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.

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    $\begingroup$ See also mathoverflow.net/questions/123438/…. $\endgroup$ Nov 19, 2013 at 20:48
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    $\begingroup$ How can the outer automorphism of $SL_n$ be realized as a conjugation inside of $GL_n$? $\endgroup$
    – ya-tayr
    Nov 19, 2013 at 22:02
  • $\begingroup$ @ya-tayr My bad. Indeed, it can not, otherwise the matrix transpose would be inner. $\endgroup$ Nov 19, 2013 at 22:46
  • $\begingroup$ @Andrei: Your question needs more detail, I think. The Berman-Moody construction aims mainly to capture the diagonal automorphisms of the Chevalley Lie algebra using an "extended" group, but I don't see how the diagram automorphisms would come in. (I wrote longer comments but realized they weren't enough.) Anyway, for type $E_6$ you have the "natural" faithful 27-dimensional representation. But again the extended Berman-Moody group only gives the diagonal automorphisms. $\endgroup$ Nov 20, 2013 at 11:30
  • $\begingroup$ Clickable paper link: Berman and Moody - Extensions of Chevalley groups (MSN). $\endgroup$
    – LSpice
    May 29, 2019 at 20:11

2 Answers 2


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.)
  • $\begingroup$ This is a much better way to interpret the underlying question. The notion of "extended group" worked out by Berman-Moody is not helpful here. (Maybe the special case $D_4$ should also be mentioned?) $\endgroup$ Nov 21, 2013 at 18:28
  • $\begingroup$ @JimHumphreys: I added D4 to the SO_2n example. $\endgroup$
    – Skip
    Nov 21, 2013 at 18:58

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


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