5

Usually the first example of a K3 surface presented to us is the Fermat quartic $x_0^4+x_1^4+x_2^4+x_3^4=0$ in $\Bbb{P}_\Bbb{C}^3$.

But I've just found out that actually any smooth quartic in $\Bbb{P}^3$ is K3, and I'm trying to understand why.

I know that since $S\subset\Bbb{P}^3$ is a smooth quartic, then in terms of linear equivalence we have $S\sim 4H$, where $H\subset \Bbb{P}^3$ is a hyperplane. By the adjunction formula: $$K_S=(K_{\Bbb{P} ^3}+S)\big|_S\sim(-4H+4H)\big|_S=0$$

Now, to prove that $h^1(S,\mathcal{O}_S)=0$, I have no idea what to do.

I imagine that the Fermat quartic has nothing special, and we should be able to prove $h^1=0$ for any smooth quartic without additional difficulties. But I really don't know how to do that.

rmdmc89
  • 10,709
  • 3
  • 35
  • 94
  • 3
    Do you know how to compute the cohomology of projective space? Did you consider the cohomology of the short exact sequence $0 \rightarrow \mathcal{O}_X(-4) \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_S \rightarrow 0$ where $X$ is $\mathbb{P}^3$? – user760870 Apr 04 '20 at 15:51
  • 3
    To be a K3 surface, the quartic has to be smooth. – Sasha Apr 04 '20 at 16:12
  • @Sasha, thanks for that! I've just corrected it – rmdmc89 Apr 04 '20 at 17:59
  • @user760870, no. Do you recommend some introductory reference for cohomology? I've looked it up in Shafarevich and Hartshorne, but they made me much more confused. – rmdmc89 Apr 04 '20 at 18:57

2 Answers2

4

First note that we have the short exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^3}(-4) \rightarrow \mathscr{O}_{\mathbb{P}^3} \rightarrow \mathscr{O}_S \rightarrow 0.$$ You get what you want then by considering the long exact sequence of cohomology groups plus the fact that $H^1(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}) = H^2(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}(-4)) = 0$.

Edit:

Since you were asking for some references, I would recommend these lecture notes or their newer (and different) version. In general I can only recommend the two lecture notes full notes (older notes) and very recent full notes by Gathmann. I think everything is well motivated, explained and contains detailed examples etc. He will also explain all the things you need to understand my answer, i.e. the long exact cohomology sequence associated to a short exact sequence and the computation of the cohomology of the twisted sheaves on projective spaces.

Con
  • 9,119
  • 3
    Best proof: very elementary, works in characteristic $p$ and does not assume smoothness ! – Georges Elencwajg Apr 04 '20 at 16:28
  • 3
    @GeorgesElencwajg I am honored to hear that from you as I am usually the one being very fond of your answers on SE. Yes, indeed. Here we do not need these assumptions to compute the irregularity (we still need smoothness for the adjunction formula, i.e for $K_S = 0$ though). – Con Apr 04 '20 at 16:37
  • @TMO, I'm aware of what the $\mathcal{O}{\Bbb{P}^3}$ e $\mathcal{O}_S$ mean, but I don't know how you buit the sequence and why $H^1(\Bbb{P}^3,\mathcal{O}{\Bbb{P}^3})=H^2(\Bbb{P}^3,\mathcal{O}_{\Bbb{P}^3}(-4))=0$ – rmdmc89 Apr 04 '20 at 22:44
  • @TMO actually, I don't even know a readable reference to learn these things. Hartshorne, Shafarevich and Beauville only confused me. If you have a recommendation, I'd be very happy – rmdmc89 Apr 04 '20 at 22:45
  • 2
    @rmdmc89 I edited the answer to include the most readable/understandable references I know where you can learn the things you need. – Con Apr 04 '20 at 23:08
2

Use Hodge theory $$H^1(S,\mathcal{O})=H^{0,1}(S)\subset H^1(S,\mathbb C),$$

together with Lefschetz hyperplane theorem $$H^1(S,\mathbb Z)\cong H^1(\mathbb P^3,\mathbb Z)=0.$$

AG learner
  • 4,846
  • 2
  • 18
  • 39