If $\alpha$ is a bounded form in a complex complete manifold $X$ (i.e $\sup_X|\alpha (x) |<\infty$, then $d\alpha$ is it also bounded? Rq: if $d\alpha$ is bounded then \alpha is not necessary bounded, take for exemple $dx_1\wedge dx_2...\wedge dx_{2n} =d\Gamma$ on $\mathbb{C}^n.$
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$\begingroup$ You are assuming that $X$ has a complete Riemannian metric? Or Finsler? Hermitian? Kaehler? $\endgroup$– Ben McKayCommented Jun 14, 2020 at 13:50
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$\begingroup$ X is a hermitian manifold $\endgroup$– MarouaniCommented Jun 14, 2020 at 13:53
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On $\mathbb{C}$, try $\omega=\sin(|z|^2) \, dz$.
answered Jun 14, 2020 at 13:55