Lemma 10.32 of Lee's Introduction to Smooth Manifolds

Question

I am reading over the proof of Lemma 10.32 (Local Frame Criterion for Subbundles) in Lee's Introduction to Smooth Manifolds.

The lemma says

Let $\pi: E \rightarrow M$ be a smooth vector bundle and suppose that for each $p\in M$ we are given an M-dimensional linear subspace $D_p \subseteq E_p$. Then $D = \cup_{p \in M} D_p \subseteq E$ is a smooth subbundle of $E$ iff each point of $M$ has a neighborhood $U$ on which there exist smooth local sections $\sigma_1, \cdots, \sigma_m: U \rightarrow E$ with the property that $\sigma_1(q), \cdots, \sigma_m(q)$ form a basis for $D_q$ at each $q \in U$.

Overall I understand the proof of this lemma, besides the part where we need to show that $D$ is an embedded submanifold with or without boundary of $E$. Professor Lee's proof says that

it suffices to show that each $p \in M$ has a neighborhood $U$ such that $D \cap \pi^{-1}(U)$ is an embedded submanifold (possibly with boundary) in $\pi^{-1}(U) \in E$.

It is not very obvious to me why it is sufficient by showing this. May someone explain the logic to me?

Edit: Here's my attempt to reason it: By Theorem 5.8, if $D ∩ \pi^{-1}(U)$ is an embedded submanifold in $\pi^{-1}(U)$, it satisfies the local k-slice condition. Now because $D$ is a union of $D ∩ \pi^{-1}(U)$ over different neighborhoods of $p \in M$, it satisfies the local k-slice condition as well, and hence again by Theorem 5.8, $D$ is an embedded submanifold.

Please let me know if anything is wrong and how it can be corrected.

Thank you very much.

Here's a screenshot of the Lemma and its (partial) proof:

Answer 1

Suppose that there is an open cover $\{U_i\}$ of $E$ such that $D\cap U_i\hookrightarrow E$ is a smooth embedding and $D\cap U_i$ is open in $D$. I claim that $D\hookrightarrow E$ must be a smooth embedding.

Because $D\cap U_i$ is open in $D$, it follows that $D\hookrightarrow E$ is locally an immersion, so it must be an immersion globally. It is left to show that the inclusion is also a topological embedding.

It suffices to show that there is an open cover $\{V_i\}$ on $D$ which also forms an open cover on $D\subseteq E$ (subspace topology) on which the inclusion is a topological embedding. We were given that the inclusion is a topological embedding on sets of the open cover $\{D\cap U_i\}$. Since each $U_i$ is open in $E$, each $D\cap U_i$ is open in $D$ with the subspace topology induced by $E$.

Answer 2

Unfortunately, Theorem 5.8 only works for smooth manifolds without boundary. But here $E$ probably has boundary. A better theorem is Theorem 5.51，but it also requires that $M$ is a smooth manifold without boundary there.

Answer 3

This question has been a while but I'm leaving this answer in case anyone else is wondering about this as well. I believe the statement in the text is misleading, or at least unclear. We actually need to use the Vector bundle chart lemma in the book. So we are given $$\Phi_\alpha : \pi^{-1}(U_\alpha)\to U_\alpha \times \mathbb{R}^k : \Phi_\alpha(s^1 \sigma_1(q)+\cdots +s^k \sigma_k(q))=(q,(s^1,\dots, s^k)).$$

And using this by restricting the last $k-m$ coordinates we get the smooth local trivialization candidate for $D$ given by $$\Psi_\alpha: D \cap \pi^{-1}(U_\alpha)=\pi_D^{-1}(U_\alpha) \to U \times \mathbb{R}^m: \Psi_\alpha(s^1 \sigma_1(q)+\cdots +s^m \sigma_m(q))=(q,(s^1,\dots, s^m)).$$

From $\Phi_\alpha$ we do get, as Lee writes, $D\cap \pi^{-1}(U_\alpha)$ as an embedded submanifold of $\pi^{-1}(U_\alpha)$, and from this we get that the map $\Psi_\alpha$ below is a diffeomorphism. But we do not yet have a smooth structure on $D$ or even a topology. So we need to find it using these maps in the Vector Bundle Chart Lemma then naturally $D$ will be a subspace of $E$.

For the Vector Bundle Chart Lemma, the first two conditions are easy. We are already given the cover $\{U_\alpha\}_{\alpha \in A}$ of $M$. And the maps $\Psi_\alpha: \pi^{-1}_D (U_\alpha)\to U_\alpha \times \mathbb{R}^m$ constructed above are bijective whose restriction to each $D_p$ is a vector space isomorphism from $D_p$ to $p \times R^m$. Note that all of this follows from its parent map $\Phi_\alpha$.

So we just need to check the third condition. For each $\alpha,\beta \in A$ with $U_\alpha \cap U_\beta \neq \emptyset$, a smooth map $\tau:U_\alpha \cap U_\beta \to GL(k,R)$ such that the map $\Phi_\alpha \circ \Phi_\beta^{-1}$ from $(U_\alpha \cap U_\beta) \times R^k$ to itself has the form $$\Phi_{\alpha} \circ \Phi_\beta^{-1}(p,v)=(p,\tau(p)v).$$

This is what we get from the vector bundle property of $E$, and note that by the uniqueness of the theorem this holds. Now we just obtain the transition map for $D$ from the transition map for $E$. This is obtained simply by taking the first $m \times m$ submatrix of $\tau$. It is still invertible, and smooth since it is just a composition of $\tau$ with the projection onto the first $m\times m$ coordinates.

Moreover, it satisfies the desired property because, note that the maps $\Phi_\alpha$ takes $D \cap \pi^{-1}(U_\alpha)$ to the subset $\{(q,(s^1,\dots, s^m,0,\dots,0))\} \subset U \times R^k$ which is basically $U\times R^m \times \{0\}^{k-m} \approx U \times R^m$. And all these maps are bijective, so when we are actually considering the transition maps inside $\pi^{-1}(U_\alpha \cap U_\beta) \cap D$, we are limiting ourselves to the vector space where the last $k-m$ components are $0$. Also by bijectivity, the range should have the last $k-m$ components $0$. That means we are multiplying a matrix $A$ by a vector $v$ of the form $v=(v^1,\dots,v^k,0,\dots,0)$ and expecting a response also of the form $w=(w^1,\dots,w^k,0,\dots,0)$. So the result is unaffected by removing the last $k-m$ rows and columns of $\tau(p)$. This is how we define $\tau_{\alpha,\beta}(p)$. The point is that $\Phi_\alpha \circ \Phi_\beta^{-1}$ and $\Psi_\alpha \circ \Psi_\beta^{-1}$ must agree on $(U_\alpha \cap U_\beta) \times R^m$ and this is achieved by setting $\tau_{\alpha,\beta}$, the transition maps for $\Psi_\alpha$ and $\Psi_\beta$ as above. And as mentioned above it is smooth because it is just a composition with a projection map.

Therefore, we have a unique topology and smooth structure making $D$ into a smooth manifold with or without boundary and a smooth rank-$m$ vector bundle over $M$ with $\pi_D$ as its projection and $\{(U_\alpha,\Psi_\alpha)\}$ as smooth local trivializations.

Next, note that from the manifold chart lemma, that the open sets constructed from this lemma are given by $((\phi_p \times Id_{R^k}) \circ \Phi_\alpha)^{-1}(\hat{V_p}\times W)$ for $E$ and $((\phi_p \times Id_{R^m}) \circ \Psi_\alpha)^{-1}(\hat{V_p} \times W)$ for $D$ where $(V_p,\phi_p)$ is a smooth chart for $M$ containing $p$ contained in $U_\alpha$ and $\phi_p(V_p)=\hat{V_p}$, and $W$ are open sets in $R^k$ and $R^m$.

If $D$ has the subspace topology then basis open sets of $D$ are given by the sets $((\phi_p \times Id_{R^k}) \circ\Phi_\alpha )^{-1}(\hat{V_p}\times W) \cap D$. Let's see how this corresponds to the basis in $D$ given from the chart lemma. Now let $W$ be an open set in $R^m$ (not $R^k$). Then the basis is really just $\Psi_\alpha^{-1}(V_p \times W)$. But the image of $\Psi_\alpha$ on $D \cap \pi^{-1}(U)=\pi_D^{-1}(U)$ was just obtained by deleting the last $k-m$ coordinates in $R^k$ from $\Phi_\alpha(\pi^{-1}(U))$ as the last $k-m$ coordinates are all $0$. As the maps are bijective, all that $\Phi_\alpha$ does by restricting the domain to $D$ is fixing the last $k-m$ coordinates to $0$. This means that if $\Phi_\alpha$ maps a set to $V_p \times (W \times R^{k-m})$ then restricting the set to $D$, $\Phi_\alpha$ maps this restricted set to $V_p \times (W \times \{0\}^{k-m})$. Thus by bijectivity, this must be equal to $\Psi_\alpha^{-1}(V_p \times W)$. And hence, $D$ has the subspace topology.

Hence we have shown that $D$ constructed by the Vector Bundle Lemma using $\Psi_\alpha$ as smooth local trivializations has the subspace topology of $E$.

Finally we have to show that the inclusion map $i:D \to E$ is a smooth injective immersion. This is the part that follows from $D \cap \pi^{-1}(U)$ is embedded in $\pi^{-1}(U)$, I believe. So $\pi^{-1}(U)$ is embedded in $E$, as an open subset of $E$, hence $D \cap \pi^{-1}(U)$ is embedded in $E$. And we have $\pi_D: D \to M$ smooth (continuous) so $\pi_D^{-1}(U)= \pi^{-1}(U)\cap D$ is open in $D$. So at every point of $D$ we have a neighborhood in which the inclusion map $i$ is an embedding to $E$, thus $i$ is a smooth immersion, clearly injective. As it has the subspace topology, it is a topological embedding. Hence $D$ is embedded in $E$.