I've posted this question already on math.stackexchage. Unfortunately, it did not receive any answers even though there was a bounty on it. So maybe somebody of you could help me. I apologize if this does not fit the scope of this site.

Let $M$ a complex surface and $\omega\in H^0(\Omega_M^2,M)$ a non degenerate holomomorphic form.
I've read somewhere (without proof), that then the first chern class of the symplecitc manifold $(M,Re~ \omega)$ vanishes.

Why is this true? As far as I know, the Chern class of $(M, Re~ \omega)$ is the chern class of any complex vector bundle with almost complex structure compatible with $Re ~\omega$. My first guess was that the original complex structure is compatible with $Re ~\omega$. But this is not true, as then $Re ~\omega$ would be of type $(1,1)$ (with repect to the original almost complex structure).

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    $\begingroup$ Is it not a consequence of the fact that thwe canonical line bundle $K_M$ is trivial (since the symplectic form $\omega$ gives a natural isomorphism between $T_M$ and $\Omega^1_M$)? $\endgroup$ – Francesco Polizzi Feb 21 at 14:31
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    $\begingroup$ @FrancescoPolizzi I am aware that $K_M$ is trivial. I argued that $\omega$ is a section without zeros, so essentially the same way as you did. So its clear that $c_1(M)=-c_1(K_M)=0$, which is the chern class with respect to the almost complex structure on $TM$ we chose such that $K_M$ is trivial. But this is not that I want to calculate. I want the chern class with respect to the symplectic structure $Re~\omega$. $\endgroup$ – MrXYZ Feb 21 at 15:03
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    $\begingroup$ Welcome new contributor. There is a twistor family of pairs $(\omega,J)$ on this manifold parameterized by the $2$-sphere $\mathbb{S}^2$. The Chern classes are discrete invariants in $H^2(M;\mathbb{Z})$. As you rotate in the twistor family, the values of the Chern classes vary continuously (e.g., by Chern-Weil theory). Thus, the values are independent of the parameter in $\mathbb{S}^2$. $\endgroup$ – Jason Starr Feb 22 at 11:37
  • $\begingroup$ @JasonStarr Thank you for your comment. Are you saying that $(..., original cx str)$ and $(Re~\omega,...)$ (where ... stands for the fitting partner), are in the same family and hence their chern classes are equal? Also, what is a twistor family? $\endgroup$ – MrXYZ Feb 22 at 14:00
  • $\begingroup$ @MrXYZ Yes, that is what I am saying. I am not certain what is the best source for twistor families. I believe there is a discussion in the K3 surfaces chapter of the updated version of Barth-Hulek-Peters-van de Ven. $\endgroup$ – Jason Starr Feb 22 at 20:28

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