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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 $\Sigma$.

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    $\begingroup$ Using the word "implicitly" are you referring to the implicit function theorem, i.e. do you want $0$ to be a regular value of $f$? $\endgroup$
    – Ryan Budney
    Commented Jul 14, 2022 at 23:10
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    $\begingroup$ In your setting, one can make the submanifold to consist of regular points of the defining function (since orientable 3-manifolds have trivial tangent bundle). $\endgroup$
    – Moishe Kohan
    Commented Jul 15, 2022 at 4:45
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    $\begingroup$ Thanks, ThiKu has answered your question provided the Riemann manifold is complete. $\endgroup$
    – Ryan Budney
    Commented Jul 16, 2022 at 0:55
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    $\begingroup$ The function $f^2$ is smooth near the manifold $\Sigma$, but it could potentially have some non-smooth points far from $\Sigma$. These could be smoothed-away using bump functions. $\endgroup$
    – Ryan Budney
    Commented Jul 16, 2022 at 8:36
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    $\begingroup$ You want to smooth $f^2$, that way you do not need to modify the function near $\Sigma$. $\endgroup$
    – Ryan Budney
    Commented Jul 19, 2022 at 15:23

2 Answers 2

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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 $x\in{\Bbb R}^n$.

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  • $\begingroup$ Thank you for your clever answer, but I edited the question to require that $f$ be smooth. I think the function you described is not smooth if more than one path minimises the distance. $\endgroup$
    – dennis
    Commented Jul 19, 2022 at 10:52
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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).

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