In 1971 Jacobowitz proved that it is always possible to locally isometrically embed a surface in $\mathbb R^4$ (thanks to @doetoe for the reference). Not all surfaces can be locally isometrically embedded in $\mathbb R^3$, but some can. What is the condition a surface must obey to be isometrically embedded in $\mathbb R^3$?
For example. a segment of a 2-sphere can be isometrically embedded in $\mathbb R^3$. However, a segment of 2-torus cannot (see the comment of @QiaochuYuan here: Embedding a torus). My intuition is that a 2-torus has an extrinsic curvature in 2 orthogonal non-intersecting planes, which are possible in $\mathbb R^4$, but not in $\mathbb R^3$.
What condition or limitation must be put on a surface (e.g on its metric or curvature or anything else) to allow an isometric embedding this surface in $\mathbb R^3$?
EDIT - based on the comment of @MoisheCohen (thanks!):
Please disregard my example of embedding of a 2-torus in $\mathbb R^3$ at the moment.
The degree of smoothness is that the surface is twice differentiable in every point.
Both necessary and sufficient conditions. If I have a differential equation for surfaces, then, out of all possible solutions, I need only and all of those, which can be isometrically embedded.
The embedding is local (e.g as pictured below).
