6
$\begingroup$

There is a basis question which puzzles me for a while. The question is the following:

Let $X\ge 2,$ and $\lambda(n)$ be the $n$-th Fourier coefficient of a $GL(2)$ newform of prime level $N>1$, with $N\ll X$ and the trivial nebentypus.

If one could show $$\sum_{p\le X} \lambda^2(p)\gg X^{1-\varepsilon}, \quad \tag{1}$$ where the implied constant does not depend on the level $N$?

My understanding is that, note that $\lambda^2(p)=1+\lambda(p^2)$ if $p\nmid N$, so that the sum in (1) becomes $$\pi(X)+\frac{1}{N}-1+\sum_{p\le X} \lambda(p^2),$$ while this guy, $\lambda(p^2)$, may be viewed as the Fourier coefficient of the the symmetric-square lift of $f$, i.e., $\text{sym}^2f$, which is however a $GL(3)$ Maass form by Gelbart and Jacquet's theory. One may thus show, by appealing to Theorem 5.13 of I-K's book, the following $$\sum_{p\le X}\lambda(p^2) =\pi(X)+O\left\{X\exp\left(- \frac{c\log X}{\sqrt{\log X}+\log N}\right) \right \} $$ for some computable constant $c>0$, where the implied $O$-constant is absolute. Thus one may deduce that the sum in (1) equals $$\pi(X)+\left\{X\exp\left(- \frac{c\log X}{\sqrt{\log X}+\log N}\right) \right \}\quad \tag{2}.$$ But it seems one fails to show that the error-term in (2) is $\gg X^{1-\varepsilon}$; for example, if one takes $N=X^{\delta}$ for some $\delta<1$, we find the error-term is $\gg X$. It seems one cannot achieve a power-saving.

If any expert leans something on this topic, please show a guide. Thanks in advance! And thanks for your time.

Improve this question
$\endgroup$
6
  • $\begingroup$ @Will Sawin Dear Prof. Will Swain, thanks for comments. Much obliged! $\endgroup$
    – hofnumber
    Commented Jul 23, 2021 at 14:23
  • 1
    $\begingroup$ Sorry, I realized my earlier comments were not so relevant and deleted. Since you found them helpful, let me give more detail instead of deleting. I'm pretty sure one can't improve the error term in (2) to be at most $X^{1-\epsilon}$ as doing so would give a zero-free region for the symmetric square $L$-function (and maybe even the zeta function too). However, this doesn't imply that one can't improve the dependence of the error term on $N$, which would also help for what you want. I'm not an expert on that direction. $\endgroup$
    – Will Sawin
    Commented Jul 23, 2021 at 14:34
  • 2
    $\begingroup$ It also doesn't show that you can't have a method focused on proving a lower bound for the whole sum rather than an upper bound for the main term. For example, it might be possible to show that, if $\sum_p \lambda(p)^2$ is too small, then $\sum_n \lambda(n)^2$ is also somewhat small, and that can be estimated with much better error terms. $\endgroup$
    – Will Sawin
    Commented Jul 23, 2021 at 14:35
  • $\begingroup$ @Will Sawin Dear Prof. Will Swain, thanks for the heuristic guide. I will try to pursue this vein. $\endgroup$
    – hofnumber
    Commented Jul 24, 2021 at 7:43
  • $\begingroup$ In your post, $X/\log X$ should be $\pi(X)$. Also, the formula $\lambda^2(p)=1+\lambda(p^2)$ is only valid for $p\nmid N$ and when the nebentypus is trivial. Finally, you seem to assume an arithmetic normalization $\lambda(1)=1$, which is in general not available (as the first Fourier coefficient of a cusp form can vanish). $\endgroup$
    – GH from MO
    Commented Jul 27, 2021 at 22:57

2 Answers 2

Reset to default
6
$\begingroup$

Assuming that the spectral parameter is absolutely bounded (which you seem to implicitly state), the best that one can achieve with existing tools is the following: There exist absolute and effectively computable constants $c_1,c_2,c_3>0$ such that if $x>N^{c_1}$, then

$\displaystyle c_2 \frac{x}{\log x} \leq \sum_{p\leq x}|\lambda(p)|^2\leq c_3 \frac{x}{\log x}$.

In some regards, this is a direct generalization of Linnik's bound on the least prime in an arithmetic progression. As such, the only way I'd know to prove this would involve using a log-free zero density estimate (as Linnik did). This result is proved in a significantly broader context here.

ADDED: In view of GH from MO's comment, I should specify that I am assuming that the cusp form under consideration is in fact a newform. If you want to assume that $x$ is at least a polynomial in $N$ (and not an even larger function of $N$), then it is unclear (at least to me) what one can say otherwise.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Many thanks! A naive question is whether or not the constants $c_2,c_3$ respectively depend on $c_1$? Following P. Humphries-J. Thorner's paper, it looks like the constant $c_6$ in the final estimate in Theorem 2.1 depends on the former given constants $c_1,c_2,c_3,c_4,c_5$? By the way, the estimate $c_2 \frac{x}{\log x}\le \sum_{p\le x} |\lambda(p)|^2\le c_3\frac{x}{\log x} $ definitely does not depend the form $f$ anymore. $\endgroup$
    – hofnumber
    Commented Jul 24, 2021 at 7:37
  • $\begingroup$ So for any given $c_1$, whether or not one could has $\displaystyle c_2 \frac{x}{\log x} \leq \sum_{p\leq x}|\lambda(p)|^2\leq c_3 \frac{x}{\log x},$ where $c_2,c_3$ are computable depending on $c_1$? $\endgroup$
    – hofnumber
    Commented Jul 24, 2021 at 7:48
  • $\begingroup$ @hofnumber Not "for any given $c_1$", but "there exists $c_1$." If you don't want to assume GRH, then the least admissible value of $c_1$ will be large. And yes, the $c_2$ and $c_3$ can be effectively computed in terms of $c_1$. But I stated earlier, each of these constants are all absolute. $\endgroup$
    – 2734364041
    Commented Jul 24, 2021 at 8:05
  • $\begingroup$ So it is urgent to figure out the minimum value of $c_1$ to ensure that $c_2 \frac{x}{\log x} \le \sum_{p\le x}|\lambda(p)|^2 \le c_3 \frac{x}{\log x}$. $\endgroup$
    – hofnumber
    Commented Jul 24, 2021 at 8:25
  • 1
    $\begingroup$ "Urgent"? I don't know what you mean. These constants $c_1,c_2,c_3$ are all absolute and effectively computable. If you want to determine them for yourself, you are welcome to do so. $\endgroup$
    – 2734364041
    Commented Jul 26, 2021 at 6:51
1
$\begingroup$

Old results of Iwaniec-Kohnen-Sengupta give non-vanishing of $\lambda_f(p)$ for $p < N^{1/2 - \delta}$ and some small $\delta > 0$ so this is the best range that you could hope for give the current technology.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Do you mean "for some $p < N^{1/2-\delta}$"? $\endgroup$
    – 2734364041
    Commented Jul 24, 2021 at 0:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .