)(C+\text{sgn}(k)S)\left(\frac{\pi s}{2}\right)\mathrm{d}s\,\mathrm{d}u\,\mathrm{d}v.\end{split}$$