# 2024-10-17-12-49-00

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Let $S, S'$ be non-ruled surfaces, then any birational map $S\birational S'$ is an isomorphism.
Factoring as a composition of $n$ blowups and $m$ blowdowns, we saw that if $n=0$ this is true.
For $n\geq 1$, we showed existence of a curve $C$ with $K_S.C \leq -2$ and $C^2 \geq 0$, which implied $P_m(S) = 0$, and in particular $P_2(S) = 0$.
There were two cases:

- $h^{1, 0}(S) = 0 \implies S$ is rational, hence ruled, a contradiction, or
- $h^{1, 0}\neq 0$.

In the latter case, we have a natural map $\psi: S\to \Alb(S)$, which is an abelian variety.
The image $\psi(S)$ is a genus $g$ curve where $g = h^{1, 0}(S)$.
Note that a priori, one could have $\dim \psi(S) = 0,1,2$, so we argue that $\dim \psi(S)\neq 0, 2$.
If $\psi(S)$ is a point, the universal property of the Albanese yields a contradiction.
If $\psi(S)$ is a surface, then $\psi$ is generically finite and $P_2(S) = P_2(\Alb(S))\neq 0$.
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## Albanese varieties

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Let $X$ be any smooth projective variety, then there is an abelian variety $\Alb(X)$.
This can be written as $H^0(\Omega_X)\dual/\iota(H_1(X; \ZZ))$ where $\iota: H_1(X;\ZZ)\to H^0(\Omega_X)\dual$ is defined by $\gamma \mapsto (\omega\mapsto \int_\gamma \omega)$.
It is a theorem that the image is always a lattice $\Gamma$, so $\Gamma \tensor \CC = H^0(\Omega_X)\dual$.
The proof uses deep facts from Hodge theory.
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:::{.remark}
Recall that $h^{1, 0} = \dim_\CC H^0(\Omega_X) \da n$, so $H^0(\Omega_X)\cong \CC^n \cong \RR^{2n}$.
Writing $H_1(X;\ZZ) = \ZZ^{b_1} + T$ where $T$ is torsion, note that $\iota(T) = 0$.
Note also that $b_1 = h^{1, 0} + h^{0, 1}$.
Moreover $\iota(\ZZ^{b_1})\cong \ZZ^{b_1}$.
Thus $\Alb(X) \cong \RR^{2n}/\ZZ^{2n}\cong (S^1)^{2n}$.
There is a natural complex structure on $\Alb(X)$, and moreover is smooth and projective.
The proof again uses Hodge theory.
By GAGA, $\Alb(X)$ is the unique variety with this underlying complex manifold structure.
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:::{.remark}
Recall $\Pic^0(X)$ is the connected component of $\Pic(X)$ containing the trivial line bundle.
A theorem of Grothendieck shows that $\Pic^0(X)$ is an abelian variety of dimension $h^{1, 0}$.
We also have $\Alb(X) = \Pic^0(X)\dual$ -- more generally, for any abelian variety $A$, one defines $A\dual \da \Pic^0(A)$ since $\Pic^0(\Pic^0(A)) = A$.
We thus regard $\Alb(X)$ as the "smallest" abelian variety associated to $X$.
The map $\psi: X\to \Alb(X)$ can be given by $x\mapsto (\omega\mapsto \int_x^p \omega)$ where $p$ is a fixed choice of point on $X$.
The universal property is that for any abelian variety $A$, any map $X\to A$ factors as $X\to \Alb(X)\to A$.
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