isometric embedding of a sphere Hi,
Is there a way to find a function $F : \mathbb S^2 \rightarrow \mathbb R^3$ of class $\mathcal C^1$, minimizing 
$$\int_{\mathbb S^2\times\mathbb S^2} (d(F(x),F(y)) - \delta(x,y))^2 ~dx ~dy$$
, where $d$ stands for the euclidean distance in $\mathbb R^3$ and $\delta$ the geodesic distance on the sphere $\mathbb S^2$?  
I tried to perform a Multi-Dimensional Scaling to get this least square solution for a finite set of point, but it seems that the solution was just the identity... is that right? 
Thanks!
 A: Although I cannot answer your question precisely, I thought I would suggest a possible direction
to pursue: embeddings of finite metric spaces with low distortion.
With those keywords you will hit a rich literature.
Perhaps the place to start is this Handbook article by Piotr Indyk and Jiri Matousek:

"Low distortion embeddings of finite metric spaces,"
  Handbook of Discrete and Computational Geometry,
  177-196, CRC, 2004. (Google books link)

For example, Bourgain's embedding theorem say that any $n$-point metric space can 
be embedded in $\ell_2$ with $O( \log n )$ distortion 
(where distortion is defined
by a factor times the source distance $\delta(x,y)$ bounding the target distance—not quite your
least squares, but a reasonable measure).
Unfortunately this embedding might be into a rather high
dimension, which is not what you want.
Matousek proved that there are $n$-point metric spaces that
require distortion $\Omega(n^{1/2})$ for embedding into $\ell^3_2$ (i.e., $\mathbb{R^3}$),
which does not bode well for your problem.
Unfortunately, negative results abound.  Here is one, not directly relevant (because both spaces have $n$ points), but its more recent references may help:

"Hardness of Embedding Metric Spaces of Equal Size,"
  Subhash Khot and
  Rishi Saket,
  Proceedings of the 10th International Workshop on Approximation, 2007.

To skirt these negative results, you might have to somehow exploit the fact that your
source distances are geodesics on a sphere. 
A: Let's say that the sphere is with the center at $(0,0,0)$ and has the radius $1$. Let's take the points $p_1=(0,0,1)$, $p_2=(0,0,-1)$, $p_3=(1,0,0)$, $p_4=(0,1,0)$. The distances are $\delta(p_1,p_3)=\delta(p_1,p_4)=\delta(p_3,p_4)=\delta(p_2,p_3)=\delta(p_2,p_4)=\pi/2$, and $=\delta(p_1,p_2)=\pi$. Now let's try to find in $\mathbb R^3$ four points $q_i$ at the same distances. $q_1,q_3,q_4$ and $q_2,q_3,q_4$ are equilateral triangles, with the length of the edges equal to $\pi_2$. The distance $d(q_1,q_2)\leq \pi\sqrt 3/2$. But we need to have $d(q_1,q_2)=\pi$, and this cannot happen.

Edit:
It seems that the question has changed. I'll let this counterexample anyway.
