regular element of a Lie algebra

An element $X\in 𝔤$ of a Lie algebra is called regular if the dimension of its centralizer ${\zeta }_{𝔤}(X)=\{Y\in 𝔤\mid [X,Y]=0\}$ is minimal among all centralizers of elements in $𝔤$.

Regular elements clearly exist and moreover they are Zariski dense in $𝔤$. The function $X\mapsto dim{\zeta }_{𝔤}(X)$ is an upper semi-continuous function $𝔤\to {\mathbb{Z}}_{\ge 0}$. Indeed, it is a constant minus $\mathrm{rank}(a{d}_X)$ and $X\mapsto \mathrm{rank}(a{d}_X)$ is lower semi-continuous. Thus the set of elements whose centralizer dimension is (greater than or) equal to that of any given regular element is Zariski open and non-empty.

If $𝔤$ is reductive then the minimal centralizer dimension is equal to the rank of $𝔤$.

More generally if $V$ is a representation for a Lie algebra $𝔤$,an element $v\in V$ is called regular if the dimension of its stabilizer is minimal among all stabilizers of elements in $V$.

Examples

1. In $𝔰{𝔩}_nℂ$ a diagonal matrix $X=\mathrm{diag}(s_1,\mathrm{\dots },s_n)$ is regular iff $(s_i-s_j)\ne 0$ for all pairs . Any conjugate of such a matrix is also obviously regular.

2. In $𝔰{𝔩}_nℂ$ the nilpotent matrix

   is regular. Moreover, it’s adjoint orbit contains the set of all regular nilpotent elements. The centralizer of this matrix is the full subalgebra oftrace zero, diagonal matricies.