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If $\alpha(s)$ is a simple closed regular plane curve, the tangent circular image $t: [0, L] \to S^1$ is onto.

Here, $\alpha$ is a curve with unit speed, and $L$ is the period. We have that, by the rotation index theorem, that the rotation index of a simple closed plane curve $\alpha$ is $\pm 1$. We may assume that the index is $1$ and that $\alpha (0)$ is minimum. Here, we use the rotation index defined as $\frac{\theta(L)-\theta(0)}{2 \pi}$ where $\theta(s) := \theta_0 + \int_0^s x'(s)y''(s)-x''(s)y'(s) ds$ where $\theta_0$ is such that $\cos \theta_0 = x(0)$ and $t(s) = (x'(s), y'(s))$ is the standard unit tangent curve.

I am not exactly sure what I would do here. Certainly, this seems to be true, in a sense that the tangent curve is related to $\theta(s)$ in such a way that it rotates completely counterclockwise. Is there a way trigorously prove this? This post mentions the Jordan-Schonflies theorem, but I am specifically looking for a theorem that uses the rotation index theorem.

Squirrel-Power
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  • Nice problem, *+1*; but what does $\theta_0 = \arccos \alpha'(0)$ mean? Since $\alpha(t)$ is a vector-valued function, how do we take $\arccos \alpha(t)$? Cheers! – Robert Lewis Oct 17 '18 at 04:23

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Since the rotation index is $+1$ (or $-1$), the difference $\theta(L)-\theta(0)$ is equal to $2\pi$ (or $-2\pi$). By the intermediate value theorem, for every angle $\theta$ between $\theta(0)$ and $\theta(L)=\theta(0)+2\pi$, there is an $s\in[0,L]$ such that $\theta(s)=\theta$. Note that $\theta(s)$ is the angle of the tangent vector with a fixed direction (usualy the positive $x$-axis). Hence, the map $t$ is onto.

Remark: the result and the same proof hold for all closed curves with non-zero rotation index. The result doesn't hold for closed curves with rotation index zero. The lemniscate of Bernouili gives a counterexample.

Ernie060
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