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Let $\gamma:[a,b]\to \mathbb{R^2}-\{0\}$ be a closed curve of class $C^1$. Define the 1-form $d\theta$ in $\mathbb{R^2}-\{0\}$ as $d\theta=\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$ and the winding number of $\gamma$ to be $n(\gamma)=\frac{1}{2\pi}\int_\gamma d\theta$. It is a standard result that $n(\gamma)$ is always an integer number.

The question is: if $\gamma:[a,b]\to \mathbb{R^2}-\{0\}$ is a simple (i.e. nonintersecting) closed curve of class $C^1$. Is it always the case that $n(\gamma)=-1$, $0$, or $1$? It intuitively looks to be true, however I can't prove it or find a counterexample.

Zero
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  • It is more subtle than I realized and had to do with Jordan curve theorem. So,intuitive, but not that easy to show. Great question though. – orangeskid May 10 '18 at 21:05

2 Answers2

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Yes, it's true, and you can prove it by applying Green's Theorem. Can you figure out how? :)

Ted Shifrin
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HINT:

A closed curve can be written as $$\gamma(t) = r(t) e^{i \theta(t)}$$ where $r\colon [0,1]\to (0, \infty)$, $r(0)= r(1)$ and $\theta\colon [0,1]\to \mathbb{R}$, $\theta(1) = 2\pi k$, where $k$ is the winding number of $\gamma$. Assume that $k>1$. We must show that there exist $0\le t_1< t_2\le 1$ so that $r(t_1) = r(t_2)$ and $\theta(t_2) - \theta(t_1) = 2\pi$. Hm... this looks like a similar problem that I have seen on this site. I leave it here for now.

orangeskid
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  • Could you elaborate please? I'm trying to use intermediate value theorem but it's not working. – Zero Sep 12 '17 at 18:09
  • I think this question is relevant https://math.stackexchange.com/questions/2424924/a-question-about-continuous-curves-in-mathbbr2/2424954#2424954 – orangeskid Sep 12 '17 at 18:37