Finite Order Automorphisms on Complex Simple Lie Algebras

Question

Let $L$ be a finite dimensional complex simple Lie algebra, and let $F(L)$ be the set of all finite order automorphisms on $L$. Suppose that we declare $f,h \in F(L)$ to be equivalent if there exists an $L$-automorphism $\phi$ such that $f = \phi h \phi^{-1}$. Then the resulting equivalence classes of $F(L)$ have been classified by Kac. Now let us impose a stronger equivalence relation, by requiring the above $\phi$ to be inner (thus $L$ is of type $A$, $D$ or $E_6$). How are such equivalence classes of $F(L)$ (with respect to this stronger equivalence relation) classified?

Answer 1

The periodic inner automorphisms are, as you say, classified by Kac diagrams, that is, copies of the affine Dynkin diagram with non-negative weights attached to each vertex. More specifically, given a Kac diagram we construct an inner automorphism of $L$; and every inner automorphism of $L$ is conjugate to one of these automorphisms. (The outer automorphisms are classified via different diagrams, but the question you ask isn't an issue there except in type $D_4$.)

Let's look at the $E_6$ case, since this is easiest to explain what's happening. The affine Dynkin diagram has $S_3$-symmetry since there are three branches of length two emanating from the central vertex. Two Kac diagrams determine conjugate automorphisms in $Aut\; L$ if and only if each diagram can be obtained from the other via a symmetry of the affine Dynkin diagram. But conjugacy in the inner automorphism group holds if and only if each Kac diagram can be obtained from the other via a cyclic permutation of the three branches.

Something similar happens in types $A_n$ and $D_n$ where we also have outer automorphisms. The affine Dynkin diagram in type $A_n$ is a circle (or rather an $(n+1)$-gon), and so the group of symmetries of the diagram is a dihedral group, consisting of rotations and reflections; then two Kac diagrams give us conjugate automorphisms in the inner automorphism group if and only if they can be obtained from each other via rotations.

In type $D_n$ for $n\geq 5$ the affine Dynkin diagram has two forks at either end, and inner conjugacy corresponds - I think - to symmetries of the affine Dynkin diagram which induce even permutations of the four outer vertices. (In other words, we can reflect in the central line of symmetry, or we can flip the forks at both ends, but we can't flip just one fork.) In type $D_4$ the symmetry group of the affine Dynkin diagram is $S_4$, and the outer automorphism group (i.e. $Aut\; L/Int\; L$) is $S_3$, so you can guess which symmetries of the affine Dynkin diagram correspond to inner conjugacy: the permutations $(1\; 2)(3\; 4)$, $(1\; 3)(2\; 4)$ and $(1\; 4)(2\; 3)$.

Finally, the case usually denoted $^{2}D_4$ (i.e. outer automorphisms of $\mathfrak{so}_8$ whose square is inner) is quite complicated. There are no symmetries of the (twisted) affine Dynkin diagram here, but each conjugacy class of these automorphisms in $Aut\; L$ splits into three conjugacy classes for the inner automorphism group. And similarly, each conjugacy class of automorphisms of type $^3D_4$ splits into two conjugacy classes for the inner automorphism group.

Answer 2

A comment and a technical point (too long to post as a comment).

The classification of automorphisms up to inner automorphism, rather than outer, is in fact quite important in the study of real forms of complex groups. For one thing, an outer automorphism of the Lie algebra may fail to exponentiate to the group (if it is not simply connected or adjoint). Consequently the fixed points of two such automorphisms may not be isomorphic, and it is necessary to distinguish these groups. For example in $\mathfrak s\mathfrak o(12,\mathbb C)$ there are two subalgebras isomorphic to $\mathfrak g\mathfrak l(n,\mathbb C)$ which are conjugate by an outer (but not an inner) automorphism. If $G=Spin(12,\mathbb C)/\langle z\rangle$ where $z^2=1$ but $z$ is not fixed by the outer automorphism, then $G$ has two non-isomorphic subgroups locally isomorphic to $GL(6,\mathbb C)$: one is $GL(6,\mathbb C)$, and the other is disconnected.

Another point is that the correct formulation of Vogan duality requires this distinction, in the first place to make some counting come out right. For example if $G(\mathbb C)$ is adjoint of rank $n$, the total number of discrete series representations of all real forms of $G(\mathbb C)$ (with fixed infinitesimal character) is $2^n$. For $PSO(2n,\mathbb C)$ this requires counting $PSO^*(2n)$ twice (these two subroups are related by an outer, but not an inner, automorphism).

Here is a technical correction on the automorphisms of the extended diagram. The center $Z(G_{sc})$ of the simply connected group acts on the extended Dynkin diagram. Automorphisms of $\mathfrak g$, up to inner automorphism, are parametrized by labellings, modulo the action of $Z(G_{sc})$. This only matters in $D_n$: $Z(G_{sc})$ is $\mathbb Z/2\times\mathbb Z/2$ ($n$ even) or $\mathbb Z/4$ ($n$ odd). Consequently all $GL(n,\mathbb C)$ subgroups are conjugate if $n$ is odd, but not if $n$ is even. (For classification up to $Aut(\mathfrak g)$ also throw in automorphisms of the non-extended diagram).

Answer 3

It may be worth adding (belatedly) that the subgroup of affine diagram automorphisms mapping to Inn(L) has a natural description. Namely, the (co)weight lattice acts by translation on the affine root system. Since the affine Weyl group contains the (co)root lattice acting by translations, the quotient group acts on the affine diagram. As is well known, this quotient is isomorphic to the center Z of the simply connected group of L. In this way, the automorphism group of the affine diagram is an extension of Out(L), the automorphism group of the Dynkin diagram, by Z.