Mapping sphere surface to a vector space such that distances are preserved? I have a unit radius sphere (say in 3D) centered on the origin. Thus the shortest distance between two points on the sphere is the geodesic. Is there a transformation (linear or non-linear) on the points on the sphere to a higher-dimensional space (or lower if it exists), such that the distance between the points on the sphere is preserved in the transformed space as well.
To make it more rigorous: let $x_1$ and $x_2$ be the two points on the sphere. Let $y_1=T(x_1)$ and $y_2=T(x_2)$ be the transformed points. Let $d_s(x_1,x_2)$ be the shortest distance between $x_1$ and $x_2$ on the sphere's surface. Is there any T(.) such that $$d_s(x_1,x_2)=||y_1-y_2||_2^2$$ (the Euclidean distance) in the transformed space holds.
Any reference to relevant research or literature is welcome as well. Any engineering approaches are welcome too. 

When $d=1$ it is of course possible to embed large chunks of $S^d$  isometrically in ${\mathbb R}^n$.
When $d\geq2$ it is not possible to embed even tiny pieces of $S^d$  isometrically in ${\mathbb R}^n$.
Proof. Take any three points $x_1$, $x_2$, $x_3\in S^d$ forming a small equilateral triangle in the metric of $S^d$. This then is an "ordinary" spherical triangle $\triangle$  with side length $s>0$ on the $2$-sphere $S^d\cap\langle x_1,x_2,x_3\rangle$. A map $f$ of the required kind will map the $x_i$ to the vertices $y_i$ of an equilateral triangle $\triangle'$ embedded in ${\mathbb R}^n$, and having side length $s$ as well. The distance from a vertex $y_i$ to the midpoint of the opposite side in $\triangle'$ amounts to ${\sqrt{3}\over2}s$, which is certainly different from the corresponding distance in $\triangle$.
