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May 22, 2015 at 14:27 history edited Francesco Polizzi CC BY-SA 3.0
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May 22, 2015 at 14:26 comment added Francesco Polizzi 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.
May 22, 2015 at 10:29 comment added Creg 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?
May 21, 2015 at 21:34 comment added Daniel Loughran 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.
May 21, 2015 at 18:56 answer added Francesco Polizzi timeline score: 5
May 21, 2015 at 15:52 history asked Creg CC BY-SA 3.0