Timeline for Counterexample to mostow rigidity theorem
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2023 at 20:40 | answer | added | Ryan Budney | timeline score: 3 | |
| Nov 7, 2021 at 18:13 | comment | added | Moishe Kohan | @mme: It's easy already in the 2d case: one-hole torus and 3-holed sphere. In 3d this is also not hard, say, trivial and nontrivial interval bundles over genus 2 surface. | |
| Nov 7, 2021 at 16:38 | comment | added | mme | (Obviously in the comment above I meant for both manifolds to have the same fundamental group, or it has a tautological answer.) | |
| Nov 7, 2021 at 16:01 | comment | added | mme | Thank you for verifying. The next interesting question would be to construct non-diffeomorphic manifolds which both carry infinite-volume hyperbolic metrics. I am not sure if this is possible or not. I believe link complements are out: I half-remember that the only link complements with infinite volume complete hyperbolic metrics are the complements of the unknot and the Hopf link. | |
| Nov 7, 2021 at 15:18 | comment | added | Moishe Kohan | @mme: You are correct. | |
| Nov 7, 2021 at 15:14 | comment | added | mme | I am ignorant of the subject, so perhaps this is silly. But a loxodromic transformation $\varphi$ of hyperbolic 3-space acts without fixed points, as does a parabolic transformation $\psi$. These isometries are not conjugate. This should imply that the resulting quotients $\Bbb H^3/\langle \varphi\rangle$ and $\Bbb H^3/\langle \psi\rangle$ should be both diffeomorphic to $S^1 \times \Bbb R^2$ but non-isometric. Have I made an error here? | |
| Nov 7, 2021 at 14:48 | comment | added | GSM | @Wojowu afaik S^2 x R cannot have an hyperbolic complete structure. | |
| Nov 7, 2021 at 14:38 | comment | added | Wojowu | $\mathbb R^3$ and $S^2\times\mathbb R$? | |
| Nov 7, 2021 at 14:31 | history | asked | GSM | CC BY-SA 4.0 |