Let $K/k$ be an extension of fields, not necessarily algebraic; let $G$ and $H$ be split, reductive groups over $K$; and let $f : H \to G$ be an embedding of groups.
Do there exist split, reductive groups $G'$ and $H'$ over $k$, an embedding $f' : H' \to G'$ of groups, and isomorphisms $G'_K \cong G$ and $H'_K \cong H$ such that $f'_K$ is identified to $f$?
If it helps, $k$ and $K$ can both be assumed algebraically closed. (I would not be surprised if this assumption is necessary, but I would also not be surprised if just being split is enough.)
(I could ask this question with fixed $k$-groups $H'$ and $G'$ at the beginning, consider a morphism $f : H'_K \to G'_K$, and then ask for $f'$, but in that case the answer is ‘no’; for example, take $H' = \operatorname{GL}_1$ and $G' = \operatorname{SL}_2$, and let $f$ be any embedding of $H'_K$ as a maximal split torus in $G'_K$ that is not defined over $k$. If I did not require that $G$ and $H$ be split over $K$, then the answer would be ‘no’ just because one or both of them might not admit a $k$-form.)
This seems like it's in the spirit of Borel and Tits – Homomorphismes “abstraits” …, but I couldn't find it there or deduce it from the results of that paper.
Best Answer
In positive characteristic the answer to the question is negative. The reason for that is that there is exists a semisimple groups $H'/k$ admitting a family of finite dimensional representations $\rho_t:H'\to GL(n,k)$, $t\in\mathbb A^1$, whose members are pairwise non-isomorphic. This family then defines a representation $\rho:H_K\to GL(n,K)$ with $K=\overline{k(t)}$ which is certainly not defined over $k$.
To construct such a family, I would look for two simple $H'$-modules $U$ and $W$ with $\dim\mathrm{Ext}^1(U,W)\ge2$. (Maybe some expert can help out with an example.) Then let $c_0$ and $c_1$ be two cocycles which stay linearly independent in $\mathrm{Ext}^1(U,W)$ and let $c_t:=(1-t)c_0+tc_1$. Then $c_t$ defines a representation on $V=U\oplus W$ depending on $t$ such that no two are isomorphic.