Parker Glynn-Adey

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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  1. […] a way to a simple but important case of the waist inequality which I’ve mentioned before. The classical isoperimetric inequality on the sphere says: Any curve of length on the sphere […]


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