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 afreeglobal (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 aglobalisometric 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.