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.


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.

  • $\begingroup$ 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. $\endgroup$ – user24328 Jun 11 '12 at 13:03
  • $\begingroup$ 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 $H$. $\endgroup$ – user24328 Jun 11 '12 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.