# Monday, April 05


:::{.remark}
Last time: line bundles are of the form $\OO(D)$ for $D$ a divisor, and the extremely important SES
\[
0 \to \OO_S(-D) \to \OO_S \to \OO_D \to 0
.\]
We now want to discuss an alternative characterization of the intersection form on an algebraic surface.
The next result comes from Beauville's "Complex Algebraic Surfaces":
:::

:::{.proposition title="Formula for computing intersection numbers between complex curves"}
Let $S \in \Mfd^2(\CC)^\cpt$, then the intersection number between complex curves $C, D$ can be computed in the following ways:
\[
C\cdot D = \deg \OO_S(C) \ro{}{D} = \sum_{p\in C \intersect D } \len_p(C \intersect D )
,\] 
where we'll define $\len_p$ soon.
:::

:::{.remark}
This will count intersection points after a small perturbation.
Note that not every two curves will intersect transversely: consider $\PP_2$ with a line $C$ and a tangent conic $D$:

\begin{tikzpicture}
\fontsize{41pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_13-57.pdf_tex} };
\end{tikzpicture}

:::

:::{.proof title="?"}
We have the first equality because
\[
C\cdot D = \int_S [C] \capprod [D] = \int_C i^* [D]
,\]
where $i: C\injects S$ is the inclusion.
This equality holds because if \( \alpha\in \Omega^2 \) is a 2-form, 
\[
\int_S [C] \cdot \alpha = \int \ro{ \alpha}{C}
.\]
Using the pullback commutes with taking Chern classes, we can write the 
\[
\int_C i^* [D] = \int_C i^* (c_1 (\OO(D))) = \int_C c_1( i^* \OO(D) ) = \int_C \OO(D)\ro{}{C} = \deg \OO(D) \ro{}{C}
.\]
Note that this formula was symmetric, so we could have done this the other way to obtain $\deg \OO_S(C)\ro{}{D} = \deg \OO_S(D) \ro{}{C}$.

For the second equality, consider the following 4-term exact sequence: 
\[
0 \to \OO_S(-C -D) 
\underset{p_1}{ \injectsvia{\tv{ s_D, s_C}} } \OO_S(-C) \oplus \OO_S(-D) 
\underset{p_2}{ \mapsvia{\tv{s_D, -s_C}^t} } \OO_S 
\underset{p_3}{\to} \OO_{C \intersect D} 
\to 0
.\]
For the first map, we have
\[
\ts{ \text{Functions vanishing on } C+D} \injects
\ts{ \text{Functions vanishing on } C}
\oplus
\ts{ \text{Functions vanishing on } D}
.\]

Locally we can write $C = V(f)$ and $D = V(g)$ for some holomorphic functions $f,g\in \Hol(U, \CC)$.
We have the following picture:

\begin{tikzpicture}
\fontsize{43pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_14-06.pdf_tex} };
\end{tikzpicture}

We have $s_C \in H^0(S; \OO_S(C))$ and $s_D \in H^0(S; \OO_S(D))$ as global sections where $V(s_c) = C, V(s_D) = D$.
In a local trivialization, we can assume $\ro{s_C}{U} = f$ and $\ro{s_D}{U} = g$.
So the first map is $(s_D, s_C)$.
The next map is $\tv{s_C, -s_D}^t$ as a column vector, i.e. given a section we map it in the following way:
\[ 
(\varphi_1, \varphi_2) \in H^0(U, \OO_S(-C) \oplus \OO_S(-D) ) \mapsto \phi_1 \cdot s_D - \phi_2 \cdot s_C
.\]
Why is this exact?
Considering the composition, we have
\[
\phi \mapsvia{p_1}
(\phi s_D, \phi s_C)
\mapsvia{p_2}
(\phi s_D) s_C - (\phi s_C) s_D = 0
.\]

So we get $\im p_1 \subseteq \ker p_2$.
Why do we have the reverse containment for exactness?
Looking locally, given a pair $\phi_1, \phi_2 \in \Hol(U; \CC)$ such that $\phi_1 \varphi- \phi_2 g = 0$ and locally $(\phi_1, \phi_2) \in \ker p_2$, we want to show that $\phi_1 = g \phi, \phi_2 = f\phi$ for some $f,g \in \Hol(U; \CC)$.
Equivalently, we want to show that 
\[
\phi_1 f = \phi_2 g \implies g\divides \phi_1
.\]
If this is true, then we can set $\phi \da {\phi_1 \over g}$, since this would yield $g\phi = \phi_1$ and $f\phi = {f\phi_1 \over g} = \phi_2$.
Note that we can divide here because the ring $\Hol(U;\CC)$ is a domain (i.e. it has no zero divisors) on small sets.

:::{.question}
Is $\Hol(U, \CC)$ a PID in general?
:::

:::{.answer}
No! Take $U \subseteq \CC^2$ a ball around $z=0$, then $\gens{x, y}$ is not principal.
:::

However, this will form a UFD, which is weaker but still enough here.
This is not obvious, but can be proved using the Weierstrass preparation theorem.
This should be believable since $R$ a UFD implies $R[x]$ is a UFD, and $\CC[x, y] \subsetneq \Hol(U; \CC) \subsetneq \CC[[x, y]]$, and the latter is a UFD.
So we do get exactness at this position.

For exactness at the next position $\OO_S(-C) \oplus \OO_S(-D) \to \OO_S$, locally we have \( (\varphi_1, \varphi_2 ) \mapsto \varphi_1 f- \varphi_2 g \) where $V(f) = C \intersect U$ and $V(g) = D \intersect U$.
We can write $\phi_1 f- \phi_2 g = \gens{f, g}$ locally, so 
the cokernel sheaf of $p_2$ is given by 
\[
\coker p_2(U) \da {\OO_S(U) \over \im p_2} = { \OO_S(U) \over \gens{f, g}}
.\]
By definition, this is equal to $\OO_{V(f, g)} = \OO_{C \intersect D}$, and if $C \intersect D \intersect U = \emptyset$ then $\OO_{C \intersect D}(U) = 0$.
So let $p \in \OO_{C \intersect D}$ and let $U_p \ni p$ which contains no other points $q\in C \intersect D$, since the set of intersection points is isolated (and thus finite).
Note that compactness here prevents accumulation of intersection points.
In this case, $\OO_{C \intersect D}(U_p)$ will be a finite-dimensional vector space $\CC^d$, and we'll define $\len_p(C \intersect D) \da d$.
:::

:::{.example title="?"}
Let $U=\CC^2$ and take $f=y$ so $C\da V(f)$ is the x-axis, and set $g = y-x^2$ so $D\da V(g)$ is a parabola.
We're then considering 
\[ 
{ \Hol(\CC^2) \over y \Hol(\CC^2) + (y-x^2) \Hol(\CC^2)} = {\Hol(\CC^2) \over \gens{ y, x^2 }  }
.\]
Elements in the ideal can be expanded as power series of the form $a_{0,1}y + a_{2, 0}x^2 + a_{1, 1} xy + a_{2,2} y^2$, where there is no $a_{1, 0} \sim x^1 y_0$ coefficient, nor any $a_{0, 0} \sim x^0 y^0$ coefficient.
So this quotient is isomorphic to $\CC 1 \oplus \CC x$, which is 2-dimensional, so $\len_{(0, 0)} V(y) \intersect V(x) = 2$.
Geometrically we have the following, where this is picking up the multiplicity 2 intersection:

\begin{tikzpicture}
\fontsize{43pt}{1em}3
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_14-35.pdf_tex} };
\end{tikzpicture}

:::


:::{.remark}
What's the payoff of this algebraic work?
We can compute the Euler characteristic as 
\[
\chi( \OO_{C \intersect D}) = h^0(\OO_{C \intersect D}) = \sum_{p\in C \intersect D} \len_p(C \intersect D )
.\]
But by additivity of $\chi$ over exact sequences, we also have
\[
\chi(\OO_{C \intersect D}) 
&= \chi(\OO_S) - \chi(\OO_S(-C)) - \chi(\OO_S(-D)) + \chi( \OO_S( -C -D))\\
&\equalsbecause{HRR}
\int_S \qty{ \ch(\OO_S) - \ch(\OO_S(-C)) - \ch( \OO_S(-D)) + \ch( \OO_S(-C-D)) } \td(S) \\
&= c_1( \OO_S(-C)) \cdot c_1(\OO_S(-D)) \\
&= \qty{ -[C] } \cdot \qty{ -[D] } \\
&= C\cdot D
.\]

:::

:::{.remark}
Next time: adjunction formula that allows computing genus for surfaces.
:::