How to compute the Picard rank of a K3 surface? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:55:56Z http://mathoverflow.net/feeds/question/107960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107960/how-to-compute-the-picard-rank-of-a-k3-surface How to compute the Picard rank of a K3 surface? 36min 2012-09-24T06:34:19Z 2012-09-24T20:23:57Z <p>I'm curious about the following question:</p> <blockquote> <p>Given a K3 surface, how does one proceed to compute its rank?</p> </blockquote> <p>Of course the answer may depend on the form of the input, i.e. how the K3 is "given". So</p> <blockquote> <p>For a given way of writing down a K3 surface, (e.g. quartics in $\mathbb{P}^3$) <br> How does one compute the Picard rank of the K3 surface?</p> </blockquote> <p>(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.)</p> http://mathoverflow.net/questions/107960/how-to-compute-the-picard-rank-of-a-k3-surface/107963#107963 Answer by Ari Shnidman for How to compute the Picard rank of a K3 surface? Ari Shnidman 2012-09-24T07:52:32Z 2012-09-24T07:52:32Z <p>There are some papers of <a href="http://www.math.leidenuniv.nl/~rvl/papers.html" rel="nofollow">van Luijk</a>, 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.)</p> <p>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 $\mathbb{Q}$. </p> <p>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... </p> http://mathoverflow.net/questions/107960/how-to-compute-the-picard-rank-of-a-k3-surface/107987#107987 Answer by Olivier Benoist for How to compute the Picard rank of a K3 surface? Olivier Benoist 2012-09-24T15:31:18Z 2012-09-24T20:23:57Z <p>In Theorem 6 of the following paper : <a href="http://arxiv.org/abs/1111.4117" rel="nofollow">http://arxiv.org/abs/1111.4117</a>, 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. </p> <p>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 $X$.</p>