Timeline for Can a covering space of the $p$-adic disc split over the circle?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2020 at 9:43 | vote | accept | Piotr Achinger | ||
| May 26, 2020 at 20:18 | answer | added | SashaP | timeline score: 6 | |
| May 26, 2020 at 19:48 | comment | added | Piotr Achinger | @AriyanJavanpeykar thank you, this answers the question, at least if "becomes disconnected" is replaced with "splits completely". To record the argument here: let $D' = {\rm Sp}\, \mathbb{C}_p\langle x^{-1}\rangle$ so that $\{D, D'\}$ is an admissible covering of $\mathbb{P}^1$ and $D\cap D'$ is the unit circle. Given a finite etale cover which splits completely over $D\cap D'$, we can extend it to a finite etale cover of $\mathbb{P}^1$ by pasting in disjoint copies of $D'$. As $\mathbb{P}^1$ is (algebraically) simply connected, the extended cover is trivial, and hence so is the original one. | |
| May 26, 2020 at 17:58 | comment | added | Alex Youcis | @AriyanJavanpeykar Does the statement itself not imply the claim since the Gauss point is contained in the unit circle and the usual fact about fundamental groups :"surjective iff remains connected upon pullback"? | |
| May 26, 2020 at 17:14 | comment | added | Ariyan Javanpeykar | Does the proof of 7.5 here pdfs.semanticscholar.org/4f28/… show that the answer is no? | |
| May 26, 2020 at 15:00 | comment | added | Alex Youcis | Just to point out something obvious, but maybe could be useful to someone trying to construct a counter example: I'm pretty sure the cover needs to not come from the special fiber of a normal integral model. In particular, any of the strange covers of $\mathbb{A}^1_{\overline{F}_p}$ won't help. | |
| May 26, 2020 at 11:37 | history | asked | Piotr Achinger | CC BY-SA 4.0 |