Timeline for Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2022 at 0:03 | vote | accept | Michael Albanese | ||
| Mar 30, 2022 at 13:38 | comment | added | YCor | @DavidESpeyer Actually there exists a Cantor subset in $\mathbf{S}^3$ whose complement is not simply connected (Antoine necklace). I don't know what can be done if the complement is required to be simply connected. (At the positive side all Cantor subsets in the line or plane are topologically equivalent.) | |
| Mar 30, 2022 at 11:39 | comment | added | David E Speyer | But I did not know that $S^n \setminus K$ was not defined up to homeomorphism! How do you see that? | |
| Mar 30, 2022 at 11:24 | comment | added | David E Speyer | It seems to me that $S^n$ minus a cantor set should be doable. Choose three disjoint, nonnested, $(n-1)$-spheres $S_1$, $S_2$, $S_3$ inside $S^n$. Let $r_i$ be inversion in $S_i$ en.wikipedia.org/wiki/Inversion_in_a_sphere . Let $G$ be the group generated by the $r_i$ and let $H$ be the index two subgroup of orientation preserving elements of $G$. Then $H$ should act freely on the complement of a cantor set and the quotient by $H$ should be a compact manifold. (As an abstract group, $H$ is the free group on $2$ generators.) | |
| Mar 30, 2022 at 8:24 | comment | added | YCor | 2 ends can be achieved (universal cover of $\mathbf{S}^{n-1}\times \mathbf{S}^1$). References for the Freudenthal-Hopf theorem can be found on Wikipedia: en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups And of course 1 end (universal covering $\mathbf{R}^n$) and 0 end (universal covering $\mathbf{S}^n$) can be achieved too. I'm not sure about $\mathbf{S}^n$ minus Cantor (by the way this is not uniquely defined up to homeomorphism — let's say, minus a Cantor that fits inside a line). | |
| Mar 30, 2022 at 8:22 | comment | added | Geva Yashfe | @YCor Thanks! I did forget the case of 2 ends last night. The correct form is just now typed in an answer... But I did not remember it is due to Freudenthal. | |
| Mar 30, 2022 at 8:21 | answer | added | Geva Yashfe | timeline score: 18 | |
| Mar 30, 2022 at 8:21 | comment | added | YCor | The theorem that the number of ends is $0,1,2$ or $\infty$ is due to Freudenthal and Hopf, decades before Stallings. By the way, this "ends" result says that the space of ends, if infinite, is a Cantor. Hence, for every $n\ge 3$ and every totally disconnected compact space $K$ with at least 3 elements and at least one isolated point, $S^n-K$ is not homeomorphic to the universal cover of any compact space. | |
| Mar 30, 2022 at 0:18 | comment | added | Geva Yashfe | For homeomorphic/diffeomorphic the answer should be no, by the Stallings theorem on ends of groups. The fundamental group is finitely generated; the assumption implies it has $k+1>1$ ends; Stallings' theorem tells you it has either $1$ or infinitely many. (I am sorry for answering in a comment. It is late, and I don't have time to check that I am not making a stupid error or to write anything in more detail.) | |
| Mar 30, 2022 at 0:02 | history | asked | Michael Albanese | CC BY-SA 4.0 |