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I have a question adressing the "abstract" definition of a submanifold and the version in the special case of $\mathbb{R}^n$. This is what I mean:

I looked up these two definitions of a $k$-dimensional submanifold of $\mathbb{R}^n$:

  1. A subset $M⊂\mathbb{R}^n$ is called a $k$-dimensional submanifold if for every $a∈ M$ there is an open neighborhood $U∈\mathbb{R}^n$ and smooth functions $f_1,\dots,f_{n-k}\colon U\to\mathbb{R}$ such that
  • $M\cap U = \{x∈ U : f_1=\dots f_{n-k}(x)=0\}$
  • $rank \begin{pmatrix} \frac{∂ f_1}{∂ x_1} & \dots & \frac{∂ f_1}{∂ x_n}\\ \vdots & & \vdots\\ \frac{∂ f_{n-k}}{∂ x_1} & \dots & \frac{∂ f_{n-k}}{∂ x_n}\\ \end{pmatrix} = n-k$
  1. (Loring Wu, p. 100) A subset $M$ of a manifold $N$ of dimension $n$ is a regular submanifold of dimension $k$ if for every $p∈ M$ there is a coordinate neighborhood $(U,x^1,\dots,x^n)$ of $N$ such that $U\cap M$ is defined by the vanishing of $n-k$ of the coordinate functions.

So the second definition is more general than the first one by taking $N=\mathbb{R}^n$. What I don't understand where the rank condition in the first definition comes from. The "vanishing of the functions part" of the second definition is the same as in the first one. I am wondering how to imply the second condition. In short: Are these definitions equivalent by taking $N=\mathbb{R}^n$ and if yes, why?

Thank you for your help!

RobRTex
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    No it’s not as simple as setting $N=\Bbb{R}^n$. To prove the equivalence, you need to use the inverse/implicit function theorem. See here for four equivalent definitions and the proof of their equivalence (unfortunately I didn’t prove them in the same order, so you might need to do some extra work, but you should be able to extract a proof for the specific directions you need). – peek-a-boo Nov 07 '23 at 20:52
  • Similar to an earlier question Equivalence of definitions of a submanifold? – peterwhy Nov 07 '23 at 20:54
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    Not "similar". Identical. They seem to have been tagged at the exact same time, down to the second, so it's probably some kind of glitch. You should remove the other one. – Arturo Magidin Nov 07 '23 at 20:55
  • That's Loring Tu, not Wu. Yes, the definitions are equivalent, because you choose a chart on a neighborhood of a point in $N$, thereby making $N=\Bbb R^n$. – Ted Shifrin Nov 08 '23 at 00:26

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