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A convex surface is a connected open subset of the boundary of a convex body in $\mathbb{R}^3$.

An "infinitesimal bending" of a convex surface $S$ is a deformation of $S$ given by a velocity field $v:S\rightarrow \mathbb{R}^3$ such that the length of every curve on the surface is preserved. Vector field $v$ is called the "bending field." A bending field is "trivial" if it corresponds to a simple translation or rotation of the surface. A surface is called "rigid" if all of it's bending fields are trivial.

A regular convex surface is one where every point has a neighborhood which can be parametrized by $r(u, v)$ with $r_u \times r_v \neq 0$.

Blaschke proved that a closed regular convex surface is rigid. Cauchy proved that all convex polyhedra (which are irregular) are rigid.

I've begun dabbling in A.V. Pogorelov's "Extrinsic Geometry of Convex Surfaces," where he proves and discusses both of these results in one of the chapters. However, the translation was published in 1973, so I'm curious about more recent research in this area. Namely, what is known about the rigidity of 1) regular convex surfaces which are not closed and 2) irregular convex surfaces which are not polyhedral?

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    $\begingroup$ In fact, Cauchy stated and proved global, rigidity of convex polyhedra, but his argument can easily be adapted to a proof of the infinitesimal rigidity. The first to formulate and prove the infinitesimal rigidity was Dehn. $\endgroup$
    – Ivan Izmestiev
    Commented Nov 5, 2018 at 14:08
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    $\begingroup$ The first proof of the rigidity of smooth closed surfaces was due to Cohn-Vossen, not Blaschke. I am not aware of any proof of this due to Blaschke. $\endgroup$
    – Mohammad Ghomi
    Commented Nov 19, 2018 at 21:49
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    $\begingroup$ Blaschke proved the infinitesimal rigidity (as defined by the OP) in his 1912 article "Ein Beweis für die Unverbiegbarkeit geschlossener konvexer Flächen". His proof is reproduced in the Spivak's volume V, before Cohn-Vossen's proof of the global rigidity. $\endgroup$
    – Ivan Izmestiev
    Commented Nov 20, 2018 at 8:25
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    $\begingroup$ @Ivan: OK, I see. The use of the term "rigid" by OP, as opposed to "infinitesimally rigid", is confusing. $\endgroup$
    – Mohammad Ghomi
    Commented Nov 20, 2018 at 12:18

3 Answers 3

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Answer to question 1: proper compact subsets of convex surfaces are infinitesimally flexible.

In his book Pogorelov proves infinitesimal flexibility of the graph of a convex function over a strictly convex compact subset of the plane. (Basically the idea is that in order to find an infinitesimal bending of a surface with positive Gauss curvature one has to solve an elliptic PDE.) This implies infinitesimal flexibility of any proper compact subset of a strictly convex surface. Indeed, infinitesimal rigidity is projectively invariant (an amazing property given that the definition is in terms of distance). If you send to infinity a plane that cuts the surface in a neighborhood of one point, then the complement of this neighborhood becomes a graph of a convex function.

Answer to question 2: closed convex surfaces without flat pieces are infinitesimally rigid.

This is proved in the cited book of Pogorelov (Theorem 4 in Section 4.7). Clearly, any flat piece allows infinitesimal deformations: take any vector field supported inside this piece and orthogonal to its plane.

As to recent research in the field, one may consult a series of surveys by Ivanova-Karatopraklieva and Sabitov. A good textbook on the topic is the first half of volume 5 of Spivak's "Comprehensive introduction...".

One direction in which one can generalize rigidity results for convex surfaces is rigidity of hyperbolic manifolds with convex boundary. (Indeed, rigidity of a convex surface translates as rigidity of a Euclidean ball enclosed by this surface.) In this article:

Schlenker, Jean-Marc, Hyperbolic manifolds with convex boundary, Invent. Math. 163, No. 1, 109-169 (2006). ZBL1091.53019.

infinitesimal rigidity of hyperbolic manifolds with convex boundary is proved (using Pogorelov's theorem, by the way).

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OP: "I'm curious about more recent research in this area." Here are two relatively recent papers. Ivan visits MO, so he may answer more definitively.

Izmestiev, Ivan. "Infinitesimal rigidity of convex surfaces through the second derivative of the Hilbert-Einstein functional II: Smooth case." arXiv:1105.5067. (2011).

"The paper is centered around a new proof of the infinitesimal rigidity of smooth closed surfaces with everywhere positive Gauss curvature."

Martinez-Maure, Yves. "Rigidity and Bellows-type Theorem for hedgehogs." (2011). Author's link.

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Answer to question 1: If a convex surface is not closed, then generally it is far from rigid as it might admit infinitely many isometric deformations; however, if the surface has $\mathcal{C}^{2,1}$ regularity and positive curvature, then it becomes rigid as soon as one fixes an arbitrarily small curve segment on that surface. This has been proved very recently in a joint work with Joel Spruck, which has been accepted for publication in International Math. Research Notices (IMRN):

Rigidity of nonnegatively curved surfaces relative to a curve, arXiv:1805.02580.

This result may also be extended to nonnegatively curved surfaces under some additional conditions. The proof uses Hormander's unique continuation principle for elliptic PDEs. Our methods also yield a very short proof of Cohn-Vossen's rigidity theorem for smooth closed convex surfaces, via Hopf's maximum principle, which is included in the appendix to the paper.

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  • $\begingroup$ Dear Pr Ghomi, I looked at your very intersting paper with Joel Spruck, but in the short appendix A, I don't understand why the equation $(9)$ satisfied by $\phi$ is elliptic. If I'm not mistaken, $\nabla_{ij}u$ becomes degenerate (and in fact vanishes) at points where the unit vector $e$ is tangent to the image of $f$, since then $\nabla u=e$ and Darboux's equation $(8)$ gives the result. This is not a problem in the main result, because there is an open cone of $e$ which avoids all the tangent planes of $f$ and $\tilde f$ at points of $\Omega$. Am I mistaken ? $\endgroup$
    – BS.
    Commented Nov 22, 2019 at 10:34
  • $\begingroup$ @BS. Thanks for your question. The argument in Appendix A is indeed a bit more subtle than has been indicated. Specifically, it is enough for $\phi$ to be elliptic near $\partial\Omega$ which we may assume is the case. Then $\phi$ will vanish near $\partial Ω$ by the unique continuation principle, and consequently $f$ and $\tilde f$ will coincide near $p$. So the main result of the paper will yield that $f$ and $\tilde f$ coincide everywhere. $\endgroup$
    – Mohammad Ghomi
    Commented Nov 22, 2019 at 20:14
  • $\begingroup$ Thanks for your answer. So this uses $C^{2,1}$ regularity, and there is still no "shorter proof" than Herglotz's one (which seems to require more than $C^2$). May I ask if you plan to post a new version on arXiv ? $\endgroup$
    – BS.
    Commented Nov 29, 2019 at 9:07
  • $\begingroup$ Yes, we will update the arXiv posting soon. The observation in the appendix, and in fact the main result of the paper, hold in $C^2$ category if, instead of Hormander's unique continuation, one uses results in 2-dimensions due to Bers-Nirenberg. See Strong unique continuation for general elliptic equations in 2D by Giovanni Alessandrini, J. Math. Anal. Appl. 386 (2012) 669–676. $\endgroup$
    – Mohammad Ghomi
    Commented Nov 29, 2019 at 14:24
  • $\begingroup$ Thank you for this answer. $\endgroup$
    – BS.
    Commented Nov 29, 2019 at 19:31

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