Let $X$ be a nonsingular surface in $P^3$ of degree $d$. Then the Picard number does not exceed $h^{1,1}\approx 2/3 d^3$, whereas the maximal number attained that I know is that of a Fermat surface, which is $\approx 3d^2$. What is the true maximum? Is there any literature on this subject?
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asked Jan 2, 2014 at 11:34
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3$\begingroup$ Probably you have already read: math.unice.fr/~beauvill/pubs/Picmax.pdf ? $\endgroup$– Benjamin DickmanCommented Jan 2, 2014 at 12:23
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2$\begingroup$ I believe that already for smooth quartic surfaces, Segre found surfaces that beat the Fermat surface. $\endgroup$– Jason StarrCommented Jan 2, 2014 at 13:30
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3$\begingroup$ Just to clarify, are you working in characteristic $0$? In positive characteristic, there are Shioda's unirational surfaces, which I believe are all supersingular. $\endgroup$– Jason StarrCommented Jan 2, 2014 at 13:33
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3$\begingroup$ For quintic surfaces, an example with maximum Picard number $\rho = 45$ can be found at arxiv.org/pdf/0812.3519.pdf $\endgroup$– Francesco PolizziCommented Jan 2, 2014 at 14:29
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5$\begingroup$ @Jason Starr Smooth quartics are K3, and (in characteristic zero) their Picard number attains its maximum of $h^{1,1} = 20$. Segre's quartic has more lines than the Fermat quartic but both have maximal Picard number. [Segre's surface is $T(X,Y) = T(X',Y')$ where $T$ is a homogeneous quartic whose zeros in ${\bf P}^1$ have tetrahedral symmetry, such as $T(X,Y) = X^4 + XY^3$; the surface has $64$ lines, whereas Fermat's $D(X,Y) = D(X',Y')$ has only $48$, with $D$ being a quartic such as $X^4-Y^4$ whose roots have $8$-element dihedral symmetry.] $\endgroup$– Noam D. ElkiesCommented Jun 29, 2014 at 3:00
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