0
$\begingroup$

Let $M$ be a smooth manifold and $G$ a Lie group acting properly on $M$. Let $k$ be the codimension of a maximal dimensional $G$-orbit in $M$.

Is it always possible to construct $k$ functions $f_1, \dots, f_k \colon M \to \mathbb{R}$, such that $f_1, \dots , f_k$ are $G$-invariant and functionally independent on an open and dense subset of $M$?

If it is not possible, under which further assumptions could it be true?

$\endgroup$

1 Answer 1

2
$\begingroup$

This can be done locally. Suppose that the manifold $M$ and the Lie group $G$ are compact. Then one can say that the type of the orbits of $x$ and $y$ coincide if and only if the stabilizer of $x$ and $y$ are conjugated. There is particular type called the principal type: The type of the orbit of $x$ is principal if and only if the canonical representation of the stabilizer $G_x$ on $T_xM/V_x$) where $T_xM$ is tangent space of $M$ at $x$ and $V_x$ is the subspace of $T_xM$ tangent to the orbit) is trivial. A principal orbit has a maximal dimension and the set of principal orbit is a dense open subset.

The slice theorem of Koszul allows then to identify equivariantly a neighborhood of the orbit of $x$ to $G/G_x\times W_x$ where is an open subset of $T_xM/V_x$. So you can find functions $f_1,...,f_k$ that are independent functionally and invariant by $G$ given by the coordinates of $W_x$.

$\endgroup$
2
  • $\begingroup$ Thank you very much. So that way the functions $f_1, \dots f_k$ are defined locally on a slice. Is it now possible to use a partition of unity, such that these locally defined functions can be patched together to extend them to "global" functionally independent functions? (so open and dense?) $\endgroup$
    – Olorin
    Mar 16, 2016 at 17:27
  • $\begingroup$ You can extend them with partition of unity, but I am not sure they will be still invariant by $G$. $\endgroup$ Mar 16, 2016 at 17:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .