Timeline for Is there any Riemannian manifold of zero dimensional isometry group such that
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2020 at 16:30 | history | edited | C.F.G | CC BY-SA 4.0 |
deleted 10 characters in body
|
| Oct 1, 2019 at 11:04 | comment | added | Dmitri Panov | Sorry, I meant of course $(x^2+y^2+z^2)+...$ - a tiny perturbation of the unite sphere | |
| Sep 23, 2019 at 18:34 | vote | accept | C.F.G | ||
| Apr 9, 2021 at 20:37 | |||||
| Sep 23, 2019 at 18:02 | comment | added | Dmitri Panov | You can take the following surface in $\mathbb R^3$: $(x^2+y^2+y^2)+10^{-10}(x^3+2y^3+3z^3)=1$. It's group of isometries is trivial. | |
| Sep 22, 2019 at 18:30 | comment | added | C.F.G | @SebastianGoette: A perturbed Ricci-positive metric may admits no more isometries? really? do you have any concrete example in your mind? | |
| Sep 22, 2019 at 16:20 | comment | added | Sebastian Goette | If you take any Ricci-positive metric and perturb the metric just a little in $C^2$, then the resulting metric is still Ricci-positive and generically admits no more isometries. You could change the question, replacing "of zero-dimensional isometry group" by "which does not admit a smooth effective action by a compact Lie group of dimension $\ge 1$". If a manifold $M$ admits a smooth action by a compact Lie group, then there exists a Riemannian metric on $M$ for which the action is isometric. | |
| Sep 21, 2019 at 16:33 | answer | added | Tsemo Aristide | timeline score: 6 | |
| Sep 21, 2019 at 16:28 | history | asked | C.F.G | CC BY-SA 4.0 |