From Alexandrov's work we know that any metric on the sphere with lower curvature bound $\kappa$ (in the sense of Alexandrov) can be realized as a closed convex surface (i.e. boundary of a compact convex domain) in the $3$-space form of constant curvature $\kappa$.

For $\kappa=0$, the surface is unique up to isometry of the surrounding space $\mathbb{R}^3$ due to Pogorelov. Is the same true for the elliptic and hyperbolic case? In other words, are two closed convex surfaces in these spaces congruent when they are isometric with respect to their inner metrics?

I asked this question here on Math SE, but did not get an answer.


1 Answer 1


Yes, closed convex surfaces in spaces of constant curvature are rigid. See Chapter V of Pogorelov's book Extrinsic Geometry of Convex Surfaces.

As described at the beginning of the chapter on p. 270-271, the main idea is to reduce the problem to the Euclidean case. Pogorelov says that this reduction is similar for elliptic and hyperbolic cases, and only considers the elliptic case, which is established in Thm. 1 on p. 321.

  • $\begingroup$ Thank you for your answer. I am not sure how easy it is to transfer the idea of the proof from elliptic case to hyperbolic case. For the hyperbolic case, a proof can be found in the paper "Univalent determination of general closed convex surfaces in Lobachevskij spaces" by Anatoliy Milka. However, the paper is in Russian and I was not able to find it anywhere. $\endgroup$
    – Pete
    Commented Jul 26, 2023 at 15:44
  • $\begingroup$ The paper by A.Milka that you cite was published in Volume 23 of "Ukrainian geometric collection". This volume can be found here: geometry.karazin.ua/en/ukraiinsky-geometrychny-zbirnyk.html $\endgroup$ Commented yesterday

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