\chapter{The Riemann-Roch theorem}
\label{ch:riemann_roch}

\section{Motivation}

Recall a basic fact in complex analysis:
\begin{quote}
	A holomorphic $\CC \to \CC$ function is uniquely determined by its Taylor series expansion at
	the origin.
\end{quote}

Compared to the case of real smooth function, this is already very rigid --- the value of the
function in a small neighborhood of the origin determines the value of the function everywhere ---
but, in order to specify a function, you still need infinitely many coordinates!

Meanwhile, we have Liouville's theorem:
\begin{quote}
	A bounded holomorphic $\CC \to \CC$ function is constant.
\end{quote}
As we have learnt earlier, this theorem, when phrased in terms of Riemann surfaces, can be more
elegantly rephrased to the following:
\begin{quote}
	A holomorphic $\CC_\infty \to \CC$ function is constant.
\end{quote}
In other words, in order to specify a holomorphic $\CC_\infty \to \CC$, you only need a \emph{single
complex number}! That is, the $\CC$-vector space $\Hom(\CC_\infty, \CC)$ has dimension $1$.

Naturally, you may ask, ``is there anything inbetween''? There is! And the Riemann-Roch theorem is
the main ingredient to understand how these things work.

So, how are we going to define this? If you compare the two situations above,
a holomorphic $\CC \to \CC$ function is a meromorphic $\CC_\infty \to \CC$ function, which
is allowed to have a pole at $\infty$, and nowhere else.

So,
\begin{moral}
	By smoothly interpolate between ``allow pole of arbitrary order'' and ``must be holomorphic'',
	we can produce many interesting spaces of functions.
\end{moral}

Conveniently, back in \cref{ch:meromorphic_fn}, we have defined the \vocab{multiplicity} of a
zero and the \vocab{order} of a pole of a meromorphic function.
So, the natural point between these two extremes is to allow a pole of order at most $d$.

For notational convenience, we defines:
\begin{definition}[Order of a meromorphic function]
	Let $f$ be meromorphic at $p$. We define $\ord_p(f)$ to be:
	\begin{itemize}
		\ii $d$, if $f$ has a zero of multiplicity $d$ at $p$;
		\ii $-d$, if $f$ has a pole of order $d$ at $p$;
		\ii $0$, otherwise.
	\end{itemize}
\end{definition}

\begin{example}[The space of functions with pole of order at most $4$ on $\CC_\infty$]
	Let $L(4 \cdot \infty)$ be the set of meromorphic $\CC_\infty \to \CC$ function, being
	holomorphic everywhere except $\infty$, and has a pole of order at most $4$ at $\infty$ ---
	in other words,
	\[
		L(4 \cdot \infty) = \{ \text{$f$ meromorphic on $\CC_\infty$} \mid \text{$f$ defined on
		$\CC_\infty \setminus \{ \infty \}$}, \ord_\infty(f) \geq -4 \}.
	\]
	(The notation $L(-)$ will be explained later.)

	Obviously, this forms a natural $\CC$-vector space.

	Consider the Taylor series of any $f \in L(4 \cdot \infty)$ at the origin:
	\[
		f(z) = \frac{c_{-m}}{z^m} + \frac{c_{-m+1}}{z^{m-1}} + \cdots + \frac{c_{-1}}{z} + c_0 +
		c_1 z + \cdots
	\]
	Obviously, because $f$ is defined at the origin, it cannot have any nonzero coefficient $c_{-m}$
	for $m > 0$.
	But more importantly, it cannot have any nonzero coefficient $c_m$ for $m > 4$
	either!\footnote{The reason is actually not very straightforward, but you can see for yourself
		why it is true: if there are only finitely many nonzero terms, then the order of the pole at
		$\infty$, $(-\ord_\infty(f))$, is precisely the degree of the highest nonzero coefficient.}

	Did you see what happened here? We start with requiring the function to be analytic and does not
	blow up too badly, and we end up with just the \emph{algebraic} function --- the polynomials!

	In particular, $L(4 \cdot \infty)$ consists of the polynomials of degree $\leq 4$, and
	\[
		\dim L(4 \cdot \infty) = 5
	\]
	as a $\CC$-vector space.
\end{example}

\begin{example}[More complicated $L(-)$ spaces]
	There's no reason why we should restrict ourselves to considering only the functions that blow
	up at $\infty$ --- as we will see, more general meromorphic functions can be considered, as long
	as we restrict the order of the poles.

	Let $L(-1 \cdot 3 + 4 \cdot i + 5 \cdot \infty)$ be the set of meromorphic functions $f \colon
	\CC_\infty \to \CC$ that are:
	\begin{itemize}
		\ii holomorphic everywhere in $\CC_\infty$,
		possibly with the exception of the points $3$, $i$, and $\infty$;
		\ii at $3$, it must have a root of order $\geq 1$;
		\ii at $i$, it cannot have a pole of order more than $4$;
		\ii at $\infty$, it cannot have a pole of order more than $5$.
	\end{itemize}
	So, for example, $(z \mapsto z-3)$, $(z \mapsto \frac{(z-3)^3}{(z-i)^2})$, or $(z \mapsto
	(z-3)^4)$ are functions in the set, but not $(z \mapsto (z-3)^2+1)$ or $(z \mapsto (z-3)^7)$.

	As before, this is a $\CC$-vector space, and furthermore, it is also finite-dimensional!
	What should its dimension be?

	Well, note that there is a $1$-$1$ bijection between functions $f \in L(-1 \cdot 3 + 4 \cdot i +
	3 \cdot \infty)$ and functions $g \in L(-1 \cdot 3 + 7 \cdot \infty)$ by
	\[ g = \Phi(f) = (z \mapsto f(z) \cdot (z-i)^4), \]
	where, as you could probably have guessed by now, $L(-1 \cdot 3 + 7 \cdot \infty)$ is the space
	of meromorphic functions that has at least a zero at $3$ and at most a pole of order $7$ at
	$\infty$.

	Using that information, it shouldn't be too hard for you to see that the dimension should be
	$7$.
\end{example}

For another piece of motivation:
Later on, we will also define the concept of \vocab{divisors} and \vocab{line bundles}.
If you have learned about these concepts in algebraic geometry context, you might be interested to
learn what they are actually about;
otherwise,
it is still very surprising that these theorems can be naturally generalized to completely algebraic
settings, and \emph{your intuition from
the case of analytic manifold will mostly work verbatim} --- in fact, you can even define the genus
of a number field, like $\QQ[\sqrt 2]$!

\section{Divisors}
\prototype{$(-3) \cdot i + (-4) \cdot \infty$ is a divisor on $\CC_\infty$.}

We start with defining a convenient notation for the above concepts.

First, observe that the condition ``$f$ must have a zero of order at most $4$ at the origin'' can be
conveniently written as
\[ z^4 \mid f.  \]
In other words, $z^4$ must be a \vocab{divisor} of $f$.

This notation works if $f$ is a polynomial, since we already know what it means for two polynomials
to divide each other.

Generalizing, we could say ``$f$ cannot have a pole of order more than $3$ at the point $i$,
and $f$ cannot have a pole of order more than $4$ at the point $\infty$'' by
\[ (z-i)^{-3} \cdot (z-\infty)^{-4} \mid f.  \]

Of course, at this point, the notation is purely formal --- there is no interpretation as
``functions'' that could be assigned to the expression $(z-\infty)$, for instance.

Those objects are, appropriately enough, called \vocab{divisors}. So we come to the formal
definition:
\begin{definition}[Divisors]
	Let $X$ be a Riemann manifold, then a divisor $D$ on $X$ is a function $D \colon X \to \ZZ$,
	which is nonzero on a discrete set of points.
\end{definition}

The formal objects $(z-i)^{-3} \cdot (z-\infty)^{-4}$ above, from now on, we will consider it as a
function $D \colon \CC_\infty \to \ZZ$ by
\[ D(z) = \begin{cases} -3 & z = i \\ -4 & z = \infty \\ 0 & \text{otherwise}. \end{cases} \]

\begin{abuse}
	For a point $p \in X$, we write $p$ to mean the divisor that takes value $1$ at $p$, and $0$
	elsewhere.
\end{abuse}

Because divisors are integer-valued functions, we can add two divisors together or multiply a
divisor with an integer, the result is an integer. So,
\begin{example}[$(z-i)^{-3} \cdot (z-\infty)^{-4}$ as a divisor]
	The divisor $D$ that corresponds to the formal object $(z-i)^{-3} \cdot (z-\infty)^{-4}$ above
	can be written as $(-3) \cdot i + (-4) \cdot \infty$.
\end{example}

\section{Degree of a divisor}
\prototype{$\deg ((-3) \cdot i + (-4) \cdot \infty) = -7$.}

If the surface $X$ is compact, any discrete set of points is finite. Thus, a divisor $D$ on $X$ has
finite support.

This allows us to define the degree of a divisor:
\begin{definition}[Degree of a divisor]
	For a divisor $D$ on a compact surface $X$, its degree is $\sum_{p \in X} D(p)$.
\end{definition}
Of course, the sum is well-defined because only finitely many terms are nonzero.

\section{The principal divisor of a meromorphic function}
\prototype{$\Div \frac{(z-3)^2}{z-i} = 2 \cdot 3 + (-1) \cdot i + (-1) \cdot \infty$ has degree
$0$.}
After defining a divisor, we want a convenient notation to formalize our fuzzy notation earlier of a
divisor ``divides'' a function.

\begin{definition}[Divisor of a meromorphic function]
	Let $f$ be meromorphic on a Riemann surface $X$. Then the divisor of $f$, $\Div(f)$, is such
	that
	\[
		\Div(f)(p) = \ord_p(f).
	\]
\end{definition}

We can also write it as a formal sum: $\Div(f) = \sum_p \ord_p(f) \cdot p$ --- by the abuse of
notation above, this would be an actual sum if $f$ has finitely many roots and poles.

If a divisor $D$ is equal to $\Div(f)$ for some $f$, we call $D$ a \vocab{principal divisor}.
(Compare this with a principal ideal, being an ideal generated by one element!)

Looking at the prototype example of this section, you might have guessed the following for the
Riemann sphere. In fact, much more is true:
\begin{proposition}
	If a divisor $D$ on a compact surface $X$ is principal, then $\deg D = 0$.
\end{proposition}

Let us not forget our goal of defining a convenient notation to talk about the space of functions
with bounded poles.
With the notation defined above, if $f = z$, $\Div f = 1 \cdot 0 + (-1) \cdot \infty$,
and we want to say $f$ ``divides'' the divisor $(-1) \cdot \infty$. The natural definition would be:
\begin{definition}[The partial ordering of divisors]
	We write $D \geq 0$ if $D(p) \geq 0$ for all $p \in X$.

	For two divisors $D_1$ and $D_2$, we write $D_1 \geq D_2$ if $D_1-D_2 \geq 0$.
\end{definition}

And finally,
\begin{definition}
	Let $D$ be a divisor on a Riemann surface $X$.
	Then the space of meromorphic functions with poles bounded by $D$ is
	\[
		L(D) = \{ \text{$f$ meromorphic on $X$} \mid \Div(f) \geq -D \}.
	\]
\end{definition}

\begin{exercise}
	This exercise is just for you to get familiar with the notation. Show the following:
	\begin{itemize}
		\ii For two divisors $D_1 \leq D_2$, then $L(D_1) \subseteq L(D_2)$.
		\ii If $X$ is compact, then $L(0) \cong \CC$.
		\ii If $X$ is compact and $\deg D < 0$, then $L(D) = \{ 0 \}$.
	\end{itemize}
\end{exercise}

\section{The Riemann-Roch theorem}

\todo{Explain canonical divisor}

\begin{theorem}[The Riemann-Roch theorem]
	Let $D$ be a divisor on an algebraic curve $X$ of genus $g$, and $K$ be a canonical divisor on
	$X$. Then
	\[ \dim L(D) - \dim L(K-D) = \deg(D) + 1 - g. \]
\end{theorem}

\todo{List some applications}
