3

I'm reading Huybrechts' Lectures on K3 Surfaces and I got stuck reading example 2.3.9, which shows that

any K3 surface $X$ with $\operatorname{Pic}(X)=\mathbb{Z}\cdot L$ and such that $(L)^2=4$ can be realized and a quartic $X \subset \mathbb{P}^3$.

His argument starts as follows:

We may assume that $L$ is ample (after passing to its dual if necessary).

Why is this true?

There's yet another fact that I dont' understand:

All curves in |L| are automatically irreducible, as $L$ generates $\operatorname{Pic}(X)$.

Hints are appreciated.

WindUpBird
  • 417

1 Answers1

7

Since the intersection form is quadratic, if we put $M = -L$, we also have $M \cdot M = 4$. Now since the Picard group is one-dimensional, and the variety is projective, either $L$ or $M$ has to be ample (because some bundle out there is very ample, and it's either a positive tensor power of $L$ or of $M$). Therefore, we can simply replace $L$ with $M$ if $L$ turns out not to be ample -- all the hypotheses of the theorem remain true.

For the second question, now that we're assuming $L$ is positive, if we take a curve in its linear system and it decomposes as $C + D$, then we can write $C = mL$ and $D = nL$ for positive integers $m$ and $n$, which is absurd.

hunter
  • 32,629
  • Can I use a similar argument if $X$ is just a complex K3 surface (I mean, not necessarily projective)? – WindUpBird Apr 04 '24 at 20:29
  • 1
    @WindUpBird I don't know! Now that I read the section in Huybrechts, I'm guessing that was your original question. Huybrechts points us the note at the bottom of page 16, which doesn't seem to help, except to point out that we still have an isomorphism of Pic(X) = Num(X) (because the self-intersection isn't 0). In chapter 8, Huybrechts asserts that the Nakai-Moizheson criterion applies even in the non-algebraic case (p. 144, remark 1.3), which would certainly be enough to give either $L$ or $-L$ ample (and hence $X$ algebraic all along), you would have to track his reference for this. – hunter Apr 04 '24 at 21:20
  • 1
    Ah, see also footnote 9 on page 17 (I suspect this is where he meant to point when he said "see Page 16," since there's another pointer to "page 16" in chapter 8 that seems to be pointing to this footnote). If there is a positive self-intersection line bundle, the surface is automatically projective. – hunter Apr 04 '24 at 21:22
  • I had totally missed the (very helpful) footnote you mentioned. Thanks a lot! – WindUpBird Apr 05 '24 at 08:47