Notes on the Adelic Zeta Function

Published March 2, 2023｜4 minute read

This post motivates the complete zeta function $\zeta_{\mathbb{A}}(z) = \pi^{-z/2} \, \Gamma(z/2) \, \zeta(z)$, explaining how it arises in a natural way by including contributions from all possible completions of $\mathbb{Q}$. The argument is from Andreas Knauf’s excellent lecture notes; I just fill in some details.

By Ostrowski's theorem, the only possible absolute values on $\mathbb{Q}$ are the usual absolute value $|\cdot| = |\cdot|_\infty$ and the $p$-adic absolute value $|\cdot|_p$. The completion of $\mathbb{Q}$ with respect to these absolute values give the real numbers $\mathbb{R} = \mathbb{Q}_\infty$ and the $p$-adic rationals $\mathbb{Q}_p$, respectively.

One performs Fourier analysis over $\mathbb{R}$ by executing a change of basis to the eigenfunctions $\chi_{u, \infty} \defeq \mathrm{exp}(2\pi iux)$ of the translation group $\mathbb{R}/\mathbb{Z}$. Here the frequency $u$ takes values in $\mathbb{Q}_\infty$. An analogous basis exists for the $p$-adics; we may take $\chi_{u, p} \defeq \mathrm{exp}(2\pi i [ux]_p)$, where $u \in \mathbb{Q}_p$ and $[ux]_p$ denotes the “fractional part.”

In either case, we may define the Fourier transform

$$\mathcal{F}_v f(u) \defeq \int_{\mathbb{Q}_v} f(x) \, \chi_{u, v}(x) \, \dd{x}_v.$$

It is natural to ask about the $\mathcal{F}$-invariant functions. In the case of $\mathbb{Q}_\infty$, the answer is the familiar Gaussian $\mathrm{exp}(-\pi x^2)$. For $\mathbb{Q}_p$, a possible solution is the closed ball

$$\gamma_p(x) \defeq \begin{cases} 1, &|x|_p \leq 1, \\ 0, &|x|_p > 1 \end{cases}$$

of radius 1. Indeed, we may write out

$$\begin{aligned} \mathcal{F}_p \gamma_p(u) &= \int_{\mathbb{Q}_p} \mathbf{1}_{|x|_p \leq 1} \mathrm{exp}(2\pi i [ux]_p) \dd{x}_p \\ &= \int_{\mathbb{Z}_p} \mathrm{exp}(2\pi i [ux]_p) \dd{x}_p. \end{aligned}$$

Here the measure $\dd{x}_p$ satisfies

$$\dd{x}_p(a + p^n \mathbb{Z}_p) = p^{-n}$$

for integers $a$ and $n \geq 0$. If $|u|_p \leq 1$, then all $ux \in \mathbb{Z}_p$, so the integral reduces to

$$\begin{aligned} \int_{\mathbb{Z}_p} \dd{x}_p &= \dd{x}_p(\mathbb{Z}_p) \\ &= 1 \\ &= \gamma_p(u), \end{aligned}$$

as claimed. If $|u|_p = p^k > 1$, then the integral becomes

$$\begin{aligned} &\sum_{a=0}^{p^k} \int_{a + p^k \mathbb{Z}_p} \mathrm{exp}(2\pi i [ux]_p) \dd{x}_p \\ &= \sum_{a=0}^{p^k} \mathrm{exp}(2\pi i [ua]_p) \, \dd{x}_p(a + p^k \mathbb{Z}_p) \\ &= p^{-k} \sum_{a=0}^{p^k} \mathrm{exp}(2\pi i a/p^k) \\ &= 0 \\ &= \gamma_p(u), \end{aligned}$$

again as claimed, where the second equality involves reshuffling the values of $[ua]_p$.

Now define the function

$$\zeta_v(s) \defeq \int_{\mathbb{Q}_v} \gamma_v(x) \, |x|_v^s \dd{x}^*_v,$$

where we now use the multiplicatively-invariant normalized Haar measure

$$\dd{x}_\infty^* \defeq \frac{\dd{x}_\infty}{\abs{x}_\infty} \qquad \mathrm{and} \qquad \dd{x}_p^* \defeq \frac{p}{p-1} \frac{\dd{x}_p}{\abs{x}_p},$$

since the main object of integration $|x|_p^s$ transforms multiplicatively (i.e. $|x|_p^s |y|_p^s = |xy|_p^s$). For the case of $\mathbb{Q}_\infty$, we straightforwardly obtain the result

$$\begin{aligned} \int_{\mathbb{R}} e^{-\pi x^2} |x|^s \dd{x}_\infty^* &= 2 \int_0^\infty e^{-\pi x^2} x^{s-1} \dd{x} \\ &= \pi^{-s/2} \Gamma(s/2). \end{aligned}$$For the case of $\mathbb{Q}_p$, we obtain$$\begin{aligned} &\int_{\mathbb{Q}_p} \mathbb{1}_{|x|_p \leq 1} |x|_p^s \dd{x}_p^* \\ &= \frac{p}{p-1} \int_{\mathbb{Z}_p} |x|_p^{s-1} \dd{x}_p \\ &= \frac{p}{p-1} \sum_{n=0}^\infty \int_{|x|_p = p^{-n}} |x|_p^{s-1} \dd{x}_p \\ &= \frac{p}{p-1} \sum_{n=0}^\infty (p^{1-s})^n \dd{x}_p(p^n \mathbb{Z}_p - p^{n+1} \mathbb{Z}_p) \\ &= \frac{p}{p-1} \sum_{n=0}^\infty (p^{1-s})^n \, \frac{p-1}{p^{n+1}} \\ &= \frac{1}{1-p^{-s}}, \end{aligned}$$

where we have evaluated

$$\begin{aligned} &\dd{x}_p(p^n \mathbb{Z}_p - p^{n+1} \mathbb{Z}_p) \\ &= \sum_{a=1}^{p-1} \dd{x}_p(a p^n + p^{n+1} \mathbb{Z}_p) \\ &= \frac{p-1}{p^{n+1}}. \end{aligned}$$

It is then natural to combine all the $\zeta_v$ to obtain an adelic zeta function

$$\zeta_{\mathbb{A}}(s) \defeq \prod_v \zeta_v(s) = \pi^{-s/2} \, \Gamma(s/2) \, \zeta(s),$$

which obeys the simple functional equation $\zeta_{\mathbb{A}}(s) = \zeta_{\mathbb{A}}(1-s)$.