TITLE: Rational automorphisms of semisimple algebraic groups
QUESTION [5 upvotes]: 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.''

REPLY [5 votes]: 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.