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Suppose $M$ is a closed $k-$manifold in $\mathbb R^n$ without boundary, can we always find a smooth function $f:\mathbb R^n\to\mathbb R^{n-k}$ such that $M$ is the level set where $f=0$?

jack
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    If $M$ is the inverse image of a regular value of $f:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n-k}$ then $M$ must be orientable. But your question is much more general, do you want 0 to be a regular value or not? – Studzinski Apr 27 '15 at 00:45
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    First you need to assume that $M$ is closed. – Jack Lee Apr 27 '15 at 00:58
  • This question is not a duplicate of the question it is being voted closed as a duplicate of. – PVAL-inactive Apr 27 '15 at 04:23

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In fact, much more is true. Every closed subset of $\Bbb R^n$ is the zero set of some function $\Bbb R^n \rightarrow \Bbb R$. See Every closed subset $E\subseteq \mathbb{R}^n$ is the zero point set of a smooth function

Community
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PVAL-inactive
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  • In fact, I mean smooth function $f$, such that f has rank $n-k$, it is the converse of a useful theorem which is used to quickly justify whether some map forms a smooth manifold. – jack Apr 27 '15 at 15:01