Timeline for What is the generator of $\pi_9(S^2)$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2020 at 12:25 | vote | accept | Alex Gavrilov | ||
| May 29, 2019 at 10:16 | comment | added | მამუკა ჯიბლაძე | The question you link to now has two fresh very relevant answers | |
| May 29, 2019 at 5:44 | answer | added | Ian Agol | timeline score: 8 | |
| May 28, 2019 at 17:36 | answer | added | Charles Rezk | timeline score: 12 | |
| May 28, 2019 at 10:50 | history | edited | Alex Gavrilov | CC BY-SA 4.0 |
added 847 characters in body
|
| May 28, 2019 at 2:31 | comment | added | Ryan Budney | I think people are mis-understanding the question. You might need to add some details of what an answer should look like, Alex. | |
| May 28, 2019 at 1:38 | comment | added | skd | The Hopf fibration implies that $\pi_9(S^2) = \pi_9(S^3)$, so the map is just the composite $S^9 \to S^3 \xrightarrow{\eta} S^2$, where the first map is the generator of $\pi_9(S^3) \cong \mathbf{Z}/3$. | |
| May 27, 2019 at 22:50 | comment | added | მამუკა ჯიბლაძე | @RyanBudney Well I would say at least the first isomorphism is quite straightforward; the problem is, I believe, in locating a nontrivial element. Essentially it is the Moore normalization of the free simplicial group $F(S^1)\simeq\Omega S^2$ on the minimal simplicial circle $S^1$ (with a single 1-simplex). $\pi_9(S^2)\cong\pi_8(F(S^1))$ consists of simplices $\sigma\in F(S^1)$ with $d_0(\sigma)=d_1(\sigma)=...=d_8(\sigma)=1$, modulo $d_9(\tau)$ for simplices $\tau$ with $d_0(\tau)=d_1(\tau)=...=d_8(\tau)=1$. So any such $\sigma$ in principle encodes an explicit map from $S^8$ to $\Omega S^2$ | |
| May 27, 2019 at 18:01 | comment | added | Denis T | It's pretty complicated in terms of Wu's isomorphism as well; for example, there's a number of easy-to-obtain elements corresponding to iteration of nontrivial stable element in $\pi^s_1$, which can be obtained as $\alpha_1 = [x_1, x_2]$ = Hopf map, $\alpha_2 = [[x_1, x_2], [x_1, x_3]] = \pi_4(S^2)$ and so on; as someone may know, fifth iterate $S^7 \to S^2$ becomes trivial, and you have to find less obvious elements in the centre of $G(n)$ to represent generator of $\pi_9$. | |
| May 27, 2019 at 17:53 | comment | added | Ryan Budney | As far as I know the isomorphism Jie Wu is using is rather indirect, making it difficult to see what precisely the map $S^9 \to S^2$ is. | |
| May 27, 2019 at 15:41 | history | edited | Arun Debray |
added some tags
|
|
| May 27, 2019 at 15:38 | comment | added | მამუკა ჯიბლაძე | You might find some algebraic descriptions in the work of Jie Wu. For example, facts listed on his home page imply that $\pi_9(S^2)$ is the center of the group $G(n)$ with generators $x_1$, ..., $x_9$ and relations $x_1\cdots x_9=1$ and moreover all commutators involving every generator trivial. Or also, $\pi_9(S^2)$ is the $9$th homology group of the group of Brunnian braids. | |
| May 27, 2019 at 14:55 | history | asked | Alex Gavrilov | CC BY-SA 4.0 |