# Is compliment of a zero section of a vector bundle a submanifold?

Let $$\pi:E\rightarrow M$$ be a smooth vector bundle. Let $$S:M\rightarrow E$$ be it's zero section. Let $$M'=E-S(M)$$.

Is $$M'$$ a smooth submanifold of $$M$$ ?

It is clear that $$S$$ is a smooth injective immersion. So we know that $$S(M)$$ is a submanifold of $$M$$. I wanted to use this fact to prove that $$M'$$ is a submanifold of $$M$$ but could not proceed much.

In Loop Spaces, Characteristic Classes and Geometric Quantization by Jean-Luc Brylinski in the section 2.1 (classification of Line bundles) it is mentioned that given a complex line bundle $$P:L\rightarrow M$$ the corresponding principle $$C^*$$- bundle is $$P':L^{+}\rightarrow M$$ where $$L^{+}$$ = $$L- \sigma(M)$$ where $$\sigma$$ is the zero section of $$P$$ and $$C^*$$ is the non zero group of complex numbers under multiplication.

My question is how do we know that $$L^{+}$$ is a submanifold of $$L$$ so that we can talk about principle bundle?

• In total generality, an open subset of a manifold is always a manifold. Sep 14, 2019 at 21:11
• @TedShifrin Yes Sir.Thank you sir for the comment. I am aware of this fact. I was actually confused in proving $L^{+}$ is indeed an open subset of $L$. But now I got it. I wrote a simple answer for it below. Sep 14, 2019 at 21:15
• The short answer: Smoothness is a local property, and bundles are locally trivial. (Second-countability or similar is often also required, but that's easy to check.) Sep 18, 2019 at 21:53
• @anomaly Sorry I didn’t get .. Can you please elaborate how your comment is related to my question? Sep 18, 2019 at 21:57

Sketch Pick a local trivialization of $$\pi : E \to M$$, say, $$\Phi : U \times \Bbb V \to \pi^{-1}(U)$$, where $$\Bbb V$$ is a model fiber. Since $$\{ 0 \}$$ is closed in $$\Bbb V$$, $$U \times \{0\}$$ is closed in $$U \times \Bbb V$$ and $$\Phi(U \times \{0\}) = \pi^{-1}(U) \cap S(M)$$ is closed in $$\Phi(U \times \Bbb V) = \pi^{-1}(U)$$. Varying $$U$$ over a cover of $$E$$ by local trivializations we conclude that $$S$$ is closed in $$E$$, so $$E \setminus S(M)$$ is open in (and hence is a smooth submanifold of) $$E$$.

• Thanks I got it. I wrote another answer to this question. Please check once at your leisure. Sep 14, 2019 at 19:12

An open subset of a manifold is a manifold, just take the intersections with the charts.

• How do you know that the compliment of the zero section is an open subset? Sep 14, 2019 at 17:48
• @Insearchforinfinity take a point in the complement. Then there is some family of open sets in M that trivializes the bundle, so the point lies in $\mathbb{R}^n\times U\setminus \{0\}\times U$, and our point has the form $(v,x)$. Take an open set $V$ about $v$ not containing zero, so that $V\times U$ contains our point and not the zero section. We can do this for all points, so our set is open. Sep 14, 2019 at 17:53
• thanks for the comment. Sep 14, 2019 at 19:13

Let $$l\in L^{+}$$. Consider $$\pi(l)\in M$$. Let $$(U,\phi)$$ be a trivialization around $$\pi(l)$$. Hence $$l\in \pi^{-1}(U)- \sigma(U) \subseteq L^{+}$$ Now $$\phi(\pi^{-1}(U)- \sigma(U))$$ is diffeomorphic and hence homeomorphic to $$J=(U \times C - U\times \lbrace 0\rbrace$$) which is open in $$U \times C$$ as $$U\times \lbrace 0\rbrace$$ is closed in $$U \times C$$. As $$\phi$$ is a homeomorphism hence ($$\pi^{-1}(U)- \sigma(U)$$) is an open subset of $$\pi^{-1}(U)$$ and hence an open subset of $$L$$ contained in $$L^{+}$$. Therefore $$l\in L^{+}$$ is an interior point of $$L^{+}$$. But $$l$$ is an arbitrary point of $$L^{+}$$. Hence $$L^{+}$$ is an open subset of $$L$$ and hence an open submanifold of $$L$$. (Hence Proved)

Alternatively it can be shown that the image set of any section of a vector bundle is a closed set in the the top space $$L$$

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 18, 2019 at 20:32
• Thanks, @AloizioMacedo. Sep 18, 2019 at 20:32
• Thanks @AloizioMacedo Sep 18, 2019 at 20:33
• @Travis I edited my answer.. Thanks for the discussion. Sep 18, 2019 at 21:34