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My question is based on this one and is prompted by my attempt to understand the constant rank theorem.

Specifically, suppose I have a smooth map $F : M \rightarrow R^k$ where $M \subset R^n$ is an $m$-dimensional manifold and $F$ has rank one at all points of $M$. Here by rank I mean the rank of the Jacobian matrix of $F$. I want to show that the image of $F$ is a curve in $R^k$. The referenced question uses a proof found in Buck's "Advanced Calculus" that is based on the implicit function theorem to show that, for every $x \in M$ (here $M \subset R^3$) there is a neighborhood $U \subset M$ containing $x$ in which one of the variables can be expressed as a function of the others.

Could we not use the following argument: Given $x \in M$ the constant rank theorem tells us there are smooth local charts $(\phi,U)$ centered at $x$ and $(\psi,V)$ centered at $y = F(x)$ with $F(U) \subset V$ and such that in $U$ $F$ has the coordinate representation:

$\tilde{F} (x_1,\ldots,x_m) = (x_1,0,\ldots,0)$

Another way of putting this is that $F = \psi^{-1} \circ \tilde{F} \circ \phi$. Hence $F(U) = \{ \psi^{-1} (t), t\in R \}$ (here I'm assuming $\phi(U) = R^m$) which is the definition of a curve, since $\psi$ is a diffeomorphism. Of course this is only a local argument, but I'd like to know what, if any, mistakes I'm making before proceeding.

user167131
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  • Your understanding of the original proof is evidently flawed, because your claim that "one of the variables can be expressed as a function of the others" is not at all sufficient to prove the result. The actual proof has to be similar to yours (at very least to provide the "variables" you speak of). The implicit function theorem can be considered a consequence of the constant rank theorem, so yes, you can use that theorem instead to prove anything proved with the implicit function theorem. – Paul Sinclair Aug 30 '23 at 15:12
  • @Paul Sinclair Probably, I wasn't careful enough in my interpretation. I simply linked the original post which references Buck's proof directly. My main concern is whether or not my proof is correct and, if not, where it fails. It actually seems simpler to me than the Advanced Calculus one. Thank you for posting. – user167131 Aug 30 '23 at 16:52

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