holomorphic sections on elliptic K3 surface - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:00:45Z http://mathoverflow.net/feeds/question/91831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91831/holomorphic-sections-on-elliptic-k3-surface holomorphic sections on elliptic K3 surface Jay 2012-03-21T15:51:21Z 2012-04-04T20:22:01Z <p>Hi all,</p> <p>I want to ask something about the holomorphic sections on elliptic K3:</p> <p>Is there any obstruction for an ellptic K3 (as an elliptic fibration) to have holomorphic sections? Is that always some number as 240? For example, how about E(2) or Fermat's quartic?</p> <p>Thanks a lot! :)</p> http://mathoverflow.net/questions/91831/holomorphic-sections-on-elliptic-k3-surface/91853#91853 Answer by Csar Lozano Huerta for holomorphic sections on elliptic K3 surface Csar Lozano Huerta 2012-03-21T19:46:44Z 2012-03-21T19:46:44Z <p>Hope this helps, it doesn't give a definite answer, but it tells about where to find an obstruction. First off, the existence of multiple fibers in an elliptic fibration is an obstruction to the existence of a differentiable section (over $\mathbb{C}$). On the other hand we can always get rid of the multiple fibers by passing to an \'etale cover.</p> <p>Now a more elaborate answer, given an elliptic fibration with no sections $X\rightarrow B$ we can associate a fibration $J\rightarrow B$ which has a section and a rational map $\phi:J\times_BX\rightarrow B$ that commutes with projections to $B$ and has certain properties. The family $J$ is called <em>jacobian family</em>.</p> <p>The elliptic fibrations are classified by their jacobian fibrations and here comes the change of quantifiers. One can introduce a group structure on the set $I(J)$ of elliptic fibrations with a given jacobian fibration. Hence if the class of the fibration $X\rightarrow B$ in $I(J)$ is not zero, (the obstruction) the fibration $X$ has no differentiable sections. This works the same way the the first Chern class of a line bundle $L$ gives and obstruction for $L$ to be trivial.</p>