(Lee's) Definition of smooth maps on manifolds, in particular vector fields

Question

In John Lee's Introduction to Smooth Manifolds, a map $F$ from a subset $A$ of a smooth manifold $M$ into a smooth manifold $N$ is, as on p. 45, smooth if every point $p$ in $A$ has a neighborhood $U$ and a smooth map $G$ from $U$ into $N$ such that $G$ and $F$ agree on $A\cap U$. In the errata from his website, this definition is modified so as to allow maps such as the identity on $R^1$ restricted to $H^1$ into $H^1$ to still be considered smooth as a map from a subset of $R^1$ into $H^1$ -- instead of requiring that $G$ be a smooth map into $N$, all that is needed is a smooth chart $\psi$ around $F(p)$ such that $\psi\circ F$ has a smooth extension around $p$.

The lemma that immediately follows this definition is the extension lemma for smooth functions: for every smooth function $f$ from a closed subset $A$ of $M$ into $R^n$ and for every neighborhood $U$ of $A$, there is a smooth extension $\tilde{f}$ from $M$ into $R^n$ of $f$ and supported in $U$. In this case, since the target space $R^n$ has no boundary, the original definition for a smooth map on a subset agrees with the modified definition, and the proof relies on the original definition, so no worries.

Fast-forward to chapter 8: vector fields. A smooth vector field along a subset $A$ of $M$ is defined, on p. 176, to be a map from $A$ into $TM$ such that every point $p$ in $A$ has a neighborhood $U$ and a smooth vector field $Y$ from $U$ into $TM$ such that $X$ and $Y$ agree on $A\cap U$. Of course, this is just a special case of his original definition of a smooth map on a subset. The lemma that immediately follows this definition is the extension lemma for vector fields: for every smooth vector field $X$ from $A$ into $TM$, and for every neighborhood $U$ of $A$, there is a global vector field $\tilde{X}$ on $M$ that agrees with $X$ on $A$ and is supported in $U$. Now $TM$ has boundary when $M$ does, so it isn't clear to me that the modified and original definitions agree. The proof is left as an exercise, and one can easily adopt the proof for the lemma for smooth functions, if in fact the original and modified definitions do agree in the case of smooth vector fields. Do they? I thought about this for a few days, but I haven't been able to get an extension of the chart/vector field composite to land in the closed upper half plane..

Answer 1

Good question.

First let me note that the definition of smoothness of a vector field on a subset $A\subseteq M$ is not exactly a special case of the definition for arbitrary smooth maps. The difference is that, to say that a vector field $X$ on $A$ is smooth, I require that each $p\in A$ has a neighborhood $V$ in $M$ on which $X$ extends to a smooth vector field $\widetilde X$ on $V$, not just to a smooth function $\widetilde X\to TM$. (I'm not sure this distinction is crucial for what I'm about to say, but I wanted to point it out for completeness.)

It turns out that the original definition of smoothness on a subset works fine in the case of vector fields on manifolds. The problem in the case of a map $F\colon M\to N$ between manifolds arose when the subset $A\subseteq M$ contained a point $p\in \partial A$ such that $F(p)\in\partial N$. It could then be the case that the derivatives of $F$ at $p$ would force any extension to "leak outside of $N$." This is what happens in the example you mentioned: if $M=\mathbb R$, $A=[0,\infty)\subseteq M$, $N=[0,\infty)$, and $F\colon A\to N$ is the identity map, then no smooth extension of $F$ to an open subset of $\mathbb R$ could map into $N$, because the extension would have a local minimum at $0$ but have nonzero derivative there.

But for vector fields, that can't happen, because in local coordinates a vector field just looks like a Cartesian product of the identity map with a smooth map into $\mathbb R^n$: $$ X(x^1,\dots,x^n) = (x^1,\dots,x^n,v^1(x),\dots,v^n(x)). $$ Extending this to a smooth vector field on an open set just involves extending $(v^1,\dots,v^n)$ as a mapping into $\mathbb R^n$ (which has no boundary).