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.
 A: 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:
$$
\begin{array}{rcl}
\varphi(NS(X)) & = & \varphi_{\mathbb{C}}(H^{1,1}(X)\cap H^2(X;\mathbb{Z}))\\
& = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \varphi_{\mathbb{C}}(H^2(X;\mathbb{Z}))\\
& = & \varphi_{\mathbb{C}}(H^{1,1}(X))\cap \Lambda\\
& \supset & V_{\mathbb{C}}\cap \Lambda\\
& = & V.
\end{array}
$$
