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Algebraic category. Ground field $\Bbb{C}$.

This is a naive question: are all smooth quartic surfaces in $\Bbb{P}^3$ isomorphic ?

The answer is NO if and only if there is a smooth quartic in $\Bbb{P}^3$ containing some (-1)-curve.

Heitor Fontana
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1 Answers1

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A smooth quartic in $\mathbb P^3$ is a K3 surface, and they are typically non-isomorphic. If you quotient out by the obvious projective symmeteries, you get a 19-dimensional family, and the K3's honestly depend on these 19 moduli. (There is a Torelli theorem to this effect, I think.)

[But, by adjunction, there are no -1 curves on a K3. I don't really understand your (-1)-curve remark. Details: Adjunction says that $2g - 2 = C \cdot C - C\cdot K,$ but on a K3 we have $K = 0$, so $2g-2 = C \cdot C.$ The left hands side is even, and so a curve on K3 cannot have self-intersection $-1$.]

Matt E
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  • I thought that by Castelnuovo's theorem absence of -1 curves is equivalent to $S$ being minimal; but minimal (non ruled) surfaces are all isomorphic, is this wrong? maybe I implicitly and incorrectly assumed that smooth quartics are all birational? – Heitor Fontana May 22 '13 at 22:55
  • @Heitor: Dear Heitor, Smooth quartics are not at all all birational; why would you think so? (I have added the adjunction calculation to my answer.) Regards, – Matt E May 23 '13 at 02:53
  • Dear Matt. This is what I don't understand: I learned that a surface is minimal (i.e. each birational map from it is an iso) if and only if it does not contain -1 curves.... so I am a bit puzzled: K3 are all minimal? not possible because for non ruled surface all minimal models are isomorphic (Enriques). Where am I wrong?? Help (btw I only studied smooth surfaces) – Heitor Fontana May 23 '13 at 07:57
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    @Heitor: the things you are saying are true, but you are drawing the wrong conclusion from them. The more careful version of the uniqueness statement you make is the following: if $X$ is a surface, not ruled, then any two minimal models of $X$ are isomorphic. (By the way, in the case at hand, this doesn't give any new information: a smooth $K3$ surface is already minimal, so it is its own minimal model.) –  May 23 '13 at 14:34
  • got it now! Thank you @AsalBeagDubh. My mistake was then the assumption that K3 surfaces are birationally equivalent! – Heitor Fontana May 23 '13 at 14:46
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    @Heitor: Dear Heitor, That's exactly my point. K3 surfaces come in 19 dim'l moduli spaces. The space of quartic surfaces mod projective equivalence is one of these moduli space. So basically, two smooth plane quartics are not isomorphic (equivalently, not birationally equivalent) unless they're projectively equivalent. (Actually, this may not quite be true, I'm not sure --- the point is that really one is getting a moduli space of polarized K3's, and I'm wrongly ignoring the polariation in the above formulation of things. But I think it's at least morally correct, and should be ... – Matt E May 24 '13 at 02:01
  • ... true for a typical quartic surface, which will admit a unique degree 4 polarization). Regards, – Matt E May 24 '13 at 02:01