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Setup:

I was reading about slices and principal orbit theorems (Theorem 3.4.6) from these notes.

Let the Lie group $G$ act on a complete Riemannian manifold $(M,g)$ isometrically on $M$, i.e. $\phi^{*}g= g \forall \phi \in G,$ i.e. the Riemannian metric $g$ is $G$-invariant. Also assume the action is proper but not necessarily free. Let the quotient be $Q:=M/G, \pi:M\to Q$ be the quotient map. Equip $Q$ with the quotient metric: $d_Q(q_1, q_2):=min_{\{(p_1,p_2)\in M \times M: \pi(p_1)=q_1, \pi(p_2)=q_2\} } d(p_1, p_2).$

When $G$ is compact, two things happen:

  1. Existence of global slice for $G$-action: By Theorem 3 of these notes, there is a global slice $S_p$ through every point $p\in M.$ (I hope I understood it correctly? I haven't gone through the proof yet) since the $M$ is equipped with a $G$-invariant metric, as $G$ acts on $M$ isometrically.

  2. Existence of principal orbits: There are open dense subsets $M_0 \subset M, Q_0 \subset Q$, such that $M_0\to Q_0$ is a Riemannian submersion and that $M_0$ is the union of all the principal (maximal) orbits, i.e. the union of orbits with minimal isotropy types.

My questions, connecting the above two, are:

  1. Assume $G$ is compact. Then Then are there any sufficient conditions on the $G$-action so that $M_0$ can be expressed as a disjoint union of global slices in $M_0$, i.e. given $p \in M_0,$ does there exist a global slice $S_p \subset M_0$ so that $M_0= \bigcup_{g \in G} S_{g.p}, G.S_p=M_0.$ (because of globality).

My question is motivated by simple examples of $G:=\operatorname{SO}(2)=\mathbb{S}^1$ action on $M:=\mathbb{R}^2,$ where $M_0=\mathbb{R}^2 \setminus \{0\}$ is a union of the principal orbits $\mathbb{S}^1.p$, $p \ne 0$, but $M_0$ is also the disjoint union of the global slices $S_p:=\{cp: c>0\}$ in $M_0$. Similar examples for the $G:=\operatorname{SO}(2)=\mathbb{S}^1$ action on $M:=\mathbb{S}^2$, where $M_0= M\setminus \{(0,0,\pm 1)\}$, can be decomposed into slices $S_p$, that is the "unique open longitudinal line on the sphere $\mathbb{S}^2$ passing through $p$."

EDIT: following the comment of Moishe Kohan, is it possible to have a counterexample where it's not the case?

  1. Same question for proper $G$ action, $G$ not necessarily compact.
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    $\begingroup$ In (1) it should be $M_0$, not $M$. Also, Riemannian submersion makes no sense until you equip the quotient with a smooth structure and a Riemannian metric. However, it plays no role in the question. The answer to your question is trivial: if and only if on $M_0$ you get a trivial principal $G$-bundle (assuming an effective $G$-action). I do not see what else is there to say. $\endgroup$
    – Moishe Kohan
    Commented Mar 26 at 20:05
  • $\begingroup$ @MoisheKohan Thanks for spotting the typo, I corrected the question. I also made an edit referring to your comment, where I asked for a counterexample. $\endgroup$
    – Learning math
    Commented Mar 26 at 20:09
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    $\begingroup$ Do you know any nontrivial principal bundles? For instance the Hopf bundle, the universal cover of the projective plane.... $\endgroup$
    – Moishe Kohan
    Commented Mar 26 at 20:11
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    $\begingroup$ You have to know all this if you want to have a chance to study group actions on manifolds. Tangent bundles are not exactly relevant. If you do not know what the Hopf bundle is, consider $H=SU(2), G=U(1)< SU(2)$ and the principal $G$-bundle $H\to H/G\cong S^2$. Another example to consider is the action of $G=\mathbb Z_2$ on $S^2$ generated by the antipodal map. The quotient $S^2\to S^2/G=RP^2$ is a nontrivial principal $G$-bundle. It is a good exercise to see why in both examples the $G$-action is free and a global slice does not exist. $\endgroup$
    – Moishe Kohan
    Commented Mar 26 at 20:43
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    $\begingroup$ Why don't you read some published books on the subject instead of some unfinished notes without detailed proofs, with nonstandard definitions, etc...? $\endgroup$
    – Moishe Kohan
    Commented Mar 28 at 17:08

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