Let $V$ be a complex vector space of dimension 6 and let $G\subset {\mathbb P}^{14}\simeq {\mathbb P}(\Lambda^2V)$ be the image of the Plucker embedding of the Grassmannian $Gr(2, V)$.


1. Why the degree of $G$ is 14? or in general, how to calculate the degree of a Plucker embedding?


Let ${\mathbb P}^8\simeq L\subset {\mathbb P}(\Lambda^2V)$ be a generic 8-plane and $S$ be the intersection of $L$ with $G$.


1. How to prove that $S$ is a K3 surface?


Another question: the paper said that this construction depends on 19 parameters. I know that this is the dimension of the deformation family of the polarized K3 we get here. But I think that in this statement 19 is coming from varying the generic 8-plane. How can we obtain this number?



K3 surface of genus 8
To be able calculate the degree it is worth to read a bit of Griffiths-Harris about Grassmanians (chapter 1 section 5). To prove that $S$ is $K3$ one needs to caluclate the canonical bundle of $G$, use simple facts about Plucker embedding, use adjunction formula and finally the fact that a simply connected surface with $K\cong O$ is $K3$. 


I will make the second bit of calculation, that proves that $S$ is a $K3$ (so I don't calculate the degree $14$).


First we want to calculate the canonical bundle of $G$. Denote by $E$ the trivial $6$-dimensional bundle over $G$, and by $S$ the universal (tautological) rank $2$ sub-bundle. Then the tangent bundle to $G$ is $TG=S^\* \otimes (E/S)$ . It follows from the properties of $c\_1$ that $c\_1(TG)=6c\_1(S^\*) $. Similarly for the canonical bundle $K\_G$ we have the expression $K\_G\cong (detS^\*)^{\otimes -6}$.


Now we will use the (simple) statement from Griffiths-Harris that under the Plucker embedding we have the isomorphism of the line bundles $det S^\*=O(1)$. Using the previous calculation we see $K\_G\cong O(-6)$. 
Finally, the surface $S$ is an iterated (6 times) hyperplane section of $G$. So by Lefshetz theorem it has same fundamental group as $G$, i.e., it is simply-connected. It suffices now to see that its canonical bundle is trivial. This is done using the adjunction formula $K\_D=K\_X+D|\_D$. Every time we cut $G$ by a hyper-plane we tensor the canonical by $O(1)$, but $O(-6)\otimes O(6)\cong O$.


Added. The calculation of the degree is done by Andrea


Added. The number 19 is obtaied in the following way. The dimension of the grassmanian of $8$-planes in $CP^{14}$ is $6\cdot 9=54$. At the same time the Grassmanian of $2$-planes in $\mathbb C^6$ has symmetires, given by $SL(6,\mathbb C)$, whose dimension is 35. We should quotient by these symmetries and get $54-35=19$


