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I am studying complex structures $I$ on [real] 4-manifolds $M$, and in particular uniqueness properties. Through one example, it seems that this property (written below) guarantees the existence (and uniqueness) of a particular $I$:

There exists $\omega$ (closed 2-form) satisfying $\omega\cdot\omega =0$ and $\omega\cdot\overline{\omega}$ nonzero everywhere.

Is this sufficient to say that there exists a [unique] $I$ such that $\omega$ is a holomorphic 2-form?

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I've been meaning to work out the details of this for some time. It seems to be linked to K3 surfaces - there are semi-vague references to this fact in Buchdahl [springerlink.com/content/h2517445047r421r/] and Huybrechts [math.uni-bonn.de/people/huybrech/HKhabmod.ps]. Sorry I can't do better than that. –  Gunnar Magnusson May 26 '11 at 9:47
    
This is how it's supposed to work: take a 4-manifold $M$ with the topology of a K3 surface, and a 2-form $\omega$ on $M$. Let $T = \Ker \omega \subset T_M$ be a subspace of the tangent bundle. Then $\omega \wedge \omega = 0$ and $\omega \wedge \overline \omega$ ensure that $T$ defines an almost complex structure on $M$ (i.e. $T$ is the space of $(0,1)$ or $(1,0)$ vectors). Then $d \omega = 0$ is supposed to be equivalent to integrability. -- Like I said, there are details to work out, and I haven't had time to do them yet. –  Gunnar Magnusson May 26 '11 at 9:53
    
Oh, yes, and under these conditions $\omega$ will be a holomorphic symplectic form on $(M,T)$ - i.e. a non-degenerate holomorphic 2-form, which then trivializes the canonical bundle of $M$. –  Gunnar Magnusson May 26 '11 at 9:55

1 Answer 1

This seems to be an old remark by Andreotti and appears as exercise 2.6.10 in the book Complex Geometry by D. Huybrechts. The tangent bundle of $M$ is complexified and $\omega$ is viewed as a complex $2$-form. An almost complex structure is equivalent to a decomposition $TM^{\mathbf C}=T^{1,0}M\oplus T^{0,1}M$, where $T^{0,1}M=\overline{T^{1,0}M}$. If $\omega$ is to be holomorphic, then it must vanish on $T^{0,1}M$. This will give uniqueness.

The conditions $\omega\wedge\omega=0$ and $\omega\wedge\bar\omega$ a volume form imply that the kernel of $\omega$ is $2$-dimensional, so can define $\ker\omega=T^{0,1}M$, and this will give existence. In fact, by skew-symmetry the rank of $\omega$ is even; the first condition implies that $\omega$ is not symplectic so the kernel is non-trivial; the second condition implies that it cannot be $4$-dimensional, hence the claim.

Next we let $T^{1,0}M$ be the complex conjugate of $T^{0,1}M$, that is, $\ker\bar\omega$. The condition $\omega\wedge\bar\omega$ never zero says that $T^{1,0}M\cap T^{0,1}M=0$, so we have the desired decomposition.

Finally, to check the integrability, it suffices to see that $T^{0,1}M$ is involutive. Let $X$, $Y$ be sections of $T^{0,1}M$ and $Z$ be an arbitrary section of $TM^{\mathbf C}$. Recall the formula $d\omega(X,Y,Z) = X\omega(Y,Z)-Y\omega(X,Z)+Z\omega(X,Y) - \omega([X,Y],Z)+\omega([X,Z],Y)-\omega([Y,Z],X) $.

We have $d\omega=0$ and all the terms on the RHS vanish by our choice of $X$, $Y\in\ker\omega$ except $d\omega([X,Y],Z)$. Hence this term also vanish and this means that $[X,Y]$ lies in $\ker\omega$.

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