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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 21}
\author{Ravi Vakil}
\date{Tuesday, November 23,  1999.}
\maketitle
\tableofcontents

{\parskip=12pt % closing bracket is just before the bibliography 

{\bf Problem set info!}  PS8 back Tuesday November 30.
PS9 due Tuesday November 30.  PS10 due Thursday December 2.
(PS11 due Thurs. Dec. 9.  PS12 due Monday Dec. 13.)

Brand new topic:

\section{Line bundles and invertible sheaves}

Invertible sheaves are fundamental objects.  They are essentially the
same thing as line bundles, which are geometrically easy to visualize.
So I'll tell you intuitively about line bundles first, and then introduce
invertible sheaves.  (I won't discuss motivation just yet, so I hope
you believe me when I say how important this concept is!)

\subsection{Line bundles, classically}

Line bundle on a complex manifold (with classical topology).  Draw a picture.  Precisely, given
a complex manifold $X$, a line bundle is given by the data of a cover
$U_i$ ($i \in I$ some index set), and trivializations $U_i \times
\com$, and transition functions $f_{ij}$ on $U_i \cap U_j$ that is an
analytic  function, nowhere zero, satisfying $f_{ij} f_{jk} =
f_{ik}$ (the {\em cocyle condition}), from which $f_{ii}=1$ and $f_{ij} = f_{ji}^{-1}$.
What this means: if $u \in U_i \cap U_j$, then the point $(u,z ) \in U_j
\times \com$ is identified with the point $(u,f_{ij} z ) \in U_i \times \com$.
{\em (I think I said this incorrectly in class.)}
If you've seen Cech cohomology before, you can check that
the data of a line bundle is the same as an element of $H^1(X,
\oh^*)$, where $\oh^*$ is the sheaf of invertible (=nowhere zero)
analytic functions. 

In particular, the total space $L$ of a line bundle is also a complex manifold (of dimension one higher
than that of $X$), with a morphism $L \rightarrow X$.

A {\em section} of a line bundle is the data of maps $g_i: U_i \rightarrow
\com$ (or if you prefer, $U_i \rightarrow U_i \times \com$),
satisfying $g_i(p) = f_{ij}(p) g_j (p) $ for points $p \in U_i
\cap U_j$.  (Draw a picture of a section of $L \rightarrow X$.)

Note that there is always a {\em zero-section} given by $g_i(p) = 0$
for all $i$, $p \in U_i$.

Two line bundles $L_1 \rightarrow X$, $L_2 \rightarrow X$ are
isomorphic if there is an isomorphism $L_1 \rightarrow L_2$ commuting with
the morphisms to $X$.

{\em Example.} The trivial line bundle $X \times \com \rightarrow X$.
Global sections are just analytic functions.

At this point you might think that all line bundles are trivial (which
isn't true).  Indeed there are line bundles that seem nontrivial that
are trivial.  {\em Exercise.} Suppose $U_i$ is a finite open cover of $X$,
and for each $i$, $h_i$ is an invertible function on $U_i$.  Prove that the data
of the open cover $U_i$ and the transition functions $f_{ij} = h_i/h_j$ defines
a line bundle, and prove that this is isomorphic to the trivial line bundle.
(In fancy language, this shows that a zero-element of $H^1(X, \com^*)$ indeed
induces a trivial line bundle.)

{\em Example.} Inside $\com^{n+1}$, we know that $\proj^n$
parametrizes lines through the origin.  This will give
a line bundle on $\proj^n$ (as we'll soon see rigorously):  ``above''
each point of $\proj^n$, we have the line corresponding to that point.
(We will later see that this is not trivial, if $n>0$.)

{\em Example.} Let $X$ be a one-dimensional complex manifold.  Then
there is a {\em tangent line bundle} (and similarly a {\em cotangent
line bundle}).  (Describe it pictorially.)  This will be tricky to
describe precisely.

\subsection{Line bundles on varieties (and schemes)}

We can use the same definition for varieties, except this time we use
the Zariski topology, and the functions should be invertible regular
functions.  (Experts in Cech cohomology can later check that line
bundles are parametrized be $H^1(X, \oh_X^*)$, where $\oh_X^*$ is the
{\em sheaf of invertible functions} and the topology on $X$ is of
course the Zariski topology; in fact, if you've seen Cech cohomology
of sheaves in any other setting, you know a whole lot about cohomology
of sheaves in algebraic geometry.)

\subsection{Sheaf of sections of a line bundle, and the correspondence with line bundles}

Note that the sections actually form a {\em sheaf}.  (Sketch
intuitively first.) Here, explicitly, is how.  Suppose the line bundle
$L$ on our variety $X$ is defined by an open cover $U_i$ and transition functions
$f_{ij}$ (on $U_{ij} = U_i \cap U_j$).  Then the sheaf of sections
$\cL$ is defined by:

Elements of $\cL(U)$ are functions $h_i$ on $U \cap U_i$ such that
$h_i = f_{ij} h_j$ on $U \cap U_i \cap U_j$.

Notice that this sheaf $\cL$ is an $\oh_X$-module.  What this means
is: for each open $U$, $\cL(U)$ is an $\oh_X(U)$-module (and this
action commutes with restriction: if $U \subset V$, and $a \in \cL(V)$
and $f \in \oh_X(V)$, then the restriction of $(fa)$ to $U$ is the
same as the restriction of $f$ to $U$ acting on the restriction of $a$
to $U$.

We can now define an invertible sheaf.

{\bf Definition.}  An invertible sheaf on a variety $X$ is an $\oh_X$-module $\cL$
such that there is an open cover $U_i$ of $X$ satisfying:
\begin{itemize}
\item there are isomorphisms $\phi_i:  \cL|_{U_i} \rightarrow \oh_{U_i}$,
\item there are invertible functions $f_{ij}$ on $U_i \cap U_j$ 
such that for $z \in \cL(U)$ for any $U \subset U_i \cap U_j$,
$\phi_i(z)$ (which is a function on $U$) equals $f_{ij} \phi_j(z)$ (the product of two
functions on $U$).
\end{itemize}
Notice that the cocycle condition $f_{ik} = f_{ij} f_{jk}$ on $U_i
\cap U_j \cap U_k$ follows from this definition.

Notice that if you have a line bundle, its sheaf of sections is an invertible sheaf.
If you have an invertible sheaf, you can cook up a line bundle.  And these
constructions commute.

{\em Philosophy:} Invertible sheaves are easier to deal with (if less
intuitive to picture) than line bundles, so we'll deal with them in
the future.  Here's why this fits into the the philosophy of algebraic
geometry.
Classically, people would study varieties by looking
at them as topological spaces.  In algebraic geometry, we study
varieties by looking at {\em functions} on them (in some sense a dual
point of view).  In the same way, we will study line bundles by
instead looking at their {\em sheaf of sections}. 

\subsection{Examples}

(I'm going to discuss this example again later on.)
Let's bring this back down to earth.
I'll now give some examples.  Try to think of them analytically at the
same time as algebraically.

Define a {\em meromorphic section} (or ``rational section'') to be a section 
of a line bundle over some non-empty open set.

{\em Example.}  The trivial line bundle  $X \times \A^1 \rightarrow X$.
Global sections are just regular functions on $X$.

A specific case: let $X = \proj^1$.  Then the meromorphic sections of
the trivial line bundle are just rational functions.  They are (in
affine coordinates) of the form $(t-a_1)^{n_1} \dots (t-a_r)^{n_r}$,
where the $n_r$ are integers, possibly negative.  Define the order of
vanishing of a meromorphic function (at $a_i$ to be $n_i$).  What's
the order of vanishing of this meromorphic function at $\infty$?  Answer:
$-\sum n_i$.  Hence {\em the sum of the order of vanishings of any non-zero meromorphic
section of the trivial line bundle on $\proj^1$ is 0.}  (Remark:  
whenever you are talking about orders of vanishing, keep in mind that there
is DVR theory lurking in the background.)

{\em Example: The ``tautological'' line bundle $\oh_{\proj^n}(-1)$ on
$\proj^n$.} Let $\proj^n$ have coordinates $(x_0; \dots; x_n)$.  This
will be the line bundle that assigns to a point of $\proj^n$ the
corresponding line through the origin in $\A^{n+1}$.  Perhaps ask
about section of $\oh_{\proj^1}(-1)$ corresponding to $x_0=1$ here.  We'll define this on
patches.  On the patch $U_0$ (i.e. the points $(1; y_1; \dots; y_n)$,
so we can take coordinates $y_i$ on $U_0$), the line is of the form
$(s, s y_1,
\dots, s y_n)$ in $\A^{n+1}$, with coordinate $s$.  

On $U_1$ (i.e. the points $(z_0; 1; z_2; \dots; z_n)$), the line is of the form
$(t z_0; t; tz_2; \dots; t z_n)$.   How do we change coordinates from $s$ to $t$ on $U_0 \cap U_1$?
$s = (x_0/ x_1) t$.  The best way to see this is by example.  Consider the point
$(1, 2, 3, ..., n+1) \in \A^{n+1}$.  This is a point on the line above $(1;2;\dots; n+1)$.
Its $s$-coordinate is 1, and its $t$-coordinate is 2.  Indeed $s/t = x_0/x_1$.
Precisely:  $f_{10} = x_0/x_1 = 1/y_0 = z_0$.  $f_{01} = x_1/x_0 = y_0 = 1/z_0$.

In general, you can define coordinates $s_0$ on $U_0$, \dots, $s_n$ on
$U_n$, and we have transition functions $f_{ij} = x_j/x_i$.  Note that
$f_{ij}$ is indeed an invertible function on $U_i \cap U_j$.

{\bf Proposition.}  $\oh_{\proj^1}(-1)$ on $\proj^1$ is not isomorphic to the trivial sheaf.

(I'll restate this proof, and give a second proof next day.)

{\em Proof.} (Advantages: foreshadows idea of ``degree of a line
bundle''.)  We construct a meromorphic section corresponding to the
plane $x_0=1$ in $\A^n$ (draw picture).  It you think about it, you
might convince yourself that it blows up at the point $x_0=0$ in
$\proj^n$, and nowhere else (and it has no zeroes).

Cover $\proj^1$ with two standard opens, $U_0$ and $U_1$.
On the first standard open, we have the section $s_0 = 1$; it has no zeroes.
On the second standard open, with coordinates $(z_0; 1)$, 
the coordinates on the line is $s_1$, and points on the line are of the
form $(z_0 s_1, s_1)$.  Thus the section $(1,0)$ gives (in this coordinate system)
$s_1 = 1/z_0$.  This has a simple pole at $z_0=0$.

In short, this nonzero meromorphic section of $\oh_{\proj^1}(-1)$ has a single simple pole, so the
sum of its orders of vanishing is -1.  But we earlier checked that any nonzero meromorphic
section of the trivial sheaf has sum of orders of vanishing 0.  So they are not the same.
\epf

{\bf Coming next:} The line bundle $\oh_{\proj^n}(m)$.  Maps to
projective space correspond to a vector space of sections of a line
bundle.  The canonical invertible sheaf, genus.  Riemann-Roch Theorem:
statement (no proof) and applications. Riemann-Hurwitz.  
} % end of parskip; it started just before the introduction
\end{document}
