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I'm reading several books and articles concerning Yau's proof of the Calabi conjecture. I want to have a deep understading of how and why such proof actually works, but most articles are aimed at specialists and give little or no detail to several steps.

For example, in Moroianu's Lectures on Kähler Geometry or Besse's Einstein Manifolds, one starts with a compact connected Kähler manifold $(X^n, \omega)$, where $\omega\in \mathcal{A}^{1, 1}(X)$ is its Kähler form. Then, if $\nu_g$ is the Riemannian measure associated to $X$, one defines $$\mathcal{K}:= \left\{ \varphi\in \mathcal C^\infty(X, \mathbb R): \omega + dd^c\varphi>0, \int_X \varphi \, \mathrm{d}\nu_g=0\right\}. $$ Here, $\omega + dd^c\varphi>0$ means that $\omega + dd^c\varphi$ is a Kähler form. More generally, an $(1, 1)-$form $\alpha= \mathrm{i}\, \alpha_{jk}dz^j\wedge d\bar{z}^k\in \bigwedge^{1, 1}(V^*)$ is said to be positive if for all $\xi\in\mathbb C^n$ one has $ \alpha_{jk}\xi^j \bar\xi^k\geq0 $.

After some considerations, one also defines the set $$\mathcal K':=\left\{ f\in \mathcal C^\infty(X, \mathbb R): \int_X e^f\cdot \omega^{(n)}= \int_X \omega^{(n)} \right\},$$ where $\omega^{(n)}$ is the $n$-fold wedge product of $\omega$. It turns out that proving Calabi's conjecture is equivalent to proving that the map $\mathrm{Cal}\colon \mathcal{K}\to \mathcal{K}'$ is a diffeomorphism, where $$\mathrm{Cal}(\varphi):=\log\left( \frac{(\omega+dd^c\varphi)^{(n)}}{\omega^{(n)}} \right).$$ Actually this is a quite standard procedure which appears in almost every place dealing with this topic.

Now, the notation bugs me a lot, what is it meant by the quotient $\displaystyle\frac{(\omega+dd^c\varphi)^{(n)}}{\omega^{(n)}}$?, and how does it make sense to take the "logarithm" of such thing?

We also know that $\omega^{(n)}= n! \mathrm{vol}$, where $\mathrm{vol}$ is the volume form of $X$

And please, don't be so harsh on me, I'm still an undergraduate student and I'm actually learning all these things on my own. I want an answer as clear and detailed as possible, no matter if it is a trivial thing after all. Thanks in advance!

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    $\begingroup$ $(\omega+dd^c\varphi)^{(n)}$ is an $n$ form and is thus of the form $f\cdot\omega^{(n)}$ for a well defined, smooth function $f$. That would be my guess. $\endgroup$
    – Olivier Bégassat
    Apr 30, 2019 at 4:46
  • $\begingroup$ Yeah, you're right, it makes sense in that way. Let's wait for someone to confirm it just to be sure :-) $\endgroup$
    – EternalBlood
    Apr 30, 2019 at 5:05
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    $\begingroup$ Indeed, and since $\varphi \in \mathcal{K}$, the function Olivier calls $f$ will be positive, so $\log f$ is well-defined. (To be precise, $(\omega+dd^c \varphi)^{(n)}$ is a $(n,n)$-form.) $\endgroup$
    – Richard Lärkäng
    Apr 30, 2019 at 7:21

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