MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '10 at 18:41
Yes. This is my question. – Hong Apr 1 '10 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 '12 at 10:50

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 $G_x$.

share|cite|improve this answer
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. – Hong Apr 1 '10 at 20:35

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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