Under what conditions can an orientable Riemannian 3-manifold be defined implicitly? Under what conditions can an orientable Riemannian 3-manifold $\Sigma$ be defined implicitly?
What I mean by implicitly is that there exists a smooth function $f:\mathbb{R}^n\to \mathbb{R}^m$, such that $\Sigma$ is diffeomorphic to $f^{-1}(0)$, and the Euclidean metric on $\mathbb{R}^n$ pulled back to $f^{-1}(0)$ is equal to the metric on $\Sigma$.
 A: By the Nash embedding theorem every Riemannian manifold $M$ embeds isometrically into some ${\Bbb R}^n$. You may then take $f(x)=dist(x,M)$ for $x\in{\Bbb R}^n$.
A: This is too long for a comment.
If you want $0$ to be a regular value of $f$, then this should be possible iff there is a smooth isometric embedding to $R^n$ with trivial normal bundle (for any dimensional manifold). To see that this is necessary, pull back a frame from $TR^{n-3}$ to the normal bundle to give a trivialization.
This will hold for a 3-manifold after crossing with $R^3$ (since the tangent bundle is trivial, the normal bundle in $R^{n+3}$  will be that of $R^n$ ). A tubular neighborhood of the manifold with trivial normal bundle in $R^{n+3}$ will be isomorphic to the unit disk in the normal bundle (for compact M). Take the map to the unit disk in $R^n$ induced by the trivialization and extend smoothly to get a defining function in which 0 is a regular value.
Remarks: I think that the argument should work more generally for manifolds with a proper isometric smooth embedding, but the tubular neighborhood might not have uniform radius. In the general dimension case, a manifold will have stably trivial normal bundle iff the tangent bundle is stably trivial (since these add trivially in K-theory).
