Let $f$ a weight $k$ holomorphic Hecke cusp form with $\|f\|^2=\langle f,f\rangle=1$ with fourier expansion $$f(z)=\sum_{r\geq 1}a_f(r)e(rz)$$ Let $\displaystyle\lambda_f(r)=\frac{a_f(r)r^{(-k+1)/2}}{a_f(1)}$, then $\lambda_f(r)$ are the eigenvalues of the normalized Hecke operators $n^{(1-k)/2}T_n$. We define $$L(s,\operatorname{sym}^2(f))=\zeta(2s)\sum_{n=1}^\infty \frac{\lambda_f(n^2)}{n^s}$$ I am trying to prove the following (this appears in different papers like this one in Luo-Sarnak)


What I was able to prove so far is that $$a_f(1)=\frac{\zeta(2s)}{L(s,\operatorname{sym}^2 (f))}\frac{L(k+s-1,\operatorname{sym}^2 (f))}{\zeta(2(k+s-1))}$$ If I put $s=1$, the problem reduces to find $L(1,\operatorname{sym}^2(f))$ and $L(k,\operatorname{sym}^2(f))$, but I don't think this is a good approach. Any ideas or references would be appreciated.

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    $\begingroup$ What have you tried so far? The standard proof is to consider $\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} f(z)^2 y^k E(z,s) \, d\mu(z)$, where $E(z,s)$ is a real-analytic Eisenstein series, and take the residue at $s = 1$. $\endgroup$ – Peter Humphries May 7 at 19:13
  • $\begingroup$ Thank you, I could figure it out after your comment, it had to be $|f(z)|^2$ in your integral though. $\endgroup$ – Julian Mejia May 8 at 18:38
  • $\begingroup$ @PeterHumphries given $ E(s,z)$ and its generalizations in $GL_n(\Bbb{A_Q}),\chi$ can be defined in term of representation theoretic objects, is there a spectral reason for $z \mapsto Res_{s=1} E(s,z)$ being constant ? $\endgroup$ – reuns May 8 at 19:17
  • $\begingroup$ @reuns actually I don't know a good spectral reason for this off the top of my head, but I imagine (or hope?) that this is written up somewhere in some notes of Cogdell or Garrett. $\endgroup$ – Peter Humphries May 8 at 19:20

Thanks to @Peter Humphries in telling me the approach in his comment. Denote $\Gamma=SL_2(\mathbb{Z})$ and consider $$F(s)=\int_{\Gamma\setminus\mathbb{H}}|f(z)|^2y^kE(z,s)d\mu(z)$$ where $E(z,s)=\sum_{\gamma\in\Gamma_\infty\setminus\Gamma}(\operatorname{Im}(\gamma z))^s$ is the usual Eisenstein series.

First, since $\operatorname{Res}_{s=1}(E(z,s))=\frac{3}{\pi}$, we have $$\operatorname{Res}_{s=1}F(s)=\frac{3}{\pi}\int_{\Gamma\setminus\mathbb{H}}|f(z)|^2y^kd\mu(z)=\frac{3}{\pi}\langle f,f\rangle=\frac{3}{\pi}$$ Now, let's compute this residue in a different manner:

By modularity, it's easy to check that $|f(z)|^2y^k=|f(z)|^2(\operatorname{Im}(z))^k$ is $\Gamma$-invariant, i.e. $|f(z)|^2(\operatorname{Im}(z))^k=|f(\gamma z)|^2(\operatorname{Im}(\gamma z))^k$. So, we can unfold the integral obtaining $$F(s)=\int_{\Gamma_\infty\setminus\mathbb{H}}|f(z)|^2y^ky^sd\mu(z)=\int_{0}^\infty\int_{0}^1|f(z)|^2y^ky^sy^{-2}dxdy$$ Using the Fourier expansion of $f(z)$ and integrating with respect to the $x$ variable, we have $$\int_{0}^1|f(z)|^2 dx=\sum_{r}|a_{f}(r)|^2e^{-4\pi ry}$$ So, $$F(s)=\sum_{r}|a_f(r)|^2\int_{0}^\infty y^{k+s-2}e^{-4\pi ry}dy$$ After a change of variable $t=4\pi ry$, we can see that $\int_{0}^\infty y^{k+s-2}e^{-4\pi ry}dy=\frac{\Gamma(k+s-1)}{(4\pi r)^{k+s-1}}$. Then, \begin{align*}F(s)&=\sum_{r}|a_f(r)|^2\frac{\Gamma(k+s-1)}{(4\pi r)^{k+s-1}}\\ &=|a_f(1)|^2\frac{\Gamma(k+s-1)}{(4\pi )^{k+s-1}}\sum_{r}\frac{\lambda_f(r)^2}{r^s}\\ &=|a_f(1)|^2\frac{\Gamma(k+s-1)}{(4\pi )^{k+s-1}}\frac{\zeta(s)}{\zeta(2s)}L(s,\operatorname{sym}^2f)\end{align*} Then $$\operatorname{Res}_{s=1}F(s)=|a_f(1)|^2\frac{\Gamma(k)}{(4\pi )^{k}}\frac{1}{\zeta(2)}L(1,\operatorname{sym}^2f)$$ Equating $$\frac{3}{\pi}=|a_f(1)|^2\frac{\Gamma(k)}{(4\pi )^{k}}\frac{1}{\zeta(2)}L(1,\operatorname{sym}^2f),$$ we get our result.

  • $\begingroup$ That is actually a fact that I have seen before, here www-users.math.umn.edu/~garrett/m/mfms/notes_c/simplest_eis.pdf is stated in [1.3], I haven't read the proof though. If you can get a better reference it would be helpful. $\endgroup$ – Julian Mejia May 8 at 18:50
  • $\begingroup$ About $\operatorname{Res}_{s=1}(E(z,s))$ not depending on $z$ and $=\frac{3}{\pi}$ : $\Gamma_\infty$ is the subgroup generated by $z \mapsto z+1$, $2E(z,s) = \sum_{c,d, gcd(c,d)=1} \frac{y^s}{|cz+d|^{2s}}$ so $y^{-s} \zeta(2s) 2 E(z,s) = \sum_{(c,d) \ne (0,0)} |cz+d|^{-2s}$ whose residue at $s=1$ is the same as $\int_{u \in \Bbb{R}^2, \|u\| > 1} \|M u\|^{-2s}du = \int_{v \in \Bbb{R}^2, \|M^{-1} v\| > 1} \|v\|^{-2s}d(M^{-1} v),M= \pmatrix{1 & 0 \\ Re(z) & Im(z)}$ ie. $4 \det|M|^{-1} =4 y^{-1}$. $\endgroup$ – reuns May 8 at 19:05

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