Timeline for Manifolds as simultaneous coset spaces
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Dec 24, 2015 at 10:23 | comment | added | Sebastian Goette | After thinking about the motivation of this question for a while, I guess that among the various equivalent conditions that there might be, the one that $H_1\le H_2$ is probably the easiest to check in practice. Of course I assume that we know what $G$ is and how $G$ acts. If this is not the case, please give some motivation, so one can understand what kind of description you are looking for. | |
| Dec 24, 2015 at 1:38 | comment | added | Claudio Gorodski | The existence a $G$-equivariant map $F:X\to Y$, namely, one such that $F(gx)=gF(x)$ for all $g\in G$, $x\in X$, is a necessary and sufficient condition. | |
| Dec 23, 2015 at 11:12 | comment | added | octopus | Thanks for your answer Sebastian. Why is the map locally trivial in the finite-dimensional case? I'm not worried about $X$ and $Y$ being manifolds that don't admit transitive actions by Lie groups since I am assuming that $X$ and $Y$ do. (I should point out that my definition of a Lie group is based on my definition of a manifold, which only allows finite-dimensional manifolds.) | |
| Dec 23, 2015 at 10:03 | comment | added | Sebastian Goette | If your assertion is satisfied, then you automatically get a map $X=G/H_1\twoheadrightarrow G/H_2=Y$ sending $gx$ to $gy$, where $x$ has stabiliser $H_1$ and $y$ has stabiliser $H_2$. In the finite-dimensional case, this map is automatically locally trivial, that is, a fibre bundle. On the other hand, there are manifolds that do not admit transitive actions by finite-dimensional Lie groups, so the fibre bundle condition alone is not sufficient. | |
| Dec 22, 2015 at 23:31 | comment | added | octopus | I think I do need $G$ to be finite-dimensional, but your idea is interesting nonetheless. Do you know if the fibre bundle condition is necessary as well as sufficient? | |
| Dec 22, 2015 at 16:09 | comment | added | Allen Knutson | Do you want $G$ to be finite-dimensional? If not, I suspect that the condition is simply that $X$ be a fiber bundle over $Y$, with $G$ the group of diffeomorphisms of $X$ taking fibers to fibers. | |
| Dec 21, 2015 at 23:00 | history | asked | octopus | CC BY-SA 3.0 |