Timeline for How to "lift" a transitive group action on a manifold?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2014 at 4:46 | vote | accept | Giovanni Moreno | ||
| Sep 30, 2014 at 9:57 | comment | added | Ben McKay | @G_infinity: you cannot always embed every Lie algebra into the Lie algebra of vector fields on every manifold. For example, you cannot embed the Lie algebra of $SU(2)$ into the vector fields on the circle, or else $SU(2)$ would have a nontrivial action on the circle, and by compactness would have a compact orbit, so the circle itself, with stabilizer of a point a closed subgroup of codimension 1, i.e. dimension 2. But the Lie algebra of $SU(2)$ is cross product, so you can picture the cross product on a plane in 3-dimensions: no invariant plane. So no subgroup of dimension 2. | |
| Sep 30, 2014 at 9:35 | comment | added | Giovanni Moreno | @BenMcKay: incidentally, can you always realise $\frak{g}$, the Lie algebra of $G$, as an algebra of vector fields on $M$, i.e., can you embed it into the (infinite-dimensional) Lie algebra of vector fields on $M$? | |
| Sep 30, 2014 at 9:15 | comment | added | Giovanni Moreno | @VítTuček: yes, actually in the cases I'm interested in, $\widetilde{M}$ is compact. | |
| Sep 30, 2014 at 6:18 | comment | added | Ben McKay | You only need the Lie algebra action to be complete upstairs, which is automatic because it is complete downstairs and intertwines with the covering map. | |
| Sep 29, 2014 at 20:35 | comment | added | Vít Tuček | Don't you need $\widetilde{M}$ to be compact? | |
| Sep 29, 2014 at 15:43 | comment | added | Ben McKay | The fundamental group of $G$ lies inside the universal covering group of $G$. The group $\Gamma$ you want is the subgroup of $\pi_1(G)$ which maps trivially to $\pi_1(M)$ when a loop in $G$ carries a base point of $M$ around. | |
| Sep 29, 2014 at 15:39 | comment | added | Ben McKay | The only extra condition you need on $\tilde{M}$ is that it is connected. | |
| Sep 29, 2014 at 15:28 | comment | added | Giovanni Moreno | Indeed. This is more or less what I suggested in my last lines. I'm aware of Lie-Palais theorem, though you need to require some extra topological conditions from $\widetilde{M}$. Still, I'd like to see some universal property characterising $\widetilde{G}$, like, e.g., "it is the unique group admitting $G$ as a factor by covering transformations", or something like that, and/or some algebraic way to construct it out of the available data! | |
| Sep 29, 2014 at 14:57 | history | answered | Ben McKay | CC BY-SA 3.0 |