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Let us call a cap the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$.

QUESTION. Given a metric on a closed 2-dimensional disk which has non-negative curvature in the sense of Alexandrov. Can the disk be isometrically imbedded into $\mathbb{R}^3$ as a cap?

A reference would be helpful.

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    $\begingroup$ If you replace the convex cap with a polyhedral cap, and replace the given metric by a polyhedral metric, then I believe Alexandrov's results in Convex Polyhedra answers Yes. Perhaps your result can be obtained as a limit of polyhedral caps? $\endgroup$
    – Joseph O'Rourke
    Commented Dec 2, 2021 at 18:46

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Take doubling of the disc, we obtain a metric on the sphere. By Perelman's theorem it had nonnegative curvature in the sense of Alexandrov. Therefore, by Alexandrov's theorem, it is isometric to a convex surface in the Euclidean space. This convex surface is unique up to congruence (Pogorelov's theorem). Therefore, the involution of our sphere extends to a reflection of the space. In particular, the boundary of the disc lies in one plane.

Postscript. You may also proceed as suggested by Joseph O'Rourke --- approximate the disc by polyhedral space, apply Alexandrov's theorem for polyhedral surfaces (which has uniqueness for free) and pass to the limit.

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    $\begingroup$ Yes, concerning "uniqueness for free," Alexandrov's Thm.3 p.184 says: Isometric polyhedral caps are congruent. $\endgroup$
    – Joseph O'Rourke
    Commented Dec 2, 2021 at 19:16
  • $\begingroup$ First, you probably misspelled Perelman. Second, what theorem of Perelman do you refer to? $\endgroup$
    – Alexandre Eremenko
    Commented Dec 2, 2021 at 22:08
  • $\begingroup$ If you have a metric of positive curvature on the (open) disk, and double it, in general you do not get a metric of positive curvature on the sphere. Some condition on the metric on the boundary is required (which is absent in the question). $\endgroup$
    – Alexandre Eremenko
    Commented Dec 2, 2021 at 22:12
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    $\begingroup$ @AlexandreEremenko nonnegative curvature in the sense of Alexandrov implies convexity of the boundary. $\endgroup$
    – Anton Petrunin
    Commented Dec 2, 2021 at 22:21
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    $\begingroup$ @AlexandreEremenko see p.28 here: anton-petrunin.github.io/papers/alexandrov/… $\endgroup$
    – Anton Petrunin
    Commented Dec 3, 2021 at 20:47

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