Let $G$ be a connected reductive group defined over a perfect field $F$. The *split component* $A$ of $G$ is the unique maximal $F$-split subtorus of the radical of $G$. For an algebraic group $H$ over $F$, let $X(H)$ denote the abelian group of rational characters $H \rightarrow \mathbf{G}\_m$, and let $X(H)\_F$ be the subgroup of $X(H)$ of characters which are defined over $F$. The book I'm reading ("Eisenstein Series and Automorphic L-functions" by Freydoon Shahidi, pg. 9) mentions that 


$$\textrm{Hom}\_{\mathbb{Z}}(X(G)\_F,\mathbb{R}) \cong \textrm{Hom}\_{\mathbb{Z}}(X(A),\mathbb{R})$$


as real vector spaces. I'm having trouble seeing this. I have worked a couple of examples to try and see what is going on:


**Example 1: $G = \textbf{GL}\_3$**


In this case, $X(G) = X(G)\_F \cong \mathbb{Z}$, since the rational characters of $G$ consist of powers of the determinant function, which is defined over $F$. Here we have $A = Z(G)$, the center of $G$, which is a one dimensional torus. Hence $X(A) \cong \mathbb{Z}$. $\blacksquare$


**Example 2: $F = \mathbb{R}$, $G = U(2,1)$**


As an algebraic group over $\mathbb{C}$, $G$ is equal to $\textrm{GL}\_3$, but the $\mathbb{R}$-structure on $G$ is different: if $x \in G$, and $\overline{x}$ denotes the entrywise application of complex conjugation to $x$, then 


$$G(F) = \{ x \in G : w ^t \overline{x}^{-1}w = x \}$$


where $^t \overline{x}^{-1}$ denotes the conjugate inverse transpose of $x$, and $w = \begin{pmatrix} & & 1 \\ & -1 & \\ 1 & & \end{pmatrix}$.


Here a maximal $F$-split subtorus of $G$ is $$A\_0 = \{ \begin{pmatrix} t & & \\ & 1 & \\ & & t \end{pmatrix} : t \in \mathbb{C}^{\ast} \}$$ and so the split component of $G$ is trivial. Hence $X(A) = 0$. 


On the other hand, the nontrivial element $\sigma$ of the Galois group $\Gamma = \textrm{Gal}(\mathbb{C}/\mathbb{R})$ acts on $G$ as $\sigma(x) = w ^t \overline{x}^{-1} w$, and a morphism of $F$-varieties $f: G \rightarrow Y$ is defined over $F$ if and only if $f$ preserves the action of $\sigma$. Since $\sigma$ acts on $\textbf{G}\_m$ as $t \mapsto \overline{t}$, we see that the determinant function $D$ is not defined over $F$: 


$$D(\sigma(x)) = D(w ^t \overline{x}^{-1}w) = \overline{D(x))}^{-1}$$ $$\sigma(D(x)) = \overline{D(x)}$$
hence also $X(G)\_F = 0$. $\blacksquare$


Any insight as to why we always have $\textrm{Hom}\_{\mathbb{Z}}(X(G)\_F,\mathbb{R}) \cong \textrm{Hom}\_{\mathbb{Z}}(X(A),\mathbb{R})$ would be appreciated.



Rational Characters of a reductive group have the same rank as split component
This is much easier than it looks. The point is that any reductive group $G$ is isogenous to the product of its radical, which is its centre $Z(G)$, and its commutator subgroup, which is a semisimple group. Since Hom into $\mathbb{R}$ will kill all torsion, it is sufficient to prove the statement when $G$ is either semisimple, or a torus.


The semisimple case is trivial because both spaces are obviously 0. The torus case is slightly more delicate, but it amounts to the statement that the natural map from the maximal $\operatorname{Gal}(\overline{F} / F)$-invariant subspace of $X(T)$ to the maximal invariant quotient is an isomorphism (which is obvious, since the action of $\operatorname{Gal}(\overline{F} / F)$ factors through a finite group, and real representations of finite groups are completely reducible).


(PS In your second example something is a little fishy since the torus $A\_0$ is neither $F$-split nor contained in the radical of $G$. But it is true that the maximal $F$-split torus in $Z(G)$ is trivial.)


