Root system automorphisms as inner automorphisms of extended Chevalley group

Question

For each automorphism $\sigma$ of a root system $\Phi$ there is a unique automorphism of the Chevalley group $G(\Phi,R)$ such that $\sigma(x_\alpha(t))=x_{\sigma\alpha}(t')$. While conjugating by certain elements of the torus normalizer realizes the action of the Weyl group of $\Phi$, the automorphisms corresponding to the symmetries of the Dynkin diagram are not inner. However, it is sometimes possible to realize them as inner automorphisms of certain extended groups, e.g. $O_{2n}$ for $SO_{2n}$ (edit: none for $SL$). It can be seen explicitly in matrices for the above cases, but when working with the group of type $\mathsf{E}_6$, it becomes difficult to work with matrices. So I wonder if there is an invariant description?

There is a general construction for extended groups, see [Berman, Moody, Extensions of Chevalley groups. Israel J. Math. 22 (1975), no. 1], but it is not clear, how to describe the desired inner automorphisms in its terms.

Answer 1

Let me rephrase your question. Fix a representation $\rho: G \to GL(V)$ of the Chevalley group $G$. You want a criterion for whether there is an element $x \in GL(V)$ so that $\rho(\sigma(g)) = x\rho(g)x^{-1}$ for all $g \in G$.

In your example of $G = SO_{2n}$ and $V$ the natural representation, you can take $\sigma$ to be an outer automorphism of $SO_{2n}$ and $x$ to be an isometry of determinant $-1$.

In case $V$ is irreducible, here is the desired criterion. Fix a set $\Delta$ of simple roots for the root system $\Phi$ of $G$, and put $\lambda$ for the highest weight of $V$. Multiplying the given automorphism $\sigma$ of $\Phi$ by an element of the Weyl group (as we may), we can assume that $\sigma(\Delta) = \Delta$ (and $\sigma$ sends dominant weights to dominant weights). Now the criterion is: Such an $x$ exists if and only if $\sigma(\lambda) = \lambda$.

To see this, note that such an $x$ exists if and only if the representations $\rho$ and $\rho \sigma$ are equivalent (definition of equivalent representation), which is true if and only if $\sigma(\lambda) = \lambda$ by the classification of irreducibles in terms of highest weights.

You can express this criterion as an exact sequence of groups, see Proposition 2.2 in this paper.

To go back to the examples that have already been mentioned:

• the natural representation of $SO_{2n}$ has highest weight $\omega_1$ in Bourbaki's numbering, and for $n \ge 5$ this is fixed by the nontrivial automorphism of the Dynkin diagram, so the criterion says you can find an $x$, as you know. For $n = 4$, the Dynkin diagram has automorphism group $S_3$, but only one of the three transpositions fixes $\omega_1$, so you can find an $x \in GL_8$ realizing that transposition, but not any of the other 4 nontrivial elements of $S_3$.
• the natural representation of $SL_n$ has highest weight $\omega_1$ (viewing $SL_n$ as having type $A_{n-1}$) and this is not fixed by the nontrivial automorphism of the Dynkin diagram for $n \ge 3$, i.e., you cannot write inverse transpose on $SL_n$ as conjugation by an invertible matrix in $GL_n$ for $n \ge 3$. (But for $n = 2$, $SL_n$ has type $A_1$ and there is no outer automorphism of the Dynkin diagram so of course inverse transpose is an inner automorphism.)
• For a minuscule 27-dimensional representation of $E_6$, the highest weight is moved by the nontrivial automorphism $\sigma$ of the Dynkin diagram, so there is no $x$ in $GL_{27}$ that realizes $\sigma$.
• The adjoint representation has highest weight the highest root, which is fixed by every automorphism of the Dynkin diagram. So for every $\sigma$ there is a corresponding $x$. (Assume that the characteristic is very good for $G$ so that the adjoint representation is irreducible.)

Answer 2

If it helps, you can realize an outer automorphism of $E_6$ inside $E_7$ (or an outer automorphism of $SL_n$ inside $SO_{2n}$).