Timeline for Is there a smooth manifold which admits only rigid metrics?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2016 at 7:39 | vote | accept | Asaf Shachar | ||
| Apr 19, 2016 at 12:30 | comment | added | HJRW | @studiosus -- that seems to do it! Thanks! | |
| Apr 18, 2016 at 18:40 | comment | added | Moishe Kohan | @HJRW: The isometry group of a compact Riemannian manifold is a compact Lie group. A compact Lie group is either finite or contains a copy of $U(1)$. The latter contains finite subgroups $Z/nZ$ for every $n$. Thus, the isometry group of a compact Riemannian manifold is either trivial or contains nontrivial finite order elements. | |
| Apr 18, 2016 at 18:06 | history | edited | Andy Putman | CC BY-SA 3.0 |
added 603 characters in body
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| Apr 18, 2016 at 18:00 | comment | added | Andy Putman | @HJRW: Whoops, you're right! For instance, flat tori have infinite-order isometries. However, using a little more technology one can show that my examples still work. I'll edit the answer accordingly. | |
| Apr 18, 2016 at 17:49 | comment | added | HJRW | I'm missing the other direction of the equivalence. It seems like you need an argument to show that an isometry is necessarily finite order... Or is it obvious? | |
| Apr 18, 2016 at 17:41 | history | answered | Andy Putman | CC BY-SA 3.0 |