Let $(X, \omega) $ be a $n$-dimensional complex Kähler manifold such that $\omega^{n-1}=d\alpha $.
Does $\partial\alpha^{n-1,n-2} =0$ (resp. $\bar\partial\alpha^{n-2,n-1} =0$) Where $\alpha^{n-1,n-2}$ is the (n-1,n-2)-component of $\alpha $?
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$\begingroup$ If $n = 2$, then $\omega = d\alpha = \partial\alpha^{1,0} + (\bar{\partial}\alpha^{1,0} + \partial\alpha^{0,1}) + \bar{\partial}\alpha^{0,1}$. Given that $\omega$ has type $(1,1)$, we see that $\partial\alpha^{1,0} = 0$ and $\bar{\partial}\alpha^{0,1} = 0$. For $n > 2$, the same argument shows that $\partial\alpha^{n-1, n-2} + \bar{\partial}\alpha^{n, n-3} = 0$ and $\partial\alpha^{n-3, n} + \bar{\partial}\alpha^{n-2, n-1} = 0$. $\endgroup$– Michael AlbaneseCommented Jun 14, 2023 at 13:18
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$\begingroup$ Thank Michael, But i could not find any example such that $\partial \alpha^{n-1,n-2} \neq 0$ $\endgroup$– SamirCommented Jun 14, 2023 at 14:53
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