Suppose $G$ is a connected reductive algebraic group over an arbitrary field $K$; let $Z$ be the center of $G$. The inner automorphisms of $G$ are given by $\operatorname{Inn}(G) = G / Z = G^{\operatorname{ad}}$. Set $\operatorname{Out}(G)$ to be the quotient $\operatorname{Aut}(G) / G^{\operatorname{ad}}.$ The forms of $G$ are parameterized by $\operatorname{H}^1(K, \operatorname{Aut}(G))$, and the inner forms are those in the image of
$$\operatorname{H}^1(K,G^{\operatorname{ad}}) \rightarrow \operatorname{H}^1(K,\operatorname{Aut}(G))$$ So we can recast the classification of inner forms of $G$ to:

What conditions can we put on $G$ to guarantee that the map $$\operatorname{H}^0(K,\operatorname{Aut}(G)) \rightarrow \operatorname{H}^0(K,\operatorname{Out}(G))$$ is surjective?

I'm primarily interested in the connected reductive case here, but I would be curious about the more general case as well.

On a related note, I've frequently seen the claim that for quasisimple $G$, the group $\operatorname{Out}(G)$ is given by the automorphism group of the Dynkin diagram of $G$. This holds for some reductive $G$ (such as $\operatorname{GL}_n$) and not others (most nontrivial tori will have a nontrivial outer automorphism group and a trivial dynkin diagram).

How can we extend the description of $\operatorname{Out}(G)$ from the case of quasi-simple $G$ to connected reductive $G$?


Split redictive groups are classified by a combinatorial structure called root datum. If your $G$ is split, the outer automorphism group is the constant group scheme of automorphisms of this structure preserving simple roots. If $G$ is semisimple simply connected or adjoint it is indeed just the automorphism group of the Dynkin diagram, but in the other semisimple cases it can be a proper subgroup of the latter; for general reductive groups it is finitely generated but not always finite. If $G$ is not split, you can get a twisted form of this constant scheme. SGA 3 Exp. XXIV may help.

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  • $\begingroup$ By the way, $GL_n$ has more outer automorphisms, so called central: $g\mapsto det(g)^mg$ for any integer $m$. $\endgroup$ – Victor Petrov Feb 21 '11 at 20:03
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    $\begingroup$ These maps aren't automorphisms if $m \ne 0$ and $1+mn \ne -1$: they aren't injective on the centre. In the other cases they are conjugate to $g \mapsto {}^t g^{-1}$. $\endgroup$ – fherzig Feb 21 '11 at 21:13
  • $\begingroup$ It might be helpful to note that any pinning (i.e. choice of a non-trivial element in each simple root subgroup) gives a canonical splitting of the sequence $1 \to Inn(G) \to Aut(G) \to Aut (based root datum) \to 1$. See e.g. Prop. 2.13 in Springer's notes in Corvallis (over alg. closed fields): ams.org/publications/online-books/pspum331-index (first link). This is important in the construction of L-groups (see Borel's notes in Corvallis, volume 2). $\endgroup$ – fherzig Feb 21 '11 at 21:29

I want to address your first question, regarding what conditions you can put on $G$ to guarantee that the map $Aut(G)(K) \rightarrow Out(G)(K)$ is surjective. I am only going to talk about the case where is $G$ semisimple. If $G$ is quasi-split, then the map has a section so it is surjective, as you can see from the references given by fherzig or Victor.

But if $G$ is not quasi-split, there can be obstructions coming from both the Tits index and Tits algebras. One example to consider is $G = Spin(6,2)$ (of type $D_4$) over the real numbers. In that case $Out(G)(K)$ is the symmetric group on 3 letters, but the image of your map has order 2, corresponding to the outer automorphism given by a hyperplane reflection.

If $G$ is absolutely simple, you can hope that the Tits algebras provide the only obstruction to the surjectivity of your map. But this seems to be open. For more details on these obstructions and some positive results (that in some cases the Tits algebras are the only obstruction), see section 2 of my recent paper `Outer automorphisms...'.

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