Let $S$ be a projective K3 surface. Then is there always a smooth projective 3-fold $V$ that has $S$ as its anticanonical section?
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1$\begingroup$ One comes close with the following example: Consider $V = S \times \mathbb{P}^1$. Here each anticanonical divisor is isomorphic to either two copies of $S$ or $S$ doubled. $\endgroup$– Daniel LoughranCommented May 21, 2015 at 21:34
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$\begingroup$ It seems that it is not a easy question. Let me try to simplify the question. How about the case that $S$ has Picard number one? $\endgroup$– CregCommented May 22, 2015 at 10:29
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$\begingroup$ If $S \subset V$ is not ample, Lefschetz Hyperplane Theorem in general fails and so, at least in principle, we have no control on the Picard number of $V$, even if the Picard number of $S$ is $1$. This makes the question quite tricky. $\endgroup$– Francesco PolizziCommented May 22, 2015 at 14:26
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The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.
If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective (with torsion free cokernel).
Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance
Picard group of a K3 surface generated by a curve
Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.
On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).
Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.
Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.
Remark 2. As observed in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.
answered May 21, 2015 at 18:56
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2$\begingroup$ Maybe I'm missing something easy--why does the threefold have to be Fano? Obviously the anticanonical is effective by hypothesis, but why ample? (That said +1, this is a very nice answer.) $\endgroup$ Commented May 21, 2015 at 19:03
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1$\begingroup$ I must still be missing something. First of all, to conclude an equality on restriction to S implies anything on V, you need amplitude of S to use Lefschetz. But that's what we're trying to prove. Second of all, you only show K_V=-S|_S, again, why is S|_S ample at all? $\endgroup$ Commented May 21, 2015 at 19:13
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1$\begingroup$ Sorry, you are right. I misread the question, and I have only proved that $S$ cannot be a hyperplane section of a threefold (in this case the threefold is clearly Fano). $\endgroup$ Commented May 21, 2015 at 19:15
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2$\begingroup$ Can't you just take a 3-fold where $-K_X$ has a smooth section (e.g., is bpf) but not ample and let $S\in |-K_X|$? For example, blow-up $P^3$ along the 16 nodes of a Kummer surface. $\endgroup$ Commented May 21, 2015 at 20:14
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1$\begingroup$ Let $S \subset \mathbb P^3$ be a smooth quartic and let $X \to \mathbb P^3$ be the blow up a bunch of disjoint curves in $S$ (we can find as many as we want, eg if $S$ is elliptic). The strict transform of $S$ on $X$ is anticanonical, but won't be ample if you blow up enough curves. $\endgroup$– user47305Commented May 21, 2015 at 20:16