Timeline for Example of homeomorphism of $3$-manifolds
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2020 at 13:45 | comment | added | Kyle Hayden | @MaximUlyanov Expanded below. As for the SnapPy calculation, the point is that isometries can reverse orientation. For example, reflection of the real line is an isometry! | |
| Jul 7, 2020 at 13:11 | comment | added | user160180 | Now, I'm confused. You mean that Golla's proof were true, then it would imply “truly” cosmetic surgeries on $4_1$, right? On the other hand, SnapPy says that $S^3_{1/3}(4_1)$ and $S^3_{-1/3}$ are even isometric: M = Manifold('4_1') M.dehn_fill((1,3)) N = Manifold('4_1') N.dehn_fill((-1,3)) M.is_isometric_to(N) | |
| Jul 7, 2020 at 13:01 | comment | added | Kyle Hayden | @MaximUlyanov A bit confused by your reply. It is known that there are no “truly” cosmetic surgeries on 4_1 (i.e. surgeries with different slopes on the same knot that are homeomorphic as oriented manifolds). There may be easier arguments, but see the discussion here: arxiv.org/pdf/math/0512253.pdf | |
| Jul 7, 2020 at 9:05 | comment | added | user160180 | @KyleHayden Yes. Golla presented the existence of desired homeomorphism. At first, I realized that the last manifold is $S^3_{-1}(8_1)$ not $S^3_{-1}(m(8_1))$. I checked the Theorem 1.2 of Ni and Wu: arxiv.org/pdf/1009.4720.pdf This result is compatible with their theorem. | |
| Jul 7, 2020 at 9:01 | vote | accept | CommunityBot | ||
| Jul 7, 2020 at 2:43 | comment | added | Kyle Hayden | This also should let you rule out the existence of an orientation-preserving homeomorphism: Mirroring everything above should give an orientation-preserving homeo $S^3_{1/3}(4_1) \cong S^3_{+1}(8_1)$ (because $m(4_1)=4_1$). If there was an orientation-preserving homeo $S^3_{-1/3}(4_1) \cong S^3_{+1}(8_1)$, then $S^3_{-1/3}(4_1)$ and $S^3_{1/3}(4_1)$ would be homeomorphic with orientation. But I believe that's ruled out; see existing work on the "cosmetic surgery conjecture". | |
| Jul 7, 2020 at 2:39 | comment | added | Kyle Hayden | @MaximUlyanov I believe Marco exhibited an orientation-preserving homeomorphism $S^3_{-1/3}(4_1) \cong S^3_{-1}(m(8_1))$, where $m(8_1)$ is the mirror of the knot $8_1$ that you drew above. Now recall that, for any knot $K \subset S^3$ and $r \in \mathbb{Q}$, there's an orientation-reversing homeomorphism from $S^3_r(K)$ to $S^3_{-r}(m(K))$. So that means there's an orientation-reversing homeomorphism from $S^3_{-1/3}(4_1)$ to $S^3_{+1}(8_1)$. | |
| Jul 3, 2020 at 22:13 | comment | added | Marco Golla | Ugh. I miscounted crossings, I suppose... | |
| Jul 3, 2020 at 17:21 | comment | added | user160180 | Ops! What I'm looking for is +1 surgery on 8_1. But you proved the left-hand side homeomorphic to +1 surgery on 7_2 or -1 surgery on 8_1. | |
| Jul 1, 2020 at 12:40 | vote | accept | CommunityBot | ||
| Jul 3, 2020 at 20:12 | |||||
| Jul 1, 2020 at 12:37 | history | answered | Marco Golla | CC BY-SA 4.0 |