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

• 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. – Kevin Buzzard May 19 '19 at 22:39
• It just means $\prod\limits_{v \in S} G(k_v) \prod\limits_{v \not\in S} G(\mathcal O_v)$ – D_S May 19 '19 at 22:50

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).