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Let $F:M\to \mathbb R^N$ be an embedding. This embedding induces a metric $g_F=dF\cdot dF$ on $M$, that turns $F$ into an isometric embedding. Probably the hardest part of the proof of the Nash Embedding Theorem is to prove a deformation result of the following type (see e.g. Thm 1.1 here):

Thm: If $F:M\to\mathbb R^N$ is a free global (analytic) embedding and $h$ is a $C^k$ $(2,0)$-tensor in a neighborhood of zero, then there exists a $C^k$ map $V:M\to\mathbb R^N$ such that $$g_{F+V}=g_F+h, \quad (*)$$ i.e., $F+V$ is a global isometric embedding of $(M,g_F+h)$ into $\mathbb R^N$.

The free condition means that the vectors $\partial F/\partial x_i$ and $\partial^2 F/\partial x_i\partial x_j$ are linearly independent at every point. The use of the above result to prove the Nash Embedding Theorem is when the ambient dimension $N$ is large, more precisely, $N\geq n(n+3)/2$. In this case, the free condition implies that the linearized version of $(*)$ is solvable, and then the problem follows by, e.g., the Nash-Moser method. My interest, however, is when $N< n(n+3)/2$. In this case, by free we mean that the vectors $\partial F/\partial x_i$ and $\partial^2 F/\partial x_i\partial x_j$ span a subspace with largest possible dimension (although, in the particular case I'm interested, although they do span such a subspace, some of them are identically zero!). In this case, the linearized system is overdetermined, and may not admit solutions...

My question is if (regardless of the linearized system being solvable or not) there are any known conditions under which the Thm above also holds for free embeddings with $N< n(n+3)/2$. In other words, when can one deform (global) isometric embeddings of $(M,g_0)$ in low codimension to (global) isometric embeddings for all metrics near $g_0$?

The analogous question for local isometric embeddings is dealt with in a paper by Bryant, Griffiths and Yang (Duke Math, 1983), but I do not know of any results (or counter-examples) regarding the global problem.

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It holds for compact surfaces in $\mathbb R^3$. In this case the embedding is free if and only if the surface is convex, so everything follows from Alexandrov's embedding theorem. –  Anton Petrunin Jun 28 '12 at 10:26
@Anton : I don't understand, the definition of free in the question seems to be in this case that the second fundamental form does'nt vanish entirely at a point. Where am I wrong ? –  BS. Jul 4 '12 at 11:30

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