Timeline for Freely Periodic map of $(S^{3} , K) $ and a fixed loop in the induced isomorphism of $\pi_{1} ( S^{3} \backslash K )$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2014 at 17:32 | answer | added | Ian Agol | timeline score: 1 | |
| Jan 11, 2014 at 16:27 | comment | added | Arnaud Mortier | The point of Marco and Ryan is that if $f$ does not fix the base point (wherever it is) then $f$ does not induce a homomorphism at the level of the fundamental group. | |
| Dec 14, 2013 at 15:16 | comment | added | user5604 | With the Dehn presentation, we can choose the base point to be any point above the region $\mathbb{R}^{2} \cup \infty$ where we have placed the regular projection. I want to understand whether or not an isomorphism of the fundamental group of a link induced by an orientation preserving periodic homeomorphism can distinguish between a periodic homeomorphism that has $S^{1}$ as its fixed point set or one that is freely periodic. | |
| Dec 14, 2013 at 14:29 | review | Close votes | |||
| Dec 16, 2013 at 9:49 | |||||
| Dec 14, 2013 at 14:15 | comment | added | Ryan Budney | It sounds like your question isn't well-stated as you don't have a basepoint. | |
| Dec 14, 2013 at 10:55 | comment | added | user5604 | Consider a regular link projection of $K$ with base-point above the link and consider then the Dehn presentation. If $f$ fixes the base-point, then we know that $f$ cannot be freely periodic and must have fixed point set homeomorphic to $S^{1} $ disjoint from $K$ with knot type the trivial knot by the Smith conjecture. | |
| Dec 14, 2013 at 6:44 | comment | added | Marco Golla | Where is the basepoint? And how does $f$ act on it? | |
| Dec 14, 2013 at 5:06 | review | First posts | |||
| Dec 14, 2013 at 6:59 | |||||
| Dec 14, 2013 at 4:47 | history | asked | user5604 | CC BY-SA 3.0 |