On the balanced metric on complex manifold

Question

Let $(M,h)$ be a Hermitian manifold and denote the Kahler form of the Hermitian metric by $\omega$. From the definition of Kahler manifold we know that $M$ is a Kahker manifold if $d\omega=0$. A balanced metric is defined as follows: We say that Hermitian metric $h$ is balanced if $d\omega^{n-1}=0$ where $n$ is the complex dimension of $M$. A proposition is that $d\omega^{n-k}=0$ for any $k>1$ implies that $\omega$ is closed, i.e. $d\omega=0$. So my problem is how to prove it?

Answer 1

I assume you mean $d\omega ^{n-k}=0$ for some $k>1$. This can be written $d\omega \wedge \omega ^{n-k-1}=0$, hence it implies $d\omega \wedge \omega ^{n-3}=0$. Now mutiplication by $\omega ^{n-3}$ induces an isomorphism of vector bundles $\wedge^3T^*_M\cong \wedge^{2n-3}T^*_M$ (just look on the fiber at each point), hence $d\omega \wedge \omega ^{n-3}$ implies $d\omega =0$.