An isomorphism on space of smooth sections

Question

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.

Answer 1

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 $[y,Z] = [z_Y(y),zZ]$.

1. The subspace $Y \times S^1 \subset Y \times {\bf C}$ is stable under the action of $S^1$, thus it defines a sub-bundle 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.

2. 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.

Answer 2

Given $f\in A$, observe that $z\mapsto f(z)z$ is constant on fibres of $L^\times$ and so gives a well-defined 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 vector-valued functions on its frame bundle. In your case, $L^\times$ is the frame bundle of $L$.

Answer 3

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.

Answer 4

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 $s(m)\in L_m$.