# Embeddings of reductive groups over algebraically closed fields

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.

• The following might work. Choose a finitely generated $k$-subalgebra $R\subset K$, choose split reductive group schemes $\mathcal{G}$ and $\mathcal{H}$ over $S:=\mathrm{Spec}$ $R$, and choose a closed immersion $\mathcal{H}\to \mathcal{G}$ which agrees with $f:H\to G$ after basechange along $\mathrm{Spec} K\to S$. Since $k$ is algebraically closed, there is a point $s$ in $S(k)$. Put $H' = \mathcal{H}_s$ and $G':= \mathcal{G}_s$. My guess is that if $s$ is chosen general enough, then $H'_K = H$ and $G'_K = G$. Oct 3, 2022 at 20:06
• This exists by "spreading out" of closed immersions. First choose $\mathcal{H}$ and $\mathcal{G}$. Now use that there is a closed immersion $\mathcal{H}_{L}\to \mathcal{G}_L$ for some finitely generated extension $L$ of $K(S)$ (contained in $K$). Oct 5, 2022 at 13:33
• I denote the function field of an integral scheme $S$ by $K(S)$. The scheme $S$ is the affine scheme defined by the ring $R$. The ring $R$ is obtained by adjoining to $k$ all the "data" necessary to write down $\mathcal{G}$, $\mathcal{H}$, and so on. Think of it this way: Writing down ${G}$ over $K$ requires only finitely many data (polynomials with finitely many coefficients), so once you choose this data to define $G$ you get an algebraic group over some subring $R\subset K$ with $R$ finitely generated over $k$ (i.e. $R =k[a_1,\ldots,a_n]$ for some well-chosen $a_i$ in $K$). Oct 6, 2022 at 8:04
• I would recommend Bjorn Poonen's book on rational points for "spreading out". The idea is very simple. Consider, for example, the polynomial $f(x,y) = y^2 - x^3+10$. This defines an affine scheme $X$ over $\mathbb{Z}$. This scheme is smooth over $\mathbb{Q}$, i.e., $X_{\mathbb{Q}} = \mathrm{Spec} \mathbb{Q}[x,y]/(y^2-x^3+10)$. In particular, by spreading out of smoothness, it is smooth over a dense open of $\mathrm{Spec} \mathbb{Z}$. (What this dense open is precisely can be computed in this case. But in general, spreading out only gives you "some dense open".) Oct 6, 2022 at 8:07
• Another example of spreading out: Consider the morphism $x\mapsto 2x$ from the affine line to itself. This is an honest isomorphism of schemes over $\mathbb{Q}$. Thus, by spreading out of isomorphisms, it spreads out to an isomorphism over some dense open of $\mathrm{Spec}$ $\mathbb{Z}$. (In this case, the dense open is $\mathrm{Spec}$ $\mathbb{Z}[1/2]$.) Oct 6, 2022 at 8:09

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.
• There are many examples, but here is one: take $G$ simple of type $D_n$ with $n \geq 4$ even, over an algebraically closed field of characteristic two. In this case the adjoint module for $G$ has a unique non-trivial composition factor $U$, and $\dim Ext_G^1(U,k) = 2$. By the way, for $G = SL_2$ we have $\dim Ext_G^1(U,W) \leq 1$ for all simple $G$-modules $U$ and $W$. Oct 15, 2022 at 13:11
• My interest was primarily (har har) in positive characteristic. (This sort of counterexample definitionally doesn't work, and it seems reasonable that the result itself has a chance of being true, for a linearly reductive group, so for all reductive groups in characteristic $0$.) Thanks! Oct 15, 2022 at 14:58
• Good point. I think one can argue as follows: It follows from Chevalley's isogeny theorem that finite normal subgroup schemes form a discrete set. Therefore, all representations $\rho_t$ have the same kernel $E$ with possibly finitely many exceptions. Now replace $H$ by $H/E$ und you have embeddings. Oct 16, 2022 at 8:22
• $\DeclareMathOperator\SL{SL}$I believe it is straightforward that, if one can realise every $\SL_2$-embedding over $k$, then one can realise every (connected) reductive-group embedding over $k$. @MikkoKorhonen's comment shows that the failure for $\SL_2$ can't be for the same reason as in this answer … but doesn't it also outline a simpler counterexample? Since there is a representation $V$ of $\SL_{2, k}$ such that $\operatorname{Ext}^1(V_K, V_K) = K$, can't we choose an extension of $V_K$ by $V_K$ coming from an element of $K \setminus k$? Nov 23, 2022 at 20:13