a question about the isotropy subgroup of circle action on manifolds with isolated fixed point 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?
 A: 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. 
A: 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'.  
