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Let $M$ be a submanifold of $\mathbb{R}^n$. I was wondering can one always find (perhaps finite) smooth functions that cut out $M$? I was wondering about analogies between manifolds and (affine) varieties and was not sure if this holds or not. Any comments would be appreciated. Thank you.

Johnny T.
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    What do you mean by `cut out $\mathbb{R}^d$'? Do you mean if there are finitely many smooth functions that cut out $M$ inside of $\mathbb{R}^n$? – Pol van Hoften Jul 07 '19 at 08:18
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    It also matters if M is required to be a regular level set of the corresponding vector function. – Moishe Kohan Jul 07 '19 at 08:34
  • Keep in mind that every closed subset of $R^n$ is the zero level set of a smooth function. – Moishe Kohan Jul 07 '19 at 09:46
  • @user45878 yes, it is fixed. thanks – Johnny T. Jul 07 '19 at 09:50
  • @MoisheKohan It is? Is there an easy way to see this? – Johnny T. Jul 07 '19 at 09:51
  • @MoisheKohan I don't quite understand what you mean by your first comment. Could you possibly elaborate on that? – Johnny T. Jul 07 '19 at 09:53
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    This is a standard exercise in real analysis was discussed at MSE many times, e.g. https://math.stackexchange.com/questions/791248/every-closed-subset-e-subseteq-mathbbrn-is-the-zero-point-set-of-a-smooth – Moishe Kohan Jul 07 '19 at 15:57
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    More for my comment: If you know the implicit function theorem or the submersion theorem, you know that if $F: R^n\to R^k$ is smooth and $0$ is a regular value of $F$ then $F^{-1}(0)$ is a submanifold of codimension $k$ (analogous to a complete intersection in AG). Not every codimension $k$ submanifold is obtained this way. This was also discussed many times at MSE. – Moishe Kohan Jul 07 '19 at 16:00
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    To elaborate slightly on @MoisheKohan's comments: Ordinarily "cut out by smooth functions" is used to mean precisely that one has the proper number of functions (i.e., the codimension) and that their differentials are linearly independent along the level set. It then follows that the submanifold must have trivial normal bundle. This condition is sufficient, as well. – Ted Shifrin Jul 07 '19 at 18:22
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    It also depends on what you mean by `submanifold'. For example the torus of infinite genus (in $\mathbb{R}^3$) cannot be cut out by finitely many equations (I think), but this is not second countable. – Pol van Hoften Jul 07 '19 at 20:01
  • I guess I have much to learn.. thank you! – Johnny T. Jul 07 '19 at 20:36

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