Isometric embedding

I am confused by the term "Isometric Embedding". To my knowledge, this refers to a distance preserving map from a space to another (a mapping $f:(E,d_1) \to (F,d_2)$ such that $d_2(f(x_1), f(x_2)) = d_1(x_1, x_2) )$. But I have the following problem :

On one side, I see papers saying that an isometric embedding of a sphere (with its geodesic distance) to an euclidean space cannot exist; e.g., see The Sphere is not Flat by P. L. Robinson.

On the other side, I see the Nash embedding theorem which says that any surface can be embedded into $R^n$ for some $n$.

What didn't I understand ?

Thanks!

The Nash embedding theorem uses a different $d_2$. When you have a submanifold of a Riemann manifold there's an induced Riemann metric, which comes from the inner product on all the tangent spaces of the ambient manifold. Call this metric $d_3$. In the Nash embedding theorem, in your statement above, replace $(F,d_2)$ with $(f(E), d_3)$. That's the kind of isometric embedding this theorem refers to.
• I am still a little confused. So when Nash talks about isometric embedding, Does he really mean that 2 points on a Riemannian manifold $M$ of dimension $m$, separated a (geodesic) distance $D$ before the embedding, will be now separated the same distance $D$ on $\mathbb{R}^n$, but following a straight line? Thanks Braco – user66877 Mar 15 '13 at 14:06
The usual 2-sphere exists naturally in $\mathbb R^3$, and in general the usual definition of $S^n$ is as a particular subset of $\mathbb R^{n+1}$ with the induced metric. In that case, the identity map is a locally metric-preserving embedding into $\mathbb R^2$, but it doesn't preserve the global distance. To wit, two diametrically opposed points have distance $2$ in $\mathbb R^3$ but distance $\pi$ along geodesics in the sphere itself.