## 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?

-
 What is "transversal slice"? Is it a small submanifold of $X$ whose tangent space at $x$ complements that of the orbit? Does "$G_x$-stable" mean "invariant under the action of $G_x$"? – Sergei Ivanov Apr 1 2010 at 18:41 Yes. This is my question. – xiyu Apr 1 2010 at 19:39 Of course, Palais'slice theorem tells you that the answer is affirmative if the $G$-action is proper (see R. Palais, On the existence of slices for actions of non-compact Lie groups, Ann. of Math. (2) 73 (1961)). – Alain Valette Apr 9 2012 at 10:50

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.

-

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 $G_x$.

-
 Thanks a lot. Do you know the situation in the algebraic geometry? Probably we need to require G is reductive, but usually the stabilizer is not reductive. – xiyu Apr 1 2010 at 20:35