MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S\subset\mathbb{P}^g$ be a smooth polarized K3 surface of genus $g$. I am interested in the existence of certain cuspidal curves in the linear system. We know a general hyperplane section $H\cap S$ is smooth, and to have a nodal singularity is a codimension 1 condition (Let's say the dual variety of $S$ is a divisor in $(\mathbb{P}^g)^{\lor}$). However, I am interested in the question of having a hyperplane section of multiple cuspidal singularities. The main case is when $g=8$ and I need a hyperplane of at least $3$ cusps. I could not find a proof or disproof for myself. Any input is very helpful. Thanks!

I guess I should add that I only want the existence for some K3 surfaces.

share|cite|improve this question
[edited only to add the "K3 surfaces" tag – Noam D. Elkies Sep 10 '11 at 19:20
up vote 5 down vote accepted

Xi Chen has a theorem that says that rational curves on K3's in a linear system of dimension $>3$ are nodal. I suppose you don't need this curve to be rational, but his techniques might help you in your quest. At least his theorem tells you that you cannot expect too many cusps.

I couldn't find the paper online, I include the MathSciNet review below.

EDIT/Addendum I just realized that it might be easy to get cuspidal curves if you don't care about the genus of the curve. A degeneration of a family of nodal curves is usually a cuspidal curve. More precise statements in this regard can be found in Kollár's book on rational curves, e.g., Miyaoka's lemma. Also, Kebekus's JAG paper on singular rational curves has techniques of this kind. These are all aimed at studying rational curves on varieties, but I think some of the deformation theoretic techniques could be useful. These will (might) give you one cusp, but then you can try to consider families with $m$ cusps and try to degenrate them to $m+1$ cusps. Of course, there are obstructions to doing thiss, but you might be able to pull it off. end of addendum

MR1675158 (2000d:14057) Chen, Xi(1-UCLA) Rational curves on K3 surfaces. J. Algebraic Geom. 8 (1999), no. 2, 245–278. 14N10 (14J28)

The paper is essentially divided into two parts. The first part proves existence of rational curves in the linear system |OS(d)| on a general K3 surface S in Pn (n≥3 and d>0). The rest of the paper is devoted to the following conjecture: For n>3, all rational curves in the linear system |OS(1)| on a general K3 surface are nodal. The conjecture is proven in the cases n≤9 and n=11 and hence justifies Yau-Zaslow's beautiful counting formula, ∑∞g=1n(g)qg=q/Δ(q) for g≤9 and g=11. A basic ingredient in the proof is the degeneration of the K3 surface to a trigonal K3 surface, that is, a K3 surface with Picard lattice congruent to (2n−2330). The moduli space of trigonal K3 surfaces consists of countably many irreducible components (of dimension 18), and the author considers three of these in order to prove the conjecture in the cases described above. Various reformulations and generalizations of the conjecture are also introduced.

share|cite|improve this answer
I think the paper you cite is majorized by Xi Chen's more recent paper: – Jim Bryan Jun 24 '11 at 5:04
Jim, right. And I actually have seen that paper, but apparently I am getting forgetful. :) – Sándor Kovács Jun 24 '11 at 5:50
Dear S$\acute{a}$ndor, Thank you very much for your input. I am sorry I did not say it clearly that I only need the cuspidal curves on some K3 surface. Chen's results is for a general K3, so I am hoping one could just construct some special K3 explicitly to do the job. When $g=8$, one can always get a 2-cuspidal section passing through two general point of $S$, because for each point, it is 3 conditions on $(\mathbb{P}^8)^{\lor}$ to get a node (i.e $H$ should contain $T_pS$) and one more conditions to get a cusp, so one can always find a 2-cusp. I will try your degeneration argument. – Jie Wang Jun 24 '11 at 14:16

Only now I notice your question. In case you are still interested in the problem, I think that the paper completely answers to your question.

share|cite|improve this answer

I know you mentioned $S\subset\mathbb{P}^g$, but in case this is not a strict requirement, you could consider the hyper-elliptic case, i.e., when $S$ is a branched double cover of a rational surface $Q$ in $\mathbb{P}^g$ and every curve in the linear system is hyper-elliptic. In this case it should be easy to find a curve in $S$ with three cusps: you just need the branch locus of a hyper-elliptic curve to contain three points of multiplicity three, which should be obtainable by choosing the branch locus of $S\rightarrow Q$ appropriately.

In the non-hyper-elliptic case, I would expect curves with three cusps to occur in codimension six in the linear system, the existence of each cusp being a codimension two condition. To find an explicit curve with three cusps, you might try to use Mukai's description of K3 surfaces of genus $g$, $6\leq g\leq 10$ in "Curves, K3 surfaces and Fano 3-folds of genus $\leq 10$". Namely, they are all complete intersections in a certain homogeneous space $G/P$.

share|cite|improve this answer
Thank you very much. I think the example you give would work. Could you please say a few words about why it is a codimension six locus to have 3 cusps? Jie – Jie Wang Sep 11 '11 at 14:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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