# 2024-11-19-12-46-23: K3 Surfaces

:::{.remark}
Let $S$ be a minimal surface with $p_g = 1, q=0, \chi(\OO_S) = 2$ with $\kappa = 0$.
We proved $K_S \sim 0$.
By the Noether formula, $\chi(\OO_S) = {K^2 + \chi(S)\over 12}$, so $\chi(S) = 24$.
We have the following Hodge diamond:

\[\begin{tikzcd}
	&& {h^{2,2}=1} \\
	& {h^{1,2}=0} && {h^{2,1}=0} \\
	{h^{0,2} = h^0(\omega_S) = 1} && {h^{1,1} = 20} && {h^{2,0}=1} \\
	& {h^{0,1}=q=0} && {h^{1,0}=0} \\
	&& {h^{0,0} = 1}
\end{tikzcd}\]
> [Link to Diagram](https://q.uiver.app/#q=WzAsOSxbMiwwLCJoXnsyLDJ9PTEiXSxbMSwxLCJoXnsxLDJ9PTAiXSxbMywxLCJoXnsyLDF9PTAiXSxbMCwyLCJoXnswLDJ9ID0gaF4wKFxcb21lZ2FfUykgPSAxIl0sWzIsMiwiaF57MSwxfSA9IDIwIl0sWzQsMiwiaF57MiwwfT0xIl0sWzEsMywiaF57MCwxfT1xPTAiXSxbMywzLCJoXnsxLDB9PTAiXSxbMiw0LCJoXnswLDB9ID0gMSJdXQ==)

:::

:::{.example}
Let $S = X_4 \subseteq \PP^3$ be a quartic, then $K_S = K_{\PP^3} + S\mid_S = -4H + 4H = 0$.
Take the SES $\OO_{\PP^3}(-4) \to \OO_{\PP^3}\to \OO_S$ given by restriction of functions.
Note that $H^i(\OO_{\PP^n}(d)) = 0$ unless $d=0, n$, so by the LES, $H^1(\OO_S) = 0$.
:::

:::{.example}
Let $X_{3,2} \subseteq \PP^4$ be a complete intersection, then $K_X = K_{\PP^4} + D_3 + D_2\mid_X = -5H + 3H + 2H = 0$.
Computing $H^1(\OO_X)$ proceeds similarly.
A similar example is $X_{2,2,2} \subseteq \PP^4$.
:::

:::{.remark}
Each of these defines a K3 with an ample line bundle $L \da \OO_X(-1)$, the restriction of $\OO_{\PP^n}(-1)$ to $X$.
In these cases:

- $L^2 = H^2 X = 4$
- $L^2 = 2\cdot 3 = 6$
- $L^2 = 2\cdot 2\cdot 2 = 8$.

All of these yield a pair $(X, L)$, and we define the degree of the pair as $d \da L^2$.
These are the only complete intersections which are K3s.
:::

:::{.lemma}
On a K3 surface, for any divisor $D$, one has $D^2$ even.
:::

:::{.remark}
By Riemann-Roch, $\chi(D) = {D^2 - DK\over 2}$, and $DK = 0$.
This in fact holds for all minimal surfaces with $\kappa = 0$, since some multiple of $K$ is zero.
:::

:::{.remark}
Suppose $d=2$.
Then $L$ can not be ample.
There is a double cover $f:X\to \PP^2$ branched over $B = \ts{f_6 = 0}$, a smooth curve given by an equation of degree 6.
By Riemann-Hurwitz, $K_X \sim_\QQ f^*(K_{\PP^2} + {1\over 2}B)\sim 0$, which forces $B$ to be a sextic.
Clearing denominators yields $2K_X \sim f^*(2K_{\PP^2} + B)$.
We generally have $f_* \OO_X = \OO_{\PP^2} \oplus F\inv$ where $F^2 = \OO(B)$.
This is the decomposition into the $\pm 1$ eigenspaces of $f_* \OO_X$ respectively under an involution.
There is a grading, this is given by a multiplication morphism $F\inv\tensor F\inv \to \OO_{\PP^2}$, or equivalently a map $\OO \to F^2$ where $1\mapsto s$.
The cover has local coordinates $z^2 = s(x, y)$, and $(s) = B$ cuts out the ramification locus, so $s\in \OO_{\PP^2}(B)$ and $F^2 = \OO_{\PP^2}(B)$ is necessary.
We have $f_* \omega_X = \omega_{\PP^2} \oplus \omega_{\PP^2}(F)$.
Moreover $H^1(\OO_X) = H^1(\OO_{\PP^2}) \oplus H^1(\OO_{\PP^2}(-3)) = 0$, so $q=0$.
Similarly, $H^0(\omega_X) = H^0(\OO_{\PP^2}(-3)) \oplus H^0(\OO_{\PP^2}(-3+3)) \cong 0 \oplus \CC$, so $\omega_X$ has a section and this forces $K_X \sim 0$.
:::

:::{.example}
Let $f:X\to \PP^1\times \PP^1$ with $B \subseteq \abs{\OO_{\PP^1\times \PP^1}(2, 2)}$, noting that $K_{\PP^1\times \PP^1} = \OO(2, 2)$.
A similar calculation shows that this is a K3 surface.
One sets $L = f^* \OO(1, 1)$, where $\OO(1, 1)$ is very ample and embeds $\PP^1\times \PP^1\embeds \PP^3$ as a conic.
Then $L^2 = 2\cdot 2 = 4$, yielding a degree 4 K3.
Note that $X \to \PP^3$ is not an embedding in this case; it covers a degree 2 surface.
:::

:::{.remark}
One could generalize this to consider double covers of other del Pezzo surfaces branched over $B\in \abs{-2K}$.
:::

:::{.example}
Elliptic K3s: by analogy, these are the $d=0$ case.
These come with a line bundle which is not ample.
They are elliptic surfaces: fibrations $f: S\to C$ with generic fiber $F_g$ of genus 1.
In the K3 case, $C = \PP^1$ since there are no differential forms.
We have $\chi(S) = \chi(C)\chi(F_g) + \sum_F (\chi(F) - \chi(F_g))$, and since $\chi(F) =2-2g = 0$, all contributions come from the excess term.
The simplest singular fibers are nodal curves, which normalize to $\CP^1\cong S^2$.
So there are generically 24 singular fibers, and there may or may not be a section.
:::

:::{.remark}
Minimal surfaces with $\kappa = 1$ are elliptic, but not every elliptic surface satisfies $\kappa = 1$.
If the fibration admits a section, the fibers are genus 1 curves with a point and the fibration is called Jacobian.
The fibers have Weierstrass form $y^2 = x^3 + Ax + B$ where $A(t_0, t_1), B(t_0, t_1)$ are homogeneous in the coordinates $t_i$ on $\PP^1$.
One checks $\Delta_{24} = 4A_8^3 + 27 B_{12}^2$.
:::

:::{.remark}
Note $L = f^* \OO_{\PP^1}(1)$ yields $L^2 = 0$.
One can choose other polarizations, e.g. $L = s + df$.
Then $L^2 = s^2 + 2dsf + f^2 = -2 + 2d + 0 = 2d-2$, where $s^2 = -2$ follows from the genus formula $g(s) = {s^2 + sK\over 2} + 1$ with $sK = 0$.
For generic choices, $L$ will be ample, and $L^2 = 2d-2$ achieves every even integer.
:::

:::{.remark}
Let's count moduli for the examples given above.

- $X_4$: there are $7\choose 4$ monomials, minus $\dim \PGL_4 = ?$, minus 1 for projectivizing equations.
    This yields $35-16 = 19$.
- $X_{2, 3}$: a more delicate count yields $19$.
- $X_{2, 2, 2}$: one gets 19 by computing the dimension of a Grassmannian.
- $f: X\to \PP^2$: the number of sextics in $\PP^2$, minus $\dim \PGL_2$, yields ${8\choose 2} - 9 = 19$.
- $f:X\to \PP^1\times \PP^1$: counting yields 18.
- Elliptic: count possibilities for $A_8$ and $B_{12}$, modulo $\PGL_2$, yields 18.

Why the difference: $\rank \Pic(X_4) = 1$ generically, while $\rank \Pic(X) = 2$ for e.g. $X\to \PP^1\times \PP^1$.
So larger Picard rank yields fewer moduli; in general one has $20 - \rank \Pic(X)$ moduli.
:::

:::{.remark}
In every even degree $d\geq 4$, there exist K3s with a very ample line bundle.
This comes from studying things like $X_4$ which also admit an elliptic fibration, and setting $\tilde L \da L + kF$ for $F$ a fiber class.
Then $\tilde L^2 = L^2 + 2kLF$, and since $L$ is very ample and $F$ is basepoint free, $\tilde L$ is again very ample.
:::

:::{.remark}
Suppose $X$ is a K3 and $C \subseteq X$ is a smooth curve with $g(C) \geq 2$.
Note $C^2 = 2g-2$ by the genus formula, since $K=0$.
Then $\abs{C}$ is basepoint free, and $\phi_{\abs{L}}: X\to \im(X) \subseteq \PP^3$ is degree either 1 or 2.
In this case we say $X$ is a hyperelliptic K3.
:::

:::{.proof}
Consider the SES $\OO_X \to \OO_X(C)\to \OO_C(C)$.
By adjunction $K_C = K_X + C\mid_C = \OO_C(C)$, so $\OO_C(C) = \OO_C(K_C)$.
The result follows from the theory of curves.
:::

:::{.remark}
Next time: Hodge theory, deformations, Enriques surfaces.
Last class: elliptic surfaces.
:::

:::{.remark}
Note $\rho \leq 20$ in characteristic zero, but possibly $\rho = 22$ in positive characteristic.
:::