# Do we have $G(\mathbb A_S) G(k) = G(\mathbb A)$ for sufficiently large $S$?

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 $$\mathbb A_S^{\ast}$$.

• 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 $I$ – reuns May 19 at 22:09
• I don't think $\mathbb A^{\ast}$ is ambiguous, the ideles are exactly the units of the adeles. – D_S May 19 at 22:51