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Intuition regarding the "rank theorem" in Rudin's Principle of Mathematical Analysis has been asked several times in this site:

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I'm trying to understand how this theorem applies to the following example. Consider $F:{\bf R}^2\to{\bf R}$ with $$ F(x,y)=x^2+y^2 $$ So here we have $n=2$ and $m=r=1$. Take $a=(1,0)$. Then $A=[2,0]$ and thus $Y_1={\bf R}$ and $Y_2=\{0\}$.

Here are my questions:

  • What should be $V$, $H$ and $\varphi$ for this example?
  • What does the rank theorem really say about this example?
  • For Q2, your example is a bit too trivial to illustrate the power of this theorem, since the target space $\Bbb R$ is just one-dimensional... A more illustrative example might be $$f:\Bbb R^2\to \Bbb R^3, \quad (x,y)\mapsto (\cos x, \sin x\cos y, \sin x\sin y).$$ – Vim Aug 05 '17 at 15:39
  • @Vim: Or just something more simple because they'll need to invert the function. Something like $F(x,y) = (2x, y) = (u,v)$ and so $H(u,v) = (u/2, v)$ is it's inverse. – Faraad Armwood Aug 05 '17 at 15:52
  • @FaraadArmwood in the setting of the rank theorem, I don't think it's necessary (or possible in general) to inverse the function. It need not have full rank. – Vim Aug 05 '17 at 15:54
  • @Vim: of course not since it is not necessarily square. I'm inverting the restriction of the function to the codimension $r$ submanifold say $M$ i.e, $$ f|_{U \subset M}: U \to f(U):=V$$ is a map from $\mathbb{R}^r$ to $\mathbb{R}^r$ and so the determinant will be a square matrix. – Faraad Armwood Aug 05 '17 at 15:57

1 Answers1

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This is the constant rank theorem. You can find it in many differential topology texts. It's just telling you that any point $p$ s.t $Df(p)$ has constant rank $r$ is a submanifold of dimension $n-k$.

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See $A\textbf{x}$ will have dimension $k$ and so you can represent it with $k$ coordinates, but $\phi$ is in the null-space, so you fill in the remaining coordinates with $0's$. So in your example,

$$ Df(\textbf{x}) = \begin{pmatrix} 2x & 2y \end{pmatrix}$$

which has rank $1$ outside of $(x,y) = (0,0)$ and so for all $c \not = 0$, we have that $f^{-1}(c)$ is a smooth submanifold of $\mathbb{R}^2$. This is certainly true since $f^{-1}(c) = \{x^2 + y^2 = c\}$ which are circles of radius $c$. In the above, $H$ is the inverse of $f$. And the open set $V$ you get by the inverse function theorem. Since $Df(p)$ where $p \not = 0$, then $f$ is a local diffemorphism on some open subset of $p$ and we have the relation $(Df(p))^{-1} = Df^{-1}(f(p))$.

Faraad Armwood
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  • Thanks for your answer. How would you answer my first question? –  Aug 05 '17 at 15:15
  • I made an edit for that part. Since you have something of rank $r$ which is potentially less than the rank of the image space, when you map that object of rank $r$ via a linear isomorphism i.e $F'(a) = DF(a)$ then it maintains its rank. However, since it is in a space which has potentially bigger dimension, if we want to write coordinates for any vector in this image, the difference $\textbf{range dimension} - r$ will be the number of $0'$s in the coordinate representation for any one of these vectors. – Faraad Armwood Aug 05 '17 at 15:19
  • Thanks. I'm slowly digesting your answer. It seems that it might be good to translate (66) in Rudin to (11.1) in your post first. I will come back later. –  Aug 05 '17 at 16:09
  • The set $V$ is a coordinate neighbourhood of $F(p)$ in your notation, which is different from the set $V$ in Rudin's statement. (Rudin's $V$ plays the role of $U$ in Theorem 11.1). –  Aug 05 '17 at 17:54