For any $x \in \mathfrak{g}$, is it true that the closure of its $G$-orbit is given by $$\overline{G \cdot x} = G \cdot \left(x_s + \overline{{C_Gx_s}^{\circ} \cdot x_n}\right)?$$
Notation: Let $G$ be a connected reductive algebraic group, defined over an arbitrary algebraically closed field $\mathbb{K}$ with characteristic $p \geq 0$. For any $x \in \mathfrak{g} = \mathrm{Lie}\,G$, let $x = x_s + x_n$ denote its Jordan decomposition. For $g \in G$ and $x \in \mathfrak{g}$, let $g \cdot x = \mathrm{Ad}(g)(x)$ denote the adjoint action. For any algebraic subgroup $H \subseteq G$, and any subset $X \subseteq \mathfrak{g}$, let $H \cdot X = \left\{ h \cdot x \mid h \in H, x \in X\right\}$ denote the $H$-saturation of $X$. Let $C_Gx_s = \left\{ g \in G \mid g \cdot x_s = x_s\right\}$ denote the stabiliser of $x_s$, and let ${C_Gx_s}^{\circ}$ denote the connected component $C_Gx_s$ which contains the identity. For any $X \subseteq \mathfrak{g}$, let $\overline{X}$ denote its closure with respect to the Zariski topology on $\mathfrak{g}$.
This feels like the kind of result that should be known, but I cannot find it anywhere - even with further restrictions on the group $G$ or the characteristic $p \geq 0$.
The following are related (and potentially useful) facts that I already know, for any $x \in \mathfrak{g}$:
- If $x = x_n$ is nilpotent then the result is true.
- If $x = x_s$ is semisimple then the result is true.
- We have the equalities $$x_s + \overline{{C_Gx_s}^{\circ} \cdot x_n} = \overline{x_s + {C_Gx_s}^{\circ} \cdot x_n} = \overline{{C_Gx_s}^{\circ} \cdot x},$$ which means we can rewrite the right hand side.
- This result is true if and only if the right hand side is a closed subset of $\mathfrak{g}$.
