When is a $k$-form a $(p, q)$-form?

Solution 1:

This is secretly a linear algebra question. First, let me set up some notation. Let $\mu$ be a complex-valued $k$-form on a real vector space $E$, but where $E$ is equipped with a complex structure $J$. (Worded differently: $E$ is a complex vector space, but we are seeing it as a vector space over $\mathbb{R}$ and then seeing the multiplication by $i$ as an endomorphism.) Then, the complex numbers act as linear maps on $E$ by saying that $(a+ib)\star v = a v + bJv$.

Now, $\mathbb{C}$ also acts on $k$ forms in the following way: Let $z$ be a complex number. Then I define $z\star \mu = \mu(z \cdot, z \cdot, \dots, z\cdot)$. I.e. $z \star \mu$ eats $k$-vectors, multiplies each one by $z$ and then feeds that to $\mu$.

My claim is that the $(p,q)$ forms are precisely those for which $z \star \mu = z^p \bar{z}^q \mu$.

The proof of this claim follows since $\mathcal{E}^{p,q}$ is the span of the wedges of $p$ many $(1,0)$ forms with $q$ many $(0,1)$ forms. This property I described is clearly linear and holds true for the $1$-forms by the observation you made.

Then, on your complex manifold, you replace this property for all $z \in \mathbb{C}$ by looking instead at all functions $f \colon X \to \mathbb{C}$, and asking that $f \star \mu = f^p \bar{f}^q \mu$.