## The Spherical Isoperimetric Inequality

Posted in Math by pgadey on 2013/03/28

Here is an application of the spherical isoperimetric inequality.

Fact You can’t cut up a beach ball into equal parts with a path that is too short.

Let’s take a closed compact surface ${S}$ with a Riemannian metric, which induces an inner metric ${d}$. Let ${D(S) = \sup_{x,y \in S} d(x,y)}$ be the inner diameter and ${L(S) = \inf_{\gamma} \{ Length(\gamma) : S \setminus \gamma = A \sqcup B, |A| = |B| \}}$, where ${\gamma : S^1 \rightarrow S}$ is a closed loop. The loops in the infimum are bisecting curves which split the sphere into two regions of equal area.

Theorem 1 The round unit sphere ${S^2_\circ}$ has ${L/D = 2}$.

The round unit sphere is nothing other than ${\{(x,y,z) \in {\mathbb R}^3 : x^2 + y^2 + z^2 = 1 \}}$. The equator ${\gamma = \{(x,y,0) : x^2 + y^2 = 1\}}$ is clearly a bisecting curve and ${Length(\gamma) = 2\pi}$. The sphere has diameter ${D(S^2_\circ) = \pi}$. This shows that ${L/D \leq 2}$. Now we need to show that we hit the equality. The spherical isoperimetric inequality says:

Theorem 2 For a simple smooth curve ${\gamma}$ on ${S^2_\circ}$ let ${Area(\gamma)}$ be the smaller of the two regions it encloses and let ${Length(\gamma)}$ be its length. We have:

$\displaystyle [Length(\gamma)]^2 \geq Area(\gamma)(4\pi - Area(\gamma))$

Suppose that ${\gamma}$ divides the sphere into two regions of equal area. We have

$\displaystyle [Length(\gamma)]^2 \geq 2\pi (4\pi - 2\pi) = (2\pi)^2$

and thus ${Length(\gamma) \geq 2\pi}$. This inequality says that if we want to enclose half the area of the sphere then we can’t do better than using ${2\pi}$ of length. That means that the equator was optimal and thus ${L/D = 2}$.

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