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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3$\begingroup$ Please read mathoverflow.net/howtoask and revise this question. $\endgroup$– Theo Johnson-FreydJan 13, 2013 at 5:08
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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.
answered Jan 13, 2013 at 14:45
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1$\begingroup$ 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. $\endgroup$ Jan 13, 2013 at 15:05
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2$\begingroup$ 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. $\endgroup$– Ben Webster ♦Jan 13, 2013 at 17:03
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1$\begingroup$ @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. $\endgroup$ Jan 13, 2013 at 19:56
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$\begingroup$ @Dick You're right, I erased by previous comment. I had a misconception of the notion, which is definitely global. $\endgroup$– YCorJan 14, 2013 at 18:53