= Solution
Write $g=\operatorname{Re}h$ for the underlying <Riemannian metric> of the <Hermitian manifold>. Its <fundamental form of a Hermitian manifold> is
$$
\omega(u,v)=g(Ju,v).
$$
It is real and skew-symmetric. In a unitary coframe it is $\omega=\sum_{j=1}^n e^j\wedge Je^j$, which also shows that it has type $(1,1)$ and that
$$
\operatorname{vol}_g=\frac{\omega^n}{n!}.
$$
The <Hodge star operator> is characterized, after complex-linear extension, by
$$
\alpha\wedge *\overline\beta=\langle\alpha,\beta\rangle\operatorname{vol}_g.
$$
Expanding in the same unitary coframe gives
$$
*\omega=\frac{\omega^{n-1}}{(n-1)!}.
$$
The <Hodge Laplacian> and <Dolbeault Laplacian> are
$$
\Delta_d=dd^*+d^*d,
\qquad
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.
$$
The <Dolbeault Hodge decomposition on a compact Hermitian manifold> says that every Dolbeault class has a unique $\Delta_{\bar\partial}$-harmonic representative and
$$
\Omega^{p,q}=\mathcal H^{p,q}_{\bar\partial}\oplus\bar\partial\Omega^{p,q-1}\oplus\bar\partial^*\Omega^{p,q+1}.
$$
If $\Delta_{\bar\partial}\eta=0$, then
$$
0=\langle\Delta_{\bar\partial}\eta,\eta\rangle
=\|\bar\partial\eta\|_2^2+\|\bar\partial^*\eta\|_2^2,
$$
so $\eta$ is $\bar\partial$-closed and $\bar\partial^*$-closed. If also $\eta=\bar\partial\xi$, then $\|\eta\|_2^2=\langle\bar\partial^*\eta,\xi\rangle=0$, hence $\eta=0$.
Now suppose $X$ is compact and <Kähler manifold>[Kähler]. With $d^c=i(\bar\partial-\partial)$ and $d^*=\partial^*+\bar\partial^*$, the <Kähler identities> make the mixed anticommutators vanish and give $\Delta_\partial=\Delta_{\bar\partial}$. Consequently
$$
d^cd^*+d^*d^c=0.
$$
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 $d^{c*}$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the <Green operator of the Hodge Laplacian>, then
$$
\alpha=\Delta_dG\alpha=dd^*G\alpha,
$$
where the $d^*dG\alpha$ term vanishes because $dG\alpha=Gd\alpha=0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved gives
$$
d^*G\alpha=d^*d^cG\gamma=-d^cd^*G\gamma.
$$
Therefore, for the $(k-2)$-form $\beta=-d^*G\gamma$,
$$
\boxed{\alpha=dd^c\beta.}
$$
This is the <d d c lemma> in the form required here.
Back to article page
