Let $\mathbb{A}_\mathbb{Q}^f$ be the subring of the adeles ring with $x_\infty=0$, is every open compact subgroup of $GL_2(\mathbb{A}_\mathbb{Q}^f)$ included in a conjugacy class of $GL_2(\widehat{\mathbb{Z}})$ ? Thanks in advance

## 1 Answers

$\newcommand{\GL}{\mathrm{GL}}$$\newcommand{\A}{\mathbb{A}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Q}{\mathbb{Q}}$Yes, this is true.

If $K$ is a compact open subgroup of $\GL_2(\mathbb{A}_f)$ then by there exists (either by definition or some small lemma depending on how you define $\GL_2(\A_f)$) a subset of $K$ of the form $\displaystyle \prod_{v\in S}U_v\times \prod_{v\notin S}\GL_2(\Z_v)$ where $U_v$ is open and $S$ is a finite set. Since each $\GL_2(\Q_p)$ is locally profinite we may assume that each $U_v$ is an open subgroup of $\GL_2(\Q_v)$.

Note in particular that if $\pi_v$ denotes the projection map, then $\pi_v(K)$ contain either $U_v$ or $\GL_2(\Z_v)$ and thus itself is an open compact subgroup of $\GL_2(\Q_v)$ (since if you contain an open subgroup you're open). But, recall that the maximal compact open subgroups of $\GL_2(\Q_v)$ are precisely the conjugates of $\GL_2(\Z_v)$. In particular, for $v\notin S$ since $\pi(K)\supseteq \GL_2(\Z_v)$ we see that $\pi(K)=\GL_2(\Z_v)$. For $v\in S$ there exists some $g_v\in\GL_2(\Q_v)$ such that $\pi_v(K)\subseteq g_v \GL_2(\Z_v)g_v^{-1}$.

Set $g:=(g_v)$ where

$$g_v=\begin{cases}g_v & \mbox{if}\quad v\in S\\ 1 & \mbox{if}\quad v\notin S\end{cases}$$

then $g\in\GL_2(\A_f)$ and $gKg^{-1}\subseteq\GL_2(\widehat{\Z})$ since, by design, we have that $\pi_v(gKg^{-1})\subseteq\GL_2(\Z_v)$ for all $v$.

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