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 ideles 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) set of finite 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 $\mathbb A_S^{\ast}$.

  • $\begingroup$ The notation $\Bbb{A}^*$ is ambiguous, the ideles are $\prod_v x_v$ with $x_v\in K_v^*$ and $x_v \in O_v^\times$ for all but finitely many $v$, the adeles are $\prod_v x_v$ with $x_v \in O_v$ for all but finitely many $v$, the motivation is the sequences in $O_K$ that converge in $O_K/I$ for every ideal $I$ $\endgroup$ – reuns May 19 at 22:09
  • $\begingroup$ I don't think $\mathbb A^{\ast}$ is ambiguous, the ideles are exactly the units of the adeles. $\endgroup$ – D_S May 19 at 22:51

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