# 2024-10-15-12-51-37

:::{.remark}
Recall that a *minimal* surface is a smooth projective surface which does not contain any smooth rational $(-1)$-curves.
Castelnuovo's contraction theorem implies that any smooth projective surface is birational to a minimal surface.
Goal: classify all smooth projective surfaces up to birational equivalence.
A related question: in any birational equivalence class, is there a unique minimal surface?
Or can two minimal surfaces be birational?
Answer:

- No for rational surfaces,
- Yes for non-ruled surfaces.

Recall that a ruled surface is birational to $C\times \PP^1$ for $C$ a smooth projective curve.
Note that rational implies ruled, since $\PP^2 \birational \PP^1\times \PP^1$ which is ruled.
Every non-ruled surface admits a unique minimal model up to isomorphism.
:::

:::{.remark}
Let $S, S'$ be non-ruled surfaces and $\phi: S'\birational S$ be a birational map. 
We want to show that $S' \cong S$.
Note that $\phi$ is a composition of blowups and blowdowns, so there exists some $\hat S$ with $f: \hat S\to S$ and $\eps: \hat{S}\to S'$ where $f, \eps$ are compositions of blowups.
Choose a diagram such that $\eps$ is composed of a minimal number $n\in \ZZ_{\geq 0}$ of blowups.
By cases: if $n=0$, then $S'\cong \hat S$, and $f:S'\to S$ is a composition of blowups.
The last blowup would introduce an exceptional curve, contradicting the fact that $S'$ is minimal.
Thus $f$ is an isomorphism.

Suppose now that $n\neq 0$; we'll show that this can never happen.
Let $E$ be the exceptional curve of the blowup $\eps$ in $\hat S$.
Consider $f(E)$, then either

- $f(E)$ is contracted, so $f(E)$ is a point, or
- $f(E)$ is a curve.

The first case contradicts the minimality of $n$ by removing the blowup/blowdown of $E$ from the diagram.
In the second case, we'll show such a curve can not exist, reaching a contradiction.
:::

:::{.lemma}
Let $X$ be any surface and $f: \hat X\to X$ is the blowup at a point.
Let $\hat \Gamma \subseteq \hat X$ be an irreducible curve which is not contracted by $f$, so $\Gamma \da f(\hat \Gamma) \subseteq X$ is a curve.
Then
\[
K_{\hat X}.\hat \Gamma \geq K_X . \Gamma
.\]

:::

:::{.proof}
As long as $\hat \Gamma \not\subseteq E$, it is not contracted to a point when $E$ is blown down.
So this yields some curve $\Gamma \subseteq X$, possibly singular.
Let $m \da E.\hat \Gamma < \infty$, which is finite because $\hat \Gamma$ is not contained in $E$.
Note $K_{\hat X} = f^* K_X + E$.
We can write $\hat \Gamma = f^* \Gamma + mE$.
Recall that $f^* D_1 . f^* D_2 = D_1. f_* f^* D_2$, so $f^* K_{X}. f^* \Gamma = K_X. f_* f^* \Gamma = K_X.\Gamma$,
so
\begin{align*}
K_{\hat X}. \hat \Gamma
&= (f^* K_X + E)(f^* \Gamma - mE) \\
&= f^* K_X. f^* \Gamma + E.f^* \Gamma -m f^* K_X.E - mE^2 \\
&= K_X.\Gamma + 0 + 0 - 1
.\end{align*}

Moreover, $f^* D_1.D = D_1.f_* D$, so we have
\[
E.f^* \Gamma = f^* \Gamma.E = \Gamma.f_* E = 0
\]
since $f_* E$ is a point and not a divisor.
This proves the lemma.
Note that equality is attained iff $m=0$, so $\hat \Gamma$ does not intersect $E$.
:::

:::{.remark}
We now apply this lemma to the previous proof.
Recall that $\eps: \hat{S}\to S$ produces an exceptional curve $E = \hat\Gamma$ where the last blowup $\eps_n$ produces $E$.
Let $C = f(E)$.
Apply the lemma to each blowup $f_1,\cdots, f_m$ by setting $\hat\Gamma = E$.
By the adjunction formula, $2g(E)-2 = E.(E+K_{\hat S})$, and since $g(E) = 0$ one obtains $E.K_{\hat S} = -1$.
Thus $C.K_S \leq -1$ by the lemma.
Claim: $C.K_S \neq -1$, so the inequality is strict.
If $K_S.C = -1$, then $K_{\hat S}.E = K_S.C$ and this only happens if $E$ does not intersect the other exceptional divisor $E$.
So $f(E) \cong C \cong E$, so $C\cong \PP^1$ with $C^2 = -1$, but this contradicts minimality of $S$.
Thus $C.K_S \leq -2$.

By adjunction, 
\[
2g(C)-2 = C.(C+K_S) = C^2 + C.K_S \implies C^2 = 2g-2-C.K_S \implies C^2\geq 0
\]
since $C.K_S \leq -2$.

:::

:::{.lemma}
We thus have two properties:

- $C.K_S \leq -2$,
- $C^2 \geq 0$.

If such a curve exists, all plurigenera $P_m(S)$ vanish.
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:::{.proof}
Note that $P_m(S) = 0$ iff $mK_S$ is not effective.
If $D = mK_S$ is effective, then $D = D' + a C$ where $a\geq 0$ such that $D'$ does not contain $C$.
Then $D.C = D'.C + aC^2 \geq 0$ by assumption, but this implies $mK_S.C \geq 0$, a contradiction.
:::

:::{.remark}
Hence $P_m(S) = 0$, and in particular $P_2(S) = 0$.
Cases:

- $h^{1, 0}(S) = 0$, which implies $S$ is rational by Castelnuovo, hence ruled, a contradiction.
- $h^{1, 0}(S) \neq 0$, which we claim also can not happen.

Assume the second case.
Black box: for any surface $S$, there is a natural map $\psi: S\to \Alb(S)$ where $\psi(S)$ is a curve of genus $h^{1, 0}$.
Moreover, $\psi$ is generically finite, and $\Alb(S)$ is generally a genus $g$ abelian variety.
Since $C$ is rational, it must be a fiber of $\psi$.
One can show $S\cong \Alb(S) \times \PP^1$, showing $S$ is ruled, a contradiction.
This forces $n=0$ and $S\cong S'$.
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:::{.remark}
Next time: the Albanese variety.
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