Let $\omega=\sum_{j=1}^ndx_j\wedge dy_j$ be the standard Kähler form on $\mathbb{C}^n$.
I'm trying to prove that $*\frac{\omega^k}{k!}=\frac{\omega^{n-k}}{(n-k)!}$, where $*$ is the Hodge star operator.
I've read the following argument here (beginning of page 5):
"We see that $\left|\frac{\omega^k}{k!}\right|^2=\frac{n!}{k!(n-k)!}$ and $\text{vol}_{\mathbb{C}^n}=\frac{\omega^n}{n!}$, so $*\frac{\omega^k}{k!}=\frac{\omega^{n-k}}{(n-k)!}$."
The argument is clear once I accept that $\left|\frac{\omega^k}{k!}\right|^2=\frac{n!}{k!(n-k)!}$, but I don't know how to prove it.
Sorry if this is trivial, I'm still at level zero in Hodge theory. Thank you!
This is just algebraic manipulations.
On $\mathbb{C}^n$, the standard Kahler form (with the usual $z_j=x_j+iy_j$ identification of $\mathbb{C}^n\cong\mathbb{R}^{2n}$) is $$\omega=\sum_{j=1}^n \frac{i}2 \mathrm{d}z_j\wedge\mathrm{d}\bar{z}_j=\sum_{j=1}^n\mathrm{d}x_j\wedge\mathrm{d}y_j$$ Note that $\mathrm{d}x_j,\mathrm{d}y_j$ forms an orthonormal basis of covectors at every point, so $\omega$ is the sum of $n$ orthonormal basis elements of $\bigwedge^2\mathbb{C}^n$. Taking the $k$-th (wedge) power gives $$ \omega^k=\sum_{1\leq j_1<j_2<\dots<j_k\leq n} k!\bigwedge_{\ell=1}^{k}(\mathrm{d}x_{j_\ell}\wedge\mathrm{d}y_{j_\ell}). $$ So $$ \frac{\omega^k}{k!}=\sum_{\substack{\{j_1,\dots,j_k\}\in [n]^{(k)}\\j_1<j_2<\dots<j_k}} \mathrm{d}x_{j_1}\wedge\mathrm{d}y_{j_1}\wedge\dots\wedge\mathrm{d}x_{j_k}\wedge\mathrm{d}y_{j_k} $$ is the sum of $\binom{n}{k}$ orthonormal basis elements of $\bigwedge^{2k}\mathbb{C}^n$. So $$\left\lvert\frac{\omega^k}{k!}\right\rvert^2=\binom{n}{k}.$$