# Friday, March 19

:::{.remark}
Recall Serre duality: let $C\in \Mfd_\CC(\cpt,\oriented)$ and $L\to C \in \Bun(\Hol)$.
Then
\[
h^1(L) = h^0(L\dual \tensor K_C)
.\]
We also have Riemann-Roch, a very important tool:
\[
h^0(L) - h^1(L) = \deg L + 1 - g(C)
,\]
where $\deg L = \int_C c_1(L)$, which is also equal to \( \deg [ \ts{ s = 0 }]  = \deg(\div s) \).
Note that $c_1$ is the most important Chern class to know, thanks to the splitting principle.
How was it defined?
There are several definitions:

1. $L$ defines an element of 
\[
H^1(C, \OO\units) 
&= \ts{ t_{UV}: U \intersect V \to \CC\units \st t_{UV} t_{UW}\inv t_{VW} = 1 } / \bd \ts{ h_u: U\to \CC\units } \\
&= \ker \bd^1 / \im \bd^0
\]
in Čech cohomology.
  By definition \( \bd \ts{ h_U \st U\in \mathcal{U} } = \ts{ h_u h_v\inv \st U, V \in \mathcal{U} } \), where $\bd^2 = 1$ since 
  \[ 
  (h_U h_V)\inv \qty{h_U h_W \inv }\inv (h_V h_W \inv) = 1 && \text{on } U \intersect V \intersect W 
  .\]
  By assigning $L$ to its transition functions, we get a map $L\to H^1$.
  We have the exponential exact sequence:
  \[
  0 \to \constantsheaf{\ZZ} \to \OO \mapsvia{\exp} \OO\units \to 1
  ,\]
  which induces a map
  \[
  H^1(C, \OO\units) &\to H^2(C, \ZZ) \\
  L &\mapsto c_1(L)
  .\]

2. $L$ defines an element $\Fr L \in \Prinbun(\CC\units)$ (which only works for line bundles), which is defined by \( \Fr L = L \sm s_0 \)  where $s_0$ is the zero section of $L$.
  By topology, we get a classifying map 
  \[
C \mapsvia{\phi_L}  B\CC\units = \CP^\infty = (\CC^{\infty} \smz) / \CC\units
  .\]
  There is a universal $c_1\in H^2(\CP^{\infty}; \ZZ)$, so we take the pullback to define $c_1(L) \da \phi_L^*(c_1)$.
  We can use that there is a cell decomposition \( \CP^{\infty } = \CC^0 \union \CC^1 \union \CC^2 \union \cdots \), and so there is a unique generator in its $H^2$.

3. Consider a smooth section $s\in C^{\infty }(L)$, then we can define $c_1(L) \da [ \ts{ s = 0 } ]$ by taking the fundamental class, assuming that $s$ is transverse to the zero section $s_z$ of $L$.
  Here we view the zero set as an oriented submanifold.
  See picture: in this case $[\ts{ s = 0 } ] = [p] - [q] + [r]$.

  \todo[inline]{Add picture.}
:::

:::{.remark}
Applying Serre duality to the left-hand side in Riemann-Roch yields the dimension of the space of holomorphic sections of some *other* bundle, $L\dual \tensor K$.
:::

:::{.example title="The structure sheaf"}
Applying Riemann-Roch to $L \da \OO$, we get
\[
\chi(\OO) = h^0(\OO) - h^1(\OO) = 0 + 1 - g
,\]

which is equal to $h^0(\OO) - h^0(K)$. 
But the only holomorphic functions on $\CC$ are constant, so $h^0(\OO) = 1$.
In particular, $h^0(K) = g$, so any Riemann surface of genus $g$ has a $g\dash$dimensional space of holomorphic 1-forms.
:::

:::{.example title="The Canonical Bundle"}
Applying Riemann-Roch to $L\da K$, we get
\[
\chi(K) = h^0(K) - h^0(K\dual \tensor K) = \deg(K) + 1 - g
.\]
Since $K\dual \tensor K = \OO$, we obtain $g-1 = \deg(K) + 1 - g$, so $\deg(K) = 2g-2$.

We also proved this using that $K$ was the dual of holomorphic vector fields, i.e. $\int_C c_1(K) = -\int_C c_1(T)$, which by Gauss-Bonnet equals $-\chi_\Top(C) = -(2-2g) = 2g-2$.
:::

:::{.example title="Genus 2 Riemann Surfaces"}
Taking $C$ of genus 2, we have $h^0(K_C) = g= 2$, so $\deg K_C = 2(2) - 2 = 2$.
Thus there exist linearly independent sections $s, t \in H^0(K_C)$, i.e. two linearly independent holomorphic 1-forms.
We can take the ratio $s/t$, which defines a map
\[
{s\over t}: C\to \PP^1
.\]
Locally we have $s = f(z) \dz$ for $z$ a local holomorphic coordinate on $C$ and $f\in \Hol(C, \CC)$, and similarly $t = g(z) \dz$.
So $s/t = f(z) / g(z)$ is meromorphic in this chart.
Choosing a new coordinate chart $w$, this yields a transition function $z(w)$ -- not of $L$, but from the atlas on $C$.
We can write $s =f(z(w)) \, d(z(w)) = f(z(w)) z'(w) \, dw$ by the chain rule.
Thus 
\[
{s\over t}(z) = {f(z(w)) z'(w) \dw \over g(z(w)) z'(w) \dw} = {s \over t}(w)
.\]
So although $s/t$ was only defined in a coordinate chart, it winds up being independent of coordinates.
This works in general for any holomorphic line bundle: for $s, t\in H^0(L)$, there is a map \( {s\over t}: C\to \PP^1 \) since writing $s_V = \varphi_{UV} s_U, t_V = \varphi_{UV} t_U$ where \( \varphi_{UV} \) is the transition function for $L$.

:::{.fact}
Important fact: we can take these ratios to get maps to $\PP^1$.
:::

:::{.slogan}
The canonical bundle is the line bundle whose transition functions are the Jacobians of the change of variables for the atlas.
:::

:::{.question}
What is the degree of this map generically?
I.e. given \( [x_0: x_1] \in \PP^1 \) fixed, what is the size of the inverse image \( \qty{s\over t}\inv \qty{ [x_0: x_1] } \)?
:::

:::{.answer}
Writing $s/t = x_1/x_0$, we have $x_0 s - x_1 t=0$.
This is in $H^0(K_C)$, and we computed \( \deg K_C = 2 \), meaning there are two zeros of this function.
Thus is $g(C) = 2$, there is a generically 2-to-1 map \( C \to \PP^1 \), a degree 2 meromorphic function.
Note that this section could have a double zero.
:::

:::

:::{.example title="?"}
Consider the curve $y^2 = (z-1)(z-2)\cdots (z-5)$, where we think of $z, y\in \PP^1$.
This has roots $z=1,\cdots, 5$, and is equal to $\infty$ if $z=\infty$.
These are the only points of $\PP^1$ with just one square root, all other points have two square roots.

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