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Suppose $G$ is a semisimple algebraic group defined over a field $k$. Let $\mathrm{Aut}(G)$ and $\mathrm{Inn}(G)$ denote the groups of automorphisms and inner automorphisms (respectively) of $G$. Given a Borel subgroup $B\subset G$ and a maximal torus $T \subset G$, one has an associated system of simple roots $\Delta(B,T)$, and there is a well-known short exact sequence $$ 1 \longrightarrow \mathrm{Inn}(G) \longrightarrow \mathrm{Aut}(G) \longrightarrow \mathrm{Aut}(\Delta(B,T)) \longrightarrow 1. $$ All of the groups in the sequence are defined over $k$. (This is true even if $B$ and $T$ are not defined over $k$.) So are the maps between them. If we choose a pinning for $(G,B,T)$, then this choice gives rise to a splitting for the sequence, thus realizing $\mathrm{Aut}(G)$ as a semidirect product.

Question: Can we always choose a pinning so that the splitting is defined over $k$?

Discussion: If we could, then we could decompose $\mathrm{Aut}_k(G)$ as a semidirect product. While the standard textbooks all give the short exact sequence above, I have not noticed that any of them have much to say about $\mathrm{Aut}_k(G)$, suggesting that such a decomposition isn't valid.

Can anyone point me to a proof or counterexample?

Partial result: Josh Lansky and I can show that everything works if $G$ is $k$-quasisplit. I believe that I can reduce the general case to the case where $G$ is $k$-anisotropic. If $k$ is $p$-adic, then we know what the $k$-anisotropic groups look like, and everything works.

Motivation: At one point, Josh and I needed to know that we could lift $k$-automorphisms of $\Delta(B,T)$ to $k$-automorphisms of $G$. While this ability would make some aspects of our work more explicit, it is no longer essential. But as Titchmarsh said of the irrationality of $\pi$, ``if we can know, it surely would be intolerable not to know.''

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You say "pinning" but don't assume $T$ is split, so the Question is unclear. The functor $S \mapsto {\rm{Aut}}_{S-{\rm{gp}}}(G_S)$ is represented by a smooth affine $k$-group ${\rm{Aut}}_{G/k}$ with identity component $G^{\rm{ad}}$. The identity component can fail to split off as a semi-direct product since projection from ${\rm{Aut}}_{G/k}$ to the component group need not be surjective on $k$-points when $G$ isn't quasi-split. A counterexample is $G = {\rm{SL}}(D)$ for a central division algebra that isn't 2-torsion in ${\rm{Br}}(k)$. So it fails in anisotropic cases over every local $k$! –  grp Sep 4 '12 at 5:16
    
True about $\mathrm{SL}(D)$. I must have had quaternion algebras on the brain. BTW, the notion of a pinning is not actually a problem, since I don't assume that it is preserved by the action of the Galois group. –  Jeff Adler Sep 4 '12 at 21:21
    
As I wrote above, Lansky and I can prove that everything works when $G$ is $k$-quasisplit. But if this fact is well known, then I'd rather just point to a reference. Is this in the literature somewhere? –  Jeff Adler Sep 5 '12 at 2:04
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It's well-known; see SGA3, XXIV, 3.9-3.10. Here's the concrete proof. Let $(G_0,B_0,T_0)$ be the split form. Use a pinning to define a section to the map from ${\rm{Aut}}_{G_0/k}$ onto its finite constant component group ${\rm{Out}}_{G_0/k}$, valued in the $k$-subgroup of automorphisms preserving $(B_0, T_0)$. Doing twists by hand, the map ${\rm{H}}^1(k,{\rm{Out}}_{G_0/k}) \rightarrow {\rm{H}}^1(k,{\rm{Aut}}_{G_0/k})$ has image consisting of quasi-split forms whose Aut-sequence is a semi-direct product and meeting every class of inner forms. Now use uniqueness of quasi-split inner forms. –  grp Sep 5 '12 at 6:33
    
Thanks for the reference and outline. –  Jeff Adler Sep 5 '12 at 15:16

1 Answer 1

up vote 4 down vote accepted

I think ${\rm Aut}(\Delta(B,T))$ doesn't always split: there are some very serious obstructions for that.

According to a recent result of Skip Garibaldi ad Philippe Gille on algebraic groups with few subgroups, published in PLMS, there exist groups $G$ of type ${\rm D}_4$ whose all proper connected subgroups defined over the base field are abelian (such groups are necessarily anisotrpic). If the extension in question splits then $\rm Aut(G)$ contains an involution which acts on the Dynkin graph by interchanging two nodes and fixing the rest. The fixed point group of such an involution would contain a connected reductive subgroup or semisimple rank $3$ defined over the base field. However, such subgroups cannot exist in the forms of type ${\rm D}_4$ constructed by Garibaldi and Gille.

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The phenomenon as in the middle of your 2nd paragraph happens already in type A (for more elementary reasons), as noted in my other comment. –  grp Sep 4 '12 at 11:54

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