Does the Kähler form $\omega$ satisfy $d^*\omega=0$?

Question

Let $X$ be a compact Kähler manifold with Kähler form $\omega$, then from Kodaira & Spencer's paper on deformations III, 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. $\omega\in \ker\Delta$?

Answer 1

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.

Answer 2

Choose a constant function $f=1$ on the Kähler manifold, using the Kähler identity $[L,\bar\partial^*]=-i\partial$ (see p.120 of Huybrechts' Complex Geometry: An Introduction for example), we have $$[L,\bar\partial^*]1=-i\partial1,$$ by simple computations, we get $$\bar\partial^*\omega=0,$$ so, $\omega\in\text{ker }\Delta=\text{ker }\Delta_{\bar\partial}$.