Isometric but differently shaped surfaces in $\mathbb{R}^3$

Question

We have the following chain of inclusions for surfaces in $\mathbb{R}^3$ $M_1,M_2$:

$M_1,M_2$ have the same shape, i.e. are related by an ambient isometry
&lrarr; $M_1,M_2$'s first and second fundamental forms agree

→ $M_1,M_2$ are isometric
&lrarr; $M_1,M_2$'s first fundamental forms agree

→ $M_1,M_2$ have the same Gaussian curvatures

→ $M_1,M_2$ have the same genus (for closed surfaces)

In a more catchy way: shape → metric → curvature → genus

I know the standard examples of isometric but differently shaped surfaces in $\mathbb{R}^3$: plane, cone, cylinder.

I am looking for (other) examples of

isometric but differently shaped surfaces (preferrably closed ones)

non-isometric surfaces with the same curvature

I assume there are no differently shaped surfaces isometric to the sphere, are there?

But what about other convex surfaces (with strictly positive but not constant curvature)? Or arbitrary surfaces homeomorphic to the sphere? Or to the torus?

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

Regarding your question of shape vs. metric, see Nash–Kuiper theorem (e.g. http://en.wikipedia.org/wiki/Nash_embedding_theorem) that implies that every Riemannian manifold has very “strange/non-standard” isometric $C^1$-embeddings. — Yury, Feb 27 '13 at 16:59
To complement the comment by @Yury, "reasonable" embeddings of convex surfaces indeed have the same shape. Pogorelov's uniqueness theorem applies to surfaces with "finite total extrinsic curvature", but unfortunately I forgot what this means, if I ever knew. It's mentioned in the Wikipedia article on Cauchy's theorem. — , Feb 27 '13 at 17:14
In 1, what do you mean by "first and second fundamental forms agree"? I'm not sure how to interpret it. — Jason DeVito - on hiatus, Feb 27 '13 at 18:07
@JasonDeVito I read it as "there is a diffeomorphism between two (embedded) surfaces that preserves both forms". — , Feb 27 '13 at 18:48
@5PM: Then isn't 1. false? Consider a long cylinder with spherical end caps (attached smoothly) and a short cylinder with the same radius with spherical end caps (attached in the same way) — Jason DeVito - on hiatus, Feb 27 '13 at 19:49
@JasonDeVito These surfaces don't have the same area, so a diffeomorphism between them cannot preserve the first fundamental form. — , Feb 27 '13 at 20:19
@5PM: Of course! I was thinking of curvature and second fundamental form! — Jason DeVito - on hiatus, Feb 27 '13 at 20:27
I was about to offer a bounty for examples involving closed surfaces, but Robert Young and Deane Yang's comments on the MathOverflow question have covered all the cases I can think of. — , Mar 01 '13 at 09:56

score 5 · Answer 1 · edited Mar 09 '17 at 17:30

5

There is a continuous, isometric deformation between a catenoid and a helicoid.

A parametrization of such a deformation is given by the system $$\begin{align} x(u,v) &= \cos \theta \,\sinh v \,\sin u + \sin \theta \,\cosh v \,\cos u\\ y(u,v) &= -\cos \theta \,\sinh v \,\cos u + \sin \theta \,\cosh v \,\sin u\\ z(u,v) &= u \cos \theta + v \sin \theta \, \end{align}$$

for $(u,v) \in (-\pi, \pi] \times (-\infty, \infty)$, with deformation parameter $-\pi < \theta \le \pi$, where $\theta = \pi$ corresponds to a right-handed helicoid, $\theta = \pm \pi / 2$ corresponds to a catenoid, and $\theta = 0$ corresponds to a left-handed helicoid.

In fact, there are lots of such families of isometric minimal surfaces.

edited Mar 09 '17 at 17:30

Community

1

answered Feb 27 '13 at 18:42

Thanks, these are good examples. But what about closed surfaces, do you know of any? – Hans-Peter Stricker Feb 28 '13 at 07:52
Sorry, I'm afraid not! Well, if you allow polyhedral surfaces, you can make a "house" by attaching a pyramid to the top face of a cube, and then get an isometric polyhedron by pushing the "roof" in. But this is probably not what you want. – Feb 28 '13 at 08:01
Also, if I had clicked on the first link in your question sooner — instead of just now — I would have realized that you were probably already aware of the examples I gave. Sorry again. If you want, I can delete this answer. – Feb 28 '13 at 08:10
Please, don't delete the answer, and thank you anyway. Would you go so far to assume that there are no isometric but differently shaped smooth closed surfaces? Would this be an interesting result? – Hans-Peter Stricker Feb 28 '13 at 08:17
I don't know nearly enough differential geometry to even guess, honestly! – Feb 28 '13 at 08:26
In general, a closed smooth 2-manifold without a boundary can have many embeddings into ${\mathbb R}^3$. However, if its Gaussian curvature is positive then it has only one smooth embedding into ${\mathbb R}^3$ by the Cohn-Vossen theorem. – Yury Feb 28 '13 at 21:34

Isometric but differently shaped surfaces in $\mathbb{R}^3$

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