I would like to know if there is a polyhedral analog to this beautiful theorem of Hong:

Theorem 11.0.1. Any smooth positive disk $(\bar{D},g)$ with a positive geodesic curvature along $\partial D$ admits a smooth isometric embedding in $\mathbb{R}^3$ that maps $\partial D$ to a planar curve. Moreover, the embedding is unique within rigid motions.

This is proved in the book by Qing Han and Jia-Xing Hong, Isometric Embedding of Riemannian Manifolds in Euclidean Spaces, American Mathematical Society Monograph, Volume 130, 2006, p.225.

I would like to replace the disk by a convex polyhedral cap $C$ with a convex boundary $\partial C$. $C$ is a connected subset of the surface of a convex polyhedron, and $\partial C$ has at every point $p \in \partial C$ at most $\pi$ angle incident to $p$ within $C$:
Alexandrov's book Convex Polyhedra includes an article in the Appendix written by L.A. Shor "On Flexibility of Convex Polyhedra with Boundary," which describes (rather complex) conditions under which flexing can occur, but does not seem to address flexing to achieve planarity of $\partial C$. It appears that Alexandrov's and Shor's results imply that it cannot always be possible, because (for example) if $\partial C$ contains no vertices of $C$, then it must be rigid (but not necessarily planar).

But perhaps there are conditions under which some polyhedral analog of Hong's theorem holds? Thanks for pointers!

Addendum. My concentration on the conditions for "flexing" above was misplaced, as Sergei Ivanov's answer demonstrates: the analog isometrically embeds a convex cap $C$ with $\partial C$ planar, but not by hinging $C$'s faces as rigid plates, but rather through an embedding which in general alters the facial structure of $C$ while maintaining isometry.

  • $\begingroup$ I'm reasonably familiar with the subject of isometric embeddings of a smooth Riemannian manifold in Euclidean space, but not at all with isometric embeddings of a polyhedron, with or without bounary. The latter, in my view, belongs to a different subject, presumably discrete geometry. Has there really been no work on the subject since Alexandrov? There is, I know, beautiful work by Connelly on flexing of non-convex polyhedron. I've always wanted to know if that has a smooth analogue to it or, if not, why not. $\endgroup$
    – Deane Yang
    Sep 26, 2011 at 14:31
  • $\begingroup$ I just looked at Bob Connelly's web page, math.cornell.edu/~connelly, and it appears to me that he is still quite interested in rigidity of geometric structures. You might want to consult him. $\endgroup$
    – Deane Yang
    Sep 26, 2011 at 14:35
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    $\begingroup$ I believe that the theorem you cite on isometric realization of convex caps is very close to one obtained by Pogorelov a while ago, see thm 4 on p. 104 of "Extrinsic geometry of convex surfaces" books.google.com/… $\endgroup$ Sep 27, 2011 at 11:31
  • $\begingroup$ @Jean-Marc: Thanks for the pointer! The theorem concludes that "there exists a regular convex cap realizing the metric," but does not appear to claim that the boundary $\gamma$ can be embedded as a planar curve. $\endgroup$ Sep 27, 2011 at 12:49
  • $\begingroup$ @ Joseph: this is included in the definition of a convex cap that Pogorelov uses (I should have made this precise, sorry). See his uniqueness thm in the book mentioned above, on p. 78: books.google.com/… So I still believe that the existence and uniquess thms of Pogorelov, together, seem to be very close to the Hong thm you cite. (for some reason I can't make the google books link go directly to p. 78...) $\endgroup$ Sep 27, 2011 at 16:31

1 Answer 1


Yes the polyhedral analog is true. Just consider the doubling of $C$, i.e., attach an isometric copy $C'$ of $C$ along the boundary, and apply Alexandrov's embedding theorem to the doubling. The common boundary of $C$ and $C'$ will go to a plane automatically.

Indeed, Alexandrov's theorem says that the doubling $S=C\cup C'$ (with its natural intrinsic metric) admits a unique (up to a rigid motion) isometric embedding $f:S\to\mathbb R^3$ as a surface of a convex polyhedron. The intrinsic metric of $S$ has a self-isometry $i$ which swaps $C$ and $C'$ and whose set of fixed points is their common boundary. Since $f$ is unique up to a rigid motion, $f\circ i$ extends to a rigid motion of $\mathbb R^3$. The set of fixed points of this rigid motion is an affine subspace, and $f(\partial C)$ is contained in this subspace. Hence $f(\partial C)$ is contained in a plane (and that rigid motion is the reflection in that plane).

The uniqueness also follows from the uniqueness part of Alexandrov's theorem as long as you require that the image is a convex half-polyhedron.

The usual warning attached to Alexandrov's embeddings applies here: you get an isometric embedding of the intrinsic metric, but its facial structure may differ from the original one.

  • $\begingroup$ Ah, great! Yes, I see now, that is Alexandrov's lemma on p.214 of the 2005 English translation: the doubling has a symmetry plane that includes the common glued boundary. I should have seen this! Thanks, Sergei! $\endgroup$ Sep 26, 2011 at 16:55
  • $\begingroup$ I also see that my concern with rigidity was misplaced, because, as you note, Alexandrov's theorem alters the facial structure. $\endgroup$ Sep 26, 2011 at 18:26
  • $\begingroup$ I find this very interesting. If I understand correctly (and I probably don't), the classical Cohn-Vossen rigidity theorem, as well as the work described above, allow for deformations where the shape as well as the number of the faces can change continuously? On the other hand, the famous example by Connelly of a flexible non-convex polyhedron does indeed fix the shape and number of faces and deforms the body only along the edges. I'm inferring this from the fact that there is a rigid metal version of Connelly's polyhedron with hinges along the edges. $\endgroup$
    – Deane Yang
    Sep 26, 2011 at 20:32
  • $\begingroup$ @Deane: Yes, in the intrinsic geometry you can see vertices but edges depend on the embedding. You can bend a planar quadrangle along any diagonal, so that one face becomes two faces (in two different ways). $\endgroup$ Sep 26, 2011 at 21:03
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    $\begingroup$ By the way, the described idea goes back to Olovyanishnikov mathnet.ru/links/69cc538ab2dd99be0c0994097345e25f/sm6287.pdf $\endgroup$ Sep 26, 2011 at 22:53

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