I am reading about the scalar curvature $S(\omega)$ of Kahler manifold $(X^n,\omega)$ and they use the following notation:
$$ S(\omega) = n\frac{Ric(\omega)\wedge \omega^{n-1}}{\omega^n}. $$
What does it mean to "divide" two differential forms? I haven't found a definition for this, I would appreciate your help.
On
When working with elements of top-level exterior powers $\alpha$, $\beta$, sometimes people write $$f=\frac{\,\,\alpha\,\,}{\beta}$$ to mean "$f$ is the defined to be the unique scalar such that $\alpha = f\cdot \beta$". Example: $$\det\mathsf{A}\stackrel{\text{def}}{=}\frac{\mathsf{A}\mathbf{e}_1\wedge\mathsf{A}\mathbf{e}_2\wedge\cdots\wedge\mathsf{A}\mathbf{e}_n}{\mathbf{e}_1\wedge\mathbf{e}_2\wedge\cdots\wedge\mathbf{e}_n}$$
Note that since $M$ is orientable, the top exterior bundle $\wedge^{2n}$ is a trivial bundle. Indeed all sections $s$ in this bundle can be written as $s = f \omega^n$ for some function $f$ on $M$. Thus $s/\omega^n$ is by definition just the function $f$.
In particular, assume if we calculate at one point $z = (z^1, \cdots, z^n)$ so that $$\omega = \frac{\sqrt{-1}}{2} \sum_i dz^i\wedge d\bar z^i$$ and $$\operatorname{Ric}_{i\bar j} =\frac{\sqrt{-1}}{2} \sum_i a_i dz^i\wedge d\bar z^i.$$ Then
\begin{equation} \begin{split} \omega^n &= \left(\frac{\sqrt{-1}}{2}\right)^n n! dz^1\wedge d\bar z^1 \wedge \cdots \wedge dz^n \wedge d\bar z^n \\ \operatorname{Ric} \wedge \omega^{n-1}&= \left(\frac{\sqrt{-1}}{2}\right)^n (n-1)! (a_1 + \cdots +a_n) dz^1\wedge d\bar z^1 \wedge \cdots \wedge dz^n \wedge d\bar z^n \end{split} \end{equation}
This imples
$$ \frac{\operatorname{Ric} \wedge \omega^{n-1}}{\omega^n} =\frac{(n-1)! (a_1 + \cdots +a_n)}{n!} = \frac 1n (a_1 + \cdots +a_n) = \frac 1n S(\omega),$$
which is the formula you have.