The Jacobson-Morozov theorem that any nilpotent element $e$ in the Lie algebra of a simple algebraic group $G$ can be embedded in an $\mathfrak{sl}\_2$-triple, has a restriction (in terms of the Coxeter number) on the characteristic of the underlying field (assumed to be algebraically closed). This restriction is also required for the "uniqueness" of the triple, up to $C\_G(e)$-action. (This result is due to Kostant.) In his 1980 paper, Pommerening had removed the restriction on the characteristic in Jacobson-Morozov's theorem, up to very small exceptions (*i.e.*, characteristic is "bad"). Does the uniqueness as in Kostant's result also hold with this weaker restriction? If it does, then where does Jacobson-Morozov along with uniqueness result of Kostant fail in positive characteristics?



Failure of Jacobson-Morozov in positive characteristics
The uniqueness can break down very badly in positive characteristic. Supose $G=SL\_p$ where $p$ is the characteristic of the base field. Take a regular nilpotent element $e$ in $\mathfrak{g}=\mathfrak{sl}\_p$. Then there is a nilpotent element $f\in\mathfrak{g}$ such that $e$, $f$ and $h=[e,f]$ form an $\mathfrak{sl}\_2$-triple with the property that $h^p=h$. Note that the identity matrix $I$ is in 
$\mathfrak{g}$. It is easy to see that there is $f\_0\in\mathfrak{g}$ such that $[e,f\_0]=I$ (many lecturers find this fact useful when explaining that Lie's theorem can fail in characteristic $p$). 
Let $\lambda$ be a scalar such that $\lambda^p\ne \lambda$.
Since $h$ commutes with $I$ and $ad\ h$ is semisimple, we may assume further that $[h,f\_0]=-2f\_0$.
Then $(e,h+\lambda I, f+\lambda f\_0)$ is another $\mathfrak{sl}\_2$-triple containing $e$. If the spans $\mathfrak{s}\_1$ and $\mathfrak{s}\_2$ of the triples are conjugate under $G$, then restricting the $p$-dimensional vector representaion of 
$\mathfrak{sl}\_p$ to $\mathfrak{s}\_1$ and $\mathfrak{s}\_2$ we would get equivalent representations of $\mathfrak{sl}\_2$.
However, the representation we get from $\mathfrak{s}\_1$ is restricted whereas the one we get from
$\mathfrak{s}\_2$ is not. So the triples are not conjugate under $C\_G(e)$.
One can replicate this example inside any Lie algebra of a reductive group $\widetilde{G}$ whch contains $G$ as a closed subgroup. 


