Volume form of a Kähler manifold Let $(X,\omega)$ be a Kähler manifold with $\dim_{\Bbb{C}}X=n$, that is, $X$ is a $n$-dimensional complex manifold, and $h$ is a positive Hermitian metric on $TX$, and $\omega$ is the real $(1,1)$-form corresponding to $h$, such that $\mathrm{d}\omega=0$.
I want to show $\frac{\omega^n}{n!}$ is the volume form of $X$ seen as a Riemannian manifold, with Riemannian metric $\operatorname{Re}h$.
Maybe it's a quite basic problem, but as a beginner I don't know how to deal with it. Thanks in advance for any help!
 A: This is really a pointwise computation, the Kähler condition does not really come into play. Let's begin by fixing a point $p\in X$ and a frame for the (real) tangent space $T_pX$, orthonormal with respect to the Riemannian metric $g$ defined by $h$ on $X$, call it $x_1,y_1,\dots,x_n,y_n$. Denote by $\chi^1,\upsilon^1,\dots,\chi^n,\upsilon^n$ its coframe, and let $$z_a:=\frac{1}{2}\left(x_a-\mathrm{i}\,y_a\right),\quad\zeta^a:=\chi^a+\mathrm{i}\upsilon^a.$$ Then, using the orthonormality of the base we get $$g_p=\frac{1}{2}\sum_a\zeta^a\bar{\zeta}^a\mbox{ and }\omega_p=\frac{\mathrm{i}}{2}\sum_a\zeta^a\wedge\bar{\zeta}^a.$$
The Riemannian volume form is just the wedge product of all the elements of the coframe (see for example this question), $$\mathrm{vol}_p=\chi^1\wedge\upsilon^1\wedge\dots\wedge\chi^n\wedge\upsilon^n,$$ and we should compare this with $\omega^n$. The direct computation gives $$\omega^n_p=n!\frac{i^n}{2^n}\zeta^1\wedge\bar{\zeta}^1\wedge\dots\wedge\zeta^n\wedge\bar{\zeta}^n$$ and by definition $$\zeta^a\wedge\bar{\zeta}^a=-2\mathrm{i}\,\chi^a\wedge\upsilon^a$$ so we get $\omega^n_p=n!\mathrm{vol}_p$, as requested.
