I've been recently doing some problems in John Lee's book Introduction to Smooth Manifolds (Second Edition), and I wanted to ask how one would approach the following problem.
Let $M$ be a smooth manifold with boundary. Show that there exists a global smooth vector field on $M$ whose restriction to $\partial M$ is everywhere inward-pointing, and one whose restriction to $\partial M$ is everywhere outward-pointing.
After doing some digging in the book, I found the definition (see Page 118) that Lee is referring to when he says "inward/outward-pointing", but it's not necessarily for vector fields.
Definition: If $p\in M$, a vector $v\in T_pM\backslash T_p\partial M$ is said to be inward pointing if for some $\epsilon>0$ there exists a smooth curve $\gamma:[0,\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$, and it is outward pointing if there exists such a curve whose domain is $(-\epsilon,0]$.
My first question is how to adapt this to a vector field, and make sure that it is inward/outward-pointing everywhere. Here is my proposed definition (as far as I'm aware, it's not in the book).
Definition (Own): Let $M$ be a smooth manifold with boundary, and let $X$ be a smooth vector field on $M$. Furthermore, let $A\subseteq \partial M$ be a closed subset. Then $X$ is said to be inward-pointing everywhere on $A$ if each $X_p\in T_pM$ is inward-pointing for all $p\in A$. We say that $X$ is outward-pointing everywhere on $A$ if $X_p\in T_pM$ is outward-pointing for all $p\in A$.
I assume that an outward/inward-pointing vector field is defined similarly. So now that all of the above groundwork has been laid out, here is my ultimate question.
Question(s): Is the above definition the correct interpretation of the problem? If so, I was thinking of just applying Lemma 8.6, or trying to use Proposition 8.7 to prove the claim, but I was unsure if this was the correct approach. For those who don't have the book, these say the following.
(Lemma 8.6). Let $M$ be a smooth manifold with or without boundary, and let $A\subseteq M$ be a closed subset. Suppose that $X$ is a smooth vector field along $A$ (meaning $X_p\in T_pM$ for each $p\in A$). Given any open subset $U$ containing $A$, there exists a smooth vector global vector field $\widetilde{X}$ on $M$ such that $\widetilde{X}|_A=X$ and $\text{supp}~\widetilde{X}\subseteq U$.
(Proposition 8.7). Let $M$ be a smooth manifold with or without boundary. Given $p\in M$ and $v\in T_pM$, there is a smooth global vector field $X$ on $M$ such that $X_p=v$.
Any ideas for how to solve this problem would be greatly appreciated!
Best Answer
Throughout my answer, I will focus on inward-pointing vectors and vector fields; to get outward-pointing ones, just put an extra minus sign in front of the vectors and vector fields. What follows is a standard partition of unity argument.
Let us make the following two observations:
Let $M_0$ denote the manifold interior of $M$ (i.e those points of $M$ for which a chart maps diffeomorphically onto an open subset of $\Bbb{R}^n$, i.e $M\setminus\partial M$). For each $p\in \partial M$, let $U_p$ be a neighborhood as above and let $X_{U_p}$ be a smooth vector field as above. Then, we have an open cover $\{M_0\}\cup\{U_p\}_{p\in\partial M}$ for $M$. Fix a smooth compactly-supported partition of unity $\{\phi_0,\phi_p\}_{p\in M}$ which is subordinate to the above open cover. Now, for each $p\in\partial M$, consider the vector field $\xi_p$ defined on all of $M$ given by \begin{align} \xi_p(q)&:= \begin{cases} \phi_p(q)\cdot X_{p}(q)&\text{if $q\in U_p$}\\ 0&\text{else} \end{cases} \end{align} This is clearly smooth on $U_p$; and if $q\notin U_p$, then this lies outside the support of $\phi_p$, so the complement $(\text{supp }\phi_p)^c$ is an open neighborhood of $q$ in $M$ which doesn’t intersect the support of $\phi_p$ (obviously), meaning that by definition $\xi_p$ vanishes on this open set, and hence is smooth in a neighborhood of $q$. So, $\xi_p:M\to TM$ is globally defined and smooth. Finally, consider the vector field $X:M\to TM$ given by \begin{align} X(q):=\sum_{p\in\partial M}\xi_p(q).\tag{1} \end{align} This looks like a funny infinite sum, but it’s really not, and in fact it is a smooth map because for each point $q$, there is a neighborhood $V$ of $q$ in $M$ such that all but finitely many $\phi$’s (hence $\xi$’s) vanish on $V$, so on $V$, the vector field $X$ is a finite sum of smooth vector fields, hence is smooth on $V$ (so by arbitrariness of $q$, it follows $X$ is globally smooth).
All that remains is proving that $X$ is inward-pointing everywhere along the boundary. So, fix a point $q\in\partial M$; since the $\phi$’s are a partition of unity, we have \begin{align} \phi_0(q)+\sum_{p\in\partial M}\phi_p(q)=1. \end{align} But now keep in mind that $\text{supp }\phi_0\subset M_0$, which is disjoint from $\partial M$, so $\phi_0(q)=0$, and thus (by non-negativity of the functions) there exists atleast one index $p_*\in \partial M$ such that $\phi_{p_*}(q)>0$. This strict inequality also implies that $q\in U_{p_*}$ and hence the vector $X_{p_*}(q)$ is inward-pointing, so the vector $\xi_{p_*}(q)=\phi_{p_*}(q)\cdot X_{p_*}(q)$ is inward-pointing. So, looking back at the definition (1), we have that $X(q)$ is a (necessarily finite) sum of terms with atleast one of them being inward-pointing (namely the $\xi_{p_*}(q)$) and with all other terms being either zero or inward-pointing; hence the sum is inward-pointing. This completes the proof.
To recap: almost everything here is standard. The existence of $(U_p,X_p)$ is ‘obvious’. From here, get a suitable open cover of $M$ and take a partition of unity subordinate to that cover. Constructing the ‘global cutoff’ $\xi_p$ and showing its smoothness is again a standard thing. Then, considering the sum, $X$, of all the $\xi$’s is also a standard thing. Finally, all that remains is to use that fact we have a partition of unity in order to show that the constructed object $X$ has the desired property (here we just needed a strict inequality at one place, but some other times, we need the full force of the fact that the sum is equal to $1$).