Is it true that for any $d \ge 4$, there exists a smooth, degree $d$ surface $X$ in $\mathbb{P}^3$ with maximal Picard rank i.e., Picard rank of $X$ equals $h^{1,1}(X)$?
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2$\begingroup$ mathoverflow.net/questions/153375/… this seems to be an open problem $\endgroup$– ssxCommented Jul 24, 2019 at 19:21
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$\begingroup$ The "restating" doesn't make sense, $H^2(X,\mathbb{Z})$ is a lattice and $H^2(X,\mathcal{O}_X)$ a vector space. As for the first question, it might be true, but it is known only for $d=4$ or $6$ -- in both cases the Fermat surface does the job. You might have a look at this paper. $\endgroup$– abxCommented Jul 24, 2019 at 19:21
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$\begingroup$ @abx: Thanks. I have corrected the mistake in the question. $\endgroup$– user45397Commented Jul 24, 2019 at 19:29
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$\begingroup$ @abx and $d=5$ arxiv.org/pdf/1308.2525.pdf $\endgroup$– ssxCommented Jul 24, 2019 at 20:29
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2$\begingroup$ @Mere Scribe: The OP is asking for smooth surfaces. Schütt examples have rational singularities. $\endgroup$– abxCommented Jul 25, 2019 at 4:39
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