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Nov 16, 2013 at 18:57 vote accept Serge Lvovski
Nov 16, 2013 at 9:32 comment added Francesco Polizzi Well, regarding my precedent comment, since all the singularities of Naruki's quartic $S$ are Rational Double Points, by the simultaneous resolution of RDP it follows that the desingularization of $S$ is diffeomorphic to a smooth quartic, i.e. a $K3$ surface. So I think that $S$ itself is simply connected.
Nov 16, 2013 at 8:30 history edited Francesco Polizzi CC BY-SA 3.0
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Nov 16, 2013 at 7:50 comment added Francesco Polizzi Nice observation. Naruki's quartic is obtained by taking the double cover of $\mathbb{P}^2$ branched over a configuration of three irreducible conics in special position. On the other hand, Kummer's branch locus is given by six distinct lines, all tangent to the same conic. It does not seem to me that this can degenerate to Naruki's branch, but perhaps there is some degeneration between the two surfaces not coming from the double cover construction. Maybe, in order to have more insight, one should compute some topological invariant of Naruki's quartic (for instance, the fundamental group).
Nov 16, 2013 at 7:36 comment added Sasha As $3\cdot 3 + 7 = 16$ one could imagine this surface is a degeneration of Kummer surfaces. Is it really the case?
Nov 16, 2013 at 7:35 history edited Francesco Polizzi CC BY-SA 3.0
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Nov 16, 2013 at 7:14 history answered Francesco Polizzi CC BY-SA 3.0