# 2024-10-24-12-40-57

## Chapter 5 Recap

:::{.remark}
For minimal surfaces, there is a division:

- Non-ruled surfaces: $\exists !$ minimal model in every birational class.
    We showed there is a sequence of contractions $S\to\to\to S_{\min}$ where $S_{\min}$ has no $(-1)$-curves.
    Therefore our next step will be classifying such minimal surfaces.

- Ruled surfaces: minimal models are never unique.
    There are two cases:
    - $q\da h^1(\OO_S) = 0$: regular surfaces.
    Minimal surfaces are $\PP^2$ and $\FF_n$ for $n\neq 1$, thus rational.
    - $q\neq 0$: non-regular surfaces.
    Minimal surfaces are $\PP_C(\mce)$ with $\rank\mce = 2$ and $\mce$ a vector bundle, i.e. geometrically ruled surfaces (every fiber is $\PP^1$).
    Here $g(C) = q$ and $p: S\to C$ is the Albanese map.

Last time: suppose $S$ is a rational surface, we considered minimal rational curves $C$ on $S$, i.e. 
\[
\ts{C\cong \PP^1 \subseteq S \st C^2 \geq 0} \contains \ts{C\st C^2 = m \text{ is minimal }} \contains \ts{C \st \deg(C) \da C.H \text{ is minimal} }
.\]
Then every $D\in \abs{C}$ is smooth, irreducible, and $D\cong \PP^1$, i.e. curves in $\abs{C}$ do not split.
For example, conics in $\PP^2$ degenerate into unions of lines and thus do split.
:::

:::{.proof}
Suppose minimal just means $C.H$ is minimal, dropping the condition that $C^2 = m$ is minimal.
Write $D \in \abs{C}$ as $D = \sum n_i C_i$.
By the genus formula, $0 = g(C) = {C^2 + CK\over 2} + 1$, and so $C^2 + CK = 2g-2 = -2$ and thus $CK = -2 - C^2 < 0$ since $C^2 \geq 0$.
Moreover $DK = CK < 0$.
There exists an $i$ such that $C_i K < 0$, and if $C_i^2 \geq 0$ then $C$ is not minimal since $\deg C_i \leq \deg C$.
If $C_i^2 < 0$, combine this with $C_i K < 0$ and substitute into the genus formula to conclude $C_i^2 = -1$ and $g(C) = 0$, this $C_i$ is a rational $(-1)$-curve, contradicting minimality of $S$.
:::

:::{.remark}
This argument generalizes to higher dimensions.
Note that $\abs{C}$ is a covering family of $S$ of rational curves with minimal degree.
See e.g. Mori's Annals paper proving Hartshorne's conjecture, which started the Mori program, or papers that use the bend-and-break technique.
E.g. if $X$ is a smooth projective Fano variety, so $-K_X$ is ample, then $X$ is covered by rational curves and such covering families exist.
:::

:::{.remark}
Recall that ruled non-regular surfaces ($q\neq 0$) are birational to $C\times \PP^1$ where $g(C) = q$.
First factor $S\to C\times \PP^1$ as $S\to\to\to S'\to\to\to C\times \PP^1$ as a composition of blowdowns and blowups.
Note that blowing up points in fibers of $S\to C\times \PP^1$ yields unions of $\PP^1$s.
The claim that $E_i$ must be contained in a fiber, and thus not horizontal.
This is because $g(E_i) = 0$ but $g(C) = q > 0$, and $\PP^1$ can not map to a curve of higher genus.
One can show this using Riemann-Hurwitz, since $\deg K_C = 2g(C) - 2$ and $\deg K_{\PP^1} = -2 < 0$.
Alternatively, $f^* \Omega_C \subset \Omega_E$ which, by taking dimensions, yields $g(C) \leq g(E)$.
Thus the exceptional curves are vertical.
Since there is a fibration $S'\to C$, there are thus fibrations to $C$ after each blowdown and $S\to C$ fibers over $C$.
:::

:::{.remark}
Note that the $E_i$ are not equivalent to fibers, because the intersection form of the exceptional curves is negative semi-definite.
Next chapter: ruled, non-regular surfaces, then we move on to non-ruled surfaces.
:::

## Chapter 6: $p_g = 0, q > 0$

:::{.remark}
Recall $p_g \da h^0(K_S)$, so if $p_g = 0$ then there is not effective $D\sim K_S$.
For ruled surfaces, all $P_m \da h^0(mK_S) = 0$, so no multiple of $K_S$ is effective.
Use the useful observation: if $C^2 \geq 0$ with $mK.C < 0$ then $mK$ is not effective.
Take $C$ to be a fiber $F$, then $mK_S.F + F^2 = 2g(F) - 2$ and so $mK_S.F = -2 < 0$.
:::

:::{.remark}
Noether's formula: $\chi(\OO_S) = {K^2 + \chi(S)\over 12} = {c_1^2 + c_2\over 12}$, which was derived from Hirzebruch-Riemann-Roch,
\[
\chi(D) = \int_X e^D \Todd(\T_X)
,\]
where you multiply the corresponding power series and take the top degree terms.
We have $\chi(\OO_S) = 1-q+p_g$ in general, and note that $p_g=0$ in this case.
Note that $\chi(S) = \sum (-1)^i b_i = 2-2b_1 + b_2$ by Poincare duality, and $b_1 = h^{1, 0} + h^{0, 1} = 2q$ by the Hodge decomposition.
Substituting these facts into Noether's formula yields
\[
12(1-q) = K^2 -4q + b_2 \implies K^2 = 10-8q - b_2
.\]

:::

:::{.lemma}
In parts:

a. If $p_g = 0$ and $q > 0$ then $K^2 \leq 0$ and $K^2 < 0$ only if $q=1, b_2=2$.
b. If $S$ is minimal with $K^2 < 0$ then $S$ is ruled.

:::

:::{.proof}
We know $q\geq 1$ and $b_2\geq 1$ since there is always a hyperplane section.
If $q\geq 2$, we're done, since $K^2 < 0$ by the above formula.
If $q = 1$ then $b_2 \geq 2$: this follows from producing two linearly independent divisors in $H^2(S; \ZZ)$.
Consider the Albanese map $p: S\to \Alb(S)$.
If $S$ is a curve $C$, recall that $\Alb(C) = \Jac(C) = \Pic^0(C)$.
This map contracts rational curves.
Two cases:

- $\dim p(X) = 1$
- $\dim p(X) = 2$

It's nonzero since $p(X)$ generates $\Alb(X)$ and thus can not be a point if $q = \dim p(X) > 0$.
In the first case, the generic fiber $F$ is a curve with $F^2 = 0$ which is not a multiple of a hyperplane section $H$ since $H^2 > 0$; then $\gens{H, F} \subseteq H^2(S; \ZZ)$.
In the second case, one can pull back differential forms on $p(X) \cong \CC^g/\ZZ^{2g}$, contradicting $p_g = 0$.
:::