Welcome to MathOverflow

Question

3

1

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.''

flag	asked Sep 4 at 3:18 Jeff Adler 168●1●4

1

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 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 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 at 2:04

1

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 at 6:33

Thanks for the reference and outline. – Jeff Adler Sep 5 at 15:16

Alexander Premet · Accepted Answer · 2012-09-04 12:23:08Z UTC

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.

Rational automorphisms of semisimple algebraic groups

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.