Does every nilpotent orbit have an element supported on the simple root spaces?

Question

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.

Answer 1

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.