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Suppose a manifold $M$ admits a smooth Lie Group action $G$, and $N$ is a closed sub-manifold of $M$ such that $G$ action freely on $N$.

Q: Why in a small neighborhood of $N$, $G$ also action freely?

What I do not understand is that how to let $G$ action freely on the vertical part of $ngh(N0$.

Thanks.

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  • $\begingroup$ Local freedom is clear: look at the rank of elements of the Lie algebra vanishing at each point. Since the rank is zero along $N$, it is zero nearby. $\endgroup$
    – Ben McKay
    Commented Jul 25, 2016 at 13:23
  • $\begingroup$ I do not follow, the rank is no continuous. Why the rank of vanishing Lie algebra is zero nearby? Sorry@BenMcKay $\endgroup$
    – DLIN
    Commented Jul 25, 2016 at 13:27
  • $\begingroup$ Maybe it is easier to picture that the dimension of the $G$-orbit at a point $m \in M$ is the dimension of the set of vectors $v(m)$ for $v \in \mathfrak{g}$. This is in standard textbooks, like Abraham and Marsden. Pick as many vector fields $v_i \in \mathfrak{g}$ as you can which are linearly independent at $m$, and they remain so nearby. $\endgroup$
    – Ben McKay
    Commented Jul 25, 2016 at 13:48

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The dimension of the isotropy Lie algebra $\mathfrak g_x$ is upper semicinuous in $x$. Thus, if it is zero in some point it is zero nearby. So, local freeness is ok as Ben has pointed out. Global freeness is not ok. One of the standard counterexamples is $G=SL(2,\mathbb C)$ acting in $3$-forms ($\cong\mathbb C^4$). Then the isotropy group of $x^2y$ is trivial but it is a group of order $3$ on a dense open subset (cyclically permuting the roots of a $3$-form). If $G$ is compact or, more generally, the action is proper then (global) freeness holds by the slice theorem.

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