Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that for $X=(M,I)$ , where $I$ is the complex structure, a K3 surfaces and $\alpha \in H^2(X,\mathbb{R})$ a Kähler class, there exist a Kähler metric g and J,K complex structures such that

1) g is Kähler with respect to I,J and K

2) $\omega_I:=g(I , )$ represents $\alpha$

3) K=IJ=-JI

i know also that for each $\lambda=(a,b,c) \in S^2$, $aI+bJ+cK$ is still a complex structure for which g is Kähler. I write $\omega_\lambda$ for the Kähler form of $(M,\lambda)$ and $\sigma_\lambda$ for the generator of $H^0((M,\lambda),\Omega^2_{(M,\lambda)})$.

this article by huybrechts http://arxiv.org/abs/1106.5573 at page 16 says that the forms $\omega_\lambda$, $Re(\sigma_\lambda)$ and $Im(\sigma_\lambda)$ are contained in the 3-space generated by $\omega_I$, $Re(\sigma_I)$ and $Im(\sigma_I)$. this seems pretty obvious to Huybrechts, but i can't understand why it is..what am i missing?

share|improve this question
add comment

1 Answer 1

I don't read Huybrecht's paper you referred to, but it is known that real and imaginary part of the period $\sigma_\lambda$, and any Kähler form span a positive 3-plane in $H^2(M,\mathbb{R})$; by decomposing $\sigma_\lambda$ into real and imaginary part, it is easy to see that they span a positive 2-plane. This plane is orthogonal to Kähler form because it is in $H^{1,1}$. So these three span a positive 3-plane. On the other hand, such a positive 3-plane is unique because $H^2(M,\mathbb{R})$ has signature $(3,19)$.

share|improve this answer
    
mmm no, i belive that the fact that $H^2(M,\mathbb{R})$ has signature $(3,19)$ doesn't mean that the positive 3-space is unique, else the whole theory of the twistor lines wouldn't have sense, since there would be a unique twistor line $T_W$ associated to the unique positive space $W$ –  ciccio Jan 7 '13 at 15:29
    
Opps! You are right. Let me think for a bit. –  M. K. Jan 7 '13 at 20:18
add comment

Your Answer

 
discard

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.