3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Consider the trivial action with $X=X//G=(X//G)_{(G)}$. So $X$ should at least be smooth. $\endgroup$ – Friedrich Knop Nov 9 '17 at 14:07
  • $\begingroup$ @FriedrichKnop I was assuming the strata are smooth, but forgot to mention it. $\endgroup$ – user117030 Nov 9 '17 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 Knop Nov 9 '17 at 17:17
1
$\begingroup$

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 $X//G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.