Timeline for Link between the hairy ball theorem and the fundamental theorem of algebra
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 3, 2016 at 15:15 | comment | added | RandomTopics | I did not see any proof using fundamental theorem of algebra among these. | |
| May 3, 2016 at 1:00 | comment | added | Gerry Myerson | Perhaps one should have a look through the 52 answers at mathoverflow.net/questions/10535/… (and maybe add a 53rd). | |
| May 2, 2016 at 23:01 | answer | added | Johannes Huisman | timeline score: 5 | |
| May 2, 2016 at 18:27 | comment | added | john mangual | don't directl answer your question but here mathoverflow.net/questions/234777/… and mathoverflow.net/questions/132036/… | |
| May 2, 2016 at 18:05 | answer | added | Josh Lackman | timeline score: 3 | |
| May 2, 2016 at 17:50 | comment | added | HJRW | You probably want the slightly more sophisticated version of the Hairy Ball theorem which says that the sum of the indices of the zeros equals 2. | |
| S May 2, 2016 at 17:11 | history | suggested | Amir Sagiv |
added relevant subject tags
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| May 2, 2016 at 16:59 | review | Suggested edits | |||
| S May 2, 2016 at 17:11 | |||||
| May 2, 2016 at 11:58 | comment | added | Alex Degtyarev | Both theorems amount to $\pi_1(S^1)=\mathbb{Z}$, so in this sense they are equivalent. | |
| May 2, 2016 at 11:53 | comment | added | Vincent | Right, I didn't read your comment properly. The 'point' of the one-point compactification is that all infinities ($+\infty, -\infty, i \infty$ etc) are the same point (often just denoted $\infty$) so that the extension to the sphere IS continuous (as Thomas also says). But this extension takes values in the sphere rather than $\mathbb{C}$ as, for instance, $f(\infty) = \infty$ | |
| May 2, 2016 at 11:49 | comment | added | Thomas Rot | non-constant Polynomials are proper hence extend to the one point compactification | |
| May 2, 2016 at 11:01 | comment | added | Vincent | I guess what helps is to ASSUME your function $f$ is non-constant and nowhere zero from the start (hoping to find a contradiction) and then work with the function $1/f$. But then we still are stuck with a continuous function from the sphere to $\mathbb{R}^2$ without a canonical way to identify this $\mathbb{R}^2$ with the tangent space to the sphere in each point. | |
| May 2, 2016 at 10:22 | comment | added | RandomTopics | I see the idea, but I am not sure that it works that easy. It is not obvious to me that we can extend a polynomial into a continuous function on its one-point compactification. For instance P(z) = z tends to +\infty when $z$ real tends to $+\infty$ or $-\infty$ if $z$ tends to $-\infty$, and thus the function on the compactified space would not be continuous, isn't it ? Also, take a constant polynomial, and the approach seems to be contradictory. maybe this is the right approach, but there are some technical difficulties that I do not manage to overcome. | |
| May 2, 2016 at 9:35 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Fixed a typo in the title; added top-level tag.
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| May 2, 2016 at 9:31 | review | First posts | |||
| May 2, 2016 at 9:35 | |||||
| May 2, 2016 at 9:29 | history | asked | RandomTopics | CC BY-SA 3.0 |