# Quail Entry: Global Sections of line bundles and Hyperplane Divisors

The context is morphisms $X \xrightarrow{|D|} \mathbb{P}^n$ determined by base point free linears sysmtems.  For a long time I never really flushed out why so called hyperplane divisors $X.H$ were the same as $|D|$.

A reasonable first step is being comfortable with descriptions of global sections.

One way to see is via sections.  First recall $O$ on $\mathbb{P}V$.  There’s a nice cover $U_i \cong \mbox{Spec }k[x_0/x_i,..., x_n/x_i]$.  To simplify notation, for $i,j$ write this as $\mbox{Spec } k[\{a_l\}]$ and $\mbox{Spec } k[\{b_k\}]$ with $a_l = x_l/x_i$ etc.

On intersections $U_{ij}$ there are ring isomorphisms

$k[..a_l..]_{a_j} \xrightarrow{m_{ij}} k[..b_l..]_{b_i}$

$a_l \mapsto b_l/b_i$

Global sections (take 1) of

1. $O$ are $s_i \in O(U_i)$ such that $s_j = m_{ij}(s_i)$

2. $O(1)$ are $s_i \in O(U_i)$ such that $s_j = \frac{x_i}{x_j} m_{ij}(s_i)$

3. $O(m)$ are $s_i \in O(U_i)$ such that $s_j = \frac{x_i}{x_j}^m m_{ij}(s_i)$

Example on $\mathbb{P}^1 = \mbox{Proj } k[X_o,X_1]$. Write $x = X_o/X_1$ and $y = X_1/X_0$.  A basis for global sections

$1 \in O(U_1), y \in O(U_o)$

$x \in O(U_1), 1 \in O(U_o)$

because $y = X_1/X_om_{1o}(1)$ and $X_o/X_1 m_{o1}(y) = X_o/X_1\cdot 1/x = 1$ etc.

This is very close to the line bundle story with transition functions, but without that analogy this might seems a little awkward.  It is exactly like describing $a \in A$ as a collection of compatible elements $a_i \in A_{f_i}$.

To think of global sections as just a single “thing”, consider it as a subsheaf of the functions field.  To do this, a choice is required.

Working with $\mathbb{P}^1$ example, the choice is to consider $O(1)$ as $O([1:o]), O([o:1])$ or indeed $O(p)$ for any $p \in \mathbb{P}^1$.  As will become evident, the choice is analogous to choosing to think of $K(\mathbb{P}^1)$ as $k(x)$ or as $k(y)$.

Global Sections (Take Two)

Talking about line bundles more generally, recall the correspondence between Weil, Cartier and line bundles.

For Weil need nisr.  For Weil and Cart. to agree need nis and locally factorial.  The latter condition says that locally have spec of a UFD and all Weil div. on UFD are trivial, which means locally is the divisors of a single eqautionn

Cart. divisors $D = \{f_i \in K^*(U_i)\}$ such that $f_i/f_j \in O(U_{ij})^*$. Get a line bundles by saying sections over $U_i$ are $\frac{1}{f_i}O(U_i) \subset K(U_i)$; denote this $O(D)$

Naturally $O(D) \subset K$ where $K$ is the sheaf of total fractions.  When the underlying space is integral $K$ is constant and any line bundles embeds in $K$ via. $L \to L \otimes_O K \cong K$.

In the integral case, we can looks at sections of a line bundle as locally in $O$ or as just a single element in $K$.  Then global sections are elements $s_i \in O(U_i)$ such that image in $K$ agree i.e.

$s_i/f_i = s_j/f_j \in K$

Going back to the $\mathbb{P}^1$ example, if I say $O(1) = O([0:1])$ then on $U_1$ this is cut out by $x$ and on $U_0$ its cut out by $1$.  So sections as a subset of $K$ are

$\frac{1}{x}O(U_o) = span(\frac{1}{x},1,x,x^2,...$

$1O(U_1) = span(1, y = 1/x, y^2 = 1/x^2, ...$

So comparing both sides pick out the elements $1,1/x \in K$ as a basis.

DIVISOR OF ZEROS: Now $(1/x) = -[0:1]$  So $(1/x) + D$ is effective where $D = [1:0]$.  Or instead we could have first multiplied by $x$ then just taken the divisor of $x/x = 1$ (here talking about Weil Divisors).

The story works nicer with Cartier Divisors.  Say $s \in \Gamma(X,O(D))$ where $D = \{ f_i \in K^*(U_i) \}$.  As above $s \in K$ is just a single function.  The way $O(D)$ is constructed it follows that $sf_i \in O(U_i)$.  So consider the divisor $(s)_0 = \{sf_i \in K^*(U_i)\}$, then $(s)_0 - D = (s)$ hence $(s)_0$ is lin. equiv to $D$.

Now back to the whole point, showing that hyperplane divisors are the same as $|D|$.  The intuitive way to define them is the divisor consisting of the intersections of a hyperplane with the image of $X$.  But this is more like Weil divisors.

Via Cartier divisors, say $X \xrightarrow{|D| = \phi} \mathbb{P}^n$, then define hyperplane divisors to be divisor of zeros of $\phi^*h$ where $h \in O(1)$.  Then the desired statement becomes trivial from the isomorphism $\phi^*O(1) = O_X(D)$.

I think the easy way of seeing this is to choose an embedding $O_X(D) \to K(X)$ by picking a basis $s_o, ..., s_n \in \Gamma(O_X(D))$ and fixing $s_o = 1$.  Then the global sections are effectively $1, s_1/s_0, ..., s_n/s_o$.

The morphism $\phi$ restricted to where $s_0$ doesn’t vanish is given by $k[.. x_i/x_o ..] \to O_X(X_o)$ sending $x_i/x_o \mapsto s_i/s_o$.

Took very long to articulate this:

Line bundles are determined (up to lin equivalence) by transition functions. From the definition of $\phi = |D|$ its clear that the transition functions $t_{ij} = x_i/x_j$ map to the transition functions $t'_{ij} = s_i/s_j$.

Cartier Divisors and transition functions are determined as follows.  From the transition functions form the cartier divisor $D'$: $1 \in O_X(U_0)$ and $1 \cdot t_{oi} \in O_X(U_i)$.  With this choice $1, s_1/s_o, ... , s_n/s_o \in K(X)$ are the global sections.  And its clear that $O_X(D)\otimes O(-D') = O$ so this recovers $O_X(D)$, hence $\phi^*O(1) = O_X(D)$

All the choice made just are just a shift by a principle divisor, so up to linear equivalence everything works out.