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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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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