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

I would like to find a gap in the following observation. I found a suspicious part but cannot prove it wrong. I would appreciate your assistance.

Let $M$ be a lattice of signature $(1,t)$ and $S$ be an $M$-polarized K3 surface, i.e. $NS(S)\cong M$. Choose a general Kahler class $\omega \in M\otimes \mathbb{R}$ in the sense that $\omega^\perp \cap M=0$. Also choose a general complex structure $\Omega_I \in M^\perp\otimes \mathbb{C}$, where we think $\Omega_I$ as a nowhere vanishing 2-form. Here is my thought;

If the complex structure $\Omega_I$ is generic in the moduli space of $M$-polarized K3 surfaces, $D.\Im \Omega_I\ne0$ for any $D\in M^\perp$, where $\Im \Omega_I$ is the imaginary part of $\Omega_I$.

By a hyperKahler rotation, we get anotehr complex structure $\Omega_K:=\Im \Omega_I+i\omega$. I would like to find an algebraic curve on $S$ with this new complex structure $\Omega_K$, (denote it $S_K$ below).

Let $C$ be an algebraic curve on $S_K$. Then it must be orthogonal to $\Omega_K$; $$ 0=C.\Omega_K=C.\Im \Omega_I+iC.\omega $$ Since $\omega$ is generic, $C.\omega$ implies $C \in M^\perp$, which then implies $C.\Im \Omega_I\ne0$. Contradiction. We therefore conclude that there is no algebraic curve on $S_K$. This is not true in general.

I suspect that the shadowed part is somehow misleading in the argument above. Could anyone kindly point out my mistake?

share|cite|improve this question

The error is that one cannot take $\Omega_I$ with $D.\Im\Omega\ne0$ for $\forall D\in M^\perp\setminus 0$. This is because $\Omega_I$ lies in the period domain; $$ \Im\Omega^2=\Re\Omega^2>0, \ \ \Im\Omega.\Re\Omega=0. $$

share|cite|improve this answer

I admit that this seems a little surprising, but I don't see the contradiction. You have made genericity assumptions about everything except for the existence of a lattice polarization. A generic K3 has no algebraic curve, and even the generic member of the twistor family of complex structures associated to a given kahler K3 surface doesn't contain an algebraic curve (or else we would have that all moduli spaces of polarized K3 surfaces are uniruled, which I'm pretty sure is false).

I don't see a problem with your argument either ...

share|cite|improve this answer

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.