# Finite volume of $G(k) \backslash G(\mathbb A)$ implies that a real-valued character trivial on $G(k)$ is unramified

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.

• $k$-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. – reuns May 20 at 4:00
• How does assuming density simplify things? – D_S May 20 at 4:13

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