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Classical theorem of Cohn-Vossen: A closed convex surface in Euclidean 3-space cannot be deformed isometrically.

Robert Connelly found an example of a polyhedral surface that can be deformed isometrically. A metal hinged model of it can be found at IHES.

But what about an arbitrary not-necessarily-convex smooth closed surface? Is it necessarily rigid? Or maybe it might be possible to make a smooth version of Connelly's example? It's easy to make smooth "hinges". The real challenge is finding a smooth model of the vertices, which is where two or more hinges meet.

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On the Springer Online Encyclopedia there's a relevant article here:

It says that a theorem due to Kuiper in 1955 implies that no smooth closed surface in R^3 is C^1-isometrically rigid. I think the reference is: N.H. Kuiper, On C^1-isometric embeddings, Indag. Math. XVII, (1954) 545-556 and 683-689.

On the other hand it says that nothing is known for the C^2 case, and a book on Open Problems in Geometry also says that as 1994 it is still open, see

As for polyhedra, Schlenker has a rigidity criterion for non-convex ones, preprint here: (sadly lacks the drawings, the published reference is Discrete and Computational Geometry, 33 (2005):2, 207-221).

(I'm no expert on this, just some googling).

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By the way, the infinitesimal rigidity criterion mentioned here is still partly open. It states that any convex polyhedron in $R^3$ with vertices in convex position which is "decomposable" (can be cut in convex polyhedra with no new vertex) is infinitesimally rigid. With Ivan Izmestiev we proved that this holds under an additional hypothesis of "codecomposability". Without this additional hypothesis it's unknown whether the criterion holds. – Jean-Marc Schlenker Jul 16 '11 at 5:43

Apparently the question is still open for smooth enough surfaces and deformations (that is, at least $C^2$).

Mike Anderson wrote a preprint claiming to prove local rigidity of smooth enough surfaces, but it was later withdrawn.

Idjad Sabitov and his collaborators have been working on this question, developing for instance a theory of higher-order isometric deformations, see e.g. Sabitov, I. Kh. Local theory of bendings of surfaces [MR1039820 (91c:53004)]. Geometry, III, 179–256, Encyclopaedia Math. Sci., 48, Springer, Berlin, 1992. He conjectures that local rigidity holds for analytic surfaces.

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Thanks for these references. I did not know about Mike Anderson's attempt. – Deane Yang Jul 16 '11 at 12:25
It is worth noting that the smooth case is equivalent to the local uniqueness of a solution to a PDE of mixed type, which is an extremely difficult question. Even if you try to take a more geometric approach, you have to confront the difficulties implied by this. Basically, the way a positively curved surface bends is different from how a negatively curved surface bends, and you have to somehow match up the bending where curvature changes sign. The only way I can see around this is to find some miraculous integral identity that implies rigidity. But non-ridigity seems more interesting to me. – Deane Yang Jul 16 '11 at 12:31
Right -- but it depends how one looks at it. One possibility (used by Anderson, for instance, in the preprint mentioned above) is to consider deformations of the metric in the domain bounded by the surface, under the condition that it remains flat (or constant curvature). In this setting the PDE is elliptic, but it's the boundary condition (on the surface) which is not too well behaved. – Jean-Marc Schlenker Jul 16 '11 at 20:19
Yes, that was a cool idea by Anderson, but in the end you can't escape the non-ellipticity and its consequences. I often wish mathematicians could get at least some credit for cool ideas, even if they don't work. – Deane Yang Jul 17 '11 at 21:24
I suspect the idea is older than Anderson's preprint. It comes up naturally if you think of the infinitesimal rigidity of convex surfaces in Euclidean (or hyperbolic) 3-space and try to extend it either to hyperbolic 3-manifolds with smooth, strictly convex boundary --Thurston had conjectured that they are determined by the induced metric on the boundary, and it turned out to be true -- or in higher dimensions to Einstein metrics with convex boundary on the ball (there the question is still open I think). I think people trying to prove rigidity of hyperbolic convex cores had tried this. – Jean-Marc Schlenker Jul 18 '11 at 17:25

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