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

Complexifying the cotangent bundle of an almost complex manifold gives the $\pm i$ eigenspace decomposition $T^{\ast }X\otimes \mathbb{C}=T^{1,0\ast }X\oplus T^{0,1\ast }X$. DefineFor a form $\alpha$ of type $(p,q)$, the Dolbeault operator is the type-$(p,q+1)$ projection of $d\alpha$:

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For $(0,1)$ vector fields $V,W$, direct evaluation givesup to the harmless overall sign fixed by the exterior-derivative convention. If $J$ is integrable, $(0,1)$ fields are closed under bracket, so the right side vanishes. Conversely, if it vanishes for every smooth complex function $f$, differentials separate tangent vectors and therefore $[V,W{]}^{1,0}=0$. Involutivity of $T^{0,1}X$, equivalently vanishing of the Nijenhuis tensor, is the Newlander-Nirenberg criterion for an integrable almost complex structure. Thus $J$ is integrable exactly when ${\bar{\partial }}^{2}f=0$ for every $f$.

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For a scalar form $\alpha \in {\Omega }^{p,q}(X)$ and a section $s$ of $E$, define the Dolbeault partial connection on decomposable forms byand extend linearly. This is independent of the chosen local expression precisely because the original operator obeys its Leibniz rule.

Applying the rule twice makes the two mixed terms cancel. Since $X$ is complex, ${\bar{\partial }}^{2}f=0$, and for every smooth function $f$ and $E$-valued form $\eta$ one obtainsThus ${\bar{\partial }}_E^2$ is linear over $C^{\infty }(X,\mathbb{C})$.

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Choose a local holomorphic frame $e$ of the holomorphic line bundle $E$ and defineIf $e^{\prime }=ge$ is another holomorphic frame, then $g$ is nowhere-zero and holomorphic, so $\bar{\partial }g=0$ and the two definitions agree. They therefore glue to a well-defined partial connection. In each holomorphic frame its square is ordinary ${\bar{\partial }}^2$, hence ${\bar{\partial }}_E^2=0$.

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