Split Group Case (gpt-5.5-pro-web)

Statement

Let $G$ be a split connected reductive group over a field $k$. Let $K$ be an algebraic field extension of $k$. Let $$ \mathbb{X}_ \ast(G) := \mathrm{Hom}_ {\mathrm{group}}(\mathbb{G}_ m, G). $$ Let $G_ K$ be the base change of $G$ to $K$. Prove that $$ \mathbb{X}_ \ast(G)/G(k) \longrightarrow \mathbb{X}_ \ast(G_ K)/G(K) $$ is injective.

Proof

In fact, for a split connected reductive group the displayed map is bijective, hence injective.

Let $T \subset G$ be a split maximal torus, and let $$ W := N_ G(T)/T $$ be the Weyl group. Since $G$ is split, $W$ is the same Weyl group after any field extension, and every $w \in W$ has a representative in $N_ G(T)(k)$.

We use the following standard fact about split reductive groups:

If $H$ is split connected reductive over a field $F$, then every $F$-split torus in $H$ is $H(F)$-conjugate into a fixed split maximal torus. Moreover, any two split maximal tori of $H$ are $H(F)$-conjugate.

Now fix any field extension $F/k$. I claim that the natural map $$ X_ \ast(T)/W \longrightarrow X_ \ast(G_ F)/G(F) $$ is a bijection.

Surjectivity is clear from the standard fact: if $$ \lambda : \mathbb{G}_ {m,F} \to G_ F $$ is a cocharacter, then its image is an $F$-split torus, so it is $G(F)$-conjugate into $T_ F$. Thus every $G(F)$-orbit has a representative in $$ X_ \ast(T_ F) = X_ \ast(T). $$

For injectivity, suppose $\lambda,\mu \in X_ \ast(T)$ and that they are $G(F)$-conjugate. Thus there is some $g \in G(F)$ such that $$ \mu = g\lambda g^{-1}. $$ Let $$ C := C_ {G_ F}\bigl(\mu(\mathbb{G}_ m)\bigr) $$ be the centralizer of the image of $\mu$. Since $\mu$ lands in $T_ F$, the group $C$ is a split connected reductive subgroup of $G_ F$. Indeed, relative to $T_ F$, it is generated by $T_ F$ and the root subgroups $U_ \alpha$ for which $$ \langle \alpha,\mu\rangle = 0. $$

Both $T_ F$ and $gT_ Fg^{-1}$ are split maximal tori of $C$. Since $C$ is split connected reductive, there exists $c \in C(F)$ such that $$ c(gT_ Fg^{-1})c^{-1} = T_ F. $$ Set $n = cg$. Then $n \in N_ G(T)(F)$, and since $c$ centralizes $\mu$, $$ n\lambda n^{-1} = c(g\lambda g^{-1})c^{-1} = c\mu c^{-1} = \mu. $$ Thus $\lambda$ and $\mu$ differ by the action of the Weyl group $W$. Conversely, if $\mu = w\lambda$ for some $w \in W$, then because $G$ is split we may choose a representative $\dot w \in N_ G(T)(k) \subset G(F)$, and $$ \dot w\lambda\dot w^{-1} = \mu. $$ Hence $$ X_ \ast(T)/W \cong X_ \ast(G_ F)/G(F). $$

Apply this with $F = k$ and with $F = K$. Since $T$ is split, $$ X_ \ast(T_ K) = X_ \ast(T), $$ and the Weyl group is unchanged by base change. Therefore both orbit sets identify canonically with the same set: $$ X_ \ast(G)/G(k) \cong X_ \ast(T)/W $$ and $$ X_ \ast(G_ K)/G(K) \cong X_ \ast(T)/W. $$

Under these identifications, the base-change map $$ X_ \ast(G)/G(k) \longrightarrow X_ \ast(G_ K)/G(K) $$ is just the identity map on $X_ \ast(T)/W$. Therefore it is injective, indeed bijective.