ag.algebraic geometry - 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 $Kcong O$ 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 $K_Gcong (detS^*)^{otimes -6}$.

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 $K_Gcong O(-6)$.
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 $6cdot 9=54$. 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 $54-35=19$