It is known that smooth K3 surface can be obtained as two fold branched cover of rational elliptic surface E(1)=CP29¯CP2 along the smooth divisor 2FE(1)=6H−2E1−2E2−⋯−2E9 . My question is if one can see a pencil of genus two curves in K3 with two base points from such description of K3. It seems to me a pencil of lines in E(1) with one base point (obtained via pencil of lines in CP2 with one base point) gives rise to such pencil in K3 since the sphere H branched at 6 points gives a genus two surface in two fold cover. Also, is it possible to see the singular curves in this pencil? It seems to me 6 tangent lines to the cubic 3H−E1−E2−⋯−E9 give rise to the singular curves upstairs, but not very sure. I would appreciate any insight.
1 Answer
You will get such a pencil of genus two curves provided that the base point of the pencil of lines is not one of the nine points that you blew up to get your E(1).
But if the goal was to construct a pencil of genus two curves with two base points, this model for a K3 seems to be too complicated. Why not take a K3 which is the double cover of the plane branched at a smooth sextic? There again if you pull back a pencil of lines in the plane, you will get a pencil of genus two curves with two base points.
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Thanks Tony. No, my goal is not to construct aferomentioned pencil of genus two curves in K3. I know one can take smooth sextic (or six lines in general position) as a branch locus in CP2, blow up one point on seventh line H away from this branch locus, and take two fold cover. I wanted to see the same smooth sextic construction from different angle. Thanks again for your clarification. Jun 11, 2012 at 13:03
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What I have above is actually a genus two fibration structure on K3 blowing up twice. To get the pencil structure, we don't blow up the base point on line H. Jun 11, 2012 at 13:16