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Let $X$ be a complex manifold and denote the space of all $(p, q)$-forms on $X$ by $\mathcal{E}^{p,q}(X)$. Forgetting about the complex structure, we can consider the real differential $k$-forms on the underlying smooth manifold; let the space of all such forms be denoted by $\mathcal{E}^k(X)$.

We have the following decomposition:

$$\mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C} = \bigoplus_{p+q=k}\mathcal{E}^{p,q}(X).$$

One thing this decomposition tells us is that it is easy to tell if a $(p, q)$-form is a $k$-form (check whether $p + q = k$). What about the converse?

Given a (complexified) $k$-form on a complex manifold, is there a way to determine if it is a $(p, q)$-form (for a particular $p$ and $q$ with $p + q = k$)?

Of course, this can be done by writing the form locally, but I would prefer a more global approach.


My motivation for asking this question is that there is an easy test to tell whether a complexified $1$-form is a $(1, 0)$-form or a $(0, 1)$-form which I outline below.

Let $J$ be the almost complex structure induced by the complex structure (extended to the complexified tangent bundle). Then there is an induced map $J'$ on the complexified cotangent bundle with $J'\circ J' = -\mathrm{id}$ - explicitly, this map is given by $J'(\alpha)(v) = \alpha(J(v))$. A complexified $1$-form $\beta$ is a $(1, 0)$-form if $J'(\beta) = i\beta$, and similarly $\beta$ is a $(0, 1)$-form if $J'(\beta) = -i\beta$.

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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 $f \star \mu = f^p \bar{f}^q \mu$.

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  • $\begingroup$ Thanks Sam. Seems so obvious now. $\endgroup$ Sep 16, 2012 at 13:12
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    $\begingroup$ @MichaelAlbanese: that's how it always is with these things... impossible before you see it the right way and obvious afterwards. $\endgroup$
    – Sam Lisi
    Sep 17, 2012 at 9:43
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    $\begingroup$ A corollary is that $\mu$ is a $(p,q)$ form iff $i\star\mu = i^{p-q}\mu$, which is easier to use in proofs. $\endgroup$ Jul 13, 2020 at 21:12
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    $\begingroup$ @DanielPlácido: That is not true. For example, if $\mu$ is a $2$-form with $i \ast \mu = -\mu$, then $\mu$ could be a $(2, 0)$-form, a $(0, 2)$-form, or a sum of a $(2, 0)$-form and $(0, 2)$-form. $\endgroup$ Jun 18, 2021 at 19:42

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