Let $G$ be a complex reductive group acting linearly on a complex affine variety $X\subseteq\mathbb{C}^n$. Then, there is a stratification by orbit type of the GIT quotient $$X//G=\operatorname{Spec}{\mathbb{C}[X]^G}.$$ Namely, if $\pi:X\to X//G$ is the quotient map and $H\subseteq G$ a subgroup, then $p\in X//G$ has orbit type $(H)$ if for a point $x\in\pi^{1}(p)$ such that $G\cdot x$ is closed, the stablizer $G_x$ is conjugate to $H$ in $G$. This gives a stratification $$X//G=\bigcup_{(H)}(X//G)_{(H)}.$$ Suppose the strata $(X//G)_{(H)}$ are smooth. Is this a Whitney stratification?

$\begingroup$ Consider the trivial action with $X=X//G=(X//G)_{(G)}$. So $X$ should at least be smooth. $\endgroup$– Friedrich KnopCommented Nov 9, 2017 at 14:07

$\begingroup$ @FriedrichKnop I was assuming the strata are smooth, but forgot to mention it. $\endgroup$– user117030Commented Nov 9, 2017 at 15:15

$\begingroup$ Any variety is a quotient. It is likely that any stratification can be realized as a Luna stratification. Clearly, the corresponding $X$ is then highly singular. So, no, the Luna stratification has no special properties unless you know something about $X$ (like smoothness). $\endgroup$– Friedrich KnopCommented Nov 9, 2017 at 17:17
1 Answer
If $X$ is smooth and $G$ acts properly, then (according to these notes) Theorem 2.7.4 on page 113 in Lie Groups by Duistermaat & Kolk says the orbittype stratification of $X$ is Whitney. Since orbits of proper actions are closed $X//G=X/G$, and the orbittype stratification of $X//G$ is a stratification by submanifolds corresponding to the Whitney stratification of $X$ by submersions (see Theorem 10 in the linked notes).
Let's consider an example where the action is not proper. Let $G=\mathrm{PGL}(2,\mathbb{C})$ act by conjugation on arbitrary $2\times 2$ complex matrices, then $X=\mathbb{C}^4$, and the quotient $X//G=\mathbb{C}^2$ is smooth parametrized by the trace $t$ and determinant $d$. Now $X$ is stratified by orbittype, but the central matrices and the nondiagonalizable matrices are in two different strata that intersect in the GIT quotient (both have repeated eigenvalues). To fix this we throw out the nondiagonalizable matrices (since they do not have closed orbits). Then the corresponding orbittype stratification of $X//G$ is $\mathbb{C}^2V(t^24d)$, corresponding to the diagonalizable matrices that are not central, and $V(t^24d)\cong \mathbb{C}$ corresponding to the central matrices.
More generally, maybe the orbittype stratification is not what you want exactly. Consider a maximal compact $K\subset G$ and a corresponding KempfNess set $N\subset X$. Then $N/K$ is homeomorphic to $X//G$ and $N$ is a real algebraic set, and so stratified by smooth manifolds (remove the singular locus, then the singular locus in the singular locus, etc). The action of $K$ preserves this stratification since it acts by isomorphisms. So now on each smooth stratum $S$ we have a compact Lie group acting on a smooth manifold. Therefore the action is proper and the orbittype stratification is Whitney. It descends to the orbittype stratification of $S/K$ by smooth submanifolds corresponding to a Whitney stratification of $S$ via submersions. Collectively this gives a "refined orbittype" stratification of $N/K$ by smooth manifolds that come from Whitney stratifications. Via the homeomorphism $N/K\cong X//G$ we obtain a similar picture for $X//G$.