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In an attempt to solve an unrelated problem, I was led to the task of estimating/bounding from above sums of the form $$\sum_{m=1}^\infty\lambda(m)e\left(-\frac{am}{q}\right)h(m)$$ where $\sum_{m=1}^\infty\lambda(m)e(mz)$ is the Fourier expansion of a cusp form of weight $1$ and a certain level $N$, with respect to some Dirichlet character $\chi$, and $h$ is some smooth compactly supported function. Note that the modulus of $\chi$ is NOT (necessarily) $q$; I need to work with the above display for a general positive integer $q$.

Now, my knowledge of all things modular is rather limited, but my understanding is that sums like the above are best handled via some variation of Voronoï summation. However, I have not been able to find a specific formulation of Voronoï summation in the literature that suits my situation. For example, in https://arxiv.org/pdf/math/0304187.pdf, Theorem 4.12 only deals with cusp forms with respect to the full modular group (or at least this is my understanding - I do not master the language used at the beginning of Section 4). In Iwaniec and Kowalski's Analytic Number Theory, Exercise 9 in Chapter 4 deals with the case of modular forms twisted by characters, but still, as far as I understand, only with respect to the full modular group (e.g. level $1$) and when the modulus of $\chi$ divides $q$.

It is possible that, for experts, it is clear how to handle the sum I presented via one of these two versions, but I do not see how to do it and I would appreciate any suggestion on general principles regarding approaches to this sum.

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  • $\begingroup$ Does Lemma 2.4 of arxiv.org/pdf/1707.01576.pdf do what you need? $\endgroup$
    – m34
    Nov 17, 2022 at 17:33
  • $\begingroup$ That lemma equates the sum that the OP is studying to an even more complicated expression involving three sums and an integral. In particular, it gives no upper bound for the OP's sum. $\endgroup$
    – Alex M.
    Nov 17, 2022 at 18:18
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    $\begingroup$ In an attempt to solve an unrelated problem, I was led to... - Doesn't that mean it's not unrelated? $\endgroup$
    – Kimball
    Nov 18, 2022 at 6:07
  • $\begingroup$ What is the support of $h$ here? Is $h$ a fixed test function? Then the sum is trivially $O(1)$. Also, you need to specify which parameters ($\pi$, $N$, $q$, $C(\chi)$, etc.) are fixed and which are varying. $\endgroup$ Nov 18, 2022 at 9:48
  • $\begingroup$ @m34 At a first glance it seems to be the sort of thing I want, yes, thank you very much! Other remarks are still welcome, of course :) $\endgroup$
    – user50139
    Nov 18, 2022 at 19:22

1 Answer 1

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Consider the cuspidal representation $\pi:=\pi_f\otimes\pi_\infty$ of $\mathrm{GL}_2(\mathbb{A})$ with the central character $\omega_\chi$, the Hecke character attached to $\chi$, such that $\pi_f$ has level $N$ and $\pi_\infty$ is isomorphic to the Discrete series of weight $k$. Let $\phi\in\pi$ be a Hecke-normalized factorizable even newform with the Whittaker function $W_\phi=\otimes_{p\le\infty}W_p$ (in this case, $W_p$ is the normalized (that is $W_p(1)=1$) newvector of $\pi_p$) such that $$W_\infty\left[\begin{pmatrix}y&\\&1\end{pmatrix}\right]=f(y),$$ where $f$ is the test function mentioned in the comment. The last restriction is possible via the Theory of the Kirillov model: $$C_c^\infty(\mathbb{R}^\times)\subset\left\lbrace W\left[\begin{pmatrix}\cdot&\\&1\end{pmatrix}\right]\mid W\in\pi_\infty\right\rbrace.$$ The Fourier expansion of $\phi$ is given by $$\phi(g)=\sum_{\gamma\in\mathbb{Q}^\times}W\left[\begin{pmatrix}\gamma&\\&1\end{pmatrix}g\right].$$ If $$g=\left(1,\begin{pmatrix}1&-a/q\\&1\end{pmatrix}\begin{pmatrix}1/B&\\&1\end{pmatrix}\right)\in\mathrm{GL}_2(\mathbb{A}_f)\times\mathrm{GL}_2(\mathbb{R}),$$ then $$\phi(g) = \sum_{m=1}^\infty\frac{\lambda(m)}{\sqrt{m}}e\left(-\frac{am}{q}\right)f(m/B).$$ Note that using $\lambda(m)\ll m^{O(1)}$ one can truncate the above sum by $m\le B^{1+\epsilon}$ with $O(B^{-N})$ error. Using the bound $\|\phi\|_\infty\ll_\pi 1$ we conclude that the sum in the OP is $O(B^{1/2+\epsilon})$.

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