What is the geometric point of view of an algebraic line bundle compared to a analytic line bundle? Hi folkz,
I'm trying to learn more about line bundles, invertible sheaves and divisors on schemes. I understand the connection beweteen Cartier and Weil Divisors and the connection between Cartier Divisors and invertible sheaves and how to get from one to another (as far as possible).
But compared to my analytic imagination of a line bundle I don't see how to come from an invertible sheaf to the line bundle (apart from the fact, that these two terms coincide).
Where is 'the line' in my locally free of rank one $\mathcal{O}_X$-module?
greatz Johannes
 A: Perhaps this might help as some intuition. Instead of looking for "the line" in a locally free sheaf, let's look in the other direction. Let's start with a line bundle, and move back towards sheaves.
So take a line bundle $\pi : L \to X$. This bundle has a sheaf of sections $\mathcal{O}_L$ defined by
$$\mathcal{O}_L(U) = \{s : U \to L \mid \pi \circ s = id_U\}$$
i.e. over an open set $U$ in $X$, $\mathcal{O}_L(U)$ is the collection of all sections of $L$ over $U$. It can be shown that this is a locally free sheaf of rank one.
Now, for a vector bundle of rank $n$, all of this is true, but the locally free sheaf is now or rank $n$.
Hopefully this provides at least a little intuition for the relation between the two.
A: Let $\mathcal F$ be a locally free $\mathcal O_X$-module. Then $\mathcal R := Sym_{\mathcal O_X}(\mathcal F)$, the tensor products being over $\mathcal O_X$, is a sheaf of rings, and we can take its $\bf Spec$ to get a space over $X$. That space is the corresponding vector bundle.
$\mathcal R$'s grading is what gives the dilation action on the fibers. The map $\mathcal F \to (\mathcal F \otimes \mathcal O_X) \oplus (\mathcal O_X \otimes \mathcal F)$, $f \mapsto (f\otimes 1) + (1\otimes f)$ induces a cocommutative comultiplication $\mathcal R \to \mathcal R \otimes \mathcal R$, which gives the vector addition on ${\bf Spec}\ \mathcal R$, I think.
