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Let $E$ be an elliptic curve over $\mathbb Q$ with conductor $N$ and $E_d$ be its twisted curve by $d$, where $d$ is a fundamental discriminant with $(d,N)=1$. Let $\chi_d$ be a Dirichlet character defined by $\chi_d(n)=\left( \frac{d}{n} \right)$.

In p.152 of 'A. Perelli and J. Pomykała, Averages of twisted elliptic L-functions, Acta Arith. 80 (1997), 149–163.', the author says that

'As we have already remarked, we follow the proof of Theorem 2 of [6], due to the similarity between $L(s,E_d)$ and $L(s,\chi_d)^2$,'...

I cannot understand this sentence. What 'similarity' means?

Let $L(s,E)=\sum_{n=1}^{\infty} a(n)n^{-s}$ for $Re(s) > 3/2$. Then $$L(s,E_d)=\sum_{n=1}^{\infty} a(n)\chi_d(n)n^{-s}.$$ Owing to the Ramanujan-Petersson bound, we have $|a(n)| \leq \sqrt n \tau(n)$, so if we just plug this into the $L$-series, we get

$$ \sum_{n=1}^{\infty} \tau(n)\chi_d(n) n^{-(s-1/2)}=L(s-\frac{1}{2},\chi_d(n))^2. $$

So it must have some 'relation' between two $L$-series, but I don't know this implies some 'similarity'.

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  • $\begingroup$ This depends on your normalisation of the L-function of an elliptic curve. The “analytic” normalisation is to replace $a(n)$ with $a(n) \sqrt{n}$, in which case the analogy is more clear. $\endgroup$
    – Peter Humphries
    Commented Jul 8, 2019 at 7:21
  • $\begingroup$ Humphries//I think that's not the issue. In the vies of Ramanujan-Petersson bound it is more natural to relate 'non-noormalized' L-function since the critical lines are coincide. But I asked how 'bound' says 'similarity'. $\endgroup$
    – LWW
    Commented Jul 8, 2019 at 9:25
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    $\begingroup$ To an analytic number theorist, it is more natural to normalise a critical line of an $L$-function to be $\Re(s) = 1/2$. So replace $a(n)$ with $a(n) \sqrt{n}$, so that the $a(n)$ are Hecke eigenvalues normalised such that $a(p) \in [-2,2]$, or more generally consider $L(s,f) = \sum_{n =1}^{\infty} \lambda_f(n) \chi_d(n) n^{-s}$, where $\lambda_f(n)$ are the Hecke eigenvalues of an automorphic form... $\endgroup$
    – Peter Humphries
    Commented Jul 8, 2019 at 12:06
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    $\begingroup$ ... Then the analogy is the fact that $\tau(n)$ is the $n$-th Hecke eigenvalue of an Eisenstein series (a noncuspidal automorphic form), while $a(n)$ is the $n$-th Hecke eigenvalue of the cuspidal automorphic form corresponding to an elliptic curve. So these are both particular cases of the $L$-function of the quadratic twist of a $\mathrm{GL}_2$ automorphic form: one cuspidal, one noncuspidal. It is a general fact that $L$-functions of such noncuspidal (isobaric) automorphic forms factorise as a product of $L$-functions of lower degree. $\endgroup$
    – Peter Humphries
    Commented Jul 8, 2019 at 12:08
  • $\begingroup$ Humphries // Ah, now I see what you mean (vaguely). But not solid, due to my lack of knowledge about the L-function of an automorphic form. Could you recommend any references or textbooks? And thank you so much. $\endgroup$
    – LWW
    Commented Jul 8, 2019 at 12:24

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