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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?

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    $\begingroup$ In total generality, an open subset of a manifold is always a manifold. $\endgroup$ Sep 14, 2019 at 21:11
  • $\begingroup$ @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. $\endgroup$ Sep 14, 2019 at 21:15
  • $\begingroup$ 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.) $\endgroup$
    – anomaly
    Sep 18, 2019 at 21:53
  • $\begingroup$ @anomaly Sorry I didn’t get .. Can you please elaborate how your comment is related to my question? $\endgroup$ Sep 18, 2019 at 21:57

3 Answers 3

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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) $E$.

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    $\begingroup$ Thanks I got it. I wrote another answer to this question. Please check once at your leisure. $\endgroup$ Sep 14, 2019 at 19:12
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An open subset of a manifold is a manifold, just take the intersections with the charts.

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  • $\begingroup$ How do you know that the compliment of the zero section is an open subset? $\endgroup$ Sep 14, 2019 at 17:48
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    $\begingroup$ @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. $\endgroup$ Sep 14, 2019 at 17:53
  • $\begingroup$ thanks for the comment. $\endgroup$ Sep 14, 2019 at 19:13
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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 $L$

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

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