Question on transversal slice of Lie group Assume we have action of Lie group $G$ on a manifold $X$. Fix some orbit $\mathcal{O}$, it is known there exist transversal slice $S$ with respect to this orbit. Fix some point $x$ in $\mathcal{O}$, and let $G_x$ be the stabilizer of $x$.
My question is, can we find a transversal slice which is $G_x$-stable?
How about in the algebraic situation?
 A: In general, no. Let $G$ be the group of all upper-triangular matrices with positive diagonal entries. It acts on $\mathbb R^2$ as a subgroup of $GL(2,\mathbb R)$. Consider $x=(1,0)$. Its orbit is the coordinate ray $\{(t,0):t>0\}$. Its stabilizer $G_x$ consists of matrices whose upper-left element is 1 and the second column is arbitrary. This stabilizer acts transitively on the upper half-plane, so there are no invariant transversals to the horizontal line.
If $G$ is compact and everything is smooth, then yes. By compactness, there is a Riemannian metric on $X$ invariant under $G$. Let $Z$ be the orthogonal complement to $T_x\mathcal O$ in $T_xX$ (with respect to the Riemannian scalar product). Let $B$ be a small open ball in $Z$ (centered at the origin). Then the submanifold $\exp_x(B)$, where $\exp_x$ is the Riemannian exponential map, is invariant under $G_x$.
A: Assume the base field is of characteristic zero. If $G$ is an affine algebraic group with reductive connected component which acts by morphisms on an affine variety $X$, then Luna has shown that there exists a slice étale at $x$ for each closed orbit $G\cdot x$ in $X$. This means there exists a $G_x$-invariant locally closed affine subvariety $S$ of $X$ containing $x$ such that the morphism $\psi: G*_{G_x}S\to X$, $[g,s]\mapsto gs$ is excellent. In particular, the image of $\psi$ is a saturated open subset $V$ of $X$ and 
$\psi: G*_{G_x}S\to V$ is étale. Here $G*_{G_x}S$ denotes the homogeneous fiber bundle 
$(G\times S)//G_x$. A good, easily available reference are the notes by 
Drézet. 
