Let $M$ be a non compact smooth manifold and suppose that $\pi:E\rightarrow M$ is a vector bundle over it. Is there a compact subset $K\subset M$ such that the restricted bundle $\pi|_U:E|_U\rightarrow U$ over $U:=M-K$ becomes trivial? In other words, is it true that every vector bundle over a non compact manifold can be extended to a (continuous) vector bundle over its one point compactification $M^+:=M\cup\{\infty\}$?
$\begingroup$
$\endgroup$
3
-
10$\begingroup$ That fails already for rank $1$ vector bundles over $M = \mathbb{R}\times \mathbb{S}^1$. For the tautological rank $1$ vector bundle $F$ on $\mathbb{S}^1=\mathbb{RP}^1$, the pullback $E=\text{pr}_2^*F$ gives a counterexample. For every compact subset $K$ of $\mathbb{R}\times \mathbb{S}^1$, the image $\text{pr}_1(K)$ is a compact subset of $\mathbb{R}$, hence bounded. Thus, there exists $t\in \mathbb{R}\setminus \text{pr}_1(K)$. For the section $\sigma_t:\mathbb{S}^1\to M$ by $\sigma_t(u)=(t,u)$, the pullback $\sigma_t^*E$ equals $F$. $\endgroup$– Jason StarrCommented May 11, 2018 at 23:13
-
$\begingroup$ That's right, Jason Starr. $\endgroup$– user95283Commented May 11, 2018 at 23:39
-
$\begingroup$ @JasonStarr Could you make an answer from you comment? (to move it from unanswered questions). $\endgroup$– Anton PetruninCommented Dec 10, 2022 at 16:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As requested by user @AntonPetrunin, I am turning my comment into an answer.
That fails already for rank $1$ vector bundles over $M=\mathbb{R}\times \mathbb{S}^1$. For the tautological rank $1$ vector bundle $F$ on $\mathbb{S}^1=\mathbb{RP}^1$, the pullback $E=\text{pr}_2^*F$ gives a counterexample. For every compact subset $K$ of $\mathbb{R}\times \mathbb{S}^1$, the image $\text{pr}_1(K)$ is a compact subset of $\mathbb{R}$, hence bounded. Thus, there exists $t\in \mathbb{R}\setminus \text{pr}_1(K)$. For the section $\sigma_t:\mathbb{S}^1\to M$ by $\sigma_t(u)=(t,u)$, the pullback $\sigma_t^*E$ equals $F$.