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I am trying to prove that if $\gamma(t)=(x(t),y(t))$ ,a function from the closed interval $[0,1]$ to $\mathbb{R^2}$ is a simple closed unit speed curve such that $\gamma '(0)=\gamma '(1)$. Then the tangent vector rotates exactly by $2 \pi$.

My Try: I think I have a proof I am not sure though.I am trying to use the idea of covering spaces to prove this. So first I observe that considering the tangent vector to be $x'(t)+iy'(t)$, we get a map from $[0,1]$ to $S^1$ and by our assumption it is a closed loop.

Now I use $\mathbb{R} $ to cover $S^1$ and the covering map is $t \rightarrow (\cos (2 \pi t), \sin (2 \pi t))$. Now I lift the path $\gamma '(t)$ to $\widetilde{\gamma '(t)}$.

Now we observe that a circle $\alpha(t)$ which passes through the point $\gamma (0)$ has the property that the lift $\widetilde{\alpha '(t)}$ has its endpoint at $1$.

As suggested by Andrey Ryabichev we observe that one of the two pieces obtained because of simple closed regular curve is diffeomorphic to the disc.Now we know that $\gamma(t)$ and $\alpha (t)$ are homotopic by a differentiable homotopy $H(s,t)$. Now I differentiate this map to get a homotopy from $\widetilde{\alpha '(t)}$ to $\widetilde{\gamma '(t)}$ . This means they both should have the same endpoint. Therefore we are done. Is this proof correct?

happymath
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    It is unclear how the essential assumption that $\gamma$ is simple has been used. – Christian Blatter Aug 10 '15 at 08:43
  • @ChristianBlatter Thanks for pointing that out I had actually forgotten that. – happymath Aug 10 '15 at 08:54
  • use that one of two pieces, on which $\gamma$ cuts $\mathbb R^2$, will be diffeomorphic to a disc $D^2$. – Andrey Ryabichev Aug 10 '15 at 10:16
  • A while ago, I asked a very similar question. I got no answer, but then I found the proof in a text book, and answered my own question by writing where the proof can be found. Have a look. http://math.stackexchange.com/questions/1031635/turning-number-vs-winding-number – Amitai Yuval Aug 10 '15 at 17:50
  • @AndreyRyabichev I tired to use your idea to complete the proof So is it correct now? – happymath Aug 11 '15 at 05:33
  • @Cristian Blatter I have included where I have used the hypotheis that it is simple, so is the proof correct now? – happymath Aug 11 '15 at 05:34
  • @happymath, as i understand that is written, no. you use only existence the homotopy $H(s,t)$, but for some parameter $s_0$ the curve $H(s_0,t)$ may have zero derivative in some points, because this homotopy (without any conditions) actually can change rotation number. – Andrey Ryabichev Aug 11 '15 at 05:47
  • @AndreyRyabichev Yes I see you are right but can you please suggest a method to fix it – happymath Aug 11 '15 at 08:00

1 Answers1

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Here is my proof, that doesn't use covering circle by line at all.

By the Jordan–Schönflies theorem, there exists an isotopy $F_t:\mathbb R^2\to \mathbb R^2, t\in [0,1]$ of $F_0=\mathrm{Id}_{\mathbb R^2}$, such that $F_1(\gamma)$ is the unit circle (because $\gamma$ is simple). This isotopy doesn't change the fact $\gamma'\ne0$, so it preserves a rotation number.

Andrey Ryabichev
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  • Can you please add why rotation number is preserved – happymath Aug 11 '15 at 08:48
  • @happymath, since $F_t$ is diffeomorphism for all $t$, a derivative of the map $F_t(\gamma(,\cdot,))$ is nonzero at all the points. so at each value of $t$ rotation number is correctly determined. therefore we can consider the function $[0,1]\to\mathbb Z$, $t\mapsto{\text{rotation number at the moment } t}$, as this function is continuous, it is constant. – Andrey Ryabichev Aug 11 '15 at 08:56
  • Thanks a lot for the answer – happymath Aug 12 '15 at 04:04