# Tuesday, February 07

:::{.remark}
Last time: Čech cochains defined as $C_{\mcu}^p(X; \mcf) \da \bigoplus_{i_0 < \cdots < i_p } \mcf(U_{i_0,\cdots, i_p})$ with homology defined as $\Hc^*(X;\mcf) \da \colim_{\mcu} H^*(C_\mcu^*(X; \mcf))$, and a SES of sheaves yields a LES in homology.
Recall $\Cl(X) \iso \Pic(X)$ in our setting via $D\mapsto \OO_X(D)$ which sends $U$ to $\ts{f\in k(U) \st \div f + D \geq 0}$.
:::

:::{.claim}
\[
\Pic(X) &\iso 
H^1(X; \OO_X\units) \\
L &\mapsto [(t_{UV})]
.\]
:::

:::{.proof title="?"}
Write $C_{\mcu}^1(X; \OO_X\units) = \ts{ ( t_{UV} ) \in \OO\units( U \intersect V) \st U,V \in \mcu}$ for some open cover $\mcu \covers X$, and $Z_{\mcu}^1(X; \OO_X\units) = \ts{(t_{UV}) \in \OO\units(U \intersect V)}$ with boundary $\bd^1(t_{UV}) = (t_{VW} t_{UW}\inv t_{UV})$.
Note that $t_{UW} = t_{UV} t_{VW}$.

A line bundle $L \mapsvia{\pi} X$ has a trivialization, and we can refine $\mcu$ to a cover that trivializes $L$, so $\ro{L}{U} \cong \OO_U$ as sheaves for each $U$.
We have $\pi\inv(U) \iso U\times \CC$, and $h_V\circ h_U\inv : (U \intersect V) \times \CC \to (U \intersect V)\times \CC$ is multiplication by $t_{UV}(p)$ on the fiber over $p \in U \intersect V$.
These satisfy $t_{UV} t_{VW} = t_{UW}$, so any $L$ defines an element of $Z^1(X; \OO_X\units)$ by sending $L$ to its transition function.

Conversely, given $(t_{UV})$, one can attempt to glue these to form a line bundle, but which collections define the same bundle?
Given two line bundles, refine their trivializing covers so that they coincide.
Then any two trivializations $L_U \mapsvia{h_U, h_{U}'} \OO_U$ differ by an element $f_U \in \OO\units(U)$:

\begin{tikzcd}
	{L_U} && {\OO_U} \\
	\\
	&& {\OO_U}
	\arrow["{h_U}", from=1-1, to=1-3]
	\arrow["{\cdot f_U}", from=1-3, to=3-3]
	\arrow["{h_U'}"', from=1-1, to=3-3]
\end{tikzcd}
> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJMX1UiXSxbMiwwLCJcXE9PX1UiXSxbMiwyLCJcXE9PX1UiXSxbMCwxLCJoX1UiXSxbMSwyLCJcXGNkb3QgZl9VIl0sWzAsMiwiaF9VJyIsMl1d)

On overlaps, we have $t_{UV}\mapsto f_U\inv t_{UV}$ and $t_{VU} \mapsto f_U t_{VU}$, so at the level of tuples, $(t_{UV}) \mapsto (t_{UV}) \cdot \del^0 (f_U)$ and thus $(t_{UV})$ is uniquely defined up to $\del^0(C_\mcu^0(X; \OO_X\units))$.
:::

:::{.remark}
This works for higher rank vector bundles: one has $t_{VW} t_{UV} = t_{UW}$ in $\Hol(U \intersect V \intersect W, \GL_n(\CC))$, however for $n\geq 2$ this is a nonabelian group and order matters.
In this case we get e.g. $\del^1(t_{UV}) = ( t_{UV} t{UW}\inv t_{VW} )$.
One has
\[
\ts{\text{Vector bundles on } X} \iso H^1(X; \GL_n(\OO))
\]
where $\GL_n(\OO)(U) = \ts{\text{holomorphic functions } U\to \GL_n(\CC)}$.
:::

:::{.remark}
Recall the exponential SES; taking the LES yields $c_1: \Pic(X) \to H^2(X; \ZZ)$.
For $X$ a smooth curve, $c_1 = \deg$.
For $D\in \Div(X)$, one can define the fundamental class in $X$ by taking the fundamental class $[D]\in H_{\dim_\RR D}(D; \ZZ)\iso H_{\dim_\RR D} (X; \ZZ) \isovia{\PD} H^2(X; \ZZ)$.
:::

:::{.remark}
Why is $c_1 = \deg$ true?
Consider $\mcl = \OO_C(p)$ for $p\in C$ a point on a curve.
One can take the *point bundle construction*: let $U\ni p$ be a neighborhood of $p$ and $V$ the complement of a smaller neighborhood of $p$, so $U \intersect V$ is an annulus.
For $z: U\to \CC$ a local coordinate, one can form a Cartier divisor $\ts{(U, z), (V, 1)}$ with transition function $t_{UV} = 1/z$.
Note that $H^0(\OO_C(p)) \ni s = (s_U = z, s_V = 1)$ has a section which vanishes precisely at $p$.

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\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2023/Spring/k3surfaces/sections/figures}{2023-02-07_10-19.pdf_tex} };
\end{tikzpicture}

Refine the open cover to split $U$ into two open subsets, then 
\[
c_1(\OO_C(p)) = ((z\inv)_{U_1V}, (z\inv)_{U_2V}, (1)_{U_1 U_2} ) \in Z^1(C; \OO\units)
.\]
Lifting to $C^1(C; \OO)$ using that exponential surjects on sheaves yields $(-\log(z), -\log(z), 0) \in C^1(C; \OO)$.
Taking its boundary yields
\[
\del^1( (-\log z)_{U_1 V}, (-\log z)_{U_2 V}, (0)_{U_1 U_2} ) = ( (-\log z)_{U_2 V} + (\log z)_{U_i V} )_{U_1 U_2 V}
\]
which is $0$ on the top component of $U_1 \intersect U_2 \intersect V$ and $2\pi i$ on the bottom.
This is an element of $\ul{2\pi i \ZZ}( U_1 \intersect U_2 \intersect V) \in Z^2(C; \ul{2\pi i \ZZ}) \to H^2(X; \ul{\ZZ} )$.
Thus $c_1 (\OO_C(p)) = [p]$ is the fundamental class of $p$.
:::

:::{.definition title="?"}
\[
\NS(X) \da \im c_1, \quad
\Pic^0(X) \da \ker c_1
.\]
:::

:::{.example title="?"}
For $X = E$ an elliptic curve, $\Pic X = E\times \ZZ$ where $D \mapsto (D, \deg D)$.
Thus $\NS(E) = \ZZ$ and $E = \Pic^0(E) = \ts{[p] - [0] \st p\in E}$.
Note that $\Pic^0(X) \cong \Jac(X)$ in this case.
:::

:::{.remark}
If $X$ is smooth projective, global holomorphic functions are constant, so part of the LES breaks into an exact piece: 
\[
H^0(\ZZ) = \ZZ^n \injects H^0(\OO) = \CC^n \surjects H^0(\OO\units) = (\CC\units)^n \qquad n = \size \pi_0 X
.\]
Thus $\Pic^0(X) = H^1(X; \OO)/ H^1(X; \ZZ)$, and Hodge theory shows $\rank H^1(X; \ZZ) = 2\dim_\CC H^1(X; \OO)$ and the image of $H^1(X;\ZZ)\to H^1(X; \OO)$ is discrete.
This yields $\ZZ^{2r} \injects \CC^r$ with image $\Lambda$ a lattice and $\Pic^0 X \cong \CC^r/\Lambda$.
In particular, for $C$ a smooth genus $g$ curve, $\Pic^0 X \cong \CC^g/ \Lambda$.
Note that $H^1(C; \ZZ)$ carries the intersection pairing, which induces a symplectic form and thus a polarization.
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