Let $G$ be a connected simple complex algebraic group. Then $G$ acts on its lie algebra $\mathfrak{g}$ by the adjoint representation. Write $\mathbb{C}[\mathfrak{g}]$ for the coordinate ring of the affine algebraic variety $\mathfrak{g}$. The algebra of invariants $\mathbb{C}[\mathfrak{g}]^G\subseteq \mathbb{C}[\mathfrak{g}]$ induces a map of spaces $\pi:\mathfrak{g}\to\text{Spec}\ \mathbb{C}[\mathfrak{g}]^G\cong\mathbb{A}^{\text{rk}(G)}$, where I am using that the Duflo isomorphism says that $\mathbb{C}[\mathfrak{g}]^G\cong \mathcal{Z}(\mathfrak{g})$, where $\mathcal{Z}(\mathfrak{g})$ is the center of the universal enveloping algebra of $\mathfrak{g}$ (note- I'm not sure if this last comment is actually relevant to answering the question).
I have read this representation is $\it{visible}$, meaning that the fibers of $\pi$ have finitely many $G$-orbits (equivalently $\pi^{-1}(\pi(0))$ has finitely many $G$-orbits). I can't see the reason, does anyone have an explanation for this?
Thanks!
The zero-fibre is the nilpotent cone, so this follows from the fact that there are only finitely many conjugacy classes of nilpotent elements.
More generally, for $X\in\mathfrak{g}$ let $X=X_s+X_n$ be its Jordan decomposition. Then, $\pi(X)=\pi(X_s)$ and $\pi(X)=\pi(Y)$ if and only if $X_s$ and $Y_s$ are conjugate.
For more on this map see: