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Suppose that we have a convex cap, i.e., a convex surface in $R^3$ homeomorphic to a disk whose boundary lies in a plane. Reflect the cap through the plane of its boundary and glue it back to the original cap along the common boundary curve. If the closed surface we obtain is convex, then it is rigid by a theorem of Pogorelov, i.e., any convex surface which is isometric to it is congruent to it.

My question is: do we still have rigidity when the doubled surface is not convex? I do not know the answer even in the case where each cap is spherical. Here we want to look at isometric surfaces which are smooth except along the curve where the two pieces meet, so that we do not get trivial examples by reflecting a portion of the surface.

In general very little is known about rigidity of non-convex surfaces, but this situation is so simple that perhaps someone might have an idea about it. It seems to me that this is the simplest non-convex case.

Addendum: There is a theorem of Alexandrov and Sen'kin, see p. 181 in Pogorelov, which seems to be relevant to the above problem. This theorem states that if a pair of isometric convex surfaces lie in the upper half-space, are star-shaped and concave with respect to the origin, and their corresponding boundary points are equidistant from the origin, then they are congruent.

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    $\begingroup$ The Bricard octahedra (en.wikipedia.org/wiki/Bricard_octahedron) are flexible, and formed in a different way by gluing together two copies of the same convex cap along a planar gluing polygon. I don't see how to make anything like them from your exact construction, but it suggests the problem could be difficult. $\endgroup$ – David Eppstein May 5 '18 at 7:28
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Here is one possible approach for convex polyhedra. This is only a sketch, not a proof. Truncate a convex polyhedron below its "equator," which I leave undefined. For a regular octahedron, clip off the lowest vertex:


          OctTrunc
Prove that this convex cap with boundary is rigid as it stands. I believe Alexandrov's results in Section 5.2 of his book,1 "Flexes of Convex Polyhedra," can establish that this particular cap is rigid.

Now join two copies along their common boundary, and argue that this closed polyhedron is rigid.


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Perhaps this could establish rigidity of your double spherical cap via polyhedral approximations.

1Alexandrov, Aleksandr D. Convex Polyhedra. Springer Monographs in Mathematics, 2005. Translation of Russian edition, 1950. Section 5.2: p.246ff.

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