# About twistor space of a K3 surface

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?

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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 $(3,19)$.
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