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I'd like to know to what point is it possible to generalize this method for obtaining the functional equation for the Dedekind zeta function $\zeta_K(s)$ of a number field ?

Let $\mathfrak{C}$ be the ideal class group $$\zeta_K(s) =\sum_{I \subset \mathcal{O}_K} N(I)^{-s} = \sum_{C_j \in \mathfrak{C}} \ \ \sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s}$$ Assuming $\mathcal{O}_K^\times$ is finite, each ideal $C_j \in \mathfrak{C}$ being a rank $n=[K:\mathbb{Q}]$ free $\mathbb{Z}$-module with basis $b_{j,1}, \ldots, b_{j,n}$, letting $C_j C_j^{-1} = (d_j)$ : $$\zeta_{K,C_j}(s)=\sum_{I \subset \mathcal{O}_K,I \,\sim \, C_j} N(I)^{-s} = \frac{1}{|O_K^\times|} \sum_{a \in C_j^{-1}} N(\frac{a}{d_j} C_j)^{-s}$$ $$=\frac{1}{|O_K^\times|}\frac{N(C_j)^{-s}}{N(d_j)^{-s}}\sum_{m \in \mathbb{Z}^n \setminus (0)} N(\sum_{l} m_l b_{j,l})^{-s}$$ $$\Gamma(s) \zeta_{K,C_j}(s) = \frac{1}{|O_K^\times|} \frac{N(C_j)^{-s}}{N(d_j)^{-s}}\int_0^\infty x^{s-1} \sum_{m \in \mathbb{Z}^n \setminus (0)} e^{-x\ N(\sum_{l} m_l b_{j,l})}dx$$ Therefore it reduces to finding a Poisson summation formula for $\displaystyle\Theta_j(x) = \sum_{m \in \mathbb{Z}^n} e^{-x \, N(\sum_{l} m_l b_{j,l})}$.

  • I did it for $\mathbb{Q}(\sqrt{-5})$ whose ideal class group has two elements, obtaining that $$\Lambda(s)= 5^{s/2}\pi^{-s}\Gamma(s)4 \zeta_{\mathbb{Q}(\sqrt{-5})}(s)=\Lambda(1-s) $$ I guess this method works at least for any imaginary quadratic field ?

  • If $K$ is imaginary quadratic, $\mathcal{O}_K = \mathbb{Z}[w]$ and any ideal has the form $(k,l+ow)$ then due to the property of $2\times 2$ matrices (see how $\scriptstyle\begin{pmatrix} 2 & 1 \\ 0 & \sqrt{5}\end{pmatrix}$ is treated in the link), I'll get a functional equation $\Lambda_{K,C_j}(s) = \lambda_j^{s/2} \pi^{-s}\Gamma(s)\zeta_{K,C_k}(s)=\Lambda_{K,C_j}(1-s)$, but I don't know how to show $\lambda_j $ is the same for each ideal class $C_j$.

  • What about the other cases ($\mathcal{O}_K^\times$ infinite, cubic extension, non-monogenic field...) ?

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    $\begingroup$ Didn't Hecke do this for any number field? Isn't it in his book Algebraic Number Theory? $\endgroup$
    – Stopple
    Commented Apr 4, 2017 at 16:03
  • $\begingroup$ @Stopple : He shows the Poisson summation formula of the theta function of an ideal in the Minkowski space p.215 eq.(180) and (185) $\theta(x) = \sum_{a \in I} e^{-\pi (\sum_{m=1}^n x_m |\sigma_m(a)|^2)}$ where $\{\sigma_m\}_{m=1}^n$ is the Galois group. But he doesn't mention the functional equation for $\zeta_K(s)$ $\endgroup$
    – reuns
    Commented Apr 4, 2017 at 17:11
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    $\begingroup$ I'm writing exactly this proof up for the notes in my ongoing algebraic number theory course and will post a link when done. $\endgroup$
    – Lior Silberman
    Commented Apr 4, 2017 at 21:33
  • $\begingroup$ There are also versions in Lang's and Neukirch's ANT books and I think also in Weil's "basic number theory". $\endgroup$
    – Lior Silberman
    Commented Apr 4, 2017 at 21:40
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    $\begingroup$ @LiorSilberman Neukrich does it p.218, but his notation is horrible. $\endgroup$
    – reuns
    Commented Apr 4, 2017 at 21:56

1 Answer 1

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In any event, this is essentially Hecke's original argument, the only other truly independent known proof being that of Tate. The original paper (in german) is, I believe,

  • Erich Hecke "Über die Zetafunktion beliebiger algebraischer Zahlkörper" (1917)

The method works for arbitrary number fields and, with some extra work, for any twist by a Hecke character.

English expositions of the proof can be found at least in Neukirch and Lang's textbooks.

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  • $\begingroup$ In fact, Hecke's paper is clearer than the account in Lang, I think. $\endgroup$
    – paul garrett
    Commented Apr 8, 2017 at 13:09

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