Timeline for Which homotopy classes $S^3 \to S^2$ lift to embeddings $S^3 \to S^2 \times D^3$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2021 at 20:44 | answer | added | Ryan Budney | timeline score: 2 | |
| Jun 3, 2021 at 7:49 | comment | added | Mark Grant | @NeilStrickland Aha! So $h$ itself is not $3$-prem. I'm not quite seeing how to rule out that some $h'\simeq h$ is $3$-prem. | |
| Jun 2, 2021 at 21:15 | comment | added | Neil Strickland | @MarkGrant in particular, as $h(v)=h(-v)$ we see that the formula $k(v)=(e(v)-e(-v))/\|e(v)-e(-v)\|$ gives an antipodal map $S^3\to S^2$, which is impossible. | |
| Jun 2, 2021 at 20:10 | comment | added | Mark Grant | There's a Haefliger-style obstruction: Suppose the Hopf map $h:S^3\to S^2$ is the projection of an embedding $(h,e):S^3\to S^2\times\mathbb{R}^3$. (By the way, the literature might say that in this case $h$ is 3-prem.) Let $\Delta(h)=\{(v,w)\in S^3 \mid v\neq w, h(v)=h(w)\}$. Then there is a $\mathbb{Z}/2$-equivariant map $\Delta(h)\to S^2$, where $\mathbb{Z}/2$ acts on $S^2$ antipodally and on $\Delta(h)$ by swapping factors, given by $(v,w)\mapsto (e(v)-e(w))/\|e(v)-e(w)\|$. | |
| Jun 2, 2021 at 19:00 | answer | added | Neil Strickland | timeline score: 9 | |
| Jun 2, 2021 at 18:34 | history | edited | Ryan Budney | CC BY-SA 4.0 |
add a comment on what I suspect the answer to be.
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| Jun 2, 2021 at 17:55 | comment | added | Ryan Budney | @StevenStadnicki: Only constant maps in that case. The $2^{nd}$ homotopy group of $S^1$ is trivial, so there is only the one homotopy-class available. | |
| Jun 2, 2021 at 17:45 | comment | added | Steven Stadnicki | What happens if you go down a dimension and consider embeddings $S^2\mapsto S^1\times D^2$? It seems like the answer there is only the constant maps, but there might be something I'm missing... | |
| Jun 2, 2021 at 17:34 | history | asked | Ryan Budney | CC BY-SA 4.0 |