# Tuesday, January 24

:::{.remark}
Recall that $\WDiv(X) = \mathrm{Ca}\Div(X)$ if $X$ is smooth or if $\codim X_\sing \geq 3$, and for any $D\in \CDiv(X)$, 
\[
\OO_X(D)(U) \da \ts{f\in k(U) \st \div f + D \geq 0}\in \Pic(X)
\]
is a line bundle.
:::

:::{.example title="?"}
Let $D = V(xyz) \subseteq \PP^2$, then $D$ is 3 copies of $\PP^1$ linked in a triangle.
Consider $f\in \OO_{\PP^2}(D)(\PP^2)$, so $\div f = -L_1 -L_2 -L_3 + \sum n_p P$ for some $n_p \geq 0$.
Since $f\in k(\PP^2)\implies f = a/b$ with $a,b$ homogeneous polynomials of the same degree, one example is $f(x,y,z) = a(x,y,z)/xyz$ with $a$ homogeneous of degree 3.
Thus $\H^0(\OO_{\PP^2}(D)) \cong k[x,y,z]^{3, \homog}$.
:::

:::{.proposition title="?"}
If $D \equiv D'\in \Cl(X)$, so $D-D'= \div h$ for some $h$, then $\OO_X(D)\cong \OO_X(D')$ as sheaves.
:::

:::{.proof title="?"}
One needs a map $\OO_X(D)(U) \to \OO_X(D')(U)$ for every open $U \subseteq X$, so take the map 
\[
\OO_X(D')(U) &\to \OO_X(D)(U) \\
f' &\mapsto f\da f'\cdot h\inv
.\]
Since $\div (f' h\inv) = \div(f') - \div(h)$, we have 
\[
\div f' + D' \geq 0 \iff \div f' + D' + \div h \geq \div h \iff \div f + D \geq 0
.\]

:::

:::{.remark}
Note that if $D\geq 0$ then $\OO_X(D)(X) \ni 1$, the constant function, and this is a global section so $H^0(\OO_X(D)) > 0$.
:::

:::{.fact}
Any irreducible codimension 1 $D \subseteq \PP^n$ is of the form $V(f)$ for a single function $f$, which follows from the fact that any height 1 prime in $\kxnz$ is principal.
Thus $\Div(\PP^n) = \ts{\displaystyle\sum_{f\in k[x_0,\cdots, x_n]^{\homog, \irr} } n_f [V(f)] }$, and if $D = \sum n_f [V(f)]$ and $D' = \sum n_{f'}[V(f')]$, then $D\equiv D' \iff \sum n_f \deg f = \sum n_{f'} \deg f'$, noting $D-D' = \div\qty{\prod f^{n_f} \over \prod (f')^{n_{f'}} }$ which is a rational function on $\PP^n$.
So $\Cl(\PP^n)\isovia{\deg} \ZZ$ where $\sum n_f [V(f)] \mapsto \sum n_f \deg f$.
:::

:::{.definition title="?"}
$\Pic(X)$ is the group of line bundles on $X$ up to isomorphism, with group structure given by the following: for $L_1, L_2\in \Pic(X)$, define 
\[
(L_1\tensor L_2)(U) \da L_1(U)\tensor_{\OO_X(U)} L_2(U)
.\]
Alternatively, the transition functions on the tensor product are products of transition functions: 
\[
t_{UV}^{L_1\tensor L_2} = t_{UV}^{L_1}\cdot t_{UV}^{L_2}
.\]
The identity element is $\OO_X$, since $L_1(U) \tensor_{\OO_X(U)} \OO_X(U) = L_1(U)$.
Inverses are given by $L\inv \da \sheafhom(L, \OO_X)$, so $L\inv(U) = \Hom_{\OO_X(U)}(L(U), \OO_X(U))$ on small enough open sets, and the transition functions are given by 
\[
t_{UV}^{L\inv} = (t_{UV}^{L})\inv
.\]
It can be checked that if $L$ satisfies the cocycle condition iff $L\inv$ does, and similarly for $L_1\tensor L_2$.
:::

:::{.proposition title="?"}
If $X$ is smooth then $\Pic(X)\cong \Cl(X)$ via $D\mapstofrom \OO_X(D)$.
:::

:::{.proof title="?"}
This uses that $\OO_X(D)\cong\OO_X(D')\iff D\equiv D'$, the interesting part is to show surjectivity.
Let $L\in \Pic(X)$, then $\ro L U = \OO_U$ for some $U$, and we can consider $1\in \OO_U(U) \cong L(U)$.
In any other trivialization, $\ro L V \cong \OO_V$ and there is a transition function $t_{UV} \in k(V)$.
Since $1\in \OO_U(U)$, we have $\div(1) = D$ where the LHS is regarded as a rational section of $L$, and $L \cong \OO_X(D)$.
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:::{.definition title="Rational sections"}
A **rational section** of $L$ is a section of $L\tensor k(X)$.
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:::{.remark}
This allows for a section $s\in H^0(L)$ to have poles, and $L\cong \OO(\div(s))$ for any section $s$ of $L$.
If $s, s'$ are rational sections, then $s/s'$ is a rational *function*. 
Concretely, if $s = \ts{s_u \in k(U) \st t_{UV} s_U = s_V}$ and $s' = \ts{s'_U \in k(U) \st t_{UV} s'_U = s'_V}$.
Then $s/s' = \ts{s_U \over s_U' = s_V/s_V'} \in k(X)$, so $\div(s) = \div(s')$.
:::

:::{.remark}
The degree of any principal divisor on a curve is zero.
:::

:::{.example title="?"}
More interesting examples come from elliptic curves.
Write $X = \CC/ \Lambda$ where \( \Lambda = \ZZ \oplus \ZZ \tau \) with $\tau \in \HH$
This yields a complex manifold, since the transition functions are translations and thus holomorphic.
We can write $\Div(X)\ni D = \sum n_p [p]$ where $p\in X$ are points.
A meromorphic function is a rational function $f: X\rational \CC$ which extends to a holomorphic map $f:X\to \CP^1$ by mapping poles to $\infty$ -- note that this extension only works because $X$ is complex dimension 1, and does not work in higher dimensions.
This pulls back to $\tilde f: \CC\to \PP^1$ which satisfies $\tilde f(z+ \lambda) = \tilde f(z)$ for all $\lambda\in \Lambda$.
The Weierstrass $\wp\dash$function is defined by 
\[
\wp(z) \da {1\over z^2} + \sum_{\lambda\in \Lambda\smz}\qty{ {1\over (z - \lambda)^2 } - {1\over \lambda^2} }
,\]
which averages over the lattice and is thus periodic.
Note 
\[
{1\over (z- \lambda)^2} - {1\over \lambda^2} = { \lambda^2 - (z- \lambda)^2 \over (z- \lambda)^2 \lambda^2}
.\]
where the denominator is $\geq C \abs{ \lambda}^4$ for $\abs{z} \gg 1$ and the numerator is $\leq c \abs{ \lambda}$ for $\abs{z} \gg 1$, so 
\[
\sum_{ \lambda \in \Lambda\smz} C\abs{\lambda}^{-3} \leq C \int_{\RR^2} \abs{\lambda}^{-3} = C \iint r^{-3} r\dr\dtheta
\]
which converges.
So the extra constant ${1\over \lambda^2}$ is necessary to make the series converge.
Since translating by $\lambda\in \Lambda$ rearranges the series, $\wp(z)$ is a well-defined rational function on $X$ with a double pole at every $z\in \Lambda$, corresponding to $0\in \CC/ \Lambda$.
So $\div \wp(z) = -2 [0]$, and it induces $X\mapsvia{\pi} \PP^1$ where we have $\deg \pi = \deg \pi\inv(0) = \deg \pi\inv(\infty)$.
:::

:::{.proposition title="?"}
Let $f\in \CC(X)$ be meromorphic and let $\div f = \sum n_p [p]$.

1. $\sum n_p = 0$
2. $\sum n_p [p] \equiv 0 \mod \Lambda$

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:::{.proof title="?"}
Let $f\in \CC(X)$ with some zeros and poles.

Take the following contour:

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\end{tikzpicture}

Note that $\int_\gamma \dlog(f) = \int_\gamma {f'(z) \over f(z)} \dz = \sum_{p \text{ inside}} n_p$ by the residue theorem.
Recall that in local coordinates $w$, if $f(w) = cw^{-k} + \cdots$ then $\dlog(f) = -k {dw \over w} + h(w)$ where $h$ is holomorphic.
However, by periodicity, the edge integrals cancel and $\int_{C_1 + C_2} \dlog(f) = \sum_{p\text{ edge}} -n_p$, forcing $\sum n_p = 0$.

Now considering $\int_\gamma z\dlog(f) = \sum_{p \text{ inside}} n_p p$ and $\Res_p z\dlog(f) = n_p \cdot z(p)$.
On the other hand, $-\sum_{p \text{ edge}} n_p p+ \int_{A\sm A'} z\dlog(f)$.
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