Consider a two-dimensional sphere with a Riemannian metric of total area $4\pi$. Does there exist a subset whose area equals $2\pi$ and whose boundary has length no greater than $2\pi$?
(To avoid technicalities, let's require that the boundary is a 1-dimensional smooth submanifold.)
If that fails, does there exists a set, say, with area between $\pi$ and $2\pi$ and length of the boundary no greater than that of the spherical cap of the same area? Or at least no greater than $2\pi$?
More generally, I am interested in any results saying that the isoperimetric profile of the round metric on the sphere is maximal in some sense (among all Riemannian metrics of the same area).
Notes.
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The answer is affirmative for central symmetric metrics (i.e. if the metric admits an $\mathbb{RP}^2$ quotient). This follows from Pu's isosystolic inequality: in $\mathbb{RP}^2$ with a metric of area $2\pi$ there exists a non-contractible loop of length at most $\pi$. The lift to the sphere is a loop of length at most $2\pi$ dividing the area in two equal parts.
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One should not require that the set is bounded by a single loop. A counter-example is the surface of a small neighborhood of a tripod (formed by three long segments starting from one point) in $\mathbb R^3$. Here one can divide the area in half by two short loops, but one loop would be long. (However one can cut off 1/3 of the area by one short loop.)
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In $S^3$ the similar assertion is false, moreover the minimal area of the boundary of a half-volume set can be arbitrary large.
Best Answer
The result of F.Balacheff and S.Sabourau mentioned by Alvarez Paiva can be improved as follows:
For any Riemannian 2-sphere M there exists a tree T with vertices of degree at most 3 and continuous maps $f:M \rightarrow T$ and $g: T \rightarrow \mathbb{R}$, such that the following holds:
Preimage of an interior point $x \in T$ under $f$ is a simple closed curve; preimage of a vertex of degree 3 is a figure eight curve.
$length((f \circ g) ^{-1} (x)) < 26 \sqrt{Area(M)}$ for all $x \in \mathbb{R}$.
In particular, the sphere can be subdivided into 2 regions each of area $\geq \frac{Area(M)}{3}$ by a simple closed curve or a figure eight curve of length $< 26 \sqrt{Area(M)}$. It can be subdivided into 2 equal parts by a collection of curves of total length $< 26 \sqrt{Area(M)}$, but then there is no control over the number of curves.
The proof of this result is not written yet, but I plan to include it in my PhD thesis. The reason for constant 26 is the following: we obtain the above sweep-out by successively dividing regions of the sphere by short arcs of length $\leq 2 \sqrt{3} \sqrt{Area(A)} +\epsilon$. $A$ denotes the region and constant $2 \sqrt{3}$ comes from the paper of P.Papasoglu. Every subdivision produces new regions of area $\leq \frac{3}{4} Area(A)$. We proceed until the total area becomes small compared to injectivity radius of $M$. Then we use Besicovitch inequality to bound lengths of curves in the sweep-out of this region in terms of its boundary length. In the process of subdivision we accumulated $2 \sqrt{3} \sum_{i=0} ^{\infty} (\frac{\sqrt{3}}{2})^i \sqrt{Area(M)}+\epsilon< 26 \sqrt{Area(M)}$.
Although $26$ is better than $10^8$ in F.Balacheff and S.Sabourau's paper it seems far from the optimal and at the moment I do not see how these methods can be used to produce a much better bound. A somewhat different problem about the relationship between isoperimetric profile and the diameter is studied in Surfaces of small diameter with large width. It is shown that one can not bound the length of a cycle subdividing the surface into two parts of equal area in terms of diameter, but if it is allowed for one of the parts to be slightly smaller such a bound exists.