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I have a question on the large sieve inequality involving $GL(2)$ harmonics. Recall that one has the analog for $GL(1)$ harmonics that, for any complex numbers $\alpha_m,\beta_n$, one has $$\sum_{q\le Q} \hskip 0.5em \sideset{_{}^{}}{^{\ast}_{}}\sum\limits_{a \bmod q}\, \, \left|\sum_{n\le N}\sum_{\substack{m\le M\\(mn,q) =1}} \alpha_m \beta_n e\left(\frac{a m\overline{n}}{q}\,\,\right )\right|^2\le (Q^2+MN) \|\alpha\|^2\, \|\beta\|^2$$ (see, for example, Exercise 5 on Page 185 of Iwaniec-Kowalski's book, Analytic Number Theory).

My question is whether we have: $$\sum_{q\le Q} \hskip 0.5em \sideset{_{}^{}}{^{\ast}_{}}\sum\limits_{a \bmod q}\, \, \left|\sum_{n\le N}\sum_{\substack{m\le M\\(mn,q) =1}} \alpha_m \beta_n \frac{S(am\overline{n}\,\,,1;q) }{\sqrt{q} }\,\,\right|^2\le (Q^2+MN) \|\alpha\|^2\, \|\beta\|^2\,\,\,?,$$ where $S(m,n;c)$ denotes the classical Kloosterman sum.

If any expert here knows something on this question, please show some guides or relevant references.

Many thanks in advance.

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  • $\begingroup$ Have you tried the following: One can first make things bigger and drop the assumption $(a,q)=1$ in the $a$-sum. Then one can open the square, insert the definition of the Kloosterman sum. Then the $a$-sum should be a complete linear sum, which can be executed using orthogonality. It seems that after rewriting the result one arrives at something that looks similar to the first large sieve inequality you state. Maybe I am missing something obvious though. $\endgroup$ Commented Oct 31 at 13:04

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Let me spell out my comment made earlier. We write $$S=\sum_{q\leq Q}{\sum_{a\text{ mod }q}}^{\ast}\left\vert \sum_{n\leq N}\sum_{\substack{m\leq M,\\ (mn,q)=1}}\alpha_m\beta_n\frac{S(am\overline{n},1;q)}{\sqrt{q}}\right\vert^2.$$ We make $S$ larger by extending the $a$-sum to a sum over all congruence classes modulo $q$. Next, we open the square. This leads to $$S\leq\sum_{q\leq Q} \sum_{n_1,n_2\leq N}\sum_{\substack{m_1,m_2\leq M,\\ (m_1m_2n_1n_2,q)=1}}\alpha_{m_1}\overline{\alpha_{m_2}}\beta_{n_1}\overline{\beta_{n_2}} \underbrace{\frac{1}{q}\sum_{a\text{ mod }q}S(am_1\overline{n_1},1;q)\overline{S(am_2\overline{n_2},1;q)}}_{T_q(n_1,n_2,m_1,m_2)}.$$ The next stept is to compute the sum $T_q(n_1,n_2,m_1,m_2)$ explicitly by opening the definition of the Kloosterman sum and executing the $a$-sum: \begin{align} T_q(n_1,n_2,m_1,m_2) &= {\sum_{x,y\text{ mod }q}}^{\ast} e(\frac{\overline{x}-\overline{y}}{q})\frac{1}{q}\sum_{a\text{ mod }q} e(a\cdot \frac{m_1\overline{n_1}x-m_2\overline{n_2}y}{q}) \\ &= \sum_{\substack{x,y\text{ mod }q,\\ (xy,q)=1,\\ m_1\overline{n_1}x\equiv m_2\overline{n_2}y \text{ mod }q}} e(\frac{\overline{x}-\overline{y}}{q}) \\ &= {\sum_{y\text{ mod }q}}^{\ast} e(\frac{\overline{n_1m_2}m_1n_2\overline{y}-\overline{y}}{q}) = {\sum_{y\text{ mod }q}}^{\ast} e(\frac{m_1\overline{n_1}y-m_2\overline{n_2}y}{q}). \end{align} For the last equality we have simply made a change of variables in the $y$-sum using that $(m_1m_2n_1n_2,q)=1$. Inserting this result in our bound for $S$ gives \begin{align} S&\leq\sum_{q\leq Q} \sum_{n_1,n_2\leq N}\sum_{\substack{m_1,m_2\leq M,\\ (m_1m_2n_1n_2,q)=1}}\alpha_{m_1}\overline{\alpha_{m_2}}\beta_{n_1}\overline{\beta_{n_2}} {\sum_{y\text{ mod }q}}^{\ast} e(\frac{m_1\overline{n_1}y-m_2\overline{n_2}y}{q})\\ &= \sum_{q\leq Q}{\sum_{y\text{ mod }q}}^{\ast}\left\vert \sum_{n\leq N}\sum_{\substack{m\leq M,\\ (mn,q)=1}}\alpha_m\beta_n e(\frac{ym\overline{n}}{q})\right\vert^2. \end{align} We can now apply the first result stated in the question and obtain the desired bound $$S\leq (Q^2+MN)\Vert\alpha\Vert^2\Vert\beta\Vert^2.$$

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    $\begingroup$ I've seen this kind of argument in several places. The first one that comes to mind is a paper of Chandee and Li on the second moment of $GL(4)\times GL(2)$ $L$-functions. Specifically, on page 20 of arxiv.org/pdf/2111.05406, they extend a sum to all residues and apply orthogonality just as you do. $\endgroup$ Commented Nov 1 at 4:18

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