Is an isometric embedding of a disk determined by the boundary?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.