## K3 surface of genus 8

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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?

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## 3 Answers

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 $54-35=19$

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Many thanks to both of you.  – Guangbo Xu Feb 10 2010 at 16:09

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Guess you are reading Beauville-Donagi :-)

To compute the degree of the Grassmannian you can note that the hyperplane divisor under the Plucker embedding has class $\sigma_1$, so the degree of the Grassmannian is just $\sigma_1^8$, which you can compute by Schubert calculus. See the paragraph about Grassmannians in Griffiths-Harris.

Namely $\sigma_1^2 = \sigma_{1,1} + \sigma_2$ hence $\sigma_1^3 = 2\sigma_{2,1} + \sigma_3$ by Pieri's formula and finally $\sigma_1^4 = \sigma_4 + 3 \sigma_{3,1} + 2 \sigma_{2,2}$ again by Pieri.

Since $\sigma_4$, $\sigma_{3,1}$ and $\sigma_{2,2}$ are Poincaré dual to themselves, you find $\sigma_1^8 = 1 + 3^2 + 2^2 = 14$

The other question has already been answered by Dmitri.

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Yes, Beauville-Donagi:) – Guangbo Xu Feb 10 2010 at 16:09

1) In general, the degree of the Grassmannian $G(k,n)$ in the Plucker embedding is given by $$(k(n-k))!\prod_{i=1}^k\frac{(i-1)!}{(n-k+i-1)!}$$This is calculated by finding $\sigma_1^{k(n-k)}$ using Pieri's rule. See this link for more details.

See Dimitri's answer for question 2).

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