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.