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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 $M^+:=M\cup\{\infty\}$?

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    $\begingroup$ 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 Starr May 11 '18 at 23:13
  • $\begingroup$ That's right, Jason Starr. $\endgroup$ – user95283 May 11 '18 at 23:39

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