# A comparison question for non-positively curved disks

Let $A$ and $B$ be two closed, 2-dimensional, non-positively-curved Riemannian disks (not necessarily with convex boundary). Suppose that their boundaries $\partial A$ and $\partial B$ have the same length, and identify $\partial A$ with $\partial B$ with a length-preserving diffeomorphism of circles. Suppose also that for every pair of points $p,q \in \partial A = \partial B$, there is a distance inequality $d_A(p,q) \le d_B(p,q)$. Does it follow that the area of $A$ is less than or equal to the area of $B$, with equality only when $A$ and $B$ are isometric?

This is a variation of a question that arose in a joint paper that I am trying to finish. Another version which is more relevant to the paper is about $\mathrm{CAT}(0)$ disks which are tiled by Euclidean unit equilateral triangles.

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The answer is yes. Moreover you don't need to assume that $B$ is nonpostively curved. (And, if you are not interested in the equality case or can afford a convex boundary, the nonpositive curvature of $A$ can be replaced by a weaker assumption that the geodesics in $A$ have no conjugate points).

This follows from a filling inequality that I have proved some time ago (and quite proud of it) and boundary rigidity results of Croke, Otal, or Pestov-Uhlmann.

Inequality. There are two papers covering it:

1) S.Ivanov, On two-dimensional minimal fillings. Algebra i Analiz 13 (2001), no. 1, 26-38 (Russian); English translation in St. Petersburg Math. J., 13 (2002), no.1, 17-25. A preprint is here. In this paper the inequality $area(A)\le area(B)$ is proved under the assumptions that $A$ is free of conjugate points and has convex boundary (sorry;)) and for arbitrary $B$ such that $d_A(p,q)\le d_B(p,q)$ for all $p,q$ on the boundary.

2) S.Ivanov, Filling minimality of Finslerian 2-discs. arXiv:0910.2257, to appear in Trudy Mat. Inst. Steklov (= Proc. Steklov Inst. Math). In this paper the same thing is proved for Finslerian metrics $A$ and $B$ and without convex boundary assumptins. This is the maximum generality I can imagine. But the paper is harder to digest if you don't like Finsler metrics.

The equality of areas implies that the boundary distances of $A$ and $B$ coincide, and then you can apply boundary rigidity results mentioned below.

Rigidity. The case you are asking for is covered by C.Croke, Rigidity for surfaces of nonpositive curvature, Comment. Math. Helv. 65 (1990), no. 1, 150–169. He proves that if $A$ is a nonpositively curved and $B$ is arbitrary Riemannian such that $d_A(p,q)=d_B(p,q)$ for all $p,q$ on the boundary, then $A$ and $B$ are isometric. He carefully works through the details of non-convex boundaries.

I have not read J.-P. Otal's paper mentioned by Igor Rivin, but from mathscinet review it seems that he assumes that both $A$ and $B$ are strictly negatively curved.

For the same result without curvature assumptions but with convex boundary, see L.Pestov and G.Uhlmann, Two dimensional compact simple Riemannian manifolds are boundary distance rigid. Ann. of Math. (2) 161 (2005), no. 2, 1093–1110.

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You could check a recent paper by Burago and Ivanov, arXiv:1011.1570: Area minimizers and boundary rigidity of almost hyperbolic metrics. They prove the statement you want when $A$ is close to being hyperbolic (but in any dimension). See their Thm 1.6. They proved earlier the same statement if A is close to being Euclidean. Actually they prove a stronger rigidity statement: if the boundary distances are equal then the interior metrics are equal, not only the areas. (Somehow this stronger rigidity statement follows from the strict minimality of the area.)

Their proof is quite involved, but in the 2d case it's possible that some elementary arguments can work, for instance using the Santalo formula which they recall in their introduction.

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Your answer is so close to the mark that I suppose that I should accept it (or rather the paper that you cite) as the state of the art. In any case these authors are surely good people to ask. – Greg Kuperberg Jul 1 '11 at 19:43
The 2-dimensional case is indeed easier; it is completely solved by now - see my answer. – Sergei Ivanov Jul 2 '11 at 1:12

For your equality question, look at

MR1057248 (91i:53054) Otal, Jean-Pierre(F-PARIS11) Sur les longueurs des géodésiques d'une métrique à courbure négative dans le disque. (French) [On the lengths of the geodesics of a metric of negative curvature in the disk] Comment. Math. Helv. 65 (1990), no. 2, 334–347.

which is a beautifully written paper, and it is quite likely that the Crofton formula which he uses there can be used to prove the sort of inequality you want. I am pretty sure that for any such result, the curvature should be strictly negative, but I could be wrong.

There are many related results proved by Chris Croke (so you might want to ask him whether what you conjecture is true), and for convex disk by Gunther Uhlmann et al (but those results, which are heavily analytic, are probably not useful to you).

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I downloaded the paper by Otal and I am still reading it. But let me explain that when I said "with equality", I meant as the hypothesis that the areas are equal, not the distances. Of course if the disks are isometric, then a posteriori the distances are equal as well. – Greg Kuperberg Feb 18 '11 at 20:46
@Greg: yes, I understood the question. It is possible that under your hypotheses, the contraction on the boundary extends to a contraction on the disks (this would be a really nice result, if true...) – Igor Rivin Feb 18 '11 at 23:02
Anyway, to make things clear for any other readers here, Otal shows that if two negatively curved disks have equal distances between points on the boundary, then they are isometric. – Greg Kuperberg Feb 19 '11 at 9:01