Let $G$ be a linear algebraic group over a number field $k$. If necessary, assume $G$ is connected and reductive. Let $\mathbb A$ be the ring of adeles of $k$, and $\mathbb A_S = \prod\limits_{v \in S} k_v \prod\limits_{v \not\in S} \mathcal O_v$ for any (large) finite set of places $S$ containing the archimedean ones. Is it the case that

$$G(\mathbb A_S) G(k) = G(\mathbb A)$$

for sufficiently large $S$? This is claimed in Moeglin and Waldspurger's book on Spectral Decomposition and Eisenstein Series, in the proof that $Z(\mathbb A)G(k)$ is closed in $G(\mathbb A)$ when $G$ is connected reductive.

This is easy to see in the case $G = \operatorname{GL}_1$. We have a copy of $H = (0,\infty)$ in $G(\mathbb A) = \mathbb A^{\ast}$ by sending $\rho$ to $(\rho^{1/n}, ... , \rho^{1/n}, 1, 1, ...)$ in $\prod\limits_{v \mid \infty} k_v$, where $n = [k : \mathbb Q]$. The quotient $\mathbb A^{\ast}/H k^{\ast}$ is compact, and is covered by the images of the open sets $\mathbb A_S^{\ast}$.

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    $\begingroup$ What does $G(\mathbb{A}_S)$ mean in this generality? It seems to me that $\mathbb{A}_S$ is not a $k$-algebra in general. $\endgroup$ May 19, 2019 at 22:39
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    $\begingroup$ It just means $\prod\limits_{v \in S} G(k_v) \prod\limits_{v \not\in S} G(\mathcal O_v)$ $\endgroup$
    – D_S
    May 19, 2019 at 22:50

1 Answer 1


Yes, $G(\mathbb A_S) G(k) = G(\mathbb A)$ holds when $S$ is sufficiently large and contains the set of archimedean places $\infty$. This is because the double coset space $G(A_\infty)\backslash G(\mathbb A)/G(k)$ is finite (its cardinality is called the class number), and a set of representatives can be chosen from $G(\mathbb A_S)$ when $S\supset\infty$ is sufficiently large. For more details, see Theorem 5.1 in Platonov-Rapinchuk: Algebraic groups and number theory (Academic Press, 1994).


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