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Let $V \rightarrow \mathbb P^2$ be the blow-up at two distinct points. ($V$ is a Del Pezzo surface of degree 7.) Choose a smooth curve $C$ from the linear system $|-2K_V|$ and let $S \rightarrow V$ be the double cover, branched along $C$. Then $S$ is a K3 surface whose Picard rank is at least three.

Assume that $S$ is of Picard rank $three$.

My question is:

Can $S$ be embedded into $\mathbb P^3$?

This question is equivalent to finding a very ample divisor $H$ of $S$ with $H^2 =4$. Since we know completely the intersection form on $Pic(S)$, one can try (in fact, I have been trying) to find a primitive divisor class $L$ such that

  1. $L^2 = 4$,

  2. there is no divisor $D$ such that $D^2 = 0$ and $D \cdot L = 1, 2$.

  3. there is no divisor $E$ such that $E^2 = −2$ and $E \cdot L = 0$.

A result of Saint-Donat guarantees that $L$ or $-L$ is very ample.

My ultimate goal is to find out whether the following set is unbounded:

{$C \cdot L$ : $L$ is a very ample divisor on S with $L^2 =4$ }.

Is this set really unbounded?

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  • $\begingroup$ $S$ has three $(-2)$-curves obtained as preimages of lines on the del Pezzo surface. It seems that these curves form a basis of $NS(S)$? $\endgroup$
    – Evgeny Shinder
    Commented Jul 2, 2022 at 16:24
  • $\begingroup$ Yes, they form a basis. $\endgroup$
    – Basics
    Commented Jul 2, 2022 at 17:15
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    $\begingroup$ Then after changing the basis, the intersection form seems to be diagonal $2(x^2 - y^2 - z^2)$, and one needs to solve $x^2 - y^2 - z^2 = 2$ to find elements of square $4$. One obvious element of square $4$ is $2h - E_1 - E_2$ (pullback from the del Pezzo surface multiplies the degree by two), however it contracts the other $(-2)$-curve (pullback of $h - E_1 - E_2$). $\endgroup$
    – Evgeny Shinder
    Commented Jul 2, 2022 at 21:14
  • $\begingroup$ There are many (probably, infinitely many) elements of square 4. For example, $210 h - 183 E_1 - 103 E_2$ is of squre 4 but the condition 3 doenot hold for it. I would like to find some geometric methods, not investigating diophantine equations. $\endgroup$
    – Basics
    Commented Jul 2, 2022 at 21:28
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    $\begingroup$ I think $S$ has no embedding as a quartic in $\mathbb{P}^3$: we are looking for an integral vector $v = zh - xe_1 - ye_2$ such that $v$ intersects $e_1$, $e_2$, $h - e_1 - e_2$ positively and $v^2 = 4$. From the positivity of the intersection we get $x, y > 0$, $z > x+y$, so that $z^2 - x^2 - y^2 > 2xy$ and from the square $4$ condition $2xy \le z^2 - x^2 - y^2 = 2$, so $xy < 1$ and we don't get any solutions. $\endgroup$
    – Evgeny Shinder
    Commented Jul 3, 2022 at 19:31

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