Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 118 / 4

Write $g=\mathrm{Re}h$ for the underlying Riemannian metric of the Hermitian manifold. Its fundamental form of a Hermitian manifold isIt is real and skew-symmetric. In a unitary coframe it is $\omega ={\sum }_{j=1}^n{e}^j\wedge J{e}^j$, which also shows that it has type $(1,1)$ and thatThe Hodge star operator is characterized, after complex-linear extension, byExpanding in the same unitary coframe gives

The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold says that every Dolbeault class has a unique ${\Delta }_{\bar{\partial }}$-harmonic representative andIf ${\Delta }_{\bar{\partial }}\eta =0$, thenso $\eta$ is $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed. If also $\eta =\bar{\partial }\xi$, then $\Vert \eta {\Vert }_2^2=⟨{\bar{\partial }}^{\ast }\eta ,\xi ⟩=0$, hence $\eta =0$.

Now suppose $X$ is compact and Kähler. With $d^c=i(\bar{\partial }-\partial )$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$, the Kähler identities make the mixed anticommutators vanish and give ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Consequently

Let $\alpha =d^c\gamma$ and $d\alpha =0$. The Kähler Laplacian identity implies that the $d$-Laplacian commutes with $d^c$. Since a harmonic form is $d^{c\ast }$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the Green operator of the Hodge Laplacian, thenwhere the $d^{\ast }dG\alpha$ term vanishes because $dG\alpha =Gd\alpha =0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved givesTherefore, for the $(k-2)$-form $\beta =-d^{\ast }G\gamma$,This is the d d c lemma in the form required here.