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

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

  2. → $M_1,M_2$ are isometric
    ⇆ $M_1,M_2$'s first fundamental forms agree

  3. → $M_1,M_2$ have the same Gaussian curvatures

  4. → $M_1,M_2$ 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.)

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Regarding your question of shape vs. metric, see Nash–Kuiper theorem (e.g. 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. –  user53153 Feb 27 '13 at 17:14
    
@5pm: Thank you very much for the hint to Cauchy's theorem. –  Hans Stricker Feb 27 '13 at 17:52
    
@Yury: ... and thank you for the hint to Nash's theorem! –  Hans Stricker Feb 27 '13 at 17:53
1  
Also on MathOverflow –  user53153 Feb 28 '13 at 22:27
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1 Answer

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.

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Thanks, these are good examples. But what about closed surfaces, do you know of any? –  Hans Stricker Feb 28 '13 at 7:52
    
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. –  Rahul Feb 28 '13 at 8:01
    
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. –  Rahul Feb 28 '13 at 8:10
    
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 Stricker Feb 28 '13 at 8:17
    
I don't know nearly enough differential geometry to even guess, honestly! –  Rahul Feb 28 '13 at 8:26
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