Timeline for What is the intuition for higher homotopy groups not vanishing?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 6, 2021 at 21:49 | comment | added | Ryan Budney | I suppose one should add, $S^1$ isn't so different. $S^0$ is very similar to $S^1$ from the perspective of homotopy groups. While $S^1$ has a non-trivial $\pi_1$ only, $S^0$ has ?two? non-trivial $\pi_0$'s, given that you have two possible basepoints. | |
| Apr 11, 2020 at 14:34 | answer | added | Connor Malin | timeline score: 11 | |
| Jul 26, 2019 at 17:55 | review | Close votes | |||
| Jul 26, 2019 at 22:29 | |||||
| Jul 24, 2019 at 5:00 | answer | added | Dev Sinha | timeline score: 15 | |
| Jul 21, 2019 at 21:16 | answer | added | Connor Malin | timeline score: 13 | |
| Jul 21, 2019 at 19:49 | answer | added | Sophie Swett | timeline score: 5 | |
| Jul 21, 2019 at 11:39 | comment | added | David Roberts♦ | The higher stages of the Whitehead tower of a manifold are geometric if one allows differentiable (higher) stacks. | |
| Jul 20, 2019 at 23:32 | history | became hot network question | |||
| Jul 20, 2019 at 20:04 | answer | added | Simon Henry | timeline score: 23 | |
| Jul 20, 2019 at 18:13 | answer | added | D. Zack Garza | timeline score: 10 | |
| Jul 20, 2019 at 16:14 | comment | added | Denis Nardin | @user43326 I never said there are no aspherical manifolds, there are obviously lots (tori, hyperbolic manifolds,...). I meant to say that if you start with a manifold with non-trivial $\pi_n$ and you take the $n$-th Whitehead cover, it will usually not be the homotopy type of a manifold anymore (in fact it will usually tend to become "infinite-dimensional" in some sense). | |
| Jul 20, 2019 at 16:11 | comment | added | user43326 | @DenisNardin actually $n$-dimensional torus $(S^1)^n$ has vanishing of higher homotopy groups, and they are n-manifolds. | |
| Jul 20, 2019 at 16:00 | comment | added | skd | I'm not sure what you mean by "this non-triviality seems to keep on going"; for instance, the homotopy group pi_10 S^6 (in the stable range) vanishes. There are, however, infinitely many nontrivial homotopy groups of S^n for n>1. This is a consequence of a result known as the McGibbon-Neisendorfer theorem, which states that if X is a simply-connected finite complex which is not p-locally trivial, then pi_n X has p-torsion for infinitely many n. From this point of view, the failure of S^1 to be simply-connected is one root of the issue. | |
| Jul 20, 2019 at 16:00 | comment | added | Denis Nardin | I think it is a combination of two facts: (1) taking the universal cover is "geometric". In particular if $X$ is a $d$-dimensional manifold, so is its universal cover. (2) There aren't that many simply connected 1-manifolds. Note that taking higher Whitehead covers does not preserve being a manifold (and in fact tends to send finite-dimensional objects to infinite-dimensional objects, in some sense) | |
| Jul 20, 2019 at 15:39 | comment | added | mme | The "universal covering space" operation is simultaneously very geometric and preserves all homotopy groups except $\pi_1$. This transparently sends $S^1$ to the contractible space $\Bbb R$. There are analogues to this for higher homotopy groups, but the 'Whitehead truncation' operations are no longer so geometric; there is no real way to see visually what the Whitehead truncations of higher spheres are, and indeed they are not contractible. | |
| Jul 20, 2019 at 15:22 | history | asked | eriugena | CC BY-SA 4.0 |