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I'm curious about the following question:

Given a K3 surface, how does one proceed to compute its rank?

Of course the answer may depend on the form of the input, i.e. how the K3 is "given". So

For a given way of writing down a K3 surface, (e.g. quartics in $\mathbb{P}^3$)
How does one compute the Picard rank of the K3 surface?

(Aside: What I've seen people sometimes did is avoiding this question by nailing down a K3 surface $X$ with its $NS(X)$ together with the intersection form. Then find an embedding given by the ample class.)

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  • $\begingroup$ Somewhat similar: mathoverflow.net/questions/26438/… Are there any sofware packages for computing Picard numbers? $\endgroup$
    – joro
    Sep 24, 2012 at 8:05
  • $\begingroup$ Picard lattices of families of K3 surfaces by Belcastro, xxx.lanl.gov/abs/math/9809008, may be of interest. $\endgroup$
    – Balazs
    Sep 24, 2012 at 11:47
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    $\begingroup$ You might also try looking at some of the papers of Matthias Schütt. $\endgroup$
    – user5117
    Sep 24, 2012 at 16:04
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    $\begingroup$ Find for example a symplectic action of your K3 surface. Then you will have an lower bound of the Picard lattice. Check for example K. Hashimoto's paper, where he classified invariant lattices of such K3 surfaces. Likewise you can use non-symplectic action in some cases. $\endgroup$
    – Atsushi Kanazawa
    Sep 24, 2012 at 18:12

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There are some papers of van Luijk, where he computes the ranks of some K3s over number fields. The trick is to note that $NS(X) \hookrightarrow NS(X_p)$, where $X_p$ is the reduction of $X$ modulo a prime ideal $p$. One can determine the rank of $NS(X_p)$ by counting eigenvalues of Frobenius which differ from $q$ (the size of the residue field) by a root of unity. If you want to find rank 1 K3s, you can reduce modulo two different primes and hope to find rank 2 reductions which have lattices which are incompatible in some sense, forcing $NS(X)$ to be rank 1. (The issue here is that the rank of $NS(X_p)$ will always be even, so you can't win by using a single prime.)

I'm not sure how this works when you want to find K3s of larger rank though, unless you had a way of exhibiting linearly independent divisor classes. Anyhow, van Luijk uses this technique to find rank 1 quartics in $\mathbb{P}^3$ and I think others have done the same with genus 2 K3s defined over $\mathbb{Q}$.

I should add that the situation is much easier for Kummer surfaces. If I'm not mistaken, the rank of $X = K(A)$ ($A$ is an abelian surface) is 16 plus the Picard rank of $A$. The 16 comes from the 16 exceptional divisors you get when you blow up $A$ at its 2-torsion points. The rank of $A$ is usually not hard to figure out: a generic $A$ has rank 1, if $A$ is a product of elliptic curves then its rank is 2,3 or 4 depending on whether the curves are isogenous and whether they have CM or not, and there are a few other cases which one can probably figure out...

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    $\begingroup$ In fact, there are tricks to decide rank $1$ by reducing modulo one single prime only: arxiv.org/abs/1006.1972 - check out also the other papers of these authors circling around Picard groups of K3s $\endgroup$
    – Christian Liedtke
    Sep 25, 2012 at 10:30
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In Theorem 6 of the following paper : http://arxiv.org/abs/1111.4117, building on Van Lujik's work, François Charles explains a (theoretical) algorithm that computes the rank of a K3 surface $X$ defined over a number field. This algorithm terminates conjecturally, for instance if $X\times X$ satisfies the Hodge conjecture.

The main new feature of this article, that allows him to obtain an algorithm, is that the discrepancy between the rank of $X$ and the rank of the reduction of $X$ at a typical prime may be read off the algebra of endomorphisms of the transcendental lattice of $X$.

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    $\begingroup$ for this algorithm, we first have to find divisors on the K3 and codimension 2 cylces on the self-product of the K3 by 'going through Hilbert schemes of a suitable projective space' in order to get a lower bound on the rank. I am not an expert, but I'd be interested in how complicated this is in practice, especially, if the rank is large. $\endgroup$
    – Christian Liedtke
    Sep 25, 2012 at 10:36

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