Problem 8-1 (Extension Lemma for Smooth Vector Fields)

Question

extbf{Lemma 8.6 (Extension Lemma for Vector Fields).} Let $M$ be a smooth manifold with or without boundary, and let $A \subseteq M$ be a closed subset. Suppose $X$ is a smooth vector field along $A$. Given any of open subset $U$ containing $A$, there exists a smooth global vector field $\widetilde{X}$ on $M$ such that $\widetilde{X}|_A = X$ and $\operatorname{supp}(X) \subseteq U$.

extbf{Problem 8-1.} Prove Lemma 8.6 (the extension lemma for vector fields).

Elu Thingol · Accepted Answer

The proof strategy is identical to that of Lemma 2.26. For the sake of completeness, we provide the full proof in what follows.

\begin{proof}
Following the definition, for each $p \in A$, choose a neighborhood $V_p$ of $p$ and a smooth vector field $\widetilde{X}_p: V_p 	o TM$ that agrees with $X$ on $V_p \cap A$. Replacing $V_p$ by $V_p \cap U$, we may assume that $V_p \subseteq U$. The family of sets $\{V_p \colon p \in A\} \cup \{M \setminus A\}$ is an open cover of $M$. Let $\{\psi_p:V_p 	o \mathbb{R}\}_{p\in M} \cup \{\psi_0\}$ be the partition of unity subordinate to this cover, i.e.,
\begin{enumerate}[label=(oman*)]
    \item $0 \leq \psi_p(x) \leq 1$ for all $p \in M$ and for all $x \in M$.
    \item $\operatorname{supp}(\psi_p) \subseteq V_p$ and $\operatorname{supp}(\psi_0) \subseteq M \setminus A$. 
    \item the collection of supports $\{\operatorname{supp}\psi_p\}$ is locally finite
    \item $\sum_{p \in M} \psi_p(x) = 1$ for all $x \in M$.
\end{enumerate}

For each $p \in A$, the product $\psi_p \cdot \widetilde{X}_p$ is a smooth vector field on $V_p$. Since $\psi_p$ is supported inside $V_p$, so is $\psi_p \cdot \widetilde{X}_p$; thus, $\psi_p \cdot \widetilde{X}_p$ has a smooth extension to all of $M$ if we interpret it to be zero on $M \setminus \operatorname{supp}(\psi_p)$. Thus, we can define $\widetilde{X}: M 	o TM$ by
\begin{equation*}
    \widetilde{X}(x) := \sum_{p \in M} \psi_p(x) \cdot \widetilde{X}_p(x).
\end{equation*}
Because the collection of supports $\{\operatorname{supp}\psi_p\}$ is locally finite, this sum has only a finite number of nonzero terms in a neighborhood of any point of $M$, and therefore defines a smooth function. Furthermore, for each $x \in A$, $\psi_0(x)=0$ and $\widetilde{X}_p(x)=X(x)$ whenever $\psi_p(x) 
eq 0$; hence,
\begin{equation*}
    \widetilde{X}(x) := \sum_{p \in M} \psi_p(x) \cdot \widetilde{X}_p(x) = \left(\psi_0 + \sum_{p \in M} \psi_p(x) ight) X(x) = X(x),
\end{equation*}
so $\widetilde{X}$ is indeed an extension of $X$. It follows from Lemma 1.13(b) that
\begin{equation*}
    \operatorname{supp}(\widetilde{X}) = \overline{\bigcup_{p \in A} \operatorname{supp}(\psi_p)} = \bigcup_{p \in A} \operatorname{supp}(\psi_p) \subseteq U.
\end{equation*}
\end{proof}

Lucas MacQuarrie · Answer

\begin{proof}
By definition of smoothness, $X$ extends to a smooth vector field $Y$ defined over an open set $V$ containing $A$. Replacing $V$ by $V \cap U$, we may assume $V \subset U$. Consider a bump function $\psi$ with support $\text{supp}\{\psi\} \subset V$ and $\psi =1$ on $A$. The vector field $\psi \cdot Y$ is smooth on $V$ since $Y$ is smooth on $V$ and $\psi$ is smooth on $V$. Furthermore, since $\psi$ has support in $V$, so does $\psi \cdot Y$. $\psi \cdot Y$ can then be extended to all of $M$ by defining it to be $0$ outside $V$. Indeed, 
$$\left. \psi \cdot Y \right|_{A} = 1 \cdot X = X$$
and 
$$\text{supp}\{\psi \cdot Y\} \subset V \subset U$$
as required.

\end{proof}

Problem 8-1 (Extension Lemma for Smooth Vector Fields)

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Problem 8-1 (Extension Lemma for Smooth Vector Fields)

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