TITLE: Smooth Quartics in $\Bbb{P}^3$
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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 $-1$.]