Timeline for Proper action on product manifold
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2023 at 17:34 | vote | accept | Yushi MuGiwara | ||
| Jan 15, 2023 at 2:04 | vote | accept | Yushi MuGiwara | ||
| Jan 16, 2023 at 16:41 | |||||
| Jan 14, 2023 at 14:07 | answer | added | Moishe Kohan | timeline score: 8 | |
| Jan 14, 2023 at 13:58 | comment | added | Moishe Kohan | Just for completeness, in such questions you should specify if you are working in topological or smooth category. But for this particular question, it does not matter since the answer is negative in both. | |
| Jan 10, 2023 at 17:33 | comment | added | Yushi MuGiwara | @MarkGrant yes, I mean the maximal $n$ for which $\mathbb{R}^n$ act properly on $N$. | |
| Jan 10, 2023 at 14:50 | comment | added | Alex B. | To add to Mark's comment: for example if $M=N$, then $\mathbb{R}^n\wr (\mathbb{Z}/2\mathbb{Z})=(\mathbb{R}^n\times \mathbb{R}^n)\rtimes \mathbb{Z}/2\mathbb{Z}$ acts properly on $N\times N$. | |
| Jan 10, 2023 at 13:09 | comment | added | Mark Grant | Good question, but I'm not sure about the formulation...is there a "maximal group that can act properly" on a manifold? Do you mean to say the maximal $n$ for which $\mathbb{R}^n$ can act properly? | |
| S Jan 10, 2023 at 12:36 | review | First questions | |||
| Jan 10, 2023 at 16:01 | |||||
| S Jan 10, 2023 at 12:36 | history | asked | Yushi MuGiwara | CC BY-SA 4.0 |