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 $Re ~\omega$. I would suspect that the original complex structure is of this kind. But doing a local calculation, this seems to be only correct up to a sign.
Update: Now posted on mathoverflow
EDIT: (The calculation)
We need that $Re ~\omega(Ju,Jv)=Re~\omega(u,v)$ for all tangent vectors $u,v$. By linearity of $\omega$ it suffices to check this for a basis of each tangent space. So I check this for $\frac \partial {\partial z_i}$ and $\frac \partial {\partial \bar z_i}$.
As $\omega$ is of type $(2,0)$, it is locally of the form $\sum a_{kl}~dz_k\wedge dz_l$.
Hence the only interesting case to check is
$\omega(\frac \partial {\partial z_i}, \frac \partial {\partial z_j})$.
We have
$$\omega(J \frac \partial {\partial z_i}, J \frac \partial {\partial z_j})
= \omega(i \frac \partial {\partial z_i}, i \frac \partial {\partial z_j})
=-\omega( \frac \partial {\partial z_i}, \frac \partial {\partial z_j}), $$
$$\bar\omega(J \frac \partial {\partial z_i}, J \frac \partial {\partial z_j})= \omega(i \frac \partial {\partial z_i}, i \frac \partial {\partial z_j})=-\bar\omega( \frac \partial {\partial z_i}, \frac \partial {\partial z_j})$$
and hence
$$Re ~\omega(J \frac \partial {\partial z_i}, J \frac \partial {\partial z_j})
=\frac 1 2[\omega(J \frac \partial {\partial z_i}, J \frac \partial {\partial z_j})+\bar \omega(J \frac \partial {\partial z_i}, J \frac \partial {\partial z_j})]
=-\frac 1 2[\omega( \frac \partial {\partial z_i}, \frac \partial {\partial z_j})+\omega( \frac \partial {\partial z_i}, \frac \partial {\partial z_j})]
= -Re~ \omega( \frac \partial {\partial z_i}, \frac \partial {\partial z_j}).$$
Now that I think about it, this just seems to be the proof that Kähler forms are of type $(1,1)$.