---
abstract: 'Let $f \in S_{\kappa}(\Gamma_0(N))$ be a Hecke eigenform at $p$ with eigenvalue $\lambda_f(p)$ for a prime $p \nmid N$. Let $\alpha_p$ and $\beta_p$ be complex numbers satisfying $\alpha_p + \beta_p = \lambda_{f}(p)$ and $\alpha_p \beta_p = p^{\k-1}$. We calculate the norm of $f_{p}^{\alpha_p}(z) = f(z) - \beta_{p} f(pz)$ as well as the norm of $U_p f$, both classically and adelically. We use these results along with some convergence properties of the Euler product defining the symmetric square $L$-function of $f$ to give a ‘local’ factorization of the Petersson norm of $f$.'
address:
- |
    $^1$Department of Mathematical Sciences\
    Clemson University\
    Clemson, SC 29634
- |
    $^2$Department of Mathematics\
    Queens College\
    City University of New York\
    Flushing, NY 11367
- |
    $^3$Department of Mathematics\
    University of Connecticut\
    Storrs, CT 06269
author:
- Jim Brown$^1$
- Krzysztof Klosin$^2$
bibliography:
- 'mybib.bib'
title: 'On the norms of $p$-stabilized elliptic newforms'
---

[^1]

Introduction
============

Let $\k \geq 2$ and $N \geq 1$ be integers and $p$ an odd prime with $p \nmid N$. Let $f \in S_{\k}(\Gamma_0(N))$ be a newform. It is well-known that the Petersson norm $\langle f,f \rangle$ serves as a natural period for many $L$-functions of $f$ [@HidaInvent81; @ShimuraMathAnn77].

In this paper we focus on related periods $\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle$ (defined below) for $\alpha_p$ a Satake parameter of $f$. When $f$ is ordinary at $p$, the forms $f_p^{\alpha_p}$ arise naturally in the context of Iwasawa theory as the objects which can be interpolated into a Hida family. It is in fact in the context of ‘$p$-adic interpolation’ of some automorphic lifting procedures (between two algebraic groups, one of them being $\GL_2$) that these calculations arise (see [@BrownKlosinGSp(4)] for example); however, our results apply in a more general setup as specified below.

Let $f \in S_{\k}(\Gamma_0(N))$ be an eigenform for the $T_p$-operator with eigenvalue $\lambda_{f}(p)$. Let $\alpha_p$ and $\beta_p$ be the pair of complex numbers satisfying $\alpha_p + \beta_p = \lambda_f(p)$ and $\alpha_p \beta_p  = p^{\k-1}$. We set $f_{p}^{\alpha_{p}}(z) = f(z) - \beta_{p} f(pz)$. In the case that $f$ is ordinary at $p$, we can choose $\alpha_p$ and $\beta_p$ so that $\alpha_{p}$ is a $p$-unit and $\beta_{p}$ is divisible by $p$. In this special case $f_p^{\alpha_p}$ is the $p$-stabilized ordinary newform of tame level $N$ attached to $f$.

Since $f_p^{\alpha_p} = p^{1-\kappa}\beta_p (U_p-\beta_p)f$, calculating $\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle$ is in fact equivalent to calculating $\langle U_p f, U_p f \rangle$. While computation of any of these inner products does not present any difficulties (see Section \[Relations between the norms\]), it is an accident resulting from the relative simplicity of the Hecke algebra on $\GL_2$, where the $T_p$ and the $U_p$ operators differ by a single term. It turns out that in the higher-rank case it is the calculation of the latter inner product that provides the fastest route to computing the Petersson norm of various $p$-stabilizations. With these future applications in mind we present an alternative approach to calculating $\langle U_p f, U_p f \rangle$, this time working adelically (see Sections \[relations\] and \[sec:adeliccalc\]), as this is the method that generalizes to higher genus most readily (see [@BrownKlosinGSp(4)], where this is done for the group $\GSp_4$).

It is well-known that the Petersson norm $\langle f,f \rangle$ is closely related to the value $L(\kappa, \Sym^2 f)$ at $\kappa$ of the symmetric square $L$-function of $f$. The absolutely convergent Euler product defining this $L$-function for ${\operatorname{Re}}(s) >\kappa$ converges (conditionally) to the value $L(s, \Sym^2 f)$ when ${\operatorname{Re}}(s)=\kappa$ (this and in fact a more general result is proved in the appendix by Keith Conrad). On the other hand our computation of $\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle$ shows that this inner product differs from $\langle f,f \rangle$ by essentially the $p$-Euler factor of $L(\kappa, \Sym^2 f)$. Combining these facts we exhibit a (conditionally convergent) factorization of $\langle f,f \rangle$ into local components defined via the inner products $\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle$ (for details, see Section \[Applications to $L$-values\]).

The authors would like to thank Henryk Iwaniec, and Keith Conrad would like to thank Gergely Harcos, for helpful email correspondence.

Classical calculation of $\langle f_p, f_p \rangle$ and $\langle U_p f, U_p f \rangle$ {#Relations between the norms}
======================================================================================

Let $N$ be a positive integer. Let $\Gamma_0(N)\subset \SL_2(\bfZ)$ denote the subgroup consisting of matrices whose lower-left entry is divisible by $N$. For a holomorphic function $f$ on the complex upper half-plane $\fh$ and for $\gamma=\bmat a&b \\ c&d \emat \in \GL_2^+(\bfR)$, where $+$ denotes positive determinant, and $\kappa \in \bfZ_+$ we define the slash operator as $$(f|_{\k} \gamma)(z) = \frac{\det(\gamma)^{\kappa/2}}{(cz+d)^\kappa} f\left(\frac{az+b}{cz+d}\right).
$$ If $\kappa$ is clear from the context we will simply write $f|\gamma$ instead of $f|_{\kappa}\gamma$. We will write $S_{\kappa}(\Gamma_0(N))$ for the $\bfC$-space of cusp forms of weight $\kappa$ and level $\Gamma_0(N)$ (i.e., functions $f$ as above which satisfy $f|_{\kappa} \gamma = f$ for all $\gamma \in \Gamma_0(N)$ and vanish at the cusps - for details see [@Miyake89]).

The space $S_{\kappa}(\Gamma_0(N))$ is endowed with a natural inner product (the *Petersson inner product*) defined by $$\langle f, g \rangle_{N} = \int_{\Gamma_0(N) \backslash \fh} f(z) \overline{g(z)} y^{\k-2} dx dy$$ for $z = x+iy$ with $x,y \in \bfR$ and $y>0$. If $\Gamma \subset \Gamma_0(N)$ is a finite index subgroup we also set $$\langle f, g \rangle_{\Gamma} = \int_{\Gamma \backslash \fh} f(z) \overline{g(z)} y^{\k-2} dx dy.$$

From now on let $p$ be a prime which does not divide $N$. Set $\eta = {\left[ \begin{matrix} p & 0 \\ 0 & 1 \end{matrix} \right]}$. We have the decomposition $$\label{decomp3}
\Gamma_0(N) {\left[ \begin{matrix} 1 & 0 \\ 0 & p \end{matrix} \right]} \Gamma_0(N) = \bigsqcup_{j=0}^{p-1} \Gamma_0(N) {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]} \sqcup \Gamma_0(N) \eta.$$

Recall the $p$th Hecke operator acting on $S_{\k}(\Gamma_0(N))$ is given by $$T_p f = p^{\k/2-1}\left( \sum_{j=0}^{p-1} f|_{\k}  {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]} + f|_{\k} \eta\right)$$ and the $p$th Hecke operator acting on $S_{\k}(\Gamma_0(Np))$ is given by $$U_pf =  p^{\k/2-1}\left( \sum_{j=0}^{p-1} f|_{\k}  {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]} \right).$$ As we will be viewing $f \in S_{\k}(\Gamma_0(N))$ as an element of $S_{\k}(\Gamma_0(Np))$, we use $T_p$ and $U_p$ to distinguish the two Hecke operators at $p$ defined above.

Let $f \in S_{\k}(\Gamma_0(N))$ be an eigenfunction for $T_p$ with eigenvalue $\lambda_f(p)$. There exist (up to permutation) unique complex numbers $\alpha_p$ and $\beta_p$ satisfying $\lambda_f(p) = \alpha_p + \beta _p$ and $\alpha_p\beta_p=p^{\kappa-1}$. We consider the following two forms: $$\begin{aligned}
f_{p}^{\alpha_p}(z) &= f(z) - \beta_{p} p^{-\k/2} (f|_{\kappa}\eta)(z),\\
f_{p}^{\beta_p}(z) &= f(z) - \alpha_{p} p^{-\k/2} (f|_{\kappa}\eta)(z).\end{aligned}$$ One immediately obtains that $f_p^{\alpha_p} \in S_{\k}(\Gamma_0(Np))$ and that $f_p^{\alpha_p}$ is an eigenfunction for the operator $U_p$ with eigenvalue $\alpha_p$. Furthermore, if $f$ is also an eigenform for $T_{\ell}$ for a prime $\ell \neq p$, then so is $f_p^{\alpha_p}$ and it has the same $T_{\ell}$-eigenvalue as $f$. The analogous statements for $f_{p}^{\beta_p}$ hold as well. Note that if $f$ is ordinary at $p$, then one can choose $\alpha_p$ and $\beta_p$ so that $\ord_p(\alpha_p) = 0$ and then $f_p^{\alpha_p}$ is the $p$-stabilized newform associated to $f$, see [@WilesInventMath88] for example.

\[classform\] Let $f \in S_{\k}(\Gamma_0(N))$ be defined as above, where $p \nmid N$. We have $$\frac{\langle U_p f, U_p f \rangle_{Np}}{\langle f, f \rangle_{Np}} = p^{\k-2} + \frac{(p-1) \lambda_{f}(p)^2}{p+1}$$ and $$\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}} =  \frac{p}{p+1}\left(1 - \frac{\alpha_p^2}{p^\kappa}\right)\left(1 - \frac{\beta_p^2}{p^\kappa}\right).$$

The definition of $f_{p}^{\alpha_p}$ and the fact that $U_p f = T_{p} f - p^{\k/2-1}f|\eta$ immediately give $$\begin{aligned}
\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np} &=
  (1 + |\beta_{p}|^2 p^{-\k}) \langle f, f \rangle_{Np} - p^{-\k/2}(\beta_{p} \langle f| \eta, f \rangle_{Np} + \ov{\beta_{p} \langle f|\eta, f\rangle}_{Np})
 \end{aligned}$$ and $$\begin{aligned}
\langle U_p f, U_p f \rangle_{Np}
    &= (p^{\k-2} + \lambda_{f}(p)^2) \langle f, f \rangle_{Np} -   p^{\k/2-1}\lambda_{f}(p) (\langle f|\eta, f \rangle_{Np} + \ov{\langle f|\eta, f\rangle}_{Np}).\end{aligned}$$

Let us now compute $\langle f|\eta, f \rangle_{Np}$. Observe that by the definition of $T_p$ we have $$\langle T_p f, g \rangle_{Np} =  p^{\k/2-1}\left( \sum_{j=0}^{p-1} \left \langle f\mid {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]}, g\right \rangle_{Np} + \langle f|\eta, g \rangle_{Np}\right).$$ Using the decomposition (\[decomp3\]) we can find $a_{j}, b_{j} \in \Gamma_0(N)$ so that $a_{j} {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]} b_{j} = {\left[ \begin{matrix} 1 & 0 \\ 0 & p \end{matrix} \right]}$, and $a,b \in \Gamma_0(N)$ so that $a {\left[ \begin{matrix} 1 & 0 \\ 0 & p \end{matrix} \right]} b = {\left[ \begin{matrix} p & 0 \\ 0 & 1 \end{matrix} \right]}.$ Using this and the fact that $f, g \in S_{\k}(\Gamma_0(N))$, we have $$\begin{aligned}
p^{1-\k/2} \langle  T_p f,g \rangle_{Np}    &= \sum_{j=0}^{p-1} \left \langle f\mid a_{j} {\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]}, g|b_{j}^{-1} \right \rangle_{Np} + \langle f|\eta, g \rangle_{Np} \\
    &=\sum_{j=0}^{p-1} \left \langle f\mid a_{j}{\left[ \begin{matrix} 1 & j \\ 0 & p \end{matrix} \right]} b_{j}, g \right \rangle_{Np} + \langle f|\eta, g \rangle_{Np} \\
        &= p \left \langle f\mid {\left[ \begin{matrix} 1 & 0 \\ 0 & p \end{matrix} \right]}, g \right \rangle_{Np} + \langle f|\eta, g \rangle_{Np} \\
    &= p \left \langle f\mid a {\left[ \begin{matrix} 1 & 0 \\ 0 & p \end{matrix} \right]}, g|b^{-1} \right \rangle_{Np} + \langle f|\eta, g \rangle_{Np}\\
        &= (p+1) \langle f|\eta, g \rangle_{Np}.
 \end{aligned}$$ Thus, setting $g = f$ we obtain $$\langle f|\eta, f \rangle_{Np} =  p^{1-\k/2}\frac{\lambda_{f}(p)}{p+1} \langle f, f \rangle_{Np}.$$ We can now easily conclude that $$\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}} = 1 + |\beta_{p}|^2 p^{-\k} - p^{1-\k}(\beta_{p}+\ov{\beta_p})\frac{\lambda_{f}(p)}{p+1}$$ and $$\frac{\langle U_p f, U_p f \rangle_{Np}}{\langle f, f \rangle_{Np}} = p^{\k-2} + \frac{(p-1) \lambda_{f}(p)^2}{p+1}.$$

Using the fact that $T_p$ is self-adjoint with respect to the Petersson inner product we have $\alpha_p + \beta_p = \lambda_f(p) \in \bfR$. We note by Lemma \[Ramanujan\] below that $\ov{\alpha_{p}} = \beta_{p}$. Thus $|\alpha_p|^2=|\beta_p|^2 = |\alpha_p||\beta_p|=p^{\kappa-1}$ and $\beta_p + \ov{\beta_p}=\lambda_f(p)$. This allows us to simplify the formula for $ \frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}}$ to $$\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}} = 1 + \frac{1}{p} - p^{1-\k}\frac{\lambda_{f}(p)^2}{p+1}.$$ Again using that $\lambda_f(p) = \alpha_p+\beta_p$ we obtain $$\begin{aligned}
\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}} &= \frac{1}{p+1}\left(p+ 1+\frac{p+1}{p} - p^{1-\kappa}(\alpha_p^2+\beta_p^2 + 2p^{\kappa-1})\right)\\
&= \frac{p}{p+1}\left(1-\frac{\alpha_p^2}{p^{\kappa}}\right)\left(1-\frac{\beta_p^2 }{p^{\kappa}}\right).\end{aligned}$$

We have $$\lim_{\substack{p\to \infty\\ p\hs\textup{is prime}}} \frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{p+1}=\langle f, f \rangle_N.$$

Using Theorem \[classform\] and the fact that $\langle f,f\rangle_{Np} = (p+1)\langle f, f\rangle_N$, we have for every prime $p \nmid N$ that $$\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{p+1}= \frac{p}{p+1}\left(1-\frac{\alpha_p^2}{p^{\kappa}}\right)\left(1-\frac{\beta_p^2 }{p^{\kappa}}\right)\langle f, f \rangle_{N}.$$ Since $|\alpha_p|^2=|\beta_p|^2=p^{\kappa-1}$, we see that the first three factors on the right tend to 1 as $p$ tends to infinity.

Relation between the classical and adelic inner products {#relations}
========================================================

While the classical calculations for $\langle U_p f, U_p f \rangle$ are rather elementary, it is also useful to note that one can perform these calculations adelically. The problem of calculating $\langle U_p f, U_p f \rangle$ is one that is local in nature, so it lends itself nicely to such an approach. Moreover, in a higher genus setting such as when working with Siegel modular forms, it is the adelic approach that generalizes most readily [@BrownKlosinGSp(4)]. In this section we provide the necessary background relating the adelic and classical inner products that is needed to relate the adelic inner product calculated in Section \[sec:adeliccalc\] to the calculation given in the previous section.

In this and the following sections $p$ will denote a prime number and $v$ will denote an arbitrary place of $\bfQ$ including the Archimedean one, which we will denote by $\infty$. Let $G=\GL_2$ and fix $N \geq 1$. By strong approximation (see for example [@BumpCambridgeUniversityPress88 Theorem 3.3.1, p. 293]) we have $$\label{strapp}
G(\AQ) = G(\bfQ) G(\bfR) \prod_{p} K_{p},$$ where $K_p$ is a compact subgroup of $G(\bfQ_p)$ such that $\det K_p = \bfZ_p^{\times}$. One example would be to take $K_{p} = K_{0}(N)_p$, where $$K_0(N)_p = \left\{ \bmat a& b \\ c& d \emat \in G(\bfZ_{p}): c \equiv 0 \bmod{N}\right\}.$$ Note that $K_0(N)_p=G(\bfZ_p)$ if $p \nmid N$. We will also set $$K_0(N):= \left\{ \bmat a& b \\ c& d \emat \in G(\hat{\bfZ}): c \equiv 0 \bmod{N}\right\}=\prod_p K_0(N)_p.$$ The decomposition (\[strapp\]) implies that \[2\] G() G() = G\^+() \_[p]{} K\_[p]{}, where $+$ indicates positive determinant.

Let $Z \subset G$ denote the center. For every $p$ there is a unique Haar measure $dg_p$ on $G(\bfQ_p)$ normalized so that the volume of any maximal compact subgroup of $G(\bfQ_p)$ is one. We use the standard Haar measure on $G(\bfR)$ as defined in [@BumpCambridgeUniversityPress88 $\S$ 2.1]. Define the adelic analogue of the Petersson inner product: $$\langle \phi_1, \phi_2 \rangle = \int_{Z(\AQ) G(\bfQ) \setminus G(\AQ)} \phi_1(g) \ov{\phi_2(g)} dg,$$ where $\phi_1$ and $\phi_2$ lie in $L^2(Z(\AQ) G(\bfQ) \setminus G(\AQ))$ and have the same central character and $dg$ is the Haar measure on $Z(\AQ) G(\bfQ) \setminus G(\AQ)$ corresponding to our choice of local Haar measures.

Let $f \in S_{\kappa}(\Gamma_0(N))$ be an eigenform. For $g=\gamma g_{\infty} k\in G(\AQ)$ with $\gamma \in G(\bfQ)$, $g_{\infty}=\bmat a&b\\c&d\emat \in G^+(\bfR)$ and $k \in K_0(N)$, set \[adelicdef\] \_f(g)=f(g\_).Then $\phi_f$ is an automorphic form on $G(\AQ)$ and it is easy to see (using the bijection in [@GelbartAFAG Equation 5.13]) that one has $$\label{eqn:classicalrelation}
\langle \phi_f, \phi_{f} \rangle = \frac{1}{\left[\SL_2(\bfZ):\Gamma_0(N)\right]} \langle f, f \rangle_{N}.$$

Let $\pi_{f} \cong \otimes \pi_{f,v}$ be the automorphic representation generated by $\phi_{f}$. If $f$ is a newform, then we can write $\phi_f= \otimes_v \phi_{f,v}$ for $\phi_{f,v} \in \pi_{f,v}$ and $\phi_{f,v}$ are spherical vectors for all $v \nmid N$, $v \neq \infty$. For every $v$ we can choose a $G(\bfQ_v)$-invariant inner product $\langle\cdot, \cdot\rangle_v$ (and any two such are scalar multiples of each other) so that $\langle \phi_{f,v}, \phi_{f,v}\rangle_v=1$ for all $v \nmid N$, $v \neq \infty$. It follows that there is constant $c$ so that $$\label{eqn:factorization}\langle \phi_{f}, \phi_{f}\rangle = c \prod_{v} \langle \phi_{f,v}, \phi_{f,v} \rangle_{v}.$$ We are now in a position to relate the ratio $\frac{\langle U_p f, U_p f \rangle_{Np}}{\langle f, f \rangle_{Np}}$ to something that can be calculated locally. In fact since we are only interested in this ratio, the precise value of the constant $c$ in (\[eqn:factorization\]) will be irrelevant.

Fix $p \nmid N$. As noted above, we normalize our Haar measure so that $\vol(K_0(1)_p) = 1$. For a vector $v_p$ inside the space of $\pi_{f,p}$, we set $$T_p v_p = \int_{K_0(1)_p \bsmat p&0 \\ 0& 1 \esmat K_0(1)_p} \pi_{f,p}(g) v_{p} dg$$ and $$V_p v_{p} = \int_{\bsmat 1 & 0 \\ 0 & p \esmat K_0(1)_p} \pi_{f,p}(g) v_{p} dg.$$ Note that we have the decompositions \[dec\] K\_0(1)\_p p&0\
0& 1 K\_0(1)\_p = \_[b=0]{}\^[p-1]{} p & b\
0 & 1 K\_0(1)\_p 1&0\
0& pK\_0(1)\_p and \[dec1\] K\_0(p)\_p p&0\
0& 1 K\_0(p)\_p= \_[b=0]{}\^[p-1]{} p & b\
0 & 1 K\_0(p)\_p. The adelic operator corresponding to the $U_p$-operator acting on classical modular forms as defined in Section \[Relations between the norms\] is given by $$U_p v_{p} := T_p v_{p} - V_p v_{p}.$$

\[classtoadel\] We have $$\frac{\langle U_p^{\rm cl} f, U_p^{\rm cl} f \rangle_{Np}}{\langle f, f \rangle_{Np}} = p^{\k-2}\langle U_p \phi_{f,p}, U_p \phi_{f,p} \rangle_{p}$$ for any local inner product pairing $\langle , \rangle_{p}$ so that $\langle \phi_{f,p}, \phi_{f,p} \rangle_{p} = 1$ and $U_p^{\rm cl}$ is the classical $U_p$-operator as defined in Section \[Relations between the norms\].

If we set $U_p \phi_f := (U_p \phi_{f,p}) \otimes \otimes_{v \neq p} \phi_{f,v}$ then it follows by the same argument as the one in the proof of [@GelbartAFAG Lemma 3.7] that \[adelclas\] \_[U\_p\^[cl]{}f]{}=p\^[/2-1]{}U\_p \_f. The lemma is now immediate from (\[eqn:classicalrelation\]) and (\[eqn:factorization\]).

It only remains to calculate $\langle U_p \phi_{f,p}, U_p \phi_{f,p} \rangle$, which is done in the next section.

Local calculation of $\langle U_p \phi_{f,p}, U_p \phi_{f,p} \rangle$ {#sec:adeliccalc}
=====================================================================

We will now give a calculation that, when combined with the results of the previous section, provides a local way to calculate $\langle U_p f, U_p f \rangle$ in terms of $\langle f, f \rangle$. As in the previous section we fix $f \in S_{k}(\Gamma_0(N))$ a newform and a prime $p$ not dividing $N$. We again let $\pi_{f} = \otimes_{v} \pi_{f,v}$ be the automorphic representation associated to $f$. Note that since $p \nmid N$, we can take the principal series representation $\pi_{p}(\chi_1, \chi_2)$ to be the model for $\pi_{f,p}$ and for functions $\psi, \psi' \in \pi_p(\chi_1, \chi_2)$ define the local inner product by $$\langle \psi, \psi' \rangle_p := \int_{K_0(N)_p} \psi(g) \ov{\psi'(g)}dg,$$ where the Haar measure is normalized so that $\vol(K_0(N)_p)=1$. Then the vector $\phi_{f,p}\in \pi_p$ corresponds to the function (which we will also denote by $\phi_{f,p} \in \pi_p(\chi_1,\chi_2)$) which can be described explicitly as $$\phi_{f,p}\left( \bmat a &* \\ 0&b \emat k\right) = \chi_1(a) \chi_2(b)|ab^{-1}|_p^{1/2},$$ where $|\cdot|_p$ denotes the standard $p$-adic norm ($|p|_p=p^{-1}$) and $k \in K_0(N)_p$.

As this section is focused on the calculation of $\langle U_p \phi_{f,p}, U_p \phi_{f,p} \rangle$, we will from now on write $\phi$ for $\phi_{f,p}$ and $K_0(1)$ (resp., $K_0(p)$) for $K_0(1)_p$ (resp., $K_0(p)_p$).

We note here that the calculation which follows can also be performed using the MacDonald formula for matrix coefficients (see [@CasselmanCompMath80 $\S$ 4]). However, in the relatively simple case of $\GL_2$ the elementary approach which we present below does not add any computational difficulty and is perhaps more transparent.

Set $$\mB:= \left\{ \bmat p & b \\ 0 & 1 \emat : b\in \{0,1, \dots, p-1\}\right\}, \quad \mB':= \mB \cup \left\{ \bmat 1&0 \\ 0&p \emat \right\}.$$ If $g \in K_0(1)$ and $\beta \in \mB'$, there is a permutation $\sigma_{g}$ of $\mB'$ and elements $k(g,\beta) \in K_0(1)$ such that $g \beta = \sigma_{g}(\beta) k(g, \beta)$. Furthermore, note that if $g \in K_0(1) - K_0(p)$, then the corresponding permutation cannot fix $\bmat 1&0 \\ 0& p \emat$. This implies that for such a $g$, there exists $\beta \in \mB$ such that $\sigma_{g}(\beta) = \bmat 1&0 \\ 0& p \emat$. Since in the computation of $U_p\phi$ only matrices in $\mB$ are used, we are interested in the restriction of $\sigma$ to $\mB$. For such a $g$ there are $p-1$ matrices in the image of $\sigma_{g}$ which have $(p,1)$ on the diagonal and one that has $(1,p)$ on the diagonal. Set $\mB_1(g) = \{\beta \in \mB: \sigma_{g}(\beta) \in \mB\}$ and $\mB_2(g) = \{\beta \in \mB: \sigma_{g}(\beta) \in \mB' - \mB\}$. So for $g \in K_0(1) - K_0(p)$ we have (note that our $\phi$ is right-$K_0(1)$-invariant and $\vol(K_0(1))=1$) $$\begin{split} (U_p\phi)(g) &= \vol(K_0(1)) \sum_{\beta \in \mB} \phi(g\beta)\\
&= \sum_{\beta \in \mB} \phi(\sigma_{g}(\beta) k(g,\beta)) \\
&= \sum_{\beta \in \mB_1(g)} \phi(\sigma_{g}(\beta)) + \sum_{\beta \in \mB_2(g)} \phi(\sigma_{g}(\beta))\\
& =  (p-1)\chi_1(p) p^{-1/2} + \chi_2(p) p^{1/2}.
\end{split}$$ If $g \in K_0(p)$, then the permutation $\sigma$ fixes $\bmat 1&0 \\0 & p \emat$, hence we obtain $$(U_p\phi)(g) = \vol(K_0(1)) \sum_{\beta \in \mB} \phi(g\beta) = \sum_{\beta \in \mB} \phi(\sigma(\beta))=p\chi_1(p)p^{-1/2} = \chi_1(p)p^{1/2}.$$ Now let us compute the integral: $$\left< U_p \phi, U_p \phi \right>_{K_0(1)} = \int_{K_0(p)} U_p\phi(g) \ov{U_p\phi(g)} dg + \int_{K_0(1) - K_0(p)} U_p\phi(g) \ov{U_p\phi(g)} dg.$$ We have \[j1\]

\_[K\_0(p)]{} U\_p(g) dg &= \_[K\_0(p)]{} p |\_1(p)|\^2 dh\
&=(K\_0(p)) p |\_1(p)|\^2\
&=

and, since $\vol(K_0(1) - K_0(p))=p/(p+1)$, $$\begin{gathered}
 \label{k1} \int_{K_0(1) - K_0(p)} U_p\phi(g) \ov{U_p\phi(g)} dg\\
= \int_{K_0(1)- K_0(p)} \left[ \frac{(p-1)^2}{p} |\chi_1(p)|^2 + (p-1)\tr(\chi_1(p) \ov{\chi_2(p)})+ p|\chi_2(p)|^2\right] dg \\
= \frac{p}{p+1} \left[ \frac{(p-1)^2}{p} |\chi_1(p)|^2 + (p-1)\tr(\chi_1(p) \ov{\chi_2(p)})+ p|\chi_2(p)|^2\right].\end{gathered}$$ Putting (\[j1\]) and (\[k1\]) together we get \[for32\]

&= |\_1(p)|\^2 + |\_2(p)|\^2+ (\_1(p)).

\[Ramanujan\] We have $\chi_j(p) = p^{s_j}$ for $j=1,2$, where $s_j$ is a purely imaginary number. In particular, $|\chi_{j}(p)|=1$ for $j=1,2$. Moreover, we have $\ov{\alpha_p} = \beta_{p}$.

The first part follows from [@GelbartAFAG p. 92] and is a direct consequence of the fact that cusp forms on $\GL_2$ satisfy the Ramanujan conjecture. Observe that $\alpha_p = p^{(\k-1)/2} \chi_1(p)$ and $\beta_{p} = p^{(\k-1)/2} \chi_2(p)$. Using that $\alpha_p \beta_{p} = p^{\k-1}$, we obtain $\chi_1(p) \chi_2(p) = 1$. This, combined with the fact that $\chi_{j}(p) = p^{s_{j}}$ with $s_{j}$ purely imaginary implies $\chi_1(p) = \ov{\chi_2(p)}$. Thus $\ov{\alpha_p} = \beta_p$.

Using Lemma \[Ramanujan\] we can simplify (\[for32\]) to $$\langle U_p \phi, U_p \phi \rangle = \frac{2p^2-p+1}{p+1} + \frac{p^2-p}{p+1} \tr (\chi_1(p)\ov{\chi_2(p)}).$$ Moreover using that $\alpha_p = p^{(\kappa-1)/2}\chi_1(p)$, $\beta_p=p^{(\kappa-1)/2}\chi_2(p)$ and $\chi_1(p) = \ov{\chi_2(p)}$ we have $$\tr (\chi_1(p)\ov{\chi_2(p)}) =  (\chi_1(p) + \chi_2(p))^2 - 2\chi_1(p) \chi_2(p) = p^{1-\k}\lambda_f(p)^2 - 2.$$ Thus we obtain $$\begin{aligned}
\langle U_p \phi, U_p \phi \rangle &= \frac{p+1}{p+1} + \frac{p(p-1)p^{1-k}\lambda_f(p)^2}{p+1}\\
    &=  1+ \frac{(p-1)\lambda_f(p)^2}{p+1}p^{2-k},\end{aligned}$$ hence we see that by Lemma \[classtoadel\] this recovers the classical formula from Theorem \[classform\].

Applications to $L$-values {#Applications to $L$-values}
==========================

Let $f \in S_{\kappa}(\Gamma_0(N))$ be a newform. In this section we apply the results of the previous sections to give a ‘local’ decomposition of the Petersson norm of $f$. This depends on showing that value $L^{N}(k, \Sym^2 f)$ obtained by meromorphic continuation of $L(s,\Sym^2 f)$ can be expressed as a conditionally convergent Euler product.

Recall that the (partial) *symmetric square $L$-function* of $f$ is defined by the Euler product \[Ep\] L\^N(s, \^2 f) = \_[p N]{}, where $$L_{p}(s, \Sym^2 f):=\left(1-\frac{\alpha_{p}^2}{p^{s}}\right)\left(1-\frac{\alpha_{p} \beta_{p}}{p^{s}}\right)\left(1-\frac{\beta_{p}^2}{p^{s}}\right).$$ The product (\[Ep\]) converges absolutely for ${\operatorname{Re}}s>\kappa$. It is well-known that $L^N(s, \Sym^2 f)$ admits meromorphic continuation to the entire complex plane with possible poles only at $s = \kappa$ and $\kappa - 1$, of order at most one [@ShimuraPLMS75 Theorem 1]. In our case (since $f$ is assumed to have trivial character), the $L$-function does not have a pole at $s=\kappa$ [@ShimuraPLMS75 Theorem 2]. We will continue to denote this extended function by $L^N(s, \Sym^2 f)$. Using that $\alpha_p\beta_p=p^{\kappa-1}$ we conclude that $$\label{form1}
\frac{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}}{\langle f, f \rangle_{Np}} = \frac{p^2}{p^2-1}\frac{1}{L_p(\kappa,\Sym^2 f)} =\frac{\zeta_p(2)}{L_p(\kappa, \Sym^2 f)},$$ where $\zeta_{p}(s) = 1/(1 - 1/p^s).$

We have $$\langle f_p^{\alpha_p}, f_p^{\alpha_p}\rangle_{Np} = \langle f_p^{\beta_p}, f_p^{\beta_p}\rangle_{Np}.$$

Set $$\langle f,f\rangle^{(p)}_N:=  \frac{\langle f, f \rangle_{Np}}{\langle f_p^{\alpha_p}, f_p^{\alpha_p} \rangle_{Np}} = \frac{\langle f, f \rangle_{Np}}{\langle f_p^{\beta_p}, f_p^{\beta_p} \rangle_{Np}}.$$ We will now show that $\langle f,f\rangle^{(p)}_N$ can in some sense be regarded as a ‘local’ (at $p$) period for the symmetric square $L$-function.

\[conv1\] The value $L^N(\kappa, \Sym^2 f)$ given by the meromorphic continuation is equal to the conditionally convergent Euler product $$\prod_{p \nmid N} \frac{1}{L_{p}(\kappa, \Sym^2 f)}$$ when we order the factors according to increasing $p$.

Let $\phi_f$ be defined as in (\[adelicdef\]) and let $\chi_1(p)=\alpha_p/p^{(\kappa-1)/2}$ and $\chi_2(p)=\beta_p/p^{(\kappa-1)/2}$ be its Satake parameters for $p \nmid N$ as in Section \[sec:adeliccalc\]. For $p \nmid N$ define $$L_{p}(s, \Sym^2 \phi_f):= \left(1-\frac{\chi_1(p)^2}{p^{s}}\right)\left(1-\frac{1}{p^{s}}\right)\left(1-\frac{\chi_2(p)^2}{p^{s}}\right)$$ and note that $L_{p}(s, \Sym^2 \phi_f)=L_{p}(s+\kappa-1, \Sym^2 f)$. Thus the Euler product $$L^N(s, \Sym^2 \phi_f):=\prod_{p \nmid N}\frac{1}{L_{p}(s, \Sym^2 \phi_f)}$$ converges absolutely for ${\operatorname{Re}}s>1$ and inherits all the corresponding properties (in particular the meromorphic continuation and the lack of a pole at $s=1$) from $L^N(s, \Sym^2 f)$. As before we will continue to denote this extended function by $L^N(s, \Sym^2 \phi_f)$.

Let $\pi$ be the automorphic representation of $\GL_2(\AQ)$ associated with $\phi_f$. It is known [@GelbartJacquet78 Theorem 9.3] that there exists an automorphic representation $\sigma$ of $\GL_3(\AQ)$ such that the (partial) standard $L$-function $L^N(s, \sigma)$ coincides with $L^N(s, \Sym^2 \pi):=L^N(s, \Sym^2 \phi_f)$. Also $L^N(s, \sigma)$ does not vanish on the line ${\operatorname{Re}}s=1$ by a result of Jacquet and Shalika (see [@JacquetShalika76 Theorem 1]; see also [@KohnenSengupta00]). Finally note that by Lemma \[Ramanujan\], we have $|\chi_1(p)| = |\chi_2(p)| = 1$ if $p \nmid N$. Thus we are in a position to apply Theorem \[ethm\] in the appendix with $K=\bfQ$ and $d=3$ to $L^N(s, \Sym^2 \phi_f)$ and the theorem follows.

By Theorem \[conv1\] and (\[form1\]) we have $$\frac{L^N(\kappa, \Sym^2 f)}{\prod_{p \nmid N} \langle f,f\rangle^{(p)}_N} = \zeta^N(2),$$ where the superscript means that we omit the Euler factors at primes dividing $N$, and the product $\prod_{p \nmid N}$ (here and below) is ordered according to increasing $p$. Using [@HidaInvent81 Theorem 5.1] we have $$L^N(\kappa, \Sym^2 f) = \prod_{p\mid N}\left(1-\frac{\lambda_f(p)^2}{p^{\kappa}}\right) \times \frac{2^{2\kappa}\pi^{\kappa+1}}{(\kappa-1)! \delta(N) N \phi(N)}\langle f,f \rangle_N,$$ where $\delta(N)=2$ or 1 according as $N \leq 2$ or not. Using this we obtain the following corollary that can be viewed as a factorization of the ‘global’ period $\langle f,f\rangle_N$ in terms of the ‘local’ periods $ \langle f,f\rangle^{(p)}_N$.

We have $$\langle f,f \rangle_N = \frac{(\kappa-1)! \delta(N) N \phi(N)\zeta^N(2)}{2^{2\kappa}\pi^{\kappa+1}} \prod_{p\mid N}\frac{1}{1-\lambda_f(p)^2/p^{\kappa}}  \prod_{p \nmid N} \langle f,f\rangle^{(p)}_N.$$

Convergence of Euler products on ${\rm Re}(s) = 1$\
by Keith Conrad$^3$
===================================================

Let $K$ be a number field. A degree $d$ Euler product over $K$ is a product $$L(s) = \prod_{\mathfrak p} \frac{1}{(1 - \alpha_{{\mathfrak p},1}{{\rm N}}{\mathfrak p}^{-s})\cdots (1 - \alpha_{{\mathfrak p},d}{{\rm N}}{\mathfrak p}^{-s})},$$ where $|\alpha_{{\mathfrak p},j}| \leq 1$ for all nonzero prime ideals ${\mathfrak p}$ in the integers of $K$ and $1 \leq j \leq d$. On the half-plane ${\operatorname{Re}}(s) > 1$ this converges absolutely and is nonvanishing. Combining factors at prime ideals lying over a common prime number, $L(s)$ is also an Euler product over ${\mathbf Q}$ of degree $d[K:{\mathbf Q}]$.

We want to prove a general theorem about the representability of $L(s)$ by its Euler product on the line ${\operatorname{Re}}(s) = 1$. If $L(s)$ is the $L$-function of a nontrivial Dirichlet character, this is in [@davenport pp. 57–58], [@landau $\S$ 109], and [@mv p. 124] if $s = 1$ and [@landau $\S$ 121] if ${\operatorname{Re}}(s) = 1$.

\[ethm\] If $L(s)$ is a degree $d$ Euler product over $K$ and it admits an analytic continuation to ${\operatorname{Re}}(s) = 1$ where it is nonvanishing, then $L(s)$ is equal to its Euler product on ${\operatorname{Re}}(s) = 1$ when factors are ordered according to prime ideals of increasing norm: if ${\operatorname{Re}}(s) = 1$ then $$L(s) =
\lim_{x \rightarrow \infty} \prod_{{{\rm N}}{\mathfrak p}\leq x} \frac{1}{(1 - \alpha_{{\mathfrak p},1}{{\rm N}}{\mathfrak p}^{-s})\cdots (1 - \alpha_{{\mathfrak p},d}{{\rm N}}{\mathfrak p}^{-s})}.$$

The proof is based on the following lemma about representability of a Dirichlet series on the line ${\operatorname{Re}}(s) = 1$.

\[rthm\] Suppose $g(s) = \sum_{n \geq 1} b_n n^{-s}$ has bounded Dirichlet coefficients. If $g(s)$ admits an analytic continuation from ${\operatorname{Re}}(s) > 1$ to ${\operatorname{Re}}(s) \geq 1$, then $g(s)$ is still represented by its Dirichlet series on the line ${\operatorname{Re}}(s) = 1$.

See [@newman].

Here is the proof of Theorem \[ethm\].

We will apply Lemma \[rthm\] to a logarithm of $L(s)$, namely the absolutely convergent Dirichlet series $$(\log L)(s) := \sum_{{\mathfrak p}} \sum_{k \geq 1} \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}},$$ where ${\operatorname{Re}}(s) > 1$. The coefficient of $1/{{\rm N}}{\mathfrak p}^{ks}$ has absolute value at most $d/k \leq d$, so if we collect terms and write $(\log L)(s)$ as a Dirichlet series indexed by the positive integers, say $\sum_{n \geq 1} c_n/n^s$, then $c_n = 0$ if $n$ is not a prime power and $|c_n| \leq d[K:{\mathbf Q}]$ if $n$ is a prime power. Therefore the coefficients of $(\log L)(s)$ as a Dirichlet series over ${\mathbf Z}^+$ are bounded.

Since $L(s)$ is assumed to have an analytic continuation to a nonvanishing function on ${\operatorname{Re}}(s) \geq 1$, $(\log L)(s)$ has an analytic continuation to ${\operatorname{Re}}(s) \geq 1$, so Lemma \[rthm\] implies that $$\label{L1t}
(\log L)(s) = \sum_{{\mathfrak p}^k}  \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}$$ for ${\operatorname{Re}}(s) = 1$, where the terms in the series are collected in order of increasing values of ${{\rm N}}({\mathfrak p}^k)$.

Although a rearrangement of terms in a conditionally convergent series can change its value, one particular rearrangement of the series in (\[L1t\]) doesn’t change the sum: $$\label{apt}
\sum_{{\mathfrak p}^k} \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}} =
\sum_{{\mathfrak p}}\sum_{k \geq 1} \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}$$ when ${\operatorname{Re}}(s) = 1$, where the sum on the left is in order of increasing values of ${{\rm N}}({\mathfrak p}^k)$ and the outer sum on the right is in order of increasing values of ${{\rm N}}({\mathfrak p})$. To prove (\[apt\]), we rewrite it as $$\label{sum2}
 \sum_{{{\rm N}}({\mathfrak p}^k) \leq x}
  \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}} =
\sum_{{{\rm N}}({\mathfrak p}) \leq x} \sum_{k \geq 1}
 \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}
  + o(1)$$ as $x \rightarrow \infty$, and we will prove (\[sum2\]) when ${\operatorname{Re}}(s) > 1/2$, not just ${\operatorname{Re}}(s) = 1$. For ${\operatorname{Re}}(s) = 1$ we can pass to the limit in (\[sum2\]) as $x \rightarrow \infty$ and conclude (\[apt\]).

The sum on the right in (\[sum2\]) the sum on the left in (\[sum2\]) is equal to $$\sum_{{{\rm N}}({\mathfrak p}) \leq x} \sum_{\substack{k \geq 2\\ {{\rm N}}({\mathfrak p})^k > x}}
  \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}},$$ which is equal to $$\label{bigt}
\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x}
\sum_{k \geq 2}
 \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}
 +
\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \sum_{\substack{k \geq 3 \\ {{\rm N}}({\mathfrak p})^k > x}}
 \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}.$$ The absolute value of the first sum in (\[bigt\]) is bounded above by $$\begin{aligned}
\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x} \sum_{k \geq 2} \left|\frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}\right| & \leq &
\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x} \sum_{k \geq 2} \frac{d}{k{{\rm N}}{\mathfrak p}^{k\sigma}} \ \ \text{ where } \sigma = {\operatorname{Re}}(s) \\
& < &  \frac{d}{2}\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x} \sum_{k \geq 2} \frac{1}{{{\rm N}}{\mathfrak p}^{k\sigma}} \\
& = & \frac{d}{2}\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x} \frac{1}{{{\rm N}}{\mathfrak p}^{\sigma}({{\rm N}}{\mathfrak p}^{\sigma} - 1)} \\
& < & \frac{d}{2}\sum_{\sqrt{x} < {{\rm N}}({\mathfrak p}) \leq x} \frac{4}{{{\rm N}}{\mathfrak p}^{2\sigma}} \quad \textup{since ${{\rm N}}(\mathfrak p)^\sigma > \sqrt{2} > \frac{4}{3}$,}\end{aligned}$$ which tends to 0 as $x \rightarrow \infty$ since $\sum_{{\mathfrak p}} 1/{{\rm N}}{\mathfrak p}^{2\sigma}$ converges. The absolute value of the second sum in (\[bigt\]) is bounded above by $$\begin{aligned}
\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \sum_{\substack{k \geq 3\\ {{\rm N}}({\mathfrak p})^k > x}}  \frac{d}{k{{\rm N}}{\mathfrak p}^{k\sigma}} &<
\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \sum_{\substack{k \geq 3\\ {{\rm N}}({\mathfrak p})^k > x}}  \frac{d}{\log_{{{\rm N}}{\mathfrak p}}(x){{\rm N}}{\mathfrak p}^{k\sigma}}\\
    &=
\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \frac{d\log {{\rm N}}({\mathfrak p})}{\log x}\sum_{\substack{k \geq 3 \\ {{\rm N}}({\mathfrak p})^k > x}}  \frac{1}{{{\rm N}}{\mathfrak p}^{k\sigma}}.\end{aligned}$$ Letting $n$ be the least integer above $\log_{{{\rm N}}{\mathfrak p}}(x)$, $$\sum_{\substack{k \geq 3 \\ {{\rm N}}({\mathfrak p})^k > x}}  \frac{1}{{{\rm N}}{\mathfrak p}^{k\sigma}}  = \frac{1/{{\rm N}}({\mathfrak p})^{n\sigma}}{1 - 1/{{\rm N}}({\mathfrak p})^{\sigma}} < \frac{1/x^{\sigma}}{1/4} = \frac{4}{x^{\sigma}},$$ so $$\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \sum_{\substack{k \geq 3 \\ {{\rm N}}({\mathfrak p})^k > x}}  \frac{d}{k{{\rm N}}{\mathfrak p}^{k\sigma}} <
\sum_{{{\rm N}}({\mathfrak p}) \leq \sqrt{x}} \frac{4d\log {{\rm N}}({\mathfrak p})}{x^{\sigma}\log x}  = O\left(\frac{\sqrt{x}}{x^{\sigma}\log x}\right),$$ which tends to 0 as $x \rightarrow \infty$ since $\sigma > 1/2$.

Now that we established (\[apt\]), take the exponential of the right side: if ${\operatorname{Re}}(s) = 1$, then $$\prod_{{\mathfrak p}} \exp\left(\sum_{k \geq 1} \frac{\alpha_{{\mathfrak p},1}^k + \cdots + \alpha_{{\mathfrak p},d}^k}{k{{\rm N}}{\mathfrak p}^{ks}}\right) =
\prod_{{\mathfrak p}} \frac{1}{(1 - \alpha_{{\mathfrak p},1}{{\rm N}}{\mathfrak p}^{-s})\cdots (1 - \alpha_{{\mathfrak p},d}{{\rm N}}{\mathfrak p}^{-s})},$$ where the products run over ${\mathfrak p}$ in order of increasing norms and the last calculation is justified since $|\alpha_{{\mathfrak p},j}/{{\rm N}}({\mathfrak p})^s| \leq 1/{{\rm N}}({\mathfrak p}) < 1$. Since $L(s) = e^{(\log L)(s)}$ for ${\operatorname{Re}}(s) \geq 1$, by (\[L1t\]) and (\[apt\]) we have $$L(s) =
\prod_{{\mathfrak p}} \frac{1}{(1 - \alpha_{{\mathfrak p},1}{{\rm N}}{\mathfrak p}^{-s})\cdots (1 - \alpha_{{\mathfrak p},d}{{\rm N}}{\mathfrak p}^{-s})}$$ for ${\operatorname{Re}}(s) = 1$, where the product is in order of increasing values of ${{\rm N}}{\mathfrak p}$.

Let $L(s)$ be the $L$-function of the elliptic curve $y^2 = x^3 - x$ over ${\mathbf Q}$. For ${\operatorname{Re}}(s) > 3/2$ it has an Euler product over the odd primes of the form $$\label{L1p}
L(s) = \prod_{p \not= 2} \frac{1}{1 - a_p p^{-s} + p \cdot p^{-2s}} =
\prod_{p \not= 2} \frac{1}{(1 - \alpha_{p}p^{-s})(1 - \beta_{p}p^{-s})},$$ where $|\alpha_{p}| = \sqrt{p}$ and $|\beta_{p}| = \sqrt{p}$ for $p \not= 2$. Since $y^2 = x^3 - x$ has CM by ${\mathbf Z}[i]$, $L(s)$ is also the $L$-function of a Hecke character $\chi$ on ${\mathbf Q}(i)$ such that $|\chi((\alpha))| = |\alpha| =  |{{\rm N}}(\alpha)|^{1/2}$ for all nonzero $\alpha$ in ${\mathbf Z}[i]$ with odd norm. Therefore $L(s)$ also has an Euler product over the nonzero prime ideals of ${\mathbf Z}[i]$ of odd norm: for ${\operatorname{Re}}(s) > 3/2$, $$\label{L2p}
L(s) = \prod_{(\pi) \not= (1+i)} \frac{1}{1 - \chi(\pi)/{{\rm N}}(\pi)^s}.$$

The function $L(s)$ is entire and is nonvanishing on the line ${\operatorname{Re}}(s) = 3/2$, so $L(s + 1/2)$ fits the conditions of Theorem \[ethm\] using $K = {\mathbf Q}$ and $d = 2$ for (\[L1p\]), and $K = {\mathbf Q}(i)$ and $d = 1$ for (\[L2p\]). Therefore (\[L1p\]) and (\[L2p\]) are both true on the line ${\operatorname{Re}}(s) = 3/2$. For instance, $L(3/2) \approx .826348$, the partial Euler product for (\[L1p\]) at $s = 3/2$ over prime numbers up to 100,000 is $\approx .826290$, and the partial Euler product for (\[L2p\]) at $s = 3/2$ over nonzero prime ideals in ${\mathbf Z}[i]$ with norm up to 100,000 is $\approx .826480$.

[^1]: The first author was partially supported by the National Security Agency under Grant Number H98230-11-1-0137. The United States Government is authorized to reproduce and distribute reprints not-withstanding any copyright notation herein. The second author was partially supported by a PSC-CUNY Award, jointly funded by The Professional Staff Congress and The City University of New York.
