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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.

  • 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.

  • 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.)

  • In $S^3$ the similar assertion is false, moreover the minimal area of the boundary of a half-volume set can be arbitrary large.

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Alex Nabutovsky is the person to ask. –  Misha May 5 '13 at 20:47
    
Can elaborate why you formulated the question in terms of "area between pi and 2pi" rather than in terms of the Cheeger constant, and how this might affect the answer? –  katz May 12 '13 at 13:46
    
@katz: This is just an example of statement that might be true if dividing in half fails. It might as well be an integral inequality on the isoperimetric profile. In other words, I am more interested in a affirmative answer to a slightly different question than in a counter-example to the precise one. There is nothing magic about the constant $\pi$, I just want to avoid the trivial answer "look at a small neighborhood of a point where curvature is greater than 1". –  Sergei Ivanov May 12 '13 at 19:01
    
Just to make sure I understand what you are looking for: it seems plausible that by applying coarea to a suitable distance function one might be able to get a non-optimal bound similar to the one you asked for. Are you interested in the optimal value $2\pi$ for boundary length, or does the coarea argument get stuck on diameter issues? –  katz May 13 '13 at 7:53
    
The question came by association from a totally different problem. I am hoping that there might be a kind of rigidity result where one concludes that the metric is round (or not far from round) by looking at some rough measures of isoperimetric profile. If there is one, I could try to apply the technique of the proof in another context. –  Sergei Ivanov May 13 '13 at 10:01

3 Answers 3

I don't know if you already thought of this, but your question is related to a result of Balacheff and Sabourau (an electronic version of their paper is in Stéphane's web page). Without going into the technical definition of the diastole over $1$-cycles, their result basically says that there exists a constant $C$ (which, unfortunately, is very large) so that given a Riemannian metric on the sphere with area $4\pi$, there exists a Morse function such that the length of any of its level curves is less than $C$. This is the best that is known in this direction.

On the other hand, a related conjecture which has perhaps a shot at being true runs as follows:

Conjecture. If the area of a Riemannian two-sphere is $4\pi$, there exists a closed geodesic that is regular homotopic to a figure 8 and whose length is less than or equal to $4\pi$.

Note that twice an equator in a round sphere is regular homotopic to a figure 8.

In fact, the conjecture should be for all (reversible and non-reversible) Finsler metrics and equality should hold only for Zoll metrics. The reason is that this conjecture is a consequence of (a reasonable extension of) the Viterbo conjecture. Enough is known about this conjecture (Artstein-Avidan, Ostrover, Milman, Alvarez Paiva and Balacheff: see the first version of Contact geometry and isosystolic inequalities and the references therein) that an optimist may harbor some hope of its validity.

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I don't get the first statement in italic, as for small $C$ it precisely violate any diastolic inequality... –  Benoît Kloeckner Sep 11 '13 at 19:38
    
@Benoit: sorry. I meant "unfortunately very large"! Just edited. –  alvarezpaiva Sep 11 '13 at 19:45

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:

  1. 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.

  2. $length((f \circ g) ^{-1} (x)) < 26 \sqrt{Area(M)}$ 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.

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that's very neat !! Please send me the preprint when it's ready. –  alvarezpaiva Sep 24 '13 at 8:25
    
thank you! yes, I will send you the preprint when it's ready. –  Yevgeny Liokumovich Sep 24 '13 at 15:39

Papasoglu's 2009 paper on Cheeger constants shows that there is a closed loop of length at most $2\sqrt{3}+\epsilon$ on a sphere of area 1 (see second displayed formula on page 5146). Thus the tripod counterexample you mentioned is not really a counterexample unless you are looking for closed geodesics. This indicates that your result holds in a weaker "coarse" sense (without the optimal constant).

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Papasoglu estimates the Cheeger constant, which does not control the area bounded be the loop, only the length/area ratio. It can happen that the loop bounds a small area and is very short itself, isn't it? –  Sergei Ivanov Sep 11 '13 at 21:42
    
His paper focuses on the Cheeger constant (as the title suggests) but the result I mentioned on page 5146 works when complementary regions have area at least 1/4 of the total area. If you think otherwise please let me know so I can correct my understanding of his paper. –  katz Sep 12 '13 at 13:13

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