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.


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