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Searched-for situation: A compact connected Lie group acts effectively on a closed Riemannian manifold by isometries, such that there is only one orbit type of dimension strictly less than that of the manifold, and the orbit projection $M\to M/G$ is not a principal $G$-bundle.

I could not find or construct any examples of the situation above. However, I am quite sure there must be some. The reason is that several authors study group actions with only one orbit type in their papers, and if there were no examples as above then the situation in those papers could be simplified dramatically.

Q: Who knows an example, or a reference to an example, of the situation described above?

Background information: It is well-known (Bredon, Introduction to Compact Transformation Groups, Theorem IV.3.3) that when there is only one orbit type, corresponding to a minimal isotropy type $(H)$, the orbit projection $M\to M/G$ is a fiber bundle with fiber $G/H$ and structure group $N(H)/H$, where $N(H)$ is the normalizer of $H$. Consequently, the situation above reduces to:

A compact connected Lie group acts effectively on a closed Riemannian manifold by isometries, such that there is only one orbit type of dimension strictly less than that of the manifold, and the isotropy groups of the action are non-trivial and core-free.

Dropping the dimension hypothesis, any homogeneous space $G/H$, where $G$ is a compact connected Lie group and $H$ is a non-trivial core-free closed subgroup and $G$ acts by left multiplication, is an example. In this case, there is just one orbit type for the trivial reason that there is only one orbit in total.

Thanks in advance for your answers.

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Let $G=O(n+1)$ act on the $n$-sphere $S^n\cong O(n+1)/O(n)$. Let $M=S^1\times S^n$ and let $G$ only act on the second factor The effectiveness is the problem, but this is clear, as the action on $S^{n}$ by linearity determines the action on $R^{n+1}$.The projection is not a principal $G$-bundle, as indeed no element of $G$ acts freely.

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  • $\begingroup$ Thank you. For $n=2$, and taking $SO(3),SO(2)$ instead of $O(3),O(2)$, your example indeed solves the problem. $\endgroup$ – B K Oct 27 '14 at 13:52
  • $\begingroup$ Small typo: $N$ should be $n$. $\endgroup$ – Jim Humphreys Oct 27 '14 at 17:37

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