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May 16, 2022 at 12:16 vote accept Elio Li
Apr 4, 2022 at 9:45 answer added Ben McKay timeline score: 3
Mar 30, 2022 at 10:22 comment added Ben McKay yes, that's right.
Mar 30, 2022 at 10:15 comment added Elio Li Thanks for answering, I just started learning Kahler geometry while reading this paper. You mean that $\Omega+d \bar{\partial} \varphi$ = $\Omega+\partial \bar{\partial} \varphi$ then integraing $\Omega+\partial \bar{\partial} \varphi$ equals to integrating $\Omega$(Since $M$ has no boundary and Stokes), then similarly expand out $(\Omega+\partial \bar{\partial} \varphi)^{m}$ and integrate by parts?
Mar 29, 2022 at 19:21 history edited Daniele Tampieri CC BY-SA 4.0
Added some detail to reference paper
Mar 29, 2022 at 18:28 comment added Ben McKay I think you expand out and integrate by parts to show that the integral is the same as for $\Omega$. In fact, the homology class of $\Omega+d\bar\partial \varphi$ is clearly that of $\Omega$.
Mar 29, 2022 at 15:38 history asked Elio Li CC BY-SA 4.0