# Image of the Harish-Chandra map is of finite index?

Let $G$ be a connected, reductive group over a global or local field $k$ with absolute value $| \cdot |$. Let $X(G)_k$ be the group of rational characters of $G$ which are defined over $k$, and let $H_G: G(k) \rightarrow \operatorname{Hom}_{\mathbb Z}(X(G)_k,\mathbb R)$ be the Harish-Chandra map, defined by

$$H_G(g)(\chi) = \log |\chi(g)|$$

What can be said about the image of $H_G$? I know that it is a discrete subgroup of $\operatorname{Hom}_{\mathbb Z}(X(G)_k,\mathbb R)$? Does it span all of $\operatorname{Hom}_{\mathbb Z}(X(G)_k,\mathbb R)$?

I know that this is true when $G$ is a split torus. What about in general?

The image always spans $\operatorname{Hom}_{\mathbb Z}(X(G)_k,\mathbb R)$. The proof essentially reduces to that of when $G$ is a split torus. Let $\chi_1, ... , \chi_n$ be a basis for $X(G)_k$. Then $f_1, ... , f_n$ is a basis for $\operatorname{Hom}_{\mathbb Z}(X(G)_k,\mathbb R)$, where $f_i(\chi_j) = \delta_{ij}$.

Fix $1 \leq i \leq n$. We need to show that there exists a $g \in G(F)$ and a real number $m$ such that $m H_G(g) = f_i$.

Let $A_G$ be the split component of $G$. It is a well known result that the restriction homomorphism $X(G)_k \rightarrow X(A_G)$ is injective, and the image is of finite index in $X(A_G)$. So in the beginning, we can choose the basis $\{\chi_i\}$ of $X(G)_k$ as well as a basis $\eta_1, ... , \eta_n$ of $X(A_G)$, so that there exist nonzero integers $d_i$ such that $\chi_i|_{A_G} = d_i \eta_i$.

There exists an $a \in k^{\ast}$ and an $m \in \mathbb R$ such that $|a|^{md_i} = e$. Choose $g \in A_G$ such that $\eta_i(g) = a$, and $\eta_j(g) = 1$ for all $j \neq i$. Then for all $\chi = c_1\chi_1 + \cdots + c_n\chi_n \in X(G)_k$ ($c_i \in \mathbb Z$), we have

$$mH_G(g)(\chi) = m \log |\chi_1(g)^{c_1} \cdots \chi_n(g)^{c_n}| = m \log |\eta_1^{c_1d_1}(g) \cdots \eta_n^{c_nd_n}(g)| = m \log |a^{c_id_i}|$$

$$= c_i \log |a|^{md_i} = c_i = f_i(\chi)$$

which shows that $m H_G(g) = f_i$.