Differential form of type (p,q) - Imported Knowlpedia Content

Definition

Let $M$ be a complex manifold, with the type decomposition $T_{\mathbb C}^*M=T^{*1,0}M\oplus T^{*0,1}M$ . A differential form of type $(p,q)$ is a smooth section of

\Lambda^{p,q}T^*M =\Lambda^pT^{*1,0}M\otimes\Lambda^qT^{*0,1}M \subseteq \Lambda^{p+q}T_{\mathbb C}^*M.

Equivalently, it is locally a sum of wedge products containing exactly $p$ holomorphic covector factors and $q$ antiholomorphic covector factors. Its total degree is $p+q$ ; by convention the space is zero if either index is negative or exceeds the complex dimension of $M$ .

Type decomposition

Every complex-valued $k$ -form decomposes uniquely as

\alpha=\sum_{p+q=k}\alpha^{p,q}, \qquad \alpha^{p,q}\in\Omega^{p,q}(M).

Thus $\Omega^k(M;\mathbb C)=\bigoplus_{p+q=k}\Omega^{p,q}(M)$ . The wedge product respects bidegree: $\Omega^{p,q}\wedge\Omega^{r,s}\subseteq\Omega^{p+r,q+s}$ .

Local coordinates and conjugation

In holomorphic coordinates $z^1,\ldots,z^n$ , a $(p,q)$ -form is a sum of terms

f_{I\bar J}\,dz^{i_1}\wedge\cdots\wedge dz^{i_p} \wedge d\bar z^{j_1}\wedge\cdots\wedge d\bar z^{j_q}.

Complex conjugation exchanges bidegrees: $\overline{\Omega^{p,q}}=\Omega^{q,p}$ . Consequently, a real complex-valued form can be of pure type only when its type is fixed by this exchange or when it is zero.

Relation to complex geometry

The Dolbeault operators split the exterior derivative according to this bidegree. The pointwise decomposition itself only requires an almost-complex structure, but holomorphic coordinates and the two-term decomposition of the exterior derivative require integrability; see Voisin, vol. I, §2.1.

References

C. Voisin, Hodge Theory and Complex Algebraic Geometry I, Cambridge University Press, 2002. DOI record. Relevant: §2.1, forms of type $(p,q)$ .
R. O. Wells Jr., Differential Analysis on Complex Manifolds, 3rd ed., Springer, 2008. DOI record. Relevant: Chapter I, §§2–3.