octic K3s inside cubic 4-folds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:20:01Z http://mathoverflow.net/feeds/question/110338 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110338/octic-k3s-inside-cubic-4-folds octic K3s inside cubic 4-folds IMeasy 2012-10-22T15:00:10Z 2012-10-22T15:31:43Z <p>From the Thesis of B.Hassett I seem to understand that a smooth cubic 4-fold \$X\$ containing a \$\mathbb{P}^2\$ should contain also a octic K3, but I cannot see a natural way by which this K3 octic could appear inside \$X\$. How do you see that?</p> <p>One wild guess is: via linkage of a quartic 3-fold that contains a quadric surface contained in \$X\$... but this doesn't seem very consistent...</p> http://mathoverflow.net/questions/110338/octic-k3s-inside-cubic-4-folds/110341#110341 Answer by Yusuf Mustopa for octic K3s inside cubic 4-folds Yusuf Mustopa 2012-10-22T15:31:43Z 2012-10-22T15:31:43Z <p>The \$\mathbb{P}^{2}\$ contained in your cubic fourfold \$X\$ is cut out by linear forms (say) \$L_{1},L_{2},\$ and \$L_{3}.\$ Since the homogeneous ideal of \$X\$ is contained in the homogeneous ideal generated by \$L_{1},L_{2}\$ and \$L_{3},\$ there exist quadrics \$Q_{1},Q_{2},\$ and \$Q_{3}\$ such that \$X={L_{1}Q_{1}+L_{2}Q_{2}+L_{3}Q_{3}=0}.\$ The octic K3 cut out by \$Q_{1},Q_{2}\$ and \$Q_{3}\$ is easily seen to lie in \$X.\$</p>