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If $f$ is a Maass form and $p$-Hecke eigenvalue (i.e. Hecke eigenvalue of usual Hecke operator $T_p$) of $f$ is $\lambda_f(p)$, do we know anything about lower bound of the sum$$S(x) = \sum_{x\le p\le 2x}|\lambda_f(p)|^2?$$

To avoid Confusion $$(T_pf)(z)=\frac{1}{\sqrt{p}}\left[\sum_{b=0}^{p-1}f\left(\frac{z+b}{p}\right)+f(pz)\right]$$ Any referecne would be highly helpful.

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If $x$ is large enough, then Rankin-Selberg theory will show that $S(x) \gg x^{1-\varepsilon}$. However, if $x$ is not large enough, then it is unknown how to obtain a lower bound for $S(x)$. In particular, it is unknown how to show that $S(x) \neq 0$. A good starting point for this is Chapter 13 of Iwaniec's book, Topics in Classical Automorphic Forms.

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  • $\begingroup$ To avoid confusion, in which normalization is the above result? I mean are you using $\lambda(p)=\sqrt{p}a(p)$ where $a(n)$ are the coefficients of the Maass form. $\endgroup$
    – Subhajit Jana
    Commented Oct 1, 2014 at 22:28
  • $\begingroup$ I am assuming the normalization where the Ramanujan-Petersson conjecture implies $|\lambda(n)| \leq d(n)$, the divisor function. Alternatively, the $L$-function associated to $f$ is $\sum_{n=1}^{\infty} \lambda_f(n) n^{-s}$ which satisfies a functional equation under $s \rightarrow 1-s$. $\endgroup$
    – Matt Young
    Commented Oct 2, 2014 at 13:38
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    $\begingroup$ If the central character is trivial, then by Hecke's relation and using standard analytic techniques involving the symmetric square L-function you can get an asymptotic formula $S(x) \sim X/\log X$. $\endgroup$
    – Idoneal
    Commented Oct 10, 2014 at 11:39
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Following up on @Idoneal's comment, one can prove something a bit more precise. In particular, $S(x)\asymp\frac{x}{\log x}$, provided that $\log x\gg\log(\lambda N)$, where $f$ has level $N$ and Laplace eigenvalue $\lambda$. I'm assuming Matt Young's normalization and that the central character is unitary; also, I'm writing $f\asymp g$ to mean that both $f=O(g)$ and $g=O(f)$. This result follows from work of Motohashi https://arxiv.org/abs/1209.4140. A little extra effort is needed to get the claimed range, but all the necessary tools are in this paper.

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    $\begingroup$ I think you need that $x>(\lambda N)^c$, where $c>0$ is a sufficiently large constant. Also, I believe the result is true (and easy to prove) for any central character (always assumed to be unitary). $\endgroup$
    – GH from MO
    Commented Oct 21, 2017 at 13:34
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    $\begingroup$ I guess I should've said "$\log x\gg \log(\lambda N)$ with a sufficiently large implied constant", but your statement is clearer. $\endgroup$
    – 2734364041
    Commented Oct 23, 2017 at 10:37

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