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There is a theorem in classical differential geometry due to Cohn Vossen (1927) that compact, connected regular surfaces of everywhere positive gaussian curvature (ovaloids) are rigid - in the sense that if a regular surface in Euclidean space is isometric to a given ovaloid, then it is necessarily congruent to it. This is sometimes called the congruence theorem of ovaloids.

This has been generalised by Pogorelov and Alexandrov, and based on their work it is possible to conclude the torus is rigid as well.

I was wondering, is a hemisphere rigid in the above sense? Can anyone give an example of a regular surface in Euclidean space which is isometric to a hemisphere, but not congruent to it?

It is claimed in the book: Geometry and imagination (by Hilbert and Cohn-Vossen) on page 228, that there are pieces of the sphere which can be bent, and that it becomes bendable as soon as we remove a segment of a great circle on the sphere. However, I have not been able to find any example.

(I am not interested in infinitesimal bendings, but interested in all examples of surfaces which are isometric and not congruent)

Artin Ostal
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An important point is that by Minding's Theorem, if you have a (not necessarily compact!) surface of constant $K=1$, then it must be locally isometric to the sphere. But you can find two different families of these just among surfaces of revolution by explicit integration. Here's a sketch:

Non-round surfaces of K=1

If you're interested in working it out, see Exercise 19 in section 2.3 of my differential geometry text.

Ted Shifrin
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  • Thank you for your comment, the lecture notes seem great! I am aware of what you are saying, but are these surfaces (globally) isometric? – Artin Ostal Feb 08 '21 at 06:20
  • One would have to chase through the proof of Minding's Theorem (e.g., in DoCarmo's text). I'm not sure how far the local isometry (to a portion of a sphere) extends. – Ted Shifrin Feb 08 '21 at 06:29
  • @TedShifrin, I guess you don't want to give away answers to problems in your textbook? – Deane Feb 08 '21 at 16:31
  • @Deane No, the problem was to solve the ODE to get surfaces of revolution of constant $K$. The bendability is beyond my scope. :) Or did you have something else in mind? – Ted Shifrin Feb 08 '21 at 16:34
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    The left appears to be a surface with constant curvature $1$ and two conic singular points. So, as an abstract Riemannian manifold, it could be the region that lies between two great half-circles ending at the poles with the two half-circles glued together. The right one might be something similar but involving the sphere with caps sliced off. – Deane Feb 08 '21 at 16:45
  • In particular, the intrinsic metric would be of the form $dr^2 + a^2 (\sin \theta)^2$, where $a < 1$. Intrinsically, any nonzero $a$ works but I don’t think the case $a > 1$ is isometrically embeddable as a surface of revolution. – Deane Feb 08 '21 at 17:16