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Let $x\in M$, $M$ - finite dimensional smooth manifold. Is there an example of a finite dimensional Lie group action on $M$ with no slice at $x$?

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Please read and revise this question. – Theo Johnson-Freyd Jan 13 '13 at 5:08
up vote 8 down vote accepted

Recall that if a free action of G on M has a slice S at a point x then the natural map of G x S into M given by (g,s) maps to gs would be a diffeomorphism onto a tubular neighborhood of the orbit Gx. So for a counterexample take the action of the real line on the 2-torus given by a 1-parameter subgroup with irrational slope acting by translation.

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After thinking about my answer a bit I realized that an even simpler example is the action of the group Z of integers on the circle generated by rotation through an irrational angle. – Dick Palais Jan 13 '13 at 15:05
The action of $\mathbb{R}$ on $\mathbb{R}^2$ by $(x,y)\mapsto (x,y+ax)$ is also a classic. The y-axis is fixed, but has no slice. – Ben Webster Jan 13 '13 at 17:03
@Yves Cormulier I don't think that's so Yves. If it were then even a compact group action wouldn't have a slice at an isolated fixed point p, whereas in fact an invariant neighborhood of p is a slice at p in that case. – Dick Palais Jan 13 '13 at 19:56
@Dick You're right, I erased by previous comment. I had a misconception of the notion, which is definitely global. – YCor Jan 14 '13 at 18:53
Nice answers! Thanks ! – Yar Jan 19 '13 at 12:47

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