-2
$\begingroup$

Let $E:y^2＝x^3-x/ \Bbb{Q}(i)$ be elliptic curve and $L(E,1)$ be a special value of $L$ function of $E$ at $1$.

Let $L(ψ,1)$ be value at $1$ of Hecke $L$ function with respect to Hecke character $ψ$, It is known that $L(E,1)＝L( \bar{ψ},1)L(ψ,1)$.

In this case, why $L(ψ,1)＝1$ ?

I may forget some trivial fact about Hecke $L$ character. Thank you for your help.

Improve this question
$\endgroup$
8
  • 4
    $\begingroup$ I think the formula $L(E,1)＝L( \bar{\psi},1)L(\psi,1)$ is simply not true. $\endgroup$
    – Will Sawin
    Commented Jul 19, 2022 at 12:55
  • 2
    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 13:16
  • 3
    $\begingroup$ Please don't change the question after it was answered. In the original version of the question, $E$ was regarded over $\mathbb{Q}$, not $\mathbb{Q}(i)$. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 14:52
  • 3
    $\begingroup$ @dandelion We answered your question. There is a certain Hecke character $\psi$ of the number field $\mathbb{Q}(i)$ such that $L(s,E/\mathbb{Q})=L(s,\psi)$ and $L(s,E/\mathbb{Q}(i))=L(s,\psi)L(s,\bar\psi)$. I even gave you a precise reference. Your other question "why $L(1,\psi)=1$?" is hard to answer, because $L(1,\psi)\neq 1$. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 16:29
  • 2
    $\begingroup$ Yes, we have $L(s,E/\mathbb{Q})=L(s,\psi)=L(s,\bar\psi)$. In general, for any Hecke character $\psi$ of any number field, $L(s,\bar\psi)$ can be obtained from $L(s,\psi)$ by complex conjugating the Dirichlet coefficients. In the case of $L(s,E/\mathbb{Q})$, the Dirichlet coefficients are real by definition. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 18:01

1 Answer 1

Reset to default
3
$\begingroup$

Not clear what you mean by $\varphi$ or $\psi$. At any rate, $E$ is a CM curve, hence $L(s,E)$ equals $L(s,\psi)$ for a suitable Hecke character $\psi$ of the number field $\mathbb{Q}(i)$. You can find the details (e.g. the definition of $\psi$) in Sections 8.3-8.4 of Iwaniec's book "Topics in classical automorphic forms".

Added. As David Loeffler pointed out, if $E$ is regarded as a curve over $\mathbb{Q}(i)$, its $L$-function is not $L(s,\psi)$ but $L(s,\psi)L(s,\bar\psi)$. In my answer, $E$ was regarded as a curve over $\mathbb{Q}$, as was in the original question (cf. revision history).

Improve this answer
$\endgroup$
18
  • 2
    $\begingroup$ @dandelion Note also that elliptic curve people like to normalize $L(s,E)$ so that $s=1$ is the center, and the functional equation relates $s$ to $2-s$. This is a bad convention. The correct normalization is that $s=1/2$ is the center, and the functional equation relates $s$ to $1-s$. So, most likely, your $L(1,E)$ is $L(1/2,E)$ with the proper normalization. The two normalizations are connected by the shift $s\mapsto s+1/2$. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 14:04
  • 3
    $\begingroup$ @GHfromMO you say the character $\overline{\psi}$ is not part of the equation relating the $L$-functions of $E$ and $\psi$. While you meant to say $L(s,E)$ is not $L(s,\psi)L(s,\overline{\psi})$, in fact you could use either $\psi$ or $\overline{\psi}$ individually since $L(s,\psi) = L(s,\overline{\psi})$ despite $\psi \not= \overline{\psi}$. It is an analytic analogue of ${\rm N}(\alpha) = {\rm N}(\overline{\alpha})$ for all Gaussian integers $\alpha$ and in fact that property is part of the explanation of the equality of the $L$-functions. $\endgroup$
    – KConrad
    Commented Jul 19, 2022 at 14:25
  • 2
    $\begingroup$ @KConrad Thanks for your comment. Another way to explain this is that $L(s,E)$ is self-dual (i.e. the Dirichlet coefficients are real). $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 14:32
  • 4
    $\begingroup$ @DavidLoeffler Of course normalization is a matter of taste and convention. I take the point of view that without a uniform normalization there is confusion. As far as we know, every reasonable $L$-function (in arithmetic) can be built up from automorphic $L$-functions. For most automorphic representations the most natural way to define the $L$-function makes $s=1/2$ the center. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 14:45
  • 4
    $\begingroup$ @KConrad Yes, I meant unitary Hecke characters. For many of us, character means a (continuous) homomorphism into the unit circle. In general, my comment was about $L(s,\pi)$ where $\pi$ is an automorphic representation with a unitary central character. The functional equation for such an $L$-function should relate $s$ to $1-s$, out of respect for Riemann and to restore harmony in the universe. $\endgroup$
    – GH from MO
    Commented Jul 19, 2022 at 14:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .