# All compact surfaces $S\subseteq \mathbb{R}^3$ are rigid?

Recently I've come across a lecture in differential geometry by Fernando Codá (in portuguese!) in which he stated that the following problem is (at least, up to 2014) open:

Given $S\subseteq\mathbb{R}^3$ a smooth (i.e. $C^\infty$) compact boundaryless connected surface (therefore orientable), let $f:S\times[0,1]\to \mathbb{R}^3$ be a differentiable map with $S_t:=f(S\times\{t\})$ being a surface and $f_t:S\to S_t$ an isometry (where $f_t(p):=f(p,t)$), where $f_0\equiv \mathrm{Id}_S$. Then there are rigid motions $A_t:\mathbb{R}^3\to\mathbb{R}^3$ with $f_t(S) = A_t(S)$.

More generally, in the cases of the sphere $S^2$ (Liebmann's Theorem) and "ovaloids", i.e. surfaces with Gaussian curvature $K$ strictly positive (Cohn-Vossen's rigidity theorem), it is known that isometric embeddings of it in $\mathbb{R}^3$ are exactly the restrictions of the rigid motions from $\mathbb{R}^3$. Not all compact connected surfaces $S\subseteq\mathbb{R}^3$ satisfies this, as pointed out in the book Curves and Surfaces, by S. Montiel and A. Ros:

Anyway, it is possible to show that this isometry in particular cannot be obtained by a continuous family of isometries from the identity map to the latter, so it's not a counterexample to the conjecture.

Although this seems to me to be a very studied problem, and I imagine that there must be several approachs to it, I just couldn't find enough information about it on the internet (the Montiel & Ros book was recommended by Codá later in the same lecture linked above), for "rigidity" seems to be a very comprehensive term. So my actual question is:

• Though I don't know much about the problem you state, it is worth pointing out that there are natural generalizations of it which are false. Namely, there exist polyhedra (necessarily non-convex) in $\mathbb{R}^3$ that can be "flexed" (i.e. deformed though non-rigid motions). To relate this to what you've stated, observe that their piecewise-linear structure endows such polyhedra with path metrics that are piecewise flat, and of course those path metrics are unchanged during the flexing. See en.wikipedia.org/wiki/Flexible_polyhedron – Andy Putman Nov 28 '15 at 4:44
• @RyanBudney: Flat torus in $\mathbb R^3$ is due to Nash--Kuiper in 1954. You are probably right that one can find a nontrivial continuous deformation of such isometric $C^1$-embeddings. The question in the OP is probably intended to be about smooth isometric embeddings, which are of course much less "flexible". – John Pardon Nov 28 '15 at 6:06
• The question is open for $C^\infty$ regularity, and even real-analytic according to mathoverflow.net/a/70484/6451 . The problem was mentioned by Yau in "Review of geometry and analysis" (Asian Math J. march 2000). See also the text "Classical Open Problems in Differential Geometry" (2004) by Mohammad Ghomi (problem 11, found by googling). – BS. Nov 28 '15 at 8:59
• this question might be in the same spirit: mathoverflow.net/questions/208421/… – Manfred Weis Nov 28 '15 at 16:24
• About the text by S.T.Yau I mentioned above, he says that global rigidity doesn't hold even in the real analytic class, without any reference. I guess that he refers to results about infinitesimal non-rigidity results by Rembs (and Trotsenko) for analytic surfaces of revolution (with the remark that a non-trivial infinitesimal isometric deformation yields a couple of non-trivially isometric surfaces). But, as remarked by Spivak (see volume 5 of "A comprehensive introduction to differential geometry"), the infinitesimal deformations found cannot be analytic, so I doubt Yau's statement. – BS. Nov 28 '15 at 21:57

Also, not an answer but some comments. When one learns about the geometry of smooth surfaces in $\mathbb{R}^3$, the question of rigidity and flexibility arises quite naturally. And, at first sight, it is plausible that there should be some characterization of these properties in terms of geometric invariants, especially the second fundamental form.

However, a solution to this problem, except for a closed convex hypersurface, has proved to be quite elusive. This is closely related to the fact that the PDE associated with the isometric embedding is an extremely nasty one unless the Gauss curvature is positive, where it is elliptic. The next least nasty case is that of negative Gauss curvature, where the PDE is nonlinear hyperbolic. However, a closed surface with negative curvature cannot be isometrically embedded into $\mathbb{R}^3$. Even if it could be, then resulting PDE is nonlinear hyperbolic and difficult to analyze on a compact manifold, especially a non-periodic one as is the case here. If the Gauss curvature vanishes anywhere, the PDE's become even nastier, and even results about local isometric embeddings become quite difficult and technical. A good reference for this is the book Isometric Embedding of Riemannian Manifolds in Euclidean Spaces, by Han and Hong.

So any results about rigidity and flexibility probably needs to circumvent the use of PDE's. As far as I know, nobody has a clue about how to do this.

If the isometric embedding is assumed to be only $C^1$, then the whole situation flips, due to the results of Nash and Kuiper. Any $C^1$ isometric embedding is flexible, and any closed orientable surface, including flat and negatively curved ones, can be globally isometrically embedded into $\mathbb{R}^3$. The proof of this is actually relatively elementary and worth learning. The techniques used were generalized by Gromov into what is now known as the $h$-principle and convex integration. More recently, the ideas of Nash and Kuiper have been adapted by De Lellis, Székelyhidi, and their collaborators to make dramatic progress on the Onsager conjecture for the Euler fluid flow equation.

ONE MORE COMMENT: As Andy Putman mentions, Bob Connelly found an explicit example of a flexible polyhedron. A metal model of this can be found in the IHES lounge. It is not obvious to me that this could not be turned into a flexible smooth isometric embedding of a smooth surface that approximates the polyhedron. The edges are easily modeled by a hinge. The question is whether one can create a smooth local model of the vertices. Note that these vertices are necessarily "negatively curved".

• I dream that someday mathematicians will learn to spell my last name correctly. Sigh. – Andy Putman Nov 28 '15 at 17:57
• Andy, sorry about that. There's a slight chance I'll remember that from now on. – Deane Yang Nov 28 '15 at 18:07
• Given Connelly's comment on the possibility of a smooth flexing surface, I suspect he has already explored and failed to construct such a surface. – Joseph O'Rourke Nov 28 '15 at 18:20
• A nice explanation of the PDE: aimsciences.org/journals/pdfs.jsp?paperID=4640&mode=full – Deane Yang Nov 28 '15 at 20:18
• Joseph, yes. You can shrink the embedding slightly, so that it becomes "short" and then apply the Nash-Kuiper construction. In particular, there are non-convex $C^1$ isometric embeddings of the standard sphere. – Deane Yang Nov 29 '15 at 4:02

Not at all an answer: I once had a conversation about this question with a friend at IHES, and Gromov was hanging around, overheard us, and said: oh, that's a stupid problem!

We said (in one voice): why do you say that?

To which Gromov responded: because the question has been around for a century, and no interesting mathematics has come out of it.

Make of this what you will.

• I had a (slightly) similar conversation when asking a professor from my university about this problem. He said that this sounded like just an example of a claim that is easy to state but difficult to prove, and (in his words) spend my precious time as a graduate student studying a problem like that would be such a big waste of time. Anyway, Gromov seems to have a somewhat stronger opinion on this haha :) – Alufat Nov 29 '15 at 15:40
• @ChrisTáfula: The notion of bending of surfaces (which is what we are discussing here) is totally natural and has been studied locally and infinitesimally. The fact that a foundational question about global bending is still unresolved is of course disappoining, but it would do no good to pretend the issue is not there. Gromov has expressed similar opinions about various things including e.g. the whole subject of global Riemannian geometry. Listening to Gromov is always illuminating, but I would advise to follow your own nose in choosing what to spend time on. – Igor Belegradek Nov 29 '15 at 16:39
• @IgorBelegradek Thank you for your advice! Also, it's always funny to hear mathematicians talking about each others area. My algebraic topology teacher always says that differential geometry is just "random calculations in $\mathbb{R}^4$". – Alufat Dec 2 '15 at 5:51
• Didn't Gromov also come to consider pseudoholomorphic curves by thinking about how to conceptualize Pogorelov's proof of the rigidity of convex surfaces? – skr Apr 10 '19 at 2:43

Not an answer, just a reference, and a quote from Robert Connelly's Rigidity article in the Handbook of Convex Geometry:

(P.M. Gruber, J.M. Wills, Eds. Handbook of Convex Geometry. Vol.2. North Holland, 1993. "Rigidity," p.231.)

Here's the reference:

Han, Qing, and Marcus Khuri. "Rigidity in the class of orientable compact surfaces of minimal total absolute curvature." Differential Geometry and its Applications 29.4 (2011): 463-472. (Elsevier link to HTML version.)