2
$\begingroup$

I am having trouble figuring out an equality on dimensions in the proof below from John Lee's Introduction to Smooth Manifolds.

In the proof, how do we get $\dim T_p M - \dim T_{\Phi(p)}N=\dim T_p S$? I can't figure out why this equality holds despite working on it for a long time now. I would greatly appreciate some help.

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$

The intersection $S\cap U$ is a regular level set of $\Phi$, so the codimension of $S\cap U$ equals the dimension of $N$, by Corollary 5.13 (Submersion Level Set Theorem). In other words, $\dim M - \dim S\cap U = \dim N$. But $\dim S\cap U = \dim S$ because $S\cap U$ is open in $S$ (as $S$ is embedded in $M$ and $U$ is open in $M$), and so $$\dim T_pM - \dim T_{\Phi(p)}N = \dim M - \dim N = \dim S\cap U = \dim S = \dim T_pS.$$

$\endgroup$
2
  • $\begingroup$ This is precisely what I figured out just now after going back to the earlier theorems. One question though, why is $\dim S \cap U = \dim S$ when $S\cap U$ is open in $S$? I got the equality using the equivalence of tangent spaces. $\endgroup$ Commented Nov 29, 2023 at 19:16
  • $\begingroup$ It is a general fact: if $M$ is any manifold and $V\subseteq M$ is open, then $V$ is a manifold on its own right and $\dim V= \dim M$. Restricting the domains of charts for $M$, we get charts for $V$. $\endgroup$
    – Ivo Terek
    Commented Nov 29, 2023 at 19:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .