A closed form in kaehlerian geometry

Question

Let $(M,g,J)$ be a Kaehler manifold and $Ric$ is the Ricci curvature of $g$. Then I define:

$$\alpha (X,Y)=g((JRic+RicJ)X,Y)$$

We have $\alpha \in \Lambda^2 (TM)$. Moreover we have $d\alpha=0$, because $\nabla J=0$ and we take $\nabla Ric=0$. So, we obtain a closed form and a class of cohomology $\dot{\alpha} \in H^2 (M,{\bf R})$.

Is $\dot{\alpha}$ an invariant of the manifold $M$?