This is, in a way, a follow up question to [Unipotent orbits and intersection with Levi and pseudo-Levi subgroups](https://mathoverflow.net/questions/415890/unipotent-orbits-and-intersection-with-levi-and-pseudo-levi-subgroups#415890).


I was reading "[A generalisation of the Bala–Carter theorem for nilpotent orbits](https://doi.org/10.1155/S107379289800035X)" 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?



The identity connected component of centralizers of unipotent orbits
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.


