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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.

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    $\begingroup$ 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 $G'_K = G$. $\endgroup$
    – Ariyan Javanpeykar
    Oct 3, 2022 at 20:06
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    $\begingroup$ 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 $K$). $\endgroup$
    – Ariyan Javanpeykar
    Oct 5, 2022 at 13:33
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    $\begingroup$ 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 $K$). $\endgroup$
    – Ariyan Javanpeykar
    Oct 6, 2022 at 8:04
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    $\begingroup$ 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".) $\endgroup$
    – Ariyan Javanpeykar
    Oct 6, 2022 at 8:07
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    $\begingroup$ 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} $ $\mathbb{Z}[1/2]$.) $\endgroup$
    – Ariyan Javanpeykar
    Oct 6, 2022 at 8:09

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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.

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    $\begingroup$ 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 $W$. $\endgroup$
    – Mikko Korhonen
    Oct 15, 2022 at 13:11
  • $\begingroup$ 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! $\endgroup$
    – LSpice
    Oct 15, 2022 at 14:58
  • $\begingroup$ Sorry, it's just occurred to me--are these obviously faithful representations? For an adjoint group I know they can't have a smooth kernel, but could there be an infinitesimal kernel? If so, I guess we just use the isogeny theorem to replace by a suitable quotient. $\endgroup$
    – LSpice
    Oct 16, 2022 at 2:45
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    $\begingroup$ 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. $\endgroup$
    – Friedrich Knop
    Oct 16, 2022 at 8:22
  • $\begingroup$ $\DeclareMathOperator\SL{SL}$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 $K \setminus k$? $\endgroup$
    – LSpice
    Nov 23, 2022 at 20:13

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