In class we learned the following variant of the Implicit Function Theorem:

Suppose $f:U \to \mathbb{R}^{n-k}$, where $U \subseteq \mathbb{R}^n$, is such that $Df(p)$ has full row rank for all $p \in U$. Then there exist diffeomorphisms $\alpha, \beta$ s.t. $\alpha: U' \to U$ and $\beta: F(U) \to W$, and $\beta \circ f \circ \alpha = \pi$, the normal orthogonal projection $\mathbb{R}^n \to \mathbb{R}^{n-k}$.

My Question: Why do we need $\beta$? WLOG suppose that the first $n-k$ columns of $Df(p)$ are linearly independent. Then the function $F:(x_1, \dotsc, x_n) \mapsto (f_1, \dotsc, f_{n-k}, x_{n-k+1}, \dotsc, x_n)$ has an inverse in a neighborhood of $p$, so

$$f \circ F^{-1}: (f_1, \dotsc, f_{n-k}, x_{n-k+1}, \dotsc, x_n) \mapsto (f_1, \dotsc, f_{n-k}).$$

So why do we need $\beta$?

  • 1
    $\begingroup$ In the way you did, you don't have a projection. You should note that $\pi$ as described is the linear projection $\pi (x_1,...,x_{n-k-1},x_{n-k},x_{n-l+1},...,x_n)=(x_1,...,x_{n-k-1},x_{n-k})$, which is linear, and mucch easier to work with. This is also called the Local Form for Submersions, widely used in Differential Topology and in several other subjects. $\endgroup$ – Marra Sep 29 '13 at 1:01
  • $\begingroup$ @GustavoMarra But why is it not a projection? $f \circ F^{-1}$ preserves the first $n-k$ variables and discards the rest, right? Here $f_1, \dotsc, f_{n-k}$ are just standing for the coordinates of the image under $f$ of the neighborhood of $p$. $\endgroup$ – Eric Auld Sep 29 '13 at 3:44

You are right: $\beta$ is unnecessary. It probably got into the statement from the more general rank theorem, where it is needed.

Here is another way to see this. The statement says that $f = \beta^{-1}\circ \pi \circ \alpha^{-1}$. Informally, we shuffle the domain, then project, then shuffle the image. But whatever shuffling we want to do after the projection could just as well be done before it, by moving the fibers of the projection map. Formally speaking, $\beta^{-1}\circ \pi $ can be written as $\pi \circ \gamma$ where $\gamma$ is a diffeomorphism of the cylinder domain lying above the domain of $\beta^{-1}$. Just let $\gamma$ keep the $k$ vertical coordinates as they are.

  • $\begingroup$ That's very helpful, thank you. $\endgroup$ – Eric Auld Oct 2 '13 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.