Of all the automorphisms of a (finite-dimensional, semisimple) Lie algebra which induce a particular automorphism of its Dynkin diagram, is there a particular one which is "nicer" than the others?

The question came about while solving the following exercise in Fulton and Harris (22.24(b), on page 362 in my copy) asks

Find the invariant subalgebra for the involutions of $(A_n)$ and $(D_n)$, and for an automorphism of order three of $(D_4)$.

Here "invariant subalgebra" means invariant under an automorphism of the corresponding Lie algebra $\mathfrak{g}$ which induces a particular automorphism of the Dynkin diagram. My question is how to choose the automorphism of $\mathfrak{g}$, because the answer can be different depending on which you choose. For example, in the case of the Lie algebra $\mathfrak{g} = \mathfrak{sl_{n+1}}$ with Dynkin diagram $A_n$ (a path of length $n$), the following automorphisms of $\mathfrak{g}$ both reverse the order of the nodes.

$$\alpha: E_{i,j} \mapsto (-1)^{j-i+1} E_{n+2-j, n+2-i}$$ $$\beta: E_{i,j} \mapsto -E_{n+2-j, n+2-i}$$ $E_{i,j}$ is the unit matrix with 1 in position $(i,j$). They are both reflection across the antidiagonal, but with different sign change patterns.

In both cases, in the invariant subalgebra, the entries off the antidiagonal are paired up by reflection across the antidiagonal followed by a possible sign change. However, assuming that $n$ is odd, for $\alpha$, the invariant subalgebra has arbitrary entries on the antidiagonal, while for $\beta$, the invariant subalgebra has zero entries on the antidiagonal.

Since the invariant subalgebras of $\alpha$ and $\beta$ do not have the same dimension, it follows that $\alpha$ and $\beta$ are not conjugate in $Aut(\mathfrak{g})$.

The solution for this exercise in Fulton and Harris uses $\alpha$ to calculate the answer. So is there some reason $\alpha$ is a "better" choice of automorphism than $\beta$ ? The only reason I can think of is that the invariant algebra for $\alpha$ is bigger.

Both $\alpha$ and $\beta$ induce the involution on the Dynkin diagram (and fix the Cartan subalgebra consisting of the diagonal matrices). They differ by an inner automorphism ($\alpha \beta^{-1}$ is inner), but their invariant subalgebras can still be very different.


1 Answer 1


It is true that both automorphisms $\alpha$ and $\beta$ stabilise the Cartan subalgebra $\mathfrak{h}$ consisting of the diagonal matrices, and both stabilise the root basis

$B:= (\alpha_i : diag(c_1, ..., c_n) \mapsto c_i-c_{i+1})_{i=1,..., n}$.

But only $\alpha$ stabilises the whole épinglage $(\mathfrak{g}, \mathfrak{h}, B, (E_{i+1,i})_{\alpha_i})$ which additionally contains fixed basis vectors of the root spaces of the $\alpha_i$, here $E_{i+1, i}$. An automorphism of $\mathfrak{g}$ which stabilises all these data is then uniquely determined by its induced action on the Dynkin diagram: that is the content of Corollaire 1 to the Bourbaki proposition whose proof (but not assertion) I doubt in MSE/764696. Further, if you fix any other épinglage and take the Lie algebra automorphism $\alpha'$, corresponding to the same Dynkin diagram automorphism, with respect to that épinglage, then $\alpha = \gamma \circ \alpha' \circ \gamma^{-1}$ for a $\gamma \in Aut_0(\mathfrak{g})$, so in particular $\alpha$ and $\alpha'$ have the same invariant subalgebra. I do not know if this is what Fulton and Harris had in mind, but I think it gives a reasonable perspective.

I do not know if among the automorphisms that induce a certain automorphism of the Dynkin diagram, the ones that stabilise an épinglage have the largest invariant algebras. It sounds reasonable.


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