Let $f \in S_2(\Gamma_0(N))$ be a newform with trivial character. I want to compute the Petersson norm $\lVert f\rVert^2$ of $f$, not normalized by $1/[\operatorname{SL}_2(\mathbf{Z}):\Gamma_0(N)]$, as in Gross–Zagier.
From Numerical evaluation of the Petersson product of elliptic modular forms, I came across the formula $$\lVert f\rVert^2 = \frac{(k-1)!}{2^{2k – 1}\pi^{k + 1}}L(\operatorname{Sym}^2(f),2)$$ with $k = 2$. I implemented this in Magma. If $N$ is not square-free, I guess the correct Euler factor for the symmetric square at $p^2 \mid N$ by testing if the functional equation for the symmetric square is satisfied with $1 \pm x$ or $1 \pm px$.
However, comparing with the result of PARI/gp (which is normalized by $1/[\operatorname{SL}_2(\mathbf{Z}):\Gamma_0(N)]$, so removing that normalization), it seems that I have to multiply my result by $N$ if $N$ is square-free, the reason for which I don't understand (maybe it's a convention of the implementation of the symmetric square $L$-function in Magma?). It is even worse for $N$ not square-free, e.g. $f \in S_2(\Gamma_0(125))^+$ or $S_2(\Gamma_0(147))^{w_3,w_{49}}$, where the normalization factors seem to be $125$ and $147 \cdot 7/8$, respectively.
PARI's code is hard to read, and Petersson scalar products are not implemented for $N \neq 1$ in Sage.
Can someone please shed light on this?
Best Answer
My guess is that the formula you are trying to use is only valid for $N=1$, and thus needs correction in general.
Maybe Shimura's paper can help sort this out. https://doi.org/10.1002/cpa.3160290618
In (2.1), which Shimura writes for $\Gamma_1(N)$ but that doesn't matter when the character is trivial, his definition is $$\langle f,f\rangle={3/\pi\over [SL_2(Z):\Gamma_0(N)]}\int_\Phi |f|^2 dx dy.$$
Then in (2.5) you have $$\langle f,f\rangle={\Gamma(2)\over (4\pi)^2}\cdot\mathop{\rm res}\limits_{s=2} D(s,f,f)$$ where by the last display of Section 1 he defines $$D(s,f,f)=\sum_{n=1}^\infty {a_n^2\over n^s}.$$ Now a comparison of Euler products gives that the local factors of $D(s,f,f)\zeta(2s-2)$ and $L(s,Sym^2 f)\zeta(s-1)$ match, at least away from $p$ that divide $N$ (this discrepancy is the issue that David Loeffler raises). This Euler product comparison is mentioned in another paper of Shimura, see (0.4) of https://doi.org/10.1112/plms/s3-31.1.79
Anyway, this gives the answer up to the bad factors, namely $$\int_\Phi |f|^2 dx dy={[SL_2(Z):\Gamma_0(N)]\over 3/\pi}\langle f,f\rangle$$ $$=[SL_2(Z):\Gamma_0(N)]{\pi\over 3}{1\over (4\pi)^2}\mathop{\rm res}\limits_{s=2} D(s,f,f)$$ $$=[SL_2(Z):\Gamma_0(N)]{\pi\over 48\pi^2}{1\over\zeta(2)}L(2,Sym^2 f)\prod_{p|N} C_p$$ $$=N\prod_{p|N}(1+1/p)\cdot{1\over 8\pi^3}L(2,Sym^2 f)\prod_{p|N}C_p$$
Note that this matches your asserted formula when $N=1$ and $k=2$.
In the more general case, considering a bad prime $p|N$, the Euler factor from $\zeta(s-1)/\zeta(2s-2)$ evaluated at $s=2$ exactly cancels out factor of $(1+1/p)$ in the index formula. Meanwhile, the Euler factor of $L(s,Sym^2f)$ when $p$ exactly divides $N$ is $(1-1/p^s)^{-1}$, as is the Euler factor of $D(s,f,f)$ in this case (since $a_p^2=1$).
Finally, when $p^2|N$, the Euler factor of $D(s,f,f)$ is trivial since $a_p^2=0$, while that of $L(s,Sym^2f)$ can be known either by theory or trial-and-error computation. For the theoretical side, one can presuably work with the $p$-minimal twist of $f$ where this minimality allows twists with nontrivial Nebentypus - see 2.1 of Coates and Schmidt, particularly (2.12). https://doi.org/10.1515/crll.1987.375-376.104
I think one aspect is that if $v_p(N)$ is odd then the Euler factor of $L(s,Sym^2f)$ is trivial; while if $v_p(N)$ is even and $f$ is itself $p$-minimal then the factor is $(1+p/p^s)^{-1}$; and otherwise the Euler factor comes from that of the $p$-minimal twist (though perhaps not completely transparently, again with this $(1+p/p^s)^{-1}$ possibly appearing).