# Existence of curves of arbitrary genus on some K3 surface

Voisin uses the fact "If $$X$$ is a K3 surface with an ample line bundle $$\mathcal L$$ such that $$\mathcal L$$ generates $$\mathop{\mathrm{Pic}}(X)$$ and $$(\mathcal L^2) = 4t - 2$$, then every smooth curve $$C \in \lvert\mathcal L\rvert$$ satisfies $$K_{t, 1}(C, K_C) = 0$$." to prove the Green conjecture holds for generic curves of even genus. My question is why given an even integer $$g$$, there always exists a K3 surface $$X$$ and a smooth curve $$C \subseteq X$$ of genus $$g$$ satisfying the conditions above.

This should be a consequence of the surjectivity of the period map for K3 surfaces. I believe with this in mind the reasoning is somewhat standard, but it's useful to try and make it explicit. The underlying strategy is as follows: 1) identify a non-empty locus $$\mathcal{W}$$ in the period domain to which a K3 surface $$X$$ with the desired property could be mapped into via the period map, then 2) use surjectivity of the period map to assert existence of such $$X$$.

To this end

• Let's make the following definitions of, respectively, the K3 lattice, the moduli space of marked K3 surfaces[1], and the period domain: $$\Lambda := E_8(-1)^{\oplus 2}\oplus U^{\oplus 3}$$ $$N := \{(X,\varphi)\}/\sim$$ $$D := \{[v] \in \mathbb{P}\Lambda_{\mathbb{C}} : (v)^2 = 0\text{ and } (v,\overline{v}) > 0\}.$$

Then the period map $$\mathcal{P} : N \rightarrow D \subset \mathbb{P}\Lambda_{\mathbb{C}}$$ sending $$(X,\varphi)$$ to $$[\varphi_{\mathbb{C}}(H^{2,0}(X))]$$ is surjective (see e.g. Theorem 4.1 in "Lectures on K3 surfaces" by D. Huybrechts).

• Now suppose $$V \subset \Lambda$$ is any sublattice. Then $$V^{\perp}$$ (in particular) determines a closed slice $$D_V := D\cap\mathbb{P}(V^{\perp})_{\mathbb{C}}$$ of the period domain. If $$\mathcal{P}(X,\varphi) \in D_V$$ this then means $$\varphi_{\mathbb{C}}(H^{2,0}(X)) \subset (V^{\perp})_{\mathbb{C}}$$ which[2] implies $$\varphi_{\mathbb{C}}(H^{1,1}(X)) \supset V_{\mathbb{C}}$$ and thus[3] $$\varphi(NS(X)) \supset V.$$

• Finally, let $$D_V^{\circ} := D_V \setminus \bigcup_{V'\not\subset V}D_{V'}.$$ Then clearly if $$\mathcal{P}(X,\varphi) \in D_V^{\circ}$$ we have $$\varphi(NS(X)) \cong V.$$ Since $$D_V^{\circ}$$ is the complement of a countable union of proper closed subsets, it is non-empty.

• So for the desired result, it now suffices to take $$\mathcal{W} = D_V^{\circ}$$ where $$V = \langle \lambda \rangle$$ for $$\lambda \in \Lambda$$ such that $$(\lambda)^2 = 4t - 2$$. The existence of such $$\lambda$$ can be verified for any $$t \geq 1$$ using our explicit knowledge of the lattice $$\Lambda$$ (as indicated above) - e.g. $$\lambda = (2t-1,1) \in U$$ has the desired property, thinking of $$U \cong \mathbb{Z}^2$$ with intersection form $$\left[\begin{array}{cc}0 & 1\\1 & 0\end{array}\right]$$.

[1] if $$X$$ is a K3 surface with marking $$\varphi : H^2(X;\mathbb{Z}) \xrightarrow \cong \Lambda$$ (an isometry) then $$(X,\varphi) \sim (X',\varphi')$$ if and only if $$\varphi' = \varphi\circ f^*$$ for some isomorphism $$f : X \rightarrow X'$$.

[2]using the fact that $$\varphi$$ is an isometry and is defined over $$\mathbb{Z}$$ (and thus commutes with conjugation on $$\Lambda_{\mathbb{C}}$$), one sees this as follows: $$\begin{array}{rcl} \varphi_{\mathbb{C}}(H^{2,0}(X)) & \subset & (V^{\perp})_{\mathbb{C}}\\ \implies \varphi_{\mathbb{C}}(H^{0,2}(X)) = \varphi_{\mathbb{C}}(\overline{H^{2,0}(X)}) & \subset &(V^{\perp})_{\mathbb{C}} = \overline{(V^{\perp})_{\mathbb{C}}}\\ \implies V_{\mathbb{C}} = (V^{\perp})_{\mathbb{C}}^{\perp} & \subset & \varphi_{\mathbb{C}}(H^{2,0}(X)\oplus H^{0,2}(X))^{\perp}\\ & = & \varphi_{\mathbb{C}}((H^{2,0}(X)\oplus H^{0,2}(X))^{\perp})\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X)) \end{array}$$

[3]using the same facts as above, one sees this as follows: $$\begin{array}{rcl} \varphi(NS(X)) & = & \varphi_{\mathbb{C}}(H^{1,1}(X)\cap H^2(X;\mathbb{Z}))\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \varphi_{\mathbb{C}}(H^2(X;\mathbb{Z}))\\ & = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \Lambda\\ & \supset & V_{\mathbb{C}}\cap \Lambda\\ & = & V. \end{array}$$