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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?

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  • $\begingroup$ 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.... $\endgroup$ – Jason DeVito Mar 29 at 15:20
  • $\begingroup$ @JasonDeVito Can you explain the compact $G$ case? $\endgroup$ – Totoro Mar 29 at 15:43
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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

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