Corollary 5.39, Lee's Smooth manifold

Question

How does Corollary 5.39 follows from Proposition 5.38? Clearly it suffices to show that $v\in d\Phi_p$ iff $v\Phi^1=\cdots=v\Phi^k=0$, but it doesn't seem obvious to me. Am I missing something?

Answer 1

$S$ is a level set of $\Phi$ is equivalent to saying that $\Phi$ is a defining map of $S$, $5.38$ implies that for every $p\in S, v\in T_pM, v\in T_pS$ if and only if $d\Phi_p(v)=(d\Phi^1_p(v),...,d\Phi^k_p(v))=(v\Phi^1,...,v\Phi^k)=0$.

Usually, if $X$ is a vector defined on $M$ and $f:M\rightarrow \mathbb{R}$ a function, $Xf(p)$ is the derivation defined by $X$ at $f$ which is $df_P.X(p)$, $Xf(p)$ depends only of the value of $X$ at $p$, this enables to extend this definition to $T_pM$, for every $v\in M$, we can construct a vector field $X^v$ which support is in chart which contains $p$ such $X^v(p)=v$ by using bump functions, $V^vf(p)=v.f$.

Answer 2

I just want to develop a bit of Tsemo's answer.

Note that $\Phi$ is a global defining function for $S$. The regularity of $S$ follows from the submersion hypothesis.

Let's suppose $v \in T_pM$ is a tangent vector to $S$. Thus, by the previous proposition, $v \in \text{Ker }d\Phi_p$, namely $$v(- \circ \Phi)=0 \in T_{\Phi(p)}\mathbb{R}^k.$$

Now, Let's see $\mathbb{R}^k$ as a product manifold, and recall the identification of the $T_{\Phi(p)}\mathbb{R}^k$ with the sum of the tangent spaces to the factors (Proposition 3.14): $$w \in T_{\Phi(p)}\mathbb{R}^k \tilde{\longrightarrow} (w(-\circ \pi_1), ..., w(-\circ \pi_n)) \in \oplus_{i=1}^n T_{\Phi(p)_i}\mathbb{R}.$$

Then, for each $i$, $$v(- \circ \Phi^i)=0;$$

in particular, $0=v(Id_{\mathbb{R}}\circ \Phi^i)=v(\Phi^i).$

Conversely, let's suppose that for all $i$, $v(\Phi^i) = 0$. In order to apply the previous proposition, we want to show that $v \in \text{Ker }d\Phi_p$. But that's equivalent to proving that $v(- \circ \Phi^i)=0$ for all $i$. Using the "identity global coordinates" on $\mathbb{R}$, we get that $$v(- \circ \Phi^i) = c_i \frac{d}{dt}\Big{|}_{\Phi(p)_i},$$

and acting on the identity function as before, we get $$0=v(\Phi^i)=v(Id_{\mathbb{R}}\circ \Phi^i) = c_i \frac{d(Id_{\mathbb{R}})}{dt}\Big{|}_{\Phi(p)_i} = c_i,$$

for all $i$.