Professor Lee's Introduction to Smooth Manifolds Second Edition Lemma 10.35

Question

I'm stuck trying to verify the proof given in the text. There are parts of the hypothesis and proof that have nothing to do with where I'm stuck, so in the interests of brevity, I'll give only the part I'm having trouble with. We are given $M$, an immersed submanifold with or without boundary in $\mathbb{R}^n$, and $D$, a smooth rank-$k$ subbundle of $T\mathbb{R}^n|_M$. The proof begins as follows:

Let $p\in M$ be arbitrary, and let $(X_1,\dots,X_k)$ be a smooth local frame for $D$ over some neighborhood $V$ of $p$ in $M$. Because immersed submanifolds are locally embedded, by shrinking $V$ if necessary, we may assume that it is a single slice in some coordinate ball or half-ball $U\subseteq\mathbb{R}^n$. Since $V$ is closed in $U$, Proposition 8.11(c) shows that we can complete $(X_1,\dots,X_k)$ to a smooth local frame $(\tilde{X}_1,\dots,\tilde{X}_n)$ for $T\mathbb{R}^n$ over $U$, ...

My problem lies wholly within the above three sentences. The first sentence is fine. The second sentence starts out fine, but I don't think there is such a thing as a coordinate half-ball in $\mathbb{R}^n$ since $\mathbb{R}^n$ has empty boundary. (Please correct me if I'm wrong here.) So I am going to assume that Professor Lee meant to say "assume that it is a single slice or half-slice of some coordinate ball $U\subseteq\mathbb{R}^n$." With that change (which doesn't solve my problem), I am OK with the second sentence and with $V$ being closed in $U$.

Now Proposition 8.11(c) applies to smooth vector fields along a closed subset $V\subseteq U$. For my question, I only need to talk about one of the $X_i$, so let me just call it $X$. I have no problem composing $X$ with smooth maps and restricting codomains so that, I can consider $X$ to be a smooth vector field $X\colon V\to TU$ over $V$. Recall that $V$ is a neighborhood in $M$ and as such has a smooth structure of manifold with (possibly empty) boundary derived from that of $M$. Now let me quote the definition of a smooth vector field along a subset (I've changed the names of the variables to match the present situation):

If $U$ is a smooth manifold with or without boundary and $V\subseteq U$ is an arbitrary subset, a vector field along $V$ is a continuous map $X\colon V\to TU$ satisfying $\pi\circ X=\mathrm{Id}_V$. We call it a smooth vector field along $V$ if for each $q\in V$, there is a neighborhood $W$ of $q$ in $U$ and a smooth vector field $\tilde{X}$ on $W$ that agrees with $X$ on $W\cap V$.

My problem is that given $q\in V$, I haven't been able to find a neighborhood $W$ of $q$ in $U$ and an associated smooth vector field $\tilde{X}$ on $W$ such that $X|_{W\cap V}=\tilde{X}|_{W\cap V}$. I've got $(U,\phi)$, a smooth chart centered at $p$ for $\mathbb{R}^n$, with $\phi(U)$ being the open cube of side $r$ centered at $0_{\mathbb{R}^n}$ and with $\phi(V)=\phi(U)\cap(Z^m\times\{0_{\mathbb{R}^{n-m}}\})$ where $m=\mathrm{dim}\,M$, and $Z=\mathbb{H}$ if $p\in\partial M$ and $Z=\mathbb{R}$ otherwise.

I think I can make things work if $p$ is not a boundary point of $M$ as follows: Let $F\colon\mathbb{R}^n\to\mathbb{R}^n$ be defined by $$F(x_1,\dots,x_m,x_{m+1},\dots,x_n)=(x_1,\dots,x_m,0,\dots,0).$$ Then $F$ is smooth, and when applying $F$ to $\phi(U)$, it produces $\phi(U)\cap(\mathbb{R}^m\times\{0_{\mathbb{R}^{n-m}}\})=\phi(V)$. Thus $\tilde{X}=X\circ\phi^{-1}\circ F\circ\phi\colon U\to TU$ is smooth and equals $X$ on $V$. [EDIT: As pointed out by Tob Ernack, this may be smooth, and may be an extension, but it is not a section and therefore not a solution for the empty boundary case. So I retract the statement that I know how to make it work at an interior point.]

But this won't work when $p\in\partial M$ since then $Z=\mathbb{H}$ and $F(\phi(U))=\phi(U)\cap(\mathbb{R}^m\times\{0_{\mathbb{R}^{n-m}}\})$ which is not equal to $\phi(V)=\phi(U)\cap(\mathbb{H}^m\times\{0_{\mathbb{R}^{n-m}}\})$ and therefore cannot be composed with $X\circ\phi^{-1}$.

So, my question is, how do I show $X$ is a smooth vector field along $V$, when $p\in\partial M$?

Answer 1

I believe your remark about the second sentence is correct. Since $V$ is an embedded submanifold with or without boundary in $\mathbb{R}^n$, there is an interior slice chart or a boundary slice chart $(U, \varphi)$ for $V$ (Theorem 5.51), which by definition means that $U$ is an open subset of $\mathbb{R}^n$ whose intersection with $V$ is a $k$-dimensional slice or half-slice in $U$. So $U$ itself is not a boundary chart for $\mathbb{R}^n$.

Your argument that $X$ is a smooth vector field along $V$ in the case where the boundary is empty seems correct to me. One technical point is that we really should use the map $x \to (x, X^j \circ \phi^{-1} \circ F \circ \phi(x))$ where $X^j$ are the component functions of $X$ with respect to the global trivialization $T\mathbb{R}^n \to \mathbb{R}^n \times \mathbb{R}^n$, in order to ensure that it is indeed a vector field (it must lie above $x$, but $X \circ \phi^{-1} \circ F \circ \phi(x)$ actually lies above $\phi^{-1} \circ F \circ \phi(x)$, which is different from $x$ in $U \setminus V$).

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Now let's assume that $p \in \partial V$. Since $X$ is part of a smooth local frame for $D$ over $V$, by definition it is a smooth section of $T\mathbb{R}^n\big\vert_V$, or in other words, a smooth map $X: V \to T\mathbb{R}^n$ such that $\pi \circ X = \text{id}_V$.

We identify $T\mathbb{R}^n$ with $\mathbb{R}^n \times \mathbb{R}^n$ using the standard smooth global trivialization. So we can view $X$ as a smooth map $X: V \to \mathbb{R}^n \times \mathbb{R}^n$ given by $X(x) = \left(x, X^j(x)\right)$ where $X^j: V \to \mathbb{R}$ are the smooth component functions.

Since $(U, \varphi)$ is a boundary slice chart for $V$ centered at $p$, we obtain a smooth boundary chart $\left(U \cap V, \pi \circ \varphi\big\vert_{U \cap V}\right)$ for $V$ centered at $p$. Note that I changed the meaning of the symbol $\pi$ to represent the projection $\pi: \mathbb{R}^n \to \mathbb{R}^k$ onto the first $k$ components. Also note that I am assuming that the smooth structure on $V$ is the one given by the slice charts and boundary slice charts. This is justified by the answer here. This is applicable because $V$ is an embedded submanifold with boundary of $\mathbb{R}^n$, which itself has empty boundary.

Then by definition of smoothness for a function on a smooth manifold, the coordinate representation of $X^j$, which is $\widehat{X^j} = X^j \circ \left(\pi \circ \varphi\big\vert_{U \cap V}\right)^{-1}$, must be smooth from $(\pi \circ \varphi)(U \cap V)$ to $\mathbb{R}$.

Now $(\pi \circ \varphi)(U \cap V)$ is an open subset of $\mathbb{H}^k$ containing $0$.

By definition of smoothness for functions defined on an open subset of $\mathbb{H}^k$ (see page 27 in the book and also page 34), there must be a smooth function $F^j: W' \to \mathbb{R}$ defined on an open subset $W'$ of $\mathbb{R}^k$ containing $0$ such that $F^j$ agrees with $\widehat{X^j}$ on $W' \cap \mathbb{H}^k$.

Let $W = \varphi^{-1}(\pi^{-1}(W') \cap \varphi(U))$, which is an open subset of $U$ containing $p$.

Then define $\widetilde{X}: W \to \mathbb{R}^n \times \mathbb{R}^n$ as $\widetilde{X}(x) = \left(x, F^j \circ \pi \circ \varphi\big\vert_W(x)\right)$, which is a smooth vector field defined on $W$ that agrees with $X$ on $W \cap V$.

Answer 2

I'm not sure if I understand your question correctly. I think you are worried that the extension of the vector fields, which are initially a section into $D$ do not become vector fields in $\mathbb{R}^n$? In any case, I don't really see what is the problem here. These sections are still vector fields in $R^n$. We choose smooth local frame $(X_1,\dots, X_k)$ for $D$ over $V$ basically from the Local Frame Criterion for Subbundles. This means that each is a smooth local section from $V \to TR^n|_M \approx M \times R^n.$ But we are actually mapping to $TR^n|_V \approx V \times R^n$. (In case you worry about smoothness, $V$ is open in $M$ so the restriction to the codomain is fine as $TR^n|_V$ is just the inverse image of the open set $V$ of the projection map $TR^n|_M \to M$.) But $V$ is embedded in $R^n$, so we can see this as a smooth map into $TR^n= R^n \times R^n$. Now $V$ is closed in $U$ so use the extension lemma for vector fields.

Answer 3

Let us assume that $p\in\partial M$. We may shrink $U$ if necessary so that it is a smooth coordinate cube. Since $(U, \varphi)$ is a boundary slice chart for $V$, we have $\varphi(U\cap V) = \mathbb{H}^{m}\times\{0\}^{n-m}\cap \varphi(U)$. Since $X : U\cap V\rightarrow T\mathbb{R}^{n}|_{M}$ is smooth, it's component functions are smooth on $U\cap V$, and thus $X^{j}\circ\varphi^{-1}\circ \iota: \pi_{m}(\varphi(U\cap V))\rightarrow \mathbb{R}$ is smooth for each $1\leq j\leq n$, where $\pi_{m}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ and $\iota : \mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$ are the canonical projection and inclusion respectively. Thus, we may extend this to a smooth function $Y^{j}:\pi_{m}(\varphi(U))\rightarrow\mathbb{R}$. To conclude, we define $\widetilde{X}^{j} : U\rightarrow \mathbb{R}$ by $\widetilde{X}^{j}(q) = Y^{j}\circ\pi_{m}\circ\varphi(q)$.

Once the extensions $\widetilde{X}_{1}, \dots, \widetilde{X}_{k}$ have been constructed, we may shrink $U$ further so that they are linearly independent on $U$. Finally, (by possibly shrinking $U$ even further), we may complete $(\widetilde{X}_{1}, \dots, \widetilde{X}_{k})$ to a smooth local frame $(\widetilde{X}_{1}, \dots, \widetilde{X}_{n})$ for $T\mathbb{R}^{n}$ over $U$.

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There is an easier proof of this result, which generalizes easily to the case when $M$ is embedded in a Riemannian manifold (with or without boundary). All computations are done with the open set $V$ of $M$ without any reference to slice charts.

By possibly shrinking $V$, we may complete $(X_{1}, \dots, X_{k})$ to a smooth local frame $(X_{1}, \dots, X_{n})$ for $T\mathbb{R}^{n}|_{M}$ over $V$. (This is Proposition $10.15$(a) in Lee's book.) Since the map $$T\mathbb{R}^{n}|_{V}\ni v^{i}\partial/\partial x^{j}|_{q} \mapsto (q, v^{1}, \dots, v^{n})\in V\times \mathbb{R}^{n}$$ is a smooth local trivialization of $T\mathbb{R}^{n}|_{M}$ over $V$, the component functions $X_{i}^{j}: V\rightarrow\mathbb{R}$ are smooth for $1\leq i, j\leq n$. We may now apply the Gram-Schmidt process to obtain a smooth orthonormal frame $(E_{1}, \dots, E_{n})$ over $V$.