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. $\endgroup$ Nov 9, 2017 at 14:07
  • $\begingroup$ @FriedrichKnop I was assuming the strata are smooth, but forgot to mention it. $\endgroup$
    – user117030
    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). $\endgroup$ Nov 9, 2017 at 17:17

1 Answer 1


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 orbit-type stratification of $X$ is Whitney. Since orbits of proper actions are closed $X//G=X/G$, and the orbit-type 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 orbit-type, but the central matrices and the non-diagonalizable matrices are in two different strata that intersect in the GIT quotient (both have repeated eigenvalues). To fix this we throw out the non-diagonalizable matrices (since they do not have closed orbits). Then the corresponding orbit-type stratification of $X//G$ is $\mathbb{C}^2-V(t^2-4d)$, corresponding to the diagonalizable matrices that are not central, and $V(t^2-4d)\cong \mathbb{C}$ corresponding to the central matrices.

More generally, maybe the orbit-type stratification is not what you want exactly. Consider a maximal compact $K\subset G$ and a corresponding Kempf-Ness 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 orbit-type stratification is Whitney. It descends to the orbit-type stratification of $S/K$ by smooth submanifolds corresponding to a Whitney stratification of $S$ via submersions. Collectively this gives a "refined orbit-type" 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 $X//G$.


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