Is a smooth closed surface in Euclidean 3-space rigid? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:08:01Z http://mathoverflow.net/feeds/question/1975 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1975/is-a-smooth-closed-surface-in-euclidean-3-space-rigid Is a smooth closed surface in Euclidean 3-space rigid? Deane Yang 2009-10-22T23:23:55Z 2011-07-16T05:40:10Z <p>Classical theorem of Cohn-Vossen: A closed convex surface in Euclidean 3-space cannot be deformed isometrically.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/1975/is-a-smooth-closed-surface-in-euclidean-3-space-rigid/2321#2321 Answer by Thomas Sauvaget for Is a smooth closed surface in Euclidean 3-space rigid? Thomas Sauvaget 2009-10-24T17:24:50Z 2009-10-24T17:24:50Z <p>On the Springer Online Encyclopedia there's a relevant article here: <a href="http://eom.springer.de/T/t092810.htm" rel="nofollow">http://eom.springer.de/T/t092810.htm</a></p> <p>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. </p> <p>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 <a href="http://books.google.fr/books?id=S5CD-YceX6QC&amp;pg=PA62" rel="nofollow">http://books.google.fr/books?id=S5CD-YceX6QC&amp;pg=PA62</a></p> <p>As for polyhedra, Schlenker has a rigidity criterion for non-convex ones, preprint here: <a href="http://www.math.univ-toulouse.fr/~schlenker/texts/rcnp.pdf" rel="nofollow">http://www.math.univ-toulouse.fr/~schlenker/texts/rcnp.pdf</a> (sadly lacks the drawings, the published reference is Discrete and Computational Geometry, 33 (2005):2, 207-221).</p> <p>(I'm no expert on this, just some googling).</p> http://mathoverflow.net/questions/1975/is-a-smooth-closed-surface-in-euclidean-3-space-rigid/70484#70484 Answer by Jean-Marc Schlenker for Is a smooth closed surface in Euclidean 3-space rigid? Jean-Marc Schlenker 2011-07-16T05:31:41Z 2011-07-16T05:40:10Z <p>Apparently the question is still open for smooth enough surfaces and deformations (that is, at least \$C^2\$). </p> <p>Mike Anderson wrote a <a href="http://arxiv.org/abs/0709.3202" rel="nofollow">preprint</a> claiming to prove local rigidity of smooth enough surfaces, but it was later withdrawn. </p> <p>Idjad Sabitov and his collaborators have been working on this question, developing for instance a theory of higher-order isometric deformations, see e.g. <em>Sabitov, I. Kh. Local theory of bendings of surfaces [MR1039820 (91c:53004)]. Geometry, III, 179–256, Encyclopaedia Math. Sci., 48, Springer, Berlin, 1992.</em> He conjectures that local rigidity holds for analytic surfaces.</p>