Given a circle action on a closed, oriented smooth manifold $M^{2n}$ with isolated fixed points. My question is, does there always exist a point $p\in M$ such that the isotropy subgroup of $p$ is trivial? In other words, does there always exist a free orbit of this circle action? Moreover, if the answer is yes, can we extend this free orbit to a tubular neighborhood $S^1\times D^{2n-1}$ such that each $S^1$ in it is a free orbit?
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$\begingroup$ COnsider the action of $S^1\subseteq\mathbb C$ on $M=\mathbb C$ such that $z\in S^1$ acts by multiplication by $z^2$. ($M$ is not closed but you can easily cook up a closed example like this; say, on the Riemann sphere with the "same" action) $\endgroup$– Mariano Suárez-ÁlvarezCommented Jan 22, 2012 at 2:55
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2 Answers
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The answer is `no', but for a stupid reason: you can have an action with an ineffective kernel, meaning that (normal closed) subgroup of $G=S^1$ consisting of those $g$ which act trivially: for all $x$ in $M$, $gx = x$. For example, take a free action of $S^1$ on $M$. Define a new action $g * x = g^p x$ (I'm thinking of the circle multiplicatively). The new action's ineffective kernel consists of the pth roots of unity.
You can get rid of the ineffective kernel by taking $G$ and
dividing by the ineffective kernel. In this circle case, the resulting action of this new'
$S^1$ will be free a.e.: that is it will have principal orbit type the identity (check out, for example, Hsiang's book,Cohomology Theory of Topological Transformation Groups', p. 10-12, esp. p.12) which means, combined with the slice theorem' (p.10) or
theGleason theorem' (p. 9) the answer to your question becomes `yes'.
answered Jan 22, 2012 at 3:49
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$\begingroup$ Thank you. But what about my second question?:-)(Moreover, if the answer is yes, can we extend...) $\endgroup$– PingCommented Jan 22, 2012 at 7:52
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Closed, orientable, and even-dimensional are irrelevant, as is the condition that all fixed points are isolated. Assume that the action is effective, and that the manifold is connected. Then for every $n>1$ the fixed point set of the subgroup of order $n$ is a submanifold of positive codimension (or possibly a disjoint union of countably many submanifolds of various positive codimensions), so the union of them all cannot be the whole manifold. In other words there is a point with trivial isotropy subgroup.
answered Jan 22, 2012 at 3:57
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$\begingroup$ Thank you. But what about my second question?:-)(Moreover, if the answer is yes, can we extend...) $\endgroup$– PingCommented Jan 22, 2012 at 7:52
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$\begingroup$ By the way, the condition "even-dimensional" is of course can be deduced from the fact that the fixed points are isolated. But the conditions "closed, orientable" are relevant:-) $\endgroup$– PingCommented Jan 22, 2012 at 7:59
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$\begingroup$ Irrelevant in the sense of being unneeded hypotheses for the conclusion you are asking about. And does "the fixed points are isolated" imply that there is at least one fixed point? (Sorry.) $\endgroup$ Commented Jan 22, 2012 at 12:57
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$\begingroup$ some literatures take "isolated” as "nonempty and finitely many fixed points" while some are not. So it depends on concrete situation. $\endgroup$– PingCommented Jan 23, 2012 at 1:22
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2$\begingroup$ I believe that that Richard Montgomery was answering your second question with the slice theorem. In general, for any smooth action of compact $G$ on $M$, for every point $p\in M$ with trivial isotropy group, let $D\subset M$ be a disk transverse to the orbit $Gp$ at $p$, of dimension $dim(M)-dim(G)$. The map $G\times D\to M$, $(g,x)\mapsto gx$, is a submersion at every point in $G\times p$, therefore in a neighborhood of $G\times p$, and it's not hard to see that for a small enough disk nbhd $\Delta \subset D$ of $p$ in $D$ it gives a diffeomorphism of $G\times \Delta$ onto its image. $\endgroup$ Commented Jan 23, 2012 at 1:56