# 2024-10-29-12-44-54

:::{.remark}
Surfaces with $p_g=0$ and $q\geq 1$, i.e. ruled irregular surfaces (plus a little more).
The next main goal: minimal non-ruled surfaces.
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:::{.lemma}
In parts:

a. (A little more) $p_g=0, q\geq 1\implies K^2\leq 0$, or $K^2=0$ and $q=1,b_2=2$.
b. (Ruled irregular) If $S$ is minimal and $K^2 < 0$ then $p_g=0$ and $q\geq 1$.

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:::{.proposition}
If $S$ is minimal and $K^2 < 0$ then $S$ is ruled.
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:::{.proof}
Of lemma part (a): by Noether's formula, $K^2 = 10-8q-b_2$.
Use the Albanese map to show $b_2\geq 2$.

Of lemma part (b): if $p_g\da h^0(K_S) \neq 0$, then $K$ is effective.
Write $K = \sum n_i C_i$, then $K^2 < 0 \implies \exists C_j$ such that $KC_j < 0$ and $C_j^2 < 0$.
By the genus formula, $p_g(C_j) = {KC_j + C_j^2\over 2} + 1$, forcing $C_j$ to be a negative curve, contradicting minimality of $S$.
This argument shows that in fact $P_m = 0$ for all $m$.
Suppose that $q\neq 0$, we will then argue $S$ is rational, and hence either $\PP^2$ or $\FF_n$ by minimality.
But $K_{\PP^2}^2=9$ and $K_{\FF_n}^2 = n > 0$, contradicting $K^2 < 0$.
By Castelnuovo's criterion, $S$ is rational iff $2K_S$ is effective, which it is not (by repeating the above argument).
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:::{.remark}
Examples where $K$ is not effective but $2K$ is effective: Enriques surfaces.
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:::{.remark}
Proving the proposition, the main tool: the Albanese map $p: S\to \Alb(S)$ which is an abelian variety of dimension $q$.
Note that $p(S)$ generates $\Alb(S)$, and $p$ has connected fibers.
If $q\geq 1$ then $\dim p(S) = 1,2$.
If $p_g = 0$, then $\dim p(S) = 1$.
Any abelian variety is of the form $A = \CC^g/\ZZ^{2g}$, and the differential forms are those with constant coefficients, e.g. $dz_1,\cdots,dz_n$, $dz_1\wedge dz_2 \cdots$, etc.
These are translation invariant and thus descend to the quotient.
One then produces a differential form in $\Alb(S)$ which pulls back to $S$, yielding a nontrivial section of $\Omega^2(S)$, forcing $\dim p(S) = 1$.
In this case, $p: S\to B$ where $B$ is a smooth curve of genus $q$.
Note that if a single fiber is $\PP^1$ then $S$ is ruled.
We're trying to rule out fibers with higher genus.
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:::{.proof}
Step 1: suppose $C \subseteq S$ is a curve with $KC < 0$ and $\abs{K+C}= \emptyset$.
Then $\ro{p}{C}: C\to B$ is etale (unramified), and if $q\geq 2$ then $\ro{p}{C}: C\iso B$.
Suppose $C$ is contained in a fiber, so $C$ collapses to a point.
We first claim that this fiber $F$ must be irreducible.
Toward a contradiction, write $F = C + \sum n_i C_i$.
Since the intersection form on fibers is negative definite, we have $C^2 < 0$.
But then $C$ is forced to be a $(-1)$-curve, contradicting minimality.
Suppose $F = nC$, then $C^2= 0$.
The genus formula yields $p_a(C) = {KC + C^2\over 2}+1 = 0 \implies KC = -2$, then $C \cong \PP^1$ and $S$ is ruled, a contradiction.
So $C$ is necessarily a horizontal curve.

Step 1.5: Riemann-Roch.
Write $0 = h^0(K+C)\geq \chi(K+C)$ since $h^2(K+C)=h^0(-C) = 0$ since $C$ is effective.
We have 
\[
\chi(K+C) = \chi(\OO_S) + {(K+C)C\over 2} = 1-q + (p_a(C) - 1) = p_a(C) - q \leq 0
,\]
and thus $p_a(C)\leq q$.
But this contradicts the fact that the genus decreases if $q\geq 2$, by Hurwitz: $2g(C)-2 = d(2g(B)-2)+\eps$.
The only way this can happen is if $B\cong C$, in which case $g(C) = g(B) = 1$ (they are elliptic curves) and $\eps = 0$ since the map is unramified.

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:::{.proof}
Step 2: we'll show there exists an irreducible curve $C \subseteq S$ such that $KC \leq -2$ and $\abs{K+C} = \emptyset$.
Recall that we proved the following: if $S$ is minimal and $K^2 < 0$, then there exists a $D'$ such that $KD' < -n$ is arbitrarily negative and $\abs{K+D}= \emptyset$.
Write $D = \sum_{1\leq i\leq r}n_i C_i \leq D' = \sum n_j C_j$ where all $KC_i \leq -1$, i.e. keep just those curves that intersect $K$ negatively.
We have $\abs{K+D} = \emptyset$.
Thus we are done unless 

a. there exists some $n_i \geq 2$ or 
b. $r \geq 2$, 

i.e. we want to show that $D$ is a single curve with coefficient 1.

In case (a), $\abs{K+2C_i} = \emptyset$ and thus
\begin{align*}
0 
&= h^0(K+2C_i) \geq \chi(K+2C_i) = \chi(\OO_S) + {(K+2C_i)2C_i \over 2} \\
&= (1-q + p_g) + (KC_i + 2C_i^2)\\
&= (1-q + p_g) + (2KC_i + 2C_i^2 - KC_i) \\
&= (1-q + p_g) + (2(2p_a(C_i)-2) - KC_i) \\
&= (1-q + p_g) + (2(2q-2) - KC_i) \\
&= (1-q) + (4(q-1) - KC_i) \\
&> 3(q-1) \\
&\geq 0
\end{align*}
noting that $h^2(K+2C_i)=0$.
This yields $0 > 0$, a contradiction.
:::

:::{.proof}
Step 3: get a contradiction.
Take $C$ as in step 2.

Case 1: $p:C\iso B$.
By Riemann-Roch,
\[
h^0(C) \geq \chi(\OO_S) + {C(C-K)\over 2}
= (1-q) + (q-1 - KC)
\geq 2
,\]
noting that ${C(C+K)\over 2} + 1 = q$.
Note that this implies that $C$, as a section of $p$, moves.
Thus intersecting $C$ with the general fiber $F$ yields a point that moves, thus defining a map $F\cong \PP^1$.

Case 2: $g(B) = g(C) = 1$ (elliptic curves) and $p$ is étale.
In this case, make a base change.
Write $S' = S \fiberprod{B} C$, which changes a multisection into an honest section.
Note that $q(S') \geq q(S) \geq 1$, and $p_g(S') = h^0(\omega_{S'}^2) = 0$.
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:::{.remark}
What remains to do: look at the "a little more" case.
Two cases:

- The genus of the general fiber satisfies $g(F)\geq 2$, then $p$ is a smooth map yielding a family of elliptic curves over an elliptic curve
- $g(F) = 1$, there may be multiple fibers.
    These appear in the classification of minimal models with $\kappa = 0$.
    Proofs here involve the topological Euler characteristic.
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