# 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

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you can take a trivialising open cover for your line bundle, that will give you transition functions. You can define a scheme by gluing the open patches cross $\mathbb{A}^1$ with the given transition functions. It's an exercise in hartshorne. –  Yosemite Sam Mar 29 '12 at 17:31
I think this question would be better suited to math.stackexchange.com. One reference: Shafarevich, "Basic Algebraic Geometry" Book 2. –  Artie Prendergast-Smith Mar 29 '12 at 17:37
if you prefer you can find 'the line' hiding in your sheaf by noticing that the fibres are of dim 1: if F is your sheaf, x is a (closed) point of your variety $X$, the fibre is defined as $F_x \otimes_{\mathcal{O}_{X,x} k(x)$, where $F_x$ is the stalk of $F$ at $x$, $\mathcal{O}_{X,x}$ is the local ring of your variety at $x$ and $k(x) = \mathbb{C}$ is the residue field. (this is the same as considering your point as a morphism $x: Spec \mathbb{C} \to X$ and taking the pullback $x^*F$). By assumption your sheaf is locally free of rank one, so $F_x \cong \mathcal{O}_{X,x}$, hence the fibre is C –  Yosemite Sam Mar 29 '12 at 17:41
–  Mark Grant Mar 29 '12 at 18:13
Do I see it right, that if I dont look at the fibres of a point, but at the pullback of some open subset I get the following: $F(U) \otimes_{\mathcal{O}_{X} k(x)$ and in some <i>good cases</i> (for example k algebraically closed) the residue field does not depend on a point (the subset) and we end up in $F(U) \otimes_{\mathcal{O}_{X} k$? –  Johannes Mar 29 '12 at 22:17

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.

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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.

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don't you want to write Sym somewhere? –  Yosemite Sam Apr 4 '12 at 8:48
Oops! Fixed (I had the tensor algebra before). –  Allen Knutson Apr 4 '12 at 10:18