# The shortest curve that divides the unit sphere into two regions of equal area

How to prove that if a simple closed curve $$\gamma$$ divides the unit sphere into two regions of same area, then its length is not smaller than $$2\pi$$. The hint is to consider a reflection of the sphere, then to show that must intersection with its reflection. But I don't know what to do.

• Can we assume that on a unit sphere, the shortest "distance" between two diametrically opposite points is $\pi$? – highgardener Sep 19 at 15:12

If $$\gamma$$ is the curve, then assume that $$(1)$$ there is $$x,\ -x\in \gamma$$. Clearly the length of $$\gamma$$ is greater than $$2\pi$$.

If not, then assume that $$\gamma$$ encloses a domain $$D$$, whose area is $$2\pi$$. When $$A$$ is an (isometric) antipodal map on the unit sphere, then $$A(D)$$ still has the area $$2\pi$$. When $$D$$ intersect $$A(D)$$, then $$A(\gamma),\ \gamma$$ have nonempty intersection. Hence $$D,\ A(D)$$ has empty intersection.

Hence unit sphere has area $$> 2{\rm area}\ D=4\pi$$, which is a contradiction. Hence we can assume $$(1)$$.

• Thank you for wonderful answer! – YinLu Sep 20 at 0:20

Suppose we have such a curve $$\gamma$$, with length smaller than $$2\pi$$, and denote with the same the set of points defining it. Let us denote the mutually disjoint sets of points of the sphere of equal area created by this curve by $$\sigma_1$$ and $$\sigma_2$$
Consider the reflection of all points of the sphere across the center. Let this transform $$\gamma$$ into $$\gamma'$$ and $$\sigma_1$$ into $$\sigma_1'$$. Note that if the curves $$\gamma$$ and $$\gamma'$$ intersect it would imply that $$\sigma_1$$ into $$\sigma_1'$$ are non-disjoint (and the areas bounded by $$\gamma$$ and $$\gamma'$$, between their points of intersection, will be in the intersection) and vice versa. So if we assume that $$\gamma$$ and $$\gamma'$$ do not intersect and thus $$\sigma_1$$ and $$\sigma_1'$$ are disjoint.
But then $$\sigma_1' \subseteq \sigma_2\cup\gamma$$ (since the latter is the complement of $$\sigma_1$$) and since these have the same area, $$\sigma_1'\setminus \gamma = \sigma_2$$. Thus $$\gamma$$ must be a great circle since it divides the sphere exactly into sets with diametrically opposite points. But then the length is not smaller than $$2\pi$$ so we reach a contradiction.

Thus $$\gamma$$ and $$\gamma'$$ must intersect, say they do so at point $$P$$, whose diametrically opposite point is $$P'$$ . Since $$P\in \gamma$$, we have $$P'\in\gamma'$$ and since $$P\in\gamma'$$, we have $$P'\in\gamma$$. Thus always in $$\gamma$$, there will be two "poles", diametrically opposite points, $$P$$ and $$P'$$. Thus, since $$\gamma$$ is a closed curve, we have two paths from $$P$$ to $$P'$$ with points in this curve, namely $$\Gamma$$ and $$\Gamma'$$, but the length of each of these must be $$\pi$$ or more. Thus, $$\gamma$$ must have length greater than or equal to $$2\pi$$.

• Thank you for wonderful answer! – YinLu Sep 20 at 0:21