centralizers of nilpotent element in simple Lie algebra and associated Levi subalgebra

Question

Let $\{e,h,f\}$ be a $sl_2$ triples in simple Lie algebra $\mathfrak g$ with usual relations $[h,e]=2e,~ [h,f]=-2f,~[e,f]=h$. Then the centralizer of $e$ is $\mathfrak g_e=\{b:[b,e]=0\}$ and the associated Levi subalgebra is $\mathfrak g_0=\{b:[b,h]=0\}$. Is $\mathfrak g_0$ and $\mathfrak g_e$ are isomorphic subalgebras?.

Answer 1

I think, the answer is negative. Taking $\mathfrak{sl}_3(K)$ and $(e,f,h)$ given by $$ e=\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; f=\begin{pmatrix} 0 & 0 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; h=\begin{pmatrix} 1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 0\end{pmatrix},\; $$ I compute that $\dim \ker {\rm ad}(e)=4$ and $\dim \ker {\rm ad}(h)=2$, hence the two subalgebras have different dimension. We have ${\rm ad}(h)={\rm diag}(2,1,-2,-1,-1,1,0,0)$ and $$ ad(e)=\begin{pmatrix} 0 &0 &0 &0 &0 &0 &-2 & 1 \cr 0 &0 &0 &1 &0 &0 &0 &0 \cr 0 &0 &0 &0 &0 &0 &0 &0 \cr 0 &0 &0 &0 &0 &0 &0 &0 \cr 0 &0 &0 &0 &0 &0 &0 &0 \cr 0 &0 &0 &0 &-1 &0 &0 &0 \cr 0 &0 &1 &0 &0 &0 &0 &0 \cr 0 &0 &0 &0 &0 &0 &0 &0 \end{pmatrix} $$