# 2024-10-31-12-45-31

:::{.remark}
Recall that we are considering minimal surfaces $S$ with $q\geq 1, p_g = 0$.
We had $K^2 = 10-8q-b_2$, and concluded that if $K^2 < 0$ then $S$ is ruled.
We consider now the extreme case where $K^2 = 0$, so $q=1, p_g = 0, b_2 = 2$, and $S$ is not ruled.
Consider the Albanese morphism $p: S \to B \subseteq \Alb(S)$ with $g(B) = q$; we'd like to show every fiber is irreducible.
If any fiber is reducible, this forces $b_2\geq 3$ by taking two components along with a horizontal part.
If any fiber is $\PP^1$, then $S$ is ruled and birational to $B\times \PP^1$.
Since $S$ is reduced and $p$ is nonconstant onto a smooth curve, $p$ is flat.
:::

:::{.theorem}
If $g\geq 2$ then $p$ is smooth, so every fiber is smooth and reduced (multiplicity one).
If $g=1$ then every fiber $F$ is of the form $F=nC$ for $C$ a smooth elliptic curve.
In any case, $p$ is an isotrivial family, i.e. any two fibers are isomorphic.
:::

:::{.remark}
In the $g=1$ case, the fibration $p$ is almost smooth.
Note that isotrivial families aren't necessarily cartesian products.
Moreover, we claim that $p$ is never the trivial family $S = B\times F$.
If it were, $K_S = \pi_1^* K_B + \pi_2^* K_F$ is effective, contradicting $p_g = 0$.
:::

:::{.remark}
To study this, we consider the topological Euler characteristic, defined as $\chi_{\Top}(X) = \sum_{i=0}^{\dim_\RR X} (-1)^i \dim H^i_c(X)$ for $X$ any algebraic variety over $\CC$, not necessarily compact.
Fact: if $X\contains Z$ a closed set with open complement $U \da X\sm Z$, one has $\chi_\Top(X) = \chi_\Top(Z) + \chi_\Top(U)$.
As a corollary, for a fibration like $p$, one has 
\[
\chi_\Top(S) = \chi_\Top(B)\cdot \chi_\Top(F_{\mathrm{gen}}) + \sum_{\mathrm{finite}}\chi_\Top(F_s)- \chi_\Top(F_{\mathrm{gen}})
\]
where $F_s$ is the special fiber.
Note that there is a LES $H^i_c(U)\to H^i_c(X) \to H^i_c(Z)$ given by including cycles along $Z\injects X$ and restricting cycles to the open subset $U \subseteq X$.

Note that $b_2 = 2q$ by Hodge conjugate symmetry, and so in our case
\[
\chi_\Top(S) = 1-b_1+b_2-b_3+1 = 2 - 2b_1 + b_3 = 4-4q=0
.\]
If $C$ is a smooth curve of genus $g$,
\[
\chi_\Top(C) = 1-b_1+1 = 2-2g
.\]
Note that when $C$ is smooth, $\chi(\OO_C) = h^0(\OO_C)-h^1(\OO_C) = 1-g$, and so $\chi_\Top(C) = 2\chi(\OO_C)$.
:::

:::{.remark}
Claim: $\chi_\Top(F_s) > \chi_\Top(F_{\mathrm{gen}})$, so the finite sum in $\chi_\Top(S)$ is nonzero.
Equality only holds in the $g=1$ case of the above theorem.
This will prove the first part of the theorem.
Let $F$ be a fiber, then there is a SES $\OO_S(-F) \injects \OO_S\surjects \OO_F$ given by restriction to a fiber.
Thus
\begin{align*}
\chi(\OO_F) 
&= \chi(\OO_S) - \chi(\OO_S(-F))
&= \chi(\OO_S) - \qty{\chi(\OO_S) + {F(F+K)\over 2}}
&= {-FK\over 2}
.\end{align*}

By the same argument, $\chi(\OO_C) = {-CK\over 2} = {-FK\over 2n}$ since $C^2 = 0$.
Consider the case $g(F_{\mathrm{gen}}) \geq 2$, then $\chi(\OO_C)-\chi(\OO_{F_{\mathrm{gen}}}) > 0$.
Similarly when $g=1$.
It remains to show smoothness.
:::

:::{.lemma}
If $C$ is singular, then 
\[
\chi_\Top(C) > 2\chi(\OO_C)
.\]
:::

:::{.example}
Let $C$ be a cuspidal rational curve, i.e. a cubic curve with a cusp.
Let $\eta: \tilde C\to C$ be its normalization, so $\tilde C = \PP^1$.
Note that $\eta$ is not an algebraic isomorphism, but is a homeomorphism.
Thus $\chi_\Top(C) = 2$.
There is an embedding $C\injects \PP^2$ as a plane cubic, and thus a SES $\OO_{\PP^2}(-3)\to \OO_{\PP^2}\to \OO_C$ and thus $\chi(\OO_C) = 0$.
:::

:::{.example}
Let $C$ be a rational cubic with a node, then again $\tilde C = \PP^1$.
Note that $\chi_\Top(\pt) = 1$, and by an argument deleting points in $C$ and $\tilde C$ one has $\chi_\Top(C) = 1$.
Again one can argue $\chi(\OO_C) = 0$.
:::

:::{.example}
The general case is a combination of the two above cases, unibranch (where $\eta$ is a homeomorphism) and two branches crossing.
Consider a curve with a node with $b=3$ independent tangent directions.
Take the seminormalization and then normalize, which separates the node into 3 points.
In the first step, the seminormalization is a homeomorphism, so $\chi_\Top$ doesn't change.
In the second step, it increases by $b-1$.
Let $\pi: C^{\sm}\to C$ be the seminormalization, then there is a SES
\[
\OO_C\to \pi_* \OO_{C^\sm}\to Q
\]
where $Q$ is a skyscraper sheaf, which is nonzero if $\pi$ is not an isomorphism.
Thus $\chi(\OO_C)$ increases in the first step.
Similarly, for the normalization $\eta: C^\sm \to \tilde C$, there is again a SES
\[
\OO_{C^{\sm}}\to \eta_* \OO_{\tilde C}\to Q'
\]
where $Q'$ is a skyscraper sheaf of length $b-1$.
:::

:::{.remark}
Fact: any proper non-isotrivial family of curves of genus $g\geq 1$ must degenerate, i.e. there must be a singular fiber.
A similar statement holds for abelian varieties.
For $g\geq 2$, consider $\Ag$ as a coarse space.
Adding a level structure for $m\geq 3$ yields a fine space $\Ag(m)$, where a level structure is an isomorphism $A[m] \iso (\ZZ/m\ZZ)^{2g}$.
This kills automorphisms.
Note that for $g=1$, this is easier since $M_1 = \AA^1$ and any map from a projective to an affine variety is constant.
:::