Skip to main content
2 of 4
Minor revision

For your first question, note that (let $\omega$ be the Hermitian form)

$$\int_M\Delta_c(f) \omega^n=\int_M(\mathrm{tr}_\omega \sqrt{-1}\partial \overline{\partial}f)\omega^n=n\int_M \sqrt{-1}\partial \overline{\partial}f \wedge \omega^{n-1}$$

Using Stoke's formula, we deduce that

$$\int_M \partial \overline{\partial}f \wedge \omega^{n-1}=\int_M \overline{\partial}f \wedge \mathrm{d}\omega^{n-1}=\int_M \overline{\partial}f \wedge \partial \omega^{n-1}=\int_M f \wedge \partial \overline{\partial} \omega^{n-1}$$

Then $$ \int_M\Delta_c(f) \omega^n=n \int_M f \wedge \sqrt{-1}\partial \overline{\partial} \omega^{n-1} $$

And $\int_M\Delta_c(f) \omega^n=0$ for all $f$ iff $\partial \overline{\partial} \omega^{n-1}=0$.