# How does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in\ker d\Phi_p$?

I'm confused about an immediate corollary in John Lee's Smooth Manifolds. Proposition 5.38 says

Suppose $M$ is a smooth manifold and $S\subset M$ is an embedded submanifold. If $\Phi\colon U\to N$ is any local defining map for $S$, then $T_pS=\ker d\Phi_p\colon T_pM\to T_{\Phi(p)}N$ for each $p\in S\cap U$.

It says an immediate corollary is that if $S\subseteq M$ is a level set of a smooth submersion $\Phi=(\Phi^1,\dots,\Phi^k)\colon M\to\mathbb{R}^k$, then $v\in T_pM$ is tangent to $S$ iff $v\Phi^1=\cdots=v\Phi^k=0$.

Why does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in T_pS$? By the proposition, how does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in\ker d\Phi_p$? I don't see how to relate this to the individual components though to use $v\Phi^1=\cdots=v\Phi^k=0$.

## 2 Answers

Here's one way to prove it. If you write $d\Phi_p(v) = a^i \partial/\partial x^i$ (using the summation convention), you can compute each coefficient $a^k$ by letting $d\Phi_p(v)$ act on the $k$-th coordinate function $x^k\colon \mathbb R^n\to \mathbb R$: $$a^k = d\Phi_p(v)(x^k) = v (x^k \circ \Phi) = v(\Phi^k).$$

Part of the reason you had a hard time following @Shuchang's answer is because he's probably thinking of $d\Phi^k_p$ as a covector, but in Chapter 5 I haven't yet introduced covectors. This will be easier to see after you've read the introduction to covectors in Chapter 11.

• Thanks again Professor Lee. – Clara Aug 10 '14 at 0:38

It's redundant to consider $f$. A straightforward verification that since $v\in\ker d\Phi_p$, then $0=d\Phi_p(v)=v\Phi_p$ and hence $v\Phi^k=0$ for each $k$.

• I'm not assuming $v\in\ker d\Phi_p$, only that $v\Phi^1=\cdots=v\Phi^k=0$. Also, isn't $d\Phi_p(v)$ a tangent vector in $T_{\Phi(p)}(\mathbb{R}^k)$, but $v(\Phi)$ a real number? How can they be equal? – Clara Aug 9 '14 at 2:35
• @Clara $v\in\ker d\Phi_p$ is equivalent to $v\in T_pS$ and hence $v$ is tangent to $S$. – Shuchang Aug 9 '14 at 2:39
• I know, that's the proposition, but how does $v\Phi^1=\cdots=v\Phi^k=0$ imply $v\in\ker d\Phi_p$? That's my question. – Clara Aug 9 '14 at 2:40
• @Clara $v\Phi^k_p=0$ implies $v\Phi_p=d\Phi_p(v)=0$ – Shuchang Aug 9 '14 at 2:42
• But $v\colon C^\infty(M)\to\mathbb{R}$, so isn't $v(\Phi)\in\mathbb{R}$, but $d\Phi_p\colon T_p(M)\to T_{\Phi(p)}(\mathbb{R}^k)$, so $d\Phi_p(v)\in T_{\Phi(p)}(\mathbb{R}^k)$? – Clara Aug 9 '14 at 2:43