This is, in a way, a follow up question to Unipotent orbits and intersection with Levi and pseudo-Levi subgroups.
I was reading "A generalisation of the Bala–Carter theorem for nilpotent orbits" by Eric Sommers and was considering the orbit $D_4(a_1)$ in $E_6^\text{ad}$ and there is an issue that I wasn't able to resolve regarding the semi-simple part of the centralizer of this orbit.
According to Sommers, this orbit intersects the principal orbit of the pseudo-Levi subgroup $A_2^3$ and the pair $(A_2^3,3A_2)$ (of pseudo-Levi subgroup and a nilpotent elment of $\mathfrak{e}_6$ that is distinguished in $\mathfrak{l}=\operatorname{Lie}(A_2^3$) corresponds (in the generalized Bala–Carter correspondence described there) to $(N,C)$, where $C$ is the conjugacy class in the component group $A(N)=S_3$ consisting of elements of order $3$: $C=\{(123), (132)\}$.
Now, if I am not mistaken, this pseudo-Levi $L$ of type $A_2^3$ can be realized in $E_6^\text{ad}$ as $\left( \operatorname{SL}_3(\mathbb{C}) \right)^3 / \mu_3^2$ and its center is $\mu_3$.
Being the principal orbit of $L$, the centralizer of the orbit in $L$ is the center of $L$ (generated by the elements of $C$). The subgroup $L$ has rank $6$ and so its centralizer should have rank $0$.
On the other side, in the table in Carter's book, it says that the semi-simple part of the centralizer of this orbit has type $T_2$ with component group $S_3$.
My question is (assuming my analysis of Sommers' map is correct), how do these two descriptions of the centralizer work together? Where am I missing the rank $2$ torus in the pseudo-Levi description of the orbit?
Best Answer
If you decompose the $E_6$ Lie algebra over the $3A_2$ subgroup $L$ then, in addition to the adjoint representation, there are two irreducible summands: with the right identifications these are isomorphic to $V^{\otimes 3}$ and its dual ($V$ the natural representation for ${\rm SL} _3$). So, to describe the centraliser of $N$ this way you have to also consider the (${\rm ad} \, N$) -fixed points in these two subspaces. By inspecting weights, we see that $V^{\otimes 3}$, decomposed over the (diagonally embedded) principal ${\mathfrak sl}_2$, equals $$V(6)\oplus V(4)^2\oplus V(2)^3\oplus V(0).$$ In particular, the centraliser of $N$ in each subspace is 7-dimensional, and the centraliser for the principal ${\mathfrak{sl}}_2$ is one-dimensional. This gives that the reductive part of the centraliser in the whole Lie algebra is 2-dimensional, so must be the Lie algebra of a torus.