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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?

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2 Answers 2

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; $$ \Im\Omega^2=\Re\Omega^2>0, \ \ \Im\Omega.\Re\Omega=0. $$

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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 ...

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