Timeline for discrete subgroups of Lie groups and actions on homogeneous spaces
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 4, 2013 at 8:43 | vote | accept | Dieter | ||
| Mar 4, 2013 at 8:43 | vote | accept | Dieter | ||
| Mar 4, 2013 at 8:43 | |||||
| Mar 4, 2013 at 8:42 | vote | accept | Dieter | ||
| Mar 4, 2013 at 8:43 | |||||
| Mar 2, 2013 at 17:33 | answer | added | Misha | timeline score: 10 | |
| Mar 2, 2013 at 14:14 | vote | accept | Dieter | ||
| Mar 2, 2013 at 15:13 | |||||
| Mar 2, 2013 at 13:31 | comment | added | Lee Mosher | Igor does have a point. One instance of the original version of OP's question, before the edit, is whether every finite group acts on $S^2$. The answer is yes, trivially. But the question becomes interesting if you restrict to faithful actions. | |
| Mar 2, 2013 at 12:12 | history | edited | Dieter | CC BY-SA 3.0 |
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| Mar 2, 2013 at 9:50 | history | edited | Dieter | CC BY-SA 3.0 |
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| Mar 2, 2013 at 4:21 | comment | added | YCor | @Igor: what's the point in adding extra unnatural requirements just to give trivial counterexamples? | |
| Mar 2, 2013 at 1:24 | answer | added | Lee Mosher | timeline score: 9 | |
| Mar 2, 2013 at 0:14 | comment | added | Igor Belegradek | Do you want $\tilde\Gamma$ act effectively? Say, if $\tilde\Gamma$ is the product of $\Gamma$ and a finite group, then $\tilde\Gamma$ acts properly via the projection onto $\Gamma$. On the other hand, if you want the action to be effective, there is an easy counterexample: take $X=\mathbb R$, $\Gamma=\mathbb Z$, and $\tilde\Gamma=\mathbb Z\times\mathbb Z_2$. The factors commute and $\mathbb Z_2$ has a fixed point from which one can show that $\mathbb Z_2$ must act trivially. | |
| Mar 1, 2013 at 22:47 | history | edited | Dieter | CC BY-SA 3.0 |
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| Mar 1, 2013 at 20:31 | history | asked | Dieter | CC BY-SA 3.0 |