Additive twists of Fourier coefficients of symmetric-square lifts (Q415264)

Finite sums of the Fourier coefficients of a holomorphic cusp form \(f\) twisted by additive characters of \(\mathbb R\) are known to exhibit a ``square-root'' cancellation in terms of the length of the sum: for any real \(\alpha\) and \(N \geq 1\) one has, NEWLINE\[NEWLINE \underset{n \leq N} \sum \lambda_f(n) e(\alpha n) \ll_f N^{1/2} \log 2N . \tag{1} NEWLINE\]NEWLINE Here the implied constant depends only on \(f\) and not on \(\alpha\) and \(e(z):= \exp (2 \pi i z)\).NEWLINENEWLINEIn this paper, the authors study sums of the form NEWLINE\[NEWLINE \underset{n \leq N}\sum \lambda_j(n^2) e(\alpha n),NEWLINE\]NEWLINE where \(\lambda_j(m)\) is the \(m\)-th Hecke eigenvalue of a Maass form \(\phi\) for the full modular group with Laplacian eigenvalue \(1/4 + t_j^2\).NEWLINENEWLINEIf \(F\) denotes the symmetric-square lift of \(\phi\) to a \(\mathrm{GL}_3\) Maass form, then a closely related sum to consider is NEWLINE\[NEWLINE S_F(N) := \underset{n \leq N}\sum A_F(1,n) e(\alpha n), NEWLINE\]NEWLINE where \(A_F(1,n) = \underset{ml^2=n} \sum \lambda_j(m^2)\) and NEWLINE\[NEWLINE L_F(s) = \sum_{n=1}^\infty A_F(1,n) n^{-s} NEWLINE\]NEWLINE is the \(L\) function attached to \(F\).NEWLINENEWLINEIt was proved by \textit{S. D. Miller} [Am. J. Math. 128, No. 3, 699--729 (2006; Zbl 1142.11033)] that \(S_F(N) \ll_F N^{3/4 + \varepsilon}\), with the implied constant independent of \(\alpha\), but depending on \(F\).NEWLINENEWLINEThis paper brings out explicitly this dependence. More precisely they prove the following. {Theorem. } Suppose \(F\) is the symmetric square lift of a \(\mathrm{SL}_2(\mathbb Z)\) Hecke-Maass form with \(A_F(1,1)=1\). Then assuming the Ramanujan conjecture at the finite places for \(F\), we have with \(D=1/4\), NEWLINE\[NEWLINE S_F(N) \ll N^{3/4 + \varepsilon} \lambda_F(\Delta)^{D + \varepsilon}, \tag{1} NEWLINE\]NEWLINE where \(\lambda_F(\Delta)\) is the Laplace eigenvalue of \(F\). The implies constant depends on \(\varepsilon\) only. Unconditionally, one has (1) with \(D=1/3\).