Suppose $G$ is a compact Lie group acting freely on a topological space $Q$ (about whose separation conditions I make no assumptions) and the qoutient $Q/G$ is known to be completely regular Hausdorff (or even paracompact Hausdorff). Can one deduce that $Q \to Q/G$ is a principal bundle? If $Q$ itself (rather than $Q/G$) were completely regular, this result is well known. Perhaps it follows that $Q$ must be completely regular if $Q/G$ is?
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$\begingroup$ I guess the ending not in the parenthetical remark in your title is a typo? $\endgroup$– Mariano Suárez-ÁlvarezCommented Aug 8, 2013 at 22:08
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$\begingroup$ Maybe you can use metrization theorem on the quotient and try to build a metric on $Q$ out of it and metric on $G$? $\endgroup$– Vít TučekCommented Aug 8, 2013 at 22:21
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$\begingroup$ Think of the case when $G$ is finite and acts simply transitively. $\endgroup$– MishaCommented Aug 9, 2013 at 3:27
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$\begingroup$ Did you try to adapt Palais' proof of the slice theorem? MR0126506 (23 #A3802) Palais, Richard S. On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 1961 295–323. $\endgroup$– Peter MichorCommented Aug 9, 2013 at 5:15
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1$\begingroup$ Maybe, you should add the requirement that each $G$-orbit (with subspace topology) is homeomorphic to $G$. $\endgroup$– Peter MichorCommented Aug 9, 2013 at 19:54
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1 Answer
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Take $X$ to be finite with trivial topology, let $G$ be a finite cyclic group acting simply transitively on $X$. If you equip $G$ with discrete topology then it is a Lie group. The quotient $Y=X/G$ is a single point, so it satisfies all the regularity properties you can imagine. However, the quotient map $X\to Y$ is clearly not a bundle. Thus, at the very least you need the original space to be Hausdorff.
