Complexifying the cotangent bundle of an <almost complex manifold> gives the $\pm i$ eigenspace decomposition $T^*X\otimes\mathbb C=T^{1,0*}X\oplus T^{0,1*}X$. Define
$$
\bigwedge^{p,q}T^*X=\bigwedge^pT^{1,0*}X\otimes\bigwedge^qT^{0,1*}X.
$$
For a form $\alpha$ of type $(p,q)$, the <Dolbeault operator> is the type-$(p,q+1)$ projection of $d\alpha$:
$$
\bar\partial\alpha=\operatorname{pr}_{p,q+1}(d\alpha).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 118 / 3 / a / Solution

{title2=$\bar\partial$}
{wiki=Dolbeault_operator}

The complexified cotangent bundle decomposes into $(1,0)$ and $(0,1)$ parts. The Dolbeault operator is the type-$(0,1)$ component of the exterior derivative, $\bar\partial:\Omega^{p,q}\to\Omega^{p,q+1}$, and integrability gives $\bar\partial^2=0$.


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