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It is known that smooth $K3$ surface can be obtained as two fold branched cover of rational elliptic surface $E(1) = \mathbb{CP}^2 9 \bar{{\mathbb{{CP}^2}}}$ along the smooth divisor $2F_{E(1)} = 6H - 2E_{1} - 2E_{2}- \cdots - 2E_{9}$ . 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 $\mathbb{CP}^2$ 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 $3H - E_{1} - E_{2}- \cdots - E_{9}$ give rise to the singular curves upstairs, but not very sure. I would appreciate any insight.

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up vote 2 down vote accepted

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 $\mathbb{CP}^2$, 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. –  user24328 Jun 11 '12 at 13:03
    
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 $H$. –  user24328 Jun 11 '12 at 13:16
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