How to interpret this sum in Tate's thesis?

Question

Let $f$ be a Schwartz function on the ring of adeles $\mathbb{A}$ of a number field $K$, and $d^\times x$ the multiplicative Haar measure on $\mathbb{A}^\times$. One can embed $K^\times$ diagonally in $\mathbb{A}^\times$ and can take a quotient $\mathbb{A}^\times/K^\times$. In Tate's thesis, the integral of $f$ over $\mathbb{A}^\times$ is replaced with an integral over $\mathbb{A}^\times/K^\times$ via

$$\int_{\mathbb{A}^\times}f(x)d^\times x = \int_{\mathbb{A}^\times/K^\times}\left(\sum_{\kappa \in K^\times}f(\kappa x)\right) d^\times x.$$

I am wondering how to interpret this sum. In particular, why does it converge, and how big is $\mathbb{A}^\times/K^\times$? For example, if in the real archimedean component $f_\infty(x) = e^{-\pi x^2}$, is he integrating over a function which in this component is equal to $$\sum_{\kappa \in K^\times}e^{-\pi \kappa^2 x^2} $$ and does this converge?

Also, how does this relate to the Jacobi theta function?

Answer 1

You need to work out an actual example so you can see that some of what you wrote makes no sense.

Take $K = \mathbf Q$. In the definition of Schwartz functions on $\mathbf A_\mathbf Q$, they are sums of finitely many products of (nice) local functions with most $p$-factors being the characteristic function of $\mathbf Z_p$. The simplest example is the product function $f(\mathbf x) = e^{-\pi x^2}\prod_p \chi_{\mathbf Z_p}(x_p)$, where $x = x_\infty$.

Fix an idele $\mathbf x$ in $\mathbf A_\mathbf Q^\times$. Let's see why the series $\sum_{k \in \mathbf Q^\times} f(k\mathbf x)$ converges and what it equals. We'll see it is not $\sum_{k \in \mathbf Q^\times} e^{-\pi k^2x^2}$, a sum over all nonzero rationals that converges for no real $x$.

Set $S = \{p : x_p \not\in \mathbf Z_p^\times\}$, which is finite. Let $r = \prod_{p\in S}p^{v_p(x_p)}$, so $r$ is a positive rational number depending on the idele $\mathbf x$ and $x_p/r \in \mathbf Z_p$ for all $p$ (not just $p \in S$).

When $k\in\mathbf Q^\times$, $f(k\mathbf x)=0$ unless $kx_p \in \mathbf Z_p$ for every prime $p$, due to how $f$ is defined. Rewrite "$kx_p \in \mathbf Z_p$" as $(kr)(x_p/r) \in \mathbf Z_p$. Since $x_p/r \in \mathbf Z_p^\times$ for each $p$ (not just $p \in S$, but all $p$), $kx_p \in \mathbf Z_p$ if and only if $kr \in \mathbf Z_p$. We need this for all $p$, which is the same as saying $kr \in \mathbf Z$. In other words, the only nonzero rational $k$ for which $f(k\mathbf x)\not=0$ are the $k \in (1/r)\mathbf Z - \{0\}$. Write $k = n/r$ for a nonzero integer $n$, so (using the actual definition of $f$ as a product of local functions) $$ f(k\mathbf x) = f(n(1/r)\mathbf x) = e^{-\pi (nx/r)^2} = e^{-\pi n^2x^2/r^2} $$ since $(1/r)\mathbf x \in \mathbf R^\times \times \prod_p \mathbf Z_p^\times$. Note $r$ depends on $\mathbf x$. (This $r$ is the first factor in a nice direct product decomposition: $\mathbf A_\mathbf Q^\times = \mathbf Q_{>0} \times \mathbf R^\times \times \prod_p \mathbf Z_p^\times$. When $K \not= \mathbf Q$, $\mathbf A_K^\times$ does not have anything like that tidy decomposition, which makes general idele groups and Hecke characters complicated!)

Now for each idele $\mathbf x$ of $\mathbf Q$, with its associated positive rational $r$, we can write your mystery sum at $\mathbf x$ in the integral as something less mysterious: $$ \sum_{k \in \mathbf Q^\times} f(k\mathbf x) = \sum_{n \in \mathbf Z - \{0\}} e^{-\pi n^2x^2/r^2} = \sum_{n \in \mathbf Z} e^{-\pi n^2(x/r)^2} - 1, $$ which (ignoring the term $-1$ at the end) is the classical Jacobi theta-function at $(x/r)^2$, where $x$ is the archimedean component of the idele $\mathbf x$. In particular, this sum depends only on the archimedean component of $\mathbf x$ and it obviously converges if you know about Jacobi theta-functions.

I haven't actually dealt directly here with the integral, just showing that the sum inside the integral makes sense when we take $f$ to be the simplest nonzero Schwartz function. What one should really be integrating is not that function $f$ I wrote down, but the product $f(\mathbf x)||\mathbf x||^s$, where $||\cdot||$ is the idelic norm and ${\rm Re}(s)$ is large enough. It turns out that the integral with that function converges when ${\rm Re}(s) > 1$ and (for suitable idelic Haar measure) is exactly the Riemann zeta-function. This is worked out in Section 6 of Gelbart and Miller's paper "Riemann's Zeta Function and Beyond" here.