General Case (gpt-5.5-pro-web)

Statement

Let $G$ be a (not necessarily reductive) smooth connected affine group over a (not necessarily perfect) field $k$. Let $K$ be an algebraic field extension of $k$. Let $$ \mathbb{X}_ \ast(G) := \operatorname{Hom}_ {k\text{-groups}}(\mathbb{G}_ m, G) $$ be the $k$-homomorphisms. Let $G_ K$ be the base change of $G$ to $K$. Prove (or disprove with a counterexample) that $$ \mathbb{X}_ \ast(G)/G(k) \to \mathbb{X}_ \ast(G_ K)/G(K) $$ is injective. (You can take for granted that this statement is true if $k$ is a perfect field.)

Proof

The statement is true, even slightly more generally: the same proof works for any field extension $K/k$, not just algebraic extensions.

Let $$ \lambda,\mu\in \mathbb X_ \ast(G)=\operatorname{Hom}_ {k\text{-groups}}(\mathbb G_ m,G) $$ and suppose that their base changes to $K$ are $G(K)$-conjugate. We prove that $\lambda$ and $\mu$ are already $G(k)$-conjugate.

For a cocharacter $\lambda:\mathbb G_ m\to G$, define the dynamic subgroup $$ P_ G(\lambda)=\{ g\in G:\lim_ {t\to 0}\lambda(t)g\lambda(t)^{-1}\text{ exists} \}. $$

Also define $$ C_ G(\lambda)=Z_ G(\lambda(\mathbb G_ m)). $$

For smooth connected affine $G$, the dynamic theory of Borel–Tits says that $P_ G(\lambda)$ is a smooth connected pseudo-parabolic $k$-subgroup, and that the limit map $$ \rho:P_ G(\lambda)\to C_ G(\lambda), \qquad \rho(g)=\lim_ {t\to 0}\lambda(t)g\lambda(t)^{-1}, $$ is a $k$-homomorphic retraction. In particular, $$ P_ G(\lambda)=C_ G(\lambda)\ltimes U_ G(\lambda) $$ for the corresponding contracted unipotent subgroup $U_ G(\lambda)$.

We also use the Borel–Tits rationality theorem in the following form: if $P\subset G$ is a pseudo-parabolic $k$-subgroup of a smooth connected affine $k$-group, then $$ G(k)\longrightarrow (G/P)(k) $$ is surjective.

Now we prove the key cohomological lemma.

Lemma. For every $k$-cocharacter $\lambda$, the natural map of pointed sets $$ H^1(k,C_ G(\lambda))\longrightarrow H^1(k,G) $$ has trivial kernel.

Proof. Put $$ C:=C_ G(\lambda),\qquad P:=P_ G(\lambda). $$

We have an inclusion $i:C\hookrightarrow P$ and a $k$-homomorphic retraction $$ \rho:P\to C $$ with $\rho\circ i=\operatorname{id}_ C$.

For the homogeneous space $G/P$, the standard exact sequence of nonabelian cohomology gives $$ G(k)\longrightarrow (G/P)(k)\longrightarrow H^1(k,P)\longrightarrow H^1(k,G). $$

Since $G(k)\to (G/P)(k)$ is surjective, the connecting map $(G/P)(k)\to H^1(k,P)$ has trivial image. Therefore $$ \ker\bigl(H^1(k,P)\to H^1(k,G)\bigr)=\{1\}. $$

Now let $\alpha\in H^1(k,C)$ map to the trivial class in $H^1(k,G)$. Its image $$ i_ \ast(\alpha)\in H^1(k,P) $$ also maps to the trivial class in $H^1(k,G)$, so $i_ \ast(\alpha)=1$. Applying $\rho_ \ast$ gives $$ \alpha=(\rho\circ i)_ \ast(\alpha)=\rho_ \ast(i_ \ast(\alpha))=1. $$

Thus $H^1(k,C)\to H^1(k,G)$ has trivial kernel. $\square$

Now return to $\lambda,\mu$. Assume that $\lambda_ K$ and $\mu_ K$ are $G(K)$-conjugate. Define the transporter $$ X:=\operatorname{Transp}_ G(\lambda,\mu)=\{ g\in G:\operatorname{Int}(g)\circ\lambda=\mu \}. $$

This is a closed $k$-subscheme of $G$. Since $\lambda_ K$ and $\mu_ K$ are $G(K)$-conjugate, we have $$ X(K)\neq\varnothing. $$

Let $$ C:=C_ G(\lambda). $$

The group $C$ acts on $X$ on the right by multiplication. This action is simply transitive after any field extension over which $X$ has a point: if $x,y\in X$, then $$ x^{-1}y\in C_ G(\lambda). $$

Thus $X$ is a right $C$-torsor over $k$. Let $$ [X]\in H^1(k,C) $$ be its cohomology class.

Now induce this torsor from $C$ to $G$. The associated $G$-torsor is $$ X\times^C G. $$

But there is a natural $G$-equivariant isomorphism $$ X\times^C G \xrightarrow{\sim} G, \qquad [x,a]\longmapsto xa. $$

Therefore the image of $[X]$ in $H^1(k,G)$ is trivial. By the lemma, $$ [X]=1\in H^1(k,C). $$

Hence the $C$-torsor $X$ is trivial, so $$ X(k)\neq\varnothing. $$

Choose $h\in X(k)$. By definition of $X$, $$ \operatorname{Int}(h)\circ\lambda=\mu. $$

Thus $\lambda$ and $\mu$ are $G(k)$-conjugate.

Therefore the map $$ \mathbb X_ \ast(G)/G(k)\longrightarrow \mathbb X_ \ast(G_ K)/G(K) $$ is injective.