Let $X$ be a compact Kähler manifold with Kähler form $\omega$, then from Kodaira & Spencer's paper on deformations [III](https://www.jstor.org/stable/1969879?casa_token=Lm4RJWHwjh4AAAAA%3Ahj6Mji5mA0uzR5pq4cNuPWCRZOUp24xKCZ50qXtMxR7_N3CaUW0ltkqwtMlvuoSb3oRx_D8wgolZsQPjd0UfEmFPzNk8jTXurtrtgmoObNCQLU-0MFokMg&seq=1), p.75, the authors state that the Kähler form satisfies the Laplace equation $\Delta\omega=0$.


As we know, $\Delta\omega=0\Leftrightarrow d\omega=0,d^\*\omega=0$. By definition, every Kähler form satisfies $d\omega=0$, but it is not obvious why $d^\*\omega$ should also be 0. Actually, from Kodaira & Morrow's book *Complex Manifolds* p.115, they have a proof of $\Delta\omega=0$ by showing $\bar\partial\omega=0$ and $\bar\partial^\*\omega=0$, then it's a result of the famous Kähler identities $\Delta=2\Delta\_{\bar\partial}$, but their proof seems too complicated as it involves covariant differentiation. Does anybody have a simple reason why $d^\*\omega$ should be $0$, i.e. $\omega\in \ker\Delta$?



Does the Kähler form $\omega$ satisfy $d^*\omega=0$?
Recall that $d^\* = -\ast d\ast$ and $\ast\omega = \frac{1}{(n-1)!}\omega^{n-1}$, see Example 1.2.32 of Huybrechts' *Complex Geometry: An Introduction* for example. Therefore


$$d^\*\omega = -\ast d\ast\omega = -\ast d\left(\frac{1}{(n-1)!}\omega^{n-1}\right) = 0$$


as $\omega$ is closed.


