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Let $M$ be an $n$-dimensional Hermitian manifold. That is, it is a complex manifold equipped with a smoothly varying family of Hermitian metrics $h_x, x\in M$ on the holomorphic tangent spaces. Such a metric is equivalent to giving a smoothly varying family of real alternating forms $\omega_x, x\in M$ of type $(1,1)$ on the tangent spaces viewed as real vector spaces.

The Hermitian metrics $h_x$ also induce Riemannian metrics $g_x$ by taking real parts $g_x = \text{Re}(h_x)$. Thus, $(M,g)$ is a Riemannian manifold, and if it is also given an orientation, then it admits a canonical volume form - ie a positively oriented nowhere vanishing smooth section of $\Omega^{2n}_{M,\mathbb{R}}$ of norm 1 relative to $g$.

Now, the alternating forms $\omega_x$ define a smooth section $\omega$ of $\Omega^2_{M,\mathbb{R}}$. It is often said (for example in Claire Voisin's first book on Hodge theory Lemma 3.8, or in this wikipedia article) that the canonical volume form of the Riemannian manifold $(M,g)$ is given by $\frac{\omega^n}{n!}$.

What does this mean? In the wikipedia article, they say $\omega^n$ is just $\omega$ wedged with itself $n$ times. However, doesn't wedging with yourself result in 0?

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It's true that a 1-form wedged with itself is always zero, but that doesn't hold for general differential forms. For example, consider

$$ \omega = dx \wedge dy + dz \wedge dw $$

on $\mathbb{R}^4$. In this case $\omega \wedge \omega$ is a positive multiple of the standard volume form.

In fact, the imaginary part of a Hermitian form should be a symplectic form, which comes with a nondegeneracy condition that is equivalent to requiring $\omega^n$ to be a volume form.

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