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Let $G$ be a connected, reductive group over a number field $k$, and $X(G)_k$ the group of $k$-rational characters of $G$. We define $G(\mathbb A)^1$ to be the subgroup of $g \in G(\mathbb A)$ such that $||\eta(g)|| = 1$ for all $\eta \in X(G)_k$, where $||x||$ is the idele norm.

A character $\chi: G(\mathbb A) \rightarrow \mathbb C^{\ast}$ is called unramified if it is trivial on $G(\mathbb A)^1$.

Suppose $\chi$ takes values in $(0,\infty)$ and is trivial on $G(k)$. Is $\chi$ necessarily unramified?

This is obviously true in the case where $G(k) \backslash G(\mathbb A)^1$ is compact, since there are no nontrivial compact subgroups of $(0,\infty)$. In general, $G(k) \backslash G(\mathbb A)^1$ is not compact, but it does have finite volume.

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  • $\begingroup$ $k$-rational character ? And why don't you just assume that $G(k)$ is dense in $G(k_v)$ for each $v$. If it is not the case then the relation between $G(k)$ and $G(\Bbb{A})=\prod_v' G(k_v)$ will be much more complicated. $\endgroup$ – reuns May 20 at 4:00
  • $\begingroup$ How does assuming density simplify things? $\endgroup$ – D_S May 20 at 4:13
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This is true. It follows from the more general fact that if $G$ is a unimodular locally compact Hausdorff topological group, and $H$ is a discrete subgroup of $G$ with $\operatorname{Vol}(H \backslash G) < \infty$, then every character $\chi: G \rightarrow (0,\infty)$ which is trivial on $H$ is trivial on $G$.

To prove this, suppose that $\chi$ is trivial on $H$ but not on $G$. Then there is a $g_0 \in G$ with $\rho = \chi(g_0) > 1$. Let $U$ be an open interval containing $\rho$ which is bounded away from $0$. There exists a sequence of positive integers $m_1 < m_2 < m_3 < \cdots$ such that the product sets $U\rho^{m_i} = U \chi(g_0^{m_i})$ are disjoint.

Each preimage

$$V_k = \chi^{-1}(U)g_0^{m_k}= \{ g \in G : \chi(g) \in U \rho^{m_k}\}$$

is open, and nonempty since it contains $g_0^{m_k+1}$. These preimages are mutually disjoint, and so are their images $\overline{V_k}$ in $H \backslash G$. In $H \backslash G$, the disjoint open sets $\overline{V_k} : k \in \mathbb N$ all have positive identical measure, contradicting the assumption that $H \backslash G$ has finite measure.

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