Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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 $L^\times$. –  John Dec 3 '13 at 11:29

1 Answer 1

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

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

where

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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.