MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Starting from the following inclusions for surfaces $M_1,M_2$ in $\mathbb{R}^3$:

     $M_1,M_2$ have the same shape, i.e. are related by an ambient isometry

→ $M_1,M_2$ have the same metric

→ $M_1,M_2$ have the same Gaussian curvature

the only examples of isometric but differently shaped surfaces in $\mathbb{R}^3$ I have seen so far do have boundaries:

  • cone and cylinder
  • catenoid and helicoid
  • other associate families of isometric minimal surfaces.

I wonder if there are no (examples of) isometric but differently shaped closed surfaces, and why that could be. (I am particularly interested in smooth surfaces.)

And I am still looking for (closed) surfaces with the same Gaussian curvature but different metrics.

A picture gallery would be highly welcome, because I really would like to see two such (non-)isometric surfaces.

share|cite|improve this question
Which metric are you talking about? – Ryan Budney Feb 28 '13 at 18:21
Also, what do you mean when you say the surfaces have the "same" Gaussian curvature, just a diffeomorphism that carries the Gauss curvature? If so, there's an immense variety of such surfaces -- the zero-curvature embeddings of the Moebius band gets you started, but there are similar things for hyperbolic surfaces. – Ryan Budney Feb 28 '13 at 18:25
Concerning the metric: I hoped not to have to be too specific:… (Myers-Steenrod theorem). – Hans Stricker Feb 28 '13 at 19:26
Concerning curvature: a) I have difficulties to imagine two non-isometric zero-curvature Moebius bands embedded in $\mathbb{R}^3$. b) Even if I managed: it would be a surface with a boundary. – Hans Stricker Feb 28 '13 at 19:28
Just imagine making a Moebius band out of a thin strip of paper. There's two standard bends you can put in it -- the kind of bend you have in the standard embedding of the cylinder $S^1 \times [0,1]$ in $\mathbb R^3$, and then there's the planar embedding of the cylinder, in polar coordinates $1 \leq r \leq 2$. – Ryan Budney Feb 28 '13 at 19:47

Here is the standard explict example of 2-spheres $S_1, S_2$ embedded in $R^3$ with identical curvature functions (under suitable parameterization), so that $S_1$ is not isometric (as a Riemannian manifold) to $S_2$. Both surfaces will be surfaces of revolution. Take two cylinders of revolution $C_i$, $i=1, 2$ with unit radius and different heights. Now, attach isometric rotationally-symmetric caps $C_i^\pm, i=1,2$ at the top and the bottom of the cylinders $C_i$. One can easily write explicit functions for the cups (similarly to writing equations for the bump-functions) so that the resulting surfaces $S_i$ are smooth. Now, it is clear that, under suitable parameterization, surfaces $S_1, S_2$ have the same curvature functions, but will not be isometric to each other (since their regions of zero curvature are not isometric as they have different heights). I think, you can easily draw pictures of such surfaces yourself.

share|cite|improve this answer

Let $D$ be the unit disc and take a smooth function $f:D\to \mathbb{R}$ such that $f$ and all its derivatives are zero near the boundary of $D$. Then the graph of $f$ and the graph of $-f$ are isometric, so if you construct a smooth surface which contains the graph of $f$, you can replace it with the graph of $-f$. This is a bit of an unsatisfying example, though -- I'm not sure whether there is, say, a closed smooth manifold with a continuous family of isometric deformations, like the flexihedra that Lee mentioned.

Pogorelov's monograph (in Russian) on "Unique determination of general convex surfaces" proves the theorem that the boundary of any convex body in $\mathbb{R}^3$ is rigid (i.e., its embedding in $\mathbb{R}^3$ is determined up to translation and rotation by its metric), so any examples would have to be nonconvex.

share|cite|improve this answer
@Robert. Thanks in advance, but I first have to figure out what your example tries to tell me - and how. (I can follow you until "[...] $-f$ are isometric", but I don't know how to construct your proposed smooth surface and how I can replace the graph of $f$ with the graph of $-f$.) – Hans Stricker Feb 28 '13 at 21:47
Imagine a cube with a hemisphere glued to one face. If you push the hemisphere into the cube, you get a cube with a hemispherical dimple, and both surfaces are isometric. Now, this example isn't smooth, but you can make it smooth using bump functions ( – Robert Young Feb 28 '13 at 21:56
Hans: Graphs of $f$ and $-f$ are isometric via the reflection in $xy$-plane. – Misha Feb 28 '13 at 22:08
Misha: That's the part I did understand. – Hans Stricker Feb 28 '13 at 22:39


share|cite|improve this answer
Are there smooth occurrences of flexihedra? – Hans Stricker Feb 28 '13 at 21:35
THis currently links to Google search preferences. I suspect you meant to link to something else. – Harry Altman Feb 28 '13 at 21:42
Thanks for the correction Harry. – Lee Mosher Feb 28 '13 at 21:58
I don't know about smooth examples, Hans. – Lee Mosher Feb 28 '13 at 21:59
As far as I know there is no known example of a smooth closed surface in $\mathbb{R}^3$ that has a continuous family of isometries. Robert Connelly constructed a polyhedron that does (but I'm not sure it's the same as a flexihedron). There is a metal model of it at IHES. I believe it is even a conjecture (by whom I don't know) that any such smooth closed surface is infinitesimally rigid. I've always wondered if it's possible to construct a flexible smooth approximation of Connelly's example. – Deane Yang Mar 1 '13 at 5:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.