Are the two nilpotent elements in an $\mathfrak{sl}_2$-triple conjugate

Question

As the title, I am wondering that suppose we have an $\mathfrak{sl}_2$-triple $(e,h,f)$, say in a reductive Lie algebra over $\mathbb{C}$. Is it true that $e$ and $f$ are always conjugate? A proof that does not require a case-by-case study will be great. Thanks!

Answer 1

The nilpotent elements $e=\begin{pmatrix} 0 & 0 \cr 1 & 0\end{pmatrix}$ and $f=\begin{pmatrix} 0 & 1 \cr 0 & 0\end{pmatrix}$ are conjugate (as matrices) in $\mathfrak{sl}_2(K)$, namely we have $ses^{-1}=f$ with $s=\begin{pmatrix} 0 & 1 \cr 1 & 0\end{pmatrix}\in \mathfrak{sl}_2(K)$. The elements are also conjugated over the complex numbers in the sense that $Ad_g(e)=geg^{-1}=f$, with $g=\begin{pmatrix} 0 & i \cr i& 0\end{pmatrix}\in SL_2(\mathbb{C})$, i.e., with $\det(g)=1$.

If we have an $\mathfrak{sl}_2(K)$-triple in a Lie algebra $L$, then $\mathfrak{sl}_2(K)$ is a subalgebra of $L$, hence $e$ and $f$ are conjugate (as matrices). Of course, there may not exist any $\mathfrak{sl}_2(K)$-triple in $L$. But the Jacobson-Morozov theorem says that for any nilpotent element $e$ of a semisimple Lie algebra $L$ of characteristic zero there exist $f,h\in L$ such that $e,f,h$ is a $\mathfrak{sl}_2(K)$-triple.