Let $G$ be a connected complex reductive group. Given an element $g \in G$, there is a canonical decomposition $g = s u$ such that $s$ is semisimple, $u$ is unipotent and $s$ and $u$ commute. Similarly, for $x \in \frak{g}$ in the Lie algebra, there is a canonical decomposition $x = s + n$, where $s$ is semisimple, $n$ is nilpotent and $[s,n] = 0$.
This is a generalisation of the Jordan normal form for matrices. However, in the case of matrices, the nilpotent component $x = s + n$ in the Jordan normal form has a special form, namely $n$ is block upper triangular with $1$'s on the upper diagonal.
My question is to what extent this more refined normal form generalises to reductive groups?
For example, let $n \in \frak{g}$ be a nilpotent element. Is it possible to find a maximal torus $T \subset G$, and a set of simple roots $\alpha_{1}, ..., \alpha_{k}$, such that
$n = \sum_{i} \epsilon_{i} \alpha_{i}$,
where each $\epsilon_{i} = 0$ or $1$? Maybe more precisely, whats the minimal number of simple roots required in such a decomposition (where we allow the choice of maximal torus and simple roots to vary as needed).
In fact, I this is enough to imply the general case. This can be seen as follows. Let $x = s + n \in \frak{g}$ be the canonical decomposition. The centralizer $H = C_{G}(s)$ is a connected and reductive subgroup of $G$. It also has maximal rank, because it contains any maximal torus containing $s$ (i.e. its Lie algebra contains $s$). But now $n \in \frak{h}$, and the roots of $H$ are a subset of the roots of $G$. So now we can apply any result about decomposing nilpotent elements inside of $H$, and this will be compatible with $s$.
Note: I also asked this question on math stack exchange.
