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[This question arises from a look at the paper

My problem arises from (4.1)

It said that itegrating $(\Omega+\partial \bar{\partial} \varphi)^{m}=(\exp \{F\}) \Omega^{m}(4.1)$ then we get $\int \exp \{F\}=\operatorname{Vol}(M)$ where $\Omega$ is the kahler form.

Does this mean $(\Omega+\partial \bar{\partial} \varphi)^{m}$ is also a volume form? I'm confused this step of integrating (4.1).

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    $\begingroup$ 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 $\Omega$. $\endgroup$
    – Ben McKay
    Commented Mar 29, 2022 at 18:28
  • $\begingroup$ 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? $\endgroup$
    – Elio Li
    Commented Mar 30, 2022 at 10:15
  • $\begingroup$ yes, that's right. $\endgroup$
    – Ben McKay
    Commented Mar 30, 2022 at 10:22

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Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold. On the other hand, since the Monge-Ampere equation is elliptic, scalar, determined, it is locally solvable, so $\Omega+d\bar\partial\varphi$ can achieve any multiple of any given volume form, locally, by suitable local choice of $\varphi$. So we cannot guaranteed that $\Omega+d\bar\partial\varphi$ is not zero somewhere, if we allow arbitrary choice of $\varphi$. So we can't be sure that this $(\Omega+\partial\bar\partial\varphi)^n$ is actually a volume form, i.e. a nowhere-zero top-degree form with positive integral. That requires more information.

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