# Cutting a sphere in half

I’m at the Banff International Research Station this week for a conference on metric geometry.�� I’ve listened to several nice talks already but one that stood out for me was by Yevgeny Liokumovich on the problem of cutting a sphere in half.� (It had, of course, a more official title!)

Consider the sphere $$S^2$$ with some Riemannian metric, scaled so that the total area is 1.� Is there an upper bound to the length of a geodesic loop that divides the sphere into two disks of equal area?

It seems plausible at first that the answer might be “yes”, but in fact it is “no”.� To see the counterexample, think about balloon animals: specifically a “balloon starfish” that has three thin, cylindrical arms of length $$\ell$$ emanating from a central core.

Such a “starfish” can be constructed with area 1 and arbitrarily large $$\ell$$, and the length of the shortest geodesic loop that cuts it in half is of order $$\ell$$.

The main point of the talk was to obtain a more delicate bound for the equal-division problem which involves both the area and the diameter of the given metric.� But in the course of the proof there is a very pretty lemma that I had not seen before (but which apparently goes back to Gromov’s Filling Riemannian Manifolds).

Theorem Given any Riemannian metric on the 2-sphere $$X$$, with area $$A$$ and diameter $$d$$, and any $$\epsilon>0$$, there exists a simple closed curve of length at most $$2d+\epsilon$$ that divides the sphere into two disks both of which have area at least $${\frac13}A-\epsilon$$.

The proof goes by contradiction.� Suppose not.� Triangulate $$X$$ with geodesic simplices of side $$<\epsilon$$.� Fix a base-point $$p\in X$$.� Now consider a 2-simplex $$s$$ of the triangulation.� Build three arcs, each of length $$\le d$$, from the vertices of $$s$$ to $$p$$. The three loops, each formed by two of these arcs together with an edge of $$s$$, have length $$<2d+\epsilon$$; so, by assumption, each bounds a disk of area $$\le {\frac13}A-\epsilon$$.� The union of these three disks and the simplex $$s$$ has area less than $$A$$ so it misses a point, and thus lies in a contractible subset of $$X$$.� Thus we can find a map from the cone on $$s$$ to $$X$$ that fixes $$s$$ and maps the vertex of the cone to $$p$$.� Do this construction inductively over the simplices of $$X$$ to obtain a map from the cone $$cX \cong B^3$$ to $$X\cong S^2$$ which is the identity on the boundary, i.e. a retraction. This contradicts the Brouwer fixed-point theorem.