# 2024-11-21-12-48-29 :::{.remark} K3 surfaces: $X$ with $p_g=1, q=0, K_X\sim 0, \chi(X) = 24, \chi(\OO_X) = 2$. Polarized: pairs $(X, L)$ with $L$ an ample line bundle satisfying $L^2 = 2d$. Note that any two K3 surfaces over $\CC$ are homeomorphic, and $\pi_1(X) = 1$ and thus $H^1(X; \ZZ) = 0$. By UCT, $H^2(X;\ZZ) = \Hom(H_1(X;\ZZ), \ZZ) \oplus T$, where $T$ is the torsion in $H_1$, is torsionfree is isomorphic to $\ZZ^{22}$. There is an integral bilinear form $H^2(X;\ZZ)\times H^2(X;\ZZ)\to H^4(X;\ZZ)\cong \ZZ$. Note that $(x, x)$ is even for all $x$, symmetric, and unimodular. The form has signature $(3, 19)$. Such indefinite unimodular even forms are unique, so this lattice is $II_{3, 19} \cong U^3 \oplus E_8^2$. Note that $U = II_{1, 1}$ and $E_8(-1) = II_{0, 8}$. Note that $\Pic(X) = H^1(\OO_X\units)$. Take the LES for the exponential SES $\ZZ \to \OO_X\to \OO_X\units$ to get \[ 0\to \Pic^0(X) \to \Pic(X) \to \NS(X) = \ker\qty{H^2(X;\ZZ)\mapsvia{f} H^2(\OO_X)} .\] We have $\Pic^0(X) = H^1(\OO_X)/H^1(X;\ZZ) = \CC^q/\ZZ^{2q}$, and $\Pic(X) \cong \NS(X) \subseteq H^2(X;\ZZ) = II_{3, 19}$. By the Hodge index theorem, $\NS(X)$ has signature $(1, r-1)$ for some $r$. The inclusion $\NS(X) \subseteq H^2(X;\ZZ)$ is a primitive embedding of lattices, i.e. $nv\in \NS(X)\implies v\in \NS(X)$, since $\NS(X)$ is isomorphic to a kernel. Nikulin classified which hyperbolic lattices can be primitively embedded into $II_{3, 19}$, and moreover shows they are all attained by some K3 surface. ::: :::{.remark} Understanding $\ker f$ from above: it is the composition \[ H^2(X;\ZZ) \to H^2(X;\CC) = H^{2, 0} \oplus H^{1, 1} \oplus H^{0, 2} \surjects H^{2, 0} = H^2(\OO_X) .\] Write $H^{0, 2} = H^0(\Omega^2) = H^0(K_X) = \CC \omega_X$. ::: :::{.lemma} \[ \NS(X) = H^2(X;\ZZ) \intersect H^{1, 1} = \ts{ \alpha\in H^2(X;\ZZ ) \st \alpha\cdot \omega_X = 0 } .\] ::: :::{.proof} Write \( \alpha = \alpha^{2, 0} + \alpha^{1, 1} + \alpha^{0, 2} \) where \( \bar \alpha^{2, 0} = \alpha^{0, 2} \) and \( \bar \alpha^{1, 1} = \alpha^{1, 1} \). If \( \alpha\in \ker f \) then \( \alpha^{2, 0} = 0 \) and so \( \alpha = \alpha^{1, 1} \). The second identification comes from the fact that \( \alpha\cdot \omega_X \in H^{2, 2} + H^{1, 3} + H^{0, 4} = H^{2, 2} \) and the pairing $H^{2, 0} \times H^{0, 2} \to H^{2, 2}$ is nondegenerate. ::: :::{.remark} Thus varying \( \omega_X \) amounts to varying $\NS(X)$, and one has $0\leq r\leq 20$ for $r \da \rank \NS(X)$. We define a period domain \[ D \da \ts{[\omega] \st \omega \in II_{3, 19} \tensor \CC, \omega\cdot\omega = 0, \omega\cdot \bar\omega > 0} \subseteq \CP^{21} .\] The Torelli theorem states that for any \( \omega\in D \) there exists a complex analytic K3 surface $X$ with \( \omega_X = \omega \), together with a marking $H^2(X;\ZZ)\iso II_{3, 19}$. Further, two marked K3s are isomorphic iff $\omega_X = \omega_{X'}$. :::