# 2024-11-05-12-50-39

:::{.remark}
Recall: let $S$ be a minimal non-ruled surface with $p_g=0$ and $q\geq 1$, which implies $K^2 = 0$ and $b_2 = 2$.
Then $\pi: S\to B$ is isotrivial, where $B$ is a curve with $g(B) = q$.
Either

- All fibers are smooth with multiplicity 1 and $g(B) \geq 2$, or
- $g(B) = 1$ and all fibers are either smooth or a multiple of a smooth curve.

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:::{.theorem}
Let $G$ be a finite group acting freely on $B'\times F$.
Then there is an etale cover $B'\times F \to {B'\times F\over G} = S$ where ${B'\times F\over G}\to B'/G = B$.
Thus any such surface is a quotient of a cartesian product.
Here $g(B'), g(F) \geq 1$.
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:::{.remark}
Thus there is an etale base change such that $S$ is a Cartesian product and the group action is free.
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:::{.lemma}
If $g(B') = g(F) = 1$, then $\exists N$ such that $NK_S \sim 0$. 
Otherwise, if either $g(B)\geq 2$ or $g(F)\geq 2$, there exists a sequence of integers $n_1\leq n_2\leq \cdots$ such that $P_{n_i}\to \infty$.
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:::{.remark}
In the first case, $\kappa(S) = 0$, and in the second, $\kappa(S) \geq 1$.
Note that if $G\actson B'$ by translations and $B'$ is elliptic, then $B = B'/G$ is elliptic as well.
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:::{.proof}
Let $S' \da B'\times F$ and $f: S'\to S$.
Then by Riemann-Hurwitz, $K_{S'} = f^*K_S + \eps$, and since $f$ is étale, $\eps = 0$.
Moreover $f_* K_{S'} = f_* f^* K_S = dK_S$ where $d \da \deg f = \size{G}$.
Thus $f_*(nK_S) = ndK_S$.
Note that $K_{B'\times F} = \pi_1^* K_{B'} + \pi_2^* K_F$, so if either $g(B'), g(F) \geq 2$, then the plurigenera go to infinity.
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:::{.example}
Let $S' = E_1\times E_2$ be a product of elliptic curves and let $G = \gens{\iota}$ be the group generated by an involution.
Define $\iota(x,y) = (x+a, -y)$ where $a\in E_1[2]$ is 2-torsion, so the action on the first coordinate is free.
Note $S'\to E_1' \da E_1/\iota_1$ with fibers $E_2$, and $B\da E_1'$ is again an elliptic curve.
There is also a projection $S'\to E_2' \da E_2/\iota_2$ with fibers either $E_1$ or $2E_1'$.
Analyzing $E_2 \to E_2'$, the ramification points are $y$ where $2y=0$, so this is ramified over 4 torsion points.
By Riemann-Hurwitz, the genus decreases, so $E_2'\cong \PP^1$.
Locally this can be written $y^2 = x(x-1)(x- \lambda)$ and the involution is $(x,y) \mapsto (x, -y)$ which ramifies over $0,1, \lambda, \infty \in \PP^1$.

Claim: $p_g = 0, q=1$.
Write $p_g(S') = h^0(\Omega^2_{E_1\times E_2})$.
Note that $h^0(\Omega_{E_1}) = 1$, generated by $dx$ which is translation invariant on $\CC$.
We have $H^0(\Omega^2_{E_1\times E_2}) = \gens{dx\wedge dy}_\CC$, and $H^0(\Omega^2_{E_1\times E_2/G}) = H^0(\Omega^2_{E_1\times E_2})^G$, the $G$-invariant differential forms on $S'$.
One now checks that under $(x,y)\mapsto (x+a, -y)$, we have 

- $dx\mapsto dx$
- $dy\mapsto -dy$
- $dx\wedge dy\mapsto -dx\wedge dy$.

Thus there are no invariant forms and $H^0(\Omega^2_{E_1\times E_2})^G = 0$.
This also shows $H^0(\Omega_{S'/G}) = H^0(\Omega_{S'})^G = \gens{dx}_\CC$.
Note also that $K_S \not\sim 0$ but $2K_S\sim 0$.
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:::{.remark}
Other examples: let $G\actson E_1\times E_2$ by a finite group action where $G\actson E_1$ by translation and $E_2/G \cong \PP^1$.
These are referred to as **bielliptic surfaces**.
Consider $\Aut(E)$ for $E$ an elliptic curve.
This always contain $\Aut^0(E)$ generated by translations $x\mapsto x+a$ for $a\in E$.
One can write $\Aut(E) = H\semidirect \Aut^0(E)$ where $H$ is a finite group isomorphic to $\Aut(E, 0)$, the automorphisms that fix the origin.
Generically $H = C_2$ generated by the elliptic involution.
Let $E_\tau \da {\CC\over \ZZ \oplus \ZZ \tau}$, then 
\[
H = 
\begin{cases}
C_2 & y\mapsto -y \\
C_4 & y\mapsto iy (E_i) \\
C_6 & y \mapsto \zeta_3 y (E_{\zeta_3})
\end{cases}
.\]
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:::{.remark}
There is a finite list of possible actions $G\actson E_1\times E_2$ for bielliptic surfaces:

1. $C_2$
2. $C_2\times C_2$
3. $C_4$
4. $C_4\times C_2$
5. $C_3$
6. $C_3\times C_3$
7. $C_6$

We have

- $2K_S \sim 0$ in cases 1,2,
- $4K_S\sim 0$ in 3,4,
- $3K_S \sim 0$ in 5,6,
- $6K_S \sim 0$ in 7.

Note that as a corollary, we always have $12K_S \sim 0$ in this case.
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:::{.remark}
Claim: there is a base change of $S\to B$ to $S' \to B'$ such that $S'$ is a Cartesian product.
This is case 1 from earlier.
An idea of the proof: there is a constant map $B\to \Mg \to \Ag$ at the level of coarse spaces.
Pass to fine spaces $\Mg(N) \to \Mg$, and $\Ag(N)\to \Ag$, which are quotients by finite groups.
The latter is comprised of pairs $[C, \Jac(C)_N \iso C_N^{2g}]$.
One then considers trivializing the local system $R^1\pi_* C_N$.
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:::{.remark}
The $\kappa(S) = 0$ case is particularly important case for us.
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