# 2024-10-10-12-51-51

:::{.remark}
Let $S$ be a smooth projective surface.
Then $S$ is rational iff 

- $q \da h^{1, 0} = 0$ (vanishing irregularity)
- $P_2 = h^0(\omega_S^{\tensor 2}) = 0$.

In the forward direction, this is clear since these are birational invariants and $q(\PP^2) = h^{1, 0}(\PP^2) = 0$ and $P_2(\PP^2) = 0$.
The other direction requires a difficult proof.
We'll assume $S$ is minimal, i.e. $S$ does not contain a smooth rational curve $C$ with $C^2 = -1$ (which can be contracted).
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:::{.proposition}
If $S$ is a minimal smooth projective surface with $q = P_2 = 0$, then there exists a curve $C\cong \PP^1$ on $S$ such that $C^2 \geq 0$.
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:::{.remark}
General philosophy: $C$ is given by $V(f)$ for some polynomial $f$, and since $C^2 \geq 0$ it can be deformed to $V(f')$.
One can form the pencil $\ts{uf + vf'}$.
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:::{.lemma}
Under the same conditions as the last proposition, there exists an irreducible algebraic curve $C$ with

- $K_S.C < 0$, and
- $\abs{K_S + C} = \emptyset$.

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:::{.remark}
Consider $S = \PP^2$, take $C$ to be any line in $\PP^2$.
Then $\OO(-3).[C] = -3 < 0$, and $L\da \OO(K_S + C) \cong \OO(-2)$ has no sections.
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:::{.remark}
Why the lemma implies the proposition: let $C$ be a curve satisfying the conditions of the lemma, then we claim $C\cong \PP^1$.
The proof uses adjunction and Riemann-Roch: $2g-2 = C(C+K_S)$ where $g$ is the arithmetic genus, and $\chi(\OO(D)) = {1 \over 2}D(D-K_S) + \chi(\OO)$.
Note that $\chi(\OO) = 1 - h^{1, 0} + h^{2, 0} = 1-0+0$ by assumption, so $\chi(\OO(D)) = {1\over 2}D(D-K_S) + 1$.
Take $D \da K_S + C$.
Recall that by Serre duality, $h^2(\OO(D)) = h^0(\OO(K_S-D))$.
Thus $h^2(\OO(K_S + C)) = h^0(\OO(-C)) = 0$ since $-C$ is not effective.
As a result, $\chi(K_S + C) = h^0(K_S + C)-h^1(K_S + C)$.
We know by assumption that $h^0(K_S + C)$, so $\chi(K_S +C)\leq 0$.
Substituting this into Riemann-Roch yields ${1\over 2}(K_S + C)C + 1 \leq 0$.
Solving for $g$ in the adjunction formula yields $g = {1\over 2}(K_S + C)C + 1$, so $g\leq 0$.
The arithmetic genus can not be negative for an irreducible curve, hence $g=0$ and $C\cong \PP^1$.

It remains to show that $C^2 \geq 0$.
By adjunction, $-2 = 2g-2 = C(K_S + C)$ since $g=0$.
We assumed $C.K_S < 0$, so $C^2 + C.K_S < C^2$, so $C^2 = -1$ or $C^2 \geq 0$.
The former case is ruled out because we assumed $S$ to be minimal.
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:::{.remark}
On the proof of the lemma: it is broken into the following three cases,

- $K^2 < 0$,
- $K^2 = 0$,
- $K^2 > 0$.

We'll consider the case $K^2 = 0$.
We first claim $-K_S$ is effective, i.e. $h^0(-K_S) \neq 0$.
Applying Riemann-Roch to $-K_S$ yields $\chi(-K_S) = {1\over 2}(-K)(-K-K) + 1 = -K^2 + 1= 1$ since $K^2 = 0$.
By Serre duality, $h^2(-K_S)= h^0(K_S - (-K_S)) = h^0(2K_S)$, but note $\OO(2K_S) = \OO(K_S)\tensor \OO(K_S) = \omega_S^{\tensor 2}$, so $h^0(2K_S) = P_2 = 0$ by assumption.
Thus $h^0(-K_S) = 1 + h^1(-K_S) > 0$.

Claim: let $H$ be a very ample divisor, which is in particular effective.
Then $-K_S.H > 0$ since $-K_S$ is effective, so $K_S.H < 0$.
We'll show $H + nK_S$ is not effective if $n$ is large enough.
Consider $H(H + nK_S) = H^2 + nHK_S$; sine $HK_S < 0$, this is negative for some $n$ since $H^2$ is a fixed number.
If $H + nK_S$ were effective, this would have to be positive.
So there exists a unique minimal $n_0$ such that $H + n_0 K_S$ is not effective for any $n\geq n_0$.
So take $C \da H + n_0 K_S$.
One then checks $K_S.C = K_S.H < 0$ since $H^2 = 0$ and $-K_S$ is effective with $H$ ample.
Furthermore $K_S + C= K_S + H + n_0 K_S = H + (n_0 + 1)K_S$ which is not effective by definition of $n_0$, thus $\abs{K_S + C} = \emptyset$.
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:::{.remark}
Note that we didn't show $C$ was irreducible.
If it is not, then take any irreducible component $C'$ with $C'.K_S < 0$, which must exist since $K_S.C < 0$.
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