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The Rank Theorem: Given a map, $F: M \rightarrow N$ of constant rank, r, there exist smooth charts $(U,\phi)$ and $(V, \psi)$ such that

$\psi \circ F \circ \phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$

In the book he defines, $\phi: U_0 \rightarrow \tilde{U}_0$, where $\tilde{U}_0$ is an open cube, with $\phi(x,y) = (Q(x,y), y)$ where $x$ is short hand for $x^1,...,x^r$ and $y$ is short hand for $x^{r+1},...,x^m$ which he relabels as $y^{r+1},...,y^m$, with $F(x,y) = (Q(x, y), R(x, y))$.

Then he arrives at $$ D(F \circ \phi^{-1}) = \begin{pmatrix} I & 0 \\ \frac{\partial \tilde{R^i}}{\partial x^j} & \frac{\partial \tilde{R^i}}{\partial y^j} \end{pmatrix} $$

He mentions that it is necessary for $\tilde{U}_0$ to be an open cube so that $\tilde{R}$ is independent of $(y^1,...,y^{m-r})$, based on $D(F \circ \phi^{-1})$ above, but I thought the same held based on rank arguments alone.

He also uses the fact that $\tilde{U}_0$ later on, where he says,

Because $\tilde{U}_0$ is a cube and $F \circ \phi^{-1}$ has the form (4.6), it follows that $F \circ \phi^{-1}(\tilde{U}_0) \subseteq V_0$

My question is, why is $\tilde{U}_0$ being a cube needed for the above statements?

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    $\begingroup$ You don’t need a cube but it’s easier to use one. $\endgroup$
    – Deane
    Mar 18, 2023 at 23:51

2 Answers 2

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I'm struggling with the same result too.

I agree one does not need $\tilde{U}_0$ to be a cube in order to the derivatives vanish.

For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,\psi)$ despite the compositions.

Note that $\psi(V_0)$ is a connected subset of $\mathbb{R}^n$, which I think (haven't proof) is a cube.

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  • $\begingroup$ I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $\tilde{R}(x,y) = \tilde{R}(x,0)$. $\endgroup$
    – Jeff
    Dec 15, 2018 at 2:46
  • $\begingroup$ For one, if you have something like a T-shaped region $\tilde{R}$ might not even be defined at $\tilde{R}(x,0)$. A second problem would be if $\tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $\tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems. $\endgroup$
    – Jeff
    Dec 15, 2018 at 2:54
  • $\begingroup$ You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/… $\endgroup$ Dec 27, 2018 at 2:04
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The key idea for the second statement is that an "open cube that contains zero is closed under annihilation of coordinates":

Suppose $(x,y) \in \tilde{U}_0$. Then, $$F \circ \phi^{-1} (x,y) = (x,S(x)).$$ We want $(x,S(x))$ to be in $V_0$, and that is equivalent (by definition of $V_0$) to $(x,0) \in \tilde{U}_0$; but the latter is true, for $(x,0)$ is just $(x,y)$ after annihilating the $y$-coordinates.

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