TITLE: Nash embedding theorem for 2D manifolds
QUESTION [22 upvotes]: The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$. What can we say in the special case of 2-manifolds? For example, can we always embed a 2-manifold in $R^3$?

REPLY [3 votes]: I had a conversation today with Christopher Croke of Penn., he pointed out that there was progress, extremely recent progress, on the local version, isometrically embedding a piece of surface in $\mathbf R^3.$ This is all in the $C^\infty$ category, it is always possible that more is known in $C^\omega$ and very likely less is known in, say $C^2.$
Anyway, if the curvature does not vanish in the neighborhood, then it can be isometrically embedded in  $\mathbf R^3.$ The same is true if the curvature vanishes at a point but some directional derivative does not. 
Well, from a new item on the arXiv by Qing Han and Marcus Khuri,

It is shown that if the Gaussian
  curvature vanishes to finite order and
  its zero set consists of two Lipschitz
  curves intersecting transversely at a
  point, then local sufficiently smooth
  isometric embeddings exist.

The title of the preprint is "On the Local Isometric Embedding in  $\mathbf R^3$ of Surfaces with Gaussian Curvature of Mixed Sign."
http://arxiv.org/abs/1009.6214 
Both Han and Khuri have published on this topic in the Journal of Differential Geometry, Han in 2003, Khuri in 2007.