Timeline for Quotient by a non-free action of a Lie group and manifolds with corners
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2018 at 0:44 | comment | added | Claudio Gorodski | In case $G$ is a finite group, this is a prime example of a smooth orbifold. | |
| Jul 28, 2018 at 1:15 | comment | added | David Roberts♦ | Thanks, @Igor - I was going to supply a more general reference dealing with proper Lie groupoids, where I learned of this result, but it would be easier in the specific case of a group action. | |
| Jul 27, 2018 at 17:13 | comment | added | Igor Belegradek | @WarlockofFiretopMountain, see theorem 15 in faculty.math.illinois.edu/~ruiloja/Math519/michiels.pdf. It refers to theorem 2.7.4 on page 113 of J.J. Duistermaat and J.A.C. Kolk. Lie Groups. Universitext. Springer, 2000. | |
| Jul 27, 2018 at 14:32 | history | edited | Overflowian | CC BY-SA 4.0 |
added 6 characters in body
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| Jul 27, 2018 at 11:10 | comment | added | Overflowian | @DavidRoberts interesting, any reference? | |
| Jul 27, 2018 at 10:54 | comment | added | David Roberts♦ | If the action is proper, then the quotient is a Whitney stratified space. Each stratum is a manifold, and restricted to these, the quotient is a submersion. Also these glue together well, to give what is arguably the stratified version of a submersion. | |
| Jul 27, 2018 at 10:16 | comment | added | mme | I should say manifold with boundary in the above because topologically there is no difference. | |
| Jul 27, 2018 at 10:09 | comment | added | mme | There is no uniform treatment of the non-free case; the quotient is not in any natural way a smooth manifold (when G is finite, though, it is an orbifold). When the orbit space is 1- or 2-dimensional it is topologically a manifold with corners; there is a proof in Bredon's book on transformation groups. | |
| Jul 27, 2018 at 10:06 | history | asked | Overflowian | CC BY-SA 4.0 |