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.

  • $\begingroup$ 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

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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.