Timeline for Covering of a knot complement
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2023 at 13:44 | answer | added | Steven Sivek | timeline score: 5 | |
| Mar 5, 2023 at 20:04 | history | became hot network question | |||
| Mar 5, 2023 at 14:47 | answer | added | Sam Nead | timeline score: 10 | |
| Mar 5, 2023 at 14:12 | comment | added | Carl-Fredrik Nyberg Brodda | Sorry, me saying that the answer is "vacuously yes" to your question is of course a typo and completely backwards, it should be: the answer is "always no". In the sense that for e.g. $B$ as in Gonzales-Acuna & Whitten none of the covering spaces $E$ you can construct will be knot complements (e.g. the obvious covering you mention induced by surjecting $\mathbb{Z}_n$). | |
| Mar 5, 2023 at 13:54 | comment | added | Andrey Ryabichev | @Carl-FredrikNybergBrodda I claim that for any $n$ there is an $n$-covering over $S^3\setminus K$ -- just take for its fundamental group the kernal of the composition $\pi_1(S^3\setminus K)\to H_1(S^3\setminus K)=\mathbb{Z}\to\mathbb{Z}_n$ | |
| Mar 5, 2023 at 12:52 | comment | added | Carl-Fredrik Nyberg Brodda | A result in the Gonzales-Acuna & Whitten paper is that "any covering between knot exteriors is cyclic", so this gives some restrictions for your question. For example, it shows that if $B$ is either the complement of (1) the two-bridge knot; or (2) a composite knot; or (3) a knot with a quadratic Alexander polynomial different from $t^2-t+1$ (i.e. that of the figure-eight knot); then in any of these cases there are no finite-fiber coverings $E$ as in the question. So in all those cases for $B$, the answer is (vacuously) "yes" to your question. | |
| Mar 5, 2023 at 12:03 | history | asked | Andrey Ryabichev | CC BY-SA 4.0 |