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Suppose we cut a disk out of a flat piece of paper and then manipulate it in three dimensions (folding, bending, etc.) Can we determine where the paper is from the position of the boundary circle?

More formally, let $\mathbb{D}$ be the unit disk in the Euclidean plane, viewed as a Riemannian manifold.

Is every isometric embedding $\varphi\colon\mathbb{D}\to\mathbb{R}^3$ determined by the restriction of $\varphi$ to the boundary circle?

That is, if two isometric embeddings $\varphi_1,\varphi_2\colon \mathbb{D} \to \mathbb{R}^3$ agree on the boundary circle, must they be equal?

Note: Here "isometric embedding" means isometric in the sense of differential geometry, i.e. a $C^1$ embedding that preserves the lengths of curves.

Jim Belk
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    This is false in the context of piecewise-isometric embeddings of triangulated disks (Connelly's flexahedron with one face removed), therefore, is probably also false in the smooth setting. – Moishe Kohan Jan 11 '14 at 07:55
  • I think this maybe wrong. Because we maybe can construct two surfaces which have different Second Fundmental Forms. – gaoxinge Jan 12 '14 at 08:19
  • For example, a disc in $R^3$ and a half sphere in $R^3$. – gaoxinge Jan 12 '14 at 11:41
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    @gaoxinge The surfaces are required to be _isometric_ embeddings of a Euclidean disk. – Jim Belk Jan 12 '14 at 19:10

3 Answers3

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The question, as I said in comments to the answer by user72694, is very sensitive to the degree of smoothness. In Jim's question, the map $\varphi$ is only assumed to be $C^1$. I will work with maps of the unit square $I^2$ rather than the unit disk, but it is irrelevant since one can start with a square containing a disk in its interior. Recall that a $C^1$-map $f: I^2\to R^3$ is short if it does not increase length of tangent vectors. A map is strictly short if it strictly decreases the lengths of nonzero tangent vectors. It is very easy to construct strictly short maps, take, for instance, a map which is a dilation by a factor $<1$.

The key theorem then is:

Theorem (Nash-Kuiper). Every short $C^1$ embedding $f: I^2\to R^3$ can be approximated (in topology of uniform convergence) by ($C^1$) isometric embeddings.

See for instance this paper for a modern proof, using Gromov's technique of convex integration.

Now, alter the original map $f$ on a sub-square $Q \subset Int(I^2)$ a little bit and see that the approximation construction can be performed without changing the approximating maps away from an open neighborhood of $Q$. (If $f$ was a strictly short map, then altering $f$ on $Q$ with another strictly short map is easy since strictly short maps form an open set in $C^1$-topology.)

The drawback of this answer is that to verify that it works you would have to go through a proof of the N-K theorem, but if you are interested in $C^1$-isometric embeddings, you would have to read a proof anyway.

Moishe Kohan
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  • Thank you! It's remarkable that the answer to such a straightforward geometric question would depend on the difference between $C^1$ and smooth. I will certainly look through the paper you linked to -- there must be a way to extract a simple counterexample without using the full force of the Nash-Kupier theorem. – Jim Belk Jan 13 '14 at 18:27
  • NO, there isn't. The theorem relies on noncontructive foundational material, so these are purely "theoretical" counterexamples. – Mikhail Katz Jan 14 '14 at 12:55
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    @user72694: Yes, indeed, this proof is nonconstructive. However, there might be some ways to construct such maps explicitly as there was, the the recent years, some work on "constructivization" of Nash-Kuiper's theorem, see discussion at http://mathoverflow.net/questions/31222/c1-isometric-embedding-of-flat-torus-into-mathbbr3 . – Moishe Kohan Jan 14 '14 at 14:34
  • @studiosus, very interesting, I was not aware of this work. – Mikhail Katz Jan 14 '14 at 15:56
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A Riemannian-isometric imbedding of an interval or triangle in $\mathbb{R}^3$ is called strongly isometric if the ambient distances coincide with the Riemannian distances. Notice that every Riemannian-isometric imbedding of the flat disk into $\mathbb{R}^3$ has the following property:

Every point of the disk lies either on a chord connecting two points of the boundary circle $S^1$, or in a triangle inscribed in $S^1$, such that the chord/triangle is imbedded strongly isometrically.

But the pattern of the strongly isometric imbeddings can be read off the boundary points, because the ambient distance will be less than or equal to the intrinsic distance, with equality corresponding to the case of a strongly isometric imbedding.

The uniqueness of the Riemannian-isometric imbedding of the disk now follows from the fact that a strongly isometric imbedding of an interval or a (flat) triangle is uniquely determined by the image of its boundary points.

Mikhail Katz
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  • Could you explain why do isometric embeddings of the flat disk have this property? In particular, I think, this question is quite sensitive to the degree of smoothness of the map: If the map is $C^2$ then the situation, I think, is very rigid, while in the $C^1$-setting (as in Jim's question), I think, pretty much anything is possible (in particular, the property you are stating can be violated), see Kuiper's theorem. – Moishe Kohan Jan 13 '14 at 09:03
  • @studiosus, I should have said that my answer applies in the $C^2$ setting. Showing this characterisation of flat imbeddings of the disk involves solving 2nd order PDEs. One needs at least the notion of curvature to be able to use the flatness of the image. – Mikhail Katz Jan 13 '14 at 09:08
  • @studiosus, note that in a $C^1$ setting the claim is most likely false by Nash's isometric imbedding theorem. – Mikhail Katz Jan 13 '14 at 09:23
  • @studiosus, if you have an argument using Kuiper's theorem you might as well post it as an answer. – Mikhail Katz Jan 13 '14 at 09:23
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    I just posted an answer along these lines. – Moishe Kohan Jan 13 '14 at 10:18
  • @user72694 Thanks for the great answer! Though I only asked about the unit disk, is it true that this reasoning would work for any smooth convex topological disk in the plane? – Jim Belk Jan 13 '14 at 18:30
  • I think so, so long as you are working with the flat metric. – Mikhail Katz Jan 14 '14 at 12:55
  • @JimBelk, Actually I don't think convexity is required. This should work for any imbedding of, say, a simply connected domain with smooth boundary in the plane. – Mikhail Katz Jan 18 '14 at 19:55
  • @user72694 I can't award another +50 bounty for this question, so I'm planning to award you a +50 bounty for [your answer to another question](http://math.stackexchange.com/questions/465048/mutually-densely-embedded-non-homeomorphic-topological-spaces/465068#465068). – Jim Belk Jan 18 '14 at 20:11
  • @JimBelk, thanks, that's thoughtful of you. – Mikhail Katz Jan 19 '14 at 10:29
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It is clear that the image of $\varphi_i$ is in a plane of $\mathbb{R}^3$, since the vertexs of a tetrahedron in $\mathbb{R}^3$ can not be isomeric embedded into $\mathbb{R}^2$ by $\varphi_i^{-1}$. Moreover, the images of $\varphi_1$ and $\varphi_2$ should be in the same plane since they agree on the boundary circle. Therefore, the problem can be reduced into isomeric embeddings from $\mathbb{D}$ to $\mathbb{R}^2$.

Choosing three different points from the image of the boundary that are not collinear. They should determine the position of every point of the image of $\mathbb{D}$. Therefore, I think the argument is right.

user120953
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    The image of $\varphi_i$ need not lie in a plane. It can lie on any ruled surface like that on a cylinder. If you intersect two cylinders of same radius, say the one $x^2 + y^2 = 1$ and $(1-x)^2 + z^2 = 1$, it is clear you can find a disc like region in $\mathbb{R}^2$ which have more than one isometric embedding in $\mathbb{R}^3$ using above intersection as boundary. The question is whether one can distort the cylinder to make the disc like region in $\mathbb{R}^2$ become a circular disc. I strongly suspect it is possible. – achille hui Jan 13 '14 at 07:12
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    You are confusing two notions of isometry her: Jim Belk is asking for an isometric map of Riemannian manifolds which is an infinitesimal condition, while you are thinking about distance preserving maps. The two are not the same. – Moishe Kohan Jan 13 '14 at 07:20
  • @studiosus ahh...I do misunderstood the notions. Thanks for reminding. – user120953 Jan 13 '14 at 07:47
  • Yes, my apologies. I mean isometric embedding in the sense of differential geometry, i.e. a $C^1$ embedding that preserves the lengths of curves. – Jim Belk Jan 13 '14 at 07:57
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    @achillehui Note that, for the intersection of cylinders surface that you describe, the boundary of the disk-like region has the same *image* under the two embeddings, but the two embeddings do not *agree* on the boundary -- they differ by a self-homeomorphism of the boundary curve. (This distinction isn't relevant for the disk, since any self-isometry of the boundary circle is just a rotation or reflection, and therefore extends to the whole disk.) – Jim Belk Jan 13 '14 at 08:46
  • @JimBelk good point. – achille hui Jan 13 '14 at 09:30