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When do the orbits of an action on an algebraic variety make a Whitney stratification?

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    $\begingroup$ Possible duplicate of Whitney stratification of affine GIT quotients $\endgroup$ Commented Sep 26, 2019 at 20:37
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    $\begingroup$ Always, if there are finitely many orbits. One can think of the Whitney conditions as positing some checkable "equi-singularity along the strata". The orbits of a group action are equi-singular in the strongest possible sense. Can't think of a ref right now but I can't imagine this is difficult to do by hand. $\endgroup$ Commented Sep 27, 2019 at 5:40

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The answer is indeed always when there are fnitely many orbits. Let $Y \subset X$ be an orbit, $y \in Y$, and $U \subset X$ a $G$-invariant open such that $y \in Y \cap U$, the relative open $Y \cap U$ is dense in $Y$, closed in $U$ and $U$ embedds $G$-equivariantly in some $\mathbb{C}^N$. Let us denote by: $$ C(X) = \overline{\{(x,H) \in U_{smooth} \times (\mathbb{C}^N)^*, \ \textrm{such that} \ T_{X,x} \subset H \}},$$ where the overline is the Zariski closure in $\mathbb{C}^N \times (\mathbb{C}^N)^*$. The variety $C(X)$ is known as the conormal space and is also $G$-invariant. Note that the projection map : $\pi : C(X) \longrightarrow X$ is $G$-equivariant. Let: $$ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$ be the blow-up of $C(X)$ along $\pi^{-1}(Y)$. The map $\rho$ is again $G$-equivariant, so that the composite: $$ \pi \circ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$ is $G$-equivariant. In particular, all fibers of $\pi \circ \rho$ are isomorphic and we find : $$ \dim (\pi \circ \rho)^{-1}(y) = \dim (\pi \circ \rho)^{-1}(Y) - \dim Y = N-2 - \dim Y.$$ Hence, proposition 2.3.1 in the paper Variétés Polaires II by Bernard Teissier shows that the pair $(X,Y)$ staisfies the Whitney conditions at $y$. This is true for all $y \in Y$, so that the pair $(X,Y)$ satisfies the Whitney conditions.

As hinted in the comments, if there are finitely many orbits, then you get a finite stratification of $X$ by orbits, and by the above arguments, this is a Whitney stratification.

Note however that it is certainly not the minimal stratification, as illustrated by the case $X$ smooth and $G$ acts non-trivially with finitely many orbits.

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