Let $G$ be a connected reductive algebraic group (over $\mathbb{C}$) and $\mathfrak{g}$ its Lie algebra. Let $O \subset \mathfrak{g}$ be an orbit of a nilpotent element. Let $\Pi = \{\alpha\_1, \dots ,\alpha\_r\}$ be a set of simple roots for $G$, and $\mathfrak{g}\_{\alpha\_i}$ be their corresponding root spaces. Then is there an element $x \in O$ such that
$$ x \in \sum\_{i=1}^r \mathfrak{g}\_{\alpha\_i}? $$
For example, if $G = \mathrm{GL}\_n$, $\alpha\_i$ the standard simple roots, then the I believe the answer is yes: take an element $x \in O$ in Jordan block form. I am wondering if this holds for any reductive algebraic group.



Does every nilpotent orbit have an element supported on the simple root spaces?
An orbit is regular if and only if it has a representative in the Lie algebra of the unipotent radical of some Borel subgroup whose projection on every simple root space is non-$0$. Thus, if your suggestion held, then every non-regular orbit would be non-distinguished, but this fails already for $G = \operatorname{Sp}\_6$, where, for example, the orbit corresponding to the partition $6 = 4 + 2$ is distinguished but not regular. This is remarked explicitly after the proof of Proposition 5.2.3 in Collingwood and McGovern - Nilpotent orbits in semisimple Lie algebras.


