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Let $G$ be a Hausdorff, path-connected group acting transitively on a Hausdorff space $X$. Assume the action is continous (i.e. $(g,x) \mapsto g \cdot x$ is continuous) and transitive. It follows easily that $X$ is also path connected.

Given any path $x_t$ in $X$, must there exist a path $g_t$ in $G$ such that $x_t = g_t \cdot x_0$, all $t \in [0,1]$.

I'm inclined to think the answer is "no". In this case, I'd also be very interested to hear of additional hypotheses which make the answer into "yes". Thanks.

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  • $\begingroup$ I think the answer is yes, but I'm not sure I get the picture clear. Take the quotient $G/G_{x_0}$ by the stabilizer subgroup of $x_0$, then it seems something like we have local homeomorphism $G/G_{x_0}\to X$ by the action, so the path uniquely lifts to $G/G_{x_0}$, I guess, then it can be lifted to $G$, too. $\endgroup$ – Berci May 28 '13 at 22:17
  • $\begingroup$ An inane remark: the question is equivalent to "does the group $\mathrm{Paths}(G)$ act transitively on the set $\mathrm{Paths}(X)$". Here the product of two paths is defined point-wise over the interval. Of course this is of no assistance in answering the question, but I thought it was an attractive way to formulate it. $\endgroup$ – Mike F May 29 '13 at 1:20
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If $G$ is a Lie group and the stabilizer is a closed subgroup then the map $G \rightarrow X$ sending $g \mapsto gx_0$ is a principal $H$-bundle and so satisfies path lifting.

More generally, it is a theorem of Steenrod that $G \rightarrow X$ is a fiber bundle (hence, in particular, satisfies path lifting) if there exists a local section of this map around $x_0$ and the stabilizer is a closed subgroup.

I can't seem to find literature on whether or not $G \rightarrow X$ might be a fibration without actually being a fiber bundle... so I don't know. I also can't find a counterexample for when $G \rightarrow G/H$ fails to be a fibration when $H$ is closed (the trouble seems to be coming up with groups $G$ that are far from being Lie groups.)

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  • $\begingroup$ Hello, I don't really know what you mean... In the 1st paragraph, $H$ is the stabilizer of your arbitrary basepoint $x_0$, right? What is it you are saying is a principal $H$-bundle? $\endgroup$ – Mike F May 31 '13 at 0:59
  • $\begingroup$ Yes. The map $G \rightarrow G/H$ is a principal $H$-bundle in this case. ($H$ acts on $G$ by right multiplication, the orbits of this action are the left cosets...) $\endgroup$ – Dylan Wilson May 31 '13 at 1:22
  • $\begingroup$ OK +1, more than a month later, I've at least been exposed to the language appearing in your answer :). Makes sense. This is certainly useful information. I have a couple ideas for examples which I will fiddle with. $\endgroup$ – Mike F Jul 11 '13 at 8:20
  • $\begingroup$ Hi, Can you please answer this question? I think you can help me math.stackexchange.com/q/1693969/322103 $\endgroup$ – M98 Mar 12 '16 at 11:03

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