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Let $\Gamma(M,L) $ be the space of smooth sections, then why $\Gamma(M,L) $ is isomorphic to $A=\{f:L^{\times}\to \mathbb{C}; f(cz)=c^{-1}f(z), c\in \mathbb{C}-\{0\} , z\in L^{\times}\}$ . Here $L^{\times}$ is the line bundle obtained from $L$ by removing zero section. $M$ is smooth complex manifold and $L$is complex line bundle

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What are $M$ and $L$? Some complex manifold and some line bundle? What does $L^\times$ mean? The complement of the zero-section? – Alex Youcis Dec 2 '13 at 21:33
Dear Alex, I edited it – 1234 Dec 2 '13 at 21:57
How are you defining $\Gamma(M,L)$? Wouldn't it contain the zero section which is clearly not in $A$? Am I misunderstanding something? – Dori Bejleri Dec 2 '13 at 22:58
@DoriBejleri: The zero section corresponds to the zero function on $L^\times$. – Arctic Char Dec 3 '13 at 11:29
up vote 1 down vote accepted

Consider the projection $\pi: L^\times \to M$ and the pullback bundle $\pi^*L$ on $L^\times$. First of all, $\pi^* L $ is a trivial bundle on $L^\times $, as it has a nonzero section $(x, v) \to v$. Thus

$$A: \{ f: L^\times \to \mathbb C\} \cong \Gamma(L^\times , \pi^*L),$$


$$(Af) (x,v) := f(x,v)v$$

On the other hand, $\Gamma(M, L)$ can be thought of all $s\in \Gamma(L^\times , \pi^*L)$ that can be pushed down to $L$. That is $s((x,cv)) = s((x,v)) \in L_x$ for all $x\in M$, $v\in L_x$ and $c\in \mathbb C^*$. That corresponds to all $f:L^\times \to \mathbb C$ such that $f(cz) = c^{-1}f(z)$, as

$$(Af)((x, cv)) = f(x, cv) cv = c^{-1}f(x, v) cv = f(x, v)v = (Af)(x, v)$$

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