I am referring to Tao's blog post about Tate's thesis. Introduce the adeles $\mathbb A$ of $\mathbb Q$ and the adelic Mellin transform $$Z(s) = \int_{\mathbb A^\times} = g(x) |x|^s d^\times x.$$

Here, $g = \prod_v g_v$ is a product over places $v$ of $\mathbb Q$ and $g_v$ is a self-dual function on $\mathbb Q_v$, more precisely the Gaussian $g(x) = \exp(-\pi x^2)$ for the real place and the characteristic function of integers $\mathbb 1_{\mathcal O_p}$ at finite places $p$. I would like to justify that $$Z(x) = \pi^{-s/2} \Gamma(s/2) \zeta(s), $$ as in the blog's equation (24), directly from properties of the adeles (and not the Euler product of $\zeta$). In particular, Tao emphasizes that we can use an explicit fundamental domain for $\mathbb A^\times / \mathbb Q^\times$, viz. $$\mathbb A^\times = \bigsqcup_{k \in \mathbb Q^\times} k \left( J := \mathbb R_+ \prod_p \mathbb Z_p^\times \right).$$

Cutting by classes modulo $\mathbb Q^\times$, we get $$Z(s) = \int_{\mathbb A^\times / \mathbb Q^\times} \left( \sum_{k \in \mathbb Q^\times} g(kx) \right) |x|^s d^\times x$$ where we recognize the adelic analogue of the theta function used in the derivation of the classical functional equation.

However, I do not understand how using the decomposition above we would find another way to recover $Z(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$.


$$Z(s)= \sum_{k\in \Bbb{Q}_{>0}}\int_{(k\Bbb{\hat{Z}^\times)\times R^*}} g(x)|x|^s d^\times x = \sum_{k\ge 1} \int_{(k\Bbb{\hat{Z}^\times)\times R^*}}g(x) |x|^s d^\times x $$ $$=\sum_{k\ge 1} |k|_{fin}^s \int_{\Bbb{\hat{Z}^\times\times R^*}}g(x)|x|^s d^\times x = \sum_{k\ge 1} k^{-s} \int_{-\infty}^\infty e^{-\pi t^2} |t|^s \frac{dt}{t}= \zeta(s)\pi^{-s/2}\Gamma(s/2)$$ Where $\int_{\Bbb{Z}^\times} d^\times x$ is normalized to $1$

  • $\begingroup$ Here is where I’m lost: why is your second equality true? What allows to reduce the sum over Q to a sum over integers? (otherwise I suppose there is a missing $g(x)$ in the third integral, and this one only is on the archimedean component now that you took out the finite part of the norm?) $\endgroup$ – Desiderius Severus Jan 17 at 9:28
  • $\begingroup$ Except for this justification, the answer is great, thank you! $\endgroup$ – Desiderius Severus Jan 17 at 11:46
  • 1
    $\begingroup$ $g(x)=0$ if $k$ is not an integer $\endgroup$ – reuns Jan 17 at 11:51

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