Rational Characters of a reductive group have the same rank as split component

Question

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.

Answer 1

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.)

Answer 2

The following is a mild and perhaps too-detailed variant on David Loeffler's answer. To adhere to Borel's textbook notation, I will write $X_F(G)$ for what you denote by $X(G)_F$ and will write $X_{\mathbf{Q}}$ to denote $X \otimes_{\mathbf{Z}} \mathbf{Q}$ for any $\mathbf{Z}$-module $X$.

Firstly, the intervention of $\mathbf{R}$ in these matters is a big red herring (even though it is useful for some analytic purposes that arise in your motivating context); we'll see that all of the content can be expressed using $\mathbf{Q}$ instead, and that clarifies matters (such as being more "algebraic"). Also, some good news is that $F$ may be an arbitrary field; it is unnecessary to assume it is perfect (so it can be a local or global function field, very important in number theory, even for proving results about reductive groups over $p$-adic fields such as Ngo's proof of the Fundamental Lemma).

The crux of the matter is twofold:

(1) If $G$ is a connected reductive group over an arbitrary field $F$, with $G' := \mathscr{D}(G)$ its connected semisimple derived group and $Z$ its maximal central $F$-torus, then the natural $F$-homomorphism $$Z \times G' \rightarrow G$$ defined by multiplication is an isogeny. In particular, the quotient $T := G/G'$ is an isogenous quotient of $Z$ (so it is an $F$-torus). (Note also that $A$ is the maximal split $F$-subtorus of $Z$, basically by definition.)

(2) If $\mathscr{T}$ is any $F$-torus at all, with $\mathscr{S}$ its maximal split $F$-subtorus and $\mathscr{T}_0$ its maximal anisotropic $F$-subtorus, then the natural map $\mathscr{S} \times \mathscr{T}_0 \rightarrow \mathscr{T}$ is an isogeny. Indeed, this is equivalent to the induced map between geometric character lattices inducing an isomorphism after rationalization. Well, the action of ${\rm{Gal}}(F_s/F)$ on the finite-dimensional $\mathbf{Q}$-vector space $X(\mathscr{T}_{F_s})_{\mathbf{Q}}$ factors through a finite Galois group and hence is semisimple since $\mathbf{Q}$ is a field of characteristic 0 (no need for $\mathbf{R}$ here): the maximal subspace with trivial Galois action is $X(\mathscr{S}_{F_s})_{\mathbf{Q}}$, and its unique ${\rm{Gal}}(F_s/F)$-stable complement is $X((\mathscr{T}_0)_{F_s})_{\mathbf{Q}}$.

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Coming back to $G$, $G'$ is equal to its own derived group and so has no nontrivial $F$-homomorphism to a smooth commutative affine $F$-group (such as to $\mathbf{G}_m$). Thus, we see via point (1) above that $X_F(G) = X_F(G/G') = X_F(T)$. Since the quotient map $q:Z \rightarrow T$ is an $F$-isogeny of $F$-tori, consideration of ${\rm{Gal}}(F_s/F)$-lattices given by their geometric character groups provides an $F$-homomorphism $f:T \rightarrow Z$ such that $q \circ f = [n]_T$ and $f \circ q = [n]_Z$ for some integer $n > 0$. Thus, the map $X_F(q): X_F(T) \rightarrow X_F(Z)$ between $\mathbf{Z}$-lattices has $X_F(f)$ as an "inverse up to $n$-multiplication". In particular, the natural map of $\mathbf{Q}$-vector spaces $$\theta: X_F(T)_{\mathbf{Q}} \rightarrow X_F(Z)_{\mathbf{Q}}$$ is an isomorphism.

Now let's get rid of the intervention of $\mathbf{R}$ in the initial setup. For any $F$-torus $\mathscr{T}$, ${\rm{Hom}}_{\mathbf{Z}}(X_F(\mathscr{T}), \mathbf{R})$ is the $\mathbf{R}$-linear dual of $$X_F(\mathscr{T}) \otimes_{\mathbf{Z}} \mathbf{R} = (X_F(\mathscr{T})_{\mathbf{Q}}) \otimes_{\mathbf{Q}} \mathbf{R},$$ so by applying this to $T$ and $Z$ in the role of $\mathscr{T}$ we conclude that the original question concerns exactly proving that the natural map $$X_F(T)_{\mathbf{Q}} \rightarrow X_F(A)_{\mathbf{Q}}$$ is an isomorphism. But this is exactly the composition of the isomorphism $\theta$ with the rationalization of the natural map of $\mathbf{Z}$-lattices $X_F(Z) \rightarrow X_F(A)$, so the task is precisely to show that this latter map of lattices induces an isomorphism after rationalization. So $\mathbf{R}$ has vanished from the scene.

Now the role of $G$ has been eliminated, or rather replaced with that of $Z$ which can be any $F$-torus at all: if $A$ is the maximal $F$-split subtorus of an $F$-torus $Z$, then show that $X_F(Z) \rightarrow X_F(A)$ induces an isomorphism after rationalization. This brings us to point (2) above, which gives an isomorphism of rationalizations $$X(Z_{F_s})_{\mathbf{Q}} \rightarrow X(A_{F_s})_{\mathbf{Q}} \oplus X((Z_0)_{F_s})_{\mathbf{Q}}$$ where $Z_0$ is the maximal $F$-anisotropic subtorus of $Z$. Passing to ${\rm{Gal}}(F_s/F)$-invariants on both sides, this becomes the map $X_F(Z)_{\mathbf{Q}} \rightarrow X_F(A)_{\mathbf{Q}} \oplus 0$, so we are done.

QED