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In Iwaniec, Luo and Sarnak article (precisely (4.23)), it is said that GRH for $L(s, \mathrm{sym}^2(f))$, for a holomorphic cusp newform $f$ of level $N$ and weight $k$, implies $$\sum_{p \nmid N} \frac{\lambda_f(p^2) \log p}{p} \ll \log\log kN$$

Why is that true?

It is not the first time that I use this "blackbox" of GRH giving bounds, is there a general intuition to have behind it, or a systematic formal justification (like using the explicit formula to relate zeroes and eigenvalues)?

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We have that \[-\frac{L'}{L}(s,\mathrm{sym}^2 f) = \sum_{n = 1}^{\infty} \frac{\Lambda_{\mathrm{sym}^2 f}(n)}{n^s},\] where $\Lambda_{\mathrm{sym}^2 f}(n)$ is equal to $\lambda_f(p^2) \log p$ if $n = p$ with $p \nmid N$, is essentially a bounded multiple of $\log p$ if $n = p^k$, and vanishes otherwise. (There are some minor issues at $p \mid N$ that are no big deal). In particular, this shows that the value at $1$ of this sum is basically equal to the desired sum up to a negligible error term.

Now use Theorem 5.17 of Iwaniec and Kowalski, which states that for $s = \sigma + it$ with $1/2 < \sigma \leq 5/4$ and assuming RH for $L(s,\mathrm{sym}^2 f)$ (as well as the Ramanujan conjecture, which is known via the work of Deligne), \[-\frac{L'}{L}(s,\mathrm{sym}^2 f) = \frac{r}{s - 1} + O\left(\frac{1}{2\sigma - 1} (\log \mathfrak{q}(\mathrm{sym}^2 f, s))^{2 - 2\sigma} + \log \log \mathfrak{q}(\mathrm{sym}^2 f, s)\right),\] where $r$ is the order of the pole of $L(s,\mathrm{sym}^2 f)$ at $s = 1$ and $\mathfrak{q}(\mathrm{sym}^2 f, s)$ is the analytic conductor. Note that $N$ is squarefree and $f$ has trivial nebentypus, so that $r = 0$. Taking $s = \sigma = 1$ and noting that $\log \mathfrak{q}(\mathrm{sym}^2 f, s) \ll \log kN$ yields the result.

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  • $\begingroup$ Does it rely on the holomorphic assumption, or does the implication also holds for Maass newforms? $\endgroup$ – Gory Oct 30 '17 at 10:02
  • $\begingroup$ Thanks a lot for the answer, it is almost everything I want to know. However, what will happen when N is not squarefree of the form hasn't trivial nebentypus? (i.e. can we also deduce estimates of the sum il the question with r not equal to zero? like with specific partial summations formulas/transforms?) $\endgroup$ – Wolker Oct 30 '17 at 10:08
  • $\begingroup$ @Gory, the bound is valid under the assumption of the Ramanujan conjecture, which is not known for Maass newforms. $\endgroup$ – Peter Humphries Oct 30 '17 at 13:34
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    $\begingroup$ @Wolker, I have actually been thinking about this a bit (I have an unfinished preprint where I generalise the ILS paper to arbitrary level and nebentypus). The issue is that $L(s,\mathrm{sym}^2 f)$ can have a simple pole at $s = 1$ when $f$ is a dihedral newform; see mathoverflow.net/questions/178654/… $\endgroup$ – Peter Humphries Oct 30 '17 at 13:39
  • $\begingroup$ So the sum $\sum_{p \leq x, \, p \nmid N} \frac{\lambda_f(p^2) \log p}{p}$ is in fact asymptotic to a constant multiple of $\log x$, coming from the pole of $L(s,\mathrm{sym}^2 f)$ at $s = 1$, and then there is an error term of size $\log \log kN$. $\endgroup$ – Peter Humphries Oct 30 '17 at 13:42

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