Let $M$ be a smooth complex manifold and $L$ be a complex line bundle over $M$. Let $\Gamma(M,L)$ be the space of smooth sections. Why $\Gamma(M,L)$ is it 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 structure bundle obtained from $L$ by removing the zero section.

1$\begingroup$ A tiny remark. It's not a good idea call $L^\times$ a line bundle. $\endgroup$ – Anton Fonarev Dec 2 '13 at 23:26

$\begingroup$ Thanks Anton, which word is suitable for $L^×$? $\endgroup$ – user21574 Dec 2 '13 at 23:28

4$\begingroup$ I'd say a bundle or a $\mathbb{C}^*$bundle. $\endgroup$ – Anton Fonarev Dec 2 '13 at 23:44

1$\begingroup$ I think you can forget the complex structure, this construction applies to real symplectic manifolds. The fact that the fiber of the fiber bundle $L$ is the complex line $\bf C$ is accessory. $\endgroup$ – Patrick IZ Dec 4 '13 at 13:54

$\begingroup$ Yes, your right, $\endgroup$ – user21574 Dec 4 '13 at 15:48
I want first to fix a confusion at the level of the two bundles involved in this construction. My way(*) is to consider, first of all, a $S^1$bundle over a symplectic manifold $(M,\omega)$, with curvature $\omega$ — assuming the symplectic form is integral. Let $(Y,\lambda)$ be this prequantization, and $\pi : y \mapsto x$ be the projection. Now, $S^1$, identified to the complex numbers of norm $1$, acts by multiplication on the field of complex numbers ${\bf C}$. You can then build a line bundle $p : L \to X$ as the associated bundle to $Y$ $$ L = Y \times_{S^1} {\bf C} = [Y \times {\bf C}]/S^1, $$ where $S^1$ acts diagonaly by $\underline z(y,Z)= (z_Y(y),zZ)$, $z \in S^1$, $y \in Y$, $Z \in {\bf C}$, and we have denoted by $\underline z : y \to z_Y(y)$ the action of $S^1$ on $Y$. Therefore $L$ is the space of classes $[y,Z]$ with $[y,Z] = [z_Y(y),zZ]$.
The subspace $Y \times S^1 \subset Y \times {\bf C}$ is stable under the action of $S^1$, thus it defines a subbundle in $L$ which is equivalent to $Y$ by $y \mapsto [y,1]$. This map is surjective since every $[y,z]$ with $z\in S^1$ writes $[y/z,1]$, and obviously injective. This is what, in this approach, replaces your bundle $L^\times$. And you can forget about it for now.
Next, consider an equivariant map $\Psi : Y \to {\bf C}$, that is, $\Psi(z_Y(y)) = z\Psi(y)$, that gives a map on the quotient $\psi :y \mapsto [y,\Psi(y)]$ satisfying $$ \psi(z_Y(y)) = [z_Y(y),\Psi(z_Y(y))]=[z_Y(y),z\Psi(y)]=[y,\Psi(y)] = \psi(y). $$ Thus, $\psi$ is invariant by the action of $S^1$, then there exists a function (smooth because the projections involved are submersions) $\varphi : X \to L$ such that $\varphi(x) = [y,\Psi(y)]$ for any $y$ such that $\pi(y)=x$. And by construction, $\varphi$ is a section of $L$. You can show now that conversely, any section $\varphi$ of $\pi : Y \to M$ gives you an equivariant map $\Psi$. Take the pullback by the projection $(y,Z) \mapsto [y,Z]$ of the image of the section $\varphi$.
(*) That is the Souriau's construction.

1$\begingroup$ The symplectic structure appears to play no role at all in this. $\endgroup$ – Fran Burstall Dec 4 '13 at 19:49

1$\begingroup$ @FranBurstall Sure, but I know also that our friend Hassan is working on geometric quantization (previous exchanges) I just anticipate his needs. And, as you mentioned in your reply, $S^1$ can be replaced by any group $G$, and the associated bundle built with any action of $G$ on a space $E$, the construction will be identical. We agree :) $\endgroup$ – Patrick IZ Dec 4 '13 at 19:58

$\begingroup$ @PatrickIZ: very nice explanation. Do you have a reference for this? In particular that explains how a connection on the principal $U(1)$bundle explicitly induces a connection on the associated line bundle $L$. $\endgroup$ – Bilateral Apr 10 '15 at 0:36

$\begingroup$ @Bilateral I would say that the KobayashiNomizu "Foundations of Differential Geometry " could be a good general reference. $\endgroup$ – Patrick IZ Apr 11 '15 at 7:27

$\begingroup$ @PatrickIZ: thanks. A general question: the associated complex, rank1, vector bundle to a $U(1)$bundle is not necessarily an holomorphic line bundle, right? $\endgroup$ – Bilateral Apr 11 '15 at 11:16
Given $f\in A$, observe that $z\mapsto f(z)z$ is constant on fibres of $L^\times$ and so gives a welldefined element of $\Gamma(M,L)$. That is the isomorphism.
What is really going on here is the standard isomorphism between sections of a vector bundle and equivariant vectorvalued functions on its frame bundle. In your case, $L^\times$ is the frame bundle of $L$.
Locally a section is just a smooth function $s:U\to\mathbb{C}$. Given such a function, define $f(x, t) = \frac{s(x)}{t}$. Here $x$ stands for a point on $M$ and $t$ is a local coordinate in the fiber. Conversely, given such an $f$, define $s(x)$ to be the unique solution of the equation $f(x, s(x))=1$ (if doesn't exist, put $s(x)=0$).
Now observe that this construction behaves nicely under gluing.
I think you are confusing the line bundle $L$ and its dual $L^*$. A section $s\in \Gamma (M,L)$ provides a linear map $L^*\rightarrow \mathbb{C}$, by $(m,\xi) \mapsto \langle s(m ),\xi \rangle$ for $m\in M$, $\xi \in L_m^*$. Conversely, a linear map $\ell:L^*\rightarrow \mathbb{C}$ defines for each $m\in M$ a linear form $\ell_m:L^*_m\rightarrow \mathbb{C}$, hence a unique $s(m)\in L_m$.

$\begingroup$ abx , your solution is not correct $\endgroup$ – user21574 Dec 3 '13 at 8:23

$\begingroup$ Funny, if there is a "right" answer to the question, it is yours. $\endgroup$ – Sasha Anan'in Jan 11 '14 at 13:22