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If $ \beta: (a,b)\rightarrow \mathbb{S}^2 $ is a simple closed curve such that $ \int_a^b\Vert\beta^{\prime}(t) \Vert dt<2\pi$ then there is an open hemisphere (or any rotation of this) containing the image of $\beta $

thanks!

helmonio
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2 Answers2

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Here's a constructive proof that feels like it's more complicated than necessary, but I wasn't able to find an easier argument.

Assume without loss of generality that $\beta$ has constant speed $|\dot \beta| = v$ and parametrized on $S^1 = \mathbb R / 2 \pi \mathbb Z$, so that the length condition becomes $v < 1$. Use the usual unit sphere $\mathbb S^2 \subset \mathbb R^3$.

I will prove that the "centre of mass vector"

$$ X = \int_{\mathbb S^1} \gamma(t) dt \in \mathbb R^3$$ defines a hemisphere $\{ p \in \mathbb S^2 : p \cdot X > 0 \}$ containing $\gamma$. (The integral here is to be understood component-wise in $\mathbb R^3$.)

This involves showing that for every $s$, $$\gamma(s) \cdot X = \int_{s-\pi}^{s+\pi} \gamma(s) \cdot \gamma(t) dt > 0.$$

Since $\gamma$ has speed $v$, $\gamma(t)$ makes an angle of at most $v |t-s|$ with $\gamma(s)$ and so $\gamma(s)\cdot \gamma(t) \ge \cos(v |t-s|)$. Thus

$$ \gamma(s) \cdot X \ge \int_{-\pi}^{\pi} \cos(vt)dt = \frac1v \int_{-v\pi}^{v\pi} \cos(z) dz = \frac2v \sin(v \pi) > 0$$

since $0 < v < 1$.

Anthony Carapetis
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    @helmonio: I had a small typo, it should be $v|s-t|$. Imagine $\gamma(s)$ is the north pole - then at $\gamma(t)$, time $|s-t|$ has passed while travelling at speed $v$, and thus $\gamma$ cannot have traveled a distance of more than $v|s-t|$ - i.e. it cannot be south of the line of latitude that makes angle $v|s-t|$ with the north pole. – Anthony Carapetis Jun 11 '14 at 10:05
  • making modifications to the arc length function can obtain a reparametrization with speed fixed. But it also ensures that domain parameterization has $S^1$,would appreciate me explain this last – Tirifilo Jun 12 '14 at 23:25
  • @Anthony Carapetis: also I have this doubt too :) – helmonio Jun 13 '14 at 02:13
  • @Tirifilo: once you have a constant-speed parametrization then you can easily reparametrize to $(0,2 \pi)$ via a scaling/translation on the parameter domain, which keeps the speed constant. You then have a closed curve $(0,2 \pi) \to \mathbb S^2$ which is the same thing as a regular map $\mathbb S^1 \to \mathbb S^2$. – Anthony Carapetis Jun 13 '14 at 03:04
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R.A. Horn gave an elementary proof in his article On Fenchel's theorem (American Mathematical Monthly, 78, 1971, 380-381). The argument goes as follow:

  • Let $\Gamma$ be a curve on $\mathbb{S}^2 \subset \mathbb{R}^3$ of length $<2 \pi$. Let $P,Q \in \Gamma$ cutting $\Gamma$ in two pieces $\Gamma_1, \Gamma_2$ of equal length, ie. $$\mathrm{lg}(\Gamma_1)= \mathrm{lg}(\Gamma_2)= \frac{1}{2} \mathrm{lg}(\Gamma).$$ Thanks to a rotation, we may suppose without loss of generality that $P$ and $Q$ are symmetric with respect to the North pole $N$.

  • Suppose that $\Gamma_1$ intersects the equator through a point $R$; let $R'$ and $\Gamma_2'$ denote respectively the images of $R$ and $\Gamma_1$ through the symmetry with respect to $N$. Let $\Gamma':= \Gamma_1 \cup \Gamma_2'$. Notice that $$\mathrm{lg}(\Gamma')= \mathrm{lg}(\Gamma).$$

  • Because $\Gamma'$ contains two arcs joining the antipodal points $R$ and $R'$, we deduce $$\mathrm{lg}(\Gamma)= \mathrm{lg}(\Gamma') \geq 2d(R,R')=2 \pi.$$

  • Therefore, $\Gamma_1$ cannot intersect the equator; of course, the same thing happens for $\Gamma_2$. Finally, we deduce that $\Gamma$ is contained in the upper hemisphere (after the previous rotation).

Seirios
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