Timeline for Non-zero winding number on a space curve implies a linked curve in the zero set?
Current License: CC BY-SA 4.0
8 events
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| Apr 19, 2019 at 15:42 | history | edited | Dmitri Panov | CC BY-SA 4.0 |
added 159 characters in body
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| Apr 19, 2019 at 13:39 | comment | added | Hao Chen | Wait ... Then there is a simple fix to my question. Just let $f$ be a map from $S^3$ to $R^2$. So I panicked too much facing @Wojowu's comment. I should just compacting the source of $f$, not its target. I'll update the question and acknowledge you. | |
| Apr 19, 2019 at 13:16 | comment | added | Dmitri Panov | I agree, this modification works | |
| Apr 19, 2019 at 12:32 | comment | added | Hao Chen | Here is another attempt, which I think should work. If it is, I'll open another question asking for reference. Let $f$ be a continuous map from 3-ball $\mathbb{B}^3$ to $\mathbb{R}^2$, and $C$ be a closed curve in $\partial \mathbb{B}^3$. If $f(C)$ has a non-zero winding number around $0$, then the degree theory tell me that $f^{-1}(0)$ has a non-empty intersection with any disk in $\mathbb{B}^3$ bounded by $C$. If $f$ is moreover real analytic, then $f^{-1}(0)$ contains a path-connected curve that intersects every disk bounded by $C$ (is there a word for this situation?) | |
| Apr 19, 2019 at 9:13 | comment | added | Dmitri Panov | Concerning your first question I am just saying that $\pi_3(S^2)\cong \mathbb Z$ and if we have a map $\phi: S^3\to S^2$ whose class is equal $n\in \mathbb \pi_3(S^2)\cong \mathbb Z$ then the preimages of two generic points in $S^2$ are links in $S^3$ that have linking number $n$. Concerning your second question, changing $\mathbb S^n$ to $\mathbb RP^n$ will not make any difference, I don't see how this can be fixed | |
| Apr 19, 2019 at 7:59 | comment | added | Hao Chen | Then, can it be fixed if I work not in $\mathbb{S}^n$, but in the projective space $\mathbb{R}P^n$? | |
| Apr 19, 2019 at 6:15 | comment | added | Hao Chen | Thanks! Do we have a linked curve in $f^{-1}(0) \cap f^{-1}(\infty)$? Although this is not what I am intending to. | |
| Apr 18, 2019 at 22:41 | history | answered | Dmitri Panov | CC BY-SA 4.0 |