Timeline for What is the intuition for higher homotopy groups not vanishing?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2019 at 16:27 | comment | added | Maxime Ramzi | @NajibIdrissi : nonzero homology implies noncontractibility without mentioning homotopy groups or Hurewicz; so I don't see how that's circular. The homology here is not part of the question as all spheres have "the same" homology (the formula is the same) | |
| Jul 22, 2019 at 20:55 | comment | added | Mykola Pochekai | Yes, of course, I meant that modulo Whitehead theorem you could reduce nonzeroness of higher homotopy groups to the (much more geometrically intuitive) fact of contrability of $S^2$. Sorry for inaccurate wording. | |
| Jul 22, 2019 at 14:07 | comment | added | Najib Idrissi | @kp9r4d That's circular... How do you know that $S^2$ is not contractible, if not because it has nonvanishing homology (therefore homotopy by Hurewicz) or homotopy groups? | |
| Jul 22, 2019 at 14:02 | comment | added | Mykola Pochekai | @NajibIdrissi Hmm, maybe I don't see something, I thought about argument of such type: if $S^2$ have all zero homotopy groups then embedding of point into $S^2$ is weak equivalence, this implies by Whitehead theorem that $S^2$ contractible, but we know that it is not. | |
| Jul 22, 2019 at 12:21 | comment | added | Najib Idrissi | @kp9r4d Sorry, I don't really understand what you mean. Which Whitehead theorem? The one about weak equivalences of CW complexes being homotopy equivalence, the one about embeddings...? I don't really see how either one applies. | |
| Jul 20, 2019 at 22:32 | comment | added | Mykola Pochekai | @NajibIdrissi Modulo Whitehead theorem it explains why at least one higher homotopy group is nonzero. | |
| Jul 20, 2019 at 22:07 | comment | added | Wojowu | @NajibIdrissi It does answer the question in the body: "What makes $S^1$ so fundamentally different"? | |
| Jul 20, 2019 at 20:46 | comment | added | Najib Idrissi | That's not really an explanation for why the homotopy groups of spheres are nonzero though... All you can conclude here is that $\pi_k(S^n) = \pi_k(S^n)$! | |
| Jul 20, 2019 at 18:15 | review | First posts | |||
| Jul 20, 2019 at 19:12 | |||||
| Jul 20, 2019 at 18:13 | history | answered | D. Zack Garza | CC BY-SA 4.0 |