# Monday, April 12

:::{.remark}
Last time: the Lefschetz hyperplane theorem.
Intersecting a projective variety of dimension $d\geq 3$ with a hypersurface $S$, the map $\pi_1(\PP^3) \to \pi_1(S)$ is an isomorphism.
We saw that K3 surfaces were thus simply connected, and $h^1(S; \CC) = 0$, so we could compute the Hodge diamond.
:::

:::{.example title="?"}
What is the Hodge diamond for a cubic surface $S \subseteq \PP^3$, such as $\sum x_i^3 = 0$?
We first need to compute the canonical bundle $K$, for which we have a useful tool: the adjunction formula.
This say \( K_S = \qty{K_{\PP^3} \tensor \PP_{\PP^3}(S) } \ro{}{S} = \qty{ \OO(-4) \tensor \OO(3) }\ro{}{S} = \ro{ \OO(-1)}{S} \).

:::{.proposition title="If a holomorphic line bundle has a section, its inverse doesn't"}
Let \( \bundle{L} \to X \) be a holomorphic line bundle.
If $h^0( \bundle{L} \inv ) > 0$, then either $\bundle{L} = \OO$ or $h^0(\bundle{L}) = 0$.
:::

:::{.slogan}
If a line bundle has a section, its inverse does not.
:::

:::{.proof title="?"}
Suppose that both $\bundle{L}, \bundle{L}\inv$ have a section, so $h^0(\bundle{L}), h^0( \bundle{L} ) > 0$.
Let $s, t$ be sections of each, then $st\in H^0( \bundle{L} \tensor \bundle{L}\inv ) = H^0(\OO) = \CC$.
So taking zero loci yields $\div(s) + \div(t) = 0$
Writing these as $\div(s) \da \sum n_D D, \div(t) \da \sum n_C C$, we have $n_D, n_C \geq 0$, which implies that $\div(s) = \div(t) = 0$.
So $s, t$ are nowhere vanishing, making \( \OO \mapsvia{\cdot s} \bundle{L} \) is an isomorphism.
:::

:::{.corollary title="H0 of cubic surfaces"}
For $S$ a cubic surface, $H^0(S; K_S) = 0$.
:::

:::{.proof title="?"}
This follows because $K_S = \OO_S(-1)$, so $K_S\inv = \OO_S(1)$ which has a nontrivial section: namely $\OO_{\CP^1}(1)$ which has sections vanishing along hyperplanes.

\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}{2021-04-12_14-06.pdf_tex} };
\end{tikzpicture}

Letting $H$ be a hyperplane containing $S$, there exists an $f\in H^0(\PP^3; \OO_{\CP^3}(1))$.
Since $\div(f) = H$, the restriction $\ro{f}{S}$ is a section of $\OO_S(-1) = K_S\inv$ which is not identically zero and vanishes along $H \intersect S$.

:::

We now know $h^0(S; K_S) = 0$, and this equals $h^0(S, \Omega^2) = h^{2, 0}(S)$, so we have the following Hodge diamond:

\begin{tikzcd}
	&& 1 \\
	& 0 && 0 \\
	0 && {h^{1, 1}} && 0 \\
	& 0 && 0 \\
	&& 1
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiMCJdLFszLDEsIjAiXSxbMCwyLCIwIl0sWzIsMiwiaF57MSwgMX0iXSxbNCwyLCIwIl0sWzEsMywiMCJdLFszLDMsIjAiXSxbMiw0LCIxIl1d)

We have $h^{0, 1} + h^{1, 0} = h^1 = 0$ since $S$ is simply connected.
We can now apply Noether's formula as before: $\chi(\OO_S) = {1\over 12} (K_S^2 + \chi_\Top(S))$.
We have $K_S = \OO_S(-1)$, so $K_S^2 = c_1( \OO(-1))^2$, and $\chi(\OO_S) = 1-0+1 = 1$.
We now want to compute $\int_S \qty{- c_1(\OO_S(1)) }^2$.
We know $c_1(\bundle{L}) = [ \div s]$ where $s\in H^0( \bundle{L} )$ is a section of a line bundle.
This equals $[H \intersect S]$.
On the other hand, $\int_S c_1 \qty{ \OO_S(1)}^2$ is the self-intersection number of $H \intersect S$.

\

Take $H_1 \da \ts{ x_0 = 0 }$ and $H_2 \da \ts{ x_1 = 0 }$.
Points in this intersection are of the form $[0:0:1:\zeta_6^a]$ where $a=1,3,5$ since this is in the triple intersection \( H_1 \intersect H_2 \intersect S \).
So there are exactly 3 points here, and in fact $\deg S =3$.
This is the same as integrating $\int_{\PP^3} c_1(S) c_1( \OO(1)) c_1( \OO(2))$, which contains 3 elements in $H^2$ and lands in $H^6$, so this yields a number.

\

We thus have $K_S = \OO_S(-1) \da \OO_{\CP^3}(-1) \ro{}{S}$.
Thus $\chi_\Top(S) = 9$ and $h^{1, 1} = 7$.


:::

:::{.example title="Hypersurfaces"}
Note that a degree 5 surface (a quintic) such as $x_0^5 + x_3^5 = 0$ would be harder, since $h^{2, 0} \neq 0$.
We would get $K_S = \OO(-4) \tensor \OO(5) \ro{}{S} = \OO_S(1)$, and there are nontrivial sections so $h^0(K_S) = \spanof{x_0, x_1, x_2, x_3}$.
This follows because there is a map given by restriction which turns out to be an isomorphism 
\[
0 \to H^0(\PP^3; \OO(1)) & \mapsvia{\res_S} H^0(S; \OO(1)) \to 0 \\
f &\mapsto \ro{f}{S}
.\]

Injectivity isn't difficult, surjectivity is harder.
We have a SES
\[
0 \to \OO_{\CP^3}(-S) \to \OO_{\CP^3} \to \OO_S \to 0
.\]
Tensor all of these with $\OO(1)$ to obtain
\[
0 \to \OO_\CP^3(-4) \to \OO_{\CP^3}(1) \to \OO_S(1) \to 0
.\]
Taking the associated LES yields

\begin{tikzcd}
	{H^1(\OO_{\CP^3}(-4)) =_? 0} \\
	\\
	{H^0(\OO_{\CP^3}(-4)) = 0} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0(\OO_{\CP^3}(-1)) = \CC \gens{x_0, x_1, x_2, x_3}} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0(\OO_{S}(1))} \\
	0
	\arrow[from=4-1, to=3-1]
	\arrow[from=3-1, to=3-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=3-5]
	\arrow[from=3-5, to=1-1, out=0, in=180]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwyLCJIXjAoXFxPT197XFxDUF4zfSgtNCkpID0gMCJdLFsyLDIsIkheMChcXE9PX3tcXENQXjN9KC0xKSkgPSBcXENDIFxcZ2Vuc3t4XzAsIHhfMSwgeF8yLCB4XzN9IixbMjQwLDYwLDYwLDFdXSxbNCwyLCJIXjAoXFxPT197U30oMSkpIixbMjQwLDYwLDYwLDFdXSxbMCwwLCJIXjEoXFxPT197XFxDUF4zfSgtNCkpID1fPyAwIl0sWzAsMywiMCJdLFs0LDBdLFswLDFdLFsxLDIsIiIsMCx7ImNvbG91ciI6WzI0MCw2MCw2MF19XSxbMiwzXV0=)

This gives us a way to relate things back to the cohomology of $\CP^3$.
Showing that the indicated term is zero involves computing Čech cohomology.

It turns out that $h^0(K_S) = 4$ here, and it turns out that the Hodge diamond is the following:

\begin{tikzcd}
	&& 1 \\
	& 0 && 0 \\
	4 && {h^{1, 1} = 45} && 4 \\
	& 0 && 0 \\
	&& 1
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiMCJdLFszLDEsIjAiXSxbMCwyLCI0Il0sWzQsMiwiNCJdLFsxLDMsIjAiXSxbMywzLCIwIl0sWzIsNCwiMSJdLFsyLDIsImheezEsIDF9ID0gNDUiXV0=)

Here $K_S^2 = c_1(\OO_S(1))^2 = 5$ and $\chi_\Top = 55$.

:::

:::{.example title="Products"}
Consider now a product of curves $C \cross D$ of genera $g, h$ respectively.
Computing the Hodge diamond is easy here due to the Kunneth formula:
\[
H^k(S; \CC) = \bigoplus_{i+j = k} H^i(C; \CC) \tensor H^j(D; \CC)
.\]

What is the actual map? 
Take cohomology classes \( [\alpha], [\beta] \), closed $i$ and $j$ forms respectively.
The surface has two maps:

% https://q.uiver.app/?q=WzAsMyxbMCwwLCJTIl0sWzIsMCwiQyJdLFswLDIsIkQiXSxbMCwxLCJcXHBpX0MiXSxbMCwyLCJcXHBpX0QiLDJdXQ==
\begin{tikzcd}
	S && C \\
	\\
	D
	\arrow["{\pi_C}", from=1-1, to=1-3]
	\arrow["{\pi_D}"', from=1-1, to=3-1]
\end{tikzcd}

Here we send \( [\alpha] \tensor [ \beta] \mapsto [ \pi_C^* \alpha\wedge \pi_D^* \beta] \) where we take pullbacks.
Note that $\pi_D, \pi_C$ are holomorphic maps, and pullbacks of $(p, q)$ forms are still $(p, q)$ forms.
Thus the Kunneth formula gives a decomposition 
\[
H^{p, q}(S; \CC) 
= \sum_{\substack{ i_1 + j_1 = p \\ i_2 + j_2 = q }} H^{i_1, j_1}(C) \oplus H^{i_2, j_2}(D)
.\]
So we can "tensor" the Hodge diamonds:

\begin{tikzcd}
	& 1 &&&& 1 \\
	g && g && h && h \\
	& 1 &&&& 1 \\
	\\
	&&& 1 \\
	&& {g+h} && {g+h} \\
	& gh && {2 + 2gh} && gh \\
	&& {g+h} && {g+h} \\
	&&& 1
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTcsWzEsMCwiMSJdLFswLDEsImciXSxbMiwxLCJnIl0sWzEsMiwiMSJdLFs1LDAsIjEiXSxbNCwxLCJoIl0sWzYsMSwiaCJdLFs1LDIsIjEiXSxbMyw0LCIxIl0sWzIsNSwiZytoIl0sWzQsNSwiZytoIl0sWzIsNywiZytoIl0sWzQsNywiZytoIl0sWzEsNiwiZ2giXSxbNSw2LCJnaCJdLFszLDgsIjEiXSxbMyw2LCIyICsgMmdoIl1d)

:::

:::{.remark}
Check out complete intersections.
:::