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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!


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 $v \circ g \circ \phi_\alpha f_\alpha$.

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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: $$d(fg)_p(v)=v_p(fg)=f(p)v_pg+g(p)v_pf=f(p)dg_p(v)+g(p)df_p(v).$$

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    $\begingroup$ Yes you can : I will edit my answer. $\endgroup$ – Balloon Jan 31 at 15:27
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    $\begingroup$ Yes: the differential you are talking about is also named the push-forward, sending a tangent vector to another one (the one which eats $g$ and returns $v(g\circ f)$). The differential here eats a tangent vector $v_p$ and returns a scalar $v_pf$. There is a correspondance between the two: since a tangent vector in $\mathbb R$ is written $\alpha\frac{\partial}{\partial t}$, you identify the scalar $\alpha$ with the tangent vector $\alpha\frac{\partial}{\partial t}$. Some authors write $f_*$ when they talk about the push-forward, and $df$ for the differential (for $f:M\to\mathbb R$.) $\endgroup$ – Balloon Jan 31 at 15:42
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    $\begingroup$ ... the relation being $f_*(v_p)=df_p(v_p)\frac{\partial}{\partial t}$. $\endgroup$ – Balloon Jan 31 at 15:46
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    $\begingroup$ First beware that $g$ is a function from $\mathbb R$ to $\mathbb R$, not a tangent vector. And indeed for a special $g$, namely the identity $t\mapsto t$, you can check what you want: $f_*(v_p)(\mathrm{id})=v_p(\mathrm{id}\circ f)=v_p(f)$, so $f_*(v_p)=v_p(f)\frac{\partial}{\partial t}$ (since $\frac{\partial}{\partial t}(\mathrm{id})=\frac{\partial t}{\partial t}=1$). I don't know if this answers your question? $\endgroup$ – Balloon Feb 1 at 12:46
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    $\begingroup$ Also note that for general $g$, you don't have $g \circ \sum_\alpha \phi_\alpha f_\alpha=\sum_\alpha g\circ \phi_\alpha f_\alpha$: $g$ has to be linear for example. $\endgroup$ – Balloon Feb 1 at 22:25

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