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Suppose $\pi:E\to B$ is a line bundle. Let $s:B\to E$ a non-vanishing section, i.e. for every $b\in B$ $s(b)\ne 0$ and $\pi\circ s=Id_B$.

I have to prove that the line bundle above is trivial.

Idea: Since $s$ is a section, for every $b\in B$, $0\ne E_b:=\pi^{-1}(b)=s(b)\cong\mathbb{R}$. I want to define an isomorphism between $E$ and $B\times\mathbb{R}$. Define $$ F:B\times \mathbb{R}\to E\\ (b,\lambda)\mapsto \lambda s(b) $$ How can I go on?

avati91
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1 Answers1

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First of all, $\pi^{-1}(b) \neq s(b)$; $\pi^{-1}(b) \cong \mathbb{R}$ while $s(b) \in \mathbb{R}$.

You have defined a vector bundle homomorphism $F : B\times\mathbb{R} \to E$. Now you just have to show that it is in fact an isomorphism, i.e. that the map $F_b : \{b\}\times\mathbb{R} \to E_{b}$ is an isomorphism of vector spaces. The map $F_b$ is given by $(b, \lambda) \mapsto \lambda s(b)$. Is this linear map injective? That is, if $\lambda s(b) = 0$, is $\lambda$ necessarily zero?

Michael Albanese
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    Yes, since $s$ doesn't vanish. – avati91 Aug 14 '15 at 16:30
  • Good. So $F_b$ is an injective linear map. Is it an isomorphism? – Michael Albanese Aug 14 '15 at 16:31
  • If I observe that $E$ is the disjoint union of $E_b$ where $b$ varies in $B$, then I want to define $G:E\to B\times\mathbb{R}$ by gluing together $G_b:E_b\to B\times \mathbb{R}$, $G:b(s(b)):=(b,1)$. Then for every $b\in B$ $F\circ G_b(s(b))=F(b,1)=s(b)$. Hence $G=F^{-1}$. Ok? – avati91 Aug 14 '15 at 16:33
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    The map $G_b$ that you have defined will be the inverse of $F_b$. However, it is not clear whether you understand why specifying $G_b(s(b)) = (b, 1)$ is enough to determine the map $G_b$. In particular, you have said what $G_b(s(b))$ is, but what is $G_b(e)$ for some $e \in E_b$ with $e \neq s(b)$? – Michael Albanese Aug 14 '15 at 16:47
  • Okay, define $G_b(e):=(\pi(e),1)$ – avati91 Aug 14 '15 at 16:54
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    That is not injective. The map you defined before is correct, you just hadn't explained why setting $G_b(s(b)) = (b, 1)$ completely determined the map $G_b$. The point is that because $E_b$ is one-dimensional, any $e \in E_b$ is of the form $e = \lambda s(b)$ for some $\lambda \in \mathbb{R}$. Once you have realised this, then $G_b(e) = G_b(\lambda s(b)) = (b, \lambda)$. You could have seen that $F_b$ is an isomorphism without constructing the inverse map $G_b$. Note that $F_b$ is an injective linear map between vector spaces of the same dimension, and is therefore an isomorphism. – Michael Albanese Aug 14 '15 at 17:00
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    Thank you very much for your clearness and patience Michael! – avati91 Aug 14 '15 at 19:05