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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 22}
\author{Ravi Vakil}
\date{Tuesday, November 30,  1999.}
\maketitle
\tableofcontents

{\parskip=12pt % closing bracket is just before the bibliography 
	
Problem Set 8 back at end of lecture.
PS9 due today.  

PS10 due Thursday December 2.  PS11 due Thurs. Dec. 9.  PS12 due
Monday December 13.  If you don't get a chance to do PS12 by December
13, but still want to think about it, I'll be around, so you can drop
by and we can talk.

\section{Example:  The invertible sheaves $\oh_{\proj^n}(m)$.}

{\em Caution.} I don't I think I was consistent with transition
functions last time.  I think everything is correct in the lecture notes.  If
$g_i$ is a section over $U_i$ and $g_j$ is a section over $U_j$, then
$g_i = f_{ij} g_j$.    

I'll now describe some invertible sheaves on projective space.  I think this
description will help you see how to deal with invertible sheaves in general.

Let $U_0$, \dots, $U_n$ be the standard opens of $\proj^n$.  Consider
the invertible sheaf (known as $\oh_{\proj^n}(m)$) given by the
transition functions $f_{ij} = (x_j/x_i)^m$.  Recall for this sheaf,
the sections corresponding to an open set $U$ are given by the data of
sections $g_i \in \oh(U \cap U_i)$ (i.e functions on $U \cap U_i$),
satisfying the transition information $g_i = f_{ij} g_j$ on the open
set $U \cap U_i \cap U_j$.

Note that $f_{ii} = 1$, and $f_{ik} = f_{ij} f_{jk}$.

We can talk about {\em rational} (or {\em meromorphic}) sections as well.

\subsection{Example:  $\proj^1$}

Let's do an example.  To keep with last day's example, let's let
$n=1$, and let's first take $m=-1$.  Here are the variables we'll use.

On $\proj^1$, use projective coordinates $[x_0; x_1]$.

On open set $U_0$, choose coordinate $y_1 = x_1/x_0$, so $U_0$ are
the points $[1;y_1]$.  (Note that this is equal to $[x_0;x_1]$!)

On open set $U_1$, choose coordinate $z_0 = x_0/x_1$, so $U_0$ are the
points $[z_0; 1]$.

Here's a section over $U_0$: $g_0(y_1) = y_1-1$.  Let me ask you some
stupid questions.  Where does it have zeroes or poles?  Answer: a zero
of order 1 at $y_1=1$, i.e. at $[x_0;x_1] = [1;1]$.

Let's consider this as a {\em rational} section over all of $\proj^1$.
What does this section look like over $U_1$?  It is given by a
function $g_1(z_0) = f_{10} f_0(y_0) = x_1/x_0 (1/z_0 - 1) =
(z_0-1)/z_0^2$.  Zero of order 1 at [1;1], and pole of order 2 at
[0;1].  

{\em Remark.}  Note that we calculated that this section had a zero of order 1 at
$[1;1]$ in both open neighbourhoods.  But the functions whose order of
vanishing we checked were different!  Why were we guaranteed to get
the same answer?  Answer: they differed by a factor that was
invertible (i.e. had 0 valuation).

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

{\em Proof 1.} The sums of orders of vanishing of this meromorphic
section is -1.  The sums of orders of vanishing of a meromorphic
structure sheaf is 0.  (This is the proof I gave last time. \epf

{\em Remark.} Notice that   In general, the sums of orders of
vanishing of a meromorphic section of $\oh_{\proj^1}(m)$ is $m$.  So they
are all non-isomorphic invertible sheaves.

{\em Proof 2.}  You can check that $\oh_{\proj^1}(-1)$ has no non-zero global sections.
But the structure sheaf has non-zero global sections (the non-zero constants). \epf

\subsection{Identification of global sections of $\oh_{\proj^n}(m)$ with
the degree $m$ polynomials in $x_0$, \dots, $x_n$ ($m\geq0$)}

As another example, I'll let you calculate the space of global sections
of $\oh_{\proj^1}(2)$.

Using the same variables as before, $g_0(y_1)$ is a polynomial in
$y_1$.  $g_1(z_0)$ is a polynomial in $z_0$, and $g_1(z_0) = z_0^2
g_0(1/z_0)$.  Hence $g_0$ can have degree at most 2.  Conversely, any
polynomial of degree 2 will work.  Hence the space of global sections
of $\oh_{\proj^1}(2)$ has dimension 3.  (The same argument will show
that the space of global sections of $\oh_{\proj^1}(n)$ has dimension
$n+1$ if $n \geq 0$.)

You can identify the space of global sections with the vector space of
degree 2 polynomials in $x_0$ and $x_1$.  Here's the identification.
Given a polynomial $a x_0^2 + b x_0 x_1 + c x_1^2$, to get the
corresponding section $g_0(y_1)$ over $U_0$, just divide by $x_0^2$.
Hence get $a + b (x_1/x_0) + c (x_1/x_0)^2 = a + b y_1 + c y_1^2$.  To
get the corresponding section $g_1(z_0)$ over $U_1$ just divide by
$x_1^2$, and get $a (x_0/x_1)^2 + b (x_0/x_1) + c = a z_0^2 + b z_0 +
c$.  Note that $g_0$ and $g_1$ satisfy the desired transition data:
$g_0 = (x_0/x_1)^2 g_1$.

In the same way you can prove:

{\bf Proposition.} Let $m \geq 0$.  Then the vector space of global
sections of $\oh_{\proj^n}(m)$ can be identified with the vector space
of polynomials $h(x_0, \dots, x_n)$ of degree $m$.  The identification
is this: $g_i = h/x_i^m$.

{\em Probably an exercise.}  You've actually done this in the case when $m=0$,
when you proved that the only regular functions on projective space are the constants.

This is really concrete, and to prove to you how concrete it is, let me ask you some questions.

Here's a section of $\oh_{\proj^2}(3)$:  $x^3+y^3+z^3$.  Where does it have zeroes?
How about $x y^2$?

We can extend this to rational sections.  Once again, they have to be homogeneous degree $m$.
How about $(x^3+y^3+z^3)/ (x^2 yz)$, which is a meromorphic section of $\oh_{\proj^2}(-1)$.
Where are the poles and zeroes?

\section{More background on invertible sheaves}

\subsection{Operations on invertible sheaves}

Here are some basic
things you can do with invertible sheaves.  

{\em i) Pullback.} You can pull back invertible sheaves (or line
bundles).  (Give picture first.)  Here's how.  If you have a morphism
$\pi: X \rightarrow Y$, and you have an invertible sheaf $\cL$ on $Y$
defined by open sets $U_i$ and transition functions $f_{ij}$ on $U_i
\cap U_j$, then define the {\em inverse invertible sheaf} $\pi^* \cL$
by the open sets $\pi^{-1}(U_i)$, and the functions $\pi^* f_{ij}$,
which are functions on $\pi^{-1}(U_i \cap U_j) = \pi^{-1}(U_i) \cap
\pi^{-1}( U_j)$.

You can pull back sections as well: if $g$ is a section of $\cL$ over
some open subset $U \subset Y$, then you can define a section
$\pi^{-1} g$ of $\pi^{-1} \cL$ over $\pi^{-1}(U)$ (which is also
open).  For convenience of notation, I'll show you how this works for
global sections.  If $g$ is a global section given by regular
functions $g_i$ on $U_i$ related by $g_i = f_{ij} g_j$ on $U_i \cap
U_j$, when consider the regular functions $\pi^* g_i$ on $\pi^{-1}
U_i$.  (Remember that you can pull back regular functions via
morhpisms!)  Then because $g_i = f_{ij} g_j$ on $U_i \cap U_j$, we
have $\pi^{-1} g_i = \pi^{-1} f_{ij} \pi^{-1} g_j$ on $\pi^{-1} (U_i
\cap U_j) = \pi^{-1} (U_i) \cap \pi^{-1} (U_j)$.  In other words, these
regular functions define a section of $\pi^* \cL$.

As an example of how you can use this, consider:

{\bf Proposition.}  $\oh_{\proj^n}(m_1)$ is not isomorphic to $\oh_{\proj^n}(m_2)$ if $m_1 \neq m_2$.

{\em Proof.} Recall that we've proved this for $n=1$: any non-zero
meromorphic section has ``total order of vanishing'' $m$.

Now we'll deal with the general $n$.  Take the closed immersion
$\proj^1 \hookrightarrow \proj^m$, sending $\proj^1$ to a line, e.g.
$(x_0; x_1) \mapsto (x_0; x_1; 0; \dots; 0)$.  Check that
$\oh_{\proj^n}(m_i)$ pulls back to $\oh_{\proj^1}(m_i)$. 
But $\oh_{\proj^1}(m_1)$ is not isomorphic to 
$\oh_{\proj^1}(m_2)$.
\qed

{\bf Fact.}  (We probably won't prove this, but the proof isn't difficult to follow.)  These are {\em all} the line
bundles on $\proj^m$.  Hence $\Pic \proj^n \cong \zed$, with $\oh(1)$ the generator.

{\em Example: Line bundles on curves.} This example is somewhat lame.
Consider the curve $C$ $y^2 z = x^3 + x z^2$.  We can pull back the
line bundle $\oh_{\proj^2}(2)$ to $C$.  We already can see that it has
at least 6 linearly-independent sections.  Where on the curve does the
section $x^2 + y^2$ vanish?  Answer: where $x^2+y^2=0$.  We can pull
back the line bundle $\oh_{\proj^2}(3)$ to $C$.  We can already see
that it has at least 9 linearly independent sections.  $\oh_{\proj^2}(3)$
has 10 sections.  But one of them vanishes on $C$:  the section $y^2 z - x^3 - x z^2$.

{\em ii) Tensor product of two invertible sheaves.} Take 1:  Suppose you have two
invertible sheaves $\cL$ and $\cM$ on $X$, given by the {\em same} open
cover $U_i$ and (possibly different) transition functions $l_{ij}$ and
$m_{ij}$.  Then define the tensor product invertible sheaves $\cL \otimes
\cM$ by the same open cover, and the transition function $l_{ij} m_{ij}$.
(You can immediately check that this satisfies the cocyle condition.)

Take 2: Suppose now that you have two invertible sheaves $\cL$ and
$\cM$, given with the data of {\em possibly different} open covers
$U_i$ and $V_j$ respectively, with transition functions $l_{ii'}$ and
$m_{jj'}$ respectively.  Then what you want to do is to take a
different cover {\em finer} than the $U_i$ and the $V_j$ so you can
express $\cL$ and $\cM$ using this new cover; then use the
construction above.  The following does the trick: $\cL \otimes \cM$
is defined using the open sets $U_i \cap V_j$, and the transition
functions are given by $f_{ij, i'j'} = l_{ii'} m_{jj'}$.

Things you might want to check: that this construction is independent
of the ``representation'' of the invertible sheaf.  Also, $\cL \otimes
\oh_X$ is isomorphic to $\cL$.

{\em Remark.}  If $a$ is a global section of $\cL$ (or more generally,
a section over some open set), and $b$ is a section of $\cM$, then
you can interpret $ab$ as a section of $\cL \otimes \cM$.

{\em Remark.}  You can see how this works with $\oh_{\proj^1}(m)$.
Immediately, we have $\oh_{\proj^1}(m+n) \cong
\oh_{\proj^1}(m) \otimes \oh_{\proj^1}(n)$.

{\em Remark.} If $a$ is a section of $\cL$ (over some open set) and
$b$ is a section of $\cM$ (over the same open set), then $ab$ is
naturally a section of $\cL \otimes \cM$.  For example, $x^2$ is a
global section of $\oh_{\proj^1}(2)$, and $(x+y)^3$ is a global
section of $\oh_{\proj^1}(3)$.  What is their product as a global
section of $\oh_{\proj^1}(5)$?  Answer: $x^2 (x+y)^3$.

{\em iii) Inverse invertible sheaves.} Suppose you have an invertible sheaf
$\cL$ defined by the open cover $U_i$ and transition functions
$f_{ij}$.  Then define the {\em inverse}, denoted $\cL^{-1}$, by the
same open cover, and the transition functions $f_{ij}^{-1}$.  Note
that the cocycle condition is satisfied, and also that $\cL \otimes
\cL^{-1} \cong \oh_X$.

{\bf Coming next:}  The Picard group.  Maps to projective space correspond to a vector space of sections
of an invertible sheaf.  The class group.  The canonical invertible sheaf (= the sheaf of differentials).  Genus.
Riemann-Roch Theorem: statement (no proof) and applications.
The Riemann-Hurwitz formula.  

} % end of parskip; it started just before the introduction
\end{document}
