How to show there exists a smooth local frame $(\tilde{X_1},\cdots,\tilde{X_n})$ on some neighborhood of $A$ such that $\tilde{X_i}|_A=X_i$?

Question

Let $M$ be a smooth $n$-manifold with or without boundary. If $(X_1,\cdots,X_n)$ is a linearly independent $n$-tuple of smooth vector fields along a closed subset $A\subset M$, how to show there exists a smooth local frame $(\tilde{X_1},\cdots,\tilde{X_n})$ on some neighborhood of $A$ such that $\tilde{X_i}|_A=X_i$ for $i=1,\cdots,n$?

Answer 1

This is an exercise in Lee's book on smooth manifolds. Here are some hints to get you started:

$1).\ $ the result follows from this more general result (also an exercise in the same book!): let $\pi : E \to M$ be a smooth vector bundle over a smooth manifold $M$ and $A$ a closed subset of $M$. If $\sigma: A \to E$ is a smooth section of $E|_A$ then for each open subset $A\subseteq U \subseteq M$ there exists a global smooth section $\tilde \sigma$ such that $\tilde \sigma|_A = \sigma$ and supp $\tilde \sigma\subseteq U.$ To show this, follow this sketch:

$2).\ $ Each $p\in A$ has a neighborhood $W_p\overset{wlog}\subseteq U$ such that there is a $\tilde \sigma:W_p\to E$ satisfying $\tilde \sigma _{W_p\cap A}=\sigma$.

$3).\ $ Using $2).$, find a cover of $M$ (the sets will be indexed by $p\in A$ except for one of them, $W_0$ which will be a different but obvious choice) and a partition of unity $\{\psi_0,\psi_{p}\}_p $ subordinate to it.

$4).\ $ Apply the extension theorem for functions to $\psi_p\tilde \sigma_p$.

$5).\ $ Define $\tilde \sigma$ to be the obvious sum, and check that it is the desired section.

Answer 2

From J. M. Lee's book "Introduction to smooth manifolds" we have the next proposition:

10.12 Let $\pi:E\to M$ be a (smooth) vector bundle. Let $C$ be a closed subset of $M$ and $s:C\to E$ a (smooth) section of $\pi_C:E|_C\to C$. For each open subset $U \subseteq M$ containing $C$, there exists a (smooth) section $\bar{s}$ of $U$ such that $\bar{s}|_C=s$ and $\operatorname{supp}\bar{s}\subseteq U$.

Proof. Let $U\subseteq M$ be an open subset such that $C\subseteq U$. For every $\alpha\in C$ we can find an open subset $V_\alpha\subseteq U$ and a section $s_\alpha:V_\alpha\to E$ such that $s_\alpha|_{V_\alpha\cap C}=s$. The family of subsets $\{V_\alpha\}_{\alpha\in C}\cup\{U-C\}$ is an open cover of $U$, so we can find a (smooth) partition of unity $\{\psi_\alpha\}_{\alpha\in C}\cup\{\psi_0\}$ subordinate to it with $\operatorname{supp}\psi_\alpha\subseteq V_\alpha$ and $\operatorname{supp}\psi_0\subseteq U-C$. For each $p\in C$, $\psi_0(p)=0$ and for every $p\in U$, $\psi_\alpha(p)=0$ except for a finite number of $\alpha$s and $\sum_{\alpha\in C}\psi_\alpha(p)=1$. Let $s_0$ be a (smooth) global section of $U$. Such a section exists: take the everywhere zero section. Hence the section we are looking for is $$\bar{s}:=\psi_0s_0+\sum_{\alpha\in C}\psi_\alpha s_\alpha$$ because for every $p\in C$ $$\Big{(}\psi_0s_0+\sum_{\alpha\in C}\psi_\alpha s_\alpha\Big{)}(p)=\psi_0(p)s_0(p)+\sum_{\alpha\in C}\psi_\alpha(p)s_\alpha(p)=0s_0(p)+\sum_{\alpha\in C}\psi_\alpha(p)s(p)=$$ $$\Big{(}\sum_{\alpha\in C}\psi_\alpha(p)\Big{)}s(p)=1s(p)=s(p)$$ Finally by the properties of partitions of unity the supports of the $\psi$s are locally finite, so $$\operatorname{supp}\tilde{s}=\overline{\operatorname{supp}s_0\cup\Big{(}\bigcup_{\alpha\in C}\operatorname{supp}\psi_\alpha\Big{)}}=\operatorname{supp}s_0\cup\Big{(}\bigcup_{\alpha\in C}\operatorname{supp}\psi_\alpha\Big{)}\subseteq U$$

10.15 If $C\subseteq M$ is a closed subset and $\{s_1,\dots,s_k\}$ are linearly independent (smooth) sections of $\pi_C:E|_C\to C$, there exists a (smooth) reference frame $\{\bar{s}_1,\dots,\bar{s}_k\}$ of some open subset $U$ such that $C\subseteq U$ and $\bar{s}_i|_C=s_i$ for all $i\in\{1,\dots,k\}$.

Proof. Thanks to point 10.12, for any open subset $V\subseteq M$ such that $C\subseteq V$, we can find a set $\{\bar{s}_1,\dots,\bar{s}_k\}$ of sections of $V$ such that $\bar{s}_i|_C=s_i$ for all $i\in\{1,\dots,k\}$. Since this sections restricted to $C$ are linearly independent, they are linearly independent in some open subset $U\subseteq V$ such that $C\subseteq U$ because being linearly independent is an open property. This is because if the sections are linearly independent at a point, its determinant (as a function) isn't zero at that point, so it isn't zero in some open neighbourhood of the point.