# 2024-09-19-12-49-21

:::{.remark}
Let $C$ be a curve of genus $g$ and $E$ a rank 2 vector bundle, and let $S = \PP_C(E)\to C$.
There is an exact sequence $0\to N\to p_* E\to \OO_S(1)\to 0$ over $S$ where the ranks are 1,2,1 respectively.
Let $h = c_1(\OO_S(1))\in \NS(S)$; on any fiber $f$ this restricts to $\OO_{\PP^1}(1)$.
Note that $h$ need not be effective.
:::

:::{.theorem}
Three equations:

- $\Pic(S) = p^* \Pic(C) \oplus \ZZ h$ 
- $\NS(S) = p^* \NS(C) \oplus \ZZ h \cong \ZZ \oplus \ZZ h$.
- $H^2(S; \ZZ) = p^* H^2(C; \ZZ) \oplus \ZZ h$.

Moreover:

- $f^2 = 0$
- $hf = 1$
- $h^2 = \deg E$

Finally,
\[
K_S = -2h + (\deg E + 2g-2)f
.\]

:::

:::{.proof}
Let $D\in \Pic(S)$ and consider $d\da D.f$.
Then $D' \da D-dh$ yields $D'.f = 0$, and we claim that $D' = p^* L$ for some $L\in \Pic(C)$.
It suffices to show that $h^0(D' + nf) > 0$ for some $n$.
Note that $h^2(D' + nf) = h^0(K-D'-nf) = 0$, since $(K-D'-nf)f = -2$ which follows from $(K+f)\mid_f = K_f = K_{\PP^1} = -2$.
Riemann-Roch yields
\[
\chi(D' + nf) = {1\over 2}(D' + nf)(D' + nf - K) + \chi(\OO_S) = n + c
\]
for $c$ some constant.
Let $A \sim D' + nf$, then $A.f=0$.
So $A$ is effective and intersects every fiber by zero, so every reducible component is contained in a fiber.
So $A$ is a sum of fibers with some multiplicities, which is thus the pullback of some divisor on $C$.

Note that we have a SES $\Pic^0(S) \to \Pic(S) \to H^2(S; \ZZ) \to H^2(\OO_S)$.
Note that $\Pic^0(S) = H^1(\OO_S)/H^1(S; \ZZ)$ and $H^2(S; \ZZ)=\NS(S)$ in this case.
This yields the conclusion.
:::

:::{.proof}
For $F$ a sheaf on $X$, $c_0(F) = \rank(F)[X]$, $c_1(F)$ is a divisor, and $c_2(F)$ is codimension 2.
Recall $c(N)\cdot c(\OO_S(1)) = c(p^* E)$.
This yields $(1+n)(1+h) = 1 + c_1(p^* E) + c_2(p^* E)$.
Note that $c_1(p^* E) = p^* c_1(E) = (\deg E)f$ and $c_2(p^*E) = p^* c_2(E) = p^* 0 = 0$, since $c_2$ vanishes for a curve.
Thus
\[
(1+N)(1+h) = 1 + (\deg(E))f
,\]
and so 

- $n+h = (\deg(E))f$,
- $nh = 0$.

Thus $\qty{ (\deg(E))f - h } h = 0$.
:::

:::{.remark}
Note that twisting $E$ by a line bundle changes $h$ but does not change $\PP_C(E)$.
:::

:::{.proof}
Let $s$ be a section, then $s = h + af$ since $s.h = 1$ and $s.f=a$ is unknown.
Note that $\deg K_s = 2g-2$, and we can write $K_S = -2h + bf$ for some $b$.
By adjunction, $K_s = (K_S + s)\mid_s$, and taking degrees yields
\[
2g-2 = (-2h + bf + h + af)(h+af) = (-h + (a+b)f)(h + af) = -\deg(E) + b
.\]
Note that $b$ does not depend on the section $s$, hence the cancellation of $a$.
Solving yield $b = \deg(E) + 2g-2$.
:::

:::{.remark}
Recall that $N$ was defined by varying lines in $\AA^2$ along a fiber.
Twisting by a line bundle yields $0\to N\tensor p^* L \to p^*(E\tensor L) \to \OO_S(1)\tensor p^* L \to 0$.
This yields $h' \da h + p^* L$ and $(h')^2 = h^2 + 2\deg(L)$.
:::

:::{.remark}
Basic numerical invariants:

- $q(S) = h^1(\OO_S) = h^0(\Omega_S)$,
- $p_g(S) = h^2(\OO_S) = h^0(\Omega_S^2)$,
- $P_m(S) = h^0(\OO_S(mK)) = h^0( (\Omega_S^2)^{\tensor m} )$.

Note that $P_1 = p_g$ and $P_0 = 1$.
:::

:::{.theorem}
$q, p_g, P_m$ are birational invariants for smooth projective varieties.
:::

:::{.proof}
Step 1:
Global sections of differential forms tend to be birational invariants.
Let $X$ be such a variety and $\phi: X\rational Y$ a rational map.
Then $\phi$ is undefined in codimension $\geq 2$ -- there exists $Z \subseteq X$ with $\codim_X(Z) \geq 2$ and $U \da X\setminus Z$ such that $\ro{\varphi}{U} = f: U\to Y$ is a regular map.
Let $D \subseteq X$ be a divisor cut out by $t=0$ in a local coordinate.
Embed $Y \subseteq \PP^N$, and clear denominators and rescale by powers of $t$.
We can thus pull back differential forms to $U$.

Step 2: Hartog's theorem.
A differential form on $U$ with $\codim(X\setminus U) \geq 2$ can be extended to all of $X$ when $X$ is smooth.
:::

:::{.corollary}
If $\phi: X\to Y$ is a dominant rational map then

- $q(X) \geq q(Y)$
- $p_g(X) \geq p_g(Y)$
- $P_m(X) \geq P_m(Y)$

:::

:::{.remark}
Rational maps thus yield inequalities in both directions.
:::

:::{.remark}
We now consider ruled surfaces $S \birational C\times \PP^1$.
We obtain

- $q(S) = g$,
- $p_g(S) = 0$,
- $P_m(S) = 0$.

Let $p_i: C\times \PP^1\to C, \PP^1$ be the coordinate projections.
Then $\T_{C\times \PP^1} = p^* \T_{C} \oplus p^* T_{\PP^1}$.
Take duals to get $\Omega_S = p^* \Omega_C \oplus \Omega_{\PP^1}$ and check $h^0(\Omega_S) = h^0(\Omega_C) + h^0(\Omega_{\PP^1}) = g+0 = g$.
Note that $p_g(S) = h^0(K_S)$.
Also note that $K^2$ is not a birational invariant, since blowing up decreases this by 1.
Here we have $K_S^2 = 8-8g$ by computing $(-2h + (\deg E + 2g-2))^2 = 4\deg E - 4(\deg E + 2g-2)$.
For $g=0$, we get $\PP^1\times \PP^1$, check $K_S^2 = (2h + 2f)^2 = 8$.
For $g=1$, check $K_S^2 = 0$.
:::

:::{.remark}
Note that a surface $S$ is birationally ruled iff $P_m(S) = 0$ for all $m$.
This is a classical theorem.
It is rational iff $q(S) = P_2(S) = 0$.
Recall that Kodaira dimension is defined by $P_m \sim m^{\kappa(S)}$.
For $\kappa(S) = 0$, there is a full classification in terms of the above invariants.
There is a structure theorem for $\kappa(S) = 1$, elliptic surfaces.
For $\kappa(S) = 2$, there are finitely many types where $q=p_g=0$, but there is not a full classification.

:::