# field is discrete and cocompact in its adèles

For brevity, we write $P_{f}$ for the set of finite places of $K$, and $P_{\infty}$ for the set of infinite places. We also write $\sideset{}{{}^{\prime}}{\prod}$ for a restricted direct product. Then

 $\mathbb{A}_{K}=\sideset{}{{}^{\prime}}{\prod}_{v\in P_{f}}K_{v}\times\prod_{v% \in P_{\infty}}K_{v}$
###### Theorem 1.

$K$ is discrete as a subgroup of $\mathbb{A}_{K}$.

Proof. Since $\mathbb{A}_{K}$ is a topological ring, it suffices to show that there is a neighborhood in $\mathbb{A}_{K}$ meeting $K$ in only $0$.

Let

 $U=\prod_{v\in P_{f}}\mathfrak{o}_{v}\times\prod_{v\in P_{\infty}}B\left(0,% \frac{1}{2}\right)$

Since $\mathfrak{o}_{v}$ is open in $K_{v}$ for $v$ finite, this is an open set. (Note that $\mathfrak{o}_{v}=\mathcal{O}_{K_{v}}$, the ring of algebraic integers of $K_{v}$).

Now consider an element $x\in U\cap K\subset\mathbb{A}_{K}$. If $x=(x_{v})$, then for $v$ finite, $x_{v}\in\mathfrak{o}_{v}$, and for $v$ infinite  , $x_{v}\in B\left(0,\frac{1}{2}\right)$. Assume $x\neq 0$. Then

 $\left\lvert x_{v}\right\rvert_{v}\leq\begin{cases}1&v\text{ finite}\\ \frac{1}{2}<1&v\text{ infinite}\end{cases}$

but then

 $\prod\left\lvert x\right\rvert_{v}=\prod\left\lvert x_{v}\right\rvert_{v}<1$

The above theorem is very sensitive to the fact that all places are included in $\mathbb{A}_{K}$. For example, it is clear that the image of $\mathbb{A}_{K}$ in $\prod_{v\in P_{\infty}}K_{v}$ is dense, since $K_{v}$ is characterized by an embedding   $K\hookrightarrow K_{v}\cong\mathbb{R},\mathbb{C},\mathbb{Q}_{p}$. Then by an argument familiar from Minkowski’s theorem, $\mathcal{O}_{K}$ is a full-rank lattice   in the image of $K$. But $K$ is the $\mathbb{Q}$-span of that lattice, so is dense in $K_{v}$.

Furthermore, the same is true for the finite places:

###### Proposition 2.

The image of $K$ in $\sideset{}{{}^{\prime}}{\prod}_{v\in P_{f}}\mathfrak{o}_{v}$ is dense.

Proof. Suppose $x=(x_{v})_{v}\in\sideset{}{{}^{\prime}}{\prod}_{v\in P_{f}}\mathfrak{o}_{v}$. We show that $x$ can be approximated as closely as desired by an element of $K$ by showing that for any ideal $I\subset\mathcal{O}_{K}$, there is $y\in K$ such that $y-x_{v}\in I\mathfrak{o}_{v}$ for each $v\in P_{f}$.

First multiply through by some $z$ so that everything is in $\mathcal{O}_{K}$: choose $z\neq 0$ such that $zx_{v}\in\mathfrak{o}_{v}$ for all $v\in P_{f}$. This is possible since all but finitely many $x_{v}$ are already in $\mathfrak{o}_{v}$. Thus $y-x_{v}\in I\mathfrak{o}_{v}$ is equivalent      to $zy-zx_{v}\in(zI)\mathfrak{o}_{v}\subset I\mathfrak{o}_{v}$. So assume wlog that $x_{v}\in\mathfrak{o}_{v}$ for all $v$; in the end simply divide by $z$ to recover the general case. But then the existence of $y$ is guaranteed by the Chinese Remainder Theorem    , since if $I=\prod\mathfrak{p}_{i}^{e_{i}}$, then $I\mathfrak{o}_{v}=\mathfrak{p}_{i}^{e_{i}}$ for some $i$.

It is true, though somewhat harder to prove, that $K$ is in fact dense in $\mathbb{A}_{K}$ if even one place is missing from the product   !

###### Theorem 3.

$\mathbb{A}_{K}/K$

Proof. The set

 $U=\prod_{v\in P_{f}}\mathfrak{o}_{v}\times\prod_{v\in P_{\infty}}K_{v}$

is open in $\mathbb{A}_{K}$.

Claim first that $K+U=\mathbb{A}_{K}$. Choose $(x_{v})\in\mathbb{A}_{K}$. There is a finite set  $S$ of finite places $v$ such that $x_{v}\notin\mathfrak{o}_{v}$ for $v\in S$. Using an argument identical to the approximation argument above, choose $y\in K$ such that $y-x_{v}\in\mathfrak{o}_{v},v\in S$ and $y\in\mathfrak{o}_{v},v\notin S$. Then $(x_{v}-y)_{v}$ is in $\mathfrak{o}_{v}$ for $v\in S$, is in $\mathfrak{o}_{v}$ for $v\notin S$ but finite, and is in $K_{v}$ for $v$ infinite. Thus $(x_{v}-y)_{v}\in U$ and we are done.

Claim next that $K\cap U=\mathcal{O}_{K}$. $\supset$ is obvious. To see $\subset$, note that an element of $K\cap U$ is an element of $K$ that is integral  at every finite place, so it is integral and is in $\mathcal{O}_{K}$.

Thus we get a natural map $U\hookrightarrow\mathbb{A}_{K}\twoheadrightarrow\mathbb{A}_{K}/K$. This map is surjective  since $K+U=\mathbb{A}_{K}$, and its kernel is $K\cap U$. So it suffices to show that $U/(K\cap U)=U/\mathcal{O}_{K}$ is compact. There is obviously an exact sequence induced by the projection $U\to\prod_{v\in P_{\infty}}K_{v}$,

 $\prod_{v\in P_{f}}\mathfrak{o}_{v}\to U/\mathcal{O}_{K}\to\prod_{v\in P_{% \infty}}K_{v}/\mathcal{O}_{K}\to 0$

The left-hand side is compact since each $\mathfrak{o}_{v}$ is, and the right-hand side is

 $\mathbb{R}^{r_{1}+2r_{2}}/\mathcal{O}_{K}$

which know is compact since $\mathcal{O}_{K}$ forms a full-rank lattice in $\mathbb{R}^{n}$. Thus $U/\mathcal{O}_{K}$ is also compact and we are done.

So we have shown that $\mathbb{A}_{K}$ is a locally compact ring, and that $K\subset\mathbb{A}_{K}$ is discrete and cocompact. This is analogous to two other situations with which we are familiar:

 $\displaystyle\mathbb{R}\text{ is locally compact, }\mathbb{Z}\subset\mathbb{R}% \text{ is discrete and cocompact}$ $\displaystyle\prod_{v\in P_{\infty}}K_{v}\text{ is locally compact, }\mathcal{% O}_{K}\subset\prod_{v\in P_{\infty}}K_{v}\text{ is discrete and cocompact}$

This is a useful concept because in such a situation one can do Fourier analysis. For example, if $f:\mathbb{R}\to\mathbb{R}$ is a $C^{\infty}$ function  with exponential decay (or at least integrable on all of $\mathbb{R}$), then we can define its Fourier transform   $\hat{f}$, and the Poisson summation formula

 $\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\hat{f}(n)$

relates the two. The same theory thus exists for appropriately defined functions $f:\mathbb{A}_{K}\to\mathbb{R}$, and the Poisson formula again holds with the sum over $K$ rather than over $\mathbb{Z}$. This can be used to show that the $L$-functions have analytic continuations, just as the real Poisson formula is used to show this for $\zeta$.

Title field is discrete and cocompact in its adèles FieldIsDiscreteAndCocompactInItsAdeles 2013-03-22 18:00:05 2013-03-22 18:00:05 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 11R56