The Geometry and Moduli of K3 Surfaces

Abstract

These notes will give an introduction to the theory of K3 surfaces. We begin with some general results on K3 surfaces, including the construction of their moduli space and some of its properties. We then move on to focus on the theory of polarized K3 surfaces, studying their moduli, degenerations and the compactification problem. This theory is then further enhanced to a discussion of lattice polarized K3 surfaces, which provide a rich source of explicit examples, including a large class of lattice polarizations coming from elliptic fibrations. Finally, we conclude by discussing the ample and Kähler cones of K3 surfaces, and give some of their applications.

Appendix: Lattice Theory

In this appendix we present a short description of the lattice theory that is used in the preceding article. The main reference for this section will be [49].

In this article, we use the word lattice in the following sense.

Definition 12.

A lattice is a pair $(L,\langle \cdot,\cdot \rangle )$ consisting of a finitely generated free $\mathbb{Z}$-module L and an integral symmetric bilinear form $\langle \cdot,\cdot \rangle$ on L.

Often we will suppress the bilinear form $\langle \cdot,\cdot \rangle$ and refer to a lattice simply as L. A lattice L is called non-degenerate if the $\mathbb{R}$-linear extension of the bilinear form $\langle \cdot,\cdot \rangle$ to the $\mathbb{R}$-vector space $L \otimes _{\mathbb{Z}}\mathbb{R}$ is non-degenerate. For the remainder of this appendix, we will assume that all lattices are non-degenerate.

A lattice L has signature (m, n) if, for some basis $u_{1},\ldots,u_{m+n}$ of $L \otimes _{\mathbb{Z}}\mathbb{R}$, we have

$$\displaystyle{\langle u_{i},u_{j}\rangle = \left \{\begin{array}{cl} 1 &\mathrm{if}\ i = j \in \{ 1,\ldots,m\}, \\ - 1&\mathrm{if}\ i = j \in \{ m + 1,\ldots,m + n\}, \\ 0 &\mathrm{if}\ i\neq j. \end{array} \right.}$$

If L is of signature (m, 0) we call it positive definite , and if it has signature (0, n) we say that it is negative definite . If a lattice is neither positive nor negative definite, it is called indefinite . If a lattice has signature (m, 1) we will call it hyperbolic .

Let L be a lattice and u _i a basis of L. Then the Gram matrix of L is the matrix of integers $g_{i,j} =\langle u_{i},u_{j}\rangle$ and the discriminant of L, denoted $\mathop{\mathrm{disc}}\nolimits (L)$, is the absolute value of the determinant of the Gram matrix. Obviously the Gram matrix depends upon the basis chosen, but the discriminant is independent of basis.

A lattice is called even if for every u in L,

$$\displaystyle{\langle u,u\rangle \equiv 0\bmod 2.}$$

For instance, a root lattice of ADE type is a positive definite even lattice. When dealing with K3 surfaces, all relevant lattices are even.

A lattice is called unimodular if it has discriminant 1. Up to isomorphism, there is a single even unimodular rank 2 lattice of signature (1, 1), which has Gram matrix for some basis given by

$$\displaystyle{\left (\begin{array}{*{10}c} 0&1\\ 1 &0 \end{array} \right ).}$$

This lattice is called the hyperbolic plane and, depending on the author, is denoted U or H. We will denote it by H.

Now suppose that L and M are two lattices and that L embeds into M. Then L is said to be a sublattice of M. This embedding is called primitive if the quotient M∕L is torsion-free. Similarly, an element u ∈ M is called primitive if the sublattice of M generated by u is primitively embedded in M.

Given a lattice L, we may define a second lattice L ^∗, called the dual lattice of L, as follows. Consider the tensor product $L \otimes _{\mathbb{Z}}\mathbb{Q}$, with bilinear form induced by the $\mathbb{Q}$-linear extension of $\langle \cdot,\cdot \rangle$. Then define L ^∗ to be the subgroup of $L \otimes _{\mathbb{Z}}\mathbb{Q}$ made up of elements v which satisfy $\langle v,u\rangle \in \mathbb{Z}$ for all u ∈ L, equipped with the integral binear form induced by $\langle \cdot,\cdot \rangle$. Note that L is a sublattice of L ^∗.

For even lattices L, we may use this to define a more refined version of the discriminant, called the discriminant lattice of L. This is given by the finite group

$$\displaystyle{A_{L}:= L^{{\ast}}/L.}$$

This group is equipped with a quadratic form and a bilinear form as follows: take u, v ∈ L ^∗ and let $\overline{u},\overline{v}$ be their images in A _L , then define

$$\displaystyle{q_{L}(\overline{u}) =\langle u,u\rangle \bmod 2\mathbb{Z}}$$

and

$$\displaystyle{b_{L}(\overline{u},\overline{v}) =\langle u,v\rangle \bmod \mathbb{Z}.}$$

Note that if u, v ∈ L, then the fact that L is an even lattice implies that $q_{L}(\overline{u}) = 0$ and $b_{L}(\overline{u},\overline{v}) = 0$, so q _L and b _L are well-defined. The group A _L is finite and $\vert A_{L}\vert =\mathop{ \mathrm{disc}}\nolimits (L)$.

The invariant A _L is obviously finer than just the discriminant of the lattice, but its true strength is made evident by the following proposition of Nikulin.

Proposition 5 ([49, Cor. 1.13.3]).

Let L be an even indefinite lattice of signature (m,n) and rank m + n, with discriminant lattice A _L . Let ℓ(L) denote the minimal number of generators of A _L . If ℓ(L) ≤ m + n − 2, then any other lattice with the same rank, signature and discriminant lattice is isomorphic to L.

1.1 Overlattices

Now assume that L and M are two even lattices of the same rank, such that L embeds inside of M. Then we say that M is an overlattice of L. If we begin with a lattice M, then it is easy to compute all possible sublattices of maximal rank of L, but the problem of computing all possible overlattices of L is more subtle. It is solved by the following theorem:

Theorem 20 ([49, Prop. 1.4.1]).

Let L be an even lattice. Then there is a bijection between subgroups G of A _L on which the form q _L satisfies q _L (u) = 0 for all u ∈ G and overlattices L _G of L.

Furthermore, the discriminant form of the lattice L _G associated to the subgroup G is given by the form q _L restricted to G ^⊥ ∕G, where orthogonality is measured with respect to b _L .

The main practical use of this proposition is to determine when a specific lattice is primitively embedded in another. In particular, if a lattice L has no non-trivial overlattices, then any embedding of L into another lattice M must be primitive. However, it can also be used to explicitly compute the possible overlattices of a given lattice, as illustrated by the next example.

Example 19.

Let L be the lattice H ⊕ (−E ₇) ⊕ (−E ₇) ⊕ (−A ₃). Then A _L is isomorphic to $\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/4$ with generators u, v, w respectively. It may be checked explicitly that

$$\displaystyle{q_{L}(u) = q_{L}(v) = \frac{1} {2}}$$

and

$$\displaystyle{q_{L}(w) = \frac{5} {4},}$$

and that u, v, w are mutually orthogonal with respect to b _L .

One checks easily that the only nontrivial element Q in A _L with q _L (Q) = 0 is Q = u + v + 2w, which has order 2. Thus L has a unique overlattice L _G of index 2, corresponding to the subgroup G of A _L generated by Q.

One may construct L _G concretely in the following way: let $\hat{Q}$ be some element of L ^∗ whose image in A _L is Q, then L _G can be identified as the sublattice of $L \otimes _{\mathbb{Z}}\mathbb{Q}$ spanned by $\hat{Q}$ and the image of L in $L \otimes _{\mathbb{Z}}\mathbb{Q}$.

However, it is often simpler to use Proposition 5 to identify the overlattice L _G . The subgroup G of A _L generated by Q has orthogonal complement generated by Q and v + w. Modulo G, this group is cyclic of order four and

$$\displaystyle{q_{L}(v + w) = -\frac{1} {4}.}$$

Therefore, the overlattice L _G of L associated to G has rank 19, signature (1, 18) and discriminant group of order 4 with a generator satisfying q _L (v + w) = −1∕4.

Now, the lattice $M = H \oplus (-E_{8}) \oplus (-E_{8}) \oplus \langle -4\rangle$ also has rank 19, signature (1, 18) and discriminant group of order 4 with generator e satisfying q _M (e) = −1∕4 so, by Proposition 5, the overlattice L _G of L must be isomorphic to the lattice M.

Remark 15.

Note that if we replaced the lattice H ⊕ (−E ₇) ⊕ (−E ₇) ⊕ (−A ₃) with the lattice (−E ₇) ⊕ (−E ₇) ⊕ (−A ₃) then we could not use Proposition 5 here, since the second lattice is not indefinite.