# Smooth point of the quotient space obtained by Lie group action

Let $$G$$ be a Lie group acting smoothly and effectively on a smooth manifold $$M$$ and $$\pi: M \to M/G$$ the quotient map. Can we find a point $$p \in M$$ such that an open neighborhood of $$\pi(p)$$ is smooth?

• Certainly if $G$ is compact - then $M$ contains an open dense set of such points. I'm afraid I have no idea what happens when $G$ is non-compact.... – Jason DeVito Mar 29 at 15:20
• @JasonDeVito Can you explain the compact $G$ case? – Totoro Mar 29 at 15:43

When you say "open neighborhood of $$\pi(p)$$ is smooth", I'm assuming you mean "open neighborhood of $$\pi(p)$$ can be given the structure of a manifold for which $$\pi$$ is smooth in a neighborhood of $$p$$"
First, if $$G$$ is non-compact, then there may not exist such a point. For example, consider the $$G=\mathbb{R}$$ action on $$M = T^2 = \mathbb{R}^2/\mathbb{Z}^2$$ given by $$t(\theta, \phi) = (\theta + t, \phi + \sqrt{2} t)$$. This action is smooth because it descends from a linear action on $$\mathbb{R}^2$$ and it is effective because $$\sqrt{2}$$ is irrational.
The orbits are dense, and so $$M/G$$ is has the property that each point is dense. In particular, no point of $$M/G$$ can have a Euclidean neighborhood.
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On the other hand, if $$G$$ is compact, then there must exist a $$p\in M$$ for which there is an smooth open neighborhood of $$\pi(p)$$. In fact, there is an open dense set of points of such $$p$$ in $$M$$. This is essentially the content of Theorem 3.4.6 of these notes: https://www.math.upenn.edu/~wziller/math661/LectureNotesLee.pdf