Constructing the Kummer K3 of an Abelian surface $A$, we have an obvious 22-dimensional collection of classes in $H^2(K3, \mathbb{Z})$ given by the 16 (-2)-curves (which by construction do not intersect each other), and the pushforward-and-pullback of the six classes generating $H^2(A, \mathbb{Z})$. However, this is clearly not all of the classes that I need to find; first of all, the intersection form is wrong---it is certainly not unimodular.

Secondly, there are a few other classes that can be constructed geometrically which are missing---for example, since $\sum_{i=1}^{16} E_i$, the sum of the exceptional divisors, is the branch locus of a 2-1 cover of K3, it must be divisible by two, which my naive description misses.

Ultimately what I am hoping to do is to produce a generating function summing over all effective curve classes in the K3 (with coefficients determined by some GW-invariants), but as stated above, I am missing some classes whose description I do not know. I've been looking through Barth, Peters, Van de Ven, and the best statement I can find is Proposition VIII 3.7:

The set of effective classes on a Kahler K3-surface is the semigroup generated by the nodal classes and the integral points in the closure of the positive cone.

That being said, is there a nice concrete description of these somewhere?


2 Answers 2


The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$.

The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $K$ containing it is computed as follows. Consider a linear combination $F=\frac12\sum a_i E_i$ with $a_i=0,1$. Recall that $E_i$ are labeled by the 2-torsion points of the torus $A$, i.e. the elements of the group $A[2]$.

Then $F$ is in $K$ $\iff$ the function $a:A[2]\to \mathbb F_2$, $i\mapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5. (The element $\frac12\sum E_i$ in your example corresponds to the constant function 1, which is affine linear). Thus, $K$ has index $2^5$ in $16A_1$ and its determinant is $2^{16}/(2^5)^2=2^6$.

$K$ is called the Kummer lattice. By the above, it is a concrete negative-definite lattice of rank 16 with determinant $2^6$. Nikulin proved that a K3 surface is a Kummer surface iff $Pic(X)$ contains $K$.

The orthogonal complement $K^{\perp}$ of $K$ in $L_{K3}$ is $H^2(A,\mathbb Z)$ but with the intersection form multiplied by 2. As a lattice, it is isomorphic to $3U(2)$. It has determinant $2^6$, the same as $K$. The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is recovered from the primitive orthogonal summands $K$ and $K^{\perp}$.

However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r-1)$, not $(3,19)$. For a Kummer surface, it contains Kummer lattice $K$ described above, and its intersection with $K^{\perp}$ is the image of the Picard group of $A$. For a Kummer surface one has $r=17,18,19$ or 20.

For the Mori-Kleiman cone of effective curves, which you would need for Gromov-Witten theory, the description you put in a box is already the best possible.

  • If you want Kummer surfaces completely dissected, look at Kummer's quartic surface, by R.W.H.T. Hudson., specifically chapters XIII, XIV.

  • Sometimes, the easiest thing to do is to project the Kummer surface (which leaves naturally in P3) from one of the nodes to P2. The tangent cone to the node projects to a quadric, the 6 lines through the node project to 6 lines tangent to the quadric - they are the ramification locus of the projection. The 15 other nodes project to the 15 intersection points of the 6 lines. In order for a curve to lift to the Kummer surface it has to intersect with even multiplicity the ramification locus at all intersection points.

  • $\begingroup$ Well, that first source might be useful, but the language it uses is excruciatingly out of date, and unfortunately my library doesn't have a copy of it (though I suppose I can just download the whole book in chunks). Is there perhaps a more modern treatment that you know of? It seems like there should be some further literature on this subject... $\endgroup$
    – Simon Rose
    Feb 17, 2010 at 19:38
  • $\begingroup$ To the best of my knowledge nobody rewrote this stuff in modern language. However, unlike some classical authors, Hudson is actually very readable once you twist your mind around the notation. $\endgroup$ Feb 17, 2010 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.