Why doesn't ever smooth vector bundle admits a line bundle?

Question

Let $E \to M$ be a smooth vector bundle. Consider $G = \sqcup_{p \in M} F_p$ where $F_p$ is just a 1 dimensional subspace of each fiber $E_p$. The trivialization is just coming from the restriction of the trivialization of $E$. Why is this argument wrong?

Answer 1

Well, here's a simple example. Let $M=\mathbb{R}$ and let $E$ be the trivial bundle $M\times\mathbb{R}^2$. You say to pick a 1-dimensional subspace $F_p$ of each fiber, so let's do so as follows. If $p=0$, then $F_p=\{p\}\times \{0\}\times\mathbb{R}$. If $p\neq0$, then $F_p=\{p\}\times\mathbb{R}\times\{0\}$.

You now say we can (locally) trivialize $G$ by just restricting a (local) trivialization of $E$. Well, in this case $E$ is globally trivial, so you'd be saying that $G$ is already trivial. But if we try to "restrict" our trivialization of $E$, we immediately hit a problem: there is no single subspace $V\subset\mathbb{R}^2$ such that $F_p=\{p\}\times V$ for all $p$, so there is no obvious way to restrict our trivialization. We could try to define a trivialization $M\times\mathbb{R}\to G$ that would be an isomorphism on each fiber, but such a map would not be continuous, since the fiber of $G$ "jumps" discontinuously at $0$. Indeed, $G$ is not a line bundle over $M$ at all.

Now you might say I just made a dumb choice of 1-dimensional subspaces $F_p$. It would have been a lot smarter to pick $F_p=\{p\}\times\mathbb{R}\times\{0\}$ for all $p$, instead of doing something crazy at $p=0$. Indeed, in that case $G$ would be a trivial line bundle and the obvious map $M\times\mathbb{R}\to G$ would be an isomorphism of line bundles. But, what if our original bundle $E$ was not trivial? Then we could make a "smart" choice like this for $F_p$ in each local trivialization of $E$, but our choices of $F_p$ in different local trivialization that overlap might not be the same. There's no reason to believe we can choose $F_p$ consistently for all $p$ such that $G$ really is locally trivial everywhere.

Answer 2

Some comments:

1. The family of 1-dimensional subspaces $p \mapsto F_p$ does indeed exist, by the axiom of choice.

2. The set $\sqcup_{p \in M} F_p$ does indeed exist, by basic principles of set theory.

3. There's a natural set-theoretic inclusion of $\bigsqcup_{p \in M} F_p$ into the total space $E$.

4. By composing the distinguished map $E \rightarrow M$ with the aforementioned inclusion, we get a surjective function from $\sqcup_{p \in M} F_p$ down onto the base space $M$.

5. It remains to show that the map in $(3)$ is smooth. If we can show this, then the map in $(4)$ is smooth, and we're done.

6. We can't show that the map in $(3)$ is smooth until we've chosen a manifold structure on $\sqcup_{p \in M} F_p.$

7. Since the $F_p$ are arbitrary, getting an actual manifold structure on $\sqcup_{p \in M} F_p$ is usually going to be impossible. There's just no guarantee they'll fit together in such a way as to smoothly vary between fibers.

8. In special cases we're able to choose $p \mapsto F_p$ in a non-arbitrary way in order to prove that the particular vector bundle under question has an embedded line bundle.