Boundary defining function

Question

I'm confused about the proof of Prop 5.43 in Lee's Introduction to Smooth Manifolds

Prop 5.43 Every smooth manifold with boundary admits a boundary defining function.

A boundary defining function is defined just a few lines earlier:

If $M$ is a smooth manifold with boundary, a boundary defining function for $M$ is a smooth function $f: M \to [0, \infty)$ such that $f^{-1}(0) = \partial M$ and $df_p \neq 0$ for all $p \in \partial M$.

The proof basically lets ${(U_\alpha, \phi_\alpha)}$ be a collection of smooth charts whose domains cover $M$; defines smooth functions $f_\alpha : U_\alpha \to [0, \infty)$, and lets ${\psi_\alpha}$ be a partition of unity subordinate to this cover. He defines $f = \sum_\alpha \psi_\alpha f_\alpha$, and then has the following equation: \begin{equation} df_p(v) = \sum_\alpha (f_\alpha (p) d \phi_\alpha \rvert_p(v) + \phi_\alpha(p) d f_\alpha \rvert_p (v)), \end{equation}

My question is why does:

$f = \sum_\alpha \psi_\alpha f_\alpha \Longrightarrow df_p(v) = \sum_\alpha (f_\alpha (p) d \phi_\alpha \rvert_p(v) + \phi_\alpha(p) d f_\alpha \rvert_p (v))?$

I don't have a lot of diff geo background other than the first five chapters of Lee's Introduction to Smooth Manifolds, so if explanations could be confined to using that knowledge, it would help me out a lot. Thanks!

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This seems like something that should be easy to see, but here's what I am thinking and am stuck on:

Let $v \in T_{p} M$, so $v: C^\infty (M) \to \mathbb{R}$, and $g \in C^\infty (\mathbb{R})$. If we use the definition of a differential, we get: \begin{equation} d f_p(v)(g) = v(g \circ f) = v (g \circ \sum_\alpha \phi_\alpha f_\alpha) = v (\sum_\alpha g \circ \phi_\alpha f_\alpha) = \sum_\alpha v \circ g \circ \phi_\alpha f_\alpha, \end{equation} where in the third equality, we used the linearity of the derivation $v$.

From looking at the expression we want, it seems like we use the chain rule property of derivations, but that would apply if our expression were $v \circ \phi_\alpha f_\alpha$ rather than what we have, which is $v \circ g \circ \phi_\alpha f_\alpha$.

Answer 1

These are properties of the exterior derivative (linearity and product rule), and you can find it later in the book (Theorem 14.24). Namely, in the case of functions (0-forms) $f,g:M\to\mathbb{R}:$ $$d(f+g)=df+dg$$ and $$d(fg)=f\,dg+g\,df.$$

You can directly prove it, knowing that $df_p(v)=v_pf.$ Linearity is by definition of a tangent vector, which is a linear map $v:C^{\infty}(M)\to\mathbb R$: $$d(f+g)_p(v)=v_p(f+g)=v_pf+v_pg=df_p(v)+dg_p(v).$$ Since a tangent vector also checks the product rule $v_p(fg)=f(p)v_pg+g(p)v_pf$, the second one follows from a similar computation: $$d(fg)_p(v)=v_p(fg)=f(p)v_pg+g(p)v_pf=f(p)dg_p(v)+g(p)df_p(v).$$

Answer 2

This answer may no longer be useful to you, but I had similar concerns while working through this part of the proof. Here's an answer that only relies on techniques present in the book up to this point.

Here if $F: M \to N$ is a smooth map and $p \in M$, the notation $dF_p$ will mean the differential as defined in Lee's book, so it is a linear map $dF_p: T_pM \to T_{F(p)}N$.

Let $F, G \in C^{\infty}(M)$, so $F$ and $G$ are smooth functions $M \to \mathbb{R}$.

First of all, it doesn't seem that the statement $d(FG)_p = F(p) dF_p + G(p)dG_p$ makes much sense in general, because $dF_p$ is a map $T_pM \to T_{F(p)}\mathbb{R}$ while $dG_p$ is a map $T_pM \to T_{G(p)}\mathbb{R}$ and $d(FG)_p$ is a map $T_pM \to T_{(FG)(p)}\mathbb{R}$. In general, $F(p)$, $G(p)$ and $F(p)G(p)$ may all be distinct so the codomains are not even the same.

However, if we assume that $F(p) = 0$ we can prove the following version of the "product rule" for differentials:

Theorem 1: Let $F, G \in C^{\infty}(M)$ and $p \in M$. Suppose $F(p) = 0$. Then $d(FG)_p = G(p)dF_p$.

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Proof of Theorem 1

Let $v \in T_pM$ and $f \in C^{\infty}(M)$. Then $d(FG)_p(v)f = v(f \circ (FG))$. By Lemma 2 below, $v(f \circ (FG)) = f'((FG)(p)) v(FG)$. Then by the product rule for derivations, we obtain $$d(FG)_p(v)f = f'(0)G(p)vF = G(p)f'(F(p))vF = G(p) v(f \circ F) = G(p) dF_p(v)f$$ where we used Lemma 2 again. This holds for any $v$ and $f$ so $d(FG)_p = G(p) dF_p$.

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In the case we are interested in, we want to find $d(\psi_\alpha f_\alpha)_p$.

For each $\alpha$ such that $p \in U_\alpha$ (where $U_\alpha$ is the open cover that the partition of unity is subordinate to), we have $f_\alpha(p) = 0$ so Theorem 1 shows that $d(\psi_\alpha f_\alpha)_p = \psi_\alpha(p) d(f_\alpha)_p$.

For the other $\alpha$, we have $\psi_\alpha(p) = 0$ (because $U_\alpha$ must then be an interior chart) and so Theorem 1 shows that $d(\psi_\alpha f_\alpha)_p = f_\alpha(p) d(\psi_\alpha)_p$. But for any $v \in T_pM$ and $f \in C^{\infty}(R)$ we have $$d(\psi_\alpha)_p(v)f = v(f \circ \psi_\alpha) = v(0) = 0$$ since $\psi_\alpha = 0$ on a neighborhood of $p$ when $p$ is not in $U_\alpha$ (assume that $\overline{U_\alpha} \cap ∂M$ is empty, which can be done by shrinking the $U_\alpha$ if necessary). Hence $d(\psi_\alpha f_\alpha)_p = 0$.

So we have $$df_p = \sum\limits_{\alpha\colon p \in U_\alpha} \psi_\alpha(p) d(f_\alpha)_p$$

Note also that $f(p) = 0$ since $f_\alpha(p) = 0$ for each $\alpha$ such that $p \in U_\alpha$ and the codomains of each differential in the sum are the same, $T_0\mathbb{R}$).

Then $$df_p(v) = \sum\limits_{\alpha\colon p \in U_\alpha} \psi_\alpha(p) d(f_\alpha)_p(v)$$

This result is good enough to complete the proof of Lee's Proposition.

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Lemma 2 provides an analogue of the chain rule for derivations:

Lemma 2: Suppose $F \in C^{\infty}(M)$ and $f \in C^{\infty}(\mathbb{R})$. Let $p \in M$ and $v \in T_pM$. Then $dF_p(v)f = v(f \circ F) = f'(F(p)) vF$.

Proof of Lemma 2:

Let $(U, \varphi)$ be a smooth chart for $M$ containing $p$. With respect to this chart, we can write $v$ as a linear combination in terms of the standard basis for $T_pM$:

$$v = \sum\limits_{i=1}^n v^i\frac{\partial}{\partial x^i}\bigg\vert_p$$

Then we compute

$$v(f \circ F) = \sum\limits_{i=1}^n v^i\frac{\partial(f \circ F)}{\partial x^i}\bigg\vert_p = \sum\limits_{i=1}^n v^i\frac{\partial(f \circ F \circ \varphi^{-1})}{\partial x^i}\bigg\vert_{\varphi(p)} = \sum\limits_{i=1}^n v^i f'(F(p))\frac{\partial(F \circ \varphi^{-1})}{\partial x^i}\bigg\vert_{\varphi(p)}$$ by the ordinary chain rule $$ = f'(F(p))\sum\limits_{i=1}^n v^i \frac{\partial(F \circ \varphi^{-1})}{\partial x^i}\bigg\vert_{\varphi(p)} = f'(F(p))\sum\limits_{i=1}^n v^i \frac{\partial F}{\partial x^i}\bigg\vert_p = f'(F(p)) vF$$

Answer 3

I just noticed this question. This was a big oversight on my part -- when I wrote that equation for $df_p$, I was thinking of $df_p$ as a covector, conveniently blanking out the fact that covectors aren't introduced until Chapter 11.

Instead, I should have phrased this in terms of the action of vectors on functions. Here's what I should have written, starting two lines above the displayed formula for $df_p$ on page 119:

For each $\alpha$ such that $p\in U_\alpha$, we have $f_\alpha(p)=0$ and $v(f_\alpha) = v^n>0$ by Proposition 5.41. Thus $$ v(f) = \sum_\alpha \big( f_\alpha(p) v(\psi_\alpha) + \psi_\alpha(p) v(f_\alpha)\big). $$ For each $\alpha$, the first term in parentheses is zero and the second is nonnegative, and there is at least one $\alpha$ for which the second term is positive. Thus $v(f)>0$, which implies that $df_p(v) = (vf) d/dt\big|_{f(p)}\ne 0$, where $t$ is the standard coordinate on $\mathbb R$.

EDIT: I just added this as a correction in my online list.