The d, partial and dbar identities

Write a complex form $\xi ={\xi }_1+i{\xi }_2$ with real forms ${\xi }_1,{\xi }_2$. The coordinate formula defining $d$ on complex coefficients is the complex-linear extension of the real exterior derivative, so $d^2\xi =d^2{\xi }_1+i{d}^2{\xi }_2=0$ by [F2].

The real graded product rule [F3] extends to complex forms: write each complex form as real part plus $i$ times imaginary part, expand the wedge product by complex bilinearity, and apply [F3] to each real pair. The coordinate definition of $d$ in [F1] is complex-linear, so the resulting identity is the same signed rule for complex forms.

For a pure type form $\eta \in {\mathrm{\Omega }}^{p,q}(U)$, [F1] gives $d\eta =\mathrm{\partial }\eta +\bar{\mathrm{\partial }}\eta$ and hence $0=d^2\eta ={\mathrm{\partial }}^2\eta +(\mathrm{\partial }\bar{\mathrm{\partial }}+\bar{\mathrm{\partial }}\mathrm{\partial })\eta +{\bar{\mathrm{\partial }}}^2\eta$ by step 1.1. These three terms have respective bidegrees $(p+2,q)$, $(p+1,q+1)$, and $(p,q+2)$; the direct sum uniqueness in [F1] forces each component to vanish, including when an endpoint component is zero by convention.

Let $\eta \in {\mathrm{\Omega }}^{p,q}(U)$ and $\theta \in {\mathrm{\Omega }}^{r,s}(U)$. Wedge products add the two bidegrees (and vanish if a repeated differential occurs). In the complex graded-derivation identity from step 1.2, the terms involving $\mathrm{\partial }$ have bidegree $(p+r+1,q+s)$ and those involving $\bar{\mathrm{\partial }}$ have bidegree $(p+r,q+s+1)$. Projecting onto these distinct summands yields the two displayed Leibniz identities.

Every smooth complex form is a finite sum of its bidegree components, and both operators and wedge product are additive. Applying step 2.2 componentwise proves the graded Leibniz rules for all homogeneous forms; applying step 2.1 componentwise proves all three square and anticommutation identities for arbitrary forms.