# 2024-11-12-12-46-21

:::{.remark}
Goal: classify surfaces with $\kappa(S) = 0$.
Let $S$ be a minimal non-ruled surface, so $P_m = 0,1$ for all $m$ and $P_m = 1$ for at least one $m$.
We know $K_S$ is nef and thus $K_S^2 = 0$.
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:::{.theorem}
There are four possibilities for $S$:

1. $p_g = 0, q = 0$ and $\chi(\OO_S) = 1-q+p_g = 1$ ($K\neq 0, 2K=0$: Enriques),
2. $p_g = 0, q = 1$ and $\chi(\OO_S) = 0$ (Bielliptic),
3. $p_g = 1, q = 0$ and $\chi(\OO_S) = 2$ ($K=0$: K3 surfaces),
4. $p_g = 1, q = 1$ and $\chi(\OO_S) = 1$ (not possible),
5. $p_g = 1, q = 2$ and $\chi(\OO_S) = 0$ ($K=0$: abelian surfaces).

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:::{.corollary}
$4K_S = 0$ or $6K_S = 0$, so $12 K_S = 0$ in any case.
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:::{.remark}
Note that if $X$ is a projective variety of dimension $n$ and $D$ is a nef divisor, then $D^n \geq 0$.
If $A$ is an ample divisor, then Kleiman's criterion $D + \eps A$ is ample $\forall \eps > 0$.
Ampleness implies $(D + \eps A)^n > 0$ since this is the degree of $X$.
Then take $\eps\to 0$.
Note that torsion in $\Pic(X)$ is numerically zero.

Kleiman's criterion: consider $N_1(X) = \ts{\sum n_i C_i}/\sim$, sums of curves modulo numerical equivalence.
Similarly $N^1(X) = \ts{\sum n_j D_j}/\sim$, sums of divisors modulo numerical equivalence.
There is an intersection product $N_1(X)\times N^1(X) \to \ZZ$, and numerical equivalence means they intersect the same way.
This is a perfect pairing, and $\rank_\RR N_1(X)\tensor \RR = \rank \NS(X)$, so $N^1(X) = \NS(X)/\tors$.
There is a cone $\NE_1(X) \subseteq N_1(X)\tensor \RR$, the Mori-Kleiman cone (or Mori cone).
Any divisor $E$ gives a linear function $f_E$ on $\NE_1(X)$.
The Kleiman criterion states that a divisor $E$ is ample if $f_E > 0$ on $\overline{\NE(X)}\smz$.
Thus nef divisors are limits of ample divisors.

Mori proved that if one partitions $\NE_1(X)$ by the hyperplane $K_X$, the negative part is polyhedral, generated by extremal rays. 
For Fano varieties, the entire cone is polyhedral, while for abelian surfaces it is the round cone.
More cones of K3 surfaces are partly polyhedral, partly round.
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:::{.lemma}
$\chi(\OO_S) \geq 0$.
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:::{.remark}
Note $\chi(\OO_S) = 1-q +p_g$, and by Noether's formula, $\chi(\OO_S) = {c_1^2 + c_2\over 12} = {K^2 + \chi(S)\over 12}$.
If $K^2=0$, then $12 \chi(\OO_S) = \chi(S) = \sum b_i = 1-2q+b_2-2q+1 = 2-4q+b_2$. 
Note $b_2 = h^{0, 2} + h^{1,1}+h^{2, 0} = 2p_g + h^{1,1}$, thus $b_2 \geq 2p_g$ and
\[
12\chi(\OO_S) = 2-4q+b_2 \geq 2-4q + 2p_g
.\]
There is a problem if $q$ is large.
Note that $-4\chi(\OO_S) = -4+4q-4p_g$, and adding this to the above equation yields 
\[
8\chi(\OO_S) = -2-4p_g + b_2 \geq -2 - 2p_g
.\]
Note that $-4p_g = 0, -4$, but if $-4p_g=-4$ then $\chi(\OO_S) < 0$ would contradict this inequality.
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:::{.remark}
Thus the possibilities for $\chi(\OO_S) = 1-q+p_g$ are 

- $p_g = 0 \implies \chi(\OO_S) = 1-q\geq 0$,
- $p_g = 1 \implies \chi(\OO_S) = 2-q\geq 0$.

These yield the four possibilities in the theorem.
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:::{.lemma}
Suppose $\kappa(S) = 1$.
Let $m, n > 0$ and $d = \gcd(m, n)$.
If $P_m = P_n = 1$ then $P_d = 1$.
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:::{.remark}
Consider $\mcs \da \ts{m \st \abs{mK_S}\neq \emptyset} \subseteq \ZZ$; this forms a semigroup.
We will show that $\mcs$ is in fact a group.
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:::{.proof}
Case 1: $mK\sim 0$, which happens iff $nK\sim 0$.
This happens if $K.H = 0$ for $H$ a hyperplane.
Then for all $a,b\in \ZZ$, we have $(am + bn)K = 0$, so $dK = 0$.

Case 2: $mK = C_m \neq 0$ is a nonzero effective curve iff $nK = C_n\neq 0$ is as well.
This happens if $K.H > 0$.
Write $m = m' d, n=n' d$ where $\gcd(m', n') = 1$.
Write $mK = \sum n_i C_i$ and $nK = \sum b_j D_j$.
Note that $dm'n' K \sim \sum n_i n' C_i \sim \sum b_j m' D_j$, but in fact these are equal divisors (not just linearly equivalent) since $P_m = 0, 1$ forces uniqueness of effective divisors in $\abs{mK}$ when they exist.
Note that $\sum {a_i \over n'}C_i$ and $\sum {b_j \over n'}D_j$ are integral divisors, and we claim these divisors are in $\abs{dK}$.
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:::{.remark}
We now discuss the proof of the theorem.
Case 1: $p_g=0, q=0, \chi(\OO_S) = 1$.
By Riemann-Roch, $\chi(D) = \chi(\OO_S) + {D(D-K)\over 2} = h^0(D) - h^1(D) + h^0(K-D)$, and so $h^0(D) + h^0(K-D)\geq \chi(D)$.
Applying this to $D = mK$ yields $h^0(mK) + h^0((1-m)K) \geq \chi(\OO_S)$.
We'll show a positive multiple and a negative multiple of $K$ is effective, which can only happen if the multiple if zero.
If $\chi(\OO_S) = 1$, we have $h^0(3K) + h^0(-2K)\geq 1$.
If $h^0(-2K) = 1$, then $-2K \sim 0$.
Consider the case where $h^0(3K) \geq 1$.
This implies $p_3 = 1$.
In this case we have $h^0(2K)\geq 1$ since $q=0$ and $S$ is not rational.
This implies $p_2 = 1$, which implies $p_1 = p_g = 1$, contradicting $p_g=0$.
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:::{.remark}
Case 2: we have already considered these in a previous chapter.
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:::{.remark}
Case 3: $p_g=1, q=0, \chi(\OO_S) = 2$, and we want to conclude $K\sim 0$.
By Riemann-Roch, $h^0(2K) + h^0(-K) \geq 2$.
We know $h^0(2K) = 1$, thus $h^0(-K)\geq 1$ and $-K$ is effective.
Since $K$ is effective, we have $K\sim 0$.
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:::{.remark}
Case 4: $p_g = 1, q=1, \chi(\OO_S) = 1$, we want to show this is impossible.
Recall the SES $\Pic^0(S)\injects \Pic(S)\surjects \NS(S)$, and note $\Pic^0(S) \cong \CC^q/\ZZ^{2q} = \CC/\ZZ^{2}$ in this case.
This is a complex torus, which has a 2-torsion point $\eps \in \Pic^0(S)$.
By Riemann-Roch, $h^0(\eps) + h^0(K-\eps) \geq 1$.
Since $\eps$ can not be effective since $2\eps = 0$ (intersect with a hyperplane), we have $h^0(\eps) = 0$.
So $h^0(K-\eps)\neq 0$.
We thus have $\exists D\in \abs{K}$ and $\exists E\in \abs{K-\eps}$.
Note that $2D=2E \in \abs{2K}$ since there is a unique divisor in this linear system.
Either this is zero, or effective and nonzero.
We want to conclude $D=E$, and thus $\eps = D-E = 0$, a contradiction.

> Called case 3.5 in class.

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:::{.remark}
Case 5: the hardest case.
Use the Albanese map $\alpha: S\to \Alb(S)$, an abelian variety of dimension $q=2$.
Claim: $\alpha$ is surjective and étale.
Fact: if $X\to Y$ is an etale map between projective varieties and $Y$ is an abelian variety, then $X$ is an abelian variety.
I.e. any etale cover of an abelian variety is again an abelian variety.
We have to prove the image of $\alpha$ is not just a curve, and that $\alpha$ is unramified.

> Called case 4 in class.

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