# Counting lattice points in adelic spaces

Let $$\mathbb{A}$$ denote the ring of adeles of $$\mathbb{Q}$$, let $$\mu$$ be the Haar measure of $$\mathbb{A}$$, and let $$\|\cdot\|_{\infty}$$ denote the sup-norm of the components in the Archimedean component of $$\mathbb{A}^m$$ (that is $$\mathbb{R}^m$$), $$m\ge 1$$.

For any subset $$E\subseteq \mathbb{A}^m$$ and any $$B>0$$, we denote $$E(B):=E\cap \{\mathbf{x}\in\mathbb{A}^m\,:\,\|\mathbf{x}\|_{\infty}\le B\}$$.

Consider also $$\mathbb{Z}^m$$ embedded diagonally in $$\mathbb{A}^m$$, and let $$\Lambda(B):=\mathbb{Z}^m\cap E(B)$$.

In analogy to what happens in Euclidean spaces, since we still have a natural (uniquely) ergodic action, I would expect that if $$E$$ is somehow "nicely shaped" then one has $$|\#\Lambda(B)-\mu^m(E(B))|=o(\mu^m(E(B)))$$ as $$B\to \infty$$.

Are you aware of any sufficient (not obvious) condition on $$E$$ to ask to make this happen?

• From some viewpoints, the thing like a "lattice" in $\mathbb A^n$ would be $\mathbb Q^n$ here... ??? And for a finite-covolume discrete subgroup $\Gamma$ of a unimodular topological group $G$, a measure-theoretic analogue of Minkowski's lemma holds. Of course, $\mathbb Z^n$ is not finite covolume... – paul garrett Jun 29 '19 at 17:15